Fluid Mechanics SI Units [2 ed.] 9780134649290, 1292247304, 9781292247304, 9781292247397

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Fluid Mechanics Second Edition in SI Units

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FLUID MECHANICS SECOND EDITION IN SI UNITS

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FLUID MECHANICS SECOND EDITION IN SI UNITS

R. C. HIBBELER SI Conversion by Kai Beng Yap

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Pearson Education Limited KAO Two KAO Park Hockham Way Harlow Essex CM17 9SR United Kingdom and Associated Companies throughout the world. Visit us on the World Wide Web at: www.pearsonglobaleditions.com. © 2021 by R. C. Hibbeler. Published by Pearson Education, Inc. or its affiliates. The rights of R. C. Hibbeler to be identified as the author of this work have been asserted by him in accordance with the Copyright, Designs and Patents Act 1988. Authorized adaptation from the United States edition, entitled Fluid Mechanics, ISBN 978-0-13-464929-0, by R. C. Hibbeler, published by Pearson Education, Inc., © 2018, 2015. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a license permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. Many of the designations by manufacturers and seller to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages with, or arising out of, the furnishing, performance, or use of these programs. This eBook is a standalone product and may or may not include all assets that were part of the print version. It also does not provide access to other Pearson digital products like MyLab and Mastering. The publisher reserves the right to remove any material in this eBook at any time. British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. ISBN 10: 1-292-24730-4 ISBN 13: 978-1-292-24730-4 eBook ISBN 13: 978-1-292-24739-7 Typeset by Integra Software Services Pvt Ltd.

To the Student With the hope that this work will stimulate an interest in Fluid Mechanics and provide an acceptable guide to its understanding.

P RE FA C E It is hoped that this book will provide the student with a clear and thorough presentation of the theory and applications of fluid mechanics. In order to achieve this objective, I have incorporated many of the suggestions and comments from the book’s reviewers, university colleagues who have e-mailed me, and from my students. Some of the more important improvements are listed below.

New to This Edition • Rewriting of Text Material. The purpose here is to achieve a further clarification of the material throughout the book, either with an expanded discussion, or by deleting material that seemed not relevant to a particular topic. • Expanded Topic Coverage. The material in Chapters 1, 3, 5, 7, 10, and 11 have enhanced discussion on some topics, including additional data in the tables in Chapters 10, 11, and 12. • New Example Problems. The addition of new examples to Chapters 7 and 10 further illustrate the applications of the theory. Many of the examples throughout the book have been expanded for clarification. • New Fundamental Problems. These problems have been added to Chapter 10 to enhance the student’s understanding of the theory and its applications, and to provide a review of the material that can be used to study for various engineering exams. • New Photos and Enhanced Art. Many new photos and figures have been added to help the student obtain a better understanding of the subject matter. • New Equation Summaries. These are found at the end of Chapters 7, 11, 12, and 13 so that students may use formula sheets when taking exams that require application of some key equations. Besides the new features mentioned, other outstanding features that define the contents of this book include the following.

Organization and Approach. Each chapter is organized into welldefined sections that contain an explanation of specific topics, illustrative example problems, and at the end of the chapter, a set of relevant homework problems. The topics within each section are placed into subgroups defined by boldface titles. The purpose of this organization is to present a structured

P R E FA C E

method for introducing each new definition or concept, and to make the book a convenient resource for later reference and review.

Procedures for Analysis. This unique feature provides the student with a logical and orderly method to follow when applying the theory that has been discussed in a particular section. The example problems are then solved using this outlined method in order to clarify its numerical application. Realize, however, that once the relevant principles have been mastered, and enough confidence and judgment has been obtained, the student can then develop his or her own procedures for solving problems.

Important Points. This feature provides a review or summary of the most important concepts in a section, and highlights the most significant points that should be remembered when applying the theory to solve problems. A further review of the material is given at the end of the chapter.

Photos. The relevance of knowing the subject matter is reflected by the realistic applications depicted in the many photos placed throughout the book. These photos are often used to show how the principles of fluid mechanics apply to real-world situations. Fundamental Problems. These problem sets are selectively located just after the example problems. They offer students simple applications of the concepts and therefore provide them with the chance to develop their problem-solving skills before attempting to solve any of the standard problems that follow. Students may consider these problems as extended examples, since they all have complete solutions and answers given in the back of the book. Additionally, the fundamental problems offer students an excellent means of preparing for exams, and they can be used at a later time to prepare for various engineering exams. Homework Problems. The majority of problems in the book depict realistic situations encountered in engineering practice. It is hoped that this realism will both stimulate interest in the subject, and provide a means for developing the skills to reduce any problem from its physical description to a model or symbolic representation to which the principles of fluid mechanics may then be applied. Throughout the book, all problems use SI units. Furthermore, in any set, an attempt has been made to arrange the problems in order of increasing difficulty. Except for every fourth problem, indicated by an asterisk (*), the answers to all the other problems are given in the back of the book.

Conceptual Problems. Throughout the text, usually at the end of most chapters, there is a set of problems that involve conceptual situations related to the application of the principles contained in the chapter. These problems are intended to engage students in thinking through a real-life situation as depicted in a photo. They can be assigned after the students have developed some expertise in the subject matter and they work well either for individual or team projects.

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P R E FA C E

Accuracy. Apart from my work, the accuracy of the text and problem solutions have all been thoroughly checked by other parties. Most importantly, Kai Beng Yap, Kurt Norlin of the Bittner Development Group, as well as Pavel Kolmakov and Vadim Semenenko at Competentum. The SI edition has been checked by three additional reviewers.

CONTENTS The book is divided into 14 chapters. Chapter 1 begins with an introduction to fluid mechanics, followed by a discussion of units and some important fluid properties. The concepts of fluid statics, including constant accelerated translation of a liquid and its constant rotation, are covered in Chapter 2. In Chapter 3, the basic principles of fluid kinematics are covered. This is followed by the continuity equation in Chapter 4, the Bernoulli and energy equations in Chapter 5, and fluid momentum in Chapter 6. In Chapter 7, differential fluid flow of an ideal fluid is discussed. Chapter 8 covers dimensional analysis and similitude. Then the viscous flow between parallel plates and within pipes is treated in Chapter 9. The analysis is extended to Chapter 10 where the design of pipe systems is discussed. Boundary layer theory, including topics related to pressure drag and lift, is covered in Chapter 11. Chapter 12 discusses open channel flow, and Chapter 13 covers a variety of topics in compressible flow. Finally, turbomachines, such as axial and radial flow pumps and turbines are treated in Chapter 14.

Alternative Coverage. After covering the basic principles of Chapters 1 through 6, at the discretion of the instructor, the remaining chapters may be presented in any sequence, without the loss of continuity. If time permits, sections involving more advanced topics, may be included in the course. Most of these topics are placed in the later chapters of the book. In addition, this material also provides a suitable reference for basic principles when it is discussed in more advanced courses.

ACKNOWLEDGMENTS I have endeavored to write this book so that it will appeal to both the student and instructor. Through the years many people have helped in its development, and I will always be grateful for their valued suggestions and comments. In particular, along with R. Sultana, California State University, Long Beach, J. Karl, University of South Florida, and C. Dreyer, Colorado School of Mines, the following individuals have contributed important reviewer comments relative to preparing this work: H. Gallegos, California State University, LA K. Lamb, California State Polytechnic University, Pomona A. Polebitski, University of Wisconsin-Platteville N. Lu, Colorado School of Mines M. Widdowson, Virginia Tech A. Shalaby, Howard University L. Grega, The College of New Jersey

P R E FA C E

R. Hotchkiss, Brigham Young University S. Kumpaty, Milwaukee School of Engineering J. Liburdy, Oregon State University S. K. Venayagamoorthy, Colorado State University J. Chen, Temple University K. Anderson, California State Polytechnic University, Pomona T. Olenik, New Jersey Institute of Technology K. Hodge, Mississippi State University E. Petersen, Texas A&M W. Rahmeyer, Utah State University D. Corti, Purdue University N. Moore, North Carolina State University S. Hilgenfeldt, University of Illinois—Urbana Champaign There are a few people that I feel deserve particular recognition. A longtime friend and associate, Kai Beng Yap, was of great help in checking the entire manuscript, and helping to further check all the problems. And a special note of thanks also goes to Kurt Norlin, and Viktoria Sirotova and all those at Competentum, for their diligence and support in this regard. During the production process I am also thankful for the support of my long time Production Editor, Rose Kernan, and my Managing Editor, Scott Disanno. My wife, Conny, has been a big help with the proofreading and typing needed to prepare the manuscript for publication. Lastly, many thanks are extended to all my students and colleagues that have e-mailed me their suggestions and comments. Since this list is too long to mention, it is hoped that those who have helped in this manner will accept this anonymous recognition. I value your judgment as well, and would greatly appreciate hearing from you if at any time you have any comments or suggestions that may help to improve the contents of this book. Russell Charles Hibbeler [email protected]

GLOBAL EDITION The publishers would like to thank the following for their contribution to the Global Edition: Contributor for the Second Edition in SI Units Kai Beng Yap is currently a registered professional engineer who works in Malaysia. He has BS and MS degrees in civil engineering from the University of Louisiana, Lafayette, Louisiana; and has done further graduate work at Virginia Tech in Blacksburg, Virginia. He has taught at the University of Louisiana and worked as an engineering consultant in the areas of structural analysis and design, as well as the associated infrastructure. Reviewers for the Second Edition in SI Units Haluk Aksel, Middle East Technical University Serter Atabay, American University of Sharjah Suresh Babu, Pondicherry University

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your answer specific feedback

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P R E FA C E

RESOURCES FOR INSTRUCTORS • Mastering Engineering. This online Tutorial Homework program allows you to integrate dynamic homework with automatic grading. Mastering Engineering allows you to easily track the performance of your entire class on an assignment-by-assignment basis, or the detailed work of an individual student. • Instructor’s Solutions Manual. An instructor’s solutions manual was prepared by the author. The manual includes homework assignment lists and was also checked as part of the accuracy checking program. The Instructor Solutions Manual is available at www.pearsonglobaleditions.com. • Presentation Resource. All art from the text is available in PowerPoint slide and JPEG format. These files are available for download from the Instructor Resource Center at www.pearsonglobaleditions.com. If you are in need of a login and password for this site, please contact your local Pearson representative. • Video Solutions. Developed by Professor Garret Nicodemus, video solutions which are located on the companion website offer step-by-step solution walkthroughs of homework problems from each section of the text. Make efficient use of class time and office hours by showing students the complete and concise problem solving approaches that they can access anytime and view at their own pace. The videos are designed to be a flexible resource to be used however each instructor and student prefers. A valuable tutorial resource, the videos are also helpful for student self-evaluation as students can pause the videos to check their understanding and work alongside the video. Access the videos at www.pearsonglobaleditions.com and follow the links for the Fluid Mechanics text.

RESOURCES FOR STUDENTS • Mastering Engineering. Tutorial homework problems emulate the instructor’s office-hour environment. • Companion Website. The companion website, located at www. pearsonglobaleditions.com, includes opportunities for practice and review, including access to animations and video solutions offering complete, step-by-step solution walkthroughs of representative homework problems from various sections of the text.

C O N T E NTS 1

2

Fundamental Concepts 19 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10

Fluid Statics

Chapter Objectives 19 Introduction 19 Characteristics of Matter 21 The International System of Units 22 Calculations 25 Problem Solving 27 Some Basic Fluid Properties 29 Viscosity 34 Viscosity Measurement 39 Vapor Pressure 43 Surface Tension and Capillarity 44

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14

61

Chapter Objectives 61 Pressure 61 Absolute and Gage Pressure 64 Static Pressure Variation 66 Pressure Variation for Incompressible Fluids 67 Pressure Variation for Compressible Fluids 69 Measurement of Static Pressure 72 Hydrostatic Force on a Plane Surface— Formula Method 80 Hydrostatic Force on a Plane Surface— Geometrical Method 86 Hydrostatic Force on a Plane Surface— Integration Method 91 Hydrostatic Force on an Inclined Plane or Curved Surface Determined by Projection 94 Buoyancy 101 Stability 104 Constant Translational Acceleration of a Liquid 107 Steady Rotation of a Liquid 112

3

4

Kinematics of Fluid Motion

Conservation of Mass 189

3.1 3.2 3.3 3.4 3.5

153

Chapter Objectives 153 Types of Fluid Flow 153 Graphical Descriptions of Fluid Flow Fluid Flow Descriptions 161 Fluid Acceleration 168 Streamline Coordinates 175

4.1 157 4.2 4.3 4.4

Chapter Objectives 189 Volumetric Flow, Mass Flow, and Average Velocity 189 Finite Control Volumes 194 The Reynolds Transport Theorem 196 Conservation of Mass 200

14

CONTENTS

5

7

Work and Energy of Moving Fluids

Differential Fluid Flow 359

5.1 5.2 5.3 5.4 5.5

231

Chapter Objectives 231 Euler’s Equations of Motion 231 The Bernoulli Equation 235 Applications of the Bernoulli Equation 238 Energy and Hydraulic Grade Lines 251 The Energy Equation 260

7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13

Chapter Objectives 359 Differential Analysis 359 Kinematics of Differential Fluid Elements 360 Circulation and Vorticity 364 Conservation of Mass 369 Equations of Motion for a Fluid Particle 371 The Euler and Bernoulli Equations 373 Potential Flow Hydrodynamics 377 The Stream Function 377 The Potential Function 383 Basic Two-Dimensional Flows 387 Superposition of Flows 396 The Navier–Stokes Equations 409 Computational Fluid Dynamics 414

8 6 Fluid Momentum 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8

301

Chapter Objectives 301 The Linear Momentum Equation 301 Applications to Bodies at Rest 304 Applications to Bodies Having Constant Velocity 313 The Angular Momentum Equation 318 Propellers and Wind Turbines 326 Applications for Control Volumes Having Accelerated Motion 331 Turbojets and Turbofans 332 Rockets 333

Dimensional Analysis and Similitude 435 8.1 8.2 8.3 8.4 8.5

Chapter Objectives 435 Dimensional Analysis 435 Important Dimensionless Numbers 438 The Buckingham Pi Theorem 441 Some General Considerations Related to Dimensional Analysis 450 Similitude 451

CONTENTS

9

11

Viscous Flow within Enclosed Conduits

Viscous Flow over External Surfaces

9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8

475

Chapter Objectives 475 Steady Laminar Flow between Parallel Plates 475 Navier–Stokes Solution for Steady Laminar Flow between Parallel Plates 481 Steady Laminar Flow within a Smooth Pipe 486 Navier–Stokes Solution for Steady Laminar Flow within a Smooth Pipe 490 The Reynolds Number 492 Fully Developed Flow from an Entrance 497 Laminar and Turbulent Shear Stress within a Smooth Pipe 499 Steady Turbulent Flow within a Smooth Pipe 502

10 Analysis and Design for Pipe Flow 521 10.1 10.2 10.3 10.4 10.5

Chapter Objectives 521 Resistance to Flow in Rough Pipes 521 Losses Occurring from Pipe Fittings and Transitions 535 Single-Pipeline Flow 541 Pipe Systems 548 Flow Measurement 554

15

575

Chapter Objectives 575 The Concept of the Boundary Layer 575 Laminar Boundary Layers 581 The Momentum Integral Equation 590 Turbulent Boundary Layers 594 Laminar and Turbulent Boundary Layers 596 11.6 Drag and Lift 602 11.7 Pressure Gradient Effects 604 11.8 The Drag Coefficient 609 11.9 Drag Coefficients for Bodies Having Various Shapes 613 11.10 Methods for Reducing Drag 620 11.11 Lift and Drag on an Airfoil 624 11.1 11.2 11.3 11.4 11.5

12 Open-Channel Flow 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9

655

Chapter Objectives 655 Types of Flow in Open Channels 655 Open-Channel Flow Classifications 657 Specific Energy 658 Open-Channel Flow over a Rise or Bump 666 Open-Channel Flow under a Sluice Gate 670 Steady Uniform Channel Flow 674 Gradually Varied Flow 681 The Hydraulic Jump 688 Weirs 693

16

CONTENTS

13

14

Compressible Flow 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 13.10 13.11 13.12 13.13

715

Chapter Objectives 715 Thermodynamic Concepts 715 Wave Propagation through a Compressible Fluid 724 Types of Compressible Flow 727 Stagnation Properties 731 Isentropic Flow through a Variable Area 738 Isentropic Flow through Converging and Diverging Nozzles 743 The Effect of Friction on Compressible Flow 752 The Effect of Heat Transfer on Compressible Flow 762 Normal Shock Waves 768 Shock Waves in Nozzles 771 Oblique Shock Waves 776 Compression and Expansion Waves 781 Compressible Flow Measurement 786

Turbomachines

807

Chapter Objectives 807 Types of Turbomachines 807 Axial-Flow Pumps 808 Radial-Flow Pumps 815 Ideal Performance for Pumps 818 Turbines 824 Pump Performance 831 Cavitation and the Net Positive Suction Head 834 14.8 Pump Selection Related to the Flow System 836 14.9 Turbomachine Similitude 838 14.1 14.2 14.3 14.4 14.5 14.6 14.7

Appendix A Physical Properties of Fluids 856 B Compressible Properties of a Gas (k = 1.4) 859 Fundamental Solutions

870

Answers to Selected Problems Index

899

886

FLUID MECHANICS SECOND EDITION IN SI UNITS

1

Eric Middelkoop/Fotolia

CHAPTER

Fluid mechanics plays an important role in the design and analysis of pressure vessels, pipe systems, and pumps used in chemical processing plants.

FUNDAMENTAL CONCEPTS

CHAPTER OBJECTIVES ■

To define fluid mechanics and indicate its various branches.

■

To discuss the system of units for measuring fluid quantities, and establish proper calculation techniques.

■

To define some important fluid properties.

■

To describe some of the behavior characteristics of fluids.

1.1

INTRODUCTION

Fluid mechanics is a study of the behavior of a gas or liquid that is either at rest or in motion. Because fluids are so common, this subject has important applications in many engineering disciplines. For example, aeronautical and aerospace engineers use fluid mechanics to study flight and to design propulsion systems. Civil engineers use it to design channels, water networks, sewer systems, and water-resisting structures such as dams and levees. Fluid mechanics is used by mechanical engineers to design pumps, compressors, control systems, heating and air conditioning equipment, wind turbines, and solar heating devices. Chemical and petroleum engineers apply this subject to design equipment used for filtering, pumping, and mixing fluids. And finally, engineers in the electronics and computer industry use fluid mechanics principles to design switches, screen displays, and data storage equipment. Apart from the engineering profession, the principles of fluid mechanics are also used in the field of biomechanics, where they play a vital role in the understanding of the circulatory, digestive, and respiratory systems; and in meteorology to study the motion and effects of tornadoes and hurricanes.

19

20

Chapter 1

F u n d a m e n ta l C o n C e p t s

Fluid Mechanics

1

Hydrostatics

Kinematics

Study of fluids at rest or flow with constant velocity

Study of the geometry of fluid motion

Fluid Dynamics Study of the forces that cause accelerated motion

Fig. 1–1

Branches of Fluid Mechanics. The principles of fluid mechanics are based on Newton’s laws of motion, the conservation of mass, the first and second laws of thermodynamics, and laws related to the physical properties of a fluid. The subject is divided into three main categories, as shown in Fig. 1–1. In this book, hydrostatics is presented in Chapter 2, kinematics is introduced in Chapter 3, and the study of fluid dynamics is presented throughout the rest of the book.

Historical Development. A fundamental knowledge of the principles of fluid mechanics has been of considerable importance throughout the development of human civilization. Historical records show that through the process of trial and error, early societies, such as the Roman Empire, used fluid mechanics in the construction of their irrigation and water supply systems. In the middle of the 3rd century B.C., Archimedes discovered the principle of buoyancy, and then much later, in the 15th century, Leonardo Da Vinci developed principles for the design of canal locks and other devices used for water transport. However, many important discoveries of basic fluid mechanics principles occurred in the 17th century. It was then that Evangelista Torricelli designed the barometer, Blaise Pascal formulated the law of static pressure, and Isaac Newton developed his law of viscosity to describe the nature of fluid resistance to flow. In the 18th century, Leonhard Euler and Daniel Bernoulli pioneered the field of hydrodynamics, which deals with the motion of an idealized fluid, that is, one having a constant density and providing no internal frictional resistance. Unfortunately, this study has limited application since not all physical properties of the fluid are taken into account. The need for a more realistic approach led to the development of hydraulics. This field uses empirical equations found from fitting curves to data determined from experiments, primarily for applications involving water. In the 19th century, contributors included Gustave de Coriolis, who developed water turbines, and Gotthilf Hagen and Jean Poiseuille, who studied the resistance to water flowing through pipes. In the early 20th century, hydrodynamics and hydraulics were essentially combined through the work of Ludwig Prandtl, who introduced the concept of the boundary layer while studying aerodynamics. Through the years, many others have also made important contributions to this subject, and we will discuss many of these throughout the book.* *References [1] and [2] provide a more complete description of the historical development of this subject.

1.2

1.2

CharaCteristiCs oF matter

21

CHARACTERISTICS OF MATTER

In general, matter can be classified by the state it is in—as a solid, a liquid, or a gas.

Solid. A solid, such as steel, aluminum, or wood, maintains a definite shape and volume, Fig. 1–2a. It maintains its shape because the molecules or atoms of a solid are densely packed and are held tightly together, generally in the form of a lattice or geometric structure. The spacing of atoms within this structure is due in part to large cohesive forces that exist between molecules. These forces prevent any relative movement, except for any slight vibration of the molecules themselves. As a result, when a solid is subjected to a load it will not easily deform, but once in its deformed state, it will continue to support the load.

Liquid. A liquid, such as water, alcohol, or oil, is a fluid that is composed of molecules that are more mobile than those in a solid. Their intermolecular forces are weaker, so liquids do not hold their shape. Instead, they flow and take the shape of their container, forming a horizontal free surface at the top, Fig. 1–2b. Although liquids can easily deform, their molecular spacing allows them to resist compressive forces when they are confined.

Gas. A gas, such as helium, nitrogen, or air, is a fluid that flows until it fills the entire volume of its container, Fig. 1–2c. Gases are composed of molecules that are much farther apart than those of a liquid. As a result, the molecules of a gas are free to travel away from one another until a force of repulsion pushes them away from other gas molecules, or from the molecules on the surface of a container. Continuum. Studying the behavior of a fluid as in Fig. 1–3a by analyzing the motion of all its many molecules would be an impossible task. Fortunately, however, almost all engineering applications involve a volume of fluid that is much greater than the very small distance between adjacent molecules of the fluid, and so it is reasonable to assume that the fluid is uniformly dispersed throughout this volume. Under these circumstances, we can then consider the fluid to be a continuum, that is, a continuous distribution of matter leaving no empty space, Fig. 1–3b. This assumption allows us to use average properties of the fluid at any point throughout its volume. For those special situations where the molecular distance does become important, which is on the order of a billionth of a meter, the continuum model does not apply, and it is necessary to employ statistical techniques to study the fluid flow, a topic that will not be considered here. See Ref. [3].

Solids maintain a constant shape (a)

Liquids take the shape of their container (b)

Gases fill the entire volume of their container (c)

Fig. 1–2

Actual fluid (a)

Continuum model (b)

Fig. 1–3

1

22

Chapter 1

F u n d a m e n ta l C o n C e p t s

1.3

1

THE INTERNATIONAL SYSTEM OF UNITS

We can describe a fluid and its flow characteristics using combinations of units based on five basic quantities, namely, length, time, mass, force, and temperature. However, because length, time, mass, and force are all related by Newton’s second law of motion, F = ma, the units used to define the size of each of these quantities cannot all be selected arbitrarily. The equality F = ma is maintained when three of these units are arbitrarily defined, and the fourth unit is then derived from the equation. The International System of units is a modern version of the metric system that has received worldwide recognition. The SI system specifies length in meters (m), time in seconds (s), and mass in kilograms (kg). The unit of force, called a newton (N), is derived from F = ma, where 1 newton is equal to the force required to give 1 kilogram of mass an acceleration of 1 m > s2 1 N = kg # m > s2 2 , Fig. 1–4a. To determine the weight of a fluid in newtons at the “standard location,” where the acceleration due to gravity is g = 9.81 m>s2, and the mass of the fluid is m (kg), we have W (N) = [m (kg)]19.81 m>s2 2

(1–1)

And so a fluid having a mass of 1 kg has a weight of 9.81 N, 2 kg of fluid has a weight of 19.62 N, and so on. 1 ms2 1N

1 kg

The newton is a unit of force (a)

Fig. 1–4

1.3

the international system oF units

Temperature. The absolute temperature is the temperature measured from a point at which the molecules of a substance have so-called “zero energy” or no motion.* Its unit in the SI system is the kelvin (K). This unit is expressed without reference to degrees, so 7 K is stated as “seven kelvins.” Although not officially an SI unit, an equivalent-size unit measured in degrees Celsius (°C) is often used. This measurement is referenced from the freezing and boiling points of pure water, where the freezing point is 0°C (273 K) and the boiling point is 100°C (373 K), Fig. 1–4b. For conversion, TK = TC + 273

(1–2)

We will use Eqs. 1–1 and 1–2 in this book, since they are suitable for most engineering applications. However, for more accurate work, use the exact value of 273.15 in Eq. 1–2. Also, at the “standard location,” the more exact value g = 9.807 m>s2 or the local acceleration due to gravity should be used in Eq. 1–1.

373 K

1008C (Steam)

273 K

08C (Ice)

–2738C (zero energy)

0K

The Kelvin and Celsius scales (b)

Fig. 1–4 (cont.)

*This is actually an unreachable point according to the law of quantum mechanics.

23

1

24

1

Chapter 1

F u n d a m e n ta l C o n C e p t s

Prefixes. In the SI system, when a numerical quantity is either very large or very small, the units used to define its size should be modified by using a prefix. The range of prefixes used for problems in this book is shown in Table 1–1. Each represents a multiple or submultiple of a unit that moves the decimal point of a numerical quantity either forward or backward by three, six, or nine places. For example, 5 000 000 g = 5000 kg (kilogram) = 5 Mg (Megagram), and 0.000 006 s = 0.006 ms (millisecond) = 6 μs (microsecond). As a general rule, the units of quantities that are multiplied together are separated by a dot to avoid confusion with prefix notation. Thus, m # s is a meter-second, whereas ms is a millisecond. And finally, the exponential power applied to a unit having a prefix refers to both the unit and its prefix. For example, ms2 = (ms)2 = (ms)(ms) = 1 10-3 s 2 1 10-3 s 2 = 10-6 s2.

TABLE 1–1 Prefixes Exponential Form

Prefix

SI Symbol

0.001

10 - 3

milli

m

0.000 001

10 - 6

micro

μ

0.000 000 001

10 - 9

nano

n

1 000 000 000

109

Giga

G

1 000 000

106

Mega

M

1 000

103

kilo

k

Submultiple

Multiple

1.4

1.4

CalCulations

25

CALCULATIONS

Application of fluid mechanics principles often requires algebraic manipulations of a formula followed by numerical calculations. For this reason it is important to keep the following concepts in mind.

Dimensional Homogeneity. The terms of an equation used to describe a physical process must be dimensionally homogeneous, that is, each term must be expressed in the same units. Provided this is the case, then all the terms of the equation can be combined when numerical values are substituted for the variables. For example, consider the Bernoulli equation, which is a specialized application of the principle of work and energy. We will study this equation in Chapter 5, but it can be expressed as p V2 + + z = constant g 2g Using SI units, the pressure p is expressed in N>m2, the specific weight g is in N>m3, the velocity V is in m>s, the acceleration due to gravity g is in m>s2, and the elevation z is in meters, m. In the form stated, each of the three terms is in meters, as noted if you cancel the units in each fraction. N>m2 N>m3

+

1 m>s 2 2 m>s2

+ m

Regardless of how the equation is algebraically arranged, it will maintain its dimensional homogeneity, and as a result, a partial check of the algebraic manipulation of any equation can be made by checking to be sure all the terms have the same units.

Rounding off Numbers. Rounding off a number is necessary so that the accuracy of the result will be the same as that of the problem data. As a general rule, any numerical figure ending in a number greater than five is rounded up and a number less than five is rounded down. For example, if 3.558 is to be rounded off to three significant figures, then because the fourth digit (8) is greater than 5, the third number is rounded up to 6, so the number becomes 3.56. Likewise 0.5896 becomes 0.590 and 9.387 becomes 9.39. If we round off 1.341 to three significant figures, because the fourth digit (1) is less than 5, then we get 1.34. Likewise 0.3762 becomes 0.376 and 9.873 becomes 9.87. There is a special case for any number that ends in exactly 5. If the digit preceding the 5 is an even number, than this digit is not rounded up. If the digit preceding the 5 is an odd number, then it is rounded up. For example, 75.25 rounded off to three significant figures becomes 75.2, 0.1275 becomes 0.128, and 0.2555 becomes 0.256.

1

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Calculation Procedure. When performing numerical calculations,

1

it is best to store the intermediate results in the calculator. In other words, do not round off calculations until reporting the final result. This procedure maintains precision throughout the series of steps to the final solution. In this text we will generally round off the answers to three significant figures, since most of the data in fluid mechanics, such as geometry and fluid properties, may be reliably measured to this accuracy. Also, when using SI units, first represent all the quantities in terms of their base or derived units by converting any prefixes to powers of 10. Then do the calculation, and finally express the result using a single prefix. For example, [3 MN](2 mm) = 3 3 1 106 2 N 4 3 2 1 10-3 2 m 4 = 6 1 103 2 N # m = 6 kN # m. In the case of fractional units, with the exception of the kilogram, the prefix should always be in the numerator, as in MN>s or mm>kg.

Complex flows are often studied using a computer analysis; however, it is important to have a good grasp of the principles of fluid mechanics to be sure reasonable predictions have been made. (© CHRIS SATTLBERGER/Science Source)

1.5

1.5

problem solving

27

PROBLEM SOLVING

At first glance, the study of fluid mechanics can be rather daunting, because there are many aspects of this field that must be understood. Success at solving problems, however, will depend on your attitude and your willingness to both focus on class lectures and to carefully read the material in the book. Aristotle once said, “What we have to learn to do, we learn by doing,” and indeed your ability to solve problems in fluid mechanics depends upon a thoughtful preparation and neat presentation. In any engineering subject, it is very important that you follow a logical and orderly procedure when solving problems. In the case of fluid mechanics this should include the sequence of steps outlined below.

GENE R AL PR OCEDU R E F O R ANALYSIS Fluid Description. Fluids can behave in many different ways, and so at the outset it is important to identify the type of fluid flow and specify the fluid’s physical properties. Knowing this provides a means for the proper selection of equations used for an analysis. Analysis. This generally involves the following steps:

• Tabulate the problem data and draw any necessary diagrams. • Apply the relevant principles, generally in mathematical form. When substituting numerical data into any equations, be sure to include their units, and check to be sure the terms are dimensionally homogeneous.

• Solve the equations, and report any numerical answers to three significant figures.

• Study the answer with technical judgment and common sense to determine whether or not it seems reasonable. When applying this procedure, do the work as neatly as possible. Being neat generally stimulates clear and orderly thinking.

1

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I MPO RTA N T PO I N T S • Solids have a definite shape and volume, liquids take the shape of their container, and gases fill the entire volume of their container. 1

• For most engineering applications, we can consider a fluid to be a continuum, and therefore use its average properties to model its behavior.

• Weight is measured in newtons in the SI system and is determined from W (N) = m (kg)(9.81 m>s2).

• Certain rules must be followed when performing calculations and using prefixes in the SI system of units. First convert all numerical quantities with prefixes to their base units, then perform the calculations, and finally choose an appropriate prefix for the result.

• The derived equations of fluid mechanics are all dimensionally homogeneous, and thus each term in an equation has the same units. Careful attention should therefore be paid to the units when entering data and then solving an equation.

• As a general rule, perform calculations with sufficient numerical accuracy, and then round off the final answer to three significant figures.

EXAMPLE

1.1 Evaluate 1 80 MN>s 2 1 5 mm 2 2, and express the result with SI units having an appropriate prefix. SOLUTION We first convert all the quantities with prefixes to powers of 10, perform the calculation, and then choose an appropriate prefix for the result.

1 80 MN>s 2 1 5 mm 2 2

=

3 80 1 106 2 N>s 4 3 5 1 10-3 2 m 4 2

3 801106 2 N>s 4 3 25110 - 6 2 m2 4

= 2 1 103 2 N # m2 >s = 2 kN # m2 >s =

Ans.

1.6

EXAMPLE

29

some basiC Fluid properties

1.2

Convert a fluid flow of 24 000 liters>h to m3 >s.

1

SOLUTION Using 1 m3 = 1000 liters and 1 h = 3600 s, the factors of conversion are arranged in the following order so that cancellation of units occurs. a 24 000

liters 1 m3 1h ba ba b h 1000 liters 3600 s

= 6.67110 - 3 2m3 >s

Ans.

Notice here that the result is rounded off to three significant figures. For engineering work, the results are expressed as a multiple of 10, having an exponential power in multiples of three as in 1 103 2 , 1 106 2 , 1 10−9 2 , etc.

1.6

SOME BASIC FLUID PROPERTIES

A fluid has several important physical properties used to describe its behavior. In this section, we will define its density, specific weight, specific gravity, and bulk modulus. Then at the end of this section, we will use the ideal gas law to discuss how an ideal gas behaves.

Density. The density r (rho) refers to the mass of the fluid that is

contained in a unit of volume, Fig. 1–5. It is measured in kg>m3 and is determined from r =

m V

V m

(1–3)

Here m is the mass of the fluid, and V is its volume.

Liquid. Through experiment it has been found that a liquid is practically incompressible, that is, the density of a liquid varies little with pressure. It does, however, have a slight but greater variation with temperature. For example, water at 4°C has a density of rw = 1000 kg>m3, whereas at 100°C its volume will expand, and so rw = 958.1 kg>m3. If the temperature range is small, we can, for most practical applications, consider the density of a liquid to be essentially constant, meaning its volume does not change, and the fluid is then referred to as incompressible.

Density is massvolume

Fig. 1–5

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Gas. Unlike a liquid, temperature and pressure can markedly affect the density of a gas, since it has a higher degree of compressibility. For example, air has a density of r = 1.23 kg>m3 when the temperature is 15°C and the atmospheric pressure is 101.3 kPa [1 Pa (pascal) = 1 N>m2]. But at this same temperature, and at twice the pressure, the density of air will double and become r = 2.46 kg>m3. Appendix A lists typical values for the densities of common liquids and gases. Included are tables of specific values for water at different temperatures, and air at different temperatures and atmospheric elevations.

1

W V

Specific Weight. The specific weight g (gamma) of a fluid is its weight per unit volume, Fig. 1–6. It is measured in N>m3. Thus,

g =

W V

(1–4)

Specific weight is weightvolume

Fig. 1–6

Here W is the weight of the fluid, and V is its volume. Since weight is related to mass by W = mg, then substituting this into Eq. 1–4, and comparing this result with Eq. 1–3, the specific weight is related to the density by

g = rg

(1–5)

Specific Gravity. The specific gravity S of a substance is a dimensionless quantity that is defined as the ratio of its density or specific weight to that of some other substance that is taken as a “standard.” It is most often used for liquids, and water at an atmospheric pressure of 101.3 kPa and a temperature of 4°C is taken as the standard. Thus,

S =

r g = rw gw

(1–6)

The density of water for this case is rw = 1000 kg>m3. So, for example, if an oil has a density of ro = 880 kg>m3, then its specific gravity will be So = 0.880.

1.6

Bulk Modulus. The bulk modulus of elasticity, or simply the bulk modulus, is a measure of the amount by which a fluid offers a resistance to compression. To define this property, consider the cube of fluid in Fig. 1–7, where each face has an area A and is subjected to an incremental force dF. The intensity of this force per unit area is the incremental pressure, dp = dF >A. As a result, the original volume V of the cube will decrease by dV, and when the incremental pressure is divided by the decrease in volume per unit volume, dV > V, it defines the bulk modulus, namely, EV = -

dp dV>V

(1–7)

dF

1 dF Final volume

dF Original volume

Bulk modulus

Fig. 1–7

The minus sign is included to show that the increase in pressure (positive) causes a decrease in volume (negative). The units for EV are the same as for pressure, that is, force per unit area, since the volume ratio is dimensionless. Typical units are N>m2 or Pa.

Liquid. Because the density of a liquid changes very little with pressure, its bulk modulus is very high. For example, sea water at atmospheric pressure and room temperature has a bulk modulus of about EV = 2.20 GPa.* If we use this value and consider the deepest region of the Pacific Ocean, where the water pressure is increased by 110 MPa, then Eq. 1–7 shows that the percent of compression of water is only ∆V>V = 3 110 1 106 2 Pa 4 > 3 2.20 1 109 2 Pa 4 = 0.05 or 5%. Similar results occur for other liquids, and so for this reason, we can assume that for most practical applications, liquids can be considered incompressible, and, as stated previously, their density remains constant.** Gas. A gas, because of its low density, is thousands of times more compressible than a liquid, and so its bulk modulus will be much smaller. For a gas, however, the relation between the applied pressure and the volume change depends upon the process used to compress the gas. Later, in Chapter 13, we will study this effect as it relates to compressible flow, where changes in pressure become significant. However, if the gas flows at low velocities, that is, less than about 30% the speed of sound in the gas, then only small changes in the gas pressure will occur, and so, even with its low bulk modulus, at constant temperature a gas, like a liquid, can in this case also be considered incompressible.

*Of course, solids can have much higher bulk moduli. For example, the bulk modulus for steel is 160 GPa. **The compressibility of a flowing liquid must, however, be considered for some types of fluid analysis. For example, “water hammer” is created when a valve on a pipe is suddenly closed. This causes an abrupt local change in density of the water near the valve, which generates a pressure wave that travels back up the pipe and produces a hammering sound when the wave encounters a bend or other obstruction in the pipe. See Ref. [7].

31

some basiC Fluid properties

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Ideal Gas Law. In this book we will consider every gas to behave as an ideal gas.* Such a gas is assumed to have enough separation between its molecules so that the molecules have no attraction to one another. Also, the gas must not be near the point of condensation into either a liquid or a solid state. From experiments, mostly performed with air, it has been shown that ideal gases behave according to the ideal gas law. It can be expressed as

1

p = rRT

(1–8)

Here p is the absolute pressure, or force per unit area, referenced from a perfect vacuum, r is the density of the gas, R is the gas constant, and T is the absolute temperature. Typical values of R for various gases are given in Appendix A. For example, for air, R = 286.9 J>(kg # K), where 1 J (joule) = 1 N # m. The volume, pressure, and temperature of the gas in this tank are related by the ideal gas law.

I MPO RTA N T PO I N T S • The mass of a fluid is often characterized by its density r = m>V, and its weight is characterized by its specific weight g = W>V, where g = rg.

• The specific gravity is a ratio of the density or specific weight of a liquid to that of water, defined by S = r>rw = g>gw. Here rw = 1000 kg>m3.

• The bulk modulus of a fluid is a measure of its resistance to compression. Since this property is very high for liquids, we can generally consider liquids as incompressible fluids. Provided a gas has a low velocity of flow—less than 30% the speed of sound in the gas—and has a constant temperature, then the pressure variation within the gas will be low, and we can, under these circumstances, also consider it to be incompressible.

• For many engineering applications, we can consider a gas to be ideal, and can therefore relate its absolute pressure to its absolute temperature and density using the ideal gas law, p = rRT.

*Nonideal gases and vapors are studied in thermodynamics.

1.6

some basiC Fluid properties

33

EXAMPLE 1.3 Air contained in the tank, Fig. 1–8, is under an absolute pressure of 60 kPa and has a temperature of 60°C. Determine the mass of the air in the tank.

4m

1.5 m

Fig. 1–8

SOLUTION We will first find the density of the air in the tank using the ideal gas law, Eq. 1–8, p = rRT. Then, knowing the volume of the tank, we can determine the mass of the air. The absolute temperature of the air is TK = TC + 273 = 60°C + 273 = 333 K

From Appendix A, the gas constant for air is R = 286.9 J>(kg # K). Then, p = rRT 60(10 ) N>m2 = r(286.9 J>kg # K)(333 K) r = 0.6280 kg>m3 3

The mass of air within the tank is therefore m r = V m 3 0.6280 kg>m = 3 p 1 1.5 m 2 2(4 m) 4 m = 17.8 kg

Ans.

Many people are often surprised by how large the mass of a gas contained within a volume can be. For example, if we repeat the calculations for the mass of air in a typical classroom that measures 4 m by 6 m by 3 m, at a standard room temperature of 20°C and pressure of 101.3 kPa, the result is 86.8 kg. The weight of this air is 851 N! It is no wonder then that the flow of air can cause the lift of an airplane and structural damage to buildings.

1

34

Chapter 1

F u n d a m e n ta l C o n C e p t s

EXAMPLE 1.4 An amount of glycerin has a volume of 1 m3 when the pressure is 120 kPa. If the pressure is increased to 400 kPa, determine the change in volume of this cubic meter. The bulk modulus for glycerin is EV = 4.52 GPa.

1

SOLUTION We must use the definition of the bulk modulus for the calculation. First, the pressure increase applied to the cubic meter of glycerin is ∆p = 400 kPa - 120 kPa = 280 kPa Thus, the change in volume is EV = 4.52 1 109 2 N>m2 = -

∆p ∆V>V

280 1 103 2 N>m2

∆V = -61.9 1 10-6 2 m3 ∆V>1 m3

Ans.

This is indeed a very small change! Notice that since ∆V is directly proportional to the change in pressure, doubling the pressure change will then double the change in volume.

1.7 U F

Fig. 1–9

VISCOSITY

The reason liquids and gases are classified as fluids is that they continuously deform or flow when subjected to a shear or tangential force. This deformation is shown by the movement of thin layers of fluid in Fig. 1–9, when a small force F is applied to the plate, causing the fluid to be dragged along. As one would expect, the resulting deformation will occur at different rates for different types of fluid. For example, thin layers of water or gasoline will slide faster than tar or syrup. The property of a fluid that is used to measure this resistance to flow is called viscosity. The higher the viscosity, the more difficult it is to move through the fluid. For example, it is easier to move through air compared to water, because the viscosity of air is many times smaller than that of water. Viscosity is a very important property in fluid mechanics, because it causes internal friction within the fluid, which produces an energy loss that must be taken into account when designing vehicles or conduits such as pipes and channels.

1.7

35

visCosity

Fast

A B

1 Slow

Fig. 1–10

Physical Cause of Viscosity. The resistance that gives rise to viscosity in a fluid can be understood by considering the two layers of fluid in Fig. 1–10 sliding relative to one another. Since the molecules composing the fluid are always in continuous motion, then if molecule A in the faster top layer happens to travel down to the slower bottom layer, it will have a component of relative motion to the right. Collisions that occur with any slower-moving molecule of the bottom layer will cause it to be pushed along due to a momentum exchange with A. The reverse effect occurs when molecule B in the bottom layer migrates upward. Here this slower-moving molecule will retard a faster-moving molecule through their momentum exchange. On a grand scale, both of these effects cause resistance or viscosity.

U F

Distortion of fluid elements due to shear (a) y U

U F

Newton’s Law of Viscosity. Let us again consider a fluid that is confined between a fixed surface and a very wide horizontal plate, Fig. 1–11a. When a very small force F is applied to the plate, it will cause elements of the fluid to deform as shown. After a brief acceleration, the viscous resistance of the fluid will bring the plate into equilibrium, such that the plate will begin to move with a constant velocity U. During this motion, the molecular adhesive force between the fluid particles in contact with the bottom fixed surface and the particles contacting the bottom of the plate creates a “no-slip condition,” such that the fluid particles at the fixed surface remain at rest, while those on the plate’s bottom surface move with the same velocity as the plate.* In between these two surfaces, very thin layers of fluid are dragged along so that the velocity profile u across the thickness of the fluid will be parallel to the plate, and will vary, as shown in Fig. 1–11b.

*Recent findings have confirmed that this “no-slip condition” does not always hold. A fastmoving fluid flowing over an extremely smooth surface develops no adhesion. Also, surface adhesion can be reduced by adding soap-like molecules to the fluid, which coats the surface, thereby making it extremely smooth. For most engineering applications, however, the layer of fluid molecules adjacent to a solid boundary will adhere to the surface, and so these special cases with slipping at the boundary will not be considered in this book. See Ref. [11].

u 1 Du u u Velocity distribution within a thin fluid layer (b)

Fig. 1–11

36

Chapter 1

F u n d a m e n ta l C o n C e p t s

y U

U

F

u 1 Du

1

u

Shear Stress. The motion just described is a consequence of the shear within the fluid caused by the moving plate. This effect subjects each element of fluid to a force per unit area or shear stress t (tau), Fig. 1–11c, that is produced by a tangential force ∆F that acts on the top and bottom area ∆A of the element. This stress is measured as a force per unit area, and so in the limit,

u Velocity distribution within a thin fluid layer

t = lim S ∆A

0

∆F dF = ∆A dA

Shear Strain. Since the fluid will flow, the shear stress will cause each

(b) t Da dx

element to deform into the shape of a parallelogram, Fig. 1–11c, and during a short time ∆t, the resulting deformation is defined by the shear strain, specified by the small angle ∆a (alpha), where

Dy

∆a ≈ tan ∆a = Shear stress

(1–9)

causes

Shear strain

(c)

Fig. 1–11 (cont.)

dx ∆y

Unlike a solid that would hold this angle under load, a fluid element will continue to deform, and so in fluid mechanics, the time rate of change in this shear strain (angle) becomes important. Since the top of the element moves ∆u faster than its bottom, Fig. 1–11b, then dx = ∆u ∆t. Substituting this into the above equation, the time rate of change of the shear strain becomes ∆a> ∆t = ∆u> ∆y. And in the limit, as ∆t S 0, da du = dt dy The term on the right is called the velocity gradient, because it indicates the change in velocity u with respect to y. In the late 17th century, Isaac Newton proposed that the shear stress in a fluid is directly proportional to this shear strain rate or velocity gradient. This is often referred to as Newton’s law of viscosity, and it can be written as

t = m

du dy

(1–10)

The constant of proportionality m (mu) is a physical property of the fluid that measures the resistance to fluid movement.* Although it is sometimes called the absolute or dynamic viscosity, we will refer to it simply as the viscosity. From the equation, m has units of N # s>m2.

*If you studied mechanics of materials you will find a similar equation applies to an elastic solid, where the shear stress t is proportional to the shear strain ∆a using the shear modulus of elasticity, G, i.e., t = G∆a.

1.7

Newtonian Fluids. Any fluid that maintains the same viscosity

37

visCosity

t Crude oil

between the applied shear stress and the velocity gradient, Eq. 1–10, is referred to as a Newtonian fluid. A plot showing how the shear stress and shear strain rate (velocity gradient) behave for some common Newtonian fluids is shown in Fig. 1–12. Notice how the slope (viscosity) increases, from air, which has a very low viscosity, to water, and then to crude oil, which has a much higher viscosity. As previously stated, the higher the viscosity, the more resistant the fluid is to flow.

Mercury Water

mo mHg

Gasoline mg Air ma

Inviscid and Ideal Fluids. Many applications in engineering

involve fluids that have very low viscosities, such as water and air, 1.00 1 10 - 3 2 N # s>m2 and 18.1 1 10 - 6 2 N # s>m2, at 20°C, and so sometimes we can approximate them as inviscid fluids. By definition, an inviscid fluid has zero viscosity, m = 0, and as a result it offers no resistance to shear stress, Fig. 1–13. In other words, it is frictionless. Hence, if the fluid in Fig. 1–11a is inviscid, then when the force F is applied to the plate, it will cause the plate to continue to accelerate, since no shear stress can be developed within an inviscid fluid to offer a restraining frictional resistance to the bottom of the plate. If in addition to being inviscid, the fluid is also assumed to be incompressible, then it is called an ideal fluid.

du dy

The higher the viscosity, the more difficult it is for a fluid to flow

Fig. 1–12

t

Ps eu do -p la

Non-Newtonian Fluids. Fluids whose very thin layers exhibit a nonlinear behavior between the applied shear stress and the shear strain rate are classified as non-Newtonian fluids. There are basically two types, and they behave as shown in Fig. 1–13. For each of these fluids, the slope of the curve for any specific shear strain rate defines the apparent viscosity for that fluid. Those fluids that have an increase in apparent viscosity (slope) with an increase in shear stress are referred to as shearthickening or dilatant fluids. An example would be water with a high concentration of sugar. Many fluids, however, exhibit an opposite behavior and are called shear-thinning or pseudo-plastic fluids. Examples include ketchup, blood, gelatin, and milk. As noted, these substances flow slowly at low applications of shear stress (large slope), but rapidly under a higher shear stress (smaller slope), but rapidly under a higher shear stress (smaller slope). For example, quicksand is pseudoplastic. If anyone falls into it, it is best to move very slowly because of its high apparent viscosity (large slope). Moving fast will loosen the quicksand and pull a person downwards, because of its low apparent viscosity (small slope). Finally, there exist other classes of substances that have both solid and fluid properties. For example, paste and wet cement hold their shape (solid) for small shear stress, but can flow (fluid) when the shear stress is larger. These substances, as well as other unusual solid–fluid substances, are studied in the field of rheology, not in fluid mechanics. See Ref. [8].

1

mw

uid cfl sti id

an

flu

ni to w e m

N

1

Di

lat

a

fl nt

ui

d

Inviscid and ideal fluids

Fig. 1–13

du dy

38

Chapter 1

F u n d a m e n ta l C o n C e p t s

Pressure and Temperature Effects. Through experiment it has

4

2 Mercury

1023 8 6

at er

W

4

ud Cr eo

2

il

Viscosity, (N · sm2 )

1

1024 8 6 4

Air xide bon dio

2 Car

1025

5 210

n

Hydroge

been found that the viscosity of a fluid is actually increased by pressure; however, this effect is quite small, and so it is generally neglected for most engineering applications. Temperature, however, affects the viscosity of fluids to a much greater extent. In the case of a liquid, an increase in temperature will decrease its viscosity, as shown in Fig. 1–14 for mercury, water, and crude oil, Ref. [9]. This occurs because a temperature increase will cause the molecules of the liquid to have more vibration or mobility, thus breaking their molecular bonds and allowing the layers of the liquid to “loosen up” and slip more easily. If the fluid is a gas, an increase in temperature has the opposite effect, that is, the viscosity will increase as noted for air, carbon dioxide, and hydrogen, Ref. [10]. Since gases are composed of molecules that are far apart, their intermolecular attraction to one another is small. When the temperature increases the molecular motion of the gas will increase, and this will increase the momentum exchange between successive layers. It is this additional resistance, developed by molecular collisions, that causes the viscosity to increase. Attempts have been made to use empirical equations to fit experimental curves of viscosity versus temperature for various liquids and gases such as those shown in Fig. 1–14. For liquids, the curves can be represented using Andrade’s equation. m = BeC>T (liquid) And for gases, the Sutherland equation works well.

0 10 40 60 100 Temperature (8C) Temperature increase causes an increase in the viscosity of a gas, but a decrease in the viscosity of a liquid

Fig. 1–14

m =

BT 3>2 (gas) (T + C)

In each of these cases T is the absolute temperature, and the constants B and C can be determined if specific values of m are known for two different temperatures.*

Kinematic Viscosity. Another way to express the viscosity of a fluid is to represent it by its kinematic viscosity, n (nu), which is the ratio of the fluid’s dynamic viscosity to its density: n =

m r

(1–11)

The units are m2 >s.** The word “kinematic” is used to describe this property because force is not involved in the dimensions. Typical values of the dynamic and kinematic viscosities are given in Appendix A for some common liquids and gases, and more extensive listings are also given for water and air. *See Probs. 1–40 and 1–42. **In the standard metric system (not SI), grams and centimeters (100 cm = 1 m) are used. In this case the dynamic viscosity m is expressed using a unit called a poise, where 1 poise = 1 g>(cm # s), and the kinematic viscosity n is measured in stokes, where 1 stoke = 1 cm2 >s.

1.8

1.8

39

visCosity measurement

VISCOSITY MEASUREMENT

The viscosity of a Newtonian liquid can be measured in several ways. One common method is to use a rotational viscometer, sometimes called a Brookfield viscometer. This device, shown in the photo, consists of a solid cylinder that is suspended within a cylindrical container, Fig. 1–15a. The liquid to be tested fills the small space between these two cylinders, and as the container is forced to rotate with a very slow constant angular velocity v, it causes the inner cylinder to twist the suspension wire a small amount before the inner cylinder comes to rest. By measuring the angle of twist of the wire, the torque M in the wire can be calculated using the theory of mechanics of materials. Once this torque is known, we can then find the viscosity of the fluid using Newton’s law of viscosity. To show this, for equilibrium we require M, the torque in the wire, to balance the moment of the resultant shear force the liquid exerts on the inner cylinder’s surface about the axis of the cylinder, Fig. 1–15b.* This gives Fs = M>ri. Since the area of the surface is (2pri)h, the shear stress acting on the surface is

1

Brookfield viscometer

M

ro

Rotating container

ri

M>ri Fs M t = = = A 2prih 2pr i2h

h Fixed cylinder

The angular rotation of the container causes the liquid in contact with its inner wall to have a speed of U = vro, Fig. 1–15c. Since the inner cylinder is held stationary by the twisted wire, then the velocity gradient or slope of the velocity profile across the thickness t of the liquid becomes

v (a) u

vro du = dr t Now, using Newton’s law of viscosity,

vro Fixed cylinder

M

t = m

vro M = m t 2pr i2h

du ; dr

Fs (b)

Solving for m in terms of the measured properties, the viscosity is then m =

Mt 2pv r i2roh

*An extended analysis should include the frictional resistance of the liquid on the cylinder’s bottom surface. See Probs. 1–45 and 1–48.

Rotating container r

t

ri

t

(c)

Fig. 1–15

40

Chapter 1

F u n d a m e n ta l C o n C e p t s

Besides a rotational viscometer, the viscosity of a liquid can also be obtained by using other methods. For example, W. Ostwald invented the Ostwald viscometer shown in the photo. Here the viscosity is determined by measuring the time for a liquid to flow through the short, small-diameter tube between the two bulbs, and then correlating this time with the time for another liquid of known viscosity to flow through this same tube. The unknown viscosity is then determined by direct proportion. If the liquid is transparent and has a high viscosity, such as honey, then its viscosity can be determined by measuring the speed of a small sphere as it falls through the liquid. It will be shown in Sec. 11.8 how this speed can be related to the viscosity of the liquid using Stokes’ equation. In addition, many other devices have been developed to measure viscosity, and the details on how they work can be found in books related to this subject. For example, see Ref. [14].

1

I MPO RTA N T PO I N T S Ostwald viscometer

• The shear resistance of a fluid is measured by its viscosity, m. The higher the viscosity, the greater the resistance to flow caused by shear.

• A Newtonian fluid, such as water, oil, or air, develops shear stress within successive thin layers of the fluid that is directly proportional to the velocity gradient that occurs between the fluid layers, t = m (du>dy).

• A non-Newtonian fluid has an apparent viscosity. If this viscosity increases with an increase in shear stress, then the fluid is a dilatant fluid, and if this viscosity decreases with an increase in shear stress, then it is a pseudo-plastic fluid.

• An inviscid fluid has no viscosity, m = 0, and an ideal fluid is both inviscid and incompressible; that is, m = 0 and r = constant.

• The viscosity will increase only slightly with an increase in pressure; however, for increasing temperature, m will decrease for liquids, but it will increase for gases.

• The kinematic viscosity n is the ratio of the two fluid properties m and r, where n = m>r.

• It is possible to obtain the viscosity of a liquid in an indirect manner by using a rotational viscometer, an Ostwald viscometer, or by several other methods.

1.8

EXAMPLE

41

1.5

The plate in Fig. 1–16 rests on top of the thin film of water, which is at a temperature of 25°C. If a pressure difference occurs between A and B, and a small force F is applied to the plate, the velocity profile across the thickness of the water can be described as u = 1 40y - 800y2 2 m>s, where y is in meters. Determine the shear stress acting on the fixed surface and on the bottom of the plate. y 0.32 ms F

A

10 mm

B u

Fig. 1–16

SOLUTION Fluid Description. Water is a Newtonian fluid, and so Newton’s law of viscosity applies. The viscosity of water at 25°C is found from Appendix A to be m = 0.897 1 10 - 3 2 N # s>m2.

Analysis. Before applying Newton’s law of viscosity, we must first obtain the velocity gradient. du d = 1 40y - 800y2 2 m>s = (40 - 1600y) s - 1 dy dy Therefore, at the fixed surface, y = 0, du t = m ` = 3 0.897 1 10-3 2 N # s>m2 4 140 - 02 s - 1 dy y = 0 t = 35.88 1 10-3 2 N>m2 = 35.9 mPa And, at the bottom of the moving plate, y = 0.01 m, t = m

visCosity measurement

Ans.

du ` = 30.897 1 10-3 2 N # s>m2 4[40 - 1600(0.01)] s - 1 dy y = 0.01 m

t = 21.5 mPa Ans. By comparison, the larger shear stress develops on the fixed surface rather than on the bottom of the plate, since the velocity gradient or increase in velocity, du, with height dy is larger at the fixed surface. Both of these slopes, du>dy, are indicated by the short dark lines in Fig. 1–16. Also, notice that the equation for the velocity profile satisfies the boundary condition of no slipping, i.e., at the fixed surface y = 0, u = 0, and with the movement of the plate at y = 10 mm, u = U = 0.32 m>s.

1

42

Chapter 1

EXAMPLE

F u n d a m e n ta l C o n C e p t s

1.6 P

0.2 ms

308

1 (a)

The 100-kg plate in Fig. 1–17a is resting on a very thin film of SAE 10W-30 oil, which has a viscosity of m = 0.0652 N # s>m2. Determine the force P that must be applied to the center of the plate to slide it over the oil with a constant velocity of 0.2 m>s. Assume the oil thickness is 0.1 mm, and the velocity profile across this thickness is linear. The bottom of the plate has a contact area of 0.75 m2 with the oil. SOLUTION Fluid Description. The oil is a Newtonian fluid, and so Newton’s law of viscosity can be applied.

100(9.81) N P 308

Analysis. First we draw the free-body diagram of the plate in order to relate the shear force F caused by the oil on the bottom of the plate to the applied force P, Fig. 1–17b. Because the plate moves with constant velocity, the force equation of equilibrium in the horizontal direction applies. + S ΣFx

F

F - P cos 30° = 0

= 0;

F = 0.8660P

N

The effect of this force on the oil is in the opposite direction, and so the shear stress on the top of the oil acts to the left. It is

(b)

t =

F 0.8660P = (1.155P) m-2 = A 0.75 m2

Since the velocity profile is linear, Fig. 1–17c, the velocity gradient is constant, du>dy = U>t, and so U 5 0.2 ms

t 5 130 Pa

t 5 0.1 mm

du

dy

t = m (1.155P ) m-2 =

du U = m dy t

1 0.0652 N # s>m2 2 c

P = 113 N

0.2 m>s 0.1(10 - 3) m

d Ans.

(c)

Fig. 1–17

Notice that the constant velocity gradient will produce a constant shear-stress distribution across the thickness of the oil, which is t = m(U>t) = 130 Pa, Fig. 1–17c.

1.9

1.9

vapor pressure

43

VAPOR PRESSURE

When a liquid is contained within a closed tank, as in Fig. 1–18, its temperature will cause continuous thermal agitation of its molecules. As a result, some of the molecules near the surface will acquire enough kinetic energy to break their molecular bonds between adjacent molecules, and will leave the surface or evaporate into the vacuum space of the tank. Eventually a state of equilibrium will be reached, such that the number of molecules that evaporate from the liquid will equal the number of molecules that condense back to it. The vacuum space is then said to be saturated. By bouncing off the walls of the tank and the liquid surface, the evaporated molecules create a pressure within the tank. This pressure is called the vapor pressure, pv . Any increase in liquid temperature will increase the rate of evaporation, and also the kinetic energy of the vapor’s molecules, so higher temperatures will cause higher vapor pressures. Knowing the vapor pressure is important because the liquid will begin to boil when the absolute pressure at its surface is at or lower than its vapor pressure. For example, if water at sea level is brought to a temperature of 100°C, then at this temperature its vapor pressure will equal the atmospheric pressure, which is 101.3 kPa, and so the water will begin to evaporate quickly and boil. Specific values of the vapor pressure for water at various temperatures are given in Appendix A. Notice that as the temperature increases, so does the vapor pressure due to the increase in the thermal agitation of its molecules.

pv

The vapor pressure pv will form within the top space of the closed tank that was originally a vacuum

Fig. 1–18

Cavitation. When engineers design pumps, turbines, or piping systems, it is important that they do not allow the liquid at any point within the flow to be subjected to a pressure equal to or less than its vapor pressure. As stated above, if this occurs rapid evaporation or boiling will occur within the liquid. The resulting bubbles will migrate to regions of higher pressure, and then suddenly collapse, creating a phenomenon known as cavitation. Essentially, a very high-speed jet of liquid enters a collapsing bubble, Fig. 1–19, and the result produces an extremely large localized pressure and temperature, along with a shock wave that strikes the surface. Repeated interactions caused by this effect result in pitting of the surface and can eventually wear it down, causing damage to the surface of propeller blades or pump casings, or at the crest of a spillway or dam. Later, in Chapter 14, we will study the significance of cavitation in greater detail, and show how it can be avoided when designing turbo-machinery.

Cavitation is the result of bubble implosion near a surface

Fig. 1–19

1

44

Chapter 1

F u n d a m e n ta l C o n C e p t s

Resultant cohesive force is vertically downward

1

Resultant cohesive force is zero (a)

s s

Surface tension is the force per unit length needed to separate the molecules on the surface (b)

Fig. 1–20

1.10

SURFACE TENSION AND CAPILLARITY

A liquid maintains its form because its molecules are attracted to one another by a force known as cohesion. It is this force that enables liquids to resist surface tension. Liquid molecules can also be attracted to those of a different substance, such as those that make up the liquid’s container. This force of attraction is known as adhesion. Both cohesion and adhesion play an important role in studying the effects of a liquid passing through small spaces and thereby producing the effects of surface tension and capillarity.

Surface Tension. The phenomenon of surface tension can be explained by visualizing the cohesive forces acting on two molecules (or particles) in a liquid, shown in Fig. 1–20a. The molecule located deep within the liquid has the same cohesive forces acting on it due to its attraction to all the surrounding molecules. Consequently, there is no resultant cohesive force acting on it. However, the molecule located on the surface of the liquid has cohesive forces that attract the molecules that are next to it on the surface and also from those below it. This will produce a net resultant cohesive downward force, and the effect of all such forces will tend to produce a contraction of the surface. The attractive force towards the surface molecules is similar to the same effect that occurs when molecules tend to restore a stretched elastic film, and so we refer to this effect as a surface tension, s (sigma), which is measured as a force per unit length in any direction along the surface, Fig. 1–20b.* Surface tension has units of N >m, and for any liquid, its value depends primarily upon the temperature. The higher the temperature, the more thermal agitation occurs, and so the surface tension becomes smaller. For example, for water at 10°C, s = 74.2 mN >m, whereas at the higher temperature of 50°C, s = 67.9 mN >m. Values of s, such as these, are very sensitive to impurities, so care should be taken when using published values. Typical values of s for some common liquids are given in Appendix A.

Like rain drops, the water ejected from this fountain forms spherical droplets due to the cohesive force of surface tension. *Surface tension can also be thought of as the amount of free-surface energy or work per increase in the surface area of the liquid. See Prob. 1–63.

1.10

R p

surFaCe tension and Capillarity

45

pa

1

s

Fig. 1–21

Liquid Drops. Surface tension is responsible for the formation of liquid droplets that naturally form when a liquid is sprayed into the atmosphere. This force minimizes the surface of any water droplet, and so it forms a sphere. We can determine the pressure that surface tension causes within a droplet by considering the free-body diagram of half the drop as shown in Fig. 1–21. If we neglect gravity and the effects of atmospheric drag as the droplet falls, then the only forces acting are those due to atmospheric pressure, pa, on its outside surface; surface tension, s, around the surface of the drop where it is sectioned; and the internal pressure, p, on the sectioned area. As will be explained in the next chapter, the resultant horizontal forces due to pa and p are determined by multiplying each pressure by the projected area of the drop, that is, pR2, and the resultant force of the surface tension is determined by multiplying s by the circumferential distance around the drop, 2pR. For horizontal equilibrium, we therefore have + S ΣFx

= 0;

p 1 pR2 2 - pa 1 pR2 2 - s(2pR) = 0

2s + pa R For example, mercury at a temperature of 20°C has a surface tension of s = 486 mN>m. If the mercury forms into a 2-mmdiameter droplet, its surface tension will create an internal pressure of pst = 2(0.486 N>m)>(0.001 m) = 972 Pa within the droplet, in addition to the pressure pa caused by the atmosphere. p =

Mercury is a nonwetting liquid, as shown by the way its edge curls inward.

46

Chapter 1

F u n d a m e n ta l C o n C e p t s

Capillarity. The capillarity of a liquid depends upon the comparison

1

Wetting liquid

Nonwetting liquid

(a)

(b)

Fig. 1–22

between the forces of adhesion and cohesion. If the force of a liquid’s adhesion to the molecules on the surface of its container is greater than the force of cohesion between the liquid’s molecules, then the liquid is referred to as a wetting liquid. In this case, the meniscus or surface of the liquid, such as water in a narrow glass container, will be concave, Fig. 1–22a. If the adhesive force is less than the cohesive force, as in the case of mercury, then the liquid is called a nonwetting liquid. The meniscus forms a convex surface, Fig. 1–22b. Wetting liquids will rise up along a narrow tube, Fig. 1–23a, and we can determine this height h by considering a free-body diagram of the portion of the liquid suspended in the tube, Fig. 1–23b. Here the free surface or meniscus makes a contact angle u with the sides of the tube. This angle defines the direction of the surface tension s of the liquid as it holds the liquid surface up against the wall of the tube. The resultant force is vertical, and acts around the inner circumference of the tube. It is therefore s(2pr) cos u. The other force is the weight of the suspended liquid, W = rg V, where V = pr 2h. For equilibrium, we require + c ΣFy = 0;

s(2pr) cos u - rg1pr 2h2 = 0

2s cos u rgr Experiments have shown that the contact angle between water and glass is u ≈ 0°, and so for water the meniscus surface, shown in Fig. 1–23a, actually becomes somewhat hemispherical. By carefully measuring h, the above equation can then be used with u = 0° to determine the surface tension s for water at various temperatures. In the next chapter we will show how to determine pressure within an open or closed conduit by measuring the height of a liquid contained within a small glass tube, placed within the conduit. When it is used for this purpose, however, errors due to the additional height h caused by capillarity within the tube can occur. To minimize this effect, notice that h in the above result is inversely proportional to the density of the liquid and the radius of the tube. The larger they are, the smaller h becomes. For example, for a 3-mm-diameter tube containing water at 20°C, where s = 72.7 mN>m and r = 998.3 kg>m3, we have h =

2r

u

u

h

(a)

s

s

s s

s

s u

u

h

W (b)

Fig. 1–23

h =

1 998.3 kg>m3 2 1 9.81 m>s2 2 (0.0015 m) 2(0.0727 N>m) cos 0°

= 9.90 mm

This is somewhat significant, and so for experimental work it is generally preferable to use tubes having a diameter of 10 mm or greater, since at 10 mm, h ≈ 3 mm, and the effect of capillarity is minimized. Throughout our study of fluid mechanics, we will find that, for the most part, the forces of cohesion and adhesion will be small compared to the effects of gravity, pressure, and viscosity. Surface tension generally becomes important, however, when we want to study phenomena related to bubble formation and growth, design nozzles for producing mist or fine sprays, examine the movement of liquids through porous media such as soil, or consider the effects of a liquid film applied to a surface.

problems

47

IMPORTANT POIN T S • A liquid will begin to boil at a specific temperature when the pressure within it, or at its surface, is equal to its vapor pressure at that temperature.

• Consideration must be given to the possibility of cavitation when designing mechanical or structural elements operating within a fluid environment. This phenomenon will occur when the pressure within the liquid is equal to or less than the vapor pressure, causing boiling, migration of the resulting bubbles to a region of higher pressure, and then their sudden collapse.

• Surface tension in a liquid is caused by molecular cohesion. It is measured as a force per unit length acting on the liquid’s surface. It becomes smaller as the temperature rises.

• Capillarity of a wetting liquid, such as water in a narrow glass tube, creates a concave surface since the force of adhesion to the walls of the tube will be greater than the force caused by the cohesion of the liquid. For a nonwetting liquid, such as mercury, the surface is convex since the force of cohesion will be greater than that of adhesion.

References 1. G. A. Tokaty, A History and Philosophy of Fluid Mechanics, Dover

Publications, New York, NY, 1994. 2. R. Rouse and S. Ince, History of Hydraulics, Iowa Institute of

Hydraulic Research, Iowa City, IA, 1957. 3. Handbook of Chemistry and Physics, 62nd ed., Chemical Rubber

Publishing Co., Cleveland, OH, 1988. 4. Handbook of Tables for Applied Engineering Science, Chemical

Rubber Publishing Co., Cleveland, OH, 1970. 5. V. L. Streeter, and E. Wylie, Fluid Mechanics, 8th ed., McGraw-Hill,

New York, N.Y., 1985. 6. D. Blevins, Applied Fluid Dynamics Handbook, Van Nostrand

Reinhold, New York, NY, 1984. 7. P. R. Lide, W. M. Haynes, eds., Handbook of Chemistry and Physics,

90th ed., CRC Press, Boca Raton, FL.

P R OBLEMS The answers to all but every fourth problem are given in the back of the book.

SEC. 1.1–1.6 1–1. Evaluate each of the following to three significant figures, and express each answer in SI units using an appropriate prefix: (a) 749 mm>63 ms, (b) (34 mm)(0.0763 Ms)>263 mg, (c) (4.78 mm)(263 Mg).

1–2. Represent each of the following quantities with combinations of units in the correct SI form, using an appropriate prefix: (a) mm # MN, (b) Mg>mm, (c) km>ms, (d) kN>(mm)2.

1

48

Chapter 1

F u n d a m e n ta l C o n C e p t s

1–3. Evaluate each of the following to three significant figures, and express each answer in SI units using an appropriate prefix: (a) 34.86(106)4 2 mm, (b) (348 mm)3, (c) (83 700 mN)2.

*1–8. The tank contains air at a temperature of 18°C and an absolute pressure of 160 kPa. If the volume of the tank is 3.48 m3 and the temperature rises to 42°C, determine the mass of air that must be removed from the tank to maintain the same pressure.

1 1–5. The container is filled with water at a temperature of 25°C and a depth of 2.5 m. If the container has a mass of 30 kg, determine the combined weight of the container and the water.

1–9. The tank contains 4 kg of air at an absolute pressure of 350 kPa and a temperature of 18°C. If 0.8 kg of air is added to the tank and the temperature rises to 38°C, determine the resulting pressure in the tank.

*1–4. Convert the following temperatures: (a) 250 K to degrees Celsius, (b) 43°C to kelvin.

1m

2.5 m

Prob. 1–5 1–6. Determine the change in the density of oxygen when the absolute pressure changes from 345 kPa to 286 kPa, while the temperature remains constant at 25°C. This is called an isothermal process. 1–7. The 8-m-diameter spherical balloon is filled with helium that is at a temperature of 28°C and an absolute pressure of 106 kPa. Determine the weight of the helium contained in the balloon. The volume of a sphere is V = 34pr 3.

Prob. 1–7

Probs. 1–8/9 1–10. Determine the mass of carbon tetrachloride that should be mixed with 80 kg of glycerin so that the combined mixture has a density of 1450 kg>m3. 1–11. Gasoline is mixed with 3 m3 of ethyl alcohol so that the volume of the mixture in the tank becomes 5.5 m3. Determine the density and the specific gravity of the mixture at standard temperature and pressure.

Prob. 1–11

49

problems *1–12. The tank contains a liquid having a density of 1.3 Mg>m3. Determine the weight of the liquid when it is at the level shown.

1–15. Water in the swimming pool has a measured depth of 3.03 m when the temperature is 5°C. Determine its approximate depth when the temperature becomes 35°C. Neglect losses due to evaporation.

0.25 m 1.5 m

9m

3m

4m

2m

Prob. 1–12 1–13. The bottle tank contains nitrogen having a temperature of 60°C. Plot the variation of the pressure in the tank (vertical axis) versus the density for 0 … r … 5 kg>m3. Report values in increments of ∆p = 50 kPa.

1

Prob. 1–15 *1–16. The tanker carries 858(103) barrels of crude oil in its hold. Determine the weight of the oil if its specific gravity is 0.940. Each barrel contains 159 liters.

Prob. 1–16 1–17. Dry air at 25°C has a density of 1.23 kg>m3. But if it has 100% humidity at the same pressure, its density is 0.65% less. At what temperature would dry air produce this same smaller density? 1–18. Determine the specific weight of hydrogen when the temperature is 85°C and the absolute pressure is 4 MPa. Prob. 1–13 1–14. If air within the tank is at an absolute pressure of 680 kPa and a temperature of 70°C, determine the weight of the air inside the tank. The tank has an interior volume of 1.35 m3.

Prob. 1–14

1–19. The bottle tank has a volume of 0.35 m3 and contains 40 kg of nitrogen at a temperature of 40°C. Determine the absolute pressure in the tank.

Prob. 1–19

50

Chapter 1

F u n d a m e n ta l C o n C e p t s

*1–20. A volume of 8 m3 of oxygen initially at 80 kPa of absolute pressure and 15°C is subjected to an absolute pressure of 25 kPa while the temperature remains constant. Determine the new density and volume of the oxygen.

1–27. The viscosity of SAE 10 W30 oil is m = 0.100 N # s>m2. Determine its kinematic viscosity. The specific gravity is So = 0.92.

1–21. The rain cloud has an approximate volume of 3 1 30 km and an average height, top to bottom, of 125 m. If a cuboid container collects 58 mm of water after the rain falls out of the cloud, estimate the total mass of rain that fell from the cloud.

*1–28. If the kinematic viscosity of glycerin is n = 1.15(10-3) m2 >s, determine its viscosity. At the temperature considered, glycerin has a specific gravity of Sg = 1.26.

125 m

SEC. 1.7–1.8

1–29. If a force of P = 2 N causes the 30-mm-diameter shaft to slide along the lubricated bearing with a constant speed of 0.5 m>s, determine the viscosity of the lubricant and the constant speed of the shaft when P = 8 N. Assume the lubricant is a Newtonian fluid and the velocity profile between the shaft and the bearing is linear. The gap between the bearing and the shaft is 1 mm.

50 mm

0.5 ms 1.5 m P

1.5 m

Prob. 1–21 Prob. 1–29 1–22. A 2-kg mass of oxygen is held at a constant temperature of 50°C and an absolute pressure of 220 kPa. Determine its bulk modulus. 1–23. At a point deep in the ocean, the density of seawater is 1035 kg/m3. Determine the absolute pressure at this point if at the surface the density is 1035 kg/m3 and the absolute pressure is pa = 101.3 kPa. Take EV = 2.42 GPa.

1–30. The Newtonian fluid is confined between the plate and a fixed surface. If its velocity profile is defined by u = (8y - 0.3y2) mm>s, where y is in mm, determine the force P that must be applied to the plate to cause this motion. The plate has a surface area of 15(103) mm2 in contact with the fluid. Take m = 0.482 N # s>m2.

*1–24. Water at 20°C is subjected to a pressure increase of 44 MPa. Determine the percent increase in its density. Take EV = 2.20 GPa. 1–25. If the change in pressure required to reduce the volume of water by 0.3% is 6.60 MPa, determine the bulk modulus of the water. 1–26. The bulk modulus and density of a liquid are EV = 1.44 GN/m2 and r = 880 kg/m3 respectively. Determine the increase in its density if it is subjected to a pressure increase of 900 kPa.

P 4 mm

u y

Prob. 1–30

51

problems 1–31. The Newtonian fluid is confined between a plate and a fixed surface. If its velocity profile is defined by u = (8y - 0.3y2) mm>s, where y is in mm, determine the shear stress that the fluid exerts on the plate and on the fixed surface. Take m = 0.482 N # s>m2.

1–34. When the force P is applied to the plate, the velocity profile for a Newtonian fluid that is confined under the plate is approximated by u = (4.23y1>3) mm>s, where y is in mm. Determine the minimum shear stress within the fluid. Take m = 0.630(10-3) N # s>m2. P

P

10 mm y

4 mm

u

u

Prob. 1–34 1–35. When the force P is applied to the plate, the velocity profile for a Newtonian fluid that is confined under the plate is approximated by u = (4.23y1>3) mm>s, where y is in mm. Determine the shear stress within the fluid at y = 5 mm. Take m = 0.630(10-3) N # s>m2.

y

. Prob. 1–31

*1–32. The plate is moving at 0.6 mm>s when the force applied to the plate is 4 mN. If the surface area of the plate in contact with the liquid is 0.5 m2, determine the approximate viscosity of the liquid, assuming that the velocity distribution is linear.

0.6 mms 4 mN

4 mm

Prob. 1–32

1–33. An experimental test using human blood at T = 30°C indicates that it exerts a shear stress of t = 0.15 N>m2 on surface A, where the measured velocity gradient is 16.8 s -1. Since blood is a non-Newtonian fluid, determine its apparent viscosity at A.

P 10 mm y

u

Prob. 1–35 *1–36. The tank containing gasoline has a long crack on its side that has an average opening of 10 μm. The velocity through the crack is approximated by the equation u = 10 1 109 2 3 10 1 10-6 2 y - y2 4 m>s, where y is in meters, measured upward from the bottom of the crack. Find the shear stress at the bottom, y = 0, and the location y within the crack where the shear stress in the gasoline is zero. Take mg = 0.317 1 10-3 2 N # s>m2.

1–37. The tank containing gasoline has a long crack on its side that has an average opening of 10 μm. If the velocity profile through the crack is approximated by the equation u = 10 1 109 2 3 10 1 10-6 2 y - y2 4 m>s, where y is in meters, plot both the velocity profile and the shear stress distribution for the gasoline as it flows through the crack. Take mg = 0.317 1 10-3 2 N # s>m2. 10 mm

A

Prob. 1–33

Prob. 1–36/37

1

52

Chapter 1

F u n d a m e n ta l C o n C e p t s

1–38. A plastic strip having a width of 0.2 m and a mass of 150 g passes between two layers A and B of paint having a viscosity of 5.24 N # s>m2. Determine the force P required to overcome the viscous friction on each side if the strip moves upwards at a constant speed of 4 mm>s. Neglect any friction at the top and bottom openings, and assume the velocity 1 profile through each layer is linear. 1–39. A plastic strip having a width of 0.2 m and a mass of 150 g passes between two layers A and B of paint. If force P = 2 N is applied to the strip, causing it to move at a constant speed of 6 mm>s, determine the viscosity of the paint. Neglect any friction at the top and bottom openings, and assume the velocity profile through each layer is linear.

1–43. The viscosity of water can be determined using the empirical Andrade’s equation with the constants B = 1.732(10-6) N # s>m2 and C = 1863 K. With these constants, compare the results of using this equation with those tabulated in Appendix A for temperatures of T = 10°C and T = 80°C. *1–44. Determine the torque T required to rotate the disk with a constant angular velocity of v = 30 rad>s as a function of the oil thickness t. Plot your results of torque (vertical axis) versus the oil thickness for 0 … t … 0.15(10-3) m for values every 0.03 (10-3) m. Assume the velocity profile is linear, and m = 0.428 N # s>m2.

P 8 mm

6 mm

150 mm A

T

B

0.30 m

Prob. 1–44

Probs. 1–38/39 *1–40. Determine the constants B and C in Andrade’s equation for water if it has been experimentally determined that m = 1.00(10-3) N # s>m2 at a temperature of 20°C and that m = 0.554(10-3) N # s>m2 at 50°C. 1–41. The constants B = 1.357(10-6) N # s>(m2 # K1>2) and C = 78.84 K have been used in the empirical Sutherland equation to determine the viscosity of air at standard atmospheric pressure. With these constants, compare the results of using this equation with those tabulated in Appendix A for temperatures of T = 10°C and T = 80°C.

1–42. Determine the constants B and C in the Sutherland equation for air if it has been experimentally determined that at standard atmospheric pressure and a temperature of 20°C, m = 18.3(10-6) N # s>m2, and at 50°C, m = 19.6(10-6) N # s>m2.

1–45. Determine the torque T required to rotate the disk with a constant angular velocity of v = 30 rad>s. The oil has a thickness of 0.15 mm. Assume the velocity profile is linear, and m = 0.428 N # s>m2.

150 mm

T

Prob. 1–45

53

problems 1–46. The read–write head for a hand-held music player has a surface area of 0.04 mm2. The head is held 0.04 μm above the disk, which is rotating at a constant rate of 1800 rpm. Determine the torque T that must be applied to the disk to overcome the frictional shear resistance of the air between the head and the disk. The surrounding air is at standard atmospheric pressure and a temperature of 20°C. Assume the velocity profile is linear.

*1–48. The tube rests on a 1.5-mm thin film of oil having a viscosity of m = 0.0586 N # s>m2. If the tube is rotating at a constant angular velocity of v = 4.5 rad>s, determine the torque T that must be applied to the tube to maintain the motion. Assume the velocity profile within the oil is linear. 1 v 5 4.5 rad s 40 mm T

80 mm 8 mm

T

Prob. 1–46

1–47. The tube rests on a 1.5-mm thin film of oil having a viscosity of m = 0.0586 N # s>m2. If the tube is rotating at a constant angular velocity of v = 4.5 rad>s, determine the shear stress in the oil at r = 40 mm and r = 80 mm. Assume the velocity profile within the oil is linear.

v 5 4.5 rad s 40 mm

Prob. 1–48 1–49. The very thin tube A of mean radius r and length L is placed within the fixed circular cavity as shown. If the cavity has a small gap of thickness t on each side of the tube, and is filled with a Newtonian liquid having a viscosity m, determine the torque T required to overcome the fluid resistance and rotate the tube with a constant angular velocity of v. Assume the velocity profile within the liquid is linear.

T T t

t

A

r

80 mm

L

Prob. 1–47

Prob. 1–49

54

Chapter 1

F u n d a m e n ta l C o n C e p t s

1–50. The conical bearing is placed in a lubricating Newtonian fluid having a viscosity m. Determine the torque T required to rotate the bearing with a constant angular velocity of v. Assume the velocity profile along the thickness t of the fluid is linear.

1–55. A boat propeller is rotating in water that has a temperature of 15°C. What is the lowest absolute water pressure that can be developed at the blades so that cavitation will not occur? *1–56. As water at 20°C flows through the transition, its pressure will begin to decrease. Determine the lowest absolute pressure it can have without causing cavitation.

v

1 T R

u

t

Prob. 1–56 Prob. 1–50

SEC. 1.9–1.10 1–51. How hot can you boil the water to make a cup of tea if you climb to the peak of Mt. Everest (8848 m)? *1–52. The city of Denver, Colorado is at an elevation of 1610 m above sea level. Determine how hot one can prepare boiling water to make a cup of tea. 1–53. Water at 28.5°C is flowing through a garden hose. If the hose is bent, a hissing noise can be heard. Here cavitation has occurred in the hose because the velocity of the flow has increased at the bend, and the pressure has dropped. What would be the highest absolute pressure in the hose at this location?

1–57. The triangular glass rod has a weight of 0.3 N and is suspended on the surface of the water, for which s = 0.0728 N>m. Determine the vertical force P needed to pull the rod free from the surface. 1–58. The triangular glass rod has a weight of 0.3 N and is suspended on the surface of the water. If it takes a force of P = 0.335 N to lift it free from the surface, determine the surface tension of the water.

1–54. Water at 25°C is flowing through a garden hose. If the hose is bent, a hissing noise can be heard. Here cavitation has occurred in the hose because the velocity of the flow has increased at the bend, and the pressure has dropped. What would be the highest absolute pressure in the hose at this location?

P

608

608 608

Probs. 1–53/54

Probs. 1–57/58

80 mm

55

problems 1–59. For water falling out of the tube, there is a difference in pressure ∆p between a point located just inside and a point just outside of the stream due to the effect of surface tension s. Determine the diameter d of the stream at this location.

1–62. The marine water strider, Halobates, has a mass of 0.36 g. If it has six slender legs, determine the minimum contact length of all of its legs combined to support itself in water having a temperature of T = 20°C. Take s = 72.7 mN>m, and assume the legs to be thin cylinders with a contact angle of u = 0°.

d

Prob. 1–59

Prob. 1–62

*1–60. The glass tube has an inner diameter d and is immersed in water at an angle u from the horizontal. Determine the average length L to which water will rise along the tube due to capillary action. The surface tension of the water is s and its density is r. 1–61. The glass tube has an inner diameter of d = 2 mm and is immersed in water. Determine the average length L to which the water will rise along the tube due to capillary action as a function of the angle of tilt, u. Plot this relationship of L (vertical axis) versus u for 10° … u … 30°. Give values for increments of ∆u = 5°. The surface tension of the water is s = 75.4 mN>m, and its density is r = 1000 kg>m3.

1–63. Because cohesion resists any increase in the surface area of a liquid, it actually tries to minimize the size of the surface. Separating the molecules and thus breaking the surface tension requires work, and the energy provided by this work is called free-surface energy. To show how it is related to surface tension, consider the small element of the liquid surface subjected to the surface tension force F. If the surface stretches dx, show that the work done by F per increase in area is equal to the surface tension in the liquid.

d

L u

Dy

Dx

F dx

Probs. 1–60/61

Prob. 1–63

1

56

Chapter 1

F u n d a m e n ta l C o n C e p t s

*1–64. Determine the distance h that the column of mercury in the tube will be depressed when the tube is inserted into the mercury at a room temperature of 20°C. Set D 5 5.5 mm. 1–65. Determine the distance h that the column of mercury in 1 the tube will be depressed when the tube is inserted into the mercury at a room temperature of 20°C. Plot this relationship of h (vertical axis) versus D for 1  mm … D … 6 mm. Give values for increments of ∆D = 1 mm. Discuss this result.

1–67. Sand grains fall a short distance into a tank of benzene. Determine the largest average diameter of a grain that will float on the benzene with a contact angle of 180°. Take rs = 2650 kg/m3 and s = 28.9 MN/m Assume each grain has the shape of a sphere, where V = 43 pr 3.

D

h

Prob. 1–67

40°

Prob. 1–64/65

* 1–68. Water is at a temperature of 30°C. Plot the height h of the water as a function of the gap w between the two glass plates for 0.4 mm … w … 2.4 mm. Use increments of 0.4 mm. Take s = 0.0718 N>m. w

1–66. Steel particles are ejected from a grinder and fall gently into a tank of kerosene. Determine the largest average diameter of a particle that will float on the kerosene with a contact angle of u = 180°. Take rst = 7850 kg/m3 and s = 26.8 MN>m. Assume that each particle has the shape of a sphere, where V = 34 pr 3.

h D

Prob. 1–68

ConCeptual problems

57

CONCEPTUAL PROBLEMS P1–1. The air pressure for the bicycle tire is 220 kPa. Assuming the volume of air in the tire remains constant, find the pressure difference that occurs between a typical summer and winter day. Discuss the forces that are responsible for supporting the tire and the bicycle when a rider is on the bicycle.

P1–3. If a drop of oil is placed on a water surface, then the oil will tend to spread out over the surface as shown. 1 Explain why this happens.

P1–2. Water poured from this pitcher tends to cling to its bottom side. Explain why this happens and suggest what can be done to prevent it.

P1–4. The shape of this municipal water tank has the same shape as a drop of water resting on a nonabsorbent surface, such as wax paper. Explain why engineers design the tank in this manner.

58

Chapter 1

F u n d a m e n ta l C o n C e p t s

CHAP T ER R EV IEW

1

Matter can be classified as a solid, which maintains its shape; a liquid, which takes the shape of its container; or a gas, which fills its entire container. SI units measure length in meters, time in seconds, mass in kilograms, force in newtons, and temperature in degrees Celsius or in kelvins.

Density is a measure of mass per unit volume.

r =

m V

Specific weight is a measure of weight per unit volume.

g =

W V

Specific gravity is the ratio of either a liquid’s density or its specific weight to that of water, where rw = 1000 kg>m3.

S =

r g = rw gw

The ideal gas law relates the absolute pressure of the gas to its density and absolute temperature.

p = rRT

The bulk modulus is a measure of a fluid’s resistance to compression.

EV = -

dp dV>V

t Crude oil Mercury

The viscosity of a fluid is a measure of its resistance to shear. The higher the viscosity, the higher the resistance.

Water

mo mHg

Gasoline

mw mg Air ma

du dy

Chapter review

For Newtonian fluids, the shear stress in a fluid is directly proportional to the time rate of change in its shear strain, which is defined by the velocity gradient, du>dy. The constant of proportionality, m, is called the dynamic viscosity or simply the viscosity. The kinematic viscosity is a ratio of the fluid’s viscosity to its density.

t = m

du dy 1

n =

m r

The viscosity is measured indirectly by using a rotational viscometer, an Ostwald viscometer, or other device.

pv

A liquid will boil if the pressure above or within it is equal to or below its vapor pressure. This can lead to cavitation, which produces bubbles that can migrate to regions of higher pressure and then suddenly collapse.

s

Surface tension on a liquid surface develops because of the cohesive (attractive) forces between its molecules. It is measured as a force per unit length.

The capillarity of a liquid depends on the comparative forces of adhesion and cohesion. Wetting liquids have a greater force of adhesion to their contacting surface than the liquid’s cohesive force. The opposite effect occurs for nonwetting liquids, where the cohesive force is greater than the adhesive force.

59

s

2

Arco Images GmbH/Alamy Stock Photo

CHAPTER

These gates have been designed to resist the hydrostatic loadings within this canal. Along with the other set, they form a “lock,“ controlling the level of water within the confined region, thus allowing boats to pass from one elevation to another.

FLUID STATICS

CHAPTER OBJECTIVES ■

To discuss pressure and show how it varies within a static fluid.

■

To present the various ways of measuring pressure.

■

To show how to calculate a resultant hydrostatic force and find its location on a submerged surface.

■

To present the topics of buoyancy and stability.

■

To show how to calculate the pressure within a liquid subjected to a constant acceleration, and to a steady rotation about a fixed axis.

2.1

PRESSURE

In general, fluids can exert both normal and shear forces on the surfaces of contact. However, if the fluid is at rest relative to the surface, then the viscosity of the fluid will not exert a shear force on the surface. Instead, only a normal force acts on the surface, and the intensity of this force is called pressure. It is the result of the impulses exerted by the fluid molecules as they collide with and then bounce off the surface. Pressure is defined as the force acting normal to an area divided by this area. If we assume the fluid to be a continuum, then at a point within the fluid the area can approach zero, Fig. 2–1a, and so the pressure becomes ∆F dF = (2–1) ∆A dA If the surface has a finite area and the pressure is uniformly distributed over this area, Fig. 2–1b, then the average pressure is p = lim S ∆A

DF

p DA (a) F

pavg

0

pavg =

F A

Pressure can have the unit of pascals Pa (N>m2).

(2–2)

A Average pressure (b)

Fig. 2–1

61

62

Chapter 2

F l u i d S ta t i C S

Ds

Dz 5 Ds sin u

u

Dx

Dy 5 Ds cos u

2

(a)

z

1 Dx Dy Dz) g (–– 2 p (Dx Ds) u

y x

Pascal’s Law. In the 17th century the French mathematician Blaise Pascal was able to show that the intensity of the pressure acting at a point in a fluid is the same in all directions. This statement is commonly known as Pascal’s law. To prove it, consider the equilibrium of a small triangular element located within the fluid that has the dimensions shown in Fig. 2–2a. Provided the fluid is at rest (or moving with constant velocity), the only forces acting on its free-body diagram are due to pressure and gravity, Fig. 2–2b. The force of gravity is the product of the specific weight g of the fluid and the volume of the element, 1 12 ∆y∆z 2 ∆x. The force created by the pressure is determined by multiplying the pressure by the area of the element face upon which it acts. In the y—z plane there are three pressure forces. Assuming the inclined face has a length ∆s, then the dimensions of the other faces are ∆y = ∆s cos u and ∆z = ∆s sin u, Fig. 2–2a, and so applying the force equations of equilibrium in the y and z directions, we have

py (Dx Dz)

Dz

ΣFy = 0;

u

Dx

Dy pz (Dx Dy) Free-body diagram (b)

Fig. 2–2

ΣFz = 0;

py(∆x)(∆s sin u) - 3p(∆x∆s)4 sin u = 0 pz(∆x)(∆s cos u) - 3p(∆x∆s)4 cos u

1 - g c ∆x(∆s cos u)(∆s sin u) d = 0 2

Dividing by ∆x∆s and letting ∆s S 0, we obtain py = p pz = p By a similar argument, the element can be rotated 90° about the z axis, and ΣFx = 0 can be applied to show that px = p. Since the angle u of the inclined face is arbitrary, this indeed shows that the pressure at a point is the same in all directions for any fluid that has no relative motion between its adjacent layers.* Actually, Pascal’s law seems intuitive, since if the pressure at the point were larger in one direction than in the opposite direction, the imbalance would cause movement or agitation of the fluid, something that is not observed. Since the pressure at a point is transmitted to each of its neighboring points by action, equal but opposite reaction, then it follows from Pascal’s law that any change in pressure ∆p at one point in the fluid will cause the same change at all the other points within the fluid. This principle has widespread application for the design of hydraulic machinery, as noted in the following example. *It is also possible to show that Pascal’s law applies even if the fluid is accelerating. See Prob. 2–1.

2.1

EXAMPLE

15 mm 280 mm

B

A

Fig. 2–3

SOLUTION Fluid Description.

The weight of the air can be neglected.

Analysis. Due to equilibrium, the force created by air pressure at A is equal and opposite to the weight of the car and lift. The average pressure at A is therefore FA ; AA

25(103)N p(0.140 m)2

= 406.00(103) Pa

Since the weight of the air is neglected, the pressure at each point is the same in all directions (Pascal’s law), and so this same pressure is transmitted to B. Therefore, the force at B is pB =

FB ; AB

63

2.1

A pneumatic jack is used in a service station as shown in Fig. 2–3. If the car and lift weigh 25 kN, determine the force that must be developed by the air compressor at B to raise the lift at a constant velocity. Air is in the line from B to A. The air line at B has an inner diameter of 15 mm, and the post at A has a diameter of 280 mm.

pA =

preSSure

406.00(103) N>m2 = FB = 71.7 N

FB p(0.0075 m)2 Ans.

This small force will therefore be sufficient to lift the 25 kN load, even though the pressure at A and B is the same. NOTE: The principles on which this example is based also extend to many

hydraulic systems, where the working fluid is oil. Typical applications include jacks, construction equipment, presses, and elevators. The pressures in these systems often range from 8 MPa when used with small vehicles, all the way to 60 MPa for hydraulic jacks.

2

64

Chapter 2

F l u i d S ta t i C S

2.2

2

ABSOLUTE AND GAGE PRESSURE

If a fluid such as air were removed from its container, a vacuum would exist and the pressure within the container would be zero. This is commonly referred to as zero absolute pressure. Any pressure that is measured above this value is referred to as the absolute pressure, pabs. For example, standard atmospheric pressure is the absolute pressure that is measured at sea level and at a temperature of 15°C. Its value is patm = 101.3 kPa Any pressure measured above or below atmospheric pressure is called the gage pressure, pg, because gages are often used to measure pressure relative to the atmospheric pressure. The absolute pressure and the gage pressure are therefore related by pabs = patm + pg

(2–3)

Realize that the absolute and atmospheric pressures are always positive; however, the gage pressure can either be positive or negative, Fig. 2–4. For example, if the absolute pressure is pabs = 301.3 kPa, then the gage pressure becomes pg = 301.3 kPa - 101.3 kPa = 200 kPa. Likewise, if the absolute pressure is pabs = 51.3 kPa, then the gage pressure is pg = 51.3 kPa - 101.3 kPa = - 50 kPa, a negative value producing a suction, since it is below atmospheric pressure. In this book we will always measure the gage pressure relative to standard atmospheric pressure; however, for greater accuracy the local atmospheric pressure should be used, and from that the absolute pressure can be determined. Also, unless otherwise stated, all pressures reported in the book and in the problems will be considered gage pressures. If absolute pressure is intended, it will be specifically stated or denoted as, for example, 50 kPa (abs.).

pabs 5 patm1 pg 1pg patm (pg = 0) 2pg pabs 5 patm 2 pg

pabs 5 0 Pressure scale

Fig. 2–4

2.2

EXAMPLE

abSolute and GaGe preSSure

65

2.2

The air pressure within the bicycle tire is determined from a gage to be 70 kPa, Fig. 2–5. If the local atmospheric pressure is 104 kPa, determine the absolute pressure in the tire.

Fig. 2–5

SOLUTION Fluid Description.

The air remains static under constant pressure.

Analysis. Before the tire was filled with air, the pressure within it was atmospheric, 104 kPa. Therefore, after the tire is filled, the absolute pressure in the tire is pabs = patm + pg pabs = 104 kPa + 70 kPa = 174kPa

Ans.

A point to remember is that a newton is about the weight of an apple, and so if you imagine this weight distributed over a square meter, you will realize that a pascal is actually a very small pressure 1Pa = N>m2 2. For this reason, for engineering work, pressures measured in pascals are almost always accompanied by a prefix such as k or M.

2

66

Chapter 2

F l u i d S ta t i C S

2.3

2

Dy pDA ∂p (p + — Dy) DA ∂y

STATIC PRESSURE VARIATION

In this section we will determine how the pressure varies within a static fluid due to the weight of the fluid. To do this, we will consider small, slender horizontal and vertical fluid elements having cross-sectional areas ∆A, and lengths of ∆y and ∆z, respectively. The free-body diagrams, showing only the forces acting in the y and z directions on each element, are shown in Fig. 2–6. For the element extending in the z direction, the weight is included. It is the product of the fluid’s specific weight g and the volume of the element, ∆V = ∆A∆z. The gradient or change in pressure from one side of each element to its opposite side is assumed to increase in the positive y and z directions, and is expressed as (0p > 0y)∆y and (0p > 0z)∆z, respectively.* If we apply the equation of force equilibrium to the horizontal element, Fig. 2–6a, we obtain

DA y

p(∆A) - a p +

ΣFy = 0;

0p = 0

(a)

∂p (p + — Dz) DA ∂z DA

z

g(DAdz) Dz

0p ∆yb ∆A = 0 0y

This same result will also occur in the x direction, and since this change in pressure is zero, it indicates that the pressure remains constant in the horizontal plane. In other words, the pressure will only be a function of z, p = p(z). From Fig. 2–6b,

ΣFz = 0;

p(∆A) - a p +

dp ∆zb ∆A - g(∆A∆z) = 0 dz

dp = -gdz

pDA

(b)

(2–4)

The negative sign indicates that the pressure will decrease as one moves upwards in the fluid, in the positive z direction. The above two results apply to both incompressible and compressible fluids. In the following sections we will treat each of these types of fluids separately.

Fig. 2–6

*This is the result of a Taylor series expansion about a point, for which we have omitted 2 2 1 0 p 1 0 p a b ∆y 2 + g and a b ∆z2 + c, because they will 2 2 2 0y 0 z2 drop out as ∆y S 0 and ∆z S 0. Also, the partial derivative is used here because the pressure is assumed to be changing in every coordinate direction, i.e., the pressure is assumed to be different at each point, and so p = p(x, y, z).

the higher-order terms,

2.4

2.4

preSSure Variation For inCompreSSible FluidS

67

PRESSURE VARIATION FOR INCOMPRESSIBLE FLUIDS

If the fluid is assumed to be incompressible, as in the case of a liquid, then its specific weight g is constant. If we establish the reference at the liquid suface, where the pressure is p0, and then measure +z downward, Fig. 2–7, then Eq. 2–4 becomes dp = gdz. Integrating from the surface to the depth z = h, we have

p

Lp0

dp = g

L0

h

dz

p = p0 + gh

(2–5)

When the pressure at the surface is the atmospheric pressure, p0 = patm, then the term gh represents the gage pressure in the liquid, and so

p = gh

(2–6)

Incompressible fluid

Like diving into a swimming pool, this result indicates that the weight of the water above the diver will cause the gage pressure to increase linearly as one descends deeper into the water.

p0 (z 5 0)

z5h

p z

Pressure increases with depth p 5 gh

Fig. 2–7

2

Water tanks are used in many municipalities in order to provide a constant high pressure within the water distribution system. This is especially important when demand is high in the early morning and early evening.

68

Chapter 2

F l u i d S ta t i C S

Pressure Head. If we solve Eq. 2–6 for h, we obtain

Water

h = 2

p g

(2–7)

Here h is referred to as the pressure head, since it indicates the height of a column of liquid that produces the (gage) pressure p. For example, if the gage pressure is 50 kPa, then the pressure heads for water 1gw = 9.81 kN>m3 2 and mercury 1gHg = 133 kN>m3 2 are

5.10 m

hw =

Mercury 0.376 m

hHg =

p 5 50 kPa

501103 2 N>m2 p = 0.376 m = gHg 1331103 2 N>m3

As shown in Fig. 2–8, there is a significant difference in these pressure heads, since the densities (or specific weights) of these two liquids are so different.

Pressure heads

Fig. 2–8

EXAMPLE

501103 2 N>m2 p = = 5.10 m gw 9.811103 2 N>m3

2.3 The tank and drainpipe in Fig. 2–9 are filled with gasoline and glycerin to the depths shown. Determine the pressure on the drain plug at C. Report the answer as a pressure head in meter of water. Take rga = 726 kg/m3 and rgl = 1260 kg/m3. SOLUTION Fluid Description. Each of the liquids is assumed to be incompressible.

2m A Gasoline

1m

B Glycerin 1.5 m

C

Analysis. Notice that the gasoline will “float” on the glycerin since it has a lower specific weight. To obtain the pressure at C, we need to find the pressure at depth B caused by the gasoline, and then add to it the additional pressure from B to C caused by the glycerin. The gage pressure at C is therefore pC = gga hAB + ggl hBC = 1726 kg>m3 2(9.81 m>s2)11 m2 + 11260 kg>m3 2(9.81 m>s2)11.5 m2 = 25.661103 2 Pa = 25.66 kPa This result is independent of the shape or size of the tank; rather, it only depends on the depth of each liquid. In other words, in any horizontal plane, the pressure is constant. Since the specific weight of water is gw = rwg = 11000 kg>m3 2 (9.81 m>s2) = 9.811103 2 N>m3, the pressure head in feet of water at C is h =

Fig. 2–9

25.661103 2N>m2 pC = 2.62 m = gw 9.811103 2 N>m3

Ans.

Therefore, the tank would have to be filled with water to this depth to create the same pressure at C as that caused by both the gasoline and glycerin.

2.5

2.5

69

preSSure Variation For CompreSSible FluidS

PRESSURE VARIATION FOR COMPRESSIBLE FLUIDS

When a fluid is compressible, as in the case of a gas, then its specific weight g will not be constant throughout the fluid. Therefore, to obtain the pressure, we must integrate Eq. 2–4, dp = -gdz. This requires that we express g as a function of p. Using the ideal gas law, Eq. 1–12, p = rRT, where g = rg, we have g = pg>RT. Then dp = -g dz = -

2

pg dz RT

or z

g dp = dz p RT

p

Remember that here p and T must represent the absolute pressure and absolute temperature. Integration can now be carried out, provided we can express T as a function of z.

p0 z z0

Constant Temperature. If the temperature throughout the gas remains constant (isothermal) at T = T0, then assuming the pressure at a reference location z = z0 is p = p0, Fig. 2–10, we have

Fig. 2–10

p

z g dp dz = Lp0 p Lz0 RT0

z (km)

g p ln = 1z - z0 2 p0 RT0

50 47

32.2

or

p = p0 e

-a

g RT0

b(z - z0)

Stratosphere

20.1

(2–8)

11 Troposphere

We can use this equation to calculate the pressure within the lowest region of the stratosphere. As shown on the graph of the U.S. standard atmosphere, Fig. 2–11, this region starts at an elevation of about 11.0 km and reaches an elevation of about 20.1 km. Here the temperature is practically constant, at –56.5°C (216.5 K).

T (8C) 256.58 244.58

22.58

158

Approximate temperature distribution in the U.S. standard atmosphere

Fig. 2–11

70

Chapter 2

EXAMPLE

2

F l u i d S ta t i C S

2.4 The natural gas in the storage tank is contained within a flexible membrane and held under constant pressure using a weighted top that is allowed to move up or down as the gas enters or leaves the tank, Fig. 2–12a. Determine the required weight of the top if the (gage) pressure at the outlet A is to be 60 kPa. The gas has a constant temperature of 20°C. W 10 m

pB B (b)

30 m

A

(a)

Fig. 2–12

SOLUTION Fluid Description. We will compare the results of considering the gas to be both incompressible and compressible. Analysis. Here the pressure at A is a gage pressure. There are two forces acting on the free-body diagram of the top, Fig. 2–12b. They are the resultant force of the pressure pB of gas in the tank and the weight of the top W. We require + c ΣFy = 0;

p BA B - W = 0

pB 3p(5 m)2 4 - W = 0 W =

3 78.54 pB 4 N

(1)

2.5

preSSure Variation For CompreSSible FluidS

Incompressible Gas. If the gas is considered incompressible, the pressure at the outlet A can be related to the pressure at B using Eq. 2–5. From Appendix A for natural gas, rg = 0.665 kg>m3, and since gg = rgg, we have pA = pB + g g h

601103 2 N>m2 = pB + 10.665 kg>m3 219.81 m>s2 2(30 m) pB = 59 804 Pa

Substituting into Eq. 1 yields W =

3 78.54(59 804) 4 N

= 4.697 MN

Ans.

Compressible Gas. If the gas is assumed to be compressible, then since its temperature is constant, Eq. 2–8 applies. From Appendix A, for natural gas, R = 518.3 J>(kg # K), and the absolute temperature is T0 = 20 + 273 = 293 K. Thus pB = pAe - a RT0 b(zB - zA) g

= 60(103)e - a [518.3(293)] b(30 - 0) 9.81

= 59 884 Pa From Eq. 1, W =

3 78.54(59 884) 4 N

= 4.703 MN

Ans.

By comparison, there is a difference of less than 0.13% between these two results. Furthermore, notice that the pressure difference between the top B and bottom A of the tank is actually very small. For incompressible gas, (60 kPa - 59.80 kPa) = 0.2 kPa, and for compressible gas, (60 kPa - 59.88 kPa) = 0.12 kPa. For this reason, it is generally satisfactory to neglect the change in pressure due to the weight of the gas, and consider the pressure within any gas to be essentially constant throughout its volume. If we do this, then pB = pA = 60 kPa, and from Eq. 1, W = 4.71 MN.

71

2

72

Chapter 2

F l u i d S ta t i C S

2.6

2

MEASUREMENT OF STATIC PRESSURE

There are several ways in which engineers measure the absolute and gage pressures at points within a fluid. Here we will discuss some of the more important ones.

Barometer. Atmospheric pressure can be measured using a simple device called a barometer. It was invented in the mid-17th century by Evangelista Torricelli, using mercury as a preferred fluid, since it has a high density and a very small vapor pressure. In principle, the barometer consists of a closed-end glass tube that is first entirely filled with mercury. The tube is then submerged in a dish of mercury and then turned upside down, Fig. 2–13. Doing this causes a slight amount of the mercury to empty from the closed end, thereby allowing a small amount of mercury vapor to accumulate in this region. For moderate seasonal temperatures, however, the vapor pressure created here is practically zero, so that at the mercury surface the absolute pressure is pA = 0.* As the atmospheric pressure, patm, pushes down on the surface of the mercury in the dish, it causes the pressure at points B and C to be the same, since they are at the same horizontal level. If the height h of the mercury column in the tube is measured, the atmospheric pressure can be determined by applying Eq. 2–5 between A and B. pB = pA + gHgh patm = 0 + gHg h = gHgh Normally the height h is stated in millimeters of mercury. For example, standard atmospheric pressure, 101.3 kPa, will cause the mercury column 1gHg = 133 550 N>m3 2 to rise h ≈ 760 mm in the tube.

A

pA 5 0

h B

C

patm

Simple barometer

Fig. 2–13 *To increase the accuracy, the vapor pressure of the mercury that exists within this space should be determined at the temperature recorded when the measurement is made.

2.6

73

meaSurement oF StatiC preSSure

Manometer. A manometer consists of a transparent tube that is used to determine the gage pressure in a liquid. The simplest type of manometer is called a piezometer. The tube is open at one end to the atmosphere, while the other end is inserted into a vessel, where the pressure of a liquid is to be measured, Fig. 2–14. Any pressure at the top of the vessel will push the liquid a distance h up the tube. If the liquid has a specific weight g, and the pressure head h is measured, then, for example, the pressure at point A is pA = g(h + d). Piezometers do not work well for measuring large gage pressures, since h would be large. Also, they are not effective at measuring high negative (suction) gage pressures, since air may leak into the vessel through the point of insertion. When negative gage pressures or moderately high pressures are to be found, a simple U-tube manometer, as shown in Fig. 2–15, can be used. Here one end of the tube is connected to the vessel containing a fluid of specific weight g, and the other end is open to the atmosphere. To measure relatively high pressures, a liquid with a high specific weight g′, such as mercury, is placed in the U-tube. The pressure at point A in the vessel is the same as at point B in the tube, since both points are on the same level. The pressure at C is therefore pC = pA + ghBC. This is the same pressure as at D, again, because C and D are on the same level. Finally, since pC = pD = g′hDE, then

h

2

d A

Piezometer

Fig. 2–14

g E hDE

B A

hBC

g′hDE = pA + ghBC

C

D g9

or Simple manometer

pA = g′hDE - ghBC

In particular, if the fluid in the vessel is a gas, then its specific weight will be very small compared to that of the manometer liquid, so g ≈ 0 and the above equation becomes pA = g′hDE. To reduce errors when using a manometer, it is best to use a manometer fluid that has a low specific gravity, such as water, when the anticipated pressure is low. This way the fluid is pushed higher in the manometer so the reading of the pressure head is more sensitive to pressure differences. Also, as noted in Sec. 1.10, errors can be reduced in reading the level of the meniscus, that is, the curved surface caused by capillary attraction, if the tube generally has a diameter of 10 mm or greater. Finally, for very sensitive work, a more precise specific weight of the fluid should be determined, based on its temperature.

Fig. 2–15

74

Chapter 2

F l u i d S ta t i C S

Manometer Rule. The previous result can also be determined in a g

more direct manner by using the manometer rule, something that works for all types of manometers. It can be stated as follows:

E hDE

B A

hBC

2

D

C

g9

Simple manometer

Start at a point in the fluid where the pressure is to be determined, and proceed to add to it the pressures algebraically from one vertical fluid interface to the next, until you reach the liquid surface at the other end of the manometer.

As with any fluid system, a pressure term will be positive if it is below a point since it will cause an increase, and it will be negative if it is above a point since it will cause a decrease. Thus, for the manometer in Fig. 2–15, we start with pA at point A, then add ghBC, and finally subtract g′hDE. This algebraic sum is equal to the pressure at E, which is zero; that is, pA + ghBC - g′hDE = 0. Therefore, pA = g′hDE - ghBC, as obtained previously. As another example, consider the manometer in Fig. 2–16. Starting at A, and realizing that the pressure at C is zero, we have

Fig. 2–15 (repeated)

pA - ghAB - g′hBC = 0 so that pA = ghAB + g′hBC

C C

C

hBC

C g9

B

B

B

B

hAB A

A

g

Increasing the pressure by squeezing the bulb at A will cause the elevation difference BC to be the same in each tube, regardless of the shape of the tube.

Fig. 2–16

2.6

75

meaSurement oF StatiC preSSure

Differential Manometer. A differential manometer is sometimes used to determine the difference in pressure between two points in a closed conduit when the fluid is flowing through it. For example, the differential manometer in Fig. 2–17 measures the difference in static pressure between points A and D. Following the path through the manometer from A to B to C to D, and summing the pressures as outlined by the manometer rule, we have pA + ghAB - g′hBC - ghCD = pD

2

g

∆p = pD - pA = ghAB - g′hBC - ghCD

A

D hCD

hAB

Since hBC = hAB - hCD, then

C

hBC

B

∆p = (g - g′)hBC

g9

Fig. 2–17

This result represents the difference, or in this case the drop in either the absolute or the gage pressure between points A and D. Notice that if g′ is chosen close to g, then their difference (g - g′) will be rather small, and a large value of hBC will occur, making it more accurate to determine ∆p. Small differences in pressure can also be determined by using an inverted U-tube manometer as in Fig. 2–18. Here the manometer fluid has a smaller specific weight g′ than that of the contained fluid g. An example would be oil 1g′2 versus water 1g2. Starting from point A and going to point D, we have

g9 B

pA - ghAB + g′hBC + ghCD = pD ∆p = pD - pA = -ghAB + g′hBC + ghCD

hBC hAB

C hCD

g A

Since hBC = hAB - hCD, then Fig. 2–18

∆p = (g′ - g)hBC If air is used as the lighter fluid, then it can be pumped into the top portion of the tube and the valve closed to produce a suitable liquid level. In this case, g′ ≈ 0 and ∆p = -ghBC. There have been several other modifications to the U-tube manometer that have improved its accuracy for measuring pressures or their differences. For example, one common modification is to make one of the tubes inclined, the principles of which are discussed in Example 2.7. Another is to use a micro-manometer, as discussed in Prob. 2–42.

D

76

Chapter 2

F l u i d S ta t i C S

Bourdon Gage. If the gage pressures are very high, then a manometer may not be effective, and so the measurement can be made using a Bourdon gage, Fig. 2–19. Essentially, this gage consists of a coiled metal tube that is connected at one end to the vessel where the pressure is to be measured. The other end of the tube is closed so that when the pressure in the vessel is increased, the tube begins to uncoil and respond elastically. Using the mechanical linkage attached to the end of the tube, the dial on the face of the gage gives a direct reading of the pressure, which can be calibrated in various units, such as kPa.

2

Pressure Transducers. An electromechanical device called a pressure transducer can also be used to measure pressure. It has the advantage of producing a quick response to changes in pressure, and providing a continuous digital readout over time. Figure 2–20 illustrates the way it works. When end A is connected to a pressure vessel, the fluid pressure will deform the thin diaphragm. The resulting strain in the diaphragm is then measured using the attached electrical strain gage. Essentially, the changing length of the thin wires composing the gage will change their resistance, producing a change in electric current. Since this change in current is directly proportional to the strain caused by the pressure, the current can be converted into a direct reading of the gage pressure in the vessel, provided region B is open to the atmosphere. Pressure transducers can also be used to find the absolute pressure, provided the volume B behind the diaphragm is sealed so it is in a vacuum. Finally, if regions A and B are connected to two different vessels, then the difference in pressure between them can be recorded.

10

5

Bourdon gage

Fig. 2–19

Diaphragm

B

A

+ –

Strain gage

Pressure transducer

Fig. 2–20

Other Pressure Gages. Apart from the gages discussed above, there are several other methods for measuring pressure. One of the more accurate gages available is a fused quartz force-balance Bourdon tube. Within it, pressure causes elastic deformation of a coiled tube that is detected optically. The tube is then restored to the original position by a magnetic field that is measured, and this is correlated to the pressure that caused the deformation. In a similar manner, piezoelectric gages, such as a quartz crystal, can change their electric potential when subjected to small pressure changes, and so the pressure can be correlated to this potential change and presented as a digital readout. The same type of gage can also be constructed using thin silicon wafers. A sudden change in pressure will cause them to deform, which causes a measured change in capacitance or vibrational frequency. Further details on these gages and others like them, along with their specific applications, can be found in the literature or from Refs. [5]–[11].

2.6

meaSurement oF StatiC preSSure

77

IMPORTANT POIN T S • The pressure at a point in a fluid is the same in all directions, provided the fluid has no relative motion. This is Pascal’s law. As a consequence, any increase in pressure ∆p at one point in the fluid will cause the same increase in pressure ∆p throughout the fluid.

• Absolute pressure is the pressure above that of a vacuum. Standard atmospheric pressure, which is measured at sea level and a temperature of 15°C, is 101.3 kPa.

• Gage pressure is measured as the pressure that is above (positive) or below (negative) atmospheric pressure.

• When the weight of a static fluid is considered, the pressure in the horizontal direction is constant; however, in the vertical direction it increases with depth.

• If a fluid is essentially incompressible, as in the case of a liquid, then its specific weight is constant, and the (gage) pressure can be determined using p = gh.

• If a fluid is considered compressible, as in the case of a gas, then the variation of the fluid’s specific weight (or density) with pressure must be taken into account to obtain an accurate measurement of pressure.

• For small changes in elevation, the static gas pressure in a tank, vessel, manometer, pipe, and the like can be considered constant throughout its volume, since the specific weight of a gas is very small.

• The pressure p at a point can be represented by its pressure head, which is the height h of a column of fluid needed to produce the pressure, h = p>g.

• Atmospheric pressure can be measured using a barometer. • Manometers can be used to measure small pressures in pipes or tanks, or differential pressures between points in two pipes. Pressures at any two points in the manometer can be related using the manometer rule.

• High pressures are generally measured using a Bourdon gage or pressure transducer. Besides these, many other types of pressure gages are available for specific applications.

2

78

Chapter 2

EXAMPLE

F l u i d S ta t i C S

2.5 The funnel in Fig. 2–21 is filled with oil and water to the levels shown, while portion CD of the tube contains mercury. Determine the distance h the mercury level is from the top of the oil surface. Take ro = 880 kg>m3, rw = 1000 kg>m3, rHg = 13 550 kg>m3.

A

2 0.3 m

h B

SOLUTION D

Fluid Description. The fluids are liquids, so we will consider them to be incompressible.

0.4 m

Analysis. We can treat the system as a “manometer” and write an equation using the manometer rule from A to D, noting that the (gage) pressures at A and D are both zero. We have

C

0 + roghAB + rwghBC - rHgghCD = 0

Fig. 2–21

0 + 1880 kg>m3 219.81 m>s2 2(0.3 m) + 11000 kg>m3 219.81 m>s2 2(0.4 m) - 113 550 kg>m3 219.81 m>s2 2(0.3 m + 0.4 m - h) = 0

Thus,

h = 0.651 m

EXAMPLE

2.6 Determine the difference in pressure between the centerline points A and B in the two pipelines in Fig. 2–22, if the manometer liquid CD is at the level shown. The density of the liquid in AC and DB is r = 800 kg>m3, and in CD, rCD = 1100 kg>m3.

D 40 mm 60 mm

65 mm C

A

Ans.

250 mm

SOLUTION

30 mm

Fluid Description. The liquids are assumed incompressible. B

Analysis. Starting at point B and moving through the manometer to point A, using the manometer rule, we have pB - rghBD + rCDghDC + rghCA = pA

Fig. 2–22

pB - 1800 kg>m3 219.81 m>s2 2(0.250 m) + 11100 kg>m3 219.81 m>s2 2(0.065 m) + 1800 kg>m3 219.81 m>s2 2(0.03 m) = pA

Thus,

∆p = pA - pB = -1.03 kPa

Ans.

Since the result is negative, the pressure at A is less than that at B.

2.6

EXAMPLE

meaSurement oF StatiC preSSure

79

2.7

The inclined-tube manometer shown in Fig. 2–23 is used to measure small pressure changes. Determine the difference in pressure between points A and E if the manometer liquid, mercury, is at the level shown. The pipe at A contains water, and the one at E contains natural gas. For mercury, rHg = 13 550 kg/m3.

E A

400 mm D B 100 mm

208 C

700 mm

Fig. 2–23

SOLUTION Fluid Description. We will consider the fluids to be incompressible and neglect the specific weight of the natural gas. As a result, the pressure at E is essentially the same as that at D. Analysis. Applying the manometer rule from point A to point D, we have pA + gwhAB + gHghBC - gHghCD = pE

pA + 11000 kg>m3 219.81 m>s2 210.4 m2 + 113 550 kg>m3 219.81 m>s2 2 10.1 m2 - 113 550 kg>m3 219.81 m>s2 210.7 sin 20° m2 = pE

pA - pE = 14.611103 2 Pa = 14.6 kPa

Ans.

Notice that any slight pressure change will cause the distance ∆ CD along the tube to be significantly altered, since the elevation change ∆hCD depends on the factor (sin 20°). In other words, ∆ CD = ∆hCD >sin 20° = 2.92∆hCD. In practice, angles less than about 5° are not practical for inclined-tube manometers. This is because at such small angles the exact location of the meniscus is hard to detect, and the effect of surface tension will be magnified if there are any surface impurities within the tube.

2

80

Chapter 2

F l u i d S ta t i C S

2.7 HYDROSTATIC FORCE ON A PLANE SURFACE—FORMULA METHOD

x

When designing gates, vessels, dams, or other bodies that are submerged in a liquid, it is important to be able to obtain the resultant force caused by the pressure loading of the liquid, and to specify the location of this force on the body. In this section we will show how this is done on a plane surface by using a derived formula. To generalize the development, we will consider the surface to be a flat plate of arbitrary shape that is submerged in the liquid and oriented at an angle u with the horizontal, Fig. 2–24a. The origin of the x, y coordinate system is located at the surface of the liquid so that the positive y axis extends downward, along the plane of the plate.

x dF

2

u

y

y

h

dA

dF

u

h

y

Resultant Force. The resultant force on the plate can be found by first considering the differential area dA that lies at a depth h = y sin u from the liquid surface. Because the pressure at this depth is p = gh, the force acting on this area is

dA

Side view

dF = p dA = (gh) dA = g (y sin u) dA

(a)

The resultant force acting on the plate is equivalent to the sum of all these forces. Therefore, integrating over the entire area A, we have

x

x FR

u y

C

h

xP

y

FR = ΣF; yP

P

u

FR

g y sin u dA = g sin u y dA = g sin u (yA) LA LA

The integral 1 y dA represents the “moment of the area” about the x axis. Here it has been replaced by its equivalent moment yA, where y is the distance from the x axis to the centroid C, or geometric center of the area, Fig. 2–24b.* Since the depth of the centroid is h = y sin u, the above equation then becomes

h y C P

y

FR =

yP

Side view FR 5 gh A FR acts through the center of pressure P (b)

Fig. 2–24

FR = gh A

(2–9)

where FR g h A

= = = =

the resultant force on the plate the specific weight of the liquid the depth to the centroid of the submerged area the submerged area of the plate

The force has a direction that is perpendicular to the plate, since the entire pressure distribution acts in this direction. *See Engineering Mechanics: Statics, R. C. Hibbeler, Pearson Education.

2.7

hydroStatiC ForCe on a plane SurFaCe—Formula method

81

Location of the Resultant Force. The resultant force of the pressure distribution acts through a point on the plate called the center of pressure, P, shown in Fig. 2–24b. The location of this point, (xP, yP), is determined by a balance of moments that requires the moment of the resultant force about the x axis and about the y axis, Fig. 2–24b, to equal the moment of the entire pressure loading about each of these axes, Fig. 2–24a.

2

The yP Coordinate. We require 1MR 2 x = ΣMx;

y dF LA

yP FR =

Since FR = g sin u (yA) and dF = g (y sin u) dA, then yP 3g sin u (yA)4 =

y 3g ( y sin u ) dA4 LA

Canceling g sin u, we get yP yA =

y2 dA LA

The integral represents the area moment of inertia Ix for the area about the x axis.* Thus, yP =

Ix yA

Values for the area moment of inertia are normally referenced from an axis passing through the centroid of the area, referred to as I x . Examples for some common shapes, along with the location of their centroids, are given on the inside back cover. With these values we can then use the parallel-axis theorem* to obtain Ix, that is, Ix = I x + Ay2, and write the above equation as yP = y +

Ix yA

(2–10)

Here yP = the distance along the plate to the center of pressure y = the distance along the plate to the centroid of the submerged area Ix = the moment of inertia of the submerged area about the centroidal axis A = the submerged area of the plate Notice that the term I x >(yA) will always be positive, and so the location of the center of pressure P will always be below the centroid of the plate 1yP 7 y2, Fig. 2–24b. *Ibid.

The formula method can be used to determine the resultant pressure force acting on a surface, such as the endplates of this water trough.

82

Chapter 2

F l u i d S ta t i C S

The xP Coordinate. The lateral position of the center of pressure, xP, can be determined by a balance of moments about the y axis, Figs. 2–24a and 2–24b. We require

x

x dF

2

u

y

h

dA

y

1MR 2 y = ΣMy;

x dF LA Again, using FR = g sin u( yA) and dF = g (y sin u) dA, we have -xP FR = -

xP [g sin u( yA)] =

x [g( y sin u) dA] LA

Canceling g sin u gives

xy dA LA This integral is referred to as the product of inertia Ixy for the area.* Thus, Ixy xP = yA If we apply the parallel-plane theorem,** Ixy = Ixy + A x y, where x and y locate the area’s centroid, then this result can also be expressed as Ixy xP = x + (2–11) yA Here xP = the location of the center of pressure x, y = the coordinate distance to the centroid of the submerged area Ixy = the product of inertia of the submerged area relative to the centroidal axes A = the submerged area of the plate xP y A =

dF

u

h

y

dA

Side view (a)

x

x Symmetrical FR

u y

C

h

xP

y

yP

P

Plate. For most engineering applications, the submerged area will be symmetrical about an axis through the centroid, as in the case of the rectangular plate in Fig. 2–25; then Ixy = 0 and xP = 0, which simply indicates that the center of pressure P lies on the centroidal y axis, as shown in the figure. p1 x

u

FR

FR

h y C P

p2

y yP

C

yP

h

P y

y

Side view

h 2 y

FR 5 gh A FR acts through the center of pressure P

h 2

FR 5 gh A FR acts through the center of pressure

(b)

Fig. 2–24 (repeated) **Ibid.

Fig. 2–25

2.7

83

hydroStatiC ForCe on a plane SurFaCe—Formula method

IMPORTANT POIN T S • A liquid creates a pressure loading that acts perpendicular to a submerged surface. Since the liquid is assumed to be incompressible, then the pressure intensity increases linearly with depth, p = g h.

• The resultant force of the pressure on a plane surface having an area A and submerged in a liquid can be determined from FR = g h A, where h is the depth of the area’s centroid C, measured from the liquid’s surface.

• The resultant force acts through the center of pressure P, determined from xP = x + Ixy >( yA) and yP = y + Ix >( yA). If the submerged area has an axis of symmetry along the y axis that passes through the centroid, then Ixy = 0, and so xP = x = 0. Here P is located on the centroidal axis of the area.

EXAMPLE

2.8

Determine the force that water pressure exerts on the inclined side plate of the storage tank, and find its location measured from AB, Fig. 2–26a.

1.5

m

A

SOLUTION

B x

Fluid Description. rw = 1000 kg/m3.

We consider water to be incompressible, where

m 1.5 y5 yP

3m C

2.5 m

P

Analysis. The centroid of the plate’s area is located at its midpoint, y = 1.5 m from its top AB, Fig. 2–26a. The depth of the water at this point is h = 1.25 m. Thus,

E D y (a)

FR = gw hA = 11000 kg>m 2(9.81 m>s )(1.25 m)[(1.5 m)(3 m)] 3

2

= 55.18 1103 2N = 55.2 kN

Ans.

1 From the inside back cover, for a rectangular area, Ix = ba 3. Here 12 b = 1.5 m and a = 3 m. Thus, 1 (1.5 m)(3 m)3 Ix 12 yP = y + = 1.5 m + = 2m yA (1.5 m)[(1.5 m)(3 m)]

xP = x +

yA

= 0 + 0 = 0

The results are shown from a side view of the plate in Fig. 2–26b.

3m

yP 5 2 m

P D

Ans.

Since the rectangle is symmetrical about its centroidal y axis, Fig. 2–26a, then Ixy = 0 and Eq. 2–11 gives Ixy

A

Ans.

(b)

Fig. 2–26

FR 5 55.2 kN

2

84

Chapter 2

EXAMPLE

F l u i d S ta t i C S

2.9 Determine the magnitude and location of the resultant force that the water pressure exerts on the circular plate in Fig. 2–27a.

2

Water surface 3m

x yP 5 3.08 m h53m

2m

C P

FR

y (a)

(b)

Fig. 2–27

SOLUTION Fluid Description. We assume the water to be incompressible. For water, rw = 1000 kg>m3. Analysis.

From Fig. 2–27b, the resultant force is

FR = gw h A

= 11000 kg>m3 219.81 m>s2 2(3 m)[p(1 m)2] = 92.5 kN

Ans.

Using the table on the inside back cover for a circle, the location of the resultant force is determined from p (1 m)4 Ix 4 = 3.08 m yP = y + = 3m + yA (3 m)[p(1 m)2]

Ans.

Since Ixy = 0, due to symmetry,

xP = x +

Ixy yA

= 0 + 0 = 0

Ans.

2.7

EXAMPLE

hydroStatiC ForCe on a plane SurFaCe—Formula method

2.10

Determine the magnitude and location of the resultant force acting on the triangular end plate of the settling tank in Fig. 2–28a. The tank contains kerosene.

2 0.5 m

1m

B x

yP 5 1 m

P

2m 2m

A

FR 5 5.32 kN

y

(a)

(b)

Fig. 2–28

SOLUTION Fluid Description. The kerosene is considered an incompressible fluid for which gk = rkg = (814 kg/m3)(9.81 m/s2) = 7985 N/m3 (Appendix A). Analysis.

From the inside back cover, for a triangle, 1 (2 m) = 0.6667 m 3 1 1 Ix = ba 3 = (1 m)(2 m)3 = 0.2222 m4 36 36 y = h =

Therefore,

1 FR = gk h A = 17985 N>m3 2(0.6667) c (1 m)(2 m) d 2 = 5323.56 N = 5.32 kN

yP = y +

85

Ix = 0.6667 m + yA

Ans.

0.2222 m4 = 1.00 m Ans. 1 (0.6667 m) c (1 m)(2 m) d 2

The triangle is symmetrical about the y axis, so Ixy = 0. Thus, Ixy xP = x + = 0 + 0 = 0 yA These results are shown in Fig. 2–28b.

Ans.

86

Chapter 2

F l u i d S ta t i C S

2.8 HYDROSTATIC FORCE ON A PLANE SURFACE—GEOMETRICAL METHOD x

2

p1 5 gh1

x

dF u

p

y

h

dV dA

Resultant Force. If an area element dA of the plate is at a depth h,

y

where the pressure is p, then the force on this element is dF = p dA. However, as shown in the figure, this force geometrically represents a differential volume element dV of the pressure distribution. Realize that this volume has units of force, because it has a height p (N>m2) and base dA (m2), and so dF = p dA = dV. The resultant force can be obtained by integrating these elements over the entire volume enclosed by the pressure distribution. We have

p2 5 gh2 (a)

Centroid of volume

FR = ΣF ;

FR =

x

FR

yP

CV y

Rather than using the equations of the previous section, the resultant force and its location on a flat submerged plate can also be determined using a geometrical method. To show how this is done, consider the flat plate shown in Fig. 2–29a.

p dA = dV = V LA LV

(2–12)

Therefore, the magnitude of the resultant force is equal to the total volume of the “pressure prism.” The base of this prism is the area of the plate, and the height varies linearly from p1 = gh1 to p2 = gh2, Fig. 2–29a.

P

xP

Location. To locate the point of application of the resultant force on Center of pressure

FR is equal to the volume of the pressure distribution, and it passes through the centroid CV of this volume (b)

Fig. 2–29

the plate, we require the moment of the resultant force about the y axis and about the x axis, Fig. 2–29b, to be equal to the moment created by the entire pressure distribution about each of these axes, Fig. 2–29a; that is, 1MR 2 y = ΣMy; 1MR 2 x = ΣMx;

xP FR =

L

x dF

yP FR =

L

y dF

Since FR = V and dF = dV, we have

xP =

x p dA LA p dA LA

x dV LV = V

yP =

y p dA LA p dA LA

y dV LV = V

(2–13)

These equations locate the x and y coordinates of the centroid CV of the pressure prism volume. In other words, the line of action of the resultant force will pass through both the centroid CV of the volume of the pressure prism and the center of pressure P on the plate, Fig. 2–29b.

2.8

Centroid of pressure volume

hydroStatiC ForCe on a plane SurFaCe—GeometriCal method

p1

h1 FR

FR

Centroid of pressure area

w1

u

h2

u

CV

p2

87

2 w2

P

CA P

h2 b FR equals the volume of the pressure distribution, and it passes through the centroid CV of this volume

FR equals the area of the w diagram, and it passes through the centroid CA of this area (b)

(a)

h1

Fig. 2–30

Plate Having Constant Width. As a special case, if the plate has a constant width b, as in the case of a rectangle, Fig. 2–30a, then the pressure along this width at depth h1 and at depth h2 is constant. As a result, the loading may be viewed along the side of the plate, in two dimensions, Fig. 2–30b. The intensity w of this distributed load is measured as a force>length, and varies linearly from w1 = p1b = (g h1)b to w2 = p2b = (g h2)b. The magnitude of FR is then equivalent to the trapezoidal area defining the distributed load, and FR has a line of action that passes through both the centroid CA of this area and the center of pressure P on the plate. Of course, these results are equivalent to finding the trapezoidal volume of the pressure prism, FR, and its centroidal location CV , as shown in Fig. 2–30a.

IM PORTANT POINT S • The resultant force on a plane surface can be determined graphically by finding the volume V of the pressure prism, FR = V. The line of action of the resultant force passes through the centroid of this volume. It intersects the surface at the center of pressure P.

• If the submerged surface has a constant width, then the pressure prism can be viewed from the side and represented as a planar distributed load w. The resultant force equals the area of this loading diagram, and it acts through the centroid of this area.

88

Chapter 2

EXAMPLE

2

F l u i d S ta t i C S

2.11 The tank shown in Fig. 2–31a contains water to a depth of 3 m. Determine the resultant force and its point of application that the water pressure creates on side ABCD of the tank and also on its bottom. B A

C

3m

2m D 1.5 m (a)

Fig. 2–31

SOLUTION Fluid Description. The water is considered to be incompressible, with rw = 1000 kg>m3. ANALYSIS I Loading. The pressure at the bottom of the tank is p = rwgh = 11000 kg>m3 219.81 m>s2 2(3 m) = 29.43 kPa Using this value, the pressure distribution acting on the side and bottom of the tank is shown in Fig. 2–31b. Resultant Forces. The magnitudes of the resultant forces are equal to the volumes of the pressure prisms. 1FR 2 s =

1 2

(3 m)129.43 kN>m2 2(2 m) = 88.3 kN

1FR 2 b = 129.43 kN>m2 2(2 m)(1.5 m) = 88.3 kN

Ans. Ans.

These resultants act through the centroids of their respective volumes, and define the location of the center of pressure P for each plate, Fig. 2–31.

2.8

89

hydroStatiC ForCe on a plane SurFaCe—GeometriCal method

z

2m

2

(FR)b P 3m

zP

xP CV

CV

(FR)s

2m

P

1.5 m

29.43 kPa

x

y

xP

yP

29.43 kPa

x (b)

Location. Using the table on the inside back cover, for the side plate, zP in Fig. 2–31b is determined for a triangle to be 13 a, so that xP = 1 m 1 zP = (3 m) = 1 m 3 For the bottom plate, due to symmetry, xP = 1 m yP = 0.75 m

Ans. Ans.

3m

P CV

Ans. Ans.

zP 5 1 m 58.86 kNm

ANALYSIS II Loading. Since the side and bottom plates in Fig. 2–31a both have a constant width of b = 2 m, the pressure loading can also be viewed in two dimensions. The intensity of the loading at the bottom of the tank is w = 1rwgh2b

= 11000 kg>m3 219.81 m>s2 2(3 m)(2 m) = 58.86 kN>m

CV P

The distributions are shown in Fig. 2–31c.

Resultant Forces. Here the resultant forces are equal to the areas of the loading diagrams. 1 (3 m)(58.86 kN>m) = 88.3 kN 2 (FR)b = (1.5 m)(58.86 kN>m) = 88.3 kN (FR)s =

(FR)b 58.86 kNm

yP = 0.75 m 1.5 m

Ans.

(c)

Ans.

Fig. 2–31 (cont.)

Location. These results act through the centroids of their respective areas as shown in Fig. 2–31c.

(FR)s

90

Chapter 2

EXAMPLE

F l u i d S ta t i C S

2.12

A 0.75 m B 1.5 m

SOLUTION C

Fluid Description. Both the water and the oil are assumed to be incompressible.

(a) A

Loading. Since the side of the tank has a constant width, the intensities of the distributed load at B and C, Fig. 2–32b, are

0.75 m B

8.277 kNm

1.5 m

wB = roghABb = 1900 kg>m3 219.81 m>s2 2(0.75 m)(1.25 m) = 8.277 kN>m

wC = wB + rwghBCb = 8.277 kN>m + 11000 kg>m3219.81 m>s2 2(1.5 m)(1.25 m) = 26.67 kN>m

Resultant Force. The resultant force can be determined by adding the two shaded triangular areas and the rectangular area shown in Fig. 2–32c.

C 26.67 kNm (b)

FR = F1 + F2 + F3

A

1 1 (0.75 m)(8.277 kN>m) + (1.5 m)(8.277 kN>m) + (1.5 m)(18.39 kN>m) 2 2 = 3.104 kN + 12.42 kN + 13.80 kN = 29.32 kN = 29.3 kN Ans. =

0.75 m B

F1 5 3.104 kN 8.277 kNm

1.5 m

F2 5 12.42 kN F3 5 13.80 kN

Location. As shown, each of these three parallel resultants acts through the centroid of its respective area.

C 8.277 kNm 26.67 kNm 2 8.277 kNm5 18.39 kNm (c) A

2 (0.75 m) = 0.5 m 3 1 y2 = 0.75 m + (1.5 m) = 1.5 m 2 2 y3 = 0.75 m + (1.5 m) = 1.75 m 3 y1 =

=

2

The storage tank contains oil and water at the depths shown in Fig. 2–32a. Determine the resultant force that both of these liquids together exert on the side ABC of the tank, if the side has a width of b = 1.25 m. Also, determine the location of this resultant, measured from the top surface of the oil. Take ro = 900 kg>m3, rw = 1000 kg>m3.

The location of the resultant force is determined by equating the moment of the resultant force about A, Fig. 2–32d, to the sum of the moments of all the component forces about A, Fig. 2–32c. We have

yP

FR 5 29.32 kN

(d)

Fig. 2–32

& yP FR = ΣyF;

yP (29.32 kN) = (0.5 m)(3.104 kN) + (1.5 m)(12.42 kN) + (1.75 m)(13.80 kN) yP = 1.51 m

Ans.

2.9

91

hydroStatiC ForCe on a plane SurFaCe—inteGration method

2.9 HYDROSTATIC FORCE ON A PLANE SURFACE—INTEGRATION METHOD If the boundary of the flat plate in Fig. 2–33a can be defined in terms of its x and y coordinates as y = f(x), then the resultant force FR and its location P on the plate can be determined by direct integration.

2

Resultant Force. If we consider a differential area strip dA of the plate that is located at a depth h, where the pressure is p, then the force acting on this strip is dF = p dA, Fig. 2–33a. The resultant force on the entire area is therefore FR = ΣF;

LA

FR =

(2–14)

p dA

Location. Here we require the moment of FR about the y and x axes to equal the moment of the pressure distribution about these axes. Since ∼, ∼ dF passes through the center (centroid) of dA, having coordinates (x y ), then from Fig. 2–33a and Fig. 2–33b, 1MR 2 y = ΣMy; 1MR 2 x = ΣMx;

xP FR =

∼ x dF LA

yP FR =

∼ y dF LA

The hydrostatic force acting on the elliptical back plate of this water truck can be determined by integration.

Or, written in terms of p and dA, we have

xP =

∼ x p dA LA

LA

yP =

p dA LA

∼ y p dA (2–15)

p dA LA

Application of these equations is given in the following examples.

x

x

dF

FR

u

p

yP

y h

˜˜ dA (x,y)

P dy

y

y

xP

y 5 f(x) (a)

(b)

Fig. 2–33

92

Chapter 2

EXAMPLE

2

F l u i d S ta t i C S

2.13 Determine the magnitude and location of the resultant force that the water pressure exerts on the circular plate in Fig. 2–34a.

0.75 m

SOLUTION y

3m

x2 1 y2 5 1 x

Fluid Description. We assume the water to be incompressible. For water, rw = 1000 kg>m3. Resultant Force. We can determine the resultant force on the plate by using integration, since the circular boundary can be defined from the center of the plate in terms of the x, y coordinates shown in Fig. 2–34a. The equation is x 2 + y2 = 1. The horizontal rectangular strip shown in Fig. 2–34b has an area

1m

(a) y

x yP

dA = 2x dy = 211 - y2 2 1>2dy

3m

It is at a depth h = 3 - y, where the pressure is p = gwh = gw(3 - y). Applying Eq. 2–14 to obtain the resultant force, we have

x

h

dF p

y

dy x

P

F=

LA

1m

p dA =

L- 1 m

c (1000 kg>m3)19.81 m>s2 2(3 - y) d (2)11 - y2 2 1>2 dy 1m

= 19 620

L-1 m

c 3(1 - y2)1>2 - y(1 - y2)1>2 d dy

= 92.5 kN

(b)

Ans.

Location. The location of the center of pressure P, Fig. 2–34b, can be determined by applying Eq. 2–15. Here dF = p dA is located at ∼ x = 0 and ∼ y = 3 - y and so

Fig. 2–34

xP = 0 (3 - y)(1000 kg>m3) 1 9.81 m>s2 2 (3 - y)(2) 1 1 - y2 2 1>2dy L-1 m 1m

LA

(3 - y)p dA

yP =

Ans.

LA

p dA

=

(1000 kg>m ) 1 9.81 m>s 2 (3 - y)(2) 1 1 - y L-1 m 1m

3

2

2

2 1>2

= 3.08 m

dy

For comparison, this problem was also worked in Example 2.9.

Ans.

2.9

EXAMPLE

93

hydroStatiC ForCe on a plane SurFaCe—inteGration method

2.14

Determine the magnitude and location of the resultant force acting on the triangular end plate of the settling tank in Fig. 2–35a. The tank contains kerosene.

2

SOLUTION

1m

Fluid Description. The kerosene is considered an incompressible fluid for which gk = rkg = (814 kg/m3)(9.81 m/s2) = 7985 N/m3 (Appendix A). Resultant Force. The pressure distribution acting on the end plate is shown in Fig. 2–35b. Using the x and y coordinates, and choosing the differential area strip shown in the figure, we have

B

2m A

dF = p dA = (gk y)(2x dy) = 15.971103 2yx dy

(a)

The equation of line AB can be determined using similar triangles:

0.5 m

2m - y x = 0.5 m 2 x = 0.25(2 - y)

Location.

LA

p dA =

L0

15.971103 2y[0.25(2 - y)] dy

dF

A y (b)

Ans.

Because of symmetry about the y axis, ∼ x = 0 and so xP = 0

x

Ans.

Applying Eq. 2–15, with ∼ y = y, we have LA

∼ y p dA

yP =

dy 2m (2 m 2 y)

2m

= 5.3241103 2 N = 5.32 kN

x

y

Hence, applying Eq. 2–14 and integrating with respect to y, from y = 0 to y = 2 m, yields F =

p dA LA

=

L0

y[15.97110 2y)][0.25(2 - y)]dy

yP 5 1.00 m

P

2m

3

5.3241103 2

x

B

= 1.00 m

The results are shown in Fig. 2–35c. They were also obtained in Example 2.10.

Ans. y (c)

Fig. 2–35

CV FR 5 5.32 kN

94

Chapter 2

F l u i d S ta t i C S

2.10

2

HYDROSTATIC FORCE ON AN INCLINED PLANE OR CURVED SURFACE DETERMINED BY PROJECTION

If a submerged surface is curved, then the pressure acting on the surface will change not only its magnitude but also its direction, since it must always act normal to the surface. For this case, it is generally best to determine the horizontal and vertical components of the resultant force caused by the pressure, and then use vector addition to find the resultant. The method for doing this will now be described with reference to the submerged curved plate in Fig. 2–36a.

Horizontal Component. The force shown acting on the differential element dA in Fig. 2–36a is dF = p dA, and so its horizontal component is dFh = (p dA) sin u, Fig. 2–36b. If we integrate this result over the entire area of the plate, we will obtain the resultant’s horizontal component Fh =

p sin u dA LA

Since dA sin u is the projected differential area onto the vertical plane, Fig. 2–36b, and from Pascal’s law the pressure p at a point must be the same in all directions, then the above integration over the entire area of the plate can be interpreted as follows: The resultant horizontal force component acting on the plate is equal to the resultant force of the pressure loading acting on the area of the vertical projection of the plate, Fig. 2–36c. Since this vertical area is flat or “planar,” any of the methods of the previous three sections can be used to determine Fh and the location of its application on this projected area.

dF 5 p dA

dFv u

dF

FR

dA

p dA

CV

dFh

h

u

P

dA cos u Horizontal projected area (a)

(b)

Fig. 2–36

dA sin u Vertical projected area

2.10

hydroStatiC ForCe on an inClined plane or CurVed SurFaCe determined by projeCtion

95

Fv

Projected area

2

FR

Fh 5 the resultant pressure loading on the vertical projected area

CV P

Centroid of pressure distribution

Fv 5 the weight of the volume of liquid above the plate

CV Fh

(c)

Fig. 2–36 (cont.)

Vertical Component. The vertical component of the resultant force acting on the element dA in Fig. 2–36b is dFv = (p dA) cos u. This same result can also be obtained by noting that the horizontal projection of dA is dA cos u, and so dFv = p(dA cos u) = gh(dA cos u) Since a vertical column of liquid above dA has a height h and base dA cos u, then its volume is d V = h(dA cos u), and so dFv = gd V. The resultant’s vertical component is therefore Fv =

LV

g dV = gV

And so, the resultant vertical force component acting on the plate is equivalent to the weight of the volume of the liquid above the plate, Fig. 2–36c. This force acts through the centroid CV of the volume, which has the same location as the center of gravity for the weight of the liquid, since the specific weight of the liquid is constant. Once the horizontal and vertical components of force are known, the magnitude of the resultant force, its direction, and its line of action can be established. As shown in Fig. 2–36c, the line of action of this force will act through the center of pressure P on the plate’s surface. This method can also be used for an inclined flat plate of constant width as in Fig. 2–37. Here the horizontal component of FR acts on the area projected onto the vertical plane, and the vertical component is equal to the weight of the volume of liquid above the plate.

Fv

Fh

Fig. 2–37

96

Chapter 2

Fv

F l u i d S ta t i C S

Liquid below Plate. This same type of analysis can also be applied in

C

B

CV D

2 P

Fh FR

A

E

Fh 5 the resultant pressure loading on the vertical projected area DE Fv 5 the weight of the volume of imaginary liquid ADCBA above the plate

Fig. 2–38

cases where the liquid is below the plate, rather than above it. For example, consider the curved plate AD of constant width shown in Fig. 2–38. The horizontal component of FR is determined by finding the force Fh acting on the projected vertical area, DE. The vertical component of FR, however, will act upward. To see why, imagine that liquid is also present within the shaded gray volume ABCDA. Then the vertical force components on the top and bottom surfaces of the plate will be equal but opposite and must have the same lines of action. Therefore, if we determine the weight of imaginary liquid contained within the volume ABCD, and reverse the direction of this weight, we can then establish Fv acting upward on AD. Using this same method, if a flat plate has a constant width and is submerged as shown in Fig. 2–39, then the horizontal and vertical components of the resultant force are found by finding the pressure force acting on the area projected onto the vertical plane, and from the weight of fluid within the triangular block.

Gas. If the fluid is a gas, then its weight can generally be neglected, and so the pressure throughout the gas is constant. The horizontal and vertical components of the resultant force are then determined by projecting the curved surface area onto vertical and horizontal planes and determining the components as shown in Fig. 2–40.

Fv

Fh

I MPO RTA N T PO I N T S • The horizontal component of the resultant force acting on a submerged flat inclined or curved surface is equivalent to the force acting on the projection of the area of the surface onto a vertical plane. The magnitude of this component and the location of its point of application can be determined using the methods outlined in Secs. 2.7 through 2.9.

Fig. 2–39

Fv 5 pAh

p

Centroid of pressure distribution

Ah

p Av

• The vertical component of the resultant force acting on a submerged flat inclined or curved surface is equivalent to the weight of the volume of liquid above the surface. This component passes through the centroid of this volume. If the liquid confined below the flat inclined or curved surface, then the vertical component is equal but opposite to the weight of imaginary liquid located within the volume extending above the surface to the liquid level.

• Pressure due to a gas is uniform in all directions, since the weight Fh 5 pAv

Weight of gas is negligible

Centroid of pressure distribution

Gas pressure is constant

Fig. 2–40

of the gas is generally neglected. As a result, the horizontal and vertical components of the resultant force of pressure acting on a flat inclined or curved surface can be determined by multiplying the pressure by its associated vertical and horizontal projected areas, respectively. These components act through the centroids of these projected areas.

2.10

EXAMPLE

97

hydroStatiC ForCe on an inClined plane or CurVed SurFaCe determined by projeCtion

2.15

The sea wall in Fig. 2–41a is in the form of a semiparabola. Determine the resultant force acting on 1 m of its width. Also, specify the distance the line of action of each component force acts from the origin of the coordinates. Take rw = 1050 kg>m3.

y

2 2m

SOLUTION Fluid Description.

We treat the water as an incompressible fluid. 8m

Horizontal Force Component. The vertical projection of the wall is AB, Fig. 2–41b. The intensity of the distributed load caused by water pressure at point A for a 1-m width of the wall is wA = (rwgh)(1 m) = 11050 kg>m 219.81 m>s 2(8 m)(1 m) 3

x

2

(a)

= 82.40 kN>m

y

Thus, Fx =

Fy

1 (8 m)(82.40 kN>m) = 329.62 kN 2

B

x C

Using the table on the inside back cover for a triangle, from the x axis, this component acts at 1 y = (8 m) = 2.67 m Ans. 3 Vertical Force Component. The vertical force is equivalent to the weight of the water contained within the volume of the exparabolic segment ABCA, Fig. 2–41b. From the inside back cover, the area of this segment is AABCA = 13 ba. Thus, Fy = 1rwg2AABCA (1 m) =

3 11050 kg>m3 219.81 m>s2 2 4 c

Resultant Force.

Fx y 82.40 kNm

x

A 2m (b)

1 (2 m)(8 m) d (1 m) = 54.94 kN 3

This force acts through the centroid of the volume (area); that is, from the inside back cover, x =

8m

3 (2 m) = 1.5 m 4

Ans.

The resultant force is therefore

FR = 2(329.62 kN)2 + (54.94 kN)2 = 334 kN

Ans.

Fig. 2–41

98

Chapter 2

EXAMPLE

F l u i d S ta t i C S

2.16 The semicircular plate in Fig. 2–42a is 4 m long and acts as a gate in a channel. Determine the resultant force the water pressure exerts on the plate, and then find the components of reaction at the hinge (pin) B and at the smooth support A. Neglect the weight of the plate.

A

2

SOLUTION Fluid Description. Water is assumed to be an incompressible fluid for which gw = (1000 kg/m3)(9.81 m/s2) = 9.81(103) N>m3.

3m

ANALYSIS I

B

We will first determine the horizontal and vertical components of the resultant force acting on the plate. (a) D

Horizontal Force Component. The vertical projected area of AB is shown in Fig. 2–42b. The intensity of the distributed loading at B (or E) is

A

wB = gwhBb = [9.811103 2N>m3](6 m)(4 m) = 235.441103 2N>m

Fy

Therefore, the horizontal froce component is

d

1 [235.441103 2N>m)(6 m) = 706.321103 2N = 706.32 kN 2 1 This force acts at h = (6 m) = 2 m. 3

C

Fx =

Fx h B

E (b) D

A

D

A

+ C

C Force acting on segment CA (d) B Force acting on segment CB

Vertical Force Component. From Fig. 2–42b, note that the vertical force pushing up on segment BC is due to the water pressure under this segment. It is equal to the imaginary weight of water contained within BCDAB, Fig. 2–42c. And the vertical force pushing down on segment AC in Fig. 2–42b is due to the weight of water contained within CDAC, Fig. 2–42d. The net vertical force acting on the entire plate is therefore the difference in these two weights, namely, an upward force equivalent to the weight of water contained within the semicircular region BCAB in Fig. 2–42b. Thus, Fy = gwVBCAB = [9.811103 2N>m3]e

1 3 p(3 m)2 4 f (4 m) 2

= 176.58p1103) N = 176.58p kN

The centroid of this semicircular volume of water can be found on the inside back cover.

(c)

Fig. 2–42

d =

4(3 m) 4r 4 = = m p 3p 3p

2.10

99

hydroStatiC ForCe on an inClined plane or CurVed SurFaCe determined by projeCtion

Resultant Force.

The magnitude of the resultant force is therefore

FA

FR = 2F 2x + F 2y = 2(706.32 kN)2 + (176.58p kN)2 = 898 kN Ans.

Reactions. The free-body diagram of the plate is shown in Fig. 2–42e. Applying the equations of equilibrium, we have + c ΣFy = 0;

-By + 176.58p kN = 0

176.58 4m

kN

2

6m

706.32 kN

By = 176.58p kN = 555 kN

2m

⤿

Ans. 4 + ΣMB = 0; FA(6 m) - (706.32 kN) (2 m) - (176.58p kN ) a mb = 0 p

+ g Fx = 0 S

FA = 353.16 kN = 353 kN

Bx

Ans.

By

706.32 kN - 353.16 kN - Bx = 0 Bx = 353.16 kN = 353 kN

(e)

Ans.

ANALYSIS II We can also determine the resultant force components directly using integration. In Fig. 2–42f, notice how the pressure varies along the surface section. To simplify the analysis, we will use polar coordinates because of the circular shape of the plate. The differential strip of width b has an area of dA = b ds = (4 m)(3 du m) = 12 du m2 . The pressure acting on it is p = gwh = [9.811103 2N/m3 ]1 3 - 3 cos u2 m = 29.431103 211 - cos u2N>m2

For the horizontal component, dFx = p dA sin u, and so

h

dF = pdA

3m d

dA

O

p sin u dA = 29.431103 2 (1 - cos u)(sin u)(12 du) = 706.32 kN LA L0 p

Fx =

3m

In a similar manner, the y component can be found from dFy = p dA cos u. You may wish to evaluate this to verify our previous result for Fy.* (f)

Fig. 2–42 (cont.)

*Be aware that this method can only be used to determine the components of the resultant force. The resultant force cannot be found from FR = not account for the changing direction of the force.

LA

p dA because it does

100

Chapter 2

EXAMPLE

F l u i d S ta t i C S

2.17 The plug in Fig. 2–43a is 50 mm long and has a trapezoidal cross section. If the tank is filled with crude oil, determine the resultant vertical force acting on the plug due to the oil pressure.

2

A G

H

F

60 mm

60 mm

E

B 15 mm

15 mm C

20 mm 10 mm

D 20 mm

10 mm

10 mm (a)

10 mm (b)

Fig. 2–43

SOLUTION Fluid Description. We take the oil to be incompressible, and from Appendix A, ro = 880 kg>m3. Analysis. With reference to Fig. 2–43b, the force pushing down on the plug is due to the weight of oil contained within region ABEFA. The force pushing up occurs due to pressure along sides BC and ED and is equivalent to the weight of oil within the dark brown strips, ABCGA and FEDHF. We have + T FR = rog3VABEFA - 2VABCGA 4

= (880 kg>m3)(9.81 m>s2) c (0.06 m)(0.04 m)(0.05 m) - 2 3 (0.06 m)(0.01 m) +

= 0.453 N

1 (0.01 m)(0.015 m) 4 (0.05 m) d 2

Since the result is positive, this force acts downward on the plug.

Ans.

2.11

2.11

buoyanCy

101

BUOYANCY

The Greek scientist Archimedes (287–212 b.c.) discovered the principle of buoyancy, which states that when a body is placed in a static fluid, it is buoyed up by a force that is equal to the weight of the fluid displaced by the body. To show why this is so, consider the submerged body in Fig. 2–44a. Due to fluid pressure, the vertical resultant force acting upward on the bottom surface of the body, ABC, is equivalent to the weight of fluid contained above this surface, that is, within the volume ABCEFA. Likewise, the resultant force due to pressure acting downward on the top surface of the body, ADC, is equivalent to the weight of fluid contained within the volume ADCEFA. The difference in these forces acts upward, and is the buoyant force. It is equivalent to the weight of an imaginary amount of fluid contained within the volume of the body, ABCDA. This force Fb acts through the center of buoyancy, Cb, which is located at the centroid of the volume of liquid displaced by the body. If the density of the fluid is constant, then this force will remain constant, regardless of how deep the body is placed within the fluid. These same arguments can also be applied to a floating body, as in Fig. 2–44b. Here the displaced amount of fluid is within the region ABCA, so that the buoyant force is equal to the weight of fluid within this displaced volume, and the center of buoyancy Cb is at its centroid. If a hydrostatic problem that involves buoyancy is to be solved, then it may be necessary to investigate the forces acting on the free-body diagram of the body. This requires the buoyant force to be shown acting upward at the center of buoyancy, while the body’s weight acts downward, through its center of gravity.

E

F

A

2

This cargo ship has a uniform weight distribution and it is empty, as noticed by how high it floats in the water relative to the top of the red boundary.

C Cb B

D

A

Cb

C Floating body

B

(b)

Submerged body (a)

Fig. 2–44

The density of ice is about nine-tenths that of sea water. It is for this reason that ninety percent of an iceberg is submerged, while only ten percent remains above the surface. (Jurgen Ziewe/Shutterstock)

102

2

Chapter 2

F l u i d S ta t i C S

Sl

Sw 5 1.0

h

Fb Fb

W W (a)

(b) Hydrometer

Fig. 2–45

Hydrometer. The principle of buoyancy can be used in a practical way to measure the specific gravity of a liquid using a device called a hydrometer. As shown in Fig. 2–45a, it consists of a hollow glass tube that is weighted at one end. If the hydrometer is placed in a liquid such as pure water, it will float in equilibrium when its weight W equals the weight of displaced water, that is, when W = gwV0, where V0 is the volume of water displaced. If the stem is marked at the water level as 1.0, Fig. 2–45a, then this position can indicate water’s specific gravity, since for water, Sw = gw >gw = 1.0, Eq. 1–10. When the hydrometer is placed in another liquid, it will float either higher or lower, depending upon the liquid’s specific weight gl relative to water. If the liquid has a lower density than water, such as kerosene, then a greater volume of the liquid must be displaced in order for the hydrometer to float. Consider this displaced volume to be V0 + Ah, where A is the cross-sectional area of the stem, Fig. 2–45b. Now W = gl(V0 + Ah). If Sl is the specific gravity of the liquid, then gl = Slgw, and so for both cases, equilibrium of the hydrometer having weight W requires W = gwV0 = Sl gw(V0 + Ah) Solving for Sl yields Sl =

V0 V0 + Ah

Using this equation, for each depth h, calibration marks can be placed on the stem to indicate the specific gravities, Sl, for various types of liquids. In the past, hydrometers were often used to test the specific weight of acid in automobile batteries. When a battery is fully charged, the hydrometer will float higher in the acid than when the battery is discharged.

2.11

EXAMPLE

buoyanCy

103

2.18

The 500-N flat-bottom container in Fig. 2–46a is 600 mm wide and 900 mm long. Determine the depth the container will float in the water (a) when it carries the 200-N steel block, and (b) when the block is suspended directly beneath the container, Fig. 2–46b. Take gst = 77.0 kN>m3.

2

350 mm

d9 d 600 mm

(a) (b)

Fig. 2–46

SOLUTION Fluid Description. The water is assumed to be incompressible, and has a specific weight of gw = (1000 kg/m3)(9.81 m/s2) = 9.81(103) N/m3. Analysis. In each case, for equilibrium, the weight of the container and block must be equal to the weight of the displaced water, which creates the buoyant force. Part (a). From the free-body diagram in Fig. 2–46c, we require + c ΣFy = 0;

-(Wcont. + Wblock) + 1Fb 2 cont. = 0

-(500 N + 200 N) + [9.811103) N>m3][(0.6 m)(0.9 m)d] = 0 d = 0.1321 m = 132 mm 6 350 mm OK Ans. Part (b). In this case, Fig. 2–46d, we first have to find the volume of steel used to make the block. Since the specific weight of steel is given, then Vst = Wst >gst. From this the buoyancy force can be determined. Thus, + c ΣFy = 0;

-Wcont. - Wblock + (Fb)cont. + (Fb)block = 0

-500 N - 200 N + [9.811103 2N>m3][(0.6 m)(0.9 m)d′] + [9.81(103) N>m3 ]C

Wblock 1 Wcont.

(Fb)cont.

(c)

Wcont.

(Fb)cont.

200 N S = 0 77.01103 2 N>m3

d′ = 0.1273 m = 127 mm

(Fb)block

Ans.

Notice that here the container floats higher in the water because when the block is supported under the water, its buoyancy force reduces the force needed to support it. Also, note that the answer is independent of the depth the block is suspended in the water.

Wblock (d)

104

Chapter 2

F l u i d S ta t i C S

2.12

STABILITY

A body can float in a liquid (or gas) in stable, unstable, or neutral equilibrium. To illustrate each of these cases, consider a uniform lightweight bar with a weight attached to its end so that its center of gravity is at G. 2

This boat is used to transport cars across a river. Care must be taken so that it does not become unstable if it makes a sharp turn or is overloaded on one side.

Stable equilibrium. If the bar is placed in a liquid so that its center of gravity is below its center of buoyancy, then a slight angular displacement of the bar, Fig. 2–47a, will create a couple moment between the weight and buoyant force that will cause the bar to restore itself to the vertical position. This is a state of stable equilibrium. Unstable equilibrium. If the bar is placed in the liquid so that its center of gravity is above the center of buoyancy, Fig. 2–47b, then a slight angular displacement will create a couple moment that will cause the bar to rotate farther from its equilibrium position. This is unstable equilibrium. Neutral equilibrium. If the weight is removed so that the weight and buoyant force are balanced, then its center of gravity and center of buoyancy will coincide, Fig. 2–47c. Any rotation of a submerged bar will cause it to remain in the newfound equilibrium position. This is a state of neutral equilibrium.

Fb

Cb

Cb

G

G Cb

W

Stabilizing couple moment

G

G

W Stable equilibrium (a)

Unstable equilibrium (b)

Fb Cb

G

Cb

G

W

Neutral equilibrium (c)

Fig. 2–47

Fb Cb

Destabilizing couple moment

2.12

Although the bar in Fig. 2–47b is in unstable equilibrium, some floating bodies can maintain stable equilibrium when their center of gravity is above their center of buoyancy. For example, consider the ship in Fig. 2–48a, which has its center of gravity at G and its center of buoyancy at Cb. When the ship undergoes a slight roll, which occurs at the water level about point O, Fig. 2–48b, the new center of buoyancy Cb′ will be to the left of G. This is because a portion of the displaced water OABO is gained on the left, while an equivalent portion ODEO is lost on the right. If we establish a vertical line through Cb′ (the line of action of Fb), it will intersect the centerline of the ship at point M, which is called the metacenter. If M is above G, Fig. 2–48b, the clockwise couple moment created by the buoyant force and the ship’s weight will tend to restore the ship to its equilibrium position. Therefore, the ship is in stable equilibrium. If a ship has a large, high deck loading, as in Fig. 2–48c, M will then be below G. In this case, the counterclockwise couple moment created by Fb and W will cause the ship to become unstable and to roll over, a condition that obviously must be avoided when designing or loading any ship. Realizing this danger, however, maritime engineers design modern cruise ships so that their centers of gravity are as high as possible above their centers of buoyancy, while remaining below M. Doing this will put G close to the metacenter, and so the restoring couple moment will be small, causing the ship to roll back and forth very slowly in the water. When G and M are farther apart, the restoring moment is larger and the back-and-forth motion is faster, which can cause discomfort to the passengers. This same rule related to the location of the metacenter also applies to the bar in Fig. 2–47. Because the bar is thin, the metacenter M is on the centerline of the bar and coincides with Cb, and so when M (or Cb) is above G it is in stable equilibrium, Fig. 2–47a; when it is below G it is in unstable equilibrium, Fig. 2–47b; and when it is at G it is in neutral equilibrium, Fig. 2–47c.

G O Cb

2

(a) W

M E

G O

A

D

B C b9

F Fb

OM . OG Stable equilibrium (b)

W

IMPORTANT POIN T S • The buoyant force on a body is equal to the weight of the fluid the body displaces. It acts upward through the center of buoyancy, which is located at the centroid of the displaced volume of fluid.

G

M O

C 9b

• A hydrometer uses the principle of buoyancy to measure the specific gravity of a liquid.

Fb

• A floating body can be in stable, unstable, or neutral equilibrium. If its metacenter M is above the center of gravity G of the body, then the body will float in stable equilibrium. If M is below G, then the body will be unstable.

105

Stability

OG . OM Unstable equilibrium (c)

Fig. 2–48

106

Chapter 2

EXAMPLE

F l u i d S ta t i C S

2.19 The wooden block (cube) in Fig. 2–49a is 0.2 m on each side. A vertical force F is applied at the center of one of its sides and pushes the edge of the block to the water surface so that it is held at an angle of 20°. Determine the buoyant force on the block, and show that the block will be in stable equilibrium when the force F is removed.

F

2 208 0.2 m

SOLUTION Fluid Description. The water is assumed to be incompressible, and has a density of rw = 1000 kg>m3. 0.2 m

Analysis. To find the buoyant force, we must first find the submerged volume of the block, shown on the free-body diagram in Fig. 2–49b. It is

(a) y

V sub = 10.2 m2 3 -

(0.2 m) tan 208

F

Then W

208

0.2 m

M G

Cb

x

Fb

Fb = rwgVsub = 3 1000(9.81) N>m3 4 3 6.5441110-3 2 m3 4 = 64.2 N Ans.

This force acts through the centroid of volume Vsub (or area) having coordinates measured from the x, y axes of

y x

0.2 m (b)

Fig. 2–49

yA Σ∼ y = = ΣA = 0.0832 m

1 (0.2 m)(0.2 tan 20°)(0.2 m) = 6.5441110-3 2 m3 2

Σ∼ xA x = = ΣA

2 1 (0.2 m) a b (0.2 m)(0.2 m tan 20°) 3 2 1 10.2 m2 2 - a b (0.2 m)(0.2 m tan 20°) 2

0.1 m10.2 m2 2 -

= 0.0926 m

1 1 10.1 m210.2 m2 2 - c (0.2 m) - a b (0.2 m tan 20°) d a b (0.2 m)(0.2 m tan 20°) 3 2 1 (0.2 m)2 - a b (0.2 m)(0.2 m tan 20°) 2 This location for Fb is to the left of the block’s center of gravity (0.1 m, 0.1 m), and so the clockwise moment of Fb about G will restore the block when the force F is removed. Hence the block is in stable equilibrium. In other words, the metacenter M will be above G, Fig. 2–49b. Although it is not part of this problem, the force F and the block’s weight W can be determined by applying the vertical force and moment equilibrium equations to the block.

2.13

2.13

ConStant tranSlational aCCeleration oF a liquid

107

CONSTANT TRANSLATIONAL ACCELERATION OF A LIQUID

In this section we will discuss both horizontal and vertical constant accelerated motion of a container of liquid, and we will study how the pressure varies within the liquid for these two motions.

Constant Horizontal Acceleration. If the container in Fig. 2–50a has a constant velocity, vc, then the surface of the liquid will remain horizontal, since it is in equilibrium. As a result, the pressure exerted on the walls of the container can be determined in the usual manner using p = gh. If the container undergoes a constant acceleration ac to the right, however, then the liquid surface will begin to rotate clockwise about the center of the container and will eventually be brought to a fixed tilted position u, Fig. 2–50b. After this adjustment, all the liquid will behave as though it were a solid. No shear stress will be developed between layers of the liquid since there is no relative motion between these layers. To study the effects of this motion, a force analysis using a free-body diagram of vertical and horizontal differential elements of the liquid will now be considered. Vertical Element. For this case, the differential element extends downward a distance h from the liquid surface and has a cross-sectional area ∆A, Fig. 2–50c. The two vertical forces acting on it are the weight of the contained liquid, ∆W = g∆V = g(h∆A), and the pressure force acting upward on its bottom, p∆A. Equilibrium exists in the vertical direction since no acceleration occurs in this direction. + c ΣFy = 0;

p∆A - g(h∆A) = 0 p = gh

2

Tank cars carry a variety of liquids. Their endplates must be designed to resist the hydrostatic pressure of the liquid within the car caused by any anticipated acceleration of the car.

(2–16)

This result indicates that the pressure at any depth from the inclined liquid surface is the same as if the liquid were static.

L 2

L

vc

L 2

u

ac

h

u

DW

p DA

Constant velocity (a)

Constant acceleration open container (b)

Fig. 2–50

Constant acceleration (c)

ac

108

Chapter 2

F l u i d S ta t i C S

Horizontal Element. Here the differential element has a length x h1 h2

x

p2 DA

ac p1 DA

and cross-sectional area ∆A, Fig. 2–50d. The horizontal forces acting on it are caused by the pressure of the adjacent liquid on each of its ends. Since the mass of the element is ∆m = ∆W>g = g(x∆A)>g, the equation of motion becomes

2

+ ΣFx = max; S

Constant acceleration (d)

p2 ∆A - p1 ∆A = p 2 - p1 =

Imaginary liquid surface hA A

u

g(x ∆A) ac g gx a g c

(2–17)

B

pA 5 ghA

Using p1 = gh1 and p2 = gh2, we can also write this expression as

hC ac C

pC 5 ghC

ac h2 - h1 = x g

(2–18)

Constant acceleration of a completely filled closed container (e)

Fig. 2–50 (cont.)

As noted in Fig. 2–50d, the term on the left represents the slope of the liquid’s free surface. Since this is equal to tan u, then

tan u =

ac g

(2–19)

Slope of liquid surface

If the container is completely filled and has a closed lid on its top, as in Fig. 2–50e, then the liquid cannot pivot about the center of the container. Rather, the lid constrains the liquid, such that the upward pressure on the lid can be replaced by an “imaginary liquid” that creates the same pressure distribution on the lid. The surface of this liquid must pivot about the corner point B, since the pressure at this point must remain the same with or without an acceleration. Although this is the case, we can still find the tilt angle u using Eq. 2–19. Once the surface is established, the pressure at any point in the liquid can be determined by finding the vertical distance from this imaginary surface to the point, Eq. 2–16. For example, at A, pA = ghA. Also, at the bottom of the container at C, the pressure is pC = ghC.

2.13

ConStant tranSlational aCCeleration oF a liquid

109

ac

2 x p1 D A

p2 DA

Constant acceleration (a)

Constant Vertical Acceleration. When a container is accelerated upward at ac, the liquid surface maintains its horizontal position; however, the pressure within the liquid will change. To study this effect, we will again select horizontal and vertical differential elements and draw their free-body diagrams.

ac

Horizontal Element. Since the horizontal element in Fig. 2–51a is at the same depth in the liquid, the pressure of the adjacent liquid on each of its ends exerts forces as shown. No motion occurs in this direction, so + ΣFx = 0; S

p2 ∆A - p1 ∆A = 0 p2 = p1

h

Hence, as in the case of a static fluid, for vertical acceleration the pressure is the same at points that lie in the same horizontal plane.

Vertical Element. The forces acting on the vertical element of depth h and cross section ∆A, Fig. 2–51b, consist of the element’s weight ∆W = g∆V = g(h ∆A) and the pressure force on its bottom. Since the mass of the element is ∆m = ∆W>g = g(h ∆A)>g, application of the equation of motion yields + c ΣFy = may;

p∆A - g(h ∆A) = p = gh a 1 +

g(h ∆A) ac g ac b g

DW

p DA

Constant acceleration (b)

Fig. 2–51

(2–20)

Thus, the pressure within the liquid will increase by gh1ac >g2 when the container is accelerated upward. If it has a downward acceleration, the pressure will decrease by this amount. If free-fall occurs, then ac = -g and the (gage) pressure throughout the liquid will be zero.

110

Chapter 2

EXAMPLE

F l u i d S ta t i C S

2.20 The tank on the truck in Fig. 2–52a is filled to its top with gasoline. If the truck has a constant acceleration of 4 m>s2, determine the pressure at points A, B, C, and D within the tank.

2

Imaginary liquid surface h 5 3.262 m 4 ms2

B

A

2m

2m D

B

u

A

C

D

C

8m

8m (a)

(b)

Fig. 2–52

SOLUTION Fluid Description. We assume that gasoline is incompressible, and from Appendix A, it has a density of rg = 726 kg>m3. Analysis. When the truck is at rest or moving with a constant velocity, the (gage) pressures at A and B are zero since the gasoline surface remains horizontal. When the truck accelerates, the surface is imagined to pivot at A and tilt back, Fig. 2–52b. We can determine the height h using Eq. 2–18. ac h2 - h1 = x g 4 m>s2 h - 0 = 8m 9.81 m>s2 h = 3.262 m The top of the tank prevents the formation of this sloped surface, and so the imaginary gasoline surface exerts a pressure on the top. This pressure can be obtained using Eq. 2–16, p = gh. Thus, pA = gghA = 1726 kg>m3 219.81 m>s2 2(0) = 0

Ans.

pD = gghD = 1726 kg>m3 219.81 m>s2 2(2 m) = 14.2 kPa

Ans.

pB = gghB = 1726 kg>m 219.81 m>s 2(3.262 m) = 23.2 kPa 3

2

Ans.

pC = gghC = 1726 kg>m3219.81 m>s22(3.262 m + 2 m) = 37.5 kPa Ans.

2.13

EXAMPLE

111

ConStant tranSlational aCCeleration oF a liquid

2.21

The container in Fig. 2–53a is 1.25 m wide and is filled with crude oil to a height of 2 m. Determine the resultant force the oil exerts on the container’s side and on its bottom if a crane begins to hoist the container upwards with an acceleration of 3 m>s2.

3 m/s2

2

SOLUTION Fluid Description. Using Appendix 8.63281103) N>m3.

The oil is assumed to be incompressible. A, go = ro g = (880 kg>m3)(9.81 m>s2) =

A

2m

Analysis. The (gage) pressure at A is zero, and hence the pressure at B and C can be determined using Eq. 2–20. Since ac = +3 m>s2, we have

C

B 1.5 m

3 m>s2 ac p = goh a 1 + b = [8.63281103 2 N>m3](2 m) a 1 + b g 9.81 m>s2 3

(a) 3 m/s2

2

= 22.55110 ) N>m

Since the tank has a width of 1.25 m, the intensity of the distributed load at the bottom of the tank, Fig. 2–53b, is w = pb = [22.551103 2 N>m2](1.25 m) = 28.182 1103 2 N>m Side of Tank. we have

(FR)s =

A

2m 28.18(103) N

For the triangular distributed load acting on side AB,

1 [28.1821103 2 N>m)](2 m) = 28.1821103 2 N = 28.2 kN Ans. 2

C

28.18(103) N/m B 1.5 m (b)

Fig. 2–53

Bottom of Tank. The bottom of the tank is subjected to a uniform distributed load. Its resultant force is (FR)b = [28.1821103 2 N>m)](1.5 m) = 42.2731103 2 N = 42.3 kN Ans.

112

Chapter 2

F l u i d S ta t i C S

2.14

2

STEADY ROTATION OF A LIQUID

If a liquid is placed into a cylindrical container that begins to rotate at a constant angular velocity v, Fig. 2–54a, the shear stress developed within the liquid will cause it to rotate with the container. Eventually, no relative motion within the liquid will occur, and the system will then rotate as a solid body. When this happens, the velocity of each fluid particle will depend on its distance from the axis of rotation. Those particles that are closer to the axis will move slower than those farther away. This motion will cause the liquid surface to form the shape of a forced vortex.

Vertical Element. If the free-body diagram of a vertical differential element of height h and cross-sectional area ∆A is considered, Fig. 2–54a, then, as in the proof of Eq. 2–16, it can be shown that the pressure in the liquid will increase with depth from the free surface, that is, p = gh. This is because there is no acceleration in this direction.

Ring Element. The constant angular rotation v of the cylinder–liquid system does, however, produce a pressure difference or gradient in the radial direction due to the radial acceleration of the liquid particles. This acceleration is the result of the changing direction of the velocity of each particle. If a particle is at a radial distance r from the axis of rotation, then from dynamics (or physics), its acceleration has a magnitude of ar = v2r, and it acts toward the center of rotation. To study the radial pressure gradient, we will consider a ring element having a radius r, thickness ∆r, and height ∆h, Fig. 2–54b. Due to the rotation, the pressure p on the inside of the ring will increase to p + (0p>0r) ∆r on the outside.* v

Liquid surface level for v 5 0

H –– 2 H –– 2

R

DW

h

p DA

Forced vortex (a)

Fig. 2–54 *The partial derivative is used here since the pressure is a function of both depth and radius.

2.14

Steady rotation oF a liquid

Since the mass of the ring is ∆m = ∆W>g = g ∆V>g = g(2pr) ∆r ∆h>g, the equation of motion in the radial direction becomes ΣFr = mar; - c p + a

113

v

g(2pr)∆r∆h 2 0p b ∆r d (2pr∆h) + p(2pr∆h) = vr g 0r

R

0p gv2 = a br g 0r

r

Dh

2

p0

Dr

Integrating, we obtain p

p = a

gv2 2 br + C 2g

p1

( rp ) Dr (b)

The constant of integration can be found by realizing that the point on the vertical axis at the free surface, where r = 0, is p0 = 0, Fig. 2–54c, and so C = 0. Therefore, p = a

gv2 2 br 2g

(2–21)

To obtain the equation of the free surface of the liquid, Fig. 2–54c, set p = gh, so that h = a

v2 2 br 2g

(2–22)

This is the equation of a parabola. Specifically, the liquid as a whole forms a surface that describes a paraboloid. Since the interior radius of the container is R, the height of this paraboloid is H = v2 R2 >2g, Fig. 2–54c. Its volume is one-half its base area pR2 times its height, H. As a result, during the rotation, the high and low points of the liquid surface will be H>2 above and below the liquid surface when the liquid is at rest, Fig. 2–54c. If the rotating container has a closed lid, then an imaginary free surface of the paraboloid can be established above the lid, and the pressure at any point in the liquid is determined by measuring the depth h from this surface.

H –– 2 H –– 2

Liquid surface level when v 5 0 2 h 5 v r2 2g

( )

p0 5 0 r

R

(c)

Fig. 2–54 (cont.)

p 5 gh

114

Chapter 2

F l u i d S ta t i C S

I MPO RTA N T PO I N T S • When an open container of liquid is uniformly accelerated

horizontally, the surface of the liquid will be inclined at an angle u, determined from tan u = ac >g. The pressure varies linearly with depth from this surface, p = gh. If a lid is on the container, then an imaginary surface should be established, and the pressure at any point can be determined using p = gh, where h is the vertical distance from the imaginary surface to this point.

2

• When a container of liquid is uniformly accelerated vertically,

the surface of the liquid remains horizontal. If this acceleration is upward, the pressure at a depth h is increased by gh(ac >g); and if the acceleration is downward, the pressure is decreased by this amount.

• When a cylindrical container of liquid has a constant rotation about a fixed axis, the liquid surface forms a forced vortex having the shape of a paraboloid. Once the surface of the rotating liquid is established, using h = (v2 >2g)r 2, the pressure varies with depth from this surface, p = gh. If a lid is on the container, then an imaginary liquid surface should be established, and the pressure at any point can be determined using the vertical distance from this surface.

EXAMPLE

2.22

1.25 m

The closed cylindrical drum in Fig. 2–55a is filled with crude oil to the level shown. If the pressure of the air within the drum is atmospheric due to the hole in the center of the lid, determine the pressures at points A and B when the drum and oil attain a constant angular velocity of 12 rad>s.

0.5 m

SOLUTION

1.5 m

Fluid Description. The oil is assumed incompressible, and from Appendix A it has a specific weight of B

A

(a)

Fig. 2–55

12 rad/s

go = ro g = 1880 kg>m3 219.81 m>s2 2 = 8.63281103) N>m3

2.14

Steady rotation oF a liquid

115

Analysis. Before finding the pressure, we must define the shape of the oil surface. As the drum rotates, the oil takes the shape shown in Fig. 2–55b. Since the volume of open space within the drum must remain constant, this volume must be equivalent to the volume of the shaded paraboloid of unknown radius r and height h. Since the volume of a paraboloid is one-half that of a cylinder having the same radius and height, we require

2

Vcyl = Vparab p10.625 m2 2(0.5 m) =

0.625 m

1 2 pr h 2

r 2h = 0.3906

(1)

r Top

Also, from Eq. 2–22, for this contained paraboloid within the drum, 112 rad>s2 2 v2 h = a b r2 = J R r2 2g 219.81 m>s2 2

12 rad/s

0.5 m h9 5 2.867 m h 5 1.693 m

h = 7.3394r 2

(2)

1.5 m

Solving Eqs. 1 and 2 simultaneously, we get r = 0.4803 m,

Imaginary liquid surface

B

h = 1.693 m

Without the lid, the oil would rise to a level h′, Fig. 2–55b, which is 112 rad>s2 2 v2 h′ = a b R2 = c d 10.625 m2 2 = 2.867 m 2g 219.81 m>s2 2

A (b)

Fig. 2–55 (cont.)

Since the free surface of the oil has now been defined, the pressures at A and B are pA = ghA = [8.63281103 2 N>m3](2 m - 1.693 m) = 2.6481103 2 Pa = 2.65 kPa

Ans.

= 27.401103 2 Pa = 27.4 kPa

Ans.

pB = ghB = [8.63281103 2 N>m3][2.867 m + 12 m - 1.693 m2]

Although it is not part of this problem, realize that if a cap were placed on the hole, and the pressure of the air within the drum were increased to 30 kPa, then this pressure would simply be added to the pressures at A and B.

116

Chapter 2

F l u i d S ta t i C S

References 1. I. Khan, Fluid Mechanics, Holt, Rinehart and Winston, New York,

NY, 1987. 2. A. Parr, Hydraulics and Pneumatics, Butterworth-Heinemann,

2

Woburn, MA, 2005. 3. The U.S. Standard Atmosphere, U.S Government Printing Office,

Washington, DC. 4. K. J. Rawson and E. Tupper, Basic Ship Theory, 2nd ed., Longmans,

London, UK, 1975. 5. S. Tavoularis, Measurements in Fluid Mechanics, Cambridge

University Press, New York, NY, 2005. 6. R. C. Baker, Introductory Guide to Flow Measurement, John Wiley,

New York, NY, 2002. 7. R. W. Miller, Flow Measurement Engineering Handbook, 3rd ed.,

McGraw-Hill, New York, NY, 1996. 8. R. P. Benedict, Fundamentals of Temperature, Pressure, and Flow

Measurement, 3rd ed., John Wiley, New York, NY, 1984. 9. J. W. Dally et al., Instrumentation for Engineering Measurements, 2nd

ed., John Wiley, New York, NY, 1993. 10. B. G. Liptak, Instrument Engineer’s Handbook: Process Measurement

and Analysis, 4th ed., CRC Press, Boca Raton, FL, 2003. 11. F. Durst et al., Principles and Practice of Laser-Doppler Anemometry,

2nd ed., Academic Press, New York, NY, 1981.

F UNDAMEN TAL PRO B L EM S The solutions to all fundamental problems are given in the back of the book.

SEC. 2.1–2.5 F2–1. Water fills the pipe AB such that the absolute pressure at A is 400 kPa. If the atmospheric pressure is 101 kPa, determine the resultant force the water and surrounding air exert on the cap at B. The inner diameter of the pipe is 50 mm.

F2–2. The container is partially filled with oil, water, and air. Determine the pressures at A, B, and C. Take rw = 1000 kg>m3, ro = 830 kg>m3.

Air 1.25 m

1m

Oil B

B 0.3 m

1.25 m Water A

1.25 m

0.4 m C

A

Prob. F2–1

Prob. F2–2

0.25 m

117

Fundamental problemS

F2–5. The air pressure in the pipe at A is 300 kPa. Determine the water pressure in the pipe at B.

SEC. 2.6 F2–3. The U-tube manometer is filled with mercury, having a density of rHg = 13 550 kg>m3. Determine the differential height h of the mercury when the tank is filled with water.

2 A

0.5 m

0.3 m

2m

0.2 m

3m

h

0.4 m B

Prob. F2–3

0.35 m

Prob. F2–5

F2–4. The tube is filled with mercury from A to B, and with water from B to C. Determine the height h of the water column for equilibrium.

F2–6. Determine the absolute pressure of the water in the pipe at B if the tank is filled with crude oil to the depth of 1.5 m. patm = 101 kPa.

0.5 m

C

A 308

1.5 m 0.6 m

h

0.1 m B 0.4 m

0.3 m

0.5 m B

0.4 m

Prob. F2–4

Prob. F2–6

118

Chapter 2

F l u i d S ta t i C S

SEC. 2.7–2.9 F2–7. The bin is 1.5 m wide and is filled with water to the level shown. Determine the resultant force on the side AB and on the bottom BC.

F2–9. The 2-m-wide container is filled with water to the depth shown. Determine the resultant force on the side panels A and B. How far does each resultant act from the surface of the water?

2

A

0.9 m A 0.6 m

2.5 m B 0.6 m 2m 0.6 m

B

C

0.6 m

2m

Prob. F2–9

Prob. F2–7

F2–10. Determine the resultant force of the water acting on the triangular end plate A of the trough. Neglect the width of the opening at the top. How far does this force act from the surface of the water? F2–8. The bin is 2 m wide and is filled with oil to the depth shown. Determine the resultant force acting on the inclined side AB. Take ro = 900 kg>m3.

1.2 m

B

A

3m 608 A

Prob. F2–8

0.3 m 0.3 m

Prob. F2–10

119

Fundamental problemS F2–11. Determine the resultant force of the water acting on the circular glass plate that is bolted to the side panel of the tank. Also, determine the location of the center of pressure along this inclined side, measured from the top.

F2–13. The 0.5-m-wide inclined plate holds water in a tank. Determine the horizontal and vertical components of force and the moment that the fixed support at A exerts on the plate. 2 3m 308

3m

A

Prob. F2–13 2m

1m

F2–14. Determine the resultant force the oil exerts on the semicircular surface AB. The tank has a width of 3 m. Take ro = 900 kg>m3. 1208 3m

Prob. F2–11 1.5 m

A

B 0.5 m

1m

F2–12. The tank is filled with water and kerosene to the depths shown. Determine the total resultant force the liquids exert on side AB of the tank. The tank has a width of 2 m. Take rw = 1000 kg>m3, rk = 814 kg>m3.

1m

Prob. F2–14

F2–15. Determine the resultant force the water exerts on side AB and on side CD of the inclined wall. The wall is 0.75 m wide.

0.5 m B

A

C

1m 2m 3m 608 A

Prob. F2–12

608

458 B

Prob. F2–15

D

120

Chapter 2

F l u i d S ta t i C S

SEC. 2.10

SEC. 2.11–2.12

F2–16. The tank has a width of 2 m and is filled with water. Determine the horizontal and vertical components of the resultant force acting on plate AB.

F2–19. The cylindrical cup A of negligible weight contains a 2-kg block B. If the water level of the cylindrical tank is h = 0.5 m before the cup is placed into the tank, determine h when the cup floats in the water.

2 1m 0.5 m

C

B

A

B

1.5 m 608

0.2 m

A

h

Prob. F2–16

F2–17. Determine the horizontal and vertical components of the resultant force the water exerts on plate AB and on plate BC. The width of each plate is 1.5 m.

A

0.4 m

Prob. F2–19

C

F2–20. The 3-m-wide cart is filled with water to the level of the dashed line. If the cart is given an acceleration of 4 m>s2, determine the angle u of the water surface and the resultant force the water exerts on the wall AB.

2m 458 B

Prob. F2–17

F2–18. The plate ABC is 2 m wide. Determine the angle u so that the normal reaction at C is zero. The plate is supported by a pin at A.

4 ms2 A u

A

B 1.5 m

5m

C

B

u C

2m

Prob. F2–18

Prob. F2–20

Fundamental problemS F2–21. The closed tank is filled with oil and given an acceleration of 6 m>s2. Determine the pressure on the bottom of the tank at points A and B. ro = 880 kg>m3.

121

F2–23. If the open cylindrical container rotates at v = 8 rad>s, determine the maximum and minimum pressure of the water acting on the bottom of the container.

2 v 2

6 ms

0.5 m

A

B 2m 1.5 m

Prob. F2–21 1m

1m

Prob. F2–23 F2–22. The open cylindrical container is filled with water to the level shown. Determine the smallest angular velocity that will cause the water to spill over the sides.

F2–24. The closed drum is filled with crude oil. Determine the pressure on the lid at A when the drum is rotating at 4 rad>s.

v

1.5 m A

3m 2m

1m

1m

Prob. F2–22

4 rads

Prob. F2–24

122

Chapter 2

F l u i d S ta t i C S

P R OB L EMS

2

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3. Also, assume all pressures are gage pressures unless stated otherwise.

SEC. 2.1–2.5

2–5. The oil derrick has drilled 5 km into the ground before it strikes a crude oil reservoir. When this happens, the pressure at the well head A becomes 25 MPa. Drilling “mud” is to be placed into the entire length of pipe to displace the oil and balance this pressure. What should be the density of the mud so that the pressure at A becomes zero?

2–1. Show that Pascal’s law applies within a fluid that is accelerating, provided there are no shearing stresses acting within the fluid. 2–2. In 1896, S. Riva-Rocci developed the prototype of the current sphygmomanometer, a device used to measure blood pressure. When it was worn as a cuff around the upper arm and inflated, the air pressure within the cuff was connected to a mercury manometer. If the reading for the high (or systolic) pressure is 120 mm and for the low (or diastolic) pressure is 80 mm, determine these pressures in pascals.

A

2–3. If the absolute pressure in a tank is 140 kPa, determine the pressure head in mm of mercury. The atmospheric pressure is 100 kPa.

5 km

*2–4. If the height of water in the piezometer is 475 mm, determine the absolute pressure at point A. Compare this pressure with that using kerosene. Take rw = 1000 kg/m3 and rkc = 814 kg>m3.

Prob. 2–5 475 mm

125 mm A

Prob. 2–4

2–6. Oxygen in a tank has an absolute pressure of 130 kPa. Determine the pressure head in mm of mercury. The atmospheric pressure is 102 kPa.

problemS 2–7. The structure shown is used for the temporary storage of crude oil at sea for later loading into ships. When it is not filled with oil, the water level in the stem is at B (sea level). Why? As the oil is loaded into the stem, the water is displaced through exit ports at E. If the stem is filled with oil, that is, to the depth of C, determine the height h of the oil level above sea level. Take ro = 900 kg>m3, rw = 1020 kg>m3.

123

2–11. The closed tank is filled with glycerin. If the space A is a vacuum and the valve at B is closed, determine the pressure near valve B where h = 6 m. Can the tank be drained if the valve at B is opened? 2

*2–8. If the water in the structure in Prob. 2–7 is displaced with crude oil to the level D at the bottom of the cone, then how high h will the oil extend above sea level? Take ro = 900 kg>m3, rw = 1020 kg>m3.

A

h

A h B

G B

Prob. 2–11

40 m

1m C D

5m 1m

E 10 m

Probs. 2–7/8 2–9. The soaking bin contains ethyl alcohol used for cleaning automobile parts. Suction pump C is to vary the pressure of the trapped air in enclosure B. If h = 1.75 m, determine the pressure developed at point A and in the trapped air. Take rea = 789 kg>m3 . 2–10. The soaking bin contains ethyl alcohol used for cleaning automobile parts. Suction pump C is to vary the pressure of the trapped air in enclosure B. If the pressure in the trapped air is pB = -5.50 kPa, determine the pressure developed at point A and the height h of the ethyl alcohol level in the bin. Take rea = 789 kg>m3 .

*2–12. The field storage tank is filled with water to the level shown. The standpipe is connected to the tank at B, and the system is open to the atmosphere at C. Determine the maximum pressure in the tank and the height h to which the water will rise in the stand pipe if the pressure of the trapped air in the tank (space A) is pA = 65 kPa. 2–13. The field storage tank is filled with oil to the level shown. The stand pipe is connected to the tank at B and open to the atmosphere at C. Determine the maximum pressure in the tank and the pressure of the trapped air in the tank (space A) if the water rises to a level h = 8 m in the stand pipe.

C

C

B A

h

2.5 m

h 2m

0.5 m

A B

Probs. 2–9/10

Probs. 2–12/13

124

2

Chapter 2

F l u i d S ta t i C S

2–14. At the bottom of a bottle a bubble within carbonated water has a diameter of 0.2 mm. Determine the bubble’s diameter when it reaches the surface. The temperature of the water and bubbles is 10°C, and the atmospheric pressure is 101 kPa. Assume that the density of the water is the same as that of pure water.

2–17. The density r of a fluid varies with depth h, although its bulk modulus EV can be assumed constant. Determine how the pressure varies with h. The density of the fluid at the surface is r0. 2–18. A heavy cylindrical glass is inverted and then placed down to the bottom of a swimming pool. Determine the height ∆h of water within the glass when it is at the bottom. Assume the air in the glass remains at the same temperature as in the atmosphere. Hint: Account for the change in volume of air in the glass due to the pressure change. The atmospheric pressure is patm = 101.3 kPa.

300 mm 4m F

Prob. 2–14 2–15. The tank is filled with water and gasoline at a temperature of 20°C to the depths shown. If the absolute air pressure at the top of the tank is 200 kPa, determine the gage pressure at the bottom of the tank. Would the results be different if the tank had a flat bottom rather than a curved one? The atmospheric pressure is 101 kPa.

0.3 m Dh

0.15 m

Prob. 2–18 2–19. Determine the temperature at an elevation of z = 15 km. Also, what is the pressure at this elevation? Assume the stratosphere begins at z = 11 km (see Fig. 2–11). *2–20. In the troposphere, which extends from sea level to 11 km, it is found that the temperature decreases with altitude such that dT>dz = -C, where C is the constant lapse rate. If the temperature and pressure at z = 0 are T0 and p0, determine the pressure as a function of altitude.

0.8 m 0.2 m

Prob. 2–15 *2–16. Due to its slight compressibility, the density of water varies with depth, although its bulk modulus EV = 2.20 GPa (absolute) can be considered constant. Accounting for this compressibility, determine the pressure in the water at a depth of 300 m, if the density at the surface of the water is r0 = 1000 kg>m3. Compare this result with water that is assumed to be incompressible.

2–21. A liquid has a density that varies with depth h, such that r = 11.75h + 8252 kg>m3, where h is in meters. Plot the variation of the pressure due to the liquid (vertical axis) versus depth for 0 … h … 30 m. Give values for increments of 5 m. 2–22. A liquid has a density that varies with depth h, such that r = 11.75h + 8252 kg>m3, where h is in meters. Determine the pressure due to the liquid when h = 25 m. 2–23. Using the data in Appendix A, make a graph showing how the atmospheric pressure p in kPa (vertical axis) varies with elevation h in meters. Plot values of p every 1000 m for 0 … h … 6 000 m.

125

problemS *2–24. As the weather balloon ascends, measurements indicate that the temperature begins to decrease linearly from T0 at z = 0 to Tf at z = h. If the absolute pressure of the air at z = 0 is p0, determine the pressure as a function of z.

2–26. Determine the maximum pressure in the system shown. Also, what is the pressure at B? The containers are filled with water. 2–27. Determine the pressure of the trapped air in the closed container. The containers are filled with water. 2

Air C

3m

1.5 m B 0.5 m A

Probs. 2–26/27 z

*2–28. Determine the difference in pressure pB - pA between the centers A and B of the pipes, which are filled with water. The mercury in the inclined-tube manometer has the level shown. Take SHg = 13.55. Water

Prob. 2–24 2–25. As the balloon ascends, measurements indicate that the temperature begins to decrease at a constant rate, from T = 15°C at z = 0 to T = 8.50°C at z = 1000 m. If the absolute pressure of the air at z = 0 is p = 101.3 kPa, plot the variation of pressure (vertical axis) versus altitude for 0 … z … 1000 m. Give values for increments of ∆z = 200 m.

A

B 100 mm C

250 mm

Mercury D

40°

Prob. 2–28 2–29. Butyl carbitol, used in the production of plastics, is stored in a tank having the U-tube manometer. If the U-tube is filled with mercury to level E, determine the pressure in the tank at point B. Take SHg = 13.55, and Sbc = 0.957.

A 300 mm

E 50 mm C

D

250 mm

100 mm

z

B

Prob. 2–25

120 mm

Prob. 2–29

Mercury

126

Chapter 2

F l u i d S ta t i C S

SEC. 2.6 2–30. The 150-mm-diameter container is filled to the top with glycerin, and a 50-mm-diameter pipe is inserted within it to a depth of 300 mm. If 0.00075 m3 of kerosene is then poured into the pipe, causing the displaced glycerin to over2 flow, determine the height h to which the kerosene rises from the top of the glycerin. 2–31. The 150-mm-diameter container is filled to the top with glycerin, and a 50-mm-diameter pipe is inserted within it to a depth of 300 mm. Determine the maximum volume of kerosene that can be poured into the pipe while causing the displaced glycerin to overflow, so the kerosene does not come out from the bottom end. How high h does the kerosene rise above the glycerin?

2–34. Determine the level h′ of water in the tube if the depths of oil and water in the tank are 0.6 m and 0.8 m, respectively, and the height of mercury in the tube is h = 0.08 m. Take ro = 900 kg>m3, rw = 1000 kg>m3, and rHg = 13 550 kg>m3.

0.6 m

50 mm h h 0.8 m h9

300 mm

100 mm

Prob. 2–34 Probs. 2–30/31 *2–32. Water in the reservoir is used to control the water pressure in the pipe at A. If h = 200 mm, determine this pressure when the mercury is at the elevation shown. Take rHg = 13 550 kg>m3. Neglect the diameter of the pipe. 2–33. If the water pressure in the pipe at A is to be 25 kPa, determine the required height h of water in the reservoir. Mercury in the pipe has the elevation shown. Take rHg = 13 550 kg>m3. Neglect the diameter of the pipe.

2–35. The pipes at A and B contain oil and the inclinedtube manometer is filled with oil and mercury. Determine the pressure difference between A and B. Take ro = 920 kg>m3 and rHg = 13 550 kg>m3.

E

D

h

A

300 mm 608 B

A

200 mm

400 mm 200 mm B

C 150 mm 100 mm 200 mm Mercury

Probs. 2–32/33

350 mm 150 mm

Prob. 2–35

127

problemS *2–36. Determine the difference in pressure pA - pB between the centers A and B of the closed pipes, which are filled with kerosene. The mercury in the inclined-tube manometer has the level shown. Take SHg = 13.55 and Sk = 0.82.

A

B

2–38. The two tanks A and B are connected using a manometer. If waste oil is poured into tank A to a depth of h = 0.6 m, determine the pressure of the entrapped air in tank B. Air is also trapped in line CD as shown. Take ro = 900 kg>m3, rw = 1000 kg>m3. 2–39. The two tanks A and B are connected using a 2 manometer. If waste oil is poured into tank A to a depth of h = 1.25 m, determine the pressure of the trapped air in tank B. Air is also trapped in line CD as shown. Take ro = 900 kg>m3, rw = 1000 kg>m3.

100 mm D 400 mm

A h

Air Air

Oil

D

C E 2m

C

B

1m

1.25 m

Water

608

1.25 m Water

0.5 m

1.5 m

Prob. 2–36 Probs. 2–38/39

2–37. Determine the height h of the mercury in the tube if the level of water in the tube is h′ = 0.3 m and the depths of the oil and water in the tank are 0.6 m and 0.5 m, respectively. Take ro = 900 kg>m3, rw = 1000 kg>m3, and rHg = 13 550 kg>m3.

*2–40. The inverted U-tube manometer is used to measure the difference in pressure between water flowing in the pipes at A and B. If the top segment is filled with air, and the water levels in each segment are as indicated, determine the pressure difference between A and B. rw = 1000 kg>m3. 2–41. Solve Prob. 2–40 if the top segment is filled with an oil for which ro = 800 kg>m3.

75 mm

A

C

0.6 m

D

D

225 mm

B

A

h

300 mm

0.5 m 150 mm C

Prob. 2–37

h9 5 0.3 m

B

Probs. 2–40/41

128

Chapter 2

F l u i d S ta t i C S

2–42. The micro-manometer is used to measure small differences in pressure. The reservoirs R and upper portion of the lower tubes are filled with a liquid having a specific weight of gR, whereas the lower portion is filled with a liquid having a specific weight of gt, Fig. (a). When the liquid flows through the venturi meter, the levels of the liquids with respect to the 2 original levels are shown in Fig. (b). If the cross-sectional area of each reservoir is AR and the cross-sectional area of the U-tube is At, determine the pressure difference pA - pB. The liquid in the venturi meter has a specific weight of gL. A

A

B

h1

*2–44. A solvent used for plastic manufacturing consists of cyclohexanol in tank A and ethyl lactate in tank B. Determine the pressure on the top of tank A if the mercury in the manometer is at the level shown, where h = 0.75 m. Take Sc = 0.953, SHg = 13.55 and Sel = 1.03. 2–45. A solvent used for plastic manufacturing consists of cyclohexanol in tank A and ethyl lactate in tank B. If the pressure on the top of tank A is PT = 130 kPa as indicated by the pressure gauge, determine the required level h of the mercury in the manometer shown. Take Sc = 0.953, SHg = 13.55, and Sel = 1.03.

B

gL R

R

d d

B 4.5 m

gR

h2

e

1.5 m h A

gt

(a)

mercury

(b)

Prob. 2–42

Probs. 2–44/45

2–43. The Morgan Company manufactures a micromanometer that works on the principles shown. Here there are two reservoirs filled with kerosene, each having a cross-sectional area of 300 mm2. The connecting tube has a cross-sectional area of 15 mm2 and contains mercury. Determine h if the pressure difference pA = pB = 40 Pa. Take rHg = 13 550 kg>m3, rke = 814 kg>m3. Hint: Both h1 and h2 can be eliminated from the analysis.

SEC. 2.7–2.9 2–46. The storage tank contains oil and water acting at the depths shown. Determine the resultant force that both of these liquids exert on the side ABC of the tank if the side has a width of 1.25 m. Also, determine the location of this resultant, measured from the top surface of the oil. Take ro = 900 kg>m3.

B A

h2

A 0.75 m B h1

h

1.5 m

C

Prob. 2–43

Prob. 2–46

129

problemS 2–47. Seepage is assumed to occur beneath the concrete wall, producing a linear distribution of hydrostatic pressure as shown. Determine the resultant force on a 1-m wide portion of the wall and its location, measured to the left and upward from point A.

2–49. The vertical pipe segment has an inner diameter of 100 mm and is capped at its end and suspended from the horizontal pipe as shown. If it is filled with water and the pressure at A is 80 kPa, determine the resultant force that must be resisted by the bolts at B in order to hold the flanges together. Neglect the weight of the pipe but not the 2 water within it.

A

6m 2m

A

B

3m 2m

Prob. 2–47 C

100 mm

*2–48. Nitrogen in the chamber is at a pressure of 300 kPa. Determine the total force the bolts at joints A and B must resist to maintain the pressure. There is a cover plate at B.

Prob. 2–49

2–50. A swimming pool has a width of 4 m and a side profile as shown. Determine the resultant force the water exerts on walls AB and DC, and on the bottom BC.

1.5 m B

D A

1m

A C

2.5 m

2.5 m B 8m

Prob. 2–48

Prob. 2–50

130

Chapter 2

F l u i d S ta t i C S

2–51. Determine the critical height h of the water level that causes the concrete gravity dam to be on the verge of tipping over due to water pressure. The density of concrete is rc = 2.40 Mg>m3. Hint: Work the problem using a 1-m width of the dam.

2–53. The uniform control gate AB is pinned at A and rests on the smooth surface at B. If the gate has a mass of 8.50 Mg, determine the maximum depth of water h in the reservoir that will cause the gate to be on the verge of opening. The gate has a width of 1 m.

2

18 m 3m

h A

308

h B

B

A 2m 6m

Prob. 2–51 Prob. 2–53 *2–52. Determine the smallest thickness b of the concrete gravity dam that will prevent the dam from overturning due  to water pressure acting on the face of the dam. The density of concrete is rc = 2.40 Mg>m3. Hint: Work the problem using a 1-m width of the dam.

2–54. The uniform rectangular relief gate AB has a mass of 20 Mg and a width of 2 m. Determine the minimum depth h of water needed to open it. The gate is pinned at B and rests on a rubber seal at A.

3m

6m

B 3m

h 45°

A A

b

Prob. 2–52

Prob. 2–54

131

problemS 2–55. The tide gate opens automatically when the tide water at B subsides, allowing the marsh at A to drain. For the water level h = 4 m, determine the horizontal reaction at the smooth stop C. The gate has a width of 2 m. At what height h will the gate be on the verge of opening? *2–56. The tide gate opens automatically when the tide water at B subsides, allowing the marsh at A to drain. Determine the horizontal reaction at the smooth stop C as a function of the depth h of the water level. Starting at h = 6 m, plot values of h for each increment of 0.5 m until the gate begins to open. The gate has a width of 2 m.

2–58. The bin is used to store carbon tetrachloride, a cleaning agent for metal parts. If it is filled to the top, determine the magnitude of the resultant force this liquid exerts on each of the two side plates, AFEB and BEDC, and the location of the center of pressure on each plate, measured from BE. Take rct = 1590 kg>m3.

D E 0.6 m C F 1m

B 2m

D

A

B

0.6 m 0.6 m 0.6 m

A

Prob. 2–58 6m h

3.5 m

C

Probs. 2–55/56

2–57. The uniform rectangular relief gate AB has a mass of 200 kg and a width of 1.5 m. Determine the components of reaction at the pin A and the normal reaction at the smooth support B.

2–59. The pressure of the air at A within the closed tank is 200 kPa. Determine the resultant force acting on the plates BC and CD caused by the water. The tank has a width of 1.75 m.

A

2m

B

3m 30°

A

C

1.5 m D

B 1.25 m

Prob. 2–57

Prob. 2–59

2

132

Chapter 2

F l u i d S ta t i C S

*2–60. Determine the resultant force that the water and oil together exert on the wall ABC. The wall has a width of 2 m. Also, determine the location of this resultant measured from the top of the tank. Take ro = 900 kg>m3.

2–62. The gate is 0.5 m wide and is pinned at A and held in place by a smooth latch bolt at B that exerts a force normal to the gate. Determine this force caused by the water and the resultant force on the pin for equilibrium.

2

1m A

A 2m

1.5 m B

B 2m

3m

C

1.5 m Prob. 2–60

Prob. 2–62

2–61. Determine the critical height h of the water level that causes the concrete gravity dam to be on the verge of tipping over. Assume water also seeps under the base of the dam and produces a uniform pressure under the dam. The density of concrete is rc = 2400 kg>m3. Hint: Work the problem using a 1-m width of the dam.

2–63. Determine the critical height h of the water level where the concrete gravity dam starts to tip over. The density of concrete is rc = 2.40 Mg>m3. Hint: Work the problem using a 1-m width of the dam. *2–64. Determine the critical height h of the water level before the concrete gravity dam starts to tip over. Assume water also seeps under the base of the dam and produces a uniform pressure under the dam. The density of concrete is rc = 2.40 Mg>m3. Hint: Work the problem using a 1-m width of the dam.

0.75 m

18 m 4m

h

h

B

A 6m

Prob. 2–61

B

A 1.5 m

Probs. 2–63/64

problemS 2–65. Determine the placement d of the pin on the 0.5-mwide rectangular gate so that it is on the verge to rotate clockwise (open) when waste water reaches a height h = 2.5 m. What is the resultant force acting on the gate? 2–66. Determine the placement d of the pin on the 1-m-diameter circular gate so that it is on the verge to rotate clockwise (open) when waste water reaches a height h = 2.5 m.What is the resultant force acting on the gate? Use the formula method.

133

*2–68. The gate is 2 m wide and is pinned at B and rests on a smooth support at A. Determine the horizontal and vertical components of reaction at B and the normal reaction at the smooth support A for equilibrium. The fluid is water. 2

1m A

4m

A B

d C

h 1m 3m

B

Prob. 2–68

0.5 m

Probs. 2–65/66 2–67. The uniform plate, which is hinged at C, is used to control the level of the water at A to maintain its constant depth of 6 m. If the plate has a width of 1.5 m and a mass of 30  Mg, determine the required minimum height h of the water at B so that seepage will not occur at D.

2–69. The tank is filled to its top with an industrial solvent, ethyl ether. Determine the resultant force acting on the plate ABC, and its location on the plate measured from edge AB of the tank. Use the formula method. Take ree = 715 kg>m3. 2–70.

Solve Prob. 2–69 using the integration method.

A

2m

C

B

B

A 4m

4m h

608 D

C

1.5 m 3m

Prob. 2–67

1.5 m

Probs. 2–69/70

134

Chapter 2

F l u i d S ta t i C S

2–71. Determine the resultant force acting on the triangular plate A and the location of the center of pressure, measured from the free water level in the tank. Solve the problem using the formula method. *2–72.

Solve Prob. 2–71 using the integration method.

*2–76. Ethyl alcohol is pumped into the tank, which has the shape of a four-sided pyramid. When the tank is completely filled, determine the resultant force acting on each side, and its location measured from the top A along the side. Use the formula method. Take rea = 789 kg>m3.

2

A 5m 6m 1.8 m A B C 2m 2m 1m

4m

B

Prob. 2–76

1m

Probs. 2–71/72

2–73. If the tank is filled with vegetable oil, determine the resultant force that the oil exerts on plate A, and its location measured from the bottom of the tank. Use the formula method. Take rvo = 932 kg>m3. 2–74. If the tank is filled with vegetable oil, determine the resultant force that the oil exerts on plate B, and its location measured from the bottom of the tank. Use the formula method. Take rvo = 932 kg>m3. 2–75.

2–77. The tapered settling tank is filled with oil. Determine the resultant force the oil exerts on the trapezoidal clean-out plate located at its end. How far from the oil surface does this force act on the plate? Use the formula method. Take ro = 900 kg>m3. 2–78. The tapered settling tank is filled with oil. Determine the resultant force the oil exerts on the trapezoidal cleanout plate located at its end. How far from the oil surface does this force act on the plate? Use the integration method. Take ro = 900 kg>m3.

Solve Prob. 2–74 using the integration method.

4m 4m 0.75 m

0.75 m 1m

1.5 m

4m

2.5 m 1m

5m 2m

B

A 1m

Probs. 2–73/74/75

Probs. 2–77/78

135

problemS 2–79. The tank is filled to its top with lubricating oil. Determine the resultant force acting on the semicircular plate ABC, and its location on the plate measured from base B of the tank. Use the formula method. Take go = 880 kg>m3. Solve Prob. 2–79 using the integration method.

*2–80.

2–82. The control gate ACB is pinned at A and rests on the smooth surface at B. Determine the minimum mass of the counter weight that should be placed at C in order to maintain a reservoir depth of h = 2 m. The gate has a width of 0.6 m. Neglect its weight. 2–83. The control gate ACB is pinned at A and rests on 2 the smooth surface at B. If the mass of counterweight C is 1200 kg, determine the maximum depth of water h in the reservoir before the gate is on the verge to open. The gate has a width of 0.6 m. Neglect its weight.

4.5 m

1.5 m

h 1.5 m

A

C C

A 608

0.75 m B

B

Probs. 2–79/80 Probs. 2–82/83

2–81. The open wash tank is filled to its top with butyl alcohol, an industrial solvent. Determine the magnitude of the resultant force on the end plate ABCD and the location of the center of pressure, measured from AB. Solve the problem using the formula method. Take rba = 800 kg>m3.

0.6 m

0.6 m

0.6 m

*2–84. The tank is filled with water. Determine the resultant force acting on the trapezoidal plate C and the location of the center of pressure, measured from the top of the tank. Solve the problem using the formula method. 2–85.

Solve Prob. 2–84 using the integration method.

3.5 m

B 1.5 m

1m A C D

3m

Prob. 2–81

1m1m

1m

Probs. 2–84/85

136

Chapter 2

F l u i d S ta t i C S

2–86. The uniform plate, which is hinged at C, is used to control the level of the water at A to maintain its constant depth of 6 m. If the plate has a width of 4 m and a mass of 80 Mg, determine the minimum height h of the water at B so that seepage will not occur at D.

2–89. The tank truck is filled to its top with water. Determine the magnitude of the resultant force on the elliptical back plate of the tank, and the location of the center of pressure measured from the top of the tank. Solve the problem using the formula method. 2–90.

2

Solve Prob. 2–89 using the integration method. y y2 1 x2 5 1 (0.75 m)2

A

2m

0.75 m C

x B

1m

4m h D 3m

Probs. 2–89/90

Prob. 2–86

2–87. The trough is filled to its top with carbon disulfide. Determine the magnitude of the resultant force acting on the parabolic end plate, and the location of the center of pressure measured from the top. Solve the problem using the formula method. Take rcd = 1260 kg>m3. *2–88.

Solve Prob. 2–87 using the integration method.

2–91. The tank truck is half filled with water. Determine the magnitude of the resultant force on the elliptical back plate of the tank, and the location of the center of pressure measured from the x axis. Solve the problem using the formula method. Hint: The centroid of a semi-ellipse measured from the x axis is y = 4b>3p. *2–92.

Solve Prob. 2–91 using the integration method. y y2 1 x2 5 1 (0.75 m)2

y

0.75 m

y 5 5x2 0.6 m

x 1m

1.8 m x

Probs. 2–87/88

Probs. 2–91/92

137

problemS

SEC. 2.10 2–93. Determine the magnitude and direction of the resultant hydrostatic force the water exerts on the parabolic face AB of the wall if it is 3 m wide.

2–95. The 5-m-wide overhang is in the form of a parabola. Determine the magnitude and direction of the resultant force on the overhang.

2

y y

3m

y 5 4x½ B

1 y 5 — x2 3

3m

8m

x A

Prob. 2–95

x 4m

Prob. 2–93

*2–96. Water is confined in the vertical chamber, which is 2 m wide. Determine the resultant force it exerts on the arched roof AB.

2–94. The bent plate is 1.5 m wide and is pinned at A and rests on a smooth support at B. Determine the horizontal and vertical components of reaction at A and the vertical reaction at the smooth support B. The fluid is water.

6m

1m A

A

B 4m

4m

2m

2m

B 2m

3m

Prob. 2–94

Prob. 2–96

138

Chapter 2

F l u i d S ta t i C S

2–97. The gate is 1.5 m wide, is pinned at A, and rests on the smooth support at B. Determine the reactions at these supports due to the water pressure. A

*2–100. Determine the horizontal and vertical components of reaction at the hinge A and the horizontal normal reaction at B caused by the water pressure. The gate has a width of 3 m.

2 3m

B 3m

Prob. 2–97 B

2–98. Determine the resultant force that water exerts on the overhanging sea wall along ABC. The wall is 2 m wide.

3m

1.5 m

A

B A 2m

Prob. 2–100 C

2.5 m

Prob. 2–98 2–99. The wall is in the form of a parabola. Determine the magnitude and direction of the resultant force on the wall if it is 2 m wide.

2–101. Determine the resultant force the water exerts on the quarter-circular wall AB if it is 3 m wide.

y

4m

y5

1 2 x 4

A

4m

3m

B x

Prob. 2–99

Prob. 2–101

139

problemS 2–102. A quarter-circular plate is pinned at A and rests on a smooth support at B. If the tank and plate are 1.5 m wide, determine the horizontal and vertical components of reaction at A, and the normal reaction at B due to the water pressure.

*2–104. The semicircular gate is used to control the flow of water over a spillway. If the water is at its highest level as shown, determine the torque T that must be applied at the pin A in order to open the gate. The gate has a mass of 8 Mg with center of mass at G. It is 4 m wide. 2–105. The semicircular gate is used to control the flow of 2 water over a spillway. If the water is at its highest level as shown, determine the horizontal and vertical components of reaction at pin A and the normal reaction at B. The gate has a weight of 8 Mg with center of mass at G. It is 4 m wide. Take T = 0.

2m

B

C 3m 2m

T

G A

A 2m

B

Prob. 2–102

2–103. Plate AB has a width of 1.5 m and a radius of 3 m. Determine the horizontal and vertical components of reaction at the pin A and the vertical reaction at the smooth stop B due to the water pressure.

Probs. 2–104/105 2–106. The parabolic and flat plates are pin connected at A, B, and C. They are submerged in water at the depth shown. Determine the horizontal and vertical components of reaction at pin B. The plates have a width of 4 m.

y 3m

A

B

3 y 5 16 (16 – x2)

3m 3m

B

A

C 4m

Prob. 2–103

Prob. 2–106

x

140

Chapter 2

F l u i d S ta t i C S

2–107. The 5-m-wide wall is in the form of a parabola. If the depth of the water is h = 4 m, determine the magnitude and direction of the resultant force on the wall.

2

*2–108. The 5-m-wide wall is in the form of a parabola. Determine the magnitude of the resultant force on the wall as a function of depth h of the water. Plot the results of force (vertical axis) versus depth h for 0 … h … 4 m. Give values for increments of ∆h = 0.5 m.

2–110. The Tainter gate for a water channel is 1.5 m wide and in the closed position, as shown. Determine the magnitude of the resultant force of the water acting on the gate. Also, what is the smallest torque T that must be applied to open the gate if its weight is 30 kN and its center of gravity is at G. 2–111. Solve the first part of Prob. 2–110 by the integration method using polar coordinates.

2m y

2m T

O

208 G 208 1.5 m y2 5 4x h

Probs. 2–110/111

x

Probs. 2–107/108

*2–112. The steel half cylinder has a density of 7850 kg>m3 and acts as a plug for the 0.5-m-long slot in the tank. Determine the resultant vertical force the bottom of the tank exerts on the half cylinder when the water in the tank is at a depth of h = 1.5 m.

h 0.3 m

2–109. The cylindrical tank is filled with gasoline and water to the levels shown. Determine the horizontal and vertical components of the resultant force on its hemispherical end. Take rg = 726 kg>m3.

A

B

0.45 m

Prob. 2–112 2–113. The steel half cylinder has a density of 7850 kg>m3 and acts as a plug for the 0.5-m-long slot in the tank. Determine the resultant vertical force the bottom of the tank exerts on the cylinder when the water in the tank just covers the top of the half cylinder, h = 0.

z 0.5 m

h 1m

0.3 m x

1m A

B

0.45 m

Prob. 2–109

Prob. 2–113

141

problemS 2–114. The sluice gate for a water channel is 2 m wide and in the closed position, as shown. Determine the magnitude of the resultant force of the water acting on the gate. Also, what is the smallest torque T that must be applied to open the gate if its mass is 6 Mg with its center of mass at G?

*2–116. The 2-m-wide Tainter gate in the form of a quartercircular arc is used as a sluice gate. Determine the magnitude and direction of the resultant force of the water on the Tainter gate. What is the moment of this force about the bearing O? 2 A 4m

3m

458

D 458

458

458

O

3m O

4m

4m

T

B

308 G 308 2m

Prob. 2–116

Prob. 2–114

2–115. Determine the horizontal and vertical components of reaction at the hinge A and the horizontal reaction at the smooth surface B caused by the water pressure. The plate has a width of 1.2 m.

2–117. The Tainter gate is used to control the flow of water over a spillway. If the water is at its highest level as shown, determine the torque T that must be applied at the pin A in order to open the gate. The gate has a mass of 5 Mg and a center of mass at G. It is 3 m wide. 2–118. The Tainter gate is used to control the flow of water over a spillway. If the water is at its highest level as shown, determine the horizontal and vertical components of reaction at pin A and the vertical reaction at the smooth spillway crest B. The gate has a mass of 5 Mg and a center of mass at G. It is 3 m wide. Take T = 0.

2.7 m C

4m T

A

G

B

1.8 m B O

Prob. 2–115

Probs. 2–117/118

608 4.5 m

A

142

Chapter 2

F l u i d S ta t i C S

SEC. 2.11–2.12 2–119. The hot-air balloon contains air having a temperature of 80°C, while the surrounding air has a temperature of 20°C. Determine the volume of hot air if the total mass of the balloon and the load is 770 kg.

2–121. The solid ball is made of plastic having a density of rp = 48 kg>m3. Determine the tension in the cable AB if the ball is submerged in the water at the depth shown. Will this force increase, decrease, or remain the same if the cord is shortened? Why? Hint: The volume of a ball is V = 43pr 3.

2

0.9 m

0.6 m

A

Prob. 2–119 *2–120. A boat having a mass of 80 Mg rests on the bottom of the lake and displaces 10.25 m3 of water. Since the lifting capacity of the crane is only 600 kN, two balloons are attached to the sides of the boat and filled with air. Determine the smallest radius r of each spherical balloon that is needed to lift the boat. What is the mass of air in each balloon if the air and water temperature is 12°C? The balloons are at an average depth of 20 m. Neglect the mass of the air and the balloon. The volume of a sphere is V = 34pr 3.

Prob. 2–121

2–122. The hollow spherical float of mass 150 g controls the level of water within the tank. If the water is at the level shown, determine the horizontal and vertical components of the force acting on the supporting arm at the pin A, and the normal force on the smooth support B.

600 kN

r

B

r B A 75 mm 450 mm

Prob. 2–120

Prob. 2–122

100 mm

143

problemS 2–123. The container of water has a mass of 20 kg. Block B has a density of 7840 kg>m3 and a mass of 30 kg. Determine the total compression or elongation of each spring when the block is fully submerged in the water.

2–125. Water in the container is originally at a height of h = 1 m. If a block having a density of 800 kg>m3 is placed in the water, determine the new level h of the water. The base of the block is 600 mm square, and the base of the container is 1.2 m square. 2

kC 5 5 kNm

C

600 mm 600 mm

B

h

D

D

kD 5 2 kNm

Prob. 2–123 Prob. 2–125 *2–124. The container of water has a mass of 20 kg. Block B has a density of 7840 kg>m3 and a mass of 30 kg. Determine the total compression or elongation of each spring when the block is fully submerged in the water.

2–126. The raft consists of a uniform platform having a mass of 2 Mg and four floats, each having a mass of 120 kg and a length of 4 m. Determine the height h at which the platform floats from the water surface. Take rw = 1 Mg>m3. C

kC 5 5 kNm

B h

D

kD 5 2 kNm

E

kE 5 5 kNm 0.25 m

Prob. 2–124

Prob. 2–126

144

Chapter 2

F l u i d S ta t i C S

2–127. Determine the height at which the oak block will float above the water surface. The density of the oak is roak = 770 kg>m3.

2–129. When loaded with gravel, the barge floats in water at the depth shown. If its center of gravity is at G, determine whether the barge will restore itself when a wave causes it to roll slightly at 9°.

2

1.5 m

G O

h

2m

1.5 m

3m

1m 6m

Prob. 2–127

Prob. 2–129

*2–128. A glass having a diameter of 50 mm is filled with water to the level shown. If an ice cube with 25-mm sides is placed into the glass, determine the new height h of the water surface. Take rw = 1000 kg>m3 and rice = 920 kg>m3. What will the water level h be when the ice cube completely melts?

2–130. When loaded with gravel, the barge floats in water at the depth shown. If its center of gravity is at G, determine whether the barge will restore itself when a wave causes it to tip slightly.

25 mm 25 mm G O h

2m

1.5 m

100 mm

50 mm

50 mm

Prob. 2–128

6m

Prob. 2–130

145

problemS

SEC. 2.13–2.14 2–131. The can of gasoline rests on the floor of a hoist. Determine the maximum pressure developed at the base of the can if the hoist is moving upward with (a) a constant velocity of 3 m>s, and (b) a constant upward acceleration of 2 m>s2. Take rg = 726 kg>m3.

2–133. The truck carries an open container of water. If it has a constant deceleration 1.5 m>s2, determine the angle of inclination of the surface of the water and the pressure at the bottom corners A and B. 2

0.5 m

a

1.5 m A

B

4m

Prob. 2–133 1.5 m

Prob. 2–131

*2–132. The closed rail car is 2 m wide and filled with water to the level shown. Determine the pressure that acts at A and B when the car has a constant acceleration of 4 m>s2.

2–134. If the truck has a constant acceleration of 2 m>s2, determine the water pressure at the bottom corners A and B of the water tank.

a 6m

3m 1m

a

2m 1m A

B

A

B

T

2m

Prob. 2–132

3m

Prob. 2–134

146

Chapter 2

F l u i d S ta t i C S

2–135. The open rail car is 2 m wide and filled with water to the level shown. Determine the pressure that acts at point B both when the car is at rest, and when the car has a constant acceleration of 3 m>s2. How much water spills out of the car?

2–137. The cart is allowed to roll freely down the inclined plane due to its weight. Show that the slope of the surface of the liquid, u, during the motion is u = f.

2

a 8m u

A a 3m 2m B

T

f

Prob. 2–137

Prob. 2–135

*2–136. If the truck has a constant acceleration of 2 m>s2, determine the water pressure at the bottom corners B and C of the water tank. There is a small opening at A.

2–138. The cart is given a constant acceleration a up the inclined plane. Show that the lines of constant pressure within the liquid have a slope of tan u = (a cos f)>(a sin f + g).

a u A a

1m 1m C

B

f 2m

3m

Prob. 2–136

Prob. 2–138

problemS 2–139. A large container of benzene is transported on the truck. Determine the level in each of the vent tubes A and B if the truck accelerates at a = 1.5 m>s2. When the truck is at rest, hA = hB = 0.4 m.

147

2–143. The glass is filled with water to a height of d = 0.1 m. To what height d = d′ does the water rise against the wall of the glass when the platform has an angular velocity of v = 15 rad>s?

*2–140. A large container of benzene is being transported by the truck. Determine its maximum constant acceleration so that no benzene will spill from the vent tubes A or B. When the truck is rest, hA = hB = 0.4 m.

2 0.1 m

0.1 m

0.2 m 0.2 m

3m

0.2 m d

a

hA

A 0.7 m

B

hB

0.05 m 0.05 m v

Prob. 2–143

Probs. 2–139/140

2–141. A woman stands on a horizontal platform that is rotating at a constant angular velocity v. If she is holding a cup of tea, and the center of the cup is 0.9 m from the axis of rotation, determine v if the slope angle of the liquid’s surface is 5°. Neglect the size of the cup. 2–142. Determine the maximum height d the glass can be filled with water so that no water spills out when the glass is rotating at 15 rad>s.

0.1 m

*2–144. The sealed tube assembly is completely filled with water, such that the pressures at C and D are zero. If the assembly is given an angular velocity of v = 15 rad>s, determine the difference in pressure between C and D. 2–145. The sealed tube assembly is completely filled with water, such that the pressures at C and D are zero. If the assembly is given an angular velocity of v = 15 rad>s, determine the difference in pressure between A and B.

D

C

0.1 m

0.6 m 0.2 m d

A B 0.5 m

0.05 m 0.05 m v v

Prob. 2–142

Probs. 2–144/145

148

Chapter 2

F l u i d S ta t i C S

2–146. The tube is filled with oil to the level h = 0.4 m. Determine the angular velocity of the tube so that the pressure at O becomes -15 kPa. Take So = 0.92.

2–147. The drum has a hole in the center of its lid and is filled to a height d with a liquid having a density r. If the drum is then placed on the rotating platform and it attains an angular velocity of v, determine the inner radius ri of the liquid where it contacts the lid.

2 ro

ri

v

h 5 0.4 m

O

d do

0.8 m

0.8 m v

Prob. 2–146

Prob. 2–147

ConCeptual problemS

149

CONCEPTUAL PROBLEMS P2–1. By moving the handle up and down on this hand pump, one is able to pump water from a reservoir. Do some research to explain how the pump works, and show a calculation that indicates the maximum height to which the pump can lift a column of water.

P2–3. Ice floats in the glass when the glass is filled with water. Explain what happens to the water level when the ice 2 melts. Does it go up, go down, or remain the same?

P2–4. The beaker of water rests on the scale. Will the scale reading increase, decrease, or remain the same if you put your finger in the water? Explain. P2–2. In 1656 Otto von Guericke placed the two halves of a 300-mm-diameter hollow sphere together and pumped the air out of the inside. He tied one rope at A to a tree and the other to a team of eight horses. Assuming a perfect vacuum was developed within the sphere, do you think the horses could pull the hemispheres apart? Explain. If he used sixteen horses, eight on each side, would this make a difference? Explain.

B

A

150

Chapter 2

F l u i d S ta t i C S

CHAPTER R EV IEW

2

Pressure is a normal force acting per unit area. At a point in a fluid, it is the same in all directions. This is known as Pascal’s law. The absolute pressure is equal to the atmospheric pressure plus the gage pressure. In a static fluid, the pressure is constant at points that lie in the same horizontal plane. If the fluid is incompressible, then the pressure depends on the specific weight of the fluid, and it increases linearly with depth.

pabs = patm + pg

p = gh

If the depth is not great, the pressure within a static gas can be assumed constant. Atmospheric pressure is measured using a barometer. A manometer can be used to measure the gage pressure in a liquid. The pressure is determined by applying the manometer rule. The pressure can also be measured using other devices such as a Bourdon gage or a pressure transducer.

The resultant hydrostatic force acting on a plane surface area has a magnitude of FR = ghA, where h is the depth of the centroid of the area. The location of FR is at the center of pressure P(xP, yP).

The resultant hydrostatic force acting on a plane surface area can also be determined by finding the volume of its pressure prism. If the surface has a constant width, one can then view the pressure prism perpendicular to its width and find the area of the load distribution that is caused by the pressure. The resultant force acts through the centroid of the volume or area.

Direct integration of the pressure distribution can also be used to determine the resultant force and its location on a plane surface area.

xP = x +

I xy

yA Ix yP = y + yA

Centroid of plate area

x

p 5 gh

x

FR y

xP

P C

h

Center of pressure

y yP

151

Chapter reView

If the surface is inclined or curved, the resultant hydrostatic force can be determined by first finding its horizontal and vertical components.

Fy B

The horizontal component is found by projecting the surface onto the vertical plane and finding the force acting on this projected area.

CV

The vertical component is equal to the weight of the volume of liquid above the inclined or curved surface. If the liquid is below this surface, then the weight of imaginary liquid above the surface is determined. The vertical component then acts upward on the surface because it represents the equivalent pressure force of the liquid below the surface.

C

2 D

P A

Fh FR

E

The principle of buoyancy states that the buoyant force acting on a body immersed in a fluid is equal to the weight of fluid displaced by the body. A floating body can be in stable, unstable, or neutral equilibrium. The body will be stable if its metacenter is located above its center of gravity. If an open container of liquid has a constant horizontal acceleration ac, the surface of the liquid will be inclined at an angle given by tan u = ac >g. If a lid is on the container, then an imaginary liquid surface should be established. In either case, the pressure at any point in the liquid is determined from p = gh, where h is the depth, measured from the liquid surface.

Imaginary liquid surface hA A

u

B

pA 5 ghA hC ac C

pC 5 ghC

If a container of liquid has a constant vertical upward acceleration ac, then the pressure within the liquid at a depth h will be increased by gh(ac >g). It is decreased by this amount if the acceleration is downward.

If a container has a constant rotation about a fixed axis, the liquid surface will form a forced vortex having the shape of a paraboloid. The surface is defined by h = (v2 >2g)r 2. If a lid is on the container, then an imaginary liquid surface can be established. The pressure at any point within the liquid is determined from p = gh, where h is the depth measured from the liquid surface.

2 h 5 v r2 2g

( )

p0 5 0 R

r

p 5 gh

3

NASA/Science Source

CHAPTER

Exhaust flow through the engines of this jet is modeled using a computer program involving computational fluid dynamics.

KINEMATICS OF FLUID MOTION

CHAPTER OBJECTIVES ■

To define the velocity and acceleration of fluid particles using either a Lagrangian or an Eulerian description.

■

To classify the various types of fluid flow and to discuss the ways in which these flow patterns are visualized experimentally.

■

To show how to express the velocity and acceleration of fluid particles in terms of their Cartesian and streamline components.

3.1

TYPES OF FLUID FLOW

In most cases, fluids do not remain static but rather they flow. In this chapter, we will consider the kinematics of this flow, that is, the study of the geometry of the flow, which provides a description of a fluid’s position, velocity, and acceleration, without considering the forces that cause the motion. Before we discuss flow kinematics, however, it is important to understand the various ways a flow can be classified. Here we will consider three of them.

153

154

Chapter 3

KinematiCs

3

of

fluid motion

Laminar flow Fluid particles follow straight-line paths since fluid flows in thin layers

Turbulent flow Fluid particles follow erratic paths which change direction in space and time

(a)

(b)

Fig. 3–1

Classification of Flow Based on Its Frictional Effects. Turbulent

Transition

Laminar

Laminar and turbulent flow from the smoke of an extinguished candle

When a highly viscid fluid such as oil flows at a very slow rate through a pipe, the paths the particles follow are uniform and undisturbed. In other words, the lamina or thin cylindrical layers of fluid are “orderly,” and so one layer slides smoothly relative to an adjacent layer. This behavior is referred to as laminar flow, Fig. 3–1a. Increase the velocity or decrease the viscosity, and the fluid particles will then follow erratic paths, which causes a high rate of mixing within the fluid. We refer to this as turbulent flow, Fig. 3–1b. Between these two types, we have transitional flow, that is, a region in which both laminar and turbulent flow coexist. In Chapter 9 we will see that one of the most important reasons for classifying the flow in this manner is to determine the amount of energy the fluid loses due to frictional effects. Obtaining this energy loss is necessary when designing pumps and pipe networks for transporting the fluid. The velocity profile for laminar flow and the average velocity profile for turbulent flow between two surfaces are shown in Fig. 3–2. Notice how laminar flow is shaped entirely by the viscosity of the fluid’s sliding layers, whereas turbulent flow “mixes” the fluid both horizontally and vertically, which causes its average velocity profile to flatten out, or become more uniform.

Velocity profile for laminar flow

Average velocity profile for turbulent flow

Fig. 3–2

3.1

Classifications of Flow Based on Dimension. A flow can also be classified by how many spatial coordinates are needed to describe it. If all three space coordinates are required, then it is called threedimensional flow. Examples include the flow of water around a submarine and the airflow around an automobile. Three-dimensional flows are rather complex and therefore difficult to analyze. They are often studied using a computer, or experimentally using models. For many problems in engineering, we can simplify the analysis by assuming the flow to be two-, one-, or even nondimensional. For example, the flow through the converging pipe in Fig. 3–3a is twodimensional flow. The velocity of any particle depends only on its axial and radial coordinates x and r. A further simplification can be made in the case of unchanging flow through a uniform straight pipe, Fig. 3–3b. This is a case of one-dimensional flow. Its velocity profile changes only in the radial direction. Finally, if we consider the unchanging flow of an ideal fluid, where the viscosity is zero and the fluid is incompressible, then its velocity profile at each location is constant, and therefore independent of its coordinate location, Fig. 3–3c. It becomes nondimensional flow.

r

x Two-dimensional flow Velocity is a function of x and r (a)

r

One-dimensional flow Velocity is a function of r (b)

Nondimensional flow Velocity is constant (c)

Fig. 3–3

types of fluid flow

155

3

156

3

Chapter 3

KinematiCs

of

fluid motion

Steady uniform flow An ideal fluid maintains the same velocity at all times (steady) and at every point (uniform)

Steady nonuniform flow The velocity remains constant over time (steady), but it is different from one location to the next (nonuniform)

(a)

(c)

Time t

Time t

Time t 1 Dt Time t 1 D t Unsteady uniform flow The valve is slowly opened and so at any instant the velocity of an ideal fluid is the same at all points (uniform), but it changes over time (unsteady)

Unsteady nonuniform flow The valve is slowly opened, and because of the changing cross section of the pipe, the velocity will be different at each point (nonuniform) and at each time (unsteady)

(b)

(d)

Fig. 3–4

Classification of Flow Based on Space and Time. When the velocity of a fluid at each point does not change with time, we refer to the flow as steady flow, and when the velocity does not change from one position to the next, it is referred to as uniform flow. In general, there are four possible combinations of these two flows, and an example of each is given in Fig. 3–4. Most engineering applications of fluid mechanics involve steady flow, and fortunately, it is the easiest to analyze. Furthermore, it is generally reasonable to assume that any unsteady flow that occurs only for a short time may, over the long term, be considered a case of steady flow. For example, flow through the moving parts of a pump will be unsteady, but the pump operation is cyclic or repetitious, so we may consider the flow at the inlet and outlet of the pump to be on average steady flow. It may also be possible to establish steady flow relative to a moving observer, while it may appear unsteady to a fixed observer. Consider the case of a car passing through still, smoky air. The airflow will appear steady to the driver when the car moves through it at constant speed on a straight road. However, to an observer standing along the roadway, the air would appear to have unsteady flow during the time the car passes by.

3.2

3.2

GraphiCal desCriptions of fluid flow

157

GRAPHICAL DESCRIPTIONS OF FLUID FLOW

The flow of any real fluid is generally very complex, and so a mathematical analysis of any particular problem is often backed up by an experiment in order to visualize the flow. Also, flow visualization plays an important role in the analysis of numerical solutions. For an analysis, this includes using streamlines or streamtubes, and for experimental work, pathlines and streaklines and optical methods are often used. We will now give each of these separate treatment.

Streamlines. As a fluid flows, each of the particles will have a unique velocity. In principle we could draw this “velocity field” of, say, water flowing around a cylinder, as in Fig. 3–5a; however, this would not produce a very clear representation of the flow. A better way to illustrate the flow at any instant would be to sketch a series of streamlines. Each streamline only contains adjacent particles that have velocity vectors tangent to the streamline. As a result, no two streamlines can cross each other, because the velocity of any particle at the point of intersection would not have a unique direction. An example of streamlines for the flow around a cylinder is shown in Fig. 3–5b. In particular, notice the streamline at the center of the flow. It intersects the cylinder at A. This point is called a stagnation point, because here the velocity of any particle is momentarily reduced to zero when it strikes the surface before traveling around the cylinder. It is important to keep in mind that streamlines are used to represent the flow field at an instant of time. In the example just given, the direction of the streamlines is maintained as time passes; however, sometimes the streamlines can be a function of both space and time. For example, this occurs when fluid flows through a rotating pipe. Here the streamlines (and the pipe) change position from one instant to the next.

A

Velocity field around a cylinder

Streamlines for ideal fluid flow around a cylinder

(a)

(b)

Fig. 3–5

3 3

158

Chapter 3

KinematiCs

of

fluid motion

3 Streamtube

Fig. 3–6

slow

fast

Fluid flows faster within a narrow streamtube (c)

Fig. 3–5 (cont.)

Streamtubes. For some types of analysis, it is convenient to consider slow

a bundle of streamlines that surround a region of flow, Fig. 3–6. Such a circumferential grouping is called a streamtube. Here the fluid flows through the streamtube as if it were contained within a curved conduit. In two dimensions, streamtubes are formed between two streamlines. For example, consider the streamtube in Fig. 3–5c for flow around the cylinder. Notice that a small element of fluid traveling along this streamtube will move slower when the streamlines are farther apart (or the streamtube is wider), and move faster when they are closer together (or the streamtube is narrower). This is a consequence of the conservation of mass, something we will discuss in the next chapter.

Equation of a Streamline. Since the particle’s velocity is always tangent to its streamline, then if the velocity field is known, the equation that defines a streamline at a particular instant can always be determined. For example, in the case of two-dimensional flow, Fig. 3–7, the particle’s velocity at any point will have two components, u in the x direction and in the y direction. We require dy v = u dx

(3–1)

Integrating this equation will give the equation of the streamline, and the method for finding this equation is demonstrated in Examples 3.1 and 3.2. y

dy V

dx y 5 f(x)

v u Streamline

x The velocity is always tangent to the streamline

Fig. 3–7

3.2

GraphiCal desCriptions of fluid flow

159

Pathlines. The pathline defines the “path” a single particle travels over a period of time. To obtain the pathline experimentally, a particle can be released within the flow stream and a time exposed photograph taken. The line on the photograph then represents the pathline for this particle, Fig. 3–8a. This method can be used, for example, to determine the average velocity on the surface of a liquid. If several particles of powdered aluminum are sprinkled on the surface, then a short time exposed photograph will show the distance each particle travels, and from this their average velocity can be obtained. We can obtain the equation of the pathline if we know a fluid particle’s velocity components, which in two dimensions are

dx dt dy v = dt

3 The pathline shows the path of a single particle using a time exposure photograph for 0 < t < t1 (a)

u =

(3–2) The streakline shows the path of many particles at the instant t 5 t1

Integrating each, we obtain x = x1t2 and y = y1t2. Eliminating the term t between them, we obtain the result y = f1x2.

(b)

Fig. 3–8

Streaklines. If smoke is released continuously in a gas, or colored dye is released in a liquid, a “trace” of all the particles will be carried along with the flow. This resulting succession of marked particles that have all come from the same point of origin is called a streakline. It can be identified by taking an instantaneous photograph of the trace, or “streak,” of all the particles, Fig. 3–8b. Realize that if the flow is steady, the streamlines, pathlines, and streaklines will always coincide. For example, this occurs for the steady flow of water ejected from the fixed nozzle, Fig. 3–9. Here a streamline will maintain its same direction from one instant to the next, and so every particle coming from the nozzle will follow this same streamline, thereby producing a coincident pathline and streakline.

This photo shows the streaklines of water particles ejected from a water sprinkler at a given instant. The streamlines, pathlines, and streaklines all coincide for steady flow

Fig. 3–9

160

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of

fluid motion

Optical Methods. For a transparent fluid, such as air or water, the

3

A schlieren photograph showing a ball that is suspended by an airstream produced by a jet of hot air. (© Ted Kinsman/Science Source)

flow can be visualized indirectly using a shadowgraph. This is the result of refracted or bent light rays that interact with the fluid and then cast a shadow on a screen in close proximity. Perhaps you have noticed the shadow of a heated air plume rising from a candle, or the shadow of the exhaust from a jet engine cast against a background? In both of these cases the heat being produced changes the local density of the air, and the light passing through this air is bent. The greater the change in the fluid density, the more the light rays are bent. Shadowgraphs have been used in industry to visualize the air flow around jet aircraft and rockets, as well as to study the flow around devices that produce heat. One other optical technique that is used to visualize the flow of a transparent fluid is schlieren photography. It also is based on detecting the density gradients that the flow produces within a fluid. However, in this case, the light that shines on the object is focused with a lens, and a knife edge is placed perpendicular to the beam at its focal point. This blocks about half the light and so the brightness of a uniform flow will be diminished. When the flow is disturbed, then the density variations within it will cast lighter and darker regions on a nearby background surface. An example is shown in the accompanying photo of a heated source generating a flow of hot air used to suspend a ball. Schlieren photography has been used extensively in aeronautical engineering to visualize the formation of shockwaves and expansion waves formed around jet aircraft and missiles. Further details on these two optical techniques are given in Ref. [5].

Computational Fluid Dynamics. Although the experimental

A CFD printout showing the flow around a model of a car. Based on the results, the model can be reshaped to enhance the design. (© Hank Morgan/Science Source)

techniques just discussed have played an important role in studying complicated flow patterns, they have been increasingly replaced with numerical techniques that apply the laws of fluid dynamics using high-speed computers. This is referred to as computational fluid dynamics (CFD). There are many types of commercially available CFD programs, and with some of them the engineer can extend an analysis to include heat transfer and multiphase changes within the fluid. Generally the output includes a plot of the streamlines or pathlines, with the velocity field displayed in a color-coded fashion. Digital printouts of steady flow can be made, or if the flow is unsteady, then a video can be produced. We will discuss this important field at greater length in Sec. 7.13.

3.3

3.3

161

fluid flow desCriptions

FLUID FLOW DESCRIPTIONS

The behavior of a fluid is quite different from that of a solid, because the particles within a fluid can move in all directions, whereas the particles in a solid maintain a fixed position relative to one another. This difference in behavior can complicate any analysis of a fluid flow unless we have a proper way to describe it. To do this, we must first identify a specific quantity of fluid, called a system, that is enclosed within a region of space apart from the fluid particles outside this region, which is called the surroundings, Fig. 3–10a. Provided we know the flow pattern of the fluid system, we can then establish, for example, the pressure or forces that act on a structure or machine submerged within the fluid. To completely define this flow pattern, it is necessary to specify the velocity of each fluid particle at each point within the system, and at each instant of time. In fluid mechanics there are two ways for doing this.

z Surroundings System

y x (a) z

Lagrangian Description—System Approach. The flow within a fluid system can be defined by “tagging” each fluid particle, and then specifying its position r as a function of time as the particle moves from one position to the next, Fig. 3–10b. This method is referred to as a Lagrangian description, named after the Italian mathematician Joseph Lagrange. Provided r = r1t2 is known, then its time derivative yields the velocity of the particle, that is,

r 5 r(t)

y x x

V = V1t2 =

dr1t2 dt

If i, j, k are the unit vectors that define the positive directions of x, y, z axes, then V can be expressed in terms of its Cartesian components, V1t2 = Vx 1t2i + Vy 1t2j + Vz 1t2k. Here the velocity is only a function of time because it is defined as the time rate of change of the particle’s position. The velocity is not a function of the particle’s position, because the position itself is a function of time, r = r1t2, Fig. 3–10b. In other words, time is the independent variable, since the x, y, z components of both r and V only depend upon t. This description works well in particle or rigid body dynamics because the body maintains a fixed shape, and so the location and motion of the particles composing the body can be readily specified with respect to one another. In fluid mechanics, however, it is very difficult to account for the position of each particle in the system from one instant to the next, and then measure the velocities of all the particles as they move about and change the shape of the system. For this reason, the Lagrangian description has very limited use in fluid mechanics, although for some applications, it may provide insight regarding the changing geometry of the system.

V 5 V(t)

z

y

Lagrangian description of motion follows a single fluid particle as it moves about within the system (b)

Fig. 3–10

3

162

Chapter 3

KinematiCs

of

Eulerian Description—Control Volume Approach. The

z

velocity of the fluid particles within a system can also be described by considering a specific point (x0, y0, z0) in the system, surrounded by a differential volume of space, Fig. 3–10c. The velocity of all the particles that pass through this point or volume can then be measured at this point. This method is named after the Swiss mathematician Leonhard Euler, and is referred to as an Eulerian description. The volume of space through which the particles flow is called a control volume, and the boundary of this volume is the control surface. An Eulerian description therefore requires knowing the velocity field for the system. Here the velocity of a particle depends upon the location x, y, z of the control volume in space and the time t, that is,

Control volume

3

V 5 V(x0, y0, z0, t) z0 y x0 x

fluid motion

y0

Eulerian description of motion specifies a point or region within the system, and it measures the velocity of the particles that pass through this point or control volume

(3–3)

V = V1x, y, z, t2

(c)

Fig. 3–10 (cont.)

An example of this velocity field is shown in Fig. 3–11. Notice that because V is a function of position (x, y, z), the flow is nonuniform, and because it is a function of time, the flow is unsteady. To summarize the above two descriptions, imagine the system to be a classroom of students. For a Lagrangian description, the velocity of a single student, V1t2, is measured from a fixed location within the classroom, such as the doorway. An Eulerian description requires specifying a location in the classroom, such as near a desk, and then measuring the velocity of each student that passes this desk, V = V1x, y, z, t2. In the first case the motion of only one student in the class is investigated, and in the second case the motion of several students is considered.

z

z

y

y

x

x time t2

time t1 Velocity field

Fig. 3–11

3.3

fluid flow desCriptions

163

IMPORTANT POIN T S • Fluid flow can be described in various ways. It can be classified as the result of viscous friction, that is, either laminar, transitional, or turbulent. It can be classified as nondimensional, or as one-, two-, or three-dimensional. And finally, for a space–time classification, the flow is steady if it does not vary with time, and uniform if it does not vary with location.

• Experimentally, the flow of a single marked particle can be visualized using a pathline, which shows the path taken by the particle using a time-lapse photograph. Streaklines are formed when smoke or colored dye is released into the flow from the same point, and an instantaneous photo is taken of the trace of many particles.

• Optical methods, such as shadowgraphs and schlieren photographs, are useful for observing flow in transparent fluids caused by heat, or to visualize shock and expansion waves created in high-speed compressible flows.

• Computational fluid dynamics uses numerical analysis to apply the laws of fluid dynamics, and thereby produces data to visualize complex flow.

• There are two ways to describe the motion of fluid particles within a flow. A Lagrangian description, or system approach, has limited use since it requires tracking the location of each particle in the flow and reporting its motion. An Eulerian description, or control volume approach, is more practical, since it considers a specific region or point in the flow, and measures the motion of any fluid particles passing through this region or point.

• A streamline identifies the direction in which particles move through a flow, since their velocities are tangent to a streamline at each instant. If the flow is steady, then the particles move along fixed streamlines. However, if unsteady flow causes the direction of the streamlines to change, then the particles will move along streamlines that have a different orientation from one instant to the next.

3

164

Chapter 3

EXAMPLE

KinematiCs

of

3.1 The velocity for the two-dimensional flow shown in Fig. 3–12 is defined by V = 56yi + 3j6m>s, where y is in meters. Determine the equation of the streamline that passes through point (1 m, 2 m).

y

3

fluid motion

1m 3 ms V 2m

12 ms

u

SOLUTION fast

slow x

Stre

amt

ube

y2 5 x 1 3

Fig. 3–12

fast

Fluid Description. Since the velocity is described only by its spatial coordinates and time is not involved, we have steady nonuniform flow, where u = 16y2 m>s and v = 3 m>s. Also, we have an Eulerian description of the motion, where V defines the velocity field for all the fluid particles. Analysis. Eq. 3–1,

To find the equations of the streamlines, we must use

dy v 3 1 = = = u dx 6y 2y Separating the variables and integrating yields L

2y dy =

L

dx

y2 = x + C This is the equation for a parabola. Each selected constant C will produce a unique streamline. For the one passing through the point (1 m, 2 m), we require 122 2 = 1 + C, or C = 3. Therefore, y2 = x + 3

Ans.

A plot of this equation (streamline) is shown in Fig. 3–12. Here particles that pass through a differential size control volume located at the point (1 m, 2 m) will have the velocity V = 512i + 3j6 m>s, as indicated. By selecting other points to evaluate the integration constant C, we can plot the streamlines for other particles and thereby establish the graphical representation of the entire flow field, a portion of which is also shown in Fig. 3–12. Once this is established, notice how the fluid element constrained in the selected streamtube travels fast, slows down as it passes the x axis, and then speeds up again.

3.3

EXAMPLE

fluid flow desCriptions

165

3.2

The velocity components of a particle in a flow are defined by u = 3 m>s and v = 16 t2 m>s, where t is in seconds. Plot the pathline for the particle if it is released from the origin when t = 0.

y

y5

1 2 3x

Pathline

x

Fig. 3–13

SOLUTION Fluid Description. Here we have a Lagrangian description of the motion because we are following the motion of a single particle, where the velocity is a function of time. Pathline. The pathline describes the location of the particle at various times. Since the particle is at (0, 0) when t = 0, then from Eqs. 3–2, dy dx u = = 3 v = = 6t dt dt L0

x

t

y

t

6t dt L0 L0 L0 x = 13t2 m y = 13t 2 2 m (1) Eliminating the time t between these two parametric equations, we obtain our result x 2 1 y = 3a b or y = x 2 Ans. 3 3 The pathline or path taken by the particle is a parabola, shown in Fig. 3–13. dx =

3 dt

dy =

3

166

Chapter 3

EXAMPLE

3

KinematiCs

of

fluid motion

3.3 The velocity of gas particles flowing along the center of the pipe in Fig. 3–14 is defined for x Ú 1 m by the velocity field V = 1t>x2 m>s, where t is in seconds and x is in meters.* If a particle is at x = 1 m when t = 0, determine its velocity when it is at x = 2 m.

V x

x

Fig. 3–14

SOLUTION Fluid Description. Because the velocity is a function of time, the flow is unsteady (it changes with time), and because it is a function of position, it is nonuniform (it changes with position). We have an Eulerian description of the motion. Analysis. To find the velocity of the particle at x = 2 m, we must first find the time for the particle to travel from x = 1 m to x = 2 m. To do this, we must find x = x(t). This is a Lagrangian description since we are following the motion of this single particle. The particle’s position as a function of time can be determined from the velocity field, by realizing that V =

dx t = x dt

Separating the variables and integrating yields L1

x

x dx =

L0

t

t dt

x2 1 1 - = t2 2 2 2

x = 2t 2 + 1

(1)

*Notice that substituting t in seconds and x in meters gives units of s>m; however, here there is a constant 1 m2 >s2 (not shown) that converts these units to velocity, m>s, as stated.

3.3

fluid flow desCriptions

167

Now applying the first of Eqs. 3–2, 3

dx 1 V = = 1t 2 + 12 -1>2 12t2 dt 2 2t + 1 t

=

(2)

2

We are now able to follow the particle as it moves along the path during each instant of time. Its position at any instant is determined from Eq. 1, and its velocity at this instant is determined from Eq. 2. Therefore, when the particle is located at x = 2 m, the time is 2 = 2t 2 + 1 t = 1.732 s

And at this time, the particle is traveling at

V =

211.7322 2 + 1 1.732

V = 0.866 m>s

Ans.

We can check this result using the Eulerian description. Since we require the particle to be located at x = 2 m, when t = 1.732 s its velocity is

V =

t 1.723 = = 0.866 m>s x 2

Ans.

as expected. In this example the velocity field has only one component, V = u = t>x, v = 0, w = 0. In other words, it is one-dimensional flow, and so the streamline does not change direction; rather, it remains in the x direction even though the flow is unsteady.

NOTE:

168

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of

fluid motion

3.4

3

FLUID ACCELERATION

Once the velocity for a fluid system is established, it is then possible to determine the acceleration for the flow. Doing this is important, because we can then apply Newton’s second law of motion, F = ma, to relate the acceleration of the fluid particles to the forces acting on them. By definition, the acceleration is the rate of change in the velocity. For a Lagrangian description, we follow the particle along its path, and so the velocity is only a function of time, V = V1t2. We can therefore determine the acceleration of the particle using the derivative to find the time rate of change in the velocity, i.e., a =

dV dt

(3–4)

For an Eulerian description, the changes made in V must be determined relative to a selected control volume as the particle moves through it. And so calculating these changes requires a different approach. To show how this is done, we will consider the unsteady nonuniform flow of a fluid through the nozzle in Fig. 3–15. A control volume is selected on the center streamline, and particles move through it with a velocity of V = V1x, t2. When a particle is at position x, where the first open control surface to the control volume is located, it will have a velocity that will be smaller than when it is at position x + ∆x, where it exits the control volume’s other open control surface. This is because the nozzle constricts the flow and therefore causes the particle to have a greater speed at x + ∆x. Thus the velocity of the particle will change due to its change ∆x in position (nonuniform flow). If the valve is gradually opened, then the velocity of the particle can also change within the control volume due to a change over time ∆t (unsteady flow). As a result, the total change in the particle’s velocity will be ∆V =

0V ∆t 0t

+

0V ∆x 0x Change of V with position (nonuniform flow)

Change of V with time (unsteady flow)

Control volume x

x Dx x 1 Dx

Fig. 3–15

3.4

Since ∆x is the distance covered in the time ∆t, then dividing this equation by ∆t and taking the limit, the acceleration of the particle becomes ∆V 0V 0V dx a = lim = + ∆t S 0 ∆t 0t 0x dt We can also determine this result by using the chain rule of calculus. Since V = V1x, t2, we must use partial derivatives to find each time rate of change in V, that is, a = 0V>0t1dt>dt2 + 0V>0x1dx>dt2. Realizing that dx>dt = V, we get a =

DV 0V 0V = + V Dt 0t 0x Local accel.

(3–5)

Convective accel.

The notation D12 >Dt is referred to as a material derivative, because it gives the time rate of change of a fluid property (in this case the velocity) as the fluid particle (material) passes through the control volume. Let’s summarize our result.

Local Acceleration. The first term on the right side, 0V>0t, indicates the time rate of change of the velocity of the particle, within the control volume. For this reason it is called the local acceleration. A further opening of the valve in Fig. 3–15 causes the flow to increase, producing this local change. For steady flow this term will be zero, because the flow will not change with time as we follow it through the control volume. Convective Acceleration. The second term on the right side, V10V>0x2, is referred to as the convective acceleration, since it measures the change in the particle’s velocity as the particle changes its location, that is, it moves from the entrance of the control volume to its exit. The conical shape of the nozzle in Fig. 3–15 causes this change. Only when the flow is uniform, as in the case of a pipe with a constant cross section, will this term be zero.

Fluid particles moving upward from this water sprinkler have a decrease in their velocity magnitude. Also the direction of their velocity is changing. Both of these effects produce acceleration.

fluid aCCeleration

169

3

170

Chapter 3

KinematiCs

z

of

fluid motion

Three-dimensional Flow. Now let’s generalize these results using the material derivative and consider the case of three-dimensional flow, Fig. 3–16. Here each particle in the velocity field has the velocity

V

V1x, y, z, t2 = u1x, y, z, t2i + v1x, y, z, t2j + w1x, y, z, t2k (3–6) 3

z y x x

y

where u, , w, are the x, y, z components of velocity. As we follow each particle through the control volume located at the general position (x, y, z), its velocity can increase (or decrease) due to either a change in time, dt, or a change in the particle’s position, dx, dy, dz. To find these changes, we must use the chain rule to obtain a result similar to Eq. 3–4. We have

Fig. 3–16

a =

DV 0V 0V dx 0V dy 0V dz = + + + Dt 0t 0x dt 0y dt 0z dt

Since u = dx>dt, v = dy>dt, and w = dz>dt, then

a =

DV 0V 0V 0V 0V = + au + v + w b Dt 0t 0x 0y 0z

Total accel.

Local accel.

(3–7)

Convective accel.

If we substitute Eq. 3–6 into this expression, the expansion results in the x, y, z components of acceleration, that is

ax =

0u 0u 0u 0u + u + v + w 0t 0x 0y 0z

0v 0v 0v 0v + u + v + w 0t 0x 0y 0z 0w 0w 0w 0w az = + u + v + w 0t 0x 0y 0z ay =

(3–8)

Besides the velocity field V = V1x, y, z, t2, other fluid properties can be described using an Eulerian description. For example, while a liquid is being heated in a boiler, there will be an uneven temperature rise of the liquid at each point. This creates a scalar temperature field T = T1x, y, z, t2 that changes with both position and time. In a similar manner, the pressure and density of the liquid within the boiler can also be described by scalar pressure and density fields, p = p1x, y, z, t2 and r = r1x, y, z, t2. Like the velocity field, the time rate of change of these fields produces local and convective changes within the fluid system at each control volume.

3.4

EXAMPLE

fluid aCCeleration

3.4

As the valve in Fig. 3–17 is being gradually closed, oil particles flowing through the nozzle along the center streamline have a velocity of V = 3 611 + 0.4x 2 211 - 0.5t2 4 m>s, where x is in meters and t is in seconds. Determine the acceleration of an oil particle when it arrives at x = 0.25 m when t = 1 s.

x

x

0.3 m

Fig. 3–17

SOLUTION Fluid Description. The flow along the streamline is nonuniform and unsteady because its Eulerian description of velocity is a function of both x and t. Analysis. have a =

=

Here V = u. Applying Eq. 3–5 or the first of Eqs. 3–8, we

0V 0V 0 + V = 3 611 + 0.4x 2 211 - 0.5t2 4 0t 0x 0t

3 611

171

+ 3 611 + 0.4x 2 211 - 0.5t2 4

+ 0.4x 2 210 - 0.52 4 +

3 611

0 3 611 + 0.4x 2 211 - 0.5t2 4 0x

+ 0.4x 2 211 - 0.5t2 4 3610 + 0.412x2211 - 0.5t24

Evaluating this expression at x = 0.25 m, t = 1 s, we get

a = -3.075 m>s2 + 1.845 m>s2 = -1.23 m>s2

Ans.

The local acceleration component 1-3.075 m>s2 2 is decreasing the velocity of the particle within the control volume at x = 0.25 m since the valve is being closed to decrease the flow. The convective acceleration component 11.845 m>s2 2 is increasing the velocity of the particle as it passes through the control surfaces, since the nozzle constricts as x increases, and this causes the velocity to increase. The net result, however, is that the particle decelerates at 1.23 m>s2.

3

172

Chapter 3

EXAMPLE

KinematiCs

of

fluid motion

3.5 The velocity for a two-dimensional flow is defined by V = 52xi - 2yj6m>s, where x and y are in meters. Plot the streamlines for the flow field, and determine the magnitudes of the velocity and acceleration of a particle located at the point x = 1 m, y = 2 m.

3

SOLUTION

y

Flow Description. The velocity does not depend on time, so the flow is steady and the streamlines will remain in fixed positions. Analysis. Here u = 12x2 m>s and v = 1-2y2 m>s. To obtain the equations of the streamlines, we must use Eq. 3–1. dy -2y v = = u dx 2x

xy 5 2 a5 8.94 ms2

Separating the variables and integrating yields dy dx = L x L y

2m V 5 4.47 ms

x

1m

ln x = - ln y + C ln1xy2 = C xy = C′

The arbitrary integration constant is C′. Using various values of this constant, the above equation represents a family of streamlines that are hyperbolas, and so the flow field is shown in Fig. 3–18a. To find the streamline that passes through point (1 m, 2 m), we require 112122 = C′, so that xy = 2.

Velocity. The velocity of a particle passing through the control volume located at point (1 m, 2 m) has components of (a)

u = 2112 = 2 m>s v = -2122 = -4 m>s

Fig. 3–18

Therefore,

V = 212 m>s2 2 + 1-4 m>s2 2 = 4.47 m>s

Ans.

Based on the direction of its components, this velocity is shown in Fig. 3–18a. It indicates the direction of the flow along its hyperbolic streamline. Flow along the other hyperbolas can be determined in the same manner, that is, by selecting a point and then showing the vector addition of the velocity components of a particle at this point.

3.4

173

fluid aCCeleration

y

y

3

x Flow striking a fixed surface

x Flow along two perpendicular surfaces

(b)

(c)

It is interesting to note that we can use a portion of this flow pattern to describe, for example, a flow that strikes a fixed surface, as in Fig. 3–18b, a flow within a corner, Fig. 3–18c, or a flow constrained between a corner and a hyperbolic boundary defined by one of the streamlines, Fig. 3–18d. In all of these cases there is a stagnation point at the origin (0, 0), since the velocity is zero at this point, that is, u = 2102 = 0 and v = -2102 = 0. As a result, if the fluid contained debris, it would tend to accumulate within the region around this stagnation point as shown in each figure. Acceleration. The components of acceleration are determined using Eqs. 3–8. Due to the steady flow there is no local acceleration, just convective acceleration. 0u 0u 0u ax = + u + v = 0 + 2x122 + 1 -2y2102 0t 0x 0y = 4x ay =

0v 0v 0v + u + v = 0 + 2x102 + 1 -2y21 -22 0t 0x 0y

= 4y The particle at point (1 m, 2 m) will therefore have acceleration components ax = 4112 = 4 m>s2 ay = 4122 = 8 m>s2 The magnitude of the particle’s acceleration, shown in Fig. 3–18a, is therefore a = 214 m>s2 2 2 + 18 m>s2 2 2 = 8.94 m>s2 Ans.

y

x Flow between a corner and a hyperbolic surface (d)

Fig. 3–18 (cont.)

174

Chapter 3

EXAMPLE

KinematiCs

of

fluid motion

3.6 The velocity of particles of gas flowing along the center of the pipe in Fig. 3–19 is defined for x Ú 1 m by the velocity field V = 1t>x2 m>s, where t is in seconds and x is in meters. If a particle within this flow is at x = 1 m when t = 0, determine its acceleration when it is at x = 2 m.

3

SOLUTION Fluid Description. As noted, since V = V1x, t2, the flow is unsteady and nonuniform. Analysis. In Example 3.3 we have obtained the Lagrangian description of the particle’s position and velocity, namely x = 2t 2 + 1

V x

and V =

x

Fig. 3–19

t t = 2 x 2t + 1

Also, when the particle arrives at x = 2 m, the time was found to be t = 1.732 s. From the Lagrangian point of view, the acceleration is simply the time derivative of the velocity, dV a = = dt

1t 2 + 12 1>2 112 - tc

Thus, when t = 1.732 s, a =

1 2 1t + 12 - 1>2 12t2 d 2 1 = 2 2 t + 1 1t + 12 3>2 1

3 11.7322 2 + 14 3>2

= 0.125 m>s2

Ans.

Now let’s check our work by taking the material derivative of the velocity field (Eulerian description). Since this is one-dimensional flow, applying Eq. 3–5, we have a =

DV 0V 0V 1 t t = + V = + a- 2 b x x x Dt 0t 0x

At the control volume, located at x = 2 m, the particle will appear when t = 1.732 s. Its acceleration will then be a =

1 1.732 1.732 + ab = 0.125 m>s2 2 2 22

which agrees with our previous result.

Ans.

3.5

3.5

175

streamline Coordinates

STREAMLINE COORDINATES

s

When a curved path or streamline for the fluid particles is known, such as when the flow is through a curved pipe, then streamline coordinates can be used to describe the motion. To show how these coordinates are established, consider fluid particles moving along the streamline in Fig. 3–20a. The origin of the coordinate axes is placed at the point on the streamline where the fixed control volume is located. The s axis is tangent to the streamline at this point, and is positive in the direction in which the particles are traveling. We will designate this positive direction with the unit vector u s. The normal n axis is perpendicular to the s axis at the control volume, with the positive sense directed toward the center of curvature O′ for an arc ds along the path, Fig. 3–20b. This positive direction, which is always on the concave side of the curve, will be designated by the unit vector u n. Once the s and n axes are established in this manner, we can then express the velocity and acceleration of a particle passing through the control volume in terms of these coordinates.

V

Control volume O

us un

s

n

Reference point

3

O9 Streamline coordinates (a) O9 O9 s R n

R

R

ds n R R

R

ds

Velocity. Since the direction of the particle’s velocity V is always tangent to the path, that is, in the positive s direction, Fig. 3–20a, we have

n

O9 s

ds s

(3–9)

V = Vu s

n–s axes at various points along a streamline (b)

where V = V1s, t2.

Acceleration. The acceleration of the particle is the time rate of change of the velocity, and so to determine it using an Eulerian description, we must account for both the local and the convective changes to the velocity as the particle moves through the control volume, Fig. 3–20c. Local Change. If a condition of unsteady flow exists, then local changes can occur to the particle’s velocity within the control volume. This change ∆V has both tangential and normal components ∆Vs and ∆Vn, and so in the limit the local change produces the acceleration components as  = a local

0V b 0t s

and

an  = a local

V

Control volume Ds (c)

Fig. 3–20

0V b 0t n

For example, a local streamline (or tangential) component of acceleration 10V>0t2 s can occur if the speed of the flow in the rotating sprinkler arm in Fig. 3–21 is increased or decreased by opening or closing the valve. Here the magnitude of the particle’s velocity within the control volume increases or decreases as a function of time. Also, a local normal component of acceleration 10V>0t2 n can occur in the control volume because the rotating arm will cause the direction of the velocity of the particle within the control volume to change with time.

Fig. 3–21

V + DV

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V + DV

V

Control volume Ds

3 (c)

of

fluid motion

Convective Change. The velocity of the particle can also change as the particle moves ∆s from the entrance to the exit control surface, Fig. 3–20c. Here ∆Vs in Fig. 3–20d represents the convective change in the magnitude of V. It indicates whether the fluid particle speeds up, as when it moves through a converging pipe (or nozzle), Fig. 3–22, or slows down, as when it moves through a diverging pipe. Both of these cases represent nonuniform flow. To obtain this convective acceleration component in the s direction, we have

DVn (due to change in direction) DVs (due to change in magnitude) DV V Du

as  = lim S conv ∆t

V + DV (d) V + DV

V

Ds R

R n

Du

0

∆Vs ∆s ∆Vs 0V = lim = V S ∆t 0 ∆t ∆t ∆s 0s

The normal component ∆Vn in Fig. 3–20d is due to the change in the direction of V, since it indicates how the velocity vector “swings” as the particle moves or is convected through the control volume. Since V is always tangent to the path, the change in angle ∆u between V and V + ∆V, Fig. 3–20d, must be the same angle ∆u shown in Fig. 3–20e. And, because ∆u is very small, then ∆u = ∆s>R (Fig. 3–20e) and also ∆u = ∆Vn >V (Fig. 3–20d). Equating these results gives ∆Vn = 1V>R2∆s. Therefore the convective acceleration component in the n direction becomes an  = lim a S

O9

conv

(e)

∆t

0

∆Vn V ∆s V2 b = lim = ∆t conv R ∆t S 0 ∆t R

Fig. 3–20 (cont.)

A typical example of this acceleration component occurs in the curved pipe in Fig. 3–22, because the direction of the velocity of the particle will change as the particle moves from the entrance to the exit control surface.

Resultant Acceleration. If we now combine both the local and the convective changes using the above results, the streamline and normal acceleration components become s

n

as = a

an = a

0V 0V b + V 0t s 0s

0V V2 b + 0t n R

(3–10) (3–11)

Fig. 3–22

To summarize, the first terms on the right in these two equations are the local changes in the velocity’s magnitude and direction, caused by unsteady flow, and the second terms are convective changes in the velocity’s magnitude and direction, caused by nonuniform flow.

3.5

streamline Coordinates

177

IMPORTANT POIN T S • A normal time derivative is used to determine the acceleration •

•

• •

•

of a particle when we are using a Lagrangian description of the flow. Its velocity is defined by V = V1t2. Here a = d V>dt. The material time derivative is used to determine the acceleration of a particle when we are using an Eulerian description of the flow provided the velocity field V = V1x, y, z, t2 is known. It consists of two parts, the local or time change in the velocity within the control volume, the result of unsteady flow; and the convective or position change of velocity as the particle moves into and out of the control surfaces, the result of nonuniform flow. Streamline coordinates are located at a point on a streamline. They consist of an s coordinate axis that is tangent to the streamline, positive in the direction of flow, and an n coordinate axis that is normal to the s axis. It is positive if it acts towards the center of curvature of the streamline at the point. The velocity of a fluid particle is always in the +s direction. The s component of acceleration of a particle consists of the magnitude change in the velocity. This is the result of the local time rate of change, 10V>0t2 s (unsteady flow), and convective change, V10V>0s2 (nonuniform flow). The n component of acceleration of a particle consists of the directional change in the velocity. This is the result of the local time rate of change, 10V>0t2 n (unsteady flow), and convective change, V 2 >R (nonuniform flow).

Using particles of smoke, streamlines provide a method of visualizing the flow of air over the body of this automobile. (© culture-images GmbH/Alamy Stock Photo)

3

178

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EXAMPLE

KinematiCs

of

fluid motion

3.7

n

3

s

s 5 0.6 m

Fig. 3–23

A

As a fluid flows through the fixed curved conduit in Fig. 3–23, the velocity of particles is described by V = 10.4s2 2e -0.4t m>s, where s is in meters and t is in seconds. Determine the magnitude of the acceleration of the fluid particle located at point A, where s = 0.6 m, and the time of arrival is t = 1 s. The radius of curvature of the streamline at A is R = 0.5 m. SOLUTION Fluid Description. Since this Eulerian description of the motion is a function of space and time, the flow is nonuniform and unsteady. Streamline Acceleration Component. Applying Eq. 3–10, to determine the magnitude change in velocity, we have

as = a =

0V 0V b + V 0t s 0s

0 3 10.4s2 2e - 0.4t 4 + 0t

3 10.4s2 2e - 0.4t 4

0 3 10.4s2 2e - 0.4t 4 0s

= 0.4s2 1-0.4e -0.4t 2 + 10.4s2 2e -0.4t 10.8s e -0.4t 2

= 0.410.6 m2 2 3-0.4 e -0.411 s24 + 10.4210.6 m2 2e -0.411 s2 30.810.6 m2e -0.411 s24 = -0.00755 m>s2

Normal Acceleration Component. Because the pipe is not rotating, then the streamline is not rotating, and so the n axis remains in a fixed direction at A. Therefore, there is no local change in the direction of velocity within the control volume along the n axis. We only have a convective change. Applying Eq. 3–11, we get an = a

0V V2 b + = 0 + 0t n R

3 0.410.6 m2 2e - 0.411 s2 4 2 0.5

= 0.01863 m>s2 Acceleration.

The magnitude of acceleration is therefore

a = 2a 2s + a 2n = 21 -0.00755 m>s2 2 2 + 10.01863 m>s2 2 2 = 0.0201 m>s2 = 20.1 mm>s2

Ans.

fundamental problems

179

References 1. D. Halliday, et. al, Fundamentals of Physics, 7th ed., J. Wiley and Sons,

Inc., N.J., 2005. 2. W. Merzkirch, Flow Visualization. 2nd ed., Academic Press, New

3

York, NY, 1897. 3. R. C. Baker, Introductory Guide to Flow Measurement, 2nd ed., John Wiley, New York, NY, 2002. 4. R. W. Miller, Flow Measurement Engineering Handbook, 3rd ed., McGraw-Hill, New York, NY, 1997. 5. G. S. Settles, Schlieren and Shadowgraph Techniques: Visualizing Phenomena in Transport Media, Springer, Berlin, 2001.

F UN DAMEN TAL PRO B L EM S The solutions to all fundamental problems are given in the back of the book.

SEC. 3.1–3.3 F3–1. A two-dimensional flow field is defined by u = 10.25x2 m>s and v = 12t2 m>s, where x is in meters and t is in seconds. Determine the position (x, y) of a particle when t = 2 s if the particle passes through point (2 m, 6 m) when t = 0.

F3–2. A two-dimensional flow field is defined by u = 12x 2 2 m>s and v = 18y2 m>s, where x and y are in meters. Determine the equation of the streamline passing through the point (2 m, 3 m).

y

y

6m 3m

x 2m

Prob. F3–1

x 2m

Prob. F3–2

180

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SEC. 3.4

SEC. 3.5

F3–3. The flow of water through the nozzle causes the velocity of particles along the center streamline to be V = 1200x 3 + 10t 2 2 m>s, where t is in seconds and x is in meters. Determine the acceleration of a particle at 3 x = 0.1 m if it arrives when t = 0.2 s.

F3–6. Fluid flows through the curved pipe at a steady rate. If the velocity of particles moving along the center streamline is V = 120s2 + 42 m>s, where s is in meters, determine the magnitude of the acceleration of a particle when it is at point A. A s

0.5 m 458

x

Prob. F3–3 F3–4. The velocity of particles of dioxitol along the x axis is given by u = 31x + 42 m>s, where x is in meters. Find the acceleration of a particle located at x = 100 mm. Also, what is the position of a particle when t = 0.02 s if it starts from x = 0 when t = 0?

Prob. F3–6 F3–7. A fluid flows through the curved pipe with a constant average speed of 3 m>s. Determine the magnitude of acceleration of water particles on the streamline along the centerline of the pipe.

0.5 m

x 3 ms

x

Prob. F3–7

Prob. F3–4 F3–5. A flow field is defined by u = 13x + 2t 2 2 m>s and v = 12y3 + 10t2 m>s, where x and y are in meters and t is in seconds. Determine the magnitudes of the local and convective accelerations of a particle at point x = 3 m, y = 1 m if it arrives when t = 2 s.

F3–8. Fluid flows through the curved pipe such that fluid particles move along the center streamline with a velocity of V = 120s2 + 1000t 3>2 + 42 m>s, where s is in meters and t is in seconds. Determine the magnitude of the acceleration of a particle at point A, where s = 0.3 m, if it arrives when t = 0.02 s.

s y A

1m

R 5 0.5 m x 3m

Prob. F3–5

Prob. F3–8

181

problems

P ROBLEMS The answers to all but every fourth problem are given in the back of the book. 3

SEC. 3.1–3.3 *3–1. Particles travel within a flow field defined by V = 5 2y2i + 4j 6 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (1 m, 2 m), and find the velocity of a particle located at this point. Draw this streamline.

3–2. A flow field is defined by u = 10 m>s and v = - 3 m>s. If metal flakes are released into the flow at point (2 m, 1 m), draw the streamline, streakline, and pathline for these particles.

3–3. A two-dimensional flow field for a liquid can be described by V = 5 1 5y2 - x 2 i + (3x + y)j 6 m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at (5 m, - 2 m), and its direction measured counterclockwise from the x axis.

3–7. A balloon is released into the air from the origin and carried along by the wind, which blows at a constant rate of u = 0.5 m>s. Also, buoyancy and thermal winds cause the balloon to rise at a rate of v = 10.8 + 0.6y2 m>s, where y is in meters. Determine the equation of the streamline for the balloon, and draw this streamline. *3–8. A balloon is released into the air from point (1 m, 0) and carried along by the wind, which blows at a rate of u = 10.8x2 m>s, where x is in meters. Also, buoyancy and thermal winds cause the balloon to rise at a rate of v = 11.6 + 0.4y2 m>s, where y is in meters. Determine the equation of the streamline for the balloon, and draw this streamline. y

v

*3–4. A flow field is defined by u = 12x 2 + 12 m>s and v = 1xy2 m>s, where x and y are in meters. Determine the equation of the streamline that passes through point (3 m, 1 m), and find the velocity of a particle located at this point. Draw this streamline.

3–5. A two-dimensional flow field for a fluid is defined by V = 314x2i - 12y + 1x2j4 m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at 11 m, 2 m2, and its direction measured counterclockwise from the positive x axis.

3–6. A two-dimensional flow field for a fluid can be described by V = {(2x + 1)i - (y + 3x)j} m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at (2 m, 3 m), and its direction measured counterclockwise from the x axis.

u

x

Probs. 3–7/8 3–9. A two-dimensional flow field for a liquid is defined by V = 31x + 2y2 2i + 12x 2 - 3y2j4 m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at 1- 2 m, 2 m2, and its direction measured counterclockwise from the positive x axis. 3–10. Particles travel within a flow field defined by 1 3 V = 3 16 y i + 6j 4 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (3 m, 4 m), and find the velocity of a particle located at this point. Draw this streamline.

182

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Chapter 3

KinematiCs

of

fluid motion

3–11. A flow field for a fluid is defined by u = 12 + y2 m>s and v = 12y2 m>s, where y is in meters. Determine the equation of the streamline that passes through point (3 m, 2 m), and find the velocity of a particle located at this point. Draw this streamline.

3–21. A fluid has velocity components of u = 13x 2 + 12 m>s and v = 14txy2 m>s, where x and y are in meters and t is in seconds. Determine the streamlines that pass through point (1 m, 3 m) at times t = 1 s and t = 1.5 s. Plot these streamlines for 0 … x … 5 m.

The circulation of a fluid is defined by the velocity = 16 - 3x2 m>s and v = 2 m>s, where x is in Plot the streamline that passes through the origin x 6 2 m.

3–22. A fluid has velocity components of u = [30>12x + 12] m>s and v = 12ty2 m>s, where x and y are in meters and t is in seconds. Determine the pathline that passes through point (2 m, 6 m) at time t = 2 s. Plot this pathline for 0 … x … 4 m.

*3–12. field u meters. for 0 …

3–13. A particle travels along a streamline defined by y3 = 8x - 12, where x and y are in meters. If its speed is 5 m>s when it is at x = 1 m, determine the two components of its velocity at this point. Sketch the velocity on the streamline. 3–14. The velocity for an oil flow is defined by V = 10.5y2i + 2j2 m>s, where y is in meters. What is the equation of the streamline that passes through point (1 m, 3 m)? If a particle is at this point when t = 0, at what point is it located when t = 2 s? 3–15. A two-dimensional flow field for benzene is defined by V = 31y2 - 62i + 12x + 32j4 m>s, where x and y are in meters. Determine a streamline that passes through point 12 m, 4 m2 and the velocity at this point. Sketch the velocity on the streamline. *3–16. field u meters. for 0 …

The circulation of a fluid is defined by the velocity = 1 3 - 21x 2 m>s and v = 4 m>s, where x is in Plot the streamline that passes through the origin x 6 6 m.

3–17. A flow field for gasoline is defined by u = 1 8y 2 m>s, and v = 12x2 m>s where x and y are in meters. Determine the equation of the streamline that passes through point (2 m, 4 m). Draw this streamline.

3–18. A particle travels along the streamline defined by y2 + 3 = 2x, where x and y are in meters. If its speed is 3 m>s when it is at x = 3 m, y = 13 m, determine the x and y components of its velocity at this point. Sketch the velocity on the streamline. 3–19. A flow of water is defined by u = 5 m>s and = 8 m>s. If metal flakes are released into the flow at the origin (0, 0), draw the streamline and pathline for these particles. *3–20. A stream of water has velocity components of u = - 2 m>s, v = 3 m>s for 0 … t 6 10 s; and u = 5 m>s, v = -2 m>s for 10 s 6 t … 15 s. Plot the pathline and streamline for a particle released at point (0, 0) when t = 0.

3–23. A flow field is defined by u = 10.9x2 m>s and = 11.8y2 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (1 m, 2 m). Draw this streamline.

SEC. 3.4

*3–24. A fluid flow is defined by u = 13x 2 - 4y2 m>s and v = 16xy2 m>s, where x and y are in meters. Determine the velocity and acceleration of particles passing through point (1 m, 2 m).

3–25. A fluid flow is defined by V = 52xi + 5j6 m>s, where x is in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point 12 m, 1 m2. Find the equation of the streamline passing through this point, and show the velocity and acceleration at the point on this streamline.

3–26. A velocity field for oil is defined by u = 13y2 m>s, v = 10.6t 2 + 22 m>s, where t is in seconds and y is in meters. If a particle is at the origin when t = 0, determine its acceleration and position when t = 2 s. 3–27. A fluid flow is defined by u = 1 14y2 2 m>s and 1 2 v = 1 16 x y 2 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (2 m, 3 m). Also, what is the acceleration of a particle at this point? Is the flow steady or unsteady?

*3–28. Oil flows through the reducer such that particles along its centerline have a velocity of V = 160xt + 32 m>s, where x is in meters and t is in seconds. Determine the acceleration of a particle at x = 400 mm if it arrives when t = 0.3 s. 500 mm

x

Prob. 3–28

problems 3–29. A fluid flow is defined by u = 10.5y2 m>s and v = 1 - 0.3x2 m>s, where x and y are in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point 14 m, 3 m2. Find the equation of the streamline passing through this point, and show the velocity and acceleration at the point on this streamline.

183

*3–36. Oil flows through the reducer such that particles along its centerline have a velocity of V = 14x + 20t + 22 m>s, where x is in meters and t is in seconds. Determine the acceleration of a particle at x = 400 mm if it arrives when t = 0.2 s. 3

3–30. The velocity for the flow of a gas along the center streamline of the pipe is defined by u = 110x 2 + 200t + 62 m>s, where x is in meters and t is in seconds. Determine the acceleration of the particle at A if it is just leaving the nozzle when t = 0.01 s.

600 mm

x

Prob. 3–36

A x 0.6 m

3–37. Airflow through the duct is defined by the velocity field u = 12x 2 + 82 m>s and v = 1 - 8x2 m>s, where x is in meters. Determine the acceleration of a fluid particle at the origin (0, 0) and at point (1 m, 0). Also, sketch the streamlines that pass through these points.

Prob. 3–30 y

3–31. A fluid flow is defined by u = 10.4x 2 + 2t2 m>s and v = 10.8x + 2y2 m>s, where x and y are in meters and t is in seconds. Determine the velocity and acceleration of a particle passing through point x = 2 m, y = 1 m if it arrives when t = 3 s.

*3–32. A fluid flow is defined by u = 16x 2 - 3y2 2 m>s and v = 14xy + y2 m>s, where x and y are in meters. Determine the magnitudes of the velocity and acceleration of a particle at point (2 m, 2 m). 3–33. A fluid has velocity components of u = 1 x 2yt 2 m>s and v = 14x - 2t2 m>s, where x and y are in meters and t is in seconds. Determine the magnitude of acceleration of a particle passing through the point (0.3 m, 0.4 m) if it arrives when t = 2 s.

3–34. Air flows uniformly through the center of a horizontal duct with a velocity of V = 1 14t 3 + 3 2 m>s, where t is in seconds. Determine the acceleration of the flow when t = 3 s.

3–35. A fluid flow is defined by u = 13xy2 m>s and v = 12y2 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (2 m, 1 m). Also find the acceleration of a particle located at this point. Is the flow steady or unsteady?

x

x51m

Prob. 3–37

3–38. The velocity of gasoline, along the centerline of a tapered pipe, is given by u = 14tx2 m>s, where t is in seconds and x is in meters. Determine the acceleration of a particle when t = 0.8 s if u = 0.8 m>s when t = 0.1 s. 3–39. A fluid flow is defined by u = 12y2 2 m>s and v = 18xy2 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (1 m, 2 m). Also, what is the acceleration of a particle at this point? Is the flow steady or unsteady?

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fluid motion

*3–40. As the valve is closed, oil flows through the nozzle such that along the center streamline it has a velocity of V = 310.5 + 0.8x 2 216 - 2t2 m>s, where x is in meters and t is in seconds. Determine the acceleration of an oil particle at x = 0.2 m if it arrives when t = 2 s. 3 A B

3–45. A fluid flow is defined by u = 18t 2 2 m>s and v = 17y + 3x2 m>s, where x and y are in meters and t is in seconds. Determine the velocity and acceleration of a particle passing through point (1 m, 1 m) if it arrives when t = 2 s.

3–46. A fluid flow is defined by u = 12x 2 - y2 2 m>s and v = 1 -4xy2 m>s, where x and y are in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point 11 m, 1 m2. Find the equation of the streamline passing through this point, and show the velocity and acceleration at the point on this streamline.

x

0.3 m

Prob. 3–40 3–41. A fluid flow is defined by u = 15y2 2 m>s and v = 14x - 12 m>s, where x and y are in meters. Determine the equation of the streamline passing through point 11 m, 1 m2. Find the components of the acceleration of a particle located at this point, and sketch the acceleration on the streamline. 3–42. Air flowing through the center of the duct decreases in speed from VA = 8 m>s to VB = 2 m>s in a linear manner. Determine the velocity and acceleration of a particle moving horizontally through the duct as a function of its position x. Also, find the position of a particle as a function of time if x = 0 when t = 0. B

A VB 5 2 ms

VA 5 8 ms

*3–44. A fluid flow is defined by V = 5 4xi + 2j 6 m>s, where x is in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point 11 m, 2 m2. Find the equation of the streamline passing through this point, and show the velocity and acceleration at the point on this streamline.

3–47. A fluid flow is defined by u = 12x2 m>s and v = 12y2 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (0.6 m, 0.3 m). Also find the acceleration of a particle located at this point. Is the flow steady or unsteady?

SEC. 3.5 *3–48. Fluid particles have velocity components of u = 12x2 m>s and v = 14y2 m>s, where x and y are in meters. Determine the acceleration of a particle located at point (2 m, 1 m). Determine the equation of the streamline passing through this point.

3–49. A fluid has velocity components of u = 14xy2 m>s and v = 12y2 m>s, where x and y are in meters. Determine the magnitudes of the streamline and normal components of acceleration of a particle located at point (2 m, 1 m).

3–50. A particle moves along the circular streamline, such that it has a velocity of 3 m>s, which is increasing at 3 m>s2. Determine the acceleration of the particle, and show the acceleration on the streamline. 3 ms

x 3m

Prob. 3–42

3–43. A fluid flow is defined by V = 5 4yi + 2xj 6 m>s, where x and y are in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point 12 m, 1 m2. Find the equation of the streamline passing through this point, and show the velocity and acceleration at the point on this streamline.

4m

Prob. 3–50

problems 3–51. As water flows steadily over the spillway, one of its particles follows a streamline that has a radius of curvature of 16 m. If its speed at point A is 5 m>s, which is increasing at 3 m>s2, determine the magnitude of acceleration of the particle.

185

3–54. Air flows around the front circular surface. If the steady-stream velocity is 4 m>s upstream from the surface, and the velocity along the surface is defined by V = 116 sin u2 m>s, determine the magnitudes of the streamline and normal components of acceleration of a particle located on the surface at u = 30°.

V 4 ms 16 m A

u

0.5 m

Prob. 3–51

Prob. 3–54 *3–52. A particle located at a point within a fluid flow has velocity components of u = - 2 m>s and v = 3 m>s, and acceleration components of ax = - 1.5 m>s2 and ay = - 2 m>s2. Determine the magnitudes of the streamline and normal components of acceleration of the particle. 3–53. Water flows into the drainpipe such that it only has a radial velocity component V = 1 -3>r2 m>s, where r is in meters. Determine the acceleration of a particle located at point r = 0.5 m, u = 20°. At t = 0, r = 1 m.

3–55. The motion of a tornado can, in part, be described by a free vortex, V = k>r, where k is a constant. Consider the steady motion at the radial distance r = 3 m, where V = 18 m>s. Determine the magnitude of the acceleration of a particle traveling on the streamline having a radius of r = 9 m.

r59m s

r 5 0.5 m u

Prob. 3–53

Prob. 3–55

3

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CHAP T ER R EVIEW Laminar flow occurs when the fluid flows in thin layers so that the fluid particles follow smooth paths. 3

Turbulent flow is very erratic, causing mixing of the fluid particles, and therefore more internal friction is produced than within laminar flow. Steady flow occurs when the flow does not change with time. Uniform flow occurs when the flow does not change with location.

y dy V

A streamline is a curve that indicates the direction of the velocity of particles located on it at a particular instant of time.

dy y —5 — u dx

dx

v u Streamline x

A pathline shows the path of a particle during a specified time period. It is visualized using a timelapse photograph.

A streakline is formed by a succession of marked particles that all come from the same point of origin. For example, streaklines are formed when smoke or colored dye is released into the flow from the same point, and an instantaneous photograph is taken of the trace of all the marked particles. If the flow is steady, the streamlines, pathlines, and streaklines will all coincide.

Chapter review

187

z

A Lagrangian description follows the motion of a single particle as the particle moves through the flow field.

3 3

r 5 r(t)

V 5 V(t)

z

y x y

x z

Control volume

An Eulerian description considers a specific region (or control volume) in the flow, and it measures the motion, or a fluid property, of all the particles that pass through this region.

V 5 V(x0 , y0 , z0 , t) z0 y x0 y0

x

If an Eulerian description is used to define the velocity field V = V1x, y, z, t2, then the acceleration will have both local and convective components. The local acceleration accounts for the time rate of change in velocity within the control volume, and the convective acceleration accounts for its spatial change, from the point where the particle enters a control surface, to the point where it exits another surface.

a =

DV = Dt Total accel.

0V 0V 0V 0V + au + v + w b 0t 0x 0y 0z

Local accel.

Convective accel.

s V

Streamline coordinates s and n have their origin at a point on a streamline. The s coordinate is tangent to the streamline and it is positive in the direction of flow. The normal coordinate n is positive towards the streamline’s center of curvature at the point.

us O

Reference point

s

un n

O9

4

Maximilian Stock Ltd./Science Source

CHAPTER

The analysis of flow through the various ducts and vessels in this chemical processing plant depends upon the conservation of mass.

CONSERVATION OF MASS

CHAPTER OBJECTIVES ■

To show how to determine the average velocity and the volumetric and mass flow in a conduit.

■

To show how a Lagrangian and an Eulerian description of fluid behavior can be related using the Reynolds transport theorem.

■

To use the Reynolds transport theorem to derive the equation for the conservation of mass.

4.1

VOLUMETRIC FLOW, MASS FLOW, AND AVERAGE VELOCITY

In this chapter, we will expand our use of a control volume, and discuss its application as it relates to the conservation of mass. Before we do this, however, we must first define ways to describe the flow in terms of volume or mass per unit time. We will now formally define each of these quantities.

189

190

4

Chapter 4

C o n s e r va t i o n

of

Mass

Volumetric Flow. The rate at which a volume of fluid flows through a cross-sectional area A is called the volumetric flow, or simply the flow or discharge. It can be determined provided we know the velocity profile for the flow through the area. For example, consider the one-dimensional flow of a viscous fluid through a pipe, where its velocity profile has the axisymmetric shape shown in Fig. 4–1. If particles passing through the differential area dA have a velocity , then during the time dt, a volume element of fluid of length v dt will pass through this area. Since this “volume” is dV = (v dt)(dA), then the volumetric flow dQ through the area is determined by dividing the volume by dt, which gives dQ = dV>dt = v dA. If we integrate over the entire cross-sectional area A, we have

Q =

v dA LA

Here Q can be measured in m3 >s. Integration is possible if the velocity can be expressed as a function of the coordinates describing the area. For example, we will show in Chapter 9 that the velocity profile in Fig. 4–1 is a paraboloid, provided the flow is laminar. Sometimes, however, the velocity profile must be determined experimentally, as for turbulent flows, in which case a graphical integration of the velocity profile may be performed. In either case, the integral in the equation geometrically represents the volume within the velocity diagram shown in Fig. 4–1.

dA v

A

Volumetric flow is represented by the volume under the velocity profile

Fig. 4–1

4.1

When calculating Q, it is important to remember that the velocity must be normal to the cross section through which the fluid flows. If this is not the case, as in Fig. 4–2, then we must use the velocity’s component cos u for the calculation. If the outward normal direction is defined by the unit vector n, then the area dA can be expressed in vector form as dA = (dA)n. Using the dot product,* v # dA = v cos u dA, to express the integral in the previous equation, and writing it in a more general form, we have Q =

v # dA LA

191

voluMetriC flow, Mass flow, and average veloCity

n n

u v

dA

4 Fig. 4–2

(4–1)

Average Velocity. When the fluid is considered ideal, then its viscous or frictional effects can be neglected. As a result, the velocity profile of the fluid over the cross section will be uniform as shown in Fig. 4–3. This type of profile also closely resembles the case of turbulent flow, where we have seen in Fig. 3–2 how turbulent mixing of the fluid tends to flatten the velocity profile to be somewhat uniformly distributed. For the case of v = V, Eq. 4–1 gives Q = V#A

(4–2)

Here V is the average velocity and 𝚨 = An is the outward vector that represents the area of the cross section. For any real fluid, where viscosity occurs, the average velocity can be determined by requiring the flow to be equivalent for both the actual and the average velocity distributions, Fig. 4–1 and Fig. 4–3, so that v # dA LA

Q = VA = Therefore, the average velocity is

dA

v # dA

V =

LA

A

V

(4–3)

Most often, however, the average velocity of the flow is determined if we know Q through the cross-sectional area A. Combining Eqs. 4–1 and 4–3, it is V =

Q A

A

Inviscid or ideal fluids produce a uniform velocity distribution

Fig. 4–3

(4–4)

Throughout the book, and for the problems, it is intended that when the velocity is given or asked for, we will assume this to mean the average velocity. *Recall that for the dot product, the angle u 10° … u … 180°2 is measured between the tails of the vectors.

Streamline

192

Chapter 4

C o n s e r va t i o n

of

Mass

4

The mass flow of air through this duct must be determined by using the open area of the duct and the velocity component that is perpendicular to this area.

Mass Flow. Since the fluid mass passing through the element of area dA in Fig. 4–1 is dm = r dV = r(v dt)dA, the mass flow or mass discharge of the fluid through the entire cross section becomes dm # m = = rv # dA (4–5) dt LA Measurements of the mass flow can be made in kg>s. If the fluid is incompressible, then r is constant, and if V is the average velocity, Eq. 4–5 gives # m = rV # A

(4–6)

I MPO RTA N T PO I N T S from Q = 1Av # dA, where v is the velocity of each fluid particle passing through the area. The dot product is used because the calculation requires the velocity to be perpendicular to the area. Here Q can have the unit of m3 >s.

• The volumetric flow or discharge through an area is determined

• In many problems, the average velocity V can be used. If the flow •

is known, then this velocity can be determined from V = Q>A. # The mass flow is determined from m = 1 rv # dA, or for an # incompressible fluid having an average velocity V, m = rV # A. # Here m can have the unit of kg>s.

4.1

EXAMPLE

voluMetriC flow, Mass flow, and average veloCity

4.1

The velocity profile for the steady laminar flow of water through a 0.4-m-diameter pipe is defined by v = 3 1 1 - 25r 2 2 m>s, where r is in meters, Fig. 4–4a. Determine the volumetric flow through the pipe and the average velocity of the flow. v 5 3 (1 2 25 r 2) ms 0.2 m

4

dA dr

r

r

3 ms

0.2 m

(b)

(a)

Fig. 4–4

SOLUTION Fluid Description.

Here one-dimensional steady flow occurs.

Analysis. The volumetric flow is determined using Eq. 4–1. A differential ring element of thickness dr is selected, Fig. 4–4b, so that dA = 2p r dr. Thus,

Q =

v # dA =

LA

L0

311 - 25r 2 2 2pr dr

0.2 m

= 6pc

r2 25r 4 0.2 m d 2 4 0

= 0.1885 m3 >s = 0.188 m3 >s

Ans.

To avoid this integration, Q can also be determined by calculating the volume under the “velocity paraboloid” of radius 0.2 m and height 3 m>s. It is

Q =

193

1 1 p r 2h = p(0.2 m)2 13 m>s2 = 0.188 m3 >s 2 2

Ans.

The average velocity is determined from Eq. 4–4.

V =

0.1885 m3 >s Q = 1.5 m>s = A p(0.2 m)2

Ans.

194

Chapter 4

C o n s e r va t i o n

of

Mass

4.2

4 Open control surface

Control volume

Fluid system

Closed control surface Open control surface Fluid surroundings

Fig. 4–5

Fixed control volume (a)

FINITE CONTROL VOLUMES

In Sec. 3.3 we defined a control volume as a selected volume of space within a system of particles. Recall that the boundary of this volume is the control surface. A portion of the surface of this volume may be open, where the fluid particles flow into or out of the control volume, and the remaining portion is closed. Although in the previous chapter we considered a fixed differential size control volume, we can also consider control volumes that have a finite size, such as the one shown in Fig. 4–5. Actually, depending upon the problem, a control volume can be fixed, it can move, or it can change shape. Also, it may include solid parts of an object within its boundary. For example, a fixed control volume within the pipe in Fig. 4–6a is indicated by the red boundary. The control volume that outlines the rocket engine and the fluid within it, Fig. 4–6b, moves with the rocket as it travels upward. Finally, the control volume within the inflatable structure in Fig. 4–6c changes its shape as air is pumped into its open control surface. As we have seen, using a control volume is fundamental when applying an Eulerian description of the flow, since here we must follow the flow as it enters and exits the control volume through its open control surfaces. In this chapter we will use this approach to solve problems that involve the conservation of mass. Then later we will extend its application to problems that involve energy, Chapter 5, and momentum, Chapter 6. In all these cases, it will be necessary to clearly define the boundaries of a selected control volume, and to specify the size and orientation of its open control surfaces.

Moving control volume (b)

Fig. 4–6

Changing control volume (c)

4.2

Open Control Surfaces. The open control surfaces of a control volume will have a cross-sectional area that either lets fluid flow into the control volume, Ain, or lets fluid flow out of it, Aout. In order to properly identify these areas, we will express each as a vector, where its direction is normal to the area and always directed outward from a control surface. For example, if the fixed control volume in Fig. 4–7 is used to study the flow into and out of the tee connection, then the direction of each open control surface is defined by the outward normal unit vector u, and so the area vectors are AA = AAu A, AB = ABu B, and AC = AC u C. Velocity. When using a control volume, we will also have to specify the velocity of flow both into and out of each control surface. Since the outward normal for each control surface is positive, then flow into a control surface will be negative, and flow out of a control surface will be positive. By this convention, VA in Fig. 4–7 is in the negative direction, and VB and VC are in the positive direction. Steady Flow. In some problems we can simplify a fluid analysis by observing the fluid as having steady flow. For example, consider the blade moving with a velocity Vb shown in Fig. 4–8a. For a fixed observer, the flow at A will appear to have a velocity Vf at time t; however, when the blade advances, then at t + ∆t the flow will change its direction and then the velocity will become Vf′. This is a case of unsteady flow since it changes with time. If we select a control volume that contains an amount of fluid on the blade, and then move this control volume and the observer with the blade at Vcv = Vb, then the flow at A relative to the control surface will appear as steady flow, Fig. 4–8b.

195

finite Control voluMes

VB uB AB B

4 AC

AA VA

VC

uA

uC A

C

Vectors directed outward from a control surface are positive

Fig. 4–7

Vf Vb Vf

A y

Vfcs

Time t

Vcv 5 Vb A

x

Vcs 5 Vb

Vfcs 5 Vf 2 Vcs

V9f A

Steady flow relative to an observer attached to the moving control volume

Vb

(b)

Time t 1 D t Unsteady flow relative to a fixed observer (a)

Fig. 4–8

196

Chapter 4

C o n s e r va t i o n

of

Mass

4.3

4

THE REYNOLDS TRANSPORT THEOREM

Fluid behavior must satisfy the conservation of mass, the principle of work and energy, and the principle of impulse and momentum. These laws were originally formulated for a particle, and they were described using a Lagrangian approach. However, in fluid mechanics it is best to apply them using an Eulerian approach. For this reason, we must have a means to convert these laws from their Lagrangian description to an Eulerian description. For a system of particles this conversion is done using the Reynolds transport theorem. In this section we will formalize this theorem, and then we will use it in the next section to develop the conservation of mass, and in the later chapters to develop the energy and momentum equations. Before we establish this theorem, however, we will first discuss how to best describe a fluid property in terms of the fluid’s mass and volume.

Fluid Property Description. Any fluid property that depends on the amount of mass in a system is called an extensive property, N, because the mass “extends” throughout the system. For example, momentum is an extensive property since it represents mass times velocity, N = mV. Fluid properties that are independent of the system’s mass, such as temperature, are called intensive properties, h (eta). We can represent any extensive property N as an intensive property h simply by expressing it per unit mass, that is, h = N>m. Since momentum is N = mV, then h = V. Likewise, kinetic energy is N = (1>2)mV 2, and so h = 11>22V 2. Since mass is related to volume by m = rV, then in general, the relation between an extensive and intensive property for a system of fluid particles, expressed in terms of either the system’s mass or its volume, is N =

h dm = hr dV Lm LV

(4–7)

The integrations are necessary since the property may vary over the entire mass of the system or the volume it covers.

The Reynolds Transport Theorem. We are now ready to relate the time rate of change of any extensive property N for a fluid system (Lagrangian description) to its time rate of change as seen from the control volume (Eulerian description). To do this we will consider the system of particles in Fig. 4–9a that is passing through the fixed conduit. For a Lagrangian description, we must follow the entire system as it moves through the conduit, from time t to time t + ∆t, Fig. 4–9a and Fig. 4–9b. As it does so, the fluid’s extensive property N will change, and this rate of change is found from the time derivative. Nt + ∆t - Nt dN = lim ∆t S 0 dt ∆t

(4–8)

4.3

197

the reynolds transport theoreM

For an Eulerian description, we must first establish a control volume, here fixed to the conduit and outlined in red, and then find the change of N relative to the control volume, as the system passes through it during the time ∆t. At time t, Fig. 4–9a, the entire system of particles is within and coincident with the control volume (CV), Fig. 4–9a. At time t + ∆t, a portion of this system has exited the open control surface and is now in region Rout, outside the CV, Fig. 4–9b. This will leave a void Rin within the control volume. In other words, the system went from occupying the CV at time t to occupying 3CV + 1Rout - Rin 24 at time t + ∆t. To find the time rate of change in N measured relative to the control volume, we must therefore use the material derivative.

System

4 Control volume

Time t

[1Ncv 2 t + ∆t + 1∆Nout - ∆Nin 2] - 1Ncv 2 t DN = lim c d S ∆t 0 Dt ∆t

(a)

1Ncv 2 t + ∆t - 1Ncv 2 t ∆Nout ∆Nin = lim c d + lim c d - lim c d (4–9) S S ∆t S 0 ∆t 0 ∆t 0 ∆t ∆t ∆t

System Rout

The first term on the right represents the local derivative, since it is the change of N within the control volume with respect to time. Using Eq. 4–7, to express this result in terms of the corresponding intensive property h, we have lim c S

∆t

0

1Ncv 2 t + ∆t - 1Ncv 2 t 0Ncv 0 d = = hr dV ∆t 0t 0t L

Rin

(4–10) Time t + Dt

cv

The partial derivative is used here since Ncv is a function of both space (location) and time. The second term on the right side of Eq. 4–9 is a convective derivative of the extensive property as the system exits the control surface. Since ∆Nout > ∆t = h∆m>∆t, and ∆m = r∆Vout, then ∆Nout > ∆t = hr(∆Vout >∆t). In Fig. 4–9c, the rate at which a small volume of fluid particles will flow out of the control surface, having an area ∆Aout, is (∆V)out > ∆t = 3 1Vf>cs 2 out cos uout 4 ∆Aout, where Vf>cs is the velocity of the fluid measured relative to the control surface. Since ∆Aout = ∆Aout n, then using the dot product we can also write ∆Vout > ∆t = 1Vf>cs 2 out # ∆Aout. Therefore, our result becomes (∆N> ∆t)out = hr1Vf>cs 2 out # ∆Aout. For the entire exit control surface, ∆Nout lim a b = hr1Vf>cs 2 out # dAout ∆t S 0 ∆t L The same arguments apply for the last term in Eq. 4–9, so that lim a

∆t S 0

∆Nin b = hr1V f>cs 2 in # dAin ∆t L

Here the dot product will produce the required negative result in Eq. 4–9, since 1Vf>cs 2 in is inward and dAin = dAinn is outward. In other words, 1Vf>cs 2 in # dAin = 1Vf>cs 2 in dAin cos uin, where uin 7 90°, Fig. 4–9c.

(b) (Vf/cs)out uout

System

DAout

DAin (Vf/cs)in uin n (c)

Fig. 4–9

n

198

Chapter 4

C o n s e r va t i o n

of

Mass

If we combine these two terms, and express the sum as a “net” flow through the control volume, and use Eq. 4–10, then the Eulerian time rate of change of N in the control volume, Eq. 4–9, can be related to the Lagrangian time rate of change of N in the system, Eq. 4–8. The result is 4

dN 0 = hr dV + hrVf>cs # dA dt 0t Lcv Lcs Local change

(4–11)

Convective change

This equation is referred to as the Reynolds transport theorem, since it was first developed by the British scientist Osborne Reynolds. In summary, it relates the time rate of change of any extensive property N of a system of fluid particles, defined from a Lagrangian description, that is, by following the system, to the changes of N defined from an Eulerian description, that is, by finding the changes relative to the control volume. Doing it this way, the first term on the right side is the local change, since it represents the time rate of change of the intensive property h within the control volume. The second term on the right is the convective change of h, since it represents the net flow of the intensive property through the control surfaces.

Applications. When applying the Reynolds transport theorem, it is first necessary to specify the control volume that contains a selected portion of the fluid system. Once this is done, the local changes of the fluid property within the control volume can be determined, as well as the convective changes that occur through its open control surfaces. A few examples will illustrate how this is done.

• As shown in Fig. 4–10a, an incompressible fluid flows through the pipe transition at a steady rate. If we take the volume outlined in red as the fixed control volume, then there will be no local changes of the fluid within it because the flow is steady and the fluid mass within the control volume remains constant. Convective changes occur at each of the two open control surfaces because there is a flow of fluid mass across these surfaces.

(a)

Fig. 4–10

4.3

the reynolds transport theoreM

199

4 (b)

• Air is being pumped into the tank in Fig. 4–10b. If the control volume is taken as the entire volume within the tank, then local changes occur because the mass of the air in the tank is increasing with time. Also, convective changes occur at the open control surface or pipe connection as the fluid mass flows through it. • As air flows at a steady rate through the pipe in Fig. 4–10c, it is being heated. If the region outlined in red is considered as the control volume, then the heating will decrease the density of the air within the control volume. However, if the heating occurs at a constant rate, then no local changes will occur. We have nonuniform flow because the expanded air will increase the velocity as the air exits the pipe. Realize, however, that if the heating is increased (or decreased) then the air will further expand (or contract), causing unsteady flow. This will then produce a local change of the fluid properties within the control volume. • An incompressible liquid leaks out of the moving cart in Fig. 4–10d. The control volume that contains this liquid in the cart is both moving and deformable. Local changes occur because the mass in the control volume is decreasing with time. Convective change occurs at the open control surface (outlet). • The liquid on the moving blade in Fig. 4–10e is taken as a control volume. If we observe the motion from the blade, the flow will be steady, and so no local changes occur to the fluid mass within this control volume. Convective changes occur through the open control surfaces.

Further examples of how to select a control volume, and specify the local and convective changes that occur, will be presented throughout the book whenever we use the Reynolds transport theorem as it relates to the conservation of mass, the principle of work and energy, and the principle of impulse and momentum.

(c)

(d)

(e)

Fig. 4–10 (cont.)

200

Chapter 4

C o n s e r va t i o n

of

Mass

I MPO RTA N T PO I N T S • A Lagrangian description is used to describe the motion of 4

fluid particles within a system of particles, whereas an Eulerian description uses a fixed, moving, or deformable control volume to describe the motion of a system of particles that enter and exit its open control surfaces.

• Fluid properties that depend upon mass, such as energy and momentum, are called extensive properties, N. Those properties that are independent of mass, such as temperature and pressure, are called intensive properties, h. Any extensive property can be made intensive by dividing it by the fluid’s mass, h = N>m.

• The Reynolds transport theorem provides a means for relating the time rate of change of an extensive property N of a fluid as the system moves along, dN>dt, to the time rate of change of N as measured from a control volume, DN>Dt. This change consists of two parts: a local change that measures the change of the property within the control volume, and a convective change that measures the net amount of this property that enters and exits the open control surfaces. This net amount must be measured relative to the control surfaces if they are moving.

4.4

CONSERVATION OF MASS

The conservation of mass states that within a region, apart from any nuclear process, matter can neither be created nor destroyed. From a Lagrangian point of view, the mass of all the particles within a system of particles must be constant over time, and so we require the change in the mass to be 1dm>dt2 syst = 0. In order to develop a similar statement that relates to a control volume of particles, we must use the Reynolds transport theorem, Eq. 4–11. Here the extensive property N = m, and so the corresponding intensive property is mass per unit mass, or h = m>m = 1. Therefore, the conservation of mass requires 0 r dV + r Vf>cs # dA = 0 0t Lcv Lcs Local mass change

Convective net mass flow

(4–12)

4.4

This equation is often called the continuity equation. It states that the local rate of change of mass within the control volume, plus the net convective rate at which mass enters and exits the open control surfaces, must equal zero. For example, consider fluid passing into and out of the boiler in Fig. 4–11a. Local changes of the fluid mass occur within the boiler (control volume) if the temperature of the burner is increasing since it will further expand the fluid and produce unsteady flow. If the temperature remains constant then steady flow occurs and no local change is produced. Convective changes occur at the inlet and outlet. As another example, if the balloon in Fig. 4–11b is deflating, then the control volume of the air within the balloon is decreasing. This causes a local change in the mass. Also, a convective change occurs as mass flows out of the stem or open control surface.

A

B

4 (a)

Special Cases. Provided we have a fixed control volume that is completely filled with fluid, then for steady flow there will be no local change of the fluid mass within the control volume. In this case, the first term in Eq. 4–12 is zero, and so the net mass flow into and out of the open control surfaces must be zero. In other words, “what flows in must flow out.” Thus for steady flow, L cs

# # rVf>cs # dA = Σmout - Σmin = 0

201

Conservation of Mass

(b)

Fig. 4–11

(4–13)

Steady flow

If we consider Vf>cs to be the average velocity through each control surface, then Vf>cs will be constant or uniform over the area, and so integration yields # # ΣrVf>cs # A = Σmout - Σmin = 0

(4–14)

VC Fixed control volume VA

nC C AC

nA

Steady uniform flow AA

A

Finally, if the fluid is incompressible, then the density can be factored out, and we have

B nB

ΣVf>cs # A = ΣQout - ΣQin = 0

(4–15)

AB

VB

Incompressible steady uniform flow

A conceptual application of this equation is shown in Fig. 4–12. By our sign convention, notice that whenever a fluid exits a control surface, Vf>cs and Aout = Aout n will both be directed outward, and their dot product will be positive. If the fluid enters a control surface, Vf>cs is directed inward and Ain = Ain n is directed outward (always), and so their dot product will be negative.

SV · A 5 0 2VA AA 1VB AB 1 VC AC 5 0 Steady uniform flow of an incompressible fluid

Fig. 4–12

202

Chapter 4

C o n s e r va t i o n

of

Mass

I MPO RTA N T PO I N T S • The continuity equation is based on the conservation of mass, which states that the mass of all the particles within a system remains constant.

4

• If a control volume is fixed and completely filled with fluid,

The mass flow of air through this duct assembly must be calculated in order to balance the airflow out of each of the room vents.

then if the flow is steady, there will be no local change within the control volume as the fluid flows through its open control surfaces. The continuity equation then requires there be no net mass flow or convective change through its open control surfaces.

PRO CE DUR E FOR A N A LY S I S The following procedure can be used when applying the continuity equation. Fluid Description.

• Classify the flow as steady or unsteady, uniform or nonuniform. Also, determine if the fluid can be assumed inviscid and>or incompressible. Control Volume. • Establish the control volume and determine what type it should be. Fixed control volumes are generally used for objects that are at rest and have a fixed amount of fluid passing through them, like pipes. Moving control volumes are used for pump or turbine blades, and changing control volumes can be used for tanks that contain a variable amount of fluid. Care should be taken to orient the open control surfaces so that their planar areas are perpendicular to the flow. Also, locate these surfaces in a region where the flow is uniform and well established. Continuity Equation. • Consideration must be given both to the rate of change of mass within the control volume and to the rate at which mass enters and exits each open control surface. For application, we will always write the fundamental equation, Eq. 4–12, and then show how this equation reduces to a specific case. For example, if the control volume does not deform and is fixed and completely filled with fluid, then for steady flow, the local change of the mass within the control volume will be zero, so “what flows in must flow out,” as indicated by Eqs. 4–13 through 4–15. Remember that planar areas of open control surfaces are defined by unit vectors n that are always directed outward, normal to the control surface, A = An. Thus, flow into a control surface will be negative since Vf>cs and A will be in opposite directions, whereas flow out of a control surface is positive since both vectors are outward. If a control surface is moving, then for steady flow, the velocity of flow into or out of a control surface must be measured relative to the moving surface, Vf>cs.

4.4

EXAMPLE

203

Conservation of Mass

4.2

Water flows into the 250-mm-diameter barrel of the fire hydrant at QC = 0.15 m3 >s, Fig. 4–13a. If the velocity out the 75-mm-diameter nozzle at A is 12 m>s, determine the discharge out the 100-mm-diameter nozzle at B.

4

SOLUTION Fluid Description. This is a condition of incompressible uniform steady flow, where the water is considered an ideal fluid. Therefore, average velocities will be used. Control Volume. We will take the control volume to be fixed and have it enclose the volume of water within the fire hydrant and extended portions of hose as shown in the figure. There is no local change within the control volume because the flow is steady. Convective changes occur through the three open control surfaces. Since the flow at C is known, the average velocity there is VC =

0.15 m3 >s QC 9.6 = = m>s 2 p AC p (0.125 m)

Continuity Equation.

0 - a

12 ms

Control volume

A

nA 75 mm

250 mm

For incompressible steady flow,

C

0 r dV + r Vf>cs # dA = 0 0t Lcv Lcs

(1)

0 - VC AC + VAAA + VB AB = 0

(2)

9.6 m>s b 3 p 1 0.125 m 2 2 4 + (12 m>s) 3 p 1 0.0375 m 2 2 4 + p

nC VC (a)

VB 3 p 1 0.05 m 2 2 4 = 0 VB = 12.35 m>s

The discharge at B is therefore QB = VB AB =

1 12.35 m>s 2 3 p 1 0.05 m 2 2 4

= 0.0970 m3 >s

Ans.

Notice that since QC = VC AC and QB = VB AB, it really is not necessary to calculate VC and VB as an intermediate step. But be careful to recognize that if QC is used in Eq. 2, then it is a negative term, i.e., Eq. 2 or application of Eq. 4–15 becomes 0 - QC + VAAA + QB = 0. NOTE: If the viscosity of the water was considered, and a specific velocity

profile, like the one shown in Fig. 4–13b, was known at C, then the solution would require integration to get QC (see Example 4–1).

(b)

Fig. 4–13

B

VB

100 mm

nB

204

Chapter 4

EXAMPLE

C o n s e r va t i o n

of

Mass

4.3 Air flows into the gas heater in Fig. 4–14 at a steady rate, such that at A its absolute pressure is 203 kPa, its temperature is 20°C, and its velocity is 15 m>s. When it exits at B, it has an absolute pressure of 150 kPa and a temperature of 75°C. Determine its velocity at B.

4

SOLUTION

150 mm VB B nB 100 mm nA 15 ms A

Fig. 4–14

Fluid Description. As stated, we have steady flow. We will neglect viscosity and use average velocities through the pipe. Due to the pressure and temperature changes within the heater, the air changes its density, and so we must include the effects of compressibility. Control Volume. As shown, the control volume is fixed and consists of the portion of pipe within the heater, along with short extensions from it. Here there are no local changes of the mass within the control volume because the flow remains steady. Convection occurs through the open control surfaces at A and B. Continuity Equation. The pressure and the temperature affect the air density at the open control surfaces. The flow of air in at A is negative. We require the net convective change to be zero. 0 r dV + r Vf>cs # dA = 0 0t Lcv Lcs 0 - rAVAAA + rBVBAB = 0

-rA 115 m>s23p(0.05 m)2 4 + rBVB 3p(0.075 m)2 4 = 0 VB = 6.667a

rA b rB

(1)

Ideal Gas Law. The ratio of the densities at A and B is obtained from the ideal gas law. We have pA = rARTA; 2031103 2 N>m2 = rAR(20 + 273) K pB = rBRTB; 1501103 2 N>m2 = rBR(75 + 273) K

The value of R for air is tabulated in Appendix A, although here it can be eliminated by division. rA 1.607 = rB Substituting this into Eq. 1 yields VB = 6.667(1.607) = 10.7 m>s

Ans.

4.4

EXAMPLE

205

Conservation of Mass

4.4

The tank in Fig. 4–15 has a volume of 1.5 m3 and is being filled with air, which is pumped into it at a rate of 8 m>s through a hose having a diameter of 10 mm. As the air enters the tank, its temperature is 30°C and its absolute pressure is 500 kPa. Determine the rate at which the density of the air within the tank is changing at this instant.

4

SOLUTION Fluid Description. We will assume that due to mixing the air has a uniform density within the tank. This density is changing because the air is compressible. The flow into the tank is steady. Control Volume. We will consider the fixed control volume to be the air contained within the tank. Local changes occur within this control volume because the mass of air within the tank is changing with time. This change is directly proportional to the density change because V is constant. The velocity of the air stream will be considered uniform at the open control surface at A. Here there is a convective change since there is a net mass flow into the control volume. Continuity Equation. Applying the continuity equation, realizing that the control volume (tank) has a constant volume, we have 0 r dV + r Vf>cs # dA = 0 0t Lcv Lcs 0ra V - rAVA AA = 0 0t

A

(1)

Ideal Gas Law. The density of air flowing into the tank is determined using the ideal gas law. From Appendix A, R = 286.9 J>(kg # K), and so pA = rARTA;

5001103 2 N>m2 = rA 3 286.9 J>(kg # K) 4 (30 + 273) K rA = 5.752 kg>m3

Therefore,

0ra 11.5 m3 2 - 3 15.752 kg>m3 218 m>s2 4 3 p(0.005 m)2 4 = 0 0t 0ra = 2.41110 - 3 2 kg> 1m3 # s2 Ans. 0t

This positive result indicates that the density of the air within the tank is increasing, which is to be expected.

Fig. 4–15

nA

206

Chapter 4

EXAMPLE

C o n s e r va t i o n

nB

Mass

4.5

4 nC

of

80 ms

B

nA C

The rocket sled in Fig. 4–16 is propelled by a jet engine that burns fuel that enters the engine at C at a rate of 60 kg>s. The air duct at A has an opening of 0.2 m2 and takes in air having a density of 1.20 kg>m3. If the engine exhausts the gas relative to the nozzle at B with a velocity of 300 m>s, determine the density of the exhaust. The sled is moving forward at a constant rate of 80 m>s, and the nozzle has a crosssectional area of 0.35 m2.

A

SOLUTION Fig. 4–16

Fuel Description. The air–fuel system is compressible, and so its density will be different at the inlet A and exhaust B. We will use average velocities. Control Volume. The control volume is represented by the enclosed region within the engine that accepts the air and fuel, burns it, and exhausts it. We will assume it moves with the rocket. From this viewpoint (as a passenger), the flow is steady, so there is no local time rate of change of the air–fuel mass within the control volume. No net convective change occurs since the mass flow of air and fuel into the control volume is equal to that which passes out of it. Assuming the outside air is stationary, then the relative velocity of airflow at the intake A is + VA = Vcs + VA>cs S 0 = 80 m>s + VA>cs VA>cs = -80 m>s = 80 m>s d At B, VB>cs = 300 m>s because the velocity of the exhaust gas is measured relative to the moving control surface. # Continuity Equation. The mass flow of fuel, mf = 60 kg>s, will be negative because it enters into an open control surface, opposite to nC in Fig. 4–16. Since there is no local change, we have 0 r dV + rVf>cs # dA = 0 0t Lcv Lcs # 0 - raVA>cs AA + rgVB>cs AB - mf = 0

-1.20 kg>m3 180 m>s210.2 m2 2 + rg 1300 m>s210.35 m2 2 - 60 kg>s = 0 rg = 0.754 kg>m3

Ans.

Notice that if the selected control volume were in a fixed location in space, and the rocket sled were passing this location, then local changes would occur within this fixed control volume as the rocket sled passed by. In other words, the flow through the control volume would appear as unsteady flow.

4.4

EXAMPLE

207

Conservation of Mass

4.6

The 1-m-diameter tank in Fig. 4–17a is being filled with water using a 0.5-m-diameter pipe, which has a discharge of 0.15 m3 >s. Determine the rate at which the level of water is rising in the tank. Neglect the effect of gravity on the falling water.

4

nA A 0.5 m

SOLUTION Fluid Description. r is constant.

We assume the water to be an ideal fluid so that 3m

Control Volume I. We will choose a deformable control volume, which consists only of the volume of water that enters the tank, Fig. 4–17a. Although we have steady flow into this control volume, local changes occur because the control volume is changing. In other words, the amount of mass within the control volume changes with time. Convective changes occur since there is a net mass flow into the control volume through the open control surface at A. To calculate the volume of water within the control volume, we have assumed that as the water falls it maintains a constant diameter of 0.3 m as shown.*

y

1m (a)

Fig. 4–17

Conservation of Mass. Applying the continuity equation, realizing that QA = VAAA, we have 0 r dV + r Vf>cs # dA = 0 0t Lcv Lcs Since the density is constant, and the volume is only a function of time, then we can use a total derivative, i.e., r

dV - r QA = 0 dt

Here V is the total volume of water within the control volume at the instant the depth is y. Factoring out r, we have

d 3 p(0.5 m)2y + p(0.25 m)2(3 m - y) 4 - 1 0.15 m3 >s 2 = 0 dt d p (0.1875y + 0.1875) = 0.15 dt dy 0.15 0.1875 + 0 = p dt dy 0.8 = m>s = 0.255 m>s Ans. p dt

*Even if this water column disperses, the same mass flow will be maintained throughout the column.

208

Chapter 4

C o n s e r va t i o n

of

Mass

VA nA A

4 0.5 m

VB

3m2y

nB 3m B y

1m (b)

Fig. 4–17 (cont.)

Control Volume II. We can also work this problem by considering a fixed control volume consisting only of the water within the tank at the instant when its depth is y, Fig. 4–17b. In this case no local changes occur because the water within this control volume is considered incompressible. Also, no net convective change occurs, because the mass flow of water into the control surface of area p(0.25 m)2 at A is equal to the mass flow of water out of the control surface at B, which has an area of 3p(0.5 m)2 - p(0.25 m)2 4. Conservation of Mass. Since QA = VAAA, we have 0 r dV + rwV # dA = 0 0t Lcv w Lf/cs 0 - VAAA + VBAB = 0

- 1 0.15 m3 >s 2 + VB 3 1 p(0.5 m)2 - p(0.25 m)2 2 4 = 0 VB =

dy 0.8 = m>s = 0.255 m>s p dt

Ans.

References 1. ASME, Flow Meters, 6th ed., ASME, New York, NY, 1971. 2. S. Vogel, Comparative Biomechanics, Princeton University Press,

Princeton, NJ, 2003. 3. S. Glasstone and A. Sesonske, Nuclear Reactor Engineering, D. van

Nostrand, Princeton, NJ, 2001.

fundaMental probleMs

209

F UN DAMEN TAL PR O B L EM S SEC. 4.1–4.2 F4–1. Water flows into the tank through a rectangular tube having a width of 60 mm. If the average velocity of the flow is 16 m>s, determine the mass flow. Take rw = 1000 kg>m3.

F4–3. Water has an average velocity of 8 m>s through the pipe. Determine the volumetric flow and mass flow. 4

8 ms

16 ms

300 mm

50 mm 608

Prob. F4–3

F4–4. Crude oil flows through the pipe at 0.02 m3 >s. If the velocity profile is assumed to be as shown, determine the maximum velocity umax of the oil and the average velocity.

0.2 m

r u max

Prob. F4–1 u 5 umax (1 2 25r 2) ms

Prob. F4–4

F4–2. Air flows through the triangular duct at 0.7 kg>s when the temperature is 15°C and the gage pressure is 70 kPa. Determine its velocity. Take patm = 101 kPa.

F4–5. Determine the mass flow of air having a temperature of 20°C and pressure of 80  kPa as it flows through the circular duct with a velocity of 3 m>s.

400 mm 3 ms 0.3 m

0.3 m

Prob. F4–2

Prob. F4–5

210

Chapter 4

C o n s e r va t i o n

of

Mass

F4–6. If the velocity profile for a very viscous liquid as it flows through a 0.5-m-wide rectangular channel is approximated as u = 16y2 2 m>s, where y is in meters, determine the volumetric flow.

F4–8. A liquid flows into the tank at A with a velocity of 4 m>s. Determine the rate, dy>dt, at which the level of the liquid is rising in the tank. The cross-sectional area of the pipe at A is AA = 0.1 m2.

4

y

3m

0.5 m

y

2m

u 5 (6y2) ms

Prob. F4–6 A 4 m/s

Prob. F4–8

SEC. 4.4 F4–7. The velocity of the steady flow at A and B is indicated. Determine the velocity at C. The pipes have cross-sectional areas of AA = AC = 0.1 m2 and AB = 0.2 m2.

F4–9. As air exits the tank at 0.05 kg>s, it is mixed with water at 0.002 kg>s. Determine the velocity of the mixture as it exits the 20-mm-diameter pipe if the density of the mixture is 1.45 kg>m3.

6 ms

A

C

B

VC

Prob. F4–7

2 ms

Prob. F4–9

211

probleMs

P ROBLEMS SEC. 4.1–4.2 4–1. Determine the average velocity V of the fluid if it has the velocity profile shown. The channel is 2 m wide.

4–5. Water flows along the semicircular trough with an average velocity of 3.6 m>s. Determine the volumetric 4 discharge.

4–2. Determine the mass flow of the fluid if it has the velocity profile shown. The channel is 2 m wide. Take r = 1600 kg>m3.

0.1 m 0.4 m 3.6 ms 308

y 0.9 ms

0.6 m

u

Prob. 4–5

y

Probs. 4–1/2

4–3. The velocity profile of a liquid flowing through the pipe is approximated by the truncated conical distribution as shown. Determine the average velocity of the flow. Hint: The volume of a cone is V = 31 pr 2h.

4–6. Water flows along the triangular channel having the cross section shown. If the average velocity is 3 m>s, determine the volumetric discharge.

0.4 m 0.1 m 0.4 m

3 ms 308

3 m/s

100 mm 200 mm 100 mm

Prob. 4–6 Prob. 4–3

*4–4. The velocity profile of a liquid flowing through the pipe is approximated by the truncated conical distribution as shown. Determine the mass flow if the liquid has a density of r = 880 kg>m3. Hint: The volume of a cone is V = 13 pr 2h.

3 m/s

100 mm 200 mm 100 mm

Prob. 4–4

4–7. A fluid flowing between two plates has a velocity profile that is assumed to be parabolic, where 4umax u = (hy - y2). Determine the average velocity and h2 volumetric discharge in terms of umax. The plates have a width of w. y

h –– 2

umax

h –– 2

Prob. 4–7

u

212

Chapter 4

C o n s e r va t i o n

of

Mass

*4–8. The liquid in the rectangular channel has a velocity profile defined by u = 13y1>2 2 m>s, where y is in meters. Determine the volumetric discharge if the width of the channel is 2 m.

*4–12. Air enters the turbine of a jet engine at a rate of 40 kg>s. If it is discharged with an absolute pressure of 750 kPa and temperature of 120°C, determine its velocity at the exit. The exit has a diameter of 0.3 m.

4–9. The liquid in the rectangular channel has a velocity 1>2 4 profile defined by u = (3y ) m>s, where y is in meters. Determine the average velocity of the liquid. The channel has a width of 2 m.

0.3 m

y

1

Prob. 4–12

u 5 (3y 2 ) ms

0.9 m

u

Probs. 4–8/9

4–10. Determine the volumetric flow through the 50-mm-diameter nozzle of the fire boat if the water stream reaches point B, which is R = 24 m from the boat. Assume the boat does not move.

4–13. The 30-mm-diameter nozzle ejects water such that the stream strikes the wall at B, where h = 4 m. Determine the volumetric flow through the nozzle.

B

A

h

308 1.2 m 12 m

308 B

A

Prob. 4–13

3m

4–14. Determine the mass flow of air in the duct if it has a velocity of 16 m>s. The air has a temperature of 15°C, and the (gage) pressure is 40 kPa. The atmospheric pressure is 101.3 kPa.

R

Prob. 4–10 4–11. Determine the volumetric flow through the 50-mm-diameter nozzle of the fire boat as a function of the distance R of the water stream. Plot this function of flow (vertical axis) versus the distance for 0 … R … 25 m. Give values for increments of ∆R = 5 m. Assume the boat does not move.

4–15. Air flows through the duct at a velocity of 16 m>s. If the temperature is maintained at 15°C, plot the variation of the mass flow (vertical axis) versus the (gage) pressure for the range of 0 … p … 50 kPa. Give values for increments of ∆p = 10 kPa. The atmospheric pressure is 101.3 kPa.

500 mm

300 mm

308 B

A

3m

R

Prob. 4–11

Probs. 4–14/15

213

probleMs *4–16. A fluid flowing between two plates has a velocity profile that is assumed to be linear as shown. Determine the average velocity and volumetric discharge in terms of umax. The plates have a width of w.

4–19. The radius of the circular duct varies as r = [0.06(1 - 1x)4 m, where x is in meters. The flow of water at A is Q = 0.003 m3 >s at t = 0, and it is increasing at dQ>dt = 0.003 m3 >s2. If a fluid particle is originally located at x = 0 when t = 0, determine the time for this particle to arrive at x = 0.3 m. Use average velocities at the cross 4 section for the solution.

b –– 2

r umax

b –– 2

0.4 m

60 mm

x

Prob. 4–16 B x

A

4–17. Determine the mass flow of air in a 300-mmdiameter duct if it has a velocity of 8 m>s. The air has a temperature of 24°C, and the gage pressure is 60 kPa. The atmospheric pressure is 101.3 kPa.

300 mm

Prob. 4–19 *4–20. The radius of the circular duct varies as r = 30.06(1 - 1x)4 m, where x is in meters. If the flow of water at A is Q = 0.003 m3 >s at t = 0, and it is increasing at dQ>dt = 0.003 m3 >s2, determine the velocity and acceleration of a water particle at x = 0.3 m when t = 0.4 s. Use average velocities at the cross section for the solution. r 0.4 m

60 mm

x

Prob. 4–17 B x

A

4–18. Oxygen gas flows through the 400-mm-diameter duct. If it has a velocity of 12 m>s and the gage pressure is maintained at 120 kPa, plot the variation of mass flow (vertical axis) versus temperature for the temperature range 0°C … T … 50°C. Give values for increments of ∆T = 10°C. The atmospheric pressure is 101.3 kPa.

Prob. 4–20 4–21. At two specific instants during a heartbeat, the velocity profiles of blood passing through a straight portion of the aorta can be modeled by a cosine curve as shown. Determine the volumetric flow through the aorta for each instant. r

r

400 mm R

u 5 a cos(

Prob. 4–18

p r) 2R

u5

Prob. 4–21

7p a [2 cos( r) 1 3 ] 2 6R

214

4

Chapter 4

C o n s e r va t i o n

of

Mass

4–22. The human heart has an average discharge of 0.1110-3 2 m3 >s, determined from the volume of blood pumped per beat and the rate of beating. Careful measurements have shown that blood cells pass through the capillaries at about 0.5 mm>s. If the average diameter of a capillary is 6 μm, estimate the number of capillaries that must be in the human body.

4–25. Kerosene flows through the nozzle at 0.25 m3 >s. Determine the velocity and acceleration of a particle on the x axis at x = 0.25 m. Use average velocities at the cross section for the solution.

300 mm 25 mm 100 mm x

x

Prob. 4–25

Prob. 4–22

4–23. Kerosene flows through the nozzle at 0.25 m3 >s. Determine the time it takes for a particle on the x axis to travel from x = 0 to x = 0.2 m. Plot the distance-versustime graph for the particle, if x = 0 at t = 0. Use average velocities at the cross section for the solution.

4–26. Rain falls vertically with an average speed of 6 m>s and accumulates in the bin. It is observed that the level of water in the bin rises at the rate of 2.5 mm>min. Determine the amount of falling rainwater in a cubic meter of air. Also, if the average diameter of a drop of rain is 2.65 mm, determine the number of raindrops in a cubic meter of air. Hint: The volume of a drop is V = 34pr 3.

300 mm 25 mm 100 mm x

x

Prob. 4–23 *4–24. Air flows through the gap between the vanes at 0.75 m3 >s. Determine the velocity of the air passing through the inlet A and the outlet B. The vanes have a width of 400 mm and the vertical distance between them is 200 mm.

1m

208 408

200 mm VB

VA

408

B

208

0.5 m

A

Prob. 4–24

Prob. 4–26

215

probleMs

SEC. 4.3 4–27. Air flows through the tapered duct, and during this time heat is being added at an increasing rate, changing the density of the air within the duct. Select a control volume that contains the air in the duct, and indicate the open control surfaces. Show the positive direction of their areas, and also indicate the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume the air is incompressible.

4–30. Air flows through the transition, and its temperature decreases as it flows from inlet B to outlet A. The average velocities are indicated. Select a control volume that contains the air in the duct. Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume the air is 4 incompressible.

2 ms 7 ms

1.5 ms

9 ms B A

Prob. 4–27 Prob. 4–30 *4–28. The average velocities of water flowing steadily through the nozzle are indicated. Outline the control volume to be the entire nozzle and the water inside it. Also, select another control volume to be just the water inside the nozzle. In each case, indicate the open control surfaces, and show the positive direction of their areas. Specify the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible. 2.5 ms A

B

8 ms

Prob. 4–28 4–29. The average velocities of water flowing steadily through the nozzle are indicated. If the nozzle is glued onto the end of the hose, outline the control volume to be the entire nozzle and the water within it. Also, select another control volume to be just the water within the nozzle. In each case, indicate the open control surfaces, and show the positive direction of their areas. Specify the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible. 3 ms

4–31. The flat plate is moving to the right at 2 m>s. Water is ejected from the nozzle at A at an average velocity of 4 m>s. Outline a moving control volume that contains the water on the plate. Indicate the open control surfaces, and show the positive direction of their areas through which flow occurs. Also, indicate the directions of the relative velocities through the control surfaces and determine the magnitude of the relative velocity through the inlet control surface. Identify the local and convective changes that occur. Assume water to be incompressible. For the analysis, why is it best to consider the control volume as moving?

4 ms 2 ms A

A 12 ms

B

Prob. 4–29

Prob. 4–31

216

4

Chapter 4

C o n s e r va t i o n

of

Mass

*4–32. The water in the tank is being drained at A with the average velocity shown. Select a control volume that includes only the water in the tank. Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible.

A

4–35. The hemispherical bowl is suspended in the air by the water stream that flows into and then out of the bowl at the average velocities indicated. Outline a control volume that contains the bowl and the water entering and leaving it. Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible.

4 ms

Prob. 4–32 4–33. Water flows steadily through the pipes with the average velocities shown. Outline the control volume that contains the water in the pipe system. Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible.

3 ms 3 ms

1.5 ms

Prob. 4–35 B 2.25 ms C A

0.75 ms

Prob. 4–33 4–34. Air is pumped into the tank, and at the instant shown it has an average velocity of 9 m>s. Select a control volume that contains the air in the tank. Indicate the open control surface, and show the positive direction of its area. Also, indicate the direction of the velocity through this surface. Identify the local and convective changes that occur.Assume the air to be compressible.

9 ms

A

*4–36. The jet engine is moving forward with a constant speed of 800 km>h. Fuel from a tank enters the engine and is mixed with the intake air, burned, and exhausted with an average relative velocity of 1200 km>h. Outline a moving control volume as the jet engine and the air and fuel within it. Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the magnitudes of the relative velocities and their directions through these surfaces. Identify the local and convective changes that occur. Assume the fuel is incompressible and the air is compressible. For an analysis, why is it best to consider this control volume to be moving?

800 kmh

Prob. 4–34

Prob. 4–36

217

probleMs 4–37. The blade on the turbine is moving to the left 6 m>s. Water is ejected from the nozzle at A at an average velocity of 2 m>s. Outline a moving control volume that contains the water on the blade. Indicate the open control surfaces, and show the positive direction of their areas through which flow occurs. Also, indicate the directions of the relative velocities through the control surfaces and determine the magnitude of the relative velocity through the inlet control surface. Identify the local and convective changes that occur. Assume water to be incompressible. For the analysis, why is it best to consider the control volume as moving?

SEC. 4.4 4–39. A liquid flows through the drainpipe such that it has a parabolic velocity profile of u = 10(1 - 400r 2) m>s, where r is in meters. Determine the time needed to drain the cylindrical tank from a depth of h = 2 m at t = 0 to a depth of 1 m. The tank has a diameter of 1.5 m. 4

1.5 m

h

2 ms

6 ms

A r

Prob. 4–37 100 mm

4–38. Compressed air is being released from the tank, and at the instant shown it has an average velocity of 3 m>s. Select a control volume that contains the air in the tank. Indicate the open control surface, and shown the positive direction of its area. Also, indicate the direction of the velocity through this surface. Identify the local and convective changes that occur. Assume the air to be compressible.

Prob. 4–39

*4–40. Water flows through the nozzle at 0.02 m3 >s. Determine the velocity V of a particle as it moves along the centerline as a function of x.

4–41. Water flows through the nozzle at 0.02 m3 >s. Determine the acceleration of a particle as it moves along the centerline as a function of x. 3 ms

50 mm 20 mm

x 80 mm

Prob. 4–38

Probs. 4–40/41

218

Chapter 4

C o n s e r va t i o n

of

Mass

4–42. The unsteady flow of glycerin is such that at A it has a velocity of VA = (5t 1>2) m>s, where t is in seconds. Determine the acceleration of a fluid particle located at x = 0.3 m when t = 1 s. Hint: Determine V = V(x, t), then find the material derivative. 4 4–43. The unsteady flow of glycerin is such that at A it has a velocity of VA = (0.8t + 2) m>s, where t is in seconds. Determine the acceleration of a fluid particle located at x = 0.3 m when t = 0.5 s. Hint: Determine V = V(x, t), then find the material derivative.

4–45. The unsteady flow of glycerin through the reducer is such that at A its velocity is VA = (0.8 t 2) m>s, where t is in seconds. Determine its velocity at B, and its average acceleration at A, when t = 2 s. The pipes have the diameters shown.

0.1 m 0.3 m B A

Prob. 4–45

0.4 m

0.1 m

0.2 m B A x

Probs. 4–42/43

*4–44. Oil flows into the pipe at A with an average velocity of 0.2 m>s and through B with an average velocity of 0.15 m>s. Determine the maximum velocity Vmax of the oil as it emerges from C if the velocity distribution is parabolic, defined by vC = Vmax(1 - 100r 2), where r is in meters measured from the centerline of the pipe.

4–46. The cylindrical plunger traveling at Vp = 10.004t 1>2 2 m>s, where t is in seconds, injects a liquid plastic into the mold to make a solid ball. If d = 50 mm, determine the amount of time needed to do this. The volume of the ball is V = 43pr 3. 4–47. The cylindrical plunger traveling at Vp = 10.004 t 1>2 2 m>s, where t is in seconds, injects a liquid plastic into the mold to make a solid ball. Determine the time needed to fill the mold as a function of the plunger diameter d. Plot the time needed to fill the mold (vertical axis) versus the diameter of the plunger for 10 mm … d … 50 mm. Give values for increments of ∆d = 10 mm. What distance must the plunger travel when d = 10 mm? Is this realistic? The volume of the ball is V = 34 pr 3. Vp

d y Vmax 10 mm

200 mm C 300 mm

200 mm

A

75 mm

B 0.15 ms

0.2 ms

Prob. 4–44

Probs. 4–46/47

219

probleMs *4–48. The jet engine draws in air at 1800 kg>s and jet fuel at 0.916 kg>s. If the density of the air–fuel mixture at the exhaust is 3.36 kg>m3, determine the velocity of the exhaust relative to the plane. The exhaust nozzle has a diameter of 0.8 m.

4–51. Nitrogen flows into the tank at A at VA = 3 m>s, and helium flows in at B at VB = 8 m>s. Both enter at a gage pressure of 260 kPa and a temperature of 150°C. Determine the steady mass flow of the mixed gas at C. The atmospheric pressure is patm = 101.3 kPa. 4

VC 0.4 m C

Prob. 4–48 4–49. Water flows through the pipe at A at 60 kg>s, and then out of B with a velocity of 4 m>s. Determine the average velocity at which it flows in through C.

0.3 m

B

A

0.2 m

VA

VB

VA A 100 mm B

Prob. 4–51 VB 60 mm

*4–52. Nitrogen flows into the tank at A at VA = 3 m>s, and helium flows in at B at VB = 8 m>s. Both enter at a gage pressure of 260 kPa and a temperature of 150°C. Determine the velocity of the mixed gas leaving the tank at a steady rate at C. The mixture has a density of r = 2.520 kg>m3. The atmospheric pressure is patm = 101.3 kPa.

160 mm C

VC

Prob. 4–49 VC

4–50. If water flows in the wye through pipes at A and C with a mass flow of 60 kg>s and 20 kg>s, respectively, determine the velocity of the flow out of the wye through the pipe at B.

0.4 m C

VA A 100 mm 0.3 m

B

B

A VA

VB

VB 60 mm

160 mm C

VC

Prob. 4–50

0.2 m

Prob. 4–52

220

4

Chapter 4

C o n s e r va t i o n

of

Mass

4–53. The flat strip is sprayed with paint using the six nozzles, each having a diameter of 2 mm. They are attached to the 20-mm-diameter pipe. The strip is 50 mm wide, and the paint is to be 1 mm thick. If the speed of the paint through the pipe is 1.5 m>s, determine the required speed V of the strip as it passes under the nozzles.

1.5 ms

4–55. Drilling fluid is pumped down through the center pipe of a well and then rises up within the annulus. Determine the diameter d of the inner pipe so that the speed of the fluid remains the same in both regions. Also, what is this average velocity if the discharge is 0.02 m3 >s? Neglect the thickness of the pipes. *4–56. Drilling fluid is pumped down through the center pipe of a well and then rises up within the annulus. Determine the velocity of the fluid forced out of the well as a function of the diameter d of the inner pipe, if the velocity of the fluid forced into the well is maintained at Vin = 2 m>s. Neglect the thickness of the pipes. Plot this velocity (vertical axis) versus the diameter for 50 mm … d … 150 mm. Give values for increments of ∆d = 25 mm.

V

Vout

2.5 m

Vin

d

200 mm

Prob. 4–53

4–54. The flat strip is sprayed with paint using the six nozzles, which are attached to the pipe. The strip is 50 mm wide and the paint is to be 1 mm thick. If the speed of the paint through the pipe is 1.5 m>s, determine the required speed V of the strip as it passes under the nozzles as a function of the diameter of the pipe. Plot this function of speed (vertical axis) versus diameter for 10 mm … D … 30 mm. Give values for increments of ∆D = 5 mm.

Probs. 4–55/56 4–57. Gasoline flows into the tank through two pipes. At A the velocity is 4 m>s. Determine the rate at which the level of gasoline is rising in the tank as a function of the flow through the inlet pipe B. Plot this rate (vertical axis) versus the discharge for 0 … QB … 0.1 m3 >s. Give values for increments of ∆QB = 0.02 m3 >s.

1.5 m

1.5 ms

V

120 mm B

200 mm A

2.5 m

Prob. 4–54

Prob. 4–57

221

probleMs 4–58. Gasoline flows into the cylindrical tank through two pipes. At A the velocity is 4 m>s and at B it is 6 m>s. Determine the rate at which the level of gasoline is rising in the tank.

*4–60. Water enters the conical tank from the bottom with a velocity of 1.5 m>s. Determine the rate at which the water level at the surface is rising in terms of h.

4

1.5 m

3m

h 120 mm B

200 mm A

100 mm 1.2 m

Prob. 4–58 Prob. 4–60

4–59. Water flows into the tank through two pipes. At A the flow is 7500 liters>min, and at B it is 5500 liters>min. Determine the rate at which the level of water is rising in the tank.

4–61. The cylindrical syringe is actuated by applying a force on the plunger. If this causes the plunger to move forward at 20 mm>s, determine the velocity of the fluid passing out of the needle. 1.5 m

20 mms 150 mm A

0.5 mm

10 mm

200 mm B

Prob. 4–59

Prob. 4–61

222

4

Chapter 4

C o n s e r va t i o n

of

Mass

4–62. Oil flows into a mixing tank through pipe A with an average velocity of 6 m>s, and glycerin flows into the tank through pipe B at 3 m>s. Determine the average density at which the mixture flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m3 tank. Take ro = 880 kg>m3 and rgly = 1260 kg>m3.

*4–64. The pressure vessel of a nuclear reactor is filled with boiling water having a density of rw = 850 kg>m3. Its volume is 185 m3. Due to failure of a pump needed for cooling, the pressure release valve A is opened and emits steam having a density of rs = 35 kg>m3 and velocity of V = 400 m>s. If it passes through the 40-mm-diameter pipe, determine the time needed for all the water to escape.Assume that the temperature of the water and the velocity at A remain constant.

80 mm

100 mm A

A

B

V

C 120 mm

Prob. 4–64 Prob. 4–62 4–63. Oil flows into the mixing tank through pipe A at 0.05 m3 >s, and glycerin flows into the tank through pipe B at 0.015 m3 >s. Determine the average density of the two fluids as the mixture flows out from the tank through the exit pipe at C. Assume uniform mixing of the fluids occurs within the 4@m3 tank. Take ro = 880 kg>m3 and rgly = 1260 kg>m3.

80 mm

100 mm A

4–65. The pressure vessel of a nuclear reactor is filled with boiling water having a density of rw = 850 kg>m3. Its volume is 185 m3. Due to failure of a pump, needed for cooling, the pressure release valve is opened and emits steam having a density of rs = 35 kg>m3. If the steam passes through the 40-mm-diameter pipe, determine its speed through the pipe as a function of the time needed for all the water to escape. Plot the speed (vertical axis) versus the time for 0 … t … 3 h. Give values for increments of ∆t = 0.5 h. Assume that the temperature of the water remains constant.

B A

C 120 mm

Prob. 4–63

Prob. 4–65

V

223

probleMs 4–66. The tank contains air at a temperature of 20°C and absolute pressure of 500 kPa. Using a valve, the air escapes with an average speed of 120 m>s through a 15@mm-diameter nozzle. If the volume of the tank is 1.25 m3, determine the rate of change in the density of the air within the tank at this instant. Is the flow steady or unsteady?

*4–68. Water flows into the rectangular tank through pipes A and B, with velocities of 2.5 m>s and 1.5 m>s, respectively. It exits at C with a velocity of 0.75 m>s. Determine the rate at which the surface of the water is rising or dropping. The base of the tank is 2 m by 1.5 m. Ignore the effect of gravity on the falling water. 4–69. Water flows into the 1.5-m-diameter cylindrical tank 4 through pipes A and B with velocities of 2.5 m>s and 1.5 m>s, respectively. It exits at C with a velocity of 0.75 m>s. Determine the time required to fill the tank if y = 0 when t = 0. Ignore the effect of gravity on the falling water. 1.5 m/s

100 mm B

2.5 m/s

A

200 mm

2m

300 mm 0.75 m/s

C

Prob. 4–66

y

1.5 m

4–67. With every breath, air enters the trachea, its flow split equally into two main bronchi, and then passes through about 150 000 bronchial tubes before entering the alveoli. If the air flow into the 18-mm-diameter trachea is 12 liter>min., determine the velocity of the air in the trachea and the main bronchi, which have a diameter of 12 mm. Note: The diameter of the alveoli is about 250 μm, and because there are so many of them, the flow is reduced to practically zero, so gaseous exchange is carried out by diffusion.

Probs. 4–68/69 4–70. The cylinder is pushed down into the tube at a rate of V = 5 m>s. Determine the velocity of the liquid as it rises in the tube. 4–71. Determine the speed V at which the cylinder must be pushed down into the tube so that the liquid in the tube rises with an average velocity of 4 m>s.

V

150 mm

y

200 mm

Prob. 4–67

Probs. 4–70/71

224

Chapter 4

C o n s e r va t i o n

of

Mass

*4–72. The natural gas (methane) and crude oil mixture enters the separator at A at 1200 liters>min and passes through the mist extractor at B. Crude oil flows out at 4800 liters>min through the pipe at C, and natural gas leaves the 50-mm-diameter pipe at D at VD = 75 m>s. Determine the density of the mixture that enters the separator at A. The process takes place 4 at a constant temperature of 20°C. Take r = 880 kg>m3, o rme = 0.665 kg>m3.

*4–76. At the instant considered, the tank contains air at a temperature of 30°C and absolute pressure of 480 kPa. If the air escapes through the nozzle at 0.05 m3 >s, and the volume of the tank is 6 m3, determine the rate of change in the density of the air within the tank at this instant. Is the flow steady or unsteady?

4–73. The natural gas (methane) and crude oil mixture having a density of 353 kg>m3 enters the separator at A at 1200 liters>min, and crude oil flows out through the pipe at C at 4800 liters>min. Determine the velocity of the natural gas that leaves the 50-mm-diameter pipe at D. The process takes place at a constant temperature of 20°C. Take ro = 880 kg>m3, rme = 0.665 kg>m3.

VD

D B

Prob. 4–76

A

C

Probs. 4–72/73 4–74. Water flows into the rectangular tank through pipes A and B, with velocities of 4 m>s and 2 m>s, respectively. Determine the rate at which the surface of the water is rising. Ignore the effect of gravity on the falling water. 4–75. Water flows into the rectangular tank of base 2 m by 1 m through pipes A and B, with velocities of 4 m>s and 2 m>s, respectively. Determine the time required to fill the tank if y = 0 when t = 0. Ignore the effect of gravity on the falling water. 2 m/s

150 mm

100 mm A

B

4–77. Gasoline flows through the pipe at A with an average velocity of 3 m>s, and kerosene flows through the pipe at B with an average velocity of 2 m>s. Determine the required velocity VC of the mixture leaving the tank at C so that the level of the mixture within the tank remains constant at y = 1.5 m. The tank has a width of 1 m. What is the density of the mixture leaving the tank at C? Take rg = 726 kg>m3 and rke = 814 kg>m3.

1.5 m 4 m/s

50 mm VA 3m

y

A 100 mm

75 mm VB

C

B

2m

Probs. 4–74/75

Prob. 4–77

VC

y

225

probleMs 4–78. Gasoline flows through the pipe at A with an average velocity of 3 m>s, and kerosene flows through the pipe at B with an average velocity of 2 m>s. If the mixture is allowed to leave the tank at C with VC = 2.5 m>s only after the tank is filled to a level y = 1.5 m, determine the rate at which the level in the tank is changing. The tank has a width of 1 m. Is the level rising or falling? What is the density of the mixture leaving the tank at C? Take rg = 726 kg>m3 and rke = 814 kg>m3.

*4–80. The block with a equilateral triangular base is pushed down into the tank having square cross-section with a velocity of Vb = 0.5 m>s. Determine the velocity of the liquid as it rises in the tank. 4–81. Determine the speed Vb at which the square-base block must be pushed down into the tank with a square cross-section 4 so that the liquid in the tank rises with a velocity of 0.2 m>s.

Vb

1.5 m

50 mm VA

0.4 m

A 100 mm

75 mm VB

y

VC

C

B

y

1m

Probs. 4–80/81 Prob. 4–78

4–79. As air flows over the plate, frictional effects on its surface tend to form a boundary layer in which the velocity profile changes from that of being uniform to one that is parabolic, defined by u = 3 1000y - 83.331103 2y2 4 m>s, where y is in meters, 0 … y 6 6 mm. If the plate is 0.2 m wide and this change in velocity occurs within the distance of 0.5 m, determine the mass flow through the sections AB and CD. Since these results will not be the same, how do you account for the mass flow difference? Take r = 1.226 kg>m3.

4–82. The water in the cylindrical storage tank is being drained using a pump. If the water is pumped out of the tank at 0.2 m3 >s, determine the rate at which the level in the tank is descending. 4–83. The water in the cylindrical storage tank is being pumped through a pipe having a diameter d. Determine the rate at which the level is descending as a function of d if the liquid is pumped out of the tank with an average velocity of 6 m>s. Plot this rate (vertical axis) versus the diameter for 0 … d … 0.3 m. Give values for increments of ∆d = 0.05 m. 3m Pumped

3 ms

3 ms

u A

C h

y

B

d

6 mm D

0.5 m

Prob. 4–79

Probs. 4–82/83

226

4

Chapter 4

C o n s e r va t i o n

of

Mass

*4–84. The cylindrical tank is filled with nitrogen using a hose having an inside diameter of 6.5 mm. If, at the instant considered, the nitrogen enters the tank with an average velocity of 10 m>s and has a density of 3.50 kg>m3, determine the rate of change in the density of the nitrogen within the tank at this instant

4–86. Hydrogen is pumped into the closed cylindrical tank such that the mass flow through the tube is # m = (0.5r1>2) kg>s, where r is the density of the gas in the tank in kg>m3. Determine the density of the hydrogen within the tank when t = 4 s from the time the pump is turned on. Assume that initially there is 0.2 kg of hydrogen in the tank. 0.5 m

0.6 m

1.5 m

3m

Prob. 4–86 Prob. 4–84 4–85. Hydrogen is pumped into the closed cylindrical tank such that the mass flow through the tube is # m = (0.5r-1>3) kg3 >s. Determine the density of the hydrogen within the tank when t = 4 s from the time the pump is turned on. Assume that initially there is 0.2 kg of hydrogen in the tank.

4–87. Brine or concentrated salt water in the cylindrical tank has an initial density of rb = 1250 kg>m3. Pure water is pumped into the tank at A at 0.02 m3>s and mixes with the salt water. If an equal flow of the diluted solution exits at B, determine the amount of water that must be added to the tank so that the density of the solution is reduced by 10% of its original value. The tank remains full throughout the mixing process. 1m

0.5 m

1.5 m 3m

B A

Prob. 4–85

Prob. 4–87

probleMs *4–88. A part is manufactured by placing molten plastic into the trapezoidal container and then moving the cylindrical die down into it at a constant speed of 40 mm>s. Determine the speed at which the plastic rises in the container as a function of yc. The container has a width of 200 mm.

227

4–91. As oxygen is pumped into the closed cylindrical tank, # the mass flow through the tube is m = (4.50r-1>2) kg>s, 3 where r is in kg>m . Determine the density of the oxygen within the tank when t = 5 s from the time the pump is turned on. Assume initially there is 5 kg of oxygen in the tank. *4–92. As air is pumped into the closed cylindrical tank, 4 # the mass flow through the tube m = (6.50r-3>2) kg>s, where 3 r is in kg/m . Determine the density of the air within the tank when t = 5 s from the time the pump is turned on. Assume initially there is 6 kg of air in the tank.

40 mms

1m

608

yc

608 100 mm

300 mm

2m

Prob. 4–88

4–89. Oil flows into the trapezoidal container at a constant rate of 1800 kg/min. Determine the rate at which the oil level is rising when y = 1.5 m. The container has a constant width of 0.5 m. Take ro = 880 kg>m3. 4–90. Oil flows into the conical frustum at a constant rate of 1800 kg>min. Determine the rate at which the level is rising when y = 1.5 m. Hint: The volume of a cone is V = 13 pr 2h. Take ro = 880 kg>m3.

y

Probs. 4–91/92

4–93. Water in the circular conical funnel is at a depth of y = 900 mm. If the drain is opened at the bottom, and water flows out at a rate of V = (5.02y1>2) m>s, where y is in meters, determine the time needed to fully drain the funnel. The hole at the bottom has a diameter of 50 mm.

308

308

A

y 5 900 mm 608

608

3m

Probs. 4–89/90

Probs. 4–93

228

4

Chapter 4

C o n s e r va t i o n

of

Mass

4–94. Water in the circular conical funnel is at a depth of y = 900 mm. If the drain is opened at the bottom, and water flows out at a rate of V = (5.02y1>2) m>s, where y is in meters, determine the time needed for the water to reach a depth of y = 300 mm. The hole at the bottom has a diameter of 50 mm.

4–95. The cylindrical tank in a food-processing plant is filled with a concentrated sugar solution having an initial density of rs = 1400 kg/m3. Water is piped into the tank at A at 0.03 m3 >s and mixes with the sugar solution. If an equal flow of the diluted solution exits at B, determine the amount of water that must be added to the tank so that the density of the sugar solution is reduced by 10% of its original value. 1m A

308

308

2m

y 5 900 mm

B

Probs. 4–94

Prob. 4–95

CONCEPTUAL PROBLEMS P4–1. Air flows to the left through this duct transition. Will the air accelerate or decelerate through the transition? Explain why.

P4–2. As water falls from the opening, it narrows as shown and forms what is called a vena contracta. Explain why this occurs, and why the water remains together in a stream.

P4–1

P4–2

229

Chapter review

CHAP TER R EV IEW A control volume is used for an Eulerian description of the flow. Depending on the problem, this volume can be fixed, be moving, or have a changing shape.

Surroundings

Open control surface

4 System

Using the Reynolds transport theorem, we can relate the time rate of change of a fluid property N for a system of particles to its time rate of change as measured from a control volume. Determining the control volume change requires measuring both a local change within the control volume, and a convection change, as the fluid passes through its open control surfaces.

Closed control Control surface volume Open control surface

The volumetric flow or discharge Q through a planar area A is determined by finding the velocity of the flow perpendicular to the area. If the velocity profile is known, then integration must be used to determine Q. If the average velocity V is used, then Q = V # A.

Q =

LA

The mass flow depends on the density of the fluid and on the velocity profile passing through the area. If the average # velocity V is used, then m = rV # A.

# m =

LA

The continuity equation is based on the conservation of mass, which requires that the mass of a fluid system remain constant with respect to time. In other words, its time rate of change is zero.

v # dA

rv # dA

0 r dV + r Vf>cs # dA = 0 0t Lcv Lcs

We can use a fixed, a moving, or a changing control volume to apply the continuity equation. In particular, if the control volume is fixed and completely filled with an incompressible fluid and the flow is steady, then no local changes will occur within the control volume, so only convective changes need be considered.

Oftentimes if the control volume is attached to a body moving with constant velocity, then steady flow may occur. Vf

Vfcs Vcv 5 Vb A

Vcs 5 Vb

5

Kyrylo Glivin/Alamy Stock Photo

CHAPTER

The design of fountains requires application of the principle of work and energy. Here the velocity of the flow out of the nozzles is transformed into lifting the water to its maximum height.

WORK AND ENERGY OF MOVING FLUIDS

CHAPTER OBJECTIVES ■

To develop Euler’s equations of motion and the Bernoulli equation for streamline coordinates, and to illustrate some important applications.

■

To show how to construct the energy grade line and hydraulic grade line for a fluid system.

■

To develop the energy equation, and to show how to solve problems that include pumps, turbines, and friction loss.

5.1

EULER’S EQUATIONS OF MOTION

In this chapter, we will analyze how the pressure, velocity, and elevation of a fluid are related from one location to another. We begin by using a Lagrangian description to study the motion of a single fluid particle as it travels along a streamline with steady flow, Fig. 5–1a. We will consider motion in the vertical plane and use a streamline coordinate s that is in the direction of motion and tangent to the streamline, and a normal coordinate n that is directed positive toward the streamline’s center of curvature. The particle has a length ∆s, height ∆n, and width ∆x. Since the flow is steady, the streamline will remain fixed, and since it happens to be curved, the particle will have two components of acceleration. Recall from Sec. 3.5 that the tangential or streamline component as measures the time rate of change in the magnitude of the particle’s velocity. It is determined from as = V1dV>ds2. The normal component an measures the time rate of change in the direction of the velocity. It is determined from an = V 2>R, where R is the radius of curvature of the streamline at the point where the particle is located.

n an as

Streamline s u Horizontal

Dn Ds Dx Ideal fluid particle (a)

Fig. 5–1 231

232

Chapter 5

u

DW 5 rg( s n x) Dx

Ds

5

and

energy

z

n

(p 1 dp –– Dn)Ds Dx dn

Work

s

–– Ds) Dn Dx (p + dp ds Dn

of

Moving fluids

The free-body diagram of the particle is shown in Fig. 5–1b. If we assume the fluid is inviscid, then the shearing forces of viscosity are not present, and only the forces caused by weight and pressure act on the particle. Notice that the pressure on two sides of the particle is considered to increase by 1dp>ds2∆s and by 1dp>dn2∆n in the positive s and n directions.* Finally, the weight of the particle is ∆W = rg∆V = rg 1∆s∆n∆x2, and its mass is ∆m = r∆V = r1∆s∆n∆x2.

s Direction. Applying the equation of motion ΣFs = mas in the s

p Dn Dx p Ds Dx

direction, we have

p∆n∆x - a p +

Free-body diagram (b)

dp ∆s b ∆n∆x - rg1∆s∆n∆x2 sin u ds

Net pressure force

z

dz

u

(c) n

z

= r1∆s∆n∆x2V a

dV b ds

Mass-Acceleration

s

ds

Weight

Dividing through by the mass, r1 ∆s∆n∆x2, and rearranging the terms yields 1 dp dV + Va b + g sin u = 0 (5–1) r ds ds As noted in Fig. 5–1c, sin u = dz>ds. Therefore,

u dz

dn

dp + V dV + g dz = 0 r

(5–2)

n Direction. Applying the equation of motion ΣFn = man in the n direction, Fig. 5–1b, we have (d)

Fig. 5–1 (cont.)

p∆s∆x - a p +

dp ∆n b ∆s∆x - rg1∆s∆n∆x2 cos u dn

Net pressure force

Weight

= r1∆s∆n∆x2 a

V2 b R

Mass-Acceleration

Using cos u = dz>dn, Fig. 5–1d, and dividing by the volume, 1∆s∆n∆x2, this equation reduces to -

dp rV 2 dz - rg = dn dn R

(5–3)

* For this increase, we have considered only the first term in a Taylor series expansion at this point, since the higher-order terms will cancel out as the particle’s size becomes infinitesimal. (See the footnote on page 361.)

5.1

233

euler’s equations of Motion

Equations 5–2 and 5–3 are differential forms of the equations of motion that were originally developed by the Swiss mathematician Leonhard Euler. For this reason they are often called Euler’s differential equations of motion. They apply only in the s and n directions to the steady flow of an inviscid fluid particle that moves along a streamline. We will now consider a few important applications.

Steady Horizontal Flow of an Ideal Fluid. Shown in Fig. 5–2 are straight horizontal open and closed conduits, where an ideal liquid is flowing at constant velocity. In both cases, the pressure at A is pA, and we wish to determine the pressure at points B and C. Since A and B lie on the same streamline, identified by the s axis, we can apply Euler’s equation in the s direction, and integrate it from point A to point B. Here VA = VB = V, and because there is no elevation change, dz = 0. Also, since the fluid density is constant, we have pB V dp 1 + V dV + g dz = 0; dp + V dV + 0 = 0 r r LpA LV 1 V 1 pB - pA 2 + 0 + 0 = 0 or pB = pA r Thus, for an ideal fluid, the pressure along both the open and closed conduits remains constant in the horizontal direction. This result is to be expected because for an ideal fluid no viscous friction forces have to be overcome by the pressure to push the fluid forward. Although point C is not on the same streamline as A, Fig. 5–2, it is on the n axis, which has its origin at A. Since the radius of curvature of the horizontal streamline at A is R S ∞, Euler’s equation in the n direction becomes V dp rV 2 dz - rg = = 0; -dp - rg dz = 0 dn dn R Integrating from A to C, noting that C is at z = -h from A, we have pC

n, z

A

B

h

s

C Open conduit n, z

A h

C Closed conduit

-h

dp - rg dz = 0 LpA L0 -pC + pA - rg1 -h - 02 = 0 pC = pA + rgh This result indicates that in the vertical direction, the pressure is the same as if the fluid were static. And so for the open conduit, pA represents the pressure at A caused by the weight of fluid above A; and for the closed conduit, pA must also include the internal pressure of the fluid within the conduit. The term “static pressure” is often used here since it is a measure of the pressure relative to the flow, in which case the fluid will appear to be at rest. For example, if you were submerged and flowing along a river (moving horizontally) or lake (stationary) at the same depth, you would feel the same pressure. Now if the conduit were curved, then we would not obtain the above result, because particles at A and C would move along streamlines having different radii of curvature, and so the time rate of change in the direction of their velocities would be different. The following example should help to illustrate this point. -

5

Fig. 5–2

B

s

234

Chapter 5

EXAMPLE

5

Work

and

energy

of

Moving fluids

5.1 A tornado has winds that essentially move along horizontal circular streamlines, Fig. 5–3. Within the eye, 0 … r … r0, the wind velocity is V = vr, which represents a forced vortex, that is, flow rotating at a constant angular rate v as described in Sec. 2.14. Determine the pressure distribution within the eye of the tornado as a function of r, if at r = r0 the pressure is p = p0. s p0 V

r0

r n

1r

v s

Fig. 5–3

SOLUTION Fluid Description. We have steady flow, and we will assume the air is an ideal fluid, that is, it is inviscid and has a constant density r. Analysis. The streamline for a fluid particle having a radius r is shown in Fig. 5–3. To find the pressure distribution as a function of r (positive outward), we must apply Euler’s equation in the n direction (positive inward). dp rV 2 dz - rg = dn dn R Since the path is horizontal, dz = 0. Also, for an arbitrarily chosen streamline, R = r and dn = -dr. Since the velocity of the particle is V = vr, the above equation becomes -

r1vr2 2 dp - 0 = r dr dp = rv2r dr Notice that the pressure increases, +dp, as we move farther away +dr from the center. Since p = p0 at r = r0, then Lp

p0

dp = rv2

Lr

r0

r dr

rv2 2 1r 0 - r 2 2 2 We will extend this analysis in Example 7.9. p = p0 -

Ans.

5.2

5.2

the Bernoulli equation

235

THE BERNOULLI EQUATION

As we have seen, Euler’s equations represent the application of Newton’s second law of motion to the steady flow of an inviscid fluid particle, expressed in terms of the streamline coordinates s and n. Since particle motion only occurs in the s direction, we can integrate Eq. 5–2 along a streamline, and thereby obtain a relationship between the motion of the particle and the pressure and gravitational forces that act upon it. dp V dV + g dz = 0 + L r L L Provided the fluid density can be expressed as a function of pressure, integration of the first term can be carried out. The most common case, however, is to consider the fluid to be ideal, that is, both inviscid and incompressible, i.e., r is constant. For this case, p V2 + + gz = const. r 2

(5–4)

Here z is the elevation of the particle, measured from an arbitrarily chosen fixed horizontal plane or datum, Fig. 5–4. For any particle above this datum, z is positive, and for any particle below it, z is negative. For a particle on the datum, z = 0. Equation 5–4 is referred to as the Bernoulli equation, named after Daniel Bernoulli, who stated it around the mid-18th century. Sometime later, though, it was expressed mathematically by Leonhard Euler. When it is applied between any two points, 1 and 2, located on the same streamline, Fig. 5–4, it can be written in the form* p1 p2 V 12 V 22 + + gz1 = + + gz2 r r 2 2

(5–5)

Steady flow, ideal fluid, same streamline

p2 2

p p1

V2

V V1

1 z1

z

z2 Datum

Fig. 5–4

*Later in Sec. 7–6 we will show that the Bernoulli equation can also be applied between any two points on different streamlines provided the flow is “irrotational.”

5

236

Chapter 5

Work

and

energy

of

Moving fluids

n

5

Weight produces gravitional potential energy DW s

Velocity produces kinetic energy V 1

2

Net pressure force produces flow work z

Datum

Fig. 5–5

This integrated form of Newton’s second law can be interpreted as a statement of the principle of work and energy as it applies to a unit mass of fluid moving from point 1 to point 2 along a streamline, Fig 5–5. In other words, each term has units of energy or work (J) per unit mass (kg). In the form p>r + V 2 >2 + gz = const., the equation states that the sum of the flow work associated with pressure 1p>r2, plus the kinetic energy 1V 2 >22, and the gravitational potential energy (gz), has the same constant value at all points along the streamline, provided no friction or external heat or energy is added to or drawn from the fluid. It is sometimes convenient to divide the Bernoulli equation by g and substitute g = rg into the equation. It then has the form p1 p2 V 12 V 22 + + + z1 = + z2 g g 2g 2g

(5–6)

steady flow, ideal fluid, same streamline

In this book, we will apply the Bernoulli equation in this form, and then later in Sec. 5.4, we will show how to represent the terms graphically in order to plot the energy and hydraulic grade lines for a network of open and closed conduits.

5.2

Limitations. It is very important to remember that the Bernoulli equation can only be applied when we have steady flow of an ideal fluid—one that is incompressible (r is constant) and inviscid 1m = 02. As developed here, its application is between any two points lying on the same streamline. If these conditions cannot be justified, then application of this equation will produce erroneous results. With reference to Fig. 5–6, the following is a list of some common situations in which the Bernoulli equation should not be used.

237

the Bernoulli equation

Heat added

Rotational flow eddies

Energy added

A B

C D

E

• Many fluids, such as air and water, have rather low viscosities, so in some situations they may be assumed to be ideal fluids. However, realize that in certain regions of the flow, the effects of viscosity can not be ignored. For example, viscous flow will always predominate near a solid boundary, such as the walls of the pipe. This region is called the boundary layer, and we will discuss it in more detail in Chapter 11. Here the fluid friction, or shear, which causes boundary layer formation, will produce heat and thereby draw energy from the flow. The Bernoulli equation cannot be applied within the boundary layer, because of this resulting energy loss.

• Sudden changes in the direction of a solid boundary can cause the

•

boundary layer to thicken and result in flow separation from the boundary. This can occur along the inside wall of the pipe at the bend. Here turbulent mixing of the fluid not only produces frictional heat loss, but greatly affects the velocity profile and causes a severe pressure drop. The streamlines within this region are not well defined, and so the Bernoulli equation does not apply. Furthermore, flow separation and the development of turbulent mixing within the flow also occur in valves and at connections such as the pipe–tube interface A. Hence the Bernoulli equation cannot be applied across this interface. Energy changes within the flow occur within regions of heat removal or heat application such as between B and C. Also, pumps or turbines can supply or remove energy from the flow such as between D and E. The Bernoulli equation does not account for these energy changes, and so it cannot be applied within these regions.

• If the fluid is a gas, then its density will change as the speed of the flow increases. Normally, as a general rule for engineering calculations, a gas can be considered incompressible provided its speed remains below about 30% the speed at which sound travels within it. For example, at 15°C the sonic speed in air is 340 m>s, so a limiting value would be 102 m>s. For faster flows, compressibility effects will cause heat loss, and this becomes important. For these high speeds, the Bernoulli equation will not give acceptable results.

Viscous friction

Flow separation

Places where the Bernoulli equation does not apply

Fig. 5–6

5

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5.3

Moving fluids

APPLICATIONS OF THE BERNOULLI EQUATION

In this section we will provide some basic applications of the Bernoulli equation to show how to determine either the velocity or the pressure at different points on a streamline.

5

Flow from a Large Reservoir. When water flows from a tank or reservoir through a drain, Fig. 5–7, the flow is actually unsteady. This happens because when the depth h is large, then due to the greater water pressure at the drain, the water level will drop at a faster rate than when h is small. However, if the reservoir has a large volume and the drain has a relatively small diameter, then the movement of water within the reservoir is very slow and so, at its surface, VA ≈ 0, Fig. 5–7. Under these circumstances it is reasonable to assume steady flow through the drain. Also, for small-diameter openings, the elevation difference between C and D is small, and so the average velocity can be used since VC ≈ VD ≈ VB. Furthermore, the water stream is in free fall just outside the opening, and so pressure at the centerline B of the opening is atmospheric, as it is at C and D. If we assume that water is an ideal fluid, then the Bernoulli equation can be applied between points A and B that lie on the selected streamline in Fig. 5–7. Setting the gravitational datum at B, and using gage pressures, where pA = pB = 0, we have

A

h

pA pB VA 2 VB 2 + + + z = + zB A g g 2g 2g C Datum B

D

0 + 0 + h = 0 +

VB 2 + 0 2g

VB = 22gh Flow from a reservoir

Fig. 5–7

This result is known as Torricelli’s law, since it was first formulated by Evangelista Torricelli in the 17th century. As a point of interest, it can be shown that VB as determined above is the same as that obtained by a fluid particle that is simply dropped from rest at the same height h, even though the time of travel for this freely falling particle is much shorter than for the one flowing through the tank.

5.3

Datum

A

appliCations of the Bernoulli equation

239

B

5

Stagnation point

Fig. 5–8

Flow around a Curved Boundary. As a fluid flows around an obstacle, the energy of the fluid will be transformed from one form into another. For example, consider the horizontal streamline that intersects the front of the curved surface in Fig. 5–8. Since B is a stagnation point, fluid particles moving from A to B must decelerate, so that their velocity at B momentarily becomes zero, before the flow begins to separate and then move along the sides of the surface. If we apply the Bernoulli equation between points A and B, we have pA pB VA 2 VB 2 + + + zA = + zB g g 2g 2g pA pB VA 2 + + 0 + 0 + 0 = g g 2g pB Stagnation pressure

=

pA Static pressure

+ r

VA 2 2

Dynamic pressure

This pressure, pB, is referred to as the stagnation pressure because it represents the total pressure exerted by the fluid at the stagnation point. As noted in Sec. 5–1, the pressure pA is a static pressure because it is measured relative to the flow, like what a submerged swimmer would feel in a flowing river. Finally, the increase in pressure, rVA 2 >2, is sometimes called the dynamic pressure because it represents the additional pressure required to bring the fluid to rest at B. Structural engineers use this pressure to determine the wind loads on buildings.

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and

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of

Moving fluids

Flow in an Open Channel. One method of determining the

5

h

d A

B

Pitot tube

Fig. 5–9

Due to dynamic pressure Due to static pressure Datum

velocity of a moving liquid in an open channel, such as a river, is to immerse a bent tube into the stream and observe the height h to which the liquid rises within the tube, Fig. 5–9. Such a device is called a stagnation tube or pitot tube, named after Henri Pitot, who invented it in the early 18th century. To show how it works, consider the two points A and B located on the horizontal streamline. Point A is upstream within the fluid, where the velocity of flow is VA and the (static) pressure is pA = gd. Point B is at the opening of the tube. It is the stagnation point, since the velocity of flow has momentarily been reduced to zero due to its impact with the liquid within the tube. The liquid at this point is subjected to both a static pressure, because of its depth d, and a dynamic pressure, because the flow has forced additional liquid farther up the vertical segment to a height h above the liquid surface. Thus the total pressure of the liquid at B is pB = g1d + h2. Applying the Bernoulli equation with the gravitational datum on the streamline, we have pA pB VA 2 VB 2 + + + zA = + zB g g 2g 2g g1d + h2 gd VA 2 + + 0 + 0 + 0 = g g 2g VA = 22gh

Hence, by measuring h on the pitot tube, the velocity of the flow can be determined.

Flow in a Closed Conduit. If liquid is flowing in a closed conduit or pipe, Fig. 5–10a, then it will be necessary to use both a piezometer and a pitot tube to determine the velocity of the flow. The piezometer measures the static pressure at A. In this case, this pressure is caused by the internal pressure in the pipe, gh, and the hydrostatic pressure gd, caused by the weight of the fluid. The static or total pressure at A is therefore g1h + d2. The total pressure at the stagnation point B will be larger than this, due to the dynamic pressure that depends upon the velocity of the fluid, V 2A >2g. If we apply the Bernoulli equation at points A and B on the streamline, using the measurements h and 1l + h2 from these two tubes, the velocity VA can be obtained. pA pB VA 2 VB 2 + + + zA = + zB g g 2g 2g g1h + d2 g1h + d + l2 VA 2 + 0 = + + 0 + 0 g g 2g VA = 22gl

5.3

appliCations of the Bernoulli equation

Rather than use two separate tubes in the manner just described, a more elaborate single tube called a pitot-static tube is often used to determine the velocity of the flow in a closed conduit. It is constructed using two concentric tubes as shown in Fig. 5–10b. Like the pitot tube in Fig. 5–10a, the stagnation pressure at B can be measured from the pressure tap at E in the inner tube. Downstream from B there are several open holes D on the outer tube. This outer tube acts like the piezometer in Fig. 5–10a so that the static pressure can be measured from the pressure tap at C. Using these two measured pressures and applying the Bernoulli equation between points A and B, neglecting any small elevation difference between C and E, we have

5

pA pB VA 2 VB 2 + + + zA = + zB g g 2g 2g pE + rgh pC + rgh VA 2 + 0 = + + 0 + 0 rg rg 2g VA =

2 1p - pC 2 Ar E

In actual practice, the difference in pressure 1pE - pC 2 can be determined either by using a manometer, attached to the outlets C and E, and measuring the differential height of the manometer liquid; or by using a pressure transducer, Fig. 2–20. For accuracy, corrections are sometimes made to the readings since the flow may be slightly disturbed at the side inlet holes D, due to its movement around the front of the tube at B and past the sides of its vertical segment.

Static pressure Dynamic pressure

A

l

h

Due to pipe internal pressure

d

Due to weight of fluid

C E

h D

Datum

B A

B

D

Piezometer and pitot tubes

Pitot-static tube

(a)

(b)

Fig. 5–10

241

Datum

242

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and

energy

of

Moving fluids

Venturi meter. A venturi meter is a device that can also be used to measure the average velocity or the flow of a fluid through a pipe, Fig. 5–11. It was conceived of by Giovanni Venturi in 1797, but the principle was not applied until about one hundred years later, by the American engineer Clemens Herschel. This device consists of a reducer followed by a venturi tube or throat, and then a gradual transition segment back to the original pipe. As the fluid flows through the reducer, the flow accelerates, causing a higher velocity and lower pressure to be developed within the throat. Applying the Bernoulli equation along the center streamline, between point 1 in the pipe and point 2 in the throat, gives

5

d1

Datum

d2 2

1 V1 h9

r h r0 Venturi meter

p1 p2 V 12 V 22 + z1 = + z2 + + g g 2g 2g p1 p2 V 12 V 22 + 0 = + 0 + + rg rg 2g 2g

In addition, since the mass of the fluid is conserved, the continuity equation can be applied using a control volume between points 1 and 2, Fig. 5–11. For steady flow we have

Fig. 5–11

0 r dV + rVf>cs # dA = 0 0 t Lcv Lcs 0 - V1 c pa

d 12 d 22 b d + V2 c pa b d = 0 4 4

Combining these two results and solving for V1, we get V1 =

21p1 - p2 2 >r A 1d1 >d2 2 4 - 1

The static pressure difference 1p1 - p2 2 is often measured using a manometer or pressure transducer. For example, if a manometer is used, as in Fig. 5–11, and r is the density of the fluid in the pipe, and r0 is the density of the fluid in the manometer, then applying the manometer rule, we have p1 + rgh′ - r0gh - rg1h′ - h2 = p2 p1 - p2 = 1r0 - r2gh

Once the measurement for h is made, the result 1p1 - p2 2 is then substituted into the above equation to obtain V1. The flow can then be determined from Q = V1 A1.

5.3

appliCations of the Bernoulli equation

243

IMPORTANT POIN T S • The Euler differential equations of motion apply to a fluid particle that moves along a streamline. These equations are based on the steady flow of an inviscid fluid. Because viscosity is neglected, the flow is affected only by pressure and gravitational forces. In the s direction these forces cause a change in magnitude of a fluid particle’s velocity, giving it a tangential acceleration; and in the n direction these forces cause a change in direction of the velocity, thereby producing a normal acceleration.

• When the streamlines for the flow are straight horizontal lines, the Euler equations show that for an ideal (frictionless) fluid having steady flow, the pressure p0 in the horizontal direction is constant. Furthermore, since the velocity does not change its direction, there is no normal acceleration. Consequently, in a horizontal open or closed conduit, the pressure variation in the vertical direction is hydrostatic. In other words, this is a measure of static pressure because when the pressure is measured relative to the moving fluid, it is the same as if the fluid is at rest.

• The Bernoulli equation is an integrated form of Euler’s equation in the s direction. It is applied at two points located on the same streamline, and it requires steady flow of an ideal fluid. This equation cannot be applied to viscous fluids, or at transitions where the flow separates and becomes turbulent. Also, it cannot be applied between points where fluid energy is added or withdrawn by external sources, such as pumps and turbines, or in regions where heat is applied or removed. then z is constant, and so p>g + V 2 >2g = const. Therefore, through a converging duct or nozzle, the velocity will increase and the pressure will decrease. Likewise, for a diverging duct, the velocity will decrease and the pressure will increase. (See the photo on the next page.)

• The Bernoulli equation indicates that if the flow is horizontal,

in an open channel. The flow creates a dynamic pressure rV 2 >2 at the tube’s stagnation point, which pushes the fluid up the tube. For a closed conduit, both a piezometer and a pitot tube must be used to measure velocity. A device that combines both of these is called a pitot-static tube.

• A pitot tube can be used to measure a fluid’s velocity at a point

• A venturi meter can be used to measure the average velocity or volumetric flow of a fluid through a closed duct or pipe.

5

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Moving fluids

PROCEDURE FOR ANALYSIS The following procedure provides a means for applying the Bernoulli equation. 5

Fluid Description. • Be sure that the fluid can be assumed to be an ideal fluid, that is, incompressible and inviscid. Also, steady flow must occur. Bernoulli Equation. • Select two points on the same streamline within the flow, where some values of pressure and velocity are known. The elevation of these points is measured from an arbitrarily established fixed datum. At pipe outlets to the atmosphere where the fluid begins to free fall, and at open surfaces, the pressure is atmospheric, that is, the gage pressure is zero. • The velocity at each point can be determined if the volumetric flow and cross-sectional area of a conduit are known, V = Q>A. • Tanks or reservoirs that drain slowly have liquid surfaces that are essentially at rest; that is, V ≈ 0. • When the ideal fluid is a gas, elevation changes, measured from the datum, can generally be neglected. • Once the known and unknown values of p, V, and z have been identified at each of the two points, the Bernoulli equation can be applied. When substituting the data, be sure to use a consistent set of units. • If more than one unknown is to be determined, consider relating the velocities using the continuity equation, or relating the pressures using the manometer equation if it applies.

As noted by the water level in the piezometers, the pressure of the water flowing through this pipe will vary in accordance with the Bernoulli equation. Where the diameter is small, the velocity is high and the pressure is low; and where the diameter is large, the velocity is low and the pressure is high.

5.3

EXAMPLE

245

appliCations of the Bernoulli equation

5.2

The jet plane in Fig. 5–12 is equipped with a piezometer that indicates an absolute pressure of 47.2 kPa, and a pitot tube B that reads an absolute pressure of 49.6 kPa. Determine the altitude of the plane and its speed.

A

5

B Datum

SOLUTION The piezometer measures the static pressure in the air, and so the altitude of the plane can be determined from the table in Appendix A. For an absolute pressure of 47.2 kPa, the altitude is approximately h = 6 km Ans.

Fig. 5–12

Fluid Description. We will assume that the speed of the plane is slow enough so that the air can be considered incompressible and inviscid—an ideal fluid. Doing so, we can apply the Bernoulli equation, provided we observe steady flow. This can be realized if we view the motion from the plane.* Thus, the air at A, which is actually at rest, will have the same speed as the plane when observed from the plane, VA = Vp. The air at B, a stagnation point, will appear to be at rest when observed from the plane, VB = 0. From Appendix A, for air at an elevation of 6 km, ra = 0.6601 kg>m3. Bernoulli Equation. Applying the Bernoulli equation at points A and B on the horizontal streamline, we have pA pB VA 2 VB 2 + + + zA = + zB g g 2g 2g 47.21103 2 N>m2

10.6601 kg>m3 219.81 m>s2 2

+

V p2

219.81 m>s2 2

Vp = 85.3 m>s

+ 0 =

(© Minerva Studio/Shutterstock)

49.61103 2 N>m2

10.6601 kg>m3 219.81 m>s2 2 Ans.

NOTE: We will show in Chapter 13 that this speed is about 25% the speed

of sound in air (343 m>s), and since 25% 6 30%, our assumption of air being incompressible is valid. Most planes are equipped with either a piezometer and a pitot tube, or a combination pitot-static tube. The pressure readings are directly converted to altitude and air speed and are shown on the instrument panel. If more accuracy is needed, corrections are made by factoring in a reduced density of the air at high altitudes. Although this device has served the aviation industry well for many years, it is important that for proper operation, the opening of any pitot tube must always be free of debris such as caused by nesting insects or ice formation. *If the flow is observed from the ground, then the flow is unsteady since the velocity of the air particles will change with time as the plane flies past these particles. Remember that the Bernoulli equation does not apply for unsteady flow.

+ 0 + 0

246

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EXAMPLE

Work

and

energy

of

Moving fluids

5.3

C

5

hdyn

Determine the average velocity of the flow of water in the pipe in Fig. 5–13, and the static and dynamic pressure at point B. The water level in each of the tubes is indicated. Take rw = 1000 kg>m3.

hB 5 150 mm hA 5 90 mm A

B

Fig. 5–13

Datum

SOLUTION Fluid Description. We have steady flow. Also, we will consider the water to be an ideal fluid. Bernoulli Equation. At A the total pressure is the static pressure, which is the result of both the weight of fluid and the internal pressure within the pipe. It is found from pA = rwghA.At B the total (or stagnation) pressure is a combination of static and dynamic pressures found from pB = rwghB. Knowing these pressures, we can determine the average velocity of flow VA using the Bernoulli equation, applied at points A and B, where B is a stagnation point on the streamline. We have pA pB VA 2 VB 2 + zA = + zB + + g g 2g 2g rwghB rwghA VA 2 + 0 = + + 0 + 0 r wg r wg 2g

VA = 22g1hB - hA 2 = 2219.81 m>s2 210.150 m - 0.090 m2 VA = 1.085 m>s = 1.08 m>s

Ans.

The static pressure at both A and B is determined from the piezometric head.

1pA 2 static = 1pB 2 static = rwghA = 11000 kg>m3 219.81 m>s3 210.09 m2 = 883 Pa

Ans.

The dynamic pressure at B is determined from

11.085 m>s2 2 VA 2 = 11000 kg>m32 = 589 Pa 2 2 This value can also be obtained by noticing that rw

Ans.

hdyn = 0.15 m - 0.09 m = 0.06 m so that

1pB 2 dyn = rwghdyn = 11000 kg>m3 219.81 m>s2 210.06 m2 = 589 Pa

Ans.

5.3

EXAMPLE

247

appliCations of the Bernoulli equation

5.4

A transition is placed in a rectangular air duct as shown in Fig. 5–14. If 2 kg>s of air flows steadily through the duct, determine the pressure change that occurs between the ends of the transition. Take ra = 1.23 kg>m3. SOLUTION

450 mm A

5

300 mm B

Fluid Description. We have steady flow. At slow velocities the air passing through the duct will be considered an ideal fluid, that is, incompressible and inviscid. Analysis. To solve this problem, we will first use the continuity equation to obtain the average velocity of flow at A and at B. Then we will use the Bernoulli equation to determine the pressure difference between A and B. Continuity Equation. We will consider a fixed control volume that contains the air within the duct, Fig. 5–14. Thus, for steady flow, 0 r dV + rVf>cs # dA = 0 0t L L cv

150 mm

Fig. 5–14

cs

0 - VAAA + VBAB = 0 Q = VAAA = VBAB But Q = so that VA = and VB =

# 2 kg>s m = 1.626 m3 >s = r 1.23 kg>m3

1.626 m3 >s Q = = 12.04 m>s AA (0.45 m)(0.3 m)

1.626 m3 >s Q = = 24.09 m>s AB (0.45 m)(0.15 m)

Bernoulli Equation. streamline, we have

Selecting points at A and B on the horizontal

pA pB VA 2 VB 2 + + zA = + + zB ga ga 2g 2g pA (1.23 kg>m3)(9.81 m>s2)

+

1 12.04 m>s 2 2 2(9.81 m>s2)

+ 0 =

pA - pB = 267.66 Pa = 0.268 kPa

pB (1.23 kg>m3)(9.81 m>s2) Ans.

This small drop in pressure, or the low velocities, will not significantly change the density of the air, and so here it is reasonable to have assumed the air to be incompressible.

+

1 24.09 m>s 2 2 2(9.81 m>s2)

+ 0

Datum

248

Chapter 5

EXAMPLE

5

Work

and

energy

of

Moving fluids

5.5 The gas tank contains a 0.6-m depth of gasoline and a 0.2-m depth of water as shown in Fig. 5–15. Determine the time needed to drain the water if the drain hole has a diameter of 25 mm. The tank is 1.8 m wide and 3.6 m long. The density of gasoline is rg = 726 kg>m3, and for water, rw = 1000 kg>m3.

1.8 m

A 0.6 m B h 5 0.2 m

Datum

C 25 mm

Fig. 5–15

SOLUTION Fluid Description. The gasoline is on top of the water because its density is less than that of water. Because the tank is large relative to the drain hole, we will assume steady flow. Also, we will consider the two fluids to be ideal. Bernoulli Equation. Here we will select the vertical streamline containing points B and C, Fig. 5–15. At any instant the level of water is h, as measured from the datum, and so the pressure at B is due to the weight of the gasoline above it, that is, pB = gghAB = (726 kg>m3)(9.81 m>s2)(0.6 m) = 4.273(103) N>m2 To simplify the analysis for using the Bernoulli equation, we will neglect the velocity at B since VB ≈ 0, and so VB 2 will be even smaller. Since C is open to the atmosphere, pC = 0. Thus, pB pC VC 2 VB 2 + + zB = + + zC gw gw 2g 2g 4.273(103) N>m2 3

2

(1000 kg>m )(9.81 m>s )

+ 0 + h = 0 +

VC 2 2(9.81 m>s2)

VC = 4.4292h + 0.4356

+ 0 (1)

5.3

appliCations of the Bernoulli equation

Continuity Equation. The continuity of flow at B and C will allow us to relate the actual nonzero VB to VC . We will choose a control volume that contains all the water up to the depth h. Since VB is downward and h is positive upward, then at the top control surface, VB = -dh>dt. Thus, 0 r dV + rVf>cs # dA = 0 0 t Lcv Lcs 0 - V BA B + V C A C = 0

0 - a-

dh b 3(1.8 m)(3.6 m)4 + VC 3 p 1 0.0125 m 2 2 4 = 0 dt dh = -75.752 1 10-6 2 VC dt

Now, using Eq. 1,

dh = -75.752 1 10-6 2 1 4.4292h + 0.4356 2 dt

or dh = -0.3355 1 10-3 2 2h + 0.4356 dt

(2)

Notice that when h = 0.2 m, VB = dh>dt = 0.268110-32 m>s, which is very slow compared to VC = 3.53 m>s, as determined from Eq. 1, and so indeed we were justified in neglecting VB 2 in Eq. 1. If td is the time needed to drain the tank, then separating the variables in Eq. 2 and integrating, we get 0

= -0.3355 1 10 - 3 2

L0.2 m 2h + 0.4356

1 22h

dh

+ 0.4356 2 2

0 0.2 m

L0

td

dt

= -0.3355 1 10-3 2 td

Evaluating the limits, we get

-0.2745 = -0.3355 1 10-3 2 td

td = 818.06 s = 13.6 min.

Ans.

249

5

250

Chapter 5

EXAMPLE

5

Work

and

energy

of

Moving fluids

5.6 Water flows up through the vertical pipe that is connected to the transition, Fig. 5–16. If the volumetric flow is 0.02 m3 >s, determine the height h to which the water will rise in the pitot tube. The level of water in the piezometer at A is indicated.

200 mm

h D

B

400 mm

SOLUTION Fluid Description. The flow is steady, and the water is assumed to be an ideal fluid, where rw = 1000 kg>m3. Bernoulli Equation. at A is

165 mm Datum

CA

From the piezometer reading, the pressure

pA = rwghA = 11000 kg>m3 219.81 m>s2 210.165 m2 = 1618.65 Pa

This total pressure is caused by the static pressure in the water. In other words, it is the pressure within the enclosed pipe at this level. Since the flow is known, the velocity at A can be determined.

100 mm

Fig. 5–16

0.02 m3 >s = VA 3 p10.05 m2 2 4

Q = VAAA;

VA = 2.546 m>s

Also, since B is a stagnation point, VB = 0. We can now determine the pressure at B by applying the Bernoulli equation at points A and B on the vertical streamline in Fig. 5–16. The datum is placed at A, and so pA pB VA2 VB 2 + + zA = + + zB gw gw 2g 2g

1618.65 N>m2

11000 kg>m3 219.81 m>s2 2

+

12.546 m>s2 2

219.81 m>s2 2

+0=

pB

11000 kg>m3 219.81 m>s2 2

+ 0 + 0.4 m

pB = 936.93 Pa

Since B is a stagnation point, this total pressure is caused by both static and dynamic pressures at B. For the pitot tube we require h =

pB 936.93 Pa = = 0.09551 m = 95.5 mm Ans. gw 11000 kg>m3 219.81 m>s2 2

it is not part of this problem, the pressure at D can be obtained 1pD = 734 Pa2 by applying the Bernoulli equation along the streamline CD, where pC = pA. First, though, the velocity VD = 0.6366 m>s must be obtained by applying Q = VD AD. NOTE: Although

5.4

5.4

energy and hydrauliC grade lines

251

ENERGY AND HYDRAULIC GRADE LINES

To analyze the flow within a system of pipes or open channels, the terms of the Bernoulli equation are often represented graphically. If we write the equation as V2 2g Velocity head

+

p g Pressure head

+

z Elevation head

= H

(5–7)

Total head

Hydraulic head

then each term has a unit of length, m. The first term on the left represents the kinetic or velocity head, which indicates the vertical distance a fluid particle must fall from rest to attain the velocity V. The second term represents the static pressure head, which is the height of a fluid column supported by a pressure p acting at its base. And finally, the third term is the elevation head, which is the height a fluid particle is placed above (or below) a selected datum. As noted, the sum of the pressure and elevation head forms the hydraulic head, and this added to the velocity head gives the total head, H. A plot of the total head measured from an arbitrary fixed selected datum is called the energy grade line (EGL). Although each of the terms on the left side of Eq. 5–7 may change, their sum H will always remain constant at every point along the same streamline, provided there are no friction losses and there is no addition or removal of energy by an external source such as a pump or turbine. Experimentally, H can be obtained at any point using a pitot tube, as shown in Fig. 5–17. For problems involving the design of pipe systems or channels, it is often convenient to plot the energy grade line and also its counterpart, the hydraulic grade line (HGL). This line shows how the hydraulic head p>g + z will vary along the pipe (or channel). Here a piezometer can be used to experimentally obtain its value, Fig. 5–17. By comparison, the HGL will always lie below the EGL by a distance of V 2 >2g.

5

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Pitot tube Piezometer Pitot tube V12 2g

V22 2g

Piezometer

HGL

p2 g 1 z2

EGL

p V2 H5 g 1 1z 2g

5 p1 g 1 z1

Datum Energy and hydraulic grade lines

Fig. 5–17

As an example of how to draw the energy and hydraulic grade lines, consider the pipe in Fig. 5–18, which drops in elevation and also changes its diameter. Because of continuity, the flow must maintain its velocity V1 through section ABCD. Also, continuity requires the flow to be faster, V2 7 V1, through section DE because the diameter of this length is smaller. The velocity heads for these two lengths are V1 2 >2g and V2 2 >2g. At all points, the elevation head z is measured from the selected datum to the centerline of the pipe.* Since the pressure at E is atmospheric and frictional effects within the pipe are neglected, then both at E and along DE, p = 0, and so the HGL is at p>g + z = z2, Fig. 5–18, and the EGL is at V2 2 >2g + z2. In particular, notice how the pressure and elevation heads trade off along the inclined section, where the pressure head increases in proportion to a drop in the elevation head.

EGL V12 2g V1

1

V22 2g

p1 g B p g

A

C

z1

D

2

V2 HGL

z

E z2 Datum

Fig. 5–18

*Actually, there will be a small hydrostatic difference in pressure along the vertical diameter of the pipe, as noted on p. 233, because the fluid within the pipe is supported by the pipe. Here we will neglect this effect since the diameter will generally be much smaller than z.

5.4

energy and hydrauliC grade lines

253

IM PORTANT POIN T S • The Bernoulli equation can be expressed in terms of the total head H of fluid. This head is measured in units of length and remains constant along a streamline provided no friction losses occur, and no energy is added to the fluid or withdrawn from it due to external sources. H = V 2 >2g + p>g + z = const.

• A plot of the total head H along a pipe or channel is called the energy grade line (EGL). Provided there are no pumps and turbines and friction losses are not considered, then this line will always be horizontal, and its value can be calculated from any point along the flow where p, V, and z are known.

• The hydraulic grade line (HGL) is a plot of the hydraulic head, p>g + z, along the pipe. If the EGL is known, then the HGL will always be V 2 >2g below the EGL.

This cropland is being watered using these siphon pipes. Throughout the length of each pipe the flow is essentially constant, and so the hydraulic grade line remains constant. As a result, the pressure within the pipe decreases as the elevation increases, and vice versa. (© Jim Parkin/Alamy Stock Photo)

5

254

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EXAMPLE

5

Work

and

energy

of

Moving fluids

5.7 Water flows through the 100-mm-diameter pipe at 0.025 m3 >s, Fig. 5–19a. If the pressure at A is 225 kPa, determine the pressure at C, and draw the energy and hydraulic grade lines from A to D. Neglect friction losses and take gw = 9.81 kN>m3. 2m C

100 mm A B

D 100 mm

30° 4m

Datum

2m (a)

Fig. 5–19

SOLUTION Fluid Description. We have steady flow, and we will assume water is an ideal fluid. 0.025 m3 >s = V 3 p 1 0.05 m 2 2 4 10 V = m>s p Since the pipe has a constant diameter over its length, this velocity must remain constant in order to satisfy the continuity equation. The pressure at A and B is the same, since segment AB is horizontal and we have neglected the effects of viscous friction. We can find the pressure at C (and D) by applying the Bernoulli equation at points B and C, which lie on the same streamline. With the gravitational datum through AB, noting that VB = VC = V, we have Bernoulli Equation. The average velocity of flow through the pipe is Q = VA;

pB pC VC 2 VB 2 + + zB = + + zC gw g 2g 2g 2 10 a m>s b 3 2 225(10 ) N>m p + + 0 = 3 3 9.81(10 ) N>m 2(9.81 m>s2) 2 10 a m>s b p pC + + (4 m) sin 30° 3 3 9.81(10 ) N>m 2(9.81 m>s2) pC = pD = 205.38(103) Pa = 205 kPa

Ans.

The pressure at B (225 kPa) is higher than the pressure at C (205 kPa) because the pressure at B has to do work to lift the water to C.

5.4

Head (m)

V2 5 0.516 m 2g

23.452 m 22.936 m

A

B

Elevation head D

C

5 Length (m)

(b)

Fig. 5–19 (cont.)

EGL and HGL. The total head remains constant, because there are no friction losses. This head can be determined from the conditions at any point along the pipe. Using point B, we have

H =

a

2 10 m>s b p

225(10 ) N>m pB + + + zB = + 0 g 2g 9.81(103) N>m3 2 1 9.81 m>s2 2 3

V B2

2

= 23.452 m = 23.5 m

The EGL is located as shown in Fig. 5–19b. The location of the HGL along AB is 225(103) N>m2 pB + zB = + 0 = 22.936 m = 22.9 m g 9.81(103) N>m3 or along CD it is 205.38(103) N>m2 pC + zC = + (4 m) sin 30° = 22.936 m = 22.9 m g 9.81(103) N>m3 Along BC the elevation head will increase, and the pressure head will correspondingly decrease 1pC >g 6 pB >g2. The velocity throughout the pipe is constant, and so the velocity head is always a

2 10 m>s b p

255

EGL HGL

pC g zC

pB g

energy and hydrauliC grade lines

V2 = = 0.516 m 2g 2 1 9.81 m>s2 2

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EXAMPLE

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Moving fluids

5.8 Water flows out of the large tank and through the pipeline shown in Fig. 5–20. Draw the energy and hydraulic grade lines for the pipe if friction losses are neglected.

A

EGL

V 2B9 2g 5 0.5625 m 5m 200 mm B9 8.44 m

B

V 2E 2g 5 9 m

C 100 mm

4m Datum

z

100 mm D E

HGL

Fig. 5–20

SOLUTION Fluid Description. We will assume the water level in the tank remains essentially constant, so that VA = 0 and steady flow will be maintained. The water is assumed to be an ideal fluid. Energy Grade Line. We will establish the datum through DE. At A the velocity and pressure heads are both zero, and so the total head is equal to the elevation head, which is at a level of H =

pA VA 2 + + z = 0 + 0 + 14 m + 5 m2 = 9 m g 2g

The EGL remains at this level since the fluid is ideal, and so there are no energy losses due to friction. Notice that if the Bernoulli equation is applied at points A and B, then VB = 0 because we have assumed VA = 0. This result, although not quite correct, requires the fluid to accelerate only within the pipe, from rest at B to a velocity VB′ at B′. Here we will assume these two points are rather close together, although actually they are somewhat separated. See Sec. 9.6.

5.4

energy and hydrauliC grade lines

Hydraulic Grade Line. Since the (gage) pressure at both A and E is zero, the velocity of the water exiting the pipe at E can be determined by applying the Bernoulli equation on a streamline that contains these two points. 2

2

pA pE VA VE + + zA = + + zE g g 2g 2g 0 + 0 + 9m = 0 +

VE 2

219.81 m>s2 2

+ 0

VE = 13.29 m>s

The velocity of the water through pipe section B′C can now be determined from the continuity equation, considering the fixed control volume to contain the water within the entire pipe. We have 0 r dV + rVf>cs # dA = 0 0 t Lcv Lcs 0 - VB′AB′ + VEAE = 0

-VB′ 3 p10.1 m2 2 4 + 113.29 m>s2 3 p10.05 m2 2 4 = 0 VB′ = 3.322 m>s

The HGL can now be established. It is located below the EGL, a distance defined by the velocity head V 2 >2g. For pipe segment B′C this head is 13.322 m>s2 2 VB′2 = = 0.5625 m 2g 219.81 m>s2 2

The HGL is maintained at 9 m - 0.5625 m = 8.44 m until the transition at C changes the velocity head along CDE to 113.29 m>s2 2 VE 2 = = 9m 2g 219.81 m>s2 2

This causes the HGL to drop to 9 m - 9 m = 0. Along CD, z is always positive, Fig. 5–20, and therefore, a corresponding negative pressure head -p>g must be developed within the flow to maintain a zero hydraulic head, i.e., p>g + z = 0. If this negative pressure becomes large enough, it can cause cavitation, something we will discuss in the next example. Finally, along DE both p>g and z are zero.

257

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EXAMPLE

Work

and

Moving fluids

The siphon in Fig. 5–21a is used to draw water from the large open tank. If the absolute vapor pressure for the water is pv = 1.23 kPa, determine the shortest drop length L of the 50-mm-diameter tube that will cause cavitation in the tube. Draw the energy and hydraulic grade lines for the length of the tube.

B

SOLUTION

0.5 m A

of

5.9

5

0.2 m

energy

Fluid Description. As in the previous example, we will assume water to be an ideal fluid, and its level in the tank to remain essentially fixed, so that we have steady flow. gw = 9810 N>m3.

D

L 5 9.90 m

Bernoulli Equation. To obtain the velocity at C, we will apply the Bernoulli equation at points A and C. With the datum at C, we have

Datum C

(a)

Fig. 5–21

pA pC VC 2 VA 2 + + zA = + + zC gw gw 2g 2g

0 + 0 + 1L - 0.2 m2 = 0 +

VC 2

219.81 m>s2 2

+ 0

VC = 4.42921L - 0.2 m2

(1)

This result is valid provided the pressure at any point within the tube does not drop to or below the vapor pressure. If this were to happen, the water would boil (cavitate), causing a “hissing” noise and energy loss. This, of course, would invalidate the application of the Bernoulli equation. Since the flow is assumed to be steady, and the tube has a constant diameter, then due to continuity, V 2 >2g is constant throughout the tube, and so the hydraulic head 1p>g + z2 must also be constant. Using standard atmospheric pressure of 101.3 kPa, the gage vapor pressure for the water is 1.23 kPa - 101.3 kPa = -100.07 kPa. To find the shortest drop length L, we will assume this negative pressure develops at B, where z, measured from the datum, is a maximum. Applying the Bernoulli equation at points B and C, realizing that VB = VC = V, we have pB pC VC 2 VB 2 + + zB = + + zC gw gw 2g 2g

-100.071103 2 N>m2

V2 V2 + 1L + 0.3 m2 = 0 + + 0 2g 2g 9810 N>m3 L + 0.3 m = 10.20 m L = 9.90 m Ans. From Eq. 1 the critical velocity is +

VC = 4.429219.90 m - 0.2 m2 = 13.80 m>s

5.4

energy and hydrauliC grade lines

If L is equal to or greater than 9.90 m, cavitation will occur in the siphon at B, because the pressure at B will then be equal to or lower than -100.07 kPa. Notice that we can also obtain these results by applying the Bernoulli equation, between A and B to obtain VB, and between B and C to obtain L. EGL and HGL. Throughout the tube, the velocity head is 113.80 m>s2 2 V2 = = 9.70 m 2g 219.81 m>s2 2

The total head can be calculated from C. It is pC VC 2 H = + + zC = 9.70 m + 0 + 0 = 9.70 m g 2g Both the EGL and the HGL for segment DBC are shown in Fig. 5–21b. Here the HGL remains at zero. We can determine the pressure head within the tube at D by applying the Bernoulli equation between D and C. pD pC VC VD2 + + + zD = + zC g g 2g 2g pD + 9.70 m + 19.90 m - 0.2 m2 = 0 + 9.70 m + 0 g pD = -9.70 m g Therefore, the pressure head decreases from –9.70 m at D to p>g = [ -100.071103 2 N>m2]>9810 N>m3 = -10.2 m at B, while the elevation head z increases from 9.70 m to 9.70 m + 0.5 m = 10.2 m, Fig. 5–21b. After rounding the top of the pipe at B, the pressure head increases, while the elevation head decreases by a corresponding amount. Head (m) 10.2 m Elevation head z 9.70 m

A D

29.70 m

EGL

C

B

p Pressure head g

210.2 m (b)

Fig. 5–21 (cont.)

Flow through the tube V2 5 9.70 m is at constant speed 2g HGL Length (m) Datum

259

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Moving fluids

dA ds Flow work out p

5

Turbine work out

dWturb dWpump

z out dQ

Pump work in z in

Datum

Heat energy in

Fig. 5–22

5.5

THE ENERGY EQUATION

In this section, we will expand our application of work and energy methods beyond the limitation of the Bernoulli equation, and will include the effects of heat and viscous fluid flow, along with work input from a pump and work output to a turbine. To do this we will consider an Eulerian approach and follow a system of fluid particles as they pass through a control volume such as the one in Fig. 5–22. Before we begin, however, we will first discuss the various forms of energy that a fluid system can have.

System Energy. At any instant, the total energy E of the fluid system consists of three parts: Kinetic energy. This is energy of motion that depends upon the macroscopic speed of the particles, as measured from an inertial reference frame. Gravitational potential energy. This is energy due to the elevation of the particles, measured from a selected datum. Internal energy. Internal energy refers to the vibrational or microscopic motion of the atoms and molecules that compose the fluid system. It also includes any stored potential energy within the atoms and molecules that cause the binding of the particles due to nuclear or electrical forces. The total of these three energies, E, is an extensive property of the system, since it depends upon the amount of mass within the system. However, it can be expressed as an intensive property e by dividing E by the mass. In this case, the above three energies are then expressed as energy per unit mass, and so for the system, we have 1 e = V 2 + gz + u (5–8) 2

5.5

the energy equation

261

Let us now consider the various forms of heat energy and work done by the fluid system in Fig. 5–22.

Heat Energy. Heat energy dQ can be added or drawn through a control surface by conduction, convection, or radiation. It increases the total energy of the system within the control volume if it flows in (system heated), and decreases the total energy if it flows out (system cooled). Work. Work dW can be done by the enclosed system on its surroundings through a control surface. Work decreases the total energy of the system when it is done by the system, and it increases the total energy of the system when it is done on the system. In fluid mechanics we will be interested in three types of work. Flow Work. When a fluid system is under pressure, it can push a volume dV of the system’s mass outward from an open control surface. This is flow work, dWp. To calculate it, consider the (gage) pressure p within the system at the outlet pushing on a small volume of mass in the system in Fig. 5–22. Since dA is the cross-sectional area of this volume, then the force exerted by the system is dF = p dA. If the volume extends outward a distance ds, then the flow work for this small volume, dV = dA ds, is dWp = dF ds = p1dA ds2 = p dV. Shaft Work. If work is performed on a turbine by the fluid system, then the work will subtract energy from the system, Fig. 5–22. However, it is also possible for work to be performed on the system by a pump, thereby adding external energy to the fluid. In both cases, the work is called shaft work, because a shaft is used to supply or extract the work. Shear Work. The viscosity of any real fluid will cause shear stress t to develop tangent to the control volume’s inner surface. Because of the no-slip condition on a fixed control surface, no work can be done by the shear stress since it does not move along the surface. Also, we will always select the open control surfaces so that they are perpendicular to the flow of the fluid coming into and passing out of the control volume. When this is the case, the viscous shear stress will do no work.*

*Normal viscous stress can occur if the flow through an open control surface is nonuniform; however, any work done by this stress will be zero provided the streamlines are all parallel. We will consider this to be the case in this book. Also, see the footnote on p. 409.

5

262

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Energy Equation. Now that we have considered the various forms of work and energy for a fluid system, we can apply the conservation of energy for the system. This is formalized as the first law of thermodynamics and represents the conservation of energy for the system. It states that # the time rate at which heat is added or transferred into the system, Qin, plus the rate of work added to the system is equal to the time rate of change of the total energy within the system. # # dE Qin + Win = dt

(5–9)

For an Eulerian description, the term on the right can be converted to the rate of change of energy relative to the control volume using the Reynolds transport theorem, Eq. 4–11, where h = e. We have # # Qin + Win =

0 er d V + erVf>cs # dA 0 t Lcv Lcs

The first term on the right indicates the local rate of change of energy per unit mass within the control volume, and the second term is the net convective energy per unit mass passing through the open control surfaces. If we assume the flow is steady, then this first term will be equal to zero because the time derivative is zero. Substituting Eq. 5–8 for e into the last term yields # # Qin + Win = 0 +

1 a V 2 + gz + u b rVf>cs # dA Lcs 2

(5–10)

# The time rate of input work, Win, can be represented by the rates of flow work and shaft work. As stated previously, the flow work is caused by pressure, where dWp = p1dA ds2, and thus the rate of flow work is out (negative) through a control surface. dWp # ds Wp = = p a dA b = - pVf>cs # dA dt dt Lcs Lcs A pump will supply shaft work (positive), dWpump, and a turbine will draw shaft work (negative), dWturbine. The negative sign indicates that the system does this work. Therefore, the time rate of total work (power) transferred into the system can be expressed as # Win = -

Lcs

# # pVf>cs # dA + Wpump - Wturbine

5.5

the energy equation

263

Substituting this result into Eq. 5–10 and rearranging the terms gives # # # Qin + Wpump - Wturbine =

c

p 1 + V 2 + gz + u d rVf>cs # dA (5–11) r 2 Lcs 5

The integration must be carried out over the outlet (out) and inlet (in) control surfaces. For our case, we will assume the flow is uniform and one-dimensional, and so average velocities will be used.* Also, we will assume that the pressure p and elevation z are constant over any opening, Fig. 5–22. Finally, the conservation of mass requires the mass flow through the inlet to be equal to the mass flow through the outlet, # so m = rinVin Ain = routVout Aout. Therefore, after integrating over each control surface, we have # # # Qin + Wpump - Wturbine = ca

pout Vout2 pin Vin2 # + + gzout + uout b - a + + gzin + uin b d m rout rin 2 2

(5–12)

This is the energy equation for one-dimensional steady flow, and it applies to both compressible and incompressible fluids.

Incompressible

Flow. If we further assume the flow is # incompressible, then rin = rout = r. Then if Eq. 5–12 is divided by m, and the terms are rearranged, we obtain 2 pin pout V out V in2 + + gzin + wpump = + + gzout + wturbine + 1uout - uin - qin 2 r r 2 2

Here each term represents energy or work per unit mass, J>kg. Specifically, wpump and wturbine are the shaft work per unit mass, performed by the pump and turbine, respectively; and the term qin is the heat energy per unit mass that is transferred into the system that is contained within the control volume.

*Actually, the flow through pumps and turbines is cyclic as the fluid passes through the machine. However, the cycles are very fast, and provided the time considered for observing the flow is larger than that of a single cycle, then the time average of a quantity of flow through an open control surface a short distance away from the inlet and outlet of a pump or turbine can be considered as quasi-steady flow.

264

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The viscous effects of the fluid will produce turbulence or eddies within the fluid, and this will increase the fluid’s internal or thermal energy per unit mass, i.e., 1uout - uin 2 7 0. This friction loss is normally much greater than the heat gain per unit mass, qin, from any external source. As a result, we will express the collective terms 1uout - uin - qin 2 as a friction loss and represent it as fl. Therefore, for our purposes, a general expression of the energy equation becomes pin pout V out2 V in2 + + gzin + wpump = + + gzout + wturbine + fl (5–13) r r 2 2 Here the total available energy per unit mass passing through the inlet control surfaces, plus the work per unit mass that is added to the fluid within the control volume by a pump, is equal to the total energy per unit mass that passes through the outlet control surfaces, plus the energy removed from the fluid within the control volume by a turbine, plus the energy losses that occur within the control volume. If Eq. 5–13 is divided by g, then the terms represent energy per unit weight or “head of fluid.” Occurs within control volume

pin pout Vout2 Vin2 + + + zin + hpump = + zout + hturbine + hL g g 2g 2g

(5–14)

Occurs at open control surfaces

The last term is called the head loss, hL = fl>g, and the terms hpump and hturbine are referred to as the pump head and turbine head, respectively. Therefore, this form of the energy equation requires that the total input head plus the pump head equals the total output head plus the turbine head plus the head loss. Although Eq. 5–14 will reduce to the same form as the Bernoulli equation if there is no shaft work or no change in the fluid’s internal energy, realize that there is a striking difference in the derivation of these two equations. The Bernoulli equation was derived using a Lagrangian method of analysis. It is an integrated form of Newton’s second law of motion applied to a single fluid particle moving with steady flow from one point to another along the same streamline. In other words, it is a statement of the principle of work and energy as it applies to the fluid particle. By comparison, the energy equation applies to a finite volume of fluid. It was derived from the first law of thermodynamics (conservation of energy) using an Eulerian or control volume approach. It accounts for the changes in energy within the control volume and the energy that is transported across its control surfaces.

5.5

Compressible Fluid. For compressible gas flow, it is often convenient to express the energy of the gas as the sum of its internal energy u and the energy created by pressure.This combination of fluid energy is called enthalpy, h. Pressure can cause fluid movement, and since the flow energy or flow work for a volume of fluid is p dV, then for a unit mass of fluid, p dV>dm = p>r. Therefore, h = p>r + u

(5–15)

If we substitute this into Eq. 5–12, we get # # # Vout2 Vin2 # Qin + Wpump - Wturbine = ca hout + + gzout b - a hin + + gzinbd m 2 2 (5–16) Application of this equation will become important in Chapter 13, where we discuss compressible flow.

Power and Efficiency. The# power of a turbine or a pump is defined as its time rate of doing work, W = dW>dt. In the SI system, power is measured in watts 1W = J>s2 and sometimes, it is also measured in horsepower, where 1 hp = 746 W. We can express the power in terms of the pump head or turbine head, where either is referred to as the shaft head, hs, by noting that from the derivation of Eq.# 5–14, hs = ws# >g. And since the shaft work per unit # # mass is ws = Ws >m , then Ws = mghs . Also, since the mass flow # m = rQ = gQ>g, then the power imparted to the fluid by a pump or extracted from it by a turbine is # # Ws = mghs = Qghs

(5–17)

Since pumps (and turbines) have friction losses, they will never be 100% efficient, and so we define their efficiency by a power ratio. For example, for pumps, mechanical # efficiency e is the ratio of mechanical power delivered to the fluid, (Ws)out, to the# electrical power supplied to the shaft that is required to run the pump, (Ws)in. Thus, # 1Ws 2 out e = # 1Ws 2 in

0 6 e 6 1

(5–18)

the energy equation

265

5

266

Chapter 5

Work

and

energy

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Moving fluids

Nonuniform Velocity. If a velocity profile for the flow at any inlet and outlet control surface is nonuniform, as it is in all cases of viscous flow, then the velocity profile must be known in order to carry out the integration in Eq. 5–11. One way of representing this integration is to use a dimensionless kinetic energy coefficient a, and express the integration of the velocity profile in terms of the average velocity V of the profile, as determined from Eq. 4–3, that is, V = 1 1 v dA2 >A. In other words, the # velocity term in Eq. 5–11 can be written as 1cs 12V 2rVf>cs # dA = a 21V 2m, so that

5

1 a = # 2 V 2rVf>cs # dA mV Lcs

Velocity profile for laminar flow (a)

Mean velocity profile for turbulent flow (b)

Fig. 5–23

(5–19)

Therefore, in cases where it may be necessary to consider the nonuniformity of the velocity at a control surface, we can substitute the terms involving V 2 in the energy equation with aV 2. For example, it will be shown in Chapter 9 that for laminar flow through a pipe the velocity profile will be a paraboloid, Fig. 5–23a, and so for this case, integration will yield a = 2.* However, in practice almost all flows are turbulent, and because of this it is sufficient to take a = 1 since turbulent mixing of the fluid will cause the velocity profile to become approximately uniform, Fig. 5–23b.

I MPO RTA N T PO I N T S • The energy equation is based upon the first law of thermodynamics, which states that the time rate of change of the total energy within a fluid system is equal to the rate at which heat is added to the system plus the rate at which work is done on the system.

• In general, the system’s total energy E within a control volume consists of the kinetic energy of all the fluid particles, their potential energy, and their atomic or molecular internal energies.

• The work done on or by a system can be flow work due to pressure, shaft work due to a pump or turbine, or shear work caused by viscous friction. Shear work is not considered here, since the flow will always be perpendicular to any open control surface.

*See Prob. 5–90.

5.5

the energy equation

267

P ROCEDUR E FOR A N A LY S I S The following procedure can be used when applying the various forms of the energy equation.

• Fluid Description. As developed here, the energy equation applies to one-dimensional steady flow of either compressible or incompressible fluids.

• Control Volume. Select the control volume that contains the fluid, and indicate the open control surfaces. Be sure that these surfaces are located in regions where the flow is uniform and well defined.

• Energy Equation. Write the energy equation, and below it substitute in the numerical data using a consistent set of units. Heat energy dQin is positive if heat flows into the control volume, and it is negative if heat flows out. Establish a fixed datum to measure the elevation (potential energy) of the fluid moving into and out of each control surface. If the fluid is assumed inviscid or ideal, then its velocity profile is uniform as it passes through an open control surface. Also, this uniform or average velocity profile can be assumed for turbulent flow of a viscous fluid. If the velocity profile is known, then the coefficient a can be determined using Eq. 5–19, and aV 2 is used instead of V 2. The average velocities “in” and “out” of the open control surfaces can be determined if the volumetric flow is known, Q = VA. Reservoirs, or large tanks that drain slowly, have liquid surfaces that are essentially at rest, V ≈ 0.

The terms 1p>g + z2 in Eq. 5–14 represent the hydraulic head at the “in” and “out” control surfaces. This head remains constant over each control surface if the streamlines are all parallel, and so it can be calculated at any point on this surface. Once this point is selected, its elevation z is positive if it is above a datum and negative if it is below a datum.

• If more than one unknown is to be determined, consider relating the velocities using the continuity equation or relating the pressures using the manometer equation if it applies.

5

268

Chapter 5

EXAMPLE

Work

and

energy

of

Moving fluids

5.10

A

5 Penstock Generator

60 m

0.3 m Datum

Fig. 5–24

Turbine B

The turbine in Fig. 5–24 is used in a small hydroelectric plant. If the discharge through the 0.3-m-diameter draft tube at B is 1.7 m3 >s, determine the amount of power that is transferred from the water to the turbine blades. The frictional head loss through the penstock, turbine, and draft tube is 4 m. SOLUTION Fluid Description. This is a case of steady flow. Here viscous friction losses occur within the fluid. We consider the water to be incompressible, where g = 9810 N>m3. Control Volume. A portion of the reservoir, along with water within the penstock, turbine, and draft tube, is selected to be the fixed control volume. The average velocity at B can be determined from the discharge. Q = V BA B;

1.7 m3 >s = VB 3p10.15 m2 2 4 VB = 24.05 m>s

Energy Equation. Applying the energy equation between A (in) and B (out), with the gravitational datum set at B, we have pA pB V A2 V B2 + + + zA + hpump = + zB + hturbine + hL g g 2g 2g 0 + 0 + 60 m + 0 = 0 +

124.05 m>s2 2

219.81 m>s2 2

+ 0 + hturbine + 4 m

hturbine = 26.52 m

As expected, the result is positive, indicating that energy is supplied by the water (system) to the turbine. Power. Using Eq. 5–17, the power transferred to the turbine is therefore # Ws = Qghs = 11.7 m3 >s219810 N>m3 2126.52 m2 = 442 kW

By comparison, the power lost due to the effects of friction is # WL = QghL = 11.7 m3 >s219810 N>m3 214 m2 = 66.7 kW

Ans.

5.5

EXAMPLE

5.11

The pump in Fig. 5–25 discharges water at 80 (103) liters>h. The pressure at A is 150 kPa, whereas the pressure in the pipe at B is 500 kPa. As the water passes through the filter, it causes the internal energy of the water to increase by 50 J>kg due to the effect of friction, while at the same time there is a heat conduction loss from the water of 250 J>s. Determine the horsepower that is developed by the pump. SOLUTION Fluid Description. We have steady flow into and out of the pump. The water is considered incompressible, but viscous friction losses occur. r = 1000 kg>m3. Control Volume. The fixed control volume contains the water within the pump, filter, and pipe extensions. Since there is 1 liter in 1000 cm3 of water, the volumetric and mass flows are Q = c 80(103)

and

liter 1000 cm3 1m 3 1h da ba b a b = 0.02222 m3 >s h 1 liter 100 cm 3600 s

# m = rQ = (1000 kg>m3)(0.02222 m3 >s) = 22.22 kg>s

Therefore, the velocities at A (in) and B (out) are

Q = VAAA; 0.02222 m3 >s = VA[p(0.075 m)2]; VA = 1.258 m>s Q = VBAB; 0.02222 m3 >s = VB[p(0.025 m)2]; VB = 11.32 m>s

# Energy Equation. Since there is heat conduction loss, Qin is negative; that is, the heat flows out. Also, there is no elevation change in the flow from A to B. For this problem, we will apply Eq. 5–12. # # # pB VB 2 Qin + Wpump - Wturbine = c a + gzB + uB b + r 2 - a

# - 250 J>s + Wpump - 0 = ca - a

269

the energy equation

500(103) N>m2 1000 kg>m3

150(103) N>m2 1000 kg>m3

+

+

(11.32 m>s)2 2

(1.258 m>s)2 2

pA VA2 # + gzA + uA b d m + r 2

+ gz + 50 J>kgb

+ gz + 0b d (22.22 kg>s)

# Wpump = 10.54(103) W = 10.5 kW

Ans.

The positive result indicates that indeed energy is added to the water within the control volume using the pump.

5

150 mm A

50 mm B

Fig. 5–25

Datum

270

Chapter 5

EXAMPLE

5

Work

and

energy

of

Moving fluids

5.12 Water is flowing with an average velocity of 6 m>s when it comes down the spillway of a dam as shown in Fig. 5–26. Within a short distance a hydraulic jump occurs, which causes the water to transition from a depth of 0.8 m to 2.06 m. Determine the head loss caused by the turbulence within the jump. The spillway has a constant width of 2 m.

D Vout 1m 6 ms A 0.5 m

F 1.06 m

0.3 m C

Datum E

B

Fig. 5–26

SOLUTION Fluid Behavior. Steady flow occurs before and after the jump. The water is assumed to be incompressible. Control Volume. Here we will consider a fixed control volume that contains the water within the jump and a short distance on either side of it, Fig. 5–26. Steady flow passes through the open control surfaces because these surfaces are removed from the regions within the jump where the flow is not well defined. Continuity Equation. Since the cross sections of AB and DE are known, we can determine the average velocity out of DE using the continuity equation. 0 r dV + rVf>cs # dA = 0 0t Lcv Lcs 0 - 11000 kg>m32 16 m>s210.8 m212 m2 + 11000 kg>m3 2Vout 12.06 m212 m2 = 0 Vout = 2.3301 m>s

5.5

Energy Equation. We will place the datum at the bottom of the control volume, Fig. 5–26. From this we can determine the hydraulic head 1p>g + z2 at each open control surface. For example, if we select points A and D, then since pin = pout = 0, we get pin + zin = 0 + 0.8 m = 0.8 m g pout + zout = 0 + 2.06 m = 2.06 m g If instead we take points B and E, then since p = gh, we again have g10.8 m2 pin + zin = + 0 = 0.8 m g g g12.06 m2 pout + zout = + 0 = 2.06 m g g Finally, if we use intermediate points C and F, then once again g10.5 m2 pin + zin = + 0.3 m = 0.8 m g g g11 m2 pout + zout = + 1.06 m = 2.06 m g g In all cases we obtain the same results, regardless of which pair of points on the control surface we choose. For application, we will choose points A and D. Since no shaft work is done, we have pin pout V out2 V in2 + + + zin + hpump = + zout + hturbine + hL g g 2g 2g

0 +

16 m>s2 2

219.81 m>s2 2

+ 0.8 m + 0 = 0 +

12.3301 m>s2 2 219.81 m>s2 2

hL = 0.298 m

+ 2.06 m + 0 + hL

Ans.

This head loss is due to turbulence and frictional heating within the jump.

the energy equation

271

5

272

Chapter 5

EXAMPLE

Work

and

energy

of

Moving fluids

5.13 D

The irrigation pump in Fig. 5–27a is used to supply water to the pond at B at a rate of 0.09 m3 >s. If the pipe is 150 mm in diameter, determine the required power of the pump. Assume the frictional head loss per meter length of pipe is 0.1 m>m. Draw the energy and hydraulic grade lines for this system.

B

5

4m

SOLUTION

A' C

A

Datum

3.5 m (a)

Fluid Description. Here we have steady flow. The water is assumed to be incompressible, but viscous friction losses occur. g = 9.81(103) N>m3. Control Volume. We will select a fixed control volume that contains the water within the reservoir A, along with that in the pipe and pump. For this case, the velocity at A is essentially zero, and the pressure at the inlet and outlet surfaces A and B is zero. Since the volumetric flow is known, the average velocity at the outlet is

Fig. 5–27

Q = V BA B;

0.09 m3 >s = VB 3p 1 0.075 m 2 2 4 VB =

16 m>s p

Energy Equation. Establishing the gravitational datum at A, and applying the energy equation between A (in) and B (out), we have pA pB VA 2 VB 2 + + + zA + hpump = + zB + hturbine + hL g g 2g 2g 2 16 a m>s b p 0 + 0 + 0 + hpump = 0 + + 4 m + 0 + (0.1 m>m)(7.5 m) 2 1 9.81 m>s2 2 hpump = 6.072 m

This positive result indicates the pump head of water is transferred into the system by the pump. Power. Using Eq. 5–17, the pump must therefore have a power of # Ws = Qgwhpump = (0.09 m3 >s)[9.81(103) N>m3](6.072 m)

= 5.361(103) W = 5.36 kW Ans. Of this amount, the power needed to compensate for the frictional head loss is # WL = QgwhL = (0.09 m3 >s)[9.81(103) N>m3][(0.1 m>m)(7.5 m)] = 662.18 W

5.5

273

the energy equation

EGL and HGL. Recall the EGL is a plot of the total head H = p>g + V 2 >2g + z along the pipe. The HGL lies V 2 >2g below and parallel to the EGL. Here the velocity head is a

5

2 16 m>s b p

5.32 m

V2 = = 1.322 m 2g 2 1 9.81 m>s2 2

4m

It remains constant along the pipe, since the pipe has the same diameter throughout its length. Point A′ in Fig. 5–27 is where the water reaches its velocity V, having been accelerated from rest at the entrance of the pipe. At A′, C, and D, the pressure head is pA′ V A2′ pA V A2 + + + zA + hpump = + zA′ + hturb + hL g g 2g 2g pA′ A 0 + 0 + 0 + 0 = + 1.322 m + 0 + 0 + 0 g pA′ = -1.322 m 21.32 m g A9

Energy added by pump

D B

C 20.35 m

Friction loss along pipe

HGL

20.75 m

21.67 m

pA pC V C2 V A2 + + + zA + hpump = + zC + hturb + hL g g 2g 2g pC 0 + 0 + 0 + 0 = + 1.322 m + 0 + 0 + (0.1 m/m)(3.5 m) g pC = -1.672 m g pA pD V A2 V D2 + + + zA + hpump = + zD + hturb + hL g g 2g 2g pD 0 + 0 + 0 + 0 = + 1.322 m + 4 m + 0 + (0.1 m/m)(7.5 m) g pD = -6.072 m g The signs indicate a negative pressure caused by the suction of the pump. The total head at A′, C, D, and B is therefore H = HA′ HC HD HB

= = = =

p V2 + + z g 2g -1.322 m + 1.322 m + 0 = 0 -1.672 m + 1.322 m + 0 = -0.35 m -6.072 m + 1.322 m + 4 m = -0.75 m 0 + 1.322 m + 4 m = 5.32 m

Stretching the pipe out, these values are plotted in Fig. 5–27b to produce the EGL. The HGL lies 1.322 m below and parallel to the EGL.

Datum

EGL

22.07 m

(b)

Fig. 5–27 (cont.)

274

Chapter 5

Work

and

energy

of

Moving fluids

References 1. D. Ghista, Applied Biomedical Engineering Mechanics, CRC Press,

Boca Raton, FL, 2009. 5

2. A. Alexandrou, Principles of Fluid Mechanics, Prentice Hall, Upper

Saddle River, NJ, 2001. 3. I. H. Shames, Mechanics of Fluids, McGraw Hill, New York, NY,

1962. 4. L. D. Landau, E. M. Lifshitz, Fluid Mechanics, Pergamon Press,

Addison-Wesley Pub., Reading, MA, 1959.

F UNDAMEN TAL PRO B L EM S SEC. 5.2–5.3 F5–1. Water flows through the pipe at A at 6 m>s. Determine the pressure at A and the velocity of the water as it exits the pipe at B.

F5–3. The fountain is to be designed so that water is ejected from the nozzle and reaches a maximum elevation of 2 m. Determine the required water pressure in the pipe at A, a short distance AB from the nozzle exit.

6 ms

A 3m

C

B 2m

Prob. F5–1 F5–2. Oil is subjected to a pressure of 300 kPa at A, where its velocity is 7 m>s. Determine its velocity and the pressure at B. Take ro = 940 kg>m3.

10 mm B

120 mm

7 ms

80 mm

50 mm A B

A

Prob. F5–2

Prob. F5–3

275

fundaMental proBleMs F5–4. Water flows through the pipe at 8 m>s. Determine the pressure at C if the pressure at A is 80 kPa.

SEC. 5.4 F5–7. Water flows from the reservoir through the 100-mm-diameter pipe. Determine the discharge at B. Draw the energy grade line and the hydraulic grade line for the flow from A to B.

0.2 m 8 ms A 0.3 m

C A

Prob. F5–4 6m

F5–5. The tank has a square base and is filled with water to the depth of y = 0.4 m. If the 20-mm-diameter drain pipe is opened, determine the initial volumetric flow of the water and the volumetric flow when y = 0.2 m.

B

VB

Prob. F5–7 A

y

F5–8. Crude oil flows through the 50-mm-diameter pipe such that at A its velocity is 4 m>s and the pressure is 300 kPa. Determine the pressure of the oil at B. Draw the energy grade line and the hydraulic grade line for the flow from A to B.

B 2m

Prob. F5–5

F5–6. Air at a temperature of 80°C flows through the pipe. At A, the pressure is 20 kPa, and the velocity is 4 m>s. Determine the pressure at B. Assume the air is incompressible. 4 ms

A B

B

A

VB

2m 1.5 m

50 mm 200 mm

Prob. F5–6

Prob. F5–8

5

276

Chapter 5

Work

and

energy

of

Moving fluids

F5–9. Water at A has a pressure of 400 kPa and a velocity of 3 m>s. Determine the pressure and velocity at B and C. Draw the energy grade line and the hydraulic grade line for the flow from A to C.

F5–11. Water is supplied to the pump at a pressure of 80 kPa and a velocity of VA = 2 m>s. If the discharge is required to be 0.02 m3 >s through the 50-mm-diameter pipe, determine the power that the pump must supply to the water to lift it 8 m. The total head loss is 0.75 m.

5

150 mm

3 ms

100 mm

A

B

50 mm B

C

Prob. F5–9 8m

2 ms

SEC. 5.5 F5–10. Water from the reservoir flows through the 150-mlong, 50-mm-diameter pipe into the turbine at B. If the head loss in the pipe is 1.5 m for every 100-m length of pipe, and the water exits the pipe at C with a velocity of 8 m>s, determine the power output of the turbine. The turbine operates with 60% efficiency.

A

Prob. F5–11

F5–12. The jet engine takes in air and fuel having an enthalpy of 600 kJ>kg at 12 m>s. At the exhaust, the enthalpy is 450 kJ>kg and the velocity is 48 m>s. If the mass flow is 2 kg>s, and the rate of heat loss is 1.5 kJ>s, determine the power output of the engine.

A

40 m

B

50 mm C

48 ms 12 ms B A

Prob. F5–10

Prob. F5–12

proBleMs

277

P ROBLEMS SEC. 5.1 5–1. Air at 20°C flows through the horizontal tapered duct. Determine the acceleration of the air if on a streamline the pressure is 101.3 kPa and 15 m away the pressure is 100.6 kPa.

5–5. If the volumetric flow of water through the pipe is 0.05 m3 >s, determine the pressure difference between points A and B. The flow occurs on the horizontal plane. 5 Assume the velocity is constant over the cross section.

5–2. Air at 20° C flows through the horizontal tapered duct. Determine the average change in pressure in 15 m, so that the air has an acceleration of 50 m>s2.

100 mm 15 m

A

B

Probs. 5–1/2

300 mm

5–3. Determine the required average change in pressure if the water flows from A to B with an acceleration of 15 m>s2 along a horizontal streamline. Take rw = 1000 kg>m3.

A

B

Prob. 5–5 0.8 m

Prob. 5–3 *5–4. An ideal fluid having a density r flows with a velocity V through the horizontal pipe bend. Plot the pressure variation within the fluid as a function of the radius r, where ri … r … ro and ro = 2ri. For the calculation, assume the velocity is constant over the cross section.

SEC. 5.2–5.3 5–6. Determine the velocity of water through the pipe if the manometer contains mercury held in the position shown. Take rHg = 13 550 kg>m3.

V

ri

r

V ro

A

B 100 mm 50 mm 50 mm

200 mm

Prob. 5–4

Prob. 5–6

278

5

Chapter 5

Work

and

energy

of

Moving fluids

5–7. A fire hydrant supplies water under a pressure of 250 kPa. If a 100-mm-diameter hose is connected to it at A and the hose extends 40 m to the fire truck inlet at B, determine the pressure of the water as it arrives at B. The friction loss is 1.2 m for every 10 m of hose. The inlet at B is 0.5 m higher than the hydrant outlet.

5–9. Air is drawn into the 200-mm-diameter cylinder through the opening at A. If the piston is moving downward at a constant velocity of 10 m>s, determine the average pressure within the cylinder and the force required to move the piston. Take ra = 1.23 kg>m3. 200 mm

A

B B A

Prob. 5–7 F

Prob. 5–9 *5–8. The level of mercury in the manometer has the reading shown. Determine the velocity of the water flowing from the nozzle at B. Neglect any head losses. Take rHg = 13 550 kg>m3.

50 mm

5–10. A fountain is produced by water that flows up the tube at 0.08 m3 >s, and then radially through two cylindrical plates before exiting to the atmosphere. Determine the velocity and pressure of the water at point A. 5–11. A fountain is produced by water that flows up the tube at 0.08 m3 >s, and then radially through two cylindrical plates before exiting to the atmosphere. Determine the pressure of the water as a function of the radial distance r. Plot the pressure (vertical axis) versus r for 200 mm … r … 400 mm. Give values for increments of ∆r = 50 mm.

B 200 mm 200 mm

5 mm

1.5 m A

A 0.3 m

0.45 m

V

Prob. 5–8

Probs. 5–10/11

279

proBleMs *5–12. The jet airplane is flying at 80 m>s in still air, A, at an altitude of 3 km. Determine the absolute stagnation pressure at the leading edge B of the wing.

5–17. Water flows out of a faucet at A at 6 m>s. Determine the velocity of the water just before it strikes the ground at B.

5–13. The jet airplane is flying in still air, A, at an altitude of 4 km. If the air flows past point C near the wing at 90 m>s, measured relative to the plane, determine the difference in pressure between the air near the leading edge B of the wing and point C.

A

5

6 ms

80 ms A 1.75 m

C B

Probs. 5–12/13 5–14. Drainage under a canal is provided using a 500-mm-diameter drainpipe. Determine the flow through the pipe. Neglect any head losses.

B

Prob. 5–17 A 3m

2m B

5–18. The water in an open channel drainage canal flows with a velocity of VA = 1.5 m > s into the drainpipe that crosses a highway embankment. Determine the volumetric discharge through the pipe. Neglect any head losses.

500 mm

Prob. 5–14 5–15. Water flows through the 30-mm-diameter pipe at 0.002 m3 >s and is ejected from the 10-mm-diameter nozzle at B. Determine the velocity and pressure of the water at point A. *5–16. Water flows through the 30-mm-diameter pipe and is ejected with a velocity of 25 m>s at B from the 10-mmdiameter nozzle. Determine the pressure and the velocity of the water at A.

A

B

900 mm

1m 1.5 m/s

3.5 m B

Prob. 5–18 5–19. Heavy rain has caused reservoir A to reach a height of 3 m above the pipe at B. Determine the flow through the concrete culvert buried beneath the embankment. Neglect any head losses. A

300 mm 3m

500 mm B

Probs. 5–15/16

A

Prob. 5–19

280

Chapter 5

Work

and

energy

of

Moving fluids

*5–20. A fountain ejects water through the two nozzles A and B, which have inner diameters of 10 mm. If the velocity of the flow at point C in the 50-mm-diameter pipe is 2 m>s, determine the pressure in the pipe at this point and the velocity of the water through each nozzle.

5–23. Water from a nozzle tapers from a diameter of 12 mm to 6 mm after falling 600 mm. Determine the mass flow in kg>s.

5 5–21. A fountain ejects water through the two nozzles A and B, which have inner diameters of 10 mm. Determine the velocity of the water stream passing through each of the nozzles and the pressure at point C if the flow through the 50-mm-diameter pipe is 0.005 m3 >s.

12 mm

A

600 mm

B 6 mm B

2m

A 1.2 m

Prob. 5–23 C

50 mm 2 ms

Probs. 5–20/21

*5–24. In order to determine the flow in a rectangular channel, a 75-mm-high bump is added on its bottom surface. If the measured depth of flow at the bump is 1.25 m, determine the volumetric discharge. The flow is uniform, and the channel has a width of 1.5 m.

5–22. Water from a nozzle tapers from a diameter of 12 mm to 6 mm after falling 600 mm. Determine the velocity of the water at A and at B.

B 1.25 m

A 1.5 m

75 mm

Prob. 5–24

12 mm

5–25. Air at a temperature of 40°C flows into the nozzle at 6 m>s and then exits to the atmosphere at B, where the temperature is 0°C. Determine the pressure at A.

A

600 mm

300 mm 100 mm

6 mm B

6 ms B A

Prob. 5–22

Prob. 5–25

281

proBleMs 5–26. Water flows through the transition at 0.3 m3 >s, which causes the water to rise within the piezometer at A to a height of hA = 350 mm. Determine the height hB. 5–27. If the water in piezometers A and B rises to hA = 250 mm and hB = 950 mm, respectively, as it flows through the transition, determine the volumetric flow.

5–30. A river has an average width of 5 m. Just after its flow falls 2 m to the lower elevation, the depth becomes h = 0.8 m. Determine the volumetric discharge. 5–31. A river has an average width of 5 m and flows with an average velocity of 6 m>s at A. Determine its depth h just after the flow falls 2 m.

1.2 m A hB hA 2m

B

A

B

h 250 mm 400 mm

Probs. 5–30/31 Probs. 5–26/27

*5–28. Air at 20°C flows through the circular duct such that the absolute pressure is 100.8 kPa at A, and 101.6 kPa at B. Determine the volumetric discharge through the duct. 5–29. Air at 20°C flows through the circular duct such that the absolute pressure at A is 100.8 kPa and the velocity is 40 m>s. Determine the absolute pressure and the velocity of the air at B.

*5–32. Air enters the circular nozzle with a velocity of 30 m>s and then exits to the atmosphere at B. Determine the pressure at A. The density of air is assumed to be constant at ra = 1.20 kg>m3.

1.2 m

B

A

0.4 m A

250 mm 150 mm

Probs. 5–28/29

B

Prob. 5–32

5

282

Chapter 5

Work

and

energy

of

Moving fluids

5–33. Water flows horizontally past the pitot tube such that mercury within the manometer is displaced as shown. Determine the mass flow if the pipe has a diameter of 100 mm. Take rHg = 13 550 kg>m3. 5

5–35. If the velocity of water changes uniformly along the transition from VA = 10 m>s to VB = 4 m>s, determine the pressure difference over the distance x. *5–36. If the velocity of water changes uniformly along the transition from VA = 10 m>s to VB = 4 m>s, find the pressure difference between A and x = 1.5 m.

2m x A B 50 mm 10 ms

4 ms A

100 mm

75 mm

B

Probs. 5–35/36 Prob. 5–33

5–37. Air is pumped into the top of the tank so that the pressure at B is 40 kPa. Determine the discharge of water through the drainpipe when h = 5 m. 5–34. Water flows through the pipe transition with a velocity of 6 m>s at A. Determine the difference in the level of mercury within the manometer. Take rHg = 13 550 kg>m3.

5–38. Air is pumped into the top of the tank so that the pressure at B is 40 kPa. Determine the discharge of the water as a function of the height h. Plot the discharge (vertical axis) versus h for 1 m … h … 6 m. Give values for increments of ∆h = 1 m.

150 mm 75 mm

B B

A

h A

h

0.15 m

Prob. 5–34

Probs. 5–37/38

283

proBleMs 5–39. A 6-m-high chimney has a circular opening A at its base. If air flows into it at 3.75 m>s, determine the speed of the air as it exits at the top B. Also, what is the pressure difference between A and B? Take ra = 1.20 kg>m3.

5–41. The pipe assembly is mounted vertically. Determine the pressure at A if the velocity of the water ejected from B is 0.75 m>s.

5 200 mm B

600 mm B

1.2 m

3.75 ms 6m

A

200 mm A 80 mm

Prob. 5–41

Prob. 5–39 *5–40. The sewage siphon regulates the level of water in the large holding tank A. Determine the flow through the 100-mm-diameter pipe when the water levels are as shown. Neglect any head losses. A

5–42. When the valve at A is opened, the initial discharge of oil is 0.0125 m3 >s. Determine the depth h of oil in the tank. Take rke = 814 kg>m3 and rg = 726 kg>m3.

1m

C 3m

0.5 m

B

1.5 m

Kerosene

h

B

50 mm

Prob. 5-40

Gasoline

A

Prob. 5–42

284

Chapter 5

Work

and

energy

of

Moving fluids

5–43. If the manometer contains mercury, determine the volumetric flow of air through the circular duct. Take rHg = 13 550 kg>m3 and ra = 1.23 kg>m3.

5–45. Water from the large closed tank is to be drained through the pipes at A and B. When the valve at B is opened, the initial discharge is QB = 0.025 m3 >s. Determine the pressure at C and the initial volumetric discharge at A if this valve is also opened. 5–46. Water from the large closed tank is to be drained through the pipes at A and B. When the valve at A is opened, the initial discharge is QA = 0.055 m3 >s. Determine the pressure at C and the initial volumetric discharge at B when this valve is also opened.

5

V

A

B

C 10 mm

2m

1.5 m A

Prob. 5–43

B

75 mm

50 mm

Probs. 5–45/46 *5–44. One method of producing energy is to use a tapered channel (TAPCHAN), which diverts sea water into a reservoir as shown. As a wave approaches the shore through the closed tapered channel at A, its height will begin to increase until it begins to spill over the sides and into the reservoir. The water in the reservoir then passes through a turbine in the building at C to generate power and is returned to the sea at D. If the speed of the water at A is VA = 2.5 m>s, and the water depth is hA = 3 m, determine the minimum height hB at the back B of the channel to prevent the water from entering the reservoir.

5–47. Air is pumped into the top A of the tank. Determine the pressure at A if the water strikes the ground at C, where d = 4 m. Consider the tank to be a large reservoir. *5–48. Air is pumped into the top A of the tank such that it has a pressure of 10 kPa. Determine the distance d at which the water strikes the ground at C. Consider the tank to be a large reservoir.

A

3m

B B A

C

C

1m

VA 5 2.5 ms D d

Prob. 5–44

Probs. 5–47/48

285

proBleMs 5–49. Water drains from the fountain cup A to cup B. Determine the depth h of the water in B in order for steady flow to be maintained. Take d = 25 mm. 5-50. Water drains from the fountain cup A to cup B. If the depth in cup B is h = 50 mm, determine the velocity of the water at C and the diameter d of the opening at D so that steady flow is maintained.

*5–52. The solution is ejected from the 20-mm-diameter syringe through a 0.4-mm-diameter needle. Determine the velocity of the solution through the needle and the maximum height h max the solution rises in the air, each as a function of the force F applied to the plunger. Assume the pressure within the syringe is constant and the velocity through the needle is much greater than through the syringe. Plot the velocity and maximum 5 height (vertical axis) as a function of the force for 0 … F … 20 N. Give values for increments of ∆F = 5 N. Take r = 1135 kg>m3.

A 100 mm

hmax 50 mm

C 20 mm

B

20 mm

h D

d

Probs. 5–49/50

5–51. The solution is ejected from the 20-mm-diameter syringe through a 0.4-mm-diameter needle. If the pressure developed within the syringe is assumed to be constant at 50 kPa, determine the velocity of the solution through the needle and the maximum height the liquid rises in the air. Take r = 1135 kg>m3. Assume the velocity through the needle is much greater than through the syringe.

F

Prob. 5–52 5–53. Determine the velocity of water at B and C if the pressure of the water in the 150-mm-diameter pipe at A is 150 kPa and the velocity at this point is 4 m>s. The pressure at B is measured to be 20 kPa. 150 mm

hmax 50 mm

A

C 120 mm

20 mm 3m

50 mm

B

F

Prob. 5–51

Prob. 5–53

286

Chapter 5

Work

and

energy

of

Moving fluids

5–54. The 300-mm square base container is to be used to slowly water the tree. If it is filled to the top with water, determine the time for the bucket to empty if there is a 2-mmdiameter hole drilled into its bottom. Assume the velocity through the hole is much greater than at the water surface.

*5–56. If the difference in the level of mercury within the manometer is 80 mm, determine the volumetric flow of the water. Take rHg = 13 550 kg>m3.

5 100 mm 40 mm

B

300 mm

A

80 mm

300 mm 600 mm

Prob. 5–56 Prob. 5–54

5–55. Determine the velocity out of the pipes at A and B if water flows into the Tee at 8 m>s and under a pressure of 40 kPa. The system is in the vertical plane.

5–57. The open cylindrical tank is filled with linseed oil. A crack having a length of 50 mm and average height of 2 mm occurs at the base of the tank. How many liters of oil will slowly drain from the tank in eight hours? Take ro = 940 kg>m3.

4m

0.5 m B 30 mm

5m

8 ms 3m

C 50 mm 3m A 30 mm

Prob. 5–55

Prob. 5–57

287

proBleMs 5–58. As the air flows through the transition from A to B, its absolute pressure drops from 101.8 kPa to 101.3 kPa. If the temperature of the air remains constant at 20°C, determine the mass flow of the air through the duct.

5–62. Water flows in a rectangular channel over the 1-m drop. If the width of the channel is 1.5 m, determine the volumetric flow in the channel.

5–59. Air flows through the transition with a velocity of 8 m>s at A. If the absolute pressure at A is 101.8 kPa, determine the pressure at B. The temperature of the air remains constant at 20°C.

5

A VA

0.5 m

400 mm 200 mm

1m

B 0.2 m

VB

B A

Prob. 5–62

Probs. 5–58/59

*5–60. Determine the velocity of the flow at B and C if the pressure of the water in the 100-mm-diameter pipe at A is 120 kPa and water flows past this point at 3 m>s. 5–61. If the pressure in the 100-mm-diameter pipe at A is 120 kPa, and the velocity of water discharge from B is 12 m>s, determine the pressure change from A to C and velocity of the water in the pipe at C.

5–63. If the pressure at A is 325 kPa, and the velocity of the water at this point is 2.25 m>s, determine the pressure in the pipe at B if the pressure at C is 175 kPa. The flow occurs in the horizontal plane. *5–64. If the pressure at A is 215 kPa, and the velocity of the water at this point is 2.25 m>s, determine the pressure in the pipe at B if the water is discharged into the atmosphere at C.

B

C 5m

20 mm

20 mm

A

C

A 80 mm B 40 mm

100 mm

Probs. 5–60/61

Probs. 5–63/64

288

Chapter 5

Work

and

energy

of

Moving fluids

5–65. Determine the average velocity and the pressure in the pipe at A if the height of the water column in the pitot tube is 450 mm. and the height in the piezometer is 50 mm. 5

5–67. Water is contained within the bowl, which has a surface defined by y = (8r 2) m, where r is in meters. If the 5-mm-diameter drain plug is opened when y = 500 mm, determine the time needed to drain the water to the level of the hole 1y = 200 mm2. y r

200 mm

B

y

450 mm

Prob. 5–67

50 mm B

A C

40 mm 80 mm

Prob. 5–65

*5–68. Determine the discharge of water from the funnel at the instant y = 50 mm. For the solution, consider the volume of water to be a cone. Note: For a cone, V = 31pr 2h. 5–69. Determine the rate at which the surface level of water is dropping in the funnel as a function of the depth y. Assume steady flow. Plot this rate (vertical axis) versus the depth y for 20 mm 6 y 6 120 mm. Give values for increments of 20 mm. Consider the volume of water to be a cone. Note: For a cone, V = 31pr 2h.

5–66. Carbon dioxide at 20°C passes through the expansion chamber, which causes mercury in the manometer to settle as shown. Determine the velocity of the gas at A. Take rHg = 13 550 kg>m3.

50 mm

50 mm

A

200 mm

300 mm

A

y C

B 150 mm

150 mm 308

40 mm

B

10 mm

100 mm

Prob. 5–66

Probs. 5–68/69

289

proBleMs

SEC. 5.4–5.5 5–70. Water at a pressure of 90 kPa and a velocity of 4 m>s at A flows through the transition. Determine the velocity and the pressure at B Draw the energy and hydraulic grade line with reference to the datum set through B.

5–73. When the valve at A is closed, the pressure at A is 230 kPa, and at B it is 180 kPa. When the valve is open, the water flows at 6 m>s, and the pressure at A is 210 kPa and at B it is 145 kPa. Determine the head loss in the pipe between A and B.

5–71. Water at a pressure of 90 kPa and a velocity of 4 m>s at A flows through the transition. Plot the pressure head and the elevation head from A to B with reference to the datum set through B.

5

B

100 mm A

C

h

0.6 m

5m

A

Prob. 5–73 B 0.5 m

5–74. Water is drawn into the pump, such that the pressure at the inlet A is -30 kPa and the pressure at B is 120 kPa. If the discharge at B is 0.15 m3 >s determine the power output of the pump. Neglect friction losses. The pipe has a constant diameter of 100 mm. Take h = 1.5 m.

Probs. 5–70/71

*5–72. Determine the power delivered to the turbine if the water exits the 400-mm-diameter pipe at 8 m>s. Draw the energy and hydraulic grade lines for the pipe using a datum at point C. Neglect all losses.

5–75. Draw the energy and hydraulic grade lines for the pipe ACB in Prob. 5–74 using a datum at A.

B

A

20 m

h

B 8m C C

A

Prob. 5–72

Probs. 5–74/75

290

Chapter 5

Work

and

energy

of

Moving fluids

*5–76. Water in the reservoir flows through the 300-mmdiameter pipe at A into the turbine. If the discharge at B is 18 000 liters>min, determine the power output of the turbine. Assume the turbine runs with an efficiency of 75%. Neglect frictional losses in the pipe.

5–79. Water flows into the pump at 3000 liters>min and has a pressure of 30 kPa. It exits the pump at 130 kPa. Determine the power output of the pump. Neglect friction losses.

5 C

12 m 200 mm A 3m

B

150 mm B

A

300 mm

Prob. 5–76

Prob. 5–79

5–77. Water in the reservoir flows through the 300-mmdiameter pipe at A into the turbine. If the discharge at B is 18 000 liters>min, determine the power output of the turbine. Assume the turbine runs with an efficiency of 75% and there is a head loss of 2 m through the pipe.

C

12 m A 3m

B 300 mm

*5–80. Water flows through the constant-diameter pipe such that at A the pressure is 80 kPa, and the velocity is 4 m>s. Determine the pressure and velocity at B. Draw the energy and hydraulic grade lines from A to B with reference to the datum set through A. 5–81. Water flows through the constant-diameter pipe such that at A the pressure is 80 kPa, and the velocity is 4 m>s. Plot the pressure head and the elevation head from A to B with reference to the datum set through A.

Prob. 5–77 5–78. The 6 kW pump has an efficiency of e = 0.8 and produces a flow of 3000 liters>min through the pipe. If the frictional head loss within the system is 2 m, determine the difference in the water pressure between A and B. 10 m 200 mm

5

150 mm B

A

3

4

4 ms

Prob. 5–78

B

A

Probs. 5–80/81

proBleMs 5–82. Water is drawn into the pump, such that the pressure at the inlet A is - 35 kPa and the pressure at B is 120 kPa. If the discharge at B is 0.08 m3 >s, determine the power output of the pump. Neglect friction losses. The pipe has a constant diameter of 100 mm. Take h = 2 m. 5–83. Draw the energy and hydraulic grade lines for the pipe ACB in Prob. 5–82 using a datum at A.

291

5–86. The turbine removes energy from the water in the reservoir such that it has a discharge of 0.3 m3 >s through the 300-mm-diameter pipe. Determine the power delivered to the turbine. Draw the pressure and elevation heads for the flow using a datum at point C. Neglect friction losses. 5

C B A h 5m B D 2m

A

Probs. 5–82/83

C

*5–84. The hose is used to siphon water from the river. Determine the smallest pressure in the hose and the volumetric discharge at C. The hose has an inner diameter of 50 mm. Draw the energy and hydraulic grade lines for the hose with reference to the datum set through C.

Prob. 5–86

5–85. The hose is used to siphon water from the river. Determine the pressure in the hose at point A′. The hose has an inner diameter of 50 mm. Draw the energy and hydraulic grade lines for the hose with reference to the datum set through point B. 5–87. A pump is used to deliver water from a large reservoir to another large reservoir that is 20 m higher. If the friction head loss in the 200-mm-diameter, 4-km-long pipeline is 2.5 m for every 500 m of pipe length, determine the required power output of the pump so the flow is 0.8 m3 >s. The inlet of the pipe is submerged in the one reservoir and the exit is just above the surface of the other reservoir.

B A

2m A9

8m

C

Probs. 5–84/85

*5–88. A 300-mm-diameter horizontal oil pipeline extends 8 km, connecting two large open reservoirs having the same level. If friction in the pipe creates a head loss of 3 m for every 200 m of pipe length, determine the power that must be supplied by a pump to produce a flow of 6 m3 >min through the pipe. The inlet of the pipe is submerged in the one reservoir and the exit is just above the surface of the other reservoir. Take ro = 880 kg>m3.

292

Chapter 5

Work

and

energy

of

Moving fluids

5–89. Determine the kinetic energy coefficient a if the velocity distribution for turbulent flow in a smooth pipe is defined by Prandtl’s one-seventh power law, u = Umax 11 - r>R2 1>7 . 5

*5–92. The pump is connected to the 80-mm-diameter hose. If the average velocity of the water at C is 8.50 m>s, determine the required power output of the pump. Neglect friction losses. 5–93. Solve Prob. 5–92 by including frictional head losses in the hose of 0.75 m for every 10 m of hose. The hose has a total length of 18 m.

r R

C

Prob. 5–89 5m 1m

B

5–90. Determine the kinetic energy coefficient a if the velocity distribution for laminar flow in a smooth pipe is defined by u = Umax 11 - 1r>R2 2 2.

2m A

Probs. 5–92/93 r R

5–94. The pump discharges water at B at 0.015 m3 >s. If the friction head loss between the intake at A and the outlet at B is 1.5 m, and the power input to the pump is 2 kW, determine the difference in pressure between A and B. The efficiency of the pump is e = 0.8.

Prob. 5–90

5–91. The pump is connected to the 80-mm-diameter hose. If the pump supplies a power of 4 kW, determine the discharge at C. Neglect frictional losses. B 75 mm C

5m B

3m

1m 2m

A

A 100 mm

Prob. 5–91

Prob. 5–94

293

proBleMs 5–95. A 4.5 kW pump with a 100-mm-diameter hose is used to drain water from a large cavity at B. Determine the discharge at C. Neglect friction losses and the efficiency of the pump. *5–96. The pump is used with a 100-mm-diameter hose to draw water from the cavity. If the discharge is 0.075 m3 >s, determine the required power developed by the pump. Neglect friction losses.

*5–100. A piezometer and a manometer containing mercury are connected to the venturi meter. If the gage reading is 35 kPa and the level of mercury is as indicated, determine the volumetric flow of water through the meter. Draw the energy and hydraulic grade lines. Take rHg = 13 550 kg>m3. 5

5–97. Solve Prob. 5–96 by including frictional head losses in the hose of 0.75 m for every 10 m of hose. The hose has a total length of 50 m. C B

B

1m

C

A

3m A

100 mm

300 mm

25 mm 100 mm

150 mm

Probs. 5–95/96/97 5–98. The siphon spillway provides an automatic control of the level in the reservoir within a desired range. In the case shown, flow will begin when the water level in the reservoir rises above the crown C of the conduit. Determine the flow through the siphon if h = 1.5 mm. Also, draw the energy and hydraulic grade lines for the siphon conduit, with reference to the datum set through B. The siphon has a diameter of 300 mm, and the water is at a temperature of 30°C. Neglect any head loss. The atmospheric pressure is 101.3 kPa. 5–99. What is the maximum permissible height h of the crown in the siphon conduit in Prob. 5–98 before it affects the flow? A

Prob. 5–100

5–101. The turbine at C draws a power of 68.5 kW. If the pressure at the intake B is PB = 380 kPa and the velocity of the water at that point is 4 m>s, determine the pressure and velocity of the water at the exit A. Neglect frictional losses between A and B.

C 3m h A9

400 mm

9m 150 mm 1m B A

C B

Probs. 5–98/99

Prob. 5–101

294

Chapter 5

Work

and

energy

of

Moving fluids

5–102. Determine the power that the pump supplies to the water if the velocity of the water at A is 6 m>s and the pressures at A and B are 175 kPa and 350 kPa, respectively. Neglect friction losses. 5

*5–104. The pump is used to transfer carbon tetrachloride in a processing plant from a storage tank A to the mixing tank C. If the head loss due to friction in the pipe and in the pipe fittings is 1.8 m, and the diameter of the pipe is 50 mm, determine the power developed by the pump when h = 3 m. The velocity at the pipe exit is 10 m>s. The storage tank is opened to the atmosphere. Take rct = 1590 kg>m3.

150 mm

B

B A

3m

6m

h

C

100 mm

Prob. 5–104

A

Prob. 5–102

5–103. The water pressure at the inlet and exit portions of the pipe are indicated for the pump. If the flow is 0.1 m3 >s, determine the power that the pump supplies to the water. Neglect friction losses.

5–105. The pump at C produces a discharge of water at B of 0.035 m3 >s. If the pipe at B has a diameter of 50 mm and the hose at A has a diameter of 30 mm, determine the power output supplied by the pump. Assume frictional head losses within the pipe system are determined from 3VB2 >2g.

B 300 kPa 50 mm 30 m 200 kPa

B C

2m

75 mm

A A

Prob. 5–103

Prob. 5–105

295

proBleMs 5–106. The pump delivers water at 4200 liters>min from the river to the irrigation stream. If the frictional head loss in the 80-mm-diameter hose is hL = 3 m, determine the power output of the pump.

*5–108. The flow of air at A through a 200-mm-diameter duct has an absolute inlet pressure of 180 kPa, a temperature of 15°C, and a velocity of 10 m>s. Farther downstream a 2-kW exhaust system increases the outlet velocity at B to 25 m>s. Determine the change in enthalpy of the air. Neglect heat transfer through the pipe. B

4m A B

20 m

30 m

Prob. 5–108

A

Prob. 5–106

5–107. A fire truck supplies 750 liters >min of water to the third story of a building at B. If the friction loss through the 20-m-long, 60-mm-diameter hose is 1.5 m for every 10 m of hose, determine the required pressure developed at the outlet A of the pump located within the truck close to the ground. Also, what is the average velocity of the water as it is ejected through a 30-mm-diameter nozzle at B?

5–109. Crude oil is pumped from a test separator at A to the stock tank using a pipe that has a diameter of 100 mm. If the total pipe length is 60 m, and the volumetric flow at A is 2400 liters> min, determine the power supplied by the pump. The pressure at A is 30 kPa, and the stock tank is open to the atmosphere. The frictional head loss in the pipe is 45 mm>m, the head loss at the pipe discharge into the tank is 1.01V 2 >2g2, and for each of the four elbows it is 0.91V 2>2g2, and V is the velocity of the flow in the pipe. Take ro = 880 kg>m3.

8m

B

A

Prob. 5–109 5–110. The pump is used to transport water at 2700 liters>min from the stream up the 10 m embankment. If frictional head losses in the 60-mm-diameter pipe are hL = 2.5 m, determine the power output of the pump.

B

B

10.8 m

10 m A

A

Prob. 5–107

Prob. 5–110

5

296

Chapter 5

Work

and

energy

of

Moving fluids

5–111. Water from the reservoir passes through a turbine at the rate of 0.75 m3 >s. If it is discharged at B with a velocity of 3 m>s, and the turbine draws 80 kW, determine the head loss in the system. 5

*5–112. The pump is used to deliver the water from the pond at A to the channel at B with a flow of 0.025 m3 >s. If the hose has a diameter of 100 mm and friction losses within it can be expressed as 5V 2 >g, where V is the velocity of the flow, determine the power the pump supplies to the water.

A

20 m

B

B 7.5 m

1.5 m A

Prob. 5–111

Prob. 5–112

ConCeptual proBleMs

297

CONCEPTUAL PROBLEMS P5–1. The level of coffee is measured by the standpipe A. If the valve is pushed open and the coffee begins to flow out, will the level of coffee in the standpipe go up, go down, or remain the same? Explain.

P5–3. When air flows through the hose, it causes the paper to rise. Explain why this happens.

P5–1 P5–2. The ball is suspended in the air by the stream of air produced by the fan. Explain why it will return to its original position if it is displaced slightly to the right or left.

P5–2

P5–3

5

298

Chapter 5

Work

and

energy

of

Moving fluids

CHAP T ER R EV IEW

5

For an inviscid fluid, the steady flow of a particle along a streamline is caused by pressure and gravitational forces. The Euler differential equations describe this motion. Along the streamline or s direction, the forces change the magnitude of the particle’s velocity, and along the normal or n direction, they change the direction of its velocity.

dp + V dV + g dz = 0 r -

rV 2 dp dz - rg = dn dn R

Heat added

Rotational flow eddies

The Bernoulli equation is an integrated form of Euler’s equation in the s direction. It applies between any two points on the same streamline for steady flow of an ideal fluid. It cannot be used between points where energy losses occur, or where fluid energy is added or withdrawn by external sources. When applying the Bernoulli equation, remember that points at atmospheric openings have zero gage pressure and that the velocity is zero at a stagnation point, and can be assumed to be zero at the top surface of a large reservoir.

Energy added

A B

C D

Viscous friction

E

Flow separation

Places where the Bernoulli equation does not apply

p1 p2 V 21 V 22 + + gz1 = + + gz2 r r 2 2 steady flow, ideal fluid, same streamline

Pitot tubes can be used to measure the velocity of a liquid in an open channel. To measure the velocity of a liquid in a closed conduit, it is necessary to use a pitot tube along with a piezometer, which measures the static pressure in the liquid. A venturi meter can also be used to measure the average velocity or the volumetric flow.

299

Chapter revieW

Pitot tube V1 2g

Pitot tube

V22 2g

2

The Bernoulli equation can be expressed in terms of the total head H of the fluid. A plot of the total head is called the energy grade line, EGL, which will always be a constant horizontal line, provided there are no friction losses or pumps or turbines. The hydraulic grade line, HGL, is a plot of the hydraulic head, p>g + z. This line will always be below the EGL by an amount equal to the kinetic head, V 2 >2g.

Piezometer

Piezometer

p2 g 1 z2

HGL

EGL

5

p V2 H5 g 1 1z 2g

p1 g 1 z1

Datum Energy and hydraulic grade lines

H =

p V2 + z = const. + g 2g

# # # Qin + Wpump - Wturbine = When the fluid is viscous and>or energy is added to or removed from the fluid, the energy equation should be used. It is based on the first law of thermodynamics, and a control volume must be specified when it is applied. It can be expressed in various forms.

Power is the rate of doing shaft work.

c ahout +

Vout2 Vin2 # + gzout b - ahin + + gzin b d m 2 2

pin pout Vout2 Vin2 + + gzin + wpump = + + gzout + wturbine + fl r r 2 2 pin pout Vout2 Vin2 + + zin + hpump = + + zout + hturbine + hL g g 2g 2g

# # Ws = m ghs = Qghs

6

Spaces Images/Blend Images/Getty Image

CHAPTER

Impulse and momentum principles play an important role in the design of both windmills and wind turbines.

FLUID MOMENTUM

CHAPTER OBJECTIVES ■

To develop the principles of linear and angular impulse and momentum for a fluid, so that the loadings the fluid exerts on a surface can be determined.

■

To illustrate specific applications of the momentum equation to propellers, wind turbines, turbojets, and rockets.

6.1

THE LINEAR MOMENTUM EQUATION

The design of many hydraulic structures, such as floodgates and flow diversion blades, as well as pumps and turbines, depends upon the forces that a fluid flow exerts on them. In this section we will obtain these forces by using a linear momentum analysis, which is based on Newton’s second law of motion, written in the form ΣF = ma = d(mV)>dt. Here V represents the velocity of the mass center for the system of particles that compose the fluid. For application, it is important to measure the time rate of change of V from an inertial or nonaccelerating frame of reference, that is, a reference that is either fixed or moves with constant velocity.

301

302

Chapter 6

Fluid MoMentuM

For a fluid, a control volume approach works best for this type of analysis, and so we will apply the Reynolds transport theorem to describe the rate of change of the momentum relative to a control volume. Linear momentum is an extensive property of a fluid, where N = mV, and so H = mV>m = V. Therefore, Eq. 4–11 becomes 6

dN 0 = Hr dV + Hr V # dA dt 0t Lcv Lcs d(mV) 0 = Vr dV + Vr V # dA dt 0t Lcv Lcs

Now, substituting this result into Newton’s second law of motion, we obtain our result, the linear momentum equation.

ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

(6–1)

The first term on the right represents the local change of the momentum within the control volume. It can occur if the flow is unsteady, as when the fluid is accelerating. The second term represents the net convective change of the momentum as the fluid enters and exits the open control surfaces. Notice that in the equation the velocity V appears three times, and in each case it has a special meaning. In the first term it represents the velocity of the fluid system at points within the control volume. In the second term the first V is the velocity of the fluid passing through the open control surfaces, and so it has components along the coordinate axes. The second V is involved in the dot product operation, rV # dA. This represents the mass flow through the open control surfaces, which is a scalar quantity. To distinguish the use of V for this purpose, we will represent it as the velocity of the fluid measured relative to the control surface, Vf>cs. Using this notation we can therefore write the linear momentum equation as

ΣF = Sum of forces

0 Vr dV + Vr Vf>cs # dA 0t Lcv Lcs Local momentum change

Convective momentum change

(6–2)

6.1

the linear MoMentuM equation

303

Steady Flow. If the flow is steady, then no local change of momentum will occur within the control volume, and the first term on the right of Eq. 6–2 will be equal to zero. Therefore ΣF =

Lcs

VrVf>cs # dA

(6–3)

Steady flow

Furthermore, if we have an ideal fluid, then r is constant and viscous friction is zero. Thus the velocity will be uniformly distributed over the open control surfaces, and so integration of Eq. 6–3 gives ΣF = ΣVrVf>cs # A

(6–4)

Steady ideal fluid flow

This equation is often used to obtain the fluid forces acting on various types of surfaces that deflect or transport the flow. As an application, consider the steady flow of an ideal fluid into and out of the two control surfaces of the fixed control volume contained within the pipe shown in Fig. 6–1a. In the x direction, the x components of Vin and Vout must be used for V. They both act in the + x direction. When writing the expression for the mass flow, rVf>cs # A, we have to follow our positive sign convention, that is, Ain = Ainnin and Aout = Aoutnout are both positive out, Fig. 6–1a. Therefore, after performing the dot products, the x components of Eq. 6–4 become + ΣFx = ΣVxrVf>cs # A = (Vin)x( -rVinAin) + (Vout)x(rVoutAout) S

y

Vout Aout

nin

nout (Vout)x

Aout

Vin 5 (Vin)x Inertial coordinate system x Control volume (a)

Fig. 6–1

6

304

Chapter 6

Fluid MoMentuM

Pressure force (out) Shear force

6

Pressure force (in) Weight

Normal force

Free-body diagram (b)

Fig. 6–1 (cont.)

Free-Body Diagram. When the momentum equation is applied, there will generally be four types of external forces ΣF that can act on the fluid system contained within the control volume. As shown on the freebody diagram of the fluid within the pipe, Fig. 6–1b, they are the resultant normal and shear reactions from solid boundaries on the closed control surfaces, the pressure forces that act on open control surfaces, and the weight that acts through the center of mass of fluid within the pipe (control volume). For analysis, we will generally represent the resultant shear and normal forces acting on the closed control surface by a single resultant force. The opposite of this resultant is the effect of the fluid system on this surface. It is referred to as the dynamic force.

6.2 APPLICATIONS TO BODIES AT REST On occasion a vane, pipe, or other type of conduit will be subjected to fluid forces when it changes the direction of the flow. These forces can be determined using the following procedure.

The attachments for this cover plate must resist the force caused by the change in momentum of the flow of water coming out of the opening and striking the plate.

6.2

appliCations to Bodies at rest

305

PROCE DUR E FOR A N A LY S I S Fluid Description. • Identify the type of flow as steady or unsteady and uniform or nonuniform. Also specify whether the fluid is compressible or incompressible and viscous or inviscid. Control Volume and Free-Body Diagram. • Select the control volume so its free-body diagram includes the unknown forces that are to be determined. This volume can include both solid and fluid parts. Open control surfaces should be located in a region where the flow is uniform and well established. These surfaces should be oriented so that their planar areas are perpendicular to the flow. For inviscid and ideal fluids, the velocity profile will be uniform over the cross section. The free-body diagram of the control volume should be • drawn to identify all of the external forces acting on it. These forces generally include the weight of the contained fluid and the weight of any solid portions of the control volume, the resultant of the frictional shear and pressure forces or their components that act on a closed control surface, and the pressure forces acting on the open control surfaces. Note that the pressure forces will be zero if the control surface is open to the atmosphere; however, if an open control surface is contained within a closed region of fluid, then the pressure on this surface may have to be determined by using the Bernoulli equation. Linear Momentum. • If the volumetric flow is known, then the average velocity at an open control surface is determined using V = Q>A. Also, a velocity can be determined using the continuity equation, the Bernoulli equation, or the energy equation. • Establish the inertial x, y coordinate system, and apply the linear momentum equation along the x and y axes, using the components of the velocity shown on the open control surfaces and the forces shown on the free-body diagram. Remember, the term rVf>cs # dA in the equation is a scalar quantity that represents the mass flow through the area A of each open control surface. The product rVf>cs # dA will be negative for mass flow into a control surface since Vf>cs and A are in opposite directions, whereas rVf>cs # A will be positive for mass flow out of a control surface since then Vf>cs and A are in the same direction.

6

306

Chapter 6

EXAMPLE

6

Fluid MoMentuM

6.1 The end of a pipe is capped with a reducer as shown in Fig. 6–2a. If the water pressure within the pipe at A is 200 kPa, determine the shear force that glue along the sides of the pipe exerts on the reducer to hold it in place.

B

25 mm B

pB 5 0

A

A

pA 5 200 kPa FR –– 2

FR –– 2 100 mm (b)

(a)

Fig. 6–2

SOLUTION Fluid Description. We have steady flow, and we will assume water is an ideal fluid, where rw = 1000 kg>m3. Control Volume and Free-Body Diagram. We will select the control volume to represent the reducer along with the portion of water within it, Fig. 6–2a. The reason for this selection is to show or “expose” the required shear force FR on the free-body diagram of the control volume, Fig. 6–2b. Also shown is the water pressure pA, developed on the inlet control surface A. There is no pressure at the outlet control surface since the (gage) pressure is pB = 0. Excluded is the symmetric horizontal or normal force of the wall of the pipe on the reducer since it produces a zero resultant force. We have also excluded the weight of the reducer, along with the weight of water within it, since here they can be considered negligible.

6.2

appliCations to Bodies at rest

307

Continuity Equation. Before applying the momentum equation, we must first obtain the velocity of the water at A and B. Applying the continuity equation for steady flow, we have 0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

6

0 - rVAAA + rVB AB = 0

-VA 3 p(0.05 m)2 4 + VB 3 p(0.0125 m)2 4 = 0 VB = 16VA

(1)

Bernoulli Equation. Since the pressures at A and B are known, the velocities at these points can be determined by using Eq. 1 and applying the Bernoulli equation to points on a vertical streamline passing through A and B.* Neglecting the elevation difference from A to B, we have pA pB VA2 VB 2 + + + zA = + zB g g 2g 2g 200(103) N>m2

11000 kg>m3 219.81 m>s2 2

+

VA2

219.81 m>s2 2

+ 0 = 0 +

VA = 1.252 m>s

116VA 2 2

219.81 m>s2 2

+ 0

VB = 16(1.252 m>s) = 20.04 m>s Linear Momentum. To obtain FR, we can now apply the momentum equation in the vertical direction. ΣF =

0 Vr dV + VrVf>cs # dA 0t Lcv Lcs

Since the flow is steady, no local change in momentum occurs, and so the first term on the right is zero. Also, since the fluid is ideal, rw is constant and average velocities can be used (Eg. 6–4). Thus, + c ΣFy = 0 + VB 1rwVB AB 2 + VA 1 -rwVA AA 2

3 2001103 2 N>m243p(0.05 m)2 4 - FR = 11000 kg>m3 2 3120.04 m>s2 2(p)(0.0125 m)2 - (1.252 m>s)2(p)(0.05 m)24 + c ΣFy = rw(VB2 AB - VA2 AA) FR = 1.39 kN

Ans.

This positive result indicates that the shear force acts downward on the reducer (control surface), as initially assumed. *Actually a fitting such as this will generate friction losses due to turbulence within it, and so the energy equation should be used. This will be discussed in Chapter 10.

308

Chapter 6

EXAMPLE

Fluid MoMentuM

6.2 Water is discharged from the 50-mm-diameter nozzle of a fire hose at 15 liters>s, Fig. 6–3a. The flow strikes the fixed surface such that 3>4 flows along B, while the remaining 1>4 flows along C. Determine the x and y components of the resultant force exerted on the surface. Assume steady flow in the vertical plane.

6

VC

608

C 50 mm VA

A 458

B VB (a)

Fig. 6–3

SOLUTION Fluid Description. We consider the water as an ideal fluid, for which rw = 1000 kg>m3.

Fx Fy

(b)

Control Volume and Free-Body Diagram. We will select a fixed control volume that contains the water from the nozzle and on the surface, Fig. 6–3a. The pressure distributed over the surface creates resultant horizontal and vertical reaction components, Fx and Fy, on the closed control surface. We will neglect the weight of the water within the control volume. Also, because the (gage) pressure on all the open control surfaces is zero, no force is acting on these surfaces. If we apply the Bernoulli equation to a point on the fluid stream along A, and another point on the fluid stream along B (or C), neglecting elevation changes between these points, and realizing that the (gage) pressure is zero, then p>g + V 2 >2g + z = const. shows that the speed

6.2

appliCations to Bodies at rest

of the water onto and along the fixed surface is the same everywhere. That is, the Bernoulli equation shows VA = V B = V C and so the surface only changes the direction of the velocity. Linear Momentum. The velocity can be determined from the flow at A, because here the cross-sectional area is known. QA = a

15 liters 1000 cm3 1m 3 ba b = 0.015 m3 >s ba s 1 liter 100 cm

VA = VB = VC =

0.015 m3 >s QA 24 = = m>s 2 p AA p(0.025 m)

For this case of steady flow, the linear momentum equation becomes 0 Vr dV + VrVf>cs # dA 0t Lcv Lcs ΣF = 0 + VB 1rVB AB 2 + VC 1rVC AC 2 + VA 1 -rVA AA 2 ΣF =

Notice that the last term is negative because the flow is into the control surface at A. Because Q = VA, then ΣF = r1QBVB + QCVC - QAVA 2

Now when we resolve this equation in the x and y directions, we get + ΣFx = r1QBVBx + QCVCx - QAVAx 2 S 1 24 -Fx = (1000 kg>m3) e 0 + (0.015 m3 >s) c a m>s b cos 60° d p 4 - (0.015 m3 >s) c a

24 m>s b cos 45° d f p Ans.

+ c ΣFy = r 3 QB 1 -VBy 2 + QCVCy - QA 1 -VAy 2 4 Fx = 66.70 N = 66.7 N d

Fy = (1000 kg>m3) e

3 24 1 24 1 0.015 m3 >s 2 a - p m>s b + 1 0.015 m3 >s 2 c a p m>s b sin 60° d 4 4 -

1 0.015 m3 >s 2 c a - p

24

Fy = 19.89 N = 19.9 N c

m>s b sin 45° d f

Ans.

These components produce a resultant force of 69.6 N that acts on the water. The equal but opposite dynamic force acts on the surface. As a comparison, if the nozzle is directed perpendicular to a flat surface, a similar calculation shows that the force is 115 N. This can certainly become quite dangerous if the stream is directed at someone!

309

6

310

Chapter 6

EXAMPLE

Fluid MoMentuM

6.3 When the sluice gate G in Fig. 6–4a is in the open position as shown, water flows out from under the gate at a depth of 1 m. If the gate is 3 m wide, determine the resultant horizontal force that must be applied by its supports to hold the gate in place. Assume the channel behind the gate maintains a constant depth of 5 m.

6

1

G VA 5m A 2 B V B

1m Datum

(a)

Fig. 6–4

SOLUTION Fluid Description. Since the channel depth is assumed constant, the flow will be steady. We will also assume water to be an ideal fluid so that it will flow with an average velocity through the gate. Here r = 1000 kg>m3.

W

FG FA FB

N (b)

Control Volume and Free-Body Diagram. To determine the force on the gate, the fixed control volume will include a control surface along the left face of the gate, and a volume of water on each side of the gate, Fig. 6–4a. There are three horizontal forces acting on this control volume, as shown on the free-body diagram, Fig. 6–4b. They are the unknown pressure force resultant of the gate, FG, and the two hydrostatic pressure force resultants from the water on the control surfaces at A and B. The viscous frictional forces acting on the closed control surfaces on the gate and on the ground are neglected since we have assumed the water to be inviscid. (Actually, these forces will be comparatively small.)

6.2

appliCations to Bodies at rest

Bernoulli and Continuity Equations. We can calculate the average velocities at A and B by applying the Bernoulli equation (or the energy equation) and the continuity equation, Fig. 6–4a. When a streamline is chosen through points 1 and 2, the Bernoulli equation gives* p1 p2 V12 V22 + + + z1 = + z2 g g 2g 2g 0 +

2 1 9.81 m>s2 2 V A2

+ 5m = 0 +

2 1 9.81 m>s2 2 V B2

+ 1m

V 2B - V 2A = 78.48

(1)

Continuity requires 0 r dV + rVf>cs # dA = 0 0t Lcv Lcs 0 - VA(5 m)(3 m) + VB(1 m)(3 m) = 0 VB = 5VA

(2)

Solving Eqs. 1 and 2 yields VA = 1.808 m>s and VB = 9.042 m>s Linear Momentum ΣF =

0 Vr dV + V rVf>cs # dA 0t Lcv Lcs

+ ΣFx = 0 + VB 1rVBAB 2 + VA 1 -rVAAA 2 S = 0 + r1VB 2AB - VA2AA 2

Referring to the free-body diagram, Fig. 6–4b, 1 2 [(1000

kg>m3)(9.81 m>s2)(5 m)](3 m)(5 m) - 12[(1000 kg>m3)(9.81 m>s2)(1 m)](3 m)(1 m) - FG

= (1000 kg>m3)[(9.042 m>s)2 (3 m) (1 m) - (1.808 m>s)2 (3 m) (5 m)] FG = 156.96 (103) N = 157 kN

Ans.

Notice that if no flow occurs, then a much larger hydrostatic force acts on the gate. (FG)st = 21(rghb)h =

1 2

[(1000 kg>m3)(9.81 m>s2)(5 m - 1 m)(3 m)](5 m - 1 m)

= 235.44(103) N = 235 kN *All the particles on the surface will eventually pass under the gate. Here we have chosen the particle at point 1 that happens to wind up at point 2.

311

6

312

Chapter 6

EXAMPLE

Fluid MoMentuM

6.4 6 ms 0.9 ms2

6 100 mm A

Oil flows through the pipe AB in Fig. 6–5a such that at the instant shown, it has a velocity of 6 m>s at A, which is increasing at 0.9 m>s2. Determine the pressure at B that creates this flow if the pressure in the pipe at A is 60 kPa. Take ro = 900 kg>m3. SOLUTION

0.75 m

Fluid Description. Because of the acceleration, we have a case of unsteady flow. We will assume the oil is an ideal fluid.

B

Control Volume and Free-Body Diagram. Here we will consider a fixed control volume that contains the oil within the vertical section AB of the pipe, Fig. 6–5a. The forces shown on its free-body diagram are the weight of oil within the control volume, Wo = goVo, and the pressures at A and B, Fig. 6–5b.* Linear Momentum. Because an ideal fluid produces average velocities, the momentum equation becomes

(a)

pA

ΣF = Wo

+ c ΣFy = pB (b)

Fig. 6–5

0 Vr dV + VrVf>cs # dA 0t Lcv Lcs

0 Vr dV + VA 1rVA AA 2 + VB 1 -rVB AB 2 0t Lcv

Since r is constant and AA = AB, the continuity equation requires VA = VB = V = 6 m>s. As a result, the last two terms will cancel each other. In other words, the flow is uniform, and so there is no net convective effect. The unsteady flow term (local effect) indicates that the momentum changes within the control volume due to the time rate of change in the flow velocity (unsteady flow). A local change does occur within the control volume due to unsteady flow. Here the velocity field for the fluid system V = V(x, y, z, t) is not a function of position since the fluid (oil) is assumed incompressible, and so there is no relative

*The horizontal pressure caused by the sides of the pipe on the control volume surface is not included here since it will produce a zero resultant force. Also, friction along the sides is excluded because the oil is assumed to be inviscid.

6.3

appliCations to Bodies having Constant veloCity

movement between the fluid particles. In other words, at every point within the control volume all the particles move together and so the fluid system only has a time rate of change, V = V(t). Therefore, the above equation becomes + c ΣFy =

dV rV; dt

-pA AA + pB AB - goVo =

dV rV dt o o

6

(1)

[ -60(103) N>m2][p(0.05 m)2] + pB[p(0.05 m)2] - (900 kg>m3)(9.81 m>s2)[p(0.05 m)2](0.75 m) = (0.9 m>s2)(900 kg>m3)[p(0.05 m)2](0.75 m) pB = 67.23(103) Pa = 67.2 kPa

Ans.

Notice that Eq. 1 is actually the application of ΣFy = may, Fig. 6–5b.

6.3

APPLICATIONS TO BODIES HAVING CONSTANT VELOCITY

For some problems, a blade or vane may be moving with constant velocity, and when this occurs, the forces on the blade can be obtained by selecting a control volume of constant shape and size that moves with the body. If this is the case, then the velocity and the mass flow in the momentum equation are measured relative to each control surface, Fig. 6–6. Therefore, for the convective term, V = Vf>cs, and so Vf

0 ΣF = Vr dV + Vf>cs rVf>cs # dA 0t Lcv Lcs

Vb

Also, since the flow will appear to be steady flow relative to the control volume, the local change or first term on the right side of this equation will be zero. Using the procedure for analysis outlined in Sec. 6.2, the following examples illustrate application of the momentum equation for a body moving with constant velocity, and in Sec. 6.5 we will consider application of this equation to propellers and wind turbines.

313

Vfb 5 Vf 2 Vb Steady flow measured relative to the blade

Fig. 6–6

314

Chapter 6

EXAMPLE

6

Fluid MoMentuM

6.5 The truck in Fig. 6–7a is moving to the left at 5 m>s into a 50-mm-diameter stream of water, which has a discharge of 8 liter>s. Determine the dynamic force the stream exerts on the truck if it is deflected off the windshield in the vertical plane.

408

B

5 ms

A

(a)

Fx

Fy (b)

Fig. 6–7

SOLUTION Fluid Description. In this problem the driver will observe steady flow, and so we will fix the x, y inertial coordinate system to the truck, so that it moves with constant velocity. We will assume water to be an ideal fluid, and so average velocities can be used because friction is negligible. Here r = 1000 kg>m3. Control Volume and Free-Body Diagram. We will consider the moving control volume to include the portion AB of the water stream that is in contact with the truck, Fig. 6–7a. As shown on its free-body diagram, Fig. 6–7b, only the horizontal and vertical components of the resultant force caused by the truck on the control volume will be considered significant. (The pressure at all the open control surfaces is atmospheric and the weight of the water is neglected.)

6.3

appliCations to Bodies having Constant veloCity

315

The nozzle velocity of the stream is determined first. Q = VA;

a

8 liter 10-3 m3 b = V3p(0.025 m)2 4 ba s 1 liter

V = 4.074 m>s 6

Relative to the control volume (or driver), the velocity of the water at A is therefore + Vf>A = Vf - VA S Vf>A = 4.074 m>s - ( -5 m>s) = 9.074 m>s Application of the Bernoulli equation will show that this relative average speed is maintained as the water flows off the windshield at B (neglecting the effect of elevation). Also, the size of the cross-sectional area at B (open control surface) must remain the same as at A to maintain continuity, Vf>AAA = Vf>BAB, although its shape will certainly change. Linear Momentum. ΣF =

For steady incompressible flow we have

0 Vr dV + VrVf>cs # dA 0 t Lcv Lcs

ΣF = 0 + Vf>B 3rVf>BAB 4 + Vf>A [ -rVf>AAA] Applying this equation in the x and y directions yields + ΣFx = 0 + 3Vf>B cos 40°4 3rVf>BAB 4 - Vf>A 3rVf>AAA 4 S

-Fx = 3(9.074 m>s)cos 40°4 3 11000 kg>m3 2(9.074 m>s)3p(0.025 m)2 4 4 - (9.074 m>s) 3 11000 kg>m3 2(9.074 m>s)3p(0.025 m)2 4 4 Fx = 37.83 N

+ c ΣFy = 0 + 3Vf>B sin 40°4 3rVf>BAB 4 - 0

Fy = 3(9.074 m>s) sin 40°4 3 11000 kg>m3 2(9.074 m>s)3p(0.025 m)2 4 4 - 0 = 103.9 N

Thus,

F = 2(37.83 N)2 + (103.9 N)2 = 111 N

Ans.

This (dynamic) force acts on the truck, but in the opposite direction.

316

Chapter 6

EXAMPLE

6

Fluid MoMentuM

6.6 The jet of water having a cross-sectional area of 2110-3 2 m2 and a velocity of 45 m>s strikes the vane of a turbine, causing it to move at 20 m>s, Fig. 6–8a. Determine the dynamic force of the water on the vane, and the power output caused by the water.

45 ms

A Fy

Fx

20 ms

B (b) 308 (a)

Fig. 6–8

SOLUTION Fluid Description. To observe steady flow, the x, y inertial reference is fixed to the vane. We assume the water is an ideal fluid, where r = 1000 kg>m3. Control Volume and Free-Body Diagram. We will take the control volume to contain the water on the vane from A to B, Fig. 6–8a. As shown on its free-body diagram, Fig. 6–8b, the force components of the vane on the control volume are denoted as Fx and Fy. The weight of the water is neglected, and the pressure on all the open control surfaces is atmospheric or zero gage pressure. Linear Momentum. surface at A is

The velocity of the water relative to the control

+ Vf>A = Vf - VA S Vf>A = 45 m>s - 20 m>s = 25 m>s

6.3

appliCations to Bodies having Constant veloCity

With negligible elevation change, the Bernoulli equation will show that water leaves the vane at B with this same speed, Vf>B = 25 m>s, and so to satisfy continuity, AA = AB = A. For steady ideal fluid flow, the momentum equation becomes

ΣF =

0 Vrd V + VrVf>cs # dA 0t Lcv Lcs

ΣF = 0 + Vf>B 3rVf>BAB 4 + Vf>A 3 -rVf>AAA 4 As in the previous example, the velocity Vf>cs has components in this equation. The mass flow terms rVf>cs A are positive or negative scalars depending upon the direction of Vf>cs and A. Therefore, + ΣFx = 3 -Vf>B cos 30°4 3rVf>BAB 4 + Vf>A 3 -rVf>AAA 4 S

-Fx = 3 -(25 m>s) cos 30°4 3 11000 kg>m3 2(25 m>s)12110-3 2 m2 2 4 + (25 m>s) 3 - 11000 kg>m3 2(25 m>s)12110-3 2 m2 2 4 Fx = 2333 N

+ c ΣFy = 3 -Vf>B sin 30°4 3rVf>BAB 4 - 0

-Fy = 3 -(25 m>s) sin 30°4 3 11000 kg>m3 2(25 m>s)12110-3 2 m2 2 4 Fy = 625 N

F = 2(2333 N)2 + (625 N)2 = 2.41 kN

Ans.

The equal but opposite force acts on the vane. This represents the dynamic force. Power. By definition, power is work per unit time, or the product of a force and the parallel component of velocity. Here only Fx produces power, since Fy does not displace up or down and hence does no work. Since the vane is moving at 20 m>s, # W = F # V;

# W = (2333 N)(20 m>s) = 46.7 kW

Ans.

317

6

318

Chapter 6

Fluid MoMentuM

6.4

THE ANGULAR MOMENTUM EQUATION

In some cases it is necessary to obtain the torque produced by a flow. This typically occurs when analyzing turbomachinery having blades that rotate about a fixed axis, as in Fig. 6–9. Examples include pumps, turbines, fans, and compressors. In addition, a fixed blade or structure may be used to redirect a flow, as in Fig. 6–10, and for this case the moment reaction at the support must be obtained. A fluid analysis of both of these types of problems can be accomplished by applying the angular momentum equation. For a fluid particle of mass m, angular momentum is defined as the moment of the particle’s linear momentum mV about a point O, Fig. 6–11. If we consider a control volume consisting of a system of particles, then the angular momentum for the system is Σ1r * mV2, where r is the position vector extending from O to each particle. Since Newton’s second law of motion for a particle is F = ma = d(mV)>dt, then it follows that the sum of the moments about point O of the external forces acting on the system,* Σ(r * F), is equal to the time rate of change of the system’s angular momentum, that is,

6

Fig. 6–9

ΣMO = Σ(r * F) =

d Σ(r * mV) dt

Since we are seeking an Eulerian description, we must use the Reynolds transport theorem to obtain the material derivative of r * mV. Hence, applying Eq. 4–11, where, for the particles in the system, N = r * m V so that H = r * V, we have

Fig. 6–10

dN 0 = Hr dV + Hr Vf>cs # dA dt 0t Lcv Lcs

Control volume

z

d 0 1r * mV2 = (r * V)r dV + (r * V)rVf>cs # dA dt 0t Lcv Lcs Therefore, the angular momentum equation becomes

mV

ΣMO =

r

y

O x Angular momentum r 3 mV

Fig. 6–11

Moment of forces about O

0 (r * V) r dV + (r * V) rVf>cs # dA 0t Lcv Lcs Local angular momentum change about O

(6–5)

Convective angular momentum change about O

*Moments of the internal forces within the system will cancel, since they occur in equal but opposite collinear pairs.

6.4

the angular MoMentuM equation

319

Steady Flow. If we consider steady flow, the first term on the right will be zero, since there will be no local changes within the control volume. Also, for an ideal fluid, which has a constant density and uniform velocity through the open control surfaces, the convective change or second term on the right can be integrated, and we obtain 6

ΣMO = Σ(r * V)rVf>cs # A

(6–6)

Steady flow

This final result is often used for the design of blades used in turbomachinery, as will be shown in Chapter 14.

P ROCEDUR E FOR A N A LY S I S Application of the angular momentum equation follows the same procedure as that for linear momentum. Fluid Description. Define the type of flow, whether it is steady or unsteady and uniform or nonuniform. Also define the type of fluid, whether it is viscous, compressible, or if it can be assumed to be an ideal fluid. For an ideal fluid, the velocity profile will be uniform, and the density will be constant. Control Volume and Free-Body Diagram. Select the control volume so its free-body diagram includes the unknown forces and couple moments or torques that are to be determined. The forces on the free-body diagram include the weight of the fluid and any solid portion of the body, the pressure forces at the open control surfaces, and the components of the resultant normal and shear forces acting on closed control surfaces. Angular Momentum. If the flow is known, the average velocities through the open control surfaces can be determined using V = Q>A, or by applying the continuity equation, the Bernoulli equation, or the energy equation. Set up the inertial x,y,z coordinate axes, and apply the angular momentum equation about a fixed point or axis so that a selected unknown force or moment can be obtained.

The impact of water flowing onto and off the blades of this waterwheel causes the wheel to turn.

320

Chapter 6

EXAMPLE

6

Fluid MoMentuM

6.7 Water flows out of the port of the fire hydrant in Fig. 6–12a at 120 liters>s. Determine the reactions at the fixed support necessary to hold the symmetric 30-kg fire hydrant in place.

B

30(9.81) N B

100 mm 0.75 m A

A

Fx MA

125 mm

pAAA

Fy (b)

(a)

Fig. 6–12

SOLUTION Fluid Description. This is a case of steady flow. The water will be assumed to be an ideal fluid, where r = 1000 kg>m3. Control Volume and Free-Body Diagram. We will consider the entire fire hydrant and water contained within it as a fixed control volume. Since the support at A is fixed, three reactions act on its freebody diagram, Fig. 6–12b. Also, the pressure force pAAA acts on the open control surface at A. Since atmospheric pressure exists at B, there is no pressure force at B. Here we will neglect the weight of the water within the fire hydrant. Bernoulli Equation. Before applying the momentum equations, we must first determine the pressure at A. Here, Q = a 120

liters 1000 cm3 1m 3 ba ba b = 0.12 m3 >s s 1 liter 100 cm

1 0.12 m3>s 2

Then, the velocities at A and B are Q = VAAA; Q = V BA B;

1 0.12 m3>s 2

= VA 3p(0.0625 m )2 4;

= VB 3p(0.05 m)2 4;

30.72 m>s p 48 VB = m>s p VA =

6.4

the angular MoMentuM equation

321

Thus, with the datum at A, pA pB VA2 VB 2 + + + zA = + zB g g 2g 2g pA (1000 kg>m3)(9.81 m>s2)

2 30.72 a m>s b p

+

2(9.81 m>s2)

+ 0 = 0 +

2 48 a m>s b p

2(9.81 m>s2)

6

+ 0.75 m

pA = 76.27(103) Pa Linear and Angular Momentum. The force reactions at the support are obtained from the linear momentum equation. For steady flow, ΣF =

0 Vr dV + V rVf>cs # dA 0t Lcv Lcs

ΣF = 0 + VB 1rVBAB 2 + VA 1 -rVAAA 2

Considering the x and y components of the velocities, we have + ΣFx = VBx 1rVBAB 2 + 0 S

Fx = a

48 48 m>s b (1000 kg>m3) a m>s b [p(0.05 m)2] p p Fx = 1833.46 N = 1.83 kN

+ c ΣFy = 0 + 1VAy 21 -rVAAA 2

[76.27(103) N>m2][p(0.0625 m)2] - 30(9.81) N - Fy = a

Ans. 30.72 30.72 m>sb e -(1000 kg>m3) a m>sb [p(0.0625 m)2] f p p

Fy = 1815.09 N = 1.82 kN

Ans.

We will apply the angular momentum equation about point A, in order to eliminate the force reactions at this point. ΣMA =

0 (r * V)r dV + (r * V) rVf>cs # dA 0t Lcv Lcs

⤿

+ ΣMA = 0 + 1rVB 21rVBAB 2

Here the vector cross product becomes a scalar moment of VB about point A. Therefore, MA = c (0.75 m) a

48 48 m>s b d e (1000 kg>m3) a m>s b [p(0.05 m)2] f p p

= 1375.10 N # m = 1.38 kN # m

Ans.

322

Chapter 6

EXAMPLE

Fluid MoMentuM

6.8 The arm of the sprinkler in Fig. 6–13a rotates at a constant rate of v = 100 rev>min. This motion is caused by water that enters the base at 3 liter>s and exits each of the two 20-mm-diameter nozzles. Determine the frictional torque on the shaft of the arm that keeps the rate of rotation constant.

6

z

300 mm 300 mm W v 5 100 revmin

C M pAC

A VA 5 3.141 ms Vƒ/A 5 4.775 ms

(b)

(a)

Fig. 6–13

SOLUTION Fluid Description. As the arm rotates, the flow will be quasi steady, that is, it is cyclic and repetitive. We also will assume the water to be an ideal fluid, where r = 1000 kg>m3. Control Volume and Free-Body Diagram. Here we will choose a control volume that is a fixed disk and contains the moving arm and the water within it, Fig. 6–13a.* We have (quasi) steady tangential flow through the side control surface of this disk. As shown on the free-body diagram, Fig. 6–13b, the forces acting on this control volume include the weight W of the arm and the water within it, the pressure force pCAC from the water supply, and the frictional torque M on the base of the arm at the axle. The fluid exiting the nozzles on the side of the disk is at atmospheric pressure, and so there are no pressure forces here. *If we chose a rotating control volume consisting of the arm and the fluid within it, then the coordinate system, which rotates with it, would not be an inertial reference. Its rotation would complicate the analysis since it produces additional acceleration terms that must be accounted for in the momentum analysis. See Ref. [2].

6.4

the angular MoMentuM equation

323

Velocity. Due to symmetry, the discharge through each nozzle is equal to half the flow. Thus, the velocity of the water passing through each nozzle is Q = VA;

1 3 liter 10-3 m a bca ba b d = Vf>A 3 p 1 0.01 m 2 2 4 s 2 1 liter Vf>A = 4.775 m>s

6

The rotation of the arms causes each nozzle to have a velocity of VA = vr = a 100

rev 2p rad 1 min ba ba b (0.3 m) = 3.141 m>s rev min 60 s

Therefore, the tangential exit velocity of the water from each nozzle through the open control surface as seen by a fixed observer looking down on the fixed control volume (disk), Fig. 6–13a, is Vf = -VA + Vf>A

(1)

Vf = -3.141 m>s + 4.775 m>s = 1.633 m>s Angular Momentum. If we apply the angular momentum equation about the z axis, then no angular momentum enters at the control inlet surface C, since the velocity of the flow is directed along the z axis. The (tangential) velocity Vf of the fluid stream from both nozzles as it exits the side control surface will, however, produce angular momentum about the z axis. The vector cross products for the two moments of Vf can be written in terms of the scalar moments of these velocities. Since they are equal, and the flow is steady, we have ΣM =

0 (r * V)r dV + (r * V)rVf>cs # dA 0t Lcv Lcs

ΣMz = 0 + 2rAVf 1rVf>AAA 2

M = 2(0.3 m)(1.633 m>s)11000 kg>m3 2 (4.775 m>s) 3 p(0.01 m)2 4 M = 1.47 N # m

(2)

Ans.

It is interesting to note that if this frictional torque on the shaft were equal to zero, there would be an upper limit to the rotation v of the arm. To determine this, note that VA = v(0.3 m), and so Eq. 1 then becomes Vf = -v(0.3 m) + 4.775 m>s 0 = 2(0.3 m)3 - v (0.3 m) + 4.775 m>s4 11000 kg>m3 2 (4.775 m>s) 3 p(0.01 m)2 4

Substituting this result into Eq. 2 yields

v = 15.92 rad>s = 152 rev>min

324

Chapter 6

EXAMPLE

Fluid MoMentuM

6.9 The axial-flow pump in Fig. 6–14a has an impeller with blades having a mean radius of rm = 80 mm. Determine the average torque T that must be applied to the impeller, in order to create a flow of water through the pump of 0.1 m3 >s while the impeller rotates at v = 120 rad>s. The open cross-sectional area through the impeller is 0.025 m2. The flow of water is delivered onto each blade along the axis of the pump, and it exits the blade with a component of velocity of 5 m>s that is tangent to the circular motion of the blade, as shown in Fig. 6–14b.

6

T v 5 120 rads

V2

608

2 1

Va 5 4 ms (Vt)2 5 5 ms

80 mm

Flow off the blade

V1 5 Va 5 4 ms Flow onto the blade

4 ms

(b)

(a)

Fig. 6–14

SOLUTION Fluid Description. Flow through the pump slightly away from the impeller can be considered mean steady flow. The water is considered an ideal fluid, where r = 1000 kg>m3. pexit T

pent

(c)

Control Volume and Free-Body Diagram. As in the previous example, we will consider a fixed control volume that includes the impeller and the water surrounding yet slightly removed from it, Fig. 6–14a. The torque on the impeller shaft is shown on the free-body diagram, Fig. 6–14c. Not shown is the weight of the water and blades, along with the force of the pressure distribution around the rim of the closed control surfaces. These forces, along with the forces produced by pent and pexit on the open control surfaces, create no torque about the shaft. Since we are looking for the torque developed by the pump on the shaft, we must apply the angular momentum equation.

6.4

the angular MoMentuM equation

325

Continuity Equation. Flow through each open control surface is determined using the axial component of V. Since the areas are equal, then 0 Vr dV + rVf>cs # dA = 0 0t Lcv Lcs

6

0 - rVa1A + rVa2A = 0 Va1 = Va2 Therefore, flow in the axial direction is constant, Fig. 6–14b, and so 0.1 m3 >s = Va 1 0.025 m2 2

Q = Va A;

Va = 4 m>s

Angular Momentum. Applying the angular momentum equation about the axis of the shaft, for steady flow we have 0 (r * V)r dV + (r * V)rVf>cs # dA 0t Lcv Lcs

ΣM =

T = 0 +

Lcs

rm Vt rVa dA

(1)

Here the vector cross product is replaced by the moment of the tangential component Vt of the water’s velocity V. Only this component produces a moment about the axis of the shaft, Fig. 6–14b. Also, notice that the flow through the open control surfaces is determined only from the axial component of V, that is, rVf>cs # dA = rVa dA. Integrating over the two areas, we get T = rm 1Vt 2 2 1rVa A2 + rm 1Vt 2 1 1 -rVa A2

(2)

As shown in Fig. 6–14b, Vt1 = 0, because the flow is delivered to the blade in the axial direction at the rate of Va = 4 m>s. Due to the torque, at the top of the blade the impeller has given the water a velocity V2 = Va + 1Vt 2 2, but as stated, only the tangential component creates angular momentum. Substituting the data into Eq. 2, we therefore have T = (0.08 m)(5 m>s) 3 11000 kg>m3 2(4 m>s)10.025 m2 2 4 - 0 = 40 N # m

Ans.

A more thorough analysis of axial-flow pumps is covered in Chapter 14.

326

Chapter 6

Fluid MoMentuM

*6.5 PROPELLERS AND WIND TURBINES

6

V1

V2

V4 V

V3 V

2 4

3 1

(a) p

1 pa

2

p4

p3

Pressure variation (b) V

V2

V V1

V3 V

V4 V

Velocity variation (c)

R F

Free-body diagram

Propellers and wind turbines both act like a screw by using several blades mounted on a rotating shaft. In the case of a boat or airplane propeller, applying a torque causes the linear momentum of the fluid in front of the propeller to increase as it flows towards and then through the blades. This change in momentum creates a reactive force on the propeller, which then pushes it forward. A wind turbine works the opposite way, extracting fluid energy or developing a torque from the wind, as the wind passes through the propeller. The design of both of these devices is actually based on the same principles used to design airfoils (or airplane wings). See, for example, Refs. [3] and [5]. In this section, however, we will provide some insight into how these devices operate, using a simplified analysis. First we will discuss the propeller and then the wind turbine. In both cases we will assume the fluid to be ideal.

Propeller. In order for the flow to appear as steady flow, we will observe it relative to the center of the propeller, which we will assume to be stationary.* The fixed control volume excludes the propeller but includes the outlined portion of the fluid slipstream that passes through it. Fluid at the left control surface, 1, is moving towards the propeller with a velocity V1. The fluid within the control volume from 1 to 3 is accelerated because of the reduced pressure (or increased suction) within this region, Fig. 6–15b. If we assume the propeller has many thin blades, then the velocity V is essentially constant as the fluid passes through the propeller from 3 to 4, Figs. 6–15a and 6–15c. The increased pressure that occurs on the right side of the propeller, Fig. 6–15b, pushes the fluid and further accelerates it from 4 to 2. Finally, because of continuity of mass flow, the control surface or slipstream boundary narrows, because the velocity at the far right control surface, 2, is increased to V2, Fig. 6–15a. Realize that this description of the flow is somewhat simplistic, since it neglects the interactive effects between the fluid and the housing to which the propeller is attached. Also, the boundary at the upper and lower closed control surface, from 1 to 2 in Fig. 6–15a, has a discontinuity between the outside still air and the flow within the control volume, when actually there is a smooth transition between the two. Finally, in addition to the axial motion we are considering, the propeller will also impart a rotary motion, or whirl, to the air. In the analysis that follows, we will neglect these effects. Linear Momentum. If the linear momentum equation is applied in the horizontal direction to the fluid within the control volume, then the only horizontal force acting on the control volume’s free-body diagram is that caused by the force of the propeller on the fluid, Fig. 6–15d. (During operation, the pressure on all outside control surfaces remains constant and equal to the atmospheric pressure of the undisturbed fluid. In other words, the gage pressure is zero.) Therefore,

(d)

Fig. 6–15

*We can also consider the center of the propeller to be moving to the left at V1, since the analysis is the same for both cases.

6.5

propellers and Wind turBines

327

0 Vr dV + V rVf>cs # dA 0t Lcv Lcs F = 0 + V2(rV2A2) + V1( -rV1A1)

ΣF =

Since Q = V2A2 = V1A1 = VA = VpR2, where R is the outer radius of the propeller, then F = r 3 V1pR2 2 4 1V2 - V1 2

(6–7)

p4 - p3 = rV1V2 - V1 2

(6–8)

6

As noted in Fig. 6–15c, the velocities V3 = V4 = V, and so no momentum change occurs between sections 3 and 4. Therefore, the force F of the propeller can also be expressed as the difference in pressure that occurs on sections 3 and 4, that is, F = ( p4 - p3)pR2, Fig. 6–15b. Thus, the above equation becomes

It is now necessary to express V in terms of V1 and V2.

Bernoulli Equation. The Bernoulli equation, p>g + V 2 >2g + z =

const. can be applied along a horizontal streamline between points at 1 and 3 and between points at 4 and 2.* Realizing that the (gage) pressures p1 = p2 = 0, we have 0 +

p3 V 12 V2 + 0 = + + 0 g 2g 2g

and p4 V 22 V2 + + 0 = 0 + + 0 g 2g 2g If we add these equations and solve for p4 - p3, where g = rg, we obtain p 4 - p3 =

1 r1V 22 - V 12 2 2

Finally, equating this equation to Eq. 6–8, we get V =

V1 + V2 2

(6–9)

This result is known as Froude’s theorem, after William Froude, who first derived it. It indicates that the velocity of the flow through the propeller is actually the average of the upstream and downstream velocities. If it is substituted into Eq. 6–7, then the force or thrust developed by the propeller on the fluid becomes F =

rpR2 1V 22 - V 12 2 2

(6–10)

*This equation cannot be applied between points at 3 and 4 because energy is added to the fluid by the propeller within this region. Furthermore, within this region the flow is unsteady.

The rotation v of the propeller will cause points along its length to have a different velocity in accordance with the equation = vr. To maintain a constant angle of attack with the air stream, the blade is given a noticeable angle of twist.

328

6

Chapter 6

Fluid MoMentuM

V1

V2

V4 V

V3 V

2 3

4

1 (a)

Fig. 6–15 (repeated)

Power and Efficiency. The power output of the propeller is caused by the rate of work done by the thrust F. If we think of the fluid ahead of the propeller as being at rest, Fig. 6–15a, and the propeller fixed to an airplane and moving forward at V1, then the power output produced by F is # Wo = FV1

(6–11)

The power input is the rate of work needed to maintain the increase of velocity of the slipstream from V1 to V2. Since this requires the fluid to have a velocity V through the propeller, then # Wi = FV

(6–12)

Finally, the ideal efficiency h (eta) is the ratio of the power output to the power input. Using Eq. 6–9, we have hprop

# Wo 2V1 = # = V1 + V2 Wi

(6–13)

In general, a propeller’s actual efficiency will increase as the speed of the aircraft or boat to which it is attached is increased, although it can never be equal to 1 (or 100%) because of friction losses. As the speed increases, though, there is a point at which the efficiency will start to drop off. This can occur on aircraft propellers when the tips of the blades reach or exceed the speed of sound. When this occurs, the drag force on the propeller will significantly increase due to compressibility of the air. Also, in the case of boats, efficiency is reduced because cavitation can occur if the pressure at the tips of the blades reaches the vapor pressure. Experimental tests on airplane propellers show that their actual efficiencies are in the range of about 60% to 80%. Boats, with their smaller-diameter propellers, have lower efficiencies, often in the range of 40% to 60%.

6.5

329

propellers and Wind turBines

Wind Turbine.

Wind turbines and windmills extract kinetic energy from the wind. The flow pattern for these devices is opposite to that of a propeller, and so it looks like that shown in Fig. 6–16, where the slipstream gets larger as it passes through the blades. Using a similar analysis as that for propellers, we can show that Froude’s theorem also applies, that is, V =

V1 + V2 2

6

(6–14)

Power and Efficiency. Using a derivation that compares with Eq. 6–10, the force of the wind on the blades is F =

rpR2 1V 12 - V 22 2 2

Here the parcel of air passing through the# blades has an area of A = pR2 and a velocity V, so the power output is Wo = FV, which becomes # 1 Wo = rVA1V 12 - V 22 2 2

(6–15)

This result actually represents the time rate of loss in the kinetic energy of the wind as it passes through the blades. It is customary to measure the power input as the time rate of change of the kinetic energy of the wind passing through the area swept by the blades, pR2, but without the blades being present to disturb the velocity # # # V1. Since m = rAV1, then Wi = 12mV 12 = 12(rAV1)V 12. Therefore, the efficiency of a wind turbine is # 1 2 2 V1V 12 - V 22 2 Wo 2 rVA1V 1 - V 2 2 hturbine = # = = 1 2 V 13 Wi 2 1rAV1 2V 1

V1

V2 1 2 Wind Turbine

Fig. 6–16

Substituting Eq. 6–14, and simplifying, we get hturbine =

V 22 V2 1 c1 - a 2b d c1 + a b d 2 V1 V1

(6–16)

If hturbine is plotted as a function of V2 >V1, we can show that the curve passes through a maximum of hturbine = 0.593 when V2 >V1 = 1>3.* In other words, a wind turbine can extract a maximum of 59.3% of the total power of the wind. This is known as Betz’s law, named after the German physicist Albert Betz, who derived it in 1919. Wind turbines have a rated power at a specified wind speed, but those currently used in the power industry base their performance on a capacity factor. This is a ratio comparing the actual power output for a year to its rated continuous power output for the year. For modern-day wind turbines, the capacity factor is usually between 0.30 and 0.40. See Ref. [6]. Also, for efficient operation, wind interference between neighboring turbines is minimized when they stand at least a distance of ten blade lengths apart. *See Prob. 6–73.

Wind turbines are gaining in popularity as a device for harnessing energy. Like an airplane propeller, each blade acts as an airfoil. The simplified analysis presented here indicates the basic principles; however, a more complete analysis would include treating each propeller as a wing.

330

Chapter 6

EXAMPLE

Fluid MoMentuM

6.10 The motor on a small boat has a propeller with a radius of 60 mm, Fig. 6–17a. If the boat is traveling at 2 m>s, determine the thrust on the boat and the ideal efficiency of the propeller if it discharges water through it at 0.04 m3 >s.

6

SOLUTION Fluid Description. We consider the flow relative to the boat to be steady flow, and the water to be incompressible, where r = 1000 kg>m3. (© Carver Mostardi/Alamy)

Analysis. The average velocity of the water flowing through the propeller is 0.04 m3 >s = V3p(0.06 m)2 4

Q = VA;

V = 3.537 m>s

60 mm

2 m/s

V2

V1

To achieve steady flow, the control volume moves with the propeller, and so the flow into it is V1 = 2 m>s, Fig. 6–17a. We can now obtain the flow out, V2, from Eq. 6–9. V =

V1 + V2 ; 2

V2 = 2 (3.537 m>s) - 2 m>s = 5.074 m>s

Applying Eq. 6–10 to find the thrust on the boat, Fig. 6–17b, we have

(a)

F =

rpR2 (V 22 - V 12) 2 (1000 kg>m3) (p) (0.06 m)2

=

2

F

= 122.94 N = 123 N (b)

Fig. 6–17

3 (5.074 m>s)2

- (2 m>s)2 4

Ans.

The power output and input are determined from Eqs. 6–11 and 6–12, respectively. # Wo = FV1 = (122.94 N) (2 m>s) = 245.88 W # Wi = FV = (122.94 N) (3.537 m>s) = 434.82 W Thus, the ideal efficiency of the propeller is # Wo 245.88 W h = # = = 0.5654 = 56.5% 434.82 W Wi This same result can also be obtained from Eq. 6–13. h =

2(2 m>s) 2V1 = = 0.5654 V1 + V2 (2 m>s) + (5.074 m>s)

Ans.

6.6

6.6

appliCations For Control voluMes having aCCelerated Motion

331

APPLICATIONS FOR CONTROL VOLUMES HAVING ACCELERATED MOTION

For some problems, it will be convenient to choose a control volume of fixed size and shape that is accelerating.* If Newton’s second law is written for a fluid system that is within an accelerating control volume, then from a Lagrangian description, we have ΣF =

d(mV) dt

(6–17)

If the fluid velocity V is measured from a fixed inertial reference, then it is equal to the velocity of the mass center of the control volume, Vcv, plus the velocity of the fluid measured relative to the mass center, Vf>cv, that is, V = Vcv + Vf>cv Substituting this into Eq. 6–17, where the first term represents the mass times the acceleration of the control volume, we have ΣF = m

d(mVf>cv) dVcv + dt dt

(6–18)

The time rate of change indicated by the second term can be expressed as an Eulerian description using the Reynolds transport theorem, with H = mVf>cv >m = Vf>cv. Therefore, d(mVf>cv) dt

=

0 V r dV + Vf>cs rVf>cs # dA 0t Lcv f>cv Lcs

Substituting this into Eq. 6–18, we get

ΣF = m Sum of forces

dVcv 0 + V r dV + Vf>cs rVf>cs # dA dt 0t Lcv f>cv Lcs

Mass times acceleration of control volume mass center

Relative local change in momentum

(6–19)

Relative convective change in momentum

This result indicates that the sum of the external forces acting on the accelerating control volume is equal to the inertial effect of the entire mass contained within the control volume, plus the rate at which the momentum of the fluid system relative to the control volume’s mass center is changing, plus the convective rate at which momentum relative to the control surfaces is leaving or entering. We will consider important applications of this equation in the following two sections.

6

332

Chapter 6

Fluid MoMentuM

*6.7

Turbine

6

Fuel Injector

Axial Compressors

Turbojet Engine (a) Outer Airflow

Fuel Injector

Turbine

Fan

Axial Compressors

Turbofan Engine (b)

Fig. 6–18 y x Vcv ma Ve ma 1 mf

(a)

y x FD

u W (b)

Fig. 6–19

TURBOJETS AND TURBOFANS

A turbojet or turbofan engine is used mainly for the propulsion of aircraft. As shown in Fig. 6–18a, a turbojet operates by taking in air at its front, and then increasing its pressure by passing it through a series of fans called a compressor. Once the air is under high pressure, fuel is injected and ignited in the combustion chamber. The resulting hot gases expand and then move at a high velocity through a turbine. A portion of the kinetic energy of the gas is used to turn the turbine, which is connected to a shaft, thereby turning the compressor. The remaining energy of the gas is used to propel the aircraft, as the gas is ejected through the exhaust nozzle. Like a propeller, the efficiency of a turbojet increases as the speed of the aircraft increases. A turbofan engine works on the same principle, except that a fan is added on the front of the axial compressors and turns with the shaft to supply additional incoming air by propelling some of it through a duct surrounding the turbojet, and thereby developing an increased thrust, Fig. 6–18b. To analyze the motion and thrust of a turbojet (or turbofan), we will consider the engine represented in Fig. 6–19a.* Here the control volume surrounds the engine and contains the gas (air) and fuel within the engine. If the engine is initially at rest, then the control volume will accelerate with the engine, and so we must apply Eq. 6–19, where measurements are made from a fixed (inertial) coordinate system. For simplicity, we will consider one-dimensional, incompressible steady flow relative to the engine. The forces acting on the free-body diagram of the control volume, Fig. 6–19b, consist of the weight of the engine W and the atmospheric drag FD. The gage pressure acting near the intake and exit control surfaces is zero, since here the air is at atmospheric pressure. Thus, for steady flow, Eq. 6–19, applied along the engine, becomes dVcv (+Q ) -W cos u - FD = m + 0 + Vf>cs r Vf>cs # dA dt Lcs The last term on the right can be evaluated by noting that at the intake the air flow has a relative velocity of (Vf>cs)i = -Vcv, and at the exhaust the fuel and air mixture has a relative velocity of (Vf>cs)e = Ve. Therefore, using our sign convention, dVcv -W cos u - FD = m + 1 -Vcv 21 -raVcv Ai 2 + 1 -Ve 21ra + f Ve Ae 2 dt # # # At the intake, ma = raVcvAi, and at the exhaust, ma + mf = (ra+f Ve Ae). Therefore, dVcv # # # -W cos u - FD = m + maVcv - 1ma + mf 2Ve (6–20) dt The thrust of the engine must overcome the two forces on the left side of this equation, along with the inertia caused by the mass of the engine and its contents, that is, m(dVcv >dt). In other words, the thrust is the “effect” on the engine caused by the mass flow, represented by the last two terms on the right side of Eq. 6–20. # # # T = (ma + mf)Ve - maVcv (6–21) It is for this reason that we have not shown this force on the free-body diagram of the control volume. *In this discussion, the engine is not attached to an airplane or a fixed support.

6.8

6.8

333

roCkets

ROCKETS

A rocket engine burns either solid or liquid fuel that is carried within the rocket. A solid-fueled engine is designed to produce a constant thrust, which is achieved by forming the propellant into a shape that provides uniform burning. Once it is ignited, it cannot be controlled. A liquidfueled engine requires a more complex design, involving the use of piping, pumps, and pressurized tanks. It operates by combining a liquid fuel with an oxidizer as the fuel enters a combustion chamber. Control of the thrust is then achieved by adjusting the flow of the fuel. The performance of either of these types of rockets can be studied by applying the linear momentum equation in a manner similar to that used for the turbojet. If we consider the control volume to be the entire rocket, as shown in Fig. 6–20a, then Eq. 6–20 can be applied, except here the # weight is vertically downward, Fig. 6–20b, and ma = 0 since only fuel is used. Therefore, for a rocket we have (+ c)

-W - FD = m

dVcv # - mfVe dt

6

Turbofan engine for a commercial jetliner.

FD

(6–22)

A numerical application of this equation is given in Example 6.12. W

Vcv

IMPORTANT POIN T S Ve

mf

• Propellers are propulsion devices that act as a screw, causing the linear momentum of the fluid in front of the propeller to increase as the fluid flows towards and then past the blades. Wind turbines act in an opposite manner, removing energy from the wind, and thus decreasing the wind’s momentum. A simple analysis of the flow can be made for both of these devices using the linear momentum and Bernoulli equations.

(a)

(b)

Fig. 6–20

• If the control volume is selected to have accelerated motion, then the momentum equation must include an additional term, m1dVcv >dt2, which accounts for the inertia of the mass within the control volume. This term is required because Newton’s second law forms the basis of the momentum equation, and measurements must be made from a nonaccelerating or inertial reference frame.

• The momentum equation can be used to analyze the motion of turbojets and rockets. The thrust produced is not shown on the free-body diagram because it is the result of the terms involving the mass flow out of the engine.

(© AF archive/Alamy)

334

Chapter 6

EXAMPLE

6

Fluid MoMentuM

6.11 The jet plane in Fig. 6–21a is in level flight at a constant speed of 140 m>s. Each of its two turbojet engines burns fuel at a rate of 3 kg>s. Air, at a temperature of 15°C, enters the intake, which has a crosssectional area of 0.15 m2. If the exhaust has a velocity of 700 m>s, measured relative to the plane, determine the drag acting on the plane. W

140 ms FD

FL (a)

(b)

Fig. 6–21

SOLUTION Fluid Description. This is a case of steady flow when it is measured relative to the plane. Within the engine the air is compressible; however, in this problem we do not have to account for this effect since the resulting exhaust velocity is known. From Appendix A, at T = 15°C, ra = 1.23 kg>m3. Analysis. To determine the drag we will consider the entire plane, its two engines, and the air and fuel within the engines as the control volume, Fig. 6–21a. This control volume moves with a constant speed of Vcv = 140 m>s, so the mass flow of the air delivered to each engine is # ma = raVcv A = 11.23 kg>m3 21140 m>s210.15 m2 2 = 25.83 kg>s

Conservation of mass requires that this same mass flow of air pass through the engine. The free-body diagram of the control volume is shown in Fig. 6–21b. Applying Eq. 6–20, realizing there are two engines, we have + ) -W cos u - FD = m (d

dVcv # + maVcv dt

1 m# a

# + mf 2 Ve

0 - FD = 0 + 23(25.83 kg>s)(140 m>s) - (25.83 kg>s + 3 kg>s)(700 m>s)4 FD = 33.131103 2 N = 33.1 kN

Ans.

Because the plane is in equilibrium, this force is equivalent to the thrust provided by the engines.

6.8

EXAMPLE

roCkets

6.12

The rocket and its fuel in Fig. 6–22 have an initial mass of 5 Mg. When the rocket takes off from rest, the 3 Mg of fuel is consumed at the rate of # mf = 80 kg>s, and is expelled at a constant speed of 1200 m>s, relative to the rocket. Determine the maximum velocity of the rocket. Neglect the change in gravity due to altitude and the drag resistance of the air.

6

V

SOLUTION Fluid Description. The flow is steady when it is measured relative to the rocket. Compressibility effects of the air on the rocket are not considered here, since the drag resistance is not considered. Analysis. The rocket and its contents are selected as an accelerating control volume, Fig. 6–22a. The free-body diagram is shown in Fig. 6–22b. For this problem, Eq. 6–20 becomes dVcv # ( + c ) -W = m - mfVe dt Here Vcv = V, and at any instant t during the flight, the mass of the rocket is m = (5000 - 80t) kg. Since W = mg, we have dV - 3(5000 - 80t) kg4 19.81 m>s2 2 = 3(5000 - 80t) kg4 - (80 kg>s)(1200 m>s) dt In this equation, the thrust of the engine is represented by the last term. Separating the variables and integrating, where V = 0 when t = 0, we have L0

V

dV =

L0

t

a

V = -1200 ln(5000 - 80t) - 9.81t `

W

5000 b - 9.81t 5000 - 80t

t 0

Maximum velocity occurs at the instant all the fuel has been exhausted. The time t′ needed to do this is # mf = mf t′; 3(103) kg = (80 kg>s)t′ t′ = 37.5 s

Vmax

(a)

80(1200) - 9.81b dt 5000 - 80t

V = 1200 ln a

Therefore,

335

5000 = 1200 ln a b - 9.81(37.5) 5000 - 80(37.5) Vmax = 732 m>s

(b)

Fig. 6–22

Ans.

If we include the effect of air resistance, then the solution will become more complicated, as we will discuss in Chapter 13.

336

Chapter 6

Fluid MoMentuM

References 1. J. R. Lamarch and A. J. Baratta, Introduction to Nuclear Engineering,

Prentice Hall, Inc., Upper Saddle River, NJ, 2001. 2. J. A. Fry. Introduction to Fluid Mechanics, MIT Press, Cambridge,

6

MA, 1994. 3. National Renewable Energy Laboratory, Advanced Aerofoil for Wind

Turbines (2000), DOE/GO-10098-488, Sept. 1998, revised Aug. 2000. 4. M. Fremond et al., “Collision of a solid with an incompressible fluid,”

Journal of Theoretical and Computational Fluid Dynamics, London, UK. 5. D. A. Griffin and T. D. Ashwill “Alternative composite material for

megawatt-scale wind turbine blades: design considerations and recommended testing,” J Sol Energy Eng 125:515–521, 2003. 6. S. M. Hock, R. W. Thresher, and P. Tu, “Potential for far-term advanced wind turbines performance and cost projections,” Sol World Congr Proc Bienn Congr Int Sol Energy Soc 1:565–570, 1992. 7. B. MacIsaak and R. Langlon, Gas Turbine Propulsion Systems, American Institute of Aeronautics and Astronautics, Ruston, VA, 2011. 8. V. Streeter, et al., Fluid Mechanics, 9th edition, Mc Graw Hill, N. Y.

F U NDAMEN TAL PRO B L EM S SEC. 6.1–6.2 F6–1. Water is discharged into the atmosphere at B through the 40-mm-diameter elbow at 0.012 m3 >s. If the pressure at A is 160 kPa, determine the resultant force the elbow exerts on the water.

F6–2. The shield of negligible weight is held at an angle of 60° to deflect the 40-mm-diameter water stream, which is discharged at 0.02 m3 >s. If the guide on the shield deflects 30% of the discharge upwards, determine the resultant force of the shield on the water. Assume all the water flows in the vertical plane.

B

A

608

Prob. F6–1

Prob. F6–2

337

FundaMental proBleMs F6–3. Water is flowing at 10 m>s from the 50-mm-diameter smooth open pipe AB. If the flow is increasing at 3 m>s2, determine the pressure in the pipe at A.

F6–5. The table fan develops a slipstream that has a diameter of 0.25 m. If the air is moving horizontally at 20 m>s as it leaves the blades, determine the horizontal friction force that the table must exert on the fan to hold it in place. Assume the air has a constant density of 1.22 kg>m3, and that the air just to the right of the blade is essentially at rest.

0.2 m

0.25 m A

B

Prob. F6–3

F6–4. Crude oil flows into the open air at the same rate out of each branch of the horizontal wye fitting. If the pressure at A is 80 kPa, determine the resultant force at A to hold the fitting to the pipe.

Prob. F6–5

SEC. 6.3 F6–6. As water flows out the 20-mm-diameter pipe, it strikes the vane, which is moving to the left at 1.5 m>s. Determine the resultant force on the vane needed to deflect the water 90° as shown.

B

C 458

20 mm 458

20 mm

A 30 mm 20 ms 1.5 ms 6 ms

Prob. F6–4

Prob. F6–6

6

338

Chapter 6

Fluid MoMentuM

P R OB L EMS SEC. 6.1–6.2 6

6–1. Flow through the circular pipe is turbulent, and the velocity profile can be modeled using Prandtl’s one-seventh power law, u = Vmax (1 - r>R)1>7. If r is the density, show that the momentum of the fluid per unit time passing through the pipe is (49>72)pR2rV 2max. Also, show that Vmax = (60>49)V, where V is the average velocity of the flow, and that the momentum per unit time is (50>49)pR2rV 2.

*6–4. Oil flows through the 100-mm-diameter pipe with a velocity of 8 m>s. If the pressure in the pipe at A and B is assumed to be 60 kPa, determine the x and y components of force the flow exerts on the elbow. The flow occurs in the horizontal plane. Take ro = 900 kg>m3.

8 ms

A 308 100 mm B

r R

100 mm

Prob. 6–1

Prob. 6–4

6–2. Determine the linear momentum of a mass of fluid in a 0.2-m length of pipe if the velocity profile for the fluid is a paraboloid as shown. Compare this result with the linear momentum of the fluid using the average velocity of flow. Take r = 800 kg>m3.

6–5. The steady jet of water flows from the 100-mmdiameter pipe at 4 m>s. If it strikes the fixed vane and is deflected in the vertical plane as shown, determine the volumetric flow towards A and towards B if the tangential component of the force that the water exerts on the vane is zero.

u 5 4 (1 2 100 r 2) ms A

458

100 mm

0.1 m r

C

0.2 m

4 ms B

Prob. 6–2 Prob. 6–5 6–3. Oil flows through the 200-mm-diameter pipe with a velocity of 8 m>s. If it discharges into the atmosphere through the nozzle, determine the total force the bolts must resist at the connection AB to hold the nozzle onto the pipe. Take ro = 920 kg>m3. 200 mm

6–6. The steady jet of water flows from the 100-mmdiameter pipe at 4 m>s. If it strikes the fixed vane and is deflected in the vertical plane as shown, determine the normal force the jet exerts on the vane.

B

A 100 mm

458

100 mm

8 m/s C 4 ms A

Prob. 6–3

B

Prob. 6–6

proBleMs 6–7. The boat is filled with water using a 50-mm-diameter hose. If the velocity of flow out of the hose is VA = 8 m>s in the direction shown, determine the force in the tie rope needed to hold the boat stationary. Assume the level of water within the boat remains horizontal and the rate at which it rises can be neglected.

339

6–9. Water is ejected from the hose at A with a velocity of 8 m>s. Determine the force that the water exerts on the ceiling. Assume the water does not splash back off the ceiling.

6 C

5m 8 ms A

B

10 m

308 3m

B

1.5 m

A

Prob. 6–7 60 mm

*6–8. Water flows out of the reducing elbow at 0.05 m3 >s. Determine the horizontal and vertical components of force that are necessary to hold the elbow in place at A. Neglect the size and weight of the elbow and the water within it. The water is discharged to the atmosphere at B.

Prob. 6–9

6–10. A volumetric discharge of 1.25 m3 >s passes out the pipe. Determine the horizontal force the water exerts on the contraction.

B

60 mm 30°

0.9 m 0.6 m A

120 mm

Prob. 6–8

Prob. 6–10

340

Chapter 6

Fluid MoMentuM

6–11. Water flows out of the reducing elbow at 0.03 m3 >s. Determine the horizontal and vertical components of force that are necessary to hold the elbow in place. The pipe and elbow and the water within have a total mass of 80 kg. The water is discharged to the atmosphere at B.

6–15. The 500-g hemispherical bowl is held in equilibrium by the vertical jet of water discharged through the 10-mm-diameter nozzle. Determine the height h of the bowl as a function of the volumetric flow Q of the water through the nozzle. Plot the height h (vertical axis) versus Q for 0.5 1 10-3 2 m3 >s … Q … 1 1 10-3 2 m3 >s. Give values for increments of ∆Q = 0.1 1 10-3 2 m3 >s.

6 100 mm

30°

6–14. The hemispherical bowl of mass m is held in equilibrium by the vertical jet of water discharged through a nozzle of diameter d. If the volumetric flow is Q, determine the height h at which the bowl is suspended. The water density is rw.

A

1m 30°

h B

50 mm

Prob. 6–11 *6–12. The pipe AB has a diameter of 40 mm. If water flows through it at a rate of 0.015 m3 >s, determine the horizontal and vertical components of force that must be supported at A to hold the pipe in place. The pipe and water within it have a total mass of 9 kg. 6–13. Pipe AB has a diameter of 40 mm. Determine the maximum volumetric flow through the pipe if the vertical tensile force developed at A cannot exceed 300 N. The pipe and water within it have a total mass of 9 kg.

Probs. 6–14/15 *6–16. The nozzle has a diameter of 40 mm. If it discharges water with a velocity of 20 m>s against the fixed blade, determine the force exerted by the water on the blade. The blade divides the water evenly in the vertical plane at an angle of u = 45°. 6–17. The nozzle has a diameter of 40 mm. If it discharges water in the vertical plane with a velocity of 20 m>s against the fixed blade, determine the force exerted by the water on the blade as a function of the blade angle u. Plot this force (vertical axis) versus u for 0° … u … 75°. Give values for increments of ∆u = 15°. The blade divides the water evenly.

B

2m 40 mm

C u A u B

A

Probs. 6–12/13

Probs. 6–16/17

proBleMs 6–18. Water flows with a velocity of 6 m>s through the pipe at A. If the pressure at A is 65 kPa, determine the resultant horizontal shear force developed along the seam at C that holds the cap to the larger pipe.

341

6–22. Water flows through the elbow with a velocity of 4 m>s. Determine the horizontal and vertical components of force the support at C exerts on the elbow. The pressure within the pipe at A and B is 200 kPa. The flow occurs in the horizontal plane. Assume there is no support at A and B.

300 mm 200 mm 6 ms

A

6

B

4 ms C 75 mm

Prob. 6–18 75 mm

6–19. The disk valve is used to control the flow of 0.008 m3 >s of water through the 40-mm-diameter pipe. Determine the force F required to hold the valve in place for any position x of closure of the valve.

A B 608 458

40 mm

C

x F

Prob. 6–19 *6–20. The water jet is ejected from the 80-mm-diameter pipe with a steady velocity of 9 m>s. If it strikes the surface and is deflected in the vertical plane as shown, determine the normal force the jet exerts on the surface.

Prob. 6–22

6–23. Air flows through the 500-mm-wide rectangular duct at 45 000 liters>min. Determine the vertical force acting on the end plate B of the duct. Take ra = 1.202 kg>m3.

6–21. The water jet is ejected from the 80-mm-diameter pipe with a steady velocity of 9 m>s. If it strikes the surface and is deflected in the vertical plane as shown, determine the volumetric flow towards A and towards B if the tangential component of the force that the water exerts on the surface is zero.

200 mm B

A 9 m/s

80 mm B

A 30°

Probs. 6–20/21

800 mm

Prob. 6–23

342

Chapter 6

Fluid MoMentuM

*6–24. Water flows through the pipe A at 6 m>s. Determine the x and y components of force the water exerts on the horizontal assembly. The pipe has a diameter of 100 mm at A, and at B and C the diameter is 60 mm. The water is discharged into the atmosphere at B and C.

6–26. The 20-mm-diameter pipe ejects water towards the wall as shown. Determine the normal force the water stream exerts on the wall.

6

6m s

B 100 mm

VA

A

1.5 m 30°

A

60 mm

8m

Prob. 6–26

C 5 4 3

B 60 mm

Prob. 6–24

6–25. The 300-kg circular craft is suspended 100 mm from the ground. For this to occur, air is drawn in at 18 m>s through the 200-mm-diameter intake and discharged to the ground as shown. Determine the pressure that the craft exerts on the ground. Take ra = 1.22 kg>m3.

6–27. Water flows through the 40-mm-diameter pipe assembly such that the velocity at A is 6 m>s and the pressure is 120 kPa. Determine the horizontal and vertical components of force at the support C, assuming there is no support resistance at A and B. The assembly and the water within it have a mass of 54 kg.

40 mm

2m A

1.5 m

6 m/s

1.5 m 200 mm C 2m

C

A

B

100 mm B

Prob. 6–25

Prob. 6–27

343

proBleMs *6–28. Water flows through the pipe C at 4 m>s. Determine the x and y components of force necessary to hold the horizontal pipe assembly in equilibrium. The pipe has a diameter of 60 mm at C, and at A and B the diameter is 20 mm.

4 5 3

6–30. A nuclear reactor is cooled with liquid sodium, which is transferred through the reactor core using the electromagnetic pump. The sodium moves through a pipe at A having a diameter of 90 mm, with a velocity of 6 m>s and pressure of 150 kPa, and passes through the rectangular duct, where it is pumped by an electromagnetic force giving it a 9-m pumphead. If it emerges at B through a 6 60-mm-diameter pipe, determine the restraining force F on each arm needed to hold the pipe in place. Take rNa = 850 kg>m3.

B

A A F

6 m/s 90 mm F

4 ms

60 mm

B

D C

Prob. 6–28 Prob. 6–30 6–29. Crude oil flows through the horizontal tapered 45° elbow at 0.02 m3 >s. If the pressure at A is 300 kPa, determine the horizontal and vertical components of the resultant force the oil exerts on the elbow. Neglect the size of the elbow.

6–31. A speedboat is powered by the jet drive shown. Seawater is drawn into the pump housing at the rate of 0.35 m3 >s through a 120-mm-diameter intake A. An impeller accelerates the water and forces it out horizontally through an 80-mm-diameter nozzle B. Determine the horizontal and vertical components of thrust exerted on the speedboat. The density of seawater is rsw = 1030 kg>m3.

A

50 mm

1358

A

B

30 mm B

Prob. 6–29

Prob. 6–31

608

344

6

Chapter 6

Fluid MoMentuM

*6–32. Water flows through the elbow at 3 m>s. Assuming the pipe connections at A and B do not offer any resistance on the elbow, determine the resultant horizontal force F that the support must exert on the elbow in order to hold it in equilibrium. The pressure within the pipe at A and B is 80 kPa.

6–34. The 80-kg man stands on the scale. If the bucket of water has a mass of 10 kg, and water is flowing into it at 0.001 m3 >s from a 20-mm-diameter hose, determine the reading on the scale at this instant. Assume the bucket is large so that the rate at which the level of water in the bucket is rising can be neglected.

20 mm

60°

60° 100 mm 3 m/s A

B

Prob. 6–34

F

6–35. Pipe AB has a diameter of 40 mm. If water flows through it at 0.015 m3 >s, determine the horizontal and vertical components of force that must be supported at A to hold the pipe in place. The pipe and the water it contains have a total mass of 12 kg.

Prob. 6–32

6–33. A stilling basin is 5 m wide and is used to confine the flow over a spillway. To retard the flow, 12 baffle blocks are used. Determine the horizontal force on each block if the flow is 80 m3 >s.

*6–36. Pipe AB has a diameter of 40 mm. If the tensile force developed in the pipe at point A cannot exceed 250 N, determine the maximum allowable volumetric flow through the pipe. The pipe and the water it contains have a total mass of 12 kg.

0.5 m B

3m

2.5 m 1.2 m 608 A

Prob. 6–33

Probs. 6–35/36

345

proBleMs

SEC. 6.3 6–37. A 25-mm-diameter stream flows at 10 m>s against the blade and is deflected 180° as shown. If the blade is moving to the left at 2 m>s, determine the horizontal force F of the blade on the water.

*6–40. The water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling friction, the cart moves to the right with a constant velocity of 2 m>s. The discharge from the 50-mm-diameter nozzle is 0.04 m3 >s. One-fourth of the discharge flows down the incline, and three-fourths flows up the incline. Assume 6 steady flow, all within the vertical plane.

B 2 ms F 10 ms B 25 mm

A

A

Prob. 6–37 2m s

608

6–38. Solve Prob. 6–37 if the blade is moving to the right at 2 m>s. At what speed must the blade be moving to the right to reduce the force F to zero?

C

Prob. 6–40

B 2 ms F 10 ms

25 mm

A

Prob. 6–38 6–39. The boat is powered by the fan, which develops a slipstream having a diameter of 1.25 m. If the fan ejects air with a velocity of 40 m>s, measured relative to the boat, and the boat is traveling with a velocity of 8 m>s, determine the force the fan exerts on the boat. Assume that the air has a constant density of ra = 1.22 kg>m3 and that the air entering at A is essentially at rest relative to the ground.

6–41. Water flows through the hose with a velocity of 3 m>s. Determine the force F on the semicylindrical cup if it is moving to the right at 2.5 m>s. 6–42. Water flows through the hose with a velocity of 3 m>s. Determine the force F needed to keep the semicylindrical cup moving to the left at 1.5 m>s.

C

1.25 m 50 mm A B

20 mm 3 ms

F B A D

Prob. 6–39

Probs. 6–41/42

346

6

Chapter 6

Fluid MoMentuM

6–43. A plow located on the front of a truck horizontally scoops up a liquid slush at the rate of 0.55 m3 >s and throws it off vertically, perpendicular to its motion, i.e., u = 90°. If the truck is traveling at a constant speed of 5 m>s, determine the resistance to motion caused by the shoveling. The density of the slush is rs = 1150 kg>m3. *6–44. The truck is traveling forward at 5 m>s, shoveling a liquid slush that is 0.25 m deep. If the slush has a density of 125 kg>m3 and is thrown upward at an angle of u = 60° from the 3-m-wide blade, determine the traction force of the wheels on the road necessary to maintain the motion. Assume that the slush is thrown off the shovel at the same rate as it enters the shovel.

6–46. Determine the power required to keep the vane moving to the left with a velocity of 4 m>s. The discharge from the 50-mm-diameter nozzle is 0.02 m3 >s. Two-thirds of the discharge flows up the incline and one-third flows down and around.

B 50 mm 458 A

4 m/s C u

Prob. 6–46

B

A

6–47. The car is used to scoop up water that is in a trough between the tracks. Determine the force needed to pull the car forward at constant velocity v for each of the three cases. The scoop has a cross-sectional area A and the density of water is rw.

Probs. 6–43/44

6–45. A 20-mm-diameter stream flows at 8 m>s against the blade and is deflected 120° as shown. If the blade is moving to the right at 3 m>s, determine the force F of the blade on the water.

v

v

F1

F2

(a)

B

(b) v

308 8 ms

20 mm

Prob. 6–45

A

F3 F (c)

Prob. 6–47

347

proBleMs *6–48. The vane is moving at 8 m>s when a jet of water having a velocity of 20 m>s enters at A. If the cross-sectional area of the jet is 650 mm2, and it is diverted as shown, determine the power developed by the water on the vane.

SEC. 6.4 6–50. Water flows through the pipe with a velocity of 3 m>s. Determine the horizontal and vertical components of force and the moment at A needed to hold the elbow in place. Neglect the weight of the elbow, the water within it, and the 6 elevation change.

40 mm B 608 8 m/s

300 mm

308 A

A

Prob. 6–48

3 m/s

80 mm

300 mm

Prob. 6–50

6–49. The large water truck releases water at the rate of 2400 liters>min through the 80-mm-diameter pipe. If the depth of the water in the truck is 1.5 m, determine the frictional force the road has to exert on the tires to prevent the truck from rolling.

6–51. The chute has a cross-sectional area of 0.03 m2. If the flow is 0.4 m3 >s, determine the horizontal and vertical force components at the pin A, and the horizontal force at the roller B, necessary for equilibrium. Neglect the weight of the chute and the water on it.

B

4m 1.5 m

A

3m

Prob. 6–49

Prob. 6–51

348

Chapter 6

Fluid MoMentuM

*6–52. Air enters into the propeller tube at A with a mass flow of 3 kg>s and exits at the ends B and C with a velocity of 400 m>s, measured relative to the tube. If the tube rotates at 1500 rev>min, determine the frictional torque M on the tube. 6

6–55. Water flows through the 200-mm-diameter pipe bend with a velocity of 8 m>s and discharges into the atmosphere at B. Determine the horizontal and vertical components of force and the moment acting on the coupling at A. The mass of the pipe bend and the water within it is 25 kg, having a center of gravity at G.

B

0.5 m 600 mm M A

A

1500 revmin

608

0.5 m

G B 450 mm

C 200 mm

Prob. 6–52 Prob. 6–55

6–53. The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water is supplied through the hose at 0.006 m3 >s and is ejected through the four nozzles into the atmosphere. Determine the torque required to keep the arms from rotating. 6–54. The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water is supplied through the hose at 0.006 m3 >s and is ejected through the four nozzles into the atmosphere. Determine the constant angular velocity of the arms. Neglect friction at the vertical axis.

*6–56. The waterwheel consists of a series of flat plates that have a width b and are subjected to the impact of water to a depth h, from a stream that has a velocity V. If the wheel is turning at v, determine the power supplied to the wheel by the water.

300 mm

308 308

308

R v

308

h V

Probs. 6–53/54

Prob. 6–56

349

proBleMs 6–57. The fan blows air at 120(103) liters>min. If the fan has a mass of 15 kg and a center of gravity at G, determine the smallest diameter d of its base so that it will not tip over. Assume the airstream through the fan has a diameter of 0.5 m. The density of the air is ra = 1.202 kg>m3.

6–59. Water flows into the bend fitting with a velocity of 3 m>s. If the water exits at B into the atmosphere, determine the horizontal and vertical components of force and the moment at C needed to hold the fitting in place. Neglect the weight of the fitting and the water within it. 6

A G

0.5 m 200 mm

150 mm

150 mm A 3 ms 0.15 m

30°

C B

1.5 m

150 mm

Prob. 6–59

d 2

d 2

Prob. 6–57 6–58. If the propeller tube rotates at a constant rate of 60 rad>s when 1.75 kg>s of air enters at A, determine the frictional torque M acting on the tube. The density of air is ra = 1.23 kg>m3. The diameter of the nozzles at B, C, D, and E is 100 mm.

*6–60. The bend is connected to the pipe at flanges A and B as shown. If the diameter of the pipe is 160 mm and the volumetric flow is 0.18 m3 >s, determine the horizontal and vertical components of force and the moment exerted at the fixed base D of the support. The total mass of the bend and the water within it is 120 kg, and the mass center is at point G. The pressure of the water at A is 120 kPa. Assume that no force is transferred to the flanges at A and B.

100 mm B

A 1m

0.5 m

1m

M

E 0.5 m

A D

D

60 rads

30°

G

0.5 m 1.5 m

C 0.5 m

Prob. 6–58

0.5 m

Prob. 6–60

B

350

6

Chapter 6

Fluid MoMentuM

6–61. Water is discharged into the atmosphere from the pipe at 10 m>s. Determine the horizontal and vertical components of force and the moment that is developed at the fixed support A in order to hold the pipe in equilibrium. Neglect the resistance provided by the pipe at C, and the weight of the pipe and the water within it. 6–62. Water flows into the Tee fitting at 3.6 m>s. If a pipe is connected to B and the pressure in the pipe at B is 75 kPa, determine the horizontal and vertical components of force and the moment that must be exerted on the fixed support at A to hold the pipe in equilibrium. Neglect the resistance provided by the pipe at B and C, and the weight of the pipe and the water within it.

100 mm

*6–64. If water flows into the horizontal bend fitting at A with a velocity of 6 m>s, and exits at B into the atmosphere, determine the x and y components of force and the moment at C needed to hold the fitting in place. 6–65. If water flows into the horizontal bend fitting at A with a velocity of 6 m>s, and exits at B into a tank having a gage pressure of 50 kPa, determine the x and y components of force and the moment at C needed to hold the fitting in place.

A

C

6 ms

150 mm

200 mm 400 mm

C

308 B B 300 mm

Probs. 6–64/65

60 mm 150 mm A

Probs. 6–61/62

6–66. If the velocity through the pipe is 4 m>s, determine the horizontal and vertical components of force and the moment exerted on the fixed support D. The total mass of the bend and the water within it is 20 kg, with a mass center at point G. The pressure of the water at A is 50 kPa. Assume that no force is transferred to the flanges at A and B.

6–63. The 5-mm-diameter arms of a rotating lawn sprinkler have the dimensions shown. Water flows out relative to the arms at 6 m>s, while the arms are rotating at 10 rad>s. Determine the torsional resistance at the bearing A, and the speed of the water as it emerges from the nozzles, as measured by a fixed observer.

100 mm B G 608 A

10 rads

200 mm 50 mm

100 mm 600 mm 100 mm 150 mm

608

A D

Prob. 6–63

Prob. 6–66

proBleMs

SEC. 6.5–6.8 6–67. Air is drawn into the jet engine at A at 25 kg>s with a velocity of VA = 125 m>s, while the fuel is burned at 0.75 kg>s. Determine the horizontal force that the jet engine exerts on the supports. The exhaust velocity at B is VB = 550 m>s.

351

6–69. The jet is traveling at a velocity of 400 m>s in still air, while consuming fuel at the rate of 1.8 kg>s and ejecting it at 1200 m>s measured relative to the plane. If the engine consumes 1 kg of fuel for every 50 kg of air that passes through the engine, determine the thrust produced by the engine and the efficiency of the engine.

400 ms

Prob. 6–69

A

B 550 ms

125 ms

Prob. 6–67

6–70. The plane has a mass of 6.5 Mg and is propelled using two jet engines. If the plane lands with a horizontal touchdown velocity of 60 m>s when the braking deflector is engaged, determine the velocity of the plane 4 s after it has landed. The exhaust from the engine measured relative to the plane is 900 m>s, and the steady mass flow is 24 kg>s. Neglect rolling resistance of the landing gear and the rate at which the fuel is consumed.

*6–68. The jet is flying at 750 km>h. Still air enters its engine nacelle at A having a cross-sectional area of 0.8 m2. # Fuel is mixed with the air at me = 2.5 kg>s and is exhausted with a velocity of 900 m>s, measured relative to the plane. Determine the force the engine exerts on the plane. Take ra = 0.850 kg>m3.

308

A

308

Prob. 6–70 6–71. The airplane is flying at 250 km>h through still air as it discharges 350 m3 >s of air through its 1.5-m-diameter propeller. Determine the thrust on the plane. Take ra = 1.007 kg>m3. A

1.5 m

Prob. 6–68

Prob. 6–71

6

352

Chapter 6

Fluid MoMentuM

*6–72. The propeller of a boat discharges 1.85 m3 >s of water as the boat travels at 40 km>h in still water. If the diameter of the propeller is 400 mm, determine the thrust developed by the propeller on the boat. 6–73. Plot Eq. 6–16 and show that the maximum efficiency of 6 a wind turbine is 59.3%, as stated by Betz’s law. 6–74. The 12-Mg helicopter is hovering over a lake as the suspended bucket collects 5 m3 of water used to extinguish a fire. Determine the power required by the engine to hold the filled water bucket over the lake. The horizontal blade has a diameter of 14 m. Take ra = 1.23 kg>m3.

*6–76. The wind turbine has an efficiency of 48% in an 8 m>s wind. If the air is at a standard atmospheric pressure with a temperature of 20°C, determine the thrust on the blade shaft and the power withdrawn by the blades. 6–77. The wind turbine has an efficiency of 48% in an 8 m>s wind. If the air is at a standard atmospheric pressure with a temperature of 20°C, determine the difference between the pressures just in front and just behind the blades. Also find the average velocity of the air passing through the blades.

8 ms

35 m

Probs. 6–76/77 Prob. 6–74 6–75. The jet boat takes in water directly through its bow at 0.03 m3 >s while traveling in still water with a velocity of 10 m>s. If the water is ejected from a pump through the stern at 30 m>s, measured relative to the boat, determine the thrust developed by the engine. What would be the thrust if the water were taken in along the sides of the boat, perpendicular to the direction of motion? If the efficiency is defined as the work done per unit time divided by the energy supplied per unit time, then determine the efficiency for each case. 10 ms

6–78. Determine the largest speed of the breeze that can be generated by the 8-kg fan so that it does not tip over, assuming that slipping will not occur. The blade has a diameter of 400 mm. The mass center is at G. Take ra = 1.20 kg>m3.

G

300 mm

100 mm

Prob. 6–75

Prob. 6–78

75 mm

proBleMs 6–79. The fan is used to circulate air within a large industrial building. The blade assembly of mass 100 kg consists of 10 blades, each having a length of 1.5 m. Determine the power that must be supplied to the motor to lift the assembly off its bearings and allow it to freely turn without friction. What is the downward air velocity for this to occur? Neglect the size of the hub H. Take ra = 1.23 kg>m3.

353

6–82. The balloon has a mass of 20 g (empty) and it is filled with air having a temperature of 20°C. When it is released, it begins to accelerate upwards at 8 m>s2. Determine the initial mass flow of air from the stem. Assume the balloon is a sphere having a radius of 300 mm. Note V = 43pr 3. 6

8 ms2

5 mm 1.5 m

H

Prob. 6–82 Prob. 6–79 *6–80. Air is exhausted from the fixed jet engine at 20 kg>s with a velocity of 480 m>s while it is being tested with an attached brake deflector. Determine the force in each of the two supporting links needed to hold the deflector in the position shown. 308 158

6–83. The balloon carries a bucket containing 200 kg of water. If it is ascending with a constant velocity of 4 m>s and it then releases 80 kg>s of water through a 100-mmdiameter opening, determine the initial upward acceleration of the balloon as the water is being released. The balloon and empty bucket have a combined a mass of 1.5 Mg.

158 308 a

Prob. 6–80 6–81. The jet engine on a plane flying at 160 m>s in still air draws in air at standard atmospheric temperature and pressure through a 0.5-m-diameter inlet. If 2 kg>s of fuel is added and the mixture leaves the 0.3-m-diameter nozzle at 600 m>s, measured relative to the engine, determine the thrust provided by the engine.

160 ms

Prob. 6–81

Prob. 6–83

354

Chapter 6

Fluid MoMentuM

*6–84. The jet has a speed of 1260 km>h when it is flying horizontally. Air enters the intake scoop S at the rate of 52 m3 >s. If the engine burns fuel at the rate of 1.85 kg>s, and the gas (air and fuel) is exhausted relative to the plane with a speed of 2880 km>h, determine the drag exerted on the plane by air resistance. The plane has a mass 6 of 8 Mg. Assume that the air has a constant density of ra = 1.112 kg>m3.

6–87. The jet has a speed of 1200 km>h when it is flying horizontally. Air enters the intake scoop S at the rate of 80 m3 >s. If the engine burns fuel at the rate of 2.50 kg>s, and the gas (air and fuel) is exhausted relative to the plane with a speed of 3000 km>h, determine the drag exerted on the plane by air resistance. The plane has a mass of 10 Mg. Assume that air has a constant density of 0.909 kg>m3.

S

Prob. 6–84 S

6–85. The rocket has an initial total mass m0, including the # fuel. When it is fired, it ejects a mass flow of me with a velocity of e measured relative to the rocket. As this occurs, the pressure at the nozzle, which has a cross-sectional area Ae, is pe. If the drag on the rocket is FD = ct, where t is the time and c is a constant, determine the velocity of the rocket if the acceleration due to gravity is assumed to be constant. 6–86. The rocket has an initial mass m0, including the fuel. If the fuel is expelled from the rocket at e measured relative to the rocket, determine the rate at which the fuel should be consumed to maintain a constant acceleration a0. Neglect air resistance, and assume that the gravitational acceleration is constant.

a0

Prob. 6–87

*6–88. The jet is traveling at 900 km>h, 30° above the horizontal. If the fuel is being burnt at 5- 20 kg>s, and the engine takes in air at 450 kg>s, whereas the exhaust gas (air and fuel) has a speed of 1500 m>s measured relative to the plane, determine the acceleration of the plane at this instant. The drag of the air is FD = (2.75v2) N, where the speed is measured in m>s. The jet has a mass of 5 Mg.

900 km/h

308

Probs. 6–85/86

Prob. 6–88

proBleMs 6–89. The rocket is traveling upwards at 300 m>s and discharges 50 kg>s of fuel with a velocity of 3000 m>s measured relative to the rocket. If the exhaust nozzle has a cross-sectional area of 0.05 m2, determine the thrust of the rocket.

*6–92. The second stage B of the two-stage rocket of mass 1.2 Mg (empty) is launched from the first stage with a relative velocity of 5400 km>h. The fuel in the second stage has a mass of 350 kg. If it is consumed at the rate of 35 kg>s, and ejected with a relative velocity of 2000 m>s, determine the acceleration of the second stage B just after its engine is fired. What is the rocket’s acceleration just before all the fuel is 6 consumed? Neglect the effect of gravity and air resistance.

B

Prob. 6–89 6–90. If the rocket consumes 7200 kg of solid fuel at a rate of 175 kg>s and ejects it with a velocity of 1800 m>s measured relative to the rocket, determine the velocity and acceleration of the rocket at the instant just before all the fuel has been consumed. Neglect air resistance and the variation of gravity with altitude. The rocket has a mass of 20 Mg at lift off and starts from rest. 6–91. Determine the constant rate at which the fuel in the rocket must be burned, so that its thrust gives the rocket a speed of 500 m>s, 30 s after lift off. The fuel is expelled from the rocket at a speed of 3000 m>s measured relative to the rocket. The rocket has a total mass of 21.5 Mg, including 6.5 Mg of fuel. Neglect air resistance and the variation of gravity with altitude.

v

Probs. 6–90/91

355

A

Prob. 6–92

356

Chapter 6

Fluid MoMentuM

CONCEPTUAL PROBLEMS

6

P6–1. The water cannon ejects water from this tug in a characteristic parabolic shape. Explain what effect this has on the tug.

P6–2. Water flows onto the buckets of the waterwheel causing the wheel to turn. Have the blades been designed in the most effective manner to produce the greatest angular momentum of the wheel? Explain.

P6–1

P6–2

Chapter revieW

357

CHAP TER R EV IEW The linear and angular momentum equations are often used to determine resultant forces and couple moments that a body or surface exerts on a fluid in order to change the fluid’s direction.

Pressure force (out) Pressure force

Shear force Weight

Application of the momentum equations requires identifying a control volume, which can include both solid and fluid parts. The forces and couple moments acting on the control volume are shown on its free-body diagram.

Since the momentum equations are vector equations, they can be resolved into scalar components along x, y, z inertial coordinate axes.

Pressure force (in)

ΣF =

ΣMO =

If the control volume has constant velocity, then the velocity entering and exiting each open control surface is measured relative to the control surface.

Shear force

Pressure force

0 Vr dV + V rVf>cs # dA 0t Lcv Lcs

0 (r * V)r dV + (r * V) rVf>cs # dA 0t Lcv Lcs

ΣF =

0 Vr dV + Vf>cs rVf>cs # dA 0t Lcv Lcs

A propeller acts as a screw, which causes the linear momentum of the fluid to increase as it flows toward, through, then past the blades. Wind turbines decrease the linear momentum of the flow and thereby remove energy from it. A simple analysis of the flow through these two devices can be developed using the linear momentum and Bernoulli equations.

If the control volume has accelerated motion, as in the case of turbojets and rockets, then the momentum equation must account for the acceleration of the mass of the control volume.

ΣF = m

dVcv 0 + V r dV + Vf>cs rVf>cs # dA dt 0t Lcv f>cv Lcs

6

7

World Perspectives/The Image Bank/Getty Image

CHAPTER

Hurricanes are a combination of free and forced vortices, referred to as a combined vortex. Their motion can be analyzed using the equations of differential fluid flow.

DIFFERENTIAL FLUID FLOW CHAPTER OBJECTIVES ■

To introduce basic kinematic principles as they apply to a differential fluid element.

■

To establish the differential form of the equation of continuity and Euler’s equations of motion using x, y, z coordinates.

■

To present the idea of the stream function and the potential function, and to show how they can be used to describe various types of ideal fluid flow.

■

To develop the Navier–Stokes equations, which apply to a differential element of a viscous incompressible fluid.

■

To explain how complex fluid flow problems are solved using computational fluid dynamics.

7.1

DIFFERENTIAL ANALYSIS

In the previous three chapters, we applied the conservation of mass and the energy and momentum equations to a system of fluid particles to analyze problems involving fluid flow. There are situations, however, where we may have to determine the pressure and shear-stress variations over a surface, or find the fluid’s velocity and acceleration profiles within a closed conduit or channel. To do this, we will want to consider a differential-size element of fluid, since the variations we are seeking will have to come from the integration of differential equations.

359

360

Chapter 7

time t

Differential fluiD flow

For any real fluid, this differential flow analysis can become quite complex because of the effects of the fluid’s viscosity and its compressibility. As a result, the differential equations must be solved using a numerical analysis with a computer in order to obtain a solution. In some situations, however, we can consider the fluid to be ideal, and when this is the case, then the differential equations that describe the flow will become more manageable, and their solution will provide valuable information for many common types of engineering problems. Before we begin to analyze a differential fluid flow, however, we will first discuss some important aspects of its kinematics.

time t 1 Dt

7 Element translation and linear distortion (Ideal fluid)

Fig. 7–1 time t

7.2

time t 1 D t

U

KINEMATICS OF DIFFERENTIAL FLUID ELEMENTS

In general, the forces acting on an element of fluid while it is flowing will tend to cause it to undergo a “rigid-body” displacement, as well as a distortion or change in its shape. The rigid-body motion consists of a translation and rotation of the element; the distortion causes elongation or contraction of its sides, as well as changes in the angles between them. For example, translation and linear distortion can occur when an ideal fluid flows through a converging channel, Fig. 7–1. And translation and angular distortion can occur if the fluid is viscous and the flow is steady, Fig. 7–2. Of course, in more complex flows, all these motions can occur simultaneously. To better understand this general motion, we will first analyze each motion and distortion separately, and then show how they are related to the velocity gradient that causes them.

Element translation and angular distortion (Viscous fluid)

Fig. 7–2

Translation. Consider the differential fluid element shown in Fig. 7–3a that moves in three dimensions. The rate of translation of this element is defined by its velocity V = ui + vj + wk, and this velocity gives the element a translation. For example, in the x direction, the left face of the element in Fig. 7–3b will move a distance u∆t during the time ∆t. V

z

z y

uDt

x y x

Translation (b)

(a)

Fig. 7–3

7.2

KinematiCs of Differential fluiD elements

361

Linear Distortion. An elongation or contraction of the element will occur if the flow is nonuniform. As shown in Fig. 7–3c, during the time ∆t, the element’s left face will translate in the x direction by an amount u∆t, whereas the right face will move* 3u + 1 0u> 0x2∆x4 ∆t. In other words, the right face moves farther by10u> 0x2∆x∆t. Here we have expressed the velocity gradient as a partial derivative because, in general, u will be a function of both the element’s location in the flow and the time, that is, u = u1x, y, z, t2. The result of the movement is therefore a translation of u∆t and a linear distortion or dilatation of 310u>0x2 ∆x4 ∆t. In the limit, the change in the volume of the element due to this distortion becomes dVx = 310u>0x2dx4 dy dz dt. Since we must also consider the effect of velocity components in the y direction, and w in the z direction, similar results are obtained. Thus, a general volume change of the element becomes dV = c

0u 0v 0w + + d 1dx dy dz2 dt 0x 0y 0z

The rate at which dV per unit volume changes is called the volumetric dilatation rate. It can be expressed as dV>dV =

dt

0u 0v 0w + + 0x 0y 0z

(7–1)

0u)Dx]Dt [u 1 (–– 0x Dz Dx

Dy uDt

0u Dx]Dt [ –– 0x

Linear distortion (c)

Fig. 7–3 (cont.) *Throughout this chapter and elsewhere, we will neglect the second and higher-order terms 2 in a Taylor series expansion, such as 1 00xu2 2 2!1 1∆x2 2 + c, since in the limit, as ∆t S 0, they will all be very small compared to the first-order term.

7

362

Chapter 7

Differential fluiD flow

uDy)Dt ( –– y

u Dy)Dt ( u + –– y u Dt

b

7

90º–( 1 )

Dy v Dx)Dt (v + –– x v Dt

Dy

v Dx)Dt ( –– x Dx

a

Dx Angular rotation (a)

(b)

Fig. 7–4

Rotation. In fluid mechanics, we define the rotation of a fluid element at a point as the average angular velocity of two perpendicular line segments passing through the point. To establish this criterion, consider the rectangular fluid element shown in Fig. 7–4a, which distorts after the time ∆t. To cause this rotation, the right bottom corner of the element moves in the +y direction a distance 310v>0x2∆x4 ∆t higher than its left bottom corner. Note carefully that this is due to the change in relative to x, since this change occurs over the distance ∆x.* Therefore, as shown in Fig. 7–4b, the bottom ∆x rotates positive counterclockwise through a very small angle a = 310v>0x2 ∆x4 ∆t> ∆x.† In a similar manner, the left bottom corner moves to the right u∆t, Fig. 7–4a, and the top left corner moves to the right 3u + 10u>0y2∆y4 ∆t. Consequently, the left side ∆y in Fig. 7–4b rotates positive clockwise through the angle b = 310u>0y2∆y4 ∆t> ∆y. The average angular velocity vz (omega) of these sides is then the average counterclockwise time rate of change of a and b as ∆t S 0. It is # 1 1a - b2 1 # = 1a - b 2 vz = lim ∆t S 0 2 ∆t 2 or 1 0v 0u (7–2) vz = a b 2 0x 0y If the flow is three-dimensional, then by the same arguments, we will also have x and y components of angular velocity. Therefore, in general, 1 0w 0v vx = a b 0z 2 0y vy = vz =

0w 1 0u b a 0x 2 0z

(7–3)

1 0v 0u b a 0y 2 0x

*The change in relative to y will elongate the element, as shown in Fig. 7–3. the right-hand rule, this counterclockwise rotation directs the positive z axis out of the page.

†By

7.2

KinematiCs of Differential fluiD elements

363

90º– ( 1 )

7

Shear strain (c)

Fig. 7–4 (cont.)

Angular Distortion. The angular distortion of a fluid element is measured as a shear strain caused by viscosity. To determine it we must find the change in angle that occurs between the perpendicular sides ∆x and ∆y of the element shown in Fig. 7–4b. Since a = 310v>0x2∆x4 ∆t> ∆x, (positive counterclockwise) and b = 310u>0y2∆y4 ∆t> ∆y (positive clockwise), Fig. 7–4c, then the change in the 90° angle between ∆x and ∆y becomes gxy = 90° - 390° - (a + b)4 = a + b = ca

0v ∆t 0u ∆t b ∆x d + c a b ∆y d 0x ∆x 0y ∆y

Unlike solids, fluids flow and so we will be interested in the time rate of change in the shear strain. Therefore, dividing by ∆t, we have # 0v 0u # # gxy = a + b = + 0x 0y

(7–4)

If the flow is three-dimensional, then in a similar manner, shear-strain rates about the x and y axes will also occur. In general, then, 0v 0u # + gxy = 0x 0y 0u 0w # + gxz = 0x 0z 0w 0v # gyz = + 0y 0z

(7–5)

364

Chapter 7

Differential fluiD flow

7.3

CIRCULATION AND VORTICITY

Rotational flow is often characterized by either describing its circulation about a region, or reporting its vorticity. We will now define each of these characteristics. 7

Circulation. The concept of circulation Γ (capital gamma) was first introduced by Lord Kelvin to study the flow around the boundary of a body. It defines the net flow along any closed three-dimensional curve. For a unit depth, which then represents a two-dimensional volumetric flow, the circulation has units of m2 >s. To obtain it, we must integrate around the curve the component of the velocity that is tangent to the curve, Fig. 7–5. Formally, this is done using a line integral of the dot product V # ds = Vds cos u, so that

Γ =

C

V # ds

(7–6)

By convention, the integration is performed counterclockwise, that is, in the +z direction.

y

G u

V

ds

x

Circulation

Fig. 7–5

7.3

To show an application, let’s calculate the circulation around the boundary of a small element located at point (x, y), which is immersed in a general two-dimensional nonuniform steady flow field V = u1x, y2i + v1x, y2j, Fig. 7–6. The average velocities of the flow along each side of the element have the magnitudes and positive directions shown. Applying Eq. 7–6, accounting for the positive or negative directions as we go counterclockwise (positive) around the element, we have

CirCulation anD VortiCity

365

y 0u Dy u 1 –– 0y

7

G v Dy

0v 0u Γ = u ∆x + a v + ∆xb ∆y - a u + ∆yb ∆x - v ∆y 0x 0y

0v Dx v 1 –– 0x

(x, y)

Dx u x Circulation

which simplifies to

Fig. 7–6

Γ = a

0v 0u b ∆x∆y 0x 0y

Note that circulation about this small element, or, for that matter, any other size boundary, does not mean that individual fluid particles “circle” around the boundary. Instead, the circulation as indicated above is simply the net or resultant flow around the boundary.

Vorticity. We define the vorticity z (zeta) at a point (x, y) in the flow as the circulation per unit area around the point. For example, for the element in Fig. 7–7, dividing its area ∆x∆y into Γ, we get

z =

Γ 0v 0u = A 0x 0y

(7–7) z Dy

Comparing this result with Eq. 7–2, we see that z = 2vz. The vorticity is actually a vector. In this two-dimensional case, using the right-hand rule, it is directed along the +z axis. If we were to consider three-dimensional flow, then by the same development, using Eqs. 7–3, we would obtain

Dx Vorticity z5G A

Fig. 7–7

Z = 2V

(7–8)

366

7

Chapter 7

Differential fluiD flow

Irrotational Flow. The angular rotation, or the vorticity, provides a means of classifying the flow. If V ≠ 0, then the flow is called rotational flow; however, if V = 0 throughout the flow, then it is termed irrotational flow. Ideal fluids exhibit irrotational flow because no viscous friction forces act on ideal fluid elements, only pressure and gravitational forces. And, because these two forces are always concurrent, ideal fluid elements that are initially not rotating cannot be forced to rotate while they are in motion. The difference between rotational and irrotational flow can be illustrated by a few simple examples. No rotation occurs in the ideal fluid because the two sides of the element in Figs. 7–8a and 7–8b do not have an average rotation, i.e., a = b = 0. This is irrotational flow. However, for a viscous fluid this is different. In Fig. 7–8c, the top and bottom surfaces of the element move at different velocities, and this will cause the vertical # sides to rotate clockwise at the rate b . As a result, this produces rotational # # # # flow, vz = 1a - b 2 >2 = 10 - b 2 >2 = -b >2. Negative because vz is clockwise. If viscous flow occurs around a curved boundary such that it is more rapid at the outer boundary, Fig.# 7–8d, then the sides of the element # undergo rates # of rotation -a and b so that the average rotation is # vz = 12 1 -a - b 2.

Irrotational flow Ideal fluid (a)

Irrotational flow Ideal fluid (b)

b b a

Rotational flow Viscous fluid (c)

Rotational flow Viscous fluid (d)

Fig. 7–8

7.3

EXAMPLE

CirCulation anD VortiCity

7.1

The ideal fluid in Fig. 7–9 has a uniform velocity of U = 0.2 m>s. Determine the circulation about the triangular and circular boundaries.

7

y C (0.2 ms) sin u

u 308

0.4 m

U = 0.2 ms

308

ds

U 5 0.2 ms

du 0.1 m

A

B x

U 5 0.2 ms

U 5 0.2 ms

(a)

(b)

Fig. 7–9

SOLUTION Fluid Description. We have uniform steady flow of an ideal fluid within the x–y plane. Triangular Path. To find the circulation around the triangular path in Fig. 7–9a, we do not need to integrate; rather we determine the length of each side of the triangle, and the component of the velocity U along each side. Thus, going from C to A, A to B, and B to C, we have Γ =

C

V # ds = ΣV # s = 10210.4 m cos 30°2 + 10.2 m>s210.4 m sin 30°2

- 10.2 m>s sin 30°210.4 m2 = 0 Ans. Note that the last term is negative because the velocity component 10.2 m>s sin 30°2 causes a clockwise motion. Circular Path. The circular boundary is easily defined using polar coordinates, with u positive counterclockwise as shown in Fig. 7–9b. The component of V along positive ds = 10.1 m2 du is - 10.2 m>s2 sin u. Therefore, Γ =

C

367

V # ds =

L0

2p

- 10.2 m>s2 sin u 10.1 m2 du = 0.021cos u2 `

2p

= 0 0

Ans. Both of these cases illustrate a general point, that regardless of the shape of the path, for uniform flow an ideal fluid will not produce a circulation, and because z = Γ>A, no vorticity is produced either.

u

368

Chapter 7

EXAMPLE

Differential fluiD flow

7.2 The velocity of a viscous fluid flowing between the parallel surfaces in Fig. 7–10a is defined by U = 0.00231 - 101103 2y2 4 m>s, where y is in meters. Determine the vorticity and shear-strain rate of a fluid element located at y = 5 mm within the flow.

7

y

10 mm

y 5 5 mm

x

10 mm

(a)

SOLUTION Fluid Description. We have nonuniform steady one-dimensional flow of a real fluid. Vorticity. We must apply Eq. 7–7, where u = 0.002 31 - 1011032y24 m>s and v = 0. vz 5 0.1 rads

z =

0v 0u 0x 0y

= 0 - 0.002 3 0 - 101103 212y2 4 ` rad>s = 0.200 rad>s

Ans.

y = 0.005 m

(b)

This vorticity is a consequence of the fluid’s viscosity, and since z ≠ 0, we have rotational flow. Actually, the element has a rate of rotation of vz = z>2 = 0.1 rad>s, Fig. 7–10b. It is positive (counterclockwise) because at y = 0.005 m, the velocity profile shows that the top of the fluid element is moving slower than its bottom, Fig. 7–10a.

2b

Shear-Strain Rate. Applying Eq. 7–4, # 0v 0u # # gxy = a + b = + 0x 0y 908

908 1 b

(c)

= 0 + 0.002 3 0 - 101103 212y2 4 ` rad>s = -0.200 rad>s

Ans.

y = 0.005 m

Fig. 7–10

The result is negative because b is positive clockwise, and the shear strain is defined as the difference in the angle 90° - 190° + b2 = -b, Fig. 7–10c.

7.4

7.4

ConserVation of mass

369

CONSERVATION OF MASS

In this section we will derive the continuity equation for a fluid flowing through a fixed differential control volume that only has open control surfaces, Fig. 7–11. We will consider three-dimensional flow, where the velocity field has components u = u1x, y, z, t2, v = v1x, y, z, t2, w = w1x, y, z, t2. Point (x, y, z) is at the center of the control volume, and at this point the density is r = r1x, y, z, t2. Within the control volume, local changes to the mass can occur due to the fluid’s compressibility. Also, convective changes can occur from one control surface to another when the flow is nonuniform. For example, in Fig. 7–11 this convective change is shown in the x direction. If we apply the continuity equation, Eq. 4–12, to the control volume in the x direction, we have 0 r dV + rVf>cs # dA = 0 0t Lcv Lcs 01ru2 0r ∆x ∆y ∆z + a ru + ∆xb ∆y ∆z - ru ∆y ∆z = 0 0t 0x Local change in mass

Net convective change in mass

Dividing by ∆x ∆y ∆z, and simplifying, we get 01ru2 0r + = 0 0t 0x

(7–9)

If we include the convective changes through the control surfaces in the y and z directions, then the continuity equation becomes 01ru2 01rv2 01rw2 0r + + + = 0 0t 0x 0y 0z

0 (ru) (ru 1 –––– Dx)Dy Dz 0x (x, y, z)

ru Dy Dz Dz

Dy Dx

Fig. 7–11

(7–10)

7

370

Chapter 7

Differential fluiD flow

Two-Dimensional Steady Flow of an Ideal Fluid. Although we have developed the continuity equation in its most general form, often it is applied to two-dimensional steady flow of an ideal fluid. For this special case, r is constant, and as a result Eq. 7–10 then becomes 0u 0v + = 0 0x 0y

7

(7–11)

Steady flow incompressible fluid

As noted by Eq. 7–1, this is the same as saying the volumetric dilatation rate must be zero. For example, if a positive rate of change in length occurs in the x direction 10u>0x 7 02, then by Eq. 7–11, a corresponding negative change in length must occur in the y direction 10v>0y 6 02.

Cylindrical Coordinates. The continuity equation can also be

applied to a differential control volume having dimensions in cylindrical coordinates r, u, z, Fig. 7–12. For the sake of completeness, we will state the result here without proof, and then use it later to describe some important types of symmetrical flow. In the general case, 01rvz 2 0r 1 01rrvr 2 1 01rvu 2 + + + = 0 r r 0u 0t 0r 0z

(7–12)

If the fluid is incompressible and the flow is steady, then in two dimensions (r, u) this equation becomes vr 0vr 1 0vu + + = 0 r r 0u 0r Steady flow incompressible fluid z

Dz Du

Dr

z

r u

Cylindrical coordinates

Fig. 7–12

(7–13)

7.5

7.5

371

equations of motion for a fluiD partiCle

EQUATIONS OF MOTION FOR A FLUID PARTICLE

In this section we will apply Newton’s second law of motion to a differential fluid particle (element) and express the result in its most general form. But before we do this, we must first formulate expressions that represent the effect of a force ∆F acting on a differential area ∆A. As shown in Fig. 7–13a, ∆F will have a normal component ∆Fz and two shear components ∆Fx and ∆Fy. Stress is the result of these surface force components. The normal component creates a normal stress on the area, defined as szz = lim S ∆A

DFz DF

∆Fz 0 ∆A

DA DFx

and the shear components create shear stresses tzx = lim S ∆A

∆Fx 0 ∆A

tzy = lim S ∆A

∆Fy

1∆Fz 2 sf = a

0txy 0x

0txz 0x

+ +

0syy 0y 0tyz 0y

+ +

0tzy 0z

0szz 0z

DFy

x

y

0 ∆A

The first letter (z) in this subscript notation represents the outward normal direction that defines the direction of the area element ∆A, and the second letter represents the coordinate direction of the stress. If we now generalize this idea and consider forces acting on the six faces of a volume element of fluid, then three components of stress will act on each face of the element as shown in Fig. 7–13b. At each point in the fluid there will be a stress field that defines these stresses. And because this field changes from one point to the next, the forces these stresses produce on the fluid element must account for these changes. For example, consider the free-body diagram of the fluid particle in Fig. 7–13c, which shows the forces produced by the stress components in the x direction. The resultant surface force in the x direction is 0sxx 1∆Fx 2 sf = a sxx + ∆xb ∆y ∆z - sxx ∆y ∆z 0x 0tyx + a tyx + ∆yb ∆x ∆z - tyx ∆x ∆z 0y 0tzx + a tzx + ∆zb ∆x ∆y - tzx ∆x ∆y 0z By collecting terms this can be simplified, and in a similar manner, the resultant surface forces produced by the stresses in the y and z directions can also be obtained. We have 0tyx 0tzx 0sxx 1∆Fx 2 sf = a + + b ∆x ∆y ∆z 0x 0y 0z 1∆Fy 2 sf = a

7

z

b ∆x ∆y ∆z b ∆x ∆y ∆z

(a) z

szz tzx txz

tzy

tyz txy tyx syy

sxx x

y (b) (tzx +

tzx )DxDy z Dz

tyx DxDz sxx DyDz

Dz

Dy sxx (sxx + Dx )DyDz x

Dx (tyx + tzx DxDy

Free-body diagram (c)

Fig. 7–13

tyx Dy )DxDz y

372

7

Chapter 7

Differential fluiD flow

Apart from these forces, there is also the body force due to the weight of the particle. If ∆m is the particle’s mass, this force is ∆W = 1∆m2g = rg∆x ∆y ∆z. To further generalize this development, we will assume the x, y, z axes have some arbitrary orientation so that the weight will have components ∆Wx, ∆Wy, ∆Wz along each axis. Therefore, the sums of all the body and surface force components acting on the fluid particle are ∆Fx = a rgx + ∆Fy = a rgy + ∆Fz = a rgz +

0tyx 0tzx 0sxx + + b ∆x ∆y ∆z 0x 0y 0z 0txy 0x 0txz 0x

+ +

0syy

+

0y 0tyz 0y

+

0tzy 0z

0szz 0z

b ∆x ∆y ∆z

(7–14)

b ∆x ∆y ∆z

With these forces established, we can now apply Newton’s second law of motion to the particle. Using the material derivative to determine the acceleration, Eq. 3–7, we have

ΣF = ∆m

DV 0V 0V 0V 0V = 1r∆x ∆y ∆z2 c + u + v + w d Dt 0t 0x 0y 0z

When we substitute Eqs. 7–14, factor out the volume ∆x∆y∆z, and then use V = u i + v j + w k, the x, y, z components of this equation become

rgx + rgy +

rgz +

0sxx 0x 0txy 0x 0txz 0x

+ +

+

0tyx 0y 0syy 0y 0tyz 0y

+ +

+

0tzx 0z 0tzy 0z 0szz 0z

= ra = ra

= ra

0u 0t 0v 0t

+ u + u

0u 0x

0v 0x

+ v + v

0u 0y

0v 0y

+ w + w

0u 0z

0v 0z

b

b (7–15)

0w 0w 0w 0w + u + v + w b 0t 0x 0y 0z

To summarize, these are the three equations of motion. The terms on the left are the result of “ΣF,” and those on the right are the result of “ma.” In the next section we will apply these equations to study the motion of an ideal fluid. Then later, in Sec. 7.12, we will consider the more general case of a Newtonian fluid.

7.6

7.6

373

the euler anD Bernoulli equations

THE EULER AND BERNOULLI EQUATIONS

z

If we consider the fluid to be an ideal fluid, then the equations of motion will reduce to a simpler form. In particular, there will be no viscous shear stress on the particle (element), and the three normal stress components will represent the pressure. Since these normal stresses have all been defined in Fig. 7–13b as positive outward, and as a convention, positive pressure produces a compressive stress, then sxx = syy = szz = -p. As a result, the general equations of motion for an ideal fluid particle become

szz tzx txz

tzy

7

tyz txy tyx syy

sxx x

y (b)

Fig. 7–13 (repeated)

rgx rgy rgz -

0p 0u 0u 0u 0u = ra + u + v + w b 0x 0t 0x 0y 0z 0p 0v 0v 0v 0v = ra + u + v + w b 0y 0t 0x 0y 0z

(7–16)

0p 0w 0w 0w 0w = ra + u + v + w b 0z 0t 0x 0y 0z

These equations are called the Euler equations of motion, expressed in x, y, z coordinates. Recall that in Sec. 5.1 we derived them for streamline coordinates, s, n, where they took on a simpler form.

Two-Dimensional Steady Flow. In many cases we will have steady two-dimensional flow, and the z component of velocity w = 0. If we orient the x and y axes so that g = -gj, then Euler’s equations (Eqs. 7–16) become

1 0p 0u 0u = u + v r 0x 0x 0y

(7–17)

1 0p 0v 0v - g = u + v r 0y 0x 0y

(7–18)

-

The velocity components u and and the pressure p at any point within the fluid can now be determined, provided we can solve these two partial differential equations along with the continuity equation, Eq. 7–11.

374

7

Chapter 7

Differential fluiD flow

The Bernoulli Equation. In Sec. 5.2 we derived the Bernoulli equation by integrating the streamline component of the Euler equation. There it was shown that the result applies at any two points on the same streamline. However, if a condition of irrotational flow exists, meaning V = 0, then the Bernoulli equation can also be applied between any two points that are on different streamlines. To show this, assume we have irrotational two-dimensional flow so that vz = 0 or 0u>0y = 0v>0x, Eq. 7–2. If we substitute this condition into Eqs. 7–17 and 7–18, we get -

1 0p 0u 0v = u + v r 0x 0x 0x

-

1 0p 0u 0v - g = u + v r 0y 0y 0y

Since 01u22>0x = 2u10u>0x2, 01v22>0x = 2v10v>0x2, 01u22>0y = 2u10u>0y2, and 01v2 2 >0y = 2v10v>0y2, then the equations become -

1 0p 1 01u2 + v2 2 = r 0x 2 0x

-

1 0p 1 01u2 + v2 2 - g = r 0y 2 0y

Integrating the first equation with respect to x, and the second equation with respect to y, yields* -

p 1 1 + f1y2 = 1u2 + v2 2 = V 2 r 2 2

p 1 1 - gy + h1x2 = 1u2 + v2 2 = V 2 r 2 2

Here V is the fluid particle’s velocity found from its components, V 2 = u2 + v2. Equating these two results, it is then necessary that f1 y2 = -gy + h1x2. This solution indicates that h1x2 = Const., since x and y can vary independent of one another. As a result, the unknown function f1 y2 = -gy + Const. Substituting this and h1x2 = Const. into the above equations, we obtain in either case the Bernoulli equation, that is, p V2 + + y = Const. g 2g

(7–19)

Steady irrotational flow, ideal fluid

Therefore, if the flow is irrotational, then the Bernoulli equation can be applied between any two points 1x1, y1 2 and 1x2, y2 2 that are not necessarily on the same streamline. Of course, as noted, we must also require the fluid to be ideal and the flow to be steady.

*Remember that p is a function of both x and y, and so, for example, when we integrate 11>r20p>0x, we get p>r + f1y2 since f1y2 does not change in the x direction.

7.6

the euler anD Bernoulli equations

375

IM PORTANT POIN T S • In general, when a differential element of fluid is subjected to forces, it tends to undergo “rigid-body” translation and rotation, as well as linear and angular distortions.

• The rate of translation of a fluid element is defined by the velocity field.

• Linear distortion is measured by the change in volume per unit volume of the fluid element. The rate at which this change occurs is called the volumetric dilatation rate.

• Rotation of a fluid element is defined by the average angular velocities of its two sides. Rotation can also be specified by the vorticity Z = 2V.

• If V = 0, then the flow is termed irrotational flow, that is, no rotation occurs. This type of flow always occurs in an ideal fluid, because viscous shear forces are not present to cause rotation.

• Angular distortion is defined by the rate of change in shear strain, or the rate at which the 90° angle between adjacent sides of the fluid element will change. These strains are caused by shear stress, which is the result of the fluid’s viscosity. Ideal or inviscid fluids have no angular distortions.

• Since an ideal fluid is incompressible, the continuity equation for steady flow states that the rate of change in volume per unit volume for a fluid element must be zero.

• Euler’s equations relate the pressure and gravitational forces acting on a differential fluid particle of an ideal fluid to its acceleration. If these equations are integrated and combined, then for steady irrotational flow, they produce the Bernoulli equation.

• The Bernoulli equation can be applied between any two points not located on the same streamline, provided the fluid is ideal and the steady flow is irrotational, that is, V = 0.

7

376

Chapter 7

EXAMPLE

Differential fluiD flow

7.3 The velocity field V = 5 -6xi + 6yj6m>s defines the two-dimensional ideal fluid flow in the horizontal plane between two walls, Fig. 7–14. Determine the volumetric dilatation rate and the rotation of a fluid element located at point B(1 m, 2 m). If the pressure at point A(1 m, 1 m) is 250 kPa, what is the pressure at point B? Take r = 1200 kg>m3.

y

7

SOLUTION B

2m

Fluid Description. Since the velocity is not a function of time, the flow is steady. The fluid is an ideal fluid. Volumetric Dilatation Rate. Applying Eq. 7–1, where u = 1-6x2 m>s, v = 16y2 m>s, and w = 0, we have

A 1m

dV>dV

x

0t

1m

=

0u 0v 0w + + = -6 + 6 + 0 = 0 0x 0y 0z

Ans.

The result confirms that there is no change in the volume of the fluid element at B, because an ideal fluid is incompressible.

Fig. 7–14

Rotation. The angular velocity of the fluid element at B is defined by Eq. 7–2. vz =

1 0v 0u 1 a b = 10 - 02 = 0 2 0x 0y 2

Ans.

Therefore, the fluid element will not rotate about the z axis. Actually, the above two results apply at all points since they are independent of x and y. This is to be expected since we have an ideal fluid. Pressure. Since the flow is irrotational and steady, we can apply the Bernoulli equation at two points not located on the same streamline, Fig. 7–14. The velocities at A and B are VA = 23 -61124 2 + 361124 2 = 8.485 m>s VB = 23 -61124 2 + 361224 2 = 13.42 m>s

Since motion is in the horizontal plane, we have

2501103 2 N>m2

11200 kg>m3 219.81 m>s2 2

+

18.485 m>s2 2

219.81 m>s2 2

pA pB VA2 VB2 + + + zA = + zB g g 2g 2g + 0 =

pB

11200 kg>m3 219.81 m>s2 2 pB = 185 kPa

+

113.42 m>s2 2

219.81 m>s2 2

+ 0 Ans.

7.8

7.7

377

POTENTIAL FLOW HYDRODYNAMICS

It was mentioned in Sec. 7.1 that the study of fluid flow including the effects of viscosity and compressibility will complicate any analysis. A simple, yet still somewhat accurate approach is to neglect these effects and assume the fluid to be ideal. This mathematical study of ideal fluid flow was developed primarily in the late 19th century in an effort to provide theoretical results that are intended to approximate many types of flow, without the need for any experimental measurements, apart from knowing the fluid’s density. In the early 1900s, Ludwig Prandtl discovered that the fluid’s viscosity is of great importance when studying the flow near a boundary. It was determined that this “boundary layer” is very thin, provided the flow is fast or the viscosity is small, as in the case of air and water. Away from this boundary layer, it is often reasonable to assume the fluid is ideal, and there the more theoretical methods of hydrodynamics can be used to model any real fluid flow. The following sections provide an introduction to this approach in order to give some insight as to how the methods are applied for an engineering analysis.* We will begin by showing how to obtain the stream and potential functions for a two-dimensional flow field. Once these are obtained, we will then be able to determine the velocity and pressure within the flow. A few elementary types of flow will be analyzed, and these will be superimposed or algebraically added together to demonstrate how to describe flows around objects having various shapes.

7.8

the stream funCtion

THE STREAM FUNCTION

In two dimensions, one method for satisfying the equation of continuity is to replace the two unknown velocity components u and by a single unknown function, thus reducing the number of unknowns, and thereby simplifying the analysis of an ideal fluid flow problem. In this section we will use the stream function as a means for doing this, and in the next section we will consider its counterpart, the potential function. The stream function c (psi) is the equation that represents all the equations of the streamlines. In two dimensions, it is a function of x and y, and for the equation of each streamline it is equal to a specific constant c1x, y2 = C. You may recall in Sec. 3.2 we developed the technique for finding the equation of a streamline as it relates to the velocity components u and . Here we will review this procedure, and extend its usefulness.

*An extended discussion of this method is mathematically rigorous, requiring considerable skill at solving partial differential equations. See, for example, Ref. [8]–[11].

7

378

Chapter 7

Differential fluiD flow

dx

V

dy

v c (x,y) 5 C

Velocity Components. By definition, the velocity of a fluid particle is always tangent to the streamline along which it travels, Fig. 7–15. As a result, we can relate the velocity components u and v to the slope of the tangent by proportion. As shown in the figure, dy>dx = v>u, or

u

7

u dy - v dx = 0

(7–20)

Now, if we take the total derivative of the streamline equation c1x, y2 = C, which describes the streamline in Fig. 7–15, we have Velocity is always tangent to streamline

dc =

Fig. 7–15

0c 0c dx + dy = 0 0x 0y

(7–21)

Comparing this with Eq. 7–20, the two components of velocity can be related to c. We require

u =

0c 0c , v = 0y 0x

(7–22)

Therefore, if we know the equation of any streamline, c1x, y2 = C, then the velocity components of a particle that travels along it can be obtained by using these equations. By obtaining the velocity components in this way, we can show that for steady flow the stream function automatically satisfies the equation of continuity. This occurs because by direct substitution of Eqs. 7–22 into Eq. 7–11, we find 0u 0v + = 0; 0x 0y

0c 0 0c 0 a b + a- b = 0 0x 0y 0y 0x

V vr vu

r c(r, u) 5 C u

Polar coordinates

Fig. 7–16

02c 02c = 0 0x 0y 0y 0x

Later we will see that in some problems, it will be convenient to express the stream function and the velocity components in terms of their polar coordinates, r and u, Fig. 7–16. Without proof, if c1r, u2 = C is given, then the radial and transverse velocity components are

vr =

0c 1 0c , vu = r 0u 0r

(7–23)

7.8

Volumetric Flow. The stream function can also be used to determine the volumetric flow between any two streamlines. For example, consider the triangular differential control volume in Fig. 7–17a that is located within the streamtube between the two streamlines c and c + dc. Since we have two-dimensional flow, we will consider the flow dq through this element as a measure of flow per unit depth z, that is, having units of m2 >s. This flow only occurs within this streamtube, because the fluid velocity is always tangent to the streamlines, never perpendicular to them. Continuity requires the flow into the control surface AB to be equal to the flow out of the control surfaces BC and AC. In the case of BC, since the depth is 1 unit, the flow out is u3dy1124, but in the case of AC, by convention, is positive upward, and so the flow outward is -v3dx1124. Applying the continuity equation, for steady incompressible flow, we have

379

the stream funCtion

y

7 dq

A dx

B u dy

dy C

c 1 dc

2v dx

dq

0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

c x

0 - r dq + ru3dy1124 - rv3dx1124 = 0 (a)

dq = u dy - v dx Substituting Eqs. 7–22 into this equation, the right side becomes Eq. 7–21. Therefore,

y

dq = dc Thus, the flow dq between the two streamlines c and c + dc is simply found by finding their difference, 1c + dc2 - c = dc. The flow between any two streamlines a finite distance apart can now be determined by integrating this result. For example, if c1 1x, y2 = C1 and c2 1x, y2 = C2, then dc = c2 1x, y2 - c1 1x, y2 = C2 - C1

c2

q =

Lc1

fast c2 5 C2

slow

(7–24)

Let’s summarize our results. If the stream function c1x, y2 is known, we can set it equal to various values of the constant c1x, y2 = C to obtain the streamlines, and thus visualize the flow. We can then use Eqs. 7–22 (or Eqs. 7–23) to determine the velocity components of the flow along a streamline. Also we can determine the volumetric flow between any two streamlines, such as c1 1x, y2 = C1 and c2 1x, y2 = C2, by finding the difference in their streamline constants, q = C2 - C1, Eq. 7–24. As we have discussed in Sec. 3.2, once constructed, the distance between the streamlines will also provide an indication of the relative speed of the flow. This is based on the conservation of mass. For example, note how the fluid element in Fig. 7–17b must become thinner as it grows wider when it moves through the streamtube in order to preserve its mass (or volume). Thus, at locations where the streamlines are close together, the flow is fast, and when these streamlines are farther apart, the flow is slow.

Conservation of mass (b)

Fig. 7–17

c1 5 C1

x

380

Chapter 7

EXAMPLE

Differential fluiD flow

7.4 The flow is defined by the stream function c1x, y2 = y2 - x. Draw the streamlines for c1 1x, y2 = 0, c2 1x, y2 = 2 m2>s, and c3 1x, y2 = 4 m2>s. Find the velocity of a fluid particle at y = 1 m on the streamline c2 1x, y2 = 2 m2 >s and show that the flow field satisfies the continuity equation.

7

SOLUTION Fluid Description. Since time is not involved, this is steady flow of an ideal fluid. Stream Functions. The equations for the three streamlines are

y

y2 - x = 0

c3 5 4 m2s c2 5 2 m2s

1 ms V

y2 - x = 2 y2 - x = 4

c1 5 0

2 ms

1m

x

These equations are graphed in Fig. 7–18. Each is a parabola and represents a streamline for the constant that defines it.

1m

Velocity. u = 4m

Fig. 7–18

The velocity components along each streamline are

0c 0 = 1y2 - x2 = 12y2 m>s 0y 0y

0c 0 = - 1y2 - x2 = - 1 -12 = 1 m>s 0x 0x Continuity requires Eq. 7–11 be satisfied. v = -

0u 0v + = 0; 0 + 0 = 0 0x 0y For the streamline y2 - x = 2, at y = 1 m, then x = -1 m, so at this point, u = 2 m>s and v = 1 m>s. These two components produce the resultant velocity of a fluid particle at this location, Fig. 7–18. It is V = 212 m>s2 2 + 11 m>s2 2 = 2.24 m>s

Ans.

Notice that the directions of the velocity components also provide a means for establishing the direction of the flow, as indicated by the small arrows on this streamline, Fig. 7–18. Although it is not part of this problem, imagine the streamlines for which c1 = 0 and c3 = 4 m2 >s represent solid boundaries for a channel, Fig. 7–18. Then from Eq. 7–24, the volumetric flow per unit depth within this channel (or streamtube) would be q = c3 - c1 = 4 m2 >s - 0 = 4 m2 >s

7.8

EXAMPLE

the stream funCtion

381

7.5

Uniform flow occurs at an angle u with the y axis, as shown in Fig. 7–19. Determine the stream function for this flow.

y

7

SOLUTION Fluid Description. is constant.

We have steady uniform ideal fluid flow, since U

U x

Velocity.

The x and y components of velocity are u = U sin u

and

v = -U cos u

Stream Function. If we relate the velocity component u to the stream function, we have 0c 0c u = ; U sin u = 0y 0y Integrating with respect to y, to obtain c, yields

c = 1U sin u2y + f1x2

(1)

Here f(x) is an unknown function that is to be determined. The component of velocity is related to the stream function using v = -

0c ; 0x

0 3 1U sin u )y + f1x24 0x 0 -U cos u = -0 3 f1x24 0x

-U cos u = -

Integrating,

Substituting into Eq. 1,

f1x2 = 1U cos u2 x + c

c1x, y2 = 1U sin u2y + 1U cos u2x + c

If we select the reference streamline to pass through the origin so that c = c0 = 0, then at x = 0, y = 0, and so c = 0. Therefore, the stream function is c(x, y) = 1U sin u2y + 1U cos u2x

Ans.

Since the velocity components are u = 0c>0y = U sin u and v = -0c>0x = -U cos u, as expected, the resultant velocity of fluid particles on each streamline is V = 21U sin u2 2 + 1 -U cos u2 2 = U

as noted in Fig. 7–19.

c 5 2 m2s c 5 1 m2s c50 u c 5 21 m2s Uniform flow

Fig. 7–19

382

Chapter 7

EXAMPLE

Differential fluiD flow

7.6 The streamlines for steady ideal fluid flow around the 90° corner in Fig. 7–20 are defined by the stream function c1x, y2 = 15xy2 m2 >s. Determine the velocity of the flow at point x = 2 m, y = 3 m. Can the Bernoulli equation be applied between any two points within this flow?

7

SOLUTION y

Fluid Description. As stated, we have steady ideal fluid flow. Velocity. 10 ms

The velocity components are determined using Eqs. 7–22. u =

c 5 30

15 ms

v = -

V

3m

0c 0 = 15xy2 = 15x2 m>s 0y 0y

0c 0 = - 15xy2 = 1 -5y2 m>s 0x 0x

At point x = 2 m, y = 3 m, x

u = 5122 = 10 m>s 2m

v = -5132 = -15 m>s

Flow around a 908 corner

Fig. 7–20

The resultant velocity has a magnitude of V = 2110 m>s2 2 + 1 -15 m>s2 2 = 18.0 m>s

Ans.

Its direction is tangent to the streamline that passes through point (2 m, 3 m), as shown in Fig. 7–20. To find the equation that defines this streamline, we require c1x, y2 = 5122132 = C = 30 m2 >s. Thus c1x, y2 = 5xy = 30, or xy = 6. An ideal fluid has irrotational flow. To check this we apply Eq. 7–2.

vz =

015x2 1 0v 0u 1 01 -5y2 a b = c d = 0 2 0x 0y 2 0x 0y

Therefore, yes the Bernoulli equation can be used to find the pressure differences between any two points within the flow, provided we know the velocities at these points and the vertical distance between them.

7.9

7.9

383

the potential funCtion

THE POTENTIAL FUNCTION

In the previous section we related the velocity components to the stream function, which describes the streamlines for the flow. Another way of relating the velocity components to a single function is to use the potential function f (phi), where f = f1x, y2. The velocity components are determined from f1x, y2 using the following equations.

u =

0f 0f , v = 0x 0y

7

(7–25)

The resultant velocity is therefore V = ui + vj =

0f 0f i + j 0x 0y

(7–26)

The potential function f only describes irrotational flow. To show this, substitute Eqs. 7–25 into Eq. 7–2. This gives

vz = =

1 0v 0u a b 2 0x 0y

02f 1 0 0f 0 0f 1 02f c a b a bd = c d = 0 2 0x 0y 0y 0x 2 0x 0y 0y 0x

Thus, if the flow can be defined by a potential function f1 x , y2, then the flow will be irrotational because this function automatically satisfies the condition vz = 0. Another characteristic of f1x, y2 is that the velocity will always be perpendicular to any equipotential line f1x, y2 = C′. As a result, any equipotential line will be perpendicular to any intersecting streamline c1x, y2 = C. This can be shown by taking the total derivative of f1x, y2 = C′, which gives df =

0f 0f dx + dy = 0 0x 0y

y

c5C V

= u dx + v dy = 0

dy dx

Or,

y

u u

dy u = v dx Graphically, this indicates that the slope u of the tangent of the streamline c = C, which is dy>dx = v>u, Eq. 7–20, is the negative reciprocal of the slope of the equipotential line f = C′, Fig. 7–21. Therefore, as shown, streamlines are always perpendicular to equipotential lines.

u u

f 5 C9

y dy dx x

Fig. 7–21

384

Chapter 7

Differential fluiD flow

Finally, if polar coordinates are used to describe the potential function, then, without proof, the velocity components r and u are related to the potential function by

vr =

7

c far apart velocity is slow

c defines fixed boundary

c close together velocity is fast

c5C c 5 C 1 DC c 5 C 1 2DC f 5 C9

f 5 C91 2DC9 f 5 C9 1 DC9

Flow net for ideal flow through a transition

Fig. 7–22

0f 1 0f , vu = r 0u 0r

(7–27)

Flow Net. A family of streamlines and equipotential lines for various values of the constants C and C′ makes up a flow net, which can serve as a graphical aid for visualizing the flow. An example of a flow net is shown in Fig. 7–22. Here the streamlines show the direction of the velocity, and the equipotential lines are perpendicular to them. The flow net is constructed so that these lines are spaced at the same incremental distance ∆C and ∆C′. As noted, where the streamlines are closer together the velocity is high (fast flow), and vice versa. For convenience, a computer can be used to construct a flow net by first plotting the equations for c1x, y2 = C and f1x, y2 = C′, and then incrementally increasing the constants by ∆C and ∆C′ and plotting these results.

I MPO RTA N T PO I N T S • A flow described by the stream function c1x, y2 satisfies the continuity condition. If c1x, y2 is known, then it is possible to determine the velocity components at any point within the flow using Eqs. 7–22. Also, the flow between any two streamlines c1x, y2 = C1 and c1x, y2 = C2 can be determined by finding the difference between the streamline constants, q = C2 - C1. The flow can either be rotational or irrotational, and its velocity components must always satisfy the equation of continuity.

• If a flow can be described by a potential function f1x, y2, then it satisfies the conditions of irrotational flow. If f1x, y2 is known, the velocity components at any point within the flow can be determined using Eqs. 7–25.

• Equipotential lines are always perpendicular to streamlines at their points of intersection, and a set of both of these “lines” forms a flow net. particular point 1x1, y1 2 can be determined by first obtaining the constants from c1x1, y1 2 = C1 and f1x1, y1 2 = C1 ′, and then plotting c1x, y2 = C1 and f1x, y2 = C1 ′.

• The streamline and equipotentional line passing through a

7.9

EXAMPLE

the potential funCtion

385

7.7

A flow has a velocity field defined by V = 54xy2i + 4x 2yj6 m>s. Is it possible to establish a potential function for this flow, and if so, what is the equipotential line passing through point x = 1 m, y = 1 m? SOLUTION Fluid Description. time.

We have steady flow since V is not a function of

Analysis. A potential function can be developed only if the flow is irrotational. To find out if it is, we apply Eq. 7–2. Here u = 4xy2 and v = 4x 2y, so that

1 0v 0u 1 a b = 18xy - 8xy2 = 0 2 0x 0y 2 Since we have irrotational flow, the potential function can be established. Using the x component of velocity, 0f 0f u = ; u = = 4xy2 0x 0x vz =

Integrating, f = 2x 2y2 + f1y2

(1)

Using the y component of velocity, 0f 0 v = ; 4x 2y = 32x 2y2 + f1y24 0y 0y 0 4x 2y = 4x 2y + 3 f1y24 0y Integrating, f1y2 = c Therefore Eq. (1) becomes f = 2x 2y2 + c For the reference equipotential line that passes through the origin, f = f 0 = 0, for x = y = 0, and so c = 0. Therefore, the potential function is f1x, y2 = 2x 2y2 To find the equipotential line passing through point (1 m, 1 m), we require f1x, y2 = 2112 2 112 2 = 2. Thus, 2x 2y2 = 2 or xy = {1

Ans.

7

386

Chapter 7

EXAMPLE

Differential fluiD flow

7.8 The potential function for a flow is defined by f1x, y2 = 10xy. Determine the stream function for the flow.

7

SOLUTION Fluid Description. This is steady fluid flow, and because it is defined by a potential function, the flow is also irrotational. Analysis. To solve, we will first determine the velocity components, and then from this, obtain the stream function. Using Eqs. 7–25, we have

y c50

f50

u =

c1 5 C1 c2 5 C2

f50

x

0f = 10y 0x

0c ; 0y

u =

c5C

Integrating with respect to y gives

c50

10y =

v = c2 5 C2

0c 0y

c = 5y2 + f1x2 Using the second of Eqs. 7–22, for

c1 5 C1

0f = 10x 0y

As expected, these velocity equations will satisfy the continuity equation, Eq. 7–11. From the first of Eqs. 7–22 for u, we have

f 5 C9

(a)

v =

0c ; 0x

10x = -

(1)

we have

0 35y2 + f1x24 0x

10x = -0 -

0 3f1x24 0x

Integrating, f(x) = -5x 2 + c Substituting this into Eq. 1, and choosing the reference streamline passing through the origin, c = c0 = 0, where x = 0, y = 0, we get c = 0. Therefore the stream function becomes Flow through a channel (b)

Fig. 7–23

c1x, y2 = 51y2 - x 2 2

Ans.

For other streamlines, the flow net can be plotted by setting c1x, y2 = 51y2 - x 2 2 = C and f1x, y2 = 10xy = C′, and then plotting these equations for different values of the constants C and C′. When this is done, the flow net will look like that shown in Fig. 7–23a. If we select two streamlines, say c1 = C1 and c2 = C2, to define the boundary of a channel, Fig. 7–23b, then our solution can be used to study the flow within the channel, assuming, of course, the fluid is ideal.

7.10

7.10

BasiC two-Dimensional flows

387

BASIC TWO-DIMENSIONAL FLOWS

Every flow must satisfy the continuity condition, but an ideal flow must also satisfy the conditions for incompressible and irrotational flow. It was stated previously that the stream function c automatically satisfies continuity; however, to ensure that it also satisfies irrotationality, we require vz = 21 10v>0x - 0u>dy2 = 0. Therefore, when we substitute the velocity components u = 0c>0y and v = -0c>0x into this equation, we get

02c 0x 2

+

02c 0y2

= 0

(7–28)

In a similar manner, since the potential function f automatically satisfies the irrotational flow condition, then to satisfy continuity, 10u>0x2 + 10v>0y2 = 0, we substitute the velocity components, u = 0f>0x and v = 0f>0y, into this equation and require

02f 0x

2

+

02f 0y2

= 0

(7–29)

The above two equations are a form of Laplace’s equation. A solution for the stream function c in Eq. 7–28, or the potential function f in Eq. 7–29, represents the flow field for an ideal fluid, and once this solution is obtained, then the flow net can be established and the velocity components u and can be obtained using Eqs. 7–22 or 7–25. Through the years, many investigators have determined c or f for various types of ideal flow, either directly by solving the above equations, or indirectly by knowing the velocity components for the flow. What follows are the solutions for c and f that involve five basic flow patterns. Once these flows have been introduced, we will then use the results to show how they can be superimposed with one another in order to represent other types of flow.

7

388

Chapter 7

Differential fluiD flow

y y f 5 22 f52 f50 f53 f 5 21 f 5 1

7

U

U

c 53 c52

2 –– U

c51 c50

1 –– U

q 5 1 m2 s x

x

c 5 21

1 –– U

c 5 22

2 –– U Uniform flow Equipotential lines (b)

Uniform flow Streamlines (a)

Fig. 7–24

Uniform Flow. If the flow is uniform and has a constant velocity U along the x axis, as shown in Fig. 7–24a, then its velocity components are u = U v = 0 We can obtain the stream function from Eqs. 7–22. Starting with the velocity component u, 0c 0c u = ; U = 0y 0y Integrating with respect to y, we get c = Uy + f1x2 (1) We now use the velocity component , 0c 0 v = - ; 0 = - 3Uy + f1x24 0x 0x 0 0 = -0 - 3 f1x24 0x

(2)

Integrating, f(x) = c. Therefore c = Uy + c. Choosing the reference streamline c = c0 = 0 that passes through the origin, x = 0, y = 0, then c = 0, and we get the stream function c = Uy

(7–30)

The streamlines are plotted in Fig. 7–24a by assigning constant values for c. As stated, when C = 0, then Uy = 0, which represents the streamline y = 0 passing through the origin. Also, if c = 1 m2 >s, then y = 1>U, and if c = 2 m2 >s, then y = 2>U, etc. Note that the flow between, say, c = 1 m2 >s and c = 2 m2 >s can be determined using Eq. 7–24, that is, q = 2 m2 >s - 1 m2 >s = 1 m2 >s, Fig. 7–24a.

7.10

BasiC two-Dimensional flows

389

y

f

7

c

x

Uniform flow net (c)

Fig. 7–24 (cont.)

In a similar manner, using Eqs. 7–25, u = 0f>0x and v = 0f>0y, we can obtain the potential function. Try following the same procedure as for c and obtain (7–31) f = Ux Equipotential lines are obtained by assigning constant values to f. For example, f = 0 corresponds to x = 0, and f = 1 m2 >s corresponds to x = 1>U, etc. As expected, these lines when plotted will be perpendicular to the streamlines described by c, Fig. 7–24b. Together all the lines form the flow net, Fig. 7–24c. Also, notice that, as required, both c and f will satisfy Laplace’s equation, and, of course, they satisfy the continuity and irrotational flow conditions.

River flow around the end of a bridge pier

390

Chapter 7

Differential fluiD flow

Line Source Flow. In two dimensions, a flow q is defined from a line source along the z axis, from which the fluid flows radially outward, uniformly in all directions in the x–y plane. Such a flow would approximate water emerging from a vertical pipe and flowing between two horizontal plates, as shown in Fig. 7–25a. Here q measures the flow per unit depth, along the z axis (line), so it has units of m2 >s. Due to the angular symmetry, it is convenient to use polar coordinates r, u to describe this flow. If we consider an arbitrary disk of radius r and unit depth in Fig. 7–25b, then the flow through the rim of the disk must pass through an area of A = 2pr112. Since q = vr A, we have

7

q = vr 12pr2112

The radial component of velocity is therefore q vr = 2pr

z

And due to symmetry, the transverse component is vu = 0 For polar coordinates, the stream function is obtained using Eqs. 7–23. For the radial velocity component, we have vr =

1 0c ; r 0u

q 1 0c = r 0u 2pr q 0c = 0u 2p

Integrating with respect to u, c =

(a)

q u + f1r2 2p

Now, considering the transverse velocity component, c5C

r

vu = -

0c ; 0r

f 5 C9

0 q c u + f1r2d 0r 2p

0 = -0 -

u

q

0 = -

0 [ f1r2] 0r

Integrating, f1r2 = c q u + c. We will choose the reference streamline 2p c = c0 = 0 so that at u = 0, c = 0. Therefore

Therefore c = Line source (b)

Fig. 7–25

c =

q u 2p

(7–32)

7.10

391

BasiC two-Dimensional flows

Hence, streamlines for which c equals any constant are radial lines at each coordinate u as expected, Fig. 7–25b. As stated, when C = 0, then 1q>2p2u = 0 or u = 0, which represents the horizontal radial line. Likewise, if c = 1, then u = 2p>q, which defines the angular position of the radial streamline for c = 1, and so on. In a similar manner, the potential function is determined using Eqs. 7–27. vr =

0f ; 0r

q 0f = 2pr 0r

vu =

1 0f ; r 0u

0 =

7

1 0f r 0u

Show that this yields f =

q ln r 2p

(7–33)

Equipotential lines for which f equals any constant are circles having a center at the source. For example, f = 1 defines the circle of radius r = e2p>q, etc., Fig. 7–25b. Notice that the source is actually a mathematical singularity, since vr = q>2pr approaches infinity as r approaches zero. The flow net we have established in Fig. 7–25, however, is still valid at distances away from the source.

Line Sink Flow. When the flow is radially inward toward a line source (z axis), then the strength of the flow q is negative and the flow is termed a line sink flow. This type of flow is similar to the draining of a shallow depth of water confined between two horizontal surfaces, Fig. 7–26a. Here the velocity components are vr = -

(a)

q 2pr

c5C

vu = 0 r

And the stream and potential functions become q c = u 2p f = -

q ln r 2p

The flow net for these functions is shown in Fig. 7–26b.

u

q f 5 C9

(7–34)

(7–35)

Line sink flow net (b)

Fig. 7–26

392

Chapter 7

Differential fluiD flow

Free-Vortex Flow. A free vortex is circular, and so it is typically f 5 C9

r

7

u

formed in tornadoes and hurricanes. Here the streamlines are concentric circles, and the equipotential lines are radial lines, Fig. 7–27a. We can represent this flow using the potential function of a line source, Eq. 7–32, to be the stream function for the vortex, that is, c = -k ln r

vu

c5C

where k = q>2p is a constant. Now equating the expressions for u from Eqs. 7–23 and Eqs. 7–27, -0c>0r = 11>r2 10f>0u2, we can obtain the potential function. The result is f = ku

Free-vortex

Applying Eqs. 7–23, the velocity components are 1 0c vr = ; vr = 0 r 0u

(a)

vu = -

b5a

a 1 ? 50 v 5 — (a? 2 a) 2

0c ; 0r

vu =

k r

(7–37)

(7–38) (7–39)

Note that u becomes larger as r becomes smaller, and the center, r = 0, is a singularity since u becomes infinite, Fig. 7–27a. This flow is irrotational since a potential function can be used to describe it. Consequently, fluid elements within the flow will have an angular distortion in such a way that they do not rotate. In other words, the two adjacent sides of a fluid element will rotate through equal but opposite angles, so that the average rate of rotation of both sides is equal to zero, Fig. 7–27b. Finally, note that this vortex is counterclockwise. To obtain a description of a clockwise vortex, the signs must be changed in Eq. 7–38 and Eq. 7–39.

Circulation. It is also possible to express the stream and potential functions for a free vortex in terms of its circulation Γ, defined by Eq. 7–6. If we choose circulation around the streamline (circle) having a radius r, then

(b)

Fig. 7–27

(7–36)

Γ =

C

V # ds =

L0

2p

k 1r du2 = 2pk r

Using this result, Eqs. 7–36 and 7–37 become c = -

f =

Γ ln r 2p

(7–40)

Γ u 2p

(7–41)

We will use these results in the next section to study the effect of fluid pressure acting on a rotating cylinder.

7.10

393

BasiC two-Dimensional flows

rdu du r

vu

u

7 v vu

Forced-vortex flow (a)

Forced-Vortex Flow. A forced vortex is so named because an external torque is required to start or “force” the motion, Fig. 7–28a. Once it is started, the viscous effects of the fluid will eventually cause it to rotate as a rigid body; that is, the fluid elements maintain their shape and rotate about a fixed axis. As they do this the two adjacent sides rotate . through the same angle a, producing the average rate of rotation a, Fig. 7–28b. A typical example would be the rotation of a real fluid in a container as discussed in Sec. 2.14. Since the fluid elements are “rotating,” a potential function cannot be established. For this case, like the case discussed in Sec. 2.14, the velocity components are vr = 0 and u = vr, where v is the angular velocity of the fluid, Fig. 7–28a.

b 5 2a

vr = 0 vu = -

a

0c = vr 0r

1 ? 5 a? v 5 — (a? 2 (2a)) 2 Rotational flow (b)

The stream function is therefore

1 c = - vr 2 2

Fig. 7–28

(7–42)

394

Chapter 7

EXAMPLE

Differential fluiD flow

7.9 A tornado consists of a whirling mass of air such that the winds essentially move along horizontal circular streamlines, Fig. 7–29a. Determine the pressure distribution within the tornado as a function of r.

7

Forced vortex

r0

Eye Free vortex

(a)

Fig. 7–29 (© Jim Zuckerman/Alamy)

SOLUTION Fluid Description. We will assume the air is an ideal fluid that has steady flow. Translational motion of the tornado will be neglected. Free Vortex. If we assume the motion takes the form of a free vortex, then from Eqs. 7–38 and 7–39, the velocity components are k vr = 0 and vu = (1) r Because the flow within a free vortex is steady irrotational flow, the Bernoulli equation can be applied to two points, each lying on different streamlines. If we choose a point within the tornado, and another at the same elevation but remote, where the air velocity V = 0 and the (gage) pressure is p = 0, then using Eq. 7–19, we have p1 p2 V 12 V 22 + + + z1 = + z2 g g 2g 2g p k2 + z = 0 + 0 + z + rg 2gr 2 rk 2 (2) p = - 2 2r Here k is a constant, as yet to be determined. The negative sign in this equation indicates that a suction develops, and both this suction and the velocity will intensify as r becomes smaller. A free vortex such as this cannot actually exist in a real fluid, since the velocity and pressure would have to approach infinity as r S 0. Instead, due to the increasing velocity gradient as r becomes smaller, the viscosity of the air will eventually create enough shear stress to cause the air in the core, or “eye,” to rotate as a solid system having an angular velocity v. We will assume this transition occurs at a radius of r = r0, Fig. 7–29a.

7.10

395

BasiC two-Dimensional flows

Forced Vortex. Because of its “rigid-body” motion, the core becomes a forced vortex, which we analyzed using Euler’s equations of motion in Example 5–1. In that example, we showed that the pressure distribution is defined by rv2 p = p0 1 r 02 - r 2 2 (3) 2 where p0 is the pressure at r0. In order to fully study the pressure variation, we must match the two solutions at r = r0 and thereby determine the constant k in Eq. 1. At r0 the velocity in the forced vortex, u = vr0, must equal that in the free vortex, that is, k r0

vu = vr0 = r 1 vr 02 2 2

7

so that k = vr 02

The pressures in Eqs. 2 and 3 must also be equivalent at r = r0, so that -

2r 02

rv2 = p0 1 r 02 - r 02 2 2

p0 = -

p

rv2r 02 2

r0

Therefore, after substituting into Eq. 3 and simplifying, we have for the forced vortex, r … r0, u

= vr

p =

rv2 2 1 r - 2r 02 2 2

r

1 2 — rv2r02 2

p 52

2rv2r02

p5

rv2r04 2r2

rv2 2 (r 2 2r02) 2

And for the free vortex, r Ú r0,

vr 02 r rv2r 04 p = Ans. 2r 2 Using these results, a plot of both the velocity and pressure variations is shown in Fig. 7–29b. Notice that the largest suction (negative pressure) occurs at the center of the forced vortex, r = 0, and the highest velocity occurs at r = r0. It is the combination of this low pressure, which can lift houses, cars, and debris off the ground, and the high velocity that makes tornadoes so destructive. In fact tornadoes often reach wind speeds over 322 km>h. The type of vortex we have considered here, that is, a combination of a forced vortex surrounded by a free vortex, is sometimes called a compound vortex. It not only occurs in tornadoes, but also forms in your bathtub as the water drains from the bottom, or in a river as the water flows off a boat oar or around a bridge pier. vu =

yu 1 yu 5 vr02 — r

y u 5 vr

r0

(b)

Fig. 7–29 (cont.)

r

396

Chapter 7

Differential fluiD flow

7.11

y U A x

q

P

7

A9

(a)

SUPERPOSITION OF FLOWS

We noted in the previous section that for any ideal fluid flow, both the stream function and the potential function must satisfy Laplace’s equation. Because the second derivatives of c and f are of the first power in this equation, that is, they are linear, then several different solutions can be superimposed, or added together, to form a new solution. For example, c = c1 + c2 or f = f 1 + f 2. In this way, complex flow patterns can be established from a series of basic flow patterns. The following examples will illustrate how this is done.

Uniform Flow Past a Half Body. If the streamline and potential functions for a uniform flow and a line source flow are added together, we obtain

U A

q q u + Uy = u + Ur sin u 2p 2p q q f = ln r + Ux = ln r + Ur cos u 2p 2p c =

A9

(b)

Asymptote

U r u P

h x

O

vr =

q 0f = + U cos u 0r 2pr

(7–45)

vu =

1 0f = -U sin u r 0u

(7–46)

h r0

(7–44)

Here we have used the coordinate transformation equations, x = r cos u and y = r sin u, to represent the results in polar coordinates. The velocity components can be determined from Eqs. 7–27 (or Eqs. 7–23). We have

y c 5 q2

(7–43)

Asymptote (c) Flow past a half body

Fig. 7–30

The resultant flow looks like that shown in Fig. 7–30a. Since no flow can cross a streamline, then any one of the streamlines can be selected or thought of as the boundary for a solid object that fits within this flow pattern. For example, streamlines A and A′ form the boundary of an infinitely extended body of the shape shown shaded in Fig. 7–30b. Here, however, we will consider a body having a shape formed by the streamline that passes through the stagnation point P, Fig. 7–30c. This point occurs where the velocity of flow from the line source q cancels the uniform flow U, Fig. 7–30a. The stagnation point is located at r = r0, where the two components of velocity must be equal to zero, just before the flow begins to divide equally and pass around the body. For the transverse component, vu = 0, and we find 0 = -U sin u u = 0, p

7.11

superposition of flows

397

The root u = p gives the location of P at r = r0. Also, since vr = 0 at this location, we have 0 =

q + U cos p 2pr0

r0 =

q 2pU

(7–47)

As one would expect, this radial position of P depends upon the magnitude of the uniform flow velocity U and the strength of the line source q. The boundary of the body can now be specified by the streamline that passes through the point r = r0, u = p. From Eq. 7–43, the constant for this streamline is c =

q q q p + Ua b sin p = 2p 2pU 2

Therefore, the equation of the boundary is q q u + Ur sin u = 2p 2 To simplify, we can solve for q in Eq. 7–47 and substitute it into this equation. This gives r =

r0 1p - u2 sin u

(7–48)

Since the body extends an infinite distance to the right, its top and bottom surfaces approach asymptotes, and so it has no closure. For this reason, it is referred to as a half body. The half width h can be determined from Eq. 7–48, by noting that y = r sin u = r0 1p - u2, Fig. 7–30c. As u approaches 0 or 2p, then y = {h = {pr0. Using Eq. 7–48, h =

q 2U

(7–49)

By selecting appropriate values for U and q, we can use the half body to model the front shape of a symmetrical object, such as the front surface of an airfoil (wing) subjected to uniform flow U. Realize, however, that this has its limitations. By assuming the fluid to be ideal, we then have a value for velocity at the boundary of the body, even though all real fluids require a “no-slip” zero velocity due to their viscosity. As stated previously, this viscous effect is generally limited only to a very thin layer near the boundary that is formed when fluids of relatively low viscosity, such as air and water, flow at high velocities. Outside this region the flow can generally be described by the analysis presented here, and indeed, the results have been found to agree rather closely with experimental results.

7

398

Chapter 7

Differential fluiD flow

y

x r1

r2

y u2

u1

x

7 a

a

x2a

x1a

(a)

Fig. 7–31

Doublet. When a line source and a line sink come close to one another and then combine, they form a doublet. This produces an unusual flow pattern; however, it will be used later to represent different types of flow around a cylinder. To show how to formulate the stream function and potential function for this case, consider the equal-strength source and sink shown in Fig. 7–31a. Using Eqs. 7–32 and 7–34, with u1 and u2 as the variables for the source and sink, respectively, we have q c = 1u - u2 2 2p 1 If we rearrange this equation and take the tangent of both sides, and then use the angle addition formula for the tangent, we get tan a

2pc tan u1 - tan u2 b = tan1u1 - u2 2 = q 1 + tan u1 tan u2

(7–50)

From Fig. 7–27a, the tangents of u1 and u2 can be expressed in terms of x and y. tan a or

3 y> 1x + a2 4 - 3 y> 1x - a2 4 2pc b = q 1 + 3 1y> 1x + a221y> 1x - a22 4 c =

-2ay q tan - 1 a 2 b 2p x + y2 - a 2

As the distance a becomes smaller, the angles u1 S u2 S u, and so the difference in the angles 1u1 - u2 2 becomes smaller. When this happens, the tangent of the difference will approach the difference itself, that is, tan 1u1 - u2 2 S 1u1 - u2 2, so tan–1 in the above equation can be eliminated. If we then convert our result to polar coordinates, where r 2 = x 2 + y2, and y = r sin u, we get c = -

qa r sin u a b p r2 - a2

7.11

superposition of flows

399

c(r, u) 5 C

f(r, u) 5 C9

7

Doublet (b)

Fig. 7–31 (cont.)

If a = 0, the flows from the source and sink would cancel each other. However, we can consider the strength q of both the source and sink as increasing to q S ∞ as a S 0, in such a way that the product qa remains constant. For convenience, if we define the strength of this doublet as K = qa>p, then in the limit, the stream function becomes

c =

-K sin u r

(7–51)

We can obtain the potential function in a similar manner. It is

f =

K cos u r

(7–52)

The flow net for a doublet consists of a series of circles that all intersect at the origin, as shown in Fig. 7–31b. With it, we will show on p. 399 how it can be superimposed with a uniform flow to represent flow around a cylinder.

400

Chapter 7

Differential fluiD flow

y

y

x

U

7

r2 r u2

u

c50

U r1

y u1

x

x

a

a

(a)

(b)

Fig. 7–32

Uniform Flow around a Rankine Oval. When uniform flow is superimposed upon a line source and sink of equal strength, each located a distance a away from the origin, Fig. 7–32a, the streamlines so produced will look like those in Fig. 7–32b. Using the established coordinates, we have q q q u2 u1 = Ur sin u + 1u - u1 2 2p 2p 2p 2 q q q r2 f = Ux ln r1 + ln r2 = Ur cos u + ln 2p 2p 2p r1 c = Uy +

(7–53) (7–54)

We can also express these functions in Cartesian coordinates, in which case c = Uy f = Ux +

2ay q tan-1 ¢ 2 ≤ 2p x + y2 - a 2 21x + a2 2 + y2 q ln 2p 21x - a2 2 + y2

(7–55)

(7–56)

The velocity components are then

u =

v =

q 0f x + a x - a = U + c d 2 2 0x 2p 1x + a2 + y 1x - a2 2 + y2 y y q 0f = c d 2 2 0y 2p 1x + a2 + y 1x - a2 2 + y2

(7–57) (7–58)

7.11

superposition of flows

401

y

U

c50

7 h x h

a

a

b

b

Rankine oval (c)

Fig. 7–32 (cont.)

Setting c = 0 in Eq. 7–53, we obtain the body having the shape shown in Fig. 7–32c. It passes through two stagnation points and forms a Rankine oval, named after the hydrodynamist William Rankine, who first developed this idea of combining flow patterns. To find the location of the stagnation points, we require u = v = 0. Thus from Eq. 7–58, with v = 0, we get y = 0. Then with u = 0 at x = b, y = 0, Eq. 7–57 gives b = a

1>2 q a + a2 b Up

(7–59)

This dimension defines the half length of the body, Fig. 7–32c. The half width h is found as the point of intersection of c = 0 with the y axis 1x = 02. From Eq. 7–55, q 2ah 0 = Uh c tan-1 ¢ 2 ≤d 2p h - a2 Rearranging the terms yields h =

h2 - a 2 2pUh tan a b q 2a

(7–60)

A specific numerical solution for h in this transcendental equation will require a numerical procedure, as will be demonstrated in Example 7–10. In general, though, when seeking a solution, begin by choosing a number slightly smaller then q>2U, because the half width of a Rankine oval will be somewhat less than the half width of a corresponding half body. By proper adjustment of h and b, the above results can then be used to study the flow around, for example, an airplane strut or a bridge pier.

402

Chapter 7

Differential fluiD flow

y

c50 U

7

r u a

x

Uniform flow around a cylinder Ideal fluid (a)

Fig. 7–33

Uniform Flow around a Cylinder. If a source and sink of equal strength are placed at the same point, producing a doublet, and this is superimposed with a uniform flow, we will produce the flow around a cylinder, Fig. 7–33a. For this case, we use Eqs. 7–30 and 7–31, with x = r cos u and y = r sin u, and Eqs. 7–51 and 7–52. The stream functions and potential functions therefore become K sin u r K cos u f = Ur cos u + r If we set c = 0, which passes through the two stagnation points, u = 0 and u = p, it will then define the streamline that produces the boundary of the cylinder having a radius r = a, Fig. 7–33a. For c = 0, 1Ua - K>a2 sin u = 0, so the doublet strength must be K = Ua 2. Therefore, a2 c = Ur a 1 - 2 b sin u (7–61) r a2 f = Ur a 1 + 2 b cos u (7–62) r c = Ur sin u -

And so the velocity components become 0f a2 = U a 1 - 2 b cos u 0r r 1 0f a2 vu = = -U a 1 + 2 b sin u r 0u r vr =

(7–63) (7–64)

7.11

superposition of flows

403

(p 2 p0)

U a

7

Pressure distribution Ideal fluid (b)

Fig. 7–33

Notice that when r = a, and u = 0° or 180°, then vr = 0 and vu = 0 as required. Since the streamlines are close together at the top (or bottom) of the cylinder 1u = 90° and 270°), Fig. 7–33a, the maximum velocity occurs here. It is 1vu 2 max = 2U. For an ideal fluid, we can have this finite value for u, but remember the viscosity of any real fluid would actually cause this velocity at the boundary to be zero, due to the no-slip condition for viscous flow. The pressure at a point on or off the cylinder can be determined using the Bernoulli equation, applied at the point and at another one far removed from the cylinder where p = p0 and V = U. Neglecting gravitational effects, we have p p0 V2 U2 + + = g g 2g 2g or since g = rg, p = p0 +

1 r1U 2 - V 2 2 2

Since V = vu = -2U sin u along the surface, r = a, then when we substitute this into the above equation, we find that the pressure on the surface is p = p0 +

1 rU 2 11 - 4 sin2 u2 2

(7–65)

A graph of (p – p0) is shown in Fig. 7–33b. By inspection, this pressure distribution is symmetrical, and so it creates no net force on the cylinder. This is perhaps to be expected for an ideal fluid, since the flow does not include the effect of viscous friction. In Chapter 11, however, we will consider this effect, and show how this alters the pressure distribution.

404

Chapter 7

Differential fluiD flow

7

G

Uniform and Free-Vortex Flow around a Cylinder. If a counterclockwise rotating cylinder is placed in the uniform flow of a real fluid, then the fluid particles in contact with the cylinder’s surface will stick to the surface, and because of viscosity, move with the cylinder. We can approximately model this type of flow by considering the cylinder as immersed in an ideal fluid and subjected to a uniform flow superimposed with a free vortex, written in terms of its circulation Γ, Eqs. 7–40 and 7–41. Adding these flows results in the stream and potential functions c = Ur a 1 -

a

G , 4 pUa Two stagnation points (a)

f = Ur a 1 +

a2 Γ b sin u ln r 2 2p r a2 Γ b cos u + u 2p r2

(7–66) (7–67)

The velocity components are therefore vr =

0f a2 = U a 1 - 2 b cos u 0r r

(7–68)

1 0f a2 Γ = -U a 1 + 2 b sin u + (7–69) r 0u 2pr r From these equations, notice that the velocity distribution around the surface of the cylinder, r = a, has components of vu =

G

vr = 0 Γ (7–70) 2pa It may be of interest to show that indeed the circulation around the cylinder is Γ. For this case, = u is always tangent to the cylinder, and since ds = a du, we have vu = -2U sin u +

G 5 4 pUa One stagnation point (b)

Γ =

G

C

V # ds =

L0

= a 2aU cos u +

2p

a -2U sin u +

2p Γ ub ` = Γ 2p 0

The location of the stagnation points can be determined by setting vu = 0 in Eq. 7–70. We get sin u =

G . 4 pUa No stagnation point on surface of cylinder (c)

Fig. 7–34

Γ b 1a du2 2pa

Γ 4pUa

(7–71)

If Γ 6 4pUa, then two stagnation points will occur on the cylinder, since Eq. 7–71 will have two roots for u, Fig. 7–34a. When the rotation increases and Γ = 4pUa, the points will merge and be located at u = 90°, Fig. 7–34b. Finally, if a further increase of rotation occurs, so that Γ 7 4pUa, then no root exists, and the flow will not stagnate on the surface of the cylinder; rather, it occurs at a point off the cylinder, Fig. 7–34c.

7.11

The pressure distribution around the cylinder can be determined by applying the Bernoulli equation in the same manner as the previous case. We get p = p0 +

2 1 2 Γ rU c 1 - a -2 sin u + b d 2 2pUa

The general shape of this distribution, 1p - p0 2, is shown in Fig. 7–34d. Integrating this distribution over the surface of the cylinder in the x and y directions, we can find the components of the resultant force that the (ideal) fluid exerts on the cylinder per unit depth. This yields

Fx = -

L0

2p

405

superposition of flows

y

U

7

dA 5 a du (p 2 p0) a

u x

1p - p0 2 cos u 1a du2 = 0

Fy = - 1p - p0 2 sin u1a du2 L0 2p

1 = - raU 2 2 L0

2p

c 1 - a -2 sin u + Fy = -rU Γ

Pressure distribution (d) y

2 Γ b d sin u du 2pUa

U

(7–72) G x

Because of the symmetry about the y axis, the result Fx = 0 indicates there is no “drag” or retarding force on the cylinder in the x direction.* Only a vertical downward force Fy exists, Fig. 7–34e. Because this force is perpendicular to the uniform flow, it is referred to as “lift.” Throwing a spinning ball into the air will produce lift in this way and cause it to curve. This is called the Magnus effect, and we will discuss it further in Chapter 11.

Other Applications. We can extend these ideas of superimposing ideal fluid flows to form other closed bodies having a variety of different shapes. For example, a body having the approximate shape of an airfoil or wing is shown in Fig. 7–35. It is formed from a streamline that results from the superposition of a uniform flow, in combination with a single line source and a row of sinks having the same total strength as the source, but decreasing linearly in intensity.

Fy 5 rUG (e)

Fig. 7–34 (cont.)

U

Superposition of a uniform flow with a line source and a row of in-line sinks *Historically, this has been referred to as d’Alembert’s paradox, named after Jean le Rond d’Alembert, who in the 1700s was unable to explain why real fluids cause drag on a body. The explanation came later, in 1904, when Ludwig Prandtl developed the concept of the boundary layer, which we will discuss in Chapter 11.

Fig. 7–35

406

Chapter 7

Differential fluiD flow

I MPO RTA N T PO I N T S • The stream function c1x, y2 or the potential function f1x, y2 7

for a given flow, which satisfies both the continuity and the irrotational flow conditions, can be determined by integrating Laplace’s differential equation. The various solutions form the basis of hydrodynamics and are valid only for ideal fluids.

• When the viscosity of a fluid is small and the flow has a high velocity, then a thin boundary layer will form on the surface of a body placed within the flow. Here the viscous effects can be limited to the flow within this boundary layer, and the flow beyond it can often be considered to be that of an ideal fluid.

• The basic solutions for c1x, y2 and f1x, y2 have been presented for uniform flow, flow from a line source or to a line sink, a doublet, and free-vortex flow. From these solutions, one can obtain the velocity at any point within the flow, and also the pressure at a point can be obtained using the Bernoulli equation.

• Since a forced vortex is rotational flow, it has a solution only for c1x, y2, not for f1x, y2.

• Laplace’s equation is a linear differential equation for c1x, y2 or f1x, y2, and so any combination of basic ideal flow solutions that satisfy this equation can be superimposed or added together to produce more complicated flows. For example, flow around a half body is formed by superimposing a uniform flow and a line source flow. Flow around a Rankine oval is formed by superimposing a uniform flow and an equal-strength line source and line sink that are not concurrent. And flow around a cylinder is formed by superimposing a uniform flow and a doublet.

• A drag is never produced on a symmetric body immersed in an ideal fluid because there are no viscous forces acting on the body that would produce drag.

7.11

EXAMPLE

407

superposition of flows

7.10

Water flows around a concrete bridge pier that has a cross section in the shape of a Rankine oval, formed by superimposing a line source and line sink placed 2 m apart, and each having a strength of 8 m2 >s, Fig. 7–36. If the uniform flow around the body is U = 10 m>s, determine the required dimensions b and h for the pier.

y 10 ms

7 8 m2s

28 m2s

h x

SOLUTION

h

Fluid Description. ideal fluid. Analysis. we have

We assume steady flow and assume water is an

The stagnation points are at x = {b, and so using Eq. 7–59, b = a

1>2 q a + a2 b Up 1>2 8 m2 >s = c 11 m2 + 11 m2 2 d 110 m>s2p = 1.12 m

Ans.

2y 4 tan-1 2 p x + y2 - 1

Setting c = 0 gives the boundary of the pier, since it contains the stagnation points. 2y 4 c = 10y - tan-1 ¢ 2 ≤ = 0 p x + y2 - 1 tan 12.5py2 =

1m

b

b

Fig. 7–36

The stream function for the flow is defined by Eq. 7–55. Here it becomes 2ay q c = Uy tan-1 2 2p x + y2 - a 2 = 10y -

1m

2y x + y2 - 1 2

To find the half width of the pier, set x = 0, y = h in this equation, Fig. 7–36. This gives h2 - 1 h = tan 12.5ph2 2 This same result can also be obtained by applying Eq. 7–60 with a = 1 m, U = 10 m>s, and q = 8 m2 >s. To solve, we note that for a half body, its half width is defined by Eq. 7–49, q>2U = 18 m2 >s2 >[2110 m>s2] = 0.4 m. And so if we start with a slightly smaller value, say h = 0.35 m, for a Rankine oval, then we can increase or decrease it until it satisfies the equation. We find that h = 0.321 m Ans.

408

Chapter 7

EXAMPLE

Differential fluiD flow

7.11 The boat in Fig. 7–37a is propelled by the vertical cylinder, which has a height of 3 m, a diameter of 0.5 m, and is rotating at 55 rad>s. If the boat is traveling forward at 3 m>s, and the wind speed against the side of the boat is 4 m>s, determine the force the cylinder exerts on the boat. Take ra = 1.22 kg>m3.

7 0.5 m 55 rads

3m

SOLUTION 3 ms

Fluid Description. We have steady flow of an ideal fluid. 4 ms

Analysis. The force per unit height acting on the rotating cylinder is defined by Eq. 7–72, F = rU Γ. Since the boat (cylinder) is moving, we must find the velocity of the wind relative to the boat, Fig. 7–37b. It is Vw>b = Vw - Vb

Vw>b = 5 -4j - 3i6 m>s

(a) z

The magnitude of the velocity is then Vw>b = 21 -4 m>s2 2 + 1 -3 m>s2 2 = 5 m>s

Vwb 3 ms 4 ms x

y

And the circulation of the flow around the cylinder is Γ =

(b)

C

Vwb

V # ds = 155 rad>s210.25 m232p10.25 m24 = 21.60 m2 >s

Since the cylinder has a height of 3 m, the force acting on the cylinder is, therefore, y

F x 55 rads

F = r Vw>b Γh = 11.22 kg>m3 215 m>s2121.60 m2 >s213 m2 F = 395 N

(c)

Fig. 7–37

Ans.

This force acts perpendicular to Vw>b as shown in Fig. 7–37c, and so it is the x component of this force that propels the boat forward. This idea for propulsion, based on the Magnus effect, was first employed by Anton Flettner in the 1920s. His “Flettner rotor ship” had two rotating cylinders, each 15 m high, having a diameter of 3 m, and driven by an electric motor.

NOTE:

7.12

7.12

the naVier–stoKes equations

409

THE NAVIER–STOKES EQUATIONS

In Sec. 7.6 we derived the Euler equations, which apply to ideal fluids, where only the forces of gravity and pressure influence the flow. These forces form a concurrent system on each fluid particle or element, and so irrotational flow occurs. Real fluids, however, are viscous, so a more precise set of equations used to describe the flow should include the viscous forces as well. The general differential equations of motion for a fluid, determined from Newton’s second law of motion, were developed in Sec. 7.5 as Eqs. 7–15. To seek a solution we will express these equations in terms of the velocity components by relating the stress components to the viscosity of the fluid and the velocity gradients. Recall that for the one-dimensional flow of a Newtonian fluid, the shear stress and velocity gradient are related by Eq. 1–14, t = m1du>dy2. However, for three-dimensional flow, similar expressions are more complicated. For the special case where the density is constant and we have a Newtonian fluid at constant temperature (isothermal), so the viscosity remains constant, then both the normal and shear stresses are linearly related to their associated strain rates. It can be shown, see Ref. [16], that the stress–strain rate relationships, known as Stokes’ law of viscosity for normal stress, then become* sxx = -p + 2m

0u 0x

syy = -p + 2m

0v 0y

szz = -p + 2m

0w 0z

txy = tyx

0u 0v = ma + b 0y 0x

tyz = tzy = ma

tzx = txz = ma

(7–73)

0v 0w + b 0z 0y

0u 0w + b 0z 0x

*Notice that the normal stresses are the result of both a pressure p, which represents the average normal stress on the fluid element, that is, p = - 13 1sxx + syy + szz 2, and a viscous shear stress term that is caused by the motion of the fluid. When the fluid is at rest (u = v = w = 0), then sxx = syy = szz = - p, a consequence of Pascal’s law. Also, when the streamlines for the flow are all parallel, and directed, say, along the x axis (onedimensional flow), then v = w = 0, and so syy = szz = - p. Additionally, with v = w = 0, for an incompressible fluid, the continuity equation, Eq. 7–11, becomes 0u>0x = 0, and so also sxx = - p.

7

410

Chapter 7

Differential fluiD flow

If we substitute these equations for the stress components into the equations of motion, and use the equation of continuity for incompressible flow, then after simplification, we obtain

7

ra

ra ra

0p 0u 0u 0u 0u 02u 02u 02u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z 0p 0v 0v 0v 0v 02v 02v 02v + u + v + w b = rgy + ma 2 + 2 + 2 b 0t 0x 0y 0z 0y 0x 0y 0z

(7–74)

0p 0w 0w 0w 0w 02w 02w 02w + 2b +u +v + w b = rgz + ma 2 + 0t 0x 0y 0z 0z 0x 0y2 0z Equations of motion

Here, the terms on the left represent “ma,” and those on the right represent “ΣF,” caused by weight, pressure, and viscosity, respectively. These equations were first developed in the early 19th century by the French engineer Louis Navier, and several years later by the British mathematician George Stokes. It is for this reason that they are often referred to as the Navier–Stokes equations. They apply to uniform, nonuniform, steady, or nonsteady flow of an incompressible Newtonian fluid, for which m is constant.* Together with the continuity equation, 0u 0v 0w + + = 0 0x 0y 0z

(7–75)

Continuity equation

these four partial differential equations provide a means of obtaining the velocity components u, , w and the pressure p within the flow. Unfortunately, there is no general solution, simply because the three unknowns u, , w appear in all the equations, and the first three equations are nonlinear and of the second order. In spite of this difficulty, for a few problems they all reduce to a simpler form, and therefore a direct solution can be obtained. This occurs when the boundary and initial conditions are simple, and laminar flow prevails. We will show one of these solutions in the example problem that follows, while some others are given as problems. Later, in Chapter 9, we will also show how these equations can be solved for the case of laminar flow between parallel plates, and within a pipe.

* They have also been developed for compressible flow and can be generalized to include a fluid of variable viscosity. See Ref. [19].

7.12

the naVier–stoKes equations

411

z Fz Fu Fr

7

z u

r

Polar coordinates

Fig. 7–38

Cylindrical Coordinates. Although we have presented the Navier–Stokes equations in terms of Cartesian x, y, z coordinates, they can also be developed in terms of cylindrical (or spherical) coordinates. Without proof, and for later use, in cylindrical component form, Fig. 7–38, they are ra

vu 0vr vu2 0vr 0vr 0vr + vr + + vz b r 0u r 0t 0r 0z

= -

ra

0vu 0vu vu 0vu vrvu 0vu + vr + + + vz b r 0u r 0t 0r 0z

= -

ra

0p vr 02vr 1 0 0vr 1 02vr 2 0vu + rgr + mc ar b - 2 + 2 2 - 2 + d r 0r 0r 0r r r 0u r 0u 0z2

vu 1 0p 1 0 0vu 1 02vu 2 0vr 02vu + rgu + mc ar b - 2 + 2 2 + 2 + d (7–76) r 0u r 0r 0r r r 0u r 0u 0z2

0vz 0t

= -

+ vr

0vz 0r

+

0vz vu 0vz + vz b r 0u 0z

2 02vz 0p 1 0 0vz 1 0 vz + rgz + mc ar b + 2 2 + d r 0r 0r 0z r 0u 0z2

And the corresponding continuity equation, for incompressible flow, Eq. 7–12, becomes 0vz vr 0vr 1 0vu + + + = 0 r r 0u 0r 0z

(7–77)

412

Chapter 7

EXAMPLE

Differential fluiD flow

7.12 When the supply valve A is slightly opened, a very viscous Newtonian liquid in the rectangular tank overflows, Fig. 7–39a. Determine the velocity profile of the liquid as it slowly spills over the sides.

7

SOLUTION

A

(a)

Fig. 7–39

Fluid Description. We will assume the liquid is an incompressible Newtonian fluid that has steady laminar flow. Also, after falling a short distance from the top, the liquid along the sides will continue to maintain a constant thickness a, Fig. 7–39b. Analysis. With the coordinate axes established as shown, there is only a velocity u in the x direction. Furthermore, due to symmetry, u changes only in the y direction, not in the x or z direction, Fig. 7–39c. Because the flow is steady and the liquid is incompressible, the continuity equation becomes 0u 0v 0w + + = 0 0x 0y 0z 0u + 0 + 0 = 0 0x Integrating yields u = u1y2 Using this result, the Navier–Stokes equations in the x and y directions reduce to ra

0p 0u 0u 0u 0u 02u 02u 02u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z 0 + 0 + 0 + 0 = rg -

ra

0p 02u + 0 + m 2 + 0 0x 0y

(1)

0p 0v 02v 0v 0v 0v 02v 02v + u + v + w b = rgy + ma 2 + 2 + 2 b 0t 0x 0y 0z 0y 0x 0y 0z 0 + 0 + 0 + 0 = 0 -

0p + 0 + 0 + 0 0y

7.12

413

the naVier–stoKes equations

This last equation shows that the pressure does not change in the y direction; and since p is atmospheric on the surface of the liquid, it remains so within the liquid, that is, p = 0 (gage). As a result, Eq. 1 now becomes

y

7

a

rg 02u = 2 m 0y Integrating twice, we obtain rg 0u = - y + C1 m 0y u = -

rg 2 y + C1 y + C2 2m

x

(2)

(b)

(3)

To evaluate the constant C2, we can use the no-slip condition; that is, the velocity of the liquid at y = 0 must be u = 0, Fig. 7–39c. Therefore, C2 = 0. To obtain C1, we realize that there is no shear stress txy on the free surface of the liquid. Applying this condition to the fourth of Eqs. 7–73 yields

txy = ma 0 = ma

0u 0v + b 0y 0x

a y

0u + 0b 0y

umax 5

Therefore, du>dy = 0 at y = a. Substituting this into Eq. 2 we get C1 = 1rg>m2a, and so Eq. 3 now becomes rg u = 12ya - y2 2 2m

Ans.

Once it is fully developed, this parabolic velocity profile, shown in Fig. 7–39c, is maintained as the liquid flows down the edge of the tank. It is the result of a balance of downward gravity and upward viscous forces.

x (c)

Fig. 7–39 (cont.)

rg 2 a 2m

414

Chapter 7

Differential fluiD flow

7.13

7

COMPUTATIONAL FLUID DYNAMICS

In the previous section we have seen that a description of any fluid flow requires satisfying the Navier–Stokes equations and the continuity equation, along with the appropriate boundary or initial conditions for the flow. However, these equations are very complicated, and so their solution has been obtained for only a few special cases involving laminar flow. Fortunately, during the past decades there has been an exponential growth in the capacity, memory storage, and affordability of high-speed computers, making it possible to use numerical methods to solve these equations. This field of study is referred to as computational fluid dynamics (CFD), and it is now widely used to design and analyze many different types of fluid flow problems, such as those involving aircraft, pumps and turbines, heating and ventilation equipment, wind loadings on buildings, chemical processes, and biomedical implant devices, and it is even used for atmospheric weather modeling. Several popular CFD computer programs are currently available—for example, FLUENT, FLOW–3D, and ANSYS, to name a few. Other phenomena related to fluid flow, such as heat transfer, chemical reactions, and multiphase changes of the fluid, have also been incorporated into these programs. As accuracy in predicting the flow improves, the use of these programs offers potential savings by eliminating the need for sophisticated experimental testing of models or their prototypes. What follows is a brief introduction and overview of the types of methods that are used to develop CFD software. Further information about this field can be found by referring to one of the many references listed at the end of this chapter, or by taking courses or seminars in this subject, offered at many engineering schools or in the private sector. The CFD Code. There are three basic parts to any CFD code. They are the input, the program, and the output. Let’s consider each of these in turn. Input. The operator must input data related to the fluid properties, specify whether laminar or turbulent flow occurs, and identify the geometry of the boundary for the flow. Fluid Properties. Many CFD packages include a list of physical properties, such as density and viscosity, that can be selected to define the fluid. Any data not included in the list must be identified and then entered separately. Flow Phenomena. The user must select a physical model provided with the program that relates to the type of flow. For example, many commercial packages will have a selection of models that can be used to predict turbulent flow. Needless to say, it takes some experience to select one that is appropriate, since a model that fits one problem may not be suitable for another.

7.13

Computational fluiD DynamiCs

415

Geometry. The physical boundary around and within the flow must also be defined. This is done by creating a grid or mesh system throughout the fluid domain. To make this job user friendly, most CFD packages include various types of boundary and mesh geometries that can be selected to improve computational speed and accuracy. Program. For many users of a CFD program, the various algorithms and numerical techniques used to perform the calculations will be unknown. The process, however, consists of two parts. The first considers the fluid as a system of discrete particles and converts the relevant partial differential equations into a group of algebraic equations, and the second uses an iterative procedure to find solutions of these equations that satisfy the initial and boundary conditions for the problem. There are several approaches that can be used to do this. These include the finite difference method, the finite element method, and the finite control volume method. Finite Difference Method. For unsteady flow, the finite difference method uses a distance–time grid that determines the conditions at a particular point, one time step in the future, based on present conditions at adjacent points. To give some idea of the application of this approach, we will use it later to model unsteady open-channel flow in Chapter 12. Finite Element Method. As the name implies, this method considers the fluid to be subdivided into small “finite elements,” and the equations that describe the flow of each element are then made to satisfy the boundary conditions at the corners or nodes of adjacent elements. It has the advantage of having a higher degree of accuracy than the finite difference method; however, the methodologies used are more complex. Also, since the elements or grid can take any geometric form, the finite element method can be made to match any type of boundary. Finite Control Volume Method. The finite control volume method combines the best attributes of both the finite difference and the finite element methods. It is capable of modeling complex boundary conditions, while expressing the governing differential equations using relatively straightforward finite difference relations. The characteristic feature of this method is that each of the many small control volumes that make up the fluid system accounts for the local time rate of change in a flow variable such as velocity, density, or temperature within the control volume, and the net flux of this variable convected through its control surfaces. Each of these terms is converted into sets of algebraic equations, which are then solved using an iterative method. As a result of these advantages, the finite control volume method has become well established and is currently used in most CFD software.

7

416

Chapter 7

Differential fluiD flow

7 This CFD analysis of flow through the transition and elbow shows how the velocity varies over the cross section as represented by the colors.

Fig. 7–40

Output. The output is usually a graphic of the flow domain, showing the geometry of the problem, and sometimes the grid or mesh used for the analysis. Superimposed on this, the operator can choose to show contours for the flow variables, including streamlines, pathlines, and velocity vector plots, etc. Hard-copy printouts of steady flow can be made, or time animation of unsteady flows can be displayed in video format. An example of this output is shown in Fig. 7–40. General Considerations. If a realistic prediction of a complex flow is to be determined, the operator must have experience at running any particular program. Of course, it is important to have a full understanding of the basic principles of fluid mechanics, in order to define a proper model for the flow, and make a reasonable selection of the time-step size and grid layout. Once the solution is obtained, it can be somewhat checked, by comparing it with experimental data or with similar existing flow situations. Finally, always remember that the responsibility of using any CFD program lies in the hands of the operator (engineer), and for that reason, he or she must ultimately be accountable for the results.

References 1. H. Lamb, Hydrodynamics, 6th ed., Dover Publications, New York,

NY, 1945. 2. J. D. Anderson Jr., Computational Fluid Dynamics: The Basics with

Applications, McGraw-Hill, New York, NY, 1995. 3. A. Quarteroni, “Mathematical models in science and engineering,”

Notices of the AMS, Vol. 56, No. 1, 2009, pp. 10–19. 4. T. W. Lee, Thermal and Flow Measurements, CRC Press, Boca Raton,

FL, 2008. 5. T. J. Chung, Computational Fluid Dynamics, Cambridge, England, 2002. 6. J. Tannechill et al., Computational Fluid Mechnanics and Heat

Transfer, 2nd ed., Taylor and Francis, Bristol, PA, 1997.

proBlems

417

7. C. Chow, An Introduction to Computational Fluid Mechanics, John

Wiley, New York, NY, 1980. 8. F. White. Viscous Fluid Flow, 3rd ed., McGraw-Hill, New York, NY, 2005. 9. J. K. Vennard, Elementary Fluid Mechanics, John Wiley and Sons,

New York, NY, 1963. 10. J. M. Robertson, Hydrodynamics in Theory and Applications, 11. 12. 13. 14. 15. 16. 17. 18.

19. 20.

7

Prentice Hall, Englewood Cliffs, NJ, 1965. L. Milne-Thomson, Theoretical Hydrodynamics, 4th ed., Macmillan, New York, NY, 1960. R. Peyret and T. Taylor, Computational Methods for Fluid Flow, Springer-Verlag, New York, NY, 1983. I. H. Shames, Mechanics of Fluids, McGraw Hill, New York, NY, 1962. J. Tu et al., Computational Fluid Dynamics: A Practical Approach, Butterworth-Heinemann, New York, NY, 2007. A. L. Prasuhn, Fundamentals of Fluid Mechanics, Prentice-Hall, Englewood Cliffs, NJ, 1980. D. N. Roy, Applied Fluid Mechanics, John Wiley, New York, NY, 1988. J. Piquet, Turbulent Flow Models and Physics, Springer, Berlin, 1999. X. Yang and H. Ma, “Cubic eddy-viscosity turbulence models for strongly swirling confined flows with variable density,” Int J Numer Meth Fluids, 45, 2004, pp. 985–1008. T. Cebui, Computational Fluid Dynamics for Engineers, SpringerVerlag, New York, NY, 2005. D. Wilcox, Turbulence Modeling for CFD, DCW Industries, La Canada, CA, 1993.

P ROBLEMS SEC. 7.1–7.6 7–1. A flow is defined by its velocity components u = 8(x 2 + y2 2 m>s and v = 1 - 16xy2 m>s, where x and y are in meters. Determine if the flow is rotational or irrotational. What is the circulation around the triangular path OABO?

7–2. A uniform flow at 6 m>s is directed at an angle of 30° to the horizontal as shown. Determine the circulation around the rectangular path OABCO.

y 0.8 m O

B

0.6 m

x

6 ms

0.8 m C

B 0.2 m

A

O

30°

Prob. 7–1

Prob. 7–2

A

418

Chapter 7

Differential fluiD flow

7–3. As the top plate is pulled to the right with a constant velocity U, the fluid between the plates has the linear velocity distribution shown. Determine the rate of rotation of a fluid element and the shear-strain rate of the element located at y. 7

U

SEC. 7.7–7.9 7–6. The velocity profile for a liquid flowing along the channel of constant width is approximated as u = 121y2 m>s, where y is in meters. Determine the stream function for the flow, and plot the streamlines for c = 0, c = 0.5 m2 >s, and c = 1.5 m2 >s. y

u

y

u 5 (2 y ) mms

h

1.25 m

Prob. 7–3 x

*7–4. The velocity within the eye of a tornado is defined by vr = 0, vu = 10.4r2 m>s, where r is in meters. Determine the circulation at r = 100 m and at r = 200 m.

Prob. 7–6 7–7. The velocity profile of a liquid flowing along the channel of constant width is approximated as u = 121y2 m>s, where y is in meters. Is it possible to determine the potential function for the flow? If so, what is it?

100 m y

200 m

u 5 (2 y ) mms

Prob. 7–4 1.25 m

7–5. Consider the fluid element that has dimensions in polar coordinates as shown and whose boundaries are defined by the streamlines with velocities and v + dv. Show that the vorticity for the flow is given by z = - 1v>r + dv>dr2. D

C G

y 1 dy

dr

x

Prob. 7–7 *7–8. A flow is described by the stream function c = 12x - 4y2 m2 >s, where x and y are in meters. Determine the potential function, and show that the continuity condition is satisfied and that the flow is irrotational.

B y

A

r Du

Prob. 7–5

7–9. A flow has velocity components u = 21y - x2 m>s and v = 14x + 2y2 m>s, where x and y are in meters. Determine the stream function, and plot the streamline that passes through the origin. 7–10. A two-dimensional flow is defined by its components u = 13x 2 2 m>s and v = 12x 2 - 6xy2 m>s, where x and y are in meters. Determine the stream function, and plot the streamline that passes through point (2 m, 4 m).

proBlems 7–11. A flow is defined by the stream function c = 10.5x 2 - y2 2 m2 >s, where x and y are in meters. Determine if the flow satisfies the continuity condition through the triangle ABCA.

419

7–13. A two-dimensional flow has a y component of velocity of v = 16y2 m>s, where y is in meters. If the flow is ideal, determine the x component of velocity and find the magnitude of the velocity at the point x = 1 m, y = 2.5 m. The velocity of the flow at the origin is zero. 7–14. The liquid confined between the two moving plates 7 has a linear velocity distribution. Determine the stream function. Does the potential function exist?

y

1.2 ms A B

C

10 mm 3m B 0.2 ms

x

A

Prob. 7–14

4m

7–15. If the stream function for a flow is c = 18x + 6y2m2 >s, where x and y are in meters, determine the potential function, and the magnitude of the velocity of a fluid particle at point (2 m, –1.5 m).

Prob. 7–11

*7–12. Determine the stream and potential functions for the two-dimensional flow.

*7–16. A two-dimensional flow is described by the stream function c = 1 - 2x 3y + 2xy3 2 m2 >s, where x and y are in meters. Show that the continuity condition is satisfied and determine if the flow is rotational or irrotational. 7–17. The flat plate is subjected to the flow defined by the stream function c = 3 8r 1>2 sin 1u>22 4 m2 >s. Sketch the streamline that passes through point r = 4 m, u = p rad, and determine the magnitude of the velocity at this point.

y

30°

r

5 ms x

Prob. 7–12

Prob. 7–17

u

420

7

Chapter 7

Differential fluiD flow

7–18. A two-dimensional flow is defined by its components u = 12y2 m>s and v = 13x2 m>s, where x and y are in meters. Determine if the flow is rotational or irrotational, and show that the continuity condition for the flow is satisfied. Also, find the stream function and the equation of the streamline that passes through point (2 m, 6 m). Plot this streamline. 7–19. The liquid confined between two moving plates has a linear velocity distribution. If the pressure at the top surface of the bottom plate is 600 N>m2, determine the pressure at the bottom surface of the top plate. Take r = 1.2 Mg>m3. 1.2 ms A

7–22. A fluid has velocity components u = - 12x + 4y2 m>s and v = 12y - 4x2 m>s, where x and y are in meters. Determine the stream function and also the potential function if possible. 7–23. The potential function for a horizontal flow of water is f = 14x 2 - 4y2 2 m2 >s, where x and y are in meters. Determine the magnitude of the velocity at point A (2 m, 1 m). What is the difference in pressure between this point and point B (1 m, 3 m)? *7–24. A fluid has velocity components u = 12y + 32 m>s and v = 12x2 m>s, where x and y are in meters. Determine the stream function and also the potential function if possible.

7–25. Water has horizontal and vertical velocity components u = 21y2 - x 2 2 m>s and v = 14xy2 m>s, where x and y are in meters. If the pressure at point A (1 m, 4 m) is 450 kPa, determine the pressure at point B (0.5 m, –2 m). Also, what is the potential function for the flow?

10 mm

B 0.2 ms

Prob. 7–19

*7–20. A fluid has velocity components u = 31x 2 - y2 2 m>s and v = 1 - 6xy2 m>s, where x and y are in meters. Determine the stream function and also the potential function if possible. Plot the streamlines and equipotential lines that pass through point (2 m, 1 m).

7–21. A fluid has velocity components u = 18xy2 m>s and v = 41x 2 - y2 2 m>s, where x and y are in meters. Determine the potential function and find the circulation around the square path ABCDA.

7–26. A fluid has velocity components u = 41 - x + y2 m>s and v = 12x 2 + 4y2 m>s, where x and y are in meters. Find the stream function, and determine if the flow is rotational or irrotational. Find the potential function if possible. 7–27. The potential function for a flow is f = 41x 2 - y2 2 m2/s, where x and y are in meters. Determine the streamline that passes through point (1 m, 2 m) and the magnitude of the velocity of a fluid particle at this point. *7–28. The air flowing around the inclined plate is defined by its stream function c = 12r 2 sin 3u2 m2 >s. Determine the speed at the point which is located at r = 2 m, u = p>6 rad. The origin is at the corner. Sketch the streamline that passes through this point.

y

C

D

2m A

B 3m

x 1m

2m 60°

Prob. 7–21

Prob. 7–28

421

proBlems 7–29. The stream function for the horizontal flow of water near the corner is defined by c = 12xy2 m2 >s, where x and y are in meters. If the pressure at point A (0.5 m, 1.5 m) is 80 kPa, determine the pressure at point B (1 m, 2 m).

7–31. The flow has a velocity of V = 516y + 42i6 m>s, where y is vertical and is in meters. Determine if the flow is rotational or irrotational. If the pressure at point A is 20 kPa, determine the pressure at the origin. Take r = 1100 kg>m3. y

7

A y 1.5 m x

O

A

B

Prob. 7–31

x

*7–32. The potential function for a flow is f = 31x 2 - y2 2, where x and y are in meters. Determine the magnitude of the velocity of a fluid particle at point A (1 m, 0.3 m). Show that continuity is satisfied, and find the streamline that passes through point A.

Prob. 7–29

7–30. The stream function for a horizontal flow of water near the corner is c = 12xy2 m2 >s, where x and y are in meters. Determine the x and y components of the velocity and the acceleration of fluid particles passing through point (1 m, 2 m). Plot the streamline that passes through this point.

y

7–33. The velocity components for a two-dimensional flow are u = 14y2 m>s and v = 14x2 m>s, where x and y are in meters. Find the stream function and the equation of the streamline that passes through point (2 m, 1 m). Plot this streamline. 7–34. The flow around the bend in the horizontal channel can be described as a free vortex for which vr = 0, vu = (8>r2 m>s, where r is in meter. Show that the flow is irrotational. If the pressure at point A is 4 kPa, determine the pressure at point B. Take r = 1100 kg>m3. A B C 2m

r u

A

B

x 0.5 m

Prob. 7–30

Prob. 7–34

422

Chapter 7

Differential fluiD flow

7–35. The x component of velocity of a two-dimensional irrotational flow is u = (2y2 - y - 2x 2) m>s, where x and y are in meters. Find the y component of velocity if v = 0 at x = y = 0. 7

*7–36. A fluid has horizontal velocity components of u = 1y2 - x 2 2 m>s and v = 12xy2 m>s, where x and y are in meters. If the pressure at point A (3 m, 2 m) is 600 kPa, determine the pressure at point B (1 m, 3 m). Also, what is the potential function for the flow? Take g = 8 kN>m3.

7–39. If the potential function for a two-dimensional flow is f = 1xy2 m2 >s, where x and y are in meters, determine the stream function and plot the streamline that passes through the point 11 m, 2 m2. *7–40. Water flow through the horizontal channel is defined by the stream function c = 21x 2 - y2 2 m2 >s. If the pressure at B is atmospheric, determine the pressure at point A (0.5 m, 0) and the flow per unit depth in m2 >s.

7–37. The stream function for a circular flow is defined by c = 1 - 2r 2 2 m2 >s. Determine the velocity components vr and vu, and vx and vy, at the point r = 2 m, u = - 30°.

y B

y

c50 1.5 m c 5 0.5 m2 s x 2m

30° x

A 1.5 m

Prob. 7–40 Prob. 7–37 7–38. A fluid has velocity components of u = 31x 2 + y2 m>s and v = 1 - 6xy2 m>s. Determine the stream function and the circulation around the rectangle. Plot the streamlines for c = 0, c = 1 m2 >s, and c = 2 m2 >s. y

7–41. A fluid flows into the corner formed by the two walls. If the stream function for this flow is defined by c = 15 r 4 sin 4u2 m2 >s, show that continuity for the flow is satisfied. Also, plot the streamline that passes through point r = 2 m, u = 1p>62 rad, and find the magnitude of the velocity at this point.

y

1m

x 2m 458

x

Prob. 7–38

Prob. 7–41

proBlems

423

7–42. The potential function for a flow is f = 61x 2 - y2 2 m2 >s, where x and y are in meters. Determine the magnitude of the velocity of fluid particles at the point (2 m, 3 m). Show that continuity is satisfied, and find the streamline that passes through this point.

7–46. As water drains from the large cylindrical tank, its surface forms a free vortex having a circulation of Γ. Assuming water to be an ideal fluid, determine the equation z = f(r) that defines the free surface of the vortex. Hint: Use the Bernoulli equation applied to two points on the surface.

7–43. The stream function for the flow around the 90° corner is c = 8r 2 sin 2u. Show that continuity of flow is satisfied. Determine the r and u velocity components of a fluid particle located at r = 0.5 m, u = 30°, and plot the streamline that passes through this point. Also, determine the potential function for the flow.

z

y

7

r

Prob. 7–46 7–47. If a building has a flat roof and is located 20 m from the center of a tornado, determine the uplift pressure on the roof. The building is within the free vortex of the tornado, where the wind speed is 20 m>s at a distance of 40 m from its center. The density of the air is ra = 1.20 kg>m3.

r u x

Prob. 7–43 Prob. 7–47 *7–48. A free vortex is defined by its stream function c = 1 - 240 ln r2 m2 >s, where r is in meter. Determine the velocity of a particle at r = 4 m and the pressure at points on this streamline. Take r = 1.20 kg>m3.

SEC. 7.10–7.11 *7–44. Show that the equation that defines a sink will satisfy continuity, which in polar coordinates is written as 01vr r2 0r

+

r54m

0vu = 0. 0u

7–45. Combine a source of strength q with a free counterclockwise vortex, and sketch the resultant streamline for c = 0.

Prob. 7–48

424

Chapter 7

Differential fluiD flow

7–49. A source at O creates a flow from O that is described by the potential function f = 18 ln r2 m2 >s, where r is in meters. Determine the stream function, and find the velocity at point r = 5 m, u = 15°. 7

7–53. Determine the location of the stagnation point for a combined uniform flow of 8 m>s and a source having a strength of 3 m2 >s. Plot the streamline passing through the stagnation point.

308 O u y

Prob. 7–49 7–50. A source q is emitted from the wall while a flow occurs towards the wall. If the stream function is described as c = 14xy + 8u2 m2 >s, where x and y are in meters, determine the distance d from the wall where the stagnation point occurs along the y axis. Plot the streamline that passes through this point.

8 ms

x

y

Prob. 7–53

d x

Prob. 7–50

7–51. Pipe A provides a source flow of 3 m2 >s, whereas the drain, or sink, at B removes 3 m2 >s. Determine the stream function and the equation of the streamline for c = 0.25 m2 >s and c = 0.5 m2 >s.

7–54. The source and sink of equal strength 6 m2 >s are located as shown. Determine the x and y components of velocity of a particle passing through point P and the equation of the streamline passing through this point.

*7–52. Pipe A provides a source flow of 3 m2 >s, whereas the drain at B removes 3 m2 >s. Determine the potential function and the equations of the equipotential line for f = 1 m2 >s and f = 2 m2 >s.

x

y

2m

3 ms2 A

3 ms2 B

2m

2m

Probs. 7–51/52

3m

x

y

P 3m

Prob. 7–54

425

proBlems 7–55. Determine the strength q of the source, if the stagnation point P is located at ro = 0.5 m in a combined uniform flow of 4 m> s and the source. Plot the streamline passing through the stagnation point.

7–58. The Rankine oval is defined by the source and sink, each having a strength of 0.2 m2 >s. If the velocity of the uniform flow is 4 m>s, determine the longest and shortest dimensions of the oval.

y

7 4 ms 0.5 m

x

q

P

y

4 ms

x

Prob. 7–55

*7–56. Two sources, each having a strength of 3 m2 >s, are located as shown. Determine the x and y components of the velocity of a fluid particle that passes point (3 m, 4 m). What is the equation of the streamline that passes through this point?

0.5 m

0.5 m

Prob. 7–58

y

x 4m

4m

7–59. The Rankine oval is defined by the source and sink of equal strength. Determine this strength q if the length of the Rankine oval is to be 1.054 m.Also, what is its corresponding width? The velocity of the uniform flow is 4 m>s.

Prob. 7–56 7–57. The Rankine oval is defined by the source and sink, each having a strength of 0.2 m2 >s. If the velocity of the uniform flow is 4 m>s, determine the equation that defines the boundary of the oval. y

4 ms

y 4 ms

x

0.5 m

0.5 m

Prob. 7–57

x

0.5 m

0.5 m

Prob. 7–59

426

Chapter 7

Differential fluiD flow

*7–60. The leading edge of a wing is approximated by the half body. It is formed from the superposition of the uniform airflow of 150 m > s and a source having a strength of 300 m2 >s. Determine the width of the half body and the location of the stagnation point. 7

7–61. The leading edge of a wing is approximated by the half body. It is formed from the superposition of the uniform airflow of 150 m > s and a source having a strength of 300 m2 >s. Determine the difference in pressure between the stagnation point O and point A. The air is at a temperature of 20°C under standard atmospheric pressure.

7–63. Determine the equation of the boundary of the half body formed by placing a source of 0.5 m2 >s in the uniform flow of 8 m > s.

y 8 ms

x 150 m/s

A

A u 5 608

Prob. 7–63

O

Probs. 7–60/61

7–62. The half body is defined by a combined uniform flow having a velocity of U and a point source of strength q. Determine the pressure distribution along the top boundary of the half body as a function of u, if the pressure within the uniform flow is p0. Neglect the effect of gravity. The density of the fluid is r.

*7–64. A fluid flows horizontally over a half body for which U = 0.4 m>s and q = 1.0 m2 >s. Plot the half body, and determine the magnitudes of the velocity and pressure in the fluid at the point r = 0.8 m and u = 90o. The pressure within the uniform flow is 300 Pa. Take r = 850 kg>m3.

y

y

U 5 0.4 ms

U r u

r u

x

q 5 1.0 ms2

Prob. 7–62

Prob. 7–64

x

proBlems 7–65. Integrate the pressure distribution, Eq. 7–65, over the surface of the cylinder in Fig. 7–22b, and show that the resultant force is equal to zero. 7–66. Air is flowing at U = 20 m>s past the model of a Quonset hut. Find the magnitude of the velocity and the absolute pressure acting on the windows located on the roof at A, u = 120°, and at B, u = 30°. The absolute pressure within the uniform flow is p0 = 90 kPa. Take ra = 1.23 kg>m3.

427

7–70. The tube is built from two segments that are glued together at A and B. If it is exposed to a uniform air flow having a velocity of 6 m>s, determine the resultant force the pressure exerts on the top segment AB per unit length of the tube. Take r = 1.22 kg>m3. 7 6 ms

7–67. The Quonset hut is subjected to a uniform wind having a velocity of 20 m>s. Determine the resultant vertical force caused by the pressure that acts on the roof if it has a length of 10 m. Take ra = 1.23 kg>m3.

B

A 308 308

*7–68. Air is flowing at U = 20 m>s past the model of a Quonset hut. Find the magnitude of the velocity and the gage pressure at point C, where r = 4.5 m and u = 150°. Take ra = 1.23 kg>m3.

0.2 m

Prob. 7–70 7–71. The long rotating cylinder is subjected to a uniform horizontal airflow of 2 m>s. Determine the circulation around the cylinder and the location of the stagnation points if the lift per unit length is 6.56 N>m. Take ra = 1.202 kg>m3.

y

20 ms

2 ms

C

A B 4m

0.5 m

u x

Probs. 7–66/67/68 Prob. 7–71

7–69. Air flows around the cylinder as shown. If r = 1.22 kg>m3, plot the pressure on its surface for 0 … u … p>2 rad for each p>12 rad increment of u.

*7–72. The 200-mm-diameter cylinder is subjected to a uniform flow having a velocity of 6 m>s. At a distance far away from the cylinder, the pressure is 150 kPa. Plot the variation of the velocity and pressure along the radial line r, at u = 0°, and specify their values at r = 0.1 m, 0.2 m, 0.3 m, 0.4 m, and 0.5 m. Take r = 1.5 Mg>m3. 6 ms

8 ms r u u

0.1 m

0.3 m

Prob. 7–69

Prob. 7–72

428

Chapter 7

Differential fluiD flow

7–73. Water flows toward the column with a uniform speed of 1.5 m>s. Determine the pressure at point A if this point is at a depth of 1.25 m below the water surface. Take rw = 1000 kg>m3.

*7–76. The 0.5-m-diameter bridge pier is subjected to the uniform flow of water at 4 m>s. Determine the maximum and minimum pressures exerted on the pier at a depth of 2 m.

1.5 m/s

7

A 750 mm 250 mm

Prob. 7–73 7–74. The 200-mm-diameter cylinder is subjected to a uniform horizontal flow having a velocity of 6 m>s. At a distance far away from the cylinder, the pressure is 150 kPa. Plot the variation of the velocity and pressure along the radial line r, at u = 90°, and specify their values at r = 0.1 m, 0.2 m, 0.3 m, 0.4 m, and 0.5 m. Take r = 1.5 Mg>m3.

Prob. 7–76

7–77. The tall circular building is subjected to a uniform wind having a velocity of 40 m>s. Determine the location u of the window that is subjected to the smallest pressure. What is this pressure? Take ra = 1.202 kg>m3.

6 ms r u

7–78. The tall circular building is subjected to a uniform wind having a velocity of 40 m>s. Determine the pressure and the velocity of the wind on the building at u = 30°, 60°, and 90°. Take ra = 1.202 kg>m3.

0.1 m

Prob. 7–74 7–75. Water flows toward the circular column with a uniform speed of 2 m>s. If the outer radius of the column is 2 m, and the pressure within the uniform flow at a particular depth is 55 kPa, determine the pressure at point A of the same depth. Take rw = 1000 kg>m3.

3m

u

A 2 m/s

30 m

2m 40 m/s

Prob. 7–75

Probs. 7–77/78

429

proBlems 7–79. The long cylinder is rotating counterclockwise with a constant angular velocity of 80 rad>s. If the air is blowing at a constant speed of 20 m>s, determine the location of the stagnation points, and find the absolute pressure at point A. The absolute pressure within the uniform flow is 98 kPa. Take ra = 1.20 kg>m3. *7–80. The long cylinder is rotating counterclockwise with a constant angular velocity of 80 rad/s. If the air is blowing at a constant speed of 20 m>s, determine the lift per unit length on the cylinder and the maximum pressure on the cylinder. The absolute pressure within the uniform flow is 98 kPa. Take ra = 1.20 kg>m3.

7–82. If the uniform velocity of the air is 15 m>s and the pressure within the uniform flow is 100 kPa, determine the required angular velocity v of the cylinder so that the lift is 96.6 N. Take ra = 1.20 kg>m3. 7

0.6 m

15 ms

v

Prob. 7–82

SEC. 7.12 80 rads

7–83. The channel for a liquid is formed by two fixed plates. If laminar flow occurs between the plates, show that the Navier–Stokes and continuity equations reduce to 02u>0y2 = 11>m2 0p>0x and 0p>0y = 0. Integrate these equations to show that the velocity profile for the flow is u = 11>2m2 1dp>dx2 3 y2 - 1d>22 2 4 . Neglect the effect of gravity.

0.3 m 20 ms

60° 0.5 m A

y

Probs. 7–79/80 d2 x d2

7–81. The 1-m-long cylinder rotates counterclockwise at 40 rad>s. If the uniform velocity of the air is 10 m>s, and the pressure within the uniform flow is 300 Pa, determine the maximum and minimum pressure on the surface of the cylinder. Also, what is the lift force on the cylinder? Take ra = 1.20 kg>m3.

Prob. 7–83 *7–84. Liquid is confined between a top plate having an area A and a fixed surface. A force F is applied to the plate and gives the plate a velocity U. If this causes laminar flow, and the pressure does not vary, show that the Navier–Stokes and continuity equations indicate that the velocity distribution for this flow is defined by u = U1y>h2, and that the shear stress within the liquid is txy = F>A. y

0.6 m

U

10 ms

F 40 rads

h

u x

Prob. 7–81

Prob. 7–84

430

7

Chapter 7

Differential fluiD flow

7–85. A horizontal velocity is defined by u = 31x 2 - y2 2 m>s and v = 1 -6xy2 m>s. Show that these expressions satisfy the continuity equation. Using the Navier–Stokes equations, show that the pressure distribution is defined by p = C - rV 2 >2 - rgz.

7–86. The sloped open channel has steady laminar flow within its depth h. Show that the Navier–Stokes equations reduce to 02u>0y2 = - 1r g sin u2 >m and 0p>0y = -r g cos u. Integrate these equations to show that the velocity profile is u = [1r g sin u2 >2m]12hy - y2 2 and the shear-stress distribution is txy = r g sin u 1h - y2.

7–87. Fluid having a density r and viscosity m fills the space between the two cylinders. If the outer cylinder is fixed, and the inner one is rotating at v, apply the Navier–Stokes equations to determine the velocity profile assuming laminar flow.

v

ri ro

r

y

Prob. 7–87

h

u x

Prob. 7–86

Chapter reView

431

CHAP TER R EV IEW

The rate of translation of a fluid element is defined by its velocity field.

The linear distortion of a fluid element is measured as the change in its volume per unit volume. The time rate of this distortion is called the volumetric dilatation.

The rate of rotation or angular velocity of a fluid element of measured by the average angular velocity of two of its adjacent sides.

The angular distortion of a fluid element is defined by the time rate of change of the 90° angle between its adjacent sides. This is the shear-strain rate of the element.

An ideal fluid, which has no viscosity and is incompressible, can exhibit irrotational flow. We require V = 0.

V = V1x, y, z, t2

dV>dV dt

=

vz =

0u 0v 0w + + 0x 0y 0z

1 0v 0u a b 2 0x 0y

0v 0u # gxy = + 0x 0y

7

432

7

Chapter 7

Differential fluiD flow

Circulation Γ is a measure of the net flow around a boundary, and vorticity z is the net circulation about each unit area of fluid.

Γ = a

0u 0v b ∆x∆y 0x 0y

The conservation of mass is expressed by the continuity equation.

The stream function c1x, y2 satisfies the continuity equation. If c1x, y2 is known, then the velocity components of the fluid at any point can be determined from its partial derivatives.

0u 0v + = 0 0x 0y

u =

The flow per unit depth, between any two streamlines, c1x, y2 = C1 and c1x, y2 = C2, can be determined from the difference of their stream function constants.

The potential function f1x, y2 satisfies the conditions of irrotational flow. The velocity components of the fluid at any point can be determined from its partial derivatives.

Γ 0v 0u = A 0x 0y

z =

0c 0y

v = -

0c 0x

q = C2 - C1

u =

0f , 0x

v =

0f 0y

Chapter reView

433

Basic Two-dimensional Flows Superposition of Flows Uniform flow c = Uy f = Ux

Uniform flow past a half body u = U v = 0

Line source flow c = f =

q u 2p

vr =

q ln r 2p

vu = 0

c =

q q u + Uy = u + Ur sin u 2p 2p

vr =

f =

q q ln r + Ux = ln r + Ur cos u 2p 2p

vu = - U sin u

q 2pr

r0 =

q 2pU

h =

Distance from source to stagnation point

q 2U

Half height

Uniform flow around a Rankine oval

Line sink flow c = -

q u 2p

vr = -

f = -

q ln r 2p

vu = 0

q 2pr

c = Uy -

f = Ux +

2ay q tan-1 2 2p x + y2 - a 2

2 1x + a)2 + y2 q ln 2p 2 1x - a)2 + y2

b = a

u = U +

v =

1>2 q a + a2 b Up

q x + a x - a c d 2p 1x + a2 2 + y2 1x - a2 2 + y2

y y q c d 2 2 2p 1x + a2 + y 1x - a)2 + y2 h =

Half length

h2 - a 2 2pUh tan a b 2a q Half width

Uniform flow around a cylinder

Free-vortex flow Γ ln r 2p Γ f = u 2p

c = - k ln r = -

vr = 0

f = ku

vu =

Γ = 2pk Forced vortex flow 1 c = - vr 2 2

q + U cos u 2pr

k r

c = Ur a1 -

a2 b sin u r2

vr = Ua1 -

a2 b cos u r2

f = Ur a1 +

a2 b cos u r2

v0 = -Ua1 +

a2 b sin u r2

Uniform and free-vortex flow around a cylinder vr = 0 u

= vr

c = Ur a1 -

f = Ur a1 +

a2 Γ b sin u ln r 2p r2 a2 Γ b cos u + u 2p r2

vr =

v0 =

sin u =

0f a2 = Ua1 - 2 b cos u 0r r

1 0f a2 Γ = -Ua1 + 2 b sin u + r 0u 2pr r

Γ 4pUa

Location of stagnation points

7

8

Photo Researchers,Inc/Alamy Stock Photo

CHAPTER

Wind tunnels are often used to test models of prototypes such as aircraft and other vehicles. To do this, the model must be properly scaled so that the results correlate to the prototype.

DIMENSIONAL ANALYSIS AND SIMILITUDE CHAPTER OBJECTIVES ■

To discuss how different types of forces influence the fluid flow behavior, and to present an important set of dimensionless numbers that involve these forces.

■

To formalize a dimensional analysis procedure that can be used to minimize the amount of data required to experimentally study the behavior of a fluid.

■

To show how one can scale a model of a full-size structure or machine, and then use this model to experimentally study the effects of a fluid flow.

8.1

DIMENSIONAL ANALYSIS

In the previous chapters we presented many important equations of fluid mechanics, and illustrated their application to the solution of some practical engineering problems. In all these cases we were able to obtain an algebraic solution in the form of an equation that describes the flow. In some cases, however, a problem can involve a very complicated flow, and the combination of variables such as velocity, pressure, density, viscosity, etc., that describe it may not be fully understood. When this occurs, the flow is often studied by performing an experiment. Unfortunately, experimental work can be costly and time consuming, and so it becomes important to be able to minimize the amount of experimental data that needs to be obtained. The best way to accomplish this is to perform a dimensional analysis of all the relevant variables. Specifically, dimensional analysis is a branch of mathematics that is used to organize all these variables into sets of dimensionless groups. Once these groups are obtained, we can use them to obtain the maximum amount of information from a minimum number of experiments.

435

436

8

Chapter 8

D i m e n s i o n a l a n a ly s i s

anD

similituDe

The method of dimensional analysis is based on the principle of dimensional homogeneity, discussed in Sec. 1.4. It states that each term in an equation must have the same combination of units. Except for temperature, the variables in almost all fluid flow problems can be described using the primary dimensions of mass M, length L, and time T; or force F, length L, and time T.* For convenience, Table 8–1 lists the combinations of these dimensions for many of the variables in fluid mechanics. Although a dimensional analysis will not provide a direct analytical solution for a problem, it will aid in formulating the problem, so that the solution can be obtained experimentally in the simplest way possible. To illustrate this point, and at the same time show the mathematical process # involved, let us consider the problem of finding how the power output W of

TABLE 8–1 Quantity

Symbol

Area

A

L

2

L2

Volume

V

L3

L3

Velocity

V

LT -1

LT -1

Acceleration

a

LT -2

LT -2

Angular velocity

v

T -1

T -1

Force

F

MLT −2

F

Mass

m

M

FT 2L -1

Density

r

ML−3

FT 2L -4

Specific weight

g

ML -2T -2

FL -3

Pressure

p

ML -1T -2

FL -2

Dynamic viscosity

m

ML -1T -1

FTL -2

Kinematic viscosity

n

L2T -1

L2T -1

ML2T -3

FLT -1

L3T -1

L3T -1

Mass flow rate

Q # m

MT -1

FTL -1

Surface tension

s

MT -2

FL -1

Weight

W

MLT -2

Power Volumetric flow rate

Torque

# W

T

M-L-T

2

ML T

F-L-T

-2

F FL

*Recall that force and mass are not independent of each other. Instead they are related by Newton’s law of motion, F = ma. Thus in the SI system, force has the dimensions of ML>T 2 (ma).

8.1

the pump in Fig. 8–1a depends upon the pressure increase ∆p from A to B and the flow Q developed by the pump. One way to obtain this unknown relationship experimentally would be to require the pump to produce a specific measured flow Q1, and then change the power several times and measure each corresponding pressure increase. The data, when plotted, would # then give the necessary relationship W = f1Q1, ∆p2, shown in Fig. 8–1b. Repeating this process for Q2, etc., we can produce a family of lines or curves as shown. Unfortunately, without a large number # of such graphs, one for each Q, it could be difficult to obtain the value of W for any specific Q and ∆p. # An easier way to obtain the needed relationship W = f1Q, ∆p2 is to first perform a dimensional analysis of the variables. Here we must require that Q and ∆p be arranged in such a way that their combined units are the same as those that are used to describe power. Furthermore, since Q and ∆p alone do not have units of power, these variables are simply not added to or subtracted from one another; rather, they must either be multiplied together or divided one by the other. Hence the unknown functional relationship must have the form # W = CQa 1 ∆p2 b Here C is some unknown (dimensionless) constant, and a and b are unknown exponents that maintain the dimensional homogeneity for the units of power. Using Table 8–1 and the# M-L-T system, the primary dimensions for these variables are W 1ML2 >T 3 2, Q 1L3 >T 2, and ∆p 1M>LT 2 2. When these dimensions for the units are substituted into the above equation, it becomes

8

A

B

(a) W Q3

W 5 f(Q3, Dp)

ML2T -3 = 1L3T -1 2 a 1ML -1T -2 2 b = M bL3a - bT -a - 2b

Q2 W 5 f(Q2, Dp)

Since the exponents for M, L, and T must be the same on each side of this equation, then M: L: T:

1 = b 2 = 3a - b -3 = -a - 2b

Solving yields a = 1 and b = 1. In other words, the above required functional relationship, shown in Fig 8–1b, is of the form # W = CQ∆p (8–1) Of course, rather than using this formal procedure, we could also obtain this result by inspection, realizing that the dimensions of Q and ∆p, when # multiplied, cancel in such a way so as to produce the same dimensions of W. Now that this relationship has been established, we only need to perform a single experiment, # measuring the flow Q1 and pressure drop ∆p1 for a known power W1. Doing this will enable us to determine the # single unknown constant C = W1 >(Q1 ∆p1), and with this known, we can then use the above equation to calculate the power requirement for the pump for any other combination of Q and ∆p.

437

Dimensional analysis

Q1 W 5 f(Q1, Dp) Dp (b)

Fig. 8–1

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8.2

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IMPORTANT DIMENSIONLESS NUMBERS

The method of applying dimensional analysis, as in the previous example, was originally developed by Lord Rayleigh. It was then later improved upon by Edgar Buckingham, and as we will show in Sec. 8.3, his method requires combining the variables that describe the flow into a set of dimensionless ratios or “numbers.” Oftentimes these numbers are represented by a ratio of the forces that act on the fluid within the flow. And because these ratios appear frequently throughout the study of experimental fluid mechanics, we will introduce some of the more important ones here. As a matter of convention, each dimensionless ratio is made up of a dynamic or inertia force, and some other force developed by the flow, such as caused by pressure, viscosity, or gravity. The inertia force is actually a fictitious force, because it represents the inertia term ma in the equation of motion. Specifically, if we write ΣF = ma as ΣF - ma = 0, then the term –ma can be thought of as an inertia force, which is needed to produce a zero force resultant on a fluid particle. We will select the inertia force for all these dimensionless ratios because it involves the fluid particle’s acceleration, and so it plays an important role in almost every problem involving fluid flow.* The inertia force has a magnitude of ma or rVa, and so, retaining the fluid property r, it has partial dimensions of r1L3 21L>T 2 2. Since velocity V has dimensions of L>T, we can omit the time dimension and instead express this force in terms of the velocity and length, that is, rV 2L2. We will now briefly define the significance of some important dimensionless force ratios, each referred to as a “number,” and in later sections we will discuss the importance of these ratios for specific applications.

Euler Number. It is the difference in static pressure between two points in a fluid that causes the fluid to flow. The ratio of the force caused by this pressure difference to the inertia force, rV 2L2, is called the Euler number or the pressure coefficient. The pressure force ∆pA can be expressed in terms of length dimensions as ∆pL2 , and so the Euler number is Eu =

pressure force ∆p = inertia force rV 2

(8–2)

This dimensionless number controls the flow behavior when pressure and inertia forces are dominant, as when a liquid flows through a pipe. It also plays a role in the cavitation of a liquid, and in the study of drag and lift effects produced by a fluid.

*Fluid particles have acceleration because they fluctuate and thereby change the magnitude and direction of their velocity.

8.2

important Dimensionless numbers

439

Reynolds Number. The ratio of the inertia force rV 2L2 to the viscous force is called the Reynolds number. For a Newtonian fluid, the viscous force is determined from Newton’s law of viscosity, Fv = tA = m(dV>dy)A. Retaining the fluid property m, the viscous force has length dimensions of m(V>L)L2, or mVL, so that the Reynolds number becomes

Re =

rVL inertia force = m viscous force

(8–3)

This dimensionless ratio was first developed by the British engineer Osborne Reynolds while investigating the behavior of laminar and turbulent flow in pipes. The length L in Eq. 8–3 is a dimension that relates to the boundary of the flow. For pipes this “characteristic length” is chosen as the diameter of the pipe. When Re is large, the inertia forces within the flow will be greater than the viscous forces, and when this occurs, the flow will become turbulent. Using this idea, in Sec. 9.5 we will show that Reynolds was able to roughly predict when laminar flow begins to transition to turbulent flow. Apart from flow in closed conduits, such as pipes, viscous forces also affect the flow of air around slowmoving aircraft, and the flow of water around ships and submarines. For this reason, the Reynolds number has important implications in a number of flow phenomena. weight rVg = (rg)L3 becomes V 2 >gL. If we take the square root of this dimensionless expression, we have another dimensionless ratio called the Froude number. It is

Froude Number. The ratio of the inertia force rV 2L2 to the fluid’s

Fr =

inertia force V = A gravitational force 2gL

(8–4)

This number is named in honor of William Froude, a naval architect who studied surface waves produced by the motion of a ship. Its value indicates the relative importance of inertia and gravity effects on fluid flow. For example, if the Froude number is greater than unity, then inertia effects will outweigh those of gravity. This number is important in the study of any flow having a free surface, as in the case of open channels, or the flow over dams or spillways.

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Weber Number. The ratio of the inertia force rV 2L2 to the surface tension force sL is called the Weber number, named after Moritz Weber, who studied the effects of flow in capillary tubes. It is

8

We =

rV 2L inertia force = s surface tension force

(8–5)

It is important because experiments have shown that capillary flow is controlled by the force of surface tension within a narrow passageway, provided the Weber number is less than one. For most engineering applications, the surface tension force can be neglected; however, it becomes important when studying the flow of thin films of liquid over surfaces, or the flow of small-diameter jets or sprays.

Mach Number. The square root of the ratio of the inertia force

rV 2L2 to the force that causes the compressibility of the fluid is called the Mach number. This ratio was developed by Ernst Mach, an Austrian physicist who used it as a reference to study the effects of compressible flow. Recall that a fluid’s compressibility is measured by its bulk modulus, EV , which is its change in pressure divided by the volumetric strain, Eq. 1–11, EV = ∆p>(∆V>V ). Since the units of EV are the same as those of pressure, and pressure produces a force F = pA, the so-called elastic force caused by compression has length dimensions of F = E V L2. The ratio of the inertia force, rV 2L2, to the compressibility force then becomes rV 2 >EV . It will be shown in Chapter 13 that for an ideal gas, EV = r c 2, where c is the speed at which a pressure disturbance (sound) will naturally travel within a fluid medium. Substituting this into the ratio, and taking the square root, gives the Mach number.

M =

inertia force V = c A compressibility force

(8–6)

Notice that if M is greater than one, as in the case of supersonic flow, then inertial effects will dominate, and the fluid velocity will be greater than the velocity c of the propagation of the pressure disturbance.

8.3

8.3

the buCkingham pi theorem

441

THE BUCKINGHAM Pi THEOREM

In Sec. 8.1 we discussed a way to perform a dimensional analysis by attempting to combine the variables of a fluid flow as terms that all have a consistent set of units. Since this method is cumbersome when dealing with problems that have a large number of variables, in 1914 the experimentalist Edgar Buckingham popularized a more direct method that has since become known as the Buckingham Pi theorem. In this section we will formalize application of this theorem, and show how it applies in cases where four or more physical variables control the flow. The Buckingham Pi theorem states that if a flow phenomenon depends upon n physical variables, such as velocity, pressure, and viscosity, and if present within these physical variables there are m primary dimensions, such as M, L, and T, then through dimensional analysis, the n variables can be arranged into (n - m) independent dimensionless numbers or groupings. Each of these groupings is called a Π (Pi) term, because in mathematics this symbol is used to symbolize a product. The five dimensionless groupings, or “numbers,” discussed in the previous section are typical Π terms. Once the functional relationship among the Π terms is established, it can then be investigated experimentally to see how it relates to the flow behavior using models. The groupings having the most influence are retained, and those having only a slight effect on the flow are rejected. Ultimately, this process will lead to an empirical equation, where any unknown coefficients and exponents are then determined by further experiment. The proof of the Buckingham Pi theorem is rather long and can be found in books related to dimensional analysis. For example, see Ref. [1]. Here we will be interested only in its application.

The drag on this car is influenced by the density, viscosity, and velocity of the air, as well as the projected area of the car into the flow. (© Takeshi Takahara/Science Source)

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PROCEDURE FOR ANALYSIS

8

The Buckingham Pi theorem is used to find the dimensionless groupings among the variables that describe a particular flow phenomenon, and thereby establish a functional relationship between them. The following procedure outlines the steps needed to apply it. • Define the Physical Variables. Specify the n variables that affect the flow phenomena. Do not include variables that are defined in terms of others through multiplication or division, such as Q = VA. See if it is possible to form any Π terms by inspection. If this cannot be done, then determine the number m of primary dimensions M, L, T or F, L, T that are involved within the collection of all n variables.* This will result in (n - m) Π terms that can be formulated to describe the phenomena. For example, if pressure, velocity, density, and length are the suspected variables, then n = 4. From Table 8–1, the dimensions of these variables are ML -1T -2, LT -1, ML -3, and L, respectively. Since M, L, and T are represented in this collection, m = 3 and so this will result in (4 - 3) = 1 Π term. • Select the Repeating Variables. From the list of n variables, select m of them so that these m variables collectively contain all the m dimensions. As a general rule, select the ones with the simplest combination of dimensions.† All these m variables are called repeating variables. In the above example we can select length, velocity, and density (m = 3) as the repeating variables, since their collection of dimensions involves M, L, T (m = 3). • Π Terms. From the remaining list of (n - m) variables, select any one of them, called q, and multiply it by the m repeating variables. Raise each of the m variables to an unknown exponent, but keep the q variable to a known power, where for simplicity we will always use the first power. This represents the first Π term. Continue the process of selecting any other q variable from the (n - m) variables, and again write the product of q and the same m repeating variables, each raised to an unknown exponential power, producing the second Π term, etc. This is done until all (n - m) Π terms are written. In our example, pressure p would be selected as q, so that the single Π term is then Π = La V brcp. • Dimensional Analysis. Express each of the (n - m) Π terms in terms of its base dimensions (M, L, T or F, L, T) using Table  8–1, and solve for the unknown exponents by requiring the sum of the exponents for each dimension in the Π term to be equal to zero. By doing this, the Π term will then be dimensionless. Once determined, the Π terms are then collected into a functional expression f (Π1, Π2, c) = 0, or in the form of an explicit equation, and the numerical values of any remaining unknown constant coefficients or exponents are then determined from experiment. The following examples should help clarify these four steps and thereby illustrate application of this procedure. *One can use either the M-L-T or the F-L-T system to do a dimensional analysis; however, it may happen that the variables selected have a different number of primary dimensions m in each of these systems. A dimensional matrix approach can be used to identify this situation, but since it does not occur very often, we will not consider its use in this book. See Ref. [5]. †Realize that for any group of variables the set of Π terms is not unique, since it depends upon the choice of repeating variables, as well as the

power of q. However, regardless of these choices, or even the exponential power of any Π term, all the Π terms are valid. See Example 8–2.

8.3

EXAMPLE

the buCkingham pi theorem

443

8.1

Establish the Reynolds number for a fluid flowing through the pipe in Fig. 8–2 using dimensional analysis, realizing that the flow is a function of the density r and viscosity m of the fluid, along with its velocity V and the pipe’s diameter D.

8

SOLUTION Define the Physical Variables. system in Table 8–1, we have

Here n = 4. Using the M-L-T

Density, r ML -3 Viscosity, m ML -1T -1 V Velocity, V LT -1 Diameter, D L Since all three primary dimensions (M, L, T) are used here, then m = 3. Thus, there is (n - m) = (4 - 3) = 1 Π term. Select the Repeating Variables. We will choose r, m, V as the m = 3 repeating variables, because the collection of their dimensions contains the m = 3 primary dimensions. (Of course, another selection, such as m, V, D, can also be made.) 𝚷 Term, q = D. Since D has not been selected, it becomes the q variable, which will be set to the first power (exponent one). And so the Π term is Π = rambV cD. Dimensional Analysis. The dimensions for this Π term are Π = rambV cD = 1M aL -3a 21M bL -bT -b 21L cT -c 2L = M a + bL -3a - b + c + 1T -b - c Since Π is to be dimensionless, the exponents on each dimension must be set equal to zero, i.e., For M: 0 = a + b For L: 0 = -3a - b + c + 1 For T: 0 = -b - c Solving yields a = 1, b = -1, c = 1 And so, rVD ΠRe = r1m-1V 1D = Ans. m This result is the Reynolds number, Eq. 8–3, where the “characteristic length” L is the pipe diameter D. Later in the book we will give a more thorough discussion on using the Reynolds number, and will show how viscous and inertia forces play a predominant role in defining whether the flow through the pipe is laminar or transitions to turbulent.

D

Fig. 8–2

444

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8.2 The wing on the aircraft in Fig. 8–3 is subjected to a drag FD created by airflow over its surface. It is anticipated that this force is a function of the density r and viscosity m of the air, the “characteristic” length L of the wing, and the velocity V of the approaching flow. Show how the drag force depends on these variables.

8

Fig. 8–3

SOLUTION Define the Physical Variables. Symbolically, the unknown function is FD = f ( r, m, L, V ). In order to collect all the variables as an implicit function, we can rewrite this as follows:* h (FD, r, m, L, V ) = 0. So here n = 5. We will solve the problem using the F-L-T system in Table 8–1. The primary dimensions of these five variables are Drag, FD Density, r Viscosity, m Length, L Velocity, V

F FT 2L -4 FTL -2 L LT -1

Since all three primary dimensions (F, L, T) are involved in the collection of these variables, m = 3, and so we have (n - m) = (5 - 3) = 2 Π terms. Select the Repeating Variables. The density, length, and velocity will be chosen as the m = 3 repeating variables. As required, the collection of their dimensions contains the m = 3 primary dimensions. The remaining variables FD and m must be selected for q. 𝚷1 Term q = FD . We will consider FD as the q variable for the first Π term, Π1 = raLbV cFD. Dimensional Analysis. This term has dimensions Π1 = raLbV cFD = 1F aT 2aL -4a 21Lb 21LcT -c 2F = F a + 1L -4a + b + cT 2a - c

Therefore, requiring the exponents to be zero, For F: For L: For T:

0 = a + 1 0 = -4a + b + c 0 = 2a - c

Solving yields a = -1, b = -2, c = -2, so Π1 = r -1L -2 V -2FD =

FD rL2V 2

*This is like writing y = 5x + 6 as y - 5x - 6 = 0, where y = f(x) and h(x, y) = 0.

8.3

the buCkingham pi theorem

445

𝚷2 Term q = M. Finally, we will consider m as the q variable, thus creating the second Π term, Π2 = rdLeV hm. Dimensional Analysis.

This term has dimensions

Π2 = rdLeV hm = 1F dT 2dL -4d 21Le 21L hT -h 2FTL -2 = F d + 1L -4d + e + h - 2T 2d - h + 1

Therefore, For F: 0 = d + 1 For L: 0 = -4d + e + h - 2 For T: 0 = 2d - h + 1 Solving yields d = -1, e = -1, h = -1, so Π2 = r-1L -1V -1m =

m rVL

We can also replace Π2 by Π2-1 since it is also a dimensionless ratio— the Reynolds number, Re.* The unknown function f1 in terms of the variables now takes the form of f1 °

FD rL2V 2

, Re¢ = 0

Ans.

If we solve for FD >rL2V 2 in this equation, then we can specify how FD is related to the Reynolds number. It is FD = f2(Re) (1) rL2V 2 or FD = rL2V 2 3 f2(Re) 4 (2) Ans.

Later, in Chapter 11, we will show that for experimental purposes, rather than obtain f2(Re), it is convenient to express the drag in terms of the fluid’s dynamic pressure, rV 2 >2, and use an experimentally determined dimensionless drag coefficient, CD. If we do this, then we require FD = rL2V 2 3 f2(Re)4 = CD L2 1rV 2 >22, so that our unknown function becomes f2(Re) = CD >2. Also, replacing L2 in Eq. 1 by the area A of the wing, since it has the same dimensions, we can write Eq. 1 as rV 2 FD = CD A a b (3) Ans. 2 Dimensional analysis has not provided the complete solution to this problem, but as will be shown in Chapter 11, once an experiment is performed to determine CD, then we can use Eq. 3 to obtain FD. *Note that if we choose m to represent, for example, r, L, FD, and use V and m for q, then we will get Π1 ′ = r1>2LV>F D1>2 and Π2 ′ = m>r1>2F D1>2. Realize that these results are still valid since Π1 = (Π ′) - 2 and Π2 = Π2 ′ >Π1 ′.

8

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8.3 The ship in Fig. 8–4 is subjected to a drag FD on its hull, created by water passing under and around its surface. It is anticipated that this force is a function of the density r and viscosity m of the water, and since waves are produced, their weight, defined by gravity g, is important. Also, the “characteristic length” of the ship, L, and the velocity of the flow, V, influence the magnitude of the drag. Show how the drag depends on all these variables.

8

SOLUTION Fig. 8–4

Define the Physical Variables. Here the unknown function is FD = f (r, m, L, V, g), which is then expressed as h(FD, r, m, L, V, g) = 0, where here n = 6. We will solve the problem using the F-L-T system in Table 8–1. The primary dimensions of the variables are Drag, FD Density, r Viscosity, m Length, L Velocity, V Gravity, g

F FT 2L -4 FTL -2 L LT -1 LT -2

Since the three primary dimensions (F, L, T) are involved in the collection of all these variables, m = 3, and so there will be (n - m) = (6 - 3) = 3 Π terms. Select the Repeating Variables. Density, length, and velocity will be chosen as the repeating variables, since the collection of their dimensions contains the m = 3 primary dimensions. 𝚷1 Term q = FD and Dimensional Analysis. We will consider FD as the q variable for the first Π term. Π1 = raLbV cFD

= 1F aT 2aL -4a 21Lb 21LcT -c 2F = F a + 1L -4a + b + cT 2a - c

Therefore, For F: For L: For T:

0 = a + 1 0 = -4a + b + c 0 = 2a - c

Solving yields a = -1, b = -2, c = -2, so Π1 = r-1L -2 V -2FD =

FD rL2V 2

8.3

the buCkingham pi theorem

447

𝚷2 Term q = M and Dimensional Analysis. Here we will consider m as the q variable, thus creating the second Π term. Π2 = rdLeV hm

= 1F dT 2dL -4d 21Le 21L hT -h 2FTL -2 = F d + 1L -4d + e + h - 2T 2d - h + 1

For F: For L: For T:

0 = d + 1 0 = -4d + e + h - 2 0 = 2d - h + 1

Solving yields d = -1, e = -1, h = -1, so Π2 = r-1L -1V -1m =

m rVL

As in the previous example, we can replace Π2 by Π2-1 since it represents the Reynolds number, Re. 𝚷3 Term q = g and Dimensional Analysis. g as the q variable for the third Π term.

Finally, we consider

Π3 = riLjV kg

= 1F iT 2iL -4i 21Lj 21LkT -k 21LT -2 2 = F iL -4i + j + k + 1T 2i - k - 2

For F: For L: For T:

0 = i 0 = -4i + j + k + 1 0 = 2i - k - 2

Solving yields i = 0, j = 1, k = -2, so Π3 = r0L1V -2g = gL>V 2 Recognizing this term to be equal to the inverse square of the Froude number, for simplicity we will consider using Fr instead of Π3 because both are dimensionless. Thus, the unknown function f1 between the Π terms takes the form f1 °

FD rL2V 2

, Re, Fr ¢ = 0

Ans.

If we solve for FD >rL2V 2 in this equation, then it can be written symbolically as a function f2 of the Reynolds and Froude numbers. FD rL2V 2

= f2 3Re, Fr4

FD = rL2V 2 f2 3Re, Fr4

(1) Ans.

8

448

Chapter 8

EXAMPLE

8

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8.4

D L

Fig. 8–5

A pressure drop ∆p provides a measure of the frictional losses of a fluid as it flows through a pipe, Fig. 8–5. Determine how ∆p is related to the variables that influence it, namely, the pipe diameter D, its length L, the fluid density r, viscosity m, velocity V, and the relative roughness factor e>D, which is a ratio of the average size of the surface irregularities to the pipe’s diameter D. SOLUTION Define the Physical Variables. In this case, ∆p = f1(D, L, r, m, V, e>D) or f2(∆p, D, L, r, m, V, e>D) = 0, and so n = 7. Using the M-L-T system in Table 8–1, the primary dimensions of the variables are Pressure drop, ∆p ML -1T -2 Diameter, D L Length, L L Density, r ML -3 Viscosity, m ML -1T -1 Velocity, V LT -1 Relative roughness, e>D LL -1 Since all three primary dimensions are involved, m = 3, and there will be (n - m) = (7 - 3) = 4 Π terms. Select the Repeating Variables. Here D, V, r will be selected as the m = 3 repeating variables. Notice that e>D cannot be selected because it is already dimensionless. Also, we cannot select the length in place of velocity since both the diameter and the length have the same dimensions and time T would not be represented. 𝚷 Terms and Dimensional Analysis. The Π terms are constructed using our repeating variables D, V, r, along with q = ∆p for Π1, q = L for Π2, q = m for Π3, and q = e>D for Π4. Thus, for Π1, Π1 = D aV brc ∆p = 1La 21LbT -b 21M cL -3c 21ML -1T -2 2 = M c + 1La + b - 3c - 1T -b - 2

For M: 0 = c + 1 For L: 0 = a + b - 3c - 1 For T: 0 = -b - 2 Solving yields a = 0, b = -2, c = -1, and so ∆p Π1 = D0V -2r-1 ∆p = rV 2 This is the Euler number.

Next, for Π2, Π2 = DdV erhL = 1L d 21LeT -e 21M hL -3h 21L2 = M hL d + e - 3h + 1T -e

8.3

For M: For L: For T:

449

0 = h 0 = d + e - 3h + 1 0 = -e

Solving yields d = -1, e = 0, h = 0, and so L Π2 = D V r L = D -1

0 0

Now, for Π3, Π3 = DiV jrkm = 1Li 21Lj T -j 21M kL -3k 21ML -1T -1 2 = M k + 1Li + j - 3k - 1T -j - 1

For M: For L: For T:

the buCkingham pi theorem

0 = k + 1 0 = i + j - 3k - 1 0 = -j - 1

Solving yields i = -1, j = -1, k = -1, so m DVr This is the inverse of the Reynolds number for the pipe, and so we will consider rVD Π3-1 = = Re m Finally, for Π4, Π4 = DlV mrn(e>D) = 1Ll 21LmT -m 21M nL - 3n 21LL -1 2 = M nLl + m - 3n + 1 - 1T -m Π3 = D -1V -1r-1m =

For M: 0 For L: 0 For T: 0 Solving yields l

n l + m - 3n + 1 - 1 -m 0, m = 0, n = 0, and so e e Π4 = D0V 0r0 a b = D D Realize that at the start, we could have saved some time in solving this problem and determined Π2 = L>D and Π4 = e>D simply by inspection, because each is a dimensionless ratio representing length over length. Had this inspection been done at the start, then only two Π terms would have to be determined. In either case, our results indicate = = = =

f2 °

∆p rV

2

, Re,

L e , ¢ = 0 D D

This equation can be solved for the ratio ∆p>rV 2 and then written in the form L e ∆p = rV 2 f3 a Re, , b Ans. D D In Chapter 10, we will show how this relationship has important applications in the design of pipe systems.

8

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8.4

8

P2

P2 5 f(P1)

P1 (a) P2 P3 5 C3 P3 5 C2 P3 5 C1

P1 (b)

Fig. 8–6

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SOME GENERAL CONSIDERATIONS RELATED TO DIMENSIONAL ANALYSIS

The previous four examples illustrate the relatively straightforward method for applying the Buckingham Pi theorem to determine a functional relationship between a dependent variable and a series of dimensionless groupings or Π terms that compose the independent variables. The most important part of the process, however, is to clearly define the variables that influence the flow. This can be done only if one has enough experience in fluid mechanics to understand what laws and forces govern the flow. For the selection, these variables include fluid properties such as density and viscosity, the dimensions used to describe the system, and variables such as gravity, pressure, and velocity that are involved in creating the forces within the flow. If the selection does not include an important variable, then a dimensional analysis will produce an incorrect result. This will be noticed when the dependent variable is plotted versus the independent variables (Π terms) because it will show data points that will be scattered. When this occurs, further investigation must be made to find the missing variable, and then reapply dimensional analysis. Also, if irrelevant variables are selected, or the variables are related to one another, such as Q = VA, then it will result in too many Π terms, and a plot of the dependent variable versus any additional Π term will indicate it to be a constant, and thereby to be excluded. In summary, if the important variables that influence the flow are properly selected, then one will be able to minimize the resulting number of Π terms that are involved, and thereby reduce not only the time, but also the cost of any experiment needed to confirm the dimensional analysis and obtain the final result. For example, the experimental process is simple if only one Π term is involved, since then this Π term will be equal# to a constant, C. An example of this is given by Eq. 8–1, where we found W >Q∆p = C as the Π term. # Once C has been experimentally determined for a specific set of values of W, Q, and ∆p, then it will not be necessary to do further experiments for other sets of values, because this same value of C will result. If two Π terms are needed to describe a phenomenon, then a functional relationship between these Π terms must be established. This situation occurred as Eq. 1 in Example 8–2. Here an experimental graph will result between Π1 and Π2, Fig. 8–6a, and by curve fitting the data the relationship Π2 = f(Π1) can be established. Once it is found, it then becomes valid for any set of data that describes the flow phenomena. This process can also be extended to situations involving three Π terms, as in Eq. 1 of Example 8–3. However, here experimental complications can arise because a family of curves will have to be established to describe the relationship between Π1, Π2, Π3, Fig. 8–6b. To circumvent this situation, engineers will then often build models in order to experimentally study the behavior of the flow. We will discuss how this is done in the next section.

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451

IMPORTANT POIN T S • The equations of fluid mechanics are dimensionally homogeneous, which means that each term of an equation must have the same combination of units.

• Dimensional analysis is used to reduce the amount of data that must be collected from the variables in an experiment in order to understand the behavior of a flow. This is done by arranging the variables into selected groups that are dimensionless. Once this is done, then it is only necessary to find one relationship among the dimensionless groups, and not to find several relationships among all the separate variables.

• Five important dimensionless ratios of force occur frequently in fluid mechanics. All involve a ratio of the dynamic or inertia force to some other force. These five “numbers” are the Euler number for pressure, the Reynolds number for viscosity, the Froude number for weight, the Weber number for surface tension, and the Mach number for the elastic force causing compression.

• The Buckingham Pi theorem provides a systematic method for performing a dimensional analysis. The theorem indicates in advance how many unique dimensionless groups of variables (Π terms) to expect, and it provides a way to formulate a relationship among them.

8.5

SIMILITUDE

Models are often used to study the three-dimensional flow around an object having a complicated geometry, such as a building, automobile, or airplane. This is done because it may be rather difficult to describe the flow using an analytical or computational solution. Even if the flow can be described by a computational analysis, in complicated cases it should be backed up with a corresponding experimental investigation, using a model to verify the results. This is necessary simply because the assumptions made using any computational study may not truly reflect the actual situation involving the complexities of the flow. If the model and its testing environment are properly proportioned, the experiment will enable the engineer to predict how the flow will affect the prototype or the actual object for which the model is made. For example, using a model, it is possible to obtain measurements of velocity, depth of liquid flow, pump or turbine efficiencies, and so forth, and with this information, the model can be altered if necessary so that the design of the prototype can be improved. Usually the model is made smaller than the prototype; however, this may not always be the case. For example, larger models have been constructed to study the flow of gasoline through an injector, or the flow of air through the blades of a turbine used for a dental drill. Whatever its size, it is very important that the model used for an experimental study correspond to the behavior of the prototype when the prototype is subjected to the actual fluid flow. Similitude is a mathematical process of ensuring that this is the case. It requires the model and the flow around it not only to maintain geometric similarity to the prototype, but also to maintain kinematic and dynamic similarity.

Due to the complexity of flow, the effect of wind on high-rise buildings is often studied using a scaled-down model in wind tunnels.

8

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Geometric Similitude. If the model and the fluid flow are geometrically similar to the prototype, then all of the model’s linear dimensions must be in the same proportion as those of the prototype, and all its angles must be the same. Take, for example, the prototype (jet plane) shown in Fig. 8–7a. We can express this linear proportion as a scale ratio, which is the ratio of the length Lm of the model to the length Lp of the prototype.

8

Vp

Lm Lp

Fp Lp Prototype Vm Fm Lm Model (a)

Fig. 8–7

If this ratio is maintained for all dimensions, then it follows that the areas of the model and the prototype will be in the proportion L2m >L2p, and their volumes will be in the proportion L3m >L3p. The extent to which geometric similitude is achieved depends on the type of problem and on the accuracy required from its solution. For example, exact geometric similitude also requires the surface roughness of the model to be in proportion to that of the prototype. In some cases, however, this may not be possible, since the reduced size of a model might require its surface to be impossibly smooth. Also, in some modeling, the vertical scale of the model may have to be exaggerated to produce the proper flow characteristics. This would be true in the case of river studies, where the river bottom may be difficult to scale.

Kinematic Similitude. The primary dimensions defining kinematic similitude are length and time. For example, the velocity of the fluid at corresponding points between the jet model and its prototype in Fig. 8–7a must have a proportional magnitude and be in the same direction. Since velocity depends on length and time, V = L>T, then

LmTp Vm = Vp LpTm If this requirement holds true, then the length ratio Lm >Lp for geometric similitude must be satisfied, and so must the time ratio Tp >Tm. Satisfying these ratios, the acceleration will also be proportional since it involves T p2 >T m2. A typical example of having kinematic similitude would be a model of the solar system showing the relative positions or “lengths” for the planets and having the proper time scale for their orbits.

8.5

Dynamic Similitude. To maintain a similar pattern of streamlines around both the prototype and its model, Fig. 8–7b, it is necessary that the forces acting on corresponding fluid particles in both cases be proportional. As stated previously, the inertia force Fi is generally considered the most important force that influences fluid flow around an object. For this reason, it is standard convention to use this force, along with each of the other forces F that influence the flow, to produce the ratios for dynamic similitude between model and prototype. We can express each force ratio symbolically as Fp Fm = 1Fi 2 m 1Fi 2 p

The many types of forces to be considered include those due to pressure, viscosity, gravity, surface tension, and elasticity. This means that for complete dynamic similitude, the Euler, Reynolds, Froude, Weber, and Mach numbers must be the same for both the model and the prototype. Actually, for complete dynamic similitude, it is not necessary to satisfy the proportional condition for all the forces that affect the flow. Instead, if similitude for all but one force is satisfied, then the condition for this remaining force will automatically be satisfied. To show why this is so, consider a case where only the forces of pressure (pr), viscosity ( ), and gravity (g) act on two fluid particles having the same mass m and located at the same relative position on the prototype and its model in Fig. 8–7b. According to Newton’s second law, at any instant the sum of these forces on each particle must be equal to the mass of the particle times its acceleration, ma.* If we express Newton’s law as ΣF - ma = 0, and consider the inertia force Fi = -ma acting on each particle, Fig. 8–7b, then for each case, the vector addition of these four forces is graphically shown by their vector polygons. For dynamic similitude the four corresponding forces of these polygons must be proportional, that is, their magnitudes must have the same scaled lengths. However, because the polygons are closed shapes, only three of the sides (forces) must satisfy this proportionality, and if this occurs, the fourth side (force) will then automatically be proportional. For example, since the inertia force is used to formulate the Euler numbers, Reynolds numbers, and Froude numbers for the prototype and its model, then if the flow satisfies, say, Reynolds and Froude number scaling, then it will automatically satisfy Euler number scaling. Also, since the inertia forces (Fi)p and (Fi)m have dimensions of ML>T 2, then dynamic similitude automatically satisfies both geometric and kinematic similitude, since L and T must be proportional when the inertia forces Fi are proportional. Actually, it is rather difficult to obtain exact similitude between a prototype and its model, due to the inaccuracies in model construction and in testing procedures. Instead, by using good judgment gained through experience, we can consider only the similitude of the dominating forces and parameters, while those of lesser importance can be safely ignored. The following three cases will illustrate how this can be achieved when performing experiments. *The particles have acceleration because the magnitude of their velocity fluctuates, and the curved streamlines along which they travel can change the direction of their velocity.

453

similituDe

8

(Fy)p (Fpr)p

(Fg)p

(Fpr)p (Fi)p

(Fi)p

(Fy)p (Fg)p Vp Fp

Lp Prototype (Fy)m (Fpr)m

(Fpr)m

(Fg)m (Fi)m

(Fi)m

(Fy)m (Fg)m Vm Fm

Lm Model (b)

Fig. 8–7 (cont.)

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Steady Flow through a Pipe. If the flow of water through a

8

pipe is to be studied using a model, Fig. 8–8, then experience has shown that the inertia and viscous forces are important.* Therefore, for pipe flow, the Reynolds number must be the same for both the model and the prototype to achieve dynamic similitude. For this ratio, the pipe diameter D becomes the “characteristic length” L, and so

a

rVD rVD b = a b m m m p

Often the same fluid is used for both the model and the prototype, so that properties r and m will be the same. If this is the case, then

Vm Dm = Vp Dp

Therefore, for given values of Vp, Dp, and Vm, the model of the pipe must have a diameter of Dm = (VD)p >Vm in order for the flow to have the same behavior through both the actual pipe and its model.

Dp

Vp

Prototype

Vm

Dm

Model

Fig. 8–8 *The pressure force is also important, but provided the viscous-to-inertia force ratios are equal, the proportionality of pressure-to-inertia force ratios will automatically be satisfied, as we have just discussed.

8.5

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455

Open-Channel Flow. For open-channel flow, Fig. 8–9, forces caused by surface tension and compressibility can be neglected without introducing appreciable errors, and if the channel length is short, then friction losses will be negligible, and therefore viscous effects can also be neglected. As a result, the flow through an open channel is primarily governed by the inertia force and gravity. Therefore, the Froude number Fr = V> 1gL can be used to establish similitude between the prototype and its model. For this number, the depth of the fluid, h, in the channel is selected as the “characteristic length” L, and since g is the same for both cases, we have

2hm Vm

Vp

=

2hp

Besides modeling flow in channels, the Froude number is also used to model the flow over a number of hydraulic structures, such as gates and spillways. However, for rivers this type of scaling can cause problems, because scaling down the model may result in depths that are too small, thereby causing the effects of viscosity and surface tension to predominate within the flow. As previously stated, however, such forces are generally neglected, and so only approximate modeling of the flow can be achieved.

hp Prototype

hm Model

Fig. 8–9

8

Wind turbines are sometimes placed off-shore, and the wave loadings on their columns can be substantial. Using similitude, the effect is studied in a wave channel so that these supports can be properly designed.

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Ships. In the previous two cases, only one equality involving a dimensionless ratio had to be satisfied to achieve similitude. However, in the case of a ship, Fig. 8–10, the drag or resistance to forward motion is due to both friction along and under the hull (viscosity) and lifting water against the hull to produce waves (gravity). As a result, the total drag on the ship is a function of both the Reynolds and Froude numbers, as shown in Example 8.3. Similitude between the model and prototype (ship) therefore requires these numbers to be equivalent. Thus, a °

rVL rVL b = a b m m m p

2gL V

¢

m

= °

2gL V

¢

(8–7) p

Since the kinematic viscosity is n = m>r, and g is the same for both the model and the prototype, these two equations become Vp n p Lm = Vm n m Lp Vp Vm

= °

Lp Lm

¢

1>2

Equating these ratios to eliminate the velocity ratio yields np nm

= °

Lp Lm

¢

3>2

(8–8)

Since here any model will be much smaller than the ship, then the ratio Lp >Lm will be very large. As a result, to preserve the equality in Eq. 8–8, it will become necessary to test the model in a liquid having a much smaller kinematic viscosity than water, which is impractical to achieve. To get around this difficulty, a method suggested by Froude has been used to solve this problem. In Example 8–3 we showed that a dimensional analysis of all the above variables produces a functional relationship between the total drag FD and the Reynolds and Froude numbers that has the form FD = rL2 V 2f 3Re, Fr4

Vp

Vm

Lp

Lm

Prototype

Model

Fig. 8–10

8.5

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457

Realizing this, Froude assumed that the total drag on the ship is the sum of its two components, namely, skin or viscous friction drag, based only on the Reynolds number, and the drag resistance due to wave creation, based only on the Froude number. Thus, the functional dependence for both model and prototype becomes a sum of two separate unknown functions. 8

FD = rL2V 2f1 (Re) + rL2V 2f2(Fr)

(8–9)

The model is then built on the basis of Froude scaling, since the effect of wave action (gravity) is the most difficult to predict. Thus, in accordance with Eq. 8–4, the length of the model and its velocity are chosen so that they will produce the same Froude number as the ship. The total drag FD on the model can then be measured by finding the force needed to pull the model through the water at the required velocity. This, of course, represents the action of both the wave and viscous forces on the model. The viscous forces alone can be measured by doing a separate test on a fully submerged thin plate that is moving at the same velocity as the model, is made of the same material and roughness as the model, and has the same length and surface area in contact with the water. The wave force on the model is then obtained by subtracting this viscous drag from the total drag previously measured. Once the viscous and wave forces on the model are known, the total drag on the ship can then be determined. To do this, from Eq. 8–9, the gravity forces on the ship and model must satisfy

1Fp2 g 1Fm2 g

=

rp Lp2 V p2 f2 1Fr2

rm Lm2 V m2 f2 1Fr2

However, since the unknown function f2(Fr) is to be the same for both the model and the prototype, it will cancel out, and so the wave or gravity force on the ship is determined from

1Fp2 g

= 1Fm2 g °

rpLp2V p2 rmL2mV m2

¢

(gravity)

Finally, the viscous force on the ship, defined as rL2V 2 f1(Re) in Eq. 8–9, is then determined by a similar scaling using a drag coefficient that will be discussed in Chapter 11. With both of these forces calculated, their sum then represents the total drag on the ship.

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Review. As noted, it is important to have a well-established background in fluid mechanics in order to recognize the primary forces that govern a particular flow, not only to perform a dimensional analysis, but also to do experimental work. Here are some examples that show which forces are important in certain cases, and the corresponding similitude that must be achieved. • • •

•

Inertia, pressure, and viscous forces predominate within the flow around cars, slow-flying airplanes, and the flow through pipes and ducts. Reynolds number similitude. Inertia, pressure, and gravity forces predominate within the flow along open channels, over dams and spillways, or the wave action on structures. Froude number similitude. Inertia, pressure, and surface tension forces predominate within the flow of liquid films, bubble formation, and the passage of liquids through capillarity or small-diameter tubes. Weber number similitude. Inertia, pressure, and compressibility forces predominate within the flow around high-speed aircraft, or the high-speed flow of gas through jet or rocket nozzles, and through pipes. Mach number similitude.

In all these examples, remember that similitude of the pressure force (Euler number) does not have to be taken into account, because it will automatically be satisfied due to Newton’s second law.

IMPORTANT POINTS • When building and testing a model, it is important that similitude or similarity between the model and its prototype be achieved. Complete similitude occurs when the model and prototype are geometrically similar, and the flow is kinematically and dynamically similar.

• Geometric similitude occurs when the linear dimensions for the model and prototype are in the same proportion to one another, and all the angles are the same. Kinematic similitude occurs when the velocities and accelerations are proportional. Finally, dynamic similitude occurs when the forces acting on corresponding fluid particles within the flow around the model and the prototype are in specific dimensionless ratios, defined by the Euler, Reynolds, Froude, Weber, and Mach numbers.

• Often it is difficult to achieve complete similitude. Instead, engineers will consider only the dominating variables in the flow in order to save cost and time, and yet get reasonable results.

• To satisfy dynamic similitude, it is necessary that all but one of the force ratios acting on the fluid particles for both the model and the prototype be equal. Because Newton’s second law must be satisfied, the remaining force ratio (generally taken as the Euler number) will automatically be equal.

• It takes good judgment and experience to decide which forces are significant in defining the flow, so that reasonable results can be obtained when testing a model used to predict the performance of its prototype.

8.5

EXAMPLE

similituDe

459

8.5

Flow through the pipe coupling (union) in Fig. 8–11 is to be studied using a scale model. The actual pipe is 100 mm in diameter, and the model will use a pipe that is 20 mm in diameter. The model will be made of the same material and transport the same fluid as the prototype. If the velocity of flow through the prototype is estimated to be 1.5 m>s, determine the required velocity through the model.

1.5 ms

Fig. 8–11

SOLUTION A relationship between the velocities and the diameters can be obtained by realizing that the forces dominating the flow are caused by inertia and viscosity, so Reynolds number similitude must be satisfied. We require a

rVD rVD b = a b m m m p

(1)

Since the same fluid is used for both cases, VmDm = VpDp Vm =

Vp Dp Dm

(1.5 m>s)(100 mm) =

20 mm

= 7.50 m>s

Ans.

As noted, Reynolds number similitude leads to high flow velocities for the model due to this scaling factor. To lower this velocity, from Eq. 1, one can use a higher-density or lower-viscosity fluid for the model.

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8.6 A model of the car in Fig. 8–12 is to be constructed to a scale of 1>4 and is to be tested at 20°C in a water tunnel. Determine the required velocity of the water if the actual car is traveling at 30 m>s in air at this same temperature.

Fig. 8–12

SOLUTION Here viscosity creates the predominating force, and so dynamic similitude must satisfy the Reynolds number equality. Using n = m>r, for the Reynolds number, the relationship between the velocities is therefore a

VL VL b = a b n m n p

Vm = Vp °

Lp nm ¢° ¢ np Lm

The values of kinematic viscosity of air and water at 20°C are found in Appendix A. We have

Vm = (30 m>s) £ = 7.95 m>s

1.00 1 10-6 2 m2 >s 15.1 1 10

-6

2

4 §a b m >s 1 2

Ans.

8.5

EXAMPLE

similituDe

461

8.7

The spillway in Fig. 8–13 is to be built such that the estimated average flow over its crest will be Q = 3000 m3 >s. Determine the required flow over the crest of a model built to a scale of 1>25.

Fig. 8–13

SOLUTION Here the weight of the water is the most significant force influencing the flow, and so Froude number similitude must be achieved. This equality will provide the necessary relationship between the velocities and the size of the spillway. Thus, °

2gL V

¢

m

= °

2gL V

¢

p

or Lm Vm = ° ¢ Vp Lp

1>2

1 = ° ¢ 25

1>2

(1)

We can express the velocity ratio in terms of the flow ratio since Q = VA, where A is the product of the depth of the water passing over the crest, Lh, and the width of the spillway, Lw. Thus, Qm Ap Qm 1Lh 2 p 1Lw 2 p Vm Qm 25 2 = = = a b Vp Qp Am Qp 1Lh 2 m 1Lw 2 m Qp 1

Substituting this into Eq. 1, we get Qm 1 5>2 = a b Qp 25 Thus, Qm = Qp a

1 5>2 b 25

= 13000 m3 >s2 a

1 5>2 b = 0.960 m3 >s 25

Ans.

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8.8 The pump in Fig. 8–14 is used to supply water through the 100-mm-diameter pipe at 3 m>s. If the measured power loss through the pipe is 3 kW, determine the power loss under similar conditions when a 75-mm-diameter pipe is used.

100 mm

8

3 m/s

SOLUTION

# By definition, power is the product of force and velocity, W = FV. Here the pump provides a pressure on the water to push it through the pipe, and since F = ∆pA, then

Fig. 8–14

# W = ∆p(pD2 >4)V or # ∆p1 D1 2 V1 W1 ba b a b # = a ∆p2 D2 V2 W2

(1)

The relationship between the pressures and velocities is obtained from Euler number similitude, and between the velocities and the dimensions from Reynolds number similitude. For the same fluid (water), and using the pipe diameter D for the “characteristic length” L, we have

a

a

rVD rVD b = a b ; m 1 m 2

b =a 2

∆p rV

1

b ; ∆p2 = a 2

∆p rV

V2 = V1 a

2

D1 100 mm b = 13 m>s2 a b = 4 m>s D2 75 mm

4 m>s 2 V2 2 b ∆p1 = a b ∆p1 = 1.7778 ∆p1 V1 3 m>s

Substituting these results into Eq. 1, we get ∆p1 3 kW 100 mm 2 3 m>s = a ba b a b # 1.7778 ∆p1 75 mm 4 m>s W2 # W2 = 4.00 kW

Ans.

problems

463

References 1. E. Buckingham, “Model experiments and the form of empirical

equations,” Trans ASME, Vol. 37, 1915, pp. 263–296. 2. S. J. Kline, Similitude and Approximation Theory, McGraw-Hill, New

York, NY, 1965. 3. P. Bridgman, Dimensional Analysis, Yale University Press, New

Haven, CT, 1922. 4. E. Buckingham, “On physically similar systems: illustrations of the

use of dimensional equations,” Physical Reviews, Vol. 4, No. 4, 1914, pp. 345–376. 5. T. Szirtes and P. Roza, Applied Dimensional Analysis and Modeling,

McGraw-Hill, New York, NY, 1997. 6. R. Ettema, Hydraulic Modeling: Concepts and Practice, ASCE,

Reston, VA, 2000.

P R OBLEMS SEC. 8.1–8.4 8–1. Determine the F-L-T dimensions of the following terms. (a) Qp>L, (b) rV 2 >m, (c) EV >p, (d) gQL.

8–2. Determine the M-L-T dimensions of the following terms. (a) Qp>L, (b) rV 2 >m, (c) EV >p, (d) gQL. 8–3. Investigate whether each ratio is dimensionless. (a) mV>L3r, (b) m>rVL, (c) 2gL>V, (d) s>rL. *8–4. Use inspection to arrange each of the following three variables as a dimensionless ratio: (a) L, t, V, (b) s, EV, L, (c) V, g, L. 8–5. Determine the Mach number for a jet flying at 1500 km>h at an altitude of 3 km. The velocity of sound in air is c = 2kRT, where the specific heat ratio for air is k = 1.40.

8–6. Express the group of variables L, μ, r, V as a dimensionless ratio. 8–7. Show that the hydrostatic pressure p of an incompressible fluid can be determined using dimensional analysis by realizing that it depends upon the depth h in the fluid and the fluid’s specific weight g. *8–8. The speed of sound V in air is thought to depend on the viscosity m, the density r, and the pressure p. Determine how V is related to these parameters. 8–9. Show that the Weber number is dimensionless using M-L-T dimensions and F-L-T dimensions. Determine its value for water at 25°C flowing at 4 m>s for a characteristic length of 1.5 m. Take sw = 0.0726 N>m.

8

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8–10. The velocity V of the stream flowing from the side of the tank is thought to depend upon the liquid’s density r, the depth h, and the acceleration of gravity g. Determine the relation between V and these parameters.

similituDe *8–12. Establish Newton’s law of viscosity using dimensional analysis, realizing that shear stress t is a function of the fluid viscosity m and the angular deformation du>dy. Hint: Consider the unknown function as f(t, m, du, dy). 8–13. The period of oscillation t, measured in seconds, of a buoy depends upon its cross-sectional area A, its mass m, and the specific weight g of the water. Determine the relation between t and these parameters.

8

h d

Probs. 8–12/13 Prob. 8–10

8–11. The discharge Q over the weir A depends upon the width b of the weir, the water head H, and the acceleration of gravity g. If Q is known to be directly proportional to b, determine the relation between Q and these variables. If H is doubled, how does this affect Q?

8–14. The discharge Q from a turbine is a function of the generated torque T, the angular rotation v of the turbine, its diameter D, and the liquid density r. Determine the relation between Q and these parameters. If Q varies linearly with T, how does it vary with the turbine’s diameter D? Take q = Q and T. 8–15. The thickness d of the boundary layer for a fluid passing over a flat plate depends upon the distance x from the plate’s leading edge, the free-stream velocity U of the flow, and the density r and viscosity m of the fluid. Determine the relation between d and these parameters. Take q = d and m.

U

H

u d A x

Prob. 8–11

Prob. 8–15

problems *8–16. Express the group of variables p, g, D, r as a dimensionless ratio. 8–17. The average velocity V of water flow through a pipe is a function of the pipe’s diameter D, the change in pressure ∆p per unit length, ∆p> ∆x, and the viscosity of the water, m. Determine the relation between V and these parameters. D V

465

8–21. The period of time t between small waves on the surface of a liquid is thought to be a function of the wave length l, the liquid depth h, gravitational acceleration g, and the surface tension s of the liquid. Show how to obtain the two Π terms h>l and t 1g>l, where q = t and h. If the liquid is water at a temperature of 15°C, an experiment with the water depth h = 150 mm produced the following results 8 for the period for various measured wavelengths. Use this data to plot the relationship between the two Π terms.

Prob. 8–17 8–18. The pressure change that occurs in the aortic artery during a short period of time can be modeled by the equation ∆p = ca ( mV>2R)1>2, where m is the viscosity of blood, V is its velocity, and R is the radius of the artery. Determine the M, L, T dimensions for the arterial coefficient ca.

t (s)

l (mm)

0.05 0.10 0.15 0.20 0.25 0.30

25 50 75 100 125 150

Prob. 8–21 8–22. The flow Q of gas through the pipe is a function of the density r of the gas, gravity g, and the diameter D of the pipe. Determine the relation between Q and these parameters. D

Prob. 8–18

V

8–19. The force of buoyancy F is a function of the volume V of a body and the specific weight g of the fluid. Determine how F is related to V and g. *8–20. A floating body will move up and down with a period of t measured in seconds. The behavior depends upon the mass m of the body, its cross-sectional area A, the acceleration due to gravity, g, and the density r of the water. Determine the relationship between t and these parameters.

Prob. 8–22 8–23. The pressure p within the soap bubble is a function of the bubble’s radius r and the surface tension s of the liquid film. Determine the relation between p and these parameters. Compare the result with that obtained in Sec. 1.10.

r

Prob. 8–20

Prob. 8–23

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*8–24. The time t needed for a liquid to drain from the pipette is thought to be a function of the fluid’s density r and viscosity m, the nozzle’s diameter d, and gravity g. Show how to obtain the two Π terms 1 1g>d2t and m> 1rd 3>2g 1>2 2 where q = t and m. If the liquid is water at a temperature of 15°C, an experiment produced the data for the time versus 8 the diameter of various tubes. Use this data to plot the relationship between the two Π terms.

d (mm)

t (s)

0.50 1.00 1.50 2.00 2.50 3.00

305 87.6 42.2 25.15 16.9 12.1

similituDe 8–27. The diameter D of oil spots made on a sheet of porous paper depends upon the diameter d of the squirting nozzle, the height h of the nozzle from the surface, the velocity V of the oil, and its density r, viscosity m, and surface tension s. Determine the dimensionless ratios that define this process. Take q = m, s, D, and d.

d V

h

D

Prob. 8–24 8–25. The velocity c of a wave on the surface of a liquid depends upon the wave length l, the density r, and the surface tension s of the liquid. Determine the relation between c and these parameters. By what percent will c decrease if the density of the liquid is increased by a factor of 1.5? 8–26. When an underwater explosion occurs, the pressure p of the shock wave at any instant is a function of the mass of the explosive m, the initial pressure p0 formed by the explosion, the spherical radius r of the shock wave, and the density r and the bulk modulus EV of the water. Determine the relation between p and these parameters. Take q = p, r, and EV.

Prob. 8–27

*8–28. The height H of water as it passes over a small weir depends upon the volumetric discharge Q, the width b and height h of the weir, the acceleration of gravity g, and the density r, viscosity m, and surface tension s of the fluid. Determine the relation between H and these parameters. Take q = H, b, and Q.

H

r

Prob. 8–26

h

Prob. 8–28

problems 8–29. The drag FD on the automobile is a function of its velocity V, its projected area A into the wind, and the density r and viscosity m of the air. Determine the relation between FD and these parameters. Take q = FD and m. v

467

8–34. The flow of air Q through the fan is a function of the diameter D of the blade, its angular rotation v, the density r of the air, and the pressure difference ∆p on each side of the blade. Determine the relation between Q and these parameters. Take q = ∆p and Q. 8

FD

Prob. 8–29 8–30. As the liquid drains from the pipette, the exit velocity V is a function of the liquid’s specific weight g, viscosity m, the nozzle’s diameter d, and the liquid’s level h. Determine the relation between V and these parameters. Take q = h and m.

D

Prob. 8–34 h

d

V

8–35. The capillary rise of a fluid along the walls of the tube causes the fluid to rise a distance h. This effect depends upon the diameter d of the tube, the surface tension s, the density r of the fluid, and the gravitational acceleration g. Show how to obtain the two Π terms h>d and s> 1rd 2g2, where q = h and s. If an experiment is performed using water at a temperature of 20°C, and the surface tension is s = 0.0736 N>m, the following data is obtained of the height h versus the diameter d of the tube. Use the data and plot the relationship between the two Π terms.

Prob. 8–30 8–31. The discharge Q of a pump is a function of the impeller # diameter D, its angular velocity v, the power output W, and the density r and viscosity m of the fluid. Determine the relation between Q and these parameters. Take q = Q, r, and m. # *8–32. The power W supplied by a pump is thought to be a function of the discharge Q, the change in pressure ∆p between the inlet and outlet, and the density r of the fluid. # Establish a relation between W and these parameters. 8–33. The length l of small water waves is thought to be a function of the period of time t of each wave cycle, the water depth h, gravitational acceleration g, and the surface tension s of the water. Determine the relation between l and these parameters. Take q = l and t.

a h (mm) d (mm)

h

30.06 15.03 10.02 7.52 6.01 5.01

Prob. 8–35

0.5 1.0 1.5 2.0 2.5 3.0

468

Chapter 8

D i m e n s i o n a l a n a ly s i s

anD

*8–36. The head loss hL in a pipe depends upon its diameter D, the velocity of flow V, and the density r and viscosity m of the fluid. Determine the relation between hL and these parameters. Take q = hL and m. 8

8–37. The drag FD on the square plate held normal to the wind depends upon the area A of the plate and the air velocity V, density r, and viscosity m. Determine the relation between FD and these parameters. Take q = FD and m.

similituDe *8–40. The change in pressure ∆p in the pipe is a function of the density r and the viscosity m of the fluid, the pipe diameter D, and the velocity V of the flow. Establish the relation between ∆p and these parameters. Take q = ∆p and m.

D V

Prob. 8–40

FD

SEC. 8.5 Prob. 8–37

8–41. If a ship having a length of 30 m is to be designed to travel at 20 m>s, determine the speed of a 0.75-m-long model in order to maintain the same Froude number.

8–38. The torque T developed by a turbine depends upon the depth h of water at the entrance, the density of the water r, the discharge Q, and the angular velocity of the turbine v. Determine the relation between T and these parameters. Take q = T and h.

8–42. The flow of water around the structural support is 1.2 m>s when the temperature is 5°C. If it is to be studied using a model built to a scale of 1>20, and using water at a temperature of 25°C, determine the velocity of the water used with the model.

8–39. The thrust T of the propeller on a boat depends upon the diameter D of the propeller, its angular velocity v, the speed of the boat V, and the density r and viscosity m of the water. Determine the relation between T and these parameters. Take q = m, V, and T.

V

1.2 ms

T

Prob. 8–39

Prob. 8–42

problems 8–43. A model of a submarine is used to determine the drag acting on its prototype. The length scale is 1>100, and the test is run in water at 20°C, with a speed of 8 m>s. If the drag on the model is 20 N, determine the drag on the prototype if it runs in water at the same speed and temperature. This requires that the drag coefficient CD = 2FD >rV 2L2 be the same for both the model and the prototype.

469

8–47. Water and crude oil flow through the pipes with a velocity of 2 m>s and 8 m>s, respectively. Determine the diameter of the oil pipe so that both fluids have the same dynamic characteristics. The temperature of the water and oil is 20°C. 8

50 mm 8 ms

D

Prob. 8–47 Prob. 8–43

*8–44. A water pump produces a flow of 0.25 m3 >s and a power of 3 kW, using a 200-mm-diameter impeller. What is the flow and power if a 150-mm-diameter impeller is used to maintain dynamic similitude?

*8–48. The effect of drag on a model airplane is to be tested in a wind tunnel with a wind speed of 450 km>h. If a similar test is performed on the same model underwater in a channel, what should be the speed of the water to achieve the same result when the temperature is 20°C?

8–45. The resistance created by waves on a 100-m-long ship is tested in a channel using a model that is 4 m long. If the ship travels at 60 km>h, what should be the speed of the model? 8–46. A model of a plane is built to a scale of 1>15 and is tested in a wind tunnel. If the plane is designed to travel at 800 km>h at an altitude of 5 km, determine the required density of the air in the wind tunnel so that the Reynolds and Mach numbers are the same. Assume the temperature is the same in both cases and the speed of sound in air at this temperature is 340 m>s.

450 km/h

Prob. 8–48

470

Chapter 8

D i m e n s i o n a l a n a ly s i s

anD

8–49. The jet plane can fly at Mach 2 at an altitude of 3 km. What is the comparable speed the plane can fly at an altitude of 1 km? Use Eq. 13–19, c = 2kRT, where k = 1.40 for air.

similituDe 8–53. If the pressure loss along a measured length of a 50-mm-diameter pipe transporting water is 4 Pa, determine the pressure loss in the same length of a 100-mm-diameter pipe transporting kerosene flowing at a velocity of 4 m>s. The temperature of the water and kerosene is 20°C.

8

50 mm

Prob. 8–49

100 mm

8–50. The drag coefficient on an airplane is defined by CD = 2FD >rV 2L2. If the drag acting on the model of a plane tested at sea level is 0.3 N, determine the drag on the prototype, which is 15 times larger and is flying at 20 times the speed of the model at an altitude of 3 km. 8–51. Consider two similar types of hydrofoil boats having a size ratio 3:2. If the maximum lift the larger boat can produce is 8 kN, when it travels with a speed of 60 km>h, determine the maximum lift that the smaller boat can produce and the corresponding speed. Assume the water temperature is the same for both cases. This requires Euler number and Reynolds number similitude. *8–52. The velocity of water waves in a channel is studied in a laboratory using a model of the channel one-twelfth its actual size. Determine the velocity of the waves in the channel if they have a velocity of 6 m>s in the model.

6 ms

Prob. 8–52

Prob. 8–53

8–54. In order to test the flow over the surface of an airplane wing, a model is built to a scale of 1>20 and is tested in water. If the airplane is designed to fly at 850 km>h, what should be the velocity of the model in order to maintain the same Reynolds number? Is this test realistic? Take the temperature of both the air and the water to be 20°C.

850 km/h

Prob. 8–54

8–55. The model of a boat is built to a scale of 1>50. Determine the required kinematic viscosity of the water in order to test the model so that the Froude and Reynolds numbers remain the same for the model and the prototype. Is this test practical if the prototype operates in water at T = 20°C?

problems *8–56. If the water in a river is flowing at 3 m>s, and it flows at 0.75 m>s in a model, determine the depth of the water in the model if the river has an average depth of 6 m. 8–57. The flow around the jet plane flying at an altitude of 10 km is to be studied using a wind tunnel and a model that is built to a 1>15 scale. If the plane has an air speed of 800 km>h, what should be the speed of the air inside the tunnel? Is this reasonable?

471

*8–60. The optimum performance of mixing blades 0.5 m in diameter is to be tested using a model one-fourth the size of the prototype. If the test of the model in water reveals the optimum speed to be 8 rad>s, determine the optimum angular speed of the prototype when it is used to mix ethyl alcohol. Take T = 20°C.

800 kmh 0.25 m 0.25 m

Prob. 8–57 8–58. If the drag on a 15-m-long airplane is to be determined when the plane is flying at 1200 km>h at an altitude of 10 km, find the speed of the air in a wind tunnel for a 1.5-m-long model if the air has a temperature of 20°C at standard atmospheric pressure. 8–59. The flow of water around the bridge pier is to be studied using a model built to a scale of 1>15. If the river flows at 0.8 m>s, determine the corresponding velocity of the water in the model at the same temperature.

Prob. 8–60

8–61. In order to determine the formation of waves around obstructions in a river, a model having a scale of 1>20 is used. If the river flows at 2 m>s, determine the speed of the water for the model. 8–62. A 20-m-long submarine is intended to travel at 40 km>h. If it is tested in a wind tunnel using a 1-m-long model, determine the speed of the air at standard atmospheric pressure and at the same temperature of 10°C. Is this reasonable?

0.8 ms

Prob. 8–59

8–63. A ship has a length of 180 m and travels in the sea where rs = 1030 kg>m3. A model of the ship is built to a 1 > 60 scale, and it displaces 0.06 m3 of water such that its hull has a wetted surface area of 3.6 m2. When tested in a towing tank at a speed of 0.5 m>s, the total drag on the model was 2.25 N. Determine the drag on the ship and its corresponding speed. What power is needed to overcome this drag? The drag due to viscous (frictional) forces can be determined using (FD)f = 1 12 rV 2A 2 CD, where CD is the drag coefficient determined from CD = 1.328> 2Re for Re 6 106 and CD = 0.455>1log10Re2 2.58 for 106 6 Re 6 109. Take r = 1000 kg>m3 and n = 1.00110 - 6 2 m2 >s.

8

472

Chapter 8

D i m e n s i o n a l a n a ly s i s

anD

*8–64. If the jet plane flies at 1500 km > h when the temperature of the air is - 10°C, and the absolute pressure is 60 kPa, what should its speed be so that it has the same Mach number when the air temperature is - 30°C and the absolute pressure is 40 kPa? Assume the air has the same bulk modulus. Use Eq. 13–20, 8 c = 2E >r. V

similituDe 8–65. A 20-m-long “check dam” on a river provides a means of collecting debris that flows downstream. If the flow over the dam is 250 m3 >s, and a model of this dam is to be built to a scale of 1>20, determine the flow over the model and the depth of water that flows over its crest. Assume that the water temperature for the prototype and the model is the same. The volumetric flow over the dam can be determined using Q = CD 2gLH 3>2, where CD is the coefficient of discharge, g the acceleration of gravity, L is the length of the dam, and H is the height of the water above the crest. Take CD = 0.71.

20 m RES

RES

CUE

CUE

51204271

Prob. 8–64

Prob. 8–65

Chapter review

473

CHAP TER R EV IEW Dimensional analysis provides a means of combining the variables that influence a flow into groups of dimensionless numbers. This reduces the number of experimental measurements required to describe the flow. When studying flow behavior, the following important dimensionless ratios of the dynamic or inertia force to some other force often occur.

Euler number Eu =

Reynolds number Re =

pressure force inertia force

∆p =

rV 2

rVL inertia force = m viscous force

Froude number Fr =

inertia force V = A gravitational force 2gL

Weber number We =

rV 2L inertia force = s surface tension force

Mach number M =

inertia force V = A compressibility force c

The Buckingham Pi theorem provides a means for determining the dimensionless groups of variables one can obtain from a set of variables. Similitude provides a means of making sure that the flow affects the prototype in the same way that it affects its model. The model must be geometrically similar to the prototype, and the flow must be kinematically and dynamically similar.

Vp

Vm Fm

Fp Lp Prototype

Lm Model

8

9

Steve Satushek/Stone/Getty Images

CHAPTER

A pressure drop that occurs along a closed conduit such as a pipe is due to friction losses within the fluid. These losses will be different for laminar and turbulent flow.

VISCOUS FLOW WITHIN ENCLOSED CONDUITS CHAPTER OBJECTIVES ■

To discuss how the forces of gravity, pressure, and viscosity affect the laminar flow of an incompressible fluid contained between parallel plates and within a pipe.

■

To illustrate how to classify the flow using the Reynolds number.

■

To present some of the ways that are used to model turbulent pipe flow.

9.1

STEADY LAMINAR FLOW BETWEEN PARALLEL PLATES y

In this section we will study laminar flow of a Newtonian (viscous) fluid confined between two inclined parallel plates, Fig. 9–1a. Here the plates are separated by a distance a, and have a sufficient width and length so that end effects can be neglected. To generalize the solution somewhat, we will also assume the top plate is moving with a speed U relative to the bottom plate. We have one-dimensional flow since the velocity will vary only in the y direction, while it remains constant in the x and z directions for each value of y. We wish to determine the shear stress within the fluid and the velocity profile of the fluid that produce steady flow, assuming the fluid is incompressible.

Top plate moving

Steady flow through plates

U u

a x

(a)

Fig. 9–1

475

476

Chapter 9

h

VisCous Flow

y

9 y

U Dy a Dx

u

Datum

x (b)

0p (p + –– D x ) Dy Dz 0x

tDxDz

u g Dx D y Dz

0p 0t ∆x b ∆y ∆z + a t + ∆y b ∆x ∆z - t ∆x ∆z 0x 0y Net shear force

Weight

Factoring out the volume ∆x ∆y ∆z, noting that sin u = - ∆h> ∆x, Fig. 9–1d, and taking the limit as ∆x and ∆h both approach zero, we obtain, after simplifying,

2Dh

0t 0 = ( p + gh) 0y 0x

(d)

The sum in parentheses is independent of y, and is constant for each cross section.* For example, consider points 1 and 2 on the cross section shown in Fig. 9–1e. At point 1 we have p1 + gh1 , and at point 2 it is (p1 - gh′) + g(h1 + h′) = p1 + gh1. Since the pressure is only a function of x, we can use a total derivative, and so integrating the above equation with respect to y yields

2 1

p ∆y ∆z - a p +

0 Vr dV + Vr Vf>cs # dA 0t Lcv Lcs

+ g∆x ∆y ∆z sin u = 0 + 0

(c)

u

To obtain the shear-stress distribution we will apply the momentum equation, and select a differential control volume having a length ∆x, a thickness ∆y, and a width ∆z, Fig. 9–1b. The forces acting on the free-body diagram of this control volume along the x direction include the pressure forces on the open control surfaces, Fig. 9–1c, the forces caused by shear stress on the top and bottom closed control surfaces, and the x component of the weight of the fluid within the control volume. The shear forces are different on their opposite surfaces because the movement of the fluid is different on adjacent streamlines. As indicated, both the pressure and the shear stress are assumed to increase in the positive x and y directions, respectively. Since the flow is steady and the fluid incompressible, and ∆Ain = ∆Aout, then no local and convective changes take place, and so the momentum equation becomes an equilibrium equation, where

Net pressure force

Free-body diagram

Dx

enClosed Conduits

ΣFx =

0t (t + –– D y) DxD z 0y

pDy Dz

within

h9

h

Datum (e)

Fig. 9–1 (cont.)

t = c

d (p + gh) d y + C1 dx

This expression is based only on a balance of forces, and so it is valid for both laminar and turbulent flow. *This equation can also be written in the form 0t>0y = g 30>0x1p>g + h24, where p>g + h is the hydraulic head, i.e., the pressure head plus the elevation head.

9.1

steady laminar Flow between parallel plates

If we have a Newtonian fluid, for which laminar flow prevails, then we can apply Newton’s law of viscosity, t = m(du>dy), to obtain the velocity profile. Thus, du d m = c ( p + gh) d y + C1 dy dx Again, integrating with respect to y, y2 C1 1 d ( p + gh) d + y + C2 c m dx m 2

u =

(9–1)

The constants of integration can be evaluated using the “no slip” boundary conditions, namely, at y = 0, u = 0, and at y = a, u = U. After substituting and simplifying, we get t =

Um d a + c ( p + gh) d a y - b a dx 2

(9–2)

Shear-stress distribution Laminar and turbulent flow

u =

U 1 d y + c ( p + gh) d 1y2 - ay2 a 2m dx

(9–3)

Velocity profile Laminar flow

If the pressures and the elevations at any two points 1 and 2 on the same streamline are known, Fig. 9–1f, then we have p2 - p1 h2 - h1 d ( p + gh) = + g dx L L We will now consider a few special cases to better understand how the forces of viscosity, pressure, and gravity influence the flow.

L 1 p1

2 p2 h1

u h2 (f)

Fig. 9–1 (cont.)

Datum

477

9

478

Chapter 9

VisCous Flow

within

Horizontal Flow Caused by a Constant Pressure Gradient—Both Plates Fixed. In this case U = 0 and

tmax u High a pressure

9

V

t a y –– 2 tmax

enClosed Conduits

Low pressure

dh>dx = 0, Fig. 9–2, so Eqs. 9–2 and 9–3 become

x

t =

umax

dp a ay - b dx 2

(9–4)

Shear-stress distribution

Shear-stress distribution and actual and average velocity profiles with negative pressure gradient and no motion of the plates

1 dp 2 1y - ay2 2m dx

u =

Fig. 9–2

(9–5)

Velocity profile Laminar flow

In order for the flow to move to the right it is necessary that dp>dx be negative. In other words, the pressure must decrease, where the higher pressure must be applied to the fluid on the left, causing it to flow to the right. It is the shear stress caused by friction within the fluid that causes the pressure to decrease in the direction of motion. A plot of Eq. 9–4 shows that the shear stress varies linearly with y, where tmax occurs at the surface of each plate, Fig. 9–2. Since there is a smaller shear stress on the fluid within the center region between the plates, the velocity there is greater. In fact, Eq. 9–5 shows the resulting velocity distribution is parabolic, and the maximum velocity occurs at the center, where du>dy = 0. Substituting y = a>2 into Eq. 9–5 reveals that this maximum velocity is umax = -

a 2 dp 8m dx

(9–6)

Maximum velocity

The volumetric flow can be determined by integrating the velocity profile over the cross-sectional area between the plates. If the plates have a width b, then dA = b dy, and we have Q =

LA

1 dp 2 1y - ay2(b dy) L0 2m dx a 3b dp Q = 12m dx a

u dA =

(9–7)

Volumetric flow

Since the cross-sectional area between the plates is A = ab, then the average velocity, Fig. 9–2, is determined from V =

Q a 2 dp = A 12m dx

Average velocity

If we compare this equation with Eq. 9–6, we find 3 umax = V 2

(9–8)

9.1

479

steady laminar Flow between parallel plates

Horizontal Flow Caused by a Constant Pressure Gradient— Top Plate Moving. In this case dh>dx = 0, so Eqs. 9–2 and 9–3 reduce to dp Um a t = + ay - b a dx 2

y U

(9–9)

9

U

u

Shear-stress distribution

Low pressure

High pressure

and

x

U 1 dp 2 u = y + 1y - ay2 a 2m dx

Velocity profile caused by a positive pressure gradient and motion of top plate

(9–10)

(a)

Velocity profile Laminar flow y

Here the volumetric flow is

U

Q =

LA

u dA =

=

L0

a

c

U 1 dp 2 y + 1y - ay2 d b dy a 2m dx

Uab a 3b dp 2 12m dx

High pressure

u

(9–11) Velocity profile caused by a negative pressure gradient and motion of top plate

Volumetric flow

And since A = ab, the average velocity is

(b)

Fig. 9–3

V =

Q U a 2 dp = A 2 12m dx

U

(9–12)

Average velocity

Realize that the maximum velocity does not occur at the midpoint; rather, it depends upon both the speed of the top plate, U, and the pressure gradient, dp>dx. For example, if a large enough positive pressure gradient occurs, then, from Eq. 9–10, u will become negative, and a net reverse (negative) flow may take place. A typical velocity profile looks like that shown in Fig. 9–3a, where the top portion of fluid is dragged to the right, due to the motion U of the plate, and the rest of the fluid moves to the left, being pushed by the positive pressure gradient. When the pressure gradient is negative, both this gradient and the plate motion will work together and cause the velocity profile to look something like that shown in Fig. 9–3b.

Low pressure x

480

Chapter 9

VisCous Flow

within

enClosed Conduits

Horizontal Flow Caused Only by the Motion of the Top Plate. If the pressure gradient dp>dx and slope dh>dx are both zero, then the flow will be caused entirely by the moving plate. In this case, Eqs. 9–2 and 9–3 become 9

y U

U

t = u t

Um a

(9–13)

Shear-stress distribution

a y x

Shear-stress distribution and velocity profile for zero pressure gradient and motion of top plate

u =

U y a

(9–14)

Velocity profile Laminar flow

Fig. 9–4

These results indicate that the shear stress is constant, while the velocity profile is linear, Fig. 9–4. We discussed this situation in Sec. 1.7, as it relates to Newton’s law of viscosity. Fluid motion caused only by movement of the boundary (in this case a plate) is known as Couette flow, named after Maurice Couette. In general, however, the term “Couette flow” refers to either laminar or turbulent flow caused only by boundary movement.

Limitations. It is important to remember that all the velocityrelated equations developed in this section apply only for steady laminar flow of an incompressible Newtonian fluid. Hence, in order to use these equations, we must be sure that laminar flow prevails. In Sec. 9.5 we will discuss how the Reynolds number Re = rVL>m can be used as a criterion for identifying laminar flow. For parallel plates, the Reynolds number can be calculated using the distance a between the plates as the “characteristic length” L. Also, using the average velocity to calculate Re, we then have Re = rVa>m. Experiments have shown that laminar flow will occur up to a certain limit for this Reynolds number. Although it has no particular exact value, in this book we will consider that upper limit to be Re = 1400. Therefore,

Re =

rVa … 1400 m

(9–15)

Laminar flow between plates

Provided this inequality is satisfied, the results of calculating the velocity profiles using the equations presented here have been shown to closely match the velocity profiles obtained from experiments.

9.2

9.2

481

naVier–stokes solution For steady laminar Flow between parallel plates

NAVIER–STOKES SOLUTION FOR STEADY LAMINAR FLOW BETWEEN PARALLEL PLATES

Rather than using first principles, the velocity profile, Eq. 9–3, can also be obtained using the continuity equation and the Navier–Stokes equations discussed in Sec. 7.12. To show how this is done, we will establish the x, y, z axes as in Fig. 9–5. Since there is steady incompressible flow only in the x direction, then v = w = 0, and as a result, the continuity equation, Eq. 7–10, gives

h

y

9

0(ru) 0(rv) 0(rw) 0r + + + = 0 0t 0x 0y 0z 0u 0 + r + 0 + 0 = 0 0x

so that 0u>0x = 0. Symmetry in the z direction and steady flow indicate that u is not a function of z and x but, rather, it is only a function of y, that is, u = u(y). Also, from Fig. 9–5, gx = g sin u = g( -dh>dx) and gy = -g cos u. Using these results, the three Navier–Stokes equations, Eqs. 7–74, reduce to 0p 0u 02u 0u 0u 0u 02u 02u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z 0p dh d 2u 0 = rgab + m 2 dx 0x dy 0p 0v 02v 0v 0v 0v 02v 02v ra + u + v + w b = rgy + ma 2 + 2 + 2 b 0t 0x 0y 0z 0y 0x 0y 0z 0p 0 = - rg cos u + 0 0y 0p 0w 02w 0w 0w 0w 02w 02w + ra + u + v + w b = rgz + ma 2 + b 0t 0x 0y 0z 0z 0x 0y2 0z2 0p 0 = 0 + 0 0z

gx 5 g sin u a

gy 5 g cos u u

u

ra

The last equation, when integrated, shows that p is constant in the z direction, which is to be expected. Integrating the second equation gives p = -r(g cos u)y + f(x) The first term on the right shows that the pressure varies in a hydrostatic manner in the y direction. The second term on the right, f(x), shows that the pressure also varies in the x direction. This is due to the viscous shear stress. If we rearrange the first Navier–Stokes equation, using g = rg, and integrate it twice, we get 1 d d 2u = ( p + gh) m dx dy2 du 1 d = ( p + gh)y + C1 m dx dy y2 1 d u = c ( p + gh) d + C1y + C2 m dx 2

This is essentially the same result as Eq. 9–1, so the analysis proceeds as before.

x

g

u dx

Fig. 9–5

2dh

482

Chapter 9

VisCous Flow

within

enClosed Conduits

I MPO RTA N T PO I N T S • Steady flow between two parallel plates is a balance of the 9

forces of pressure, gravity, and viscosity. The viscous shear stress varies linearly along the thickness of the fluid, regardless of whether the flow is laminar or turbulent.

• The velocity profile for steady laminar flow of an incompressible Newtonian fluid is determined using Newton’s law of viscosity. Motion of the fluid is caused by the relative motion of the plates and the pressure gradient within the fluid.

• The formulations in Sec. 9.1 were developed from first principles, and in Sec. 9.2 by solving the continuity and Navier– Stokes equations. These results are in close agreement to experimental measurements. Furthermore, experiments have shown that laminar flow between parallel plates will occur up to a critical value of the Reynolds number, which we have taken to be Re = rVa>m … 1400. Here a is the distance between the plates, and V is the average velocity of flow.

PR O C E DU R E FO R A N A LY S I S The equations in Sec. 9.1 can be applied using the following procedure. Fluid Description. The flow must be steady, and the fluid must be an incompressible Newtonian fluid. Also, laminar flow must exist, so be certain to check that the flow conditions produce a Reynolds number Re = rVa>m … 1400. Analysis. Establish the coordinates and follow their positive sign convention. Here x is positive in the direction of flow; y is positive measured from the bottom plate, upward and normal to the plate, so it is perpendicular to the flow; and h is positive vertically upward, Fig. 9–1b. Finally, be sure to use a consistent set of units when substituting numerical data into any of the equations.

9.2

naVier–stokes solution For steady laminar Flow between parallel plates

483

EXAMPLE 9.1 Glycerin flows at a steady rate of 0.005 m3 >s between the two smooth plates that are 15 mm apart, as shown in Fig. 9–6. Determine the pressure gradient acting on the glycerin.

9 h

SOLUTION Fluid Description. For the analysis, we will assume the plates are wide enough (0.4 m) to neglect end effects. Also, we will assume steady, incompressible, laminar flow. From Appendix A, rg = 1260 kg>m3 and mg = 1.50 N # s>m2. Analysis. Since the flow Q is known, we can obtain the pressure gradient by first finding the velocity profile, Eq. 9–3. Here the plates are not moving relative to one another, so U = 0, and therefore, 1 d u = c ( p + gh) d 1y2 - ay2 2m dx The coordinates are established with x along the left plate edge, positive in the direction of flow (downward), and h positive upward, Fig. 9–6. Since these axes are collinear, then dh = -dx and so dh>dx = -1. Therefore, the above equation becomes 1 dp u = a - gb 1y2 - ay2 (1) 2m dx The flow is therefore a 1 dp Q = u dA = a - gb 1y2 - ay2 b dy LA L0 2m dx

0.4 m y a 5 15 mm

a b dp b dp a3 1y2 - ay2 dy = - a a - gb - gb a b 2m dx 2m dx 6 L0 Substituting the data into this equation, we get (0.015 m)3 dp 0.4 m 3 2 0.005 m3 >s = 11260 kg>m 219.81 m>s 2 d a b c 6 211.50 N # s>m2 2 dx

=

dp = -54.311103 2 Pa>m = -54.3 kPa>m Ans. dx The negative sign indicates that the pressure within the glycerin is decreasing in the direction of flow. This is to be expected due to the frictional drag caused by the viscosity. Finally, we need to check if the flow is indeed laminar by using our Reynolds number criterion. Since V = Q>A, we have Re =

11260 kg>m3 23 10.005 m3 >s2 >(0.015 m(0.4 m))4(0.015 m) rVa = m 1.50 N # s>m2 = 10.5 6 1400

(laminar flow)

x

Fig. 9–6

484

Chapter 9

EXAMPLE

VisCous Flow

enClosed Conduits

9.2

y

9

within

200 mm x B 100 mm

A

The 100-mm-diameter plug in Fig. 9–7 is placed within a pipe and supported so that oil can flow between it and the walls of the pipe. If the gap between the plug and the pipe is 1.5 mm, and the pressure at A is 400 kPa, determine the discharge of the oil through the gap. Take ro = 920 kg>m3 and mo = 0.2 N # s>m2. SOLUTION

Fig. 9–7

Fluid Description. We will assume the oil is incompressible and the flow is steady laminar flow. Also, since the gap size is very small compared to the radius of the plug, we will neglect the curvature of the pipe and any elevation difference, and assume that flow occurs between horizontal “parallel plates” that are at rest. Analysis. The discharge is determined from Eq. 9–7. The x coordinate is on the top of the plug and is positive in the direction of flow, so dp>dx = ( pB - pA)>LAB. Since pA = 400 kPa, pB = 0, and LAB = 0.2 m, we have

Q = -

(0.0015 m)3 [2 p(0.05 m)] 0 - 400(103) N>m2 a 3b dp = c d 12mo dx 0.2 m 12 (0.2 N # s>m2)

= 0.8836(10 -3) m3 >s = 0.884(10 -3) m3 >s

Ans.

To check if the flow is laminar, we must first obtain the average velocity using Eq. 9–8. Q = VA; 0.8836 (10 -3) m3 >s = V[ 2 p(0.05 m)(0.0015 m)] V = 1.875 m>s The Reynolds number is therefore

Re =

(920 kg>m3)(1.875 m>s)(0.0015 m) roVa = mo 0.2 N # s>m2 = 12.94 6 1400

(laminar flow)

NOTE: A more exact analysis of this problem can be made by accounting for the curvature of the pipe and plug. It represents steady laminar flow through an annulus, and the relevant equations are developed as part of Prob. 9–46.

9.2

EXAMPLE

485

naVier–stokes solution For steady laminar Flow between parallel plates

9.3

During a manufacturing process, a 45-mm-wide strip of paper is pulled upward at 0.6 m>s through a narrow channel from a reservoir of glue, as shown in Fig. 9–8a. Determine the force per unit length exerted on the strip when it is in the channel, if the thickness of the glue on each side of the strip is 0.1 mm. Assume the glue is a Newtonian fluid having a viscosity of m = 0.843110 - 3 2 N # s>m2 and a density of r = 735 kg>m3.

F

9 0.6 ms B

SOLUTION 0.1 mm

Fluid Description. Within the channel, steady flow occurs. We will assume the glue is incompressible and the flow is laminar. Analysis. In this problem, gravity and viscosity predominate. There is no pressure gradient throughout the glue because the pressure at A and B is atmospheric, that is, pA = pB = 0 and so ∆p = 0 or 0p>0x = 0 from A to B. The paper acts as a moving plate, and so to obtain the force per unit length on the paper, we will first apply Eq. 9–2 to obtain the shear stress on the paper, that is Um d a t = + c ( p + gh) d a y - b a dx 2 The coordinate axes are established in the usual manner from the fixed surface on the left side, Fig. 9–8b. As the strip moves upward, the glue adheres to it, but it must overcome the shear stress on its surface at y = a = 0.1 mm. Since the x and h axes are collinear, then dh = dx and so dh>dx = 1, and since 0p>0x = 0, the above equation becomes Um a t = + ga b a 2 =

(0.6 m>s)30.843110-3 2 N # s>m2 4 0.1110 2 m -3

+ 1735 kg>m3 219.81 m>s2 2 £

0.1110 - 3 2 m 2

A

(a) h, x F dh 5 dx

§

= 5.419 N>m2

This stress must be overcome on each side of the strip, and because the paper has a width of 45 mm, the force per unit length on the strip is w = 215.419 N>m2 2(0.045 m) = 0.488 N>m Ans. We must now check our assumption of laminar flow. Rather than establishing the actual velocity profile, and then finding the average velocity, here we will consider the maximum velocity, which occurs on the strip at y = 0.1 mm. It is umax = 0.6 m>s. Since umax 7 V, even at this maximum velocity, we have Re =

0.1 mm

1735 kg>m3 2(0.6 m>s)(0.0001 m) rumaxa = 52.3 6 1400 (laminar flow) = m 0.843110 - 3 2 N # s>m2

y 0.1 mm

(b)

Fig. 9–8

486

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enClosed Conduits

h

y

L

9

1

R

DA9 r DA Dx

h1

f

Datum

2

h2 x

(a)

Fig. 9–9

9.3 D A9 p DA

0p (p 1 –– 0x D x)D A f gd V Free-body diagram (b)

2Dh

Dx

STEADY LAMINAR FLOW WITHIN A SMOOTH PIPE

In this section we will determine the shear stress and velocity profile that produce steady flow of an incompressible fluid within a smooth pipe. Since the flow will be axisymmetric, it is convenient to consider a control volume element within the fluid to be a differential disk, Fig. 9–9a. To obtain the shear-stress distribution, we will apply the momentum equation to this control volume. When this is done, no convective change occurs between the back and front open control surfaces due to steady flow, and no local change occurs within the control volume. As shown on the freebody diagram of the control volume, Fig. 9–9b, the forces to be considered in the x direction are due to pressure, gravity, and viscosity. We have 0 ΣFx = Vr dV + VrVf>cs # dA 0t Lcv Lcs p∆A - a p +

f (c)

tmax

0p ∆xb ∆A + t∆A′ + g∆V sin f = 0 + 0 0x From Fig. 9–9a, the cross-sectional area of the open control surface is ∆A = pr 2, the area of the closed control surface is ∆A′ = 2pr∆x, and the control volume is ∆V = pr 2 ∆x. Substituting these results into the above equation, noting that sin f = - ∆h> ∆x, Fig. 9–9c, and taking the limit, we get t =

Shear-stress distribution for both laminar and turbulent flow (d)

r 0 ( p + gh) 2 0x

(9–16)

Shear-stress distribution Laminar and turbulent flow

The shear stress varies directly with r, being largest at the wall, r = R, and zero at the center, r = 0, Fig. 9–9d. Also, since t was determined simply from a balance of forces, this distribution is valid for both laminar and turbulent flow.

9.3

steady laminar Flow within a smooth pipe

487

If we consider the flow to be laminar, then we can relate the shear stress to the velocity gradient at any point within the fluid flow by using Newton’s law of viscosity, t = m(du>dr). Substituting this into Eq. 9–16 and rearranging terms, we have du r 0 = ( p + gh) dr 2m 0x

9

The quantity ( p + gh) remains constant over any cross section and is independent of y.* For example, consider points 1 and 2 on the cross section shown in Fig. 9–9e. At point 1 we have p1 + gh1, and at point 2 it is p1 - gh′ + g(h1 + h′) = p1 + gh1. Since the pressure is only a function of x, we can use the total derivative, and so integrating the above equation with respect to r, we get r2 d u = ( p + gh) + C 4m dx The constant of integration can be determined using the “no-slip” condition that u = 0 at r = R. Once it is obtained, the result is u = -

1R2 - r 2 2 d ( p + gh) 4m dx

(9–17)

Velocity profile Laminar flow

The velocity profile therefore takes the form of a paraboloid, Fig. 9–9f. This is because t is smaller within the central region of the pipe, Fig. 9–9d, and so the fluid has its greatest velocity there. The maximum velocity occurs at the center of the pipe, r = 0, where du>dr = 0. It is umax = -

R2 d ( p + gh) 4m dx

(9–18)

Maximum velocity

umax Velocity distribution for laminar flow (f)

Fig. 9–9 (cont.) *Since we can also write du>dr = 1r>2m2g0>0x3p>g + h4, then we also state that the hydraulic head p>g + h remains constant over the cross section.

2 1

h9 h1

Datum (e)

488

Chapter 9

VisCous Flow

within

enClosed Conduits

The volumetric flow is determined by integrating the velocity profile over the cross-sectional area. Choosing the differential ring element of area dA = 2pr dr, shown in Fig. 9–9g, we have r

2p d 1R2 - r 2 2r dr Q = u dA = u 2pr dr = ( p + gh) 4m dx LA L0 L0 R

9 umax

R

or

(g)

Q = V

pR4 d ( p + gh) 8m dx

(9–19)

Volumetric flow

Because the cross-sectional area of the pipe is A = pR2, the average velocity, Fig. 9–9h, is therefore

Average velocity distribution (h)

Fig. 9–9 (cont.)

Q R2 d = ( p + gh) A 8m dx

V =

(9–20)

Average velocity

By comparison with Eq. 9–18, we see that umax = 2V

(9–21)

To establish the term d( p + gh)>dx, consider the pipe in Fig. 9–9i. If the pressures p1 and p2 and the elevations h1 and h2 between the two points 1 and 2 on the cross sections are known, then p2 - p1 h 2 - h1 d ( p + gh) = + g dx L L

h

(9–22)

y

L

p1

1 h1 Datum

p2

f

2

h2 x

(i)

9.3

steady laminar Flow within a smooth pipe

489

L V

tmax umax 5 2V

High pressure

1

2

Low x pressure

9 tmax Shear-stress distribution

Actual velocity profile

Average velocity distribution

Laminar flow

Ideal flow

Shear-stress and velocity profiles for negative pressure gradient

Fig. 9–10

Horizontal Flow through a Circular Pipe. If the pipe is horizontal, then the force of gravity will not influence the flow since dh>dx = 0. If there is a higher pressure on the left side of the pipe than on the right side, Fig. 9–10, then over the length L, this pressure difference will “push” the fluid to the right. Realize that the pressure will decrease along the pipe in the direction of flow because of fluid friction (viscosity), and so the pressure gradient dp>dx will be negative (∆p>L 6 0) according to the sign convention. Using this negative gradient, or pressure drop, and the previous results, the maximum velocity, average velocity, and volumetric flow, expressed in terms of the inner diameter of the pipe, D = 2R, then become D2 ∆p umax = a b (9–23) 16m L D2 ∆p V = a b (9–24) 32m L Q =

pD4 ∆p a b 128m L

(9–25)

This type of flow is sometimes called Poiseuille flow, and the last equation is known as the Hagen–Poiseuille equation, since it was originally developed from experiments in the mid-1800s by the German engineer Gotthilf Hagen, and independently by a French physician, Jean Louis Poiseuille.* Shortly thereafter the analytical formulation, as developed here, was presented by Gustav Wiedemann. If we know the flow Q, we can solve the Hagen–Poiseuille equation for the pressure drop that occurs over the length L of the pipe. It is 128mLQ ∆p = (9–26) pD4 Notice that the greatest influence on pressure drop comes from the pipe’s diameter. For example, a pipe having half the diameter will experience sixteen times the pressure drop due to viscous fluid friction! This effect can have serious consequences on the ability of pumps to provide adequate water flow through pipes that may have narrowed due to the accumulation of corrosion or scale. *Poiseuille attempted to study the flow of blood using water confined to small-diameter tubes. Actually, however, veins are flexible, and blood is a non-Newtonian fluid, that is, it does not have a constant viscosity.

490

Chapter 9

VisCous Flow

within

9.4

9 r

dz f

mg

NAVIER–STOKES SOLUTION FOR STEADY LAMINAR FLOW WITHIN A SMOOTH PIPE

Rather than using first principles, we can also obtain the velocity profile within the pipe using the continuity and the Navier–Stokes equations discussed in Sec. 7.12. Due to symmetry, here we will use cylindrical coordinates, established as shown in Fig. 9–11a. For this case, steady incompressible flow is along the axis of the pipe, Fig. 9–11a, so that vr = vu = 0. Therefore, the continuity equation, Eq. 7–77, gives

u

2dh

enClosed Conduits

z

0(rvz) 1 0(rrvr) 1 0(rvu) + + = 0 r 0r r 0u 0z 0vz 0vz 0 + 0 + r = 0 or = 0 0z 0z

(a)

uu

ur

r u uz z

u gr 5 g cos f sin u

u f

gu 5 g cos f cos u r-u plane

Since the flow is steady and symmetrical about the z axis, integration indicates vz = vz(r). Carefully note from Fig. 9–11b that the cylindrical components of g are gr = -g cos f sin u, gu = -g cos f cos u, and gz = g sin f. Using these and the above result, the first Navier–Stokes equation reduces to

g cos f

g Vertical plane gz 5 g sin f

ra

vu 0vr vu2 0vr 0vr 0vr + vr + + vz b r 0u r 0t 0r 0z = -

(b)

Fig. 9–11

0p 0vr vr 02vr 1 0 1 02vr 2 0vu + rgr + mc ar b - 2 + 2 2 - 2 + d r 0r 0r 0r r r 0u r 0u 0z2 0p 0 = - rg cos f sin u + 0 0r

Integrating with respect to r gives p = -rgr cos f sin u + f(u, z) For the second Navier–Stokes equation, ra

0vu 0vu vu 0vu vrvu 0vu + vr + + + vz b r 0u r 0t 0r 0z = -

0vu vu 02vu 1 0p 1 0 1 02vu 2 0vr + rgu + mc ar b - 2 + 2 2 + 2 + d r 0u r 0r 0r r r 0u r 0u 0z2 0 = -

1 0p - rg cos f cos u + 0 r 0u

Integrating with respect to u yields p = -rgr cos f sin u + f(r, z)

9.4

491

naVier–stokes solution For steady laminar Flow within a smooth pipe

Comparing these two results, we require f(u, z) = f(r, z) = f(z) since r, u, z can vary independent of one another. From Fig. 9–11c, the vertical distance h′ = r cos f sin u, and so

f r sin u r

p = -rgh′ + f(z)

u

In other words, the pressure is hydrostatic in the vertical plane since it only depends upon the vertical distance h′. The last term, f(z), is the variation in pressure caused by viscosity. Finally, the third Navier–Stokes equation becomes 0vz vu 0vz ra + vr + + vz b r 0u 0t 0r 0z 0vz

0vz

= 0 = -

2 0vz 02vz 0p 1 0 1 0 vz + rgz + mc ar b + 2 2 + d r 0r 0z 0r r 0u 0z2

0vz 0p 1 0 + rg sin f + mc ar bd r 0r 0z 0r

From Fig. 9–11a, sin f = -dh>dz, so this equation can be rearranged and written as 0 0vz r 0p 0h ar b = c + rga b d m 0r 0r 0z 0z Integrating twice yields r

0vz 0r

=

vz =

r 2 0p 0h c + rga b d + C1 2m 0z 0z

r 2 0p 0h c + rg a b d + C1 ln r + C2 4m 0z 0z

The velocity z must be finite at the center of the pipe, and since ln r S - ∞ as r S 0, then C1 = 0. At the wall, r = R, vz = 0 because of the “no-slip” condition. Thus, C2 = -

R2 0p 0h c + rga b d 4m 0z 0z

Since 1p + gh2 only depends upon z, the final result becomes vz = -

R2 - r 2 d ( p + gh) 4m dz

This is the same as Eq. 9–17, and so a further analysis will produce the rest of the equations in Sec. 9.3.

9

h9 = r cos f sin u z

(c)

Fig. 9–11 (cont.)

492

Chapter 9

VisCous Flow

within

enClosed Conduits

A B

9

(a)

Fig. 9–12

9.5 Laminar flow (b)

Transitional flow (c)

Turbulent flow (d)

THE REYNOLDS NUMBER

In 1883, Osborne Reynolds established a criterion for identifying laminar flow within a pipe. He did this by controlling the flow of water passing through a glass tube, using an apparatus similar to the one shown in Fig. 9–12a. Here, colored dye was injected within the stream at A and the valve at B was opened. For slow rates, the flow in the tube was observed to be laminar since the dye streak remained straight and uniform, Fig. 9–12b. As the flow was increased, by further opening the valve, the streak began to break down as it underwent transitional flow, Fig. 9–12c. Finally, through a further increase in flow, turbulence occurred because the dye became fully dispersed throughout the water in the tube, Fig. 9–12d. Experiments using other liquids, as well as gases, showed this same type of behavior. From these experiments, Reynolds suspected that this change from laminar to transitional to turbulent flow was dependent upon the pressure, viscosity, and inertia forces acting on each of the fluids’ particles. For any two different experimental setups, Reynolds reasoned that a similar type of flow will occur because the forces acting on the fluid particles in one flow should be in the same ratio as those acting on the particles in some other flow, Fig. 9–13. However, as noted in Sec. 8.5, similitude between the viscous and inertia forces will automatically ensure similitude between the pressure and inertia forces because the sum of the forces in each case must satisfy Newton’s second law of motion. Realizing this, Reynolds chose to study the ratio of the inertia force to the viscous force, as shown by the dimensional analysis in Example 8–1, and so produced the dimensionless “Reynolds number,” rVD>m, as the criterion for flow similitude.

9.5

the reynolds number

493

Fv1 Fv 2 Fp1

a1

Fp2

a2

9

D2

D1

Similitude among pressure, viscous, and inertia forces for two different experimental setups

Fig. 9–13

Although any velocity and pipe dimension can be chosen for this ratio, what is important is to realize that for two different flow situations, they must correspond with each other. In practice, the velocity that has been accepted is the mean or average velocity V = Q>A, and the dimension, referred to as the “characteristic length,” is the inner diameter D of the pipe. Using V and D, the Reynolds number then becomes Re =

rVD VD = m n

(9–27)

Experiments have confirmed that the higher this number, the greater the chance for laminar flow to break down, since the inertia force will overcome the viscous forces and begin to dominate the flow. And so, from Eq. 9–27, the faster the fluid flows, the more chance the fluid has of becoming unstable. Likewise, the larger the diameter of pipe, the greater the volume of fluid passing through it, and so instability can occur more readily. Finally, a lower kinematic viscosity will cause the flow to become unstable, because then any flow disturbances will not easily be dampened out by viscous shear forces. In practice, it is very difficult to predict at exactly what specific value of the Reynolds number the flow in a pipe will suddenly change from laminar to transitional flow. Experiments as in Fig. 9–12 show that the critical velocity at which this happens is highly sensitive to any initial vibrations or disturbances to the equipment, the type of pipe inlet, the pipe’s surface roughness, or any slight adjustment that occurs when the valve is opened or closed. For most engineering applications, however, laminar flow for a Newtonian fluid begins to change to transitional flow at about Re = 2300.* This value is called the critical Reynolds number, and in this book, unless otherwise stated, we will assume it to be the limiting value for laminar flow in uniform straight pipes. Therefore, Re … 2300

Laminar flow in straight pipes

(9–28)

By using this as the critical upper bound for laminar flow, we will show in the next chapter how to estimate the energy loss or pressure drop that occurs within a pipe, and thereby determine the flow through the pipe. *Some authors use other limits, generally ranging from 2000 to 2400.

494

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enClosed Conduits

I MPO RTA N T PO I N T S • Steady flow through a pipe is a balance of pressure, gravitational, 9

and viscous forces. For this case, the viscous shear stress varies linearly from zero at the center and has its greatest value at the wall of the pipe. It does not depend upon the type of flow, whether it is laminar, transitional, or turbulent.

• The velocity profile for steady laminar flow of a Newtonian incompressible fluid within a pipe is in the shape of a paraboloid. The maximum velocity is umax = 2V, and it occurs along the centerline of the pipe. The velocity at the wall is zero since it is a fixed boundary—the “no-slip” condition.

• The formulations for laminar flow in a pipe can be developed from first principles, and also by solving the continuity and Navier–Stokes equations. These results are in close agreement with experimental data.

• Flow within a horizontal pipe is dependent upon both pressure and viscous forces. Reynolds recognized this and formulated the Reynolds number Re = rVD>m as a criterion for dynamic similitude between two different sets of flow conditions.

• Experiments have indicated that laminar flow in any pipe will

occur provided Re … 2300. This general estimate of an upper limit will be used in this book.

PR O C E DU R E FO R A N A LY S I S The equations developed in Sec. 9.3 can be applied using the following procedure. Fluid Description. Be sure that the fluid can be assumed as incompressible and the flow is steady. Since laminar flow must prevail, then the flow conditions must not exceed the Reynolds number criterion, Re … 2300. Analysis. Establish the coordinates and follow their positive sign convention. Here the longitudinal axis is positive in the direction of flow, the radial axis is positive outward from the centerline of the pipe, and the vertical axis is positive upward. Finally, be sure to use a consistent set of units when substituting numerical data into any of the equations.

9.5

EXAMPLE

the reynolds number

9.4

Oil flows through the 100-mm-diameter pipe in Fig. 9–14. If the pressure at A is 34.25 kPa, determine the discharge at B. Take ro = 870 kg>m3 and mo = 0.0360 N # s>m2.

9

x

h B

4m y A

Datum 3m

Fig. 9–14

SOLUTION Fluid Description. We will assume steady laminar flow, and that the oil is incompressible. Analysis. The discharge is determined using Eq. 9–19. The origin of the coordinates for x and h is at A, and by convention, the positive x axis is extended in the direction of flow, and the positive h axis is vertically upward. Thus, Q = = -

pR4 d ( p + gh) 8mo dx

hB - hA pR4 pB - pA c + ga bd 8mo L L

= -a

p10.05 m2 4

810.0360 N # s>m2 2

bc

0 - 34.251103 2 N>m2

= 0.001516 m3 >s = 0.00152 m3 >s

5m

+ 1870 kg>m3 219.81 m>s2 2 a

4m - 0 bd 5m

Ans.

Since this result is positive, the flow is directed from A to B. The assumption of laminar flow is checked using the mean velocity and the Reynolds number criterion. V =

0.001516 m3 >s Q = = 0.1931 m>s A p10.05 m2 2

Re =

495

1870 kg>m3 2(0.1931 m>s)(0.1 m) roVD = 467 6 2300 = mo 0.0360 N # s>m2

(laminar flow)

496

Chapter 9

EXAMPLE

VisCous Flow

within

enClosed Conduits

9.5 Determine the maximum pressure at A so that the steady flow of water through the vertical standpipe in Fig. 9–15 remains laminar. The pipe has an inner diameter of 80 mm. Take rw = 1000 kg>m3 and mw = 1.52110-3 2 N # s>m2.

x,h

9 B

SOLUTION Fluid Description. We will assume water to be incompressible. We have steady laminar flow. 3m

Analysis. For laminar flow, the maximum average velocity is based on the Reynolds number criterion.

Re = A

rwVD mw

Datum

2300 =

Fig. 9–15

(1000 kg>m3)(V)(0.08 m) 1.52 (10-3) N # s>m2

V = 0.0437 m>s We will apply Eq. 9–20 to obtain the pressure at A. Following the sign convention, positive x is in the direction of flow, which is vertically upward, and positive h is also vertically upward, Fig. 9–15. Since dh>dx = 1, we have V = -

0.0437 m>s = -

ca ) N # s>m2]

(0.04 m)2 8[1.52(10

-3

R2 d ( p + gh) 8m dx

0 - PA 3m - 0 b + (1000 kg>m3)(9.81 m>s2) a bd 3m 3m

pA = 29.430996(103) Pa = 29.4 kPa

Ans.

Because the velocity and the viscosity are very small, this pressure is essentially hydrostatic, that is, p = gh = 11000 kg>m3 219.81 m>s2 2(3 m) = 29.43(10)3 Pa. In other words, the pressure at A is mainly used to support the water column, and little is needed (0.996 Pa) to overcome the slight frictional resistance and push the water through the pipe to maintain any laminar flow.

9.6

9.6

497

Fully deVeloped Flow From an entranCe

FULLY DEVELOPED FLOW FROM AN ENTRANCE

When fluid begins to flow through the opening of a pipe or duct attached to a reservoir, it will initially accelerate and then transition to either fully laminar or fully turbulent steady flow. We will now consider each of these cases separately.

9

Laminar Flow. As the fluid, which is originally at rest, begins to accelerate within the entrance of the pipe, its velocity profile within the pipe will initially be nearly uniform, Fig. 9–16a. Then, as the fluid travels farther down the pipe, its viscosity will begin to slow down the fluid located near the wall, since at the wall the fluid particles must have zero velocity. With a further advance, the viscous layers form a boundary layer that will begin to spread outward from the wall towards the pipe’s centerline until the central core of fluid, which originally had uniform velocity, narrows down and disappears at the distance L′. Once it does, the flow becomes fully developed, that is, the parabolic velocity profile for laminar flow becomes constant. The transition or entrance length L′ is actually a function of the pipe diameter D and the Reynolds number. An estimate of this length can be made using an equation formulated by Henry Langhaar. See Ref. [2]. It is L′ = 0.06(Re)D

Laminar flow

Flow through long straight pipe will be fully developed. (Prisma Bildagentur AG/Alamy Stock Photo)

(9–29)

Using our criterion for laminar flow in pipes, that is, Re … 2300, then as an upper limit to the entrance length, L′ = 0.06(2300)D = 138D. This is a relatively long distance, for example, for a 150-mm-diameter pipe L′ = 20.7 m, and so fully developed laminar flow rarely occurs in pipes, either because the velocity will be high so that Re is close to its upper limit of 2300, or the flow will get disrupted by a valve, transition, or bend in the pipe.

Turbulent Flow. Experiments have shown that the entrance length to fully developed turbulent flow is not very dependent upon the Reynolds number; rather, it depends more upon the shape or type of inlet and upon the actual roughness of the wall of the pipe. For example, a rounded inlet, as shown in Fig. 9–16b, produces a shorter transition length to full turbulence than a sharp or 90° inlet. Also, pipes with rough walls produce turbulence at a shorter distance than those with smooth walls. Through experiments, along with a computer analysis, it has been found that fully developed turbulent flow can occur within a relatively short distance. See Ref. [3]. For example, it is on the order of 12D for a low Reynolds number, Re = 3000. Although longer transition distances occur at larger Reynolds numbers, for most engineering analysis it is reasonable to assume this transition from unsteady to mean steady turbulent flow is localized near the entrance. And as a result, engineers account for the friction or energy loss that occurs at a turbulent entrance length by using a loss coefficient, something we will discuss in the next chapter.

D Fully developed laminar flow

L9

Transition to laminar flow (a)

D L9

Fully developed turbulent flow

Transition to turbulent flow (b)

Fig. 9–16

498

Chapter 9

EXAMPLE

VisCous Flow

within

enClosed Conduits

9.6 If the flow through the 150-mm-diameter drainpipe in Fig. 9–17 is 0.0062 m3 >s, classify the flow along the pipe as laminar or turbulent if the fluid is water, and if it is oil. Determine the entrance length to fully developed flow if it is oil. Take nw = 0.898110-6 2 m2 >s and no = 0.0353110-3 2 m2 >s.

9

SOLUTION

Fluid Description. Beyond the entrance length, we consider the flow as steady. The water and oil are both assumed to be incompressible.

150 mm

Analysis. The flow is classified on the basis of the Reynolds number. The mean velocity of the flow is V =

0.0062 m3 >s Q = 0.3508 m>s = A p(0.075 m)2

Water. Here the Reynolds number is Re =

Fig. 9–17

(0.3508 m>s)(0.15 m) VD = 58.6(103) 7 2300 = nw 0.898(10 -6) m2 >s

Ans.

The flow is turbulent. By comparison, note that if Re = 2300, then the average velocity for laminar flow would have to be Re =

VD = 2300; nw

2300 =

V(0.15 m)

0.898(10 -6) m2 >s

V = 0.0138 m>s This is indeed a very small value, and so practically speaking, due mainly to its relatively low viscosity, the flow of water through a pipe will almost always be turbulent. Oil.

In this case, Re =

(0.3508 m>s)(0.15 m) VD = 1491 6 2300 = no 0.0353(10 -3) m2 >s

Ans.

Here laminar flow prevails, although it is not fully developed within the region near the entrance. Applying Eq. 9–29, the transitional length for fully developed laminar flow of the oil is L′ = 0.06 (Re) D = 0.06(1491)(0.15 m) = 13.4 m

Ans.

This is a rather long distance, but once it occurs, laminar flow is well understood, and is defined by a balance of pressure and viscous forces, as we discussed in Sec. 9.3.

9.7

9.7

laminar and turbulent shear stress within a smooth pipe

499

LAMINAR AND TURBULENT SHEAR STRESS WITHIN A SMOOTH PIPE

Pipes of circular cross section are by far the most common conduit for a fluid, and for any design or analysis it is important to be able to understand how shear stress, or frictional resistance, develops within the pipe for both laminar and turbulent flow. Here we will assume the pipe walls are smooth, and study the viscous resistance developed within the fluid. In the next chapter we will take into account the resistance caused by the roughness of the pipe wall.

Laminar Flow. In Sec. 9.3 we obtained the velocity profile for steady laminar flow through a straight pipe, assuming the fluid is viscous, Fig. 9–18. This parabolic profile requires that the fluid surrounding the wall have zero velocity, since fluid particles tend to adhere (or stick) to the wall. Layers of fluid a greater distance from the wall have greater velocities, with the maximum velocity occurring at the centerline of the pipe. As discussed in Sec. 1.7 the viscous shear stress, or frictional resistance within the fluid, is caused by the continuous exchange of momentum among the molecules of fluid, as each cylindrical layer slides over an adjacent layer.

Turbulent Flow. As the rate of flow within the pipe is increased, the laminar fluid layers will become unstable and begin to break down as the flow transitions to turbulent flow. As this occurs, the fluid particles form eddy currents or small vortices within the flow, and thereby mix the fluid throughout the pipe. These effects cause a larger loss of energy, and therefore a greater drop in pressure, compared to laminar flow.

Velocity profile Laminar flow

Fig. 9–18

9

500

Chapter 9

VisCous Flow

within

enClosed Conduits

u Fluctuating horizontal velocity component u9 Mean horizontal velocity component u

u

9

Actual velocity u 1 u9

t Horizontal velocity components of fluid particles passing through a control volume (a)

Actual velocity

Mean velocity

Instantaneous velocity profile Turbulent flow (c) Viscous sublayer near pipe wall

Velocity profile Turbulent flow (d)

Fig. 9–19

(b)

To study the effects of turbulent flow, we will consider a small, fixed control volume located at a point within the pipe, Fig. 9–19a.The velocities of all fluid particles passing through this control volume will have a random pattern; however, these velocities can be resolved horizontally into a mean velocity, u, and a random horizontal fluctuating velocity about the mean, u′, as shown on the graph in Fig. 9–19b. The fluctuations will have a very short period, and their magnitude will be small compared to the mean velocity. If this mean velocity component remains constant, then we can classify the flow as steady turbulent flow, or, more properly, mean steady flow. The “turbulent mixing” of the fluid tends to flatten the mean horizontal velocity component u within a large region around the center of the pipe, and as a result, the velocity profile will be more uniform than laminar flow. The actual velocity may have a “wiggly” profile like that shown in Fig. 9–19c, but it will average out to that shown by the dark line. Although turbulent mixing occurs readily within the central region of the pipe, it tends to diminish rapidly near the pipe’s inner wall to satisfy the boundary condition of zero velocity at the wall. This region of low velocity produces a laminar viscous sublayer near the wall, Fig. 9–19d. The faster the flow, the larger the uniform region of turbulence within the center of the pipe, and the thinner this sublayer. It is the very large velocity gradient, du>dy, near the wall within the viscous sublayer that produces a much larger shear stress on the wall compared to the lower velocity gradient for laminar flow, Fig. 9–18.

Turbulent Shear Stress. The flow characteristics of the fluid are greatly affected by turbulence. The fluid “particles” in turbulent flow are far greater in size than just the “molecules” being transferred between layers in laminar flow. For both laminar and turbulent flow, however, the same phenomenon occurs, that is, the slower-moving particles migrate to faster-moving layers, and so they tend to decrease the momentum of the faster layers. Particles migrating from faster- to slower-moving layers will have the opposite effect; they increase the momentum of the slower layer. For turbulent flow, this type of momentum transfer gives rise to an apparent shear stress that will be many times greater than the viscous shear stress that is being created by molecular exchange within the flow.

9.7

501

laminar and turbulent shear stress within a smooth pipe

u = u + u9

9 Velocity profile for turbulent flow (a)

Fig. 9–20

To show conceptually how apparent shear stress develops, consider the steady turbulent flow along two adjacent fluid layers in Fig. 9–20a. At any instant, the velocity components in the direction of flow can be written in terms of their time-averaged, u, and mean horizontal fluctuating, u′, components, as shown in Fig. 9–20a. That is, for any horizontal velocity,

y

dA u9 v9

u = u + u′ Due to turbulent fluid mixing, there will also be vertical components of velocity that fluctuate, but no mean velocity component occurs, since there is no mean flow in this direction. Thus

x (b)

v = v′ Now let’s consider the movement v′ of a group of fluid particles that pass through a small rectangular element of area dA, Fig. 9–20b. The mass flow # through the area is therefore dm = rv′dA. Since u′ is the difference in the horizontal velocity between the top and bottom of the area, or the velocity fluctuation, then the change in the horizontal momentum of the mass flow is u′(rv′dA). This is the result of a shear force dF, and since the shear stress is t = dF>dA, the apparent turbulent shear stress at this location is then tturb = ru′v′ where u′v′ is the mean product of u′v′. This apparent turbulent shear stress, as described here, is sometimes referred to as the Reynolds stress, named after Osborne Reynolds, who developed these arguments in 1886. To summarize, the shear stress within turbulent flow therefore consists of two components. The viscous shear stress is due to molecular exchange, tvisc = m du>dy, which results from the time-averaged velocity component u, and the apparent turbulent shear stress tturb, which is based upon the much larger eddy particle exchange between fluid layers. It is the result of the mean horizontal fluctuating component u′. Therefore, we can write

t

t = tvisc + tturb As noted from the previous development of Eq. 9–16, regardless of the type of flow, t will always have a linear variation throughout the flow, as shown in Fig. 9–10 or Fig. 9–20c.

Shear-stress distribution (c)

Fig. 9–20 (cont.)

502

9

Chapter 9

VisCous Flow

within

enClosed Conduits

In practice it is difficult to obtain the apparent or turbulent shear-stress component since the vertical and horizontal fluctuating components v′ and u′ will be different for each location and at each moment within the flow. In spite of this, empirical formulations for this stress, based on Reynolds’s work, were later developed by the French mathematician Joseph Boussinesq, using a concept called the eddy viscosity of flow. This was followed by the work of Ludwig Prandtl, who created a mixing-length hypothesis based on the size of the eddies formed within the flow. Although both of these efforts provide some understanding of the notion of turbulent shear stress, and its relation to velocity, they have very limited application and today are no longer used. Turbulent flow is very complex, due to the erratic motion of the particles, and this has made it practically impossible to obtain a single accurate mathematical formulation to describe its behavior. Instead, this difficulty has led to many experimental investigations, and from this, engineers have produced many different models to predict turbulent behavior, Ref. [3], [4]. Some of these have been incorporated into sophisticated computer programs used for computational fluid dynamics (CFD), as we have discussed in Sec. 7.13.

9.8

STEADY TURBULENT FLOW WITHIN A SMOOTH PIPE

Careful measurements of the velocity profile for turbulent flow within a pipe have made it possible to identify three different regions of flow within the pipe. These are shown in Fig. 9–21a, and are referred to as the viscous sublayer, the transitional region, and the turbulent flow region. Viscous sublayer Transitional region

Turbulent flow region (a) Actual velocity u 5 u 1 u9 Turbulent flow Transitional flow Laminar flow Mean velocity u (solid line)

(b) Fig. 9–21

9.8

503

steady turbulent Flow within a smooth pipe

Viscous Sublayer. For almost all fluids, the particles at the wall of the pipe have zero velocity, no matter how great the flow is through the pipe. These particles “stick” to the wall, and the fluid layers near them exhibit laminar flow because of their slow velocity. Consequently, viscous shear stress within the fluid dominates within this region, and so if the fluid is a Newtonian fluid, the shear stress can be expressed by tvisc = m(du>dy). If we integrate this equation, using the boundary condition u = 0 at y = 0, we can then relate the wall shear stress t0 (a constant) to the velocity. Since for laminar steady flow u = u, which is the time-averaged or mean velocity, indicated by the solid curve in Fig. 9–21b, we get t0 = m

u y

9

(9–30)

We can express this result as a dimensionless ratio in order to compare it with experimental results that are normally plotted in terms of “dimensionless” variables. To do this, researchers have used the factor u* = 2t0 >r. This constant has units of velocity, and so it is sometimes referred to as the friction velocity or shear velocity. If we divide both sides of the above equation by r, and use the kinematic viscosity, n = m>r, we get u*y u = n u*

(9–31)

Since u* and n are constants, then u is linearly related to y. It represents the dimensionless velocity profile within the viscous sublayer, Fig. 9–22a. Equation 9–31 is sometimes referred to as the law of the wall. On a semi-logarithmic graph it plots as a curve, Fig. 9–22b, and as noted, it fits the experimental data, originally obtained mainly by Johann Nikuradse in the 1930’s, for values of 0 … u*y>n … 5. See Ref. [6]. 30 u* y n Pipe centerline

u* y u u* 5 2.5 ln ( n ) 1 5.0

20 u u*

u* y u u* 5 2.5 ln ( n ) 1 5.0

Turbulent flow

u* y u u* 5 n

10

Actual velocity profile

Viscous sublayer

Assumed velocity profile

u* y u u* 5 n

Dimensionless velocity profile

u u*

0

1

2

5

10

2

Viscous Transitional region sublayer

(a)

5

102 u* y n (b)

Fig. 9–22

2

5

103

Turbulent flow

2

5

104

504

Chapter 9

VisCous Flow

within

enClosed Conduits

Transitional and Turbulent Flow Regions. Within these two regions, the flow is subjected to both viscous shear stress and the much larger turbulent shear stress, and so here we can express the resultant shear stress as t = tvisc + tturb = m

9

du +  ru′v′  dy

(9–32)

Recall from the discussion in the previous section, the turbulent (or Reynolds) shear stress results from the momentum exchange of large groups of particles between fluid layers. Of the two components in Eq. 9–32, the turbulent shear stress will predominate within the center of the pipe, but its effect will diminish rapidly as the flow nears the wall and enters the transitional region, where the velocity suddenly drops off, Fig. 9–21a. A velocity profile for turbulent flow has been established using a theory developed by Theodore von Kármán and Ludwig Prandtl, and experimental data again primarily found by Johann Nikuradse. The result is u*y u = 2.5 ln a b + 5.0 n u* Table 9–1 n

Re

6

4 (103)

7

1 (105)

9

1 (106)

10

3 (106)

When plotted, this results in the curve shown in Fig. 9–22a, although on a semi-logarithmic scale, it results in a straight line, Fig. 9–22b. Notice the scale in Fig. 9–22b. The viscous sublayer and the dashed transition zone extend only a very small distance, u*y>n … 30, whereas the turbulent flow region extends to u*y>n = 104. For this reason, and for most engineering applications, flow within the viscous sublayer and transition zone can be neglected. Instead, Eq. 9–33 alone can be used to model the velocity profile for the flow in the pipe. Here, of course, it is assumed that the fluid is incompressible, the flow is fully turbulent and steady in the mean, and the walls of the pipe are smooth.

r R

Power Law Approximation. A turbulent velocity profile can also be modeled using an empirical power law having the form

Turbulent 1.0

n56 n57 n59 n 5 10

0.8 0.6 0.4

Laminar

0.2 0 0

0.2

0.4

0.6

0.8

Velocity profiles

Fig. 9–23

(9–33)

1.0

u umax

u umax

= a1 -

r 1>n b R

(9–34)

Here umax is the maximum velocity, which occurs at the center of the pipe, and the exponent n depends upon the Reynolds number. A few values of n for specific values of Re are listed in Table 9–1. See Ref. [4]. The corresponding velocity profiles, including the one for laminar flow, are shown in Fig. 9–23. Notice how these profiles flatten out as n gets larger. This is due to the faster flow, or higher Reynolds number. Of these profiles, n = 7 is often used for calculations and it provides adequate results for many cases.

9.8

Since the velocity profile in a pipe is axisymmetric, Fig. 9–24, we can integrate Eq. 9–34 and determine the flow for any value of n. We have Q =

LA

u dA =

= 2pR2u max

L0

umax a 1 -

R

n2 (n + 1)(2n + 1)

r

9 dr

(9–36)

IMPORTANT POIN T S • When fluid flows into a pipe from a reservoir, it will accelerate

•

r n1 ) R

(9–35)

Besides using Eq. 9–34, there have also been efforts to predict turbulent time-average flow using various “turbulent models” that include chaotic fluctuations within the flow. Research in this important area is ongoing, and hopefully these models will continue to improve through the years.

•

u 5 umax(1 2

r 1>n b (2pr)dr R

Also, since Q = V1pR2 2, the average velocity of flow is 2n2 V = u max c d (n + 1)(2n + 1)

•

505

steady turbulent Flow within a smooth pipe

for a certain distance along the pipe before its velocity profile becomes fully developed. For steady laminar flow, this transition or entrance length is a function of the Reynolds number and the pipe diameter. For steady turbulent flow, it depends on the type of entrance, and the diameter and the surface roughness of the pipe. Turbulent flow involves erratic and complex motion of fluid particles. Small eddy currents form within the flow and cause localized mixing of the fluid. It is for this reason that shear stress and the energy losses for turbulent flow are much greater than those for laminar flow. The shear stress is a combination of viscous shear stress and a much larger “apparent” turbulent shear stress, which is caused by the momentum transfer of groups of fluid particles from one layer of fluid to an adjacent layer. The mixing action within turbulent flow tends to “flatten out” the velocity profile and make it more uniform, like an ideal fluid. This profile will always have a narrow viscous sublayer (laminar flow) near the walls of the pipe. Here the fluid must move slowly due to the boundary condition of zero velocity at the wall. The faster the flow, the thinner this sublayer becomes, and the steeper the velocity gradient at the wall, which in turn produces a larger shear stress on the wall. Since turbulent flow is so erratic, an analytical solution that describes the velocity profile cannot be obtained. Instead, it is necessary to rely on experimental methods to define this shape and then match it with empirical approximations such as Eqs. 9–33 and 9–34.

Turbulent velocity profile approximation

Fig. 9–24

506

Chapter 9

EXAMPLE

VisCous Flow

within

enClosed Conduits

9.7

50 mm

9 A

B

8m

Turbulent flow occurs within the 50-mm-diameter smooth-wall pipe shown in Fig. 9–25a. If the pressure at A is 10 kPa and at B it is 8.5 kPa, determine the shear stress acting along the wall of the pipe, and also at a distance of 15 mm from its center. Use Eq. 9–33 to determine the velocity at the center of the pipe. Also, what is the thickness of the viscous sublayer? Take rw = 1000 kg>m3 and nw = 1.08110-6 2 m2 >s. SOLUTION

(a)

Fluid Description. We have mean steady turbulent flow, and we assume water is incompressible. Shear Stress. The shear stress along the wall of the pipe is not affected by the turbulence because it is within the laminar sublayer. Therefore, it can be determined using Eq. 9–16. Since the pipe is horizontal, dh>dx = 0, we have

t0 =

25 mm

2.34 Pa t

15 mm

(b)

umax

(c)

Fig. 9–25

The shear-stress distribution within the fluid varies linearly from the pipe’s center as shown in Fig. 9–25b. Recall from Sec. 9.3 that this result comes from a balance of pressure and viscous forces, and it is valid for both laminar and turbulent flow. We can determine the shear stress at r = 15 mm by proportion. It is t 2.344 Pa = ; t = 1.41 Pa Ans. 15 mm 25 mm Velocity. To use Eq. 9–33 we first must determine u*, u* = 2t0 >rw = 212.344 N>m2 2 > 11000 kg>m3 2 = 0.04841 m>s

y

25 mm

3 2 r ∆p 0.025 m (8.5 - 10)110 2 N>m a b = †a bc d † = 2.344 Pa = 2.34 Pa Ans. 2 L 2 8m

At the centerline of the pipe, Fig. 9–25c, y = 0.025 m, and so u*y u max = 2.5 ln a b + 5.0 nw u* (0.04841 m>s)(0.025 m) u max = 2.5 ln c d + 5.0 0.04841 m>s 1.08110 - 6 2 m2 >s u max = 1.09 m>s Ans. Viscous Laminar Sublayer. The viscous sublayer extends to u*y>n = 5, Fig. 9–22b. Thus,

531.08110-6 2 m2 >s4 5nw = = 0.11154110-3 2 m = 0.112 mm Ans. u* 0.04841 m>s Remember that this result applies only to a smooth-walled pipe. If the pipe has a rough surface, then there is a good chance for protuberances to pass through this very thin layer, disrupting the flow and creating added friction. We will discuss this effect in the next chapter.

y =

9.8

EXAMPLE

steady turbulent Flow within a smooth pipe

507

9.8

Kerosene flows through the 100-mm-diameter smooth pipe in Fig. 9–26a, with an average velocity of 20 m>s. Viscous friction causes a pressure drop (gradient) along the pipe to be 0.8 kPa>m. Determine the total shear stress, and the viscous and turbulent shear-stress components within the kerosene at r = 10 mm from the centerline of the pipe. Use a power-law velocity profile, and take nk = 2110-6 2 m2 >s and rk = 820 kg>m3.

20 ms SOLUTION Fluid Description. We will assume mean steady turbulent flow. Also, the kerosene can be considered incompressible. Shear Stress. The shear-stress distribution is linear, as shown in Fig. 9–26b. The maximum shear stress is caused only by viscous effects at the wall within the laminar sublayer. The magnitude of this stress is determined from the absolute value of Eq. 9–16. With dh>dx = 0, we have -800 N>m2 r ∆p 0.05 m t0 = a b = 2a ba b 2 = 20 Pa 2 L 2 1m By proportion, the shear stress at r = 10 mm is then 20 ms 20 N>m2 t = ; t = 4 Pa Ans. 10 mm 50 mm Viscous Shear-Stress Component. The viscous shear-stress component can be determined at r = 10 mm, using Newton’s law of viscosity and the power law, Eq. 9–34. To do this we must first determine the exponent n, which depends upon the Reynolds number. (20 m>s)(0.1 m) VD Re = = = 11106 2 nk 2110 - 6 2 m2 >s From Table 9–1, for this Reynolds number, n = 9. The maximum velocity umax, Fig. 9–26a, is obtained from Eq. 9–36. 2192 2 2n2 V = u max ; 20 m>s = u max c d (n + 1)(2n + 1) (9 + 1)32(9) + 14 umax = 23.46 m>s Using Eq. 9–34 for u, and mk = rnk, Newton’s law of viscosity becomes tvisc = mk 1d u>dy2 = -mk 1d u>dr2 since y = R - r, so that dy = -dr. Thus, mkumax du d r 1>n r (1 - n)>n tvisc = -mk = -mk c u max a 1 - b d = a1 - b dr dr R nR R 3 -6 2 1820 kg>m 232110 2 m >s4(23.46 m>s) 0.01 m (1 - 9)>9 = a1 b 9(0.05 m) 0.05 m = 0.1042 Pa Ans. This is a very small contribution. Turbulent Shear-Stress Component. The turbulent shear-stress component provides the majority of the shear stress. At r = 10 mm it is t = tvisc + tturb; 4 N>m2 = 0.1042 Pa + tturb tturb = 3.90 Pa Ans.

9

100 mm umax (a)

10 mm 100 mm

t

50 mm

t0 (b)

Fig. 9–26

508

Chapter 9

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enClosed Conduits

References 1. S. Yarusevych et al., “On vortex shedding from an airfoil in low-

Reynolds-number flows,” J Fluid Mechanics, Vol. 632, 2009, pp. 245–271. 2. H. Langhaar, “Steady flow in the transition length of a straight tube,”

9

J Applied Mechanics, Vol. 9, 1942, pp. 55–58. 3. J. T. Davies. Turbulent Phenomena, Academic Press, New York, NY,

1972. 4. J. Hinze, Turbulence, 2nd ed., McGraw-Hill, New York, NY, 1975. 5. F. White, Fluid Mechanics, 7th ed., McGraw-Hill, New York, NY,

2008. 6. J. Schetz et. al., Boundary Layer Analysis, 2nd ed., American Institute

of Aeronautics and Astronautics, 2011. 7. T. Leger and S. L., Celcio, “Examination of the flow near the leading

8. 9. 10. 11. 12.

edge of attached cavitation,” J Fluid Mechanics, Cambridge University Press, UK, Vol. 373, 1998, pp. 61–90. D. Peterson and J. Bronzino, Biomechanics: Principles and Applications, CRC Press, Boca Raton, FL, 2008. K. Chandran et al., Biofluid Mechanics: The Human Circulation, CRC Press, Boca Raton, FL, 2007. H. Wada, Biomechanics at Micro and Nanoscale Levels, Vol. 11, World Scientific Publishing, Singapore, 2006. L. Waite and J. Fine, Applied Biofluid Mechanics, McGraw-Hill, New York, NY, 2007. A. Draad and F. Nieuwstadt, “The Earth’s rotation and laminar pipe flow,” J Fluid Mechanics, Vol. 361, 1988, pp. 297–308.

P R OB L EMS SEC. 9.1–9.2 9–1. The 20-kg uniform plate is released and slides down the inclined plane. If it has a terminal velocity of 2 m>s along the plane, determine the thickness of the oil film under the plate. The plate has a width of 400 mm. Take ro = 900 kg>m3 and mo = 0.0685 N # s>m2.

9–2. Glue is applied to the surface of the 200-mm-wide plastic strip by pulling the strip through the container. Determine the force F that must be applied to the tape if the tape moves at 10 mm>s. Take rg = 730 kg>m3 and mg = 0.860 N # s>m2.

10 mms

40 mm

600 mm

F 40 mm 108 300 mm

Prob. 9–1

Prob. 9–2

problems 9–3. A 2-mm-wide separation forms between the window frame and the wall of a building. If the difference in pressure between the inside and the outside of the building is 4 Pa, determine the flow of air through the crack. The wall is 150 mm thick and the temperature of the air is 20°C.

509

9–5. Determine the flow of the water and the shear stress on each plate if the pressure drop from A to B is 100 Pa. The plates are fixed, and their width is 600 mm, and the temperature of the water is 20°C. 9

4 mm

A

B

0.75 m 2 mm 900 mm

Prob. 9–5

Prob. 9–3 *9–4. The water tank has a rectangular crack on its side having a width of 100 mm and an average opening of 0.1 mm. If laminar flow occurs through the crack, determine the volumetric flow of water through the crack. The water is at a temperature of T = 20°C.

9–6. Determine the average velocity of the water if the pressure drop from A to B is 100 Pa. The plates are fixed, and the temperature of the water is 20°C.

100 mm

2m 100 mm

4 mm

A

B

900 mm

Prob. 9–4

Prob. 9–6

510

9

Chapter 9

VisCous Flow

within

enClosed Conduits

9–7. The boy has a mass of 50 kg and attempts to slide down the inclined plane. If a 0.3-mm-thick oil surface develops between his shoes and the surface, determine his terminal velocity down the incline. Both of his shoes have a total contact area of 0.0165 m2. Take ro = 900 kg>m3 and mo = 0.0638 N # s>m2.

9–10. Plates A and B having a mass of 10 kg and 4 kg, respectively, are connected with a cord and placed on the inclined plane and released. If the oil under the plates has a thickness of 0.4 mm, determine the terminal velocity of the plates and the tension in the cord. Both plates have a width of 300 mm. Take ro = 920 kg>m3 and mo = 0.18 N # s>m2.

250 mm

400 mm B

A 158

2°

Prob. 9–7

Prob. 9–10

*9–8. The flat belts are moving at the speeds shown. Plot both the velocity profile within the oil film and the shearstress distribution. The pressures at A and B are atmospheric. Take mo = 0.45 N # s>m2 and ro = 920 kg>m3. 4 mm

0.6 ms A

B

0.2 ms

Prob. 9–8

9–11. A solar water heater mounted on the roof consists of two flat plates. Water enters at A and exits at B. If the gap between the plates is a = 1.5 mm, determine the maximum allowable pressure drop from A to B so that the flow remains laminar. Assume the average temperature of the water is 45°C.

9–9. The belt is moving at a constant rate of 3 mm>s. The 2-kg plate between the belt and surface is resting on a 0.5-mm-thick film of oil, whereas oil between the top of the plate and the belt is 0.8 mm thick. Determine the plate’s terminal velocity as it slides along the surface. Assume the velocity profile is linear. Take ro = 900 kg>m3 and mo = 0.0675 N # s>m2. 3 mms

A 2m a B

58

0.8 mm

0.5 mm 300 mm

Prob. 9–9

Prob. 9–11

511

problems *9–12. The plug is pin connected to the cylinder such that there is a gap between the plug and the walls of 0.2 mm. If the pressure within the oil contained in the cylinder is 4 kPa, determine the initial volumetric flow of oil up the sides of the plug. Assume the flow is similar to that between parallel plates since the gap size is very much smaller than the radius of the plug. Take ro = 880 kg>m3 and mo = 30.5(10-3) N # s>m2.

9–14. The 100-mm-diameter shaft is supported by an oillubricated bearing. If the gap within the bearing is 2 mm, determine the torque T that must be applied to the shaft, so that it rotates at a constant rate of 180 rev>min. Assume no oil leaks out, and the flow behavior is similar to that which occurs between parallel plates, since the gap size is very much smaller than the radius of the shaft. Take 9 ro = 840 kg>m3 and mo = 0.22 N # s>m2.

200 mm

180 revmin 100 mm

T

50 mm

100 mm

2 mm

Prob. 9–14

Prob. 9–12

9–13. Use the Navier–Stokes equations to show that the velocity distribution of the steady laminar flow of a fluid flowing down the inclined surface is defined by u = 3rg sin u>(2μ)4(2hy - y2), where r is the fluid density and m is its viscosity.

9–15. A fluid has laminar flow between the two parallel plates, each plate moving in the same direction, but with different velocities, as shown. Use the Navier–Stokes equations, and establish an expression that gives the shear-stress distribution and the velocity profile for the fluid. Plot these results. There is no pressure gradient between A and B.

y y

U

h

Ut

u A

a

B

u

x x

Prob. 9–13

Ub

Prob. 9–15

512

Chapter 9

VisCous Flow

within

enClosed Conduits

*9–16. When you inhale, air flows through the turbinate bones of your nasal passages as shown. Assume that for a short length of 15 mm, the flow is passing through parallel plates, the plates having a mean total width of w = 20 mm and spacing of a = 1 mm. If the lungs produce a pressure drop of ∆p = 50 Pa, and the air has a temperature of 20°C, 9 determine the power needed to inhale air.

SEC. 9.3–9.6 9–18. Lymph is a fluid that is filtered from blood and forms an important part of the immune system. Assuming it is a Newtonian fluid, determine its average velocity if it flows from an artery into a 0.8-mm-diameter precapillary sphincter at a pressure of 120 mm of mercury, then passes vertically upwards through the leg for a length of 1200 mm, and emerges at a pressure of 25 mm of mercury. Take rl = 1030 kg>m3 and ml = 0.0016 N # s>m2. 9–19. Gasoline at a temperature of 20°C flows through the 100-mm-diameter pipe. If the pressure drop that occurs along a 50-m length is 0.5 Pa, determine the flow in the pipe.

1 mm

100 mm

Prob. 9–16

Prob. 9–19 9–17. A nuclear reactor has fuel elements in the form of flat plates that allow cooling water to flow between them. The plates are spaced 1.5 mm apart. Determine the pressure drop of the water over the length of the fuel elements if the average velocity of the flow is 0.3 m>s through the plates. Each fuel element is 900 mm long. Neglect end effects in the calculation. Take rw = 965 kg>m3 and mw = 0.318(10-3) N # s>m2.

*9–20. SAE 10W-30 oil flows through the 125-mm-diameter pipe with an average velocity of 2 m>s. Determine the drop in pressure caused by viscous friction over the 10-m-long section. Take ro = 920 kg>m3 and mo = 0.2 N # s>m2. 9–21. Determine the greatest pressure drop allowed over the 10-m-long pipe caused by viscous friction, so the flow remains laminar. The 125-mm-diameter smooth pipe is transporting SAE 10W-30 oil with ro = 920 kg>m3 and mo = 0.2 N # s>m2.

0.3 m/s

900 mm

A

B

1.5 mm 150 mm

Prob. 9–17

10 m

Probs. 9–20/21

problems 9–22. Castor oil is subjected to a pressure of 550 kPa at A and to a pressure of 200 kPa at B. If the pipe has a diameter of 30 mm, determine the shear stress acting on the wall of the pipe, and the maximum velocity of the oil. Also, what is the flow Q? Take rco = 960 kg>m3 and mco = 0.985 N # s>m2.

513

*9–24. Castor oil is poured into the funnel so that the level of 200 mm is maintained. It flows through the stem at a steady rate and accumulates in the cylindrical container. Determine the time needed for the level to reach h = 50 mm. Take ro = 960 kg>m3 and mo = 0.985 N # m>s2. 9–25. Castor oil is poured into the funnel so that the level of 9 200 mm is maintained. It flows through the stem at a steady rate and accumulates in the cylindrical container. If it takes 5 seconds to fill the container to a depth of h = 80 mm, determine the viscosity of the oil. Take ro = 960 kg>m3. A

B

12 m

200 mm

308 A 300 mm

15 mm

Prob. 9–22

h

9–23. Crude oil at 20°C is ejected through the 50-mm-diameter smooth pipe. If the pressure drop from A to B is 36.5 kPa, determine the maximum velocity within the flow, and plot the shear-stress distribution within the oil.

40 mm

Probs. 9–24/25 9–26. The 100-mm-diameter horizontal pipe transports castor oil in a processing plant. If the pressure drops 100 kPa in a 10-m length of the pipe, determine the maximum velocity of the oil in the pipe and the maximum shear stress in the oil. Take ro = 960 kg>m3 and mo = 0.985 N # s>m2.

50 mm

100 mm

B

4m

10 m

Prob. 9–26 A

Prob. 9–23

9–27. The retinal arterioles supply the retina of the eye with blood flow. The inner diameter of an arteriole is 0.08 mm, and the mean velocity of flow is 28 mm>s. Determine if this flow is laminar or turbulent. Blood has a density of 1060 kg>m3 and an apparent viscosity of 0.0036 N # s>m2.

514

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*9–28. If the reading of the mercury manometer is h = 20 mm, determine the volumetric flow of SAE 10W oil through the pipe. Take r = 13 550 kg>m3, ro = 920 kg>m3, Hg and mo = 0.182 N # s>m2. 9

*9–32. Glycerin flows through a 50-mm-diameter horizontal pipe such that the pressure at A is pA = 125 kPa, and at B it is pB = 110 kPa. Determine the shear stress that the glycerin exerts on the wall of the pipe.

9–29. Determine the mass flow of SAE 10W oil through the pipe if the mercury manometer reads h = 40 mm. Take rHg = 13 550 kg>m3, ro = 920 kg>m3, and mo = 0.182 N # s>m2.

A

B 3m

Prob. 9–32

1.5 m 80 mm A

B

9–33. The 100-mm-diameter pipe carries gasoline at a temperature of 20°C. If the pressure drops 30 kPa within the 3-m length, determine the shear stress acting along the wall of the pipe.

h

Probs. 9–28/29

9–30. Crude oil is flowing vertically upward through a 50-mm-diameter pipe. If the difference in pressure between two points 3 m apart along the pipe is 26.4 kPa, determine the volumetric flow. Take ro = 880 kg>m3 and mo = 30.2110-3 2 N # s>m2.

9–31. Most blood flow in humans is laminar, and apart from pathological conditions, turbulence can occur in the descending portion of the aorta at high flow rates as when exercising. If blood has a density of 1060 kg>m3, and the diameter of the aorta is 25 mm, determine the largest average velocity blood can have before the flow becomes transitional. Assume that blood is a Newtonian fluid and has a viscosity of mb = 0.0035 N # s>m2. At this velocity, determine if turbulence occurs in an arteriole of the eye, where the diameter is 0.008 mm.

3m 100 mm

Prob. 9–33 9–34. The cylindrical tank is filled with crude oil at 20°C. When the valve at A is opened, crude oil flows out through the 20-mm-diameter pipe. If h = 8 m at t = 0, determine the time for h = 1 m. Assume that laminar flow occurs within the pipe. 1.5 m

25 mm

8m

A

Prob. 9–31

Prob. 9–34

h

problems 9–35. The cylindrical tank is to be filled with crude oil at T = 20°C using the 60-mm-diameter supply pipe. If the flow is to be laminar, determine the shortest time needed to fill the tank from h = 0 to h = 2 m. Also, what pressure gradient should be maintained to achieve this task? Air escapes through the top of the tank.

515

9–39. The man blows air through the 3-mm-diameter straw with a velocity of 3 m>s. Determine the force his lips exert on the straw to hold it in place. Assume fully developed flow occurs along the straw. The temperature of the air is 20°C. 9

2m

h 60 mm

225 mm

Prob. 9–35 *9–36. The 100-mm-diameter oil pipe is used to transport crude oil at a temperature of 20°C. If the pressure drop over the 8-m length is 500 Pa, determine the average velocity of the flow. 9–37. The 100-mm-diameter pipe is used to transport crude oil at a temperature of 20°C. What is the maximum allowable pressure drop if the flow is to be laminar over the 8-m length of pipe? 100 mm

A

B

Prob. 9–39

*9–40. Glycerin at 20°C is ejected through the 50-mmdiameter pipe. Determine the volumetric flow if the pressure drop from A to B is 60.5 kPa. Also plot the shear stress distribution within the glycerin.

8m

Probs. 9–36/37 9–38. Water flows from a beaker into a 4-mm-diameter tube with an average velocity of 0.45 m>s. Classify the flow as laminar or turbulent if the water temperature is 10°C and if it is 30°C. If the flow is laminar, then find the length of tube for fully developed flow.

50 mm B

4m

A

L

Prob. 9–38

Prob. 9–40

9–41. Determine the shear stress along the wall of a 250-mmdiameter horizontal air duct if the air is at a temperature of 30°C and the average velocity of the flow is 0.12 m>s.

516

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enClosed Conduits

9–42. Oil and kerosene are brought together through the wye. Determine if they will mix, that is, create turbulent flow, as they travel along the 60-mm-diameter pipe. Take ro = 880 kg>m3, and rk = 810 kg>m3. The mixture has a viscosity of mm = 0.024 N # m>s2. 9

9–45. The Reynolds number Re = rVDh >m for an annulus is determined using a hydraulic diameter, which is defined as Dh = 4 A>P, where A is the open cross-sectional area within the annulus and P is the wetted perimeter. Determine the Reynolds number for water at 30°C if the flow is 0.01 m3 >s. Is this flow laminar? Take ri = 40 mm and ro = 60 mm.

40 mm 0.2 ms

60 mm 0.2 ms

9–46. A Newtonian fluid has laminar flow as it passes through the annulus. Use the Navier–Stokes equations to show that the velocity profile for the flow is r o2 - r 2i 1 dp 2 r vz = c r - r o2 - a b ln d . 4m dz ln(ro >ri) ro

40 mm

Prob. 9–42 ro

9–43. As glycerin flows through a horizontal pipe, the pressure drops 40 kPa in a 50-m length. Determine the shear stress within the glycerin at a distance of 25 mm from the pipe wall. Also find the shear stress along the centerline of the pipe and the volumetric discharge.Take rg = 1260 kg>m3 and mg = 1.50 N # s>m2.

ri

Probs. 9–45/46

200 mm

SEC. 9.7–9.8

Prob. 9–43

*9–44. Glycerin is at a pressure of 15 kPa at A when it enters the vertical segment of the 100-mm-diameter pipe. Determine the discharge at B.

A

9–47. Ethyl alcohol at T = 20°C flows through the 60-mm-diameter smooth pipe. If the pressure drops from 120 kPa to 108.5 kPa along the 6-m length, determine the shear stress along the wall of the pipe and at the center of the pipe. What is the velocity of the flow along the centerline of the pipe? The flow is turbulent. Use Eq. 9–33. *9–48. Ethyl alcohol at T = 20°C flows through the 60-mm-diameter smooth pipe. If the pressure drops from 120 kPa to 108.5 kPa along the 6-m length, determine the shear stress a distance of 10 mm from the wall of the pipe. What is the thickness of the viscous sublayer?

4m 60 mm B

100 mm

B

A

6m

Prob. 9–44

Probs. 9–47/48

517

problems 9–49. SAE 10W-30 oil flows through the 30-mm-diameter smooth pipe at 0.0095 m3 >s. Determine the velocity within the oil 5 mm from the wall of the pipe. Use the power law velocity profile, Eq. 9–34, to determine the result. Take no = 0.1(10-3) m2 >s.

30 mm

A

B

*9–52. An 80-mm-diameter smooth horizontal pipe transports kerosene at 20°C. If the pressure drops 1.25 kPa in 10 m, determine the maximum velocity of the flow. What is the thickness of the viscous sublayer? Use Eq. 9–33. 9–53. SAE 10W-30 oil flows through the 30-mm-diameter smooth pipe. If the pressure at A is 200 kPa and at B it is 9 170 kPa, determine the thickness of the viscous sublayer, and find the maximum shear stress and the maximum velocity of the oil in the pipe. Use Eq. 9–33 if the flow is turbulent. Take ro = 920 kg>m3 and no = 0.1(10-3) m2 >s.

4m 30 mm

Prob. 9–49 A

B

9–50. The smooth pipe transports water at 50°C with a maximum velocity of 4 m>s. Determine the pressure drop along the 10-m length and the shear stress on the wall of the pipe. Also, what is the thickness of the viscous sublayer? Use Eq. 9–33.

200 mm

10 m

Prob. 9–50

9–51. The smooth pipe transports water at 20°C. If the pressure at A and B is 250 kPa and 238 kPa, respectively, determine the thickness of the viscous sublayer and the velocity along the centerline of the pipe.

4m

Prob. 9–53

9–54. Kerosene at 20°C flows through the 200-mm-diameter smooth pipe. Determine its velocity 40 mm from the wall of the pipe. Also, what is the corresponding shear stress? The pressure drop throughout the pipe length is 20 kPa>m. Use Eq. 9–33 if the flow is turbulent. 9–55. Determine the viscous and turbulent shear stress components within the SAE 10W-30 oil 40 mm from the wall of the smooth pipe. The volumetric flow is 0.095 m3 >s and the pressure drop throughout the pipe length is 3 kPa>m. Use the power law velocity profile, Eq. 9–34, to determine the result. Take ro = 920 kg>m3 and mo = 0.14 N # s>m2.

200 mm

A

B 3m

Prob. 9–51

80 mm

Probs. 9–54/55

518

9

Chapter 9

VisCous Flow

within

enClosed Conduits

*9–56. Water at 250°C flows through the smooth pipe at 0.141 m3 >s. If the pressure drop from A to B is 250 Pa, determine the viscous and turbulent shear stress components within the water at r = 50 mm and r = 100 mm from the centerline of the pipe. Use the power law velocity profile, Eq. 9–34.

9–57. Experimental testing of artificial grafts placed on the inner wall of the carotid artery indicates that blood flow through the artery at a given moment has a velocity profile that can be approximated by u = 8.36(1 - r>3.4)1>n mm>s, where r is in millimeters and n = 2.3 log10 Re - 4.6. If Re = 21109 2, plot the velocity profile over the artery wall, and determine the flow at this instant.

3.4 mm B

A 200 mm 6m

Prob. 9–56

Prob. 9–57

r

Chapter reView

519

CHAP TER R EV IEW If steady flow occurs between two parallel plates, or within a pipe, then regardless if the flow is laminar or turbulent, the shear stress within the fluid varies in a linear manner such that it will balance the forces of pressure, gravity, and viscosity.

tmax

9

Shear-stress distribution for both laminar and turbulent flow.

In this book, laminar flow between two parallel plates requires Re … 1400, and for pipes, Re … 2300. If this occurs, then for Newtonian fluids, Newton’s law of viscosity can be used to determine the velocity profile and the pressure drop along these conduits. When using the relevant equations, for either plates or pipes, be sure to follow the sign convention as it relates to the established coordinates.

r

umax Velocity distribution for laminar flow

When fluid flows from a large reservoir into a pipe, it will accelerate a certain distance before it becomes fully developed steady laminar or turbulent flow. Turbulent flow within a pipe causes additional frictional losses due to the erratic mixing of the fluid. This mixing tends to even out the mean velocity profile, making it more uniform; however, along the walls of the pipe there will always be a narrow viscous sublayer having laminar flow.

Viscous sublayer Transitional region

Turbulent flow region

The velocity profile for turbulent flow cannot be studied analytically because the internal mixing of the fluid is very unpredictable. Instead, we must use the results of experiments to develop empirical equations that describe this profile.

u* y u = 2.5 lna b + 5.0 n u* u umax

= a1 -

r 1>n b R

10

Acceptphot/Alamy Stock Photo

CHAPTER

In order to design a system of pipes it is necessary to know the friction losses within the pipe, along with any losses that occur at the connections and fittings.

ANALYSIS AND DESIGN FOR PIPE FLOW

CHAPTER OBJECTIVES ■

To discuss friction losses due to surface roughness within a pipe, and describe how to use experimental data to determine these losses.

■

To show how to analyze and design pipe systems having various fittings and connections.

■

To explain some of the methods engineers use to measure flow through a pipe.

10.1

RESISTANCE TO FLOW IN ROUGH PIPES

We will now extend our discussion of the previous chapter regarding the effect of fluid viscosity, and discuss how frictional resistance within the fluid and along the rough walls of a pipe contributes to the pressure drop within the pipe. This is important when designing any pipe system, or selecting a pump that is required to maintain a specific flow. Here we will focus on straight pipes that have a circular cross section, since this shape provides the greatest structural strength for resisting pressure, and furthermore a circular cross section will transport the largest amount of fluid with the smallest frictional resistance. 521

522

C h a p t e r 10

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design

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pipe flow

D

in

out

V L

10

Fig. 10–1

In engineering practice, any energy loss due to both fluid friction and wall roughness is often referred to as a major head loss, hL, or simply a major loss. We can determine this loss by measuring the pressure in the pipe at two locations a distance L apart, Fig. 10–1. If we apply the energy equation to a control volume extending between these two points, then for steady incompressible flow, since no shaft work is done, the pipe is horizontal, zin = zout = 0, and Vin = Vout = V, we have pin pout V out2 V in2 + + + zin + hpump = + zout + hturbine + hL g g 2g 2g pin pout V2 V2 + + + 0 + 0 = + 0 + 0 + hL g g 2g 2g ∆p pin - pout = hL = (10–1) g g This head loss creates a pressure drop over the pipe’s length L, because the pressure must do work to overcome the frictional resistance that creates this loss. For this reason, the pressure at the entrance will have to be greater than at the exit. Of course, if the fluid is an ideal fluid, then hL = 0, since frictional resistance will not occur, and the pressure at both ends would be the same.*

Laminar Flow. For laminar flow, the major head loss occurs within the fluid. It is due to the frictional resistance or shear stress developed between cylindrical layers of fluid when they slide past one another with different relative velocities. For a Newtonian fluid, this shear stress is related to the velocity gradient by Newton’s law of viscosity, t = m (du>dy). In Sec. 9.3, we were able to use this expression to relate the average velocity of flow in the pipe to the pressure gradient ∆p>L. The result is Eq. 9–24, V = (D2 >32m)(∆p>L). With it, and Eq. 10–1, we can now write the head loss in terms of the average velocity as hL =

32mVL D 2g

(10–2)

Laminar flow

Notice that the loss increases as the internal diameter of the pipe decreases, since the losses vary inversely with the square of D. This head loss is due to the viscosity of the fluid and so it is produced throughout the flow. Any mild surface roughness on the wall of the pipe will generally not affect laminar flow to any appreciable degree, and so it will have little effect on the loss. *Here we are neglecting the small difference in pressure caused by the weight of the fluid. See Fig. 5–2.

10.1

resistanCe to flow in rough pipes

523

For convenience later on, we will express Eq. 10–2 in terms of the Reynolds number, Re = rVD>m, and rearrange it in the form hL = f

L V2 D 2g

(10–3)

where

10

f =

64 Re

(10–4)

Laminar flow

This term f is called the Darcy friction factor, or here we will simply call it the friction factor.* For laminar flow, it is seen to be a function only of the Reynolds number, and it does not depend upon whether the inner surface of the pipe’s wall is smooth or rough; rather, the friction loss is produced only by the viscosity of the fluid.

Turbulent Flow. Unlike laminar flow, there is no analytical means for determining the head loss in a pipe due to turbulent flow, and so it must be determined by experiment. This is done by measuring the pressure drop over the length L, either with two pressure gages, as in Fig. 10–2a, or by using a manometer, Fig. 10–2b. Such experiments have shown that this pressure drop depends upon the pipe diameter D, the pipe length L, the fluid’s density r, viscosity m, average velocity V, and the roughness or average height e of the protuberances from the pipe’s inner surface. In order to reduce the number of experiments necessary to understand how all these variables relate to ∆p, it is necessary to do a dimensional analysis. D

in

out

V L (a) L V

D

Manometer (b)

Fig. 10–2 *Some engineers use a less popular Fanning friction factor or friction coefficient, which is defined as Cf = f>4.

524

C h a p t e r 10

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Actually, this was done in Chapter 8, Example 8–4, where it was shown that ∆p is a function g1 of three dimensionless ratios, namely, ∆p = rV 2g1 a Re,

10

L e , b D D

Further experiments have shown that it is reasonable to expect the pressure drop to be directly proportional to the length of the pipe—the longer the pipe, the greater the pressure drop, and so the ratio L>D can be factored out of the function g1, and so we have ∆p = rV 2

L e g2 a Re, b D D

Finally, using this result and applying Eq. 10–1 to determine the head loss in the pipe, realizing that g = rg, yields hL =

L V2 e g a Re, b D 2g 3 D

For convenience we have incorporated the factor 2, in order to express hL in terms of the velocity head V 2>2g. In other words, our unknown function is now g3(Re, e>D) = 2g2. The fact that for a given Re the head loss is also directly proportional to the velocity head, as established here, is something that has also been confirmed by experiment. If we compare the above equation with Eq. 10–3, letting the friction factor represent f = g3 a Re,

e b D

then we can express the head loss for turbulent flow in the same form as we did for laminar flow, that is, hL = f

L V2 D 2g

(10–5)

This important result is called the Darcy–Weisbach equation, named after Henry Darcy and Julius Weisbach, who first proposed its use in the late 19th century. It was derived by dimensional analysis, and it applies to fluids having either laminar or turbulent flow. In the case of laminar flow, the friction factor is determined from Eq. 10–4; however, for turbulent flow we must determine the friction factor relationship f = g3 1Re, e>D2 through experiment. The first attempts at doing this were made by Johann Nikuradse, and then by others, using pipes artificially roughened by uniform sand grains of a specific size so that e is well defined. Unfortunately, for practical applications, commercially available pipes do not have a uniform well-defined roughness. However, using a similar approach, Lewis Moody and Cyril Colebrook were able to extend the work of Nikuradse by performing experiments using commercially available pipes.

10.1

525

resistanCe to flow in rough pipes

Moody Diagram. Moody presented his data for f = g3(Re, e>D) in the form of a graph plotted on a log–log scale. It is often called the Moody diagram, and for convenience it is shown here as Fig. 10–3, and in greater detail on the inside back cover. To use this diagram it is necessary to know the average surface roughness e of the pipe’s inner wall. The table above the Moody diagram on the inside back cover gives some typical values of e, provided the pipe is in fairly good condition [Ref. 1, 2]. However, realize that through use, pipes can become corroded, or scale can build up on their walls, and this can significantly alter the value of e or, in extreme cases, lower the value of D. It is for this reason that engineers must exercise conservative judgment when choosing a value for e. Once e is known, then the relative roughness e>D and the Reynolds number can be used to determine the friction factor f from the Moody diagram. When using this diagram, notice that it is divided into four different regions, depending upon the Reynolds number.

10

0.1 0.09 0.08

Critical zone

Laminar flow

f

0.07

Transition zone

Fully Rough Turbulent Flow

64 Re

0.05 0.04

0.06 0.03 0.05

0.02 0.015

0.04 0.01 0.008

f

0.006 0.03 0.004

e D

0.025 0.002 0.02

0.001 0.0008 0.0006 0.0004

0.015 0.0002 0.0001 0.000 05

Smooth Pipe

0.01

0.000 001 0.000 005

0.009

0.000 01

0.008

(103)

2

3 4 5 67 9

(104)

2

3 4 5 67 9

(105)

2

Re

3 4 5 67 9

(106)

VD n

Fig. 10–3

2

3 4 5 67 9

(107)

2

3 4 5 67 9

(108)

526

C h a p t e r 10

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Laminar flow

10

(a) Turbulent eddies form

Transitional flow (b)

Laminar sublayer

P

Rough surface turbulent flow

Laminar sublayer

(c)

Fig. 10–4

Laminar Flow. Experimental evidence indicates that if laminar flow is maintained, the friction factor will be independent of the roughness of the pipe and, instead, will vary inversely with the Reynolds number in accordance with Eq. 10–4, f = 64>Re. This is to be expected, since here the Reynolds number is low, and the resistance to flow is caused only by the laminar shear stress within the fluid, Fig. 10–4a.

Critical Zone and Transitional Flow. If the flow in the pipe is increased just above the Reynolds number of Re = 2300, then the f values are uncertain (critical zone) because the flow becomes unstable. Here we have transitional flow, where the flow can switch between laminar and turbulent, or be a combination of both. When this is the case, it is important to be conservative and select a somewhat high value of f. Turbulence will begin to develop within some regions of the pipe, but along the wall, the slower moving fluid will still maintain laminar flow. This laminar sublayer will become thinner as the velocity increases, and eventually some of the rough elements on the pipe wall will pass through this sublayer, Fig. 10–4b, and so the effect of surface roughness will begin to become important. And so, the friction factor will become a function of both the Reynolds number and the relative roughness.

10.1

resistanCe to flow in rough pipes

Turbulent Flow. At very large Reynolds numbers, most of the rough elements will penetrate through the laminar sublayer, and so the friction factor then depends primarily on the size e of these elements, Fig. 10–4c. For very rough pipes, high e>D, notice that the curves of the Moody diagram tend to quickly flatten out and become horizontal. In other words, the values for f become less dependent on the Reynolds number, and instead become strongly influenced by the shear stress near the wall instead of within the fluid.

Empirical Solutions. Rather than using the Moody diagram to determine f, we can also obtain this value using an empirical formula that provides a close fit to the curves. This is particularly helpful when using a computer program or spreadsheet. The Colebrook equation is most often used for this purpose, Ref. [2]. It is

2f 1

= -2.0 loga

e>D 3.7

+

2.51 Re2f

b

(10–6)

Unfortunately, this is a transcendental equation that cannot be solved explicitly for f, and therefore it must be solved using an iterative procedure, something that can be done on a pocket calculator or personal computer. A more direct approximation would be to use the following formulation developed by S. Haaland in 1983, Ref. [5].

2f 1

= -1.8 logc a

e>D 3.7

b

1.11

+

6.9 d Re

(10–7)

This equation gives a result that is within 2% to that obtained using the Moody diagram or the Colebrook equation.*

*Another set of formulas has been developed by P. K. Swamee and A. K. Jain. See Ref. [10].

527

10

528

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Whatever method is used to determine f, keep in mind that experimental approximations were made in producing the Moody diagram, such as specifying the surface roughness of the pipe, or curvefitting the data. As a result the expected accuracy of the data from this diagram will be between 10% and 15%. Also, as stated before, the surface roughness of a pipe and its diameter will change with time due to sediment and scale deposits, or corrosion. Thus, calculations based on f have a rather limited reliability. Sufficient allowance should be made for future use, by increasing any value of f using sound judgment. Once the final diameter of a pipe is determined, a chart that lists standard diameters in millimeters should be used so that the size conforms to industry standards. This size should be slightly larger than the one that is “required.” A typical chart that lists these sizes is published by the American National Standards Institute and can be found on the Internet.

Noncircular Conduits. Throughout this discussion we have only considered pipes having a circular cross section; however, the formulations can also be applied to conduits having a noncircular cross section, such as those that are oval or rectangular. In such cases the hydraulic diameter for the conduit is normally used as the “characteristic length” when calculating the Reynolds number. This “diameter” is Dh = 4A>P, where A is the crosssectional area of the conduit and P is its perimeter. For example, for a circular pipe, Dh = 34(pD2 >4)4 >(pD) = D. Once Dh is known, then the Reynolds number, relative roughness, and the Moody diagram can be used in the usual manner. The results obtained are generally within a range of accuracy that is acceptable for engineering practice, although they are not very reliable for extremely narrow shapes such as an annulus or elongated opening; Ref. [19].

Hazen–Williams Equation. Engineers who design water pipes that are used for water distribution, irrigation or sprinkler systems sometimes use an empirical equation proposed by A. Hazen and G. Williams. Because it does not depend upon the Reynolds number, suitable accuracy is obtained when it is applied to pipes having a diameter of 50 mm or greater, and the water temperature is between 5°C and 25°C. For calculating the head loss per meter of pipe, the formula is

∆hL = 10.67Q1.85C -1.85D -4.87

(10–8)

Here Q is in m3 > s, D is in meters, and the roughness coefficient C is dimensionless. Typical values, accounting for the effects of pitting of the pipe wall, or build-up of scale over time, are C = 130 for copper pipes, C = 140 for steel pipes, and C = 150 for PVC pipe. See Ref. [19] for further information on this equation.

10.1

resistanCe to flow in rough pipes

529

IMPORTANT POIN T S • Resistance to laminar flow in rough pipes is independent of the surface roughness of the pipe, since the surface conditions will not severely disrupt the flow. Instead the friction factor is only a function of the Reynolds number, and for this case it can be determined analytically using f = 64>Re.

• For fully developed steady incompressible flow, the resistance to turbulent flow in rough pipes is characterized by a friction factor f that depends upon both the Reynolds number and the relative roughness e>D of the pipe wall. This relationship, f = g3(Re, e>D), is expressed graphically by the Moody diagram, or analytically by the empirical Colebrook equation or an alternative form such as Eq. 10–7. As shown on the Moody diagram, for very high Reynolds numbers, f depends mostly on the relative roughness of the pipe wall, and not very much on the Reynolds number.

P ROCEDUR E FOR A N A LY S I S Many problems involving the head loss within a single pipe require satisfying the conditions of three equations. • The pressure drop ∆p over a length of pipe is related to the head loss hL using the energy equation. pin pout Vout2 V in2 + + + zin + hpump = + zout + hturbine + hL g g 2g 2g

• The head loss in the pipe is related to the variables f, L, D, and V by the Darcy–Weisbach equation, hL = f

L V2 D 2g

Moody diagram, which graphically represents f = g3 1Re, e>D2, or analytically by using Eq. 10–6 or 10–7. Depending upon the problem, satisfying these three equations may be very direct, as in Examples 10–1, 10–2, and 10–3, but in cases where f and hL are unknown, then a trial-and-error procedure using the Moody diagram will be required for the solution. Problems of this sort are represented by Examples 10–4 and 10–5.

• The friction factor f is related to Re and e>D either by using the

10

530

C h a p t e r 10

EXAMPLE

a n a ly s i s

and

design

for

pipe flow

10.1 The 200-mm-diameter galvanized iron pipe in Fig. 10–5 transports water at a temperature of 20°C. Determine the head loss and pressure drop in 200 m of the pipe if the flow is Q = 90 liters>s.

10

SOLUTION Fluid Description. We will assume fully developed steady flow, and the water can be considered incompressible. From Appendix A, at T = 20°C, rw = 998.3 kg>m3 and nw = 1.00110-6 2 m2 >s. In order to classify the flow, we must calculate the Reynolds number.

200 mm

V =

200 m

Fig. 10–5

Re =

Q = A

a 90

liters 1 m3 ba b s 1000 liters p(0.1 m)2

= 2.865 m>s

(2.865 m>s)(0.2 m) VD = 5.73(105) 7 2300 (turbulent) = nw 1.00(10 -6) m2 >s

Analysis. The value of e is taken from the table on top of the Moody diagram on the inside back cover for galvanized iron pipe. The relative roughness is then e>D = 0.15 mm>200 mm = 0.00075. Using this value and Re, the Moody diagram gives f = 0.019. If instead we use the Haaland equation, Eq. 10–7, we get 2f 1

= -1.8 log c a

0.00075 1.11 6.9 d; b + 3.7 5.731105 2

f = 0.0189

which is close enough to the previous value. To be consistent, however, we will use f = 0.019. Therefore, from the Darcy–Weisbach equation, the head loss is 2 L V2 200 m (2.865 m>s) hL = f = (0.019) a bc d = 7.9477 m = 7.95 m Ans. D 2g 0.2 m 2(9.81 m>s2) This represents the loss of energy along the 200 m of pipe. The resulting pressure drop can be determined from the energy equation, or Eq. 10–1. pin pout Vout2 V in2 + + + zin + hpump = + zout + hturbine + hL g g 2g 2g pin pout V2 V2 + + 0 + 0 = + + 0 + 0 + hL gw gw 2g 2g hL =

∆p ; gw

7.9477 m =

1 998.3 kg>m3 2 1 9.81 m>s2 2 ∆p

∆p = 77.83(103) Pa = 77.8 kPa

Ans.

Due to turbulence, this pressure drop is the result of the friction loss within the water due to the roughness of the pipe wall.

10.1

EXAMPLE

resistanCe to flow in rough pipes

10.2

A heavy oil flows through 3 km of cast iron pipe having a diameter of 250 mm, Fig. 10–6. If the volumetric flow is 40 liter>s, determine the head loss in the pipe. Take no = 0.120110-3 2 m2>s.

250 mm

Fig. 10–6

SOLUTION Fluid Description. We have fully developed steady flow, and we will assume the oil is incompressible. To classify the flow, we must check the Reynolds number. V =

140 liter>s211 m3 >1000 liter2 Q = = 0.8149 m>s A p10.125 m2 2

Then Re =

10.8149 m>s210.250 m2 VD = = 1698 6 2300 1laminar2 no 0.120110-3 2 m2 >s

Analysis. Rather than use the Moody diagram to obtain f, for laminar flow we can obtain f directly from Eq. 10–4. f =

64 64 = = 0.0377 Re 1698

Thus, L V2 3000 m 10.8149 m>s2 = 10.03772 a bc d D 2g 0.250 m 219.81 m>s2 2 2

hL = f

531

= 15.3 m

Ans.

Here the head loss is a consequence of the oil’s viscosity and does not depend upon the surface roughness of the pipe.

10

532

C h a p t e r 10

EXAMPLE

a n a ly s i s

10.3 200 mm

10

and

design

for

pipe flow

The fan in Fig. 10–7 is used to force air having a temperature of 20°C through the 200-mm-diameter galvanized sheet metal duct. Determine the required power output of the fan if the length of the duct is 150 m, and the flow is to be 0.15 m3 >s. Take e = 0.15 mm.

SOLUTION 150 mm

Fig. 10–7

Fluid Description. We will assume the air is incompressible and the fan maintains fully developed steady flow. From Appendix A, for 20°C air at atmospheric pressure, ra = 1.202 kg>m3 and na = 15.1110-6 2 m2 >s. The type of flow is determined from the Reynolds number. Since 0.15 m3 >s Q 15 = V = = m>s p A p(0.1 m)2 15 a m>s b (0.2 m) p VD Re = = 6.32 1 104 2 7 2300 (turbulent) = na 15.1(10 -6) m2 >s Analysis. We can determine the shaft head of the fan by applying the energy equation between the inlet and outlet of the duct, but first we must determine the head loss along the duct. Here e>D = 0.15 mm>200 mm = 0.00075. Using this value and Re, the Moody diagram gives f = 0.0228. Therefore, from the Darcy–Weisbach equation, the head loss through the duct is 2 15 a m>s b L V2 150 m p hL = f = (0.0228) a b£ § = 19.87 m D 2g 0.2 m 2 1 9.81 m>s2 2 We will select a control volume that includes the fan, the duct, and a portion of still air on the inlet side of the duct. Then the pressure pin = pout = 0, since it is atmospheric, and Vin ≈ 0, since the air is still just before it begins to accelerate within the duct as it heads towards the fan.* The fan acts like a pump and will add energy to the air. Therefore, the energy equation becomes pin pout Vout2 V in2 + + zin + hpump = + + zout + hturbine + hL g g 2g 2g

0 + 0 + 0 + hfan = 0 +

a

2 15 m>s b p

+ 0 + 0 + 19.87 m 2(9.81 m>s2) hfan = 21.03 m Notice that, because of the high velocity, most of this head is used to overcome the frictional resistance of the air (19.87 m), and the rest (1.16 m) to generate kinetic energy. The power output of the fan is therefore # Ws = gaQhfan = (1.202 kg>m3)(9.81 m>s2)(0.15 m3 >s)(21.03 m) = 37.2 W Ans. *Of course some air flow toward the duct does occur near the opening. But lacking the use of any computational methods (CFD), we assume the air is still just before the entrance. With this in mind, other choices of the control volume can also be used, and this is discussed in the next section. Also, see Example 10–7.

10.1

EXAMPLE

resistanCe to flow in rough pipes

533

10.4

Crude oil flows through the 150-mm-diameter pipe in Fig. 10–8. Determine its maximum average velocity if the head loss is not to be greater than hL = 7.5 m in 100 m of pipe. Take no = 40.0110-6 2 m2 >s and e = 0.06 mm.

10

SOLUTION Fluid Description. We assume the oil is incompressible and has fully developed steady flow. Analysis. Here we do not know the friction factor or the average velocity. However, since the head loss is given, it can be related to the velocity using the Darcy–Weisbach equation. L V2 100 m V2 hL = f ; 7.5 m = f a bc d D 2g 0.15 m 219.81 m>s2 2 so 0.220725 V = (1) A f In order to obtain the friction factor, we will use the Moody diagram, but to do so, we need to first calculate the Reynolds number. Unfortunately, it can only be expressed in terms of velocity. V(0.15 m) VD Re = = 3750V (2) = no 40.0110-6 2 m2 >s Using a trial-and-error approach, we will assume a value for Re. Let’s take an average value, say Re = 3(104), and see what happens. From the Moody diagram, for the curve representing e>D = 0.06 mm>150 mm = 0.0004, an estimate for f would be about f = 0.0245. Thus, from Eq. 1, 0.220725 = 3.00 m>s A 0.0245 And from Eq. 2, this produces a Reynolds number of V =

Re = 3750(3.00 m>s) = 1.131104 2 With this value, the Moody diagram gives a new value of f = 0.031, and with this, using Eqs. 1 and 2, V = 2.66 m>s and Re = 1.001104 2. Now using this value of Re, from the Moody diagram f = 0.0313, which is close to the previous value of 0.031. ( … 10% difference is usually adequate.) Thus, Eq. 1 gives V = 2.66 m>s NOTE: We

Ans.

can also solve this problem by expressing Re in terms of f by eliminating V from Eqs. 1 and 2, and then substituting this result into the Colebrook equation, Eq. 10–6; or use Eq. 10–7 and then solve for f using a numerical method on a calculator.

150 mm

Fig. 10–8

534

C h a p t e r 10

EXAMPLE

a n a ly s i s

10.5

10

and

design

for

pipe flow

The cast iron pipe in Fig. 10–9 is used to transport water at 0.30 m3>s. If the head loss is to be no more than 0.006 m for every 1-m length of pipe, determine the smallest diameter D of pipe that can be used. Take nw = 1.15110-6 2 m2>s.

SOLUTION

Fluid Description. We assume water to be incompressible and we have fully developed steady flow.

Fig. 10–9

Analysis. In this problem, both the friction factor f and pipe diameter D are unknown. However, since the head loss is known, we can relate f and D using the Darcy–Weisbach equation. L V2 hL = f D 2g 0.3 m3>s 2 c d (p>4)D2 1m 0.006 m = f a b D 2(9.81 m>s2) D D5 = 1.2394 f

(1)

The Reynolds number can also be expressed in terms of the pipe’s diameter as 0.3 m3>s a bD (p>4)D2 VD Re = = nw 1.15110 - 6 2 m2>s Re =

(Prisma Bildagentur AG/Alamy Stock Photo)

3.32151105 2 D

(2)

The Moody diagram relates Re to f, but we do not know these values, and so we must use a trial-and-error approach. That is, assume a value for f, then determine D and Re from Eqs. 1 and 2, and with these values, find a value for f from the Moody diagram. Then using this value, the procedure is repeated until f is close to the previous value. We can also solve the problem using, say, the Colebrook equation. For cast iron e = 0.00026 m, and therefore, using Eqs. 1 and 2,

0.00026> 11.2394f2 1 = -2.0 loga 3.7 1f

1>5

+

3 3.3215(105)>(1.2394f )1>5 4 1f 2.51

b

This equation can be solved numerically for f. The result is f ≈ 0.01777, and then from Eq. 1, D = 0.4662 m = 466 mm Ans. Here we should select a pipe with a slightly larger diameter, based on the available manufactured size. Also, as noted from the Darcy–Weisbach equation, this larger size D will produce a slight reduction in the calculated head loss.

10.2

10.2

losses oCCurring from pipe fittings and transitions

535

LOSSES OCCURRING FROM PIPE FITTINGS AND TRANSITIONS

In the previous sections, we showed how a major head loss occurs along a straight length of pipe due to the frictional effects of viscosity and pipe roughness. In addition to this, the effects of turbulence can produce head losses at pipe connections such as bends, valves, fittings, entrances, and transitions. These are called minor losses. Realize though, in cases where a system of pipes has short lengths, this term can be a misnomer since the minor losses will often be greater than the major losses in all the straight lengths of pipe. Minor losses are the result of accelerated turbulent mixing of the fluid within the connection as the fluid passes through it. The eddies or swirls that are produced are carried downstream, where they decay and generate heat before fully developed laminar or turbulent flow is restored. Although a minor loss is not necessarily localized within the connection, for application of the energy equation, we will assume it is and will express this loss in terms of the velocity head as we did in the case for a major loss. Here we will formulate it as

hL = K L

V2 2g

(10–9)

where KL is called the resistance or loss coefficient, which is determined from experiment.* Design manuals often provide such data; however, care should be taken when selecting a loss coefficient, since the reported values can vary for a particular fitting that is manufactured from different sources. See Refs. [13] and [19]. Generally, the manufacturer’s recommendations should be considered. What follows is a partial list of values for KL for some common types of fittings encountered in practice, and we will use these values for problem solving.

*Flow coefficients are sometimes used in the valve industry to report minor losses. This is particularly true for control valves. This factor is similar to the resistance coefficient, and can be related to it by the Darcy–Weisbach equation. Further details are given in Ref. [19]. At the end of this chapter we will also discuss how discharge coefficients are used to represent losses incurred in various types of nozzles and flow meters.

10

536

C h a p t e r 10

a n a ly s i s

and

design

for

pipe flow

Inlet and Exit Transitions. When fluid enters a pipe from a reservoir, it will cause a minor loss that depends upon the type of transition that is used. Well-rounded transitions, as in Fig. 10–10a, will cause the smallest loss, since they provide a gradual change in flow. The value of KL depends upon the radius r of the transition; however, as noted in the figure, if r>D Ú 0.15 then KL = 0.04. Entrance transitions that produce greater losses may have a flush entrance, KL = 0.5, Fig. 10–10b, or have a re-entrant pipe, KL = 1.0, Fig. 10–10c. These situations can cause the fluid to separate from the wall of the pipe and form a vena contracta or “necking” of the fluid near the entrance, because the fluid streamlines are not able to suddenly bend 90° around a sharp corner. Because the vena contracta constricts the flow, it will cause an increase in velocity near the entrance, which in turn lowers the pressure and creates flow separation, producing localized eddies at these locations, as shown. At the discharge end of a pipe into a large reservoir, the loss coefficient is KL = 1.0, regardless of the shape of the transition, Fig. 10–10d. Here the kinetic energy of the fluid is converted into thermal energy as the fluid exits the pipe and eventually comes to rest within the reservoir. Care must be taken when the energy equation is applied to a pipe system that involves a transition. If a large reservoir supplies fluid to an inlet, then we can assume the fluid has zero velocity at the inlet, and it begins to accelerate and create the loss within the pipe. By comparison, the loss created by a discharge pipe does not occur within the pipe, rather it occurs within the reservoir. See Example 10–7 for further clarification of this point.

10

r D

D

r $ –– D 0.15 KL 5 0.04 Well-rounded entrance (a)

KL 5 0.5 Flush entrance (b)

r r$0 KL 5 1.0 Discharge pipe (d)

Fig. 10–10

L. D 2

KL 5 1.0 Re-entrant pipe (c)

D

10.2

537

losses oCCurring from pipe fittings and transitions

Vena contracta

V1 D1

10

D2

A 0.6 d2

0.4

V1

KL 0.2 B

D1

D2 2

2

D KL 5 1 2 —12 D2 Sudden expansion (a)

0

0

0.2

0.4

0.6

0.8

1.0

2 2 D1D2

Sudden contraction V1

u

(b) D1 D2

Expansion and Contraction. Connections between pipes of

different diameters are subjected to minor losses, and the loss coefficient KL for each case applies to the velocity head (V 12 >2g) for the pipe having the smaller diameter D1 6 D2. Here we will consider three types.

D u ( ––2 5 4) D1

KL

108 208 308

0.13 0.40 0.80

Sudden Expansion. The minor loss for a sudden expansion, Fig. 10–11a, is only slightly affected by the stagnation that occurs within corners A and B of the larger pipe. Mainly it is the result of the dissipation of the fluid’s kinetic energy as it enters the larger pipe. As a result, the loss can be determined by applying the continuity, energy, and momentum equations between points upstream and downstream of the connection. The resulting equation for KL is shown in Fig. 10–11a. It has been shown to be in close agreement with experiment.

Gradual Expansion (c)

Sudden Contraction. When the fitting is a sudden contraction, a vena contracta will form within the smaller-diameter pipe as shown in Fig. 10–11b. Since this formation is not well defined because it depends upon the diameter ratios, then the loss coefficient has to be determined experimentally. A few results are indicated by the graph in Fig. 10–11b. See Ref. [3].

Gradual Expansion. If the change in flow is gradual, as in the case of a conical diffuser, Fig. 10–11c, then for rather small angles, u 6 8°, losses can be significantly reduced. Higher values of u will produce flow separation and the formation of eddies in addition to friction losses along the wall. For this case, values of KL may then be larger than those for a sudden expansion. See Ref. [17]. Some typical values of KL for this fitting are given in Fig. 10–11c. Gradual Contraction. When the flow passes through a gradual contraction, as in the case of a nozzle, then very little loss occurs since the flow is well defined, with no separation and very little eddy formation. A few experimental results are shown in Fig. 10–11d. See Ref. [21].

V1

u

D1 D2 D u ( ––2 5 4) D1

KL

108 208 308

0.02 0.04 0.04

Gradual Contraction (d)

Fig. 10–11

538

C h a p t e r 10

a n a ly s i s

and

design

for

pipe flow

Pipe Connections. Consideration should also be given to the possibility of head losses at pipe-to-pipe connections. For example, smalldiameter pipes are generally threaded, so if any burrs remain on cut sections, then they can disturb the flow and cause additional losses through the connection. Likewise, larger-diameter pipes are welded, flanged, or glued together, and these joints can also produce further head loss unless they are properly fabricated and connected.

10

Bends. As the fluid passes through a bend as shown in Fig. 10–12a, its particles will be subjected to normal or radial acceleration along their curved streamline because the velocities of the fluid particles are changing direction. As a result, the centrifugal force m(V 2 >r) that is created within the flow will be largest at the center of the pipe because the velocity at the center is largest. By comparison, near the walls this force is small because of the no-slip condition. This imbalance in the forces within the fluid will push the center portions of the fluid up towards the upper wall, and this will cause a flow back and around towards the lower wall. This swirling motion, referred to as secondary flow, combines with the flow along the pipe and creates two spiral motions within the flow. This and the possible flow separation from the inner wall produces a large viscous friction loss that eventually dissipates once the fluid moves farther down the straight section of pipe. To avoid these effects, a larger-radius bend or “long sweep” can be used, or guide vanes can be placed in sharp bends of larger pipes in order to reduce the swirl and head loss, Fig. 10–12b. See Table 10–1 for the loss coefficients of some pipe bends.

Outer wall Secondary flow Inner wall

Guide vanes on 908 bend

908 bend (a)

(b)

Fig. 10–12

10.2

losses oCCurring from pipe fittings and transitions

Valves. There are many types of valves used to control the flow of fluids. In particular the gate valve works by blocking the flow with a “gate” or plate that is perpendicular to the flow, as shown in Fig. 10–13a. Gate valves are mainly used to either permit or prevent the flow of liquids, and as result they are either fully open or fully closed. In the open position they allow little or no obstruction to the flow, and so they have a very low resistance as indicated by the loss coefficient in Table 10–1. The globe valve, shown in Fig. 10–13b, is designed to regulate flow. It consists of a stopper disk that goes up and down from a stationary ring seat. The name “globe” refers to the spherical shape of the outer housing, although modern designs are generally not fully spherical. Similar to the globe valve in the way it functions, the angle valve is used when there is a 90° transition in the flow, Fig. 10–13c. Two additional valves are also shown in Fig. 10–13, namely, the swing check valve, which prevents flow reversal, Fig. 10–13d, and the butterfly valve, Fig. 10–13e, which provides a low-cost, quick shut-off means of regulating the flow. For all these cases, the loss is greatly increased when the valves are partially opened, as shown in each figure, as opposed to when they are fully opened.

Gate valve partially opened

Globe valve partially opened

(a)

(b)

TABLE 10–1 Loss coefficients for pipe fittings

KL

Gate valve—fully opened

0.19

Globe valve—fully opened

10

Angle valve—fully opened Ball valve—fully opened Swing check valve

5 0.05 2

90° elbow (short radius)

0.9

90° long sweep elbow

0.6

45° bend (short radius)

0.4

180° return bend (short radius)

2

Tee for V along pipe run

0.4

Tee for V along branch

1.8

Typical butterfly valve

Angle valve partially opened

Swing check valve partially opened

Butterfly valve partially opened

(c)

(d)

(e)

Fig. 10–13

539

10

540

C h a p t e r 10

a n a ly s i s

and

design

for

pipe flow

10

Even for large-diameter pipe systems such as this, an assessment must be made of the head loss from filters, elbows, and tees. (© Fotolia)

Equivalent Length. Another way to describe the minor losses of valves and fittings is to use an equivalent-length ratio, Leq >D. This requires converting the friction loss within a fitting or valve to an equivalent length of pipe, Leq, that would produce the same “major loss” as that due to its “minor loss” coefficient KL. Since the head loss through a straight pipe is determined from the Darcy–Weisbach equation, hL = f 3Leq >D4(V 2 >2g), and the head loss through a valve or fitting is expressed as hL = KL 1V 2 >2g2, then by comparison, KL = f a

Leq D

b

Therefore, the equivalent length of pipe producing the same loss is

Leq =

K LD f

(10–10)

The overall head loss or pressure drop for the system is then calculated using the total length of pipe plus the equivalent lengths determined for each fitting.

10.3

single-pipeline flow

541

10

Minor head losses from valves, elbows, tees, and other fittings have to be considered when choosing a pump to be used with this pipe system. (© Aleksey Stemmer/Fotolia)

IM PORTANT POIN T S • The friction loss in a fluid flowing through a straight pipe is

expressed as a “major” head loss, which is determined from the Darcy–Weisbach equation, hL = f1L>D21V 2>2g2.

and connections. They are expressed as hL = KL(V 2>2g), where the loss coefficient KL is determined from tabular data found by experiment and published in design handbooks or manufacturers’ catalogs.

• “Minor” head losses occur in pipe fittings, entrances, transitions,

• Both major and minor head losses are responsible for a pressure drop along the pipe.

10.3

SINGLE-PIPELINE FLOW

Many pipelines consist of a single-diameter pipe with bends, valves, filters, and transitions, as in Fig. 10–14. These systems are often used to transport water for industrial and residential usage, and for hydroelectric power. They may also be used to transport crude oil or lubricants through mechanical equipment. The following procedure can be used to properly design such a system.

Tee

Gate valve

Filter Elbow

Fig. 10–14

542

C h a p t e r 10

a n a ly s i s

and

design

for

pipe flow

PR O C E DU R E FO R A N A LY S I S

10

Problems that involve flow through a single pipeline must satisfy both the continuity equation and the energy equation, and account for all the major and minor head losses throughout the system. For incompressible steady flow these two equations, referenced to points where the flow is “in” and “out,” are Q = VinAin = VoutAout and pin pout V out2 Vin 2 + + + zin + hpump = + zout + hturbine g g 2g 2g + f

L V2 V2 + ΣKL a b D 2g 2g

Depending upon what is known and unknown, this results in three basic types of problems. Determine the Pressure Drop. • The pressure drop for a pipe having a known length, diameter, elevation, roughness, and discharge can be determined directly using the energy equation. Determine the Flow. • When the pipe length, diameter, roughness, elevation, and pressure drop are all known, then a trial-and-error solution is necessary to determine the flow (or average velocity V), since the Reynolds number, Re = VD>n, is unknown, and therefore the friction factor cannot be directly determined from the Moody diagram. Determine the Length or Diameter of the Pipe. • The design of a pipe generally requires specifying the length of the pipe and its diameter. Either of these parameters can be found provided the other is known, along with the flow (or average velocity) and the allowed pressure drop or head loss. With the Moody diagram, the solution requires a trial-and-error procedure because, as in the previous case, the Reynolds number and friction factor must be obtained. The examples that follow illustrate application of each of these types of problems.

10.3

EXAMPLE

543

single-pipeline flow

10.6

When the globe valve at B in Fig. 10–15 is fully opened, water flows out through the 65-mm-diameter cast iron pipe with a velocity of 2 m>s. Determine the pressure in the pipe at A. Take rw = 996 kg>m3 and nw = 0.8 1 10-6 2 m2 >s.

4m

10 A

SOLUTION

Fluid Description. We assume steady incompressible flow. Analysis. The pressure drop can be determined using the energy equation, but first we must determine the major and minor head losses. For the major head loss, the friction factor is determined from the Moody diagram. For cast iron pipe, e>D = 0.26 mm>65 mm = 0.004. Also, Re =

(2 m>s)(0.065 m) VD = 1.625 1 105 2 = nw 0.8 1 10 - 6 2 m2 >s

6m B

C

Fig. 10–15

Thus, f = 0.029. Minor head loss for the elbow is 0.9(V 2>2g), and for the fully opened globe valve, it is 10(V 2>2g). Thus the total head loss is L V2 V2 V2 + 0.9a b + 10a b D 2g 2g 2g 2 (2 m>s) (2 m>s)2 10 m = 0.029a bc d + (0.9 + 10) c d 0.065 m 2 1 9.81 m>s2 2 2 1 9.81 m>s2 2

hL = f

= 0.9096 m + 2.222 m = 3.132 m

Comparing the two terms, notice that for this case the minor losses provide the largest contribution (2.222 m) to the total loss, even though this loss is referred to as “minor.” We will consider the control volume to contain the water in the pipe from A to C. Since the pipe has the same diameter throughout its length, continuity requires VA A = VC A or VA = VC = 2 m>s. With the gravitational datum through C, the energy equation becomes pA pC VC 2 VA 2 + + zA + hpump = + + zC + hturbine + hL gw gw 2g 2g

1 996 kg>m3 2 1 9.81 m>s2 2 pA

Solving yields

+

2 1 9.81 m>s2 2 (2 m>s)2

+ 6m + 0 = 0 +

pA = -28.02 1 103 2 Pa = -28.0 kPa

The result indicates that a suction in the pipe occurs at A.

2 1 9.81 m>s2 2 (2 m>s)2

Ans.

+ 0 + 0 + 3.132 m

Datum

544

C h a p t e r 10

EXAMPLE

a n a ly s i s

and

design

pipe flow

10.7

C

10

E D h53m 3.5 m

Datum

A

for

h54m

Water is to be pumped at 1.5 m>s from a reservoir into a storage tank using an 80-mm-diameter pipe that is 25 m long, Fig. 10–16. Determine the required power output of the pump if a) h = 3 m, and b) h = 4 m. Take the friction factor to be f = 0.025 and include minor losses in the pipe system. SOLUTION Fluid Description. We assume steady incompressible flow. rw = 1000 kg>m3 .

1m B

Part a) Fig. 10–16

h = 3 m. (The pipe at C is not submerged.)

Analysis I. We will consider a control volume that contains a portion of the water in the reservoir and the water in the entire pipe, Fig. 10–16, and apply the energy equation between A and C. The continuity equation requires the velocity throughout the pipe to remain constant since the pipe has a constant diameter. The minor head loss occurs from the re-entrant pipe at B, 1.01V 2 >2g2, and the four elbows, 430.91V 2 >2g24. With the datum at the level of the reservoir, we have pA pC VC 2 VA 2 + + zA + hpump = + + zC + hturbine + hL gw gw 2g 2g

0 + 0 + 0 + hpump = 0 +

(1.5 m>s)2 2

2(9.81 m>s )

+ 3.5 m + 0 + c (0.025) a

+ 1.0 + 4(0.9) d a

(1.5 m>s)2 2(9.81 m>s2)

b

25 m b 0.08 m

= 5.038 m Analysis II. Now consider the control volume to contain only the water within the pipe and a slight horizontal distance from the pipe at B, Fig. 10–16. In this case the water near the pipe entrance B is assumed to have zero velocity,* but as it accelerates into the pipe a re-entrant loss occurs. The energy equation applied between points B and C is therefore pB pC VC 2 VB 2 + + zB + hpump = + + zC + hturbine + hL gw gw 2g 2g (1 m)(1000 kg>m3)(9.81 m>s2) (1000 kg>m3)(9.81 m>s2)

+ 0 + ( -1) + hpump = 0 + + c0.025a

(1.5 m>s)2 2(9.81 m>s2)

+ 3.5 m + 0

(1.5 m>s)2 25 m b + 1.0 + 4(0.9)d c d 0.08 m 2(9.81 m>s2)

*We have made this assumption because the reservoir level remains at rest, and so if the energy equation were applied between points A and B, it would require VB = 0.

10.3

single-pipeline flow

545

Solving, we obtain the same value as before, namely, hpump = 5.038 m Applying Eq. 5–17, the power output is # Ws = Qghs = (1.5 m>s)[p(0.04 m)2][(1000 kg>m3)(9.81 m>s2)](5.038 m) 10

= 372.65 W = 373 W Part b)

Ans.

h = 4 m. (The pipe at C is submerged.)

Analysis III. Consider the control volume to contain a portion of the water in the reservoir and in the tank and the water in the entire pipe, Fig. 10–16. We will apply the energy equation between A and E, and so we must include the minor loss through the discharge at C, 1.0 (V 2 >2g).

pA pE VA 2 VE 2 + + zA + hpump = + + zE + hturbine + hL gw gw 2g 2g 11.5 m>s2 2 25 m 0 + 0 + 0 + hpump = 0 + 0 + 4 m + 0 + c 0.025a b + 1.0 + 4(0.9) + 1.0 d c d 0.08 m 219.81 m>s2)

Solving, hpump = 5.538 m Analysis IV. Now we will consider a control volume that contains only the water within the pipe, and apply the energy equation between B and C, Fig. 10–16. Here we exclude the minor loss of the discharge from the pipe at C because this loss occurs only within the storage tank. We have pB pC V C2 V B2 + + zB + hpump = + + zC + hturbine + hL gw gw 2g 2g 11 m2(1000 kg>m3 2(9.81 m>s2) + 0 + ( -1 m) + hpump (1000 kg>m3)(9.81 m>s2) (0.5 m)(1000 kg>m3)(9.81 m>s2) =

(1000 kg>m3)(9.81 m>s2) + c 0.025a

+

(1.5 m>s)2 2(9.81 m>s2)

+ 3.5 m + 0 +

(1.5 m>s)2 25 m b + 1.0 + 4(0.9) d c d 0.08 m 2(9.81 m>s2)

Solving for hpump, we obtain the same result as before. Therefore, # Ws = Qghs = (1.5 m>s)[p(0.04 m)2][(1000 kg>m3)(9.81 m>s2)](5.538 m) = 409.63 W = 410 W

Ans.

This example illustrates the importance of accounting for all the losses that occur within any selected control volume when applying the energy equation. Remember, inlet transitions produce minor losses within a pipe, and exit transitions produce losses within the reservoir.

NOTE:

546

C h a p t e r 10

EXAMPLE

10

a n a ly s i s

and

design

for

pipe flow

10.8 Determine the required diameter of the galvanized iron pipe in Fig. 10–17 if the initial discharge at C is to be 0.475 m3 >s when the gate valve at A is fully opened. The reservoir is filled with water to the depth shown. Also, what is the equivalent length of pipe that produces the same total head loss? Take nw = 1110-6 2 m2 >s. B 4m

6m A

C Datum

3m

8m

Fig. 10–17

SOLUTION Fluid Description. We will assume the reservoir is large so that VB ≈ 0, and therefore the flow will be steady. Also, the water is assumed to be incompressible. Analysis. Continuity requires the velocity through the pipe to be the same at all points since the pipe has the same diameter throughout its length. Therefore,

Q = VA;

0.475 m3 >s = V a V =

p 2 D b 4

0.6048 D2

(1)

The energy equation will be applied between B and C, with the datum through C, Fig. 10–17. The control volume for this case contains the water in the reservoir and pipe.

10.3

The major loss is determined from the Darcy–Weisbach equation. The minor losses through the pipe come from the flush entrance, 0.5(V 2 >2g), the two elbows, 2[0.9(V 2 >2g)], and the fully opened gate valve, 0.19(V 2 >2g). Thus, pB pC VC2 VB2 + + zB + hpump = + + zC + hturbine + hL gw gw 2g 2g V2 0 + 0 + 14 m + 6 m2 + 0 = 0 + + 0 + 0 + 2g 17 m V 2 V2 V2 V2 cf a b a b + 0.5a b + 2c 0.9a b d + 0.19a b d D 2g 2g 2g 2g or 17 V2 10 = c f a b + 3.49 d c (2) d D 219.81 m>s2 2

Combining Eqs. 1 and 2 by eliminating V, we obtain 536.40D5 - 3.49D - 17f = 0 (3) Now we must use a trial-and-error procedure with the Moody diagram. Rather than assuming a value of f and then solving this fifth-order equation for D, it is easier to assume a value of D, calculate f, and then verify this result using the Moody diagram. We begin by assuming D = 0.350 m; then from Eq. 3, f = 0.0939. From Eq. 1, V = 4.937 m>s, and so (4.937 m>s) 10.350 m2 VD = Re = = 1.731106 2 nw 1110 - 6 2 m2 >s

For galvanized iron pipe e>D = 0.15 mm>350 mm = 0.000429. Therefore, from the Moody diagram with these values of e>D and Re, we obtain f = 0.0165 ≠ 0.0939. On the next iteration, we choose a value of D that gives an f close to our value of f = 0.0165. Say D = 0.3 m, then from Eq. 3, f = 0.01508. Then with this value, V = 6.72 m>s, Re = 2.0211062, and e>D = 0.15 mm>300 mm = 0.0005. With these new values, f = 0.0168 from the Moody diagram, which is fairly close to the calculated value (0.01508). Therefore, we will use D = 300 mm Ans.

Equivalent Length of Pipe. Without the fittings in Fig. 10–17, the pipe has a length of 17 m. Using f = 0.01508 and applying Eq. 10–10, for the flush entrance, KL = 0.5, the two elbows, KL = 0.9, and the fully opened gate valve, KL = 0.19, to determine each of their equivalent lengths, we have 0.910.3 m2 0.510.3 m2 0.1910.3 m2 Leq = 17 m + + 2a b + 0.01508 0.01508 0.01508 Leq = 66.5 m Ans.

single-pipeline flow

547

10

548

C h a p t e r 10

a n a ly s i s

and

design

for

10.4

pipe flow

PIPE SYSTEMS

If several pipes, having different diameters and lengths, are connected together, they can form two types of pipe systems. In particular, if the pipes are connected in succession, as shown in Fig. 10–18a, the system is in series, whereas if the pipes cause the flow to divide into different loops, Fig. 10–18b, the system is in parallel. We will now give each of these cases separate treatment.

10

2

A

2

1

3

B

A

1

Pipes in series

Pipes in parallel

(a)

(b)

B

Fig. 10–18

Pipes in Series. The analysis of a pipe system in series is similar to that used to analyze a single pipe. In this case, however, to satisfy continuity, the flow through each pipe segment must be the same, so that for the three-pipe system in Fig. 10–18a, we require Q = Q1 = Q2 = Q3 Also, the total head loss for the system is equal to the sum of the major head loss along each segment of pipe, plus all the minor head losses for the system. Therefore, the energy equation between A (in) and B (out) becomes pA VA2 pB VB2 + + + zA = + zB + hL g g 2g 2g where hL = hL1 + hL2 + hL3 + hminor Compared to having a single pipe, as in the last section, the problem here is more complex, because the friction factor and the Reynolds number will be different for each pipe.

10.4

pipe systems

549

Pipes in Parallel. Although it is possible for a parallel system to have several loops, here we will consider a system having only one, as shown in Fig. 10–18b. If the problem requires finding the pressure drop between A and B and the flow in each of the pipes, then for continuity of flow, we require 10

QA = QB = Q1 + Q2 If the energy equation is applied between A (in) and B (out), then pA VA2 pB VB2 + + + zA = + zB + hL g g 2g 2g Since the fluid will always take the path of least resistance, the amount of flow through each branch pipe will automatically adjust, to maintain the same head loss or resistance to flow in each branch. Therefore, for each length we require hL1 = hL2 Using this, the analysis of a single looped system is straightforward and is based on satisfying the above equations. Of course, if a parallel system has more than one loop, then it becomes more difficult to analyze. For example, consider a network of pipes shown in Fig. 10–19. Such a system is representative of the type used for large buildings, industrial processes, or municipal water supply systems. Due to its complexity, the direction of flow and its rate within each loop may not be certain, and so an iterative analysis will be required for the solution. The most efficient method for doing this is based upon matrix algebra, using a computer. The details for applying it will not be covered here; rather, it is discussed in articles or books related to the flow in pipe networks. See, for example, Ref. [14].

P ROCE DUR E FOR A N A LY S I S • The solution of problems involving pipe systems in series or in

•

parallel follows the same procedure outlined in the previous section. In general, the flow must satisfy both the continuity equation and the energy equation, and the order in which these equations are applied depends on the type of problem that is to be solved. For pipes in series, the flow through each pipe segment must be the same, and the head loss is the total for all the pipes. For pipes in parallel, the total flow is the accumulation of flow from each branch in the system. Also, since the flow takes the path of least resistance, the head loss for each branch will be the same.

Pipe network

Fig. 10–19

550

C h a p t e r 10

EXAMPLE

10

a n a ly s i s

and

design

for

pipe flow

10.9 Pipes BC and CE in Fig. 10–20 are made of galvanized iron and have diameters of 200 mm and 100 mm, respectively. If the gate valve at F is fully opened, determine the initial discharge of water at E in liters>min. The filter at C has a KL = 0.7. Neglect the loss of the gradual contraction at D and take nw = 1.15110-6 2 m2 >s. A

3m B

C

2m

D 5m

F

1.5 m Datum

E

Fig. 10–20

SOLUTION Fluid Description. We assume steady incompressible flow occurs, where VA ≈0. Continuity Equation. If the average velocity through the largerdiameter pipe is V, and through the smaller-diameter pipe it is V′, then choosing a localized control volume of the water within the reducer at D, and applying the continuity equation, we have 0 rdV + rVf>cs # dA = 0 0t Lcv Lcs 0 - V 3 (p 0.1 m)2 4 + V′ 3 (p 0.05 m)2 4 = 0

V′ = 4V Energy Equation. Using this result, we will now apply the energy equation between A and E in order to obtain a relationship between the velocity and the friction factors. The control volume for this case contains the water throughout the reservoir and pipe system. pA VA2 pE VE2 + + zA + hpump = + + zE + hturbine + hL gw gw 2g 2g 0 + 0 + 6.5 m + 0 = 0 +

(4V)2 + 0 + 0 + hL 2g

(1)

10.4

pipe systems

The minor losses in the system come from the flush entrance at B, 0.51V 2 >2g2, the elbow, 0.91V 2 >2g2, the filter, 0.71V′2 >2g2, and the fully opened gate valve, 0.191V′2 >2g2. Using the Darcy–Weisbach equation for the major head loss in each pipe segment, and expressing the total head loss in the pipe system in terms of V, we have hL = f a

(4V)2 (4V)2 7 m V2 1.5 m (4V)2 V2 V2 b + f′a bc d + 0.5a b + 0.9a b + 0.7c d + 0.19c d 0.2 m 2g 0.1 m 2g 2g 2g 2g 2g

hL = (35f + 240f′ + 15.64) a

V2 b 2g

Here f and f′ are friction factors for the large and small diameter pipes, respectively. Substituting into Eq. 1 and simplifying, we get 127.53 = 135f + 240f′ + 31.642V 2 (2) Moody Diagram. For galvanized iron pipe, e = 0.15 mm, so e 0.15 mm = = 0.00075; D 200 mm Thus, Re =

Re′ =

e 0.15 mm = = 0.0015 D′ 100 mm

V 1 0.2 m 2 VD = 1.739 1 105 2 V = nw 1.15 1 10-6 2 m2 >s

4V 1 0.1 m 2 V′D = 3.478 1 105 2 V = nw 1.15 1 10-6 2 m2 >s

(3)

(4)

To satisfy the conditions of the Moody diagram, we will assume intermediate values for f and f′, say f = 0.0195 and f′ = 0.022. Thus, from Eqs. 2, 3, and 4, we get V = 1.842 m>s, Re = 3.201105 2, and Re′ = 6.411105 2. Using these results and checking the Moody diagram, we get f = 0.0195 and f′ = 0.022. These values are the same as those we have assumed*. Therefore, V = 1.842 m>s, and so the discharge can be determined by considering, say, the 200-mm-diameter pipe. It is Q = VA =

1 1.842 m>s 2 3p 1 0.1 m 2 2 4

Or, since there is 1000 liters>m3, then Q = a 0.05787

= 0.05787 m3 >s

m3 1000 liters 60 s ba b = 3472 liters>min ba s 1 min 1 m3

Ans.

*If they are not the same, then use the new values of f and f ′ and perform another iteration.

551

10

552

C h a p t e r 10

EXAMPLE

2

10

a n a ly s i s

and

design

for

pipe flow

10.10

50 mm 2m

1 A

100 mm 3m

Fig. 10–21

B

Water flows at a rate of 0.03 m3 >s through the branch piping system shown in Fig. 10–21. The 100-mm-diameter pipe has a filter and globe valve on it, and the 50-mm-diameter diverter pipe has a gate valve. The pipes are made of galvanized iron. Determine the flow through each pipe, and the pressure drop between A and B when both valves are fully opened. The head loss due to the filter is KL = 1.61V 2 >2g2. Take gw = 9810 N>m3 and nw = 1110-6 2 m2 >s. SOLUTION

Fluid Description. incompressible flow.

We will assume fully developed steady

Continuity Equation. If we consider the water in the tee at A as the control volume, then continuity requires Q = V1 A 1 + V2 A 2

0.03 m3 >s = V1 3p10.05 m2 2 4 + V2 3p10.025 m2 2 4 15.279 = 4V1 + V2

(1)

Moody Diagram. Branch 1 has flow through the two tees, the filter, and the fully opened globe valve. Using Table 10–1, the total head loss is 1hL 2 1 = f1 a

V12 V12 V12 V12 3m b a b + 210.42 a b + 1.6a b + 10a b 0.1 m 2g 2g 2g 2g V 12 = 130 f1 + 12.42 2g

(2)

Branch 2 has flow along the branch of two tees, the two elbows, and the fully opened gate valve. Therefore, 1hL 2 2 = f2 a

V22 V22 V22 V22 7m ba b + 211.82 a b + 210.92 a b + 0.19a b 0.05 m 2g 2g 2g 2g

= 1140 f2 + 5.592

V22 2g

(3)

We require the head losses to be the same in each branch, 1hL 2 1 = 1hL 2 2, so that 130 f1 + 12.42V12 = 1140 f2 + 5.592V22

(4)

10.4

Equations 1 and 4 contain four unknowns. In order to satisfy the conditions of the Moody diagram, we will assume intermediate values for the friction factors, say f1 = 0.02 and f2 = 0.025. Therefore, Eqs. 1 and 4 yield V1 = 2.941 m>s V2 = 3.517 m>s so that

1Re2 1 = 1Re2 2 =

12.941 m>s210.1 m2 V1D1 = 2.941105 2 = nw 1110 - 6 2 m2 >s

13.517 m>s210.05 m2 V2D2 = 1.761105 2 = nw 1110 - 6 2 m2 >s

Since 1e>D2 1 = 0.15 mm>100 mm = 0.0015 and 1e>D2 2 = 0.15 mm>50 mm = 0.003, then using the Moody diagram, we find f1 = 0.022 and f2 = 0.027. Repeating the calculations with these values, from Eqs. 1 and 4 we get V1 = 2.95 m>s and V2 = 3.48 m>s, which are very close to the previous values. Therefore, the flow through each pipe is Q1 = V1A1 = 12.95 m>s23p10.05 m2 2 4 = 0.0232 m3 >s

Ans.

Q2 = V2A2 = 13.48 m>s23p10.025 m2 4 = 0.0068 m >s 2

3

Note that Q = Q1 + Q2 = 0.03 m3>s, as required.

Ans.

Energy Equation. The pressure drop between A and B is determined from the energy equation. The control volume contains all the water in the system from A to B. With the datum through A (in) and B (out), zA = zB = 0, and VA = VB = V. We have pA VA2 pB VB2 + + zA + hpump = + + zB + hturbine + hL gw gw 2g 2g pA pB V2 V2 + + 0 + 0 = + + 0 + 0 + hL gw gw 2g 2g

or p A - p B = g wh L Using Eq. 2, we have

pA - pB = 19810 N>m3 233010.0222 + 12.44 c = 56.81103 2 Pa = 56.8 kPa

(2.95 m>s)2 2(9.81 m>s2)

d Ans.

Since 1hL 2 1 = 1hL 2 2 for this parallel system, we can also obtain this result using Eq. 3.

pipe systems

553

10

554

C h a p t e r 10

a n a ly s i s

and

design

10.5

for

pipe flow

FLOW MEASUREMENT

Through the years, many devices have been developed that measure the volumetric flow or the velocity of a fluid passing through a pipe or a closed conduit. Each device has a specific application, and a choice depends upon the required accuracy, the cost, the size of the flow, and the ease of use. In this section, we will describe some of the more common devices used for measuring the flow. Greater details can be found in the references listed at the end of the chapter, or on a specific manufacturer’s website.

10

p1

D1

Piezometer ring

D2

p2

Piezometer ring Venturi meter

Venturi Meter. The venturi meter was discussed in Sec. 5.3, and here its principles will be briefly reviewed. As shown in Fig. 10–22, this device provides a converging transition in flow from a pipe to a throat, and then a gradual diverging transition back to the pipe, in order to prevent flow separation from the walls, and thereby minimizes friction losses within the fluid. It was discussed in Sec. 5.3 that by applying the Bernoulli and continuity equations, we can obtain the mean velocity of flow at the throat using the equation

Fig. 10–22

V2 =

21p1 - p2 2 >r A 1 - 1D2 >D1 2 4

(10–11)

For accuracy, a venturi meter is often fitted with two piezometer rings, one located at the upstream side of the meter, and the other at its throat, Fig. 10–22. Each ring surrounds a series of holes in the pipe to which it is attached so that the average pressure is produced within the ring. A manometer or pressure transducer is then connected to these rings, to measure the average differential static pressure 1p1 - p2 2 between them. Since the Bernoulli equation does not account for any friction losses within the flow, in practice engineers modify the above equation by multiplying it by an experimentally determined venturi discharge coefficient, Cv. This coefficient represents the ratio of the actual average velocity in the throat to its theoretical velocity, that is, Cv =

1V2 2 act 1V2 2 theo

Specific values of Cv are generally reported by the manufacturer as a function of the Reynolds number for the pipe. Once Cv is obtained, the actual velocity within the throat is then 1V2 2 act = Cv

21p1 - p2 2 >r A 1 - 1D2 >D1 2 4

Noting that Q = V2 A2, the volumetric flow can then be determined from 21p1 - p2 2 >r p Q = Cv a D22b 4 A 1 - 1D2 >D1 2 4

10.5

555

flow measurement

Nozzle Meter. A nozzle meter works basically the same way as a venturi meter. When this device is inserted into the path of the flow, as shown in Fig. 10–23, the flow is constricted at the front of the nozzle, passes through its throat, and then leaves the nozzle without diverting the flow. This causes localized turbulence due to the acceleration of the flow through the nozzle, and then deceleration as the flow adjusts farther downstream. As a result, the friction losses through the nozzle will be greater than flow through a venturi meter. Measurements of the pressure drop from port 1 to port 2 are used to determine the theoretical velocity, V2, by applying Eq. 10–11. Here engineers use an experimentally determined nozzle discharge coefficient, Cn, to account for any friction losses. Therefore, the flow becomes Q = Cn a

21p1 - p2 2 >r p 2 D2 b 4 A 1 - 1D2 >D1 2 4

10

p1

p2

Values for Cn as a function of the upstream Reynolds number are provided by the manufacturer for various area ratios of the nozzle and pipe.

Nozzle meter

Fig. 10–23

Orifice Meter. Another way to measure the flow in a pipe is to constrict the flow with an orifice meter, Fig. 10–24. It simply consists of a flat plate with a hole in it. The pressure is measured upstream, and at the vena contracta, or narrowing, where the streamlines are still horizontal and parallel, and the static pressure is constant. As before, the Bernoulli and continuity equations, applied at these points, result in a theoretical mean velocity defined by Eq. 10–11. The actual flow through the meter is determined by using an orifice discharge coefficient, Co, supplied by the manufacturer, which accounts for both the friction losses in the flow and the effect of the vena contracta. Thus, Q = Co a

21p1 - p2 2 >r p 2 D2 b 4 A 1- 1D2 >D1 2 4

Of these three meters, the venturi meter is the most expensive, but it will give the most accurate measurement, because the losses within it are minimized. The orifice meter is the least expensive and is easy to install, but it is the most inaccurate since the size of the vena contracta is not well defined. Also, this meter subjects the flow to the largest head loss, or pressure drop. Regardless of which meter is selected, however, it is important that it be installed along a straight section of pipe that is long enough to establish a fully developed flow. In this way, the results should correlate well with those obtained by experiment.

D2

D1

D1

D1 2

p1

p2 D1

D2

Orifice meter

Fig. 10–24

556

C h a p t e r 10

a n a ly s i s

10 High flow

Low flow

and

design

for

pipe flow

Rotometer. A rotometer can be attached to a vertical pipe as shown in Fig. 10–25. The fluid flows in from the bottom, passes up a tapered glass tube, and back into a pipe after leaving the top. Within the tube there is a weighted float that is pushed upward by the flow. Since the crosssectional area of the tube becomes larger as the float rises, the velocity of the flow becomes smaller, and the float eventually reaches an equilibrium level, indicated by gradations on the tube. This level is directly related to the flow in the pipe, and so the reading at the level of the float will indicate the flow. For horizontal pipes, a similar device will cause an obstruction to compress a spring a measured distance, and its position can be viewed through a glass tube. Both of these meters can be made to measure flow to an accuracy of about 99%, but their application is somewhat limited, since they cannot be used to measure the flow of a very opaque fluid, such as crude oil. Turbine Flow Meter. For large-diameter pipes, such as from 38 mm

Rotometer

Fig. 10–25

Turbine flow meter

Fig. 10–26

to 305 mm, a turbine rotor can be installed within a section of the pipe so that the flow of fluid through the pipe will cause the rotor blades to turn, Fig. 10–26. For liquids these devices normally have only a few blades, but for gases more are required in order to generate enough torque to turn the blades. The greater the flow, the faster the blades will turn. One of the blades is marked, so as it turns, the rotation is detected by an electrical impulse, which is produced as the blade passes by a sensor. Turbine meters are often used to measure the flow of natural gas or water through municipal distribution systems. They can also be designed to be held in the hand, so that the blades can be turned into the wind, for example, to measure its speed. An anemometer, shown in the photo, works in a similar manner. It uses cups mounted on an axle, where measuring the rotation of the axle correlates to the wind speed.

Vortex Flow Meter. If a cylindrical obstruction, called a shedder

Anemometer

Vortex street

Shedder bar

bar, is placed within the flow, as shown in Fig. 10–27, then as the fluid passes around the bar, the disturbance it produces will generate a trail of vortices, called a Von Kármán vortex street.* The frequency f at which each vortex alternates off each side of the bar can be measured using a piezoelectric crystal, which produces a small electrical pulse for every fluctuation. This frequency f is proportional to the fluid velocity V and is related to it by the Strouhal number, St = f D>V, where the “characteristic length” is D, the diameter of the shedder bar. Since the Strouhal number will have a known constant value within specific operating limits of the meter, the average velocity V can be determined, i.e., V = fD>St. The flow is then Q = VA, where A is the cross-sectional area of the meter. The advantage of using a vortex flow meter is that it has no moving parts, and it has an accuracy of about 99%. One disadvantage, however, is the head loss created by the disruption of the flow.

Vortex flow meter

Fig. 10–27

*We will discuss this further in Chapter 11.

10.5

557

flow measurement

Thermal Mass Flow Meter. As the name implies, this device measures temperature to determine the velocity of a gas within a localized region of the flow. One of the most popular types is called a constant-temperature anemometer. It consists of a very small thin wire, usually made of tungsten and having, for example, a diameter of 0.5 μm and a length of 1 mm. When it is placed in the flow, Fig. 10–28, it is heated to a constant temperature, which is maintained electrically as the flow stream tends to cool it. The velocity of the flow can be correlated to the voltage that must be applied to the wire to maintain its temperature. Several of these sensors can even be arranged within a small region to measure the flow in two and three directions. Since the wire is very fragile, care must be taken so that particulate matter within the gas will not damage or break it. Higher-velocity flow, or gases having a large number of contaminants, can be measured with less sensitivity using a hot-film anemometer, which works on the same principle, but consists of a sensor made of a thin metallic film attached to a much thicker ceramic support.

Positive Displacement Flow Meter. One type of flow meter that can be used to determine the quantity of a liquid that flows past it is called a positive displacement flow meter. It consists of a measuring chamber, such as the volume between the lobes of two gears within the meter, Fig. 10–29. By ensuring close tolerances between the lobes and the casing, each revolution allows a measured amount of liquid to pass through. By counting these revolutions, either mechanically or through electrical pulses, the total amount of liquid can be measured.

10

Constant-temperature anemometer

Fig. 10–28

Nutating Disk Flow Meter. This meter is commonly used to measure the supply of household water or the amount of gasoline drawn from a pump. It has an accuracy of about 99%. As shown in Fig. 10–30, it consists of an inclined disk that isolates a measured volume of the liquid within the chamber of the meter. The pressure from the liquid forces the disk to nutate, or turn about the vertical axis, since the center of the disk is fixed to a ball and spindle. The contained volume of liquid thereby passes through the chamber for every revolution about the axis. Each of these nutations can be recorded, either by a magnetic fluctuation caused by a magnet attached to the rotating disk, or by a gear-and-axle arrangement attached to the spindle.

Positive displacement flow meter

Fig. 10–29

Rotating gear

Motion of nutating disk

Nutating disk flow meter

Fig. 10–30

558

C h a p t e r 10

a n a ly s i s

and

design

for

pipe flow

Magnetic coil

10

Magnetic field Electrodes Magnetic flow meter

Fig. 10–31

Magnetic Flow Meter. This type of flow meter requires very little maintenance and measures the average velocity of a liquid that can conduct electricity, such as seawater, wastewater, liquid sodium, and many types of acidic solutions. The principle of operation is based upon Michael Faraday’s law, which states that the voltage induced across any conductor (liquid), as it moves at right angles through a magnetic field, is proportional to the velocity of the conductor. To measure the flow, two electrodes are placed on opposite sides of the inner pipe wall and attached to a volt meter, Fig. 10–31. A magnetic field is established over the entire flow cross section, and the volt meter then measures the electric potential or voltage between the electrodes. This is directly proportional to the velocity of the flow. Magnetic flow meters can have an accuracy of 99% to 99.5%, and they have been used on pipes up to 305 mm in diameter. The readings are very sensitive to entrained air bubbles at the electrodes and to any static electricity present within the fluid and pipe. For this reason, the pipe must be properly grounded for best performance. Other Types. There are other types of meters that can also be used to accurately measure the velocity within a small region of the flow. A laser Doppler flow meter is based on directing a laser beam toward a targeted area, and measuring the change in frequency of the beam after it is reflected off small particles that pass through the region.* This data is then converted to obtain the velocity in a particular direction. This technique offers high accuracy, and though expensive, it can be set up to determine the velocity components of particles within a region in all three directions. Ultrasonic flow meters also work on the Doppler principle. They send sound waves through the fluid, and the changes in frequency of any waves that are reflected back are measured using a piezoelectric transducer and then converted to determine the velocity. Finally, particle image velocimetry (PIV) is a method where very small particles are released into the fluid. Using a camera and laser strobe light, the speed and direction of the flow can then be measured by tracking the illuminated particles. *The Doppler principle states that a higher frequency of a light or sound wave is produced when the source moves toward the observer, and a lower frequency is produced when it moves away. The effect is quite noticeable when one hears the siren on a moving police car or fire truck.

10.5

flow measurement

559

References 1. L. F. Moody, “Friction Factors for Pipe Flow,” Trans ASME, Vol. 66,

1944, pp. 671–684. 2. F. Colebrook, “Turbulent flow in pipes with particular reference to

3. 4. 5. 6. 7.

8. 9. 10.

11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.

the transition region between the smooth and rough pipe laws,” J Inst Civil Engineers, London, Vol. 11, 1939, pp. 133–156. V. Streeter, Handbook of Fluid Dynamics, McGraw-Hill, New York, NY. L. W. Mays, Hydraulic Design Handbook, McGraw Hill, New York, N.Y. S. E. Haaland, “Simple and explicit formulas for the friction-factor in turbulent pipe flow,” Trans ASME, J Fluids Engineering, Vol. 105, 1983. H. Ito, “Pressure losses in smooth pipe bends,” J Basic Engineering, 82 D, 1960, pp. 131–134. C. F. Lam and M. L Wolla, “Computer analysis of water distribution systems,” Proceedings of the ASCE, J Hydraulics Division, Vol. 98, 1972, pp. 335–344. H. S. Bean, Fluid Meters: Their Theory and Application, ASME, New York, NY, 1971. R. J. Goldstein, Fluid Mechanics Measurements, 2nd ed., Taylor and Francis, New York, NY, 1996. P. K. Swamee and A. K. Jain, “Explicit equations for pipe-flow problems,” Proceedings of the ASCSE, J Hydraulics Division, Vol. 102, May 1976, pp. 657-664. H. H. Brunn, Hot-Wire Anemometry—Principles and Signal Analysis, Oxford University Press, New York, NY, 1995. R. W. Miller, Flow Measurement Engineering Handbook, 3rd ed., McGraw-Hill, New York, NY, 1996. E. F. Brater et al. Handbook of Hydraulics, 7th ed., McGraw-Hill, New York, NY, 1996. R. W. Jeppson, Analysis of Flow in Pipe Networks, ButterworthHeinemann, Woburn, MA, 1976. R. J. S. Pigott, “Pressure Losses in Tubing, Pipe, and Fittings,” Trans. ASME, Vol. 73, 1950, pp. 679–688. Measurement of Fluid Flow on Pipes Using Orifice, Nozzle, and Venturi, ASME MFC-3M-2004. J. Vennard and R. Street, Elementary Fluid Mechanics, 5th ed., John Wiley and Sons, New York, NY, 1975. Fluid Meters, Their Theory and Application, ASME, 6th ed., New York, NY, 1971. Flow of Fluid through Valves, Fittings and Pipe, Technical Paper 410, Crane Co., Stamford, CT, 2011. R. D. Blevins, Applied Fluid Dynamics Handbook, Krieger Publishing, Malabar, FL, 2003. Bradshaw, P. T., et al., Engineering Calculation Methods for Turbulent Flow, Academic Press, New York, 1981.

10

560

C h a p t e r 10

a n a ly s i s

and

design

for

pipe flow

F UNDAMEN TAL PRO B L EM S SEC. 10.1

SEC. 10.2–10.3

F10–1. Glycerin at a temperature of 20°C drains from the tank through a 4-m-long commercial steel pipe having a 10 diameter of 40 mm. Determine the pressure gradient along the pipe and the maximum velocity of the glycerin along its centerline. Also, what is the head loss over the 4-m length? Neglect the minor loss. A

*F10–4. SAE 10W30 oil flows from one tank to another when the gate valve is fully opened. The pipe is made of commercial steel and has a diameter of 20 mm. Determine the pressure that should be applied to the tank so that the flow is 0.002 m3 >s. Include minor losses due to the flush entrance, fully opened gate valve, and two 90° elbows. Take ro = 920 kg>m3 and no = 0.1110-3 2 m2 >s.

1.5 m

A C

5m

2m

40 mm

4m

6m

B

B 1m

Prob. F10–1 F10–2. Water at a temperature of 25°C is pumped through the 100-mm-diameter commercial steel pipe over a distance of 500 m. Determine the power supplied by the pump if the flow is to be 0.03 m3 >s and the pressure drop over this length is 25 kPa. A

Prob. F10–4 F10–5. The tank is filled with water at a temperature of 20°C. When the gate valve at C is opened, determine the flow through the 80-mm-diameter cast iron pipe. Include minor losses at the flush entrance, fully opened gate valve, and 90° elbow.

B A 500 m

4m

1m

10 m C

Prob. F10–2 F10–3. Water at a temperature of 25°C flows through a horizontal cast iron pipe at 0.0314 m3 >s. Determine the smallest diameter pipe if the allowed pressure drop over a 200-m length is to be no more than 80 kPa.

10 m

2m B 200 m

Prob. F10–3

Prob. F10–5

561

problems

P ROBLEMS SEC. 10.1 10–1. A horizontal 50-m-long galvanized iron pipe having a diameter of 150 mm is used to transport water at a temperature of 10°C. Determine the pressure drop over its length if the velocity is 3 m>s. 10–2. Water at 25°C is delivered horizontally through a 150-m-long galvanized iron pipe to a sewage treatment plant. If the water is discharged into the atmosphere using a pump that develops a pressure at A of 650 kPa, determine the diameter of the pipe to the nearest mm if the discharge at B is to be 7500 liters>min.

10–7. Determine the greatest air flow Q through the galvanized steel duct so that the flow remains laminar. What is the pressure drop along a 200-m-long section of the duct 10 for this case? Take ra = 1.202 kg>m3, na = 15.1110-6 2 m2 >s.

200 mm 2m

150 mm

10–3. Gasoline at T = 20°C flows through a 250-mm-diameter smooth pipe at the rate of 0.185 m3 >s. Determine the head loss in a 20-m-long segment of the pipe. *10–4. A 75-mm-diameter pipe supplies water with a velocity of 3 m>s. If the pipe is made of commercial steel and the water temperature is 25°C, determine the friction factor. 10–5. Glycerin flows through a 200-mm-diameter, horizontal smooth pipe with an average velocity of 4 m>s. Determine the pressure drop in a 15-m-long segment of the pipe.

Prob. 10–7 *10–8. Air flows through the galvanized steel duct, with a velocity of 4 m>s. Determine the pressure drop along a 2-m length of the duct. Take ra = 1.202 kg>m3 , na = 15.1110-6 2 m2 >s.

10–6. If air at a temperature of 40°C flows through the smooth circular duct at 0.685 m3 >s, determine the pressure drop that occurs over a 10-m length of the duct. 200 mm 2m

150 mm

10 m

Prob. 10–8 10–9. Water in the old 450-mm-diameter concrete drain pipe runs full with a flow of 0.650 m3 >s. Determine the pressure drop from point A to point B. The pipe is horizontal. Take f = 0.07.

300 mm

Prob. 10–6 15 m

450 mm

A

B

Prob. 10–9

562

C h a p t e r 10

a n a ly s i s

and

design

for

10–10. The old 400-mm-diameter concrete drain pipe runs full of water with a flow of 0.5 m3 >s. Determine the pressure drop from A to B. The pipe is sloping upward at 5 m>100 m. Take f = 0.021. 10

pipe flow 10–13. Water is pumped from the river through a 40-mm-diameter hose having a length of 3 m. Determine the maximum volumetric discharge from the hose at C so that cavitation will not occur within the hose. The friction factor is f = 0.028 for the hose, and the gage vapor pressure for water is -98.7 kPa.

15 m

400 mm

A

B

B

C

Prob. 10–10 2m A

10–11. Determine the flow of methane at T = 20°C through a 1000-m-long horizontal 30-mm-diameter commercial steel pipe if the pressure drop is 150 Pa. Prob. 10–13 30 mm

1000 m

Prob. 10–11

*10–12. Water at 20°C flows upwards through the 50-mm-diameter cast iron pipe at 5.88 kg>s. Determine the major head loss that occurs over the 8-m-long vertical segment. Also, what is the pressure at A? The water is discharged into the atmosphere at B.

10–14. A cast iron pipe having a diameter of 100 mm is used to fill the cylindrical tank with water at T = 25°C. Determine the required power output of a pump necessary to fill the empty tank to a depth of 3 m in 6.5 min. The pipe has a total length of 50 m.

B

2m

B

12 m 8m

A A 50 mm

Prob. 10–12

Prob. 10–14

563

problems 10–15. A cast iron pipe having a diameter of 100 mm is used to fill the empty cylindrical tank with water at T = 25°C. If the power output of the pump is 4.5 kW, determine the depth h of the water in the tank 6 minutes after the pump is turned on. The pipe has a total length of 50 m.

10–17. The galvanized iron pipe is used to carry water at 20°C with a velocity of 3 m>s. Determine the pressure drop that occurs over a 4-m length of the pipe. 75 mm

10 B

2m

4m

Prob. 10–17

12 m

10–18. Water at 35°C is pumped from the well through a 50-mm-diameter pipe having a length of 5 m. Determine the maximum mass flow at C so that cavitation will not occur. The roughness of the pipe is e = 0.2 mm, and the gage vapor pressure for water is –95.7 kPa. C

B

A 4m

Prob. 10–15

A

*10–16. The 20-mm-diameter copper coil is used for a solar hot water heater. If water at an average temperature of T = 50°C passes through the coil at 9 liter>min, determine the major head loss that occurs within the coil. Neglect the length of each bend. Take e = 0.03 mm for the coil. Prob. 10–18 400 mm

10–19. Air at 30°C flows through the commercial steel duct at 0.650 m3 >s. Determine the pressure drop along a 12-m length of the duct.

400 mm 12 m 300 mm

Prob. 10–16

Prob. 10–19

564

C h a p t e r 10

a n a ly s i s

design

and

for

*10–20. If the pressure drop along the 12-m-long commercial steel duct is 12.5 Pa, determine the flow Q of 30°C air.

pipe flow 10–25. Water at T = 30°C flows through the 400-mm-diameter concrete pipe from the reservoir at A to the one at B. Determine the flow. The length of the concrete pipe is 100 m. The roughness of the concrete pipe is e = 0.8 mm.

10 A 15 m

400 mm B

12 m 300 mm

Prob. 10–25 Prob. 10–20

10–21. A horizontal 75-mm-diameter galvanized iron pipe, having a roughness of e = 0.2 mm, is used to transport water at a temperature of 60°C and with a velocity of 3 m>s. Determine the pressure drop over its 12-m length. 10–22. Determine the friction factor of a 40-mm-diameter horizontal tube if natural gas is to flow through it at 10 liters>min. The density of the natural gas is rg = 0.665 kg>m3 and the pressure drop is 15.5 Pa per 1000 m. 10–23. Drinking water at 20°C flows at 4 m>s through a horizontal galvanized iron pipe. If the pressure drop over a 20-m length is to be no more than 36 kPa, determine the smallest allowable diameter D of the pipe. *10–24. Water in the concrete detention pond is to be pumped over the berm and into the lake. If the 200-mm-diameter hose has a roughness of e = 0.16 mm, determine the power output of the pump so that the flow out of the pipe is 12 000 liters>min. The hose is 250 m long. The water temperature is 25°C.

10–26. When the valve at A is opened, methane at 20°C flows through the 200-mm-diameter commercial steel pipe at 0.095 m3 >s. Determine the pressure drop over the length AB of the pipe.

A

B

200 m

Prob. 10–26 10–27. The section AB of the 100-mm-diameter galvanized iron pipe has a mass of 15 kg. If glycerin is discharged from the pipe at 3 liter>s, determine the pressure at A and the force on the flange bolts at A.

A

B 20 m 3m A

B 100 mm

Prob. 10–24

Prob. 10–27

565

problems *10–28. A commercial steel pipe is required to carry glycerin at T = 20°C with a mass flow of 14.8 kg>s. If the pressure drop over its 100-m horizontal length is not to exceed 350 kPa, determine the required diameter of the pipe. 10–29. A 75-mm-diameter galvanized iron pipe, having a roughness of e = 0.2 mm, is to be used to carry water at a temperature of 60°C and with a velocity of 3 m>s. Determine the pressure drop over its 12-m length if the pipe is vertical and the flow is upward.

10–34. The sump pump is required to lift water from the pit to a reservoir. If a 40-mm-diameter commercial steel pipe is used, determine the required power supplied by the pump so that the flow is 960 liters>min. The pipe is 80 m long. The water temperature is at 30°C. 10 40 mm B

10–30. If a pipe has a diameter D and a friction factor f, by what percent will the pressure drop in the pipe increase if the volumetric flow is doubled? Assume that f is constant due to a very large Reynolds number.

12 m A

3

10–31. Sewage, assumed to be water where r = 998.3 kg>m , is pumped from the wet well using a 50-mm-diameter pipe. Determine the maximum discharge from the pump without causing cavitation. The friction factor is f = 0.026. The (gage) vapor pressure for water is - 98.7 kPa. Neglect the friction losses in the submerged segment of the pipe. *10–32. Sewage, assumed to be water where r = 998.3 kg>m3, is pumped from the wet well using a 50-mm-diameter pipe having a friction factor of f = 0.026. If the pump delivers 500 W of power to the water, determine the discharge from the pump. Neglect the friction losses in the submerged segment of the pipe. 1m 0.5 m

C

Prob. 10–34 10–35. Air at 30°C is transported by the fan at the rate of 0.75 m3 >s through the 400-mm-diameter galvanized iron duct. Determine the head loss over a 60-m length.

B 3m

A

B

A 400 mm

60 m

Prob. 10–35

Probs. 10–31/32 10–33. Water at 25°C flows through the 80-mm-diameter cast iron pipe. If the pressure at A is 500 kPa, determine the discharge.

*10–36. Determine the power the pump must supply in order to discharge 1500 liters>min of water at T = 25°C at B from the 80-mm-diameter hose. The 80-m-long hose is made of a material having a roughness of e = 0.16 mm.

B 30 m

30 m

B A 30° A

Prob. 10–33

Prob. 10–36

566

C h a p t e r 10

a n a ly s i s

and

design

for

10–37. Water is pumped into the truck at 300 liter>min through a 40-mm-diameter hose. If the total length of the hose is 8 m, the friction factor is f = 0.018, and the tank is open to the atmosphere, determine the power that must be supplied by the pump.

10 10–38. Water is pumped at 0.003 m3 >s into the top of the truck through a 40-mm-diameter hose. If the length of the hose from C to A is 10 m, and the friction factor is f = 0.018, determine the power output of the pump.

pipe flow 10–41. Water from the reservoir at A drains through the 50-mm-diameter pipe assembly. If commercial steel pipe is used, determine the initial discharge into the tank at B when the valve E is closed and F is opened. Take nw = 1.00110-6 2 m2 >s.

10–42. Water from the reservoir at A drains through the 50-mm-diameter pipe assembly. If commercial steel pipe is used, determine the initial flow into the pipe from reservoir A when both valves E and F are fully opened. The end C is open to the atmosphere. Take nw = 1.00110-6 2 m2 >s.

A

3m A B 4m 2m C

Probs. 10–37/38

8m F 3m

10–39. A 150-m-long horizontal commercial steel pipe having a diameter of 200 mm is used to transport water at T = 30°C. Determine the power output of a pump if the discharge through the pipe is to be 6000 liters>min and the pressure at the pump inlet is 25 kPa. The pipe is open to the atmosphere at its outlet.

D

B

E 4m

8m

6m C

Probs. 10–41/42

*10–40. A 150-mm-diameter galvanized iron pipe is used to transport water at T = 30°C with a velocity of 1.5 m>s. Determine the pressure drop over the 20-m length of the pipe.

B

10–43. Water at 20°C used for irrigation is to be siphoned from a canal onto a field using a pipe having a roughness of e = 0.02 mm. If the pipe is 120 m long, determine its required diameter so that it provides a flow of 0.02 m3 >s.

20 m

5m 458 A

Prob. 10–40

Prob. 10–43

567

problems *10–44. Kerosene at T = 20°C flows through the 100mm-diameter commercial steel pipe at 22.5 kg>s. Determine the pressure drop that occurs over the 50-m length.

10–47. The submersible pump is used to fill the cylindrical tank. If the level of water in the tank rises at 35 mm>s, when the height h = 1.5 m, determine the pressure developed at the exit A of the pump at this instant. The 60-mm-diameter cast iron pipe is 30 m long. The temperature of the water is 30°C.

B 800 mm

B h 50 m

25 m A

608

A

Prob. 10–44

10–45. The 50-mm-diameter pipe has a roughness of e = 0.01 mm. If the discharge of 20°C water is 0.006 m3 >s, determine the pressure at A. 10–46. The 50-mm-diameter pipe has a roughness of e = 0.01 mm. If the water has a temperature of 20°C and the pressure at A is 50 kPa, determine the discharge at B.

Prob. 10–47 *10–48. Determine the power extracted from the water at 15°C by the turbine at C if the discharge is 0.07 m3 >s. The galvanized iron pipe has a diameter of 150 mm. Also draw the energy grade line and the hydraulic grade line for the pipe. 20 m A

50 mm 10 m

B

B

30 m

2m

C A

Probs. 10–45/46

Prob. 10–48

10

568

10

C h a p t e r 10

a n a ly s i s

and

design

for

10–49. If the length of the 150-mm-diameter commercial steel pipe from A to the pump and from the pump to B is 8 m and 50 m, respectively, determine the discharge generated by the 25-kW pump. Also, draw the EGL and HGL for the pipe from A to B. The water temperature is at 20°C.

pipe flow 10–51. Water at 30°C is pumped from the large underground tank at A and flows through the 150-mm-diameter smooth pipe. If the discharge at B is 9000 liters>min, determine the required power output for the pump that is connected to a 28-m length of the pipe. Draw the energy grade line and the hydraulic grade line for the pipe with reference to the datum set through A.

3m B

B A

15

2m C

Prob. 10–49

m

30°

10 m A

10–50. The spillway from a reservoir consists of a 900-mm-diameter concrete pipe having a total length of 400 m. Determine the flow through the pipe. The roughness of the concrete is e = 1.8 mm, and the temperature of the water is at 20°C.

Prob. 10–51

SEC. 10.2–10.3

A

180 m

*10–52. Water flows at 6000 liters>min through the 150-mm-diameter pipe. As it passes through the filter the pressure drop is 2.50 kPa. Determine the loss coefficient for the filter.

900 mm

B

A

Prob. 10–50

B

Prob. 10–52

569

problems 10–53. The 100-mm-diameter commercial steel pipe discharges 0.025 m3 >s of water at 25°C through the 40-mm-diameter nozzle at B. Determine the pressure at A. The minor head loss coefficient for the nozzle is KL = 0.15.

*10–56. Air at a temperature of 40°C flows through the duct at A with a velocity of 2 m>s. Determine the change in pressure between A and B. Account for the minor loss caused by the sudden change in diameter of the duct. 300 mm

10–54. The 100-mm-diameter commercial steel pipe discharges water at 25°C through the 40-mm-diameter nozzle. If the pressure at A is 250 kPa, determine the discharge. The minor loss coefficient of the nozzle is KL = 0.15.

200 mm

10 B

A

Prob. 10–56 10–57. Water at 20°C flows through the 20-mm-diameter galvanized iron pipe such that it discharges at C from the fully opened gate valve B at 0.003 m3 >s. Determine the required pressure at A. Include the minor losses of the four elbows and gate valve.

40 mm B

3m 3m

A 1m 608 1m

A

1m C

100 mm B

Probs. 10–53/54

20 mm 1m 0.5 m

10–55. Air at a temperature of 40°C flows through the duct at A with a velocity of 15 m>s. Determine the change in pressure between A and B. Account for the minor loss caused by the sudden change in diameter of the duct.

Prob. 10–57 10–58. The pressure of air in a large tank at A is 300 kPa. Determine the flow of water at 20°C from the tank after the gate valve at B is fully opened. The 50-mm-diameter pipe is made of galvanized iron. Include the minor losses for the flush entrance, the two elbows, and the gate valve.

800 mm

B

300 mm A

2m

B

6m A 1.5 m

4m

Prob. 10–55

Prob. 10–58

570

C h a p t e r 10

a n a ly s i s

and

design

for

10–59. Water at 20°C flows from the large reservoir through the 80-mm-diameter galvanized iron pipe. Determine the discharge if the globe valve is fully opened. The length of the pipe is 50 m. Include the minor losses of the flush entrance, the four elbows, and the globe valve. 10

pipe flow 10–61. Water at 90°C enters the radiator at A with a velocity of 2 m>s and a pressure of 450 kPa. If each 180° bend has a minor loss coefficient of KL = 1.03, determine the pressure within the pipe at the exit B. The copper pipe has a diameter of 6 mm. Take e = 0.0012 mm. The radiator is in the vertical plane.

600 mm

A

A

5m 500 mm B 1m

B

Prob. 10–59 Prob. 10–61

*10–60. Water at 15°C is pumped from the reservoir A into the large tank B. If the power output of the pump is 3.70 kW, determine the volumetric flow into the tank when h = 8 m. The commercial steel pipe has a total length of 30 m and a diameter of 75 mm. Include the minor losses for the elbow and sudden expansion into the tank.

10–62. Water at 40°C flows at 2 m>s through the 20-mm-diameter pipe at A. If the 12-mm-diameter gate valve at B is fully opened, determine the pressure in the water at A. The minor loss coefficient is KL = 0.6 for the spout at B. Also account for the minor loss in the elbow, the tee, and the gate valve. The gate valve at E is closed. For the pipe take f = 0.016.

0.8 m

C B

0.15 m

2m h

E D A

B

0.08 m 1m A

Prob. 10–60

Prob. 10–62

571

problems 10–63. Water at 40°C flows at 2 m>s through the 20-mm-diameter copper pipe at A. As the water is flowing, it emerges from the showerhead that consists of 100 holes, each having a diameter of 1.5 mm. Determine the pressure of the water at A if the minor loss coefficient is KL = 0.45 for the showerhead. Also account for the minor loss in the three elbows, the fully opened gate valve at E, and the tee. The gate valve at B is closed. For the copper pipe take f = 0.016.

10–65. Water at 25°C flows through the 40-mm-diameter commercial steel pipe such that it exits the gate valve at 600 liters>min. If the open diameter of the gate valve is also 40 mm and the pressure at A is 20 kPa, determine the required power that must be supplied by the pump at B. Consider the minor losses at the two 90° elbows and fully opened gate valve. Draw the energy grade line and hydraulic grade line for 10 the pipe with reference to the datum set through A. 1m E

D

0.8 m

C 0.15 m

5m

2m E D

B C

B

A

0.08 m 1m 1.5 m

A

1.5 m

Prob. 10–65

Prob. 10–63

SEC. 10.4 *10–64. An automatic sprinkler system for a yard is made from a 20-mm-diameter PVC pipe, for which e = 0.0025 mm. If the system has the dimensions shown, determine the volumetric flow through each sprinkler head C and D. The faucet at A delivers water at 25°C with 275 kPa pressure. Neglect elevation changes, but include the minor losses at the two elbows and the tee. Also, the loss coefficient of the 6-mm-diameter sprinkler nozzles at C and D is KL = 0.05.

10–66. Water at 25°C is pumped into the galvanized iron pipe branch consisting of two 50-mm-diameter pipes ABD and ACD, which are 8 m and 12 m long, respectively. Determine the flow through each pipe if the pressure drop from A to D is 75 kPa. Neglect minor losses. The pipe branch is in the horizontal plane. 10–67. Water at 25°C is pumped into the galvanized iron pipe branch consisting of two 50-mm-diameter pipes ABD and ACD, which are 8 m and 12 m long, respectively. Determine the pressure drop from A to D if the discharge through A is 0.0215 m3 > s. The pipe branch is in the horizontal plane. Neglect minor losses.

A

D B

50 mm B 75 mm D

75 mm

12 m

18 m

A

15 m 18 m C

Prob. 10–64

C 50 mm

Probs. 10–66/67

572

C h a p t e r 10

a n a ly s i s

and

design

for

*10–68. The two water tanks are connected together using the 100-mm-diameter pipes. If the friction factor for each pipe is f = 0.024, determine the flow out of tank C when the valve at A is opened, while the valve at B remains closed. Neglect minor losses. 10 10–69. The two water tanks are connected together using the 100-mm-diameter galvanized iron pipes. Determine the flow out of tank C when both valves A and B are opened. Neglect minor losses. Take nw = 1.00110-6 2 m2 >s.

pipe flow 10–71. Two galvanized iron pipes branch to form a loop. Branch CAD is 120 m long and branch CBD is 60 m long. Determine the power of the pump used in branch CAD if an equal flow of water at 20°C occurs through each branch. All pipes have a diameter of 75 mm. The loop is in the horizontal plane. Neglect minor losses.

D

B

A C 3 ms

C

Prob. 10–71 7m

1m

D B 3m

A

E

10 m

8m

Probs. 10–68/69

10–70. Water at 20°C is pumped through the two commercial steel pipes having the lengths and diameters shown. If the pressure developed at pump outlet A is 215 kPa, determine the discharge. Neglect minor losses.

*10–72. The copper pipe system, which transports water at 60°C, consists of two branches. The pipe for branch ABC has a diameter of 10 mm and length of 3 m, whereas the pipe for branch ADC has a diameter of 20 mm and length of 10 m. If a pump provides an inlet flow at A of 300 liters>min, determine the flow through each branch. Take e = 0.025 mm. The system is in the horizontal plane. Include minor losses of the elbows and tees. The diameters at A and C are the same. 10–73. If the pressure at A is 400 kPa and at C it is 100 kPa, determine the flow through each branch of the pipe system described in Prob. 10–72. Include minor losses of the elbows and tees. The diameters at A and C are the same. 20 mm D

2.5 m

4m C 40 mm

B 4m 80 mm

C

A

A B 0.8 m

Prob. 10–70

10 mm

Probs. 10–72/73

573

Chapter review

CHAPTER R EVIEW The friction loss within a pipe can be determined analytically for laminar flow. It is defined by the friction factor f = 64>Re. For transitional and turbulent flow, we can determine f either from the Moody diagram or by using an empirical equation that fits the curves of the Moody diagram.

10

Once f is obtained, then the head loss, called a “major loss,” can be determined using the Darcy– Weisbach equation.

hL = f

L V2 D 2g

If a pipe network has fittings and transitions, then the head loss produced by these connections must be taken into account. These losses are called “minor losses.” Tee

Filter Elbow

Gate valve

hL = KL

Pipe systems can be arranged in series, in which case the flow through each pipe must be the same, and the head loss is the total loss for all the pipes.

A

2

1

Q = Q1 = Q2 = Q3 Pipes can be arranged in parallel, in which case the total flow is the accumulation of flow from each branch pipe in the system, and the head loss for each branch pipe is the same.

3

B

hL = hL1 + hL2 + hL3 + hminor

2

A

QA = QB = Q1 + Q2 Flow through a pipe can be measured in several different ways. These include using a venturi meter, nozzle meter, or orifice meter. Also, a rotometer, turbine flow meter, as well as several other meters can be used.

V2 2g

1

B

1hL 2 1 = 1hL 2 2

11

Colin Underhill/Alamy Stock Photo

CHAPTER

Airflow over the surface of this airplane creates both drag and lift. Analysis of these forces requires an experimental investigation.

VISCOUS FLOW OVER EXTERNAL SURFACES CHAPTER OBJECTIVES ■

To introduce the concept of the boundary layer and discuss its characteristics.

■

To show how to determine the shear stress or shear drag created by both laminar and turbulent boundary layers that form on a flat surface.

■

To determine the normal and drag forces on a body caused by the pressure of a fluid stream.

■

To discuss the effect of lift and flow separation on bodies of various shapes.

11.1

THE CONCEPT OF THE BOUNDARY LAYER

When a fluid flows over a surface, the layer of fluid particles adjacent to the surface will have zero velocity, but above the surface the fluid velocity will begin to increase until it reaches the free-stream velocity U as shown in Fig. 11–1. Here the flow behaves as if it is nonviscous, since it is uniform, resulting in little or no shear or sliding between adjacent layers of fluid. In 1904, Ludwig Prandtl recognized this difference in flow behavior, and termed the localized region where the velocity is variable the boundary layer. Using this concept, the flow above the boundary layer can often be treated as if the fluid were ideal, and a separate analysis is required to analyze the viscous effects of the flow within the boundary layer.

U

U

Boundary layer

Development of the boundary layer

Fig. 11–1 575

576

C h a p t e r 11

U1

VisCous Flow

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U1

U2

U2

11

Fast-moving fluids, U2 . U1 , or fluids with low viscosity, produce a thin boundary layer

Slow-moving fluids, or fluids with high viscosity, produce a thick boundary layer

Fig. 11–2

Understanding boundary layer formation is very important when it becomes necessary to determine the resistance to flow over the surface of a body. The design of propellers, wings, turbine blades, and other mechanical and structural elements that interact with moving fluids depends on an analysis of the flow within the boundary layer. In this chapter we will study only the effects created by a thin boundary layer, which occurs when the fluid has a low viscosity and the flow over the surface is relatively fast. As shown in Fig. 11–2, slow-moving fluids, or those having a high viscosity, produce a thick boundary layer, and understanding its effect on the flow requires a specialized experimental analysis or numerical modeling using a computer. See Ref. [5].

Boundary Layer Description. The development or growth of the boundary layer can best be illustrated by considering the steady uniform flow of a fluid over a long flat plate, which is similar to what occurs along the hull of a ship, a flat section on an airplane, or the side of a building. The basic features of the boundary layer can be divided into three regions, as shown in Fig. 11–3.

U

U

U

Viscous sublayer

Laminar

Turbulent Transition

Boundary layer over a flat plate (greatly exaggerated vertical scale)

Fig. 11–3

11.1

the ConCept oF the Boundary layer

577

Laminar Flow. As the fluid flows over the leading edge of the plate with a uniform free-stream velocity U, the particles on its surface stick to it, while those directly above it slow down and build up in smooth layers farther along the plate’s length. This will produce laminar flow. Transitional Flow. A point along the plate is reached where the laminar flow buildup becomes unstable and then tends to break down. This is a transition region, where some of the fluid particles begin to undergo turbulence, characterized by the random mixing of large groups of particles moving from one layer of fluid into another.

Turbulent Flow. The mixing of fluid that occurs causes the thickness of the boundary layer to grow rapidly, eventually forming a turbulent boundary layer. In spite of this transformation from laminar to turbulent flow, there will always remain, below the turbulent boundary layer, a very thin laminar or viscous sublayer of fluid that is “slow moving,” since the fluid will always cling along the surface of the plate.

Boundary Layer Thickness. Along the plate, the boundary layer’s thickness will continue to grow as the velocity within it approaches the free-stream velocity. Since this thickness is not well defined, engineers use three methods for specifying its value.

Disturbance Thickness. The simplest way to report the boundary layer’s thickness at each location x along the plate, Fig. 11–4, is to define it as the height d where the maximum velocity reached is equal to a certain percentage of the free-stream velocity. The agreed-upon value is u = 0.99U. This is called the disturbance thickness, because this is the height to which the fluid is disturbed in building the boundary layer.

y U

U u 5 0.99U d x x Disturbance thickness

Fig. 11–4

11

578

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VisCous Flow

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y

y U

11

U

U

dA 5 b dy

U

dA 5 b dy

u

d* Imaginary solid boundary

U2u

y x

x

Mass flow deficit in this region

Mass flow deficit

Uniform flow No boundary layer

Boundary layer (real fluid)

Displacement thickness (ideal fluid)

(a)

(b)

Fig. 11–5

Displacement Thickness. The boundary layer thickness can also be specified as a displacement thickness, d *. This refers to the distance the actual surface must be displaced, so that if we had an ideal fluid, then the mass flow with this new boundary, Fig. 11–5b, would be the same as it is for the real fluid, Fig. 11–5a. This concept has been used to design wind tunnels and the intake of a jet engine. To determine the distance d *, we must equate the solid blue shaded regions shown in Fig. 11–5. They represent the decrease in mass flow or the mass flow deficit for each case. If the surface (or plate) has a width b, then in the case of the real fluid, Fig. 11–5a, the mass flow at x through the # differential strip dA = b dy is dm = ru dA = ru(b dy). If an ideal fluid is present, then viscous effects will not occur, Fig. 11–5b, and so u = U, and # the mass flow through the area dA at x will be dm0 = rU(b dy). The mass # # flow deficit due to the viscosity is therefore dm0 - dm = r(U - u)(b dy). For the entire boundary layer, integration over its height is necessary to determine this total deficit. For the ideal fluid in Fig. 11–5b, the deficit is simply rU(bd *). Therefore, rU 1b d * 2 = Loss in uniform flow

L0

∞

r (U - u)(b dy) Loss in boundary layer

Since r, U, and b are constant, we can write this equation as d* =

L0

∞

a1 -

u b dy U

(11–1)

Therefore, to determine the displacement thickness, d*, the velocity profile u = u(y) of the boundary layer must be known. If it is, then this integral can be evaluated, either analytically or numerically, at each location x along the plate.

11.1

y

the ConCept oF the Boundary layer

579

y

U

U

U

U

dA 5 b dy

dA 5 b dy

11

U

u U2u

y

Momentum flux

x

Momentum flux

x

Uniform flow No boundary layer Boundary layer (real fluid)

Momentum thickness (ideal fluid)

(a)

(b)

Fig. 11–6

Momentum Thickness. Another way to treat the velocity disturbance brought about by the boundary layer is to consider how the actual surface should be displaced, so that the momentum flux, or rate at which the fluid’s momentum passes through an area, would be the same as if the fluid were ideal. This displacement of the surface is called the momentum thickness Θ, Fig. 11–6b. We will use this concept later in Sec. 11.3 to calculate the resistance of a fluid along the surface of a plate for various velocity profiles that are used to describe the boundary layer flow. The momentum thickness represents the loss of momentum in the boundary layer compared to having ideal flow, Fig. 11–6. To find it we must determine the rate of momentum flow deficit in each case. If the plate has a width b, then for the real fluid, Fig. 11–6a, at height y the fluid passing through the area dA has a rate of # momentum flow of dm u = r(dQ)u = r(u dA)u. Since dA = b dy, then # dm u = r(ub dy)u. Therefore, the rate of momentum flow deficit will be r[ub dy](U - u). For the case of the ideal fluid, Fig. 11–6b, the rate of momentum flow deficit through dA is r1U dA2U. Since here the total area is Θb, then we require r(UΘb)U =

L0

∞

ru (U - u)b dy

or u u a 1 - b dy U U L0 ∞

Θ =

(11–2)

In summary, we now have three definitions for the boundary layer thickness: d refers to the height to which the boundary layer disturbs the flow, up to where the velocity becomes 0.99U; and d * and Θ define the heights to which the surface must be displaced or repositioned so that if the fluid were ideal and flowing with the free-stream velocity U it would produce the same rates of mass and momentum flow, respectively, as in the case of the real fluid.

580

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VisCous Flow

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y

U

U

11

Transition to turbulent boundary layer

Laminar boundary layer

x xcr (Rex)cr 5 5 (10 5)

Fig. 11–7

Boundary Layer Classification. The magnitude of shear stress a fluid develops on the surface of a plate depends upon the type of flow within the boundary layer, and so it is important to find the location xcr where laminar flow begins to transition to turbulent flow. The Reynolds number can be used to do this, since both inertia and viscous forces play a role in boundary layer formation. For flow along a flat plate, we will define the Reynolds number on the basis of the “characteristic length” x, which is the distance downstream from the front of the plate, Fig. 11–7. Therefore,

Rex =

rUx Ux = m n

(11–3)

From experiments, it has been found that laminar flow begins to break down at about Rex = 11105 2, but it can be as high as 31106 2. The specific value when this happens is rather sensitive to the surface roughness of the plate, the uniformity of the flow, and any temperature or pressure changes occurring along the plate’s surface. See Ref. [11]. In this book, to establish a consistent value, we will take this critical value for the Reynolds number to be 1Rex 2 cr = 51105 2

(11–4)

Flat plate

For example, for air at a temperature of 20°C and standard pressure, and flowing at 25 m>s, the boundary layer will maintain laminar flow up to a critical distance of xcr = 1Rex 2 cr n>U = 51105 2[15.1110-6 2 m2 >s]> 125 m>s2 = 0.302 m from the front of the plate.

11.2

11.2

581

laminar Boundary layers

LAMINAR BOUNDARY LAYERS

In this section, we will show how the velocity varies within a laminar boundary layer, and then determine the shear stress the fluid exerts on the surface of a flat plate. To do this, it will be necessary to satisfy both the continuity and momentum equations, while analyzing for the viscous flow within the boundary layer. In Sec. 7.12, we showed that the components of the momentum equation, written for a differential fluid element, became the Navier–Stokes equations. These equations form a complex set of partial differential equations that have no known general solution, although, when they are applied within the region of the boundary layer, with certain assumptions, they can be simplified to produce a usable solution. Here we will present this solution and discuss its application to the steady incompressible flow over the flat plate in Fig. 11–8. Experiments have shown that as the fluid flows over the plate, the streamlines for the flow will begin to gradually curve upward, so that a particle located at (x, y) has velocity components u and . For high Reynolds numbers, the boundary layer is very thin, and so the vertical component will be very much smaller than the horizontal component u, Fig. 11–8. Also, because the fluid travels primarily along the plate, the changes of u and in the y direction, that is, 0u>0y, 0v>0y, and 02u>0y2, will be much greater than respective changes 0u>0x, 0v>0x, and 02u>0x 2 in the x direction. Furthermore, because the boundary layer is very thin, the pressure within it is practically constant, and becomes equivalent to the pressure impressed on it by the outer flow. Thus, 0p>0y ≈ 0. Finally, since the pressure above the boundary layer is constant, then within the boundary layer, 0p>0x ≈ 0. With these assumptions, Prandtl was able to reduce the three Navier–Stokes equations, Eq. 7–74, to just one in the x direction, and it, along with the continuity equation, becomes

y U

U Boundary layer

v u

0u 0v + = 0 0x 0y

x

Fig. 11–8

y x Rex 5 4 3 2 1

0.4

0.6

0.8

1.0 0.99

Dimensionless boundary layer velocity profile for laminar flow (a)

Fig. 11–9

To obtain the velocity distribution within the boundary layer it is necessary to solve these equations simultaneously for u and , using the boundary conditions u = v = 0 at y = 0 and u = U at y = ∞. In 1908 Paul Blasius, who was one of Prandtl’s graduate students, did this using a numerical analysis. See Ref. [16]. He presented his results in the form of a curve, shown in Fig. 11–9a, that is plotted on axes of dimensionless velocity u>U versus the dimensionless parameter (y>x) 2Rex, where Rex is defined by Eq. 11–3. For convenience, some numerical values for this curve are listed in Table 11–1. Thus, at any specified point (x, y), Fig. 11–8, the velocity u for a particle within the boundary layer can be determined from the curve or the table.

y

x

0.2

0u 0u 02u u + v = n 2 0x 0y 0y

11

Streamline

u U

582

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VisCous Flow

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TABLE 11–1 y

y

x U

11 d 5 0.707 mm

x 100 mm

2Rex

The Blasius Solution — Laminar Boundary Layer y

u>U

x

2Rex

u>U

0.0

0.0

2.8

0.81152

0.4

0.13277

3.2

0.87609

0.8

0.26471

3.6

0.92333

1.2

0.39378

4.0

0.95552

1.6

0.51676

4.4

0.97587

2.0

0.62977

4.8

0.98779

2.4

0.72899

∞

1.00000

(b)

Disturbance Thickness. This thickness, y = d, is the point where the velocity of flow u is at 99% of the free-stream velocity, that is, u>U = 0.99. From Blasius’s solution in Fig. 11–9a, it occurs when y 2Rex = 5.0 x

Fig. 11–9 (cont.)

Thus, U9 . U

U

d =

U0 . U9

2Rex 5.0

x

(11–5)

Disturbance thickness d

It is worth noting how thin a laminar boundary layer can actually be. For example, if the velocity U is such that the Reynolds number reaches its critical value of (Rex)cr = 5(105) at xcr = 100 mm, then using Eq. 11–5, the laminar boundary layer thickness at this distance is only 0.707 mm, Fig. 11–9b.

x Center core flow accelerates (a)

d*

Displacement Thickness. If Blasius’s solution for u>U is substituted into Eq. 11–1, it can be shown that after numerical integration, the displacement thickness for laminar boundary layers becomes

d

U

U

U Original surface

d

d* =

2Rex 1.721

x

(11–6)

Displacement thickness

d* x

Displaced surface Velocity profile for displaced surface

Center flow is maintained at U (b)

Fig. 11–10

Once obtained, this thickness can then be used to simulate the new location of the solid boundary over which laminar flow is considered inviscid or ideal. For example, when fluid flows between two parallel plates, Fig. 11–10a, the growing boundary layer on each plate will cause the center “core” of fluid to accelerate because of the conservation of mass. To maintain the uniform mass flow at the core of the airstream, the solid boundary at the top and bottom can be displaced outward by d - d *, Fig. 11–10b.

11.2

583

laminar Boundary layers

Momentum Thickness. To obtain the momentum thickness of the boundary layer, we must integrate Eq. 11–2, using Blasius’s solution for u>U. After a numerical integration, the result becomes Θ =

2Rex 0.664

11

(11–7)

x

Momentum thickness

Shear Stress. For the laminar boundary layer shown in Fig. 11–11a, a Newtonian fluid exerts a shear stress on the plate’s surface of

t0 = ma

du b dy y = 0

(11–8) y

The velocity gradient at y = 0 can be obtained by measuring it from the graph of Blasius’s solution, Fig. 11–9a. It is shown to be da

u b U

y d a 2Rex b x

4

= 0.332

x du large dy

y=0

For a specific location x, the Reynolds number is Rex = Ux>n. Also, since U is constant, then d(u>U) = du>U and d1y2Rex >x2 = dy2Rex >x. After rearrangement, the derivative du>dy then becomes du U = 0.332a b 2Rex x dy Substituting this into Eq. 11–8 gives the result

t0 = 0.332ma

U

U

U b 2Rex x

(11–9)

With this equation, we can now calculate the shear stress on the plate at any position x from the plate’s leading edge. Notice that this stress will become smaller, as the distance x increases, Fig. 11–11a, because the velocity gradient du>dy becomes smaller.

0

large

x

Shear stress (a)

Fig. 11–11

small du small dy 0

584

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VisCous Flow

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Since t0 causes a drag on the plate, in fluid mechanics we often express this effect by writing it in terms of the product of the fluid’s dynamic pressure, that is, 1 t0 = cf a rU 2 b 2

11 y

(11–10)

Substituting Eq. 11–9 for t0, and using Rex = rUx>m, the skin friction coefficient cf is determined from

U

cf = Ff x L

2Rex 0.664

(11–11)

Like the shear stress, cf becomes smaller as the distance x from the leading edge of the plate increases.

Friction Drag. If the resultant force acting on the surface of the plate is to be determined, then it is necessary to integrate Eq. 11–9 over the surface. This force is called the friction drag, and if the width of the plate is b and its length is L, Fig. 11–11b, we have

Friction drag (b)

Fig. 11–11 (cont.)

rUx U dx = t0 dA = 0.332ma b a b (b dx) = 0.332b U 3>2 2mr m x A LA L0 L0 2x L

FDf

L

FDf =

0.664brU 2L 2ReL

(11–12)

where ReL =

rUL m

(11–13)

Experiments have been carried out to measure the friction drag on a plate caused by laminar boundary layers, and the results have been in close agreement with those obtained from Eq. 11–12. Like t0 in Eq. 11–10, the friction drag is usually expressed in terms of the fluid’s dynamic pressure using a friction drag coefficient. It applies to both laminar and turbulent boundary layers. It is 1 FDf = CDf bL a rU 2 b 2

(11–14)

Substituting Eq. 11–12 for FDf and solving for CDf, for a laminar boundary layer, we get CDf =

2ReL 1.328

ReL 6 51105 2

(11–15)

11.2

laminar Boundary layers

585

IMPORTANT POIN T S • A very thin boundary layer will form over the surface of a flat plate when a low-viscosity, fast-moving fluid flows over it. Within this layer, the shear stress changes the fluid’s velocity profile such that it is zero at the plate’s surface and then increases to approach the free-stream velocity U of the flow above the surface.

• Due to viscosity, the thickness of the boundary layer will grow along the length of the plate. As it does, the flow within the boundary layer can change from laminar, to transitional, to turbulent, provided the plate over which it is flowing is long enough. As a convention in this book, we have defined the Reynolds number for the maximum length, xcr, of the laminar boundary layer to be (Rex)cr = 51105 2.

• One of the main purposes of using boundary layer theory is to determine the shear-stress distribution that the flow exerts on a plate’s surface and, from this, to determine the friction drag on the surface.

• At any location along the plate, we define the disturbance thickness d of the boundary layer as the height at which the flow reaches a velocity of 0.99U. The displacement thickness d * and the momentum thickness Θ are the distances to which the solid boundary surface must be displaced, or repositioned so that if the fluid were ideal and flowing with the uniform speed U, it would produce the same rates of mass and momentum flow, respectively, as in the case of the real fluid.

• A numerical solution for the thickness, velocity profile, and shear-stress distribution for a laminar boundary layer along the surface of a flat plate was determined by Blasius. His results are presented in both graphical and tabular form.

• The friction drag FDf caused by a boundary layer is generally reported using a dimensionless friction drag coefficient CDf, along with the plate area bL and the fluid’s dynamic pressure, FDf = CDf bL1 21 rU 2 2. Values of CDf are a function of the Reynolds number.

11

586

C h a p t e r 11

EXAMPLE

VisCous Flow

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e x t e r n a l s u r Fa C e s

11.1 Water flows around the plate in Fig. 11–12a with an average velocity of 0.25 m>s. Determine the shear-stress distribution and boundary layer thickness along one of its sides, and sketch the boundary layer for 1 m of its length. Take rw = 1000 kg>m3 and mw = 0.001 N # s>m2.

11

0.25 ms 1m

(a)

Fig. 11–12

SOLUTION Fluid Description. We assume steady incompressible flow along the plate. Analysis. Using Eq. 11–3, the Reynolds number for the flow in terms of x is Rex =

11000 kg>m3 210.25 m>s2x rwUx = = 2.51105 2x mw 0.001 N # s>m2

When x = 1 m, Rex = 2.51105 2 6 51105 2, so the boundary layer remains laminar. The shear-stress distribution can therefore be determined from Eq. 11–9.

y

U t0 = 0.332 mw 2Rex x

0.25 ms d 5 10 mm x 1m (b)

= 0.33210.001 N # s>m2 2 a

b Pa Ans. 2x The boundary layer thickness can be determined by applying Eq. 11–5.

t0 5 0.0415 Pa

= a

0.25 m>s b 22.51105 2x x

d =

0.0415

2Rex 5.0

x =

22.51105 2x 5.0

x = 0.0102x m

Ans.

These results are shown in Fig. 11–12b for 0 … x … 1 m. Notice that when x increases, t0 decreases and d increases. In particular, when x = 1 m, t0 = 0.0415 Pa and d = 10 mm.

11.2

EXAMPLE

laminar Boundary layers

587

11.2

The ship in Fig. 11–13a is moving slowly at 0.2 m>s through still water. Determine the thickness d of the boundary layer at a point x = 1 m from the bow. Also, at this location, find the velocity of the water within the boundary layer at y = d and y = d>2. Take nw = 1.10110-6 2 m2 >s.

11

SOLUTION Fluid Description. We will assume the ship’s hull is a flat plate, and relative to the ship the water exhibits steady incompressible flow. Disturbance Thickness. First we will check to see if the boundary layer remains laminar at x = 1 m. (0.2 m>s)(1 m) Ux Rex = = = 1.8181105 2 6 1Rex 2 cr = 511052 OK nw 1.10110-6 2 m2 >s We can now use Blasius’s solution, Eq. 11–5, to determine the boundary layer thickness at x = 1 m. d =

2Rex 5.0

x =

21.8181105 2 5.0

0.2 ms

(1 m) = 0.01173 m = 11.7 mm Ans.

Velocity. At x = 1 m and y = d = 0.01173 m, by definition, u>U = 0.99. Thus, the velocity of the water at this point, Fig. 11–13b, is u = 0.99(0.2 m>s) = 0.198 m>s

Ans.

To determine the velocity of the water at x = 1 m and y = d>2 = 5.863(10 -3) m, we must first find the value of u>U, either from the graph in Fig. 11–9a, or from Table 11–1. Here, 5.863110-3 2 m y 21.8181105 2 = 2.5 2Rex = x 1m Using linear interpolation between the values 2.4 and 2.8 given in the table, for 2.5 we obtain u>U - 0.72899 0.81152 - 0.72899 = ; u>U = 0.7496 2.5 - 2.4 2.8 - 2.4 u = 0.7496(0.2 m>s) = 0.150 m>s Ans.

(a) y

0.2 ms

Other points along the velocity profile in Fig. 11–13b can be obtained in a similar manner. Also, this method can be used to find u for other values of x as well, but realize that the maximum value for x cannot exceed xcr, that is, 10.2 m>s2xcr Uxcr 1Rex 2 cr = ; 51105 2 = ; xcr = 2.75 m nw 1.10110 - 6 2 m>s Beyond this distance the boundary layer begins to transition, and then farther along, it becomes turbulent.

0.2 ms 0.198 ms d 2 5 5.863 mm

y 5 d 5 11.73 mm 0.150 ms

x51m (b)

Fig. 11–13

x

588

C h a p t e r 11

EXAMPLE

11

VisCous Flow

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e x t e r n a l s u r Fa C e s

11.3 Air flows into the rectangular duct in Fig. 11–14a at 3 m>s. Determine the displacement thickness at the end of its 2-m length, and the uniform velocity of the core flow of air coming out of the duct. Take ra = 1.20 kg>m3 and ma = 18.1110-6 2 N # s>m2.

400 mm

300 mm

SOLUTION Fluid Description. We will assume the air is incompressible and has steady flow.

3 ms

Displacement Thickness. Using Eq. 11–3, the Reynolds number for the air as it moves through the duct is 11.20 kg>m3 2(3 m>s)x raUx = 0.19891106 2x Rex = = ma 18.1110 - 6 2 N # s>m2

(a)

y U

U

u 5 0.99 U d

d* 5 0.005457 m Real fluid Ideal fluid At x 5 2 m (b) 300 mm 2 2d*

400 mm 2 2d* At exit, d* 5 5.46 mm (c)

Fig. 11–14

x

When x = 2 m, then Rex = 3.9781105 2 6 51105 2, so we have a laminar boundary layer along the duct. Therefore, we can use Eq. 11–6 to determine the displacement thickness. 1.721 1.721 d* = x = x = 3.859110-3 2 2x m (1) 2Rex 20.19891106 2x Where x = 2 m,

d * = 0.005457 m = 5.46 mm

Ans.

Velocity. If the air were an ideal fluid, then the dimensions of the cross section of the duct at x = 2 m would have to be reduced, as shown on the right in Fig. 11–14b, in order to produce the same mass flow through the duct as in the case of the actual flow. In other words, the exit crosssectional area, shown in light shade in Fig. 11–14c, must be Aout = 30.3 m - 2(0.005457 m)4 30.4 m - 2(0.005457 m)4 = 0.1125 m2 Since a constant mass flow must pass through each cross section, the uniform portion of the air exiting the duct must have a greater velocity than the uniform airstream entering the duct. To determine its value, we require continuity of flow. 0 r dV + rVf>cs # dA = 0 0 t Lcv Lcs 0 - raUinAin + raUoutAout = 0 -(3 m>s)[(0.3 m)(0.4 m)] + Uout 10.1125 m2 2 = 0 Uout = 3.20 m>s Ans. This increase in U occurs because the boundary layer constricts the flow. In other words, the cross section is decreased by the displacement thickness. If, instead, we wanted to maintain a constant uniform speed of 3 m>s through the duct, then it would be necessary for the cross section to become divergent so that its width and height would increase by 2d * along its length, in accordance with Eq. 1.

11.2

EXAMPLE

589

laminar Boundary layers

11.4 1m

A small submarine has a triangular stabilizing fin on its stern with the dimensions shown in Fig. 11–15a. If the water temperature is 15°C, determine the drag on the fin when the submarine is traveling at 0.5 m>s.

11

0.5 ms 1m

SOLUTION Fluid Description. Relative to the submarine, this is a case of steady incompressible flow. From Appendix A, for water at 15°C, rw = 999.2 kg>m3 and nw = 1.15110-6 2 m2 >s.

(a) y

Analysis. First we will determine if the flow within the boundary layer remains laminar. Since the flow creates the largest Reynolds number at the base of the fin, where x = 1 m in Fig. 11–15b is the largest, then (Re) max

(0.5 m>s)(1 m) Ux = 4.35 1 105 2 6 5 1 105 2 = = vw 1.15 1 10 - 6 2 m2 >s

y5x

0.664brwU 2L 2ReL

d =

2

= 0.5031 1 1 - x 2 1>2 dy

£

Laminar

0.664(dy) 1 999.2 kg>m3 2 10.5 m>s2 2(1 - x) 10.5 m>s2(1 - x)

A 1.15 1 10 - 6 2 m2 >s

Since x = y, the total drag acting on all the strips composing the area of the fin becomes FDf = 0.5031

L0

1m

(1 - y)1>2 dy

1m 2 = 0.5031c - (1 - y)3>2 d = 0.335 N 3 0

1m

12x y

x

Integration is necessary here, since the length x of the fin changes with y. If we establish the x and y axes as shown in Fig. 11–15b, then an arbitrary differential strip of the fin has a length of 1 - x and a width of dy. Applying Eq. 11–12, with b = dy and L = 1 - x, the drag on both sides of the strip is therefore dFDf = 2c

dy

x

Ans.

This is indeed a very small value, the result of the submarine’s small velocity and the low kinematic viscosity of water.

1m (b)

Fig. 11–15

§

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11.3

THE MOMENTUM INTEGRAL EQUATION

In the previous section we were able to determine the shear stress distribution caused by a laminar boundary layer using the solution developed by Blasius. This analysis was possible because his solution of the velocity profile could be related to the shear stress using Newton’s law of viscosity, t = m(du>dy). However, no such relationship between t and u exists for turbulent boundary layers, and so a different approach must be used to study the effects for turbulent flow. In 1921, Theodore von Kármán proposed an approximate method for boundary layer analysis that is suitable for both laminar and turbulent flow. Rather than writing the continuity and momentum equations for a differential control volume at a point, von Kármán considered doing this for a differential control volume having a thickness dx and extending from the plate’s surface up to a streamline that intersects the boundary layer, as shown in Fig. 11–16a. Through this element the flow is steady, and due to its small height, the pressure within it is virtually constant. Over time, the x component of flow enters the open control surface at the left side, AC, and exits through the open control surface DE on the right side. No flow can cross over the closed control surface AE or the streamline boundary CD, since the velocity is always tangent along the streamline.

11

Continuity Equation. If we consider a unit width of the plate perpendicular to the page, then for steady flow, the continuity equation becomes

y

U

U

U

Boundary layer

0 r dV + rVf>cs # dA = 0 0 t Lcv Lcs

D C dl

B dy A

U U ul

dr ur

dy E

x Streamline

dx

0 -

L0

dl

# rul dy - mBC +

L0

dr

rur dy = 0

(11–16)

(a) D

C

# Here mBC is due to the constant flow U into the open control surface BC between the top of the boundary layer and the streamline.

A t0 (1dx)

E

Free-body diagram (b)

Fig. 11–16

Momentum Equation. The free-body diagram of the control volume is shown in Fig. 11–16b. Since the pressure p within the control volume is essentially constant, and the height hAC ≈ hED, because AE = dx, then the resultant forces caused by pressure on each open control surface will cancel. The only external force acting on the closed

11.3

the momentum integral equation

591

control surfaces is due to shear stress from the plate. This force is t0(1dx). Applying the momentum equation in the x direction, we have + ΣFx = S

0 Vr dV + Vr Vf>cs # dA 0t Lcv Lcs dr

dl

# rul 2 dy - U mBC

L0 L0 # Solving for mBC in Eq. 11–16, and substituting into the above equation, realizing that r is constant for an incompressible fluid, gives -t0 (1dx) = 0 +

-t0 dx = r -t0 dx = r

L0

dr

L0

dr

rur 2 dy -

ur 2dy - r

L0

dl

ul 2dy - Urc

1ur 2 - Uur 2 dy - r

L0

dl

L0

dr

ur dy -

1ul 2 - Uul 2 dy

L0

ul dy d

dl

Since the vertical sides AC and ED are a differential distance dx apart, the terms on the right represent the differential difference of the integrals, that is, -t0 dx = r d c

L0

d

1u2 - Uu2 dy d

The free-stream velocity U is constant, and so we can also write this equation in terms of the dimensionless velocity ratio, u>U, and express the shear stress as d u u a 1 - b dy dx L0 U U d

t0 = rU 2

(11–17)

Recognizing that the integral represents the momentum thickness, Eq. 11–2, we can also write t0 = rU 2

dΘ dx

(11–18)

Either of the above two equations is called the momentum integral equation for a flat plate. To apply it, we must either know the velocity profile u = u(y) at each position x, or approximate it by an equation having the form u>U = f(y>d). Some of these profiles are listed in Table 11–2. With any one of these we can evaluate the integral in Eq. 11–17, but due to the upper integration limit, the result will be in terms of d. In order to determine d as a function of x, t0 must also be related to d. The following example shows how this can be done for a laminar boundary layer using Newton’s law of viscosity. It also illustrates how the results of d, cf, and CDf reported in Table 11–2 were determined. In the next section, we will show how to use the momentum integral equation for turbulent boundary layer flow.

11

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TABLE 11–2 Velocity profile

d

Blasius y u = U d

y 2 y Parabolic u = - a b + 2a b U d d u 1 y 3 3 y = - a b + a b U 2 d 2 d

Cubic

CDf

5.00

x

2Rex

0.664

2Rex

2Rex

3.46

x

2Rex

0.578

2Rex

2Rex

5.48

x

2Rex

0.730

2Rex

2Rex

4.64

2Rex

2Rex

2Rex

11 Linear

cf

0.646

x

1.33

1.16

1.46

1.29

EXAMPLE 11.5 The velocity profile for a laminar boundary layer developed over the plate of width b and length L is approximated by the parabola u>U = -(y>d)2 + 2(y>d), as shown in Fig. 11–17. Determine, as a function of x, the thickness d of the boundary layer, the skin friction coefficient cf , and the friction drag coefficient CDf .

y U

U

u d

x

y

SOLUTION x

t0 L

Fig. 11–17

Fluid Description. Here we have steady incompressible laminar flow over the surface. Boundary Layer Thickness. Substituting the function for u>U into Eq. 11–17 and integrating, we get d u u a 1 - b dy dx L0 U U d

t0 = rU 2

t0 = rU 2

d y 2 y 2 y y d d 2 a db c- a b + 2 a b d c 1 + a b - 2 a b d dy = rU 2 dx L0 d d d d dx 15

t0 =

2 dd rU 2 15 dx

(1)

11.3

the momentum integral equation

593

To obtain d we must now express t0, which is at the plate’s surface ( y = 0), in terms of d. Since laminar flow exists, this can be done using Newton’s law of viscosity, evaluated at y = 0. t0 = m

y 2 y 2mU du d ` = mU c - a b + 2a b d = dy y = 0 dy d d y=0 d

(2)

Therefore, Eq. 1 becomes 2mU 2 dd = rU 2 d 15 dx

or

rU d dd = 15m dx

Since d = 0 at x = 0, that is, where the fluid initially contacts the plate, then integration gives rU

L0

d

d dd =

L0

x

15m dx;

d =

30mx A rU

(3)

Since Rex = rUx>m, we can also write this expression as 2Rex 5.48x

d =

Ans.

Skin Friction Coefficient. Substituting Eq. 3 into Eq. 2, the shear stress on the plate as a function of x is therefore t0 =

= 0.365 B 30mx B rU 2mU

mrU 3 mU = 0.365 1Rex x x

Using Eq. 11–10, the skin friction coefficient is mU 0.365 1Rex t0 x 0.730 = = cf = 2 2 1Rex (1>2)rU (1>2)rU

Ans.

Friction Drag Coefficient. To determine CDf , we must first find FDf. FDf =

LA

x

t0 dA =

0.365 B L0

x mrU 3 (b dx) = 0.365b 2mrU 3 12x 1>2 2 ` = 0.730b 2mrU 3x x 0

Therefore, using Eq. 11–14, CDf =

FDf >bx

11>22rU 2

1.461b 2mrU 3x =

CDf =

rU 2bx 2Rex 1.46

The three results obtained here are listed in Table 11–2.

Ans.

11

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11.4

TURBULENT BOUNDARY LAYERS

Turbulent boundary layers are thicker than laminar ones, and the velocity profile within them will be more uniform due to the erratic mixing of the fluid. In this section, we will use the momentum integral equation to determine the drag caused by a turbulent boundary layer; however, it is first necessary to express the velocity profile as a function of y. The accuracy of the result, of course, depends upon how close this function approximates the true velocity profile. Although many different formulas have been proposed, one of the simplest that works well is Prandtl’s oneseventh power law. It is

11

y 1>7 u = a b U d

(11–19)

A portion of this equation is shown in Fig. 11–18a. Notice that here the profile is flatter than the one developed for a laminar boundary layer. The flatness is necessary because, as previously stated, there is a large degree of fluid mixing and momentum transfer within turbulent flow. Also, because of this flatness, there is a larger velocity gradient near the plate’s surface. As a result, the shear stress developed on the surface will be much larger than that caused by a laminar boundary layer. Prandtl’s equation does not apply within the laminar sublayer, because the velocity gradient, du>dy = (U>7d)(y>d) -6>7, at y = 0 becomes infinite, something that cannot happen. Therefore, as in all cases of turbulent boundary layers, the surface shear stress t0 must be related to d by experiment.* An empirical formula that agrees well with the data was developed by Prandtl and Blasius. It is t0 = 0.0225rU 2 a

n 1>4 b Ud

(11–20)

We will now apply the momentum integral equation with these two equations, and thereby obtain the height d of the turbulent boundary layer as a function of its position x. Applying Eq. 11–17, we have

y

t0 = rU 2 0.0225rU 2 a

1

y 7 u 5 (–– –– ) U d d

y x

x

du large, dy

Prandtl’s one-seventh power law (a)

Fig. 11–18

d u u a 1 - b dy dx L0 U U d

U

0

large

0.0225a

d y 1>7 y 1>7 n 1>4 d a b c 1 - a b d dy b = rU 2 Ud dx L0 d d

n 1>4 d 7 7 7 dd b = c d - dd = Ud dx 8 9 72 dx d 1>4dd = 0.231a

n 1>4 b dx U

*Recall that Newton’s law of viscosity can be used for laminar boundary layers, provided, of course, that we have a Newtonian fluid.

11.4

595

turBulent Boundary layers

Disturbance Thickness. Although all boundary layers are initially laminar, as shown in Fig. 11–3, here we will assume that the front surface of the plate is rather rough, and so the boundary layer will be forced to become turbulent, practically at the outset. Therefore, integrating from x = 0, where d = 0, we have d x n 1>4 n 1>5 d 1>4 dd = 0.231a b dx or d = 0.371a b x 4>5 U U L0 L0

Using Eq. 11–3 to express the thickness in terms of the Reynolds number, we have 0.371 d = x (11–21) 1Rex 2 1>5

11

y

U

Shear Stress along Plate. Substituting Eq. 11–21 into Eq. 11–20, we obtain the shear stress along the plate as a function of x, Fig. 11–18b. t0 = 0.0225rU 2 a t0 =

n1Rex 2 1>5 1>4 b U(0.371x)

1Rex 2 1>5

x

(11–22)

L

Drag on Plate. If the plate has a length L and a width b, then the drag

Shear stress on plate

can be found by integrating the shear stress over the plate’s area, Fig. 11–18c.

(b)

FDf =

L

t0 dA =

A

L0

L

0.0288rU Ux a b n

2

1>5

(b dx) = 0.0360rU 2

1ReL 2 1>5 bL

(11–23) y

The frictional drag coefficient, Eq. 11–14, is therefore FDf >bL 0.0721 (11–24) = CDf = 11>22rU 2 1ReL 2 1>5 This result has been checked by numerous experiments, and it has been found that a slightly more accurate value for CDf can be obtained by replacing the constant 0.0721 with 0.0740. The result works well within the following range of Reynolds numbers: CDf =

0.0740

1ReL 2 1>5

51105 2 6 ReL 6 107

0.455 1log10 ReL 2 2.58

107 … ReL 6 109

U

x

(11–25)

FD L

Notice that the lower bound for ReL represents our limit for laminar boundary layers. For higher values of the Reynolds number, another empirical equation that fits relevant experimental data has been proposed by Hermann Schlichting. See Ref. [16]. It is CDf =

x

t0

0.0288rU 2

(11–26)

Keep in mind that all these equations are valid only if a turbulent boundary layer extends over the entire length of the plate.

Drag on plate (c)

Fig. 11–18 (cont.)

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11.5

LAMINAR AND TURBULENT BOUNDARY LAYERS

y Turbulent

11

Laminar x xcr Model

=

(a)

y

x All turbulent

As stated in Secs. 11.1 and 11.2, a real boundary layer on a smooth flat plate develops a laminar region, grows in height to become unstable at about (Rex)cr = 5(105), then eventually transitions to become turbulent. Therefore, to develop a more exact approach when calculating the friction drag on the plate, it is necessary to consider the effect of both the laminar and turbulent portions of the boundary layer. Through experiment, Prandtl discovered that this could be done by neglecting the friction drag within the transition region, since for fast flows it covers a very short distance along the plate. In other words, we can model the boundary layer like that shown in Fig. 11–19a. To calculate the friction drag, Prandtl first assumed that the boundary layer is completely turbulent over the plate’s entire length L, Fig. 11–19b, and so the drag can be found using Eq. 11–26 for the higher Reynolds numbers or greater distance x. Then an adjustment to this result is made by subtracting a turbulent segment of the drag up to the start of the transition zone, x = xcr, Eq. 11–25, Fig. 11–19c, and finally, adding the drag caused by the laminar portion up to this point (Blasius’s solution), Eq. 11–15, Fig. 11–19d. The friction drag on the plate is therefore

(b)

_

y

FDf =

0.455 0.0740 11>22rU 2(bL) 11>22rU 2 1bxcr 2 1>5 1log10ReL 2 2.58 1Rex 2 cr +

x xcr

+

CDf = x xcr Laminar (d)

Fig. 11–19

0.455 0.0740 xcr 1.328 xcr + 1>5 L 11Re 1log10 ReL 2 2.58 1Rex 2 cr x 2 cr L

Finally, since by proportion xcr >L = 1Rex 2 cr >ReL, and if 1Rex 2 cr = 51105 2, then to fit experimental data for values of the Reynolds number between 51105 2 … ReL 6 109, we have

(c)

y

Since CDf for the plate is defined by Eq. 11–14, the friction drag coefficient is CDf =

Turbulent segment

1.328 11>22rU 2 1bxcr 2 11Rex 2 cr

0.455 1700 2.58 ReL 1log10 ReL 2

51105 2 … ReL 6 109

(11–27)

A plot of all the relevant equations used to determine the friction drag coefficient for a range of Reynolds numbers for which they are valid is shown as solid curves in Fig. 11–20. Notice how the experimental data, obtained from several different researchers, is in close agreement with this theory. Also, realize that the curves are valid only when the transition in boundary layer flow occurs at a Reynolds number of 51105 2. If the

11.5

laminar and turBulent Boundary layers

597

0.008 0.006

Turbulent Eq 11.25

0.004 CDf

11 Eq 11.27 0.002

Eq 11.26

Laminar (Blasius Solution) Eq 11.15 0.001 10 5 2

5 10 6 2

5

10 7 2 5 ReL

10 8 2

5 10 9 2

5

Friction drag coefficient for a flat plate

Fig. 11–20

free-stream velocity is subjected to any sudden turbulence, or the surface of the plate is somewhat rough, then the transition in flow from laminar to turbulent will occur at a lower Reynolds number. When this occurs, the constant in the second term (1700) in Eq. 11–27 will have to be modified to account for this change. The calculations accounting for these effects will not be outlined here; rather, they are covered in the literature related to this subject. See Ref. [16].

IMPORTANT POIN T S • The momentum integral equation provides an approximate method for obtaining the thickness and surface shear-stress distribution caused by either a laminar or a turbulent boundary layer. To apply this equation, it is necessary to know the velocity profile u = u(y) and to be able to express the shear stress on the surface of the plate in terms of the boundary layer thickness d, t0 = f(d).

• Blasius’s solution for u = u(y) can be used for laminar boundary layers, and the shear stress t0 can be related to d using Newton’s law of viscosity. For turbulent boundary layers the flow is erratic, and so the needed relationships must be approximated from results obtained from experiments. For example, for a fully turbulent boundary layer, Prandtl’s one-seventh power law, along with a formulation by Prandtl and Blasius, seems to work well.

• A laminar boundary layer exerts a much smaller shear stress on a surface than a turbulent boundary layer. A high degree of fluid mixing occurs within turbulent boundary layers, and this creates a larger velocity gradient at the surface, and therefore a higher shear stress on the surface.

• The frictional drag coefficient on a flat surface, caused by a combination laminar and turbulent boundary layer, can be determined by superposition, neglecting the region where transitional flow occurs.

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11.6 Oil flows over the top of the flat plate in Fig. 11–21 with a free-stream velocity of 20 m>s. If the plate is 2 m long and 1 m wide, determine the friction drag on the plate due to the formation of a combination laminar and turbulent boundary layer. Take ro = 890 kg>m3 and mo = 3.40110-3 2 N # s>m2. y 20 ms

31.9 N 972 N x 0.0955 m 2m

Fig. 11–21

SOLUTION Fluid Description. We have steady flow and assume the oil is incompressible. Analysis. We will first determine the position xcr where the laminar boundary layer begins to transition to turbulent, Fig. 11–21. Here 1Rex 2 cr = 51105 2, so 1Rex 2 cr =

roUx mo

51105 2 =

1890 kg>m3 2(20 m>s)xcr 3.40110-3 2 N # s>m2

xcr = 0.0955 m 6 2 m

Also, at the end of the plate, ReL =

1890 kg>m3 2(20 m>s)(2 m) roUL = 1.0471107 2 = mo 3.40110-3 2 N # s>m2

11.5

laminar and turBulent Boundary layers

Using Eq. 11–27 to determine the drag coefficient, since this equation applies for laminar–turbulent boundary layers within the range 51105 2 … ReL 6 109, we have CDf

0.455 1700 = ReL 1log10 ReL 2 2.58

3 log10 1.047110 2 4 0.455

=

7

2.58

-

= 0.002819

1700 1.0471107 2

The total friction drag on the plate can now be determined using Eq. 11–14. 1 FDf = CDf bL a roU 2 b 2

= 0.002819(1 m)(2 m) 3 12 1890 kg>m3 2120 m>s2 2 4 = 1004 N

Ans.

The portion of this force created by the laminar boundary layer, which extends along the first 0.0955 m of the plate, can be determined from Eq. 11–12. 1FDf 2 lam = =

11Rex 2 cr

0.664broU 2xcr 0.664(1 m)1890 kg>m3 2(20 m>s)2(0.0955 m)

= 31.9 N

251105 2

By comparison, the turbulent boundary layer contributes the most friction drag. It is 1FDf 2 tur = 1004 N - 31.9 N = 972 N

599

11

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EXAMPLE

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11.7 Water has a free-stream velocity of 10 m>s over the rough surface of the plate in Fig. 11–22, causing the boundary layer to suddenly become turbulent. Determine the shear stress on the surface at x = 2 m, and the thickness of the boundary layer at this position. Take rw = 1000 kg>m3 and mw = 1.00110-3 2 N # s>m2. y

10 ms

10 ms

d t0

x

x52m

Fig. 11–22

SOLUTION Fluid Description. We have steady flow and we assume water is incompressible. Analysis.

For this case, the Reynolds number is

Rex =

11000 kg>m3 2(10 m>s)(2 m) rwUx = = 201106 2 mw 1.00110-3 2 N # s>m2

Using Eq. 11–22, since the boundary layer has been defined as fully turbulent, the shear stress on the surface at x = 2 m is t0 =

0.0288rwU 2 1Rex 2 1>5

=

0.028811000 kg>m3 2110 m>s2 2

= 99.8 Pa

3 201106 2 4 1>5

Ans.

And from Eq. 11–21, the thickness of the boundary layer at x = 2 m is d =

1Rex 2

1>5

3 201106 2 4 1>5 0.371(2 m)

0.371x =

= 0.02572 m = 25.7 mm

Ans.

Although this is a small thickness, it is much thicker than a laminar boundary layer. Also, the calculated shear stress at x = 2 m will be higher than that caused by a laminar boundary layer.

11.5

EXAMPLE

laminar and turBulent Boundary layers

11.8

Estimate the friction drag on each airplane wing in Fig. 11–23 if it is assumed to be a flat plate, having an average width of 0.9 m and length of 4 m. The plane is flying at 125 m>s. Assume the air is incompressible. Take ra = 0.819 kg>m3 and ma = 16.6110-6 2 N # s>m2.

125 ms

11 0.9 m

4m

SOLUTION Fluid Description. We have steady incompressible flow relative to the airplane. Analysis. At the trailing or back edge of the wing, the Reynolds number is

raUL ReL = = ma

601

1 0.819 kg>m3 2 (125 m>s)(0.9 m) 16.6 1 10-6 2 N # s>m2

= 5.550 1 106 2 7 5 1 105 2

Therefore, the wing is subjected to a combined laminar and turbulent boundary layer. Applying Eq. 11–27, since it applies for 51105 2 … ReL 6 109 , we have CDf =

0.455 1700 ReL 1log10 ReL 2 2.58

5 log10 3 5.550 1 10 246 0.455

=

= 0.0030001

6

2.58

-

5.550 1 106 2 1700

Since the friction drag acts on both the top and bottom wing surfaces, then using Eq. 11–14, 1 FDf = 2c CDf bL a raU 2 b d 2

1 = 2e (0.0030001)(4 m)(0.9 m) c (0.819 kg>m3)(125 m>s)2 d f 2 = 138.21 N = 138 N

Ans.

Fig. 11–23

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11.6

In almost all cases, the natural flow of a fluid will be both unsteady and nonuniform. For example, the speed of the wind varies with time and elevation, and so does the speed of water in a river or stream. Even so, for engineering applications we can often approximate the effect of these irregularities, either by averaging them out or by considering an extreme condition. Then, with either of these approximations, we can investigate the flow as if it were steady and uniform. Here and in the following sections, we will study the effect of steady and uniform flow on bodies having different shapes, including flow past an axisymmetric body such as a rocket, past a two-dimensional body such as a very tall chimney, and past a three-dimensional body such as an automobile. Being able to calculate the forces created by the flow has important applications in structural and hydraulic engineering, as well as mechanical and aeronautical engineering.

11

t

p

U

DRAG AND LIFT

dA u

Pressure and shear stress on a fixed surface (a)

y

Drag and Lift Components. If a fluid has a steady and uniform free-stream velocity U, and it encounters an obstruction such as a body having a curved surface, Fig. 11–24a, then the fluid will exert both a viscous tangential shear stress t and a normal pressure p on the surface. For a differential element dA, we can resolve the forces produced by t and p into their horizontal (x) and vertical (y) components, Fig. 11–24b. The drag is in the direction of U. When integrated over the entire surface of the body, this force becomes t cos u dA + p sin u dA LA LA The lift is the force that acts perpendicular to U. It is FD =

u t dA

p dA

dA u

x

t sin u dA p cos u dA (11–29) LA LA Provided the distributions of t and p are known, then these integrations can be carried out. The following is an example of how this can be done. FL =

Forces on an element of surface area (b)

(11–28)

Fig. 11–24

Since the loading ramps on this truck are held in the vertical position, they put a large pressure drag on the truck and reduce its fuel efficiency.

11.6

EXAMPLE

drag and liFt

603

11.9

The semicircular building in Fig. 11–25a is 12 m long and is subjected to a steady uniform wind having a velocity of 18 m>s. Assuming the air is an ideal fluid, determine the lift and drag on the building. Take r = 1.23 kg>m3.

11 18 ms

SOLUTION Fluid Description. Since the air is assumed to be an ideal fluid, there is no viscosity, and so no boundary layer or viscous shear stress acts on the building, only a pressure distribution. We will assume the building is long enough so end effects do not disturb this two-dimensional steady flow. Analysis. Using ideal fluid flow, as outlined in Sec. 7.11, we have determined that the pressure distribution over the surface of a cylinder is symmetrical, and so for a semicylinder, as shown in Fig. 11–25b, it can be described by Eq. 7–68, namely, 1 p = p0 + rU 2 11 - 4 sin2 u2 2 Since p0 = 0 (gage pressure), we have 1 p = 0 + 11.23 kg>m3 2118 m>s2 2 11 - 4 sin2 u2 2 = 199.2611 - 4 sin2 u2 The drag is the horizontal or x component of dF, shown in Fig. 11–25b. For the entire building it is FD =

LA

( p dA) cos u =

= 9564.48

L0

L0

p

(199.26)11 - 4 sin2 u2 cos u3(12 m)(4 m) du4

11 - 4 sin2 u2 cos u du

p

= 9564.48 a sin u -

p 4 3 sin u b ` = 0 Ans. 3 0 This result is to be expected since for an ideal fluid the pressure distribution is symmetrical about the y axis. The lift is the vertical or y component of dF. Here dFy is directed in the -y direction, so that p

FL = -

( p dA) sin u = - (199.26)11 - 4 sin2 u2 sin u3(12 m)(4 m)du4 LA L0

= -9564.48

L0

p

11 - 4 sin2 u2 sin u du

= -9564.48 a 3 cos u -

p 4 cos3 u b ` 3 0 3 = 31.88110 2 N = 31.9 kN Ans. The positive sign indicates that the air stream tends to pull the building up, as the word “lift” would suggest.

4m

(a) y

dF 5 p dA dA

4m

u

(b)

Fig. 11–25

x

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FD

FD

11

Friction drag caused by the viscous shear boundary layer (a)

Pressure drag caused by the change in momentum of the fluid (b)

FD FL

Combination of friction drag and pressure drag (c)

Fig. 11–26

11.7

The shape of an automobile affects both the pressure drag and the viscous friction drag developed within the boundary layer. (© Frank Herzog/culture-images GmbH/Alamy Stock Photo)

PRESSURE GRADIENT EFFECTS

In the previous section we noted that the forces of drag and lift are the result of a combination of viscous shear stress and pressure acting on the surface of a body. To further demonstrate this, consider the case of a flat plate. When its surface is aligned with the flow, Fig. 11–26a, only viscous friction drag will be produced on the plate. This is the effect of the boundary layer; however, when the flow is perpendicular to the plate, as in Fig. 11–26b, then the plate acts as a bluff body. Here only pressure drag is created. Pressure drag is caused by the change in the momentum of the fluid, as discussed in Chapter 6. The resultant shear drag is zero in this case because the shear stresses act equally up and down the plate’s front surface. Notice that in both of these cases, the lift is zero because neither effect produces a resultant force in the vertical direction, perpendicular to the flow. To achieve both lift and drag, the plate must be oriented at an angle to the flow, as shown in Fig. 11–26c. A body that has a curved or irregular surface can also be subjected to both lift and drag, and to better understand how these forces are created, let us consider the uniform flow over the surface of a long cylinder.

11.7

pressure gradient eFFeCts

Minimum pressure

Maximum velocity U

Decreasing pressure (favorable) U

Decreasing velocity

Increasing velocity

Increasing pressure (adverse)

11

B B C

A

605

A

C

Velocity

Pressure distribution

Ideal flow (a)

Ideal flow (b)

Fig. 11–27

Ideal Flow Around a Cylinder. When a cylinder is placed within a uniform flow, it will redirect the flow around the cylinder as shown in Fig. 11–27a. Provided the fluid is ideal, then the velocity of the fluid particles will increase from zero at the stagnation point A to a maximum at B. The velocity then decreases as the fluid rounds the back of the cylinder, until it reaches zero at the stagnation point C. This variation of the velocity will cause a corresponding, yet opposite variation of the pressure on the cylinder, which was discussed in Sec. 7.11 and is shown in Fig. 11–27b. From A to B the pressure will decrease. The change in pressure, or favorable pressure gradient, begins with a pushing or positive pressure on the surface, followed by a suction or negative pressure which reaches a maximum value at B. Then, due to the decrease in velocity, an increase in pressure occurs from B to C. This is referred to as an adverse pressure gradient. Since we have an ideal fluid, the pressure distribution is symmetrical around the cylinder, and so the resultant pressure drag (horizontal force) on the cylinder will be equal to zero. Furthermore, since an ideal fluid has no viscosity, there also is no viscous friction drag on the cylinder.

606

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Maximum velocity

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Minimum pressure

Decreasing velocity

Increasing pressure Constant pressure and velocity

Decreasing pressure U

11

Increasing velocity

B9

C9 Flow separation

Region of reverse flow

B9

C9

C

A FD p

Wake

Shedding vortices

A

Boundary layer separation

Pressure distribution

Viscous fluid (a)

Viscous fluid (b)

Fig. 11–28

Real Flow around a Cylinder. Unlike an ideal fluid that can

The Coanda effect, indicated by the bending of the waterstream as the water clings to the glass.

Flow separation from the sides of the post is clearly evident in this photo.

slide freely around a cylinder, a real fluid has viscosity, and as a result, the fluid will tend to form a boundary layer and, due to its adhesion, cling to the cylinder’s surface as it flows around it. (See adjacent photo.) This phenomenon was studied in the early 1900s by the Romanian engineer Henri Coanda and is called the Coanda effect. To understand this viscous behavior over a curved surface, we will again consider a long cylinder shown in Fig. 11–28a. For a real fluid, the flow starts from the stagnation point at A, and thereafter it forms a laminar boundary layer on the cylinder as the fluid begins to travel around the surface with increasing velocity, Fig. 11–28a. This produces a favorable pressure gradient (pressure decrease) within the region AB′, Fig. 11–28b. Here the flow must overcome the drag effect of viscous friction within the boundary layer, and so the maximum velocity and minimum pressure occur at point B′, which is sooner than in the case of ideal fluid flow. As the boundary layer continues to grow in thickness downstream of point B′, an adverse pressure gradient (increasing pressure) occurs due to the decrease of velocity. Point C′ marks a separation of flow from the cylinder since the velocity of the slower-moving particles slightly above the surface is finally reduced to zero. Beyond C′, the flow within the boundary layer will begin to back up and move in the opposite direction to the free-stream flow. This ultimately will form a vortex, which will shed from the cylinder, as shown in Fig. 11–28a. A series of these vortices or eddies produce a wake, whose energy will eventually dissipate as heat. The pressure within the wake is relatively constant, and the resultant of the entire pressure distribution around the cylinder produces the pressure drag FDp, Fig. 11–28b.

11.7

Flow separation

U

U

607

pressure gradient eFFeCts

Flow separation

C9

C9 Wake

C

(FDp)l

(FDp)t

Rough cylinder

Smooth cylinder

Turbulent boundary layer separation

Laminar boundary layer separation

Smaller drag force (a)

Larger drag force (b)

11 Wake

Fig. 11–29

If the flow within the boundary layer is completely turbulent, Fig. 11–29a, then separation will occur later than if the boundary layer flow is laminar, Fig. 11–29b. This happens because within a turbulent boundary layer, the fluid is stirred up and so it has more kinetic energy than in the laminar case. As a result, the adverse pressure gradient will take longer to arrest the flow, and so the point of separation is farther back on the surface. Consequently, the resultant of the turbulent pressure distribution, 1FDp 2 t , Fig. 11–29a, will create a smaller pressure drag than in the case of a laminar pressure distribution, where this resultant is 1FDp 2 l , Fig. 9–29b. Although it may seem counterintuitive, one way to reduce pressure drag is to roughen the cylinder’s front surface so as to produce a turbulent boundary layer. Unfortunately, the actual point C′ of flow separation for either laminar or turbulent boundary layers cannot be determined analytically, except through approximate methods. However, experiments have shown that, as expected, the point of transition from laminar to turbulent flow is a function of the Reynolds number, and therefore the pressure drag, like viscous or friction drag, will be a function of this parameter. For a cylinder, the “characteristic length” for finding the Reynolds number is its diameter D, so Re = rVD>m.

Whale flippers have tubercles on their leading edge as noted in this photo. Wind tunnel tests have revealed that the flow of water between each pair of tubercles produces clockwise and counterclockwise vortices, which energize the turbulent flow within the boundary layer, thus preventing the boundary layer from separating from the flipper. This gives the whale more maneuverability and less drag. (© Masa Ushioda/Stephen Frink Collection/ Alamy Stock Photo)

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Alternating force

11

von Kármán vortex street

Fig. 11–30

Vortex Shedding. When the flow around a cylinder is at a low Reynolds number, laminar flow prevails, and the boundary layer will separate from the surface symmetrically from each side, forming eddies that rotate in opposite directions as shown in Fig. 11–30. As the Reynolds number increases, these eddies will elongate, and one will begin to break away from one side of the surface before another breaks away from the other side. This stream of alternating vortex shedding causes the pressure to fluctuate on each side of the cylinder, which in turn tends to cause forced vibrations of the cylinder perpendicular to the flow. Theodore von Kármán was one of the first to investigate this effect, and the vortex stream so formed is often called a von Kármán vortex trail or vortex street, so named because the alternate vortices are placed like houses along a street. The frequency f at which the vortices are shed from each side of a cylinder of diameter D is a function of the Strouhal number, which is defined as St =

This tall thin-walled metal chimney can be subjected to severe wind loadings. At a critical wind speed, its cylindrical shape will produce a von Kármán vortex street that will shed off each side as noted in Fig. 11–30. This can cause the chimney to oscillate perpendicular to the direction of the wind. To prevent this, the spiral winding, referred to as a fence or “strake,” disturbs the flow and prevents the formation of the vortices.

fD V

This dimensionless number is named after the Czech scientist Vincenc Strouhal, who studied this phenomenon as it pertains to the “singing” noise developed by wires suspended in an airstream. Empirical values for the Strouhal number, as it relates to the Reynolds number for flow around a cylinder, have been developed and can be found in the literature. See Ref. [15]. At very high Reynolds numbers, however, the vortex shedding tends to disintegrate into uniform turbulence off both sides of the cylinder, and so the vibrations will tend to die out. In spite of this, conditions can occur where the oscillating force caused by vortex shedding produces vibrations that must be taken into account when designing structures such as tall chimneys, antennas, submarine periscopes, and even the cables of suspension bridges.

11.8

11.8

the drag CoeFFiCient

609

THE DRAG COEFFICIENT

As previously stated, the drag and lift on a body are a combination of the effects of both viscous friction and pressure, and if the shear and pressure distributions over the surface can be established, then these forces can be determined, in the manner demonstrated by Example 11.9. Unfortunately, the distributions of p and especially t are generally difficult to obtain, either through experiment or by some analytical procedure. A simpler method for finding the drag and lift is to directly measure them through experiment. In this section we will consider how this is done for drag, and later, in Sec. 11.11, we will consider lift. It has become a standard procedure in fluid mechanics to express the drag in terms of the fluid’s dynamic pressure, the projected area Ap of the body into the fluid stream, and a dimensionless drag coefficient CD. The relation is FD = CD Ap a

rV 2 b 2

(11–30)

The value of CD is determined from experiments, normally performed on a prototype or model placed in a wind or water tunnel or in a channel. For example, CD for a car is determined by placing it in a wind tunnel, and then, for each wind velocity V, a measurement is made of the horizontal force FD required to prevent the car from moving. The value of CD for each velocity is then determined from CD =

FD (1>2)rV 2Ap

Specific values of CD are generally available in engineering handbooks and industrial catalogs for a wide variety of shapes. Its value is not constant;

Acceleration occurs when the weight of these people is greater than the drag force caused by the air. As their speed increases, the drag increases until at terminal velocity, equilibrium is achieved at about 200 km >h or 120 mi >h.

11

610

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rather, experiments indicate that this coefficient depends on a number of factors, and we will now discuss some of these for several different cases.

11

Reynolds Number. In general, the drag coefficient is highly dependent on the Reynolds number. Consider, for example, objects or particles that have a very small size and weight, such as powder falling through air, or silt falling through water. They have a very low Reynolds number (Re V 1), and so no flow separation occurs from their sides. As a result, the drag is due only to viscous friction caused by laminar flow. If the object has a spherical shape, then for laminar flow the drag can be determined analytically using a solution developed in 1851 by George Stokes. He obtained his result by solving the Navier–Stokes and continuity equations, and his result was later confirmed by experiment. See Ref. [7]. It is* FD = 3pmVD Using the definition of the drag coefficient in Eq. 11–30, where the projected area of the sphere into the flow is Ap = (p>4)D2, we can now express CD in terms of the Reynolds number, Re = rVD>m. It is CD = 24>Re

(11–31)

Experiments have shown that CD for other shapes also has this reciprocal dependence on Re, provided Re … 1. See Ref. [19].

Cylinder. Experimental values of CD for flow around the sides of a long smooth and a rough cylinder as a function of the Reynolds number are shown on the graph in Fig. 11–31. Ref. [16]. Notice that as the Reynolds number increases, CD begins to decrease, until it reaches a constant value at about Re ≈ 103. From this point until Re ≈ 105, laminar flow continues around the cylinder until it begins to separate from the surface. The sudden drop in CD, at Reynolds numbers within the high range 105 6 Re 6 106, occurs because the boundary layer over the surface changes from laminar to turbulent, thereby causing the separation point for the flow to occur farther downstream on the back of the cylinder. As mentioned in the previous section, this results in a smaller region for the wake and a lower pressure drag. Although there is a slight increase in the viscous drag, the net effect will still cause the total drag to be reduced. Also, notice that the drop in CD occurs sooner for the cylinder with the rough surface, Re ≈ 61104 2, because a rough surface disturbs the boundary layer into becoming turbulent sooner than a smooth surface.

*This equation is sometimes used to measure the viscosity of a fluid when it has very high viscosity. The experiment is done by dropping a small sphere of known weight and diameter into the fluid contained in a long cylinder and measuring the sphere’s terminal velocity. Then m = FD >(3pVD). Further details are given in Example 11.11 as to how to obtain the needed value of FD.

11.8

611

the drag CoeFFiCient

400 200 100 60 40

11

20 10 6

CD

4

CD 5 24 Re

Rough cylinder

Smooth cylinder

2 1 0.6 0.4

Smooth sphere

0.2 0.1 0.06 10 –1

10 0

10 1

10 2

10 3

10 4

Re Drag coefficients for a sphere and long cylinder

Fig. 11–31

Sphere. When a fluid flows around a three-dimensional body, it behaves in much the same way as it does in a two-dimensional case. However, here the flow can go around the ends of the body as well as around its sides. This has the effect of extending the point of flow separation and thereby lessens the drag on the body. For example, the drag acting on a smooth sphere is also shown in Fig. 11–31, Ref. [18]. Note that it initially follows Stokes’ equation for Re 6 1. The shape of this curve is similar to that for the cylinder; however, for high Reynolds numbers, the values of CD for the sphere are about half of those obtained for the cylinder. As before, the characteristic drop in CD at high Reynolds numbers is attributed to the transition from a laminar to a turbulent boundary layer; and as with the cylinder, a rough surface on a sphere will also contribute a further drop in CD. This is why manufacturers put dimples on golf balls and roughen the surface of tennis balls. Such objects move at high speeds, and so, within the range of the Reynolds number where this occurs, the rough surface will produce a lower CD, and the ball will travel farther than it would if it had a smooth surface.

10 5

10 6

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11

The bulb on the lower bow of this ship significantly reduces the height of the bow wave and the turbulent flow around the bow. The reduced wave drag coefficient thereby saves fuel costs, since the total drag on the ship is reduced.

CD 1.8 Blunt body 1.6 1.4

Shock wave

1.2 1.0

oVer

e x t e r n a l s u r Fa C e s

Froude Number. When gravity plays a dominant role on drag, then the drag coefficient will be a function of the Froude number, Fr = V> 2gl. Recall from Sec. 8.5 that both the Froude and Reynolds numbers are used for similitude studies of ships. These numbers are important, since the drag is produced by both viscous friction on the hull and the lifting of water to create waves. We showed how models can be tested to study these effects independently. In particular, the wave drag is found from an experimental relationship between the wave drag coefficient (CD)wave and the Froude number. If this data is plotted for different hull geometries of a ship’s model, it is then possible to make comparisons before choosing a proper shape for design. Mach Number. When the fluid is a gas, such as air, compressibility effects may have to be considered when determining the drag on the body. As a result, the drag coefficient will then be a function of both the Reynolds number and the Mach number, since, apart from pressure, the dominant forces on the body will be caused by inertia, viscosity, and compressibility.* Although the drag is affected by both viscous friction and the pressure on the body, when compressibility becomes significant, the effect of these two components will be quite different than when the flow is incompressible. To understand why this is so, consider the variation in CD versus M for a blunt body (cylinder) and one with a tapered or conical nose, Fig. 11–32, Ref. [6]. At low Mach numbers (M 66 1), or velocities well below the speed of sound, the drag is primarily affected by the Reynolds number. For both geometries this will cause only a slight increase in CD. As sonic flow is approached there is a sharp increase in CD. At the point M = 1, a shock wave is formed on or in front of the body. Shock waves are very thin, on the order of 0.3 μm, and cause abrupt changes in the flow characteristics, as we will discuss in Chapter 13. What is important here is to realize that a sudden increase in pressure occurs across the wave, and this increase causes an additional drag on the body. In aeronautics it is referred to as wave drag. At such high speeds, it becomes independent of the Reynolds number, and instead, the pressure change over the shock wave produces the predominant amount of the drag.

0.8 0.6

Tapered body

0.4 0.2 Shock wave M 1

2

3

4

Drag coefficient for a blunt body and a tapered body

Fig. 11–32

*Recall that the Mach number is M = V>c, were c is the speed of sound measured in the fluid.

11.9

drag CoeFFiCients For Bodies haVing Various shapes

By comparison, note that the greatest reduction in drag occurs when the nose is tapered, as on a jet aircraft, since then the wave drag is confined to the small “front” region on the nose. As the Mach number increases, for both the blunt and tapered bodies, the shock wave becomes more inclined farther downstream, and the width of the wake behind the wave is thereby reduced, causing CD to decrease. Incidentally, the shape of the back end of both the blunt and tapered bodies has very little effect on reducing CD for supersonic flow, because as stated, the predominant drag is caused by the shock wave. On the other hand, for subsonic flow, no shock wave is formed, and so the viscous drag is reduced if the back end is tapered, since this discourages flow separation.

11.9 DRAG COEFFICIENTS FOR BODIES HAVING VARIOUS SHAPES The flow around bodies having a variety of different shapes follows the same behavior as flow around a cylinder and a sphere. Experiments have shown that for large Reynolds numbers, generally on the order of Re 7 104, the drag coefficient CD is essentially constant for many of these shapes, because the flow will separate at the sharp edges of the body and thereby become well defined. Typical values of drag coefficients for some common shapes are given in Table 11–3, for Re 7 104. Many other examples can be found in the literature. See Ref. [19]. Realize that in general each value of CD depends not only on the body’s shape and the Reynolds number, but also on the body’s surface roughness, and the angle at which the body is oriented relative to the flow.

Applications. If a composite body is constructed from the shapes listed in Table 11–3, and each is fully exposed to the flow, then the drag, as determined from Eq. 11–31, FD = CD Ap 1rV 2 >22, becomes a superposition of forces calculated for each shape. When using the table, or other available data for CD, some care should be exercised by considering the conditions under which it is applied. In actual practice, nearby bodies can influence the pattern of flow around the object being studied, and this can greatly affect the actual value of CD. For example, in a large city, structural engineers construct models of existing buildings surrounding the building to be designed, and then test this entire system in a wind tunnel. Occasionally, the complicated pattern of wind flow will lead to an increased pressure loading on the building being considered, and so this increase must be taken into account as part of the design loading. Finally, remember that tabulated data is normally prepared for steady uniform flow, something that actually never occurs in nature. Consequently, one should exercise good engineering judgment, based on experience and intuition about the flow, whenever any value of CD is selected.

613

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TABLE 11–3 Drag Coefficients for Simple Geometric Shapes, Re + 104 L

0.5 1 2

CD 5 2.3

11

D D

Re = UD n

1.1 0.95 0.86

Re = UD n Cylinder

Hollow semicylinder

CD 5 0.39

D

D

u

CD 5 1.4

CD

LD

CD 5 1.20

Re = UD n

CD

308 608

0.55 0.80

Re = UD n

Cone

Hollow hemisphere

u

a D

Re = Ua n

CD 5 0.43

Re = UD n

CD 5 1.1

a Cube

Solid hemisphere

b

D h

CD 5 1.1

bh

CD

, 0.1 0.5 1 5 10

1.9 2.5 1.18 1.2 1.3

Re = UD n Re = Uh n Disk

Rectangular plate

D

a CD 5 1.4

Re = Ua n Triangular cross section

CD 5 0.04

Re = UD n Streamlined body

11.9

drag CoeFFiCients For Bodies haVing Various shapes

615

IMPORTANT POIN T S • Flow can create two types of drag on an object: a tangential viscous friction drag caused by the boundary layer, and a normal pressure drag caused by the momentum change of the fluid stream. The combined effect of these forces is represented by a dimensionless drag coefficient CD, which is determined by experiment.

• When fluid flows over a curved surface, a pressure gradient is developed over the surface. As the velocity increases, the pressure becomes less, or the gradient decreases (favorable), and as the velocity decreases, the pressure becomes greater, or the gradient increases (adverse). With a further decrease in velocity, the flow will eventually separate from the surface, and form a wake or region of turbulence behind the body. Experiments have shown that the pressure within the wake is essentially constant and is about the same as it is within the free-stream flow.

• The separation of boundary layer flow from a curved surface will give rise to an unsymmetrical pressure distribution over the surface of the body, and this produces the pressure drag.

• Uniform flow of an ideal fluid around a symmetric body will produce no boundary layer, and hence no shear stress on the surface, because there is no fluid viscosity. Also, the pressure distribution around the body will be symmetric, and therefore it produces a zero force resultant. Consequently, the body will not be subjected to drag.

• Cylindrical surfaces can be subjected to boundary layer separation, which can lead to alternating vortex shedding at high Reynolds numbers. This can produce oscillating forces on the cylinder that must be considered in design.

• Experimentally determined drag coefficients, CD, for cylinders, spheres, and many other simple shapes are available in the literature. Specific values either are tabulated or are presented in graphical form. In either case, CD is a function of the Reynolds number, the body’s shape, its surface roughness, and its orientation relative to the flow. For some applications, fluid forces other than viscosity become important, and so CD may, for example, also depend upon the Froude number or the Mach number.

11

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EXAMPLE

11

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11.10 The hemispherical dish in Fig. 11–33 is subjected to a direct uniform wind speed of 15 m>s. Determine the moment of the force created by the wind on the dish about the base of the post at A. For air, take ra = 1.20 kg>m3 and ma = 18.1 (10 - 6) N # s>m2.

0.5 m

SOLUTION

3m

Fluid Description. Because of the relatively slow velocity, we will assume the air is incompressible and the flow is steady. A

Analysis. The “characteristic length” for the dish is its diameter, 1 m. Therefore, the Reynolds number for the flow is

Fig. 11–33

Re =

raVD = ma

1 1.20 kg>m3 2 (15 m>s)(1 m) 18.1(10 - 6) N # s>m2

= 9.94 1 105 2 7

1 104 2

From Table 11–3, the drag coefficient is CD = 1.4, so applying Eq. 11–30 yields FD = CD Ap a

r aV 2 b 2

= 1.43p(0.5 m) 4 £ 2

(1)

1 1.20 kg>m3 2 (15 m>s)2 2

§

= 148.44 N Notice how much the diameter and velocity affect this result, since each is squared. Due to the uniformity of the wind distribution, FD acts through the geometric center of the dish. The moment of this force about A is therefore M = (148.44 N)(3 m) = 445 N # m

Ans.

An additional moment of the wind loading on the post (cylinder) can also be included, and if it is, then CD is determined from Fig. 11–31 and its projected area is used in Eq. 1.

11.9

EXAMPLE

drag CoeFFiCients For Bodies haVing Various shapes

11.11

The 0.5-kg ball has a diameter of 100 mm and is released into the tank of oil, Fig. 11–34a. Determine its terminal velocity as it falls downward. Take ro = 900 kg>m3 and mo = 0.0360 N # s>m2.

11

SOLUTION Fluid Description. Relative to the ball we initially have unsteady flow, until the ball reaches its terminal velocity. We assume the oil is incompressible. Analysis. The forces acting on the ball include its weight, buoyancy, and drag, Fig. 11–34b. Since equilibrium occurs when the ball reaches terminal velocity, we have + c ΣFy = 0; Fb + FD - mg = 0 The buoyancy force is Fb = rogV, and the drag is expressed by Eq. 11–30. Thus r oV t 2 rogV + CD Ap a b - mg = 0 2 1900 kg>m 219.81 m>s 2 3 3

2

4 3 p(0.05

m) 4 + CDp(0.05 m) c 3

2

1900 kg>m3 2Vt 2 2

- (0.5 kg)19.81 m>s2 2 = 0 CDVt 2 = 0.07983 m2 >s2 (1) The value for CD is found from Fig. 11–31, but it depends upon the Reynolds number. 1900 kg>m3 21Vt 2(0.1 m) r oVt D = 2500Vt (2) Re = = mo 0.0360 N # s>m2 The solution will proceed using an iteration process. First, we will assume a value of CD and then calculate Vt using Eq. 1. This result will then be used in Eq. 2 to calculate the Reynolds number. Using this value in Fig. 11–31, the corresponding value of CD is obtained. If it is not close to the assumed value, we must repeat the same procedure until they are approximately equal. The iterations are tabulated. Iteration

617

CD (Assumed)

Vt (m>s) (Eq. 1)

Re (Eq. 2)

CD (Fig. 11–31)

1

1

0.2825

706

0.55

2

0.55

0.3810

952

0.50

3

0.50

0.3996

999

0.48 (okay)

Therefore, from the table, the terminal velocity is Vt = 0.3996 m>s = 0.400 m>s

Ans.

(a)

Fb FD

d

mg (b)

Fig. 11–34

618

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EXAMPLE

11

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11.12 The sports car and driver in Fig. 11–35a have a total mass of 1.5 Mg. The car is traveling at 11 m>s when the driver puts the transmission into neutral and allows the car to freely coast until after 40 s its speed reaches 10 m>s. Determine the drag coefficient for the car, assuming its value is constant. Neglect rolling and other mechanical resistance effects. The projected front area of the car is 2 m2.

(a)

Fig. 11–35

SOLUTION Fluid Description. Relative to the car we have uniform unsteady flow because the car is slowing down. We will assume the air is incompressible. At standard temperature, ra = 1.23 kg>m3. Analysis. Since the car can be considered as a rigid body, we can apply the equation of motion, ΣF = ma, and then use a = dV>dt to relate the car’s velocity to the time. Alternatively, although it amounts to the same thing, we can choose the car as a “control volume,” draw its free-body diagram, Fig. 11–35b, and then apply the momentum equation, + S ΣFx

=

0 Vr dV + VrVf>cs dA 0t Lcv Lcs

Since there are no open control surfaces, the last term is zero. The first term on the right can be simplified, noting that Vr is independent of its position within the volume of the car, and so 1cvdV = V. But rV = m, the mass of the car and driver, and therefore, the above equation becomes + S ΣFx

=

d(mV ) dV = m dt dt

11.9

drag CoeFFiCients For Bodies haVing Various shapes

619

W FD

11 NB

NA (b)

Finally, since FD is the drag on the car, we have -CD Ap a

r aV 2 dV b = m 2 dt

Separating the variables V and t, and integrating, we get t

V

1 dV CD Apra dt = -m 2 2 L0 LV0 V t 1 1 V CD A pr a t ` = m ` 2 V V0 0

1 1 1 C A r t = ma b 2 D p a V V0

Substituting the data, 1 1 1 b CD 12 m2 211.23 kg>m3 2(40 s) = [1.51103 2kg]a 2 10 m>s 11 m>s CD = 0.277

Ans.

Aerodynamic design of the 2014 C7 corvette used in this example was based on both a CFD analysis and over 700 hours of wind tunnel testing. The purpose was to achieve an optimum balance of zero lift and required airflow for mechanical cooling, while maintaining a low CD. (© General Motors Corporation)

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11.10 METHODS FOR REDUCING DRAG

11

Streamlined body

Fig. 11–36

In Sec. 11.7 it was shown that when a cylinder has a rough surface, turbulence occurs sooner in the boundary layer, thereby moving the point of flow separation farther back on the cylinder, Fig. 11–29a. As a result, pressure drag is reduced. Another way of moving the point of separation back is to streamline the body, such that it takes the form of a teardrop, as shown in Fig. 11–36. Although pressure drag is reduced, more of the surface is in contact with the fluid stream, and so the friction drag is increased. The optimal shape occurs when the total drag, which is a combination of both the pressure drag and the friction drag, is a minimum. The flow around an irregularly shaped body can be complex, so the optimal shape for any streamlined body must be determined by experiment. Also, a design that works well within one range of Reynolds numbers may not be as effective within other ranges. As a general rule, however, for low Reynolds numbers the viscous shear will create the maximum component of drag, and for high numbers the pressure drag component will dominate.

Airfoils. A common streamlined shape is an airfoil, Fig. 11–37a, and the drag acting on it depends upon its angle of attack a with the freeflow airstream, Fig. 11–37b. As shown, this angle is defined from the horizontal to the cord of the wing, that is, the length measured from the wing’s leading edge to its trailing edge. Notice that as a increases, the point of flow separation will move towards the leading edge, and this causes the pressure drag to increase, because the inclination of the wing projects a larger area of its bottom side into the airstream and in addition the pressure on the top side is diminished.

Separation point

Cord

(a)

Separation point

a

Flow separation at angle of attack a (b)

Fig. 11–37

11.10 methods For reduCing drag

621

Design. To properly design an airfoil to reduce pressure drag, the separation point should therefore be as far back from the leading edge as possible. To accomplish this, modern wings have a smooth surface on the front region of the wing, to maintain a laminar boundary layer, and then, at the point where the transition to turbulence occurs, the boundary layer is energized, either by a rough surface or by using vortex generators, which are small protruding fins on the top of the wing. Both methods allow the boundary layer to cling farther back onto the wing’s surface, and although this increases friction drag on the wing, it will decrease the pressure drag. Apart from defining an appropriate shape for an airfoil, aeronautical engineers have devised other methods for boundary layer control. Delayed separation for large angles of attack can be achieved by using slotted flaps or leading-edge slots, as shown in Fig. 11–38. These devices are designed to transfer fast-moving air from the bottom of the wing to its upper surface in an effort to energize the boundary layer. Another method produces suction of the slow-moving air within the boundary layer, either through slots or through the use of a porous surface. Both of these methods will increase the speed of the flow within the boundary layer and thus delay its separation. They also have the advantage of thinning the boundary layer, and thus delaying its transition from laminar to turbulent. Achieving these designs can be difficult, however, since it requires some ingenuity in addressing the structural and mechanical problems that can arise.

FD = CDApl a

rV 2 b 2

Leading edge slots

Fig. 11–38

0.44 0.40 Drag Coefficient, (CD)`

Airfoil Drag Coefficients. The drag on airfoils has been studied extensively by the National Advisory Committee for Aeronautics (NACA).* They, and others, have published graphs that have been used by aeronautical engineers to determine the section drag coefficient, 1CD 2 ∞ , that applies to aircraft wings of various shapes. As stated this coefficient is for section drag; that is, it assumes the wing has an infinite length so that flow around the wing tip is not considered. A typical example, for a 2409 wing profile, showing (CD) ∞ vs. the angle of attack is given in Fig. 11–39. The additional flow around the wing tip produces an induced drag on the wing, and we will study its effects in the next section. Once 1CD 2 ∞ and the induced drag coefficient 1CD 2 i are known, then the “total” drag coefficient CD = 1CD 2 ∞ + 1CD 2 i can be determined. The drag on the airfoil is then

Slotted flap

a

0.36 0.32 0.28 0.24 0.20 0.16 0.12 0.08

(11–32)

0.04 0.0 0° 4° 8° 12° 16° 20° 24° 28° 32°

Here Apl represents the planform of the wing, that is, the projected area of its top or bottom. *In 1958, this agency was incorporated into the newly created National Aeronautics and Space Administration (NASA).

Angle of attack, a Section drag coefficient (CD)` for NACA airfoil 2409 wing of infinite span

Fig. 11–39

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11

Modern trucks have rounded grills in their front, skirts along the bottom of their sides, and a roof fairing, which if properly fitted will reduce drag, and this alone will produce fuel savings of about 6%. Also, the fenders on the back side will further reduce turbulence and decrease drag.

Road Vehicles. Through the years it has become important to reduce the aerodynamic drag on cars, buses, and trucks to save on fuel consumption. Although streamlining is limited for these cases by length restrictions on the vehicle, it is possible to decrease CD for cars by redesigning their front and back profiles, rounding the forward surfaces of sideview mirrors, recessing the door handles, eliminating outside antennas, and rounding the corners of the body. By doing this, automotive engineers have been able to reduce drag coefficients from values as high as 0.60 to around 0.30. For freight trucks, drag coefficients can be as high as 1.35. However, a reduction of about 20% can be achieved by adding a roof fairing and wind deflectors that direct the airstream smoothly around the cab and along the bottom sides of the trailer. See the accompanying photos. The drag coefficient for any type of vehicle is a function of the Reynolds number; however, within the typical range of highway speeds, the value of CD is practically constant. Table 11–4 lists values for some other types of vehicles, although specific values for any particular vehicle can be obtained from published literature. For example, see Ref. [23]. Once CD is obtained, the drag can then be determined using Eq. 11–30, that is FD = CDAp a

rV 2 b 2

Here Ap is the projected area of the vehicle into the flow.

11.10 methods For reduCing drag

623

TABLE 11–4 Drag Coefficients for Vehicles Bicycle CD = 191.5 Ap = 0.42 m2 90.56 m2

11

Sports Car (Corvette C5) CD = 0.29 Ap = 2.0 m2 Automobile (Toyota Prius) CD = 0.25 Ap = 1.9 m2 Sport Utility Vehicle (SUV) CD = 0.3590.4 Ap = 2.3 m2 Utility Truck CD = 0.690.8 Ap = 3.0 m2 Freight Truck CD = 0.9591.35 Ap = 8.9 m2 Train CD = 1.892.3 Ap = 14 m2

Parachute CD = 1.291.6 p Ap = D2 4

D

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11.11 LIFT AND DRAG ON AN AIRFOIL The effect of a fluid flowing over the surface of a body not only creates drag on the body but it can also create lift. We have mentioned that the drag acts in the direction of motion of the airstream, but the lift acts perpendicular to it.

11 Higher velocity Lower pressure

Lower velocity Higher pressure

Fig. 11–40

Airfoil Lift. The phenomenon of lift produced by an airfoil or wing can be explained in different ways. Most commonly, the Bernoulli equation is used to do this.* Basically, the argument states that the higher the velocity, the lower the pressure, and vice versa 1p>g + V 2 >2g = const.2. Since the flow over the longer top of an airfoil, Fig. 11–40, is faster than that beneath its shorter bottom, the pressure on top will be lower than on the bottom, and so the resultant force of these distributions produces lift. Lift, however, can more fully be explained by using the Coanda effect, which was discussed in Sec. 11.7. To show this, consider the airstream that passes over the wing’s surface in Fig. 11–41a. Because adhesion causes the air to cling to the surface, the layers of air just above the surface begin to move faster and faster as they build and form the boundary layer, until the speed of the air eventually matches the uniform speed of the airflow relative to the wing, Va>w. The pressure difference between these moving layers causes the flow to bend in the direction of each slower-moving layer. In other words, the air is forced to follow the curved surface of the wing. This effect of redirecting the airstream propagates upward from the surface of the wing at the speed of sound, since the pressure within the airstream tends to prevent the formation of voids between fluid layers. As a result, a very large volume of air above the wing will be redirected downward and ultimately produce a Vaw

Vaw

Segment of airflow over wing as seen from wing (a)

(b)

Fig. 11–41 *This equation gives a conceptual reason why lift can occur, but it cannot be used to actually calculate the lift, because adjacent air particles that begin to move over the top and bottom of the wing do not meet at the back of the wing.

625

11.11 liFt and drag on an airFoil

“downwash” behind the wing, Fig. 11–41b. If the wing moves through calm still air with a velocity Vw, as in Fig. 11–41c, then the velocity of the air coming off the trailing edge of the wing, as observed from the wing (or by the pilot), will be Va>w. By vector addition, Fig. 11–41d, the “downwash” velocity of the air as seen by an observer on the ground will almost be vertical since Va = Vw + Va>w. In other words, when a plane flies close overhead, a ground observer will feel the airstream directed somewhat vertically downward. The bending of the air around the wing in the manner just described in effect gives the air an (almost) vertical momentum. To do this, the wing must produce a downward force on the airstream, and by Newton’s third law, the airstream will produce an equal but opposite upward force on the wing. It is this force that produces lift. Notice from the pressure distribution, Fig. 11–42a, that the greatest lift (greatest negative or suction pressure) is produced on the front top third of the wing, because here the airstream must bend the greatest amount to follow the wing’s surface. On the bottom surface there is also a component of lift, caused by a redirection of flow, although here the pressure is positive. Of course, this resultant lift will be even larger if the wing profile is somewhat curved or cambered, as in Fig. 11–42b.

Circulation. In Sec. 7.11 we showed that an ideal fluid will cause lift on a rotating cylinder when we superimpose a circulation Γ about the cylinder, while the cylinder is subjected to a uniform flow. Martin Kutta and Nikolai Joukowski have independently shown that the lift, as calculated from Eq. 7–74, L = rU Γ, also holds for any closed-shaped body subjected to two-dimensional flow. This important result is known as the Kutta–Joukowski theorem, and in aerodynamics it is often used to estimate the lift since it has been shown to be in close agreement with the lift measured experimentally for airfoils (and hydrofoils) having low angles of attack.

FL

FL Low pressure (suction)

High pressure

Pressure distribution

Pressure distribution–cambered airfoil

(a)

(b)

Fig. 11–42

Vaw Vw

11 (c) Vw Vaw Va Va 5 Vw 1 Vaw (d)

Fig. 11–41 (cont.)

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To show how this works, consider the airfoil in Fig. 11–43a when it is subjected to the uniform flow U. Here stagnation points develop on the leading edge A and back at the top B of the trailing edge. However, ideal fluids cannot support this flow, because an ideal fluid would have to round the bottom of the trailing edge, and then move up to the stagnation point at B. Physically, this is not possible, since it would take an infinite normal acceleration to change the direction of the velocity to round a sharp edge. Therefore, in order to bring the flow in line, so that it leaves the tail smoothly, in 1902 Kutta proposed adding a clockwise circulation Γ about the airfoil, Fig. 11–43b. In this way, when the two flows, Fig. 11–43a and 11–43b, are superimposed, air on top of the airfoil moves faster than air on the bottom, and the stagnation point B moves to the trailing edge, Fig. 11–43c. The faster-moving air on top then has a lower pressure, whereas the slower-moving air on the bottom has a higher pressure, and this pressure difference gives rise to the lift as calculated by L = r UΓ.

11

G B

A

Flow without circulation

Flow with circulation

(a)

(b)

A

B

Superposition of flow (c)

Fig. 11–43

11.11 liFt and drag on an airFoil

Experimental Data. Although a lift coefficient CL can be

2.0

calculated analytically using the circulation for small angles of attack, for larger angles the values of CL must be determined by experiment. Values of CL are usually plotted as a function of the attack angle a, and they look something like Fig. 11–44, which, again, is for the 2409 NACA wing section. With this data, and the planform area Apl of the wing, the lift force is then calculated using the equation

1.8

rV 2 b 2

(11–33)

It is interesting to note what happens to the lift as the angle of attack increases. As shown in Fig. 11–45a, for a properly designed airfoil, the point of separation of the boundary layer will be near the trailing edge of the wing when the attack angle is zero. As a increases, however, this forces the air within the boundary layer to move faster over the top of the leading edge, causing the point of separation to move forward. When the angle of attack reaches a critical value, a large turbulent wake over the top surface of the wing develops, which increases the drag and causes the lift to suddenly drop off. This is a condition of stall, Fig. 11–45b. In Fig. 11–44, it is the point having the maximum lift coefficient, CL ≈ 1.5, which occurs at a ≈ 20°. Obviously, a stall is dangerous for any low-flying aircraft that may not have enough altitude to recover level flight. Apart from changing the angle of attack to generate lift, modern aircraft also have moveable flaps on their leading and>or trailing edges to increase curvature of the wing, Fig. 11–46. They are used during takeoffs and landings, when the velocity is low and the need for controlling lift relative to drag is most important.

Race Cars. Airfoils, such as those described here, are also on racing cars. See the photo on p. 604. These devices, along with a unique body shape, are designed to enhance the braking or cornering of the car by creating a downward force due to the aerodynamic effects of the foil. Also, without an airfoil, lifting forces developed underneath the car may cause the tires to lose contact with the road, resulting in a loss of stability and control. Unfortunately, an airfoil used for this purpose can result in a higher-than-normal drag coefficient. For example, compare the drag coefficients for the vehicles in Table 11–4 to that of a Formula One race car, which can have a drag coefficient in the range of CD = 0.7–1.1.

Flap down

Flap up

Fig. 11–46

a

1.6 Lift Coefficient, CL

FL = CL Apl a

627

1.4 1.2

11

1.0 0.8 0.6 0.4 0.2 0.0 0° 4° 8° 12° 16° 20° 24° 28° 32° Angle of attack, a Lift coefficient CL for NACA airfoil 2409

Fig. 11–44

Separation point

(a)

Separation point

a

Stall (b)

Fig. 11–45

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The trailing vortex trail is clearly evident off the tip of the wing on this agricultural plane. It is made visible due to the condensation of water vapor within the low pressure vortex. (© NASA Archive/Alamy Stock Photo)

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Trailing Vortices and Induced Drag. Our previous discussion of drag produced by an airfoil (wing) was in reference to flow over the wing, with no consideration for the variation in the length of the wing or for conditions at the wing tips. In other words, the wing was thought to have an infinite span or to be confined between parallel walls. If we account for flow over an actual wing, then the air directed down off the trailing edge and tip will produce a swirl, called a wing vortex trail, that contributes an additional drag on the aircraft. See the adjacent photo. To show how this occurs, consider the wing in Fig. 11–47. Here the higher pressure on the bottom of the wing will cause the flow to travel up and around the trailing edge, and also around the tip of the wing. This flow off the tip will pull the flow to the left under the wing, and as it comes around, it pushes the lower pressure flow on top of the wing to the right. As a result, the cross flow off the trailing edge will form a multitude of small vortices—a vortex trail, that helps to form a much larger trailing vortex near the wing tip. The production of this disturbance requires energy, and so it places an extra burden on the lift, resulting in an induced drag, which must be taken into account when calculating the total drag and strength of the wing. Actually, the turbulence created in this manner by large aircraft can be substantial and may persist for several minutes, creating a hazard for lighter planes flying too close behind them. The induced drag on a typical jet aircraft generally amounts to 30% to 50% of the total drag, and at low speeds, such as during takeoffs and landings, this percentage is even higher. To reduce this force component, modern aircraft add split-scimitar winglets, or small turned up and down airfoils on the wing tips, as shown in the adjacent photo. Experiments in wind tunnels have shown that when winglets are used, the trailing vortices lose strength, leading to about a 5%–6% reduction in total aircraft drag at cruising speed, and to an even greater reduction during takeoffs and landings.

Most jet planes now have split-scimitar winglets at their wing tips in order to mitigate trailing vortices that produce induced drag. (© Konstantin Yolshin/Alamy Stock Photo)

Low pressure on top of wing

Vortex trail

High pressure under wing causes air to flow up and off tip and trailing edge

Fig. 11–47

Trailing vortex

629

11.11 liFt and drag on an airFoil

Induced Drag Coefficient. An airfoil of infinite length and traveling horizontally at V0 must only overcome its section drag, and so in flight it will be directed at the effective angle of attack a0 to maintain the lift, (FL)0, needed to support the weight of the airfoil (or plane), Fig. 11–48a. A finite-length wing span, however, loses some of its lift because of the disturbance created by the tip vortices, which create not only an induced drag, but also a small induced downward velocity Vi onto the wing. As a result, the air stream velocity relative to the wing becomes V = V0 + Vi, Fig. 11–48b. To provide the necessary lift, FL, which by definition must be perpendicular to V, it will therefore become necessary to change the angle of attack from a0 to a greater angle a. The difference in these angles, ai = a - a0, is very small, and so the actual lift FL needed to support the weight is only slightly greater than 1FL 2 0. Since the section drag is determined relative to the free-stream velocity of the air, V0, it is customary to resolve FL into components 1FL 2 0 and 1FD 2 i, which are parallel and perpendicular to V0. By vector addition, Fig. 11–48c, we see that the horizontal component 1FD 2 i is the induced drag, and so for small values of ai its magnitude can be related to the lift by 1FD 2 i ≈ FLai. Through both experiment and analysis, Prandtl has shown that if the air that is disturbed over the wing has an elliptical shape, as in Fig. 11–41b, which closely approximates many actual cases, then ai becomes a function of the lift coefficient CL, the length b of the wing, and its planform area Apl. His result is ai =

pb2 >Apl CL

11

(FL)

0

V0 a0 Infinite length wing (a)

FL V

(11–34) a

Since the induced drag and the lift coefficients are proportional to their respective forces (see Eqs. 11–32 and 11–33), then from Fig. 11–48c, ai =

1FD 2 i 1CD 2 i = FL CL

V0 ai V

And so, using Eq. 11–34, the induced drag coefficient is determined from 1CD 2 i =

Finite length wing

C 2L

pb2 >Apl

(b)

Notice that if the length b S ∞ , then 1CD 2 i S 0, as expected. The above equation represents the minimum induced drag coefficient for any wing shape, and if it is used, then the total drag coefficient for the wing is CD = 1CD 2 ∞ +

C 2L 2

pb >Apl

(FD)i

ai (FL)

FL

0

(11–35)

where 1CD 2 ∞ is the section drag determined from a graph such as Fig. 11–39.

(c)

Fig. 11–48

Vi

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Spinning Ball. Lift can also greatly affect the trajectory of a spinning V

11 Ideal uniform flow (a)

+ Circulation (b)

= Low pressure high velocity

High pressure low velocity Combined flow (c)

Fig. 11–49

ball, because the spin will alter the pressure distribution around the ball, thereby changing the direction of the momentum of the air. For example, consider the ball moving to the left, without spin, Fig. 11–49a. Here the air flows symmetrically around the ball, and so the ball experiences a horizontal drag but no lift. If we consider what happens only when the ball spins, we see that its surface will pull air around it, forming a boundary layer in the direction of spin, Fig. 11–49b. Adding these two effects produces the condition shown in Fig. 11–49c. That is, in both Fig. 11–49a and Fig. 11–49b, the air is flowing on top of the ball in the same direction. This increases its energy and allows the boundary layer to remain attached to the surface for a longer time. Air passing under the ball goes in the opposite direction for each case, thereby losing energy and causing early boundary layer separation. Both these effects create a net downwash behind the ball. As a reaction to this, like an airfoil, the air in turn pushes or lifts the ball upward as shown in Fig. 11–49c. This tendency of a rotating ball to produce lift as described here is called the Magnus effect, named after the German scientist Heinrich Magnus, who discovered it. Most people who play baseball, tennis, or ping-pong instinctively notice this phenomenon, and take advantage of it when throwing or striking a ball. Experimental data for the drag and lift coefficients for a smooth spinning ball is given in Fig. 11–50. It is valid for Re = VD>n … 61104 2. See Ref. [20]. Notice how, up to a point, the lift coefficient is highly dependent upon the ball’s angular velocity. Any increase in v after this point will hardly affect the lift. Further lift is possible by roughing the surface, since this causes turbulence and an increased circulation around the ball, providing an even larger pressure difference between its top and bottom. In addition, the rough surface will reduce the drag because boundary layer separation is delayed. CD ,CL 0.8

CD

0.6

0.4

CL v

0.2

U

vD2U

This rotating ball is suspended in the airstream due to the Magnus effect, that is, the lift provided by the rotation and redirection of the airstream balances the ball’s weight.

1

2

3

4

Drag and lift coefficients for a smooth spinning ball

Fig. 11–50

5

11.11 liFt and drag on an airFoil

631

IMPORTANT POIN T S • Streamlining a body tends to decrease the pressure drag on the body, but it has the effect of increasing • • • • •

friction drag. For proper design, the resultant of both of these effects should be minimized. Pressure drag is reduced on a streamlined body by extending the boundary layer over the surface, or preventing the boundary layer from separating from the surface. For airfoils, the methods include using wing slots, vortex generators, roughing the surface, and using a porous surface to draw air toward the boundary layer. Both drag and lift are important when designing an airfoil. These forces are related to their drag and lift coefficients, CD and CL, which are experimentally determined, and are represented graphically as a function of the angle of attack. Lift is produced by a wing which causes the airstream to be redirected as it passes over the wing. The force generated changes the direction of the air’s momentum, so the air flows downward. This results in an opposite force reaction of the air on the wing, which produces lift. Trailing vortices are produced off the wing tips of aircraft due to differences in pressure acting on the top and bottom surfaces of the wing. As the air cross-flows off the trailing edge and wing tips, these vortices produce an induced drag, which must be accounted for in design. The trajectory of a spinning ball will curve in a steady airstream because the spinning causes an unequal pressure distribution on the ball’s surface, and this results in a change in the direction of the air’s momentum, thereby producing lift.

EXAMPLE

11.13

The airplane in Fig. 11–51 has a mass of 1.20 Mg and is flying horizontally at an altitude of 7 km. If each wing is classified as a NACA 2409 section, with a span of 6 m and a cord length of 1.5 m, determine the angle of attack when the plane has an airspeed of 70 m>s. Also, what is the drag on the plane due to the wings, and what is the angle of attack and the speed that will cause the plane to stall?

6m

70 ms 1.5 m

Fig. 11–51

SOLUTION Fluid Description. We have steady flow when measured relative to the plane. Also, the air is considered incompressible. Using Appendix A, at 7-km altitude ra = 0.590 kg>m3.

11

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Angle of Attack. For vertical equilibrium, the lift must be equal to the weight of the plane, and so using Eq. 11–33, we can determine the required lift coefficient. Since we have two wings, FL = 2CLApl a

rV 2 b 2

1.201103 2 kg 19.81 m>s2 2 = 2CL(6 m)(1.5 m) °

10.590 kg>m3 2170 m>s2 2 2

¢

CL = 0.452 From Fig. 11–44, the angle of attack must be approximately a = 5° Ans. Drag. The section drag coefficient at this angle of attack is determined from Fig. 11–39. It is for a wing of infinite span. We have, approximately, (CD) ∞ = 0.02 The total drag coefficient for the wing is determined from Eq. 11–35. CD = (CD) ∞ +

CL2

pb2 >Apl

= 0.02 +

(0.452)2

p(6 m)2 >[(6 m)(1.5 m)]

= 0.02 + 0.0163 = 0.0363

Therefore, the drag on both wings is FD = 2CD Apl a

rV 2 b 2

= 2(0.0363)3(6 m)(1.5 m)4 °

10.590 kg>m3 2170 m>s2 2 2

FD = 944 N

¢ Ans.

Stall. From Fig. 11–44, stall will occur when the angle of attack is approximately 20° so that CL = 1.5. The speed of the plane when this occurs is then FL = 2CLApl a

rV 2 b 2

1.201103 2 kg 19.81 m>s2 2 = 2(1.5)(6 m)(1.5 m) ° Vs = 38.4 m>s

10.590 kg>m3 2Vs2 2

¢

Ans.

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References 1. T. von Kármán, “Turbulence and skin friction,” J Aeronautics and Science, Vol. 1, No. 1. 1934, p. 1. 2. W. P. Graebel, Engineering Fluid Mechanics, Taylor Francis, NY, 2001. 3. W. Wolansky et al., Fundamentals of Fluid Power, Houghton Mifflin, Boston, MA., 1985. 4. A. Azuma, The Biokinetics of Flying and Swimming, 2nd ed., American Institute of Aeronautics and

Astronautics, Reston, VA, 2006. 5. J. Schetz et al., Boundary Layer Analysis, 2nd ed., American Institute of Aeronautics and Astronautics, Reston,

VA, 2011. 6. J. Vennard and R. Street, Elementary Fluid Mechanics, 5th ed., John Wiley, 1975. 7. G. Tokaty, A History and Philosophy of Fluid Mechanics, Dover Publications, New York, NY, 1994. 8. E. Torenbeek and H. Wittenberg, Flight Physics, Springer-Verlag, New York, NY, 2002. 9. T. von Kármán, Aerodynamics, McGraw-Hill, New York, NY, 1963. 10. L. Prandtl and O. G. Tietjens, Applied Hydro- and Aeromechanics, Dover Publications, New York, NY, 1957. 11. D. F. Anderson and S. Eberhardt, Understanding Flight, McGraw-Hill, New York, NY, 2000. 12. H. Blasius, “The boundary layers in fluids with little friction,” NACA. T. M. 1256, 2/1950. 13. L. Prandtl, “Fluid motion with very small friction,” NACA. T. M. 452, 3/1928. 14. P. T. Bradshaw et al., Engineering Calculation Methods for Turbulent Flow, Academic Press, New York, NY,

1981. 15. O. M. Griffin and S. E. Ramberg, “The vortex street wakes of vibrating cylinders,” J Fluid Mechanics, Vol. 66,

1974, pp. 553–576. 16. H. Schlichting, Boundary-Layer Theory, 8th ed., Springer-Verlag, New York, NY, 2000. 17. F. M. White, Viscous Fluid Flow, 3rd ed., McGraw-Hill, New York, NY, 2005. 18. L. Prandtl, Ergebnisse der aerodynamischen Versuchsanstalt zu Göttingen, Vol. II, p. 29, R. Oldenbourg, 1923. 19. CRC Handbook of Tables for Applied Engineering Science, 2nd ed., CRC Press, Boca Raton, Fl, 1973. 20. S. Goldstein, Modern Developments in Fluid Dynamics, Oxford University Press, London, 1938. 21. J. D. Anderson, Fundamentals of Aerodynamics, 4th ed., McGraw-Hill, New York, NY, 2007. 22. E. Jacobs et al., “The characteristics of 78 related airfoil sections from tests in the variable-density wind

tunnel.” National Advisory Committee for Aeronautics, Report 460, U.S. Government Printing Office, Washington, DC. 23. A. Roshko, “Experiments on the flow past a circular cylinder at very high Reynolds numbers,” J Fluid

Mechanics, Vol. 10, 1961, pp. 345–356. 24. W. H. Huchs, Aerodynamics of Road Vehicles, 4th ed., Society of Automotive Engineers, Warrendale, PA, 1998. 25. S. T. Wereley and C. D. Meinhort, “Recent advances in micro-particle image velocimetry,” Annual Review of

Fluid Mechanics, Vol. 42, No. 1, 2010, pp. 557–576.

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P R OB L EMS SEC. 11.1–11.3 11 11–1. Air at 20°C is blowing at 2 m>s as it passes over the flat surface. Determine the disturbance and displacement thickness of the boundary layer at a distance of x = 0.5 m from the leading edge. What is the velocity of the flow at half the disturbance thickness? 2 ms

*11–4. Determine the maximum free-stream velocity of oil so that if it flows over a plate having a length of 3 m it maintains a laminar boundary layer for the entire length. Take no = 40110-6 2 m2 >s.

11–5. Water at 20°C has a free-stream velocity of 500 mm>s. Determine the drag on the plate’s surface if it has a width of 0.2 m.

11–6. Water at 20°C has a free-stream velocity of 500 mm>s. Plot the boundary layer disturbance thickness along the distance x, in increments of 0.1 m for 0 … x … 0.5 m.

x 500 mms

Prob. 11–1 11–2. A fluid has laminar flow and passes over the flat plate. If the disturbance thickness of the boundary layer at a distance of x = 0.5 m from the plate’s edge is 10 mm, determine the disturbance thickness at a distance of x = 1 m.

x

6 ms 0.5 m

Probs. 11–5/6 10 mm

x

Prob. 11–2

11–7. Wind flows along the side of the rectangular sign. If the air is at a temperature of 20°C and has a free-stream velocity of 1.5 m>s, determine the friction drag on both front and back surfaces of the sign.

11–3. Oil has laminar flow and passes over the flat plate. Plot the velocity profile a distance of x = 2 m from the leading edge of the plate. Using Table 11–1, give values of u for every 0.8 increment of 1y>x2 1Rex until u = 0.99U. Take no = 40110-6 2 m2 >s.

4m

5 ms

1.5 m/s

2m

y

x

Prob. 11–3

Prob. 11–7

635

proBlems *11–8. Water at 20°C flows over the top surface of the plate. If the free-stream velocity is 0.8 m>s, determine the boundary layer disturbance thickness and momentum thickness at the back end of the plate. 11–9. If the free-stream velocity of water at 20°C is 0.8 m>s, determine the friction drag acting on the surface of the plate. 300 mm

*11–12. If the disturbance thickness of a laminar boundary layer of oil at a distance of 0.75 m from a plate’s front edge is 10 mm, determine the free-stream velocity of the oil. Take no = 40110-6 2 m2 >s.

11–13. Castor oil flows over the surface of the 2-m-long flat plate at a free-stream speed of 2 m>s. Plot the 11 boundary layer and the shear stress versus x. Give values for every 0.5 m. Also calculate the friction drag on the plate. The plate is 0.5 m wide. Take rco = 960 kg>m3 and mco = 985110-3 2 N # s>m2.

600 mm 2 ms 0.8 m/s

Probs. 11–8/9 11–10. The boundary layer for wind blowing over rough terrain can be approximated by the equation u>U = 3y> 1y + 0.0224, where y is in meters. If the free-stream velocity of the wind is 10 m>s, determine the velocity at y = 0.2 m and at y = 0.4 m from the ground.

x 2m

Prob. 11–13

10 ms

11–14. Determine the force F that must be applied to the cable to lift the fully submerged 50-kg plate up to the surface of the glycerin at T = 20°C at a constant speed of 2 m>s. Include the effect of buoyancy.

y

Prob. 11–10

F

11–11. Determine the distance x = xcr over the flat plate to where the boundary layer for gasoline at T = 20°C begins to transition from laminar to turbulent flow. 10 mm 0.4 m/s

2m

3m x

Prob. 11–11

Prob. 11–14

636

C h a p t e r 11

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11–15. Water at 15°C, confined in a channel, flows past the diverter fin at U = 2 m>s. Determine the friction drag acting on both sides of the fin, which has a width of 900 mm. Neglect end effects.

11–18. The boat is traveling at 0.8 m>s through water having a temperature of 20°C. If the rudder can be assumed to be a flat plate having a height of 800 mm and a length of 600 mm, determine the friction drag acting on both sides of the rudder. 11–19. The boat is traveling at 0.8 m>s through still water having a temperature of 20°C. If the rudder can be assumed to be a flat plate, determine the boundary layer thickness at the trailing edge A. Also, what is the displacement thickness of the boundary layer at this point?

11

U 5 2 ms 200 mm

A

800 mm

Prob. 11–15 *11–16. Air at 60°C flows through the very wide duct. Determine the required dimension a of the duct at x = 4 m so that the central 200-mm-core flow velocity maintains the constant free-stream velocity of 0.5 m>s. 0.5 ms

200 mm

0.5 ms

a

600 mm

Probs. 11–18/19 *11–20. The laminar boundary layer for a fluid is assumed to be parabolic, such that u>U = C1 + C2(y>d) + C3(y>d)2. If the free-stream velocity U starts at y = d, determine the constants C1, C2, and C3. 11–21. The laminar boundary layer for a fluid is assumed to be cubic, such that u>U = C1 + C2(y>d) + C3(y>d)3. If the free-stream velocity U starts at y = d, determine the constants C1, C2, and C3. U

x54m d y

Prob. 11–16 11–17. Air at a temperature of 30°C flows at 3 m>s over the plate. Determine the distance x where the disturbance thickness of the boundary layer becomes 15 mm. 3 m/s

Probs. 11–20/21 11–22. Assume a laminar boundary layer for a fluid can be approximated by u>U = y>d. Determine the thickness of the boundary layer as a function of x and Rex. y U

80 mm d y x

Prob. 11–17

x

Prob. 11–22

637

proBlems 11–23. A laminar boundary layer for a fluid is approximated by u>U = sin (py>2d). Determine the thickness of the boundary layer as a function of x and Rex. *11–24. A laminar boundary layer for a fluid is approximated by u>U = sin (py>2d). Determine the displacement thickness d * for the boundary layer as a function of x and Rex.

SEC. 11.4–11.5 *11–28. The plane is flying at 450 km>h in still air at an altitude of 3 km. Determine the friction drag on the wing. Assume the wing is flat having the dimensions shown, and the boundary layer is completely turbulent. 11

U

12 m 2m

x

Probs. 11–23/24 11–25. The velocity profile for a laminar boundary layer of a fluid is approximated by u>U = 1.5(y>d) - 0.5(y>d)3. Determine the thickness of the boundary layer as a function of x and Rex. 11–26. The velocity profile for a laminar boundary layer of a fluid is approximated by u>U = 1.5(y>d) - 0.5(y>d)3. Determine the shear-stress distribution acting on the surface as a function of x and Rex. y U

d y x

Probs. 11–25/26 11–27. A boundary layer for laminar flow of a fluid over the plate is approximated by u>U = C1(y>d) + C2(y>d)2 + C3(y>d)3. Determine the constants C1, C2, and C3 using the boundary conditions when y = d, u = U; when y = d, du>dy = 0; and when y = 0, d 2u>dy2 = 0. Find the disturbance thickness of the boundary layer as a function of x and Rex using the momentum integral equation.

Prob. 11–28 11–29. An airplane is flying at 300 km>h in still air at an altitude of 2 km. If the wings can be assumed to be flat plates, each having a width of 1.5 m and a length of 5 m, determine the friction drag on each wing if the boundary layer is considered to be fully turbulent. 11–30. An airplane is flying at 300 km>h in still air at an altitude of 2 km. If the wings can be assumed to be flat plates, each having a width of 1.5 m, determine the disturbance boundary layer thickness at their midpoint and also at their trailing or back edge if the boundary layer is considered to be fully turbulent. 11–31. Assume the turbulent boundary layer for a fluid has a velocity profile that can be approximated by u = U(y>d)1>6. Use the momentum integral equation to determine the disturbance thickness as a function of x and Rex. Use the empirical formula developed by Prandtl and Blasius. *11–32. Assume the turbulent boundary layer for a fluid has a velocity profile that can be approximated by u = U(y>d)1>6. Use the momentum integral equation to determine the displacement thickness as a function of x and Rex. Use the empirical formula developed by Prandtl and Blasius. y U

U

d y x

x

Prob. 11–27

Probs. 11–31/32

638

C h a p t e r 11

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11–33. Determine the disturbance thickness of the boundary layer along the side of the trailer truck at its mid-length x = 2.5 m. The trailer travels at 90 km>h. The air is still and has a temperature of 30°C. Assume the surface is smooth and flat. 11

*11–36. The cargo ship is traveling forward at 15 km>h in still water having a temperature of 15°C. If the bottom of the ship can be assumed to be a flat plate of length 200 m and width 40 m, determine the friction drag of the water on the bottom of the ship. Consider both laminar and turbulent boundary layers.

5m 2.5 m x

15 km/h 90 km/h

200 m

Prob. 11–33 Prob. 11–36

11–34. Determine the drag acting on each side of the trailer truck when it is traveling at 90 km>h. The air is still and has a temperature of 30°C. Assume the surfaces are smooth and flat. Consider both laminar and turbulent boundary layers.

11–37. The wind tunnel operates using air at a temperature of 20°C with a free-stream velocity of 40 m>s. If this velocity is to be maintained throughout the central 1-m core of the tunnel, determine the dimension a at the exit in order to accommodate the growing boundary layer. Show that the boundary layer is turbulent at the exit, and use d * = 0.0463x>(Rex)1>5 to calculate the displacement thickness.

5m 2.5 m x

90 km/h

a

Prob. 11–34 1m

11–35. An airplane is flying at an altitude of 3 km and a speed of 750 km>h in still air. If each wing is assumed to have a flat surface of width 2 m and length 6 m, determine the friction drag acting on each wing. Consider both laminar and turbulent boundary layers.

a

6m 1m

Prob. 11–37

639

proBlems 11–38. The flat-bottom boat is traveling at 4 m>s on a lake for which the water temperature is 15°C. Determine the approximate drag acting on the bottom of the boat if the length of the boat is 10 m and its width is 2.5 m. Assume the boundary layer is completely turbulent.

11–41. An airplane has wings that each, on average, are 5 m long and 3 m wide. Determine the friction drag on the wings when the plane is flying at 600 km>h in still air at an altitude of 2 km. Assume the wings are flat plates and the boundary layer is completely turbulent. 11–42. If the wings of an airplane flying at a speed of 400 km>h 11 are assumed to have a flat surface of width 2 m, determine the disturbance thickness of the boundary layer and the shear stress at the midpoint of the wing. Assume the boundary layer is fully turbulent. The airplane flies at an altitude of 2 km.

4 ms

400 km/h A

B 2m 10 m

Prob. 11–38 11–39. The oil tanker has a smooth surface area of 4.5(103) m2 in contact with the sea. Determine the friction drag on its hull and the power required to overcome this force if the velocity of the ship is 2 m>s. Consider both laminar and turbulent boundary layers. Take r = 1030 kg>m3 and m = 1.14110-3 2 N # s>m2.

Prob. 11–42 11–43. The wings of an airplane flying at an altitude of 3 km and a speed of 450 km>h can be assumed to have a flat surface of width 2 m and length 8 m. Determine the friction drag on each wing. Assume the boundary layer is fully turbulent. 450 kmh B

A

2 ms

2m

300 m

Prob. 11–39 *11–40. A barge is being towed at 3 m>s. Determine the total friction drag of the water on its sides and bottom. The water is still and has a temperature of 15°C. The submerged depth of the barge is 1.5 m. Assume the boundary layer is completely turbulent.

Prob. 11–43 *11–44. The tail of the airplane has an approximate width of 0.6 m and a length of 1.5 m. Assuming the airflow onto the tail is uniform, plot the disturbance layer thickness d. Give values for every increment of 0.01 m for the laminar boundary layer, and every 0.1 m for the turbulent boundary layer. Also, calculate the friction drag on the tail. The plane is flying in still air at an altitude of 2 km with a speed of 600 km>h. Consider both laminar and turbulent boundary layers.

F 3 ms 600 km/h

1.5 m 20 m 6m 1.5 m 0.6 m

Prob. 11–40

Prob. 11–44

640

C h a p t e r 11

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SEC. 11.6 11–45. The windward roof on the house is 20 m wide and is subjected to the wind as shown. If the average absolute pressure on the top of the roof is 3 kPa, and under the roof, that is, within the attic, it is 2 kPa, determine the pressure 11 drag on the windward side of the roof.

*11–48. The pressure distribution of air as it passes over a cylinder is p = 36 - 16>p2u4 kPa. Determine the pressure drag acting on the cylinder for 0 … u … 180°. The cylinder has a length of 4 m.

p

10 m 30°

600 mm u

Prob. 11–45 Prob. 11–48 11–46. The front of the building is subjected to wind that exerts a pressure of p = 10.9y1>2 2 kPa, where y is in meters, measured from the ground. Determine the resultant pressure force on the windward face of the building due to this loading. 11–47. The building is subjected to a uniform wind having a speed of 35 m>s. If the temperature of the air is 10°C, determine the resultant pressure force on the front of the building if the drag coefficient is CD = 1.38.

11–49. Air pressure acting on the inclined surfaces is approximated by the linear distributions shown. Determine the resultant horizontal force acting on the surface if it is 3 m wide.

y 40 m 3 kPa 3 kPa 15 m 45°

5 kPa 20° 6m

Probs. 11–46/47

Prob. 11–49

6m

proBlems 11–50. The wall of the building is subjected to a wind that produces a pressure distribution that can be approximated by p = 1215.5ry1>2 2 Pa, where y is in meters. Determine the resultant pressure force on the wall. The air is at a temperature of 20°C, and the wall is 10 m wide.

641

*11–52. The motorcycle and passenger has a projected front area of 0.635 m2. Determine the power required to travel at a constant velocity of 108 km>h if the drag coefficient is CD = 0.72 and the air is at 30°C. 11 108 km/h

p 6m y

Prob. 11–52 11–53. The drag coefficient for the car is CD = 0.28, and the projected area into the 20°C airstream is 2.5 m2. Determine the power the engine must supply to maintain a constant speed of 160 km>h.

Prob. 11–50

160 km/h

SEC. 11.7–11.9 11–51. The antenna on the building is made from two smooth cylindrical sections as shown. Determine the restraining moment at its base to hold it in equilibrium if it is subjected to a wind having an average speed of 25 m>s. The air is at a temperature of 10°C.

200 mm

10 m

400 mm

15 m

Prob. 11–53 11–54. The uniform crate has a mass of 50 kg and rests on a surface having a coefficient of static friction of ms = 0.5. If the speed of the wind is 10 m>s, determine whether it will cause the crate to either tip over or slide. The air temperature is 20°C. Take CD = 1.06.

2m 2m

2m

Prob. 11–51

Prob. 11–54

642

C h a p t e r 11

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11–55. The rocket has a 60° nose cone and a base diameter of 1.25 m. Determine the drag of the air on the cone when the rocket is traveling at 60 m>s in air having a temperature of 10°C. Use Table 11–3 for the cone, but explain why this may not be an accurate assumption. 11 60°

11–59. A 5-m-diameter balloon is at an altitude of 2 km. If it is moving with a terminal velocity of 12 km>h, determine the drag on the balloon. *11–60. A boat tows a half-submerged 300-kg oil drum having an approximate diameter of 0.9 m. If the drag coefficient is CD = 0.82, determine the tension in the horizontal tow rope at the instant the boat is traveling at 8 m>s and has an acceleration of 1.5 m>s2. Take rw = 1000 kg>m3.

1.25 m 8 m/s

Prob. 11–60

Prob. 11–55 *11–56. Impure water at 20°C enters the retention tank and rises to a level of 2 m. Determine the shortest time needed for all spherical sediment particles having a diameter of 0.05 mm or greater to settle to the bottom. Assume the density of the particles is r = 1.6 Mg>m3 or greater. Note: The volume of a sphere is V = 34pr 3.

11–61. A periscope on a submarine has a submerged length of 2.5 m and a diameter of 50 mm. If the submarine is traveling at 8 m>s, determine the moment developed at the base of the periscope. The water is at a temperature of 15°C. Consider the periscope to be a smooth cylinder. 11–62. Determine the moment developed at the base A of the circular sign due to the 14 m>s wind. The air is at 20°C. Neglect the drag on the pole.

0.75 m 2m 5m

14 ms

2m

Prob. 11–56 11–57. A ball has a diameter of 250 mm. When it is kicked, it is given a speed of 10 m>s. Determine the initial drag acting on the ball. Does this force remain constant? The air is at a temperature of 20°C. 11–58. A solid ball has a diameter of 20 mm and a density of rb = 3.00 Mg>m3. Determine its terminal velocity if it is dropped into a liquid having a density of rl = 2.30 Mg>m3 and a kinematic viscosity of nl = 0.052 m2 >s. Note: The volume of a sphere is V = 43pr 3.

3m

A

Prob. 11–62

proBlems 11–63. The truck has a drag coefficient of CD = 1.12 when it is traveling with a constant velocity of 80 km>h. Determine the power needed to drive the truck at this speed if the average front projected area of the truck is 10.5 m2. The air is at a temperature of 10°C. *11–64. The truck has a drag coefficient of CD = 0.86 when it is traveling with a constant velocity of 60 km>h. Determine the power needed to drive the truck at this speed if the average front projected area of the truck is 10.5 m2. The air is at a temperature of 10°C.

643

11–67. A styrofoam ball having a diameter of 80 mm and a mass of 25 g is dropped from a high building. Determine its terminal velocity. The temperature of the air is 20°C. Note: The volume of a sphere is V = 43pr 3. *11–68. A raindrop has a diameter of 1 mm. Determine its approximate terminal velocity as it falls. Assume that 11 for air ra = 1.247 kg>m3 and na = 14.2110-6 2 m2 >s. Neglect buoyancy. Note: The volume of a sphere is V = 43pr 3. 11–69. The parachute has a drag coefficient of CD = 1.36 and an open diameter of 4 m. Determine the terminal velocity as the man parachutes downward. The air is at 20°C. The total mass of the parachute and man is 90 kg. Neglect the drag on the man. 11–70. The parachute has a drag coefficient of CD = 1.36. Determine the required open diameter of the parachute so the man attains a terminal velocity of 10 m>s. The air is at 20°C. The total mass of the parachute and man is 90 kg. Neglect the drag on the man.

Probs. 11–63/64

11–71. The man and the parachute have a total mass of 90 kg. If the parachute has an open diameter of 6 m and the terminal velocity is 5 m>s, determine the drag coefficient of the parachute. The air is at 20°C. Neglect the drag on the man.

11–65. A 600 mm by 600 mm square plate is held in air at 30°C, which is blowing at 12 m>s. Compare the drag on the plate when it is held normal and then parallel to the airflow. 11–66. Particulate matter at an altitude of 8 km in the upper atmosphere has an average diameter of 3 μm. If a particle has a mass of 42.5110-12 2 g, estimate the time needed for it to settle to the earth. Assume gravity is constant, and for air, r = 1.202 kg>m3 and m = 18.1110-6 2 N # s>m2. V

8 km

Prob. 11–66

Probs. 11–69/70/71

644

C h a p t e r 11

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*11–72. Determine the pressure drag acting on the side of the truck if the air is at 20°C and it is blowing at 54 km>h. Take CD = 1.3.

11–74. Each of the smooth bridge piers (cylinders) has a diameter of 0.75 m. If the river maintains an average speed of 0.08 m>s, determine the drag the water exerts on each pier. The water temperature is 20°C.

11 3m

15 m

54 km/h

0.08 ms

0.75 m 6m

Prob. 11–72

11–73. The mast on the boat is held in place by the rigging, which consists of rope having a diameter of 20 mm and a total length of 50 m. Assuming the rope to be cylindrical, determine the drag it exerts on the boat if the boat is moving forward in still air at a speed of 10 m>s. The air has a temperature of 30°C.

Prob. 11–74

11–75. A ball has a diameter of 40 mm and falls in honey with a terminal velocity of 0.3 m>s. Determine the mass of the ball. For honey, take rh = 1360 kg>m3 and nh = 0.04 m2 >s. Note: The volume of a sphere is V = 43pr 3.

Prob. 11–73

Prob. 11–75

645

proBlems *11–76. A rock is released from rest at the surface of the lake, where the average water temperature is 15°C. If CD = 0.5, determine its speed when it reaches a depth of 600 mm. The rock can be considered a sphere having a diameter of 50 mm and a density of rr = 2400 kg>m3. Note: The volume of a sphere is V = 43 pr 3.

11–79. A sprayer ejects a coating of particles at 25 m>s. Determine their velocity 8 μs after leaving the nozzle. Assume the average diameter of the particles is 0.6 μm and each has a mass of 0.8110-12 2 g. The air is at 20°C. Neglect the vertical component of the velocity. Note: The volume of a sphere is V = 34pr 3.

25 ms

600 mm

Prob. 11–76

Prob. 11–79

11–77. A smooth ball has a diameter of 43 mm and a mass of 45 g. Determine its initial deceleration when it is thrown vertically upward with a speed of 20 m>s. The temperature is 20°C. 11–78. The smooth cylinder is suspended from the rail and is partially submerged in water. If the wind blows at 8 m>s, determine the terminal velocity of the cylinder. The water and air are both at 20°C.

*11–80. The blades of a mixer are used to stir a liquid having a density r and viscosity m. If each blade has a length L and width w, determine the torque T needed to rotate the blades at a constant angular rate v. Take the drag coefficient of the blade’s cross section to be CD. Assume the body of liquid remains at rest as it is being stirred.

T v

1m 8 ms L L 0.5 m 0.25 m w

Prob. 11–78

Prob. 11–80

11

646

C h a p t e r 11

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11–81. A 2-mm-diameter sand particle having a density of 2.40 Mg>m3 is released from rest at the surface of oil. As the particle falls downward, “creeping flow” will be established around it. Determine the velocity of the particle and the time at which Stokes’ law becomes invalid, at about Re = 1. The oil has a density of ro = 900 kg>m3 and a viscosity of 11 m = 30.2110-3 2 N # s>m2. Assume the particle is a sphere, o where its volume is V = 43pr 3.

11–83. Dust particles having an average diameter of 0.05 mm and an average density of 450 kg>m3 are stirred up by an airstream and blown off the edge of the 600-mm-high desk into a horizontal steady wind of 0.5 m>s. Determine the distance d from the edge of the desk where most of them will strike the ground. Since we have “creeping flow,” the path of descent is practically a straight line. The air is at a temperature of 20°C. Note: The volume of a sphere is V = 34 pr 3.

0.5 ms 600 mm

d

Prob. 11–83 Prob. 11–81

11–82. The spherical balloon filled with helium has a mass of 9.50 g. Determine its terminal velocity of ascent. The air temperature is 20°C. Note: The volume of a sphere is V = 34 pr 3.

*11–84. The parachutist and chute has a total mass of 90 kg and is in free fall at 6 m>s when she opens her 3-m-diameter parachute. Determine the time for her speed to become 10 m>s. Also, what is her terminal velocity? For the calculation, assume the parachute to be similar to a hollow hemisphere. The air has a density of ra = 1.25 kg>m3.

150 mm

V

Prob. 11–82

Prob. 11–84

647

proBlems

SEC. 11.10–11.11 11–85. The 5-Mg plane can take off from an airport when it attains an airspeed of 180 km>h. If it carries an additional load of mass 450 kg, what must be its airspeed before takeoff at the same angle of attack?

11–87. The 5-Mg airplane is flying at an altitude of 1 km. Each wing has a span of 6 m and a cord length of 1.5 m. If each wing can be classified as a NACA 2409 section, and the plane is flying at 450 km>h, determine the total drag on the wings. Also, what is the angle of attack and the corresponding velocity at which stall occurs?

180 km/h

Prob. 11–85 6m 1.5 m

Prob. 11–87 11–86. The 5-Mg airplane is flying at an altitude of 1 km. Each wing has a span of 6 m and a cord length of 1.5 m. If each wing can be classified as a NACA 2409 section, determine the lift coefficient and the angle of attack when the plane is flying at 450 km>h.

*11–88. The glider has a constant speed of 8 m>s through still air. Determine the angle of descent u if it has a lift coefficient of CL = 0.70 and a wing drag coefficient of CD = 0.04. The drag on the fuselage can be considered negligible compared to that on the wings, since the glider has a very long wingspan.

8 ms 6m 1.5 m

Prob. 11–86

u

Prob. 11–88

11

648

C h a p t e r 11

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11–89. A 5-Mg airplane is flying at a speed of 60 m>s. If each wing can be assumed to be rectangular of span 6 m and cord distance of 1.5 m, determine the smallest angle of attack a to provide lift, assuming the wing is a NACA 2409 section. The density of air is r = 1.21 kg>m3. 11 11–90. The 4-Mg airplane has wings that are each 6 m long and 1.5 m wide. It is flying horizontally at an altitude of 2 km with a speed of 450 km>h. Determine the lift coefficient.

11–93. The glider has a weight of 160 kg. If the drag coefficient is CD = 0.316, the lift coefficient is CL = 1.20, and the total area of the wings is A = 6 m2, determine if it can land on a 1.5-km-long landing strip that is located 5 km away from where its altitude is 1.5 km. Assume the density of the air remains constant.

u

Prob. 11–93 6m

6m

Probs. 11–89/90

11–91. The glider has a mass of 180 kg. If the drag coefficient is CD = 0.456, the lift coefficient is CL = 1.20, and the total area of the wings is A = 8.52 m2 , determine the angle u at which it is descending with a constant speed.

11–94. The plane can take off at 200 km>h when it is at an airport located at an elevation of 3 km. Determine the takeoff speed from an airport at sea level.

200 km/h

Prob. 11–94

u

Prob. 11–91

11–95. A smooth ball has a diameter of 80 mm. If it is hit with a speed of 10 m>s and given an angular velocity of 80 rad > s, determine the lift on the ball. Take ra = 1.23 kg>m3 and na = 14.6110-6 2 m2 >s .

80 rads

*11–92. A 6-Mg airplane is flying at a speed of 360 km>h. If each wing can be assumed to be rectangular of span 6 m and cord distance of 2 m, determine the section drag on each wing when it is flying at the proper angle of attack a. Assume each wing is a NACA 2409 section. The density of air is ra = 1.007 kg>m3.

10 ms

Prob. 11–95

649

proBlems *11–96. A 7.5-Mg airplane has two wings, each having a span of 6 m and average cord distance of 1.5 m . If the total drag on each wing is 3.85 kN when it is flying at 324 km>h, determine the total drag on each wing when the plane is flying at the same angle of attack and the same altitude with a speed of 486 km>h. Assume an elliptical lift distribution. Take ra = 0.8820 kg>m3.

11–97. A 0.5-kg ball having a diameter of 50 mm is thrown with a speed of 10 m>s and has an angular velocity of v = 400 rad>s. Determine its horizontal deviation d from striking a target a distance of 10 m away. Use Fig. 11–50 and neglect the effect on lift caused by the vertical component of velocity. Take ra = 1.20 kg>m3 and na = 15.0110-6 2 m2 >s. y

v

d 10 m Top view

Prob. 11–97

x

11

650

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CONCEPTUAL PROBLEMS P11–1. When the cup of hot tea is stirred, top photo, the “leaves” seem to eventually settle at the bottom in the 11 center of the cup. Why do they do this, instead of accumulating along the rim? Explain.

P11–2. Which structure will best survive in a hurricane, the triangular-shaped building or the dome-shaped building? Draw the pressure distribution and streamlines for the flow in each case to explain your answer.

P11–2

P11–1 P11–3. Lift, drag, thrust, and weight act on the plane. When the plane is climbing during takeoff is the lift force less than, greater than, or equal to the weight of the plane? Explain.

P11–3

P11–4. A baseball is thrown straight up into the air. Will the time to travel to its highest point be longer, shorter, or the same as the time for it to fall to the same height from which it was thrown?

P11–4

651

Chapter reView

CHAP TER R EV IEW

11 The boundary layer is a very thin layer of fluid located in a region just above the surface of a body. Within it, the velocity changes from zero at the surface to the free-stream velocity of the fluid. The fluid within the boundary layer formed over the surface of a flat plate will be laminar up to the critical distance xcr. In this book, this distance is determined from 1Rex 2 cr = Uxcr >n = 51105 2.

The velocity profile for a laminar flow boundary layer has been solved by Blasius. The solution is given in both graphical and tabular form. Knowing this velocity profile, one can find the thickness of the boundary layer and the friction drag that the flow exerts over a flat plate.

The friction drag caused by turbulent flow boundary layers is determined by experiment. For both laminar and turbulent cases, this force is reported by using a dimensionless friction drag coefficient CDf , which is a function of the Reynolds number.

y U

U

x xcr

1 FDf = CDfAa rU 2 b 2

652

11

C h a p t e r 11

VisCous Flow

oVer

e x t e r n a l s u r Fa C e s

y

The thickness and shear-stress distribution for both laminar and turbulent boundary layers can be determined by an approximate method, using the momentum integral equation. Since turbulent flow creates a larger shear stress on a surface, compared to laminar flow, turbulent boundary layers create a larger friction drag on the surface.

U

t0 x L

Experimental drag coefficients CD, caused by both viscous friction and pressure, have been determined for the cylinder, the sphere, and bodies of many other shapes. In general, CD is a function of the Reynolds number, the body’s shape, its surface roughness, and its orientation within the flow. For some cases, this coefficient may also depend on the Froude number or the Mach number.

When designing an airfoil, both drag and lift are important. Their coefficients, CD and CL, are determined by experiment and are represented graphically as a function of the angle of attack.

FD = CDAp a

rV 2 b 2

x

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Chapter reView

Fundamental Equations for Flow over External Surfaces Boundary Layer Thickness d* =

L0

∞

a1 -

u bdy U

Θ =

Displacement thickness

L0

∞

u u a1 - bdy U U

11

Momentum thickness

Laminar Boundary Layers Rex =

rUx Ux = m n

d =

2Rex 5.0

d* =

x

Disturbance thickness

1Rex 2 cr = 51105 2

2Rex 1.721

Θ =

x

Displacement thickness

2Rex 0.664

x

Momentum thickness y x Rex

Flat plate

1 t0 = cf a rU 2 b 2

FDf

5

2Rex 0.664

cf = Shear stress

4 3

1 = CDf bL a rU 2 b 2

CDf =

Friction drag

2ReL 1.328

2 1

Momentum Integral Equation d u u a1 - b dy dx L0 U U d

t0 = rU 2

0.4

0.2

0.6

0.8

1.0 0.99

Dimensionless boundary layer

Turbulent Boundary Layers y 1>7 u = a b U d

d =

Prandtl’s law

0.371

1Rex 2

1>5

x

t0 =

0.0288rU 2 1Rex 2 1>5

Disturbance thickness

CDf =

Shear stress

0.0740

1ReL 2 1>5

CDf =

5(105) 6 ReL 6 107 0.455

1log10 ReL 2 2.58

Friction drag coefficients

Laminar and Turbulent Boundary Layers CDf =

0.455

1log10 ReL 2 2.58

-

1700 ReL

51105 2 … ReL 6 109

Friction drag coefficient

Drag and Lift FD = CDApl a

Airfoil drag

rV 2 b 2

FD = CDAp a

rV 2 b 2

Shape drag

FL = CLApl a Lift

rV 2 b 2

107 … ReL 6 109

u U

12

Olaf Speier/Alamy Stock Photo

CHAPTER

Open channels are often used for drainage and irrigation. It is important that they be designed properly so that adequate flow through them is maintained.

OPEN-CHANNEL FLOW CHAPTER OBJECTIVES ■

To categorize the flow in an open channel based on the concept of specific energy.

■

To study the flow over a rise and under a sluice gate.

■

To show how to analyze steady uniform and steady nonuniform flow through a channel.

■

To discuss the hydraulic jump and to show how to measure flow in an open channel using different types of weirs.

12.1

TYPES OF FLOW IN OPEN CHANNELS

In this chapter, we will consider flow through an open channel, that is, a conduit that has an open or free surface. Here, the main driving force for the flow is gravity, as opposed to closed conduits where the flow is driven by pressure. Open channel flow is of primary interest to civil engineers, since it forms the basis for the design of drainage systems, and the hydraulic analysis of rivers and streams. The intent here is to provide an introduction to some of the more important aspects of open-channel flow, so that it provides a suitable basis for further study in specific courses devoted to this subject. Typical open channels include rivers, canals, culverts, and flumes. Of these, rivers and streams have variable cross sections, which change over time due to erosion and deposition of earth. Canals are generally very long and straight, and are used for drainage, irrigation, or navigation. Culverts normally do not flow full, and are usually made of concrete or masonry. They are often used to carry drainage under roadways. And finally, a flume is a conduit that is supported above ground and is generally designed to carry drainage over a depression.

655

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O p e n -C h a n n e l F l O w

If the channel has a constant cross section, it is known as a prismatic channel. For example, canals and flumes are typically constructed with rectangular or trapezoidal cross sections, whereas culverts often have circular or elliptical shapes. Rivers and streams have nonprismatic cross sections; however, for approximate analysis they are sometimes modeled by a series of different-size prismatic sections, such as trapezoids and semi-ellipses.

1.5 1.0 0.5

12

Laminar and Turbulent Flow. Although laminar flow can occur Typical velocity contours, in ms, for a rectangular channel

Fig. 12–1

Uniform steady flow, constant depth (a)

in an open channel, in engineering practice it is rarely encountered. This is because the flow must be very slow to meet any Reynolds number criteria for laminar flow. Instead, open-channel flows are predominantly turbulent. The mixing of the liquid that actually takes place can be caused by the friction of the wind blowing over its surface and friction along the sides of the channel. These effects will cause the velocity profile to become highly irregular; and as a result, the maximum velocity will be somewhere near the top of the liquid surface, but normally not at the surface. A typical velocity profile for water flowing through an open rectangular channel may look something like that shown in Fig. 12–1. Hence, even though the surface may appear calm, as if the flow is laminar, beneath the surface it will be turbulent. In spite of this irregularity, we can often approximate the actual flow as being uniform one-dimensional flow, and still achieve reasonable results in estimating the flow.

Uniform and Steady Flow. Besides being either laminar or

Accelerated nonuniform flow (b)

Retarded nonuniform flow (c)

Fig. 12–2

turbulent, open-channel flow can also be classified in another way. Uniform flow occurs when the depth of the liquid remains the same along the length of the channel, because then the velocity of the liquid will not change from one location to the next. An example occurs when the channel has a small slope, so that the force of gravity causing the flow is balanced by the friction force that retards it, Fig. 12–2a. If the depth varies along the length, then the flow is nonuniform. This can happen if there is a change in slope, or where there is a change in the channel’s cross-sectional area. Accelerated nonuniform flow occurs when the depth of flow decreases downstream, Fig. 12–2b. An example would be water flowing down a chute or spillway. Retarded nonuniform flow occurs if the depth is increasing, Fig. 12–2c, such as when water in a downward-sloping channel backs up to meet the face of a dam. Steady flow in a channel occurs when the flow remains constant over time as in Fig. 12–2a, and so its depth at a specific location remains constant. This is the case for most problems involving open-channel flow. However, if a wave passes by a specific location, the depth and hence the flow will change with time, and so unsteady flow occurs.

Hydraulic Jump. Apart from the types of flow mentioned above, there Hydraulic jump

Fig. 12–3

is one other phenomenon that can occur in open channels. A hydraulic jump is a localized turbulence that rapidly dissipates kinetic energy, changing the flow from rapid to tranquil. It generally occurs at the bottom of a chute or spillway, Fig. 12–3.

12.2

12.2

657

Open-Channel FlOw ClassiFiCatiOns

OPEN-CHANNEL FLOW CLASSIFICATIONS

Later in this section it will be shown that the type of flow that occurs in an open channel can be classified by comparing the average speed of the liquid in the channel to the speed of a wave on its surface. Specifically, the speed of the wave relative to the speed of the liquid in the channel is called the wave celerity, c. One way to produce a wave within a channel is to create a surge in the flow. A positive surge occurs when a gate or other obstruction suddenly blocks the flow, causing the liquid to back up against the gate and suddenly elevate the flow as shown in Fig. 12–4a. The resulting wave front will then propagate upstream of the flow and move with a speed c. To find this speed, we will assume the channel is horizontal, has a rectangular cross section, and the liquid is an ideal fluid. Initially the flow has an average velocity V1 and the depth is y1, but because the gate creates an impulse on the fluid, it will slow it down to zero and lift it up to a new depth y2. To a fixed observer unsteady flow occurs as the wave begins to increase the depth. Realize, however, that the wave only moves the liquid up as it passes by, although it creates the illusion that the liquid contained within the wave is actually moving over the surface with a velocity c, which is not the case. For the analysis that follows, it is easier to fix a reference frame to a moving control volume that has the speed of the wave, so that for an observer on the wave, the flow will appear as steady flow, Fig. 12–4b. As a result, for one-dimensional flow, the liquid at the open control surface 1 will appear to move to the left at V1 + c, and since the flow is blocked, the liquid at the open control surface 2 will appear to move to the left at c. If the channel has a constant width b, then the continuity equation gives 0 r dV + r Vf>cs # dA = 0 0t Lcv Lcs 0 + rc(y2b) - r(V1 + c)y1b = 0 y2 V1 = c a - 1b y1 We will now apply the momentum equation, where the flow rVf>cs # A is calculated from section 1. Using the free-body diagram of the control volume shown in Fig. 12-4c, we have, 0 ΣF = VrdV + VrVf>cs # dA 0t L L cv

cs

1 1 (rgy2)y2b - (rgy1)y1b = 0 - c[r(V1 + c)y1b] - (V1 + c)[ -r(V1 + c)y1b] 2 2 1 2 1 gy1 - gy22 = -(V1 + c)V1y1 2 2 Substituting the result for V1 into this equation and solving for c, we get gy1(y1 + y2) c = A 2y2

12 c

y2

V1

y1

(a)

c

c

y2

V1 + c

Moving control volume (b)

W

r g y2

r g y1 N Free-body diagram (c)

Fig. 12–4

y1

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Provided the wave has a small height compared to the depth of the liquid, then y1 ≈ y2 ≈ y, and so c = 2gy

(12–1)

12

Notice that this result is independent of the magnitude of V1. The same speed would be obtained if a negative surge occurred, as in the case of suddenly lifting the gate in Fig. 12–4a. Also, it can be shown that c represents the speed of a wave moving over the surface of the water even if there is no flow. It is only a function of the liquid depth in the channel.

Froude Number. Like waves, the driving force for all openchannel flow is due to gravity, and so in 1871, William Froude formulated the Froude number and showed how it can be used for describing this flow. Recall that in Chapter 8 we defined the Froude number as the square root of the ratio of the inertia force to the gravitational force. Here the result is expressed as

c

V

Subcritical flow: Wave moves upstream V , c Critical flow: Stationary wave V 5 c Supercritical flow: Wave moves downstream V . c

Fr =

V V = c 1gy

(12–2)

(d)

Fig. 12–4 (cont.)

where V is the average velocity of the liquid in the channel, and y is the depth of the liquid. To show why the Froude number is important, consider, for example, the case shown in Fig. 12–4d, where a plate momentarily obstructs the steady stream having a velocity V. If Fr = 1, then from Eq. 12–2, c = V. When this occurs, the wave on the left will be stationary. This is referred to as critical flow. If Fr 6 1, then c 7 V, and this wave will propagate upstream, which is a condition of subcritical flow or tranquil flow. In other words, the gravitational force or the weight of the wave overcomes the inertia force caused by its movement. Finally, if Fr 7 1, then V 7 c, and the wave will be washed downstream. This is termed supercritical flow or rapid flow and is the result of the gravitational force being overcome by the wave’s inertia force.

12.3

SPECIFIC ENERGY

The actual behavior of the flow at each location along an open channel depends upon the total energy of the flow at that location. To find this energy, we will apply the Bernoulli equation, which assumes steady flow of an ideal liquid. If we establish the datum at the bottom of the channel,

12.3

as shown in Fig. 12–5, and choose the point on the streamline at an arbitrary depth d in the liquid, then at this point the Bernoulli equation requires

Energy line V2 2g

V

p V2 + + y′ = const. g 2g gd V2 + E = + 1y - d2 = const. g 2g

659

speCiFiC energy

d y

y9 Datum 12

(12–3) Fig. 12–5

or, E =

V2 + y 2g

(12–4)

This sum is referred to as the specific energy, E. It is independent of the chosen depth d and so indicates the amount of the kinetic and potential energy that each unit weight of the liquid has at each point on the cross section at a specific location. Stated another way, it is also called the specific head, because it has units of length, and thereby represents the vertical distance from the channel bottom to the energy line, Fig. 12–5. The specific energy can also be expressed in terms of the volumetric flow by using Q = VA. Thus, y

E =

Q2 2gA2

+ y

(12–5)

We can further express E only as a function of y by considering a unique cross section.

b Rectangular cross section (a)

Rectangular Cross Section. If the cross section is rectangular, Fig. 12–6a, and the volumetric flow per unit width is q = Q>b, then since A = by we have E =

q2 2gy2

y

+ y

q 5 0 (no motion)

(12–6)

q

Specific energy

There are two independent variables in this equation, namely, q and y. However, for a constant value of q, a plot of Eq. 12–6 has the shape shown in Fig. 12–6b. It is called a specific energy diagram. Notice that when q = 0, then E = y, as shown by the 45° sloped line. This represents a condition of the liquid having no motion or kinetic energy, only potential energy. When the liquid has a flow q, then there will be two possible depths, y1 and y2, that produce the same specific energy E = E′. Here the smaller value y1 represents low potential energy and high kinetic energy. This is rapid or supercritical flow. Likewise, the larger value y2 represents high potential energy and low kinetic energy. It is referred to as tranquil or subcritical flow.

y2 yc y1

Emin E9 Specific energy diagram for flow per unit width (b)

Fig. 12–6

Subcritical flow Fr , 1 Supercritical flow Fr . 1 E

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y q 5 0 (no motion) q

12 Subcritical flow Fr , 1 Supercritical flow Fr . 1 E

y2 yc Emin E9

y1

Specific energy diagram for flow per unit width (b)

Fig. 12–6 (repeated)

As shown by the graph, the minimum value for the specific energy, Emin, occurs at the critical depth, yc. It can be found by setting the derivative of Eq. 12–6 equal to zero and evaluating the result at y = yc. Since q is constant, -q2 dE + 1 = 0 = dy gyc3 yc = a

q2 1>3 b g

(12–7)

Substituting this into Eq. 12–6, the value of Emin is therefore Emin =

Canals having rectangular cross sections are often used for small flows or within confined spaces through crowded neighborhoods.

yc 3 2yc

2

+ yc =

3 y 2 c

(12–8)

To summarize, this is the smallest amount of specific energy the liquid can have and yet still maintain the required flow q. It occurs at the nose of the curve in Fig. 12–6b, where the flow q is at the critical depth yc. To find the critical velocity at this depth, substitute q = Vcyc into Eq. 12–7, so that V c2yc2 yc = ° ¢ g and so

Vc = 1gyc

1>3

(12–9)

Notice that when the flow is at this critical velocity, the Froude number becomes Vc Fr = = 1 1gyc

12.3

speCiFiC energy

661

Therefore, for any point on the upper branch of the curve in Fig. 12–6b, the depth of flow will exceed the critical depth, y = y2 7 yc. When this happens, V 6 Vc, and so Fr 6 1. This is subcritical, or tranquil, flow. Likewise, for any point on the lower branch of the curve, the depth of flow will be less than the critical depth, y = y1 6 yc, and then V 7 Vc and Fr 7 1. This is supercritical, or rapid, flow. These three classifications are therefore Fr 6 1, y 7 yc or V 6 Vc

Subcritical (tranquil) flow

Fr = 1, y = yc or V = Vc

Critical flow

Fr 7 1, y 6 yc or V 7 Vc

Supercritical (rapid) flow

12

(12–10)

Because of the branching of the specific energy diagram that takes place at yc, engineers do not design channels to flow at the critical depth. If they did, then stationary waves or undulations would develop on the surface of the liquid, and any slight disturbance in the depth of flow would cause the liquid to constantly adjust between subcritical and supercritical flow—an unstable condition.

Large drainage channels are often constructed in a trapezoidal shape since this shape is relatively easy to construct. Notice that engineers have placed “weep holes” along the sloped sides in order to reduce the hydrostatic pressure on the inside wall caused by groundwater absorption.

Nonrectangular Cross Section. When the channel cross section is nonrectangular, Fig. 12–7, then the minimum specific energy must be obtained by taking the derivative of Eq. 12–5, setting it equal to zero, and requiring A = Ac. This gives -Q2 dA dE = + 1 = 0 dy gAc3 dy At the top of the channel, the elemental area strip dA = btop dy, and so, rearranging, we get Ac3 Q2 = g btop

(12–11)

Provided btop and Ac can both be related by the geometry of the cross section to the critical depth yc, Fig. 12–7, then a solution for yc can be determined from this equation. To do this for an irregular cross section an initial guess is made for yc and the calculated value of the right side is compared to the constant value Q2 >g on the left. Iterations of yc are made until the equation is satisfied. See Example 12.3. To find the critical velocity of the flow, substitute Q = Vc Ac into the above equation and solve for Vc. We get Vc =

gAc A btop

(12–12)

At this speed Fr = 1, and so for any other velocity, the flow can be classified as supercritical or subcritical in accordance with Eqs. 12–10.

btop dy Ac

Critical depth in channel with arbitrary cross section

Fig. 12–7

yc

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IM PORTANT POINTS • Open-channel flow is predominantly a turbulent phenomenon due to the mixing action that takes 12

place within the liquid. Although the velocity profile is highly irregular, for a reasonable analysis we can assume the liquid to be an ideal fluid, so that the flow is one-dimensional and has an average velocity over the cross section.

• Steady flow requires the velocity profile at a particular cross section to remain constant over time. Uniform flow requires that this profile remain the same at all cross sections. When uniform flow occurs in an open channel of constant width, the depth of flow will remain constant throughout the channel.

• Open-channel flow is classified according to the Froude number, which is the square root of the ratio of the inertia force to the gravitational force. When Fr = 1, critical flow occurs at depth yc. A wave produced on the surface will remain stationary. When Fr 6 1, subcritical flow occurs at depth y 7 yc, and it is possible for a wave produced on the surface to move upstream. Finally, when Fr 7 1, supercritical flow occurs at depth y 6 yc, and all waves will be washed downstream.

• The specific energy or specific head E of the flow in an open channel is the sum of its kinetic and potential energies, as measured from a datum located at the bottom of the channel. The specific energy diagram is a plot of E = f (y) for a given flow q. It shows when the flow will either move rapidly, V 7 Vc, at a shallow depth, y 6 yc (high kinetic energy and low potential energy), or move as tranquil flow, V 6 Vc, at a deeper depth, y 7 yc (low kinetic energy and high potential energy).

• The specific energy of a given flow q is a minimum when the flow is at the critical depth yc.

EXAMPLE

12.1 Water has a velocity of 4 m>s as it flows along the rectangular channel shown in Fig. 12–8a. If the depth of the flow is 3 m, classify the flow. What is the velocity of the flow at the alternate depth that provides the same specific energy for the flow?

4 m s y53m

SOLUTION Fluid Description. The flow is uniform and steady, and water is assumed to be an ideal fluid. 2m

Subcritical flow (a)

Fig. 12–8

Analysis. To classify the flow, we must first determine the critical depth from Eq. 12–7. Since the flow per unit width is q = Vy = (4 m>s)(3 m) = 12 m2 >s, then 112 m2 >s2 2 q2 1>3 yc = a b = £ § g 19.81 m>s2 2

1>3

= 2.45 m

Here yc 6 y = 3 m, so the flow will be subcritical, or tranquil.

Ans.

12.3

663

speCiFiC energy

For the arbitrary depth y, the specific energy for the flow is determined using Eq. 12–6.

E =

112 m2 >s2 2

219.81 m>s 2y 2

2

+ y

y (m)

A plot of this equation is shown in Fig. 12–8b. At y = 3 m,

E =

q

2gy

112 m >s2 2

2 2

+ y =

12

(1)

q 5 12 m 2 s 2

219.81 m>s2 213 m2 2

+ 3 m = 3.815 m

3 Subcritical flow Supercritical flow E (m)

yc 5 2.45

To find the alternate depth that provides this same specific energy of E = 3.815 m, we must substitute this value into Eq. 1, which, after simplification, yields

2.02 Emin 5 3.67 3.815 (b)

y - 3.815y + 7.339 = 0 3

2

Solving for the three roots, we obtain 5.93 m s

y = 3 m 7 2.45 m

Subcritical (as before)

y = 2.024 m 6 2.45 m

Supercritical

y = -1.21 m

Not realistic

y 5 2.02 m

Ans.

2m

For the case of supercritical or rapid flow, Fig. 12–8c, when the depth is y = 2.024 m, the velocity must be q = Vy;

12 m2 >s = V(2.024 m) V = 5.93 m>s

Supercritical flow (c)

Fig. 12–8 (cont.)   

Ans.

The specific energy at critical flow can be determined from Eq. 12–8, Emin = 32 yc, or from Eq. 1, using yc = 2.45 m. Its value, 3.67 m, is also shown in Fig. 12–8b. To summarize, a flow having a specific energy or specific head of E = 3.815 m can be rapid at a depth of 2.02 m or tranquil at a depth of 3 m. If this same flow has some other specific energy E, then it will occur at two other depths, found from the roots of Eq. 1.

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EXAMPLE

12

O p e n -C h a n n e l F l O w

12.2 The horizontal rectangular channel in Fig. 12–9a is 2 m wide and gradually tapers so that it becomes 1 m wide. If water is flowing at 8.75 m3 >s and has a depth of 0.6 m within the 2-m section, determine the depth when it is in the 1-m section.

2m 1m

0.6 m y (a)

Fig. 12–9

SOLUTION Fluid Description. We have uniform steady flow in each region, although within the transition there is nonuniform flow. The water is assumed to be an ideal fluid.

Analysis. Since q = (8.75 m3 >s)>(2 m) = 4.375 m2 >s, within the wide section, the critical depth is yc = a

14.375 m2 >s2 2 1>3 q2 1>3 b = c d = 1.25 m g 9.81 m>s2

Since the depth y = 0.6 m 6 yc, the flow is supercritical, or rapid, in this section. Using Eq. 12–6, we can find the specific energy for the flow in this section, b = 2 m. It is

E =

q2 2gy2

+ y =

When y = 0.6 m, E = 3.310 m.

14.375 m2 >s2 2

219.81 m>s2 2y2

+ y

(1)

12.3

665

speCiFiC energy

This value of E must remain constant through the channel since the channel bottom remains horizontal, and there are no friction losses. When the width is 1 m, then q = (8.75 m3 >s)>(1 m) = 8.75 m2 >s, and Eq. 12–6 becomes q2

E =

2gy2

+ y =

18.75 m2 >s2 2

219.81 m>s2 2y2

+ y

12

(2)

Therefore, to find the depths in the 1 m section, we have 3.310 m =

18.75 m2>s2 2

219.81 m>s2 2y2

+ y

y3 - 3.310y2 + 3.902 = 0

(3) y (m)

The critical depth in this section is yc = a

18.75 m2 >s2 2 1>3 q2 1>3 b = c d = 1.98 m g 9.81 m>s2

q 5 4.375 m2 s q 5 8.75 m 2 s 2.82

Solving Eq. 3 for the depths, we obtain y = 2.82 m 7 1.98 m

Subcritical

y = 1.45 m 6 1.98 m

Supercritical

y = -0.956 m

Not realistic

1.98 1.45 1.25 0.6

b52m

b51m

Since the flow was originally supercritical, it remains in this state, and so the depth in the 1-m section will be y = 1.45 m

3.310 Emin 5 2.975

Emin 5 1.874

Ans.

If we plot Eqs. 1 and 2, Fig. 12–9b, it will help to understand why the flow of 8.75 m3 >s remains supercritical throughout the channel. In the figure, the values of Emin were found from Eq. 12–8. As the width of the channel gradually narrows from 2 m to 1 m, the water rises upward from y = 0.6 m on the curve for b = 2 m, until it reaches the point y = 1.45 m on the curve for b = 1 m. It is not possible for the water to reach the greater depth of y = 2.82 m in this section, because the specific energy must remain constant, and so it cannot decrease to Emin = 2.975 m and then increase again to the required E = 3.310 m.

(b)

E (m)

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EXAMPLE

O p e n -C h a n n e l F l O w

12.3 The channel has a triangular cross section as shown in Fig. 12–10. Determine the critical depth if the flow is 12 m3 >s.

btop 2

12

SOLUTION yc

Fluid Description. We assume steady flow of an ideal fluid. 60°

60°

Analysis. For critical flow, we require the specific energy to be a minimum, which means we must satisfy Eq. 12–11. From Fig. 12–10,

Fig. 12–10

btop = 2( yc cot 60°) = 1.1547yc Ac =

1 1 b y = (1.1547yc) yc = 0.5774yc2 2 top c 2

Thus, Ac 3 Q2 = g btop

112 m3 >s2 2 9.81 m>s2

=

(0.5774 yc 2)3 1.1547yc

Solving yields yc = 2.45 m

12.4

Ans.

OPEN-CHANNEL FLOW OVER A RISE OR BUMP

When liquid flows over a rise in a channel bed, as shown in Fig. 12–11a, it will change the depth of flow, since the increased elevation of the channel bed will decrease the specific energy in lifting the liquid mass. To investigate this effect, we will consider the flow to be one-dimensional, that is, horizontal, by assuming that the change in elevation is small and occurs over a short distance. We will also neglect any frictional effects.

12.4

Supercritical flow V1

Supercritical flow V2

y

q50 q

y2

y1

Datum for upper bed Datum for lower bed

h

Subcritical flow

12 2

yc

Supercritical flow over a rise (a)

y2 y1

1

Emin E2

Fig. 12–11

Subcritical flow V1

supercritical flow, Fig. 12–11a. As the flow passes over the rise, the liquid is lifted, and so as noted in Fig. 12–11b, the depth of flow will increase from y1 to y2. In other words, as energy is used to lift the liquid (increase in potential energy), because of the continuity of flow, the liquid will slow down (decrease in kinetic energy), although it still remains supercritical flow.* If we now consider y1 7 yc, as in the case of subcritical flow, Fig. 12–11c, after passing the rise, there is a decrease in the depth of flow (decrease in potential energy) from y1 to y2 , which causes the velocity of flow to increase (increase in kinetic energy); however, subcritical flow will prevail, Fig. 12–11d.

Bump. If a bump or hill is placed on the channel bed, Fig. 12–11e, then there will be a limit by which E can decrease in lifting the fluid. As shown in Fig. 12–11f, it is (E1 - Emin). This gives a maximum rise of the bump of (E1 - Emin) = hc. If this rise occurs, and the bump is designed properly, specific energy can follow around the nose of the curve, and the flow will thereby change, for example, from supercritical to subcritical.† In other words, just when reaching the top of the bump, the flow will be at the critical depth. Then as the bump begins to slope downward, kinetic energy will be added back to the flow. This gets converted to potential energy and raises the depth to y2 as the specific energy is returned to E1.

E1

Subcritical flow V2 y2

y1

Subcritical flow over a rise (c) q50 q

y

1 y1

2

hc

Subcritical flow

y2 yc Emin E2

E1

Supercritical flow E

(d) y

q50 q 2 Subcritical flow

y2

y2

yc

Datum for y2 Datum for y1

h

Subcritical flow

y1

Supercritical flow E

(b)

Rise. Let’s first consider the case where y1 6 yc, so the approach flow is

Supercritical flow

667

Open-Channel FlOw Over a rise Or Bump

yc Datum

(e) *The continuity of flow requires Q = V1 (y1b) = V2 (y2b), or V2 = V1 1y1 >y2 2. But y1 >y2 6 1, so V2 6 V1. †Without a proper design shape of the bump, the flow will instead be returned to supercritical flow. Also, a properly designed bump can change subcritical flow into supercritical flow. See Ref. [2] .

1 Supercritical flow E E1

y1 Emin (f)

668

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EXAMPLE

12

O p e n -C h a n n e l F l O w

12.4

y2

y1 5 1 m

Water flows through the rectangular open channel in Fig. 12–12a that is 1.5 m wide. The initial depth is 1 m and the flow is 10 m3 >s. If the bed of the channel rises 0.15 m, determine the new depth.

0.15 m Datum

SOLUTION Fluid Description. We have steady flow. The water is assumed to be an ideal fluid.

(a)

y (m)

q 5 6.667 m2 s 2.83 yc 5 1.65

Subcritical flow 21

y1 5 1

Emin 5 2.48

y2 5 1.05 (b)

Fig. 12–12

3.27 3.12

Analysis. The critical depth for the flow is determined from Eq. 12–7. Since q = 110 m3 >s2 > 11.5 m2 = 6.667 m2 >s, we have yc = a

16.667 m2 >s2 2 1>3 q2 1>3 b = c d = 1.65 m g 9.81 m>s2

And the minimum specific energy, Eq. 12–8, is 3 3 Emin = yc = (1.65 m) = 2.48 m Supercritical flow 2 2 E (m) Since originally y = 1 m 6 1.65 m, the flow is supercritical, or rapid. As the water passes over the rise, it gives up some of its specific energy because the water is being lifted. The specific energy on the left side of the rise is determined using Eq. 12–6, with y1 = 1 m. We have q2 + y1 E1 = 2g y12 E1 =

16.667 m2 >s2 2

219.81 m>s2 2(1 m)2

+ 1 m = 3.27 m

On the right side of the rise, E2 = E1 - h = 3.27 m - 0.15 m = 3.12 m. Therefore, Eq. 12–6 becomes* 3.12 m =

16.667 m2 >s2 2

219.81 m>s2 2y22

+ y2

y23 - 3.115y22 + 2.265 = 0 Solving for the three roots yields y2 = 2.83 m 7 1.65 m Subcritical y2 = 1.05 m 6 1.65 m Supercritical y2 = -0.764 m Unrealistic The flow must remain supercritical because it cannot have less specific energy than Emin = 2.48 m as it is being lifted. In other words, the specific energy E is confined between 3.12 m and 3.27 m, Fig. 12–12b. Consequently, y2 = 1.05 m *In each case, the datum is set at the bottom of the channel.

Ans.

12.4

EXAMPLE

12.5

Water flows through the channel in Fig. 12–13a that has a width of 0.9 m. If the flow is 2.5 m3>s and the depth is originally 0.75 m, show that the upstream flow is supercritical, and determine the required height h of a bump placed on the bed of the channel so that the flow is able to change to subcritical downstream from the bump.

12

0.75 m

y2

yc h

SOLUTION Fluid Description. an ideal fluid.

yc = a

(a)

Steady flow occurs. The water will be considered

Analysis. Since q = 12.5 m3 >s2 > 10.9 m2 = 2.7778 m2 >s, the critical depth for the flow is 12.7778 m2 >s2 2 1>3 q2 1>3 b = c d = 0.9231 m g 9.81 m>s2

Emin =

3 2

yc = 32 10.9231 m2 = 1.385 m

q 5 2.7778 m2 s

c

In general, the specific energy for the flow is E =

2gy2

+ y;

12.7778 m2 >s2 2

219.81 m>s2 2y2 0.3933 + y E = y2 E =

Subcritical flow

0.75

Supercritical flow Emin 5 1.385 1.449 0.0645

+ y

(b)

(1)

At the depth y1 = 0.75 m, E1 = 1.449 m, Fig. 12–13b. The depth for subcritical flow at this same specific energy can be determined by solving 0.3933 1.449 = + y y2 y3 - 1.449 y2 + 0.3933 = 0 The three roots are y2 = 1.154 m 7 0.9231 m y2 = 0.75 m 6 0.9231 m y2 = -0.455 m

Datum

y (m)

Since y1 = 0.75 m 6 yc, then before the bump the flow is supercritical. 1.154 Also, the minimum specific energy is y 5 0.9231

q2

669

Open-Channel FlOw Over a rise Or Bump

Subcritical Supercritical (as before) Unrealistic

As shown in Fig. 12–13b, to produce subcritical flow after the bump, the rise must first remove E1 - Emin = 1.449 m - 1.385 m = 0.0645 m of the specific head from the water. Then, just after the top of the bump, if the bump is designed properly, the same amount of the energy will be returned, so that subcritical flow occurs at the new depth of y2 = 1.154 m. Thus, the required height of the bump is h = 0.0645 m Ans.

Fig. 12–13

E (m)

670

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O p e n -C h a n n e l F l O w

1

y1 2

12

y2 Datum Flow under a sluice gate

Fig. 12–14

12.5

OPEN-CHANNEL FLOW UNDER A SLUICE GATE

A sluice gate is a type of underflow gate that is often used to regulate the discharge of a liquid from a reservoir into a channel.* An example is shown in Fig. 12–14, where the gate is partially opened and we have free outflow. To study the flow we will neglect friction losses through the gate and assume we have steady flow of an ideal fluid. Now we can apply the Bernoulli equation on the streamline between point 1, where there is essentially no flow, and point 2, where the flow is steady and has an average velocity V2. This gives p1 p2 V 12 V 22 + + y1 = + + y2 g g 2g 2g 0 + 0 + y1 = 0 +

V 22 + y2 2g

If the channel is rectangular having a width b, then to obtain the flow per unit width, q = Q>b, as a function of depth, we can use q = V2y2. Substituting for V2 and solving for q, we get y1 =

q2

+ y2

q = 22g 1 y22y1 - y23 2 1>2 2gy22

(12–13)

Opening and closing the gate will vary y2 and also the flow through the gate. Maximum flow can be determined by taking the derivative of the above equation with respect to y2 and setting it equal to zero. 0q = 22g 31 12 2 1 y22y1 - y23 2 - 1>2 1 2y2y1 - 3y22 24 = 0 0y2

The solution requires Sluice gates are placed at the crest of this dam in order to maintain control of the water elevation in the reservoir behind the dam.

y2 =

2 y 3 1

*The force on the gate is found using the momentum equation. See Example 6.3.

(12–14)

12.5

Open-Channel FlOw under a sluiCe gate

671

When y2 is at this critical depth, then the maximum flow per unit width is determined from Eq. 12–13. We have qmax =

8 gy 3 A 27 1

(12–15)

The Froude number at maximum flow, that is, when y2 = 23y1 and V = qmax >y2, is Fr =

=

qmax >y2 8gy13 V 1 = = 1g y2 1gy2 y2 1gy2 B 27

8 y1 3 b = 1 a B 27 23y1

Therefore, the classification of flow just past the sluice gate is Fr 6 1 Fr = 1 Fr 7 1

Subcritical flow Critical flow Supercritical flow

These results can be interpreted as follows: Provided there is subcritical flow upstream, then as the gate is opened, the velocity of the flow past the gate increases so that Fr 7 1. The velocity reaches a maximum at the critical depth y2 = 23y1 (Fr = 1). A further opening of the gate will now cause the velocity to decrease (Fr 6 1). Here the gravitational force becomes greater than the inertia force. In other words, it is more difficult for the liquid to pass under the gate, since on the other side y2 is high enough so the weight of fluid restrains any increase in the velocity. In practice, the results of this analysis are modified somewhat to account for the friction losses under the gate. This is usually done using an experimentally determined discharge coefficient that is multiplied by the flow. See Ref. [8].

IMPORTANT POIN T S • A steady flow will remain subcritical or supercritical both before and after the channel experiences either a change in width or a change in elevation (rise or drop).

• A bump can be designed to change the flow from one type to another; for example, from supercritical to subcritical. This occurs because the bump can first remove specific energy from the flow, bringing the flow to the critical depth, and then return it to its original specific energy. Doing so allows the specific energy to transfer around the “nose” of the specific energy diagram.

• The steady flow over a channel rise, over a bump, or under a sluice gate can be approximated as onedimensional flow, with no friction losses, since the length of the transition is normally short. The flow on each side of the transition is classified according to the Froude number.

• Maximum velocity under a sluice gate will occur if the gate is opened so the exit flow depth is y2 = 23y1.

12

672

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EXAMPLE

O p e n -C h a n n e l F l O w

12.6 The sluice gate in Fig. 12–15 is used to control the flow of water from a large reservoir that has a depth of 6 m. If it opens into a channel that has a width of 4 m, determine the maximum flow that can occur through the channel and the associated depth of flow.

12

SOLUTION

6m

Fluid Description. The water surface in the reservoir is assumed to be at a constant elevation so we have steady flow. Also, the water is assumed to be an ideal fluid.

y2

Analysis. Maximum flow occurs at the critical depth, which is determined from Eq. 12–14. 2 2 y2 = y1 = (6 m) = 4 m Ans. 3 3 The flow at this depth is determined from Eq. 12–15. 8 Qmax = qmaxb; Qmax = a gy 3 b b A 27 1

Fig. 12–15

= a

8 19.81 m>s2 216 m2 3 b 14 m2 A 27

= 100.23 m3 >s = 100 m3 >s

Ans.

As expected, the Froude number for this flow is Fr =

EXAMPLE 0

2gyc Vc

=

1100.23 m3 >s2 > 34 m(4 m)4 219.81 m>s2 2(4 m)

= 1

12.7 1

0.5 m/s y0

y1 5 3 m 2 y2

The flow in the 2-m-wide channel in Fig. 12–16a is controlled by the sluice gate, which is partially opened so that it causes the depth of the water near the gate to be 3 m and the mean velocity there to be 0.5 m>s. Determine the depth of the water far upstream, where it is essentially at rest, and also find its depth downstream from the gate where free outflow occurs. SOLUTION

Datum (a)

Fig. 12–16

Fluid Description. The water a distance far upstream from the sluice gate is assumed to have a constant depth so that the flow at points 1 and 2 will be steady. Also, the water is assumed to be an ideal fluid.

12.5

Open-Channel FlOw under a sluiCe gate

673

Analysis. If the Bernoulli equation is applied between points 0 and 1, located on a streamline at the water surface, we have p0 p1 V 02 V 12 + y0 = + y1 + + g g 2g 2g 1 0.5 m>s 2 2 0 + 0 + y0 = 0 + + 3m 2 1 9.81 m>s2 2 y0 = 3.01274 m = 3.0127 m

12

Ans.

The Bernoulli equation can be applied between points 1 and 2, but we can also apply it between points 0 and 2. If we do this, we have p0 p2 V 02 V 22 + y0 = + y2 + + g g 2g 2g V 22 0 + 0 + 3.01274 m = 0 + + y2 2 1 9.81 m>s2 2

y (m)

(1)

Continuity requires the flow at 1 and 2 to be the same.

q 5 1.5 m2 s 3.00

Q = V1 A 1 = V2 A 2 (0.5 m>s)[3 m (2 m)] = V2 y2(2 m) V2y2 = 1.5

Subcritical flow

Substituting V2 = 1.5>y2 into Eq. 1 yields y32 - 3.01274y22 + 0.11468 = 0

2.008 0.202

Solving for the three roots, we get y1 = 3 m Subcritical (as before) y2 = 0.2020 m Supercritical y2 = -0.1892 m Not realistic The first root indicates the depth y2 = y1 = 3 m, and the second root is the depth downstream from the gate. Thus, y2 = 0.202 m

Ans.

The specific energy for the flow can be determined either from point 0, 1, or 2 since the bed elevation of the channel is constant and friction losses through the gate have been neglected (ideal fluid). Using point 1, E =

q2 2gy22

+ y2 =

3(0.5 m>s) (3 m)4 2

2 1 9.81 m>s2 2 (3 m)2

+ 3 m = 3.01274 m

A plot of the specific energy is shown in Fig. 12–16b. Here 2 2 yc = y0 = (3.01274 m) = 2.0085 m 3 3 3(0.5 m>s)(3 m)4 2 q2 + 2.0085 m = 2.037 m Emin = + yc = 2gyc2 2 1 9.81 m>s2 2 (2.0085 m)2 As noted, subcritical flow occurs upstream near the gate, and supercritical flow occurs downstream.

2.037

Supercritical flow E (m) 3.0127 (b)

Fig. 12–16 (cont.)

674

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O p e n -C h a n n e l F l O w

12.6 STEADY UNIFORM CHANNEL FLOW A 5 yb P 5 2y + b

y

b

12 2 –– (a 2 sina) A5 R 2 P 5 aR

R a

y2 A 5 tana

y a

a

P5

2y sina

A5 y y( + b) tana

y a

a P5

b

2y +b sina

Since all open channels have a rough surface, then in order to maintain steady uniform flow in the channel, it is essential that it have a constant slope and a constant cross section and surface roughness along its length. Although these conditions rarely occur in practice, an analysis based on these assumptions is often used to design many types of channels for drainage and irrigation systems. Furthermore, this analysis is sometimes used to approximate the constant-flow characteristics of natural channels such as streams and rivers. Typical prismatic cross sections of open channels that are commonly used in engineering practice are shown in Fig. 12–17. Important geometric properties that are related to these shapes are defined as follows: Area of Flow, A. Fig. 12–17.

The area of the flow cross section. Shown as blue in

Wetted Perimeter, P. The distance around the cross section of the channel where the channel is in contact with the liquid. This does not include the distance over the free liquid surface, shown in red in Fig. 12–17. Hydraulic Radius, Rh. The ratio of the area of the flow cross section to the wetted perimeter.

Fig. 12–17

Rh =

A P

(12–16)

is generally defined as Re = VRh >n, where the hydraulic radius Rh is the “characteristic length.” Experiments have shown that laminar flow depends upon the cross-sectional geometry, but in many cases it can be specified as Re … 500. For example, a channel having a rectangular cross section of width 1 m and flow depth 0.5 m has a hydraulic radius of Rh = A>P = 31 m(0.5 m)4 > 32(0.5 m) + 1 m4 = 0.25 m. If the channel is carrying water at standard temperature, it would require the mean velocity for laminar flow to be at most

Reynolds Number. For open-channel flow, the Reynolds number

Re =

Drainage canal having a trapezoidal cross section

VRh ; n

500 =

V(0.25 m)

1.00110 - 6 2 m2 >s

V = 2.00 mm>s

This is extremely slow, and so, as mentioned previously, practically all open-channel flow is turbulent. In fact, nearly all flow occurs at very high Reynolds numbers.

12.6

675

steady uniFOrm Channel FlOw

Chézy Equation. To analyze the steady uniform flow along a sloped channel, we will apply the energy equation since its surface is rough and so a head loss will occur over its length L. If we consider the control volume of liquid shown in Fig. 12–18, then the vertical open control surfaces have the same height y. Also, Vin = Vout = V. To calculate the hydraulic head p>g + z, we will make reference to points on the liquid surface. Here pin = pout = 0.*

12 L V y

2 V out

V in2

pin pout + + + zin + hpump = + zout + hturbine + hL g g 2g 2g

y Datum

V2 V2 0 + + y + ∆y + 0 = 0 + + y + 0 + hL 2g 2g

Dy

u Slope S0 Steady uniform flow

Here ∆y = L sin u. For small slopes, or angle u, sin u ≈ tan u = S0. Then ∆y = LS0, and so

hL = LS0 We can also express this head loss using the Darcy–Weisbach equation. hL = f a

L V2 b Dh 2g

To account for channels that have a variety of cross sections, here we will use the hydraulic diameter Dh = 4Rh, where Rh is the hydraulic radius, or the ratio of the cross-sectional area to the wetted perimeter, as discussed in Sec. 10.1. If we now use the above two equations to eliminate hL and solve for V, we get V = C 2RhS0

(12–17)

where C = 28g>f . This result is known as the Chézy formula, named after Antoine de Chézy, a French engineer who deduced it experimentally in 1775. The coefficient C was originally thought to be constant; however, through later experiments it was found to depend on the shape of the channel’s cross section and on the roughness of its surface.

*This head remains constant over each open control surface, and so it can be calculated at any point on the control surface, as shown in Example 5.12.

Fig. 12–18

V

676

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TABLE 12–1 Surface Roughness Coefficient Perimeter 12

n

Earth channel Grass land Sparse vegetation Heavy weeds Firm soil Smooth soil Gravel surface Rocky surface Artificially lined channel Steel Cast iron Corrugated metal Finished concrete Unfinished concrete Precast concrete Brick surface Wood

0.02–0.04 0.05–0.1 0.07–0.15 0.025–0.032 0.017–0.025 0.02–0.035 0.035–0.050

Manning Equation. In 1891 Robert Manning, an Irish engineer, established experimental values for C by expressing it in terms of the hydraulic radius and a dimensional surface roughness coefficient, n, which has units of s>m1>3. He found C = Rh1>6 >n. Typical values of n, given in SI units for some common surface conditions encountered in engineering, are listed in Table 12–1. See Ref. [2]. Notice that for any one surface there is a range of values for n, because it has been found to vary with the channel’s shape and size and the water depth. Also, there can be a variability of surface roughness along the length of the channel and, with time, vegetation growth and the effects of erosion will alter the value of n. As a result, care should be taken when selecting a particular value for design. Expressing the average velocity in terms of n, the Manning equation is then V =

0.012–0.018 0.012–0.019 0.022–0.030 0.010–0.013 0.012–0.016 0.011–0.015 0.013–0.018 0.010–0.013

kRh2>3S 01>2 n

(12–18)

When n is selected from Table 12–1,* k = 1 (SI units)

(12–19)

This formula continues to be widely used because it gives reasonably accurate results. Since Q = VA, and the hydraulic radius is Rh = A>P, Eq. 12–18 can also be expressed in terms of the volumetric flow. It is Q =

kA5>3S 01>2 nP 2>3

(12–20)

Best Hydraulic Cross Section. It can be seen that for a given y

b

Fig. 12–19

slope S0 and surface roughness n, the flow Q will increase if the wetted perimeter P decreases. Therefore, we can obtain the maximum Q by minimizing the wetted perimeter P. Such a constant cross section is called the best hydraulic cross section since it will both minimize the amount of material needed to construct the channel and maximize the flow. For example, if the channel has a rectangular cross section of given width b and liquid depth y, Fig. 12–19, then A = by and so P = 2y + b = 2y + A>y. Thus, for a constant value of A, although the relationship between b and y is not as yet known, we have

dP d A A = a 2y + b = 2 - 2 = 0 y dy dy y b A = 2y2 = by or y = 2 Therefore, a rectangular channel flowing at a depth of y = b>2 will require the smallest amount of material used for its construction. Since this design size will also provide the maximum amount of uniform flow, it is the best cross-sectional size for the rectangle. *To simplify, we will not include the units for n whenever numerical data is substituted into any equation that involves n.

12.6

steady uniFOrm Channel FlOw

677

In a strict sense, a semicircular cross section, flowing full, is the most efficient shape; however, for very large flows this shape is generally difficult to excavate in soil and costly to construct. Instead, large channels have trapezoidal cross sections, or for shallow depths, their cross sections may be rectangular. In any case, the best hydraulic cross section for a selected shape can always be determined as demonstrated here. Once this best cross section is found, then the uniform velocity within it can be determined using Manning’s equation.

12

Critical Slope. If Eq. 12–20 is solved for the slope of the channel, and we express it in terms of the hydraulic radius Rh = A>P, then we get S0 =

Q 2 n2 k 2Rh4>3A2

Flow through a concrete channel having a trapezoidal cross section.

(12–21)

Channel slope

The critical slope for a channel of any cross section requires the depth of flow to be at the critical depth, yc, and since the critical depth is determined from Eq. 12–11, for the critical slope the above equation becomes Sc =

n2gAc 2

4>3 k btop Rhc

(12–22)

Critical slope

Here the critical area Ac and hydraulic radius Rhc are determined by setting y = yc for the section. With this equation, like the depth y, we can compare the actual slope of a channel S0 with its critical slope Sc, and thereby classify the flow. S 0 6 Sc S0 = Sc S 0 7 Sc

Subcritical flow Critical flow Supercritical flow

The examples that follow illustrate a few applications of these concepts.

IMPORTANT POIN T S • Steady uniform open-channel flow can occur provided the channel has a constant slope and surface roughness, and the liquid in the channel has a constant cross section. When this occurs, the gravitational force on the liquid will be in balance with the frictional forces along the bottom and sides of the channel.

• The Manning equation can be used to determine the average velocity within an open channel having •

steady uniform flow. The best hydraulic cross section for a channel of any given shape having a constant flow, slope, and surface roughness can be determined by minimizing its wetted perimeter.

• The type of steady uniform flow in a channel can be determined by comparing its slope S0 with the critical slope Sc.

678

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EXAMPLE

12

O p e n -C h a n n e l F l O w

12.8 The channel in Fig. 12–20 is made of finished concrete, and the bed drops 2 m in elevation for a horizontal stretch or reach of 1000 m. Determine the steady uniform flow when the depth of the water is 2 m. Take n = 0.012.

2m

1000 m 3m

2m

Fig. 12–20

SOLUTION Fluid Description. The flow is steady and uniform, and the water is assumed to be incompressible and flows with an average velocity V. Analysis. Here we will use the Manning equation, with k = 1, for SI units. The slope of the channel is S0 = 2 m>1000 m = 0.002. Also, for a 2-m depth of water, the hydraulic radius is

Rh =

(3 m)(2 m) A = = 0.8571 m P 2(2 m) + 3 m

Applying Eq. 12–18,

V =

kRh2>3S 01>2 n

(1)(0.8571 m) 2>3(0.002)1>2 Q = (3 m)(2 m) 0.012 Q = 20.18 m3 >s = 20.2 m3 >s

Ans.

12.6

EXAMPLE

679

steady uniFOrm Channel FlOw

12.9

The channel in Fig. 12–21 consists of an unfinished concrete section (n = 0.014) and overflow regions on each side that contain light brush (n = 0.050). If the bottom of the channel has a slope of 0.0015, determine the steady volumetric flow when the depth is 2.5 m, as shown.

1

SOLUTION Fluid Description. We have steady uniform flow, and we will assume the water is an incompressible fluid.

5m

2

3 1m 1.5 m

2m

5m

Fig. 12–21

Analysis. For an approximate solution, the cross section is divided into three composite rectangles, Fig. 12–21. The flow through the entire cross section is thus the sum of the flows through each composite rectangle. For the calculation, note that the wetted perimeter does not include the liquid boundary between the rectangles because they are not part of the channel’s wall or bed surface. Since n = 0.014 s>m1>3 for unfinished concrete, and for light brush n = 0.050 s>m1>3, as stated, the Manning equation, written in the form of Eq. 12–20, becomes Q = Σ

kA5>3S 01>2 nP 2>3

= (0.0015)1>2 c = 20.7 m3 >s

= (1) S 01>2 a

A15>3

n1P 12>3 3(5 m)(1 m)4 5>3

0.050(1 m + 5 m)2>3

+ +

A25>3 n2P 22>3

b n3P 32>3 3(2 m)(2.5 m)4 5>3 +

A35>3

0.014(1.5 m + 2 m + 1.5 m)2>3

+

3(5 m)(1 m)4 5>3

0.050(1 m + 5 m)2>3

Ans.

EXAMPLE

12.10 1

The triangular flume in Fig. 12–22a is used to carry water over a ravine. It is made of wood and has a slope of S0 = 0.001. If the intended flow is to be Q = 3 m3 >s, determine the depth of the flow. Take n = 0.012. SOLUTION

nP 2>3

1

(a)

2y

Analysis. If y is the depth of the flow, Fig. 12–22b, then 1 P = 212y and A = 2c ( y)( y) d = y2 2 For SI units, Q =

1 1

Fluid Description. We assume steady uniform flow of an incompressible fluid.

kA5>3S 01>2

d

;

3 m >s = 3

(1)1 y2 2 5>3(0.001)1>2 0.0121212y2 2>3 y = 1.36 m

y

y (b)

Ans.

Fig. 12–22

y

12

680

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EXAMPLE

O p e n -C h a n n e l F l O w

12.11 The channel in Fig. 12–23 has a triangular cross section and is made of unfinished concrete. If the flow is 1.5 m3 >s, determine the slope that produces critical flow. Take n = 0.014.

12

SOLUTION Fluid Description. We assume steady uniform flow of an incompressible fluid. Analysis. The critical slope is determined using Eq. 12–22, but first we need to determine the critical depth. Since Ac = 2 3 12 1 12yc 2 yc 4 = 21yc2

btop = 2 1 12yc 2 = yc

(© Andrew Orlemann/Shutterstock)

Then applying Eq. 12–11, 1 y 2 c

Ac 3 Q2 = g btop

2

yc

2 1

11.5 m3 >s2 2

1

9.81 m>s2

=

1 12yc2 2 3 yc

yc = 1.129 m

Fig. 12–23

Using this value, btop = 1.129 m

Ac = 12 11.129 m2 2 = 0.6374 m2

Pc = 22(1.129 m)2 + (1.129 m>2)2 = 2.525 m

Rhc =

Ac 0.6374 m2 = = 0.2525 m Pc 2.525 m

Thus, Sc =

n2gAc 4>3 k 2btop Rhc

=

(0.014)2 19.81 m>s2 210.6374 m2 2 (1)2(1.129 m)(0.2525 m)4>3

= 0.00680

Ans.

Therefore, when the channel has a flow of Q = 1.5 m3 >s, then any slope less than Sc will produce subcritical (tranquil) flow, and any slope greater than Sc will produce supercritical (rapid) flow.

12.7

12.7

681

gradually varied FlOw

GRADUALLY VARIED FLOW

In the previous section we considered steady uniform flow in an open channel having a constant slope. This required that the flow occurs at a unique depth, so that this depth remains constant over the length of the channel. However, when the slope or cross-sectional area of the channel gradually changes, or the flow comes to a transition, then the depth of the liquid will vary along the channel length, and steady nonuniform flow will result. To analyze this case, we will apply the energy equation between the “in” and “out” sections on the differential control volume shown in Fig. 12–24. We will select the points at the top of the liquid surface to calculate the hydraulic head terms,* p>g + z. Here pin = pout = 0, and so

12

2 Vin

2 pin pout V out V in2 + + + yin + hturbine = + yout + hpump + hL g g 2g 2g 2 V out V in2 0 + + (y′in + yin) + 0 = 0 + + (y′out + yout) + 0 + dhL 2g 2g 2 V out V in2 = ( yout - yin) + (y′out - y′in) + dhL 2g 2g

d V2 a b dx = dy - S0 dx + dhL dx 2g

If we define the friction slope Sf as the slope of the energy grade line, then dhL = Sf dx, Fig. 12–24. Substituting this for dhL into the above equation, we have, after simplifying, dy d V2 a b = S 0 - Sf + dx dx 2g

2

2g

Vout 2g

yin

yout

y9in

y9out

1

(12–23)

*Any point on the open control surfaces can be selected, as noted in Example 5.12.

Sf

S0

Datum Nonuniform flow

Fig. 12–24

The two terms on the left represent the change in the velocity head over the length dx. Also, yout - yin = dy, and y′out - y′in = -S0 dx, where S0 is the slope of the channel bed, which is positive when the channel slopes downward to the right. Therefore,

-

dx dhL Energy line

1

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Rectangular Cross Section. If the channel has a rectangular cross section of constant width b, then V = Q>by, and so dy Q2 Q2 d V2 d V 2 dy a b = a b = -2a b = a ba b gy dx 2g dx 2gb2y2 dx 2gb2y3 dx

12

Finally, we can express this result in terms of the Froude number. Since Fr = V> 1gy, then V 2 >gy = Fr 2. Therefore, dy d V2 a b = -Fr2 dx 2g dx y

For a rectangular cross section, Eq. 12–23 can now be expressed as 0 2 Sf dy 5 ,0 dx 1 2 Fr2 H2

y . yc

S 0 - Sf dy = dx 1 - Fr2

(12–24)

x S0 5 0

Fig. 12–25

Surface Profiles. Since Fr2 is a function of y, the above equation is a nonlinear first-order differential equation that requires an integration to obtain the depth y of the liquid surface. It is important to be able to determine the shape of this surface and the depth of the flow, especially when the channel bed changes slope or the flow meets an obstruction such as a dam or a sluice gate. Here there is a possibility that flooding, spillover, or some other unforeseen effect can occur. As shown in Table 12–2, there are twelve possible shapes the liquid surface can have. Each group of shapes is classified according to the channel’s slope, that is, horizontal (H), mild (M), critical (C), steep (S), or adverse (A). Also, each shape is classified by a zone, indicated by a number that depends upon the depth of the actual flow, y, compared to the depth for uniform or normal flow, yn, and the depth for critical flow, yc. Zone 1 is for high values of y, zone 2 is for intermediate values, and zone 3 is for low values. Typical examples of how these shapes or profiles can form in a channel are also shown in the table. The preliminary shape of the liquid surface for any problem can be sketched by studying the behavior of its slope as defined by Eq. 12–24. For example, for the case shown in Fig. 12–25, S0 = 0, and where y 7 yc, the flow must be subcritical or tranquil so that Fr 6 1. From Eq. 12–24, dy>dx 6 0, and so the initial slope of the water surface will decrease as shown. In other words, the depth will decrease as x increases. This is an H2 profile as noted in Table 12–2.

12.7

TABLE 12–2

683

gradually varied FlOw

Surface Profile Classification M1 y . yn . yc M1 Horizontal

yn yc

yn . yc . y

12

M2

M2

yn . y . yc

M1

M3

M3 Mild slope (M) S0 , Sc S1

Horizontal

y . yc . yn S1 yc

S1

yc . y . yn yn

S2

S2

yc . yn . y

S1

S3

S3 Steep slope (S) S0 . Sc

C1

Horizontal

y . (yn 5 yc)

C1

yn 5 yc y , (yn 5 yc) C3

C1 C3

Critical slope (C) S0 5 Sc

H2

H2 yn 5 `

y . yc H2

y , yc

yc

H2

H3 H3

Horizontal (H) S0 5 0

A2

A2

A2

y . yc A3

yn 5 ` yc

y , yc

A3

Adverse slope (A) S0 , 0

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Calculating the Surface Profile. Once the water surface profile has

12 V1 V2 y1 y2 5 y1 1 Dy S0

1

Dx

Fig. 12–26

been classified, using the values of y, yn, yc, S0, and Sc, then the actual profile can be determined by integrating Eq. 12–24, and checking the results to be sure if it has the same shape and is indeed within the bounds of yc and yn. Over the years, several procedures have been developed for doing this; however, here we will consider using a finite difference method to perform a numerical integration. To do this, we first write Eq. 12–23 in the form d( y + V 2 >2g) d V2 ay + b = S0 - Sf or dx = dx 2g S0 - Sf If the channel is divided into small finite reaches, or segments, then this equation can be written in terms of a finite difference, ( y2 - y1) + (V 22 - V 12)>2g ∆x = (12–25) S0 - Sf We start from a control point, having a known discharge Q and water elevation y1, Fig. 12–26. For small slopes, which is the usual case, the vertical depth y1 can be used to calculate the cross-sectional area A1 of the flow. Then the mean velocity V1 can be calculated using V1 = Q>A1. An increase in the water depth ∆y is assumed, and the area A2 at y2 = y1 + ∆y is then calculated. And finally the mean velocity is found from V2 = Q>A2. If we further assume the head loss along the segment is the same as that along the same segment having uniform flow, then we can use the Manning equation, Eq. 12–18, to determine the friction slope. n2V m2 Sf = 2 4>3 (12–26) k Rhm The values Vm and Rhm are average values of the mean velocity and the mean hydraulic radius for the segment. Substituting all the relevant values into Eq. 12–25, we can then calculate ∆x. The iteration is continued for the next segment until one reaches the end of the channel. Although this process is tedious to do by hand, it is rather straightforward to program into a pocket calculator or computer. Actually several software programs are also accessible for calculating the water surface profile for openchannel flow. One popular program, HEC-RAS, works well for naturally formed open channels, such as rivers and drainage canals. It also is based on an iteration process, and the details of its use can be found online, and a copy can be obtained from the Army Corps of Engineers.

IMPORTANT POINTS • When steady nonuniform flow exists, the liquid depth will gradually vary along the length of the channel. • •

The slope of the liquid surface depends upon the Froude number, Fr, the slope of the channel bed, S0, and the friction slope Sf. For nonuniform flow there are twelve classifications for the liquid surface profile, as shown in Table 12–2. To determine which profile occurs, it is necessary to compare the channel slope S0 with its critical slope Sc, and to compare the liquid depth y with the depths for uniform or normal flow, yn, and critical flow, yc. For nonuniform flow in rectangular channels, it is possible to use a finite difference method to numerically integrate the differential equation dy>dx = (S0 - Sf)>11 - Fr2 2, and thereby obtain a plot of the liquid surface.

12.7

EXAMPLE

gradually varied FlOw

685

12.12

The rectangular channel in Fig. 12–27a is made of unfinished concrete and has a slope of S0 = 0.035. When it reaches the check dam, the flow backs up as shown in Fig. 12–27b. At a specific location before the dam, the water has a depth of 1.25 m and the flow is Q = 0.75 m3 >s. Classify the surface profile for the flow. Take n = 0.014.

12

SOLUTION

1.25 m

Fluid Description. The flow is steady and nonuniform. The water is considered incompressible. Analysis. To classify the surface profile, we must determine the critical depth yc, the normal flow depth yn, and the critical slope Sc. Using Eq. 12–7, with q = Q>b = (0.75 m3 >s)>(2 m) = 0.375 m2 >s,

2m

(0.375 m2 >s)2 1>3 q2 1>3 yc = a b = c d = 0.2429 m g 9.81 m>s2

(a)

Since y = 1.25 m 7 yc = 0.2429 m, the flow is subcritical. The depth yn that will produce normal or uniform flow for Q = 0.75 m3 >s is determined from the Manning equation. Since P = (2yn + 2 m) then Eq. 12–20 becomes Q =

kA5>3S 01>2 nP 2>3

;

0.75 m3 >s = y5>3 n

(2yn + 2 m)2>3

(1)3(2 m)yn 4 5>3(0.035)1>2 0.014(2yn + 2 m)2>3

= 0.017678

S1

Solving by trial and error, or using a numerical procedure, we get yn = 0.1227 m The critical slope is now determined from Eq. 12–22, Sc =

n2gAc k 2btop Rhc4>3

=

(0.014)2 19.81 m>s2 2(2 m)(0.2429 m) (1)2(2 m) c

= 0.004118

4>3 2 m(0.2429 m) d 2(0.2429 m) + 2 m

Since y = 1.25 m 7 yc 7 yn and S0 = 0.035 7 Sc, then according to Table 12–2 the water surface in Fig. 12–27 is an S1 profile. Ans.

(b)

Fig. 12–27

686

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EXAMPLE

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12.13 Water flows from a reservoir under a sluice gate and into the 1.5-m-wide unfinished concrete rectangular channel, which is horizontal, Fig. 12–28a. At the control point 0, the measured flow is 2 m3 >s and the depth is 0.2 m. Show how to determine the variation of the water depth along the channel, measured downstream from this point. Take n = 0.014.

12

0

y

0.2 m x

SOLUTION Fluid Description. Assuming the reservoir maintains a constant level, then the flow will be steady. Just after point 0, it will be nonuniform because its depth is changing. As usual, we assume water to be an ideal fluid, and use a mean velocity profile. Analysis. First we will classify the water surface profile. Since q = Q>b = (2 m3 >s)>(1.5 m) = 1.333 m2 >s, the critical depth is

y

1.5 m (a)

Fig. 12–28

11.333 m2 >s2 2 1>3 q2 1>3 yc = a b = c d = 0.5659 m g 9.81 m>s2

Here y = 0.2 m 6 yc = 0.5659 m, so the flow is supercritical. Since the channel is horizontal, S0 = 0. Using Table 12–2, the water surface profile is H3, as shown in Fig. 12–28a. This indicates that the water depth will increase as x increases from the control point. (Note that if the channel had a slope, then the Manning equation would have to be used to find yn since this value is also needed for a profile classification.) To determine the downstream depths, we will divide the calculations into increments of depth, choosing ∆y = 0.01 m. At station 0, y = 0.2 m, and the velocity is

V0 =

2 m3 >s Q = = 6.667 m>s A0 (1.5 m)(0.2 m)

The hydraulic radius is therefore

Rh0 =

(1.5 m)(0.2 m) A0 = = 0.1579 m P0 [1.5 m + 2(0.2 m)]

These results are entered into the first line of the table shown in Fig. 12–28b.

12.7

Station

0

y

V

(m)

(m/s)

0.2

6.667

Vm

Rh

Rhm

(m/s)

(m)

(m)

0.21

6.349

0.22

6.061

0.23

(m)

(m)

5.797

687

0 0.1610

0.09479

2.116

0.1671

0.08200

2.104

0.1731

0.07144

2.089

12

2.12

0.1701 5.929

3

x

0.1641 6.205

2

x

0.1579 6.508

1

Sfm

gradually varied FlOw

4.22

0.1760

6.31

(b)

At station 1, y1 = 0.2 m + 0.01 m = 0.21 m, thus, V1 = Q>A1 = 6.349 m>s and Rh1 = A1 >P1 = 0.1641 m (third line of table). The intermediate second line reports the values of

Vm =

Rhm

6.667 m>s + 6.349 m>s V0 + V1 = = 6.508 m>s 2 2

Rh0 + Rh1 0.1579 m + 0.1641 m = = = 0.1610 m 2 2

Sfm =

∆x =

=

n V m2 4>3 k 2Rhm

2

2

(0.014) (6.508 m>s) =

(1)2(0.1610 m)4>3

0 0.20 m

1

2

0.21 m

0.22 m

= 0.09479

( y1 - y0) + 1V 12 - V 02 2 >2g S - Sfm

(0.21 m - 0.2 m) + 3(6.349 m>s)2 - (6.667 m>s)2 4 > 3219.81 m>s2 24 (0 - 0.09479)

= 2.116 m The calculations are repeated for the next two stations as shown in the table, and the results are plotted in Fig. 12–28c. They appear satisfactory, since each increase ∆y = 0.01 m produces somewhat uniform changes ∆x. Had larger changes occurred, then smaller increments of ∆y should be selected to improve the accuracy of the finite-difference method.

0.23 m x

2.12 m 4.22 m

2

3

6.31 m (c)

Fig. 12–28

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12.8

12

Hydraulic jump

Fig. 12–29

THE HYDRAULIC JUMP

When water passes down the spillway of a dam or underneath a sluice gate, it is typically supercritical flow, Fig. 12–29. The transition to subcritical flow downstream can occur, and when it does, it happens in a rather abrupt manner resulting in a hydraulic jump. This is turbulent mixing that releases some of the water’s kinetic energy, and in doing so, brings the water surface up to the depth needed for subcritical flow. Because of the rapid energy loss, engineers often design hydraulic jumps using a stilling basin, in order to mitigate the effects of a rapid flow that may otherwise damage a structure or scour the foundation upon which it rests. Regardless of how the jump is formed, it is possible to determine the difference in water depth across the jump and the energy loss within the jump. To do this we must apply the continuity, momentum, and energy equations to a control volume containing the region of the jump, Fig. 12–30a. For the analysis, we will consider the jump to occur along the horizontal bed of a rectangular channel having a width b.

Continuity Equation. Since the control volume extends just before and beyond the jump into regions of steady flow, we have

0 r dV + r Vf>cs # dA = 0 0t Lcv Lcs 0 - rV1( y1b) + rV2( y2b) = 0 (12–27)

V1y1 = V2y2

2 V2

V1 1

y2 y1 Datum (a)

Fig. 12–30

12.8

689

the hydrauliC Jump

Momentum Equation. Experiments have shown that the jump happens within a short distance, and so frictional forces acting on the fixed control surfaces at the channel’s bed and sides are negligible compared to the forces due to pressure, Fig. 12–30b. Applying the momentum equation in the horizontal direction, we get 12

ΣF =

0 Vr dV + Vr Vf>cs # dA 0t Lcv Lcs

Typical formation of a hydraulic jump

1 1 (rgy1b)y1 - (rgy2b)y2 = 0 + V2r3V2(by2)4 + V1r3 -V1(by1)4 2 2 Or, after simplifying,

W

y 12 y22 V 22 V 12 = y2 y g g 1 2 2

y2 y1 p2

p1

Using the continuity equation to eliminate V2, then dividing each side by y1 − y2, and finally multiplying both sides by 2y2 > y21, we get y2 y2 2 2V 21 = a b + gy1 y1 y1

(12–28)

y2 y2 2 b + y1 y1

Finally, using the quadratic formula, and solving for the positive root of y2 >y1, we get y2 1 = c 21 + 8Fr12 - 1 d y1 2

Free-body diagram (b)

Since the Froude number at section 1 is Fr1 = V1 > 1gy1, then 2Fr12 = a

N

(12–29)

Notice that if critical flow occurs upstream, then Fr1 = 1, and this equation shows y2 = y1. In other words, there is no jump. When supercritical flow occurs upstream, Fr1 7 1, and so y2 7 y1, as expected. Consequently, subcritical flow will occur downstream.

Fig. 12–30 (cont.)

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on top of the water surface to obtain the hydraulic head 1p>g2 + z at the open control surface, Fig. 12–30a, we have

Energy Equation. If we apply the energy equation, choosing points 2 V2

12

V1 1

y2

p1 p2 V 12 V 22 + + + z1 + hpump = + z2 + hturbine + hL g g 2g 2g 0 +

y1

V 12 V 22 + y1 + 0 = 0 + + y2 + 0 + hL 2g 2g

Datum (a)

Fig. 12–30 (repeated)

Or, written in terms of the specific energy, the head loss is hL = E 1 - E2 = a

V 12 V 22 + y1 b - a + y2 b 2g 2g

This loss reflects the turbulent mixing of the liquid within the jump, which is dissipated in the form of heat. Using the continuity equation, V2 = V1(y1 >y2), we can also write this expression as hL =

y1 2 V 12 c 1 - a b d + ( y1 - y2) y2 2g

Finally, if Eq. 12–28 is solved for V 12, and the result is substituted into the above equation, it can be shown that after simplification the head loss across the jump is Hydraulic jump in a small stream

hL =

(y2 - y1)3 4y1y2

(12–30)

For any real flow, it is always necessary for hL to be positive. A negative value would violate the second law of thermodynamics, since frictional forces only dissipate energy, they never add energy to the fluid. As shown by Eq. 12–30, hL is positive only if y2 7 y1, and so a hydraulic jump occurs only when the flow changes from supercritical flow upstream to subcritical flow downstream.

I MPO RTA N T PO I N T • Nonuniform flow can occur suddenly in a channel in the form of a hydraulic jump. This process removes energy from supercritical flow and thereby converts it to subcritical flow over a very short distance. The difference in water depth between each side of the jump, and the energy loss within the jump, can be determined by satisfying the continuity, momentum, and energy equations.

12.8

EXAMPLE

8 ms y2 y1 5 0.5 m

Fig. 12–31

SOLUTION Fluid Description. Steady uniform flow occurs before and after the jump. The water is considered incompressible, and average velocity profiles will be used. The Froude number for the flow before the jump is

Fr1 =

691

12.14

Water flows down a dam spillway and then forms a hydraulic jump, Fig. 12–31. Just before the jump, the velocity is 8 m>s, and the water depth is 0.5 m. Determine the velocity downstream in the channel.

Analysis.

the hydrauliC Jump

2gy1 V1

8 m>s

=

2(9.81 m>s2)(0.5 m)

= 3.6122 7 1

Thus the flow is supercritical, as expected. After the jump the depth of the water is y2 1 = a 21 + 8Fr12 - 1b y1 2

y2 1 = a 21 + 8(3.6122)2 - 1b 0.5 m 2 y2 = 2.316 m We can now obtain the velocity V2 by applying the continuity equation, Eq. 12–27. V1y1 = V2 y2 (8 m>s)(0.5 m) = V2(2.316 m) V2 = 1.727 m>s = 1.73 m>s

Ans.

12

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EXAMPLE

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12.15 The sluice gate in Fig. 12–32 is partially opened in a 2-m-wide channel, and water passing under the gate forms a hydraulic jump. At the low level, just before the jump, the water depth is 0.2 m, and the flow is 1.30 m3 >s. Determine the depth farther downstream and also the head loss across the jump.

12

y2 y1 5 0.2 m

Fig. 12–32

SOLUTION Fluid Description. We will assume the water is incompressible and the level in the reservoir is maintained, so steady flow occurs past the sluice gate. Analysis. Just before the jump, the Froude number is Fr1 =

1gy1 V1

=

1gy1

Q>A1

=

11.30 m3 >s2 >[(2 m)(0.2 m)] 2(9.81 m>s2)(0.2 m)

= 2.320 7 1

Thus the flow is supercritical, so indeed a jump can occur. Applying Eq. 12–29 to determine the water height after the jump, we have

y2 y2 1 1 = a 21 + 8Fr12 - 1b ; = a 21 + 8(2.320)2 - 1b y1 2 0.2 m 2 y2 = 0.5638 m = 0.564 m Ans.

At this depth, Fr2 =

1gy2

Q>A2

=

11.30 m3 >s2 >[(2 m)(0.5638 m)] 2(9.81 m>s2)(0.5638 m)

= 0.4902 6 1

The flow is subcritical, as expected. The head loss can be determined from Eq. 12–30, hL =

( y2 - y1)3 (0.5638 m - 0.2 m)3 = = 0.1068 m = 0.107 m Ans. 4y1y2 4(0.2 m)(0.5638 m)

Note that since the original specific energy of the flow is E1 =

q2 2gy12

+ y1 =

[11.30 m3 >s2 >(2 m)]2

219.81 m>s2 2(0.2 m)2

+ 0.2 m = 0.7384 m

then, after the jump, the specific energy of the flow becomes E2 = E1 - hL = 0.7384 m - 0.1068 m = 0.6316 m This produces the following percent of energy lost within the jump, EL =

hL 0.1068 m b (100%) = 14.46% * 100% = a E1 0.7384 m

12.9

12.9

weirs

693

WEIRS

A weir is a device that is placed within a channel to obstruct the flow, causing the water to back up and then flow over it. If the depth of water passing over the weir is measured, it then provides a simple but effective way to measure the flow. Generally there are two types of weirs: sharpcrested weirs and broad-crested weirs.

12

Sharp-Crested Weirs. A sharp-crested weir normally takes the form of a rectangular or triangular plate that has a sharp edge at the upstream side to minimize contact with the water, Fig. 12–33. As the water flows over the weir, it forms a vena contracta called a nappe. To maintain this shape it is necessary to provide proper air ventilation underneath the nappe so that the water will fall clear of the weir plate. This is especially true for rectangular plates that extend the entire width of the channel, as in Fig. 12–33.* Streamlines within the nappe are curved, and so the accelerations that occur here will cause nonuniform flow. Also, in the channel just downstream from the weir plate, the flow combines the effects of turbulence and vortex motion. Upstream from the weir, however, the streamlines remain approximately parallel, the pressure varies hydrostatically, and the flow is uniform. As a result, if we consider this region, we can show that the flow over the weir is a function only of the upstream depth of the liquid. Assuming steady flow and water as an ideal fluid, we will now analyze three different cases.

Weir plate

H Nappe h9 Air vent

Flow over a rectangular weir

Fig. 12–33

*It may be of interest to note that the spillway of a large dam often has the same profile as a free-falling nappe, because then the water is only in slight contact with the surface.

Flood control of water flowing from a reservoir

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Rectangle. If the weir has a rectangular shape that extends over the entire width of the channel, Fig. 12–34a, then the Bernoulli equation can be applied between points 1 and 2 on the streamline shown in Fig. 12–34b. Assuming the approach velocity V1 is small compared to V2, so that V1 can be neglected, we have 12

p1 p2 V 12 V 22 + + + z1 = + z2 g g 2g 2g p1 V 22 + 0 + z1 = 0 + + (h′ + y) g 2g b

H h

dh

Here p2 = 0 because within the nappe the liquid is in free fall, so the pressure is atmospheric. Also, note from Fig. 12–34b that the hydraulic head p1 >g + z1 = h′ + y + h. Substituting this into the above equation and solving for V2, we obtain V2 = 12gh

Since the velocity is a function of h, the theoretical discharge through the entire cross section must be determined by integration, Fig. 12–34a. We have

Rectangular weir (a)

A

p1 g z1

2 1

h H y h9 Datum

(b)

Actual inside width of channel b

Qt =

Qt =

LA

V2 dA =

L0

H

22gh (b dh) = 22g b

2 22g bH 3>2 3

Suppressed rectangular weir

Fig. 12–34

L0

H

h1>2dh

(12–32)

To account for the effects of friction loss, and for any other assumptions that were made, an experimentally determined discharge coefficient Cd is included in Eq. 12–32 to calculate the actual discharge. Its value also accounts for the smaller depth of flow over the weir, to point A in Fig. 12–34b, and the amount of horizontal contraction of the nappe. Specific values of Cd can be found in the literature related to open-channel flow. See Ref. [9]. Using Cd, Eq. 12–32 then becomes 2 Qactual = Cd 22g bH 3>2 3

(c)

(12–31)

(12–33)

To further restrict the upstream flow, a suppressed rectangular weir can be used, Fig. 12–34c. Care must be taken, however, in selecting the width, b, since the nappe also contracts along its sides for very narrow widths.

12.9

weirs

695

b x h

dh

H

u 2

u 2

12

Triangular weir

Fig. 12–35

Triangle. When the discharge is small, it is convenient to use a weir plate with a triangular opening, Fig. 12–35. The Bernoulli equation yields the same result for the velocity as before, which is expressed by Eq. 12–31. Using the differential area dA = x dh, the theoretical discharge now becomes Qt =

LA

V2 dA =

L0

H

22gh x dh

Here the value of x is related to h by similar triangles, x b = H - h H x =

b (H - h) H

so that Qt = 22g

b 4 h1>2(H - h)dh = 22g bH 3>2 H L0 15 H

Because tan (u>2) = (b>2)>H, then Qt =

8 u 22g H 5>2 tan 15 2

(12–34)

Using a discharge coefficient, Cd, determined from experiment, the actual flow through a triangular weir is therefore

Qactual = Cd

8 22g H 15

5>2

tan

u 2

(12–35)

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Broad-Crested Weirs. A broad-crested weir consists of a weir block

12

that supports the nappe of the flow over a horizontal distance, Fig. 12–36. It provides a means for finding both the critical depth yc and the flow. As in the previous two cases, we will assume the fluid to be ideal and the steady velocity of approach to be negligible. Then applying the Bernoulli equation between points 1 and 2 on the surface streamline, we have

1 2

H

y

p1 p2 V 12 V 22 + + + z1 = + z2 g g 2g 2g

Datum

0 + 0 + H = 0 + Broad-crested weir

Fig. 12–36

Thus,

V 22 + y 2g

V2 = 22g(H - y)

Provided the flow is originally subcritical, then as it begins to pass over the weir block, it will accelerate until the depth drops to the critical depth y = yc, and the specific energy will be a minimum. Here Eq. 12–9 applies. V2 = Vc = 2gyc Substituting this result into the above equation yields

yc =

2 H 3

(12–36)

If the channel is rectangular and has a width b, then using these results, the theoretical discharge is Qt = V2A = Weir used to measure the flow of a river

2 2 ga Hb c b a Hb d B 3 3

3>2 2 = b 2ga Hb 3

(12–37)

This result is rather close to the actual flow obtained by experiment; however, to bring the two into better agreement, so as to account for the assumptions made, an experimentally determined broad-crested-weir coefficient Cw is used. See Ref. [9]. Thus, 3>2 2 Qactual = Cw b 2ga Hb 3

(12–38)

12.9

weirs

697

IMPORTANT POIN T S • Sharp-crested and broad-crested weirs are used to measure the flow in an open channel. • Sharp-crested weirs are plates that have either rectangular shapes or triangular openings, and are placed 12

perpendicular to the flow in the channel.

• Broad-crested weirs support the flow over a horizontal distance. They are used to determine the critical depth and the flow.

• The flow or discharge over a weir can be determined using the Bernoulli equation, and the result is adjusted to account for friction loss and other effects by using an experimentally determined discharge coefficient.

EXAMPLE

12.16

Water in the channel in Fig. 12–37 is 2 m deep, and the depth from the bottom of the triangular weir to the bottom of the channel is 1.75 m. If the discharge coefficient is Cd = 0.57, determine the volumetric flow in the channel. 30° 30°

2m 1.75 m

Fig. 12–37

SOLUTION Fluid Description. The water level in the channel is assumed to be constant, so we have steady flow. Also, water is considered to be an ideal fluid. Analysis. Here H = 2 m - 1.75 m = 0.25 m. Applying Eq. 12–35, the flow is therefore

8 u 22g H 5>2 tan 15 2 8 = (0.57) a b 2219.81 m>s2 2 (0.25 m)5>2 tan 30° = 0.0243 m3 >s 15 Ans.

Qactual = Cd

698

C h a p t e r 12

O p e n -C h a n n e l F l O w

References 1. R. W. Carter et al., “Friction factors in open channels,” Journal

Hydraulics Division, ASCE, Vol. 89, No. AY2, 1963, pp. 97–143. 2. V. T. Chow, Open Channel Hydraulics, McGraw-Hill, New York,

12

NY, 2009. 3. R. French, Open Channel Hydraulics, McGraw-Hill, New York,

NY, 1992. 4. C. E. Kindsater and R. W. Carter, “Discharge characteristics of

rectangular thin-plate weirs,” Trans ASCE, 124, 1959, pp. 772–822. 5. R. Manning, “The flow of water in open channels and pipes,” Trans

Inst of Civil Engineers of Ireland, Vol. 20, Dublin, 1891, pp. 161–201. 6. H. Rouse, Fluid Mechanics for Hydraulic Engines, Dover Publications,

Inc., New York, N. Y. 7. A. L. Prasuhn, Fundamentals of Fluid Mechanics, Prentice-Hall,

NJ, 1980. 8. E. F. Brater, Handbook of Hydraulics, 7th ed., McGraw-Hill, New

York, NY. 9. P. Ackers et al., Weirs and Flumes for Flow Measurement, John Wiley,

New York, NY, 1978. 10. M. H. Chaudhry, Open Channel Flow, 2nd ed., Springer-Verlag, New

York, NY, 2007. 11. Roughness Characteristics of Natural Channels, U.S. Geological Survey,

Water-Supply Paper 1849.

P ROB L EMS SEC. 12.1–12.3 12–1. Water flows in a rectangular channel with a velocity of 2 m>s and depth of 4 m. What other possible depth of flow provides the same specific energy? Plot the specific energy diagram. 12–2. Water flows in a rectangular channel with a velocity of 4 m>s. If the depth is 1.5 m, what other possible velocity of flow provides the same specific energy?

12–5. A large tank contains water having a depth of 1.5 m. If the tank is on an elevator that is descending, determine the horizontal speed of a wave created on its surface if the rate of descent is (a) constant at 6 m>s, (b) accelerated at 1.5 m>s2, (c) accelerated at 9.81 m>s2. 12–6. The rectangular channel transports water at 8 m3 >s. Plot the specific energy diagram for the flow and indicate the depth y for E = 3 m.

12–3. Water flows with a velocity of 3 m>s in a rectangular channel having a width of 2 m. If the depth of the water is 0.9 m, determine the specific energy and the alternate depth that provides the same flow. *12–4. A rectangular channel has a width of 2 m. If the flow is 5 m3 >s, determine the Froude number when the water depth is 0.5 m. At this depth, is the flow subcritical or supercritical? Also, what is the critical speed of the flow?

y

1.5 m

Prob. 12–6

699

prOBlems 12–7. A canal is 2 m deep and 3 m wide, and the water flows with a velocity of 1.5 m>s. If a stone is thrown into it, determine how fast the waves will travel upstream and downstream. What is the Froude number?

12–11. The rectangular channel has a transition that causes its width to narrow from 4 m to 3 m. If the flow is 900(103) liters>min and yB = 1.25 m, determine the depth at A. 12

4m A 3m 1.5 ms B

yA

2m yB

3m

Prob. 12–7

Prob. 12–11

*12–8. A rectangular channel has a width of 2 m. If the flow is 5 m3 >s, determine the Froude number when the water depth is 1.5 m. At this depth, is the flow subcritical or supercritical? Also, what is the critical speed of the flow? 12–9. A rectangular channel having a width of 3 m is transporting water at a rate of 120(106) liters>min. If the depth is 2 m, determine the specific energy at this depth, and also the specific energy at the critical depth. 12–10. The rectangular channel passes through a transition that causes its width to narrow from 4 m to 3 m. If the flow is 900(103) liters>min and yA = 3 m, determine the depth at B.

*12–12. The rectangular channel transports water at 360(103) liters>min. Determine the critical depth yc, plot the specific energy diagram for this flow, and calculate E for y = 0.75yc and y = 1.25yc.

y

2m

Prob. 12–12 12–13. The rectangular channel transports water at 1.5 (106) liters>min. Determine the critical depth and the minimum specific energy. If the specific energy is 3 m, what are the two possible flow depths?

4m A 3m

B

yA y yB

Prob. 12–10

4m

Prob. 12–13

700

C h a p t e r 12

O p e n -C h a n n e l F l O w

12–14. The channel transports water at 8 m3 >s. If the depth is y = 1.5 m, determine if the flow is subcritical or supercritical. What is the critical depth of flow? Compare the specific energy of the flow with its minimum specific energy. 12

12–17. The rectangular channel tapers gradually from 2.5 m wide to 1.5 m wide, with a transition as shown. If water flows with a velocity of 0.75 m>s on the 2.5-m-wide side, determine the depth y2 on the 1.5-m-wide side. Also, what is the horizontal force directed downstream that the water exerts on each of the flared wing walls?

2.5 m

1.5 m y 2.5 m y2

Prob. 12–14

12–15. Water flows at 2 m3 >s in the channel, which is originally 3 m wide and then gradually narrows to b2 = 2 m. If the original depth of water is 1.5 m, determine the depth y2 after it passes through the transition. What width b2 will produce critical flow y2 = yc?

3m

1m

Prob. 12–17 12–18. Water flows within the 4-m-wide rectangular channel at 20 m3 >s. Determine the depth yB at the downstream end, and the velocity of flow at A and B. Take yA = 5 m. 12–19. Water flows within the 4-m-wide rectangular channel at 20 m3 >s. Determine the depth yB at the downstream end, and the velocity of flow at A and B. Take yA = 0.5 m.

b2

A

B

1.5 m

yB

yA

0.2 m y2

Probs. 12–18/19 Prob. 12–15 *12–16. The volumetric flow in the channel is measured using the venturi segment. If yA = 4 m and yB = 3.8 m, determine the flow through the channel. 4m

1.5 m

SEC. 12.4 *12–20. The rectangular channel has a width of 5 m and transports water at 1.20(106) liters>min. If the depth at A is 2.5 m, determine the minimum height h of the channel bed at B in order to produce critical flow at B.

A

A

B

B yA yB

Prob. 12–16

yB

2.5 m

h

Prob. 12–20

701

prOBlems 12–21. Water flows at 840(103) liters>min through a rectangular channel having a width of 3 m. If yA = 1 m, show that yB = 1.25 m is possible. What is the necessary increase in elevation h of the channel bed to create this depth? B

12–25. The rectangular channel has a width of 1.5 m, and the depth of the water is originally 2 m. If the flow is 4.50 m3 >s, show that the upstream flow is subcritical, and determine the required depth of the water, y′, at the top of the bump so that the downstream flow can be transformed to supercritical flow. What is the downstream depth?

A yB yA

y9

2m

y2 h

Prob. 12–21 12–22. The rectangular channel has a width of 4 m and transports water at 1.20(106) liters>min. If the elevation of the bed drops by 0.4 m, determine the depth y2 after the depression.

y2

3m

Prob. 12–25

12–26. The rectangular channel is 1.5 m wide, and the depth of the water is 2 m as it flows with a velocity of 2 m>s. Determine if the flow is initially subcritical or supercritical. What is the required height h of the properly designed bump so that the flow can change to another type of flow after it passes over the bump? What is the new depth y2 for this flow?

0.4 m

Prob. 12–22

2 m/s

3

12–23. Water flows at 840(10 ) liters>min through a rectangular channel having a width of 3 m. If yA = 2.5 m, show that yB = 2 m is possible. What is the necessary increase in elevation h of the channel bed to create this depth?

y15 2 m

y2 h

A B

Prob. 12–26

yB

yA

h

Prob. 12–23 *12–24. The channel is 2 m wide and transports water at 18 m3 >s. If the elevation of the bed is raised 0.25 m, determine the new depth y2 of the water and the speed of the flow. Is the new flow subcritical or supercritical?

12–27. The rectangular channel is 1.5 m wide, and the water flows with a velocity of 3.5 m>s. If y1 = 0.9 m, show that this flow is supercritical. Determine the required height h of the bump so that the flow can change to subcritical flow after it passes over the properly designed bump. What is the depth y2 for this flow?

3.5 ms

h

0.25 m

Prob. 12–24

y2

y1

y2

1.5 m

Prob. 12–27

12

702

C h a p t e r 12

O p e n -C h a n n e l F l O w

*12–28. If the depths y0 = 3 m, where the water is essentially at rest, and y2 = 1 m, determine the volumetric flow through the sluice gate and the depth y1 just before the gate.

SEC. 12.6 12–33. Determine the hydraulic radius for each channel cross section.

12–29. If the flow is 540 (10)3 liters>min and y0 = 4 m, 12 where the water is essentially at rest, determine the depth y2 and the depth y1 just before the gate.

30° 30° h

12–30. If the depth y0 = 6 m, where the water is essentially at rest, determine the volumetric flow through the channel as a function of depth y2. Plot the results of flow on the vertical axis, versus depth y2 for 0 … y2 … 5 m. Use increments of ∆y2 = 1 m.

l

b (a)

(b)

2m

30°

30°

l

l

y0

b (c) y2

Prob. 12–33

Probs. 12–28/29/30 12–31. A flow of 60 m3 >s passes under the sluice gate. If y1 = 6 m, determine the depth y2 and the type of flow downstream. Also, what is the maximum volumetric flow that can pass through the gate?

SEC. 12.5

*12–32. Determine the volumetric flow of water through the gate if y1 = 6 m, y2 = 2.5 m. What type of flow occurs?

12–34. The channel is made of wood and has a downward slope of 0.0020. Determine the volumetric flow of water if y = 0.8 m. Take n = 0.012. 12–35. The channel is made of wood and has a downward slope of 0.0020. Determine the depth of flow y that will produce the maximum volumetric flow using the least amount of wood when it is flowing full of water, that is, when y = h. What is this volumetric flow? Take n = 0.012.

4m

y y1 h y2

Probs. 12–31/32

2m

Probs. 12–34/35

703

prOBlems *12–36. The channel has a triangular cross section and is flowing full. Determine the volumetric flow of water if the sides are made of wood and the downward slope is 0.003. Take n = 0.012.

12–41. The channel is made of unfinished concrete and has a slope of 0.004. Determine the volumetric flow and the corresponding critical slope. Specify whether the flow is subcritical or supercritical. Take n = 0.012.

2m

12

0.75

0.75 1

1

1.5 m 3m 60°

60°

Prob. 12–41

Prob. 12–36 12–37. The culvert is at a downward slope S0. Determine the depth y of water that will produce the maximum volumetric flow.

12–42. The channel is made of unfinished concrete and has a downward slope of 0.006. Determine the volumetric flow if the depth is y = 3 m. Take n = 0.014.

12–38. The culvert is at a downward slope S0. Determine the depth y of water that will produce maximum velocity for the flow.

12–43. The channel is made of unfinished concrete. Determine the critical depth if the volumetric flow is 145 m3 >s. Also, what is the corresponding critical slope? Take n = 0.014.

y R

5

3

5

4

u

3 4

y 2.5 m

Probs. 12–37/38

Probs. 12–42/43

12–39. The channel is made of unfinished concrete and has the cross section shown. If the downward slope is 0.0008, determine the flow of water through the channel when y = 3 m. Take n = 0.014.

*12–44. The drainage canal has a downward slope of 0.0035. If its bottom and sides have weed growth, determine the volumetric flow of water when the depth is 2 m. Take n = 0.025.

*12–40. The channel is made of finished concrete and has the cross section shown. If the downward slope is 0.0008, determine the flow of water through the channel when y = 2.5 m. Take n = 0.012.

3m

3m 2m

y 1m

3m

4m

3m

Probs. 12–39/40

Prob. 12–44

704

C h a p t e r 12

O p e n -C h a n n e l F l O w

12–45. The channel has a triangular cross section. Determine the critical depth y = yc in terms of u and the flow Q.

12–47. The channel is made of unfinished concrete. If it is required to transport water at 12 m3 >s, determine the critical depth y = yc and the critical slope. Take n = 0.014. *12–48. The channel is made of unfinished concrete. If it is required to transport water at 12 m3 >s, determine its downward slope when the depth of the water is 2 m. Also, what is the critical slope for this depth, and what is the corresponding critical flow? Take n = 0.014.

12

y u

u 60°

60°

3m

Prob. 12–45

Probs. 12–47/48

12–46. The rectangular channel has a finished concrete surface and has a slope of 0.0015. If it is to discharge 8.75 m3 >s, determine the depth for uniform flow. Is the flow subcritical or supercritical? Take n = 0.012.

y

12–49. The sewer pipe, made of unfinished concrete, is required to carry water at 1.5 m3 >s when it is half full. If the downward slope of the pipe is 0.0015, determine the required internal radius of the pipe. Take n = 0.014.

R

2m

Prob. 12–46

Prob. 12–49

705

prOBlems 12–50. Water flows uniformly through the triangular channel having a downward slope of 0.0083. If the walls are made of finished concrete, determine the volumetric flow when y = 1.5 m. Take n = 0.012.

20°

*12–52. The drainage pipe is made of finished concrete and is sloped downward at 0.002. Determine the volumetric flow from the pipe if the center depth is y = 1 m. Take n = 0.012. 12–53. The drainage pipe is made of finished concrete and is sloped downward at 0.002. Determine the volumetric flow from the pipe if the center depth is y = 0.3 m. Take n = 0.012.

20° y 0.6 m y

Prob. 12–50 Probs. 12–52/53

12–51. A 1.5-m-diameter unfinished-concrete drain pipe is required to transport 4.25 m3 >s of water at the depth shown. Determine the required downward slope of the pipe. Take n = 0.014.

12–54. An 800-mm-diameter unfinished-concrete drain pipe is required to transport 2.25 m3 >s of water at a depth of 600 mm. Determine the required downward slope of the pipe. Take n = 0.014. 12–55. Water flows through the 1-m-diameter culvert at a depth of 0.25 m. If the culvert has an unfinished-concrete surface and has a slope of 0.002, determine the volumetric flow. Take n = 0.014.

1.125 m 0.75 m

0.5 m 0.25 m

Prob. 12–51

Prob. 12–55

12

706

C h a p t e r 12

O p e n -C h a n n e l F l O w

*12–56. The channel is made of unfinished concrete. Determine the required slope if the flow is to be 20 m3 >s when the depth is y = 2 m. Take n = 0.014.

12

12–57. The channel is made of unfinished concrete. If the slope is 0.004, determine the volumetric flow when the depth y = 3 m. Is the flow subcritical or supercritical? Take n = 0.014.

12–59. The unfinished concrete channel is intended to have a downward slope of 0.002 and sloping sides at 60°. If the flow is estimated to be 100 m3 >s, determine the base dimension b of the channel bottom. Take n = 0.014.

4m 60° 1

1

60°

y

1

1

b 3m

Prob. 12–59 Probs. 12–56/57

12–58. Show that when the depth of flow y = R, the semicircular channel provides the best hydraulic cross section.

R

*12–60. Determine the volumetric flow of water through the channel if the depth is y = 1.25 m and the downward slope of the channel is 0.005. The sides of the channel are finished concrete. Take n = 0.012.

y 2

2 1

1 2m

Prob. 12–58

Prob. 12–60

y

707

prOBlems 12–61. Determine the normal depth of water in the channel if the flow is Q = 15 m3 >s. The sides of the channel are finished concrete, and the downward slope is 0.005. Take n = 0.012.

*12–64. The finished concrete channel has a flow of 20 m3 >s. Determine the critical slope. Take n = 0.012.

12 2

2 1

y

1 2m 2

yc

Prob. 12–61

2 1

1

3m

12–62. Show that the width b = 2h(csc u - cot u) in order to minimize the wetted perimeter for a given cross-sectional area and angle u. At what angle u will the wetted perimeter be the smallest for a given cross-sectional area and depth h?

Prob. 12–64

h u

u

b

12–65. Determine the angle u and the length l of its sides so that the channel has the best hydraulic trapezoidal cross section of base b.

Prob. 12–62

12–63. Determine the length of the sides a of the channel in terms of its base b, so that for the flow at full depth it provides the best hydraulic cross section that uses the minimum amount of material for a given discharge.

120° 120° a

a

u

u

l

l b b

Prob. 12–63

Prob. 12–65

708

C h a p t e r 12

O p e n -C h a n n e l F l O w

SEC. 12.7 12–66. A rectangular channel is made of finished concrete, and it has a width of 1.25 m and a downward slope of 0.01. Determine the surface profile for the flow if it is 0.8 m3 >s and the depth of the water at a specific location is 0.3 m. 12 Sketch this profile. Take n = 0.012. 12–67. A rectangular channel is made of finished concrete, and it has a width of 4 m and a downward slope of 0.0002. Determine the type of surface profile for a flow of 7.50 m3 >s and the depth of the water at a specific location is 1.5 m. Sketch this profile. Take n = 0.012. *12–68. A rectangular channel is made of finished concrete, and it has a width of 1.25 m and a downward slope of 0.01. Determine the surface profile for the flow if it is 0.8 m3 >s and the depth of the water at a specific location is 0.6 m. Sketch this profile. Take n = 0.012.

*12–72. Water flows at 4 m3 >s along a horizontal channel made of unfinished concrete. If the channel has a width of 2 m, and the water depth at a control section A is 0.9 m, approximate the depth at the section where x = 2 m from the control section. Use increments of ∆y = 0.004 m and plot the profile for 0.884 m … y … 0.9 m. Take n = 0.014. 12–73. The unfinished concrete channel has a width of 3 m and slopes upward at 0.008. At control section A, the water depth is 2 m and the velocity is 6 m>s. Determine the depth of the water y at the section 5.5 m downstream from A. Use ∆y = 0.03 m and plot the surface profile. Take n = 0.014.

6 ms B A

y

2m

12–69. A rectangular channel is made of unfinished concrete, and it has a width of 1.5 m and a downward slope of 0.06. Determine the type of surface profile for a flow of 8.5 m3 >s if the depth of the water at a specific location is y = 2 m. Sketch this profile. Take n = 0.012. 12–70. A rectangular channel is made of unfinished concrete, and it has a width of 3 m and an upward slope of 0.008. Determine the type of surface profile for a flow of 15.25 m3 >s if the depth of the water at a specific location is y = 1.25 m. Sketch this profile. Take n = 0.014.

12–71. Water flows at 7.50 m3 >s down a rectangular channel made of finished concrete. The channel has a width of 4 m and a downward slope of 0.0002, and the water depth is 1.5 m at the control section A. Determine the distance x from A to where the depth becomes 1.25 m. Use ∆y = 0.05 m and plot the surface profile. Take n = 0.012.

5.5 m

Prob. 12–73

12–74. Water flows at 8.50 m3 >s along a horizontal rectangular channel made of finished concrete. If the channel has a width of 3 m and the water depth at the control section A is 0.5 m, determine the approximate distance x from A to where the depth becomes 0.75 m. Use ∆y = 0.05 m and plot the surface profile. Take n = 0.012.

SEC. 12.8 12–75. The hydraulic jump has a depth of 5 m at the downstream end, and the velocity is 1.25 m>s. If the channel is 2 m wide, determine the depth y1 of the water before the jump and the energy head lost during the jump.

1.25 ms

A B

1.5 m

5m

y1 x

Prob. 12–71

Prob. 12–75

709

prOBlems *12–76. Water flows down the 5-m-wide spillway, and at the bottom it has a depth of 0.75 m and attains a speed of 12 m>s. Determine if a hydraulic jump will form, and if so, determine the velocity of the flow and its depth after the jump.

12–79. The sill at A causes a hydraulic jump to form in the channel. If the channel width is 1.5 m, determine the upstream and downstream velocities of the water. What amount of energy head is lost in the jump? 12 4m 2.5 m

y2

A

0.75 m

Prob. 12–79 Prob. 12–76

12–77. Water flows under the partially opened sluice gate, which is in a rectangular channel. If the water has the depth shown, determine if a hydraulic jump forms, and if so, find the depth yc at the downstream end of the jump.

*12–80. Water runs from a sloping channel with a flow of 8 m3 >s onto a horizontal channel, forming a hydraulic jump. If the channel is 2 m wide, and the water is 0.25 m deep before the jump, determine the depth of water after the jump. What amount of energy head is lost within the jump?

A

0.25 m y2

6m

C

B

yC

2m

Prob. 12–80 Prob. 12–77

12–78. Water flows at 2.50 m3 >s through the 2-m-wide rectangular channel. Determine if a hydraulic jump will form, and if so, determine the depth of flow y2 after the jump and the energy head that is lost due to the jump.

12–81. The hydraulic jump has a depth of 3 m at the downstream end, and the velocity is 1.5 m>s. If the channel is 2 m wide, determine the depth y1 of the water before the jump and the head loss within the jump.

1.5 m/s

y2

0.5 m

Prob. 12–78

3m

y1

Prob. 12–81

710

C h a p t e r 12

O p e n -C h a n n e l F l O w

12–82. Water flows at 25 m3 >s over the 4@m wide spillway of the dam. If the depth of the water at the bottom apron is 0.5 m, determine the depth y2 of the water after the hydraulic jump.

12–85. The rectangular channel is fitted with a 60° triangular weir plate. If the upstream depth of the water is 2 m, and the bottom of the weir plate is 1.6 m from the bottom of the channel, determine the volumetric flow of water passing over the weir. Take Cd = 0.72.

12 y2 0.5 m

Prob. 12–82 60°

SEC. 12.9

2m

12–83. The flow of water over the broad-crested weir is 15 m3 >s. If the weir and the channel have a width of 3 m, determine the depth of water y within the channel. Take Cw = 0.80.

1.6 m

Prob. 12–85 y 1m

Prob. 12–83 *12–84. The rectangular channel has a width of 3 m and the depth is 1.5 m. Determine the volumetric flow of water over the rectangular sharp-crested weir if the depth of the water over the top of the weir plate is H = 1 m. Take Cd = 0.83.

12–86. Determine the volumetric flow of water passing over the broad-crested weir if it is in a channel having a width of 3 m. Take Cw = 0.87.

3m

1.5 m

1.5 m

0.75 m

Prob. 12–84

Prob. 12–86

Chapter review

711

CHAP TER R EV IEW Practically all open-channel flow is turbulent. For analysis, we assume the flow is onedimensional and use an average velocity over the cross section.

12

The characteristics of flow along a channel are classified using the Froude number, since gravity is the driving force for open-channel flow. When Fr 6 1, the flow is subcritical, and when Fr 7 1, the flow is supercritical. If Fr = 1, then critical flow occurs, in which case a standing wave can form on the surface of the liquid.

V 1gy

Fr =

y q 5 0 (no motion) q

The specific energy or specific head E of a flow in an open channel is the sum of its kinetic and potential energies. When plotted for a given flow, it indicates the depths at which the flow is subcritical, critical, or supercritical.

Subcritical flow Fr , 1 Supercritical flow Fr . 1 E

y2 yc Emin E9

y1

Specific energy diagram for flow per unit width

Both subcritical and supercritical flow can produce the same specific energy for a specified flow and channel width. Using a properly designed bump, where critical flow occurs at its peak, the flow can be changed from supercritical to subcritical, or from subcritical to supercritical.

Subcritical Supercritical yc

712

12

C h a p t e r 12

O p e n -C h a n n e l F l O w

Steady uniform flow in an open channel will occur if the channel has a constant cross section, slope, and surface roughness. The velocity of the flow can be determined using Manning’s equation, which uses an empirical coefficient n to specify the surface roughness.

If nonuniform flow occurs in a channel, then there are twelve possible types of surface profiles. The profile can be determined numerically using a finite difference method derived from the energy equation.

Nonuniform flow can occur suddenly in an open channel, causing a hydraulic jump. The jump removes energy from the flow, causing it to convert from supercritical to subcritical flow.

The flow in an open channel can be measured using a sharp-crested or broad-crested weir. In particular, a broad-crested weir can also convert the flow into flow having a critical depth.

V

V =

1>2 kR2>3 h S0

n Su 1

Chapter review

713

Fundamental Equations of Open-Channel Flow c = 2gy Wave speed

Rh =

A P

Hydraulic radius

Fr =

V 1gy

Froude number

E =

Q2 2gb2y2

+ y

Specific energy

Vc = 1gyc

V =

kRh2>3S 01>2 n

Manning equation

k = 1 (SI units)

Critical velocity Rectangular channel

Q =

2

Ac3

Q = g btop

Vc =

gAc A btop

Critical velocity Nonrectangular channel

S0 =

kA5>3S 01>2 nP 2>3 Q2n2 k 2Rh4>3A2

Channel slope

Sc =

n2gAc 4>3 k 2btop Rhc

Critical slope Steady uniform flow

q = 22g 1 y22y1 - y23 2 1>2 y2 =

qmax

2 y 3 1

8 = gy 3 A 27 1 Sluice gate

y2 1 = c 21 + 8Fr12 - 1 d y1 2 hL =

(y2 - y1)3 4y1y2

Hydraulic jump

12

13

Stocktrek Images/Getty Image

CHAPTER

Local shock waves form on the surface of this jet plane at places where the airflow approaches the speed of sound. The effect is seen because the air is humid and condensation forms on the shock, which defines a vapor cone.

COMPRESSIBLE FLOW

CHAPTER OBJECTIVES ■

To present some of the important concepts of thermodynamics that are used to analyze compressible gas flow.

■

To discuss compressible flow through a variable area, and show how the area affects the properties of a gas.

■

To show how compressible flow is affected by friction within a pipe, and by the addition or removal of heat from the surface of the pipe.

■

To study the formation of shock waves within convergent and divergent nozzles, and the formation of compression and expansion waves on curved or irregular surfaces.

■

To present some of the methods used to measure velocity and pressure for compressible flow.

13.1

THERMODYNAMIC CONCEPTS

Up to this point we have considered applications of fluid mechanics only to problems where the fluid is considered incompressible. In this chapter we will extend our study to include the effects of compressibility, something that is important in the design of jet engines, rockets, and aircraft, and the design of pipes and ducts used in industry for high speed gas flow. We will begin the chapter with a presentation of some of the important concepts in thermodynamics that are involved in compressible gas flow. Thermodynamics plays an important role in understanding compressible flow behavior because any drastic change to the kinetic energy of a gas will be converted into heat, causing large density and pressure variations throughout the gas.

715

716

13

C h a p t e r 13

Compressible Flow

Ideal Gas Law. Throughout this chapter we will consider the gas to be an ideal gas, that is, one that is composed of molecules that behave like elastic spheres. The molecules are assumed to be in a quasi-stationary state and have random motion and interact only on occasion, since there is a large distance between them compared to their size. Real gases are quite similar to ideal gases at normal pressures and temperatures, although this approximation becomes more accurate as the density of the gas becomes smaller. It has been found through experiment that the absolute temperature, absolute pressure, and the density of an ideal gas are related with high precision through a single equation, referred to as the ideal gas law. It is p = rRT

(13–1)

Here R is the gas constant, which has a unique value for each gas. The ideal gas law is referred to as an equation of state, since it relates the three state properties p, r, T of a gas when it is in a specific state or condition.

Internal Energy and the First Law of Thermodynamics. Another important state property is internal energy, and for an ideal gas this refers to the collective kinetic energy caused by the microscopic motions of its atoms and molecules, and the potential energy associated with the microscopic bonds between these particles. Provided the gas is a closed system, so that no mass can enter or exit the system, then a change in internal energy will occur if heat and work are transferred to or from the gas.* The first law of thermodynamics is a formal statement of this balance. It states that if we consider a unit mass of a gas within this system, then the change in its internal energy du, as the system moves from one state to another, is equal to the heat (thermal energy) dq transferred into the system minus the work dw done by the system on its surroundings. The work is defined as flow work, discussed in Sec. 5.5. It is caused by pressure p, and for a volume per unit mass , it can be written as dw = p dv. Thus, Heat added

du = dq - p dv Change in internal energy

(13–2)

Flow work removed

This change in internal energy is independent of the process used to move the system from one state to another; however, the values of dq and p dv depend upon this process. Typical processes include changes under constant volume, constant pressure, or constant temperature; or, if the system is also insulated, then no heat enters or leaves the system, and the process is adiabatic. *In this discussion we will exclude shaft work added by a pump or removed by a turbine.

13.1

thermodynamiC ConCepts

717

Specific Heat. The amount of heat dq that can be added to a gas is directly proportional to the change in temperature dT of the gas. The constant of proportionality is called the specific heat, c, which is a physical property of the gas. It is defined as the amount of heat needed to raise the temperature of a unit mass of the gas by one degree. 13

dq c = dT

(13–3)

This property depends upon the process by which the heat is added. It is measured in J>(kg # K). Normally, it is determined by adding the heat at constant volume, cv, or at constant pressure, cp.

Constant-Volume Process. When the volume of a gas remains constant, then dv = 0, and so no external flow work can be done. The first law of thermodynamics then states that du = dq. In other words, the internal energy of the gas is raised only because of the amount of heat supplied. Thus, Eq. 13–3 becomes cv =

du dT

(13–4)

For the range of most engineering applications, cv is essentially constant as the temperature changes, and so we can integrate this equation between any two points of state and write it as ∆u = cv ∆T

(13–5)

Therefore, if cv is known, this result provides a means of obtaining the change in internal energy for a given temperature change ∆T.

Constant-Pressure Process. For a constant-pressure process, the gas can expand, and so Eq. 13–2 states that a change in both internal energy and flow work can occur. Solving for dq, then Eq. 13–3 becomes du + p dv dT To simplify this expression, we will define another state property of a gas, called the enthalpy, h. Formally, enthalpy is defined as the sum of the internal energy u and the flow work p of a unit mass of the gas. Since the volume per unit mass is v = 1>r, then cp =

h = u + pv = u +

p r

Using the ideal gas law, the enthalpy can also be written in terms of temperature. h = u + RT To find the change in the enthalpy, we take the derivative, dh = du + dp v + p dv

718

13

C h a p t e r 13

Compressible Flow

Since the pressure is constant, dp = 0, and so dh = du + p dv. Therefore, cp now becomes dh cp = (13–6) dT This equation can be integrated between any two points of state, since cp is only a function of temperature; however, within the temperature ranges normally considered in engineering, cp, like cv, is essentially constant. Thus, ∆h = cp ∆T (13–7) Therefore, if cp is known, then if a change in temperature ∆T occurs, we can calculate the corresponding change in the enthalpy (internal energy plus flow work). If we take the derivative of h = u + RT, and substitute Eq. 13–4 and Eq. 13–6 into the result, we obtain a relation among cp, cv, and the gas constant. It is cp - cv = R

(13–8)

If we now express the ratio of specific heats as k =

cp cv

(13–9)

then, using Eq. 13–8, we can write cv =

R k - 1

(13–10)

cp =

kR k - 1

(13–11)

and also

Appendix A gives values of k and R for common gases, so the values of cv and cp can be calculated from these two equations.

Entropy and the Second Law of Thermodynamics. Entropy s is a state property of a gas, and in thermodynamics, as well as here, we will be interested in how it changes. For a closed system, we define the change in entropy as the amount of heat that is transferred per degree of temperature, as a unit mass of gas moves from one state of pressure, volume, and temperature to another. Thus, ds =

dq T

(13–12)

For example, when two objects at different temperatures are located within a closed and insulated container, to form an isolated system, then eventually the objects will reach the same temperature due to the increments of heat dq moving from the hot to the cold object. One will gain entropy, and the other will lose entropy. This process of heat flowing from the hot to the cold body is irreversible. In other words, heat never flows from a cold to a hot body, because there is more thermal agitation of the molecules or internal energy within the hot body than in the cold body.

13.1

The second law of thermodynamics is based on the change in entropy that occurs within an isolated system. It determines the time order in which physical phenomena can take place. It gives a preferred direction to time and it is sometimes called the “arrow of time.” The second law states that the entropy will always increase, since the process for change is irreversible. In a gas it is the result of viscous friction which increases the agitation of the gas, thereby causing the heat of the gas to increase. If the process is assumed to be reversible, that is, without internal friction, no change in entropy will occur. Thus, ds = 0 ds 7 0

Reversible Irreversible

(13–13)

For purposes of calculation, we can obtain a relationship between the change in entropy and the intensive properties T and r by first combining Eq. 13–12 with the first law of thermodynamics to eliminate dq. We have T ds = du + p dv Now, if we substitute v = 1>r into this equation, and use the ideal gas law, p = rRT, and the definition of cv, Eq. 13–4, we get ds = cv

dT R 1 + da b r T 1>r

Integrating, with cv remaining constant during the temperature change, yields

s2 - s1 = cv ln

r2 T2 - R ln r1 T1

(13–14)

Also, the change in entropy can be related to T and p. First, the enthalpy can be related to the entropy by substituting T ds = du + p dv into dh = du + dp v + p dv. We get T ds = dh - v dp Then using the definition of cp, Eq. 13–6, and the ideal gas law, written as p = RT>v, we have

ds = cp

dp dT - R p T

When integrated, this gives s2 - s1 = cp ln

p2 T2 - R ln p1 T1

(13–15)

thermodynamiC ConCepts

719

13

720

C h a p t e r 13

Compressible Flow

Isentropic Process. Many types of problems involving compressible

13

No heat loss No internal friction

flow through nozzles and pipe transitions occur within a localized region over a very short period of time, and when this happens, the changes in the gas can often be modeled as an isentropic process. This process involves no heat transfer to the surroundings while the gas is suddenly brought from one state to another; that is, the process is adiabatic, dq = 0. Also, it is reversible, provided friction within the system is neglected, and as a result, ds = 0, Fig. 13–1. Therefore, from our previous results, T ds = du + p dv and T ds = dh - v dp, for an isentropic process, 0 = du + p dv 0 = dh - v dp If we express these equations in terms of the specific heats, we have

Isentropic process

0 = cv dT + p dv

Fig. 13–1

0 = cp dT - v dp Eliminating dT and using Eq. 13–9, k = cp >cv, yields dp dv + k = 0 p v

Since k is constant, integrating between any two points of state gives ln

p2 2 + k ln = 0 p1 v1

a

p2 v2 k ba b = 1 p1 v1

Substituting v = 1>r and rearranging terms, the relationship between pressure and density becomes p2 r2 k = a b r1 p1

(13–16)

Also, we can express the pressure in terms of the absolute temperature by using the ideal gas law, p = r RT. We obtain p2 T2 k>(k - 1) = a b p1 T1

(13–17)

Throughout the chapter, we will use several of the equations developed in this section to describe either isentropic or adiabatic compressible flow.

13.1

thermodynamiC ConCepts

721

IM PORTANT POIN T S • For most engineering applications, a gas can be considered as ideal, that is, composed of molecules in random motion, with large distances between them compared to their size. Ideal gases obey the ideal gas law, p = rRT.

• A system in equilibrium has certain state properties, such as pressure, density, temperature, internal energy, entropy, and enthalpy.

• A change in internal energy per unit mass of a gas occurs within a closed system when heat and work are transferred to or from the gas.

• Enthalpy is a state property of a gas that is defined as the sum of the internal energy and flow work per unit mass, h = u + p>r.

• The specific heat at constant volume relates the change in internal energy of a gas to the change in its temperature, ∆u = cv ∆T.

• The specific heat at constant pressure relates the change in enthalpy of a gas to the change in its temperature, ∆h = cp ∆T.

• A change in entropy ds provides a measure of the amount of heat that is transferred per degree of temperature, ds = dq>T.

• The second law of thermodynamics states that for an isolated

system, entropy will always increase because of friction, ds 7 0. If the process is reversible, or frictionless, then the change in entropy will be zero, ds = 0.

• The specific heat cv can be used to relate the change in entropy ∆s to the change in T and r, Eq. 13–14, and cp can be used to relate ∆s to the change in T and p, Eq. 13–15.

• An isentropic process is one where no heat is gained or lost and the flow is frictionless. This means that the process is adiabatic and reversible, ds = 0.

13

722

C h a p t e r 13

EXAMPLE

13.1

0.5 m

13

Compressible Flow

pA 5 80 kPa TA 5 50°C

pB 5 20 kPa TB 5 20°C B

Air flows at 5 kg>s through the duct in Fig. 13–2, such that the gage pressure and temperature at A are pA = 80 kPa and TA = 50°C, and at B, pB = 20 kPa and TB = 20°C. Determine the changes in the enthalpy, internal energy, and entropy of the air between these points. SOLUTION

A

Fig. 13–2

Fluid Description. Because the temperature and pressure change between A and B, the density of the air will also change, and so we have steady compressible flow. For air, k = 1.4 and R = 286.9 J>kg # K. Change in Enthalpy. The change in the enthalpy is determined from Eq. 13–7. However, first we must determine the specific heat at constant pressure using Eq. 13–11. 1.41286.9 J>kg # K2 kR cp = = = 1004.15 J>kg # K k - 1 11.4 - 12 The temperature in thermodynamic calculations must be expressed in kelvins, and so we have ∆h = cp 1TB - TA 2 = (1004.15 J>kg # K)3 1273 + 202 K - 1273 + 502 K4 = -30.1 kJ>kg

Ans.

The negative sign indicates a decrease in enthalpy. Change in Internal Energy. In this case Eq. 13–5 applies. First we determine the specific heat at constant volume. 286.9 J>kg # K R cv = = = 717.25 J>kg # K k - 1 11.4 - 12 ∆u = cv 1TB - TA 2 = 1717.25 J>kg # K23 1273 + 202 K - 1273 + 502 K4 = -21.5 kJ>kg Ans. This decrease in the internal energy of the gas is caused by the gas moving from a hotter to a cooler temperature.

Change in Entropy. Since both the temperature and pressure are known at points A and B, we will use Eq. 13–15 to find ∆s. Always remember p and T must be absolute values. pB TB sB - sA = cp ln - R ln pA TA 1273 + 202 K 1101.3 + 202 kPa ∆s = 11004.15 J>kg # K2 ln - 1286.9 J>kg # K2 ln 1273 + 502 K 1101.3 + 802 kPa # = 17.4 J>kg K Ans. If needed, the velocity at A (or B) can be determined from the # mass flow, m = rAVAAA, where the density of the air rA is found from pA = rARTA. The results are rA = 1.956 kg>m3 and VA = 13.0 m>s.

NOTE:

13.1

EXAMPLE

thermodynamiC ConCepts

723

13.2

The closed container with the movable lid in Fig. 13–3 contains 4 kg of helium at a gage pressure of 100 kPa and a temperature of 20°C. Determine the initial density of the helium and its final temperature and density if the force F compresses it isentropically to a pressure of 250 kPa.

F

13

SOLUTION Fluid Description. Since we have an isentropic process, during the compression no heat or friction losses occur. Temperature. The new temperature can be determined using Eq. 13–17 since the initial and final pressures are known. From Appendix A, for helium k = 1.66, and so p2 T2 k>1k - 12 = a b p1 T1

1.66>11.66 - 12 (101.3 + 250) kPa T2 = a b 1101.3 + 1002 kPa 1273 + 202 K

T2 = 365.6 K = 366 K

Ans.

Density. The initial density of helium can be determined from the ideal gas law. Since R = 2077 J>kg # K, p1 = r1RT1

1101.3 + 1002110 2 Pa = r1 12077 J>kg # K21273 + 202 K 3

r1 = 0.3308 kg>m3 = 0.331 kg>m3

Ans.

Applying Eq. 13–16, to determine the final density, we have p2 r2 k = a b r1 p1

1.66 1101.3 + 2502 kPa r2 = a b 1101.3 + 1002 kPa 0.3308 kg>m3

r2 = 0.463 kg>m3

Ans.

This same result can also be determined by using the ideal gas law. p2 = r2R2T; 1101.3 + 25021103 2 Pa = r2 12077 J>kg # K21365.6 K2 r2 = 0.463 kg>m3

This represents about a 40% increase in the density.

Ans.

Fig. 13–3

724

C h a p t e r 13

Compressible Flow

13.2

13

WAVE PROPAGATION THROUGH A COMPRESSIBLE FLUID

If a pressure disturbance occurs within an incompressible fluid, then it will be noticed instantly at all points within the fluid; however, since all fluids are compressible, then the pressure disturbance will propagate through the fluid at a finite speed. This speed c is referred to as the speed of sound, or the sonic velocity. To determine the sonic velocity, we will consider the fluid to be a gas contained within the long open tube shown in Fig. 13–4a. If the piston moves a slight distance to the right at velocity ∆V, a sudden increase in pressure ∆p will be developed in the gas just next to the piston. The molecular collisions within this region will propagate to adjacent molecules on the right, and the momentum exchange that occurs will produce a very thin wave that will separate from the piston and travel along the tube with the sonic velocity c, where c W ∆V. If we now consider a differential control volume of this wave, Fig. 13–4b, then just behind the wave the movement of the piston has caused the density, the pressure, and the velocity of the gas to increase by ∆r, ∆p, and ∆V, respectively. In front of the wave the gas is still undisturbed, and so the density is r, the pressure is p, and the velocity is zero. If the wave is seen by a fixed observer, then the flow past the observer will appear as unsteady flow since the gas, originally at rest, will begin to change with time as the wave passes by. To simplify the analysis, we will consider the observer to be fixed to the wave and moving with the same velocity c, Fig. 13–4c. From this viewpoint we have steady flow, so that the gas appears to enter the control volume on the right with a velocity c, and leave it on the left with a velocity c - ∆V.

DV

c

(a) c c

p p 1 Dp 1D Control volume having a velocity c

Velocities relative to the control volume

(b)

(c)

c 2 DV

Fig. 13–4

13.2

wave propagation through a Compressible Fluid

725

Continuity Equation. We can relate the density of the gas to the sonic speed by applying the continuity equation to the control volume. Since the cross-sectional area A on each side remains the same, we have 0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

13

0 - rcA + 1r + ∆r21c - ∆V 2A = 0

-rcA + rcA - rA∆V + c∆rA - ∆r∆VA = 0 As ∆V and ∆r approach zero, the last term can be neglected since it is of the second order. Therefore, this equation reduces to c dr = r dV

Linear Momentum Equation. As shown on the free-body diagram, Fig. 13–4d, the only forces acting on the open control surfaces are those caused by pressure. And using the momentum equation, we can relate the pressure to the sonic velocity. For steady flow, we have

(p + ∆p)A - pA = 0 + 3 -cr( -cA) - (c - ∆V)(r + ∆r)(c - ∆V)A4

Neglecting the second- and third-order terms, in the limit, dp = 2rc dV - c 2 dr Using the continuity equation, and solving for c, we get

dp A dr

pA

Free-body diagram (d)

Fig. 13–4 (cont.)

+ ΣF = 0 S VrdV + VrVf>cs # dA 0t Lcv Lcs

c =

(p 1 Dp) A

(13–18)

Since the wave is very thin, no heat is transferred into or out of the control volume during the very short time the wave passes through it. In other words, the process is adiabatic. Also, friction losses within the control volume can be neglected, and so the pressure and density changes involve a process that is reversible. Consequently, sound waves or pressure disturbances form an isentropic process. Therefore,

726

C h a p t e r 13

Compressible Flow

we can relate the pressure to the density using Eq. 13–16, which can be written in the form p = Crk 13

where C is a constant. Taking the derivative, the change in p and r gives p>C p dp rk = Ckrk - 1 = Ck a b = Ck a b = ka b r r r dr Finally, using the ideal gas law, where p>r = RT, Eq. 13–18 now becomes c = 2kRT

(13–19)

Sonic velocity

The velocity of sound in the gas therefore depends upon the absolute temperature of the gas. For example, in air at 15°C 1288 K2, c = 340 m>s, which is very close to its value found from experiment. We can also express the speed of sound in terms of the bulk modulus and the density of a fluid (liquid or gas). Recall that the bulk modulus is defined by Eq. 1–11 as EV = -

dp dV>V

For a mass m = rV, the change in mass is dm = dr V + r dV; however, the mass is constant, dm = 0, and so -dV>V = dr>r. Therefore, EV =

dp dr>r

Finally, from Eq. 13–18, we have c =

EV A r

(13–20)

This result shows that the speed of sound, or the speed of the pressure disturbance, depends on the elasticity or compressibility (EV ) of the fluid and on its inertial property (r). The more incompressible the fluid (larger EV ), the faster a pressure wave will be transmitted through it; and the larger the fluid’s density, r, the slower this wave will travel. For example, the density of water is about a thousand times the density of air, but water’s bulk modulus is so much greater than that of air that sound travels a little over four times faster in water than in air. At 20°C, cw = 1482 m>s and ca = 343 m>s.

13.3

13.3

types oF Compressible Flow

727

TYPES OF COMPRESSIBLE FLOW

To classify a compressible flow, we will use the Mach number, M. Recall from Chapter 8 that this is a dimensionless number that represents the square root of the ratio of the inertial force to the compressible force acting on the fluid. There we showed that it can be expressed as a ratio of the velocity of the fluid, V, to the sonic velocity c produced by a pressure wave within the fluid. Using Eq. 13–19, we can therefore write the Mach number as

M =

V V = c 2kRT

13

The Concorde was a supersonic commercial aircraft that was capable of flying at speeds up to M = 2.3.

(13–21)

or, if the Mach number is known, then

V = M 2kRT

(13–22)

Let us now consider a body such as an airfoil moving through air with a velocity V, Fig. 13–5a. During the motion, its forward surface, like the piston in Fig. 13–4, will compress the air in front of it, causing pressure waves to form and move away from the front surface at the sonic velocity c. The effect produced depends upon the magnitude of V.

Subsonic Flow, M * 1. As long as the body continues to move at a subsonic velocity V, the pressure waves the body creates will always move ahead of the body with a relative speed of c - V. In a sense, these pressure disturbances signal the air in front of the body that it is advancing and enable the air molecules to adjust before the body arrives. This will produce a smooth flow over and around the body’s surface, as shown in Fig. 13–5a. As a general rule, the changes in pressure generated by the movement of the body begin to become significant when M 7 0.3, or V 7 0.3c. At the speed V = 0.3c, the compressibility of the air creates pressure changes of about 1%; and it is for this reason we have assumed in the previous chapters that speeds at or below 0.3c, or 30% of c, can be considered reasonable enough to consider the gas to be incompressible. Speeds within the range 0.3c 6 V 6 c are referred to as subsonic compressible flow.

V,c

Subsonic flow (a)

Fig. 13–5

728

C h a p t e r 13

Compressible Flow

V.c

13

Supersonic flow (b)

Sonic and Supersonic Flow, M # 1. When the body is traveling c3 c2 c1

Since c1 . c2 . c3 the waves bunch up and form a shock wave (c)

Fig. 13–5 (cont.)

at a speed V that is equal to or faster than the pressure waves it creates, the air just ahead of the body does not have enough time to sense the presence of the body advancing, and so the air does not move out of the way. Instead, the pressure waves will collect and form a very thin wave in front of the body, Fig. 13–5b. To fully understand how this occurs, consider a close-up view of the interaction between the air and the surface of the body shown in Fig. 13–5c. As the air impacts the surface, because of the increased molecular motion a temperature gradient is created within the air, such that the highest temperature develops at the surface. Recall the sonic velocity is a function of temperature, c = 2kRT, and so the pressure disturbance or wave formed closest to the surface will have the highest sonic speed as it moves away from the surface at c1. As a result it will catch up with the wave in front of it since c1 7 c2. The succession of all sonic waves coming off the surface of the body will therefore bunch up, and create an increasing pressure gradient within a localized region. Although each successive pressure disturbance or wave can be considered an isentropic process, as stated in Sec. 13.2, the collection of all these waves will reach a point where the pressure will get so large that viscous friction and heat conduction effects begin to stabilize their formation. In other words the collective process becomes nonisentropic. In standard atmospheric air, the thickness is on the order of three to five times the mean free path of the molecules, which is about 0.4 μm. It is called a shock wave, and its effect will cause an abrupt localized change in the pressure, density, and temperature as it passes through the air. If the body and the shock wave near it move at M = 1, it is sonic flow, and if M 7 1, it is termed supersonic. The motion is further classified as hypersonic if a body, such as a missile or low-flying spacecraft, is moving at M Ú 5.

13.3

729

types oF Compressible Flow

Mach Cone. It is important to realize that the formation of a shock wave is a very localized phenomenon that occurs on or near the surface of a body moving at or faster than the speed of sound. As the body moves from one position to the next, each spherical shock wave formed will move away from the body at the sonic speed c. To illustrate this, consider a jet plane that is flying horizontally at a supersonic speed V, Fig. 13–6. As shown, the wave produced when t = 0 will travel a distance ct′ when the plane travels a distance Vt′ in the time t = t′. Portions of the waves that are produced at t = t′>3 and t = 2t′>3 are also shown in the figure. If we summed all the waves produced during the time t′, they will form a conical wavefront, called a Mach cone. The energy of the waves is mostly concentrated at this cone’s surface, where the spherical waves interact, and so when this surface passes by an observer outside the cone, the large pressure discontinuity created on the surface will produce a loud “crack” or sonic boom as it passes by. The half angle a of the Mach cone in Fig. 13–6 depends upon the speed of the plane. It can be determined from the red shaded triangle drawn within the cone. It is t 5 t9

c 1 sin a = = V M

13

V5c 2t 9 c — 3

t9 c — 3

ct9

a

2t 9 t =— 3

t9 t =— 3

t=0 V5c

(13–23)

As the jet increases its speed V, then sin a, and therefore a, will become smaller.

V5c

a

Vt9

V5c

Development of a Mach cone

IMPORTANT POIN T S • The speed at which a pressure wave travels through a medium is called the sonic velocity, c, because sound is a result of changes in pressure. All pressure waves undergo an isentropic process, and for a particular gas, the sonic velocity is c = 2kRT .

• Compressible flow is classified by using the Mach number, M = V>c. Pressure waves for subsonic flow (M 6 1) will always move ahead of the body and signal the fluid that the body is advancing, thereby allowing the fluid to move out of the way. If M … 0.3, we can generally consider the fluid to be incompressible.

• Pressure waves for sonic (M = 1) or supersonic flow (M 7 1) cannot move ahead of the body, so they bunch up in front of the body and develop a shock wave near or on its surface. As the body moves from one position to the next, the series of spherical shock waves formed at each position interact and form a Mach cone having a surface that travels at M = 1.

Fig. 13–6

730

C h a p t e r 13

EXAMPLE

13

Compressible Flow

13.3 The jet plane flies at M = 2.3 when it is at an altitude of 18 km, Fig. 13–7. Determine the time for someone on the ground to hear the sound of the plane just after the plane passes overhead. Take c = 295 m>s.

M 5 2.3 a

18 km

a x

Fig. 13–7

SOLUTION Fluid Description. The plane compresses the air ahead of it since it is flying faster than M = 1. Although the speed of sound (or the Mach number) depends upon the air temperature, which actually varies with elevation, for simplicity, here we have assumed c is constant. Analysis. The Mach cone, which forms off the plane, has an angle of sin a =

c 1 = = 0.4348 V 2.3

a = 25.77°

Since this same cone angle can be extended to the ground, then from Fig. 13–7, 18 km x x = 37.28 km

tan 25.77° =

Thus, x = Vt;

37.281103 2 m = 2.31295 m>s2t t = 54.9 s

This is a rather long time, making it difficult to locate the plane.

Ans.

13.4

13.4

stagnation properties

731

STAGNATION PROPERTIES

The high-speed gas flow through nozzles, transitions, or venturis can usually be approximated as an isentropic process. This is because the flow is over a short distance through these devices, and as a result, it is reasonable to assume that no heat transfer occurs, and that frictional effects can be ignored. During the flow, the state of the gas can be described by its temperature, pressure, and density. In this section, we will show how to obtain these properties at any point in the gas, provided we know their values at some other reference point. For problems involving compressible flow, we will choose the stagnation point in the flow as this reference point, because here the kinetic and potential energies of the flow will be zero, and also because experimental measurements can conveniently be made from this point, as will be shown in Sec. 13.13.

13

Stagnation Temperature. The stagnation temperature T0 represents the temperature of the gas when its velocity is zero. For example, the temperature of a gas at rest in a reservoir is at the stagnation temperature. For either adiabatic or isentropic flow no heat is lost, and so this temperature T0 is sometimes referred to as the total temperature. It will be the same at each point in the flow after the flow is brought to rest. Realize that it is different from the static temperature T, which is measured by an observer moving with the flow. We can relate the static temperature to the stagnation temperature of a gas by considering the fixed control volume in Fig. 13–8, where point O is in a reservoir where the stagnation temperature is T0 and the velocity is V0 = 0, and some other point, located in the pipe where the static temperature is T and the velocity is V. If we apply the energy equation, Eq. 5–16, neglecting the changes in the potential energy of the gas, and assuming the flow is adiabatic, so there is no heat transfer, we have 2 # # # Vout V in2 # Qin - Wturbine + Wpump = ca hout + + gzout b - a hin + + gzin bd m 2 2

0 - 0 + 0 = ca h +

V2 # + 0b - (h0 + 0 + 0)d m 2

h0 = h +

V2 2

(13–24)

This result can be written in terms of temperature using Eq. 13–7, ∆h = cp ∆T or h0 - h = cp (T0 - T). Therefore, cpT0 = cpT + or T0 = Ta 1 +

V2 2

V2 b 2cpT

(13–25)

O

V

Fig. 13–8

732

C h a p t e r 13

Compressible Flow

To eliminate cp and also express this equation in terms of the Mach number, we can use Eq. 13–11, cp = kR>(k - 1), Eq. 13–19, c = 1kRT, and M = V>c. This gives T0 = Ta 1 + 13

k - 1 2 M b 2

(13–26)

To summarize, T is the static temperature of the gas at a point because it is measured relative to the flow, whereas T0 is the temperature of the gas after it is brought to rest at this point through an adiabatic process. Notice that when M = 0, then T0 = T as expected.

Stagnation Pressure. The pressure p in a gas at a point is referred to as the static pressure because it is measured relative to the flow. The stagnation or total pressure p0 is the pressure of the gas after the flow has been brought isentropically to rest at the point. Doing this, no heat transfer and frictional effects will change the total pressure. For an ideal gas, the temperature and pressure for an isentropic process (adiabatic and reversible) are related using Eq. 13–17. T0 k>1k - 12 b T Substituting Eq. 13–26, to relate the pressures to the Mach number, we get p0 = p a

p0 = p a 1 +

k - 1 2 k>1k - 12 M b 2

(13–27)

Although the static pressure p may change from one point to the next along a streamline, the stagnation pressure p0 will remain constant, provided the flow is isentropic. When k = 1.4, which is appropriate for air, oxygen, and nitrogen, the ratios T>T0 and p> p0 have been calculated from the above two equations for various values of M, and for convenience, they are tabulated in Appendix B, Table B–1.* If you take a moment to scan the data in this appendix, you will notice that the ratios T>T0 and p>p0 are always less than one so that the static values of T and p will always be smaller than the corresponding stagnation values of T0 and p0 .

Stagnation Density. If Eq. 13–27 is substituted into Eq. 13–16, p = Crk, we obtain a relationship between the stagnation density r0 and the static density r of the gas in terms of the Mach number. It is r0 = ra 1 +

k - 1 2 1>1k - 12 M b 2

(13–28)

As expected, like T0 and p0 , this value r0 is constant along any streamline, provided the flow is isentropic. *The many equations involving compressible flow can also be solved using programmable pocket calculators, or their calculated values can be found on websites that provide this information. Most of the examples have been solved in this manner.

13.4

stagnation properties

733

IM PORTANT POIN T S • Stagnation temperature T0 is the same at each point on a streamline, provided the flow is adiabatic (no heat transfer). The stagnation pressure p0 and density r0 remain the same if the flow is isentropic (adiabatic and no friction losses). Most often, these properties are measured at a point in a reservoir where the gas is stagnant or at rest.

13

• The static temperature T, pressure p, and density r are measured within the gas while moving with the flow.

• The value of T is related to T0 using the energy equation, assuming an adiabatic process. Since the flow of gas through a nozzle or transition is essentially isentropic, then p and r can be related to p0 and r0. For each case the corresponding values depend upon the Mach number and the ratio k of the specific heats for the gas.

EXAMPLE

13.4

Air at a temperature of 100°C is under pressure in the large tank shown in Fig. 13–9. If the nozzle is open, the air flows out at M = 0.6. Determine the temperature of the air at the exit.

T 5 100°C M 5 0.6

SOLUTION Fluid Description. The Mach number, M = 0.6 6 1, indicates that this problem involves subsonic compressible flow. We will assume the flow is steady. Analysis. The stagnation temperature is T0 = 1273 + 1002 K = 373 K since the air is at rest within the tank. Assuming the flow to be adiabatic through the nozzle, this is the same stagnation temperature throughout the flow since no heat is lost in the process. Applying Eq. 13–26 yields k - 1 2 T0 = Ta 1 + M b 2 1.4 - 1 373 K = Ta 1 + (0.6)2 b 2 T = 347.95 K = 74.9°C Ans. We can also obtain this result using the ratio T>T0 listed in Table B–1 for M = 0.6. We have T = 373 K10.93282 = 348 K This lower temperature, as measured relative to the flow, is actually the result of a drop in pressure that occurs as the air emerges from the tank.

Fig. 13–9

734

C h a p t e r 13

EXAMPLE

13

Compressible Flow

13.5 Nitrogen flows isentropically through the pipe in Fig. 13–10 such that its gage pressure is p = 200 kPa, the temperature is 80°C, and the velocity is 150 m>s. Determine the stagnation temperature and stagnation pressure for this gas. The atmospheric pressure is 101.3 kPa.

150 ms

T 5 80°C p 5 200 kPa

Fig. 13–10

SOLUTION Fluid Description. The Mach number for the flow is first determined. The speed of sound for nitrogen at T = 1273 + 802 K = 353 K is c = 2kRT = 21.401296.8 J>kg # K21353 K2 = 383.0 m>s

Thus,

M =

150 m>s V = 0.3917 6 1 = c 383.0 m>s

Here we have steady subsonic compressible flow. Stagnation Temperature.

T0 = Ta 1 +

Applying Eq. 13–26, we have

k - 1 2 1.4 - 1 M b = (353 K) a 1 + (0.3917)2 b 2 2 T0 = 363.8 K = 364 K

Ans.

13.4

stagnation properties

Stagnation Pressure. The static pressure p = 200 kPa is measured relative to the flow. Applying Eq. 13–27 and reporting the result as an absolute stagnation pressure, we have p0 = p a 1 +

k - 1 2 k>1k - 12 M b 2

p0 = 31101.3 + 2002 kPa4 a 1 +

1.4>(1.4 - 1) 1.4 - 1 10.39172 2 b 2

p0 = 334.91 kPa = 335 kPa

Ans.

Since k = 1.4 for nitrogen, these values for T0 and p0 can also be determined by using Table B–1 with linear interpolation. For example, the temperature ratio in Table B–1 is determined as follows: M = 0.39, T>T0 = 0.9705, and M = 0.40, T>T0 = 0.9690. Therefore, for M = 0.3917, 0.4 - 0.39 0.9690 - 0.9705 = 0.4 - 0.3917 0.9690 - T>T0 0.9690 - T>T0 = -0.001251 T>T0 = 0.97025 Thus, T0 =

353 K = 363.82 K = 364 K 0.97025

Because we have an isentropic process, realize there will be no change in the entropy. This can be shown by applying Eq. 13–15.

NOTE:

s - s0 = cp ln ∆s = c

p T - R ln p0 T0

1.41296.8 J>kg # K2

∆s = 0

1.4 - 1

d ln a

353 K 301.3 kPa b - (296.8 J>kg # K) ln a b 363.82 K 334.9 kPa

735

13

736

C h a p t e r 13

EXAMPLE

13

Compressible Flow

13.6 The absolute pressure within the entrance to the pipe in Fig. 13–11 is 98 kPa. Determine the mass flow into the pipe once the valve is opened. The outside air is at rest, at a temperature of 20°C and atmospheric pressure of 101.3 kPa. The pipe has a diameter of 50 mm. 50 mm T 5 20°C p 5 101.3 kPa p 5 98 kPa

Fig. 13–11

SOLUTION Fluid Description. We will assume steady isentropic flow through the entrance. The Mach number can be obtained using Eq. 13–27 since the pressures are known. Here the stagnation pressure is p0 = 101.3 kPa since the outside air is at rest. Also, it maintains this value within the pipe. Since k = 1.4 for air, we have p0 = p a 1 +

k - 1 2 k>(k - 1) M b 2

101.3 kPa = 198 kPa2 a 1 +

1.4 - 1 2 1.4>11.4 - 12 M b 2

M = 0.2180 6 1 subsonic flow This result can be obtained by interpolating in Table B–1, using p>p0 = 98 kPa>101.3 kPa = 0.9674. # Analysis. The mass flow is determined using m = rAV, and so we must first find the density and velocity of the gas. The stagnation density of the air at T = 20°C is determined from Appendix A. It is r0 = 1.202 kg>m3. Therefore, using Eq. 13–28, the density of the air in the pipe is r0 = ra 1 + 1.202 kg>m3 = ra 1 +

k - 1 2 1>(k - 1) M b 2

1>11.4 - 12 1.4 - 1 (0.2180)2 b 2

r = 1.1739 kg>m3

This same value can also be determined from Eq. 13–16, p>p0 = 1r>r0 2 k.

13.4

The velocity of the flow into the entrance is determined using Eq. 13–22, V = M 2kRT, which depends upon the temperature within the flow. The temperature can be found from Table B–1 or using Eq. 13–26 for M = 0.218. T0 = Ta 1 +

(273 + 20) K = T a 1 +

k - 1 2 M b 2 1.4 - 1 (0.2180)2 b 2

T = 290.24 K Thus,

V = M 2kRT = (0.2180) 21.41286.9 J>kg # K21290.24 K2 = 74.44 m>s

The mass flow is therefore # m = rVA = 1.1739 kg>m3 174.44 m>s23p(0.025 m)2 4 = 0.172 kg>s

Ans.

Since here M = 0.218 6 0.3, it would be suitable to assume the air to be an ideal fluid (incompressible and frictionless), with steady flow, and then use the Bernoulli equation to determine the velocity. If we did this, then

NOTE:

p0 p1 V02 V12 + = + r r 2 2 101.31103 2 N>m2 1.202 kg>m3

+ 0 =

981103 2 N>m2 1.202 kg>m3

+

V12 2

V1 = 74.10 m>s This value has an error of about 0.46% from the value V = 74.44 m>s, found by accounting for the compressibility of the air.

stagnation properties

737

13

738

C h a p t e r 13

Compressible Flow

13.5

ISENTROPIC FLOW THROUGH A VARIABLE AREA

Compressible flow analysis is often applied to gases that pass through ducts in jet engines and rocket nozzles. For these applications we will generalize the discussion, and show how the pressure, velocity, and density of the gas are affected by varying the cross-sectional area of the duct through which the gas flows, Fig. 13–12a. For short distances, we will require the flow to be steady and the process to be isentropic. Also, the cross-sectional area of the duct is assumed to change gradually, so that the flow can be considered one-dimensional and average gas properties can be used. The fixed control volume shown in Fig. 13–12a contains a portion of the gas in the duct.

13

Continuity Equation. Since the velocity, density, and cross-sectional area all change, the continuity equation gives 0 r dV + rVf>cs # dA = 0 0t Lcv Lcs 0 - rVA + 1r + ∆r21V + ∆V21A + ∆A2 = 0

After multiplying and taking the limit as ∆x S 0, the second- and thirdorder terms will drop out. Simplifying, we get rV dA + VA dr + rA dV = 0 Solving for the velocity change yields A

A 1 DA

V

dV = -V a

V 1 DV

dr dA + b r A

(13–29)

Linear Momentum Equation. As shown on the free-body diagram of the control volume, Fig. 13–12b, the surrounding gas exerts pressure on the front and back open control surfaces. Since the sides of the duct increase the cross-sectional area by ∆A, the average pressure, p + ∆p>2, will act horizontally on this increased area. Applying the linear momentum equation in the direction of flow, we have

Dx (a)

p 1 Dp D A 2

+ F = 0 Vr dV + VrVf>cs # dA Sg 0t Lcv Lcv

(p 1 Dp)(A 1 D A)

pA

pA + a p +

∆p b ∆A - 1 p + ∆p21A + ∆A2 = 2

0 + Vr1 -VA2 + 1V + ∆V)(r + ∆r21V + ∆V21A + ∆A2

Free-body diagram

Expanding, and again realizing that the higher-order terms drop out in the limit, we get

(b)

Fig. 13–12

dp = - a 2rVdV + V 2dr + rV 2

dA b A

(13–30)

13.5

isentropiC Flow through a variable area

739

We can eliminate dV in this equation by substituting its value from Eq. 13–29. This yields dp = rV 2 a

dr dA + b r A

13

Also, we can eliminate dr by substituting Eq. 13–18, dr = dp>c 2. Then expressing the result in terms of the Mach number M = V>c, we find that dp =

rV 2

dA 1 - M2 A

(13–31)

The change in the velocity can be related to the Mach number and the change in the area by equating Eq. 13–31 to Eq. 13–30, and then eliminating the term dr>r using Eq. 13–29. This gives dV = -

V dA 2 A 1 - M

(13–32)

M , 1 (subsonic flow) Divergent duct dA . 0 Pressure and density increase Velocity decreases (a)

Finally, the change in the density can be related to the Mach number and the change in the area by equating Eq. 13–32 and Eq. 13–29. The result is dr =

rM2

dA 1 - M A 2

(13–33)

Subsonic Flow. When the flow is subsonic, M 6 1, then 11 - M2 2

in the above three equations is positive. As a result, an increase in area, dA, or a divergent duct, will cause the pressure and density to increase and the velocity to decrease, Fig. 13–13a. Likewise, a decrease in area, or a convergent duct, will cause the pressure and density to decrease and the velocity to increase, Fig. 13–13b. These results for pressure and velocity are similar to those for incompressible flow, as noted from the Bernoulli equation. For example, if the pressure increases, then the velocity will decrease, and vice versa.

Supersonic Flow. When the flow is supersonic, M 7 1, then

11 - M2 2 will be negative. Now the opposite effect occurs; that is, an increase in duct area, Fig. 13–13c, will cause the pressure and density to decrease and the velocity to increase, whereas a decrease in area will cause the pressure and density to increase and the velocity to decrease, Fig. 13–13d. This seems contrary to what we might expect, but experimental results have shown that indeed it occurs. In a sense, supersonic flow behaves in a manner similar to traffic flow. When cars come to a widening of a roadway, their speed increases (higher velocity) because they begin to spread out (lower pressure and density). A narrowing of the roadway causes congestion (higher pressure and density) and a lowering in speed (lower velocity).

M , 1 (subsonic flow) Convergent duct dA , 0 Pressure and density decrease Velocity increases (b)

M . 1 (supersonic flow) Divergent duct dA . 0 Pressure and density decrease Velocity increases (c)

M . 1 (supersonic flow) Convergent duct dA , 0 Pressure and density increase Velocity decreases (d)

Fig. 13–13

740

C h a p t e r 13

Compressible Flow

M,1

M.1 M51

Subsonic

Sonic

Supersonic

Laval nozzle

13

Fig. 13–14

Laval Nozzle. From these comparisons, we see that to produce supersonic flow (M 7 1), a nozzle can be fashioned as in Fig. 13–14 so that it has an initial convergent section to increase the subsonic velocity until it becomes sonic at the throat, M = 1. Here it is not possible to have the flow move faster than M = 1 because once M = 1 is obtained a pressure at the throat cannot move upstream (against the flow) to signal an increase in flow into the nozzle. A divergent section is then added to the nozzle to further increase the now sonic velocity to supersonic speed, M 7 1. This type of nozzle is termed a Laval nozzle, named after the Swedish engineer Karl de Laval, who designed it in 1893 for use in a steam turbine. Area Ratios. We can determine the cross-sectional area at any point along a nozzle by expressing it in terms of the Mach number using the continuity equation. If sonic conditions occur at the throat, then the cross section at the throat A* can be used as a reference, where T = T*, r = r*, and M = 1. At any other point, since V = Mc = M 2kRT, continuity of mass flow requires # m = rVA = r*V*A*; r 1 M 2kRT 2 A = r* 1 12kRT* 2 A*

or

A 1 r* T* = a b A* M r AT

(13–34)

This result can also be expressed in terms of the stagnation density and temperature by introducing the appropriate ratios. A 1 r* r0 T* T0 = a ba b r A T0 A T A* M r0

(13–35)

Substituting Eqs. 13–26 and 13–28 for each ratio, and realizing that M = 1 for the ratios r*>r0 and T*>T0, we get, after simplifying, k+1

1 2 1 1 + 2 1k - 12M 21k - 12 A = £ § 1 A* M 2 1k + 12

(13–36)

13.5

isentropiC Flow through a variable area

A graph of this equation for a given value of k is shown in Fig. 13–15. Except when A = A*, there are two values for the Mach number for each value of A>A*. One value, M1, is for the area A′ in the region of subsonic flow, and the other, M2, is for A′ in the region of supersonic flow. Rather than solving Eq. 13–36, for convenience, if k = 1.4, we can use Table B–1. It is worth noting how the values within this table follow the shape of the curve in Fig. 13–15. Scan down the table and notice that as M increases, A>A* decreases, until M = 1, and then A>A* increases again. With Eq. 13–36, or Table B–1, we can now determine the required cross-sectional area A2 of a nozzle at a point where the flow must be M2, provided we know M1 where the cross-sectional area is A1. To show how to do this, take the case shown in Fig. 13–16, where M1 = 0.5, A1 = p10.03 m2 2, and M2 = 1.5. To determine A2 we must reference A1 and A2 to the throat area A*, since the ratio A>A* is used in the equation and the table. Using M1 = 0.5, the ratio A1 >A* = 1.3398 is determined from Table B–1 (or Eq. 13–36). Likewise, using M2 = 1.5, the ratio A2 >A* = 1.176 is also found from Table B–1. With these two area ratios, we can then write A1 >A* A1 1.3398 = = A2 A2 >A* 1.176

so A2 =

1 1.176 pd 22 = 3p10.03 m2 2 4 a b 4 1.3398 d2 = 56.2 mm

This method will also provide a valid solution even if flow through the throat is not at M = 1, because we can simply imagine that the nozzle has a location where the throat narrows to A*.

d2 d1 5 60 mm M2 5 1.5

M1 5 0.5 A* 1

2

Fig. 13–16

741

A A*

A9 A*

13

A 51 A*

M51 M1 Subsonic

M M2 Supersonic

Laval nozzle area ratio vs. Mach number

Fig. 13–15

742

C h a p t e r 13

EXAMPLE

13

Compressible Flow

13.7 Air is drawn through the 50-mm-diameter pipe in Fig. 13–17 and passes section 1 with a speed of V1 = 150 m>s, while it has an absolute pressure of p1 = 400 kPa and a temperature of T1 = 350 K. Determine the required area of the throat of the nozzle to produce sonic flow at the throat. Also, if supersonic flow is to occur at section 2, find the velocity, temperature, and required pressure at this location. 75 mm 50 mm 150 ms

1

2

Fig. 13–17

SOLUTION Fluid Description. We assume isentropic steady flow through the nozzle. Throat Area. The Mach number at section 1 is first calculated. For air k = 1.4, R = 286.9 J>kg # K. Therefore, M1 =

2kRT1 V1

150 m>s

=

21.4(286.9 J>kg # K)(350 K)

= 0.40

Although we can use Eq. 13–36 with M1 and A1 to determine the throat area A*, it is simpler to use Table B–1 since k = 1.4. Thus, for M1 = 0.40, we find A1 = 1.5901 A* A* =

p10.025 m2 2 1.5901

= 0.0012348 m2 = 0.00123 m2

Ans.

Properties at Section 2. Now that A* is known, the Mach number at A2 can be determined from Eq. 13–36; however, again it is simpler to use Table B–1, with the ratio p10.0375 m2 2 A2 = = 3.58 A* 0.0012348 m2

13.6

isentropiC Flow through Converging and diverging nozzles

We get M2 = 2.8232 (approximately), because supersonic flow must occur at the end of the divergent section. (The other root, M1 = 0.1645 refers to subsonic flow at the exit.) The temperature and pressure at the exit can be determined using M = 2.8230 and Eqs. 13–26 and 13–27; however, first we must know the stagnation values T0 and p0 . To find them we can again use Eqs. 13–26 and 13–27 with M1 = 0.40 and T1 and p1. A simpler method is to use Table B–1 by referencing the ratios T2 >T1 and p2 >p1 in terms of the stagnation ratios as follows: T2 >T0 T2 0.38552 = = = 0.3978 T1 T1 >T0 0.9690

p2 >p0 p2 0.03557 = = = 0.03972 p1 p1 >p0 0.8956

Therefore, without having to find T0 and p0, we have T2 = 0.3978T1 = 0.3978(350 K) = 139.23 K = 139 K

Ans.

p2 = 0.03972p1 = 0.03972(400 kPa) = 15.9 kPa

Ans.

This lower pressure at the exit is what draws the air through the 50-mm-diameter pipe at 150 m>s. The average velocity of the air at section 2 is V2 = M2 2kRT2 = 2.823221.41286.9 J>kg # K21139.23 K2 = 668 m>s

13.6

Ans.

ISENTROPIC FLOW THROUGH CONVERGING AND DIVERGING NOZZLES

In this section we will study the compressible flow through a nozzle that is attached to a large vessel or reservoir of stagnant gas, Fig. 13–18a. Here the pipe on the end of the nozzle is connected to a tank and vacuum pump. By opening the valve on the pipe, we can regulate the flow through the nozzle by pumping the air out of the tank. This will produce a so-called backpressure, pb, within the tank, and here we will be interested in how the backpressure affects the pressure along the nozzle and the mass flow through the nozzle.

743

13

744

C h a p t e r 13

V50 p0 5 constant T0 5 constant

Compressible Flow

Converging Nozzle. First we will consider attaching a converging nozzle to the reservoir, Fig. 13–18a.

Tank pb

Valve

•

When the backpressure is equal to the pressure within the reservoir, pb = p0, no flow occurs through the nozzle, as indicated by curve 1, Fig. 13–18b.

•

If pb is slightly lower than p0, then flow through the nozzle will remain subsonic. Here the velocity will increase through the nozzle, causing the pressure to decrease, as shown by curve 2.

•

Further drops in pressure pb will cause the flow at the nozzle exit to eventually reach sonic flow, M = 1. The backpressure at this point is called the critical pressure p*, curve 3. The value of p* can be determined from Eq. 13–27 with M = 1. Since p0 is the stagnation pressure, the result is

Pump

13

(a)

k>(k - 1) p* 2 = a b p0 k + 1

p 1 no flow 2 subsonic flow 3 sonic flow 4 expansion waves x

p0 p* p9

For example, for air k = 1.4, then p*>p0 = 0.5283 (also see Table B–1). Therefore, the pressure just outside the nozzle must be approximately one-half that within the reservoir to achieve sonic flow. For backpressures from pb = p0 down to pb = p* the flow can be considered isentropic without appreciable error. This is because the flow is rapid, resulting in boundary layers that are thin so as to produce minimal friction losses as the flow is accelerated through the nozzle.

(b)

m·

• m· max

4

3 2

1 p0 Mass flow vs. backpressure (c)

Fig. 13–18

pb

(13–37)

If the backpressure is lowered further, say to p′ 6 p*, then neither the pressure distribution through the nozzle nor the mass flow out of it will be affected. The nozzle is said to be choked since the pressure at the exit of the nozzle, “the throat,” must remain at p*. Outside and just beyond the nozzle exit, the pressure will suddenly decrease to the lower backpressure p′; however, this occurs only through the formation of three-dimensional expansion waves, curve 4. Within this region the isentropic process ceases, since the expansion of the gas causes an increase in entropy due to friction and heat loss.

The mass flow as a function of the backpressure for each of these four cases is shown in Fig. 13–18c.

13.6

Converging–Diverging Nozzle. Now let’s consider the same test using a converging–diverging, or Laval, nozzle, Fig. 13–19a.

V50 p0 5 constant T0 5 constant

Tank pb

Valve

•

745

isentropiC Flow through Converging and diverging nozzles

Pump

As before, if the backpressure is equal to the pressure in the chamber, pb = p0, then no flow occurs through the nozzle because the pressure through the nozzle is constant, curve 1, Fig. 13–19b. (a)

•

If the backpressure drops somewhat, then subsonic flow occurs. Through the convergent section the velocity at the throat increases to its maximum, while the pressure decreases to a minimum. Through the divergent section the velocity decreases, while the pressure increases, curve 2. p

•

When the backpressure becomes p3, the pressure at the throat drops to p* so that the flow finally reaches sonic speed, M = 1, at the throat. This is a limiting case where subsonic flow continues to occur in both the convergent and divergent sections, curve 3. Any further slight decrease in backpressure will not cause an increase in mass flow through the nozzle since the velocity through the throat is at its maximum (M = 1). Hence, the nozzle becomes choked, and the mass flow remains constant.

1 No flow p0 p3 p* p4

2 3 Isentropic subsonic flow

M51

4 Isentropic supersonic flow x (b)

Fig. 13–19

•

To further accelerate the flow isentropically within the divergent section, it is necessary to decrease the backpressure all the way until it reaches p4 , as shown by curve 4. Again, this will not affect the mass flow since the nozzle is choked.

•

Because of the branching of the pressure curves 3 and 4, isentropic flow through the nozzle only occurs for p Ú p3 and p = p4. This is because for a given ratio of A>A* (exit area versus throat area), Eq. 13–36 gives only two possible Mach numbers at the exit, as noted in Fig. 13–15 (M1 subsonic and M2 supersonic). Thus, if the backpressure is somewhere in between these isentropic exit pressures of p3 and p4, or is lower than p4, then the exit pressure will suddenly convert to this pressure only through a shock wave, formed within the nozzle 1p4 6 pb 6 p3 2 or just outside of it 1pb 6 p4 2. This is nonisentropic since a shock involves friction losses and will result in an inefficient use of the nozzle. We will discuss this phenomenon further in Sec. 13.10.

Military jet aircraft have nozzles that can be flared or closed so that they are able to alter the efficiency of their thrust.

13

746

C h a p t e r 13

Compressible Flow

IMPORTANT POINTS • Subsonic flow through a converging duct will cause the velocity to increase and the pressure to decrease. Supersonic flow causes the opposite effect; the velocity will decrease and the pressure will increase. 13

• Subsonic flow through a diverging duct will cause the velocity to decrease and the pressure to increase. Supersonic flow causes the opposite effect; the velocity will increase and the pressure will decrease.

• A Laval nozzle has a convergent section to accelerate subsonic flow to sonic speed at the throat, M = 1, and a divergent section to further accelerate the flow to supersonic speed.

• It is not possible to make a gas flow faster than sonic speed, M = 1, through the throat of any nozzle, since at this speed the lower pressure at the throat cannot be transmitted back upstream to signal an increase in flow.

• The Mach number at a cross-sectional area A of a nozzle is a function of the ratio A>A*, where M = 1 at A*, the area of the throat.

• A nozzle becomes choked provided M = 1 at the throat. When this occurs, the pressure at the throat is called the critical pressure p*. This condition provides maximum mass flow through the nozzle.

• For a Laval nozzle, when M = 1 at the throat there are two possible backpressures that produce

isentropic flow within the nozzle. One produces subsonic speeds within the divergent section, M 6 1, and the other produces supersonic speeds within this section, M 7 1. No shock wave is produced in either case.

EXAMPLE

13.8 50 mm

0 1

Fig. 13–20

Determine the required pressure at the entrance of the 50-mm-diameter pipe at section 1 in Fig. 13–20, in order to produce the maximum flow of air through the pipe. Outside the pipe, the air is at standard atmospheric pressure and temperature. What is the mass flow? SOLUTION Fluid Description. We assume steady isentropic flow through the nozzle. Analysis. The stagnation pressure, temperature, and density are equal to the “standard atmospheric values” since the outside air is at

13.6

isentropiC Flow through Converging and diverging nozzles

rest. From Appendix A, we have p0 = 101.3 kPa, T0 = 15° C, and r0 = 1.23 kg>m3. The maximum flow into the pipe occurs when M = 1 at the pipe entrance. In other words, once this occurs, the air flowing through the entrance cannot transmit a reduced pressure to the air behind it any faster than M = 1. Using Table B–1 or Eq. 13–27 to obtain this required pressure, we have p0 = p 1 a 1 +

k>(k - 1) k - 1 M12b 2

101.3 kPa = p1 c 1 + a p1 = 53.5 kPa

1.4>(1.4 - 1) 1.4 - 1 b (1)2 d 2

Ans.

# To obtain the mass flow m = rVA, we must first determine the density of the air and the velocity of the flow at the pipe entrance to produce M = 1. The density of the air can be determined from Eq. 13–28 or Eq. 13–16, p2 >p1 = (r2 >r1)k. Using Eq. 13–28, r0 = r1 a 1 +

k - 1 2 1>(k - 1) M1 b 2

1.23 kg>m3 = r1 c 1 + a

1>(1.4 - 1) 1.4 - 1 b (1)2 d 2

r1 = 0.7797 kg>m3

The velocity is a function of the air temperature at the entrance, that is, V = M 2kRT. We can obtain this temperature using Table B–1, Eq. 13–26 with M = 1, or Eq. 13–17, p2 >p1 = 1T2 >T1 2 k>1k - 12. Using Eq. 13–26, we have T0 = T1 a 1 +

k - 1 M12b 2

(273 + 15°C) K = T1 c 1 + a T1 = 240 K

1.4 - 1 b (1)2 d 2

Thus,

V1 = M1 2kRT1 = (1) 21.41286.9 J>kg # K2(240 K) = 310.48 m>s

The mass flow into the pipe is therefore # m = r 1 V1 A 1 =

1 0.7797 kg>m3 2 1310.48 m>s23p 1 0.025 m 2 2 4

= 0.475 kg>s

Ans.

747

13

748

C h a p t e r 13

EXAMPLE

Compressible Flow

13.9

13 300 mm

300 kPa

The converging nozzle on the tank in Fig. 13–21 has a 300-mm exit diameter. If nitrogen within the tank has an absolute pressure of 500 kPa and an absolute temperature of 1200 K, determine the mass flow from the nozzle if the absolute pressure in the pipe at the nozzle is 300 kPa. What pressure would be needed here to create the greatest mass flow through the nozzle? SOLUTION Fluid Description. We assume steady isentropic flow through the nozzle.

500 kPa

Fig. 13–21

Analysis. The nitrogen within the tank is at rest, and so the stagnation pressure and temperature are p0 = 500 kPa and T0 = 1200 K. Since we know both p and p0 , then p>p0 = 300 kPa>500 kPa = 0.6, and so we can determine M from Eq. 13–27 or Table B–1. We get M = 0.8864. # To obtain the mass flow, m = rVA, we must obtain the density and velocity at the exit. First the temperature is determined from Eq. 13–26 or from Table B–1 for M = 0.8864 or p>p0 = 0.6. We have T = 0.8642 T0 T = 0.864211200 K2 = 1037.00 K Therefore, the exit velocity of the nitrogen is

V = M 2kRT = 10.88642 21.41296.8 J>kg # K211037 K2 = 581.86 m>s The density can be found using the ideal gas law. The mass flow from the nozzle is therefore

3001103 2 N>m2 p # m = rVA = a b VA = c d 1581.86 m>s23p(0.15 m)2 4 RT 1296.8 J>kg # K211037.00 K2 # m = 40.1 kg>s

Ans.

For the greatest mass flow through the nozzle, it is necessary that M = 1 at the exit, and so Eq. 13–27 or Table B–1 requires p* = 0.5283 or p* = 1500 kPa210.52832 = 264 kPa p0

Ans.

13.6

EXAMPLE

isentropiC Flow through Converging and diverging nozzles

749

13.10

The Laval nozzle in Fig. 13–22 is connected to a large chamber containing air at an absolute pressure of 350 kPa. Determine the backpressure in the pipe at B that will cause the nozzle to choke and yet produce isentropic subsonic flow through the pipe. Also, what backpressure is needed to cause isentropic supersonic flow?

100 mm

50 mm

pb

p 5 350 kPa B

Fig. 13–22

SOLUTION Fluid Description. nozzle.

We assume steady isentropic flow through the

Analysis. Here we must find the two backpressures, p3 and p4 , in Fig. 13–19b, required to produce M = 1 at the throat. The area ratio for the nozzle between the exit and the throat is p10.05 m2 2 AB = = 4 A* p10.025 m2 2 If this ratio is used in Eq. 13–36, two roots for M at the exit can be determined, Fig. 13–15. However, we can also solve this problem using Table B–1. With AB >A* = 4, we get M1 = 0.1467 (subsonic flow) and pB >p0 = 0.9851. Thus, the higher backpressure at B that will cause subsonic flow is 1pB 2 max = 0.98511350 kPa2 = 345 kPa

Ans.

1pB 2 min = 0.029791350 kPa2 = 10.4 kPa

Ans.

Further in the table, the alternative solution, where AB >A* = 4, gives M2 = 2.940 (supersonic flow) and pB >p0 = 0.02979. Thus, the lower backpressure at B is for supersonic flow. It is

NOTE: Both of these pressures will produce the same mass flow through

# the nozzle, and its value can be calculated using m = r*V*A*.

13

750

C h a p t e r 13

EXAMPLE 100 mm

Compressible Flow

13.11 p 5 90 kPa d

13

M2 5 0.7 2

Air flows through the 100-mm-diameter pipe in Fig. 13–23 having an absolute pressure of p1 = 90 kPa. Determine the diameter d at the end of the nozzle so that isentropic flow occurs out of the nozzle at M2 = 0.7. The air within the pipe is taken from a large reservoir at standard atmospheric pressure and temperature.

1

Fig. 13–23

SOLUTION I Fluid Description. We assume steady isentropic flow through the nozzle. Analysis. The diameter d can be determined from continuity of the mass flow, which requires # m = r1V1A1 = r2V2A2 (1) We must first find M1 and then find the densities and velocities at 1 and 2. The stagnation values for atmospheric air are determined from Appendix A as p0 = 101.3 kPa, T0 = 15°C, and r0 = 1.23 kg > m3. Knowing p1 and p0 , we can now determine M1 at the entrance 1 of the nozzle using Eq. 13–27. p0 = p1 a 1 +

k - 1 2 k>(k - 1) M1 b 2

101.3 kPa = 190 kPa2 a 1 +

1.4>(1.4 - 1) 1.4 - 1 M12b 2

M1 = 0.4146

As expected, M1 6 M2 = 0.7. Applying Eq. 13–26 to find the temperatures at the entrance and exit, we have T0 = T1 a 1 +

(273 + 15) K = T1 a 1 +

k - 1 M1 2b 2

1.4 - 1 (0.4146)2 b 2

T1 = 278.43 K

T0 = T2 a 1 +

k - 1 M22b 2

(273 + 15) K = T2 a 1 +

1.4 - 1 (0.7)2 b 2

T2 = 262.30 K

13.6

isentropiC Flow through Converging and diverging nozzles

751

Thus, the velocities at the entrance and exit are

V1 = M1 2kRT1 = (0.4146) 21.41286.9 J>kg # K21278.4 K2 = 138.63 m>s V2 = M2 2kRT2 = (0.7) 21.41286.9 J>kg # K21262.3 K2 = 227.21 m>s

The density of the air at the entrance and exit of the nozzle is determined using Eq. 13–28. r0 = r1 a 1 +

1>(k - 1) k - 1 M12b 2

r0 = r2 a 1 +

1>(k - 1) k - 1 M22b 2

1.23 kg>m3 = r1 c 1 +

1>(1.4 - 1) 1.4 - 1 (0.4146)2 d 2

1.23 kg>m3 = r2 c 1 +

1>(1.4 - 1) 1.4 - 1 (0.7)2 d 2

r1 = 1.1304 kg>m3

r2 = 0.9736 kg>m3

Finally, applying Eq. 1, r 1 V1 A 1 = r 2 V2 A 2 d 2 11.1304 kg>m3 21138.63 m>s23p(0.05 m)2 4 = 10.9736 kg>m3 21227.21 m>s2pa b 2 d = 84.2 mm

Ans.

SOLUTION II We can also solve this problem in a direct manner using Table B–1, even though subsonic flow occurs at the end of the nozzle. To do so, we will make reference to a phantom extension of the nozzle, where M = 1 and A = A*, and then relate the area ratios A1 for M1 = 0.4146 and A2 for M2 = 0.7 to this reference.

Thus,

A2 >A* A2 = A1 A1 >A* A2 = A1 a

Using Table B–1, we therefore have

A2 >A*

A1 >A*

b

1 2 1.0944 pd = p10.05 m2 2 a b 4 1.5451 d = 84.2 mm

Ans.

13

752

C h a p t e r 13

Compressible Flow

13.7

THE EFFECT OF FRICTION ON COMPRESSIBLE FLOW

In contrast to nozzles and transitions, a long conduit or duct through which a gas flows will have a rough surface, and so frictional effects will cause heating of the gas and thereby alter the characteristics of the flow. This typically occurs in exhaust and compressed-air pipes. In this section we will consider how these fast flows will change if the conduit is a pipe having a constant cross section, and a wall friction factor f, as determined from the Moody diagram.* We will assume the gas is ideal and has a constant specific heat, and the flow is steady. Also, the heat that is generated in the gas is assumed not to escape through the walls of the duct, and so the process will be adiabatic. This type of flow is sometimes called Fanno flow, named after Gino Fanno, who was the first to investigate it. To study how the flow is affected by friction and the Mach number, we will apply the fundamental equations of fluid mechanics to the fixed differential control volume in Fig. 13–24a. The flow properties are listed at each open control surface.

13

High volumetric gas flows within industrial pipes can be studied using compressible flow analysis. (© Kodda/Shutterstock)

Continuity Equation. Since the flow is steady, the continuity equation becomes 0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

In the limit,

0 + 1r + ∆r21V + ∆V2A + r1 -VA2 = 0 dr dV + = 0 r V

V 1 DV p 1 Dp T 1 DT r 1 Dr

V p T r

(13–38)

D

Dx (a)

Fig. 13–24 *Although most duct cross sections are circular, if other geometries are to be considered, then we can replace the pipe diameter D by the hydraulic diameter of the duct, defined as Dh = 4A>P. Here A is the cross-sectional area, and P is the perimeter of the duct. Note that for a circular duct, Dh = 41p D2 >42 > 1p D2 = D, as required.

13.7

the eFFeCt oF FriCtion on Compressible Flow

753

Linear Momentum Equation. As shown on the free-body diagram, Fig. 13–24b, the friction force ∆Ff acts on the closed control surface, and is the result of wall shear stress tw, discussed in Chapter 9. r 0 It is defined by Eq. 9–16, tw = 1p + gh2. Since the fluid is a gas, its 2 0x weight can be neglected, and so we get

pA

DFf Dx

Free-body diagram (b)

∆p D tw = a b a b 4 ∆x

Fig. 13–24 (cont.)

We can eliminate ∆p and ∆x by noting that the head loss from Eq. 10–1 is ∆hL = ∆p>rg, and from Eq. 10–3, ∆hL = f1∆x>D21V 2 >2g2. If we equate the right sides of these two equations and solve for ∆p> ∆x, we get ∆p> ∆x = 1f>D21rV 2 >22. Therefore, tw = a

frV 2 f rV 2 D ba ba b = 4 D 2 8

Finally, since tw acts on the fixed control surface area pD ∆x, and since the open control surfaces have an area of A = pD2 >4, the friction force becomes ∆Ff = tw 3pD∆x4 =

fA rV 2 a b ∆x D 2

Using this result, the momentum equation for the control volume becomes + F = 0 Vr dV + VrVf>cs # dA Sg 0t Lcv Lcs

-

fA rV 2 a b ∆x - 1p + ∆p2A + pA = 0 + 1V + ∆V21r + ∆r21V + ∆V2A + Vr1 -VA2 D 2

In the limit, where ∆x S 0, neglecting the second- and third-order terms, and using Eq. 13–38, we get

-

f rV 2 a b dx - dp = rVdV D 2

(p 1 Dp)A

(13–39)

Our goal is to now use this result, along with the ideal gas law and the energy equation, to relate f dx>D to the Mach number for the flow.

13

754

C h a p t e r 13

Compressible Flow

Ideal Gas Law. This law is p = rRT, but its differential form is dp = drRT + rR dT or 13

dp = a

p dT dr bp + r T

Here dr>r can be eliminated by using the continuity equation, Eq. 13–38, so that dp dT dV = p T V

(13–40)

Energy Equation. Since the flow is adiabatic, the stagnation temperature throughout the pipe will remain constant, and so application of the energy equation produces Eq. 13–26, which is T0 = Ta 1 +

k - 1 2 M b 2

(13–41)

Taking the derivative, we get after simplification

21k - 12M dT dM = T 2 + 1k - 12M2

(13–42)

Also, since V = M 2kRT, its derivative becomes

1 kR dV = dM 2kRT + M a b dT 2 2kRT dV dM 1 dT = + V M 2 T

(13–43)

Now eliminating r in Eq. 13–39 using the ideal gas law, and realizing that V = M 2kRT, we obtain dp 1 dx dV f + + = 0 2 2 D V kM p Substituting Eqs. 13–40, 13–42, and 13–43 into this equation and simplifying the algebra gives our final result. f

11 - M2 2 d1M2 2 dx = D kM4 11 + 12 1k - 12M2 2

(13–44)

13.7

1 M1

the eFFeCt oF FriCtion on Compressible Flow

755

2 M2 p2 T2

p1 T1

p* T* (Lmax)2

L

M5 1

M1

M5 1

Critical location

Critical location

(Lmax)1

(Lmax)1

2

(a)

M2

M5 1

(Lmax)2

5 M1

M2

L (b)

Fig. 13–25

Pipe Length versus Mach Number. When Eq. 13–44 is integrated along the pipe, say, from one position to another, it will result in a complicated expression, and additional work will be required to apply it numerically. However, if the pipe is actually long enough (or imagined to be long enough), we will show later that the effect of friction will tend to change the flow until it reaches the sonic speed M = 1. This occurs at the critical location, and so we will use it as a reference point to apply the upper limit of integration. At this position, xcr = Lmax, p = p*, T = T*, and r = r*, Fig. 13–25a. Along the length Lmax the friction factor f will actually vary, because it is a function of the Reynolds number, but since the Reynolds number is generally high,* for our purposes we will use a mean value for f. Therefore, Lmax 1 11 - M2 2 d1M2 2 f dx = 1 D L0 LM kM4 31 + 2(k - 1)M2 4

3 21 1k + 124M2 fLmax k + 1 1 - M2 + = ln c d D 2k kM2 1 + 12 1k - 12M2

(13–45)

With this equation, we can now determine the length L of pipe needed to change the Mach number from, say, M1 to M2 if the pipe length L … Lmax. As shown in Fig. 13–25b, to do this we simply require f(Lmax)1 f(Lmax)2 fL = ` ` (13–46) D D D M1 M2

*High values of Re produce almost constant values for f, because the curves of the Moody diagram tend to flatten out.

Critical location

13

756

C h a p t e r 13

Compressible Flow

Temperature. We will now apply Eq. 13–41 to a position where the temperature is T and to the critical or reference location where M = 1. Realizing that the stagnation temperature remains constant because the process is adiabatic, the temperature ratio expressed in terms of the Mach number becomes 13 1 T>T0 T 2 (k + 1) = = T* T*>T0 1 + 12(k - 1)M2

(13–47)

Velocity. Relating the velocity to the Mach number, we can use Eq. 13–47 to express the velocity ratio as 1 V M 2kRT 2 (k + 12 = = M£ § V* 1 + 12(k - 12M2 (1) 2kRT*

1>2

(13–48)

Density. Applying the continuity equation, rVA = r*V*A, and using Eq. 13–48, the density ratio becomes r>r* = V *>V, or 1 2 r 1 1 + 2(k - 12M = £ § 1 r* M 2 (k + 12

1>2

(13–49)

Pressure. From the ideal gas law, p = rRT, we have p>p* = 1r>r*21T>T*2. Therefore, from Eqs. 13–47 and 13–49, we obtain the pressure ratio 1 p 1 2 (k + 1) = £ § p* M 1 + 12(k - 12M2

Ratio p p*

T T* M

1.0 Subsonic

Supersonic Fanno flow

Fig. 13–26

(13–50)

Finally, the stagnation pressure ratio will vary along the pipe since the process is nonisentropic. It can be obtained by realizing that p0 >p0* = 1p0 >p21 p>p*21 p*>p0*2. And so, using Eqs. 13–27 and 13–50, we get

V V* 1.0

1>2

p0 1 2 k - 1 2 (k + 12>32(k - 1)4 = M bd ca b a1 + p 0* M k + 1 2

(13–51)

Graphs of the variation of the ratios T>T*, V>V*, and p>p* versus M are shown in Fig. 13–26, and their numerical values can be determined from the equations, or by using a “calculator” found on the Internet, or if k = 1.4, by interpolation using Table B–2 of Appendix B.

13.7

the eFFeCt oF FriCtion on Compressible Flow

757

The Fanno Line. Although we can describe the flow completely using the previous equations, it is instructive to show how the fluid behaves by considering how the entropy will vary along the pipe as a function of temperature. To do this, we must first express the change in entropy between an initial location 1 and some arbitrary location along the pipe. We will begin with Eq. 13–14, r T s - s1 = cv ln - R ln (13–52) r1 T1 Here we will want to express r>r1 in terms of temperature. Since A is constant, the continuity equation requires r>r1 = V1 >V, and since the stagnation temperature remains constant for an adiabatic process, then from Eq. 13–25, we have V = 22cp 1T0 - T2. Using these expressions, Eq. 13–52 now becomes

13

s - s1 = cv ln T - cv ln T1 + R ln 22cp 1T0 - T2 - R ln V1

R R ln(T0 - T) + c-cv ln T1 + ln 2cp - R ln V1d (13–53) 2 2 The last three terms are constant and are evaluated at the initial location of the pipe, where T = T1 and V = V1. If we plot Eq. 13–53, it represents the Fanno line (T–s diagram) for the flow, and looks like that shown in Fig. 13–27. The point of maximum entropy is found by taking the derivative of the above expression and setting it equal to zero, ds>dT = 0. This occurs when the flow is sonic, that is, M = 1 at T = 32>(k + 1)4T0. The region above M = 1 is for subsonic flow (M 6 1), and the lower region is for supersonic flow (M 7 1). For both cases, friction increases the entropy as the gas travels down the pipe.* As expected, for supersonic flow the Mach number decreases because of friction until it reaches M = 1, where the flow becomes choked at the critical length. For subsonic flow, however, the Mach number increases until it reaches M = 1. Although this may seem counterintuitive, it happens because the pressure drops rapidly, as noted in Fig. 13–26, for M … 1. This drop cools the flow and increases its velocity more than friction can provide heat or resistance to slow the flow. = cv ln T +

T

Subsonic flow

M51

ow

ic fl

on ers

Sup

(s 2 s1)

IMPORTANT POIN T S • Ideal gas flow through a pipe or duct that includes the effect of friction along the wall of the pipe without heat loss is called Fanno flow. Using an average value for the friction factor f, the gas properties T, V, r, and p can be determined at a location along the pipe where the Mach number is known, provided these properties are known at the reference or critical location, where M = 1.

• Friction in the pipe will cause the Mach number to increase for subsonic flow until it reaches M = 1, and to decrease for supersonic flow until it reaches M = 1. *The entropy cannot be decreased, since that would violate the second law of thermodynamics.

Fanno line (T–s diagram)

Fig. 13–27

758

C h a p t e r 13

EXAMPLE

Compressible Flow

13.12 30 mm

13 153 ms 0.8 m

L9 Lmax

Fig. 13–28

Air enters the 30-mm-diameter pipe with a velocity of 153 m > s, and a temperature of 300 K, Fig. 13–28. If the average friction factor is f = 0.040, determine how long, Lmax, the pipe should be so that sonic flow occurs at the exit. Also, what is the velocity of the air in the pipe at the exit Lmax, and at the location L = 0.8 m? SOLUTION Fluid Description. We assume that adiabatic steady compressible (Fanno) flow occurs along the pipe. Maximum Pipe Length. The critical length, Lmax, is determined using Eq. 13–45 or Table B–2.* First we need to determine the initial Mach number. V = M 2kRT;

153 m>s = M1 21.41286.9 J>kg # K21300 K2

M1 = 0.4408 6 1 subsonic flow Using Eq. 13–45 or Table B–2, we get 1 f>D21Lmax2 = 1.6817, so that Lmax = a

0.03 m b 11.68172 = 1.2613 m = 1.26 m 0.040

Ans.

At this exit M = 1. The velocity of the gas is determined from Eq. 13–48 or from the tabulated ratio for M1 = 0.4408. The result is V* = 322.98 m>s = 323 m>s

Ans.

Flow Properties at L = 0.8 m. Since the equations (and table) are referenced from the critical location, we must calculate (f>D)L from this location, Fig. 13–28. Thus, f 0.04 L′ = 11.2613 m - 0.8 m2 = 0.6150 D 0.03 m

This time using the table with interpolated values of the ratio for V>V*, we have V V* = 10.613321322.98 m>s2 = 198 m>s Ans. V* As the air travels 0.8 m down the pipe, notice how the velocity has increased from 153 m>s to 196 m>s. As an exercise, show that the temperature decreases from 300 K to 293 K at 0.8 m. The values of V and T follow the trend shown by the curves in Fig. 13–26. V =

*More accurate results can be obtained from the equation or from an Internet website, rather than using linear interpolation from the table.

13.7

EXAMPLE

759

the eFFeCt oF FriCtion on Compressible Flow

13.13

If air within a large reservoir flows into the 50-mm-diameter pipe in Fig. 13–29a, at M = 0.5, determine the Mach number of the air when it exits the pipe. Take L = 1 m. Explain what happens if the pipe is extended so that L = 2 m. The average friction factor for the pipe is f = 0.030.

50 mm

13

SOLUTION Fluid Description. We assume that adiabatic steady compressible (Fanno) flow occurs within the pipe.

L (a)

L = 1 m. First we will calculate Lmax for the pipe, where sonic flow occurs. Using M = 0.5, Eq. 13–45 gives f Lmax = 1.0691; D

Lmax =

1.069110.05 m2 = 1.782 m 0.030

Since L = 1 m 6 1.782 m, then at the exit, f 0.030 L′ = 11.782 m - 1 m2 = 0.4691 D (0.05 m)

Using Eq. 13–45,

M2 = 0.606

Ans.

L = 2 m. If the pipe is extended to L = 2 m, then L 7 Lmax for M = 0.5 at the entrance. As a result, friction will cause a reduced flow into the pipe until sonic flow chokes the exit. In this case, 10.03212 m2 f Lmax = = 1.2 D 10.05 m2

M19 . 1 2m (b)

Ans.

Consider what would happen if we required supersonic flow 1M1 7 12 through the entrance of the extended 2-m-long pipe. In this case, sonic flow 1M = 12 will still occur at the pipe exit; however, a normal shock wave will form within the pipe, Fig. 13–29b. This wave will convert the supersonic flow on the left side of the wave to subsonic flow on the right side. In Sec. 13.9 we will show how to relate the Mach numbers M1 ′ and M2 ′ on each side of this wave. With this relationship, the specific location L′ of the wave can then be determined, since it must be where L′1 gives M = 1 at the exit. As the pipe is extended farther, the wave will be located farther towards the entrance, and then eventually within the supersonic supply nozzle. If it reaches the throat of this nozzle, it will then choke it (M = 1). NOTE:

M29 , 1 L9

Then using Eq. 13–45, the new Mach number at the entrance becomes M1 = 0.485

M51

Fig. 13–29

760

C h a p t e r 13

EXAMPLE

Compressible Flow

13.14 A room is at atmospheric pressure, 101 kPa, and a temperature of 293 K. If the air from the room is drawn into a 100-mm-diameter pipe isentropically, such that it has an absolute pressure of p1 = 80 kPa as it enters the pipe, determine the mass flow and the stagnation temperature and stagnation pressure at the location L = 0.9 m. The average friction factor is f = 0.03. Also, what is the total friction force acting on this 0.9-m length of pipe?

13

SOLUTION Fluid Description. We assume that adiabatic steady compressible (Fanno) flow occurs along the pipe. Mass Flow. The mass flow can be determined at the entrance to the # pipe using m = r1V1A1, but we must first determine V1 and r1. Since the flow into the pipe is isentropic, and the pressure is p1 = 80 kPa, while the stagnation pressure is p0 = 101 kPa, we can determine the Mach number of the air and its temperature at the entrance using Eq. 13–27 and Eq. 13–26.

100 mm 1

2

p1 80 kPa = = 0.792 p0 101 kPa L 5 0.9 m

L9

M1 = 0.5868

Lmax (a)

Fig. 13–30

and

T1 = 0.93557 T0

Therefore, T1 = 0.935571293 K2 = 274.12 K, and so

V1 = M1 2kRT1 = 0.586821.41286.9 J>kg # K21274.12 K2 = 194.71 m>s

Using the ideal gas law to obtain r1, we have p1 = r1 RT1;

801103 2 Pa = r1 1286.9 J>kg # K21274.12 K2 r1 = 1.0172 kg>m3

The mass flow along the pipe is then # m = r1V1A1 = 11.0172 kg>m3 21194.71 m>s23p10.05 m2 2 4 # m = 1.5556 kg>s = 1.56 kg>s Ans. Stagnation Temperature and Pressure. Because the flow is adiabatic through the pipe, the stagnation temperature remains constant at 1T0 2 2 = 1T0 2 1 = 293 K

Ans.

13.7

the eFFeCt oF FriCtion on Compressible Flow

Friction will change the stagnation pressure throughout the pipe because the flow is nonisentropic. We can determine ( p0)2 at L = 0.9 m by using Eq. 13–51 (or Table B–2). First, however, we must find the length Lmax needed to choke the flow. Using M1 = 0.5868, Eq. 13–45 gives fLmax >D = 0.03Lmax >0.1 = 0.5455, and so Lmax = 1.8183 m. At this location, Eqs. 13–48, 13–51, and 13–50 give 194.71 m>s

V1 = 0.6218; V*

V* =

1p0 2 1 = 1.2043; p0*

p 0* =

101 kPa = 83.87 kPa 1.2043

p* =

80 kPa = 44.30 kPa 1.8057

p1 = 1.8057; p*

0.6218

13

= 313.16 m>s

Since Lmax locates the critical point, then at section 2, Fig. 13–30a, fL′>D = 0.03(1.8183 m - 0.9 m)>0.1 m = 0.27548. From Eq. 13–45, M = 0.6690, and then from Eq. 13–51 the stagnation pressure at this location is 1p0 2 2 = 1.1188; p*0

1p0 2 2 = 1.1188(83.87 kPa) = 93.8 kPa

Ans.

Friction Force. The resultant friction force is obtained using the momentum equation applied to the free-body diagram of the control volume, shown in Fig. 13–30b. First we must determine the static pressure p2 and the velocity V2 at fL′>D = 0.27548, p2 = 1.5689; p* V2 = 0.7021; V* Therefore,

Ff p1A

p2A Free-body diagram (b)

p2 = 1.5689(44.30 kPa) = 69.51 kPa

Fig. 13–30 (cont.)

V2 = 0.7021(313.16 m>s) = 219.86 m>s

+ F = 0 Vr dV + VrVf>cs # dA Sg 0t Lcv Lcs . # -Ff + p1 A - p2 A = 0 + V2 m + V1 ( -m) -Ff + 3801103 2 N>m2 4 3 p(0.05 m)2 4 - 3 69.5111032 N>m2 4 3p10.05 m2 24 = 0 + 1.5556 kg>s 1219.86 m>s - 194.71 m>s2 Ff = 43.3 N

Ans.

761

762

C h a p t e r 13

Compressible Flow

13.8

13

Pipes in chemical processing plants are sometimes heated along their lengths, resulting in the conditions for Rayleigh flow. (© Eric Gevaert / Alamy)

V p T r

V 1 DV p 1 Dp T 1 DT r 1 Dr

DQ

THE EFFECT OF HEAT TRANSFER ON COMPRESSIBLE FLOW

In this section we will consider how heat transfer through the walls of a straight pipe of constant cross-sectional area will affect the steady compressible flow of an ideal gas having a constant specific heat. This type of flow can typically occur in the pipes and ducts of a combustion chamber of a turbojet engine, where heat transfer is significant and friction can be ignored. The heat can also be added within the gas itself, and not through the walls of the pipe. For example, this can occur by a chemical process or by nuclear radiation. No matter how the heat is added, this type of flow is sometimes called Rayleigh flow, named after the British physicist Lord Rayleigh. To simplify the numerical work, we will do the same as for Fanno flow, and develop the necessary equations in terms of the Mach number, making reference to the gas properties T*, p*, r*, and V* at the location in the pipe where the critical or choked condition M = 1 occurs. A differential control volume for this situation is shown in Fig. 13–31a. Here ∆Q is positive if heat is supplied to the gas, and it is negative if cooling occurs.

Dx (a)

Continuity Equation. The continuity equation is the same as Eq. 13–38, namely

(p 1 Dp)A

pA

Free-body diagram (b)

Fig. 13–31

dr dV + = 0 r V

(13–54)

Linear Momentum Equation. Only pressure forces act on the open control surfaces, as shown on the free-body diagram, Fig. 13–31b. We have + ΣF = 0 S Vr dV + VrVf>cs # dA 0t Lcv Lcs -(p + ∆p)A + pA = 0 + (V + ∆V)(r + ∆r)(V + ∆V)A + V( -rVA) Taking the limit, and eliminating dr using Eq. 13–54, we obtain dp + rVdV = 0 If we divide this equation by p and use the ideal gas law, p = rRT, and V = M 1kRT to eliminate r and then T, we get dp dV + kM2 = 0 p V

(13–55)

13.8

the eFFeCt oF heat transFer on Compressible Flow

763

Ideal Gas Law. When the ideal gas law, p = rRT, is expressed in differential form, combined with the continuity equation, we get Eq. 13–40. dp dT dV = p T V

(13–56) 13

Energy Equation. No shaft work is performed on the gas, and there is no change in its potential energy. Therefore, the energy equation becomes # # # V out2 V in 2 # Qin - Wturbine + Wpump = ca hout + + gzout b - ahin + + gzin b d m 2 2 # (V + ∆V )2 V2 # Q - 0 + 0 = c a h + ∆h + + 0b - a h + bd m 2 2

# Dividing both sides by m, then in the limit we have dQ = dh + VdV dm = d ah +

V2 b 2

At the stagnation point, h + V 2 >2 = h0, and so using Eq. 13–7 in differential form, dh = cp dT, we get dQ = d1h0 2 = cp dT0 dm

And for a finite application of heat,

∆Q = cp 3 1T0 2 2 - 1T0 2 14 ∆m

(13–57)

As expected, because we do not have an adiabatic process, the result indicates that the stagnation temperature will not remain constant; rather, it will increase as heat is applied. We will now combine, and then integrate, the above equations from a position where the speed is M to the critical position where M = 1 to show how the velocity, pressure, and temperature are related to the Mach number.

Velocity. Since V = M 2kRT, its derivative produces Eq. 13–43,

that is, dV dM 1 dT = + V M 2 T

(13–58)

764

C h a p t e r 13

Compressible Flow

If we combine this equation with Eqs. 13–55 and 13–56, we obtain dV 2 = dM V M11 + kM2 2

(13–59)

Integrating between the limits V = V, M = M and V = V*, M = 1, we get

13

M2(1 + k) V = V* 1 + kM2

(13–60)

Density. From the continuity equation, for a pipe of finite length, r*V*A = rVA or V>V* = r*>r, and so the densities are related by r 1 + kM2 = 2 r* M (1 + k)

(13–61)

Pressure. For the pressure, combining Eqs. 13–55 and 13–59, we get dp 2kM = dM p 1 + kM2 which, when integrated from p = p, M = M to p = p*, M = 1, yields p 1 + k = p* 1 + kM2

(13–62)

Temperature. Finally, the temperature ratio is determined by substituting Eq. 13–59 into Eq. 13–58. This gives 211 - kM2 2 dT = dM T M11 + kM2 2

And upon integrating from T = T, M = M to T = T*, M = 1, we obtain

Ratio

M2(1 + k)2 T = T* 11 + kM2 2 2

p p* V V*

The variation of the ratios V>V*, p>p*, and T>T* versus the Mach number is shown in Fig. 13–32, and for k = 1.4 their numerical values can be determined either by direct calculation, or from the Internet, or by interpolation using the values given in Appendix B, Table B–3.

1.0 T T* M

1.0 Subsonic flow

Supersonic flow Rayleigh flow

Fig. 13–32

(13–63)

Stagnation Temperature and Pressure. The ratios of the stagnation temperatures and pressures at a location in the pipe, and at the critical or reference location, are sometimes needed for calculations. They can be determined using Eqs. 13–63 and 13–26, M2 11 + k2 2 k - 1 2 2 T0 T0 T T* = a1 + M b £ § = 2 2 2 11 + kM 2 k + 1 T0* T T* T0* T0 = T0*

k - 1 2 M b 2 11 + kM2 2 2

21k + 12M2 a 1 +

(13–64)

13.8

765

the eFFeCt oF heat transFer on Compressible Flow

And in a similar manner, using the pressure ratios of Eqs. 13–62 and 13–27, we get p0 1 + k 2 k - 1 2 k>(k - 1) = a bca b a1 + M bd 2 p0* k + 1 2 1 + kM

(13–65) 13

For convenience these ratios are also given in Table B–3.

Rayleigh Line. To better understand Rayleigh flow, we will show, just as we did for Fanno flow, how the entropy of the gas changes with temperature. Making reference to the critical location, where M = 1, the change in entropy in terms of the temperature and pressure ratios is expressed by Eq. 13–15; that is, s - s* = cp ln

p T - R ln T* p*

The last term can be expressed in terms of the temperature by squaring Eq. 13–62 and then substituting the result into Eq. 13–63. This gives a

p 2 T b = 2 p* M T*

Finally, when we solve for M2 in Eq. 13–63 and substitute this into the above equation, the change in entropy becomes s - s* = cp ln

T k + 1 k + 1 2 T - R ln c { d a b - k T* 2 A 2 T*

When this equation is graphed, it produces the Rayleigh line (T–s diagram) shown in Fig. 13–33. If we set ds>dT = 0, then like Fanno flow, the maximum entropy occurs when M = 1. Also, like Fanno flow, the upper portion of the graph defines subsonic flow (M 6 1) and the lower portion defines supersonic flow (M 7 1). Notice that for supersonic flow, the addition of heat will cause the temperature of the gas to increase, but its Mach number will decrease until it reaches M = 1, and the flow becomes choked.* Therefore, to increase supersonic flow, it is necessary to cool the pipe rather than heat it. For subsonic flow the addition of heat will cause the gas to reach a maximum temperature, Tmax, while its speed is increasing until its Mach number is M = 1> 1k (at dT>ds = 0); then the gas temperature will drop as M approaches the limit M = 1. This is also evident from the red curve in Fig. 13–32. *Entropy cannot be decreased, since that would violate the second law of thermodynamics.

T M 5 1 k

Tmax

M51

Subsonic flow heating cooling

Supersonic flow (s 2 s*) Rayleigh flow (T–s diagram)

Fig. 13–33

766

C h a p t e r 13

Compressible Flow

I MPO RTA N T PO I N T S • Rayleigh flow occurs when heat is added or removed as the gas flows through a pipe or duct. Because the process is not adiabatic, the stagnation temperature is not constant.

13

• The gas properties V, r, p, and T, at a specific location in the pipe where M is known, can be determined, provided these properties are also known at the reference or critical location, where the flow reaches M = 1.

EXAMPLE

13.15

200 mm 1

2

75 ms 2m

Outside air is drawn isentropically into the pipe having a diameter of 200 mm, Fig. 13–34. When it arrives at section 1, it has a velocity of 75 m>s, an absolute pressure of 135 kPa, and a temperature of 295 K. If the walls of the pipe supply heat at 100 kJ>kg # m, determine the properties of the air when it reaches section 2. SOLUTION

Fig. 13–34

Fluid Description. We assume the air to be inviscid and to have steady compressible flow. Due to the heating, this is Rayleigh flow. Air Properties at Critical Location. The air properties at location 2 can be determined using the ratios in Table B–3, provided we first know the properties at the critical location, where M = 1. We can find these using the properties at section 1, but first we need the Mach number at section 1. V1 = M1 2kRT1

75 m>s = M1 21.41286.9 J>kg # K2(295 K)

M1 = 0.2179 6 1 Subsonic Using Table B–3, T* = p* =

V* =

T1 295 K = = 1227.01 K T1 >T* 0.24042

p1 135 kPa = = 59.99 kPa p1 >p* 2.2504

75 m>s V1 = = 702.03 m>s V1 >V* 0.10683

13.8

the eFFeCt oF heat transFer on Compressible Flow

Air Properties at Section 2. We can determine the Mach number at section 2 by using Eq. 13–64. However, before we can do this or use the tables, we must find the stagnation temperatures (T0)2 and T0*. First, (T0)1 can be determined using Eq. 13–26, or Table B–1 for isentropic flow. For M1 = 0.2179, we get T1 295 K = = 297.80 K T1 > 1T0 2 1 0.9906

1T0 2 1 =

Now, to determine (T0)2, we will use the energy equation, Eq. 13–57. cp =

1.41286.9 J>kg # K2 kR = = 1004.15 J>kg # K k - 1 1.4 - 1

1001103 2 J (2 m) = kg # m

∆Q = cp 3 1T0 2 2 - 1T0 2 1 4 ∆m

3 1.004151103 2 J>kg # K 4 3 1T0 2 2 1T0 2 2 = 496.97 K

- 297.80 K 4

Also, from Table B–3, for M1 = 0.2179, the stagnation temperature at the critical or reference location is therefore T0* =

1T0 2 1 297.80 K = = 1472.14 K 1T0 2 1 >T0* 0.20225

Finally, we can find M2 from the stagnation temperature ratio. 1T0 2 2 496.97 K = = 0.33758 T0* 1472.14 K Using Table B–3, we get M2 = 0.2949. The other ratios at M2 give T2 = T*a p2 = p*a V2 = V*a

T2 b = 1227.01 K(0.39810) = 488 K T*

p2 b = 59.99 kPa(2.13950) = 128 kPa p*

Ans. Ans.

V2 b = 702.03 m>s (0.18607) = 131 m>s Ans. V*

The results indicate that as the Mach number increases from M1 = 0.2179 to M2 = 0.2949, the pressure decreases from 135 kPa to 128 kPa, the temperature increases from 295 K to 488 K, and the velocity increases from 75 m>s to 131 m>s. These changes follow the trend shown by the curves for Rayleigh subsonic flow in Fig. 13–32.

767

13

768

C h a p t e r 13

Compressible Flow

13.9

13

The exhaust from this rocket is designed to pass through its nozzles at supersonic speed, ideally without forming a shock. However, as the rocket ascends, the ambient pressure will decrease, and so the exhaust will fan or flare off the sides of the nozzles, forming expansion waves. (© Valerijs Kostreckis/ Alamy) 1

2 T2, p2, r2 (high)

T1, p1, r1 (low) M1 . 1 into wave Back

M2 , 1 out of wave Front

Standing shock wave (a)

(a) p 1A

p2 A

NORMAL SHOCK WAVES

When designing any nozzle or diffuser used in supersonic wind tunnels or for high-speed aircraft or rockets, it is possible to develop a standing shock wave within the nozzle. For example, a standing shock can form at the intake of a jet plane flying at M 7 1. Here the shock will decelerate the air to subsonic flow, thereby increasing its pressure just before it enters the compressor of the engine. In this section we will study how the flow properties change across a shock wave as a function of the Mach number. To do this, we will use the equations of continuity, momentum, and energy, and the ideal gas law. In Sec. 13.3, we stated that a shock wave is a high-intensity compression wave that is very thin. If the wave is “standing,” that is, at rest, then on the downstream side the temperature, pressure, and density will be high and the velocity low, whereas the opposite effect occurs on the upstream side, Fig. 13–35a. A large amount of heat conduction and viscous friction develops within the wave due to the extreme decelerations of the gas molecules. As a result, the thermodynamic process within the wave becomes irreversible, and so the entropy across the wave will increase. This process is therefore nonisentropic. If we consider a control volume to surround the wave and extend a slight distance beyond it, then the gas system within this control volume undergoes an adiabatic process because no heat passes through the control surfaces. Instead, the changes in temperature are made within the control volume.

Continuity Equation. When the wave is standing, the flow is steady,* and so the continuity equation becomes 0 r dV + rVf>cs # dA = 0 0t Lcv Lcs 0 - r1V1 A + r2V2 A = 0

Linear Momentum Equation. As shown on the free-body diagram of the control volume, Fig. 13–35b, only pressure acts on each side of the wave, and it is the difference in this pressure that causes the gas to decelerate and thereby lose its momentum. Applying the linear momentum equation, we have + F = 0 Vr dV + Sg 0t Lcv

Free-body diagram (b)

Fig. 13–35

(13–66)

r1V1 = r2V2

Lcs

VrVf>cs # dA

p1A - p2A = 0 + V1r1 1 -V1A2 + V2 r2 1V2A2 p1 + r1V12 = p2 + r2V22

(13–67)

*If the wave is moving, then steady flow occurs if we attach our reference to the wave. This will produce the same results.

13.9

normal shoCk waves

769

Ideal Gas Law. Considering the gas to be ideal with constant specific heats, then using the ideal gas law, p = rRT, we can write Eq. 13–67 as p1 a 1 +

V1 2 V2 2 b = p2 a 1 + b RT1 RT2

Since V = M 2kRT, the pressure ratio, written in terms of the Mach numbers, becomes p2 1 + kM12 = p1 1 + kM2 2

(13–68)

Energy Equation. Since an adiabatic process occurs, the stagnation temperature will remain constant across the wave. Therefore, (T0)1 = (T0)2, and from Eq. 13–26, which was derived from the energy equation, we have T2 = T1

k - 1 M1 2 2 k - 1 1 + M2 2 2

1 +

(13–69)

The ratio of the velocities on each side of the shock can be determined from Eq. 13–22, V = M 2kRT. It is V2 M2 2kRT2 M2 T2 = = V1 M1 A T1 M1 2kRT1

(13–70)

Using Eq. 13–69, our result is 1>2 k - 1 M1 2 M2 V2 2 = ≥ ¥ V1 M1 k - 1 1 + M2 2 2

1 +

(13–71)

From the continuity equation, Eq. 13–66, we obtain the density ratio 1>2 k - 1 M2 2 r2 M1 V1 2 = ≥ ¥ = r1 V2 M2 k - 1 2 1 + M1 2

1 +

(13–72)

Relationship between Mach Numbers. We can establish a relationship between the Mach numbers M1 and M2 by first forming the temperature ratio using the ideal gas law. p2 >r2R p2 r 1 T2 = = a b p1 r 2 T1 p1 >r1R

13

770

C h a p t e r 13

Compressible Flow

If we now substitute Eqs. 13–68, 13–69, and 13–72 into this expression and equate the result to Eq. 13–69, we will be able to solve for M2 in terms of M1. Two solutions are possible. The first one leads to the trivial solution M2 = M1, which refers to isentropic flow with no shock. The other solution leads to irreversible flow, which gives the desired relationship. 13

M1 2 + M2 2 =

2 k - 1

2k M2 - 1 k - 1 1

M1 7 M2

(13–73)

Thus, if M1 is known, then M2 can be found from this equation, and then the ratios p2 >p1, T2 >T1, V2 >V1, and r2 >r1 that occur in front of and behind the shock can be determined from the previous equations. Finally, the increase in entropy occurring across the shock can be found from Eq. 13–15. p2 T2 s2 - s1 = cp ln - R ln (13–74) p1 T1 This increase will cause the stagnation pressure across the shock to decrease. To determine ( p0)2 >( p0)1 in terms of the Mach numbers, we first use Eq. 13–27 and write

k>(k - 1) k - 1 M22 1p0 2 2 p2 1p0 2 2 p2 p1 2 = a ba ba b = ≥ ¥ (13–75) p2 p1 p1 1p0 2 1 1p0 2 1 k - 1 2 1 + M1 2 If Eqs. 13–68 and 13–73 are combined and simplified, we obtain p2 2k k - 1 = M12 (13–76) p1 k + 1 k + 1

1 +

1

2 T2, p2, r2 (high)

T1, p1, r1 (low) M1 . 1 into wave Back

M2 , 1 out of wave Front

Standing shock wave (a)

Fig. 13–36

Substituting this and Eq. 13–73 into Eq. 13–75 and simplifying, we obtain our result, k>(k - 1) k + 1 M12 2 ≥ ¥ k - 1 2 1 + M1 1p0 2 2 2 = (13–77) 1p0 2 1 2k k - 1 1>(k - 1) 2 c M d k + 1 1 k + 1

For convenience, this ratio along with p2 >p1, r2 >r1, T2 >T1, and M2 can be determined using the approximate tabulated values listed in Appendix B, Table B–4, for k = 1.4. More exact results can be obtained from the equations, or using a “calculator” on an Internet website. For any specific value of k, it can be shown, using Eq. 13–73, that when supersonic flow occurs behind the shock, Fig. 13–36a, M1 7 1, subsonic flow will occur in front of the shock, M2 6 1. Realize that subsonic flow cannot occur behind the shock, M1 6 1, because Eq. 13–73 would then

771

13.10 shoCk waves in nozzles

predict that supersonic flow must occur in front of the shock, M2 7 1. This, of course, is not possible because there are friction losses within the shock. In other words, Eq. 13–74 would indicate a decrease in entropy, which is a violation of the second law of thermodynamics.

V5 0 p0 5 constant T0 5 constant

pb

13

13.10

SHOCK WAVES IN NOZZLES

Propulsion nozzles used, for example, on jet planes and rockets are subjected to changing conditions of outside pressure and temperature as the jet or rocket ascends through the atmosphere. Consequently, the thrust, which is a function of this pressure, will also be changing. Depending upon this outside pressure, this can lead to a shock wave forming within the nozzle. To understand this process, let us again review the pressure variations through a converging–diverging (Laval) nozzle in Fig. 13–36a for various changes in the backpressure. • •

When the backpressure is at the stagnation pressure, p1 = p0, there is no flow through the nozzle, curve 1 in Fig. 13–36b. Lowering the backpressure to p2 causes subsonic flow through the nozzle, with the lowest pressure and maximum velocity occurring at the throat. This flow is isentropic, curve 2.

•

When the backpressure is lowered to p3, sonic velocity (M = 1) develops at the throat, and subsonic isentropic flow continues to occur through both the convergent and the divergent sections. At this pressure, maximum mass flow with no shock occurs through the nozzle, curve 3.

•

As the backpressure is lowered to p5, a standing normal shock wave will now develop within the divergent section, Fig. 13–36c. This is nonisentropic flow. Across the shock the pressure rises suddenly from A to B, curve 5, Fig. 13–36b, causing subsonic flow to occur from the shock to the exit plane. In other words, the pressure follows the curve from B and reaches the backpressure p5 at the exit.

•

•

A further lowering of the backpressure to p6 will bring the shock wave to the exit plane, Fig. 13–36d. Here the divergent section has supersonic flow throughout its length so that it reaches a pressure of p4 to the left of the wave. At the exit the shock wave will suddenly change the pressure from p4 to p6 so that the exit flow is subsonic, curve 6. An even further lowering of the backpressure from p6 to p7 will not affect the pressure to the left of the exit plane from the nozzle. It will remain at p4 6 p7. Under these conditions, at p4 the gas molecules are farther apart than when they are at the higher pressure p7, and so the gas to the left of the exit plane is said to be overexpanded. As a result, when the gas leaves the nozzle it will develop a series of oblique compression shock waves forming a

(a) p Compression waves formed p0 p*

B A

p7

1 2 3 Shock 5 at exit 6 7 4 8 Expansion waves formed x

}

}

p8

Throat

Exit

(b)

p5

Curve 5 (c)

p4

Curve 6 (d)

p6

p4 , p6

Fig. 13–36

772

C h a p t e r 13

Compressible Flow

series of shock diamonds outside the nozzle as the gas pressure rises to equal the backpressure, p7, Fig. 13–36e.

p4

•

When the backpressure is lowered to p4, it has reached the isentropic design condition for the nozzle, with subsonic flow through the converging section, sonic flow at the throat, and supersonic flow through the divergent section, curve 4. No shock will be produced, and so no energy is lost. Flow efficiency is at a maximum.

•

A further reduction in backpressure to p8 will cause the flow within the divergent section to be underexpanded, since the pressure to the left of the exit plane of the nozzle, p4, will now be higher than the backpressure, p4 7 p8. As a result the gas will undergo a series of expansion shock waves, again forming a pattern of shock diamonds outside the nozzle until the pressure equals the backpressure, Fig. 13–36f.

p4 , p7

13

Overexpanded flow backpressure, curve 7 (e)

p4

p4 . p8 Underexpanded flow backpressure, curve 8 (f)

Fig. 13–36 (cont.)

The effect of overexpanded and underexpanded flow is discussed further in Sec. 13.12. Also, these details are treated more thoroughly in books related to gas dynamics. For example, see Ref. [3].

IMPORTANT POINTS • A shock wave is very thin. It is a nonisentropic process that causes the entropy to increase due to frictional effects within the wave. Because the process is adiabatic, no heat is gained or lost across the shock, and so the stagnation temperature on each side of the shock is the same. The stagnation pressure and density, however, will be larger in front of a standing shock, due to the change in entropy.

• If the Mach number M1 of the flow behind a standing shock is known, the Mach number M2 in front of the shock can be determined. Also, if the temperature, pressure, and density, T1, p1, r1, are known behind the shock, then the corresponding values T2, p2, r2 in front of the shock can be found.

• A standing shock wave always requires supersonic flow behind it and subsonic flow in front of it. The opposite effect cannot occur, since it would violate the second law of thermodynamics.

• A converging–diverging or Laval nozzle will be most efficient when it operates at a backpressure that produces isentropic flow, with M = 1 in the throat and supersonic flow at the exit plane, curve 4, Fig. 13–36b. This is a design condition in which no heat transfer or friction loss will occur.

• Initially when a nozzle becomes choked, then M = 1 at the throat and the flow is subsonic at the exit plane, curve 3. A further slight lowering of the backpressure will cause a standing shock wave to form within the divergent section of the nozzle, curve 5. Further lowering of the backpressure to p6 will cause this wave to move toward, and finally reach, the exit plane, curve 6.

• A further drop in backpressure to p7 results in oblique shock waves forming off the edges of the nozzle, due to overexpansion of the gas. When the backpressure is lowered to p4, then supersonic isentropic flow occurs at the exit of the nozzle. And finally, if the backpressure is lowered to p8, expansion waves will form on the edge of the nozzle, creating conditions of underexpansion.

13.10 shoCk waves in nozzles

EXAMPLE

13.16

The pipe in Fig. 13–37 transports air at a temperature of 20°C having an absolute pressure of 30 kPa and a speed of 550 m>s, measured just behind a standing shock wave. Determine the temperature, pressure, and speed of the air just in front of the wave.

13

1 550 ms Back

2 V2 Front

SOLUTION Fluid Description. Shock wave formation is an adiabatic process. Steady flow occurs in back and in front of the wave.

p1 5 30 kPa T1 5 20°C

Fig. 13–37

Analysis. The control volume contains the shock, as shown in Fig. 13–37. To use the necessary equations or Table B–4, we must first determine M1. We have M1 =

2kRT1 V1

773

550 m>s

=

21.41286.9 J>kg # K2(273 + 20) K

= 1.6032

Since M1 7 1, we expect M2 6 1 and for both the pressure and temperature to increase in front of the wave. Using M1 = 1.6032, the value of M2 in front of the shock (subsonic) and the ratios of pressure and temperature are found from Eqs. 13–73, 13–68, and 13–69. They are M2 = 0.66746 p2 = 2.8321 p1 T2 = 1.3902 T1 Thus, p2 = 2.8321(30 kPa) = 85.0 kPa

Ans.

T2 = 1.3902(273 + 20) K = 407.3 K

Ans.

The speed of the air in front of the shock wave can be determined from Eq. 13–71, or, since T2 is known, we can use Eq. 13–22.

V2 = M2 2kRT2 = 0.6674621.41286.9 J>kg # K2(407.3 K) = 270 m>s Ans.

774

C h a p t e r 13

EXAMPLE

13

Compressible Flow

13.17 A jet plane is traveling at M = 1.5, where the absolute air pressure is 50 kPa and the temperature is 8°C. At this speed, a shock forms at the intake of the engine, as shown in Fig. 13–38a. Determine the pressure and velocity of the air just to the right of the shock. M 5 1.5 Back

p1 5 50 kPa T1 5 8°C

Front M2

M1 5 1.5 1

2 (b)

(a)

Fig. 13–38

SOLUTION Fluid Description. The control volume that contains the shock moves with the engine so that steady flow occurs through the open control surfaces, Fig. 13–38b. An adiabatic process occurs within the shock.

Analysis. The velocity V2 will be obtained from V2 = M2 2kRT2, and so we must first obtain M2 and T2. Since k = 1.4, using Eqs. 13–73, 13–76, and 13–69, or Table B–4, for M1 = 1.5 (supersonic), we have M2 = 0.70109 Subsonic p2 = 2.4583 p1 T2 = 1.3202 T1 Therefore, just to the right of the shock, p2 = 2.4583(50 kPa) = 123 kPa

Ans.

T2 = 1.3202(273 + 8) K = 370.98 K Thus, relative to the engine, the velocity of the air is

V2 = M2 2kRT2 = (0.70109) 21.41286.9 J>kg # K2(370.98 K)

V2 = 271 m>s

Ans.

775

13.10 shoCk waves in nozzles

EXAMPLE

13.18

The nozzle in Fig. 13–39a is connected to a large reservoir where the absolute air pressure is 350 kPa. Determine the range of outside backpressures that cause a shock wave to form within the nozzle and just outside of it.

125 mm 250 mm

13

p 5 350 kPa

SOLUTION Fluid Description.

(a)

We assume steady flow through the nozzle.

Analysis. First we will establish the backpressures that produce subsonic and supersonic isentropic flow through the nozzle, curves 3 and 4 in Fig. 13–39b. The area ratio between the exit and the throat of the nozzle (divergent section) is A>A* = p10.125 m2 2 >p10.0625 m2 2 = 4. This ratio gives two values for the Mach number at the exit using Eq. 13–36 or Table B–1 (k = 1.4). Choosing the isentropic subsonic flow within the divergent section (curve 3) for A>A* = 4, we get M ≈ 0.1465 6 1 and p>p0 = 0.9851. Since there is no shock wave, the stagnation pressure throughout the flow is 350 kPa. Then, at the exit, p3 = 0.9851(350 kPa) = 345 kPa. In other words, this backpressure will cause M = 1 at the throat and subsonic isentropic flow of M = 0.1465 at the exit. For isentropic supersonic flow within the divergent region (curve 4), from Eq. 13–36, for A>A* = 4, we get M = 2.9402 7 1 and p>p0 = 0.02979. Thus, at the exit, p4 = 0.02979(350 kPa) =10.43 kPa = 10.4 kPa. This pressure is lower than the previous value, since it must produce the required supersonic flow at the exit of M = 2.9402. The two solutions are for isentropic flow with backpressures that produce no shock within the nozzle; however, for both conditions the nozzle is choked since M = 1 at the throat. If a standing shock is formed at the exit of the nozzle, Fig. 13–39b, curve 6, then this will occur when the pressure in the nozzle at the exit, to the left or behind the shock, is 10.43 kPa. To find the backpressure in front of the shock, that is, just outside the exit plane, we must use Eq. 13–76 or Table B–4 with M = 2.9402, in which case p6 >p4 = 9.9187, so that p6 = 9.9187, p4 = 9.9187(10.43 kPa) = 103 kPa. Thus, a normal shock is created within the divergent section (such as curve 5, which is between curves 3 and 6) when the backpressure is in the following range: 103 kPa 6 pb 6 345 kPa

Ans.

For compression waves to occur at the exit, curve 7, the backpressure (between curves 6 and 4) should be in the range 10.4 kPa 6 pb 6 103 kPa Ans. Finally, expansion waves will form if the backpressure is anywhere below curve 4 (as in curve 8), that is, pb 6 10.4 kPa Ans.

p

3 (345 kPa) 1

350 kPa p*

5 6 (103 kPa) 7 8

Throat

Exit

(b)

Fig. 13–39

4 (10.4 kPa) x

776

C h a p t e r 13

Compressible Flow

13.11 OBLIQUE SHOCK WAVES In Sec. 13.2 we showed that when a jet plane or other fast-moving body encounters the surrounding air in front of it, the pressure created by the plane pushes the air to flow around it. At subsonic speeds, M 6 1, the streamlines adjust and follow the contour of the surface, Fig. 13–40a. However, at supersonic speeds, M Ú 1, the pressure created near the front surface cannot communicate to the air upstream fast enough to move out of the way. Instead the air molecules bunch up and create an oblique shock wave. This shock begins to bend just in front of the surface, and initially it is detached from the surface, Fig. 13–40b. At higher speeds, and if the plane has a sharp nose, Fig. 13–40c, the shock can become attached to the surface, causing the wave to turn at a sharp angle b. As the speed increases further, this angle will continue to decrease. Farther removed from the surface, the effect of this shock weakens and develops into a Mach cone having an angle a that travels at M = 1 through the atmosphere, Fig. 13–41. The result of all this will alter the direction of the streamlines for the flow. Near the origin of the oblique shock, the streamlines are turned the most, as they become almost parallel to the surface of the plane.* Farther away they essentially remain unchanged when they pass through the weaker Mach cone, Fig. 13–41. Oblique shock waves can be studied in the same manner as normal shocks, although here the change in direction of the streamlines becomes important. To analyze the situation, we will use two angles to define the geometry. As shown in Fig. 13–42a, b defines the angle of the shock wave, and u defines the angle of the deflected streamline or the direction of the velocity V2 in front of the shock. Here we will reference the flow normal and tangential to the wave. Resolving V1 and V2 into their n and t components, we have

13 M,1 (a)

Detached oblique or bow shock wave (blunt-nosed body) M.1 (b)

b

Attached shock wave (sharp-nosed body)

V1n = V1 sin b

M.1

V1t = V1 cos b

(c)

M.1 b

Oblique shock M.1

Mach cone M51

Mach cone M51

Fig. 13–41

M1n = M1 sin b M1t = M1 cos b

M2n = M2 sin 1b - u2

M2t = M2 cos 1b - u2

(13–78) (13–79)

For analysis, we will consider a standing wave and select a fixed control volume that includes an arbitrary portion of the wave, having a front (on the right) and back (on the left), each with an area A, Fig. 13–42a.

M.1

M.1

V2t = V2 cos 1b - u2

Or, since M = V>c, we can also write

Fig. 13–40 b5a

V2n = V2 sin 1b - u2

*At supersonic speeds the boundary layer is very thin, and so it has little effect on the direction of the streamlines.

777

13.11 oblique shoCk waves

Continuity Equation. Since the flow is steady, measured relative to the surface, and no flow is assumed to occur through the wave in the t direction, we have

t

0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

V2n V1t

V1n

(13–80)

Momentum Equation. As shown on the free-body diagram of the

13

V2t

b2u

u (deflection angle)

b (shock angle)

control volume, Fig. 13–42b, the pressure only acts in the n direction. Also, because the wave is very thin, the flow rVf>cs # A is caused only by the normal components of V1 and V2. Therefore, applying the momentum equation in the t direction, we have

d

+ g Ft =

u

V2

b V1

0 - r1V1n A + r2V2n A = 0 r1V1n = r2V2n

Streamline

n

(a) t

0 V r dV + Vt rVf>cs # dA 0t Lcv t Lcs

n p1A

0 = 0 + V1t ( -r1V1n A) + V2t (r2V2n A)

p2 A

Using Eq. 13–80, we obtain V1t = V2t = Vt

Free-body diagram

In other words, the tangential velocity components remain unchanged on each side of the shock. In the n direction, we have

d

+ g Fn =

0 V r dV + VnrVf>cs # dA 0t Lcv n Lcs

p1A - p2 A = 0 + V1n( -r1V1n A) + V2n(r2V2n A) or p1 + r1V1n2 = p2 + r2V2n2

(13–81)

Energy Equation. Applying the energy equation, neglecting the effect of gravity, and assuming an adiabatic process occurs, we have 2 # # # Vout V in2 # Qin - Wturbine + Wpump = c a hout + + gzout b - a hin + + gzin b d m 2 2

0 - 0 = c a h2 +

V2n2 + V2t2 V1n2 + V1t2 # + 0b - a h1 + + 0b d m 2 2

(b)

Fig. 13–42

778

C h a p t e r 13

Compressible Flow

Since V1t = V2t, we get h1 +

V 1n2 V 2n2 = h2 + 2 2

(13–82)

Equations 13–80, 13–81, and 13–82 are the same as Eqs. 13–66, 13–67, and 13–24. As a result, for oblique shocks we can describe the flow in the normal direction using the normal shock equations (Table B–4) developed previously. These equations become

13

M1n2 + M2n2 =

A schlieren image showing the development of oblique shocks formed on a model that is being tested in a wind tunnel (© L. Weinstein/Science Source)

2 k - 1

(13–83)

2k M2 - 1 k - 1 1n

p2 2k k - 1 = M1n2 p1 k + 1 k + 1

1 p0 2 2 = 1 p0 2 1

(13–84)

k>(k - 1) k + 1 M1n2 2 ≥ ¥ k - 1 2 1 + M1n 2

2k k - 1 1>(k - 1) c M1n2 d k + 1 k + 1

k - 1 M 1n2 T2 2 = T1 k - 1 1 + M 2n2 2

(13–85)

1 +

(13–86)

1>2 k - 1 M2n2 r2 M1n V1n 2 = ≥ ¥ (13–87) = r1 V2n M2n k - 1 2 1 + M1n 2 If M1 and M2 on each side of the shock wave are both supersonic, and the component M1n is also supersonic, then M2n must be subsonic, so it does not violate the second law of thermodynamics. For many practical problems, the initial flow properties V1, p1, T1, r1 and the angle u will be known. We can relate the angles u and b and the Mach number M1 in the following manner. Since V1t = V2t, then by Eq. 13–22,

1 +

50° M1

Deflection angle u

40°

10 6 4 3 2.5 2 1.8 1.6 1.4 1.2

30° 20° 10° 0° 0°

u 5 umax

`

30°

60° Shock angle b

M2 5 1

Oblique deflection vs. wave angle (k 5 1.4)

Fig. 13–43

M2 . 1 M2 , 1

Strong shock Weak shock 90°

M1t 2kRT1 = M2t 2kRT2 M2t = M1t

T1 A T2

13.11 oblique shoCk waves

779

Like velocity, in Fig. 13–42a, M2n = M2t tan (b - u). Therefore, M2n = M1t

T1 T1 tan1b - u2 = M1 cos b tan1b - u2 A T2 A T2

Squaring this equation and substituting the temperature ratio, Eq. 13–86, we obtain k - 1 M2n2 1 + 2 2 2 2 M2n = M1 cos b ± ≤ tan2 1b - u2 k - 1 1 + M1n2 2 Combining this equation with Eq. 13–83 and Eq. 13–78, and noting that in Fig. 13–42a, like velocity, M1n = M1 sin b, we get our final result. tan u =

2 cot b1M12 sin2 b - 12 M12 1k + cos 2b2 + 2

(13–88)

A plot of this equation, for several different values of M1, produces curves of u vs. b, shown in Fig. 13–43. Notice, for example, that for the curve M1 = 2, when the deflection angle u = 20°, there are two values for the shock angle b. The lower value b = 53° corresponds to a weak shock, and the higher value b = 74° to a strong shock. Most often a weak shock will form before a strong shock because its pressure ratio will be smaller.* Also, notice that solutions are not possible for all deflection angles u. For M2 = 0, the maximum deflection angle is approximately umax = 25°. For an angle larger than this, the wave will become detached from the surface and instead form in front of it, producing a higher drag, Fig. 13–40b. For the extreme case of no deflection, u = 0°, Eq. 13–88 gives b = sin-1 11>M1 2 = a, which produces a Mach cone, Fig. 13–41.

IMPORTANT POIN T S • An oblique shock wave will form on the front surface of a body

traveling at M Ú 1. As the speed increases, the wave will begin to attach itself to the surface of the body, and the shock angle b will begin to decrease. Farther from the localized region of the wave, a Mach cone will form and will travel at M = 1.

• The tangential component of velocity or its Mach number for an oblique shock remains the same on each side of the wave. The normal component can be analyzed using the same equations as for normal shocks. An equation is also available to determine the deflection angle u of the streamlines that pass through the wave. *An increase in pressure can occur downstream if the flow is blocked by a sudden change in the shape of the surface. If this occurs, the pressure ratio will be higher and a strong shock will be produced.

13

780

C h a p t e r 13

EXAMPLE

Compressible Flow

13.19

845 ms

13 Back Front 40°

A jet plane is flying horizontally at 845 m>s, at an altitude where the air temperature is 10°C and the absolute pressure is 80 kPa. If an oblique shock forms on the nose of the plane, at the angle shown in Fig. 13–44a, determine the pressure and temperature, and the direction, of the air just behind the shock. SOLUTION Fluid Description. The air is considered compressible, and the shock is an adiabatic nonisentropic process. Steady flow occurs as viewed from the plane.

(a) t

Analysis. M1t

40°

M1 5 2.5063

M1n

M1 =

The Mach number for the plane must first be determined.

845 m>s V1 V1 = = 2.5063 = c 2kRT1 21.41286.9 J>kg # K2(273 + 10) K

n

(b) t 40° 2 u 40°

M2 5 1.7563

Streamline

u

M2n 5 0.6651 n

T2 = 1.3956; T2 = 1.3956(273 + 10) K = 394.95 K = 395 K Ans. T1

(c)

Streamline

From the geometry shown in Fig. 13–44b, M1 is resolved into its normal and tangential components relative to the wave. The normal component is M1n = 2.5063 sin 40° = 1.6110. We can now use Table B–4 or Eqs. 13–83, 13–84, and 13–86 to obtain the speed, temperature, and pressure in front of the shock. We get M2n = 0.6651. This is subsonic, which is to be expected so as not to violate the second law of thermodynamics. Also,

M2 5 1.76 17.7°

p2 = 2.8619; p1

p2 = 2.8619(80 kPa) = 228.90 kPa = 229 kPa Ans.

The angle u, shown in Fig. 13–44c, can be obtained directly from Eq. 13–88, since M1 and b are known. tan u = (d)

Fig. 13–44

tan u =

2 cot b1M12sin2 b - 12 M12 1k + cos 2b2 + 2

2 cot 40°3 12.50632 2 sin2 40° - 14 (2.5063)2(1.4 + cos 2(40°22 + 2 u = 17.7°

The result is shown in Fig. 13–44d.

Ans.

13.12 Compression and expansion waves

781

13.12 COMPRESSION AND EXPANSION WAVES When an airfoil or other body is moving at supersonic speed, not only will it form an oblique shock at its front surface, but in addition, if the surface is curved, the flow must follow this surface, and in doing so it may also form compression or expansion waves as the flow is redirected. For example, in Fig. 13–45a the concave surface on the jet plane causes compression waves to form, which, when extended, become concurrent and merge into an oblique shock—something we studied in the previous section. However, if the surface is convex, then the air has room to expand as it follows the surface, and so a multitude of divergent expansion waves will form, creating a “fan” of an infinite succession of Mach waves, Fig. 13–45b. This behavior also occurs at a sharp corner, as shown in Fig. 13–45c. During this expansion process, the Mach number between each succeeding wave will increase, while the Mach angle a between each wave will decrease. Because the changes in the gas properties that are made through the formation of each of these waves are infinitesimal, the process of creating each wave can be considered isentropic, and for the entire “fan” of waves it is also isentropic. Oblique shock Mach wave Mach wave

Expansion waves (b)

Compression waves (a)

M1 M2 a u

Expansion waves (c)

Fig. 13–45

13

782

Streamline

C h a p t e r 13

Compressible Flow

a

Back Front

Deflection of streamline

V

a

du

13

u Mach angle (a)

Fig. 13–46

V 1 dV a 1 du

To study this wave fan, we will isolate one wave and assume it deflects the streamline that passes through the wave by the differential turning angle du, Fig. 13–46a. Our goal is to express du in terms of the Mach number M that is just in back of the wave. Here M1 7 1 and the wave acts at the Mach angle a. Resolving the velocities V and V + dV into their normal and tangential components to the wave, and realizing that the velocity components in the tangential direction remain the same, then Vt1 = Vt2 or V cos a = (V + dV ) cos (a + du) = (V + dV )(cos a cos du - sin a sin du) Since du is small, cos du ≈ 1 and sin du ≈ du. Therefore, our equation becomes dV>V = (tan a)du. Using Eq. 13–23, sin a = 1>M, and tan a is tan a =

sin a sin a 1 = = 2 cos a 21 - sin a 2M2 - 1

Therefore, dV du = V 2M2 - 1

(13–89)

Since V = M 2kRT, its derivative is

or

1 kR dV = dM 2kRT + M a b dT 2 2kRT dV dM 1 dT = + V M 2 T

(13–90)

For adiabatic flow, the relationship between the stagnation temperature and the static temperature is determined from Eq. 13–26, that is, T0 = Ta 1 +

k - 1 2 M b 2

Taking the derivative of this expression, and rearranging the terms, we obtain 2(k - 1)M dM dT = T 2 + (k - 1)M2

(13–91)

Finally, combining Eqs. 13–89, 13–90, and 13–91, the deflection angle for the wave can now be expressed in terms of the Mach number and its change. It is du =

22M2 - 1 dM 2 + (k - 1)M2 M

783

13.12 Compression and expansion waves

For a specified finite turning angle u, we integrate this expression between the initial and final waves, each having a different Mach number. Normally, the final Mach number for the flow is unknown, so it is more convenient to integrate this expression from a reference position, where at u = 0°, M = 1. Then from this reference, we can determine the turning angle u = v, through which the flow expands, for the Mach number to change from M = 1 to M. We have

L0

v=

13

M

v

du =

22M2 - 1 dM 2 L1 2 + (k - 1)M M

k+1 k-1 tan - 1 a 1M2 - 12 b - tan - 1 12M2 - 12 (13–92) Ak - 1 Ak + 1

This equation is referred to as the Prandtl–Meyer expansion function, named after Ludwig Prandtl and Theodore Meyer. With it we can find the total turning angle of the flow caused by the fan of isentropic expansion waves. For example, if the flow in Fig. 13–46b has an initial Mach number of M1 and the surface or streamline deflects u, then we can determine the resulting Mach number M2 by first applying Eq. 13–92 to obtain v1 for M1. Since u = v2 - v1, then v2 = v1 + u. With this angle, v2, we can then reapply Eq. 13–92 to determine M2. Although this will require a numerical procedure for the solution, it can also be solved by selecting a “calculator” on an Internet website, or by using tabular values for Eq. 13–92. Some of these values are listed in Appendix B, Table B–5.

IMPORTANT POIN T S • As the flow passes over a curve or edge of a surface, it can cause compression or expansion waves when M Ú 1.

• The compression waves merge into an oblique shock. • The expansion waves form a “fan” of an infinite succession of Mach waves. The streamline deflection angle caused by the expansion waves can be determined using the Prandtl–Meyer expansion function.

M1 M2 v1 v u 2

M51

v50

(b)

Fig. 13–46 (cont.)

784

C h a p t e r 13

EXAMPLE

13

Compressible Flow

13.20

V1 5 900 ms V2 170°

SOLUTION Fluid Description. We have steady compressible flow that expands isentropically. The initial or upstream Mach number for the flow is

(a)

M1

Air flows over a surface at a speed of 900 m>s, where the absolute pressure is 100 kPa and the temperature is 30°C. Expansion waves occur at the sharp transition shown in Fig. 13–47a. Determine the velocity, temperature, and pressure of the flow just to the right of the wave.

M2 u 5 10°

M1 =

900 m>s V1 V1 = = 2.5798 = c 2kRT 21.41286.9 J>kg # K2(273 + 30) K

(b)

Fig. 13–47

Analysis. Since the Prandtl–Meyer expansion function, Eq. 13–92, has been referenced from M = 1, the turning angle v1 from this reference is

v1 =

k + 1 k - 1 tan - 1 1M12 - 12 - tan - 1 2M12 - 1 Ak - 1 Ak + 1

(1)

1.4 + 1 1.4 - 1 tan - 1 1(2.5798)2 - 12 - tan - 1 2(2.5798)2 - 1 A 1.4 - 1 A 1.4 + 1 = 40.96°

=

This same value (or one close to it) can also be determined from Table B–5. Since the boundary layer over the inclined surface is very thin, the deflection angle for the streamlines is defined by the same angle as the deflection of the surface, that is, 180° - 170° = 10°, Fig. 13–47b. The downstream wave must therefore have a Mach number M2 that produces a turning angle of 40.96° + 10° = 50.96° from the M = 1 reference. Rather than using this angle and trying to solve for M2 using the Prandtl–Meyer function, Eq. 13–92, we can use an Internet website, or use the table with v2 = 50.96°. Using the more exact approach, we get M2 = 3.0631

13.12 Compression and expansion waves

In other words, the expansion waves increase the Mach number from M1 = 2.5798 to M2 = 3.0631. Realize, however, that this increase does not violate the second law of thermodynamics, because only the normal component of the wave converts from supersonic to subsonic. Since we have isentropic expansion, we can determine the temperature on the right side of the waves using Eq. 13–69.

T2 = T1

k - 1 1.4 - 1 1 + M12 (2.5798)2 2 2 = = 0.8104 k - 1 1.4 - 1 2 2 1 + M2 1 + (3.0631) 2 2

1 +

T2 = 0.8104(273 + 30) K = 245.54 K = 246 K

Ans.

We can also use Table B–1 as follows:

T2 T2 T0 1 = = (0.34764) a b = 0.8104 T1 T0 T1 0.42894 so that again T2 = 0.8104(273 + 30) K = 246 K. The pressure is obtained in a similar manner using Eq. 13–17, or Table B–1. Here p2 p2 p0 1 = = (0.024775) a b = 0.47915 p1 p0 p1 0.051706 so

p2 = 0.47915(100 kPa) = 47.9 kPa

Ans.

The velocity of the flow after the expansion is V2 = M2 2kRT2 = 3.063121.41286.9 J>kg # K2(245.54 K) = 962 m>s Ans.

785

13

786

C h a p t e r 13

Compressible Flow

13.13 COMPRESSIBLE FLOW MEASUREMENT Pressure and velocity in compressible gas flows can be measured in a variety of ways. Here we will discuss a few of them. 13

Pitot Tube and Piezometer. As in the case of incompressible flow, discussed in Sec. 5.3, a pitot-static tube, such as the one shown in Fig. 13–48a, can also be used for compressible flow measurement. The static pressure p within the flow is measured at the side opening of the tube, whereas the stagnation or total pressure p0 is measured at the stagnation point, which occurs at the front opening. At the stagnation point the flow comes rapidly to a standstill, with no significant heat or friction loss, and so the process can be assumed isentropic. related by Eq. 13–27. Since V = M 1kRT, solving for M, substituting this into Eq. 13–27, and then solving for V, we get

Subsonic Flow. For subsonic compressible flow, the pressures are

V =

2kRT p0 (k - 1)>k ca b - 1d Bk - 1 p

(13–93)

Provided we know the temperature T, we can then use this equation to determine the velocity of the flow. In practice, it is generally easier to measure the stagnation temperature T0 at the stagnation point, rather than T within the flow, since the flow is easily disturbed. The relationship used for this case can be determined by substituting cp = kR>(k - 1) and Eq. 13–17, where T = T0(p>p0)(k - 1)>k, into Eq. 13–93. We obtain p (k - 1)>k b d p0

(13–94) B Substitution of the measured quantities then gives the undisturbed, freeflow velocity of the gas. V =

2cpT0 c 1 - a

p p0

p p0

Pitot-static tube Subsonic flow (a)

Fig. 13–48

787

13.13 Compressible Flow measurement

Supersonic Flow. If the gas is flowing at supersonic speeds, then just before it strikes the nose of the pitot tube, because of the sudden change in pressure it will form a curved shock, Fig. 13–48b. The shock changes the flow from supersonic, point 1, to subsonic, point 2. Since the shock is locally normal to the stagnation streamline and the flow is brought to rest isentropically across the shock, then we can use Eq. 13–68 to relate the pressures p1 and p2 across the shock. Also, the relationship between p2 and the measured stagnation pressure p0 is determined using Eq. 13–27. We can combine these two equations and express the result in terms of the upstream Mach number using Eq. 13–73. Doing this, we get

p0 = p1

a

k + 1 2 k>(k - 1) M1 b a 2

2k k - 1 M2b k + 1 1 k + 1

1>(k - 1)

p0

p1

1 2

p p0

Pitot tube and piezometer Sonic or supersonic flow (b)

(13–95)

Fig. 13–48

The pressure p1 can be measured independently, using a piezometer on the boundary of the flow, well upstream of the shock, Fig. 13–48b. With p1 and p0 known, we can then obtain the Mach number M1 7 1 for the flow from Eq. 13–95. As an alternative procedure, rather than using a pitot tube and piezometer, we can measure the velocity of the gas using a hot-wire anemometer, which was described in Sec. 10.5.

Venturi Meter. Also in Sec. 10.5, we showed that if the flow has a

low velocity V … 0.3c, then the gas can be considered incompressible, and the mass flow through a venturi meter, Fig. 13–49, can be determined from # m = Cv A 2

B 1 - 1D2 >D1 2 4 2r1 1 p1 - p2 2

D1 1

For compressible flow this equation is modified to account for the change in density of the gas as it passes through the meter. This is done using an experimentally determined expansion factor, Y, so that then we have # m = CvYA2

B 31 - 1D2 >D1 2 4 4 2r1 1 p1 - p2 2

(13–96)

Values of the expansion factor are available from graphs for various pressure ratios. See Ref. [6]. Equation 13–97 can also be used with a flow nozzle and orifice meter since the basic principles of operation are the same.

D2 2

Fig. 13–49

13

788

C h a p t e r 13

Compressible Flow

References 1. 2. 3. 4.

13

5. 6. 7. 8.

J. E. A. John, Gas Dynamics, Prentice Hall, Upper Saddle River, NJ, 2005. F. M. White, Fluid Mechanics, McGraw-Hill, New York, NY, 2011. H. Liepmann, Elements of Gasdynamics, Dover, New York, NY, 2002. P. H. Oosthuizen and W. E. Carsvallen, Compressible Fluid Flow, McGraw-Hill, New York, NY, 2003. S. Schreier, Compressible Flow, Wiley-Interscience Publication, New York, NY, 1982. A. H. Shapiro, The Dynamics and Thermodynamics of Compressible Fluid Flow, John Wiley and Sons, Inc, New York, N. Y. W. B. Brower, Theory, Table, and Data for Compressible Flow, Taylor and Francis, New York, NY, 1990. R. Vos and S. Faroldhi, Introduction to Transonic Aerodynamics, Springer, New York, NY, 2015.

P ROB L EMS SEC. 13.1 13–1. Air flows in a horizontal duct at 20°C with a velocity of 180 m>s. If the velocity increases to 250 m>s, determine the corresponding temperature of the air. Hint: Use the energy equation to find ∆h. 13–2. A closed tank contains oxygen at 200°C and under an absolute pressure of 250 kPa If the temperature decreases to 150°C, determine the changes in density and pressure, and the change per unit mass in the internal energy and enthalpy of the oxygen. 13–3. If a pipe contains helium at a gage pressure of 100 kPa and a temperature of 20°C, determine the density of the helium. Also, determine the temperature if the helium is compressed isentropically to a gage pressure of 250 kPa. The atmospheric pressure is 101.3 kPa. *13–4. Oxygen is contained in a vessel under an absolute pressure of 850 kPa. If the temperature increases from 10°C to 60°C, determine the changes in pressure and entropy. 13–5. Nitrogen is compressed from an absolute pressure of 200 kPa to 500 kPa, with no change in temperature. Determine the changes in entropy and enthalpy. 13–6. As nitrogen flows from a point A to a point B its temperature increases from 10°C to 30°C and its absolute pressure decreases from 200 kPa to 175 kPa. Determine the change per unit mass in internal energy, enthalpy, and entropy between the two sections.

13–7. Air has a temperature of 15°C and an absolute pressure of 250 kPa at A. As it passes through the transition, its temperature becomes 40°C and the absolute pressure becomes 125 kPa at B. Determine the change in the density and the enthalpy of the air. *13–8. Air has a temperature of 20°C and an absolute pressure of 600 kPa at A. If the temperature at B is 5°C and the absolute pressure is 450 kPa, determine the changes in the enthalpy and the entropy of the air.

B A

Probs. 13–7/8 13–9. A closed tank contains helium at 200°C and under an absolute pressure of 530 kPa. If the temperature is increased to 250°C, determine the changes in density and pressure, and the change per unit mass of the internal energy and enthalpy of the helium. 13–10. As oxygen flows from a point A to a point B its temperature increases from 50°C to 65°C and its absolute pressure decreases from 300 kPa to 240 kPa. Determine the change per unit mass in internal energy, enthalpy, and entropy between the two sections.

problems

SEC. 13.2–13.4 13–11. A jet plane has a speed of 900 km>h when flying at an altitude of 3 km. Determine the Mach number. *13–12. A jet plane has a speed of M = 2 at an altitude of 10 km. How long does it take for the sound to travel to an observer on the ground after the plane passes directly overhead?

M52 a a

Prob. 13–12 13–13. Water is at a temperature of 10°C. If a sonar signal takes 3 s to detect a large whale, determine the distance from the whale to the ship. Take r = 1025 kg>m3 and EV = 2.42 GPa. 13–14. The Mach cone formed on the top of a rocket has a half angle of 15°. If the air temperature is -20°C, determine the speed of the rocket. 13–15. Determine the half angle a of the Mach cone and the speed of a jet plane that flies at M = 2.2 at an altitude of 5 km. *13–16. Determine the Mach number for a race car that travels at 360 km>h in 15°C weather. 13–17. Determine the speed of sound in water and in air, both at a temperature of 20°C. Take EV = 2.24 GPa for water. 13–18. Compare the speed of sound in water and air at a temperature of 20°C. The bulk modulus of water at T = 20°C is EV = 2.2 GPa. 13–19. How fast in kilometers per hour must a jet plane fly at an altitude of 8 km in order to have a Mach number of M = 1.5? *13–20. The Mach number of the air flow out the nozzle at B is M = 0.4. If the air has a temperature of 8°C and absolute pressure of 20 kPa, determine the absolute pressure and temperature of the air within the large reservoir at A.

13–21. A 300-mm-diameter pipe transports air at M = 1.6. If the stagnation temperature is 25°C, and the mass flow through the pipe is 60 kg>s, determine the density of the air and the stagnation pressure. 13–22. Air at a temperature of 30°C and an absolute pressure of 680 kPa flows through a 200-mm-diameter duct. 13 If the Mach number is M = 0.42, determine the mass flow. 13–23. A ship is located where the depth of the ocean is 3 km. Determine the time needed for a sonar signal to reflect off the bottom and return to the ship. Assume the average water temperature is 10°C. Take r = 1030 kg>m3 and EV = 2.111109 2 Pa for sea water.

*13–24. A jet plane passes 5 km directly overhead. If the sound of the plane is heard 6 s later, determine the speed of the plane. The average air temperature is 10°C.

13–25. The absolute stagnation pressure for methane is 750 kPa when the stagnation temperature is 20°C. If the pressure in the flow is 550 kPa, determine the corresponding velocity of the flow. 13–26. A 300-mm-diameter pipe transports nitrogen at M = 0.85. If the absolute pressure is 280 kPa and the temperature is 10°C, determine the mass flow through the pipe and the stagnation density.

SEC. 13.5–13.6 13–27. The tank contains oxygen at a temperature of 70°C and absolute pressure of 800 kPa. If the converging nozzle at the exit has a diameter of 6 mm, determine the initial mass flow out of the tank if the outside absolute pressure is 100 kPa. *13–28. The tank contains helium at a temperature of 80°C and absolute pressure of 175 kPa. If the converging nozzle at the exit has a diameter of 6 mm, determine the initial mass flow out of the tank if the outside absolute pressure is 98 kPa.

6 mm

B A

Prob. 13–20

789

Probs. 13–27/28

790

C h a p t e r 13

Compressible Flow

13–29. A converging nozzle having an exit diameter of 30 mm is connected to the large tank. If the temperature of the air in the tank is 20°C and the absolute pressure is 580 kPa, determine the mass flow from the tank. The absolute pressure outside the tank is 102 kPa. 13 13–30. A converging nozzle having an exit diameter of 30 mm is connected to the large tank. If the temperature of the air in the tank is 60°C and at the exit it is 10°C, determine the absolute pressure within the tank and the mass flow of the air. The absolute pressure outside the tank is 102 kPa.

*13–32. If the fuel mixture within the chamber of the rocket is under an absolute pressure of 1.30 MPa, determine the Mach number of the exhaust if the area ratio of the exit to the throat is 2.5. Assume that fully expanded supersonic flow occurs. Take k = 1.40 and R = 286.9 J>kg # K for the fuel mixture. The atmosphere has a pressure of 101.3 kPa.

Prob. 13–32 13–33. The diameter of the exit of a converging nozzle is 50 mm. If its entrance is connected to a large tank containing air at an absolute pressure of 180 kPa and temperature of 125°C, determine the mass flow from the tank. The ambient air is at an absolute pressure of 101.3 kPa. 13–34. Determine the greatest possible mass flow through the nozzle if the throat has a diameter of 50 mm. The air in the reservoir has an absolute pressure of 400 kPa and a temperature of 30°C.

Probs. 13–29/30

13–31. If the Laval nozzle is open to the atmosphere where the absolute pressure is 102 kPa, determine the required absolute pressure of the air in the tank so that isentropic supersonic flow occurs through the divergent section of the nozzle. The nozzle has an outer diameter of 30 mm and the throat has a diameter of 20 mm.

50 mm

20 mm

30 mm

Prob. 13–31

Prob. 13–34

791

problems 13–35. Air flows at 200 m>s through the pipe. Determine the Mach number of the flow and the mass flow if the temperature is 500 K and the absolute stagnation pressure is 200 kPa. Assume isentropic flow. *13–36. Air flows at 200 m>s through the pipe. Determine the pressure within the flow if the temperature is 400 K and the absolute stagnation pressure is 280 kPa. Assume isentropic flow.

0.3 m

13–39. Air flows through the nozzle at 2 kg>s, and the absolute pressure and temperature of the air at A are 650 kPa and 350 K, respectively. Determine the pressure at B to maintain isentropic flow in the divergent section of the nozzle for both subsonic and supersonic flow at B. Also determine the Mach number at the throat C and at B. *13–40. Air flows through the nozzle at 2 kg>s, and the absolute pressure and temperature of the air at A are 650 kPa and 350 K, respectively. Determine the temperature at B to maintain isentropic flow in the divergent section of the nozzle for both subsonic and supersonic flow at B.

Probs. 13–35/36 80 mm

50 mm

13–37. The large tank contains air at an absolute pressure of 150 kPa and temperature of 20°C. Air flows out of the tank through the 5-mm-diameter nozzle at A. Determine the mass flow and the horizontal force that must be applied to the tank to prevent it from moving. The atmospheric pressure is 100 kPa.

20 mm

C A

B

Probs. 13–39/40

A

13–41. Nitrogen in the reservoir is at a temperature of 20°C and an absolute pressure of 300 kPa. Determine the mass flow through the nozzle. The atmospheric pressure is 100 kPa.

F

Prob. 13–37

13–38. If the Laval nozzle is open to the atmosphere where the absolute pressure is 102 kPa, determine the required pressure of the air in the tank so that isentropic subsonic flow occurs through the divergent section of the nozzle. The nozzle has an outer diameter of 30 mm and the throat has a diameter of 20 mm.

20 mm 10 mm 30 mm

Prob. 13–38

Prob. 13–41

13

792

C h a p t e r 13

Compressible Flow

13–42. The large tank contains air at an absolute pressure of 700 kPa and temperature of 400 K. Determine the mass flow from the tank into the pipe if the converging nozzle has an exit diameter of 40 mm and the absolute pressure in the pipe is 150 kPa. 13 13–43. The large tank contains air at an absolute pressure of 700 kPa and temperature of 400 K. Determine the mass flow from the tank into the pipe if the converging nozzle has an exit diameter of 40 mm and the absolute pressure in the pipe is 400 kPa.

13–46. The large tank contains air at an absolute pressure of 400 kPa, and the temperature is 20°C. If the pressure at the entrance A of the nozzle is 300 kPa, determine the mass flow out of the tank. 13–47. Atmospheric air at an absolute pressure of 103 kPa and temperature of 20°C flows through the converging nozzle into the tank, where the absolute pressure at A is 30 kPa. Determine the mass flow into the tank.

A 40 mm

Probs. 13–46/47

*13–48. The large cylindrical tank contains air at an absolute pressure of 1200 kPa and a temperature of 45°C. The throat of the nozzle has a diameter of 20 mm, and the exit diameter is 50 mm. Determine the absolute pressure in the pipe required to produce isentropic supersonic flow through the pipe. What is the Mach number of this flow?

Probs. 13–42/43

*13–44. Air flows into the nozzle at MA = 0.2 and isentropically out at MB = 2. If the diameter of the nozzle at A is 30 mm, determine the diameter of the throat and the diameter at B. Also, if the absolute pressure at A is 300 kPa, determine the stagnation pressure and the pressure at B.

13–49. The large cylindrical tank contains air at an absolute pressure of 1200 kPa and a temperature of 45°C. The throat of the nozzle has a diameter of 20 mm, and the exit diameter is 50 mm. Determine the absolute pressure in the pipe required to choke the nozzle, and also maintain isentropic subsonic flow through the pipe. What is the velocity of the flow through the pipe for this condition?

13–45. Air flows into the nozzle at MA = 0.2 and isentropically out at MB = 2. If the diameter of the nozzle at A is 30 mm, determine the diameter of the throat and the diameter at B. Also, if the temperature at A is 300 K, determine the stagnation temperature and the temperature at B.

d

30 mm dt

A

B

Probs. 13–44/45

Probs. 13–48/49

problems 13–50. The large tank contains air at an absolute pressure of 420 kPa and a temperature of 20°C. The nozzle has a throat diameter of 20 mm and an exit diameter of 35 mm. Determine the absolute temperature and pressure within the connected pipe so that the nozzle chokes, but also maintains isentropic subsonic flow within the divergent section of the nozzle. Also, what is the mass flow into the tank if the absolute pressure within the pipe is 200 kPa? 13–51. The large tank contains air at an absolute pressure of 420 kPa and temperature of 20°C. The nozzle has a throat diameter of 20 mm and an exit diameter of 35 mm. Determine the absolute temperature and pressure within the connected pipe, and the corresponding mass flow through the pipe, when the nozzle chokes and maintains isentropic supersonic flow within the divergent section of the nozzle.

793

13–55. Nitrogen flows through the pipe. If the stagnation temperature is 10°C, the stagnation pressure is 1500 kPa, and the absolute pressure is 1250 kPa, determine the mass flow. 13

200 mm

Prob. 13–55

*13–56. The large tank contains air at 250 K under an absolute pressure of 1.20 MPa. When the valve is opened, the nozzle chokes. The outside atmospheric pressure is 101.3 kPa. Determine the mass flow from the tank. The nozzle has an exit diameter of 40 mm and a throat diameter of 20 mm. Probs. 13–50/51 *13–52. Air at a temperature of 20°C and atmospheric pressure of 102 kPa flows through the nozzle into the pipe where the absolute internal pressure is 45 kPa. Determine the throat diameter of the nozzle so that the mass flow into the pipe is 20 g>s.

13–57. The large tank contains air at 250 K under an absolute pressure of 150 kPa. When the valve is opened, determine if the nozzle is choked. The outside atmospheric pressure is 90 kPa. Determine the mass flow from the tank. Assume the flow is isentropic. The nozzle has an exit diameter of 40 mm and a throat diameter of 20 mm.

13–53. Air at a temperature of 25°C and standard atmospheric pressure of 101.3 kPa flows through the nozzle into the pipe where the absolute internal pressure is 80 kPa. Determine the mass flow into the pipe. The nozzle has a throat diameter of d = 10 mm. 13–54. Air at a temperature of 20°C and atmospheric pressure of 102 kPa flows through the nozzle into the pipe where the absolute internal pressure is 60 kPa. Determine the mass flow into the pipe. The nozzle has a throat diameter of d = 15 mm.

d

Probs. 13–52/53/54

Probs. 13–56/57

794

C h a p t e r 13

Compressible Flow

13–58. The converging–diverging nozzle at the end of a supersonic jet engine is to be designed to operate efficiently when the absolute outside air pressure is 25 kPa. If the absolute stagnation pressure within the engine is 400 kPa and the stagnation temperature is 1200 K, determine the exit diameter and the throat diameter for the nozzle if the 13 mass flow is 15 kg>s. Take k = 1.40 and R = 256 J>kg # K. 13–59. The large tank contains air at an absolute pressure of 680 kPa and a temperature of 85°C. If the diameter at the end of the converging nozzle is 15 mm, determine the mass flow out of the tank where the standard atmospheric pressure is 101.3 kPa.

SEC. 13.7–13.8 13–62. Nitrogen having a temperature of T1 = 270 K and absolute pressure of p1 = 330 kPa flows into the pipe at M1 = 0.3. If it is heated at 100 kJ>kg # m, determine the velocity and pressure of the nitrogen when it exits the pipe at section 2. 13–63. Nitrogen having a temperature of T1 = 270 K and absolute pressure of p1 = 330 kPa flows into the pipe at M1 = 0.3. If it is heated at 100 kJ>kg # m, determine the stagnation temperatures at sections 1 and 2, and the change in entropy per unit mass between these two sections.

1

2

15 mm

4m

Probs. 13–62/63

Prob. 13–59

*13–60. Air has an absolute pressure of 400 kPa and flows through the pipe at A at M = 0.5. Determine the Mach number at the throat of the nozzle where dt = 110 mm, and the Mach number in the pipe at B. Also, what is the stagnation pressure and the pressure in the pipe at B? 13–61. Natural gas (methane) has an absolute pressure of 400 kPa and flows through the pipe at A at M = 0.1. Determine the diameter of the throat of the nozzle so that M = 1 at the throat. Also, what is the stagnation pressure, the pressure at the throat, and the subsonic and supersonic Mach numbers for isentropic flow through pipe B?

*13–64. The 80-mm-diameter pipe has an average friction factor of f = 0.005. A nozzle on the large tank A delivers air to the pipe at section 1 with M1 = 1.8, a temperature of 40°C, and an absolute pressure of 1050 kPa. Determine the mass flow. Show that a normal shock forms within the pipe if L0 = 5 m. 13–65. The 80-mm-diameter pipe has an average friction factor of f = 0.005. The nozzle on the large tank A delivers air to the pipe at section 1 with M1 = 1.8 and a temperature of 40°C. Determine the velocity and temperature of the air at L = 2 m if L0 = 3 m.

200 mm 120 mm

A

B

dt

80 mm 2

1

B

A C

L L0

Probs. 13–60/61

Probs. 13–64/65

problems 13–66. Air from a large reservoir is at a temperature of 300 K and an absolute pressure of 200 kPa. If 65 kJ>kg of heat is added to the air as it flows from 1 to 2, determine the density and pressure at section 1 if the duct is choked at section 2. The backpressure at 2 causes M1 6 1. 13–67. Air from a large reservoir is at a temperature of 300 K and an absolute pressure of 200 kPa. If 65 kJ>kg of heat is added to the air as it flows from 1 to 2, determine the temperature and pressure at 2 if it is choked there. The backpressure at 2 causes M1 6 1.

795

13–70. Air in a large room has a temperature of 24°C and absolute pressure of 101 kPa. If it is drawn into the 200-mm-diameter duct such that the absolute pressure at section 1 is 90 kPa, determine the critical length of duct Lmax where the flow becomes choked, and the Mach number, temperature, and pressure at section 2. Take the 13 average friction factor to be f = 0.002. 13–71. Air in a large room has a temperature of 24°C and absolute pressure of 101 kPa. If it is drawn into the 200-mm-diameter duct such that the absolute pressure at section 1 is 90 kPa, determine the mass flow through the duct and calculate the resultant friction force acting on the duct. Also, what is the required length Lmax to choke the flow? Take the average friction factor to be f = 0.002.

200 mm 1 1

2

2 150 m Lmax

Probs. 13–70/71

Probs. 13–66/67

*13–68. Air is drawn into the pipe at M1 = 1.85, T1 = 60°C, and an absolute pressure of p1 = 600 kPa. If it exits the pipe at M2 = 1.15, determine the amount of heat per unit mass that is absorbed or released by the air. 13–69. Air is drawn into the 200-mm-diameter pipe at M1 = 1.85, T1 = 60°C, and an absolute pressure of p1 = 600 kPa. If it exits the pipe at M2 = 1.15, determine the stagnation temperatures at sections 1 and 2, and the change in entropy per unit mass between these sections due to uniform heating along the pipe.

*13–72. A large reservoir contains air at a temperature of T = 20°C and absolute pressure of p = 300 kPa. The air flows through the 1.5-m-long, 50-mm-diameter pipe having an average friction factor of 0.03. Determine the mass flow and the corresponding velocity, pressure, and temperature at the inlet 1 and outlet 2 if the flow is choked at section 2. 13–73. A large reservoir contains air at a temperature of T = 20°C and absolute pressure of p = 300 kPa. The air flows through the 1.5-m-long, 50-mm-diameter pipe having an average friction factor of 0.03. Determine the stagnation temperature and stagnation pressure at outlet 2 and the change in entropy between the inlet 1 and outlet 2 if the pipe is choked at section 2.

50 mm 1 2

2

1 1.5 m

Probs. 13–68/69

Probs. 13–72/73

796

C h a p t e r 13

Compressible Flow

13–74. Outside air at a temperature of 25°C is drawn into the duct and then heated along the duct at 130 kJ>kg. At section l the temperature is T = 15°C and the absolute pressure is 98 kPa. Determine the Mach number, temperature, and pressure at section 2. Neglect friction.

13–78. Air having a temperature of 280 K and pressure of 320 kPa flows from a large reservoir into the duct. As it flows, 122 kJ>kg of heat is added. Determine the greatest velocity it can have at section 1. The backpressure at 2 causes M1 6 1.

13 13–75. Outside air at a temperature of 25°C is drawn into the duct and then heated along the duct at 130 kJ>kg. At section 1 the temperature is T = 15°C and the absolute pressure is 98 kPa. Determine the mass flow and the change in internal energy between sections 1 and 2.

13–79. Air having an absolute temperature of 450 K and pressure of 600 kPa flows from a large reservoir into a duct. As it flows, 150 kJ>kg of heat is added. Determine the mass flow if the backpressure causes M1 7 1 and the flow chokes at section 2.

200 mm

50 mm 1

2

1

2

2m

Probs. 13–74/75

Probs. 13–78/79

*13–76. The duct has a diameter of 150 mm. If the average friction factor is f = 0.004, and air is drawn into the duct with an inlet velocity of 125 m>s, a temperature of 285 K, and an absolute pressure of 165 kPa, determine these properties at the exit.

*13–80. The 100-mm-diameter pipe is connected by a nozzle to a large reservoir of air that is at a temperature of 40°C and absolute pressure of 450 kPa. If the backpressure causes M1 7 1, and the flow is choked at the exit, section 2, when L = 5 m, determine the mass flow through the pipe. Assume an average friction factor of 0.0085 throughout the pipe.

13–77. The duct has a diameter of 150 mm. If the average friction factor is f = 0.004, and air is drawn into the duct with an inlet velocity of 125 m>s, a temperature of 285 K, and an absolute pressure of 165 kPa, determine the mass flow through the duct and the resultant friction force acting on the 90-m length of duct.

13–81. The 100-mm-diameter pipe is connected by a nozzle to a large reservoir of air that is at a temperature of 40°C and absolute pressure of 450 kPa. If the backpressure causes M1 6 1, and the flow is choked at the exit, section 2, when L = 5 m, determine the mass flow through the pipe. Assume an average friction factor of 0.0085 throughout the pipe.

100 mm

150 mm 1

2

1

2

L 90 m

Probs. 13–76/77

Probs. 13–80/81

problems

SEC. 13.9–13.10 13–82. A standing shock occurs in the pipe. At section 1 upstream the air has an absolute pressure of p1 = 80 kPa, temperature T1 = 75°C, and velocity V1 = 700 m>s. Determine the pressure, temperature, and velocity of the air at section 2. Also, what is the Mach number at section 1 and 2?

797

13–86. The air in the large tank to the left of the nozzle is at a temperature of 80°C and an absolute pressure of 630 kPa. Determine the mass flow from the nozzle if the backpressure is 350 kPa. 13–87. The air in the large tank to the left of the nozzle is at a temperature of 80°C and an absolute pressure of 630 13 kPa. Determine the two values of the backpressure that will choke the nozzle yet produce isentropic flow. Also, what is the maximum exit velocity of the isentropic flow?

1 2

Prob. 13–82

60 mm

13–83. A Laval nozzle is connected to the large tank. If the temperature of the air in the tank is 375 K and the absolute pressure is 480 kPa, determine the range of backpressures that will cause expansion shock waves to form at the exit of the nozzle.

25 mm

*13–84. A Laval nozzle is connected to the large tank. If the temperature of the air in the tank is 375 K and the absolute pressure is 480 kPa, determine the range of backpressures that will cause oblique compression shock waves to form at the exit of the nozzle. 40 mm

75 mm

Probs. 13–83/84 13–85. The bottle tank contains 0.13 m3 of oxygen at an absolute pressure of 900 kPa and temperature of 20°C. If the exit nozzle has a diameter of 15 mm, and when it is opened, determine the time needed to drop the absolute pressure in the tank to 300 kPa. Assume the temperature remains constant in the tank during the flow and the ambient air is at an absolute pressure of 101.3 kPa.

Probs. 13–86/87 *13–88. The converging nozzle has an exit diameter of 0.25 m. If the fuel-oxidizer mixture within the large tank has an absolute pressure of 4 MPa and temperature of 1800 K, determine the mass flow from the nozzle if the atmospheric pressure is 100 kPa. The mixture has k = 1.38 and R = 296 J>kg # K. 13–89. The converging nozzle has an exit diameter of 0.25 m. If the fuel-oxidizer mixture within the large tank has an absolute pressure of 4 MPa and temperature of 1800 K, determine the mass flow from the nozzle when the backpressure is a vacuum. The mixture has k = 1.38 and R = 296 J>kg # K.

15 mm 0.25 m

Prob. 13–85

Probs. 13–88/89

798

C h a p t e r 13

Compressible Flow

13–90. A normal shock is produced at the nose of a jet plane flying with M = 2.3. If the air is at a temperature of - 5°C and an absolute pressure of 40 kPa, determine the pressure and the stagnation pressure just in front of the shock. 13–91. A 200-mm-diameter pipe contains air at a temperature 13 of 10°C and an absolute pressure of 100 kPa. If a shock forms in the pipe and the speed of the air in front of the shock is 1000 m>s, determine the speed of the air behind the shock. *13–92. The jet is flying at M = 1.3, where the absolute air pressure is 50 kPa. If a shock forms at the inlet of the engine, determine the Mach number of the air just within the engine where the diameter is 0.6 m. Also, what is the pressure and the stagnation pressure in this region? Assume isentropic flow within the engine.

13–95. A large tank supplies air at a temperature of 275 K and an absolute pressure of 560 kPa to the nozzle. Determine the backpressure that will cause the nozzle to choke. What range of backpressures will cause expansion waves to form at the exit? *13–96. A large tank supplies air at a temperature of 275 K and an absolute pressure of 560 kPa to the nozzle. Determine the range of backpressures that will cause a standing shock to form within the nozzle.

25 mm

M 5 1.3 55 mm

Probs. 13–95/96 0.4 m

0.6 m

Probs. 13–91/92 13–93. The jet engine has a converging–diverging exhaust nozzle. The absolute pressure at the inlet to the nozzle is 900 kPa, and the temperature of the fuel–air mixture is 1850 K. If the mixture flows at 125 m>s into the nozzle, determine the diameters of the throat and exit plane for isentropic flow. The inlet has a diameter of 600 mm. The outside absolute pressure is 102 kPa. The fuel mixture has k = 1.4 and R = 265 J>kg # K. 13–94. The jet engine has a converging–diverging exhaust nozzle. The absolute pressure at the inlet to the nozzle is 900 kPa, and the temperature of the fuel mixture is 1850 K. If the mixture flows at 125 m>s into the nozzle and exits with isentropic flow, determine the diameters of the throat and exit plane, and the mass flow through the nozzle. The inlet has a diameter of 600 mm. The outside absolute pressure is 50 kPa. The fuel mixture has k = 1.4 and R = 265 J>kg # K.

Probs. 13–93/94

13–97. The jet engine is tested on the ground at standard atmospheric pressure of 101.3 kPa. If the fuel–air mixture enters the inlet of the 300-mm-diameter nozzle at 250 m>s, with an absolute pressure of 300 kPa and temperature of 800 K, and exits isentropically with supersonic flow, determine the velocity of the exhaust. Take k = 1.4 and R = 249 J>kg # K. Assume isentropic flow. 13–98. The jet engine is tested on the ground at standard atmospheric pressure of 101.3 kPa. If the fuel–air mixture enters the inlet of the 300-mm-diameter nozzle at 250 m>s, with an absolute pressure of 300 kPa and temperature of 800 K, determine the required diameter of the throat dt, and the exit diameter de, so that the flow exits with isentropic supersonic flow. Take k = 1.4 and R = 249 J>kg # K.

300 mm dt

Probs. 13–97/98

de

799

problems 13–99. A jet plane creates a shock wave that forms in front of the plane in air having a temperature of 10°C and an absolute pressure of 60 kPa. If the plane travels at M = 2.3, determine the pressure and temperature just behind the shock wave. *13–100. The jet plane travels at M = 2.5 in still air at an altitude of 8 km. If a shock forms at the air inlet of the engine, determine the stagnation pressure within the engine just before the shock and the stagnation pressure a short distance within the chamber.

*13–104. Air in the large reservoir A has an absolute pressure of 450 kPa Determine the range of backpressures at B so that oblique shock waves form at the exit. 13–105. Air in the large reservoir A has an absolute pressure of 450 kPa. Determine the range of backpressures at B so that expansion shock waves form at the exit. 80 mm

60 mm

B A

Probs. 13–104/105 13–106. A shock is formed in the nozzle at C, where the diameter is 100 mm. If the air flows through the pipe at A at MA = 3.0 and the absolute pressure is pA = 15 kPa, determine the pressure in the pipe at B.

Probs. 13–99/100

150 mm

13–101. A normal shock is produced at the nose of a jet plane flying with M = 2.3. If the air is at a temperature of - 5°C and an absolute pressure of 40 kPa, determine the velocity of the air relative to the plane and its temperature just in front of the shock. 13–102. The cylindrical plug is fired with a speed of 150 m>s in the pipe that contains still air at 20°C and an absolute pressure of 100 kPa. This causes a shock wave to move down the pipe as shown. Determine its speed and the pressure acting on the plug.

C

A

B

Prob. 13–106 13–107. If the absolute pressure of air within the large tank A is 500 kPa, determine the range of backpressures at B that will cause an oblique compression shock wave to form at the exit plane. *13–108. If the absolute pressure of air within the large tank A is 500 kPa, determine the range of backpressures at B that will cause a standing normal shock wave to form within the divergent section of the nozzle.

150 ms

Prob. 13–102 13–103. Air in the large reservoir A has an absolute pressure of 450 kPa. Determine the range of backpressures at B so that a standing normal shock wave will form within the nozzle. 80 mm

100 mm

13–109. If the absolute pressure within the large tank A is 500 kPa, determine the range of backpressures at B that will cause expansion shock waves to form at the exit plane. 80 mm

60 mm

50 mm

B A

Prob. 13–103

B

Probs. 13–107/108/109

A

13

800

C h a p t e r 13

Compressible Flow

13–110. Air at a temperature of 20°C and an absolute pressure of 180 kPa flows from a large tank through the nozzle. Determine the backpressure at the exit that causes a shock wave to form at the location where the nozzle diameter is 50 mm. 13

*13–112. A jet plane is flying in air that has a temperature of 8°C and absolute pressure of 90 kPa. The leading edge of the wing has the wedge shape shown. If the plane has a speed of 800 m>s and the angle of attack is 2°, determine the pressure and temperature of the air at the upper surface A just in front or to the right of the weak oblique shock wave that forms at the leading edge.

80 mm 50 mm 20 mm

A

3°

B

3°

a 5 2°

Prob. 13–112

Prob. 13–110

13–113. Air flows at 700 m>s through a long duct in a wind tunnel, where the temperature is 15°C and the absolute pressure is 70 kPa. The leading edge of the wing in the tunnel is represented by the 8° wedge. Determine the pressure created on its top surface if the angle of attack is set at a = 1.5°.

SEC. 13.11–13.12 13–111. A jet plane is flying in air that has a temperature of 8°C and absolute pressure of 90 kPa. The leading edge of the wing has the wedge shape shown. If the plane has a speed of 800 m>s and the angle of attack is 2°, determine the pressure and temperature of the air at the lower surface B just in front or to the right of the weak oblique shock wave that forms at the leading edge.

13–114. Air flows at 700 m>s through a long duct in a wind tunnel, where the temperature is 15°C and the absolute pressure is 70 kPa. The leading edge of the wing in the tunnel is represented by the 8° wedge. Determine the pressure created on its bottom surface if the angle of attack is set at a = 1.5°. 13–115. Air flows at 700 m>s through a long duct in a wind tunnel, where the temperature is 15°C and the absolute pressure is 70 kPa. The leading edge of the wing in the tunnel can be represented by the 8° wedge. Determine the pressure created on its top surface if the angle of attack is set at a = 6.5°. *13–116. Air flows at 700 m>s through a long duct in a wind tunnel, where the temperature is 15°C and the absolute pressure is 70 kPa. The leading edge of the wing in the tunnel can be represented by the 8° wedge. Determine the pressure created on its bottom surface if the angle of attack is set at a = 6.5°.

4°

A

3°

B

3°

a 5 2°

Prob. 13–111

700 ms

a 4°

Probs. 13–113/114/115/116

problems 13–117. A jet plane is flying at M = 2.4, in air having a temperature of 2°C and absolute pressure of 80 kPa. If the leading edge of the wing has an angle of d = 16°, determine the velocity, pressure, and temperature of the air just in front or to the right of the weak oblique shock that forms on the wing. What is the angle d of the leading edge that will cause the shock wave to separate from the front of the wing?

801

13–119. The leading edge on the wing of an aircraft can be approximated by the wedge shape shown. If the plane is flying at 900 m>s in air that has a temperature of 5°C and absolute pressure of 60 kPa, determine the angle b of a weak oblique shock wave that forms on the wing. Also, determine the pressure and temperature on the wing just in 13 front or to the right of the shock.

b 5°

M 5 2.4

900 ms

5°

d b

Prob. 13–117

Prob. 13–119

13–118. A jet plane is flying upward such that its wings make an angle of attack of 15° with the horizontal. The plane is traveling at 700 m>s, in air having a temperature of 8°C and absolute pressure of 90 kPa. If the leading edge of the wing has the wedge shape shown, determine the pressure and temperature of the air just in front or to the right of the expansion waves.

*13–120. Oxygen at a temperature of 25°C and an absolute pressure of 200 kPa flows through the rectangular duct at 900 m >s. When it comes to the transition, it is redirected as shown. Determine the angle b of the weak oblique shock that forms at A, and the temperature and pressure of the oxygen just in front or to the right of the wave.

15°

B 15° 4° 4° b

15°

700 ms A

Prob. 13–118

Prob. 13–120

802

C h a p t e r 13

Compressible Flow

13–121. Oxygen at a temperature of 25°C and an absolute pressure of 200 kPa flows in the rectangular duct at 900 m>s. When it comes to the transition, it is redirected as shown. Determine the temperature and pressure just in front or to the right of the expansion waves that form in the duct at B. 13

13–122. The wing of a jet plane is assumed to have the profile shown. It is traveling horizontally at 900 m>s, in air having a temperature of 8°C and absolute pressure of 85 kPa. Determine the pressure that acts on the top surface in front or to the right of the weak oblique shock at A and in front or to the right of the expansion waves at B.

15°

B

B A

3° 3°

900 ms

b 15° A

Prob. 13–121

Prob. 13–122

3°

Chapter review

803

CHAP TER R EV IEW Ideal gases obey the ideal gas law. The first law of thermodynamics states that the change in internal energy of a system is increased when heat is added to the system, and decreased when the system does flow work.

The change in entropy ds indicates the heat energy transfer per unit mass that occurs at a specific temperature during the change in state of a gas. The second law of thermodynamics states that this change will always increase due to friction.

A pressure wave travels through a medium at the maximum velocity c, called the sonic velocity. This is an isentropic process; that is, no heat is lost (adiabatic) and the process is frictionless, so ds = 0.

Compressible flow is classified according to the Mach number, M = V>c. If M 6 1 the flow is subsonic, if M = 1 it is sonic, and if M 7 1 it is supersonic. If V … 0.3c, we can assume the flow is incompressible.

The stagnation temperature remains the same if the flow is adiabatic, and the stagnation pressure and stagnation density are the same if the flow is isentropic. These properties can be measured at a point where the gas is at rest.

Specific values of the static temperature T, pressure p, and density r of the gas, measured while moving with the flow, are obtained from formulas that relate them to the stagnation values T0, p0, r0. They depend on the Mach number and the ratio k of the specific heats for the gas.

In a converging duct, subsonic flow will cause an increase in velocity and a decrease in pressure. For supersonic flow the opposite occurs; the velocity decreases and the pressure increases.

In a diverging duct, subsonic flow will cause a decrease in velocity and an increase in pressure. For supersonic flow the opposite occurs; the velocity increases and the pressure decreases.

p = rRT 13 dn = dq - rdv

ds =

dq 7 0 T

c = 2kRT

804

C h a p t e r 13

Compressible Flow

A Laval nozzle can be used to convert subsonic flow in the convergent section to sonic speed at the throat, and then to further accelerate the flow through the divergent section to supersonic flow at the exit. 13

A Laval nozzle becomes choked when M = 1 at the throat. If this occurs, there are two backpressures that will produce isentropic flow throughout the divergent section. One produces subsonic flow downstream of the throat, and the other produces supersonic flow.

M,1

M.1 M51

Subsonic

Sonic

Supersonic

Fanno flow considers the effect of pipe friction as the gas undergoes an adiabatic process. Equations are used to determine the properties of the gas at any location, provided these properties are known at the reference or critical location, where M = 1. Rayleigh flow considers the effect of heating or cooling the gas as it flows through a pipe with no friction loss. The properties of the gas can be determined at any location, provided they are known at the reference or critical location, where M = 1. A shock wave is a nonisentropic process that occurs over a very small thickness. Because the process is adiabatic, the stagnation temperature will remain the same on each side of the shock. The Mach number, temperature, pressure, and density of a gas on each side of a shock can be related, and the results from these equations are presented in tabular form. If the backpressure does not create isentropic flow through a nozzle, then a shock can form within the nozzle or at its exit. This is an inefficient use of the nozzle. An oblique shock will form at the point where the speed over a surface is the greatest. The properties of temperature, pressure, and density on each side of the shock are related to the normal component of the Mach number, in the same manner as they are for normal shocks. On curved surfaces or at sharp corners, compression waves will merge into an oblique shock, or they can produce expansion waves. The deflection angle of the flow caused by expansion waves can be calculated using the Prandtl–Meyer expansion function. Compressible flow can be measurement using a pitot tube and piezometer, or a venturi meter. Hot-wire anemometers and other meters can also be used for this purpose.

1 T1, p1, r1 (low) M1 . 1 into wave Back

2 T2, p2, r2 (high) M2 , 1 out of wave Front

Chapter review

805

Fundamental Equations of Compressible Flow s2 - s1 = cv ln

r2 T2 - R ln r1 T1

s2 - s1 = cp ln

p2 T2 - R ln T1 p1

Entropy

p2 T2 k>(k - 1) = a b p1 T1

p2 r2 k = a b r1 p1

∆Q = cp 3 1T0 2 2 - 1T0 2 14 ∆m

M2(1 + k)2 T = T* 11 + kM2 2 2

p0 1 + k 2 k - 1 2 k>(k - 1) = a b c a b a1 + M bd p*0 k + 1 2 1 + kM2 T0 = T0*

Isentropic process

c = 2kRT

V = M 2kRT

p0 = pa1 + r0 = ra1 +

k - 1 2 M b 2

11 + kM2 2 2

Rayleigh flow

k - 1 2 M b 2

k - 1 2 M b 2

21k + 12M2 a1 +

M2(1 + k) V = V* 1 + kM2

Sonic velocity

T0 = T a1 +

p 1 + k = p* 1 + kM2

k - 1 M12 2 k - 1 1 + M22 2 p2 1 + kM12 = p1 1 + kM22

T2 = T1

k>1k - 12

k - 1 2 1>1k - 12 M b 2

Stagnation properties

1 +

1>2 k - 1 M22 r2 M1 2 = ≥ ¥ r1 k - 1 M2 1 + M12 2

1 +

k+1

1 2 A 1 1 + 2 1k - 12M 21k - 12 = £ § 1 A* M 2 1k + 12

1>2 k - 1 M12 V2 M2 2 = ≥ ¥ V1 M1 k - 1 1 + M22 2 2 M12 + k - 1 2 M2 = M1 7 M2 2k M12 - 1 k - 1

Nozzle area ratios

1 +

31k + 12 >24M2 fLmax 1 - M2 k + 1 = lnc d + D 2k kM2 1 + 12 1k - 12M2 1

T 2 (k + 1) = T* 1 + 12(k - 1)M2

p 1 = £ p* M 1 +

1 2 (k + 1) § 1 2 2 (k - 12M

tan u =

p0 1 k - 1 2 2 = ca b a1 + M bd p*0 M k + 1 2 1 V 2 (k + 12 = M£ § V* 1 + 12(k - 12M2

Fanno flow

Normal shock wave

1>2

(k + 12>2(k - 1)

2 cot b1M12 sin2 b - 12 M12 1k + cos 2 b2 + 2

Oblique shock wave

1>2

v=

k+1 k-1 tan - 1 a 1M2 - 12 b - tan - 1 12M2 - 12 Ak - 1 Ak + 1 Expansion waves

13

14

Alacatr/E+/Getty Images

CHAPTER

Pumps play an important role in transporting fluids in chemical processing plants, and distributing water as in this water treatment plant.

TURBOMACHINES CHAPTER OBJECTIVES ■ ■ ■ ■ ■

To discuss how axial-flow and radial-flow pumps and turbines operate by adding energy to or removing it from the fluid. To study the flow kinematics, torque, power, and performance characteristics of a turbomachine. To discuss the effects of cavitation and show how it can be reduced or eliminated. To show how to select a pump so that it meets the requirements of a flow system. To present some important pump-scaling laws that are related to turbomachine similitude.

14.1

TYPES OF TURBOMACHINES

Turbomachines consist of various forms of pumps and turbines, which transfer energy between the fluid and the rotating blades of the machine. Pumps, which include fans, compressors, and blowers add energy to fluids, whereas turbines remove energy. Each of these machines can be categorized by the way the fluid flows through it. If the fluid flows along its axis, it is called an axial-flow machine. Examples include the compressor and turbine for a jet engine and an axial-flow pump such as shown in Fig. 14–1a. A radial-flow machine directs the flow mainly in the radial direction through its rotating blades. This occurs in a centrifugal or radial flow pump, Fig. 14–1b. Finally, a mixed-flow machine changes the direction of axial flow a moderate amount in the radial direction, as in the case of the mixed-flow pump in Fig. 14–1c.

Axial-flow pump

Radial-flow pump

Mixed-flow pump

(a)

(b)

(c)

Fig. 14–1 807

808

C h a p t e r 14

turbomaChines

14

Positive-displacement pump

These three types of turbomachines are generally referred to as dynamic fluid devices because the flow is changed by its dynamic interaction with a series of rotating blades as it freely passes through the machine. A device called a positive-displacement pump, as shown in Fig. 14–1d, will not be discussed here. It works like your heart, transferring each quantity of fluid contained in a moving chamber. Examples include a gear pump, as in Fig. 14–1d, and a pump composed of a piston and cylinder or screw. Although these devices can produce pressures that are much higher than those produced by a dynamic fluid device, the flow through them is generally much smaller.

(d)

Fig. 14–1 (cont.)

14.2

T v

U 5 vrm 2

Stator (fixed)

1

Impeller (rotating) rm

Axial-flow pump (a)

Fig. 14–2

AXIAL-FLOW PUMPS

Axial-flow pumps can produce a high flow, but they have a disadvantage of delivering the fluid at a relatively low pressure. Like a fan or airplane propeller, the fluid enters and then exits the pump along its axis, Fig. 14–2a. Energy is added to the fluid by using an impeller consisting of a series of vanes or blades that are fixed to a rotating shaft. When the fluid exits the impeller, fixed stator vanes are often used to remove the swirl and redirect the flow in the axial direction. Stator vanes are sometimes also located at the upstream side if the fluid has an initial swirl, although in most cases straight axial flow occurs, and so initial vanes or stators are not used. To show how an axial-flow pump works, consider the impeller blade in Fig. 14–2b. As the fluid is scooped upwards by the impeller, this removal of fluid lowers the pressure in the downstream region, and so more fluid is drawn upward into the pump. As shown, when the fluid just enters the blade it is given a velocity V1, and when it is about to leave the blade it will have a greater velocity V2. The resulting higher kinetic energy converts, in part, to an increased pressure in the region off the blade, which pushes the fluid upwards. To determine the torque needed to produce a specified flow, we will assume the fluid is ideal, and that it is guided smoothly onto and over the blades of the impeller. With these assumptions, we will consider the control volume to be the contained liquid within the impeller, Fig. 14–2a, and then write the equations of continuity and angular momentum for the flow. Although within the impeller the flow is unsteady, it is cyclic, and on average it can be considered quasi-steady as it enters and then leaves the open control surfaces, which are a short distance away from the impeller.

14.2

809

axial-Flow pumps

Continuity. As the pump draws incoming axial flow through the open control surface 1 in Fig. 14–2a, it must equal the outgoing axial flow through the open control surface 2. Since these surfaces have the same cross-sectional area, applying the continuity equation for steady flow, we have

Stator

V2 Va2

0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

V1

0 - rVa1A + rVa2A = 0 Va1 = Va2 = Va

Va1

Impeller blade

Vt 1 Flow

This result is to be expected; that is, the mean velocity of the flow in the axial direction remains constant.

(b)

Angular Momentum. The torque T that is applied to the liquid by the impeller changes the liquid’s angular momentum as it passes over the blades. If we assume that the blades are relatively short, then as a first approximation, the angular momentum of the liquid can be determined using the impeller’s mean radius rm, Fig. 14–2c. Applying the angular momentum equation, Eq. 6–5, about the center axis of the control volume, ΣM =

14

Vt 2

v

rrm m

U 5 vrm

0 1r * V2rdV + 1r * V2rVf>cs # dA 0t Lcv Lcs

= 0 +

Lcs

1r * V2rVf>cs # dA

(c)

The component of V that produces the moment in the term r * V is Vt, and the component of V that contributes to the flow rVf>cs # dA is Va . Therefore, (rmVt)rVa dA (14–1) Lcs Integrating over the control surfaces at sections 1 and 2,* using Q = Va A, yields T = rmVt 2 rQ - rmVt 1 rQ or T =

T = rQrm(Vt 2 - Vt1)

(14–2)

This equation is often referred to as the Euler turbomachine equation. The terms on the right are the product of the mass flow produced by the pump, rQ, and the moment, rmVt.

*If the blades are long, then the torque on the liquid can be determined to a closer approximation by dividing the blades into small segments, each having its own small width ∆r and mean radius rm. In this way, numerical integration can be carried out.

Fig. 14–2 (cont.)

810

C h a p t e r 14

turbomaChines

Power. The power developed by a pump is often called the shaft or brake power, since it refers to the actual power transferred to the pump shaft and not the electric power supplied to the motor. In the SI system, power is measured in watts, where 1 W = 1 N # m>s = 1 J>s. The shaft power developed by the pump to the fluid can be expressed as the product of the applied torque and the angular velocity v (omega) of the impeller. Using Eq. 14–2, we have

14

# Wpump = Tv = rQrm(Vt2 - Vt1)v

(14–3)

We can also write this equation in terms of the velocity of the midpoint of the impeller, rather than the impeller’s angular velocity. At this point, the blade has a speed of U = vrm, Fig. 14–2c, and so # Wpump = rQU(Vt 2 - Vt1)

(14–4)

The above equations imply that the torque developed, or the rate of energy transferred to the liquid, is independent of the shape of the pump, or the number of blades that are on the impeller.* Instead, it only depends upon the motion of the impeller and the tangential components of the velocity of the liquid as it enters and leaves the impeller.

Flow Kinematics. It is convenient to establish velocity kinematic

V1

Va

a1

(Vrel)1 b1

U Velocity of blade’s midpoint

U Vt 1

(a)

diagrams in order to fully understand how the flow enters and then exits each impeller blade. The center of the blade has a velocity U = vrm, and the initial velocity of the liquid relative to the head of the blade, (Vrel)1, will be at the tangent angle b1, Fig. 14–3a. These two components when added together produce the resultant velocity of the liquid, V1, acting at the angle a1. Note carefully the convention used to establish a1 and b1. The angle a1 is measured between V1 and + U, whereas b1 is measured between Vrel and –U. From the parallelogram law of vector addition, we have V1 = U + (Vrel)1

(14–5)

But also note that V1 can be resolved into its rectangular components, V1 5 Va a1 5 90° U

(b)

Fig. 14–3

(Vrel)1 b1

V1 = Vt1 + Va

(14–6)

where Vt1 = Va cot a1. For the most efficient performance of the pump, the blade angle b1 should be designed so that V1 is directed upwards, so that a1 = 90°, Fig. 14–3b. When this is the case, the initial upward direction of the liquid will be maintained so that V1 = Va. More importantly, Vt1 = 0, so that the power, calculated from Eq. 14–4, will be a maximum.

*There are disadvantages to having too many blades because the flow becomes restricted and friction losses increase.

14.2

A similar situation occurs just as the liquid is directed off the impeller. Here, the relative velocity of the liquid at the tail of the impeller blade is (Vrel)2, which is directed at the blade angle b2, Fig. 14–3c. The components of V2 are therefore V2 = U + (Vrel)2

(14–7)

V2 = Vt 2 + Va

(14–8)

811

axial-Flow pumps

V2

Va

a2 U

(Vrel)2 b2

Vt 2

14

or (c)

To reduce turbulence and friction losses, proper design requires that V2 be directed tangentially onto the stator vanes at the angle a2, as shown in Fig. 14–3d.

Stator

V2 Velocity onto stator

P ROCEDUR E FOR A N A LY S I S

a2 (d)

The following procedure provides a method for analyzing the flow through the blades of an axial-flow pump. • For flow onto a blade, first establish the direction of the midpoint velocity of the blade, U. Its magnitude is determined from U = vrm. The axial velocity through the pump Va is always perpendicular to U. Its magnitude is determined from the flow Q = Va A, where A is the open cross-sectional area through the blades. The relative velocity of the flow on or off a blade, Vrel, is tangent to the blade at the design angle b, measured from the direction of –U, Figs. 14–3a and 14–3c.

• Depending on what components are known, establish the velocity of flow V1 at the angle a1 measured from +U, Fig. 14–3a.

• Analyzing flow off the blades follows the same procedure, as noted by the kinematic diagram in Fig. 14–3c.

• When U, V, Va, and Vrel are established, then the tangential component of velocity Vt can be constructed by the vector resolution of V into its rectangular components. Here V = Va + Vt, but also V = U + Vrel. Depending upon the problem, the various magnitudes and>or angles can be determined from these vector resolutions.

• Once the tangential components of velocity, Vt1 and Vt 2, are known, then the torque and power requirements for the pump can be found from Eqs. 14–2 through 14–4.

Fig. 14–3 (cont.)

812

C h a p t e r 14

EXAMPLE

14

turbomaChines

14.1 The axial-flow pump in Fig. 14–4a has an impeller that is rotating at 150 rad>s. The blades are 50 mm long and are fixed to the 50-mm-diameter shaft. If the pump produces a flow of 0.06 m3 >s, and the head blade angle is b1 = 30°, determine the velocity of the water when it is just on the blade. The average cross-sectional area of the open region within the impeller is 0.02 m2.

50 mm 50 mm

Va

150 rads (a)

Fig. 14–4

SOLUTION Fluid Description. We assume steady ideal flow, using average velocities. Kinematics. Using the mean radius to determine the velocity of the midpoint of a blade, we have U = vrm = (150 rad>s) a

0.05 m 0.05 m + b = 7.50 m>s 2 2

Also, since the flow is known, the axial velocity of the liquid through the blades is Q = VA;

0.06 m3 >s = Va 10.02 m2 2 Va = 3 m>s

The kinematic diagram for the water as it just encounters the head of a blade is shown in Fig. 14–4b. Following the convention, note carefully how a1 and b1 are established. As usual, there are two sets of components for V1. They are V1 = Vt1 + Va and V1 = U + (Vrel)1.

14.2

axial-Flow pumps

813

SOLUTION I Using trigonometry, one way we can determine V1 is as follows: Vt1 = 7.50 m>s - (3 m>s) cot 30° = 2.304 m>s tan a1 =

3 m>s 2.304 m>s

(3 ms) cot 30°

U 5 7.50 ms 30°

V1

,

a1 = 52.48°

Vt1

a1

Va

Va 5 3 ms

3 m>s = V1 sin 52.48° V1 = 3.78 m>s

Ans.

b1 5 30° (Vrel)1 (b)

SOLUTION II

Fig. 14–4 (cont.)

Another way to determine V1 is to note that tan 30° =

3 m>s

7.50 m>s - Vt1 Vt1 = 2.304 m>s

so that V1 = 21Va 2 2 + 1Vt1 2 2 = 213 m>s2 2 + 12.304 m>s2 2 = 3.78 m>s Ans. SOLUTION III Using Cartesian vectors, Fig. 14–4b, we can write V1 = Va + Vt1 = 3i + Vt1 j

(1)

V1 = U + 1Vrel 2 1 = 1Vrel 2 1 sin 30°i + 37.50 - 1Vrel 2 1 cos 30°4j

Equating these equations, then setting the i and j components of the left side equal to those on the right side, gives 3 = 1Vrel 2 1 sin 30°

Vt1 = 7.50 - 1Vrel 2 1 cos 30°

Solving, 1Vrel 2 1 = 6 m>s, Vt1 = 2.304 m>s . Therefore, from Eq. 1, V1 = 213 m>s2 2 + 12.304 m>s2 2 = 3.78 m>s

Ans.

14

814

C h a p t e r 14

EXAMPLE

turbomaChines

14.2 125 mm (mean radius)

14

1000 revmin.

The blades on an impeller of the axial-flow water pump in Fig. 14–5a rotate at 1000 rev>min. If the pump is required to produce a flow of 0.2 m3 >s, determine the required initial blade angle b1 so that the pump runs efficiently. Also, if b2 = 70°, find the average torque that must be applied to the shaft of the impeller, and the average power output of the pump. The average open cross-sectional area through the impellers is 0.03 m2. SOLUTION

(Vrel)1

(Vrel)2 b1 b2 5 708 Impeller (a)

Fluid Description. We assume steady ideal flow through the pump and we will use average velocities. Here r = 1000 kg>m3. Kinematics.

The midpoint velocity of the blades is 1000 rev 1 min 2p rad U = vrm = a ba ba b (0.125 m) = 13.09 m>s min 60 s 1 rev And the axial velocity of the flow through the impeller is Q = Va A;

U = 13.09 ms

b1

a1 5 90°

V1 5 Va 5 6.667 ms

0.2 m3 >s = Va 10.03 m2 2;

For the most efficient operation, a1 = 90°, so that Vt1 = 0. The velocity of the water onto the blades is therefore V1 = Va = 6.667 m>s. From Fig. 14–5b, we require

b1

tan b1 =

(Vrel)1

Ans.

Vt 2 = 13.09 m>s - (6.667 m>s) cot 70° = 10.664 m>s V2 6.667 ms

Vt2

13.09 m>s

At the exit, Fig. 14–5c, one way to find Vt 2 is as follows:

(b) 70°

6.667 m>s

b1 = 27.0°

Entrance

U 5 13.09 ms

Va = 6.667 m>s

a2

Torque and Power. Since the tangential components of velocity are known, Eq. 14–2 can be applied to determine the torque. T = rQrm(Vt2 - Vt1)

Va 5 6.667 ms (Vrel)2 b2 5 70° Exit (c)

Fig. 14–5

= 11000 kg>m3 210.2 m3 >s2(0.125 m)(10.664 m>s - 0) = 267 N # m Ans.

From Eq. 14–4, the power supplied to the water by the pump is # Wpump = rQU(Vt 2 - Vt1)

= (1000 kg>m3)(0.2 m3 >s)(13.09 m>s)(10.664 m>s - 0)

= 27.9 kW

Ans.

Note: If a1 6 90°, Fig. 14–5b, V1 would have a component Vt1 as in the previous example and this would decrease the power.

14.3

14.3

radial-Flow pumps

815

RADIAL-FLOW PUMPS

A radial-flow pump is probably the most common type of pump used in industry. It operates at slower speeds than an axial-flow pump, so it produces a lower flow, but at a higher pressure. Radial-flow pumps are designed so that the fluid enters the pump in the axial direction at the center of the pump, and then it is directed onto the impeller in the radial direction, Fig. 14–6a. To show how this pump works, consider the impeller blade in Fig. 14–6b. Due to its rotation, here liquid is scooped up on the back side of the blade. Just as in the axial-flow pump, this will lower the pressure in this region and draw more liquid onto the blade. As the liquid flows off each blade with an increased velocity, it then moves onto, then later off, the guide vanes, Fig. 14–6a, and accumulates around the entire circumference of the pump’s casing. Because of the way the liquid flows, this type of pump is also called a centrifugal or a volute pump.

Kinematics. The kinematics of the flow over the impeller can be analyzed in a manner similar to that used for an axial-flow pump, as described in the Procedure for Analysis given in Sec. 14.2.

r2 r2

Casing Guide vane

Impeller blade

T v

1

Guide vane

(Vrel)2

V2 b2

2

Impeller blade

b

(Vrel)1 U2 5 vr2 b1 r2

V1 r1

Centrifugal pump

U1 5 vr1

v (b)

(a)

Fig. 14–6

14

This leaf blower acts as a radial-flow pump. Notice the volute shape of the casing as it accumulates air along its axis and expels it out through the pipe to the right.

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(Vrel)1 Vr 1

b1 r2

V1 a1

14

r1 Vt 1 v

U1 =

r1

Flow entering blade (Vrel)2

Vr 2 V2

b2

a2

Continuity. Since steady flow through the open control surfaces

b2

Vt2

A typical blade is shown in Fig. 14–6c, where the front or head of the blade has a velocity of U1 = vr1, and the rear or tail has a higher velocity of U2 = vr2. Note carefully how the front and tail blade angles b are established between –U and the relative velocity Vrel, and the angles a are established between the velocity of the fluid V and + U. Similar to Eqs. 14–5 and 14–6, we can express V in terms of its components, namely, V = U + Vrel, or its tangential and radial components, V = Vt + Vr. To relate the torque to the flow, we will again assume that friction can be neglected, the fluid is incompressible, and the flow is guided smoothly over the blades of the impeller, each of which has a constant width b. The control volume contains the fluid within the impeller as shown in Fig. 14–6a.

U2 = r2

1 and 2 occurs in the radial direction, then the continuity equation applied to these surfaces becomes 0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

r2 r1

0 - rVr1(2pr1b) + rVr 2(2pr 2b) = 0

v

Vr1r1 = Vr 2r2

Flow exiting blade (c)

Fig. 14–6 (cont.)

(14–9)

Since r2 7 r1, then the radial velocity component Vr2 6 Vr1.

Angular Momentum. The torque on the shaft of the impeller can be related to the angular momentum of the fluid using the angular momentum equation. ΣM =

0 1r * V2r d V + 1r * V2rVf>cs # d𝚨 0t Lcv Lcv

The component of V that produces the moment in the factor r * V is Vt, and the component that contributes to the flow rVf>cs # dA is Vr . T = 0 + r1Vt1 3 -r(Vr1)(2pr1b)4 + r2Vt 2 3r(Vr2)(2pr2b)4 = r(Vr 2)(2pr2b)r2Vt 2 - r(Vr1)(2pr1b)r1Vt1

(14–10)

Since Q = Vr1(2pr1b) = Vr 2(2pr2b), we can also express this result as T = rQ(r2Vt 2 - r1Vt1)

(14–11)

From the impeller kinematics in Fig. 14–6c, for any r, Vt = Vr cot a. Therefore, the torque can also be expressed in terms of Vr and a. It is T = rQ(r 2Vr 2 cot a2 - r1Vr1 cot a1)

(14–12)

14.3

Power. With Eq. 14–12, the power delivered to the fluid can now be expressed in terms of the speed of the head and tail of the impeller blades. Since U1 = vr1 and U2 = vr2, we have # Wpump = Tv = rQ1U2Vr 2 cot a2 - U1Vr1 cot a1 2

V1 5 Vr1

r2

b1

v U1 Flow entering blade (d)

(14–14)

Flow within the Casing. The flow that occurs within the pump

Casing

casing can be obtained by applying the angular momentum equation to the fluid within a control volume having open control surfaces just after the tails of the guide vanes, at r2, and at a general location r within the casing, where the fluid is free flowing, Fig. 14–6e. Since here there is no torque on the fluid, T = 0, and so Eq. 14–11 becomes

0 = rVt - r2Vt 2 or Flow within casing (e)

r2 Vt 2 const. = r r

This represents a case of free-vortex flow, discussed in Sec. 7.10. It is this accumulation of the flow, as it passes off the guide vanes, that requires the casing to have the growing spiral or volute shape; hence the name “volute pump.”

Guide vane

r2

r

Vt =

14

a1 5 908 r1

Like axial-flow pumps, radial-flow pumps are often designed so that the blade angle b1 provides an initial flow V1 = U1 + 1Vrel 2 1 that is directed onto the impeller blades only in the radial direction. This requires a1 = 90°, and so the tangential component of velocity Vt1 = 0, Fig. 14–6d. When this occurs, Eq. 14–13 will provide maximum power because the last term in the above equation will be zero, i.e., cot 90° = 0. We can also express Eq. 14–13 in terms of the tangential components of velocity. Since in general Vt = Vr cot a, Fig. 14–6c, then # Wpump = rQ1U2Vt 2 - U1Vt1 2

(Vrel)1

b1

(14–13)

817

radial-Flow pumps

Fig. 14–6 (cont.)

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14.4

T v

The performance of any pump depends upon a balance of energy as the fluid flows through the impeller. For example, in the case of an axial-flow pump, shown in Fig. 14–2a, assuming we have incompressible steady flow through the open control surfaces, and neglecting friction losses, hL = 0, the energy equation, Eq. 5–14, applied between points 1 (in) and 2 (out) on the open control surfaces becomes

14 U 5 vrm 2

Stator (fixed)

1

Impeller (rotating)

IDEAL PERFORMANCE FOR PUMPS

pin pout V out2 V in2 + + + zin + hpump = + zout + hturbine + hL g g 2g 2g hpump = a

rm

pout pin V out2 V in2 + + + zout b - a + zin b g g 2g 2g

This is an ideal pump head because it neglects any friction losses that can occur. It represents the change in the total head of the fluid and it applies to both axial- and radial-flow pumps. Using Eq. 5–17, the ideal power produced by the pump is then

Axial-flow pump (a)

Fig. 14–2 (repeated)

# Wpump = Qghpump

(14–15)

hpump

b2 5 90°

Finding hpump for a pump is important, because it refers to the additional height to which the liquid can be raised after the fluid passes through the pump.

b2 , 90° Backward curved blade

Head Loss and Efficiency. If we express the ideal pump head in

2

U2 g

Q Ideal pump head versus flow

terms of the impeller’s tangential velocity components, using Eq. 14–4 for an axial-flow pump, or Eq. 14–14 for a radial-flow pump, we get

(a) (Vrel)2

hpump =

V2

b2 , 90°

U2 (Vrel)1

U(Vt 2 - Vt1) g

(14–16)

Axial-flow pump

V1

r2

a2 5 90°

U1 r1

hpump =

U2Vt 2 - U1Vt1 g

(14–17)

Radial-flow pump

v

b2 , 90°

(b)

Fig. 14–7

The actual head, (hpump)act, delivered by the pump will be less than this ideal value because of the mechanical head losses hL within the pump. These losses are a result of friction developed at the shaft bearings, fluid friction within the pump casing and the impeller, and additional fluid

14.4

flow losses that occur because of inefficient circulation into and out of the impeller. The head loss is therefore

(Vrel)2

V2

b2 5 90°

U2

hL = hpump - (hpump)act

(Vrel)1

r2

V1

The hydraulic or pump efficiency hpump (eta) is the ratio of the actual head delivered by the pump to its ideal head, that is,

hpump =

(hpump)act hpump

819

ideal perFormanCe For pumps

14 U1

r1 v

(100%)

(14–18) b2 5 90°

(c)

Head-Discharge Curve—Radial-Flow Pump. As stated earlier, radial-flow pumps generally have impeller blades designed so that the fluid has no inlet swirl, so that Vt1 = 0 since a1 = 90°, Fig. 14–7d. Also, for this case, the component Vt 2 can be related to Vr 2 at the tail of the blade by noting that Vt 2 = U2 - Vr 2 cot b 2, Fig. 14–7e. The ideal pump head then becomes

(Vrel)1 V1 5 Vr1

b1 r2

a1 5 90° r1

hpump =

U2(U2 - Vr2 cot b2) U2Vt 2 - U1Vt1 = g g

v U1 Flow entering blade (d) (Vrel)2

Since Q = Vr A = Vr 2(2pr2b), where b is the width of the blades, we get

hpump

U 22 U2Q cot b2 = g 2p r2bg

b1

Vr2 V2

b2

a2

b2

Vt 2

(14–19)

(a1 = 90°)

This equation is plotted in Fig. 14–7a for blade angles b2 6 90°, Fig. 14–7b, and b2 = 90°, Fig. 14–7c. Notice from the graph that when the blades of the impeller are curved backward, b2 6 90°, which is the usual case, then the ideal pump head will decrease as the flow Q increases. If b2 = 90°, then the tails of the blades are in the radial direction, and so, cot 90° = 0, and hpump does not depend on the flow Q. It is simply hpump = U 22 >g. Engineers generally do not design radialflow pumps with forward curved blades, for which b2 7 90°, because the flow within the pump has a tendency to become unstable and cause the pump to surge. This effect causes rapid pressure changes within the casing that make the impeller oscillate back and forth in an attempt to find its operating point.

r2 r1 v Flow exiting blade (e)

Fig. 14–7 (cont.)

U2

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IM PORTANT POINTS • The continuity equation requires that the velocity of the flow through an axial-flow pump remains constant 14

along its axis, whereas for radial-flow pumps, the radial component of velocity must decrease as the fluid flows outward.

• Blade angles for axial- and radial-flow pumps are defined in the same way; that is, a is measured between

the velocity + U of the blade and the velocity V of the flow on or off the blade; and b is measured between –U and the relative velocity Vrel of the flow on or off the blade.

• The torque and power developed by axial- and radial-flow pumps depend upon the motion of the blades, and on the tangential components of the flow as it just enters and just leaves the blades.

• Axial- and radial-flow pumps are usually designed so that the entrance flow onto the blades is only in the axial or radial direction. Therefore, Vt1 = 0 since a1 = 90°.

• The performance of axial- and radial-flow pumps is measured by the increase in energy, or the increase in head hpump produced by the pump once the fluid passes through the impeller. This head depends on the speed of the impeller’s midpoint U and the difference in the tangential components of velocity Vt of the fluid just after it enters and just before it leaves the impeller.

• The efficiency of axial- and radial-flow pumps is the ratio of the actual pump head divided by the ideal pump head.

EXAMPLE

14.3 Determine the hydraulic efficiency for the axial-flow pump in Example 14.2 if frictional head losses produced by the pump are 2.8 m. SOLUTION Fluid Description. We assume steady, incompressible flow through the pump. Pump Head. We can use Eq. 14–16 and the results of Example 14.2 to determine the ideal pump head. hpump =

U(Vt 2 - Vt 1) 13.09 m>s (10.664 m>s - 0) = 14.23 m = g 9.81 m>s2

The actual pump head is then (hpump)act = (hpump) - hL = 14.23 m - 2.8 m = 11.43 m Hydraulic Efficiency. Applying Eq. 14–18 yields hpump =

(hpump)act hpump

(100%) =

11.43 m (100%) = 80.3% 14.23 m

Ans.

14.4

EXAMPLE

821

ideal perFormanCe For pumps

14.4

The impeller blades on the radial-flow pump in Fig. 14–8a have an average inlet radius of 50 mm and outlet radius of 150 mm, and an average width of 30 mm. If the blade angles are b1 = 20° and b2 = 10°, determine the flow through the pump, and the actual pump head when the impeller is rotating at 400 rev>min. The angle b1 allows the flow to enter the impeller in the radial direction. Frictional head losses produced by the pump are 0.8 m.

14

50 mm 150 mm

SOLUTION Fluid Description. We will assume steady, incompressible flow and use average velocities. Kinematics. To find the flow, we must first determine the speed of the fluid as it moves onto the blades. We will also need the speed of the blades at its entrance and its exit.

30 mm

U1 = vr1 = a

400 rev 1 min 2p rad ba ba b (0.05 m) = 2.094 m>s min 60 s 1 rev 400 rev 1 min 2p rad U2 = vr2 = a ba ba b (0.150 m) = 6.283 m>s min 60 s 1 rev

400 revmin b2 5 10° b1 5 20°

The kinematic diagram for the flow onto the impeller is shown in Fig. 14–8b. Since V1 is in the radial direction (a1 = 90°),

50 mm

V1 = Vr 1 = U1 tan b1 = (2.094 m>s) tan 20° = 0.7623 m>s Flow.

150 mm

The flow into, as well as out of, the pump is

(a)

Q = V1A1 = V1(2pr1b1)

= 10.7623 m>s232p(0.05 m)(0.03 m)4

= 0.007184 m3 >s = 0.00718 m3 >s

a1 5 90°

Ans.

U2Q cot b2 U 22 g 2p r2bg (6.283 m>s)2

=

9.81 m>s2

-

(6.283 m>s)(0.007184 m3 >s) cot 10° 2p(0.150 m)(0.03 m)(9.81 m>s2)

= 3.10 m The actual pump head is then

1hpump 2 act = hpump - hL = 3.10 m - 0.8 m = 2.30 m

Ans.

(Vrel)1 b1 5 20°

20° U1 5 2.094 ms

Hydraulic Efficiency. The ideal pump head is hpump =

V1

(b)

Fig. 14–8

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EXAMPLE

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14.5 The impeller of a radial-flow water pump has an outer radius of 200 mm, an average width of 50 mm, and a tail angle b2 = 20°, Fig. 14–9. If the flow onto the blades is in the radial direction, and the blades are rotating at 100 rad>s, determine the ideal power. The discharge is 0.12 m3 >s.

Typical radial-flow pump

SOLUTION I Fluid Description. We assume steady, incompressible flow and will use average velocities. Here rw = 1000 kg>m3. Kinematics. The power can be determined using Eq. 14–14, with a1 = 90° (Vt 1 = 0). First we must find U2 and Vr 2. The speed of the impeller at the exit is U2 = vr2 = (100 rad>s)(0.2 m) = 20 m>s Also, the radial component of velocity Vr 2 can be found from Q = Vr2 A2. 0.12 m3 >s = Vr 2 32p(0.2 m) (0.05 m)4;

Vr 2 = 1.910 m>s

As shown in Fig. 14–9, Vt 2 = U2 - Vr 2 cot 20° = 20 m>s - (1.910 m>s) cot 20° = 14.75 m>s

14.4

ideal perFormanCe For pumps

823

Vr2 5 1.910 ms V2

U2 5 20 ms

208 a2

Vt2

V1

14 908

b2 5 208

(Vrel)2

100 rad s

U1 0.2 m

Fig. 14–9

Ideal Power # Wpump = rQ(U2Vt 2 - U1Vt 1)

= (1000 kg>m3)(0.12 m3 >s)[(20 m>s)(14.75 m>s) - 0] = 35.41(103) W = 35.4 kW

Ans.

SOLUTION II The ideal power can also be related to the ideal pump head by # Eq. 14–15, Wpump = Qghpump . First we must find hpump by using Eq. 14–19. This gives hpump =

U2Q cot b2 U 22 g 2p r2bg (20 m>s)2

=

9.81 m>s2

-

(20 m>s) 1 0.12 m3 >s 2 cot 20°

2p 1 0.2 m 21 0.05 m 21 9.81 m>s2 2

Therefore, # Wpump = Qghpump =

= 30.08 m

1 0.12 m3 >s 21 1000 kg>m3 21 9.81 m>s2 21 30.08 m 2

= 35.41(103) W = 35.4 kW

Ans.

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14.5

TURBINES

Unlike a pump that delivers energy to a fluid, a turbine is a turbomachine that receives energy from the fluid. Turbines can be classified into two different types: impulse turbines and reaction turbines. Each type removes the fluid’s energy in a specific way.

14

v

r V

Pelton wheel (a) Vfcs u

Vfcs

V

Impulse Turbines. An impulse turbine consists of a series of “buckets” attached to a wheel as shown in Fig. 14–10a. A high-velocity water jet strikes the buckets, and the momentum of the water is then converted into an angular impulse acting on the wheel. If the flow is split equally into two directions by using double-cupped buckets, as shown in Fig. 14–10b, then the device is generally referred to as a Pelton wheel, named after Lester Pelton, who designed this turbine in the late 1870’s. An impulse turbine such as this is often used in mountainous regions, where water is delivered with a high velocity and with small flow. The force created by the fluid striking the buckets of a Pelton wheel can be determined by applying the equation of linear momentum to the control volume, Fig. 14–10b, that is attached to the bucket and moving with a constant velocity U. If V is the velocity of flow from the nozzle jet, then Vf>cs = V - U is the relative velocity onto, and then off, each bucket, Fig. 14–10c. Therefore, from the free-body diagram, Fig. 14–10d, for steady flow, with Q = Vf>cs A, we have

U

(b)

ΣF =

V U

0 Vr dV + V r Vf>cs # dA 0t Lcv Lcs

Vfcs

1S+ 2

(c)

-F = 0 + Vf>cs r1 -Vf>cs A2 + 1 -Vf>cs cos u2r1Vf>csA2

F

F = rQVf>cs(1 + cos u)

(14–20)

Free-body diagram (d)

Fig. 14–10

Notice that when the exit angle 0° … u 6 90°, the cos u remains positive, thereby producing a larger force than when 90° … u … 180° so that the cos u is negative.

14.5

turbines

825

Torque. The torque developed on the wheel is the moment of this impulsive force about the axis of the wheel. Here we have a succession of buckets that receives the flow, and so for continuous turning, we have T = Fr = rQVf>cs(1 + cos u)r

(14–21)

14

Power. Since each bucket has an average speed of U = vr, Fig. 14–10a, the shaft power developed by the wheel is # Wturbine = Tv = rQVf>csU(1 + cos u)

(14–22)

Maximum power from the fluid requires the# bucket angle u = 0°, since cos 0° = 1. Also, since Vf>cs = V - U, then Wturbine = rQ(V - U )U(2). The product (V - U)U must also be a maximum, which requires

Pelton wheels are most efficient at high hydraulic heads and low flows. This one has a diameter of 2.5 m, and was used for a head of 700 m, flow of 4 m3 >s, and a rotation of 500 rpm.

d (V - U )U = (0 - 1)U + (V - U )(1) = 0 dU U =

V 2

Therefore, to generate maximum power, the relative velocity of the fluid onto the buckets is Vf>cs = V - V>2 = V>2, so as the fluid comes off the buckets, it will have a velocity of V = U + Vf>cs = V>2 - V>2 = 0. In other words, the fluid has no kinetic energy, and instead the kinetic energy of the water jet will be converted completely into the rotational kinetic energy of the wheel. Substituting these results into Eq. 14–22, we get # V2 (Wturbine)max = rQa b 2

Vfcs u

V

Vfcs U

(14–23)

This, of course, is a theoretical value, which unfortunately is not practical since the fluid coming off one bucket would actually strike and splash off the back of the next bucket, thus causing a reverse impulse. To avoid this “back splashing,” engineers normally design the exit angle u to be about 20°, Fig. 14–10b. Considering other losses due to splashing, and viscous and mechanical friction, a Pelton wheel will have an efficiency of about 85% in converting the energy of the fluid into rotational energy of the wheel.

(b)

Fig. 14–10 (repeated)

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EXAMPLE

turbomaChines

14.6 The Pelton wheel in Fig. 14–11a has a diameter of 3 m and bucket deflection angles of 160°. If the diameter of the water jet striking the wheel is 150 mm, and the velocity of the jet is 8 m>s, determine the power developed by the wheel when it is rotating at 3 rad>s.

14

V2 u

v = 3 rads

1608

V 5 8 ms

3m U

U 5 4.50 ms

Vfcs

150 mm

Vfcs

8 ms (a)

(b)

Fig. 14–11

SOLUTION Fluid Description. We have steady flow onto each bucket measured relative to the bucket, and we will assume water to be an ideal fluid for which rw = 1000 kg>m3. Kinematics. We will use Eq. 14–22 to solve the problem, but first we must find U and Vf>cs. The average speed of the buckets is U = vr = 13 rad>s2(1.5 m) = 4.50 m>s

The kinematic diagram in Fig. 14–11b shows the flow of the water onto and then off each bucket. As noted, the relative speed of the flow onto the bucket is Vf>cs = 8 m>s - 4.50 m>s = 3.50 m>s. Power. Since the water flows on all the blades, then Q = VA. With u = 20°, we have # Wturbine = rwQVf>csU(1 + cos u)

# Wturbine = 11000 kg>m3 23(8 m>s)p(0.075 m)2 4(3.50 m>s)(4.50 m>s)(1 + cos 20°) = 4.32 kW

Ans.

14.5

turbines

827

Generator Runner blades Wicket gate

Draft tube

Scroll or volute

Inlet from Guide penstock vanes

Turbine

Propeller or Kaplan turbine (a)

Draft tube

Francis turbine (b)

Fig. 14–12

Reaction Turbines. Unlike an impulse turbine, the flow delivered to blades or runners of a reaction turbine is relatively slow; however, this type of turbine can handle a large fluid flow. If the turbine is designed to receive axial flow, it is known as a propeller turbine. A Kaplan turbine, named after Viktor Kaplan, Fig. 14–12a, is a type of propeller turbine, where the blades can be adjusted to accommodate various flows. Propeller turbines are used for low flows and high hydraulic heads. If the flow through the turbine is radial or a mix of radial and axial flow, it is called a Francis turbine, named after James Francis, Fig. 14–12b. This turbine is the most common type used for hydroelectric power because it can be designed for a variety of flows and hydraulic heads.

This propeller turbine has a diameter of 4.6 m, and was used for a head of 7.65 m, flow of 87.5 m3 >s, and had a rotation of 75 rpm.

This Francis turbine has a diameter of 4.6 m, and was used for a head of 69 m, flow of 147 m3 >s, and had a rotation of 125 rpm.

14

828

14

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Kinematics. The analysis of reaction turbines follows the same methods used to analyze axial- and radial-flow pumps, discussed previously. Even the blade kinematics is the same, as outlined in the Procedure for Analysis of Sec. 14.2. As a review, consider the case of a turbine fan used on a jet engine. This is a type of propeller turbine, where constant axial flow of a mixture of hot air and fuel passes through a sequence of stationary or stator blades, then rotating or rotor blades, Fig. 14–13a. Considering one of these interactions, Fig. 14–13b, the stator directs the flow at an angle a1, so that it strikes a rotor blade with a velocity V1. The rotor blade then delivers the flow with a velocity V2 at the angle a2 tangent to the blades on the next stator, etc. Since the turbine removes energy from the flow, the fluid velocity (kinetic energy) will decrease (V2 6 V1) as the flow passes over each set of rotors. Following the same convention as for pumps, note carefully how the velocity components are established. By convention, a is measured between +U and V, and b is measured between –U and Vrel.

Rotors Stators Axial-flow turbine (a)

(Vrel)2

b2 b1 V1 a1

b2

(Vrel)1 Vt2

Va

Va

a2

Vt1

V2

V2

a1 a2

V1

Stator

U Flow onto rotor blade

U Flow off rotor blade Rotor (b)

Fig. 14–13

Stator

14.5

turbines

829

Torque. If we apply the angular momentum equation to the rotor, then we can determine the torque applied to the rotor from Eq. 14–2 or Eq. 14–11. T = -rQrm 1Vt 2 - Vt1 2

(14–24)

Propeller turbine

T = -rQ(r2Vt 2 - r1Vt1 2

(14–25)

Francis turbine

The minus signs are included here since Vt1 7 Vt 2, Fig. 14–13b, and we require the torque of the liquid on the rotor to be positive.

Power. Knowing the torque, the power produced by the turbine is then # Wturbine = Tv

(14–26)

Head and Efficiency. The ideal turbine head removed from the fluid can be expressed as a function of the power by using Eq. 14–15. # Wturbine hturbine = (14–27) Qg Finally, the actual turbine head removed from the fluid will be greater than the ideal head, because the actual head also accounts for friction losses as the fluid passes through the turbine. Thus, (hturbine)act = hturbine + hL Therefore, the turbine efficiency based on the friction losses is hturbine =

hturbine (100%) (hturbine)act

(14–28)

The example that follows provides a typical application of these equations.

IMPORTANT POIN T S • A Pelton wheel acts as an impulse turbine, where a high-velocity

• •

jet of fluid strikes the buckets on the wheel. Changing the momentum of the jet creates a torque, causing the wheel to turn, thereby producing power. Maximum power produced by a Pelton wheel occurs when the flow is completely reversed by the buckets and the wheel turns so that the velocity of the fluid leaving the buckets is zero. Reaction turbines, such as Kaplan and Francis turbines, are similar to axial-flow and radial-flow pumps, respectively. All turbines remove energy from the fluid.

14

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EXAMPLE

turbomaChines

14.7 The guide vanes of a Francis turbine in Fig. 14–14a direct water onto the 200-mm-wide runner blades at an angle of a1 = 30°. The blades are rotating at 60 rev>min, and they discharge water at 4 m3 >s in the radial direction, that is, toward the center of the turbine, a2 = 90°, Fig. 14–14b. Determine the power developed by the turbine and the ideal head loss.

14

SOLUTION Fluid Description. We have steady flow onto the blades and assume water to be an ideal fluid for which rw = 1000 kg>m3. Linkage used to operate guide vanes for a Francis turbine

Kinematics. In order to determine the power, we must first find the tangential components of velocity onto and off the blades, Fig. 14–14b. Using the flow, the radial component of velocity of the water onto each blade, at r1 = 1 m, is Q = Vr1A1

4 m >s = Vr1 32p(1 m)(0.20 m)4 3

1m

Vr1 = 3.183 m>s

A

60 revmin

308

Thus, from Fig. 14–14b, the tangential component is 0.75 m

Vt1 = (3.183 m>s) cot 30° = 5.513 m>s At the tail of the blade, Vt 2 = 0, since only radial flow occurs. Power.

(a)

Applying Eqs. 14–25 and 14–26, for a Francis turbine, we get # Wturbine = -rwQ(r2Vt 2 - r1Vt1)v

# 60 rev 1 min 2p rad Wturbine = -(1000 kg>m3)(4 m3 >s) c 0 - (1 m)(5.513 m>s) d a ba ba b min 60 s 1 rev # Wturbine = 138.6(103) W = 139 kW Water delivered to blade

(Vrel)1

(Vrel)2 V2 5 Vr 2

a1 5 308

Vt1 U2 a2 5 908

60 revmin (b)

Fig. 14–14

Head Loss.

U1 V1 Vr1 5 3.183 ms

hturbine

Ans.

The ideal head loss is determined from Eq. 14–27. # 138.6(103) W Wturbine = = Qg (4 m3 >s)(1000 kg>m3)(9.81 m>s2) = 3.53 m

Ans.

14.6

14.6

pump perFormanCe

831

PUMP PERFORMANCE

When selecting a specific pump to use for an intended application, an engineer must have some idea about its performance characteristics. For # any given flow, these include the shaft power requirements Wpump, the actual pump head (hpump)act that can be developed, and the pump’s efficiency h. In past sections, we have studied how these characteristics are calculated based on an analytical treatment; however, for any real application, where fluid and mechanical losses occur, the data must be determined experimentally. In order to give some idea as to how this is accomplished, we will consider an experimental test as it applies to a radial-flow pump.* The testing of the pump follows a standardized procedure, which is outlined in Ref. [12]. The setup is shown in Fig. 14–15a, where the pump is made to circulate water (or the intended fluid) from tank A to another tank B through a constant-diameter pipe. Pressure gages are located on each side of the pump, and the valve is used to control the flow, while a meter measures the flow before the water is returned from B to A. The pump’s impeller is turned by an electric motor and so the power input # Wpump can also be measured. The test begins with the valve closed. The pump is then turned on and its impeller begins to run at a fixed nominal speed v0. The valve is then opened slightly, and the rate of flow Q, the pressure difference across the # pump (p2 - p1), and the power supply Wpump are all measured. The actual pump head is then calculated using the energy equation, applied between points 1 and 2. Here V2 = V1 = V, and any elevation difference z2 - z1 and head losses within the pump are factored in as part of the pressure head, i.e., (hpump)act = hpump - hL. We have

A

B

Flow meter 1

2 Pump test (a)

Fig. 14–15

*A more detailed description of the methods used in engineering practice is discussed in books related to pump design. See Ref. [3, 5, 6, 7].

14

Pumps and pipe systems play an important role in refineries and chemical processing plants.

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pin pout V out2 V in2 + + + zin + hpump = + zout + hturbine + hL g g 2g 2g

14

p1 p2 V2 V2 + + + 0 + hpump = + 0 + 0 + hL g g 2g 2g (hpump)act = hpump - hL =

p 2 - p1 g

# Once (hpump)act is calculated, and Q and Wpump are measured, we can then determine the hydraulic efficiency using Eqs. 14–15 and 14–18 # written as hpump = 1Wpump 2 >Qg . We have

hpump =

(hpump)act hpump

(100%) =

Qg(hpump)act (100%) # Wpump

If we continue to change Q in increments, until the pump is running# at its maximum capacity, and plot the successive values of (hpump)act, Wpump, and hpump versus Q, we will produce three performance curves that look similar to those shown in Fig. 14–15b. Included on this graph is the straight-line ideal pump head, as established previously in Fig. 14–7a. Both this and the actual pump head (blue curve) assume the impeller blade has a backward curve, b2 6 90°, which, as we stated earlier, is most often the case.

(hpump)act, hpump, Wpump Ideal pump head b2 , 908 B A Power BEP

Actual pump head

Efficiency Qreq’d Pump performance (b)

Fig. 14–15 (cont.)

Qmax

Q

14.6

pump perFormanCe

Note that the actual pump head, Fig. 14–15b, is always below its ideal head. Several factors are responsible for this. The most important is the loss of head that occurs because the impeller has a finite, rather than an infinite, set of blades. As a result, the flow will actually leave each blade at an angle that will be slightly different from the blade’s design angle b2. Hence, the pump head will be proportionally reduced. Along with this loss, there are additional losses due to fluid friction and mechanical friction at the shaft bearings and seals, and turbulent losses due to improper flow along the impeller. The green curve for hydraulic efficiency in Fig. 14–15b shows that as Q increases, this efficiency reaches a maximum called the best efficiency point (BEP), and then it begins to decrease to zero at Qmax. If the pump is required to produce a flow of Qreq’d, then based on these curves, this particular pump should be chosen, because it will operate at maximum efficiency (BEP) for this flow. When used, it will require a power defined at point A on the red curve, and produce a head defined at point B on the blue curve.

14

Manufacturer’s Pump Performance Curves. From similar

4 D

B

2

Head (m)

0 65% 70% 75% 56 175 mm dia. 80% 86% 48 150 mm dia. C A 40 125 mm dia. 32

80%

24 16 8 kW

8 300

600

(NPSH)req’d (m)

types of experimental tests, manufacturers provide pump performance curves for many of their pumps. Normally, each pump is designed to run at a constant nominal speed v0 and is designed to accommodate several different diameter impellers within its casing. An example of these curves is shown in Fig. 14–16, which is for an impeller that runs at v0 = 1750 rpm and can be fitted with 125-mm-, 150-mm-, and 175-mm-diameter impellers as indicated by the blue curves. For a specific flow, we can obtain the actual pump head (blue), the hydraulic efficiency (green), and brake horsepower (red).* For example, note that a pump with a 175-mm-diameter impeller and running at 1750 rpm will be most efficient at 86% (point A) when it produces a flow of 1400 liters>min. It will produce a total head of about 46 m and require a power of 16 kW. 6

ω0=1750 RPM 75% 70% 65% 18 kW

15 kW 12 kW

900 1000 1200 1400 1500 1800 Flow Q (liters/min)

2100

2400

*The efficiency curves in Fig. 14–16 look different from the shape in Fig. 14–15b, because each curve is only for a constant efficiency.

833

Fig. 14–16

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14.7

14

CAVITATION AND THE NET POSITIVE SUCTION HEAD

One important phenomenon that can actually limit the performance of a pump occurs when the pressure within its casing drops below a certain limit, so that it causes cavitation. Recall from Sec. 1.9 that cavitation is the result of the pressure within a liquid falling to or below the vapor pressure pv for the liquid. For a radial-flow pump this usually occurs on the suction side, within the center or eye of the impeller, since there the pressure is the lowest. When cavitation does occur, the liquid will boil, and bubbles or cavities will form within the liquid. As these bubbles are transported along the blades of the impeller, they reach regions of higher pressure, where they suddenly collapse. This violence causes outward pressure waves that produce a repeated pounding on any adjacent hard surface, which eventually causes material fatigue and wearing away of the surface. The process of wear is further aggravated by corrosion, or other electromechanical effects. Cavitation is associated with vibration and noise, which usually resembles the sound of rocks or pebbles striking the sides of the casing, and once it occurs, the pump’s efficiency falls dramatically. For any specific pump there will be a critical suction head, developed at the suction side of the pump, where cavitation within the pump will begin to occur. Its value can be determined experimentally by maintaining a specified flow and varying the elevation of the pump above a reservoir. As the vertical length of pipe needed to draw the fluid up into the pump increases, a critical elevation will be reached where the pump’s efficiency will suddenly drop. Using the energy equation, with the datum at the pump inlet, we can obtain this critical suction head. It is the sum of the pressure and velocity heads at the inlet, 1p>g + V 2 >2g2, since z = 0. If the vapor pressure head pv >g for the liquid is subtracted from this head, we have p pv V2 + (14–29) NPSH = g g 2g This result is referred to as the net positive suction head NPSH. Since cavitation or pv usually occurs within the pump on the surface of the impellers, and not at the inlet, the NPSH is actually the additional head needed to move the fluid from the inlet to the point of cavitation within the pump. Manufacturers normally perform experiments of this sort, and for various values of the flow, the results are plotted on the same graph as the performance curves. The values are termed the required net positive suction head, or (NPSH)req’d. Notice that for the pump in Fig. 14–16, the (NPSH)req’d is about 3.50 m (point B) when the flow is 1400 liters >min and the pump with the 175-mm-diameter impeller is running at its maximum efficiency (86%). Once (NPSH)req’d is determined from the graph, it is then compared to the available net positive suction head, (NPSH)avail, which is determined from the pressure and velocity heads for a specific flow system. To prevent cavitation, it is necessary that (NPSH)avail Ú (NPSH)req’d (14–30) For further clarification, the following example provides an application of this concept.

14.7

EXAMPLE

14.8

Datum B

75 mm

h

835

Cavitation and the net positive suCtion head

A

Fig. 14–17

The pump shown in Fig. 14–17 is used to transfer 20°C sewage water from the wet well to the sewage treatment plant. If the flow through the 75-mm-diameter pipe is to be 1>60 m3 >s, determine whether cavitation occurs when the pump in Fig. 14–16 is selected. The pump turns off just after the water reaches its lowest level of h = 4 m. Take the friction factor for the pipe to be f = 0.02, and neglect minor losses. SOLUTION Fluid Description. We will assume steady flow of an incompressible fluid. From the table in Appendix A, rw = 998.3 kg> m3 for water at T = 20°C . Inlet Pressure. We can determine the available suction head at the pump’s impeller inlet by applying the energy equation. The control volume consists of the water in the vertical pipe and the well, Fig. 14–17. The largest suction at B occurs when h = 4 m. Here we will work with absolute pressures since the vapor pressure is generally reported as an absolute pressure. At A the atmospheric pressure is pA = 101.3 kPa. Since 1 3 m >s Q 60 VB = V = = = 3.7726 m>s A p 1 0.0375 m 2 2 then pA pB V A2 V B2 L V2 V2 + + zA + hpump = + + zB + h turbine + f + ΣKL g g 2g 2g D 2g 2g

101.3(103)N>m2 (998.3 kg>m3)(9.81 m>s2)

+ 0 - 4m + 0 =

pB (998.3 kg>m3)(9.81 m>s2)

+

(3.7726 m>s)2 2(9.81 m>s2)

+ 0 + 0 + 0.02 a

(3.7726 m>s) 4m d + 0 bc 0.075 m 2(9.81 m>s2) 2

pB = 47.445(103) Pa The available suction head at the pump inlet is therefore 47.445(103) N>m2 (3.7726 m>s)2 pB V B2 + = + = 5.570 m g 2g (998.3 kg>m3) (9.81 m>s2) 2(9.81 m>s2) From Appendix A, at 20°C, the (absolute) vapor pressure for water is 2.34 kPa. Therefore, the available NPSH is (NPSH)avail = 5.570 m -

2.34(103) N>m2 (998.3 kg>m3)(9.81 m>s2)

= 5.331 m

The flow in liters per minute is 1 3 m 60 1000 liters 60 s Q = £ ≥a ba b = 1000 liters>min 3 s 1 min 1m From Fig. 14–16, at Q = 1000 liters>min, (NPSH)req’d = 2.33 m (point D), and since (NPSH)avail 7 (NPSH)req’d, cavitation will not occur in the pump. Also note from Fig. 14–16 that if a 175-mm-diameter impeller is used in this pump (point C), then the pump will have an efficiency of about 77% and will have a shaft power of about 13.5 kW.

14

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14.8

PUMP SELECTION RELATED TO THE FLOW SYSTEM

A flow system may consist of reservoirs, pipes, fittings, and a pump that is used to convey the fluid. If the pump is required to provide a specific flow for the system, then this should be accomplished in the most economical and efficient way. For example, consider the system shown in Fig. 14–18a. If we apply the energy equation between points 1 and 2, we have

14

pin pout Vout2 Vin2 + + + zin + hpump = + zout + hturbine + hL g g 2g 2g

0 + 0 + z1 + 1hpump 2 req’d = 0 + 0 + z2 + 0 + hL 1hpump 2 req’d = (z2 - z1) + hL

Here (hpump)req’d is the required head to be supplied by the pump to the system. It is actually a function of Q2. To show why, note that the accumulated head loss in the above equation is of the form hL = C(V 2>2g), where C is a constant. Since V = Q>A, then we have hL = C1Q2 >2gA2 2 = C′Q2. If we plot 1hpump 2 req’d = (z2 - z1) + C′Q2, it will be a parabola and will look something like the solid black curve A shown in Fig. 14–18b. If the pump in Fig. 14–18a produces a pump head 1hpump 2 req’d, like the head performance curve (blue) in Fig. 14–15b, then the intersection of these two curves is at point O. This point represents the flow anticipated or required for the system and it also indicates the flow provided by the pump. This is the operating point for the system, and if it is close to the best efficiency point (BEP) for the pump (Fig. 14–15b), it justifies choosing this pump for this application. In doing so, realize that over time the pump’s characteristics will change. For example, the pipes in the system may corrode, causing an increased frictional head loss, and this would raise the system curve, Fig. 14–18b. Also, the pump can deteriorate, causing its performance curve to lower. Both effects would shift the operating point to O′, and lower the pump’s efficiency. As a result, for best engineering design, the consequences of these changes should be considered when choosing a pump for any particular application. hpump

2

Older system New system A

(hpump)req’d

O9 O New pump head performance

z2

1 z1

Datum

Older pump head performance Q

(z2 2 z1)

Qreq’d

Pump system curves (b)

(a)

Fig. 14–18

14.8

EXAMPLE

837

pump seleCtion related to the Flow system

14.9 B

The radial-flow pump in Fig. 14–19a is used to transfer water from the lake at A into a large storage tank B. This is done through a 100-mm-diameter pipe that is 100 m long and has a friction factor of 0.015. The manufacturer’s data for the performance of the pump is given in Fig. 14–19b. Determine the flow if this pump is selected with a 175-mm-diameter impeller to do the job. Neglect minor losses.

40 m Datum

(a)

System Equation. We can relate the actual pump head to the flow Q by applying the energy equation between points A and B. The control volume contains the water within the pipe and a portion of the water in the lake. Since the friction factor has been given, we do not need to obtain its value from the Moody diagram. pA V A2 pB V B2 + + zA + hpump = + + zB + hturbine + hL g g 2g 2g

0 + 0 + 0 + 1hpump 2 req’d = 0 +

Also,

Head (m)

SOLUTION Fluid Description. We assume steady, incompressible flow while the pump is operating.

14

A

72 64 65% 70% Eq. 1 75% 56 175 mm dia. 80% 86% 80% 48 150 mm dia. O9 O 75% 40 125 mm dia. 70% 65% 32 18 kW 24 15 kW 16 12 kW 8 8 kW 0.005 0.01 0.015 0.02 0.025 0.03 0.035

Flow Q (m3 s) v0 5 1750 rpm (b)

Fig. 14–19

V2 100 m V2 + 40 m + 0 + (0.015) a b c d 2 0.1 m 219.81 m>s 2 2(9.81 m>s2)

Q = V [p(0.05 m)2]

When the above two equations are combined, we get 1hpump 2 req’d = [40 + 13.22 (103) Q2] m (1) A plot of this equation is shown in Fig. 14–19b. It shows the pump head 1hpump 2 req’d that must be supplied by the pump, thereby providing a flow Qreq’d for the system. Along with this curve are the manufacturer’s curves for the pump. (For convenience Q is reported in m3 >s; however, in practice it is commonly in liters>min.) We see that for the pump with a 175-mm-diameter impeller, the operating point O is where the performance and system head curves intersect, graphically, at about Q = 0.022 m3 >s (Ans.) and 1hpump 2 req’d = 46 m. Therefore, if this pump is selected, then the efficiency for this flow is determined from the efficiency curves to be about h = 84%.* By comparison, this is near the best efficiency point for the pump (86%), and so the choice of this pump with a 175-mm-diameter impeller would be appropriate. It is interesting to note that if the elevation difference between the water levels in the lake and storage tank were 30 m, rather than 40 m, then Eq. 1 would plot as the dashed curve in Fig. 14–19b. In this case, the operating point O′ gives a flow of about 0.0275 m3 >s; however, then the operating efficiency for the 175-mm-diameter impeller is about h = 78%, making the pump a relatively poor choice for this condition. Instead, performance curves for other pumps would have to be considered to match a more efficient pump for the system. *As noted by the curves, use of a smaller-diameter impeller would give a lower efficiency.

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14.9

14

TURBOMACHINE SIMILITUDE

In the previous two sections we showed how to select a radial-flow pump to produce a required flow. However, if we are to select a type of pump, such as an axial-, radial-, or mixed-flow pump, that works best for a specific job, it then becomes convenient to use dimensional analysis, and express each machine’s performance parameters in terms of dimensionless groups of variables that involve its geometry and the fluid properties. We can also use these dimensionless groups to compare the performance of one type of pump with that of a similar type of pump, or if we build a model of a pump, to test its performance characteristics and thereby predict the characteristics for a prototype. In Sec. 14.6 we found that when producing performance curves, it was convenient to consider the # dependent variables for the pump to be the pump head h, the power W, and the efficiency h. From experiments, it has been found that these three variables all depend upon the fluid properties r and m, the flow Q, the rotation v of the impeller, and some “characteristic length”—usually the diameter D of the impeller. As a result, the three dependent variables are related in some way to these independent variables by the functions

gh = f1(r, m, Q, v, D) # W = f2(r, m, Q, v, D) h = f3(r, m, Q, v, D)

Here we have considered gh, the energy per unit mass, for the convenience of a dimensional analysis. If we apply the Buckingham Pi theorem, as described in Chapter 8, we can create dimensional groups of the variables in these three functions.* They are

gh

= f4 a

rvD2 b vD3 m

h = f6 a

rvD2 b vD3 m

Q

,

v2D2 # Q rvD2 W , = f a b 5 vD3 m rv3D5

*See, for example, Probs. 8–37 and 8–39.

Q

,

14.9

turbomaChine similitude

The dimensionless parameters on the left of these three equations are called the head coefficient, the power coefficient, and, as noted previously, the efficiency. On the right, Q>vD3 is the flow coefficient, and rvD2 >m is a form of the Reynolds number that considers the viscous effects within the pump. Experiments have shown that for either pumps or turbines, this number 1rvD2 >m2 does not affect the magnitudes of the three dependent variables as much as the flow coefficient. As a result, we will neglect its effect and write # gh Q Q Q W = f7 a b b h = f9 a b = f8 a 2 2 3 3 3 5 vD vD vD vD3 rv D

14

Q1 v1D13

=

Q2 v2D23

(14–31)

Efficiency, head, and power coefficients

Pump Scaling Laws. For a particular type of pump, these three functional relationships can be determined by experiment, which requires varying the flow coefficient, and then plotting the resultant head coefficient, power coefficient, and the efficiency. The resulting curves will all have shapes something like those shown in Fig. 14–20; and, because these shapes are all similar for any family of pumps (axial-, radial-, or mixed-flow), the coefficients become pump scaling laws, sometimes called pump affinity laws. In other words, they can be used to design or compare any two pumps of the same type. For example, if the flow coefficients for two pumps of the same type are to be the same, then

Efficiency

Head coefficient

Power coefficient

Flow coefficient

Likewise, the other two coefficients can be equated between, say, two radial-flow pumps, or a model and its prototype, to determine their scaled characteristics. If the same fluid is used, then r cancels, as well as g, so that h1 v12D12

=

h2 v22D22

(14–32)

Head coefficient

# W1 v13D15

=

# W2 v23D25

(14–33)

Power coefficient

h1 = h2 Efficiency

As stated previously, a radial-flow pump can often house different-diameter impellers within its casing, or it can be made to run at different angular velocities. As a result, these scaling laws can also be used to estimate Q, h, # and W for the pump when D or v changes. For example, if the pump can produce a head of h1 when the impeller diameter is D1, then from the head coefficient, the head produced by the same pump with an impeller diameter D2 would be h2 = h1 1D22 >D12 2, provided it has the same v.

839

Flow coefficient

Fig. 14–20  

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C h a p t e r 14

hpump

Mixed-flow pumps Radial-flow pumps

turbomaChines

Axial-flow pumps

14

Specific Speed. To select the type of turbomachine to use for a specific job, it is sometimes helpful to use another dimensionless parameter that does not involve the dimensions of the machine. This parameter is called the specific speed, Ns, and it can be determined either through a dimensional analysis or by simply eliminating the impeller diameter D from the ratio of the (dimensionless) flow and head coefficients. This gives

Ns Pumps (a)

Ns =

hturb Francis turbine

Propeller turbine

Impulse turbine Ns Turbines (b)

Fig. 14–21

vQ1>2 (gh)3>4

(14–34)

For each type of turbomachine we can plot Ns versus its efficiency as shown in Fig. 14–21. See Ref. [15]. Notice that maximum efficiency for a particular type of turbomachine occurs at the peak of each curve, and this peak is located at a particular value of specific speed Ns. For example, it is inherent in their design that radial-flow pumps operate at low specific speeds, so they produce low flows and large heads (high pressure), Fig. 14–21a. On the other hand, axial-flow pumps produce high flows and develop low heads (low pressure). They operate well at high specific speeds, although this does make them susceptible to cavitation. Pumps designed for mixed flow operate in the intermediate range of specific speeds. Typical profiles of the impellers used for these three types of pumps are shown in Fig. 14–22. The same trend occurs for turbines that work on the same principles as pumps, Fig. 14–21b.

impeller casing

I MPO RTA N T PO I N T S • Due to both mechanical and fluid friction losses in turbomachines,

Radial flow impeller Ns 5 700

Mixed flow impeller Ns 5 2500

Axial flow impeller Ns 5 10 000

•

Impeller selection for various values of Ns

Fig. 14–22

•

• •

the actual performance of the machine must be determined by experiment. Cavitation can occur within turbomachines wherever the liquid pressure falls below the vapor pressure for the liquid. To avoid this, it is important to select a turbomachine that has a (NPSH)avail that is greater than its required (NPSH)req’d. Performance curves of efficiency, total head, and power versus flow provide a means for selecting a proper-size pump for a particular application. The pump selected must match the required flow and total head demands for the fluid system, and it must operate with high efficiency. The performance characteristics of a pump or turbine can be compared to a geometrically similar pump or model by using the dimensionless parameters of the head, power, and flow coefficients. The selection of a type of turbomachine to be used for a specific task is based upon the machine’s specific speed. For example, radial-flow pumps are efficient for low flows and delivering high heads, whereas axial-flow pumps are efficient at high flows and delivering low heads.

14.9

EXAMPLE

turbomaChine similitude

841

14.10

A turbine for the dam operates under a hydraulic head of 90 m, producing a discharge of 50 m3 >s. If the reservoir level drops so that the hydraulic head becomes 60 m, determine the discharge from the turbine.

SOLUTION

Here h1 = 90 m and Q1 = 50 m3 >s. Because we also know h2 = 60 m, then we will eliminate the unknowns v1 and v2 from the flow and head coefficients to obtain a relationship between h and Q. Using Eqs. 14–31 and 14–32, we have Q1D23 v1 = v2 Q2D13 v12 v22

=

(1)

h1D22 h2D12

so that Q12D24 Q22D14

=

h1 h2

(2)

Since D1 = D2, Q2 = Q1

h2 A h1

= (50 m3 >s)

60 m = 40.8 m3 >s A 90 m

Ans.

14

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EXAMPLE

turbomaChines

14.11 The Francis turbine in Fig. 14–23 is rotating at 75 rev>min under a hydraulic head of 10 m and develops 85 kW with a discharge of 0.10 m3 >s. If the guide vanes remain in their fixed position, what is the rotation of this turbine when the hydraulic head is 3 m? Also, what is the corresponding discharge and estimated power of the turbine?

14

SOLUTION

# Here v1 = 75 rev>min, h1 = 10 m, W1 = 85 kW, and Q1 = 0.10 m3 >s . For h2 = 3 m, the rotation v2 can be determined from the head coefficient similitude, Eq. 14–32, h1 v12D12

=

h2 v22D22

Since D1 = D2, v

v2 = v1

h2 A h1

= (75 rev>min)

3m = 41.08 rev>min = 41.1 rev>min Ans. A 10 m

To obtain Q2 we use the flow coefficient similitude, Eq. 14–31, with D1 = D2; that is, Fig. 14–23

Q2 = Q1 a =

v2 b v1

1 0.10 m3 >s 2 a

41.08 rev>min 75 rev>min

b = 0.0548 m3 >s Ans.

# Finally, W2 is determined from the power coefficient similitude, Eq. 14–33, with D1 = D2. We have # # v2 3 W2 = W1 a b v1 = (85 kW) a

41.08 rev>min 75 rev>min

b = 14.0 kW 3

Ans.

14.9

EXAMPLE

turbomaChine similitude

14.12

The pump has an impeller diameter of 250 mm, and when operating, discharges 0.15 m3 >s of water while producing a hydraulic head of 6 m. The shaft power requirement is 9 kW. Determine the required impeller diameter of a similar type of pump that must deliver a discharge of 0.25 m3 >s and produce a 10-m head. What is the estimated shaft power requirement for this pump?

14

SOLUTION

Here we have D1 = 250 mm, Q1 = 0.15 m3 >s, h1 = 6 m, and # W1 = 9 kW. Since Q2 = 0.25 m3 >s and h2 = 10 m, we can eliminate the angular velocity ratio v1 >v2 by using Eq. 2 in Example 14.10 to determine D2. Q12D24 Q22D14 D2 = D1 a = (250 mm) a

843

h1 h2

Q2 1>2 h1 1>4 b a b Q1 h2

0.25 m3 >s

0.15 m >s 3

=

b

1>2

a

6 m 1>4 b = 284.05 mm = 284 mm Ans. 10 m

From the flow coefficient, or Eq. 1 of Example 14.10, the angular velocity ratio is v1 Q1D23 = v2 Q2D13 Thus, the power coefficient similitude, for constant r, becomes # # W1 W2 = 3 5 v13D15 v2 D 2 # W1 v1 3 D1 5 Q1 3 D2 9 D1 5 Q1 3 D2 4 # = a b a b = a b a b a b = a b a b v2 D2 Q2 D1 D2 Q2 D1 W2 Therefore,

0.25 m3 >s 3 250 mm 4 # # Q2 3 D1 4 W2 = W1 a b a b = (9 kW) a b = 25.0 kW b a Q1 D2 284.05 mm 0.15 m3 >s

Ans.

844

C h a p t e r 14

turbomaChines

References 1. 2. 14 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

W. Janna, Introduction to Fluid Mechanics, Cengage Learning, Mason, OH, 1983. F. Yeaple, Fluid Power Design Handbook, Marcel Dekker, New York, NY, 1984. R. Warring, Pumping Manual, 7th ed., Gulf Publishing, Houston, TX, 1984. O. Balje, Turbomachines: A Guide to Design, Selection and Theory, John Wiley, New York, NY, 1981. I. J. Karassick, Pump Handbook, 2th ed., McGraw-Hill, New York, NY, 1995. R. Evans et al., Pumping Plant Performance Evaluation, North Carolina Cooperative Extension Service, Publ. No.AG 452-6. G. M. Jones et al., Pumping Station Design, 3rd ed., Butterworth-Heinemann, London, 2008. I. J. Karassick, Pump Handbook, McGraw-Hill, New York, NY. Hydraulic Institute Standards, 14th ed., Hydraulic Institute, Cleveland, OH, 1983. P. N. Garay, Pump Application Desk Book, Fairmont Press, Lilburn, GA, 1990. G. F. Wislicenus, Fluid Mechanics of Turbomachinery, 2nd ed., Dover Publications, New York, NY, 1965. Performance Test Codes: Centrifugal Pumps, ASME PTC 8.2-1990, New York, NY, 1990. Equipment Testing Procedure: Centrifugal Pumps, American Institute of Chemical Engineers, New York, NY, 2002. E. S. Logan and R. Roy, Handbook of Turbomachinery, 2nd ed., Marcel Dekker, New York, NY, 2003. J. A. Schetz and A. E. Fuhs, Handbook of Fluid Dynamics and Fluid Machinery, John Wiley, New York, NY, 1996. L. Nelik, Centrifugal and Rotary Pumps, CRC Press, Boca Raton, FL, 1999.

P ROB L EMS SEC. 14.1–14.2 14–1. The axial flow pump produces a discharge of 0.20 m3 >s . If the impeller has an angular velocity of v = 150 rad>s, determine the velocity of the water when it is delivered to the stator blades. What is the relative velocity of the water as it flows off the impeller? The impeller blades have a mean radius of 80 mm and the angles shown.

14–2. The axial flow pump produces a discharge of 0.20 m3 >s . Determine the power supplied to the water by the pump when v = 150 rad>s . The impeller blades have a mean radius of 80 mm and the angles shown.

80 mm

v 80 mm

v

30° Stator

30°

Impeller

Stator Impeller

45°

45°

250 mm 250 mm

ri

ri

Prob. 14–1

Prob. 14–2

problems 14–3. If the axial-flow pump produces a discharge of 0.150 m3 >s, determine the required blade angle b1 so a1 = 90° . Also what is the power supplied to the water by the pump? The impeller blades have a mean radius of 60 mm and v = 200 rad>s.

60° b1

v ri

60 mm

845

14–6. An axial-flow pump has an impeller with a mean radius of 100 mm. It rotates at 1200 rev>min. At the exit the stator blade angle is a2 = 70°. If the velocity of the water leaving the impeller is 8 m>s, determine the tangential component of the velocity and the relative velocity of the water at this instant. 14–7. Water flows through the axial-flow pump at 4(10-3) m3 >s, while the impeller has an angular velocity of 30 rad>s. If the blade tail angle is 35°, determine the velocity and tangential velocity component of the water when it leaves the blade.

180 mm

Prob. 14–3

*14–4. The axial flow pump produces a discharge of 0.095 m3 >s . If the impeller is rotating at 80 rad>s, and it has a mean radius of 75 mm, determine the initial blade angle b1 so that a1 = 90°. Also, find the relative velocity of the water as it flows onto the blades of the impeller.

35°

30 rads

75 mm 150 mm

200 mm

b1

75 mm 45°

Prob. 14–7

80 rad/s

Prob. 14–4

14–5. The axial flow pump produces a discharge of 0.095 m3 >s . If the impeller is rotating at 80 rad>s, and it has a mean radius of 75 mm, determine the velocity of the water as it exits the blades, and the relative velocity of the water as it flows off the blades of the impeller.

SEC. 14.3–14.4 *14–8. The velocity of water flowing onto the 40-mm-wide impeller blades of the radial-flow pump is directed at 20° as shown. If the flow leaves the blades at the blade angle of 40°, determine the torque the pump shaft must exert on the impeller.

40°

200 mm 50° b1

75 mm 45° 80 rad/s

Prob. 14–5

250 mm

100 mm

V1 20° 60 rads

Prob. 14–8

14

846

C h a p t e r 14

turbomaChines

14–9. The velocity of water flowing onto the 40-mm-wide impeller blades of the radial-flow pump is directed at 20° as shown. If the flow leaves the blades at the blade angle of 40°, determine the total head developed by the pump.

14–11. The blades on the radial-flow pump rotate at 180 rad>s. If the blades are 60 mm wide, determine the discharge if the water enters each blade in the radial direction.

14

40° 30° 50°

60°

250 mm

100 mm

V1 20° 60 rads

60 mm

180 rad/s

Prob. 14–9

250 mm

Prob. 14–11 14–10. The blades on the radial-flow pump rotate at 300 rad>s. If the discharge is 30(103) liters>min, determine the ideal pump head. The blades are 60 mm wide.

30°

*14–12. The blades on the radial-flow pump rotate at 300 rad>s. If the discharge is 30(103) liters>min, determine the velocity of the water as it flows onto and off the blades. The blades are 60 mm wide.

30° 60°

60 mm

300 rad/s

250 mm

Prob. 14–10

60°

60 mm

300 rad/s

250 mm

Prob. 14–12

problems 14–13. Water flows through the impeller such that the entrance velocity is V1 = 6 m>s and the exit velocity is V2 = 10 m>s. If the discharge is 0.04 m3 >s and the width of each blade is 20 mm, determine the torque that must be applied to the pump shaft.

847

14–15. Show that the ideal head for a radial-flow pump can be determined from hpump = (U2V2 cos a2)>g, where V2 is the velocity of the water leaving the impeller blades. Water enters the blades in the radial direction. *14–16. The impeller is rotating at 1200 rev>min and produces a flow of 0.03 m3 >s. Determine the speed of the 14 water as it exits the blades, and the ideal power and the ideal head produced by the pump.

V2 a2 30° 60°

175 mm V 1 v a

50 mm

1

80 mm 1200 revmin 150 mm

Prob. 14–13

20 mm

14–14. Water flows through the pump impeller at the rate of 0.04 m3 >s. If the blades are 20 mm wide and the velocities at the entrance and exit are directed at the angles a1 = 45° and a2 = 10°, respectively, determine the torque that must be applied to the pump shaft.

Probs. 14–15/16 14–17. The radial-flow pump impeller rotates at 900 rev >min. If the width of the blades is 50 mm, and the blade head and tail angles are as shown, determine the power developed by the pump. Water is initially guided radially onto the blades.

b2 5 308

V2 a2

175 mm V 1 v a

b1 5 458

1

200 mm

80 mm 900 rev/min 100 mm

Prob. 14–14

Prob. 14–17

848

C h a p t e r 14

turbomaChines

14–18. The radial-flow pump impeller rotates at 900 rev>min. If the width of the blades is 50 mm, and the blade head and tail angles are as shown, determine the relative velocity of the water as it flows onto and off the blades. Water is initially guided radially onto the blades. 14

b2 5 308

*14–20. Air at a temperature of 30°C enters the 80-mm-wide blades of the blower in the radial direction and is discharged from the blades at an angle of b2 = 40°. Determine the power required to turn the blades at 120 rad>s, producing a discharge of 2.50 m3 >s.

b2 5 40°

b1 5 458 120 rad/s 200 mm

180 mm 900 rev/min 100 mm

300 mm

Prob. 14–18

Prob. 14–20 14–19. The radial fan is used to force air into the ducts of a building. If the air is at a temperature of 30°C, and the impeller is rotating at 80 rad>s, determine the power output of the motor. The blades have a width of 20 mm. Air enters the blades in the radial direction and is discharged with a velocity of 30 m>s at the angle shown.

14–21. The blades of the centrifugal pump are 30 mm wide and are rotating at 60 rad>s. Water enters the blades in the radial direction and flows off the blades with a velocity of 20 m>s as shown. If the discharge is 0.4 m3 >s, determine the torque that must be applied to the shaft of the pump.

30 ms 40° 20 ms

100 mm

60 rads

80 rads 250 mm 200 mm

Prob. 14–19

Prob. 14–21

problems 14–22. Water flows radially onto the blades of the impeller and exits with a velocity of 16 m>s . If the impeller is turning at 1200 rev>min and the discharge is 48 (103) liters>min, determine the tail blade angle b2 and the power supplied to the pump if it has an efficiency of h = 0.65. The blades are 80 mm wide.

849

*14–24. Water flowing at Va = 6 m>s is directed from the stator onto the blades of the axial-flow turbine at an angle of a1 = 30°, where the mean radius of the blades is 500 mm. If b2 = 15°, the turbine blades are rotating at 60 rad>s, and the flow is 10 m3 >s, determine the torque produced by the water. b2

80 mm

b1

200 mm

50 mm

a2

1200 rev/min. 50 mm a1

200 mm

Stator blades

Prob. 14–22

SEC. 14.5

Turbine blades

Stator blades

Prob. 14–24

14–23. Water flowing at Va = 6 m>s is directed from the stator onto the blades of an axial flow turbine at an angle of a1 = 30° . If the turbine blades are rotating at 60 rad>s, determine the required angle b1 of the blades so that they properly accept the flow. What should be the angle a2 of the stator so that water is delivered to it from the turbine at an angle of b2 = 15°? The turbine has a mean radius of 500 mm.

14–25. The blades of a gas turbine are rotating at 80 rad>s. If the mean radius of the blades is 400 mm, and the horizontal velocity of the flow entering the blades is Va = 8 m>s, determine the relative velocities of the gas as it flows smoothly onto and off the blades. Also, what is the required angle b1 at the entrance of the blades, and exit angle a2?

20°

b2 b1

15°

a2

8 m/s

a1 Stator Turbine

Stator blades

Turbine blades

Prob. 14–23

Stator blades

Prob. 14–25

14

850

C h a p t e r 14

turbomaChines

14–26. Water flows through the 400-mm-diameter delivery pipe at 2 m>s. Each of the four 50-mm-diameter nozzles is aimed tangentially at the Pelton wheel, which has bucket deflection angles of 150°. Determine the torque and power developed by the wheel when it is rotating at 10 rad>s. 14 2 ms

14–29. The buckets of the Pelton wheel deflect the 150-mm-diameter water jet 120° as shown. If the velocity of the water from the nozzle is 30 m>s, determine the angular velocity of the wheel to produce a torque of T = 55.7 kN # m on the shaft. How fast must the wheel be turning to maximize the power developed by the wheel?

400 mm v 4m 2.5 m 30 m/s 10 rads 120°

Prob. 14–26

14–27. Water flows from a lake through a 300-m-long pipe having a diameter of 300 mm and a friction factor of f = 0.015. The flow from the pipe passes through a 60-mm-diameter nozzle and is used to drive the Pelton wheel, where the bucket deflection angles are 160°. Determine the power and torque produced when the wheel is turning under optimum conditions. Neglect minor losses.

Probs. 14–28/29 14–30. Water enters the blades of the Francis turbine with a velocity of 16 m>s as shown. If the blades are rotating at 120 rev> min, and the flow off the blades is radial, determine the power the water supplies to the turbine. The blades are 100 mm wide. 14–31. Water enters the blades of the Francis turbine with a velocity of 16 m>s as shown. If the blades are rotating at 120 rev> min, and the flow off the blades is radial, determine the ideal head the turbine draws from the water. The blades are 100 mm wide.

A 20 m

16 ms

2m

40°

B 120 revmin

Prob. 14–27

0.8 m

*14–28. The buckets of the Pelton wheel deflect the 150 mmdiameter water jet 120° as shown. If the velocity of the water from the nozzle is 30 m>s, determine the power delivered to the shaft of the wheel if the wheel is rotating at a constant angular velocity of 30 rev>min.

Probs. 14–30/31

problems *14–32. Water is directed at a1 = 50° onto the blades of the Kaplan turbine and leaves the blades in the axial direction. Each blade has an inner radius of 200 mm and outer radius of 600 mm. If the blades are rotating at v = 28 rad>s, and the flow is 8 m3 >s, determine the power the water supplies to the turbine.

851

14–34. The blades of the Francis turbine rotate at 40 rad>s as they discharge water at 0.5 m3 >s. Water enters the blades at an angle of a1 = 30° and leaves in the radial direction. If the blades have a width of 0.3 m and the turbine operates under a total head of 3 m, determine the hydraulic efficiency. 14

v 1.2 m 40 rads 0.4 m V1 30° 0.3 m

Prob. 14–32 14–33. The blades of the Francis turbine rotate at 40 rad>s as they discharge water at 0.5 m3 >s. Water enters the blades at an angle of a1 = 30° and leaves in the radial direction. If the blades have a width of 0.3 m, determine the torque and power the water supplies to the turbine shaft.

Prob. 14–34 14–35. Water enters the 50-mm-wide blades of the turbine with a velocity of 20 m>s as shown. If the blades are rotating at 75 rev>min and the flow off the blades is radial, determine the power the water supplies to the turbine. *14–36. Water enters the 50-mm-wide blades of the turbine with a velocity of 20 m>s as shown. If the blades are rotating at 75 rev>min and the flow off the blades is radial, determine the ideal head the turbine draws from the water. 20 m/s

1.2 m

358

40 rads 0.4 m V1 30°

75 rev/min

0.3 m

0.6 m

Prob. 14–33

Probs. 14–35/36

852

C h a p t e r 14

turbomaChines

14–37. If the angles for an axial-flow turbine blade are a1 = 30°, b1 = 60° and b2 = 30°, determine the velocity of the water entering and exiting the blades if the mean radius of the blades is 500 mm. The turbine is rotating at 70 rad>s.

*14–40. The velocities on and off the 90-mm-wide blades of a Francis turbine are directed as shown. If the blades are rotating at 80 rad>s and the discharge is 1.40 m3 >s, determine the power that the turbine withdraws from the water.

14–38. The blades of a Kaplan turbine have a mean radius 14 of 0.6 m and are rotating at 50 rad>s, which provides a flow of 40 m3 >s. If the angles for each blade are a1 = 35°, b1 = 70°, and b2 = 40°, determine the ideal power supplied to the turbine.

V1 30° b1

b2 V2 80 rads

150 mm

v 275 mm

Prob. 14–40

SEC. 14.6–14.8 14–41. A radial-flow pump has a 175-mm-diameter impeller, and the performance curves for it are shown in Fig. 14–13. Determine the approximate flow it provides to pump water from the reservoir tank to the fill tank, where h = 42 m. Neglect minor losses, and use a friction factor of f = 0.02 for the 40-m-long, 80-mm-diameter pipe.

Prob. 14–38

14–39. The velocities on and off the 90-mm-wide blades of the Francis turbine are directed as shown. If V1 = 18 m>s and the blades are rotating at 80 rad>s, determine the relative velocity of the flow off the blades. Also, determine the blade angles b1 and b2.

14–42. The radial-flow pump has a 150-mm-diameter impeller and the performance curves shown in Fig. 14–13. If it is used to pump water from the reservoir tank to the fill tank, determine the approximate flow when the water level is at h = 34 m. Neglect friction losses in the 80-mm-diameter hose, but consider the minor losses to be KL = 3.5.

V1 30° b1

b2

h

V2 80 rads 150 mm 2m 275 mm

Prob. 14–39

Probs. 14–41/42

problems 14–43. Water at 20°C is pumped from a lake into the tank on the truck using a 50-mm-diameter galvanized iron pipe. If the pump performance curve is as shown, determine the maximum flow the pump will generate. The total length of the pipe is 50 m. Include the minor losses of five elbows.

hpump (m) 700 600 500 400 300 200 100

853

14–45. Water at T = 25°C is drawn from an underground detention tank and discharged into a drain using a 6-m-long pipe and pump. The pipe has a diameter of 100 mm and a friction factor of f = 0.02. Determine if cavitation occurs when the velocity through the pipe is 4 m>s and h = 3 m. Use the pump performance curves in Fig. 14–16. The 14 atmospheric pressure is 101.3 kPa. Neglect minor losses.

B

h

A

0.005 0.015 0.025 0.035

Q (m3s) B

8m A

Prob. 14–45

Prob. 14–43

*14–44. The radial-flow pump having an impeller diameter of 150 mm and the performance curves shown in Fig. 14–16 is to be used to pump water from the reservoir into the tank. Determine the efficiency of the pump if the flow is 1200 liters>min. Also, what is the maximum height h to which the tank can be filled? Neglect any losses.

14–46. Water at T = 25°C is drawn from an underground detention tank and discharged into a drain using a 6-m-long pipe and pump. The pipe has a diameter of 100 mm and a friction factor of f = 0.02. Determine if cavitation occurs when the velocity through the pipe is 4 m>s and h = 4 m. Use the pump performance curves in Fig. 14–16. The atmospheric pressure is 101.3 kPa. Neglect minor losses.

B

h

A h

Prob. 14–44

Prob. 14–46

854

C h a p t e r 14

turbomaChines

14–47. The pipe system consists of an 80-mm-diameter, 35-m-long galvanized iron pipe, a fully opened gate valve, four elbows, a flush entrance, and a pump with the pump head curve shown. If the friction factor is f = 0.022, estimate the flow and corresponding pump head generated by the pump. 14

14–50. A pump provides a discharge of 1500 liters>min when operating at a head of 40 m with its impeller rotating at 1500 rev>min. Determine the discharge and head if the same pump is used with its impeller rotating at 1000 rev>min. 14–51. A pump delivers 800 liters>min. when operating at 1500 rev>min. to supply a total head of 25 m. If the flow is held constant but the head is increased to 40 m, determine the speed at which a substitute pump should operate and yet provide the same efficiency?

B

6m

A

*14–52. The model of a water pump has an impeller with a diameter of 50 mm that discharges 150 liters>min with a pressure head of 1.5 m. Determine the diameter of the impeller of the prototype that will discharge 1000 liters> min with a pressure head of 25 m.

hpump(m)

14–53. A radial flow pump whose 100-mm-diameter impeller rotates at 120 rad>s discharges water at 0.25 m3 >s. Determine the discharge of a geometrically similar pump having a 150-mm-diameter impeller that rotates at 90 rad>s.

45 40 35 30 25 20 15 10 5 0

200 400 600 800 1000 1200 1400 1600 1800 2000

Q(liters/min)

Prob. 14–47

SEC. 14.9 *14–48. The temperature of benzene in a processing tank is maintained by recycling this liquid through a heat exchanger, using a pump that has an impeller speed of 1750 rpm and produces a flow of 3600 liters>min. If it is found that the heat exchanger can maintain the temperature only when the flow is 2400 liters>min, determine the required angular speed of the impeller. 14–49. A variable-speed pump requires 20 kW to run at an impeller speed of 1750 rpm. Determine the power required if the impeller speed is reduced to 750 rpm.

14–54. The temperature of benzene in a processing tank is maintained by recycling this liquid through a heat exchanger, using a pump that has an impeller diameter of 140 mm and produces a flow of 3600 liters >min. If it is found that the heat exchanger can maintain the temperature only when the flow is 2400 liters>min, determine the required diameter of the impeller if it maintains the same angular speed. 14–55. A radial flow pump whose 150-mm-diameter impeller rotates at 200 rad>s produces a change in ideal head of 300 mm. Determine the change in head for a geometrically similar pump that has an impeller diameter of 300 mm and operates at 90 rad>s. *14–56. A pump has a 100-mm diameter impeller that discharges 900 liters>min. If the power required is 4 kW, determine the discharge of a similar pump having an impeller diameter of 150 mm that requires a power of 6.5 kW.

855

Chapter review

CHAP TER R EV IEW (Vrel)2 V2 b2

14

Axial-flow pumps maintain the direction of the flow as it passes through the impeller of the pump. Radial-flow pumps direct the flow from the center of the impeller radially outward into a volute. The kinematic analysis of both types of pumps is similar. It depends upon the speed of the impeller and its blade angles.

(Vrel)1

U2

b1 r2

V1 r1 v

U1

The torque, the power, and the head produced by an axial- or a radial-flow pump depend on the motion of the impeller, and on the tangential components of the velocity of the flow as it enters and leaves the impeller.

A Pelton wheel acts as an impulse turbine. It produces power by changing the momentum of the flow as it strikes the buckets of the turbine wheel. v

Kaplan and Francis turbines are called reaction turbines. The analysis of these devices is similar to that of axial- and radial-flow pumps, respectively.

r V

The actual performance characteristics of any turbomachine are determined from experiment, to account for both mechanical and fluid friction losses in the machine. Cavitation within the flow can occur in a turbomachine. It can be avoided if the (NPSH)avail is greater than the required (NPSH)req’d, which is determined by experiment. A pump that is selected to work within a fluid system must produce a required flow and total head, and must operate with a high efficiency. Q1 v1D31 To compare the performance characteristics of two turbomachines of the same type, the flow, head, and power coefficients must be similar.

Turbomachines can be selected for a specific job based on their specific speed, which is a function of the angular velocity v, the flow Q, and the pump head h. Radial-flow pumps are efficient for low specific speeds, and axial-flow pumps are efficient for high specific speeds.

h1

=

v21D21

=

v31D51

=

# W1

Ns =

Q2 v2D32 h2 v22D22 # W2 v32D52

vQ1>2

1gh2 3>4

APPENDIX

A

PHYSICAL PROPERTIES OF FLUIDS Physical Properties of Liquids at Standard Atmospheric Pressure 101.3 kPa, and 20°C (SI Units)

Liquid

Density r (kg>m3)

Dynamic Viscosity m (N # s>m2)

Ethyl alcohol

789

1.19(10-3)

Gasoline

726

-3

Carbon tetrachloride Kerosene Glycerin Mercury Crude oil

1590 814 1260 13 550 880

Kinematic Viscosity n (m2 >s)

0.317(10 )

1.51(10-6)

0.0229

-6

0.0221

-6

0.0269

0.437(10 )

-3

0.958(10 )

0.603(10 )

-3

1.92(10 )

-6

0.0293

-3

0.0633

2.36(10 )

1.50

1.19(10 ) -3

-6

1.58(10 )

0.466

0.117(10 )

-3

Surface Tension s (N>m)

-3

30.2(10 )

0.0344(10 )

Physical Properties of Gases at Standard Atmospheric Pressure 101.3 kPa (SI Units)

Gas

Density r (kg>m3)

Dynamic Viscosity m (N # s>m2)

Kinematic Viscosity n (m2 >s)

Gas Constant R (J>[kg # K])

14.6(10−6)

286.9

Specific Heat Ratio k = cp >cv

15.2(10−6)

259.8

1.40

296.8

1.40

Air (15°C)

1.23

17.9(10−6)

Oxygen (20°C)

1.33

20.4(10−6)

Nitrogen (20°C)

1.16

17.5(10 )

15.1(10 )

Hydrogen (20°C)

0.0835

8.74(10−6)

106(10−6)

Helium (20°C)

0.169

Carbon dioxide (20°C)

1.84

14.9(10 )

Methane (20°C) (natural gas)

0.665

11.2(10−6)

856

−6

−6

−6

114(10 )

−6

−6

19.2(10 )

−6

4124 2077

1.40

1.41 1.66

8.09(10 )

188.9

1.30

16.8(10−6)

518.3

1.31

APPENDIX A

857

Physical Properties of Water vs. Temperature (SI Units) Temperature T (°C)

Density r (kg>m3)

Dynamic Viscosity m (N # s>m2)

Kinematic Viscosity n (m2 >s)

Vapor Pressure pv (kPa abs.)

1.80(10−6)

0.681

0

999.8

1.80(10−3)

5

1000.0

1.52(10−3)

1.52(10−6)

0.872

10

999.7

1.31(10−3)

1.31(10−6)

1.23

15

999.2

1.15(10−3)

1.15(10−6)

1.71

20

998.3

1.00(10−3)

1.00(10−6)

2.34

25

997.1

0.897(10−3)

0.898(10−6)

3.17

30

995.7

0.801(10−3)

0.804(10−6)

4.25

35

994.0

0.723(10−3)

0.727(10−6)

5.63

40

992.3

0.659(10−3)

0.664(10−6)

7.38

45

990.2

0.599(10−3)

0.604(10−6)

9.59

50

988.0

0.554(10−3)

0.561(10−6)

12.4

55

985.7

0.508(10−3)

0.515(10−6)

15.8

60

983.2

0.470(10−3)

0.478(10−6)

19.9

65

980.5

0.437(10−3)

0.446(10−6)

25.0

70

977.7

0.405(10−3)

0.414(10−6)

31.2

75

974.8

0.381(10−3)

0.390(10−6)

38.6

80

971.6

0.356(10−3)

0.367(10−6)

47.4

85

968.4

0.336(10−3)

0.347(10−6)

57.8

90

965.1

0.318(10−3)

0.329(10−6)

70.1

95

961.6

0.300(10−3)

0.312(10−6)

84.6

100

958.1

0.284(10−3)

0.296(10−6)

101

A

858

APPENDIX A

Properties of Air at Standard Atmospheric Pressure 101.3 kPa vs. Temperature (SI Units)

A

Temperature

Density

T (°C)

r (kg>m3)

- 50

1.582

- 40

1.514

- 30

1.452

- 20

1.394

- 10

1.342

0

1.292

10

1.247

20

1.202

30

1.164

40

1.127

50

1.092

60

1.060

70

1.030

80

1.000

90

0.973

100

0.946

150

0.834

200

0.746

250

0.675

Dynamic Viscosity

Kinematic Viscosity v (m2 >s)

m (N # s>m2) 14.6110 -6 2

9.21110 -6 2

15.1110 2

9.98110 -6 2

-6

15.6110 -6 2

10.8110 -6 2

17.2110 -6 2

13.3110 -6 2

16.1110 -6 2

11.6110 -6 2

16.7110 -6 2

12.4110 -6 2

17.6110 2

14.2110 -6 2

-6

18.1110 2

15.1110 -6 2

-6

18.6110 2

16.0110 -6 2

-6

19.1110 -6 2

16.9110 -6 2

19.5110 -6 2

17.9110 -6 2

20.0110 -6 2

18.9110 -6 2

20.5110 -6 2

19.9110 -6 2

20.9110 2

20.9110 -6 2

-6

21.3110 2

21.9110 -6 2

-6

21.7110 2

23.0110 -6 2

-6

23.8110 -6 2

28.5110 -6 2

25.7110 -6 2

34.5110 -6 2

27.5110 -6 2

40.8110 -6 2

Properties of Air vs. Altitude (SI Units) Altitude (km) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Temperature

Pressure

Density

T (°C)

p (kPa)

r (kg>m3)

15.00 8.501 2.004 - 4.491 - 10.98 - 17.47 - 23.96 - 30.45 - 36.94 - 43.42 - 49.90 - 56.38 - 56.50 - 56.50 - 56.50 - 56.50 - 56.50 - 56.50 - 56.50 - 56.50 - 56.50 - 55.57 - 54.58 - 53.58 - 52.59 - 51.60

101.3 89.88 79.50 70.12 61.66 54.05 47.22 41.10 35.65 30.80 26.45 22.67 19.40 16.58 14.17 12.11 10.35 8.850 7.565 6.468 5.529 4.729 4.048 3.467 2.972 2.549

1.225 1.112 1.007 0.9092 0.8194 0.7364 0.6601 0.5900 0.5258 0.4671 0.4135 0.3648 0.3119 0.2666 0.2279 0.1948 0.1665 0.1423 0.1217 0.1040 0.08891 0.07572 0.06451 0.05501 0.04694 0.04008

Dynamic Viscosity

Kinematic Viscosity

17.89110 -6 2 17.58110 -6 2 17.26110 -6 2 16.94110 -6 2 16.61110 -6 2 16.28110 -6 2 15.95110 -6 2 15.61110 -6 2 15.27110 -6 2 14.93110 -6 2 14.58110 -6 2 14.22110 -6 2 14.22110 -6 2 14.22110 -6 2 14.22110 -6 2 14.22110 -6 2 14.22110 -6 2 14.22110 -6 2 14.22110 -6 2 14.22110 -6 2 14.22110 -6 2 14.27110 -6 2 14.32110 -6 2 14.38110 -6 2 14.43110 -6 2 14.48110 -6 2

14.61110 -6 2 15.81110 -6 2 17.15110 -6 2 18.63110 -6 2 20.28110 -6 2 22.11110 -6 2 24.16110 -6 2 26.46110 -6 2 29.04110 -6 2 31.96110 -6 2 35.25110 -6 2 39.00110 -6 2 45.57110 -6 2 53.32110 -6 2 62.39110 -6 2 73.00110 -6 2 85.40110 -6 2 99.90110 -6 2 0.1169110 -3 2 0.1367110 -3 2 0.1599110 -3 2 0.1884110 -3 2 0.2220110 -3 2 0.2614110 -3 2 0.3074110 -3 2 0.3614110 -3 2

m (Pa # s)

n (m2 >s)

APPENDIX

B

COMPRESSIBLE PROPERTIES OF A GAS ( = 1.4) TABLE B–1

Isentropic Relations (k = 1.4) T T0

p

M

A

p0

A′

0 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.30

1.0000 0.9980 0.9976 0.9971 0.9966 0.9961 0.9955 0.9949 0.9943 0.9936 0.9928 0.9921 0.9913 0.9901 0.9895 0.9886 0.9877 0.9867 0.9856 0.9846 0.9835 0.9823

1.0000 0.9930 0.9916 0.9900 0.9883 0.9864 0.9844 0.9823 0.9800 0.9776 0.9751 0.9725 0.9697 0.9668 0.9638 0.9607 0.9575 0.9541 0.9506 0.9470 0.9433 0.9395

5.8218 5.2992 4.8643 4.4969 4.1824 3.9103 3.6727 3.4635 3.2779 3.1123 2.9635 2.8293 2.7076 2.5968 2.4956 2.4027 2.3173 2.2385 2.1656 2.0979 2.0351

∞

0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.40 0.41 0.42 0.43 0.44 0.45 0.46 0.47 0.48 0.49 0.50 0.51 0.52 0.53 0.54 0.55 0.56 0.57

0.9811 0.9799 0.9787 0.9774 0.9761 0.9747 0.9733 0.9719 0.9705 0.9690 0.9675 0.9659 0.9643 0.9627 0.9611 0.9594 0.9577 0.9560 0.9542 0.9524 0.9506 0.9487 0.9468 0.9449 0.9430 0.9410 0.9390

0.9355 0.9315 0.9274 0.9231 0.9188 0.9143 0.9098 0.9052 0.9004 0.8956 0.8907 0.8857 0.8807 0.8755 0.8703 0.8650 0.8596 0.8541 0.8486 0.8430 0.8374 0.8317 0.8259 0.8201 0.8142 0.8082 0.8022

1.9765 1.9219 1.8707 1.8229 1.7780 1.7358 1.6961 1.6587 1.6234 1.5901 1.5587 1.5289 1.5007 1.4740 1.4487 1.4246 1.4018 1.3801 1.3595 1.3398 1.3212 1.3034 1.2865 1.2703 1.2550 1.2403 1.2263

859

860

APPENDIX B

TABLE B–1

B

Isentropic Relations (k = 1.4) T T0

p

M

A

p0

A′

0.58 0.59 0.60 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.69 0.70 0.71 0.72 0.73 0.74 0.75 0.76 0.77 0.78 0.79 0.80 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.90 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10

0.9370 0.9349 0.9328 0.9307 0.9286 0.9265 0.9243 0.9221 0.9199 0.9176 0.9153 0.9131 0.9107 0.9084 0.9061 0.9037 0.9013 0.8989 0.8964 0.8940 0.8915 0.8890 0.8865 0.8840 0.8815 0.8789 0.8763 0.8737 0.8711 0.8685 0.8659 0.8632 0.8606 0.8579 0.8552 0.8525 0.8498 0.8471 0.8444 0.8416 0.8389 0.8361 0.8333 0.8306 0.8278 0.8250 0.8222 0.8193 0.8165 0.8137 0.8108 0.8080 0.8052

0.7962 0.7901 0.7840 0.7778 0.7716 0.7654 0.7591 0.7528 0.7465 0.7401 0.7338 0.7274 0.7209 0.7145 0.7080 0.7016 0.6951 0.6886 0.6821 0.6756 0.6691 0.6625 0.6560 0.6495 0.6430 0.6365 0.6300 0.6235 0.6170 0.6106 0.6041 0.5977 0.5913 0.5849 0.5785 0.5721 0.5658 0.5595 0.5532 0.5469 0.5407 0.5345 0.5283 0.5221 0.5160 0.5099 0.5039 0.4979 0.4919 0.4860 0.4800 0.4742 0.4684

1.2130 1.2003 1.1882 1.1767 1.1657 1.1552 1.1452 1.1356 1.1265 1.1179 1.1097 1.1018 1.0944 1.0873 1.0806 1.0742 1.0681 1.0624 1.0570 1.0519 1.0471 1.0425 1.0382 1.0342 1.0305 1.0270 1.0237 1.0207 1.0179 1.0153 1.0129 1.0108 1.0089 1.0071 1.0056 1.0043 1.0031 1.0022 1.0014 1.0008 1.0003 1.0001 1.000 1.000 1.000 1.001 1.001 1.002 1.003 1.004 1.005 1.006 1.008

1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.30 1.31 1.32 1.33 1.34 1.35 1.36 1.37 1.38 1.39 1.40 1.41 1.42 1.43 1.44 1.45 1.46 1.47 1.48 1.49 1.50 1.51 1.52 1.53 1.54 1.55 1.56 1.57 1.58 1.59 1.60 1.61 1.62 1.63 1.64 1.65 1.66 1.67 1.68 1.69

0.8023 0.7994 0.7966 0.7937 0.7908 0.7879 0.7851 0.7822 0.7793 0.7764 0.7735 0.7706 0.7677 0.7648 0.7619 0.7590 0.7561 0.7532 0.7503 0.7474 0.7445 0.7416 0.7387 0.7358 0.7329 0.7300 0.7271 0.7242 0.7213 0.7184 0.7155 0.7126 0.7097 0.7069 0.7040 0.7011 0.6982 0.6954 0.6925 0.6897 0.6868 0.6840 0.6811 0.6783 0.6754 0.6726 0.6698 0.6670 0.6642 0.6614 0.6586 0.6558 0.6530 0.6502 0.6475 0.6447 0.6419 0.6392 0.6364

0.4626 0.4568 0.4511 0.4455 0.4398 0.4343 0.4287 0.4232 0.4178 0.4124 0.4070 0.4017 0.3964 0.3912 0.3861 0.3809 0.3759 0.3708 0.3658 0.3609 0.3560 0.3512 0.3464 0.3417 0.3370 0.3323 0.3277 0.3232 0.3187 0.3142 0.3098 0.3055 0.3012 0.2969 0.2927 0.2886 0.2845 0.2804 0.2764 0.2724 0.2685 0.2646 0.2608 0.2570 0.2533 0.2496 0.2459 0.2423 0.2388 0.2353 0.2318 0.2284 0.2250 0.2217 0.2184 0.2151 0.2119 0.2088 0.2057

1.010 1.011 1.013 1.015 1.017 1.020 1.022 1.025 1.026 1.030 1.033 1.037 1.040 1.043 1.047 1.050 1.054 1.058 1.062 1.066 1.071 1.075 1.080 1.084 1.089 1.094 1.099 1.104 1.109 1.115 1.120 1.126 1.132 1.138 1.144 1.150 1.156 1.163 1.169 1.176 1.183 1.190 1.197 1.204 1.212 1.219 1.227 1.234 1.242 1.250 1.258 1.267 1.275 1.284 1.292 1.301 1.310 1.319 1.328

861

APPENDIX B

TABLE B–1

Isentropic Relations (k = 1.4) T T0

p

M

A

p0

A′

1.70 1.71 1.72 1.73 1.74 1.75 1.76 1.77 1.78 1.79 1.80 1.81 1.82 1.83 1.84 1.85 1.86 1.87 1.88 1.89 1.90 1.91 1.92 1.93 1.94 1.95 1.96 1.97 1.98 1.99 2.00 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.08 2.09 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23

0.6337 0.6310 0.6283 0.6256 0.6229 0.6202 0.6175 0.6148 0.6121 0.6095 0.6068 0.6041 0.6015 0.5989 0.5963 0.5936 0.5910 0.5884 0.5859 0.5833 0.5807 0.5782 0.5756 0.5731 0.5705 0.5680 0.5655 0.5630 0.5605 0.5580 0.5556 0.5531 0.5506 0.5482 0.5458 0.5433 0.5409 0.5385 0.5361 0.5337 0.5313 0.5290 0.5266 0.5243 0.5219 0.5196 0.5173 0.5150 0.5127 0.5104 0.5081 0.5059 0.5036 0.5014

0.2026 0.1996 0.1966 0.1936 0.1907 0.1878 0.1850 0.1822 0.1794 0.1767 0.1740 0.1714 0.1688 0.1662 0.1637 0.1612 0.1587 0.1563 0.1539 0.1516 0.1492 0.1470 0.1447 0.1425 0.1403 0.1381 0.1360 0.1339 0.1318 0.1298 0.1278 0.1258 0.1239 0.1220 0.1201 0.1182 0.1164 0.1146 0.1128 0.1111 0.1094 0.1077 0.1060 0.1043 0.1027 0.1011 0.09956 0.09802 0.09649 0.09500 0.09352 0.09207 0.09064 0.08923

1.338 1.347 1.357 1.367 1.376 1.386 1.397 1.407 1.418 1.428 1.439 1.450 1.461 1.472 1.484 1.495 1.507 1.519 1.531 1.543 1.555 1.568 1.580 1.593 1.606 1.619 1.633 1.646 1.660 1.674 1.688 1.702 1.716 1.730 1.745 1.760 1.775 1.790 1.806 1.821 1.837 1.853 1.869 1.885 1.902 1.919 1.935 1.953 1.970 1.987 2.005 2.023 2.041 2.059

2.24 2.25 2.26 2.27 2.28 2.29 2.30 2.31 2.32 2.33 2.34 2.35 2.36 2.37 2.38 2.39 2.40 2.41 2.42 2.43 2.44 2.45 2.46 2.47 2.48 2.49 2.50 2.51 2.52 2.53 2.54 2.55 2.56 2.57 2.58 2.59 2.60 2.61 2.62 2.63 2.64 2.65 2.66 2.67 2.68 2.69 2.70 2.71 2.72 2.73 2.74 2.75 2.76 2.77 2.78 2.79 2.80 2.81 2.82

0.4991 0.4969 0.4947 0.4925 0.4903 0.4881 0.4859 0.4837 0.4816 0.4794 0.4773 0.4752 0.4731 0.4709 0.4688 0.4668 0.4647 0.4626 0.4606 0.4585 0.4565 0.4544 0.4524 0.4504 0.4484 0.4464 0.4444 0.4425 0.4405 0.4386 0.4366 0.4347 0.4328 0.4309 0.4289 0.4271 0.4252 0.4233 0.4214 0.4196 0.4177 0.4159 0.4141 0.4122 0.4104 0.4086 0.4068 0.4051 0.4033 0.4015 0.3998 0.3980 0.3963 0.3945 0.3928 0.3911 0.3894 0.3877 0.3860

0.08785 0.08648 0.08514 0.08382 0.08251 0.08123 0.07997 0.07873 0.07751 0.07631 0.07512 0.07396 0.07281 0.07168 0.07057 0.06948 0.06840 0.06734 0.06630 0.06527 0.06426 0.06327 0.06229 0.06133 0.06038 0.05945 0.05853 0.05762 0.05674 0.05586 0.05500 0.05415 0.05332 0.05250 0.05169 0.05090 0.05012 0.04935 0.04859 0.04784 0.04711 0.04639 0.04568 0.04498 0.04429 0.04362 0.04295 0.04229 0.04165 0.04102 0.04039 0.03978 0.03917 0.03858 0.03799 0.03742 0.03685 0.03629 0.03574

2.078 2.096 2.115 2.134 2.154 2.173 2.193 2.213 2.233 2.254 2.273 2.295 2.316 2.338 2.359 2.381 2.403 2.425 2.448 2.471 2.494 2.517 2.540 2.564 2.588 2.612 2.637 2.661 2.686 2.712 2.737 2.763 2.789 2.815 2.842 2.869 2.896 2.923 2.951 2.979 3.007 3.036 3.065 3.094 3.123 3.153 3.183 3.213 3.244 3.275 3.306 3.338 3.370 3.402 3.434 3.467 3.500 3.534 3.567

B

862

APPENDIX B

TABLE B–1

B

Isentropic Relations (k = 1.4) T T0

p

M

A

p0

A′

2.83 2.84 2.85 2.86 2.87 2.88 2.89 2.90 2.91 2.92 2.93 2.94 2.95 2.96 2.97 2.98 2.99 3.00 3.01 3.02 3.03 3.04 3.05 3.06 3.07 3.08 3.09 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20 3.21 3.22 3.23 3.24 3.25 3.26 3.27 3.28 3.29 3.30 3.31 3.32 3.33 3.34 3.35 3.36

0.3844 0.3827 0.3810 0.3794 0.3777 0.3761 0.3745 0.3729 0.3712 0.3696 0.3681 0.3665 0.3649 0.3633 0.3618 0.3602 0.3587 0.3571 0.3556 0.3541 0.3526 0.3511 0.3496 0.3481 0.3466 0.3452 0.3437 0.3422 0.3408 0.3393 0.3379 0.3365 0.3351 0.3337 0.3323 0.3309 0.3295 0.3281 0.3267 0.3253 0.3240 0.3226 0.3213 0.3199 0.3186 0.3173 0.3160 0.3147 0.3134 0.3121 0.3108 0.3095 0.3082 0.3069

0.03520 0.03467 0.03415 0.03363 0.03312 0.03263 0.03213 0.03165 0.03118 0.03071 0.03025 0.02980 0.02935 0.02891 0.02848 0.02805 0.02764 0.02722 0.02682 0.02642 0.02603 0.02564 0.02526 0.02489 0.02452 0.02416 0.02380 0.02345 0.02310 0.02276 0.02243 0.02210 0.02177 0.02146 0.02114 0.02083 0.02053 0.02023 0.01993 0.01964 0.01936 0.01908 0.01880 0.01853 0.01826 0.01799 0.01773 0.01748 0.01722 0.01698 0.01673 0.01649 0.01625 0.01602

3.601 3.636 3.671 3.706 3.741 3.777 3.813 3.850 3.887 3.924 3.961 3.999 4.038 4.076 4.115 4.155 4.194 4.235 4.275 4.316 4.357 4.399 4.441 4.483 4.526 4.570 4.613 4.657 4.702 4.747 4.792 4.838 4.884 4.930 4.977 5.025 5.073 5.121 5.170 5.219 5.268 5.319 5.369 5.420 5.472 5.523 5.576 5.629 5.682 5.736 5.790 5.845 5.900 5.956

3.37 3.38 3.39 3.40 3.41 3.42 3.43 3.44 3.45 3.46 3.47 3.48 3.49 3.50 3.51 3.52 3.53 3.54 3.55 3.56 3.57 3.58 3.59 3.60 3.61 3.62 3.63 3.64 3.65 3.66 3.67 3.68 3.69 3.70 3.71 3.72 3.73 3.74 3.75 3.76 3.77 3.78 3.79 3.80 3.81 3.82 3.83 3.84 3.85 3.86 3.87 3.88 3.89 3.90 3.91 3.92 3.93 3.94 3.95

0.3057 0.3044 0.3032 0.3019 0.3007 0.2995 0.2982 0.2970 0.2958 0.2946 0.2934 0.2922 0.2910 0.2899 0.2887 0.2875 0.2864 0.2852 0.2841 0.2829 0.2818 0.2806 0.2795 0.2784 0.2773 0.2762 0.2751 0.2740 0.2729 0.2718 0.2707 0.2697 0.2686 0.2675 0.2665 0.2654 0.2644 0.2633 0.2623 0.2613 0.2602 0.2592 0.2582 0.2572 0.2562 0.2552 0.2542 0.2532 0.2522 0.2513 0.2503 0.2493 0.2484 0.2474 0.2464 0.2455 0.2446 0.2436 0.2427

0.01579 0.01557 0.01534 0.01512 0.01491 0.01470 0.01449 0.01428 0.01408 0.01388 0.01368 0.01349 0.01330 0.01311 0.01293 0.01274 0.01256 0.01239 0.01221 0.01204 0.01188 0.01171 0.01155 0.01138 0.01123 0.01107 0.01092 0.01076 0.01062 0.01047 0.01032 0.01018 0.01004 0.009903 0.009767 0.009633 0.009500 0.009370 0.009242 0.009116 0.008991 0.008869 0.008748 0.008629 0.008512 0.008396 0.008283 0.008171 0.008060 0.007951 0.007844 0.007739 0.007635 0.007532 0.007431 0.007332 0.007233 0.007137 0.007042

6.012 6.069 6.126 6.184 6.242 6.301 6.360 6.420 6.480 6.541 6.602 6.664 6.727 6.790 6.853 6.917 6.982 7.047 7.113 7.179 7.246 7.313 7.382 7.450 7.519 7.589 7.659 7.730 7.802 7.874 7.947 8.020 8.094 8.169 8.244 8.320 8.397 8.474 8.552 8.630 8.709 8.789 8.870 8.951 9.032 9.115 9.198 9.282 9.366 9.451 9.537 9.624 9.711 9.799 9.888 9.977 10.07 10.16 10.25

863

APPENDIX B

TABLE B–1

Isentropic Relations (k = 1.4)

M

T T0

3.96 3.97 3.98 3.99 4.00 4.01 4.02 4.03 4.04 4.05 4.06 4.07 4.08 4.09 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20 4.21 4.22 4.23 4.24 4.25 4.26 4.27 4.28 4.29 4.30 4.31 4.32 4.33 4.34 4.35 4.36 4.37 4.38 4.39 4.40 4.41 4.42 4.43 4.44 4.45 4.46 4.47 4.48

0.2418 0.2408 0.2399 0.2390 0.2381 0.2372 0.2363 0.2354 0.2345 0.2336 0.2327 0.2319 0.2310 0.2301 0.2293 0.2284 0.2275 0.2267 0.2258 0.2250 0.2242 0.2233 0.2225 0.2217 0.2208 0.2200 0.2192 0.2184 0.2176 0.2168 0.2160 0.2152 0.2144 0.2136 0.2129 0.2121 0.2113 0.2105 0.2098 0.2090 0.2083 0.2075 0.2067 0.2060 0.2053 0.2045 0.2038 0.2030 0.2023 0.2016 0.2009 0.2002 0.1994

p

A

p0

A′

0.006948 0.006855 0.006764 0.006675 0.006586 0.006499 0.006413 0.006328 0.006245 0.006163 0.006082 0.006002 0.005923 0.005845 0.005769 0.005694 0.005619 0.005546 0.005474 0.005403 0.005333 0.005264 0.005195 0.005128 0.005062 0.004997 0.004932 0.004869 0.004806 0.004745 0.004684 0.004624 0.004565 0.004507 0.004449 0.004393 0.004337 0.004282 0.004228 0.004174 0.004121 0.004069 0.004018 0.003968 0.003918 0.003868 0.003820 0.003772 0.003725 0.003678 0.003633 0.003587 0.003543

10.34 10.44 10.53 10.62 10.72 10.81 10.91 11.01 11.11 11.21 11.31 11.41 11.51 11.61 11.71 11.82 11.92 12.03 12.14 12.24 12.35 12.46 12.57 12.68 12.79 12.90 13.02 13.13 13.25 13.36 13.48 13.60 13.72 13.83 13.95 14.08 14.20 14.32 14.45 14.57 14.70 14.82 14.95 15.08 15.21 15.34 15.47 15.61 15.74 15.87 16.01 16.15 16.28

4.49 4.50 4.51 4.52 4.53 4.54 4.55 4.56 4.57 4.58

0.1987 0.1980 0.1973 0.1966 0.1959 0.1952 0.1945 0.1938 0.1932 0.1925

0.003499 0.003455 0.003412 0.003370 0.003329 0.003288 0.003247 0.003207 0.003168 0.003129

16.42 16.56 16.70 16.84 16.99 17.13 17.28 17.42 17.57 17.72

4.59 4.60 4.61 4.62 4.63 4.64 4.65 4.66 4.67 4.68 4.69 4.70 4.71 4.72 4.73 4.74 4.75 4.76 4.77 4.78 4.79 4.80 4.81 4.82 4.83 4.84 4.85 4.86 4.87 4.88 4.89 4.90 4.91 4.92 4.93 4.94 4.95 4.96 4.97 4.98 4.99 5.00 6.00 7.00 8.00 9.00 10.00

0.1918 0.1911 0.1905 0.1898 0.1891 0.1885 0.1878 0.1872 0.1865 0.1859 0.1852 0.1846 0.1839 0.1833 0.1827 0.1820 0.1814 0.1808 0.1802 0.1795 0.1789 0.1783 0.1777 0.1771 0.1765 0.1759 0.1753 0.1747 0.1741 0.1735 0.1729 0.1724 0.1718 0.1712 0.1706 0.1700 0.1695 0.1689 0.1683 0.1678 0.1672 0.1667 0.1220 0.09259 0.07246 0.05814 0.04762

0.003090 0.003053 0.003015 0.002978 0.002942 0.002906 0.002871 0.002836 0.002802 0.002768 0.002734 0.002701 0.002669 0.002637 0.002605 0.002573 0.002543 0.002512 0.002482 0.002452 0.002423 0.002394 0.002366 0.002338 0.002310 0.002283 0.002255 0.002229 0.002202 0.002177 0.002151 0.002126 0.002101 0.002076 0.002052 0.002028 0.002004 0.001981 0.001957 0.001935 0.001912 0.001890 0.0006334 0.0002416 0.0001024 0.00004739 0.00002356

17.87 18.02 18.17 18.32 18.48 18.63 18.79 18.94 19.10 19.26 19.42 19.58 19.75 19.91 20.07 20.24 20.41 20.58 20.75 20.92 21.09 21.26 21.44 21.61 21.79 21.97 22.15 22.33 22.51 22.70 22.88 23.07 23.25 23.44 23.63 23.82 24.02 24.21 24.41 24.60 24.80 25.00 53.18 104.1 190.1 327.2 535.9

B

864

APPENDIX B

Table B–2 M

B

0.0

Fanno Flow (k = 1.4) fLmax

T

V

p

p0

D

T*

V*

p*

p0*

∞

1.2000

0.0

1.1976

0.1094

0.1

66.9216

∞ 10.9435

∞ 5.8218

0.2

14.5333

1.1905

0.2182

5.4554

2.9635

0.3

5.2993

1.1788

0.3257

3.6191

2.0351

0.4

2.3085

1.1628

0.4313

2.6958

1.5901

0.5

1.0691

1.1429

0.5345

2.1381

1.3398

0.6

0.4908

1.1194

0.6348

1.7634

1.1882

0.7

0.2081

1.0929

0.7318

1.4935

1.0944

0.8

0.0723

1.0638

0.8251

1.2893

1.0382

0.9

0.0145

1.0327

0.9146

1.1291

1.0089

1.0

0.0000

1.0000

1.0000

1.0000

1.0000

1.1

0.0099

0.9662

1.0812

0.8936

1.0079

1.2

0.0336

0.9317

1.1583

0.8044

1.0304

1.3

0.0648

0.8969

1.2311

0.7285

1.0663

1.4

0.0997

0.8621

1.2999

0.6632

1.1149

1.5

0.1360

0.8276

1.3646

0.6065

1.1762

1.6

0.1724

0.7937

1.4254

0.5568

1.2502

1.7

0.2078

0.7605

1.4825

0.5130

1.3376

1.8

0.2419

0.7282

1.5360

0.4741

1.4390

1.9

0.2743

0.6969

1.5861

0.4394

1.5553

2.0

0.3050

0.6667

1.6330

0.4082

1.6875

2.1

0.3339

0.6376

1.6769

0.3802

1.8369

2.2

0.3609

0.6098

1.7179

0.3549

2.0050

2.3

0.3862

0.5831

1.7563

0.3320

2.1931

2.4

0.4099

0.5576

1.7922

0.3111

2.4031

2.5

0.4320

0.5333

1.8257

0.2921

2.6367

2.6

0.4526

0.5102

1.8571

0.2747

2.8960

2.7

0.4718

0.4882

1.8865

0.2588

3.1830

2.8

0.4898

0.4673

1.9140

0.2441

3.5001

2.9

0.5065

0.4474

1.9398

0.2307

3.8498

3.0

0.5222

0.4286

1.9640

0.2182

4.2346

865

APPENDIX B

Table B–3 M

Rayleigh Flow (k = 1.4) T

V

p

T0

p0

T*

V*

p*

T *0

p0*

0.0

0.0

0.0

2.4000

0.0

1.2679

0.1

0.0560

0.0237

2.3669

0.0468

1.2591

0.2

0.2066

0.0909

2.2727

0.1736

1.2346

0.3

0.4089

0.1918

2.1314

0.3469

1.1985

0.4

0.6151

0.3137

1.9608

0.5290

1.1566

0.5

0.7901

0.4444

1.7778

0.6914

1.1140

0.6

0.9167

0.5745

1.5957

0.8189

1.0753

0.7

0.9929

0.6975

1.4235

0.9085

1.0431

0.8

1.0255

0.8101

1.2658

0.9639

1.0193

0.9

1.0245

0.9110

1.1246

0.9921

1.0049

1.0

1.0000

1.0000

1.0000

1.0000

1.0000

1.1

0.9603

1.0780

0.8909

0.9939

1.0049

1.2

0.9118

1.1459

0.7958

0.9787

1.0194

1.3

0.8592

1.2050

0.7130

0.9580

1.0437

1.4

0.8054

1.2564

0.6410

0.9343

1.0776

1.5

0.7525

1.3012

0.5783

0.9093

1.1215

1.6

0.7017

1.3403

0.5236

0.8842

1.1756

1.7

0.6538

1.3746

0.4756

0.8597

1.2402

1.8

0.6089

1.4046

0.4335

0.8363

1.3159

1.9

0.5673

1.4311

0.3964

0.8141

1.4033

2.0

0.5289

1.4545

0.3636

0.7934

1.5031

2.1

0.4936

1.4753

0.3345

0.7741

1.6162

2.2

0.4611

1.4938

0.3086

0.7561

1.7434

2.3

0.4312

1.5103

0.2855

0.7395

1.8860

2.4

0.4038

1.5252

0.2648

0.7242

2.0450

2.5

0.3787

1.5385

0.2462

0.7101

2.2218

2.6

0.3556

1 .5505

0.2294

0.6970

2.4177

2.7

0.3344

1.5613

0.2142

0.6849

2.6343

2.8

0.3149

1.5711

0.2004

0.6738

2.8731

2.9

0.2969

1.5801

0.1879

0.6635

3.1359

3.0

0.2803

1.5882

0.1765

0.6540

3.4244

B

866

APPENDIX B

TABLE B–4

B

Normal Shock Relations (k = 1.4)

M1

M2

1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.30 1.31 1.32 1.33 1.34 1.35 1.36 1.37 1.38 1.39 1.40 1.41 1.42 1.43 1.44 1.45 1.46 1.47 1.48 1.49 1.50 1.51 1.52 1.53 1.54 1.55 1.56 1.57 1.58 1.59 1.60 1.61 1.62 1.63 1.64 1.65 1.66 1.67 1.68

1.000 0.9901 0.9805 0.9712 0.9620 0.9531 0.9444 0.9360 0.9277 0.9196 0.9118 0.9041 0.8966 0.8892 0.8820 0.8750 0.8682 0.8615 0.8549 0.8485 0.8422 0.8360 0.8300 0.8241 0.8183 0.8126 0.8071 0.8016 0.7963 0.7911 0.7860 0.7809 0.7760 0.7712 0.7664 0.7618 0.7572 0.7527 0.7483 0.7440 0.7397 0.7355 0.7314 0.7274 0.7235 0.7196 0.7157 0.7120 0.7083 0.7047 0.7011 0.6976 0.6941 0.6907 0.6874 0.6841 0.6809 0.6777 0.6746 0.6715 0.6684 0.6655 0.6625 0.6596 0.6568 0.6540 0.6512 0.6485 0.6458

p2

r2

T2

p1

r1

T1

(p0)1

1.000 1.023 1.047 1.071 1.095 1.120 1.144 1.169 1.194 1.219 1.245 1.271 1.297 1.323 1.350 1.376 1.403 1.430 1.458 1.485 1.513 1.541 1.570 1.598 1.627 1.656 1.686 1.715 1.745 1.775 1.805 1.835 1.866 1.897 1.928 1.960 1.991 2.023 2.055 2.087 2.120 2.153 2.186 2.219 2.253 2.286 2.320 2.354 2.389 2.423 2.458 2.493 2.529 2.564 2.600 2.636 2.673 2.709 2.746 2.783 2.820 2.857 2.895 2.933 2.971 3.010 3.048 3.087 3.126

1.000 1.017 1.033 1.050 1.067 1.084 1.101 1.118 1.135 1.152 1.169 1.186 1.203 1.221 1.238 1.255 1.272 1.290 1.307 1.324 1.342 1.359 1.376 1.394 1.411 1.429 1.446 1.463 1.481 1.498 1.516 1.533 1.551 1.568 1.585 1.603 1.620 1.638 1.655 1.672 1.690 1.707 1.724 1.742 1.759 1.776 1.793 1.811 1.828 1.845 1.862 1.879 1.896 1.913 1.930 1.947 1.964 1.981 1.998 2.015 2.032 2.049 2.065 2.082 2.099 2.115 2.132 2.148 2.165

1.000 1.007 1.013 1.020 1.026 1.033 1.059 1.016 1.052 1.059 1.065 1.071 1.078 1.084 1.090 1.097 1.103 1.109 1.115 1.122 1.128 1.134 1.141 1.147 1.153 1.159 1.166 1.172 1.178 1.185 1.191 1.197 1.204 1.210 1.216 1.223 1.229 1.235 1.242 1.248 1.255 1.261 1.268 1.274 1.281 1.287 1.294 1.300 1.307 1.314 1.320 1.327 1.334 1.340 1.347 1.354 1.361 1.367 1.374 1.381 1.388 1.395 1.402 1.409 1.416 1.423 1.430 1.437 1.444

1.000 1.000 1.000 1.000 0.9999 0.9999 0.9997 0.9996 0.9994 0.9992 0.9989 0.9986 0.9982 0.9978 0.9973 0.9967 0.9961 0.9953 0.9916 0.9937 0.9928 0.9918 0.9907 0.9896 0.9884 0.9871 0.9857 0.9842 0.9827 0.9811 0.9794 0.9776 0.9758 0.9738 0.9718 0.9697 0.9676 0.9653 0.9630 0.9607 0.9582 0.9557 0.9531 0.9504 0.9476 0.9448 0.9420 0.9390 0.9360 0.9329 0.9298 0.9266 0.9233 0.9200 0.9166 0.9132 0.9097 0.9061 0.9026 0.8989 0.8952 0.8915 0.8877 0.8538 0.8799 0.8760 0.8720 0.8680 0.8640

(p0)2

867

APPENDIX B

TABLE B–4

Normal Shock Relations (k = 1.4)

M1

M2

1.69 1.70 1.71 1.72 1.73 1.74 1.75 1.76 1.77 1.78 1.79 1.80 1.81 1.82 1.83 1.84 1.85 1.86 1.87 1.88 1.89 1.90 1.91 1.92 1.93 1.94 1.95 1.96 1.97 1.98 1.99 2.00 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.08 2.09 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29 2.30 2.31 2.32 2.33 2.34 2.35

0.6431 0.6405 0.6380 0.6355 0.6330 0.6305 0.6281 0.6257 0.6234 0.6210 0.6188 0.6165 0.6143 0.6121 0.6099 0.6078 0.6057 0.6036 0.6016 0.5996 0.5976 0.5956 0.5937 0.5918 0.5899 0.5880 0.5862 0.5844 0.5826 0.5808 0.5791 0.5774 0.5757 0.5740 0.5723 0.5707 0.5691 0.5675 0.5659 0.5643 0.5628 0.5613 0.5598 0.5583 0.5568 0.5554 0.5540 0.5525 0.5511 0.5498 0.5484 0.5471 0.5457 0.5444 0.5431 0.5418 0.5406 0.5393 0.5381 0.5368 0.5356 0.5344 0.5332 0.5321 0.5309 0.5297 0.5286

p2

r2

T2

p1

r1

T1

(p0)1

3.165 3.205 3.245 3.285 3.325 3.366 3.406 3.447 3.488 3.530 3.571 3.613 3.655 3.698 3.740 3.783 3.826 3.870 3.913 3.957 4.001 4.045 4.089 4.134 4.179 4.224 4.270 4.315 4.361 4.407 4.453 4.500 4.547 4.594 4.641 4.689 4.736 4.784 4.832 4.881 4.929 4.978 5.027 5.077 5.126 5.176 5.226 5.277 5.327 5.378 5.429 5.480 5.531 5.583 5.636 5.687 5.740 5.792 5.845 5.898 5.951 6.005 6.059 6.113 6.167 6.222 6.276

2.181 2.198 2.214 2.230 2.247 2.263 2.279 2.295 2.311 2.327 2.343 2.359 2.375 2.391 2.407 2.422 2.438 2.454 2.469 2.485 2.500 2.516 2.531 2.546 2.562 2.577 2.592 2.607 2.622 2.637 2.652 2.667 2.681 2.696 2.711 2.725 2.740 2.755 2.769 2.783 2.798 2.812 2.826 2.840 2.854 2.868 2.882 2.896 2.910 2.924 2.938 2.951 2.965 2.978 2.992 3.005 3.019 3.032 3.045 3.058 3.071 3.085 3.098 3.110 3.123 3.136 3.149

1.451 1.458 1.466 1.473 1.480 1.487 1.495 1.502 1.509 1.517 1.524 1.532 1.539 1.547 1.554 1.562 1.569 1.577 1.585 1.592 1.600 1.608 1.616 1.624 1.631 1.639 1.647 1.655 1.663 1.671 1.679 1.688 1.696 1.704 1.712 1.720 1.729 1.737 1.745 1.754 1.762 1.770 1.779 1.787 1.796 1.805 1.813 1.822 1.821 1.839 1.848 1.857 1.866 1.875 1.883 1.892 1.901 1.910 1.919 1.929 1.938 1.947 1.956 1.965 1.974 1.984 1.993

0.8598 0.8557 0.8516 0.8474 0.8431 0.8389 0.8346 0.8302 0.8259 0.8215 0.8171 0.8127 0.8082 0.8038 0.7993 0.7948 0.7902 0.7857 0.7811 0.7765 0.7720 0.7674 0.7627 0.7581 0.7535 0.7488 0.7442 0.7395 0.7349 0.7302 0.7255 0.7209 0.7162 0.7115 0.7069 0.7022 0.6975 0.6928 0.6882 0.6835 0.6789 0.6742 0.6696 0.6649 0.6603 0.6557 0.6511 0.6464 0.6419 0.6373 0.6327 0.6281 0.6236 0.6191 0.6145 0.6100 0.6055 0.6011 0.5966 0.5921 0.5877 0.5833 0.5789 0.5745 0.5702 0.5658 0.5615

(p0)2

B

868

APPENDIX B

TABLE B–4

B

Normal Shock Relations (k = 1.4)

M1

M2

2.36 2.37 2.38 2.39 2.40 2.41 2.42 2.43 2.44 2.45 2.46 2.47 2.48 2.49 2.50 2.51 2.52 2.53 2.54 2.55 2.56 2.57 2.58 2.59 2.60 2.61 2.62 2.63 2.64 2.65 2.66 2.67 2.68 2.69 2.70 2.71 2.72 2.73 2.74 2.75 2.76 2.77 2.78 2.79 2.80 2.81 2.82 2.83 2.84 2.85 2.86 2.87 2.88 2.89 2.90 2.91 2.92 2.93 2.94 2.95 2.96 2.97 2.98 2.99 3.00

0.5275 0.5264 0.5253 0.5242 0.5231 0.5221 0.5210 0.5200 0.5189 0.5179 0.5169 0.5159 0.5149 0.5140 0.5130 0.5120 0.5111 0.5102 0.5092 0.5083 0.5074 0.5065 0.5056 0.5047 0.5039 0.5030 0.5022 0.5013 0.5005 0.4996 0.4988 0.4980 0.4972 0.4964 0.4956 0.4949 0.4941 0.4933 0.4926 0.4918 0.4911 0.4903 0.4896 0.4889 0.4882 0.4875 0.4868 0.4861 0.4854 0.4847 0.4840 0.4833 0.4827 0.4820 0.4814 0.4807 0.4801 0.4795 0.4788 0.4782 0.4776 0.4770 0.4764 0.4758 0.4752

p2

r2

T2

p1

r1

T1

(p0)1

3.162 3.174 3.187 3.199 3.212 3.224 3.237 3.249 3.261 3.273 3.285 3.298 3.310 3.321 3.333 3.345 3.357 3.369 3.380 3.392 3.403 3.415 3.426 3.438 3.449 3.460 3.471 3.483 3.494 3.505 3.516 3.527 3.537 3.548 3.559 3.570 3.580 3.591 3.601 3.612 3.622 3.633 3.643 3.653 3.664 3.674 3.684 3.694 3.704 3.714 3.724 3.734 3.743 3.753 3.763 3.773 3.782 3.792 3.801 3.811 3.820 3.829 3.839 3.848 3.857

2.002 2.012 2.021 2.031 2.040 2.050 2.059 2.069 2.079 2.088 2.098 2.108 2.118 2.128 2.138 2.147 2.157 2.167 2.177 2.187 2.198 2.208 2.218 2.228 2.238 2.249 2.259 2.269 2.280 2.290 2.301 2.311 2.322 2.332 2.343 2.354 2.364 2.375 2.386 2.397 2.407 2.418 2.429 2.440 2.451 2.462 2.473 2.484 2.496 2.507 2.518 2.529 2.540 2.552 2.563 2.575 2.586 2.598 2.609 2.621 2.632 2.644 2.656 2.667 2.679

0.5572 0.5529 0.5486 0.5444 0.5401 0.5359 0.5317 0.5276 0.5234 0.5193 0.5152 0.5111 0.5071 0.5030 0.4990 0.4950 0.4911 0.4871 0.4832 0.4793 0.4754 0.4715 0.4677 0.4639 0.4601 0.4564 0.4526 0.4489 0.4452 0.4416 0.4379 0.4343 0.4307 0.4271 0.4236 0.4201 0.4166 0.4131 0.4097 0.4062 0.4028 0.3994 0.3961 0.3928 0.3895 0.3862 0.3829 0.3797 0.3765 0.3733 0.3701 0.3670 0.3639 0.3608 0.3577 0.3547 0.3517 0.3487 0.3457 0.3428 0.3398 0.3369 0.3340 0.3312 0.3283

6.331 6.386 6.442 6.497 6.553 6.609 6.666 6.722 6.779 6.836 6.894 6.951 7.009 7.067 7.125 7.183 7.242 7.301 7.360 7.420 7.479 7.539 7.599 7.659 7.720 7.781 7.842 7.903 7.965 8.026 8.088 8.150 8.213 8.275 8.338 8.401 8.465 8.528 8.592 8.656 8.721 8.785 8.850 8.915 8.980 9.045 9.111 9.177 9.243 9.310 9.376 9.443 9.510 9.577 9.645 9.713 9.781 9.849 9.918 9.986 10.06 10.12 10.19 10.26 10.33

(p0)2

APPENDIX B

TABLE B–5 Prandtl–Meyer Expansion (k = 1.4) M

v (degrees)

1.00 1.02 1.04 1.06 1.08 1.10 1.12 1.14 1.16 1.18 1.20 1.22 1.24 1.26 1.28 1.30 1.32 1.34 1.36 1.38 1.40 1.42 1.44 1.46 1.48 1.50 1.52 1.54 1.56 1.58 1.60 1.62 1.64 1.66 1.68 1.70 1.72 1.74 1.76 1.78 1.80 1.82 1.84 1.86 1.88 1.90 1.92 1.94 1.96 1.98 2.00 2.02 2.04 2.06 2.08 2.10 2.12 2.14 2.16 2.18 2.20

0.00 0.1257 0.3510 0.6367 0.9680 1.336 1.735 2.160 2.607 3.074 3.558 4.057 4.569 5.093 5.627 6.170 6.721 7.279 7.844 8.413 8.987 9.565 10.146 10.730 11.317 11.905 12.495 13.086 13.677 14.269 14.860 15.452 16.043 16.633 17.222 17.810 18.396 18.981 19.565 20.146 20.725 21.302 21.877 22.449 23.019 23.586 24.151 24.712 25.271 25.827 26.380 26.930 27.476 28.020 28.560 29.097 29.631 30.161 30.688 31.212 31.732

2.22 2.24 2.26 2.28 2.30 2.32 2.34 2.36 2.38 2.40 2.42 2.44 2.46 2.48 2.50 2.52 2.54 2.56 2.58 2.60 2.62 2.64 2.66 2.68 2.70 2.72 2.74 2.76 2.78 2.80 2.82 2.84 2.86 2.88 2.90 2.92 2.94 2.96 2.98 3.00 3.02 3.04 3.06 3.08 3.10 3.12 3.14 3.16 3.18 3.20 3.22 3.24 3.26 3.28 3.30 3.32 3.34 3.36 3.38 3.40 3.42 3.44 3.46 3.48 3.50

32.249 32.763 33.273 33.780 34.283 34.782 35.279 35.772 36.261 36.746 37.229 37.708 38.183 38.655 39.124 39.589 40.050 40.508 40.963 41.415 41.863 42.307 42.749 43.187 43.622 44.053 44.481 44.906 45.328 45.746 46.161 46.573 46.982 47.388 47.790 48.190 48.586 48.980 49.370 49.757 50.14 50.52 50.90 51.28 51.65 52.01 52.39 52.75 53.11 53.47 53.83 54.18 54.53 54.88 55.22 55.56 55.90 56.24 56.58 56.90 57.24 57.56 57.89 58.21 58.53

869

B

Fundamental Solutions Chapter 2

patm 5 0

F2–1. pB + rwghw = 400 1 103 2 Pa

pB + 11000 kg>m32 19.81 m>s22 (0.3 m) = 400 11032 Pa

pB = 397.06 1 103 2 Pa

+ c FR = ΣFy; FR = 3397.06 1 10 -

2 N>m 43p(0.025 m) 4 3101 1 103 2 N>m243p(0.025 m)24 3

2

= 581.31 N = 581 N

patm 5 0

patm 5 0

pB 1.25 m

1.25 m

oil

1.25 m

oil

1.25 m

water

1m

water

2

water pC

pA

Ans.

oil

0.25 m

(b)

(a)

(c)

pB

F2–3.

Referring to the figure, patm + rwghw - rHgghHg = patm

0.3 m

1 1000 kg>m3 2 1 9.81 m>s2 2 (2 + h) - 1 13 550 kg>m3 2 1 9.81 m>s2 2 h rwghw - rHgghHg = 0

patm 5 101(103) Pa

pw 5 rwghw

= 0

2000 + 1000h - 13 550h = 0

pB 5 397.06(103) Pa

400 kPa

h = 0.1594 m = 159 mm

Ans.

patm

F2–2.

The pressures at A, B and C can be obtained by writing the manometer equation. For point A, referring to Fig. a,

2m

patm + rwghw + rog ho = pA

patm

hw 5 2 1 h

water

0 + (1000 kg>m )(9.81 m>s )(1.25 m) + 3

2

hHg 5 h

(830 kg>m3)(9.81 m>s2)(1.25 m) = pA pA = 22.44(103) Pa = 22.4 kPa

Ans.

For point B, referring to Fig. b,

Hg

F2–4.

patm + rog ho - rwg hw = pB

patm + rwghw - rHgghHg = patm rwhw = rHghHg

0 + (830 kg>m3)(9.81 m>s2)(1.25 m) 3

2

(1000 kg>m )(9.81 m>s )(0.25 m) = pB pB = 7.7254(103) Pa = 7.73 kPa

Ans.

For point C, referring to Fig. c, patm + rogho + rwghw = pC 0 + (830 kg>m3)(9.81 m>s2)(1.25 m) +

pC = 19.988(10 ) Pa = 20.0 kPa

870

rHg rw

bhHg

b(0.1 m + 0.5 sin 30° m) 1000 kg>m3 h = 5.0425 m = 5.04 m Ans.

13 550 kg>m3

(h 2 0.3) m

(1000 kg>m3)(9.81 m>s2)(1 m) = pc 3

1h - 0.3 m2 = a

hw = a

0.5 sin 30° m 0.1 m

Ans.

0.3 m

871

FUNDAMENTAL SOLUTIONS F2–5. pB -

F2–6.

1 1000 kg>m 2 1 9.81 m>s 2 (0.4 m) = pB = 303.92 1 103 2 Pa = 304 kPa pB - rwghw = pA 3

2

300 1 10

3

2 N>m

wA = roghAb = 900(9.81)(3)(2) = 52.974 11032 N>m

The intensity of the distributed load is

F2–8.

2

3101 110 2 N>m 4 + 1880 kg>m 2 19.81 m>s 2 (1.1 m) patm + rcoghco + rwghw = pB

1 1000 kg>m 2 1 9.81 m>s 2 (0.9 m) 119.33 1 103 2 Pa = 119 kPa 3

2

3

+

pB =

3

2

2

= pB Ans.

352.974 1 103 24 (3.464)

Here, LAB = 3>sin 60° = 3.464 m

Ans.

FR =

1 2

= 91.8 kN

Ans.

FR = roghA = 900(9.81)(1.5) 313>sin 60°2 (2) 4 = 91.8 kN Ans. Alternatively,

patm

FR 3.464 m

hco 5 0.5 1 0.6 5 1.1 m

pco

wA 5 52.974(103) Nm hw 5 0.4 1 0.5 5 0.9 m

pw

pB

wA = rwghAb = 1000(9.81)(2.5)(1.5) = 36.7875 1 103 2 N>m

F2–7.

The intensities of the distributed loads at the bottom of A and B are

F2–9.

The intensity of the distributed load at A is

The resultant forces on AB and BC are 1 (FR)AB = 3 36.7875 1 103 2 4 (2.5) = 45.98 1 103 2 N 2 = 46.0 kN

(FR)BC = 3 36.7875 110324 (2) = 73.575 11032 N = 73.6 kN

Ans. Ans.

w1 = rwgh1b = 1000(9.81)(0.9)(2) = 17.658 1 103 2 N>m w2 = rwgh2b = 1000(9.81)(1.5)(2) = 29.43 1 103 2 N>m Then, the resultant forces are

1 (FR)A = 317.658 110324 (0.9) = 7.94611103 2 N = 7.94 kN Ans. 2 (FR)B1 = 3 17.658 1 103 2 4 (0.6) = 10.5948 1 103 2 N 1 (FR)B2 = 3 29.43 1 103 2 - 17.658 1 103 2 4 (0.6) = 3.5316 1 103 2 2 (FR)B = (FR)B1 + (FR)B2 = 10.5948 1 103 2 + 3.5816 1 103 2

(FR)AB = rwghABAAB = 1000(9.81)12.5>22 32.5(1.5)4

Alternatively,

= 45.98 1 103 2 N = 46.0 kN

= 73.575 1 103 2 N = 73.6 kN

And they act at (yp)A =

Ans.

(FR)BC = rwghBCABC = 1000(9.81)(2.5)32(1.5)4

(yp)B =

Ans.

= 14.13 1 103 2 N = 14.1 kN

2 (0.9) = 0.6 m 3

14.1264 11032

= 1.225 m

(yp)A

w1

0.9 m 0.6 m 2m 3

w1 5 17.658(10 ) Nm

(yp)B2 5 1.3 m (FR)B1

(FR)A

wA 5 36.7875(10 ) Nm

Ans.

(yp)B1 5 1.2 m

(FR)BC

3

Ans.

310.5948 110324(0.9 + 0.6>2) + 3.5316 1103 230.9 + 32 (0.6)4

(FR)AB 2.5 m

Ans.

(FR)B2 w2 5 29.43(103) Nm

872

F U N D A M E N TA L S O L U T I O N S Alternatively, the resultant force on A is = 7.9461 1 103 2 = 7.94 kN

F1 w1

(FR)A = rwghAAA = 1000(9.81)(0.45)(0.9)(2)

And acts at (yp)A =

1Ix 2 A

yAAA

1 12

+ yA =

(2) 1 0.9

3

2

Ans. F3

3m

w2

+ 0.45 = 0.6 m

0.4530.9(2)4

1m

F2

Ans.

The resultant force on B is = 14.1264 1 103 2 = 14.1 kN

(Fp)B = rwghBAB = 1000(9.81)(0.9 + 0.6>2)(0.6)(2) And acts at (yp)B =

(Ix)B yBAB

+ yB =

10.9

(2) 10.632

+ 0.6>22 (0.6)(2) 1 12

+ 10.9 + 0.6>22

= 1.225 m F2–10.

Here, yA = hA = AA =

1 2

2 3

Ans.

Fh =

1 2

=

1 2

(Ix)A yAAA

1 36

(0.6) 1 1.2

3

2

= 1000 kg>m3 19.81 m>s22 312 (3 cos 30° m)(3 sin 30° m)(0.5 m)4

Fy = rwgV

Ans. 4

= 0.0288 m . Then

0.0288 + yA = + 0.8 = 0.9 m Ans. 0.8(0.36)

y = 2 m, h = 2 sin 60° = 23 m,

∙

A = p 1 0.52 2 = 0.25p m2 p (Ix) = 1 0.54 2 = 0.015625p m4. Then 4

= 13.3 kN

= 9557.67 N = 9.558 kN + S ΣFx = 0; Ax - 5.518 kN = 0 Ax = 5.52 kN

Ans.

+ c ΣFy = 0; 9.558 kN - Ay = 0 Ay = 9.56 kN

Ans.

+ ΣMA = 0; (9.558 kN) 3 (3 cos 30° m)4 1 3

+ (5.518 kN)13(3 sin 30° m) - MA = 0

MA = 11.0 kN # m

Ans.

1 — 3 (3

0.015625p Ix + y = + 2 = 2.03125 m = 2.03 m Ans. yA 2(0.25p)

Ans.

cos 30°) m Fv

w1 = rkghkb = 814(9.81)(1 sin 60°)(2) = 13.831 1 103 2 N>m

F2–12.

= 5518.125 N

Vertical Component:

FR = rwghA = 1000(9.81) 1 23 2 (0.25p) = 13.345 1 103 2 N yp =

1 7357.58 N>m 2 (3 sin 30° m)

wAhA

= 5.518 kN

(0.6)(1.2) = 0.36 m2. Then

Also, (Ix)A =

F2–11.

= 11000 kg>m3219.81 m>s22 (3 sin 30° m)(0.5 m) = 7357.5 N>m

(1.2) = 0.8 m

= 2.83 kN

Horizontal Component:

wA = rwghAb

Ans.

FR = rwghAAA = 1000(9.81)(0.8)(0.36) = 2825.28 N

yp =

F2–13.

The intensities of the distributed loads are

w2 = w1 + rwghwb = 13.831 1 103 2 + 1000(9.81)(3 sin 60°)(2) = 64.805 1 103 2 N>m

Thus, the resultant force is 1 FR = 3 13.831 1 103 24 (1) + 13.831 1 103 2 (3) 2 1 + 3 64.805 1 103 2 - 13.831 1 103 24 (3) 2 = 124.87 1 103 2 N = 125 kN

3m Fh Ax

30°

wA

MA Ay

Ans.

1 — (3 3

sin 30°) m

FUNDAMENTAL SOLUTIONS F2–14.

The resultant force is equal to the weight of the oil block above surface AB.

Fh =

Fy = rog(AACB + AABDE)b

p = 1900 kg>m32 19.81 m>s22 c (0.5 m)2 + (1)(1.5 m) d (3 m) 2 = 50.13 1 103 2 N = 50.1 kN

Ans.

Fy = rwgV

1 1000 kg>m3 2 1 9.81 m>s2 2 c

= 14.715 1 103 2 N = 14.7 kNT

D

1.5 m

Ans.

Vertical Component

=

Fy

E

1 1 wDhD = 314.715 1 103 2 N>m4(2 m) 2 2 = 14.715 1 103 2 N = 14.7 kN d

873

1 2m a b(2 m)(0.75 m) d 2 tan 45° Ans.

Fy

C A

45°

B C

0.5 m

2m Fh

F2–15.

Side AB

45°

1 1000 kg>m 2 1 9.81 m>s 2 (2 m)(0.75 m) 14.715 1 103 2 N>m

D

Horizontal Component wB = rwghBb = =

3

1 1 Fh = wBhB = 3 14.715 1 103 2 N>m 4 (2 m) 2 2 = 14.715 1 103 2 N = 14.7 kN S

F2–16.

=

Ans.

=

1 2m = 1 1000 kg>m 2 1 9.81 m>s 2 c a b(2 m)(0.75 m) d 2 tan 60° Side CD

1 1000 kg>m3 2 1 9.81 m>s2 2 (2 m)(2 m) = 39.24 1 103 2 N>m 39.81 11032 N>m4(1.5 m) = 14.715 11032 N = 14.715 kN 1 3 3 2 3 39.24 1 10 2 N>m - 9.81 1 10 2 N>m 4 (1.5 m) 22.0725 1 103 2 N = 22.0725 kN wA = rwghAb =

2

Ans.

1 1000 kg>m3 2 1 9.81 m>s2 2 (0.5 m)(2m) 9.81 1 103 2 N>m

wB = rwghBb

Fy = rwgV

= 8495.71 N = 8.50 kN c

Plate AB Horizontal Component

Vertical Component

3

wD

2

(Fh)1 = (Fh )2 =

Fy

=

Thus,

A

Fh = (Fh)1 + (Fh)2 = 14.715 kN + 22.0725 kN = 36.8 kN d

2m Fh

Ans.

Vertical Component 60° wB

B

1 1000 kg>m3 2 1 9.81 m>s2 2 (2 m)(0.75 m) 14.715 1 103 2 N>m

Horizontal Component wD = rwghDb = =

(Fy)1 = rwgV1

1 1000 kg>m3 2 1 9.81 m>s2 2 c a

= 8.4957 1 103 2 N = 8.4957 kN =

1.5 m b(0.5 m)(2 m) d tan 60°

874

F U N D A M E N TA L S O L U T I O N S

(Fy)2 = rwgV2 =

1 1000 kg>m32 19.81 m>s22 c

Plate BC

1 1.5 m a b(1.5 m)(2 m)d 2 tan 60°

= 12.7436 1 103 2 N = 12.7436 kN

Fh =

Fy = (Fy)1 + (Fy)2 = 8.4957 kN + 12.7436 kN Ans.

(Fy)2

1 1 wBhB = 329.43 1 103 2 N>m4(2 m) 2 2

= 29.43 1 103 2 N = 29.4 kN d

Vertical Component

= 11000 kg>m3219.81 m>s22c(2 m)(2 m)(1.5 m) -

= 12.631 1 103 2 N = 12.6 kN c

wB

1.5 m

Ans.

(Fh)1 C

(Fh)2

0.5 m 60°

2m wA

1m

1 1000 kg>m3 2 1 9.81 m>s2 2 (2 m)(1.5 m) 29.43 1 103 2 N>m

=

1 1 w h = 3 29.43 1 103 2 N>m 4 (2 m) 2 B B 2 = 29.43 1 10

3

Vertical Component

2N

= 29.4 kN S

= 29.43 1 103 2 N = 29.4 kN c Fy

1 1000 kg>m3 2 1 9.81 m>s2 2 (5 m)(2 m) 98.1 1 103 2 N> m

= Fh =

1 2

=

1 2

3 98.1 1 103 2 N>m 4 (5 m) wC hC

= 245.25 kN d

Vertical Component: Fy = rwgV =

Ans.

=

= 245.25 1 103 2 N

1 1000 kg>m3 2 1 9.81 m>s2 2 c

245.25 1 103 2 tan u

N =

1 5m a b(5 m)(2 m) d 2 tan u

245.25 kN c tan u

∙

+ ΣMA = 0;

A

a

45°

2m Fh 45° wB

Horizontal Component:

=

Ans.

1 2m b(2 m)(1.50 m)d Fy = rwgV = 11000 kg>m3219.81 m>s22 c a 2 tan 45°

wB

wC = rwghCb

Horizontal Component wB = rwghBb =

Fh

B

F2–18.

Plate AB

Fh =

p (2 m)2(1.5 m)d 4

Fy

FR 5 Fv

F2–17.

Ans.

Fy = rwgV

(Fy)1

0.5 m

= 29.43 1 103 2 N>m

wB = rwghBb =

Thus,

= 21.2 kN c

1 1000 kg>m3 2 1 9.81 m>s2 2 (2 m)(1.5 m)

Horizontal Component

B

245.25 1 5m 2 kNb c a b d - (245.25 kN) c (5 m) d = 0 tan u 3 tan u 3 1 = 2, u = 35.3° Ans. tan2 u

875

FUNDAMENTAL SOLUTIONS

Ay

A Fy

1 5 — 3 tan u

Ax

u

u

hB

FR

hC

2 — (5) m 3

5m

A

B

B

wB

2m

Fh u

F2–21.

wc

tan u =

6 m>s2 ac = 0.6116 = g 9.81 m>s2

h′ = (1.5 m) tan u = (1.5 m)(0.6116) = 0.9174 m F2–19.

= 11000 kg>m321 9.81 m>s22 3p 10.1 m2 2d4 = 308.2d

Fb = rwgV b

+ c ΣFy = 0;

1 880 kg>m3 2 1 9.81 m>s2 2 (1.4174 m) 12.2364 1 103 2 Pa = 12.2 kPa Ans. 1 880 kg>m3 2 1 9.81 m>s2 2 (0.5 m) 4.3164 1 103 2 Pa = 4.32 kPa Ans.

hA = hB + h′ = 0.5 m + 0.9174 m = 1.4174 m

308.2d - 32(9.81) N4 = 0; d = 0.06366 m

pA = roghA = =

p(0.2 m) (0.5 m) = p 1 0.2 m2 2 h - p(0.1 m)2(0.0636 m) Vw = V′ - Vwb

pB = roghB =

2

h = 0.516 m

=

Ans.

a 5 6 ms2

Fb

h9 u hA

hB 5 0.5 m

d 1.5 m

F2–22.

(Vair)i = (Vair)f

W 5 2(9.81) N

p(1 m)2(3 m - 2 m) = 0.5 m

h = 2m h

5

0.0637 m

2

h =

0.2 m 0.4 m 0.4 m

v2 2 v2 r ; 2m = £ § (1 m)2 2g 2 1 9.81 m>s2 2

v = 6.26 rad>s F2–20.

tan u =

4 m>s2 ac = = 0.4077 g 9.81 m>s2 u = 22.18° = 22.2°

1m

Ans.

hB = 1.5 m + (1 m) tan 22.18°

1 1000 kg>m3 2 1 9.81 m>s2 2 (1.9077 m)(3 m) 56.145 1 103 2 N>m = 56.143 kN>m

= 1.9077 m wB = rwghBb = =

1 FR = (56.145 kN>m)(1.9077 m) = 53.56 kN = 53.6 kN Ans. 2

h

1 3 p(1 m)2 4 h 2

Ans.

876

F U N D A M E N TA L S O L U T I O N S

h =

F2–23.

h = c

v2 2 r ; 2g

2 1 9.81 m>s2 2 (8 rad>s)2

d (1 m)2

Chapter 3 F3–1.

= 3.2620 m

Since x = 2 m, y = 6 m when t = 0, x

Vw = V′ - Vpar

p(1 m) (2 m) = p(1 m)2(h0 + 3.2620 m) - 12 3p(1 m)2(3.2620 m)4

ln x ` =

1 t x 1 t ` ; ln = (t) 4 0 2 4

x

2

2

h0 = 0.3690 m

1 (t)

Thus,

x = 2e 4

h max = h + h0 = 3.2620 m + 0.3690 m = 3.6310 m dy = y = 2t; dt

1 1000 kg>m 2 1 9.81 m>s 2 (3.6310 m) 35.62 1 103 2 Pa = 35.6 kPa Ans. 1 1000 kg>m3 2 1 9.81 m>s2 2 (0.3690 m) 3.62 1 103 2 Pa = 3.62 kPa Ans.

h min = h0 = 0.3690 m pmax = rwghmax = = pmin = rwghmin = =

3

2

h

dy =

L0

t

2t dt

y

t

6

0

y = t2 + 6

When t = 2 s, the position of the particles is 1 (2)

x = 2 e4 F3–2.

= 3.30 m, y = (2)2 + 6 = 10 m

Ans.

dy 8y 4y = 2 = dx 2x 2 x

dy y = ; dx u

x y dy 1 x 4dx ; ln y ` = = -4a b ` 2 x 2m 3m L3 m y L2 m x y 1 1 ln = - 4a - b 3 x 2 y

O h0

h =

L6

y

y ` = t2 ` ; y - 6 = t2

r51m

F2–24.

t

dx 1 = dt L2 x L0 4

dx 1 = u = x; dt 4

hmax = £

v2 2 r ; 2g

2 1 9.81 m>s2 2 (4 rad>s)2

1 880 kg>m3 2 1 9.81 m>s2 2 (1.8349 m) 15.84 1 103 2 Pa = 1.8349 m

p max = rogh max = =

= 15.8 kPa

2(x - 2)

§ (1.5 m)2

y = 3e F3–3.

a =

0V 0V + V 0t 0x

0V = 20t; 0t a =

Ans.

1.5 m

3 20t

+

+ 10t 2 2 1 600x 2 2 4 m>s2

3 200 1 0.13 2

when t = 0.2 s, x = 0.1 m hmax

= 7.60 m>s2 F3–4.

a =

Ans.

0V = 600x 2 0x

1 200x3

a = 20(0.2) +

x

+ 10 1 0.22 24 3 600 1 0.12 24 Ans.

0u 0u + u 0t 0x

= 0 + 3(x + 4)(3) = 9(x + 4) m>s2 At x = 0.1 m, a = 9(0.1 + 4) = 36.9 m>s2

Ans.

877

FUNDAMENTAL SOLUTIONS x

x 1 ln(x + 4) ` = t; 3 0

x = 4e

3t

1 x + 4 lna b = t 3 4

341e

- 4; x =

When t = 0.025 s

3t

(ax)local =

- 12 4 m

an = a

an = 0 +

(ax)local = 34(2)4 m>s2 = 8 m>s2

= =

1 3x + 3 3 1 3x

2t 2 2 (3) +

1 2y3

+ 2t 2 2 4 m>s2

3 3 1 3(3)

(ay)conv

= =

+ 10t 2 (0)

1 3x + 2t 2 2 (0) + 1 2y3 + 3 6y2 1 2y3 + 10t 2 4 m>s2

3 6 1 12 243 2 1 13 2

When t = 2 s, y = 1 m. (ay)conv =

= 132 m>s2

= 100.37 m>s2

0.5

a = 2a 2s + a 2n = 2 1 111.28 m>s2 2 2 + = 150 m>s2

F3–7.

Since the speed of the particle is constant,

Therefore, an = a

a = 2a 2s + a 2n = 202 + = 18 m>s2

= 142 m>s2

1 3 m>s 2 2 0.5 m

= 18 m>s2

1 18 m>s2 2 2

Ans.

s as

1 10 m>s2 2 2

streamline an

aconv = 2(ax)2conv + (ay)2conv = 2 1 51 m>s2 2 2 + 1 132 m>s2 2 2 = 12.8 m>s2

0V V2 b + = 0 + 0t n R

Thus,

+ 10(2) 4 m>s2

alocal = 2(ax)2local + (ay)2local = 2 1 8 m>s2 2 2 +

Ans.

Since the streamline does not rotate, (0V>0t)n = 0.

10t 2 1 6y2 2

Thus

1 100.37 m>s2 2 2

as = 0

+ 2 1 22 224 m>s2 = 51 m>s2

0y = 10 m>s2 0t 0y 0y = u + y 0x 0y

(ay)local =

(7.0842)2

Then,

When t = 2 s, x = 3 m. (ax)conv =

0y b = 0, R = 0.5 m, and at A, 0t n

V = 320(0.125 p)2 + 44 m>s = 7.084 m>s

When t = 2 s,

(ax)conv

V2 0V b + 0t n R

Here, a

Ans.

0u = (4t) m>s2 0t

0u 0u = u + y 0x 0y

p radb = 0.125p m 4

as = 40(0.125p)320(0.125p)2 + 44 = 111.28 m>s2

x = 43e3(0.025) - 14 = 0.2473 m = 247 mm F3–5.

At A, s = r u = (0.5 m) a

t

dx = dt L0 3(x + 4) L0

dx = u = 3(x + 4); dt

Ans.

n

Ans. (a)

F3–6.

as = a a

0V 0V b + V 0t s 0s

0V b = 0 (steady flow) 0t s

F3–8. 0y = 40s 0s

as = 0 + (20s2 + 4)(40s) = 340s(20s2 + 4)4 m>s2

as = a a

0V 0V b + V 0t s 0s

0V 3 b = c (1000) t 1>2 d m>s2 = 0t s 2 0V = 40s 0s

1 1500 t 1>2 2 m>s2

878

F U N D A M E N TA L S O L U T I O N S

as =

3 1500 t 1>2

5 1500 1 0.021>2 2

+

1 20s2

+ 1000t 3>2 + 4 2 (40s) 4 m>s2

3 20 1 0.32 2

VdA is equal to the volume of the paraboLA loid 1 12pr 2h 2 under the velocity profile.

Also,

+ 1000 1 0.023>2 2 + 4 4

At A, s = 0.3 m and t = 0.02 s.

as =

+

3 40(0.3)46 m>s2

= 315.67 m>s2

0V V2 an = a b + 0t n R

3 20 1 0.32 2 Then,

Vavg =

+ 1000 1 0.023>2 2 + 4 4 m>s = 8.628 m>s (8.628 m>s)2

an = 0 +

1 p(0.2 m)2V0 2 V0 = 0.318 m>s

0.5 m

= 148.90 m>s

a = 2a 2s + a 2n = 2 1 315.67 m>s2 2 2 + Thus,

= 349 m>s2

0.02 m3 >s Q = = 0.159 m>s A p(0.2 m)2

p = rRT;

# m = rVA =

1 148.90 m>s2 2 2

Ans.

= 0.812 kg>s F4–6.

F4–1. # m = rwV # A = 11000 kg>m32 (16 m>s)(0.06 m)3(0.05 m) sin 60°4

Q =

LA

VdA =

1 2.0695 kg>m3 2 (V)3 12 (0.3 m)(0.3 m)4 r = 2.0695 kg>m3

F4–3.

Ans.

= 0.565 m3 >s

Ans.

Q = VA = (8 m>s)3p(0.15 m)2 4

# m = rwQ =

F4–4.

V = 7.52 m>s

1 1000 kg>m3 2 1 0.565 m3 >s 2

Q =

LA

L0

= 565 kg>s

Ans.

= 0.125 m3 >s

L0

0.5 m

umax 1 1 - 25r 2 2 (2pr dr)

Ans.

= 0.125 m3 >s

F4–7.

Ans.

0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

0 - (6 m>s) 1 0.1 m2 2 + (2 m>s) 1 0.2 m2 2 + VC 1 0.1 m2 2 = 0 0 - VA AA + VB AB + VC AC = 0

F4–8.

Ans.

Choose a changing control volume. V = (3 m)(2 m)(y) = (6y) m3;

dy dV = 6 dt dt

0 r dV + rl Vf>cs # dA = 0 0t Lcv l Lcs

0.01 r2 25r 4 0.2 m b` = umax a p 2 4 0

2510.24 2 0.01 0.22 = umax a b p 2 4

3y2dy

Thus,

0.2 m

umax = 0.318 m>s

6y2(0.5dy) =

VC = 2 m>s

udA;

0.02 m3 >s =

0.5 m

Q = 13(0.5 m)36 1 0.52 2 m>s4(0.5 m)

Ans.

(70 + 101) 1 103 2 N>m2 = r(286.9 J>kg # K)(15 + 273) K

L0

Ans.

Also the volume of the parabolic block under the velocity profile is

p = rRT;

# m = rVA; 0.7 kg>s =

1 2.1532 kg>m3 2 1 3 m>s 2 3p(0.2 m)2 4 r = 2.1532 kg>m3

2

= 41.6 kg>s

Ans.

(80 + 101) 1 103 2 N>m2 = r(286.9 J>kg # K)(20 + 273) K

F4–5.

Chapter 4

F4–2.

Ans.

The average velocity can be determine from

0V At A, a b = 0, R = 0.5 m and 0t n

V =

0.02 m3 >s =

0V - rlVAAA = 0 0t 0y 0V = VAAA; 6 = 0t 0t 0y = 0.0667 m>s 0t

rl

Ans.

1 4 m>s 2 1 0.1 m2 2 Ans.

879

FUNDAMENTAL SOLUTIONS

F4–9.

0 - 0.05 kg>s - 0.002 kg>s + 1 1.45 kg>m32 (V) 3p 10.01 m224 = 0 V = 114 m>s

0 - VA 3p(0.025 m)2 4 +

Since AB is a short distance,

Ans.

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2

Since the pipe has a constant diameter, VB = VA = 6 m>s

Ans.

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 pA 3

+

V2 + 2

1 9.81 m>s2 2 (3 m)

= 0 +

pA = - 29.43 1 103 2 Pa = - 29.4 kPa

1000 kg>m

V2 + 0 2

pA 1000 kg>m3

0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

VB = 15.75 m>s

Ans.

pA pB VA2 VB2 + + gzA = + + gzB r0 r0 2 2 Select A and B on the same horizontal streamline, zA = zB = z. +

940 kg>m3

=

ps

+

pB = 206.44 1 10 940 kg>m3

3

F5–3.

2 Pa

2

Here pC = pB = 0 and VC = 0. pB pC VC2 VB2 + + gzB = + + gzC rw rw 2 2 0 +

VB2 + 0 = 0 + 0 + 2

VB = 6.264 m>s

1 9.81 m>s2 2 (2 m)

0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

1 6.264 m>s 2 2 2

+ 0

Ans.

80 1 103 2 N>m2

+

1 8 m>s 2 2 2

+ 0 =

pB = 112 1 103 2 Pa

pB 1000 kg>m3

+ 0 + 0

The manometer equation gives pB + rwghw = pC

112 1 103 2 Pa + F5–5.

1 1000 kg>m3 2 1 9.81 m>s2 2 (0.3 m)

pC = 114.943 1 103 2 Pa = 115 kPa

= pC Ans.

0 r dV + rw Vf>cs # dA = 0 0t Lcv w Lcs 0V + rwVBAB = 0 0t

0V = VBAB 0t

+ gz

= 206 kPa

+ 0 = 0 +

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2

rw

+ gz

1 15.75 m>s 2 2 2

2

1000 kg>m3

- 1 7 m>s 2 3p(0.06 m)2 4 + VB 3p(0.04 m)2 4 = 0

1 7 m>s 2 2

1 0.2506 m>s 2 2

Here VA = 8 m>s, VB = 0 (B is a stagnation point), and zA = zB = 0 (AB is a horizontal streamline).

F5–4.

0 - VAAA + VBAB = 0

300 1 103 2 N>m2

+

pA = 19.59 1 103 2 Pa = 19.6 kPa

Ans.

The negative sign indicates that the pressure at A is a partial vacuum. F5–2.

= 0

VA = 0.2506 m>s

Chapter 5 F5–1.

1 6.264 m>s 2 3p(0.005 m)2 4

0 - VAAA + VB AB = 0

0 r dV + rVf>cs # dA = 0 0t Lcv Lcs # # 0 - ma - mw + rmVA = 0

However, V = (2 m)(2 m)y = 4y Ans.

0y 0V = 4 0t 0t Thus, 4

0y = VB 3p(0.01 m)2 4 0t

VA =

0y = 25p 1 10 - 6 2 VB 0t

pA = pB = 0, VA can be neglected since VB ⪢ VA (Eq. 1).

(1)

880

F U N D A M E N TA L S O L U T I O N S

1 10.85 m>s 2 VB 2 = 6m = 2g 2 1 9.81 m>s2 2

The velocity head is

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 0 + 0 +

1 9.81 m>s2 2 y

= 0 +

VB = 219.62 y

2

VB 2 + 0 2

Since this is constant, then so is the HGL. EGL

At y = 0.4 m,

VB = 219.62(0.4) = 2.801 m>s

Q = VBAB = = At y = 0.2 m,

6m

1 2.801 m>s 2 3p(0.01 m)2 4 0.88 1 10-3 2 m3 >s

2

VB — 56m 2g 0

=

F5–6.

1 1.981 m>s 2 3p(0.01 m)2 4 0.622 1 10-3 2 m3 >s

0 -

1 4 m>s 2 3p(0.1 m)2 4

VB = 64 m>s

pA pB VA2 VB2 + zA = + zB + + gco gco 2g 2g 300 1 103 2 N>m2

Ans.

1 880 kg>m3 2 1 9.81 m>s2 2 =

+ VB 3p(0.025 m)2 4 = 0

Between A and B, zA = zB = 0, ra = 1.000 kg>m3 at T = 80°C (Appendix A). 20 1 103 2 N>m2

+

1.000 kg>m3

=

pB 1.000 kg>m3

+

1 4 m>s 2 2

1 880 kg>m 2 1 9.81 m>s 2 pB

3

2

2

+

V2 + 1.5 m 2g Ans.

pA + zA + gco 2g

300 1 103 2 N>m2

1 880 kg>m3 2 1 9.81 m>s2 2

= 37.567 m

1 64 m>s 2 2 2

V2 + 2m 2g

VA2

=

+ 0

+

pB = 304.32 1 103 2 Pa = 304 kPa

H =

pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2

F5–7.

VA = VB = V (constant pipe diameter), zA = 2 m, zB = 1.5 m, and rco = 880 kg>m3 (Appendix A).

F5–8.

0 rdV + rVf>cs # dA = 0 0t Lcv Lcs 0 - VAAA + VB AB = 0

B

HGL

Ans.

VB = 219.62(0.2) = 1.981 m>s Q = VB AB =

A

+

2 1 9.81 m>s2 2 (4 m>s)2

+ 2m

(4 m>s)2 V2 = = 0.815 m 2g 2 1 9.81 m>s2 2

+ 0

pB = 17.69 1 103 2 Pa = 18.0 kPa

EGL

2 V — 5 0.815 m 2g

37.567 m 36.751 m

Ans.

HGL

pA = pB = 0, VA _ 0 (large reservoir), zA = 6 m and zB = 0.

p 1z g— co

pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g 0 + 0 + 6m = 0 +

+ 0

1 10.85 m>s 2 3p(0.05 m)2 4

VB = 10.85 m>s Q = VBAB =

2 1 9.81 m>s2 2 VB2

= 0.0852 m3 >s

F5–9.

Ans.

0 r dV + rVf>cs # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0

0 - (3 m>s)3p(0.075 m)2 4 + VB 3p(0.05 m)2 4 = 0 VB = 6.75 m>s

Ans.

881

FUNDAMENTAL SOLUTIONS Also,

F5–10.

- 1 3 m>s 2 3p(0.075 m)2 4 + VC 3p(0.025 m)2 4 = 0 0 - VAAA + VCAC = 0 VC = 27 m>s

Ans.

VA _ 0, pA = pC = 0, zA = 40 m, and zC = 0;

pA pC VC 2 VA2 + zA + hpump = + zC + hturb + hL + + gw gw 2g 2g 0 + 0 + 40 m + 0 = 0 +

Between A and B, 400 1 10

2 N>m

2

+

1 3 m>s 2 2 1 9.81 m>s2 2

9810 N>m

pB

=

+

+ 0

1 6.75 m>s 2 2 2 1 9.81 m>s2 2

pB = 381.72 1 103 2 Pa = 382 kPa 9810 N>m3

= 5314.43 W = 5.314 kW # # # Wo Wo e = # ; 0.6 = Wo = 3.19 kW Ans. Wi 5.314 kW

+ 0 Ans.

F5–11.

Between A and C,

+

9810 N>m3

2 1 9.81 m>s2 2 (3 m>s)2 pC

=

pC = 40.0 1 103 2 Pa = 40 kPa 3

9810 N>m

pA VA2 H = + + zA gw 2g =

400 1 103 2 N>m2 9810 N>m3

+

= 41.233 m

+

40.774 m

(27 m>s)2 2

2

80 1 103 2 N>m2

+ 0

9810 N>m3

Ans. = 0 +

41.233 m

EGL

1 2 m>s 2 2 2 1 9.81 m>s2 2

2 1 9.81 m>s2 2 (10.19 m>s)2

+ 0

+ 0 + hpump

+ 8 m + 0 + 0.75 m

There the power supplied by the pump to the water is # W = Qgwhpump = 1 0.02 m3 >s 2 1 9810 N>m3 2 (5.6793 m) F5–12.

= 1.114 1 103 2 W = 1.11 kW # Here, Qm = - 1.5 kJ>s. Then

Ans.

# # V2 V2 Qin - Wout = c ah + B + gzBb - ahA + A + gzAb d m# B 2 2

# - 1.5 11032 J>s - Wout = e c450 11032 J>kg + - c 600 1 103 2 J>kg +

# Wout = 298 kW HGL

+

hpump = 5.6793 m

2

38.911 m

4.077 m

pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g

+ 0

2 1 9.81 m>s

1 3 m>s 2 = = 0.459 m 2g 2 1 9.81 m>s2 2 1 6.75 m>s 2 2 VB2 = = 2.322 m 2g 2 1 9.81 m>s2 2 1 27 m>s 2 2 VC2 = = 37.156 m 2g 2 1 9.81 m>s2 2 VA2

pB = 0, zA = 0, zB = 8 m

1 3 m>s 2 2 2 1 9.81 m>s2 2

Velocity heads at A, B, and C are

Q = VBAB; 0.02 m3 >s = VB 3p(0.025 m)2 4 VB = 10.19 m>s

pA pC VC2 VA2 + + zA = + + zC gw gw 2g 2g 400 1 103 2 N>m2

150 b (1.5 m) 100

Q = VC AC = 1 8 m>s 2 3p(0.025 m)2 4 = 5p 1 10-3 2 m3 >s # Wi = Qgwhs = 3 5p 1 10-3 2 m3 >s41 9810 N>m3 2 (34.488 m)

2

3

+ 0 + hturb + a

hturb = 34.488 m

pA pB VA2 VB 2 + + zA = + + zB gw gw 2g 2g 3

1 8 m>s 2 2 2 1 9.81 m>s2 2

1 12 m>s 2 2 2

1 48 m>s 2 2 2

+ 0d

+ 0 d f 1 2 kg>s 2

Ans.

The negative sign indicates that the power is being added to the flow by the engine.

882

F U N D A M E N TA L S O L U T I O N S - Fy = (15.915 m>s) sin 60° 1 1000 kg>m3 2 1 0.006 m3 >s 2

+ c ΣFy = 0 + VB sin 60°(rwVBAB) + (VC sin 60°)[-(rwVCAC)]

Chapter 6 F6–1.

Q = VA; 0.012 m3 >s = V3p(0.02 m)2 4; V = 9.549 m>s

+ (15.915 m>s) sin 60° 3 - 1 1000 kg>m3 2 1 0.014 m3 >s 2 4

F = 2F 2x + F 2y = 21381.97 N2 2 + 1110.27 N2 2 = 397.57 N

VA = VB = V = 9.549 m>s and pB = 0 ΣF =

Fy = 110.27 N

0 Vr dV + VrVf>cs # dA 0t Lcv Lcs

= 398 N

110.27 N u = tan-1 a b = tan-1 a b = 16.1° Fx 381.97 N Fy

+ ΣFx = 0 + VBrw(VBAB) S

Fx = (9.549 m>s) 1 1000 kg>m3 2 1 0.012 m3 >s 2 = 114.59 N S

Fy

Ans. Ans.

B

+ c ΣFy = 0 + VA( - rw VAAA)

1 160 1 103 2 N>m2 2 1 p 1 0.02 m 2 2 2 - Fy = (9.549 m>s) 3 - 1 1000 kg>m3 2 1 0.012 m2 >s 24

C

Fy = 315.65 NT

F = 2F 2x + F 2y = 2(114.59 N)2 + (315.65)2 = 336 N Ans. u = tan -1 a

Fy Fx

b = tan -1 a

315.65 N b = 70.0° 114.59 N

F6–3.

Ans.

(FA)x = pAAA = pA 3p(0.025 m)2 4 = 0.625p 110-32 pA V0 = 3p(0.025 m)2 4(0.2 m) = 0.125p 1 10-3 2 m3 AA = AB so VA = VB

Fy

ΣF =

Fx

pAA

QA = VAAA; 0.02 m3 >s = VA 3p(0.02 m)2 4;

0.625p 110-32 pA = 13 m>s22 1 1000 kg>m3 2 3 0.125p 110-3 2 m34 pA = 600 Pa

QB = 0.3QA = 0.3 1 0.02 m3 >s 2 = 0.006 m3 >s Also, VB = VC = VA = 15.915 m>s and pA = pB = pC = 0. 0 ΣF = VrVf>cs # dA Vr dV + 0t L cv L cs + SΣF x = 0 + VA[ - (rwVAAA)] + VB cos 60°(rwVBAB) + ( -VC cos 60°)(rwVCAC) - Fx = (15.915 m>s) 3 - 1 1000 kg>m3 2 1 0.02 m3 >s 2 4

+ (15.915 m>s)(cos 60°) 1 1000 kg>m3 2 1 0.006 m3 >s 2

+ ( -15.915 m>s)(cos 60°) 1 1000 kg>m3 2 1 0.014 m3 >s 2 Fx = 381.97 N

Ans.

W

VA = 15.915 m>s

QC = 0.7QA = 0.7 1 0.02 m3 >s 2 = 0.014 m3 >s

0 Vr dV + V rVf>cs # dA 0t Lcv Lcs

dV + r V + VA[ -(rwVAAA)] + VB (rwVBAB) SΣF x = dt w 0 dV dV + r V ; where = 3 m>s2 SΣF x = dt w 0 dt

B

MA

F6–2.

Fx

A

(FA)y (FA)x

QA = VAAA = (6 m>s)3p(0.015 m)2 4 = 1.35p 1 10-3 2 m3 >s

F6–4.

QB = QC =

1 1 QA = 3 1.35p 1 10-3 2 m3 >s 4 2 2

= 0.675p 1 10-3 2 m3 >s

QB = VBAB;

0.675p 1 10-3 2 m3 >s = VB 3p(0.01 m)2 4 VB = 6.75 m>s

VC = VB = 6.75 m>s ΣF =

0 Vr dV + VrVf>cs # dA 0t Lcv Lcs

FUNDAMENTAL SOLUTIONS

883

3 2 + SΣF x = 0 + (VC cos 45°)(rcoVCAC) + ( - VB cos 45°)(rcoVB AB) - Fx = (21.5 m>s) 3 - (1000 kg>m )(21.5 m>s) p(0.01 m) 4

+ c ΣFy = 0 + 1Vf>cs 2 out 3 r1Vf>cs 2 out 1Aout 24

Fx = (VC cos 45°)rcoQC - (VB cos 45°)rcoQB

Fx = 145.2 N

Fx = 0 + c ΣFy = 0 + VA( - rcoVAAA) + VB sin 45°(rcoVBAB)

3 80 1 10 2 N>m 4 3 p 1 0.015 m 2 4 - Fy = (6 m>s) 3 - 1 880 kg>m3 2 1 1.35p 1 10-3 2 m3 >s 2 4 + 2(6.75 m>s) sin 45° 3 1 880 kg>m3 2 1 0.675p 1 10-3 2 m3 >s 2 4 + VC sin 45°(rcoVCAC)

3

2

2

Fy = (21.5 m>s) 31 1000 kg>m3 2 (21.5 m>s) p(0.01 m)2 4 F =

2F 2x

= 145.2 N

+ F y2 = 2(145.2)2 + (145.2)2 = 205 N

u = tan-1 a

Fy Fx

b = tan - 1 a

145.2 b = 45° 145.2

Ans. Ans.

Fy = 61.1 N

Since Fx = 0, Fy is the total force the fitting exerts on the oil flowing through it. F = Fy = 61.1 NT

Ans.

Fy

Chapter 10

Fy

F10–1.

Fx

F6–5.

Fx

0 VrVf>cs # dA VrdV + 0t L cv L cs

ΣF =

Re =

+ SΣF x = 0 + ( - Vout)[raVout Aout]

- Ff = ( - 20 m>s) 3 1 1.22 kg>m3 2 (20 m>s) p(0.125 m)2 4 Ff = 24.0 N

Ans.

f =

Ff Nf

= Vcv + Vf>cs

20 m>s = -1.5 m>s + Vf>cs

64 64 1.904 = = Re 33.61V V

L V2 1.904 4m V2 d = 9.7044 V = a ba bc D 2g V 0.04 m 2(9.81 m>s2) (2) Setting the datum through B, and applying the energy equation between A and B,

pA pB V 2A V 2B + + + zA + hpump = + zB + hturbine + hL ggl ggl 2g 2g 0 + 0 + 5.5 m + 0 = 0 +

Vf>cs = 21.5 m>s ΣF =

(1)

The major head loss is

hL = f

+ 2V 1S f

V(0.04 m) VD = = 33.61 V ngl 1.19(10-3) m2 >s

Assuming that laminar flow occurs in the pipe. Then,

W

F6–6.

We will consider the tank as large and the flow as steady. From the table in Appendix A, rgl = 1260 kg>m3 and n = 1.19(10-3) m2 >s at T = 20°C.

0 VrVf>cs # dA VrdV + 0t L cv L cs

+ SΣF x = 0 + (Vf>cs)in 3 - r(Vf>cs)in Ain 4

V2 + 0 + 0 + 9.7044V 2(9.81 m>s2)

V 2 + 190.4V - 107.91 = 0 V = 0.5651 m>s

884

F U N D A M E N TA L S O L U T I O N S From Eq. 1,

F 10–3.

Re = 33.61(0.5651 m>s) = 18.99 6 2300

From Appendix A, rw = 997.1 >m3 and nw = 0.898(10-6) m2 >s for water at 25°C. The average velocity of the flow is

0.0314p m3 >s Q 0.03998 = = p 2 A D2 D 4 0.03998 a b(D) 4.4521(104) VD D2 Re = = = -6 2 nw D 0.898(10 ) m >s

The flow is indeed laminar. Then from Eq. 2,

V =

hL = 9.7044(0.5651 m>s) = 5.4837 m = 5.48 m Ans. For laminar flow, the velocity along the centerline is u max = 2V = 2(0.5651 m>s) = 1.1302 m>s = 1.13 m>s Ans. pC pB VC2 VB2 + + + zC + hpump = + zB + hturbine + hL ggl ggl 2g 2g

(1)

The major head loss is

V2 V2 2 2 f + + 4m + 0 = 0 + + 0 + 0 + 5.4837 m L V2 200 m (0.03998>D ) 3 2 2g 2g (1260 kg>m )(9.81 m>s ) = fa hL = f bc d = 0.01629 5 2 pC

D 2g

D

pC = 18.3397(103) Pa

p1 p2 V12 V22 + + z1 + hpump = + + z2 + hturbine + hL gw gw 2g 2g

18.3397(103) Pa - 0 pC - pB = L 4m = 4.58 kPa>m

p1 p2 V2 V2 + + 0 + 0 = + + 0 + 0 + hL gw gw 2g 2g

Ans.

From Appendix A, rw = 997.1 kg>m3 and nw = 0.898(10-6) m2 >s.

Q = VA; 0.03 m >s = V3p(0.05m) 4 3

Re =

2

p1 - p2 = gwhL Here p1 - p2 = 80(103) N>m2. Then

V = 3.8197 m>s

80(103) N>m2 = (997.1 kg>m3)(9.81 m>s2) a 0.01629

(3.8197 m>s)(0.1 m) VD = 4.25(105) = nw 0.898(10-6) m2 >s

For commercial steel pipe, e 0.045 mm = = 0.00045. From the Moody D 100 mm diagram, f = 0.0175. Then, the major head loss is

D =

L V2 500 m (3.8197 m>s) d = 65.069 m = 0.0175 a bc D 2g 0.1 m 2(9.81 m>s2) Applying the energy equation between A and B,

pA pB V 2B + + zA + hpump = + + zB + hturbine + hL gw gw 2g 2g

- 25(103) N>m2 =

(997.1 kg>m3)(9.81 m>s2)

= 18.3 kW

f from Moody diagram

D (m); Eq. 2

e (m>m) D

1

0.02

0.1318

0.00197

3.38(105)

0.024

2

0.024

0.1367

0.00190

3.26(105)

0.024

Iteration

Re; Eq. 1

The assumed f in the second iteration is almost the same as that from the Moody diagram. Thus, D = 0.1367 m is an acceptable result, i.e., D = 137 mm F10–4.

Ans.

b

(2)

Assume f

+ 65.069 m

= 62.513 m # Ws = gwQhpump = (997.1 kg>m3)(9.81 m>s2)(0.03 m3>s)(62.513 m)

D5

f 1>5 3.4684

V 2A

pA pB V2 V2 + + z + hpump = + + z + 0 + hL gw gw 2g 2g pB - pA hpump = + hL gw

f

For cast iron, the roughness factor is e = 0.00026 m. A numerical iterative procedure is required.

2

hL = f

D

Applying the energy equation between points 1 and 2,

Thus, the pressure gradient is

F 10–2.

2(9.81 m>s )

We will consider the upper tank as large and the flow as steady. Q = VA

0.002 m3 >s = V3p(0.01 m)2 4 V = 6.3662 m>s

Ans.

885

FUNDAMENTAL SOLUTIONS The pressure head at B is pB >gw = 2 m. Set the datum through B,

(6.3662 m>s)(0.02 m) VD = = 1273.24 ne 0.1(10-3) m2 >s Since Re 6 2300, the flow is laminar. 64 64 f = = = 0.05027 Re 1273.24 Re =

pA pB V 2A V 2B + + zA + hpump = + + zB + hturbine + hL gw gw 2g 2g 0 + 0 + 15 m + 0 = 2 m +

The major head loss is

(6.3662 m>s) L V2 9m = 0.05027a bc d = 46.72 m D 2g 0.02 m 2(9.81 m>s2)

V2 + 0 + 0 2(9.81 m>s2)

2

(hL)major = f

+ 14.0163 f V 2 + 0.08104 V 2 (14.0163 f + 0.13201) V 2 - 13 = 0

The minor head loss is contributed by the flush entrance (KL = 0.5), fully opened gate valve (KL = 0.19) and 90° elbows (KL = 0.9).

e 0.26 mm = = 0.00325. D 80 mm An iterative procedure is required.

For cast iron pipe,

(6.3662 m>s)2 V2 = 30.5 + 0.19 + 2(0.9)4 c d (hL)minor = a KL 2g 2(9.81 m>s2) = 5.144 m Set the datum through B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturbine + hL go go 2g 2g pA (920 kg>m3)(9.81 m>s2) = 0 +

+ 0 + 11 m + 0

(6.3662 m>s)2 2

2(9.81 m>s )

+ 0 + 0 + (46.72 m + 5.144 m)

pA = 387 kPa F10–5.

Ans.

We treat the tank as large and the flow as steady. From Appendix A, rw = 998.3 kg>m3 and nw = 1.00(10-6) m2 >s at T = 20°C. Re =

(hL)major = f

V(0.08 m) VD = = 80(103)V nw 1.00(10-6) m2 >s

(1)

L V2 22 m V2 = fa bc d = 14.0163 f V 2 D 2g 0.08 m 2(9.81 m>s2)

The minor head loss is contributed by the flush entrance (KL = 0.5), fully opened gate valve (KL = 0.19), and 90° elbow (KL = 0.90). (hL)minor = ΣKL

V2 V2 = (0.5 + 0.19 + 0.9) c d = 0.08104V 2 2g 2(9.81 m>s2)

Iteration

Assumed f

V (m>s); Eq. 2

Re; Eq. 1

f from Moody diagram

1

0.028

4.9787

3.98(105)

0.027

2

0.027

5.0466

4.04(105)

0.027

The assumed f in the second iteration is almost the same as that from the Moody diagram. Thus V = 5.0466 m>s is acceptable. Q = VA = (5.0466 m>s)3p(0.04 m)2 4 = 0.0254 m3 >s

Ans.

Answers to Selected Problems Chapter 1 1–1.

1–2.

1–3.

1–5. 1–6. 1–7. 1–9. 1–10. 1–11. 1–13. 1–14. 1–15. 1–17. 1–18. 1–19. 1–21. 1–22. 1–23. 1–25. 1–26. 1–27. 1–29. 1–30. 1–31. 1–33. 1–34. 1–35. 1–38. 1–39. 1–41.

1–42. 1–43. 1–45. 1–46. 1–47.

886

a. 11.9 mm>s b. 9.86 Mm # s>kg c. 1.26 Mg # m a. kN # m b. Gg>m c. Mm>s d. GN>m2 a. 23.6 Gm b. 42.1110-3 2 m3 c. 7.011103 2 N2 W = 19.5 kN ∆r = - 0.762 kg>m3 W = 446 N p2 = 449 kPa Mct = 137 kg rm = 755 kg>m3, Sm = 0.755 p = (98.8r) kPa W = 91.5 N h = 3.05 m T = 26.9°C g = 26.6 N>m3 p = 10.6 MPa Mr = 13.9 Tg EV = 220 kPa p = 23.6 MPa EV = 2.20 GPa ∆r = 0.550 kg>m3 no = 109(10-6) m2 >s m = 0.849 N # s>m2, y = 2.00 m>s P = 0.0405 N tp = 2.70 Pa, tfs = 3.86 Pa ma = 8.93110-3 2 N # s>m2 tmin = 0.191 MPa t = 0.304 mPa P = 1.84 N m = 5.03 N # s>m2 Using the Sutherland equation, at T = 283 K, m = 17.9110-6 2 N # s>m2 at T = 353 K, m = 20.8110-6 2 N # s>m2 B = 1.36110-6 2 N # s> 1m2 # K1>2 2, C = 78.8 K At T = 283 K, m = 1.25110-3 2 N # s>m2 At T = 353 K, m = 0.339110-3 2 N # s>m2 T = 68.1 N # m T = 0.218 mN # m t  r = 40 mm = 7.03 Pa, t  r = 80 mm = 14.1 Pa

1–49. 1–50. 1–51. 1–53. 1–54. 1–55. 1–57. 1–58. 1–59. 1–61. 1–62. 1–65.

4pmvr 3L t pmvR4 T = 2t sin u Tboil = 70.2°C Pmax = 3.93 kPa Pmax = 3.17 kPa pmin = 1.71 kPa P = 0.335 N s = 0.0729 N>m 2s d = ∆p 0.0154 L = a bm sin u l = 24.3 mm T =

d1mm2

1

2

3

4

5

6

h1mm2 - 10.74 - 5.37 - 3.58 -2.69 - 2.15 - 1.79 1–66. 1–67.

d = 1.45 mm d = 2.58 mm

Chapter 2 2–3. 2–5. 2–6. 2–7. 2–9. 2–10.

ps = 16.0 kPa pd = 10.6 kPa hHg = 301 mm rm = 1390 kg>m3 hHg = 211 mm h = 5.33 m pA = 9.68 kPa, pB = - 5.81 kPa pA = 9.98 kPa

2–11. 2–13.

h = 1.79 m 1pB 2 g = - 27.1 kPa pmax = 78.5 kPa

2–2.

2–14. 2–15.

2–17. 2–18. 2–19.

pA = 58.9 kPa db = 0.202 mm 1pb 2 gage = 107 kPa The pressure at the bottom of the tank does not depend on its shape. r ∘ gh p = - E Vlna1 EV b ∆h = 86.9 mm TC = - 56.5°C p = 12.0 kPa

ANSWERS TO SELECTED PROBLEMS

2–21. 2–22. 2–23.

p = 30.0098110.875h2 + 825h24 kPa, where h is in meters. p = 208 kPa Elevation 0 (m)

1000 2000 3000 4000 5000 6000

p (kPa) 101.3 89.88 79.50 70.12 61.66 54.05 47.22

2–25.

2–26. 2–27. 2–29. 2–30. 2–31. 2–33. 2–34. 2–35. 2–37. 2–38. 2–39. 2–41. 2–42. 2–43. 2–45. 2–46. 2–47.

2–49. 2–50.

2–51. 2–53. 2–54. 2–55. 2–57.

2–58. 2–59. 2–61. 2–62. 2–63.

z 1m2

0

200

400

600

800

1000

p 1kPa2 101.3 98.92 96.58 94.29 92.04 89.84

pmax = 34.3 kPa pB = 29.4 kPa pc = 14.7 kPa pB = 17.8 kPa h = 135 mm hke = 164 mm, Vke = 0.912110-3 2 m3 h = 893 mm h′ = 256 mm pB - pA = 20.1 kPa h = 76.8 mm pB = 15.1 kPa pB = 20.8 kPa pB - pA = 1.32 kPa At At pA - pB = ec yt - a1 byR - a by d AR AR L h = 0.319 mm h = 0.741 m FR = 29.3 kN, yP = 1.51 m FR = 197 kN u = 26.6° x = 1m y = 2m FB = 937 N FAB = 123 kN FCD = 19.6 kN FBC = 549 kN h = 14.6 m h = 4.95 m h = 3.06 m FC = 25.3 kN, h = 3.5 m NB = 42.4 kN Ax = 25.9 kN, Ay = 26.7 kN FBEDC = 7.94 kN, d = 0.283 m FAFEB = 34.3 kN d′ = 0.576 m FBC = 480 kN, FCD = 596 kN h = 12.2 m NB = 14.3 kN, FA = 10.2 kN h = 3.90 m

2–65. 2–66. 2–67. 2–69. 2–70. 2–71. 2–73. 2–74. 2–75. 2–77. 2–78. 2–79. 2–81. 2–82. 2–83. 2–85. 2–86. 2–87. 2–89. 2–90. 2–91. 2–93. 2–94. 2–95. 2–97. 2–98. 2–99. 2–101. 2–102. 2–103. 2–105. 2–106. 2–107. 2–109. 2–110. 2–111. 2–113. 2–114. 2–115. 2–117. 2–118. 2–119. 2–121. 2–122. 2–123. 2–125. 2–126. 2–127. 2–129. 2–130.

FR = 7.36 kN, d = 0.556 m FR = 11.6 kN, d = 0.542 m h = 3.60 m FR = 48.6 kN, yP = 2.00 m FR = 48.6 kN yP = 2.00 m FR = 67.1 kN, yP = 3.85 m FR = 73.1 kN, d = 917 mm FR = 40.4 kN, d = 2.44 m FR = 40.4 kN, d = 2.44 m FR = 72.8 kN, yP = 3.20 m FR = 72.8 kN, yP = 3.20 m FR = 96.1 kN d = 0.820 m FR = 3.92 kN, yP = 0.600 m MC = 1181 kg h = 2.03 m FR = 121 kN, yP = 4.17 m h = 3.60 m 6 4 m FR = 12.8 kN, yP = 1.03 m F = 17.3 kN, yP = 0.938 m F = 17.3 kN, yP = 0.938 m FR = 3.68 kN, yP = 442 mm FR = 993 kN, u = 18.4° NB = 194 kN, Ax = 177 kN, Ay = 31.9 kN FR = 368 kN, u = 53.1° NB = 104 kN, Ay = 0, Ax = 66.2 kN FR = 179 kN FR = 262 kN, u = 53.1° FR = 144 kN, u = 23.2° NB = 71.5 kN, Ax = 88.3 kN, Ay = 0 NB = 104 kN, Ay = 0, Ax = 66.2 kN NB = 52.3 kN, Ax = 654 kN, Ay = 476 kN Bx = 235 kN, By = 275 kN FR = 472 kN, u = 33.7° Fy = 17.7 kN, Fh = 35.4 kN FR = 53.8 kN, T = 42.3 kN # m FR = 53.8 kN N = 5.34 kN FR = 253 kN, T = 140 kN # m NB = 118 kN, Ax = 0, Ay = 101 kN T = 196 kN # m NB = 50.3 kN, Ay = 55.3 kN, Ax = 298 kN V = 3812 m3 TAB = 8.45 kN NB = 32.4 N, Ay = 7.20 N, Ax = 32.4 N dC = 51.4 mm, dD = 58.4 mm h = 1.12 m h = 136 mm h = 514 mm It will restore itself. It will restore itself.

887

888

ANSWERS

TO

SELECTED PROBLEMS

2–131. a. p = 10.7 kPa b. p = 12.9 kPa 2–133. u = 8.69°, pA = 11.7 kPa, pB = 17.7 kPa 2–134. pA = 21.6 kPa, pB = 11.6 kPa 2–135. At rest: pB = 19.6 kPa, With acceleration: ∆V = 3.57 m3, pB = 29.4 kPa 2–139. h′A = 0.171 m, h′B = 0.629 m 2–141. v = 0.977 rad>s 2–142. d = 0.132 m 2–143. d′ = 0.151 m 2–145. pB - pA = 28.1 kPa 2–146. v = 7.95 rad>s 2–147.

ri = c

4gr 2o 1do - d2 2

v

d

1>4

3–33. 3–34. 3–35.

3–37. 3–38. 3–39. 3–41. 3–42. 3–43. 3–45. 3–46.

Chapter 3 3–1. 3–2. 3–3. 3–5. 3–6. 3–7. 3–9. 3–10. 3–11. 3–13. 3–14. 3–15.

3–47.

y3 = 6x + 2, V = 8.94 m/s, u = 26.6 ∘ 1 y = 10 116 - 3x2 V = 19.8 m>s, u = 40.9° V = 6.40 m>s, u = 309° V = 10.3 m>s, u = 299° y = 34 1e1.2x - 12 V = 6.32 m>s, u = 18.4° V = 7.21 m>s, u = 56.3° , y4 = 4196x - 2242 ln1y2 2 + y = 2x - 2.61, V = 5.66 m>s, u = 45° u = 3.43 m>s, v = 3.63 m>s y3 = 12x + 15, y = 7 m, x = 27.3 m y3 - 18y = 3x 2 + 9x - 38, V = 12.2 m>s, u = 35.0° y = 4e 8 1x 1

3–17. 3–18. 3–19. 3–21.

2

- 42

u = 2.60 m>s, v = { 1.50 m>s y = 1.6x 2 For t = 1 s, y = 3 3 1 3x 4+ 1 2 2>3 4 m, For t = 1.5 s, y = 3 34 13x 2 + 124 m 1x2 + x + 5422 - 4 900

3–49. 3–50. 3–51. 3–53. 3–54. 3–55.

3–27. 3–29. 3–30. 3–31.

y = 6e y = 2x 2 V = 6.40 m>s, a = 8 m>s2, y = 1 52 ln x2 + 1 2 m y = 5.60 m, x = 14.4 m, a = 13.4 m>s2, u = 10.3° 1 y2 = (x 3 + 46), a = 2.19 m>s2, u = 59.0° 6 V = 1.92 m>s, a = 0.750 m>s2, 5y2 + 3x 2 = 93 a = 339 m>s2 V = 8.41 m>s, uv = 25.3° , a = 19.4 m>s2, ua = 43.2°

0.540 m3 >s, V = 0.450 m>s 864 kg>s 1.75 m>s 0.537 m3 >s 0.166 m3 >s 2 2 Q = whumax, V = umax 3 3 V = 1.90 m>s Q = 0.0294 m3 >s

Chapter 4 4–1. 4–2. 4–3. 4–5. 4–6. 4–7. 4–9. 4–10. 4–11.

1

3–22 3–23. 3–25. 3–26.

a = 1.77 m>s2 a = 6.75 m>s2 x 2 y = y = c ln + 1 d m 3 2 a = 30.3 m>s2 u = 7.59 ∘ At (0, 0), a = 64 m>s2 T At (1 m, 0), a = 89.4 m>s2, u = 63.4° a = 100 m>s2 y = 2x, a = 286 m>s2, u = 63.4° a = 36.1 m>s2, u = 33.7° V = (8 - 2x) m>s, a = 4(x - 4) m>s2 x = 411 - e -2t 2 m V = 5.66 m>s, a = 17.9 m>s2 V = 33.5 m>s, uv = 17.4° a = 169 m>s2, ua = 79.1° V = 4.12 m>s, a = 17.0 m>s2 x y = , a = 2.68 m>s2, u = 26.6° u 2 2 2 as = 47.5 m>s , an = 7.76 m>s a = 3.75 m>s2, u = 36.9° a = 3.38 m>s2 a = 72 m>s2 as = 222 m>s2, an = 128 m>s2 a = 4 m>s2

4–13. 4–14.

Q # m V Q Q

= = = = =

Q = c

0.00502R

10.5774R + 32 1>2

Q = 0.0107 m3 >s # m = 4.10 kg>s

d m3 >s where R is in m

4–15. pg(kPa) m(kg/s) 4–17. 4–18.

0

10

20

30

40

50

2.94 3.23 3.52 3.81 4.10 4.39

# m = 1.07 kg>s Tc(°C) m(kg/s)

0

10

20

30

40

50

4.71 4.53 4.38 4.24 4.10 3.98

ANSWERS TO SELECTED PROBLEMS 4–19. 4–21. 4–22. 4–23. 4–25. 4–26. 4–31. 4–39. 4–41. 4–42. 4–43. 4–45. 4–46. 4–47. 4–49. 4–50. 4–51. 4–53. 4–54. 4–55. 4–57.

4–58. 4–59. 4–61. 4–62. 4–63. 4–65. 4–66. 4–67. 4–69. 4–70. 4–71. 4–73. 4–74. 4–75. 4–77. 4–78.

t = 0.396 s 4(p - 2) 2 Q1 = aR , Q2 = 0.991aR2 p n = 7.071109 2 t = 0.0147 s u = 56.6 m>s, a = 42.71103 2 m>s2 V = 6.94110-6 2 m3, n = 713 A moving control volume allows the flow to be treated as steady. t = 45 s 15.41103 2 a = c 2 d m>s2 p 11 - 7.50x2 5 a = 662 m>s2 a = 153 m>s2 VB = 28.8 m>s, aA = 3.20 m>s2 t = 48.5 s t = 0.893d -4>3, s = 22.5 m, unrealistic VC = 7.22 m>s VB = 3.98 m>s # mm = 0.714 kg>s V = 9.42 m>s V = 10.0236 D2 2 m>s where D is in mm d = 141 mm, V = 1.27 m>s dy QB = a + 0.0178b m>s where dt 2.25p 3 QB is in m >s dy = 0.0274 m>s dt 0y = 0.123 m>s 0t VA = 8.00 m>s rmix = 972 kg>m3 rm = 968 kg>m3 993 Vs = a b m>s where t is in h t dr = - 0.101 kg> 1m3 # s2, unsteady dt Vt = 0.786 m>s, Vb = 0.884 m>s t = 4.19 min dy = 6.43 m>s dt Vr = 3.11 m>s V = 153 m>s dy = 21.4 mm>s dt t = 1.48 min rm = 779 kg>m3, VC = 1.875 m>s dy rm = 779 kg>m3, = -3.27 mm>s dt

4–79. 4–81. 4–82. 4–83. 4–85. 4–86. 4–87. 4–89. 4–90. 4–91. 4–93. 4–94. 4–95.

889

# # mAB = 4.41 g>s,, mCD = 2.94 g>s, mass also flows through control surface AC. Vb = 1.05 m>s T dh = - 0.0283 m>s dt dh 2 = c - d 2 d m>s, where d is in meters dt 3 r = 1.90 kg>m3 r = 0.247 kg>m3 Vw = 1.63 m3 dy = 0.0538 m>s dt dy = 0.0270 m>s dt r = 9.04 kg>m3 t = 32.7 s t = 30.6 s V = 0.677 m3

Chapter 5 5–1. 5–2. 5–3. 5–5. 5–6. 5–7. 5–9. 5–10.

as = 38.8 m>s2 ∆p = - 901.5 Pa ∆p = - 12.0 kPa pB - pA = 11.7 kPa VA = 3.51 m>s pB = 198 kPa pB = -61.5 Pa, F = 1.93 N VA = 12.7 m>s, pA = - 60.8 kPa

5–11.

p = 0.5c 40.5 -

5–13. 5–14. 5–15. 5–17. 5–18. 5–19. 5–21 5–22. 5–23. 5–25. 5–26. 5–27. 5–29. 5–30. 5–31. 5–33. 5–34.

where r is in mm pB - pC = 3.32 kPa Q = 0.870 m3 >s VA = 2.83 m>s, pA = 320 kPa VB = 8.39 m>s Q = 5.36 m3 >s Q = 1.51 m3 >s VA = 32.1 m>s, VB = 31.5 m>s, pC = 525 kPa VA = 0.886 m>s, VB = 3.54 m>s . m = 0.100 kg>s pA = 1.23 kPa hB = 1.96 m Q = 0.198 m3 >s VB = 14.4 m>s, pB = 102 kPa Q = 36.8 m3 >s h = 0.789 m # m = 19.5 kg>s h = 137 mm

6.481106 2 r2

d kPa,

890 5–35. 5–37. 5–38. 5–39. 5–41. 5–42. 5–43. 5–45. 5–46. 5–47. 5–49. 5–50. 5–51. 5–53. 5–54. 5–55. 5–57. 5–58. 5–59. 5–61. 5–62. 5–63. 5–65. 5–66. 5–67. 5–69.

ANSWERS

TO

SELECTED PROBLEMS

p(x) - pA = 130x - 4.5 x 2 2kPa Q = 0.236 m3 >s Q = 30.005625 p280 + 19.62 h4 m3 >s VB = 0.417 m>s, pA - pB = 62.3 Pa pA = 1.07 kPa h = 1.62 m Q = 146 m3 >s pC = 66.3 kPa, QA = 0.05625 m3 >s pC = 62.8 kPa, QB = 0.0244 m3 >s pA = 9.81 kPa h = 41.0 mm VC = 1.40 m>s, dD = 23.8 mm V = 9.33 m>s, hmax = 4.49 m VB = 18.3 m>s, VC = 3.07 m>s t = 2.78 h VA = 14.6 m>s, VB = 7.58 m>s Vle = 18.91103 2liters # ma = 1.12 kg>s pB = 101.2 kPa VC = 0.969 m>s, ∆p = 0.580 kPa Q = 1.65 m3 >s pB = 317 kPa VA = 0.700 m>s, pA = 4.17 kPa VA = 39.3 m>s t = 24.7 min dy 19.6 y = a bm>s, dt A 6.251106 2y4 - 1 where y is in meters

VB = 16 m>s, pB = 19.05 kPa hL# = 1.53 m W# s = 24.7 kW Wout = 26.7 kW pB - pA = 73.6 kPa # 5–79. W# s = 5.14 kW 5–82. Ws = 14.0 kW 5–85. pA' = -5.89 kPa # 5–86. Ws = 17.9 kW # 5–87. Ws = 573 kW 5–89. a = 1.06 5–90. a = 2 5–91. Q = 0.0424 m3 >s # 5–93. Wout = 4.62 kW 5–94. pB - pA = 58.6 kPa 5–95. Q = 0.0773 m3 >s # 5–97. Ws = 6.91 kW 5–98. Q = 1.04 m3 >s 5–99. h = 1.89 m 5–101. VB = 28.4 m>s, pB = -153 kPa 5–70. 5–73. 5–74. 5–77. 5–78.

# 5–102. Wout = 8.95 kW # 5–103. Ws = 116 kW # 5–105. Ws = 32.5 kW # 5–106. Wout = 19.8 kW 5–107. pA = 282 kPa # 5–109. Ws = 4.59 kW # 5–110. Ws = 11.2 kW 5–111. hL = 8.67 m

Chapter 6 6–2. 6–3. 6–5. 6–6. 6–7. 6–9. 6–10. 6–11.

L = 10.1 kg # m>s F = 8.32 kN QA = 0.00460 m3 >s, QB = 0.0268 m3 >s Fn = 88.9 N T = 109 N F = 133 N F = 1.92 kN Fx = 51.5 N, Fy = 1.33 kN

6–13.

Q = 0.0151 m3 >s

6–14.

h =

6–15.

h = c

8.26(106)Q4 - 0.307(10-6)

F = c

2.55(10-3)

6–17. 6–18. 6–19. 6–21. 6–22. 6–23. 6–25. 6–26. 6–27. 6–29. 6–30. 6–31. 6–33. 6–34. 6–35. 6–37. 6–38. 6–39. 6–41. 6–42. 6–43.

8Q2 2 4

pd g

-

m2g

8rw2Q2

F = 3 160p11 + sin u2 4 N F = 1.67 kN Q2

x2

+ 25.5 d N

d m

QA = 0.0422 m3 >s, QB = 0.00303 m3 >s Fx = 197 N, Fy = 1.50 kN F = 7.61 N p = 418 Pa Fn = 36.5 N Cx = 196 N, Cy = 309 N Fx = 349 N, Fy = 419 N F = 125 N Th = 19.5 kN, Tv = 9.66 kN F = 36.4 kN N = 886 N Ax = 102 N, Ay = 239 N F = 141 N F = 62.8 N, Vh = 10 m>s F = 1.92 kN F = 166 N F = 258 N F = 3.16 kN

ANSWERS TO SELECTED PROBLEMS F# = 13.6 N, u = 30° W = 1.36 kW F1 = F2 = F3 = rwAV 2 F = 318 N Ax = 385 N, Ay = 181 N, MA = 54.3 N # m Bx = 4 kN, Ax = 1.33 kN, Ay = 5.33 kN M = 46.5 N # m v = 86.1 rad>s Ax = 1.01 kN d = 199 mm M = 13.4 N # m Cx = 21.3 N, Cy = 79.5 N, MC = 15.9 N # m Ax = 444 N, Ay = 283 N, MA = 90.7 N # m Ax = 1.03 kN, Ay = 495 N, MA = 236 N # m Vw = 5.48 m>s, M = 0.211 mN # m Cx = 2.28 kN, Cy = 1.32 kN, MC = 1.18 kN # m Dx = 778 N, Dy = 645 N, MD = 514 N # m T = 11.0 kN T = 74.2 kN, e = 0.502 V = 47.7 m>s F# = 90.7 kN W = 3.50 MW T1 = 600 N, T2 = 900 N, e1 = 0.5, e2 = 0.6 V = 6.58 m>s, ∆p = 22.4 Pa Vb = 16.1 m>s # V2 = 15.0 m>s, Ws = 7.37 kW T = 18.2 kN # me = 0.00584 kg/s ai = 0.479 m>s2 peAe m0c m0 c 6–85. V = aVe + # - # 2 blna # b + a # - gbt me m m t m me 0 e e m0 # -(a0 + g)t>Ve 6–86. mf = (a + g)e Ve 0 6–87. FD = 38.4 kN 6–89. T = 150 kN 6–90. a = 14.8 m>s2, V = 400 m>s # 6–91. mf = 167 kg>s 6–45. 6–46. 6–47. 6–49. 6–50. 6–51. 6–53. 6–54. 6–55. 6–57. 6–58. 6–59. 6–61. 6–62. 6–63. 6–65. 6–66. 6–67. 6–69. 6–70. 6–71. 6–74. 6–75. 6–77. 6–78. 6–79. 6–81. 6–82. 6–83.

rotational, Γ = 3.07 m2 >s Γ = 0 U # U vz = ,g = 2h xy h 4 3>2 c = y 3 For c = 0, y = 0. For c = 0.5 m2 >s, y = 0.520. For c = 1.5 m2 >s, y = 1.08.

Chapter 7 7–1. 7–2. 7–3. 7–6.

7–7. 7–9.

f cannot be established. c = y2 - 2x 2 - 2xy, y = 11 { 232x

7–10.

c = 3x 2y -

7–11. 7–13.

Yes u = 1 - 6x2m>s, V = 16.2 m>s

7–14. 7–15. 7–17. 7–18.

V = 10 m>s, f = (6x - 8y) m2 >s V = 2 m>s 3 rotational, c = y2 - x 2, 2 3 2 y = x + 30 A2 pA = 482 Pa

7–21.

f = 4x 2y -

7–23.

2 3 2x 3 + 128 x ,y = 3 9x 2

c = 50y2 + 0.2y, f cannot be established.

7–19.

7–22.

4 3 y ,Γ = 0 3 c = 21x 2 - y2 2 - 2xy, f = y2 - x 2 - 4xy VA = 17.9 m>s, pB - pA = 160 kPa

7–27.

2 3 x , pB = 1.05 MPa 3 2 rotational, c = 2y2 - 4xy - x 3 3 V = 17.9 m>s, xy = 2

7–29.

pB = 75.0 kPa

7–30.

xy = 2, u = 2 m>s S, v = 4m>s T ax = 4 m2 >s S , ay = 8 m2 >s c

7–25. 7–26.

891

f = 2xy2 -

7–33. 7–34. 7–35. 7–37. 7–38.

c = 2(y2 - x 2), y = { 2x 2 - 3 pB = 832 Pa v = 4xy - x vr = 0, vu = 8 m>s, vx = 4 m>s, vy = 6.93 m>s c = y13x 2 + y2 2, Γ = -12 m2 >s

7–39.

c =

7–31.

rotational, pO = 36.2 kPa

7–47.

1 2 ( y - x 2) 2 V = 160 m>s V = 43.3 m>s, xy = 6 vr = 4 m>s, vu = -6.93 m>s Γ2 z = 8p2gr 2 p = -0.960 kPa

7–49.

c = 8u, V = 1.60 m>s

7–41. 7–42. 7–43. 7–46.

7–50.

d = 22 m

892

7–51.

ANSWERS

TO

SELECTED PROBLEMS

4y 3 d f m2 >s tan-1 c 2 2p x + y2 - 4 For c = 0.25 m2 >s, x 2 + 1y + 2232 2 = 16 c = e -

For c = 0.5 m2 >s, x 2 + ay +

7–53.

u = p, r =

7–54.

u =

7–55. 7–57.

8–1.

223 2 16 b = 3 3

3 m 16p

18 p m>s 13 v = 0 x 2 + 1y + 1.252 2 = 3.252

8–2.

q = 4p m2 >s y

= tan40py

x + y2 -0.25 2

7–58. 7–59. 7–61.

L = 1.02 m, W = 0.0485 m q = 0.697 m2 >s, w = 0.157 m pO - pA = 21.4 kPa

7–62.

p = p0 -

7–63.

y = x tan3p11 - 32y24

7–66.

VA = 20 m>s

7–67. 7–69. 7–70. 7–71. 7–73.

pB = 90 kPa, pA = 89.5 kPa 1FR 2 y = 32.8 kN p = 339.04(1 - 4 sin2q)4 Pa FR = 15.2 N>m c Γ = 2.73 m2 >s, u = 12.5° and 167° pA = 12.5 kPa

7–74.

Chapter 8

rU 2 2(p - u)2

3sin2u + (p - u)sin2u4

0.1

V (m>s) p (kPa)

8–13. 0.4

0.5

12

7.5

6.67

6.375

6.24

69

135

144

147

148

u = 90° or 270°, pmin = - 2.88 kPa

V  u = 30 ∘ = 40 m>s V  u = 60 ∘ = 69.3 m>s

V  u = 90 ∘ = 80 m>s

P  u = 60 ∘ = - 1.92 kPa P  u = 90 ∘ = - 2.88 kPa

7–79.

u = 36.9° and 143°, pA = 97.4 kPa

7–81.

Fy = 1.09 kN, pmax = 350 Pa, pmin = - 802 Pa

7–82.

v = 2.37 rad>s

7–87.

vu =

a 2

- ri

8–14. 8–15. 8–17.

7–77. 7–78.

r 2o

M = 1.27 rVL m or m rVL We = 3.301105 2

8–10. 8–11.

0.3

pA = 52.8 kPa

vr 2i

8–5.

8–9.

0.2

7–75.

P  u = 30 ∘ = 0

no yes yes no

30°

r (m)

r o2 - r 2 b r

c. 1 FL d. T ML a. T3 1 b. T c. 1 ML2 d. T3

8–3.

8–6.

, VB = 34.6 m>s

F T 1 b. T

a.

8–18. 8–19. 8–22. 8–23.

V = k2gh Q = kb2gH 3, Q increases by a factor of 2.83. m t = k A gA T Q = k vrD2 d = xf1Re2 V = Cc a

D2 ∆p a bd m ∆x

M 1>2 b LT 2 F = kyV Q = k2gD5 s p = k r

s , percent decrease = 18.4% A rl

8–25.

c = k

8–26.

p = p0 f a

rr 3 E5 , b m p0

ANSWERS TO SELECTED PROBLEMS

8–27.

gaRe, We,

D d , b = 0 h h

8–29.

FD = rV 2L2 3f1Re24

8–30.

V =

8–31.

Q = D3vf a

8–33. 8–34. 8–37. 8–38. 8–39. 8–41. 8–42. 8–43. 8–45. 8–46. 8–47. 8–49. 8–50. 8–51. 8–53. 8–54. 8–55. 8–57. 8–58. 8–59. 8–61. 8–62. 8–63. 8–65.

gd 2 h fa b m d

rD5v3 D3v2m # , # b W W

9–6.

9–11. 9–14. 9–15.

g l = hf at b Ah ∆p Q = vD3f a 2 2 b rv D

9–17. 9–18. 9–19. 9–21. 9–22.

v 1>3 T = rQ v f c a b h d Q m V , T = rv2D4f a b 2 vD rvD Vm = 3.16 m>s Vm = 14.2 m>s 1FD 2 p = 200 kN Vm = 12 km>h rp = 11.0 kg>m3 D = 428 mm V2 = 673 m>s 1FD 2 p = 20.0 kN Vs = 90 km>h, 1Fmax 2 s = 8 kN 1∆p2 ke = 4.54 Pa Vm = 1126 km>h vm = 2.83110-9 2m2 >s, too low to be practical Vm = 4.97 Mm>h, not reasonable Vm = 5140 km>h Vm = 12 m>s Vm = 0.447 m>s Vm = 8672 km>h Vp = 3.87 m>s,1FD 2 p = 250 kN, # W = 967 kN Qm = 0.140 m3 >s, Hm = 0.158 m

9–23. 9–25. 9–26.

FD = rV 2Af1Re2 5>3 1>3

Chapter 9 9–1. 9–2. 9–3. 9–5.

9–7. 9–9. 9–10.

a = 0.960 mm F = 0.0258 N Q = 0.737110-3 2 m3 >s Q = 0.356110-3 2 m3 >s tt = 0.222 Pa, tb = 0.222 Pa V = 0.148 m>s

9–27. 9–29. 9–30. 9–31. 9–33. 9–34. 9–35.

9–37. 9–38. 9–39. 9–41. 9–42. 9–43.

9–45. 9–47.

9–49. 9–50.

9–51. 9–53.

893

U = 4.88 m>s Vp = 1.15 mm>s T = 0.532 N U = 0.738 m>s ∆p … 1.91 kPa T = 0.326 N # m m (Ut - Ub) Ut - Ub u = a by + Ub, txy = a a ∆p = 8.06 kPa V = 5.24 mm>s Q = 77.4(10 - 6) m3 >s ∆p … 16.4 kPa t = 183 N>m2, umax = 1.40 m>s Q = 0.494(10-3) m3 >s tmax = 6.15 Pa, umax = 2.55 m>s Q = 0.970 N # s>m2 umax = 6.35 m>s tmax = 250 N>m2 laminar # m = 16.8 kg>s Q = 0.849(10-3) m3 >s 1V2 max = 0.304 m>s, turbulence will not occur. t = 428 N>m2 t = 231 min t = 28.2 min, N>m2 ∆p = 353 L m ∆p … 610 Pa At T = 10°C, L′ = 330 mm At T = 30°C, L′ = 537 mm F = 0.307110-3 2 N t = 71.4(10 - 6) N>m2 no mixing t  r = 75mm = 30 N>m2 tr=0 = 0 Q = 0.0209 m3 >s Re = 79.141103 2 The flow is not laminar. tO = 28.75 N>m2 tr=0 = 0 umax = 4.89 m>s u = 14.1 m>s t0 = 17.4 N>m2 ∆p = 3.47 kPa y = 21.2 mm u = 8.02 m>s y = 17.7 mm t0 = 56.25 N>m2, y = 2.02 mm

894 9–54. 9–55. 9–57.

ANSWERS

10–14. 10–15. 10–17. 10–18. 10–19. 10–21. 10–22. 10–23. 10–25. 10–26. 10–27. 10–29. 10–30. 10–31. 10–33. 10–34. 10–35. 10–37. 10–38. 10–39. 10–41. 10–42. 10–43. 10–45. 10–46. 10–47. 10–49. 10–50. 10–51. 10–53. 10–54. 10–55. 10–57.

SELECTED PROBLEMS

t = 600 N>m2 u = 32.8 m>s tvisc = 1.91 N>m2 tturb = 148 N>m2 Q = 278 mm3 >s

Chapter 10 10–1. 10–2. 10–3. 10–5. 10–6. 10–7. 10–9. 10–10. 10–11. 10–13.

TO

p1 - p2 = 30.9 kPa Use D = 143-mm-diameter pipe. hL = 0.892 m p1 - p2 = 72.0 kPa p1 - p2 = 28.4 Pa Q = 0.00608 m3 >s, p1 - p2 = 0.801 Pa pA - pB = 19.5 kPa pA - pB = 13.6 kPa Q = 0.267110 - 3 2 m3 >s Q = 8.98 liter>s # W0 = 4.40 kW h = 2.80 m p#1 - p2 = 5.87 kPa m = 10.6 kg>s p1 - p2 = 11.1 Pa ∆p = 17.7 kPa f = 0.106 Use D = 100-mm-diameter pipe Q = 0.821 m3 >s pA - pB = 2.06 kPa pA = - 31.6 kPa, FA = 190 N p1 - p2 = 133 kPa percent increase = 300% Q = 0.0145 m3 >s Q# = 0.0424 m3 >s W = 56.1 kW hL = 5.66 m P# = 427 W W# = 194 W Wout = 3.17 kW Q = 0.00951 m3 >s QAD = 0.0137 m3 >s Use D = 107 mm pA = 22.9 kPa Q = 19.5 liter>s pA = 519 kPa Q = 0.126 m3 >s Q# = 11.1 m3 >s Wout = 41.8 kW pA = 250 kPa Q = 0.0250 m3 >s pB - pA = 30.6 Pa pA = 742 kPa

10–58. Q = 0.0142 m3 >s 10–59. Q = 0.00806 m3 >s 10–61. pB = 359 kPa 10–62. pA = 65.3 kPa 10–63. pA = 47.7 kPa # 10–65. Wo = 2.41 kW 10–66. QA = 0.0212 m3 >s QB = 0.0117 m3 >s QC = 0.00955 m3 >s QD = 0.0212 m3 >s 10–67. pA - pD = 76.8 kPa 10–69. Q = 0.0430 m3 >s 10–70. Q = 0.0133 m3 >s # 10–71. W = 152 W 10–73. QABC = 0.000534 m3 >s QADC = 0.00182 m3 >s

Chapter 11 11–1.

11–2. 11–5. 11–7. 11–9. 11–10. 11–11. 11–13. 11–14. 11–15. 11–17. 11–18. 11–19. 11–21. 11–22. 11–23. 11–25. 11–26. 11–27.

d = 9.71 mm d A = 3.34 mm u = 1.50 m>s d = 14.1 mm FDf = 0.0331 N FD = 0.0456 N FDf = 0.110 N u  y = 0.2 m = 9.09 m>s, u  y = 0.4 m = 9.52 m>s xcr = 546 mm FDf = 40.8 N F = 444 N FDf = 1.62 N x = 1.69 m F = 0.588 N d = 4.33 mm d * = 1.49 mm 3 1 C1 = 0, C2 = , C3 = 2 2 3.46x ∆ = 2Rex 4.80x d = 2Rex 4.64x d = 2Rex U t0 = 0.323ma b 2Rex x 3 1 C1 = , C2 = 0, C3 = 2 2 4.64x d = 2Rex

11–29. FDf = 165 N

895

ANSWERS TO SELECTED PROBLEMS 11–30. d  x = 1.5m = 23.6 mm d  x = 0.75m = 13.6 mm 0.343x 11–31. d = (Rex)1>5 11–33. 11–34. 11–35. 11–37. 11–38. 11–39. 11–41. 11–42. 11–43. 11–45. 11–46. 11–47. 11–49. 11–50. 11–51. 11–53. 11–54. 11–55. 11–57. 11–59. 11–61. 11–62. 11–63. 11–65. 11–66. 11–67. 11–69. 11–70. 11–71. 11–73. 11–74. 11–75. 11–77. 11–78. 11–79. 11–81. 11–82. 11–83. 11–85. 11–86. 11–87.

d = 44.6 mm FDf = 13.2 N FDf = 1.22 kN a = 1.02 m F = 495 N # FDf = 15.7 kN, W = 31.4 kW F = 2.14 kN d = 16.1 mm t0 = 15.5 N>m2 FDf = 652 N 1FD 2 p = 100 kN FR = 1.39 MN FD = 632 kN FPD = 43.7 kN FR = 25.4 kN MA = 18.6 kN # m # W = 36.9 kW The crate will slide. FD = 2.20 kN FD = 1.42 N FD = 17.6 N MA = 4.25 kN # m MA = 859 N # m # W = 80.5 kW Normal: FD = 33.2 N Parallel: FD = 0.119 N t = 114 days UT = 12.4 m>s U = 9.27 m>s d = 3.71 m CD = 2.08 FD = 75.7 N FD = 20.1 N m = 673 g a = - 13.7 m>s2 V0 = 0.374 m>s V = 8.98 m>s V = 16.8 mm>s, t = 2.97 ms V = 1.86 m>s d = 8.88 m U2 = 188 km>h CL = 314 a = 2.50°1approx.2 FD = 3.30 kN a = 20° Vs = 57.2 m>s

11–89. 11–90. 11–91. 11–93. 11–94. 11–95. 11–97.

a ≈ 15° CL = 0.277 u = 20.8° The glider can land. U2 = 172 km>h FL = 0.03 N d = 31.8 mm

Chapter 12 12–1. 12–2. 12–3. 12–5.

12–6. 12–7. 12–9. 12–10. 12–11. 12–13. 12–14. 12–15. 12–17. 12–18. 12–19.

12–21. 12–22. 12–23. 12–25. 12–26. 12–27. 12–29.

y = 1.01 m V2 = 3.78 m>s E = 1.36 m, y = 0.912 m a. 3.84 m / s b. 3.53 m / s c. 0 y = 2.82 m 1subcritical2 y = 0.814 m 1supercritical2 Vup = 2.93 m>s, Vdown = 5.93 m>s, Fr = 0.339 E = 2.57 m Emin = 2.48 m yB = 2.93 m yA = 0.734 m yc = 1.59 m, Emin = 2.38 m y = 0.997 m or y = 2.73 m yc = 1.01 m, subcritical, Emin = 1.52 m At y = 1.5 m, E = 1.73 m y2 = 1.49 m y2 = 0.938 m F = 2.35 kN yB = 4.80 m, VA = 1 m>s, VB = 1.40 m>s yB = 0.511 m VA = 10 m>s VB = 9.79 m>s h = 0.150 m y2 = 3.43 m bc = 0.632 m h = 0.400 m y = 0.972 m y2 = 0.540 m h = 0.438 m y2 = 0.749 m h = 18.7 mm, y2 = 1.12 m y2 = 0.547 m y1 = 3.93 m

12–30.

y2 1m2

0

1

2

3

4

5

Q1m >s2 0 19.8 35.4 46.0 50.11max2 44.3 3

896

ANSWERS

TO

SELECTED PROBLEMS

12–31. Qmax = 100 m3 >s Supercritical: y2 = 1.62 m Subcritical: y2 = 5.64 m bh 12–33. a. Rh = 2h + b 23 b. Rh = l 8 23l1l + 2b2 c. Rh = 412l + b2 12–34. y = 1 m Qmax = 4.70 m3 >s 12–35. y = 1 m, Qmax = 4.70 m3 >s 12–37. y = 1.88R 12–38. y = 1.63R 12–39. Q = 63.5 m3 >s 12–41. Q = 98.7 m3 >s, Sc = 0.00141, supercritical 12–42. Q = 145 m3 >s 12–43. yc = 3.92 m Sc = 0.00185 2Q2 tan2 u 1>2 b 12–45. a g 12–46. y = 1.82 m, subcritical 12–47. yc = 1.09 m, Sc = 0.00278 12–49. R = 798 mm 12–50. Q = 2.51 m3 >s 12–51. S0 = 0.00504 12–53. Q = 0.259 m3 >s 12–54. S0 = 0.0404 12–55. Q = 0.136 m3 >s 12–57. Q = 110 m3/s, supercritical 12–59. b = 3.08 m 12–61. y = 1.25 m 12–62. u = 60 ∘ 12–63. a = b 12–65. u = 60 ∘ , l = b 12–66. S2 12–67. M2 12–69. S1 12–70. A3 12–71. x = 1545 m 12–73. y = 2.13 m 12–74. x = 65.7 m 12–75. y1 = 0.300 m, hL = 17.3 m 12–77. yes, yc = 4.74 m 12–78. y2 = 0.586 m, hL = 0.550 m 12–79. V1 = 7.14 m>s V2 = 4.46 m>s hL = 0.0844 m 12–81. y1 = 0.404 m, hL = 3.61 m

12–82. 12–83. 12–85. 12–86.

y2 = 3.75 m y = 3.38 m Qactual = 0.0994 m3 >s Qactual = 2.89 m3 >s

Chapter 13 13–1. 13–2.

13–3. 13–5. 13–6.

13–7. 13–9.

13–10. 13–11. 13–13. 13–14. 13–15. 13–17. 13–18. 13–19. 13–21. 13–22. 13–23. 13–25. 13–26. 13–27. 13–29. 13–30. 13–31. 13–33. 13–34. 13–35. 13–37. 13–38. 13–39.

Tout = 5.01∘ C The density r will remain constant, ∆r = 0 ∆p = - 26.4 kPa ∆u = - 32.5 kJ>kg ∆h = - 45.5 kJ>kg r1 = 0.331 kg>m3, T2 = 366 K ∆h = 0, ∆s = - 272 J>kg # K ∆u = 14.8 kJ>kg ∆h = 20.8 kJ>kg ∆s = 111 J>(kg # k) ∆r = - 1.63 kg>m3 ∆h = 25.1 kJ>kg The density r will remain constant. ∆p = 56.0 kPa, ∆u = 157 kJ>kg ∆h = 261kJ>kg ∆u = 9.74 kJ>kg, ∆h = 13.6 kJ>kg ∆s = 99.2 J>kg # K M = 0.761 s = 2.30 km V = 1.231103 2m>s a = 27.04° = 27.0° V = 705 m>s Cair = 343 m>s, Cw = 1.50 km>s cair = 343 m>s, cw = 1.48 km>s V = 1663 km>h r = 1.89 kg>m3 p0 = 453 kPa # m = 36.0 kg>s t = 4.19 s V = 301 m>s # m = 68.7 kg>s, r0 = 4.67 kg>m3 # m = 0.0519 kg>s # m = 0.968 kg>s# p0 = 180 kPa, m = 0.281 kg>s P0 = 13.31103 2 # m = 0.714 kg>s # m = 1.82 kg>s M = 0.446 # m = 17.2 kg>s # m = 0.00665 kg>s, F = 1.69 N p0 = 107 kPa MC = 1 Subsonic: MB = 0.150, pB = 722 kPa Supersonic: MB = 2.92, pB = 22.6 kPa

ANSWERS TO SELECTED PROBLEMS

13–41. 13–42. 13–43. 13–45. 13–46. 13–47. 13–49. 13–50. 13–51. 13–53. 13–54. 13–55. 13–57. 13–58. 13–59. 13–61. 13–62. 13–63. 13–65. 13–66. 13–67. 13–69. 13–70. 13–71. 13–73. 13–74.

13–75. 13–77. 13–78. 13–79. 13–81. 13–82. 13–83. 13–85. 13–86. 13–87. 13–89. 13–90.

# m = 0.0547 kg>s # m = 1.78 kg>s # m = 1.77 kg>s dt = 17.4 mm, dB = 22.6 mm T0 = 302 K, TB = 168 K # m = 1.05 kg>s # m = 0.306 kg>s p3 = 1193 kPa, V = 33.2 m>s # T = 291 K, P = 409 kPa, m = 0.312 kg>s # T = 121 K, P = 19.2 kPa, m = 0.312 kg>s # m = 0.0155 kg>s # m = 42.2 g>s # m = 85.1 kg>s P = 90 6 pB, the nozzle will choke. # m = 0.120 kg>s dt = 197 mm de = 313 mm # m = 0.257 kg>s p0 = 403 kPa, dt = 49.5 mm, pt = 219 kPa MB = 0.0358, MB = 4.07 P2 = 242 kPa V2 = 310 m>s 1T0 2 1 = 275 K, 1T0 2 2 = 660 K ∆s = 943 J>kg # K T = 390 K , V = 502 m>s p1 = 156 kPa, r1 = 1.95 kg>m3 T * = 304 K, p* = 98.4 kPa 1T0 2 1 = 561 K, 1T0 2 2 = 671 K, ∆s = 265 J>1kg # K2 Lmax = 215 m, M2 = 0.565 p2 = 64.3 kPa, T2 = 279 K 1T0 2 1 = 275 K, 1T0 2 2 = 660 K ∆s = 943 J>kg # K p*0 = 231 kPa, t *0 = 293 K, ∆s = 75.1 J>kg · K M2 = 0.583 T2 = 400 K P2 = 82.5 kPa # m = 0.330 kg>s, ∆u = 80.5 kJ>kg # m = 4.46 kg>s, Ff = 806 N V1 = 165 m>s # m = 17.4 kg>s # m = 6.91 kg>m3 M1 = 1.87, M2 = 0.601 T2 = 552 K, p2 = 314 kPa, V2 = 283 m>s pb 6 17.6 kPa t = 4.28 s # m = 0.665 kg>s 1pe 2 sub = 626 kPa, 1pe 2 sup = 10.6 kPa Vmax = 699 m>s # m = 183 kg>s p2 = 240 kPa, 1p0 2 2 = 292 kPa

13–91. 13–93. 13–94. 13–95.

13–97. 13–98. 13–99. 13–101. 13–102. 13–103. 13–105. 13–106. 13–107. 13–109. 13–110. 13–111. 13–113. 13–114. 13–115. 13–117. 13–118. 13–119.

13–121. 13–122.

V2 = 261 m>s dout = 410 mm, dt = 304 mm # dout = 504 mm, dt = 304 mm, m = 64.9 kg>s to choke the nozzle, pb 6 554 kPa To form expansion shock waves Pb 6 12.4 kPa Ve = 659 m>s dt = 254 mm, de = 272 mm p2 = 360 kPa ,T2 = 551 K T2 = 522 K, V2 = 245 m>s Vs = 445 m>s, p2 = 179 kPa 250 kPa 6 pb 6 413 kPa pb 6 52.2 kPa pB = 176 kPa 30.7 kPa 6 pb 6 213 kPa pb 6 30.7 kPa pe = 40.7 kPa pB = 123 kPa, TB = 307 K 1P2 2 t = 80.7 kPa 1p2 2 b = 95.1 kPa 1p2 2 t = 60.5 kPa p2 = 131 kPa, T2 = 317 K V2 = 742 m>s, d = 57.4° T2 = 231 K, p2 = 45.3 kPa b = 25.6° p2 = 84.5 kPa T2 = 307 K 1T2 2 B = 211 K, 1p2 2 B = 59.7 kPa pA = 105 kPa, pB = 68.5 kPa

Chapter 14

1V# rel 2 2 = 5.10 m>s, V2 = 10.4 m>s Wpump = 4.48# kW b1 = 28.9°, Wpump = 14.7 kW V2 = 4.49 m>s, 1Vrel 2 2 = 5.70 m>s (Vt)2 = 2.74 m>s, Vrel = 12.4 m>s V2 = 1.29 m>s, (Vt)2 = 1.26 m>s hpump = 18.9 m hpump = 494 m Q = 0.423 m3 >s T = 54.5 N # m T# = 59.5 N # m W = 59.6 kW 1Vrel 2 1 = 13.3 m>s, 1Vrel 2 2 = 9.42 m>s # Wpump = 207 W T = 1.81 kN # m # 14–22. Win = 429 kW 14–23. b1 = 17.0° a2 = 38.3° 14–1. 14–2. 14–3. 14–5. 14–6. 14–7. 14–9. 14–10. 14–11. 14–13. 14–14. 14–17. 14–18. 14–19. 14–21.

897

898

ANSWERS

TO

SELECTED PROBLEMS

14–25. b1 = 38.6°, 1Vrel 2 1 = 12.8 m>s 1Vrel 2 2 = 30.9 m>s,# a2 = 75.0° 14–26. T = 8.21 kN # m, W# turbine = 82.1 kW 14–27. T = 2.10 kN # m, Wturbine = 10.3 kW 3 14–29. ForT # = 55.7110 2 N # m, v = 29.8 rev>min # For # W = Wmaxv = 35.8 rev>min 14–30. Wturb = 637 kW 14–31. hturbine = 12.6 m # 14–33. T = 230 N # m, Ws = 9.19 kW 14–34. h# = 62.4% 14–35. W = 167 kW 14–37. V1 = 30.3 m/s, V2 = 17.5 m/s # 14–38. Wturb = - 16.6 MW

14–39. 14–41. 14–42. 14–43. 14–45. 14–46. 14–47. 14–49. 14–50. 14–51. 14–53. 14–54. 14–55.

b1 = 54.5°, b2 = 54.0°, 1Vrel 2 2 = 20.4 m>s Q = 1210 liters>min Q = 1360 liters>min Q = 0.03375 m3 >s Cavitation will not occur. Cavitation will occur. Q = 1660 liters>min and hpump = 29.0 m # W2 = 1.57 kW Q2 = 1000 liters>min, h2 = 17.8 m v2 = 2134 rev>min Q2 = 0.633 m3 >s D2 = 122 mm h2 = 0.243 m

INDEX A Absolute pressure, 32, 58, 64–65, 150 ideal gas law and, 32, 58 standard atmospheric pressure and, 64–65, 150 zero, 64 Absolute temperature, 23, 32, 58 Absolute viscosity, 36 Accelerated motion, 331–335, 357 control volume with, 331, 357 fluid momentum and, 331–335, 357 rockets, 333–335 turbojets and turbofans, 332 Accelerated nonuniform open-channel flow, 656 Acceleration of fluids, see Fluid acceleration; Streamline coordinates Acceleration of liquid, 107–111, 151 Adhesion, 44 Adiabatic process, 720, 803 Adverse pressure gradient, 605 Air, properties of, 858 Airfoils, 620–621, 624–632, 652 angle of attack, 620, 627, 629 circulation and, 625–626 design of, 621 drag coefficients for, 621 drag on, 624–632 drag reduction for, 620–621, 652 induced drag, 621, 628–629 lift and, 624–632 race cars with, 627 split-scimitar winglets on, 628 vortex trail, 628 Andrade’s equation, 38 Anemometer, 556 Angle of attack, 620, 627, 629 Angle valve, 539 Angular distortion, 363, 431 Angular momentum, 318–325, 357, 809, 816 axial-flow pumps, 809 control volumes for, 318–319, 357 procedure for analysis of, 319 radial-flow pumps, 816 representation of, 318–325 steady flow, 319 Apparent shear stress, 500–501 Apparent viscosity, 37 Area of flow, 674 Average angular velocity, 362 Average pressure, 61

Average velocity, 191 Axial-flow machines, 807, 808–814, 855 angular momentum and, 809 continuity equation for, 809 flow categorization of, 807 flow kinematics, 810–811, 855 fluid flow through, 808–809 power by, 810 procedure for analysis of, 811 pumps, 808–814, 855

B Backpressure, 743–751, 771–772, 804 Barometer, 72 Bends, losses from, 538 Bernoulli equation, 235–252, 298–299, 327, 374–376 differential fluid flow and, 374–376 energy (EGL) and hydraulic grade lines (HGL) from, 251–252 flow work from, 236 fluid flow applications, 238–250 limitations, 237 procedure for analysis using, 244 propeller fluid momentum and, 327 streamline applications, 235–237, 243 Best hydraulic cross section, 676–677 Betz’s law, 329 Bluff body, 604 Bodies at rest, fluid momentum of, 304–313 Bodies with constant velocity, fluid momentum of, 313–317 Body force, 372 Boundary layers, 237, 497, 575–601, 651–652 development of, 575–576 displacement thickness, 578, 582 disturbance thickness, 577, 582, 595 drag and, 584, 595, 596–601 external surfaces, 575–601, 651 flat plate analysis for combined layers, 596–601 friction drag, 584, 596–601 fundamental equations for, 653 laminar, 577, 581–589, 596–601, 651–652 momentum integral equation for, 590–593, 652 momentum thickness, 579, 583 pipe fluid flow and, 237 Prandtl’s one-seventh power law for, 594

regions of, 576–577 shear stress, 583–584, 595 thickness of, 577–579 transitional flow, 577 turbulent, 577, 594–601, 651–652 viscous fully developed flow and, 497 Bourdon gage, 76 Broad-crested weir, 696 Brookfield viscometer, 39 Buckingham Pi theorem, 441–449, 473 Bulk modulus, 31, 58 Bump, open-channel flow over, 666–669 Buoyancy, 101–105, 151 center of, 101, 105 hydrometer measurement of, 102 principle of, 101, 151 stability and, 104–105 Buoyant force, 101 Butterfly valve, 539

C Calculations for fluid mechanics, 25–26 Canals, 655 Capacity factor, 329 Capillarity, 46–47, 59 Casing, flow through, 817 Cavitation, 43, 59, 834–835, 855 Centerline, pipe flow and, 252 Centrifugal pump, 815 Centroid (C), 80, 86, 150 Channel cross sections, 659, 674, 676–677, 682 best hydraulic, 676–677 geometric properties of channel shapes, 674 gradually varied flow and, 681 nonrectangular, 661 rectangular, 659–661, 682 specific energy and, 659–661 Chézy equation, open-channel flow and, 675 Choked nozzles, 744, 804 Circulation (Г), 364–365, 367, 392, 432, 625–626 free-vortex flow, 392 Kutta-Joukowski theorem for, 625–626 lift and, 625–626 rotational flow and, 364–365, 367, 432 Closed conduits, fluid flow in, 240–241 Coanda effect, 606

899

900

INDEX

Cohesion, 44 Colebrook equation, 527 Compressible flow, 715–805 continuity equation for, 725, 738, 752, 762, 768, 777 energy equation for, 754, 763, 769, 777–779 expansion and compression waves, 781–785 friction effect on, 752–761, 804 fundamental equations of, 805 heat transfer effect on, 762–767, 804 hypersonic flow, 728 ideal gas law and, 716, 754, 763, 769 isentropic flow analysis, 738–751 linear momentum equation for, 725–726, 738–739, 753, 762, 768, 777 Mach cone, 729 Mach number (M) for, 727–730, 755, 769–771, 803 measurement of, 786–787, 804 nozzles, 740, 743–751, 771–775, 804 shock waves, 728, 768–780, 804 sonic flow, 728 stagnation properties, 731–737, 803 subsonic flow, 727, 739, 786 supersonic flow, 728, 739, 787 thermodynamic concepts for, 715–723, 803 variable areas with, 738–743 wave propagation through, 724–726 Compressible fluids, 69–71, 265 Compression waves, 781 Compressor, 332 Computational fluid dynamics (CFD), 160, 414–416 Conduits, 475–519, 528, 655–713. See also Parallel plates; Pipes enclosed conduits, 475–519 hydraulic diameter, 528 noncircular conduits, 528 open channels, 653–713 Connections, losses from, 538 Conservation of energy, 262–263 Conservation of mass, 189–229, 369–370, 432 average velocity, 191 continuity equation for, 200–201, 229, 369–370, 432 control volume change and, 196–198, 229 differential fluid flow, 369–370, 432 finite control volume for, 194–195, 229 mass flow, 192, 229 procedure for analysis of, 202 Reynolds transport theorem for, 196–200, 229

special cases for, 201 volumetric flow, 190–191, 229 Constant-pressure process, 717–718 Constant temperature, compressible fluids and, 69 Constant-volume process, 717 Continuity equation, 200–201, 229, 369–370, 432, 590, 688, 725, 738, 752, 762, 768, 777, 809, 816 axial-flow pumps, 809 compressible flow and, 725, 738, 752, 762, 768, 777 cylindrical coordinates, 370 differential fluid flow, 369–370, 432 fluid flow change and, 200–201, 229 friction effects on flow and, 752 heat transfer effects on flow and, 762 hydraulic jump and, 688 momentum integral equation and, 590 radial-flow pumps, 816 shock waves and, 768, 777 two-dimensional flow, 370 wave propagation using, 725 Continuum, 21 Control surface, 162, 194–195 Control volume, 162, 168–169, 187, 194–198, 229, 302–305, 318–319, 331, 357 accelerated motion of, 331, 357 angular momentum equation and, 318–319, 357 approach to fluid flow, 162, 187 convective and local change, 198, 229 finite, 194–195 fluid acceleration and, 168–169 fluid momentum and, 302–305, 318–319, 331, 357 linear momentum equation and, 301–304, 357 open control surfaces, 195 Reynolds transport theorem and, 196–199, 229 steady flow and, 195, 229, 303 time rate of change, 196–198, 229 velocity of flow for, 195 Convective acceleration, 169, 176 Convective control volume change, 198, 229 Converging–diverging nozzle, 745 Converging nozzle, 744 Cord, wing measurement of, 620 Critical depth, 660–661 Critical flow, 658 Critical pressure, 744 Critical Reynolds number, 493

Critical slope of open channels, 677 Critical suction head, 834 Critical zone, 526 Culverts, 655 Curved boundary, fluid flow around, 239 Curved surfaces, hydrostatic force on, 94–100, 151 horizontal component, 94–95, 151 liquid below plate, 96 resultant forces on, 94–96 vertical component, 95, 151 Cylinders, 402–405, 605–607, 610–611 drag coefficient for, 610–611 ideal flow around, 605 pressure gradient effects, 605–607 real flow, 606–607 superposition of flow around, 402–405 Cylindrical coordinates, continuity equation for, 370

D d’Alembert’s paradox, 405 Darcy friction factor, 523, 573 Darcy-Weisbach equation, 524, 573 Density, 29, 58, 732, 756, 764 compressible flow and, 732, 756, 764 property of, 29, 58 friction effect on flow and, 756 heat transfer effects on flow and, 764 Derived units, 22 Differential fluid flow, 359–433 analysis for, 359–360 angular distortion of, 363, 431 Bernoulli equation for, 374–376 circulation, 364–365, 367, 392, 432 computational fluid dynamics (CFD), 414–416 conservation of mass, 369–370, 432 continuity equation for, 369–370, 432 dilatation of, 361, 431 equations of motion for fluid particles, 371–376 Euler equations for, 373, 375–376 hydrodynamics, 377–386 ideal fluid flow, 366, 370, 377–386, 431 irrotational flow, 366 kinematics of elements, 360–363 linear distortion of, 361, 431 Navier-Stokes equations for, 409–413 potential function, 383–386, 432 rotation of, 362, 431 rotational flow, 364–368

INDEX stream function, 377–382, 432 superposition of flows, 396–408, 433 translation of, 360–361, 431 two-dimensional flow, 387–395, 433 velocity components, 378, 383–384 vorticity, 365, 368, 432 Differential manometer, 75 Dilatant fluids, 37 Dilatation, 361, 431 Dimensional analysis, 434–473 Buckingham Pi theorem, 441–449, 473 dimensionless numbers for, 438–440, 473 flow considerations for, 450–451 principle of dimensional homogeneity, 436–437 procedure for analysis, 442 similitude and, 451–462, 473 Dimensional homogeneity, 25, 436–437 Dimensionless groups, 435 Dimensionless numbers, 438–440, 473 Euler number (Eu), 438, 473 Froude number (Fr), 439, 473 Mach number (Ma), 440, 473 Reynolds number (Re), 439, 473 Weber number (We), 440, 473 Dimensionless ratio, 503 Discharge, see Mass flow; Volumetric flow Discharge coefficients, 535, 554–555, 671, 694–695 flow measurement adjustment using, 554–555 losses and, 535, 671 nozzles, 555 orifice, 555 Venturi, 554 weirs, 694–695 Displacement thickness, boundary layers, 578, 582 Distortion, 361, 363, 431 Disturbance thickness, boundary layers, 577, 582 Dot product, 191 Doublet, 398–399 Drag, 456–457, 584, 595, 596–608, 624–632, 651–652 airfoils, 621, 624–632, 652 angle of attack, 620 boundary layers and, 584, 595, 596–601 components of, 602–604 direction of, 602 flat plate with, 595 friction drag, 584, 596–601, 604 fundamental equations for, 653

induced, 621, 628–629 lift and, 602–604, 624–632 pressure drag, 604–608 pressure gradient effects, 604–608 reduction of, 620–623 section, 621 similitude for, 456–457 skin friction coefficient, 584 spinning ball trajectory and, 630 streamline the body for, 620 vehicles, 622–623 vortex trail and, 628 Drag coefficient, 609–619, 621, 623, 629, 651 airfoils, 621 cylinders, 610–611 Froude number (Fr) and, 612 geometric shapes, applications for, 613–619 Mach number (M), 612–613 Reynolds number (Re) and, 610–611 spheres, 611 vehicles, 623 Dynamic fluid devices, 808. See also Turbomachines Dynamic force, 304 Dynamic pressure, 239 Dynamic similitude, 453 Dynamic viscosity, 36

E Eddy viscosity of flow, 502 Effective angle of attack, 629 Efficiency, 265, 328–329, 819, 829, 833 fluid momentum and, 328–329 hydraulic, 819 manufacturer performance curves, 833 power and, 265, 328–329 propellers, 328 turbomachines, 819, 829, 833 wind turbines, 329 Elevation head, 251 Empirical solutions for resistance in rough pipes, 527–528 Enclosed conduits, viscous flow in, 475–519 fully developed flow from an entrance, 497–498 Navier-Stokes solution for flow in, 481–485, 490–491 procedures for analysis of, 482, 494 Reynolds number for, 492–496 shear stress in a smooth pipe, 499–502

901

steady laminar flow between parallel plates, 475–485 steady laminar flow in smooth pipes, 486–491 steady turbulent flow in smooth pipes, 502–507 Energy equation, 260–273, 299, 690, 754, 763, 769, 777–779 compressible flow, 265 conservation of energy, 262–263 flow work, 261, 262–263 friction effects on flow and, 754 gravitational potential energy, 260 heat energy, 261–262 heat transfer effects on flow and, 763 hydraulic jump and, 690 incompressible flow, 263–264 internal energy, 260 kinetic energy, 260 mechanical efficiency, 265 nonuniform velocity, 266 power, 265 procedure for analysis using, 267 shaft work, 261, 262–263 shear work, 261 shock waves and, 769, 777–779 system energy, 260 work, 261–263 Energy grade line (EGL), 251–260, 299 Enthalpy, 265, 717 Entrance length, 497 Entropy, 718–719, 803 Equations of motion for fluid particles, 371–376 Equilibrium, stability and, 104–106, 151 Equivalent-length ratio, 540 Euler number (Eu), 438, 473 Eulerian description of fluid flow, 162, 187 Euler’s equations of motion, 231–234, 298, 373, 375–376 differential fluid flow and, 373, 375–376 differential forms of, 231–232 inviscid fluids, 232–233, 298 n and s directions, 232 steady horizontal flow of ideal fluid, 233 two-dimensional steady flow, 373 Expansion and contraction, losses from, 537 Expansion factor, 787

902

INDEX

Expansion waves, 781 Extensive fluid properties, 196 External surfaces, 575–653 airfoils, 621, 624–632, 652 boundary layers, 575–601, 651 drag and lift effects, 602–604, 624–632 drag coefficient for, 609–619, 621, 623, 629, 651 drag reduction, 620–623 fundamental equations for, 653 laminar boundary layers, 577, 581–589, 596–601, 651 momentum integral equation for, 590–593 pressure gradient effects, 604–608 turbulent boundary layers, 577, 594–601, 652 vehicles, 622–623 viscous flow over, 575–653

F Fanno flow, gas properties and, 864 Fanno line, 757, 804 Favorable pressure gradient, 605 Finite control volume, 194–195, 229, 415 Finite difference method, CFD, 415 Finite element method, CFD, 415 Flat plate analysis, 594–601 combined boundary layers, 596–601 drag on, 595 shear stress along, 595 turbulent boundary layers, 594–595 Floatation, see Buoyancy; Stability Flow coefficients, 535 Flow kinematics, see Kinematics of fluid motion Flow meters, 554–558, 573. See also Measurement tools Flow net, 384 Flow systems, 836–837 Flow work, 236, 261–263 Bernoulli equation and, 236 energy equation and, 261–263 pressure and, 261 rate of, 262 Fluid acceleration, 168–174, 187 control volume and, 168–169 convective acceleration, 169, 176 local acceleration, 169, 175 resultant acceleration, 176 streamline coordinates for, 175–177 time rate change in velocity, 168–169, 187 three-dimensional flow, 170 Fluid flow, 153–167, 186–187, 190–193, 235–260, 298–299, 538 See also Differential fluid flow; Pipe flow around a curved boundary, 239 Bernoulli equation for, 235–242

closed conduits, 240–241 computational fluid dynamics (CFD), 160 dimension basis of classification, 155 energy equation, 260–273, 299 energy (EGL) grade lines, 251–260, 299 Eulerian description of, 162, 187 frictional effects and, 154 gases, compression and, 237 graphical descriptions of, 157–160 hydraulic (HGL) grade lines, 251–260, 299 Lagrangian description of, 161, 187 laminar flow, 154, 186 mass flow, 192, 229 measurement of, 240–242 open channels, 240, 298 optical methods of visualization, 160 pathlines, 159, 186 pipes, 240–242, 251–260, 299 reservoirs, 238 secondary, 538 space and time basis of classification, 156 steady flow, 156, 186 streaklines, 159, 186 streamlines, 157, 186–187, 231–237, 243, 298 systems and particle behavior, 161–167, 187 transitional flow, 154 turbulent flow, 154, 186 uniform flow, 156, 186 velocity profile, 154–155, 190–193 volumetric flow, 190–191, 229 Fluid mechanics, 18–59 branches of, 20 calculations, 25–26 capillarity, 46–47, 59 characteristics of matter, 21, 58 historical development of, 20 international systems of units, 22–24, 58 problem solving, 27–29 procedure for analysis of, 27 properties of fluids, 29–34, 58 surface tension (s), 44–47, 59 vapor pressure, 43, 59 viscosity measurement, 39–42 viscosity, 34–42, 58–59 Fluid momentum, 301–357 accelerated motion and, 331, 357 angular momentum equation, 318–325, 357 bodies at rest, 304–313 bodies with constant velocity, 313–317 control volumes for, 302–305, 318–319, 331, 357 Froude’s theorem for, 327, 329 linear momentum equation, 301–317, 326–327, 357 procedures for analysis of, 305, 319

propellers, 326–329, 330, 357 rockets, 333–335 steady flow and, 303, 319 turbofans, 332 turbojets and turbofans, 332 wind turbines, 329 Fluid motion, see Kinematics of fluid motion Fluid particles, 161–167, 371–376 Bernoulli equation for, 374–376 behavior in fluid systems, 161–167 body and surface forces, 372 equations of motion for, 371–376 Euler equations for, 373, 374–375 normal and shear stresses, 371 stress field, 371 Fluid properties, 29–38 bulk modulus, 31 density, 29–30 ideal gas law, 32 incompressibility of liquids, 29 specific gravity, 30 specific weight, 30 viscosity, 34–38 Fluid statics, 61–151 absolute pressure, 64–65, 150 acceleration of a liquid, 107–111, 151 buoyancy, 101–103, 151 compressible fluids, 69–71 curved surfaces, 94–100, 151 formula method, 80–85 gage pressure, 64, 150 geometrical method, 86–90 hydrostatic force, 80–100, 150 inclined surfaces, 94–100, 151 incompressible fluids, 67–68 integration method, 91–93 Pascal’s law, 62 plane surfaces, 80–93, 150 pressure, 61–79, 150 rotation of a liquid, 112–115, 151 stability, 104–106, 151 static pressure, 66, 72–79, 150 Fluid system particle behavior, 161–167, 187 Eulerian description of, 162, 187 Lagrangian description of, 161, 187 region of, 161 surroundings of fluid particles, 161 velocity of, 161–167 Fluids, 34, 37, 231–299, 856–858 Bernoulli equation for, 235–250, 298–299 classification of liquids and gases as, 34 dilatant, 37 energy equation for, 260–273, 299 energy grade line (EGL), 251–260, 299

INDEX Euler’s equations of motion for, 231–234, 298 hydraulic grade line (HGL), 251–260 ideal, 37, 233, 235–236, 298 inviscid, 37, 232–233, 298 measurement of flow of, 240–242, 298 movement of, 231–299 physical properties of, 856–858 procedures for analysis of, 244, 267 pseudo-plastic, 37 viscous, 237, 264, 299 work by, 236, 261–269, 299 Flume, 655 Forced vortex, 112–113, 151, 393 Forces, similitude corresponding to, 458 Formula method for plane surfaces, 80–93, 150 Francis turbine, 827, 855 Free-body diagrams, 304 Free-vortex flow, 392, 404–405, 817 Friction, 752–761, 804 compressible flow, effect on, 752–761, 804 continuity equation and, 752 density and, 756 energy equation and, 754 Fanno line and, 757, 804 ideal gas law and, 754 linear momentum equation and, 753 pipe length vs. Mach number, 755 pressure and, 756 temperature and, 756 Friction drag, 584, 596–601, 604 Friction factor, 752 Friction loss, 264, 522–523 Friction slope, 681 Froude number (Fr), 439, 473, 612, 658, 711 dimensional analysis using, 439, 473 drag coefficient and, 612 open-channel flow and, 658, 711 Froude’s theorem, 327, 329 Fully developed flow from an entrance, 497–498 Fused quartz force-balance Bourdon tube, 76

G Gage pressure, 64, 150 Gas, 21, 30–32, 34, 58, 237, 265, 856–857, 859–869 bulk modulus of, 31 classification as fluid, 34 compressible flow of, 265 compressible properties of, 859–869 density of, 30 energy equation for, 265 enthalpy of, 265 Fanno flow, 864

hydrostatic force of, 96 ideal gas law, 32, 58 incompressible flow of, 237 isentropic relations of, 859–863 normal shock relations, 866–868 physical properties of, 856–857 Prandtl-Meyer expansion, 869 Rayleigh flow, 865 Gate valve, 539 Geometric shapes, drag coefficients for, 613–619 Geometric similitude, 452 Geometrical method for plane surfaces, 86–90 Globe valve, 539 Gradual expansion and contraction, 537 Gradually varied open-channel flow, 681–687 Gravitational potential energy, 236, 260

H Haaland equation, 527 Hagen-Poiseuille equation, 489 Half body, superposition of flow around, 396–397 Hazen-Williams equation, 528 Head datum for flow analysis, 251–252 Head-discharge curve, 819 Head loss, 264. See also Losses Heat energy, 261–263 Heat transfer, 762–767 compressible flow affected by, 762–767, 804 continuity equation for, 762 density, 764 energy equation for, 763 ideal gas law and, 763 linear momentum equation for, 762 pressure, 764 Rayleigh line and, 765, 804 stagnation temperature and pressure, 764–765 temperature, 764 velocity, 763–764 Hydraulic diameter, 528 Hydraulic efficiency, 819 Hydraulic grade line (HGL), 251–260, 299 Hydraulic head, 251 Hydraulic jump, 656, 688–692, 712 continuity equation for, 688 energy equation for, 690 momentum equation for, 689 Hydraulic radius, 674 Hydraulics, 20 Hydrodynamics, 20, 377–386 ideal fluid flow, 366, 370, 377–386, 431 potential flow, 377–386

903

potential function, 383–386, 432 stream function, 377–382, 432 velocity components, 378, 383–384 Hydrometer, 102 Hydrostatic force, 80–100, 150–151 curved surfaces, 94–100, 151 formula method for, 80–85 gas, effects of, 96 geometrical method for, 86–90 inclined surfaces, 94–100, 151 integration method for, 91–93 plane surfaces, 80–93, 150 resultant forces of, 80–82, 86–87, 91 Hypersonic flow, 728

I Ideal fluid flow, 366, 370, 377–386, 431 irrotational flow of, 366 rotational flow of, 366 viscous flow compared to, 366 Ideal fluids, 37, 233, 235–237, 298 Bernoulli equation for, 235–237, 298 Euler’s equations of motion for horizontal flow of, 233 low viscosity and compressibility of, 37 streamlines of, 235–236, 298 Ideal gas law, 32, 58, 716, 754, 763, 769, 803 absolute pressure and temperature of, 32, 58 compressible flow and, 716, 754, 763, 769, 803 friction effects on flow and, 754 gas behavior and, 32, 716, 803 heat transfer effects on flow and, 763 shock waves and, 769 Ideal pump head, 818 Ideal turbine head, 829 Impeller, 808 Impulse turbine, 824–826 Inclined surfaces, hydrostatic force on, 94–100, 151 gas effects on, 96 horizontal component, 94–95, 151 liquid below plate, 96 resultant forces on, 94–96, 151 vertical component, 95, 151 Inclined-tube manometer, 79 Incompressibility of liquids, 29, 31 Incompressible fluids, 67–68, 237, 263–264 energy equation for flow of, 263–264 fluid flow of gas, 237 pressure head, 68 pressure variation of, 67–68 Induced drag, 621, 628–629 Inlet and exit transitions, losses from, 536 Integration method, 91–93

904

INDEX

Intensive fluid properties, 196 Internal energy, 260, 716, 803 Inviscid fluids, 37, 232–233, 298 Irrotational flow, 366 Isentropic flow, compressible flow analysis of, 738–751 area ratios for, 740–741 backpressure, 743–751 continuity equation for, 738 converging nozzles, 744 converging-diverging nozzles, 745 Laval nozzle, 740, 745 linear momentum equation for, 738–739 subsonic flow, 739 supersonic flow, 739 variable areas of, 738–743 Isentropic process, 720, 725, 803 Isentropic relations of gas, 859–863

K Kaplan turbine, 827, 855 Kinematic similitude, 452 Kinematic viscosity (n), 38, 59 Kinematics of differential fluid elements, 360–363 angular distortion, 363, 431 dilatation, 361, 431 linear distortion, 361, 431 rotation, 362, 431 translation, 360–361, 431 Kinematics of fluid motion, 153–187, 810–811, 815–816, 828 axial-flow pumps, 810–811 computational fluid dynamics (CFD), 160 control volume, 162, 168–169, 187 Eulerian description for, 162, 187 fluid acceleration, 168–174, 187 fluid flow, 153–167, 186 fluid particles, 161–167 Lagrangian description for, 161, 187 pathlines, 159, 186 radial-flow pumps, 815–816 reaction turbines, 828 schlieren photography, 160 shadowgraphs, 160 streaklines, 159, 186 streamlines, 157–158, 175–178, 186–187 Kinetic energy, 236, 260 Kinetic energy coefficient, 266 Kinetic head, 251 Kutta-Joukowski theorem, 625–626

L Lagrangian description of fluid flow, 161, 187 Laminar boundary layers, 577, 581–589, 596–601, 651–652

Blasius solution for, 581–582 boundary layer region, 577 displacement thickness, 582 disturbance thickness, 582 flat plate analysis for combined layers, 596–601 friction drag, 584, 596–601 fundamental equations for, 652 momentum thickness, 583 shear stress, 583–584 skin friction coefficient for, 584 Laminar flow, 154, 186, 475–499, 519, 522–523, 526, 573, 656 between parallel plates, 475–485, 519 Darcy friction factor for, 523, 573 fully developed flow from an entrance, 497–498 Moody diagram for, 526 Navier-Stokes solution for, 481–485, 490–491 open channels, 656 procedures for analysis of, 482, 494 resistance in rough pipes, 522–523, 526, 573 Reynolds number for, 492–496 shear stress in a smooth pipe, 499 smooth pipes with, 486–491 viscous fluids, 475–499, 519 viscous shear stress, 499 Laminar (viscous) sublayer, 500, 577 Laplace’s equation, 387 Laser Doppler flow meter, 558 Laval nozzle, 740, 745, 804 Law of the wall, 503 Lift, 602–604, 624–632, 652 airfoils, 624–632 angle of attack and, 620, 627, 629 circulation and, 625–626 components of, 602–604 drag and, 602–604, 624–632 force of, 602 fundamental equations for, 652 Kutta-Joukowski theorem for, 625–626 Magnus effect of, 630 spinning ball trajectory and, 630 Lift coefficient, 627 Line sink flow, 391, 398–399 Line source flow, 390–391, 398–399 Linear distortion, 361, 431 Linear momentum equation, 301–317, 326–327, 357, 725–726, 738–739, 753, 762, 768, 777 bodies at rest, 304–313

bodies with constant velocity, 313–317 compressible flow and, 725–726, 738–739, 753, 762, 768, 777 control volumes for, 302–305, 357 free-body diagram for, 304 friction effects on flow and, 753 heat transfer effects on flow and, 762 procedure for analysis of, 305 propeller fluid momentum and, 326–327 representation of, 301–304 shock waves and, 768, 777 steady flow, 303 wave propagation using, 725–726 Liquid droplets, 45 Liquids, 21, 29–31, 34, 43–47, 59, 107–115, 151, 856 acceleration of, 107–111, 151 bulk modulus of, 31 capillarity of, 46–47, 59 density of, 29 forced vortex formation, 112–113, 151 incompressibility of, 29, 31 nonwetting, 46 paraboloid surface of, 113, 151 physical properties of, 856 pressure variations in, 107–111, 151 rotation of, 112–115, 151 saturation of, 43 specific gravity of, 30 surface tension, 44–47, 59 wetting, 46 Local acceleration, 169, 175 Local control volume change, 198, 229 Losses, 522–523, 574, 535–541, 573, 818–819, 829 bends causing, 538 Darcy friction factor for, 523, 573 Darcy-Weisbach equation, 524, 573 equivalent-length ratio, 540 expansion and contraction causing, 537 friction loss, 522–523 head loss, 522–524, 573, 818–819, 829 hydraulic efficiency and, 819 inlet and exit transitions causing, 536 major head loss, 522–524, 541, 573 minor head loss, 535–537, 541, 573 pipe connections causing, 538 pipe fittings and transitions causing, 535–541 pumps, 818–819 resistance in rough pipes and, 522–523, 573

INDEX resistance (loss) coefficient, 535 turbines, 829 valves causing, 539

M Mach cone, 729 Mach number (M), 440, 473, 612–613, 727–730, 755, 769–771, 803 compressible flow and, 727–730, 755, 769–771, 803 dimensional analysis using, 440, 473 drag coefficient and, 612–613 flow classification using, 727–730 friction effect on flow and, 755 pipe length vs., 755 shock waves, relationships and, 769–771 Mach wave fans, 781–786 Magnetic flow meter, 558 Magnus effect, 630 Major head loss, 522–524, 541, 573. See also Losses Manning equation, 676, 712 Manometer rule, 74 Manometers, 73–75, 79, 241 Mass, see Conservation of mass Mass flow, 191, 229 Material derivative, 169 Matter, characteristics of, 21, 58 Mean steady flow, 500 Measurement tools, 39–42, 72–79, 102, 150, 240–242, 554–558, 573, 786–787, 804 anemometer, 556 barometer, 72 Bourdon gage, 76 Brookfield viscometer, 39 buoyancy, 102 compressible flow, 786–787, 804 differential manometer, 75 flow meters, 554–558, 573 fluid flow, 240–242 fused quartz force-balance Bourdon tube, 76 hydrometer, 102 magnetic flow meter, 558 manometer rule, 74 manometers, 73–75, 241 nozzle meter, 555 nutating disk flow meter, 557 orifice meter, 555 Ostwald viscometer, 40 piezoelectric gages, 76 piezometer, 240, 786 pipe flow, 554–558, 573 pitot (stagnation) tube, 240, 298, 786 pitot-static tube, 241

positive displacement flow meter, 557 pressure transducers, 76, 241 rotational viscometer, 39 static pressure, 72–79, 150 subsonic flow, 786 supersonic flow, 787 thermal mass flow meter, 557 turbine flow meter, 556 venturi meter, 242, 554, 787 viscosity, 39–42 vortex flow meter, 556 Mechanical efficiency, 265 Meniscus, 46 Metacenter, 105 Minor head loss, 535–537, 541, 573 Mixed-flow machines, 807 Mixing-length hypothesis, 502 Models, 451, 456–457. See also Similitude Momentum, see Angular momentum; Fluid momentum; Linear momentum equation Momentum integral equation, 590–593, 652, 689 boundary layer analysis using, 590–593, 651–652 continuity equation and, 590 hydraulic jump and, 689 velocity profile for, 591–592 Momentum thickness, boundary layers, 579, 583 Moody diagram, 525–527 Motion, 231–299, 371–376. See also Kinematics of fluid motion Bernoulli equation for, 235–250, 299–299, 374–376 Euler’s equations of, 231–234, 298, 373, 375–376 fluid particles, 371–376 fluids, 231–299

N Navier-Stokes equations, 409–413, 481–485, 490–491 Cartesian coordinate form, 409–410 cylindrical coordinate form, 411 laminar flow in parallel plates, 481–485 laminar flow in smooth pipes, 490–491 procedure for analysis using, 482 viscous flow solutions using, 481–485, 490–491 Net positive suction head (NPSH), 834–835 Neutral equilibrium, 104 Newtonial fluids, 37, 59 Newton’s law of viscosity, 35, 36 Newton’s second law of motion, 168

905

No-slip condition, 35 Non-Newtonian fluids, 37 Noncircular conduits, 528 Nondimensional flow, 155 Nonuniform open-channel flow, 656, 712 Nonuniform velocity, 266 Nonwetting liquids, 46 Normal stresses, 371 Nozzle discharge coefficient, 555 Nozzle meter, 555 Nozzles, 740, 743–751, 771–775, 804 backpressure in, 743–751, 771–772, 804 choked, 744, 804 compressible flow and, 740, 743–751, 771–775 converging, 744, 803 converging-diverging, 745, 803 isentropic flow through, 740, 743–751 Laval, 740, 745, 804 shock waves in, 771–775, 804 underexpanded flow in, 772 Nutating disk flow meter, 557

O Oblique shock waves, 776–780 One-dimensional flow, 155 Open-channel flow, 240, 298, 455, 654–713 Bernoulli equation for, 240 canals, 655 Chézy equation for, 675 critical slope of, 677 cross sections, 659, 674, 676–677, 682 culverts, 655 flume, 655 Froude number (Fr) for, 658, 711 fundamental equations for, 713 geometric properties of channel shapes, 674 gradually varied, 681–687 hydraulic jump, 656, 688–692, 712 laminar, 656 Manning equation for, 676, 712 nonuniform, 656, 712 over a rise or bump, 666–669 pitot tube for measurement of, 240, 298 prismatic channel, 656 Reynolds number (Re) for, 674 similitude for, 455 slopes, 677, 681–684 specific energy and, 658–666, 711 steady, 656, 674–680 surface profile for, 682–684, 712 surface roughness coefficient for, 676

906

INDEX

Open-channel flow (Continued) turbulent, 656 under a sluice gate, 670–673 uniform, 656, 674–680 wave celerity, 657–658 weirs, 693–697, 712 Operating point, 836 Orifice discharge coefficient, 555 Orifice meter, 555 Ostwald viscometer, 40

P Paraboloid liquid surface, 113, 151 Parallel-axis theorem, 81 Parallel pipe systems, 548, 549, 573 Parallel-plane theorem, 82 Parallel plates, 475–485, 519 horizontal flow from constant pressure gradient, 478–479 horizontal flow from motion of top plate, 480 Navier-Stokes solution for flow in, 481–485 steady laminar flow between, 475–485, 519 viscous fluid flow in, 475–485 Particle image velocimetry (PIV), 558 Pascal’s law, 62, 150 Pathlines, 159, 186 Pelton wheel, 824–826, 855 Piezoelectric gages, 76 Piezometer, 73, 240, 786 Piezometer rings, 554 Pipe flow, 521–573 analysis and design, 521–573 flow measurement, 554–558, 573 losses, 522, 535–541, 573 parallel systems for, 548, 549, 573 procedures for analysis, 529, 542, 549 relative roughness, 525 resistance in rough pipes, 521–534 series systems for, 548, 573 single pipelines, 541–547 surface roughness, 525 systems for, 548–553, 573 Pipes, 240–242, 251–260, 454, 486–507, 517, 521–534, 535–541 bends, 538 centerline, 252 connections, 538 energy grade line (EGL), 251–260, 299 entrance length, 497 expansion and contraction, 537 fittings and transitions, 535–541 fluid flow in, 240–242

fully developed flow from an entrance, 497–498 head datum for flow analysis, 251–252 horizontal flow through circular, 489 hydraulic grade line (HGL), 251–260, 299 inlet and exit transitions, 536 laminar flow in, 486–499, 519 Navier-Stokes solution for flow in, 490–491 resistance in, 521–534 Reynolds number for flow through, 492–496 rough, 521–534 shear stress in, 499–502 similitude for flow in, 454 smooth, 486–491, 499–507, 517 turbulent flow in, 497–507, 517 valves, 539 viscous fluid flow in, 486–507, 517 Pitot (stagnation) tube, 240–241, 298, 786 Pitot-static tube, 241 Plane surfaces, hydrostatic force on, 80–93, 150 formula method for, 80–85 geometrical method for, 86–90 integration method for, 91–93 plates with constant width, 87 resultant force on, 80–82, 86–87, 91, 150 symmetrical plates, 82 Planform, 621 Plates, see Plane surfaces Poiseuille flow, 489 Positive displacement flow meter, 557 Potential flow, see Hydrodynamics Potential function (f), 383–386, 432 Power, 265, 299, 328–329, 810, 817, 825, 829 axial-flow pumps, 810 capacity factor, 329 efficiency and, 265, 328–329 fluid momentum and, 328–329 propeller output of, 328 radial-flow pumps, 817 rate of work, 265, 299 shaft, 810 turbines, 825, 829 wind turbine output of, 329 Power law approximation, 504–505 Prandtl-Meyer expansion, 783, 869 Prandtl’s one-seventh power law, 594 Pressure, 32, 38, 58, 61–79, 81–82, 150, 233, 239, 261, 732, 743–751, 756, 764–765 absolute, 32, 58, 64–65, 150 average, 61 backpressure, 743–751

center of, 81–82 compressible flow and, 732, 743–751, 756, 764–765 compressible fluids, 69–71 critical, 744 dynamic, 239 effects on viscosity, 38 flow work and, 261 fluid flow around curved boundary, 239 fluid statics and, 61–79 friction effect on flow and, 756 gage, 64 heat transfer effects on flow and, 764 incompressible fluids, 67–68 isentropic flow through nozzles, 743–751 Pascal’s law, 62, 150 stagnation, 239, 732, 764–765 standard atmospheric, 64–65 static, 66, 72–79, 150, 233, 239, 732 total, 239 variation of, 66–71 Pressure coefficient, 438 Pressure drag, 604–607 Pressure gradient effects, 604–608 adverse and favorable, 605 Coanda effect, 606 ideal flow around a cylinder, 605 real flow around a cylinder, 606–607 vortex shedding, 608 Pressure head, 68, 251 Pressure transducers, 76, 241 Prismatic channel, 656 Problem solving, 27–29 Propeller turbine, 828 Propellers, fluid momentum of, 326–329, 330, 357 Bernoulli equation for, 327 Froude’s theorem for, 327 linear momentum equation for, 326–327 power output and efficiency of, 328 Prototype, 451. See also Similitude Pseudo-plastic fluids, 37 Pump head, 264, 818 Pump scaling laws, 839 Pumps, 261–263, 807, 808–823, 831–837, 855 axial-flow, 808–814, 855 cavitation and, 834–835, 855 flow system and selection of, 836–837 head-discharge curve, 819 head loss, 818–819 hydraulic efficiency, 819 ideal performance for, 818–823 manufacturer performance curves, 833 net positive suction head (NPSH), 834–835

INDEX performance characteristics, 831–833 positive suction head and, 834–835 procedure for analysis of, 811 radial-flow, 815–817, 855 shaft work by, 261–263

Q Quasi-steady flow, 263

R Race cars, airfoil effects on, 627 Radial-flow machines, 807, 815–817, 819, 855 angular momentum and, 816 casing, flow through, 817 centrifugal pump, 815 continuity equation for, 816 flow kinematics, 815–816, 855 head-discharge curve, 819 power by, 817 Rankine oval, superposition of flow around, 400–401 Rapid flow, 658 Rayleigh flow, gas properties and, 865 Rayleigh line, 765, 804 Reaction turbine, 827–830 Relative roughness, 525 Reservoirs, fluid flow from, 238 Resistance (loss) coefficient, 535 Resistance in rough pipes, 521–534 critical zone, 526 Darcy friction factor for, 523 Darcy-Weisbach equation for, 524 empirical solutions for, 527–528 Hazen-Williams equation for, 528 head loss, 522–524 laminar flow, 522–523, 526, 577 losses and, 522–523 Moody diagram for, 525–527 noncircular conduits, 528 procedure for analysis of, 529 transitional flow and, 526 turbulent flow, 523–524, 527, 577 Resultant acceleration, 176 Resultant force, 80–82, 86–87, 94–96, 91, 150–151 center of pressure, 81–82 centroid (C) of, 80, 86, 150 curved and inclined surfaces, 94–96, 151 formula method and, 80–82 geometrical method and, 86–87 horizontal component, 94–95, 151 hydrostatic forces, 80–82, 86–87, 94–96, 91, 150–151 integration method and, 91

liquid below plate, 96 location of, 81–82, 86–87, 91 parallel-axis theorem for, 81 parallel-plane theorem for, 82 plane surfaces, 80–82, 86–87, 91, 150 plates with constant width, 87 symmetrical plates, 82 vertical component, 95, 151 xp coordinate location, 82 yp coordinate location, 81 Retarded nonuniform open-channel flow, 656 Reynolds number (Re), 439, 473, 492–496, 610–611, 674 critical, 493 dimensional analysis using, 439, 473 drag coefficient and, 610–611 laminar flow in pipes determined using, 492–496 open-channel flow and, 674 procedure for analysis using, 494 Reynolds stress, 501 Reynolds transport theorem, 196–200, 229 applications of, 198–199 control volume and, 196–198, 229 extensive and intensive fluid properties, 196 time rate of change and, 196–198, 229 Ring elements, rotation of liquid in, 112–113 Rise, open-channel flow over, 666–669 Rotation of differential fluid elements, 362, 431 Rotation of liquid, 112–115, 151 Rotational flow, 364–368 circulation, 364–365, 367, 432 ideal vs. viscous behavior, 366 vorticity, 365, 368, 432 Rotational viscometer, 39 Rotor blades, 828 Roughness, pipe walls, 525. See also Resistance in rough pipes Rounding off numbers, 25

S Saturation, 43 Scale ratio, 452 Schlieren photography, 160 Secondary flow, 538 Section drag, 621 Series pipe systems, 548, 573 Shadowgraphs, 160 Shaft head, 265 Shaft power, 810 Shaft work, 261–263 Sharp-crested weir, 693–695 Shear strain (∆r), 36, 363

907

Shear stress (t), 36, 371, 499–502, 583–584 apparent, 500–501 fluid particle motion and, 371 laminar boundary layers, 583–584 laminar pipe flow with, 499 Reynolds stress, 501 skin friction coefficient for, 584 turbulent pipe flow with, 499–502 viscosity and, 36 viscous, 499 Shear velocity, 503 Shear work, 261 Shedder bar, 556 Ship motion, similitude for, 456–457 Shock diamonds, 772 Shock waves, 728, 768–780, 804, 866–868 continuity equation for, 768, 777 energy equation for, 769, 777–779 gas, relations with, 866–868 ideal gas law and, 769 linear momentum equation for, 768, 777 Mach number relationships and, 769–771 normal, 768–771, 866–868 nozzles with, 771–775 oblique, 776–780 sonic and supersonic flow and, 728 standing, 768 S.I. units (International System), 58 Similitude, 451–462, 473, 838–843 dimensional analysis and, 451–462, 473 dynamic, 453 forces corresponding to, 458 geometric, 452 kinematic, 452 models and, 451, 456–457 open-channel flow, 455 prototypes and, 451 pump scaling laws for, 839 ship motion and, 456–457 specific speed, 840 steady pipe flow, 454 turbomachines, 838–843 Single pipeline flow, 541–547 Skin friction coefficient, 584 Slopes, open channels, 677, 681–684 Sluice gate, open-channel flow under, 670–673 Solids, characteristics of, 21 Sonic flow, 728 Sonic velocity, 724–726, 803 Space, fluid flow and, 156 Specific energy, 658–666, 711 critical depth for, 660–661 nonrectangular cross section channels, 661 rectangular cross section channels, 659–661

908

INDEX

Specific energy diagram, 659 Specific gravity, 30, 58 Specific heat, 717–718 Specific speed, 840 Specific weight, 30, 58 Speed of sound, 724–726 Spheres, drag coefficient for, 611 Spinning balls, drag and lift effects on, 630 Split-scimitar winglets, 628 Stability, buoyancy and, 104–106, 151 Stable equilibrium, 104 Stagnation point, 157, 239 Stagnation properties, 239, 731–737, 764–765, 803 compressible flow and, 731–737, 764–765, 803 density, 732, 764–765, 803 heat transfer effects on flow and, 764–765 pressure, 239, 732 temperature, 731–732, 764–765, 803 Stagnation tube, 240 Stall condition, 627 Standard atmospheric pressure, 64–65, 150 Standing shock waves, 768 Static pressure, 66, 72–79, 150, 233, 239, 732 Euler’s equation of motion and, 233 fluid flow around curved boundary, 239 measurement tools for, 72–79, 150 stagnant pressure vs., 732 variation, 66 Static temperature, 731 Stationary waves, 661 Stator blades, 828 Stator vanes, 808 Steady flow, 156, 186, 195, 229, 233, 303, 319, 373, 656, 674–680 angular momentum equation for, 319 Euler’s equations of motion for, 233, 373 finite control volume and, 195, 229 horizontal flow of ideal fluid, 233 linear momentum equation for, 303 open channel, 656, 674–680 two-dimensional, 373 Streaklines, 159, 186 Stream function (c), 377–383, 432 hydrodynamics and, 377–383, 432 velocity components, 378 volumetric flow, 379 Streamline coordinates, 175–178, 187 Streamlines, 157–158, 175–178, 186–187, 231–236, 243, 298 Bernoulli equation applied to, 235–236, 243, 298 equation of, 158 Euler’s equations applied to, 231–234, 298 ideal flow and, 235–236, 298

inviscid flow and, 231–234, 298 stagnation point in, 157 velocity fields as, 157, 186 Streamtubes, 158 Stress field, 371 Stresses, fluid particle motion and, 371 Strouhal number (St), 556, 608 Subcritical flow, 658, 711 Subsonic flow, 727, 739, 786 Suction heads, 834–835 Sudden expansion and contraction, 537 Supercritical flow, 658, 711 Superposition of flows, 396–408, 433 around a cylinder, 402–405 around a Rankine oval, 400–401 doublet, 398–399 fundamental equations for, 433 free-vortex flow, 404–405 past a half body, 396–397 uniform flow, 396–397, 400–405 Supersonic flow, 728, 739, 787 Surface force, 371 Surface profile for open-channel flow, 682–684, 712 Surface roughness, 525, 676 Surface tension (σ), 44–47, 59 adhesion and, 44, 46 capillarity, 46–47 cohesion and, 44, 46 liquid droplets, 45 nonwetting and wetting liquids, 46 Surfaces, see Curved surfaces; External surfaces; Inclined surfaces; Plane surfaces Sutherland’s equation, 38 Swing check valve, 539 System approach to fluid flow, 161, 187 System energy, energy equation for, 260

T Temperature, 22, 23, 32, 38, 58, 69, 731–732, 756, 764–765 absolute, 23, 32, 58 compressible flow and, 731–732, 756, 764–765 constant, 69 effects on viscosity, 38 friction effect on flow and, 756 heat transfer effects on flow and, 764 ideal gas law and, 32 SI unit, 23 stagnation, 731–732, 764–765 static, 731

Theoretical discharge, 694 Thermal mass flow meter, 557 Thermodynamics, 715–723, 803 compressible flow, concepts for, 715–723 enthalpy and, 717 entropy and, 718–719, 803 first law of, 716, 803 ideal gas law, 716, 803 internal energy and, 716, 803 isentropic process, 720, 803 second law of, 718–719, 803 specific heat, 717–718 Three-dimensional flow, 155, 170 Time rate of change, 168–169, 187, 196–198, 229 control volume and, 196–198, 229 convective and local change, 198, 229 Reynolds transport theorem and, 196–198, 229 velocity, 168–169, 187 Time, fluid flow and, 156, 157, 186 Torque, turbines and, 825, 829 Torricelli’s law, 238 Total head, 251 Total pressure, 239 Tranquil flow, 658 Transitional flow, 154, 526, 536, 577 Transitional flow region, 504 Translation of differential fluid elements, 360–361, 431 Turbine efficiency, 829 Turbine flow meter, 556 Turbine head, 264 Turbines, 261–263, 329, 807, 824–830, 855 efficiency, 829 flow kinematics, 828 Francis, 827, 855 head loss, 829 impulse, 824–826 Kaplan, 827, 855 Pelton wheel, 824–826, 855 power by, 825, 829 propeller, 828 reaction, 827–830 shaft work by, 261–263 torque of, 825, 829 wind output, 329 Turbofan engine, fluid momentum and, 332 Turbojets, fluid momentum and, 332 Turbomachines, 807–855 axial-flow machines, 807, 808–814, 855 cavitation and, 834–835, 855 efficiency of, 819, 829, 833 flow systems for, 836–837 mixed-flow machines, 807 net positive suction head (NPSH), 834–835

INDEX procedure for analysis of, 811 pumps, 807, 808–823, 831–837, 855 radial-flow machines, 807, 815–817, 855 similitude, 838–843 turbines, 807, 824–830, 855 Turbulent boundary layers, 577, 594–601, 651–652 boundary layer region, 577 disturbance thickness, 595 external surfaces, 577, 594–601, 651 flat plate analysis for combined layers, 596–601 fundamental equations for, 653 Prandtl’s one-seventh power law for, 594 viscous fluids, 577, 594–601 Turbulent flow, 154, 186, 497–507, 517, 523–524, 527, 573, 656 apparent shear stress, 500–501 Darcy-Weisbach equation for, 524, 573 fully developed flow from an entrance, 497–498 laminar viscous sublayer, 500 law of the wall, 503 mean steady flow, 500 Moody diagram for, 527 open channels, 656 power law approximation, 504–505 resistance in rough pipes, 523–524, 527, 573 shear stress in a smooth pipe, 499–502 shear velocity, 503 smooth pipes with, 502–507, 517 transitional flow region, 504 viscous fluids, 497–507, 517 viscous sublayer, 500, 503 Turbulent flow region, 504 Two-dimensional flow, 155, 370, 373, 387–395, 433 continuity equation for, 370 differential fluid flow, 370, 373, 387–395, 433 equipotential lines and, 389 Euler’s equations of motion for, 373 forced-vortex flow, 393 free-vortex flow, 392 fundamental equations for, 433 Laplace’s equation for, 387 line sink flow, 391 line source flow, 390–391 steady flow, 373 uniform flow, 388–389

U U-tube manometer, 73 Ultrasonic flow meters, 558 Underexpanded flow, 772

Undulations, 661 Uniform flow, 156, 186, 388–389, 396–397, 400–405, 656, 674–680 around a cylinder, 402–405 around a Rankine oval, 400–401 open channel, 656, 674–680 past a half body, 396–397 superposition of, 396–397, 400–405 two-dimensional, 388–389 Uniform velocity, 191 Units, 22–24, 58 Unstable equilibrium, 104 Unsteady open-channel flow, 656

V Valves, losses from, 539 Vapor pressure, 43, 59, 834 Vehicles, drag reduction for, 622–623 Velocity, 154–156, 158, 161–169, 187, 191, 195, 229, 503, 763–764 average, 191 flow classification and, 154–156 equation of the streamline and, 158 finite control volume and, 195, 229 fluid systems, 161–167 heat transfer effects on flow and, 763–764 shear, 503 time rate change in, 168–169, 187 Velocity components of differential fluid flow, 378, 383–384 Velocity field, 157, 162 Velocity gradient, 36 Velocity head, 251 Velocity kinematic diagrams, 810 Velocity profile, 154–155, 190–193, 266, 487, 591–592 average velocity, 191 flow classification using, 154–155 momentum integral equation using, 591–592 nonuniform velocity, 266 paraboloid form, 487 volumetric flow from, 190–191 uniform velocity, 191 Vena contracta, 536, 693 Venturi discharge coefficient, 554 Venturi meter, 242, 554, 787 Viscosity, 34–42, 58–59 absolute, 36 apparent, 37 dynamic, 36 kinematic (n), 38, 59 measurement of, 39–42 Newtonian fluids, 37, 59 Newton’s law of, 35, 36

909

non-Newtonian fluids, 37 physical cause of, 35, 58 pressure and temperature effects, 38 shear stress and strain, 36 Viscous flow, 366, 475–519, 575–653 boundary layers, 497, 575–601, 651 drag and lift effects, 602–604, 624–632, 652 drag coefficient for, 609–619, 621, 623, 629, 651 drag reduction, 620–623 enclosed conduits and, 474–519 external surfaces and, 575–653 fully developed flow from an entrance, 497–498 irrotational vs. rotational, 366 laminar boundary layers, 577, 581–589, 596–601, 651 laminar flow, 475–499, 519 momentum integral equation for, 590–593 Navier–Stokes solution for, 481–485, 490–491 pressure gradient effects, 604–608 procedures for analysis of, 482, 494 Reynolds number for, 492–496 shear stress in a smooth pipe, 499–502 steady laminar flow between parallel plates, 475–485, 519 steady laminar flow in smooth pipes, 486–491 steady turbulent flow in smooth pipes, 502–507, 517 turbulent boundary layers, 577, 594–601, 652 turbulent flow, 497–507, 517 Viscous fluids, 237, 264, 299 Viscous shear stress, 499 Viscous sublayer, 500, 503, 577 Volume, 86, 162 Volumetric dilatation rate, 361 Volumetric flow, 190–191, 229, 379 Volute pump, 815 Von Kármán vortex street, 556, 608 Vortex flow, 112–113, 151, 392–393, 404–405, 817 forced-vortex (rotation), 112–113, 151, 393 free-vortex (circulation), 392, 404–405, 817 Vortex flow meter, 556 Vortex shedding, 608 Vortex trail, 628 Vorticity (z), 365, 368, 432

W Water, physical properties of, 857 Wave celerity, 657–658

910

INDEX

Wave propagation, 724–726 Waves, 781–785. See also Shock waves compression and expansion, 781 Mach wave fans, 781–786 Prandtl-Meyer expansion function, 783 Weber number (We), 440, 473 Weirs, 693–697, 712 broad-crested, 696 discharge coefficient for, 694–695 rectangular openings, 694

sharp-crested, 693–695 triangular openings, 695 Wetted perimeter, 674 Wetting liquids, 46 Wind turbines, fluid momentum of, 329 Wing vortex trail, 628 Work, 236, 261–269, 298 compressible flow, 265 conservation of energy, 262–263 energy equation and, 261–269, 298

flow work, 236, 261–263 incompressible flow, 263–264 power rate of, 265, 299 shaft work, 261–263 shear work, 261

Z Zero absolute pressure, 64 Zero viscosity, 37

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Fundamental Equations of Fluid Mechanics Density and Specific Weight r =

Energy

m V

g =

W V

Bernoulli’s Equation p1 p2 V 12 V 22 + z1 = + z2 + + g g 2g 2g

Ideal Gas Law p = rRT

Steady flow, ideal fluid, same streamline

Viscosity m n = r

du t = m dy

Pressure

H =

p V2 + z + g 2g

Energy Equation

F A

pavg =

Hydraulic Head

pin pout Vout2 V in2 + zin + hpump = + zout + hturbine + hL + + g g 2g 2g

pabs = patm + pg p = p0 + gh

Incompressible fluid

Head Losses

Hydrostatic Resultant Force FR = ghA yP = y +

L V2 D 2g

hL = f Ix yA

xP = x +

hL = KL

Ixy

Power

yA

# # Ws = mghs = Qghs

V2 2g

Volumetric and Mass Flow Q = V#A

# m = rV # A

Momentum

Streamline Kinematics

Liner Momentum Equation

dy v = dx u

ΣF =

a =

Angular Momentum Equation

DV 0V 0V = + V Dt 0t 0x

as = a

0V 0V b + V 0t s 0s

0 VrdV + VrVf>cs # dA 0t Lcv Lcs

ΣMO = an = a

Conservation of Mass

0V V2 b + 0t n R

0 (r * V) r dV + (r * V)rVf>cs # dA 0t Lcv Lcs

Dimensionless Numbers Re =

rVL m

Continuity Equation

Drag

0 r dV + rVf>cs # dA = 0 0t Lcv Lcs

FD = CD Ap a

Fr =

rV 2 b 2

2gL V

M =

V c

SI Prefixes Multiple

Exponential Form

Prefix

SI Symbol

1 000 000 000

109

giga

G

1 000 000

106

mega

M

1 000

103

kilo

k

10-3

Submultiple 0.001 0.000 001 0.000 000 001

milli

m

10

-6

micro

m

10

-9

nano

n

Conversion Factors 1 atm = 101.3 kPa 1000 liters (l) = 1 m3 1 hp = 745.7 W

Geometric Properties of an Area y

y

A 5 1 pr2 2

A 5 ab

a

x

C

r

1 Ix 5 12 ba3

C

x

4r 3p

1 Iy 5 12 ab3

Ix 5 0.1098 r4 1 Iy 5 8 pr4

Semicircle b y Rectangle

2 A 5 3 ab

y a

1 A 5 2 ab a 1 a 3

C

x

x

C

Ix 5

8 ba3 175

a

1 Ix 5 21 ba3

3a 5

1 Ix 5 36 ba3 1 Iy 5 36 ab3

3b 8

1b 3

b Parabola

b y

Triangle y

1 A 5 3 ab A 5 pr

2

1 Ix 5 4 pr4 1 Iy 5 4 pr4

r x

C

C

x

3b 4 b

3 a 10

Exparabola Circle y

y

A 5 u r2

A 5 pab a x

C a

1 Ix 5 4 pba3 1 Iy 5 4 pab3

u r

C

x u

b

b

Ellipse

2 r sin u u 3

Circular sector

1 1 Ix 5 4 r 4 (u 2 2 sin 2 u) u is in radians

Roughness Factors for New Pipe Riveted steel e 5 0.9 mm 2 9 mm Commercial steel e 5 0.045 mm Drawn tubing e 5 0.0015 mm Wood stave e 5 0.5 mm

Smooth glass, plastic e 5 0 Concrete e 5 0.3 mm 2 3 mm Cast iron e 5 0.26 mm Galvanized iron e 5 0.15 mm

0.1 0.09 0.08 0.07

Critical zone

Laminar flow

f5

Fully Turbulent Flow

64 Re

0.05 0.04

0.06

0.03

0.05

0.02 0.015

0.04

(Ref. [1], Ch. 10)

Moody Diagram

0.01 0.008

f

0.006

0.03

0.004 0.025

e D

0.002

0.02

0.001 0.0008 0.0006 0.0004

0.015

0.0002 0.0001 0.000 05

Smooth Pipe

0.01

0.000 001 0.000 005

0.009

0.000 01

0.008 (103)

2

3 4 5 67 9 (104)

2

3 4 5 67 9 (105)

2

3 4 5 67 9 (106)

VD Re 5 n

2

3 4 5 67 9 (107)

2

3 4 5 67 9 (108)

GLOBAL EDITION

This is a special edition of an established title widely used by colleges and universities throughout the world. Pearson published this exclusive edition for the benefit of students outside the United States. If you purchased this book within the United States, you should be aware that it has been imported without the approval of the Publisher or Author.

Fluid Mechanics provides a clear and thorough understanding of the theory and the many applications of fluid mechanics through: • a student-friendly approach with a clear organization of chapters and a structured method to facilitate easy understanding of each new definition and concept; • a broad range of problems of varying difficulty levels to stimulate interest in the subject, assist in the preparation for various engineering exams, and help develop the ability to reduce any problem to a model to which the principles of fluid mechanics may be applied; • insightful photos depicting real-world situations; and • video solutions offering step-by-step solution walkthroughs for homework problems, available on the companion website.

Also available for this book is Mastering Engineering, an optional suite featuring an online homework, tutorial, and assessment program designed to work with this text to engage students and improve results.

First published in July 2021.

New Enterprise House St Helens Street Derby DE1 3GY UK

email: [email protected] Copyright © 2021 David Icke No part of this book may be reproduced in any form without permission from the Publisher, except for the quotation of brief passages in criticism Cover Design: Gareth Icke Book Design: Neil Hague

British Library Cataloguing-in Publication Data

A catalogue record for this book is available from the British Library eISBN 978-18384153-1-0

Dedication:

To Freeeeeedom!

Renegade:

Adjective ‘Having rejected tradition: Unconventional.’ Merriam-Webster Dictionary

Acquiescence to tyranny is the death of the spirit You may be 38 years old, as I happen to be. And one day, some great opportunity stands before you and calls you to stand up for some great principle, some great issue, some great cause. And you refuse to do it because you are afraid … You refuse to do it because you want to live longer … You’re afraid that you will lose your job, or you are afraid that you will be criticised or that you will lose your popularity, or you’re afraid that somebody will stab you, or shoot at you or bomb your house; so you refuse to take the stand. Well, you may go on and live until you are 90, but you’re just as dead at 38 as you would be at 90. And the cessation of breathing in your life is but the belated announcement of an earlier death of the spirit. Martin Luther King

How the few control the many and always have – the many do whatever they’re told ‘Forward, the Light Brigade!’ Was there a man dismayed? Not though the soldier knew Someone had blundered. Theirs not to make reply, Theirs not to reason why, Theirs but to do and die. Into the valley of Death Rode the six hundred. Cannon to right of them, Cannon to le of them, Cannon in front of them Volleyed and thundered; Stormed at with shot and shell, Boldly they rode and well, Into the jaws of Death, Into the mouth of hell Rode the six hundred Alfred Lord Tennyson (1809-1892)

The mist is li ing slowly I can see the way ahead And I’ve le behind the empty streets That once inspired my life And the strength of the emotion Is like thunder in the air ’Cos the promise that we made each other Haunts me to the end The secret of your beauty And the mystery of your soul I’ve been searching for in everyone I meet And the times I’ve been mistaken It’s impossible to say And the grass is growing Underneath our feet The words that I remember From my childhood still are true That there’s none so blind As those who will not see And to those who lack the courage And say it’s dangerous to try Well they just don’t know That love eternal will not be denied I know you’re out there somewhere Somewhere, somewhere I know you’re out there somewhere

Somewhere you can hear my voice I know I’ll find you somehow Somehow, somehow I know I’ll find you somehow And somehow I’ll return again to you The Moody Blues

Are you a gutless wonder - or a Renegade Mind? Monuments put from pen to paper, Turns me into a gutless wonder, And if you tolerate this, Then your children will be next. Gravity keeps my head down, Or is it maybe shame ... Manic Street Preachers

Rise like lions a er slumber In unvanquishable number. Shake your chains to earth like dew Which in sleep have fallen on you. Ye are many – they are few. Percy Shelley

Contents

CHAPTER 1 CHAPTER 2 CHAPTER 3 CHAPTER 4 CHAPTER 5 CHAPTER 6 CHAPTER 7 CHAPTER 8 CHAPTER 9 CHAPTER 10 CHAPTER 11 CHAPTER 12

‘I’m thinking’ – Oh, but are you? Renegade perception The Pushbacker sting ‘Covid’: The calculated catastrophe There is no ‘virus’ Sequence of deceit War on your mind ‘Reframing’ insanity We must have it? So what is it? Human 2.0 Who controls the Cult? Escaping Wetiko  

Postscript APPENDIX BIBLIOGRAPHY INDEX  

 

 

Cowan-Kaufman-Morell Statement on Virus Isolation

CHAPTER ONE I’m thinking’ – Oh, but are you? Think for yourself and let others enjoy the privilege of doing so too Voltaire

F

rench-born philosopher, mathematician and scientist René Descartes became famous for his statement in Latin in the 17th century which translates into English as: ‘I think, therefore I am.’ On the face of it that is true. Thought reflects perception and perception leads to both behaviour and self-identity. In that sense ‘we’ are what we think. But who or what is doing the thinking and is thinking the only route to perception? Clearly, as we shall see, ‘we’ are not always the source of ‘our’ perception, indeed with regard to humanity as a whole this is rarely the case; and thinking is far from the only means of perception. Thought is the village idiot compared with other expressions of consciousness that we all have the potential to access and tap into. This has to be true when we are those other expressions of consciousness which are infinite in nature. We have forgo en this, or, more to the point, been manipulated to forget. These are not just the esoteric musings of the navel. The whole foundation of human control and oppression is control of perception. Once perception is hijacked then so is behaviour which is dictated by perception. Collective perception becomes collective behaviour and collective behaviour is what we call human society. Perception is all and those behind human control know that which is

why perception is the target 24/7 of the psychopathic manipulators that I call the Global Cult. They know that if they dictate perception they will dictate behaviour and collectively dictate the nature of human society. They are further aware that perception is formed from information received and if they control the circulation of information they will to a vast extent direct human behaviour. Censorship of information and opinion has become globally Nazilike in recent years and never more blatantly than since the illusory ‘virus pandemic’ was triggered out of China in 2019 and across the world in 2020. Why have billions submi ed to house arrest and accepted fascistic societies in a way they would have never believed possible? Those controlling the information spewing from government, mainstream media and Silicon Valley (all controlled by the same Global Cult networks) told them they were in danger from a ‘deadly virus’ and only by submi ing to house arrest and conceding their most basic of freedoms could they and their families be protected. This monumental and provable lie became the perception of the billions and therefore the behaviour of the billions. In those few words you have the whole structure and modus operandi of human control. Fear is a perception – False Emotion Appearing Real – and fear is the currency of control. In short … get them by the balls (or give them the impression that you have) and their hearts and minds will follow. Nothing grips the dangly bits and freezes the rear-end more comprehensively than fear.

World number 1

There are two ‘worlds’ in what appears to be one ‘world’ and the prime difference between them is knowledge. First we have the mass of human society in which the population is maintained in coldlycalculated ignorance through control of information and the ‘education’ (indoctrination) system. That’s all you really need to control to enslave billions in a perceptual delusion in which what are perceived to be their thoughts and opinions are ever-repeated mantras that the system has been downloading all their lives through ‘education’, media, science, medicine, politics and academia

in which the personnel and advocates are themselves overwhelmingly the perceptual products of the same repetition. Teachers and academics in general are processed by the same programming machine as everyone else, but unlike the great majority they never leave the ‘education’ program. It gripped them as students and continues to grip them as programmers of subsequent generations of students. The programmed become the programmers – the programmed programmers. The same can largely be said for scientists, doctors and politicians and not least because as the American writer Upton Sinclair said: ‘It is difficult to get a man to understand something when his salary depends upon his not understanding it.’ If your career and income depend on thinking the way the system demands then you will – bar a few freeminded exceptions – concede your mind to the Perceptual Mainframe that I call the Postage Stamp Consensus. This is a tiny band of perceived knowledge and possibility ‘taught’ (downloaded) in the schools and universities, pounded out by the mainstream media and on which all government policy is founded. Try thinking, and especially speaking and acting, outside of the ‘box’ of consensus and see what that does for your career in the Mainstream Everything which bullies, harasses, intimidates and ridicules the population into compliance. Here we have the simple structure which enslaves most of humanity in a perceptual prison cell for an entire lifetime and I’ll go deeper into this process shortly. Most of what humanity is taught as fact is nothing more than programmed belief. American science fiction author Frank Herbert was right when he said: ‘Belief can be manipulated. Only knowledge is dangerous.’ In the ‘Covid’ age belief is promoted and knowledge is censored. It was always so, but never to the extreme of today.

World number 2

A ‘number 2’ is slang for ‘doing a poo’ and how appropriate that is when this other ‘world’ is doing just that on humanity every minute of every day. World number 2 is a global network of secret societies and semi-secret groups dictating the direction of society via

governments, corporations and authorities of every kind. I have spent more than 30 years uncovering and exposing this network that I call the Global Cult and knowing its agenda is what has made my books so accurate in predicting current and past events. Secret societies are secret for a reason. They want to keep their hoarded knowledge to themselves and their chosen initiates and to hide it from the population which they seek through ignorance to control and subdue. The whole foundation of the division between World 1 and World 2 is knowledge. What number 1 knows number 2 must not. Knowledge they have worked so hard to keep secret includes (a) the agenda to enslave humanity in a centrally-controlled global dictatorship, and (b) the nature of reality and life itself. The la er (b) must be suppressed to allow the former (a) to prevail as I shall be explaining. The way the Cult manipulates and interacts with the population can be likened to a spider’s web. The ‘spider’ sits at the centre in the shadows and imposes its will through the web with each strand represented in World number 2 by a secret society, satanic or semi-secret group, and in World number 1 – the world of the seen – by governments, agencies of government, law enforcement, corporations, the banking system, media conglomerates and Silicon Valley (Fig 1 overleaf). The spider and the web connect and coordinate all these organisations to pursue the same global outcome while the population sees them as individual entities working randomly and independently. At the level of the web governments are the banking system are the corporations are the media are Silicon Valley are the World Health Organization working from their inner cores as one unit. Apparently unconnected countries, corporations, institutions, organisations and people are on the same team pursuing the same global outcome. Strands in the web immediately around the spider are the most secretive and exclusive secret societies and their membership is emphatically restricted to the Cult inner-circle emerging through the generations from particular bloodlines for reasons I will come to. At the core of the core you would get them in a single room. That’s how many people are dictating the direction of human society and its transformation

through the ‘Covid’ hoax and other means. As the web expands out from the spider we meet the secret societies that many people will be aware of – the Freemasons, Knights Templar, Knights of Malta, Opus Dei, the inner sanctum of the Jesuit Order, and such like. Note how many are connected to the Church of Rome and there is a reason for that. The Roman Church was established as a revamp, a rebranding, of the relocated ‘Church’ of Babylon and the Cult imposing global tyranny today can be tracked back to Babylon and Sumer in what is now Iraq.

Figure 1: The global web through which the few control the many. (Image Neil Hague.)

Inner levels of the web operate in the unseen away from the public eye and then we have what I call the cusp organisations located at the point where the hidden meets the seen. They include a series of satellite organisations answering to a secret society founded in London in the late 19th century called the Round Table and among them are the Royal Institute of International Affairs (UK, founded in 1920); Council on Foreign Relations (US, 1921); Bilderberg Group (worldwide, 1954); Trilateral Commission (US/worldwide, 1972); and the Club of Rome (worldwide, 1968) which was created to exploit environmental concerns to justify the centralisation of global power to ‘save the planet’. The Club of Rome instigated with others the human-caused climate change hoax which has led to all the ‘green

new deals’ demanding that very centralisation of control. Cusp organisations, which include endless ‘think tanks’ all over the world, are designed to coordinate a single global policy between political and business leaders, intelligence personnel, media organisations and anyone who can influence the direction of policy in their own sphere of operation. Major players and regular a enders will know what is happening – or some of it – while others come and go and are kept overwhelmingly in the dark about the big picture. I refer to these cusp groupings as semi-secret in that they can be publicly identified, but what goes on at the inner-core is kept very much ‘in house’ even from most of their members and participants through a fiercely-imposed system of compartmentalisation. Only let them know what they need to know to serve your interests and no more. The structure of secret societies serves as a perfect example of this principle. Most Freemasons never get higher than the bo om three levels of ‘degree’ (degree of knowledge) when there are 33 official degrees of the Sco ish Rite. Initiates only qualify for the next higher ‘compartment’ or degree if those at that level choose to allow them. Knowledge can be carefully assigned only to those considered ‘safe’. I went to my local Freemason’s lodge a few years ago when they were having an ‘open day’ to show how cuddly they were and when I cha ed to some of them I was astonished at how li le the rank and file knew even about the most ubiquitous symbols they use. The mushroom technique – keep them in the dark and feed them bullshit – applies to most people in the web as well as the population as a whole. Sub-divisions of the web mirror in theme and structure transnational corporations which have a headquarters somewhere in the world dictating to all their subsidiaries in different countries. Subsidiaries operate in their methodology and branding to the same centrally-dictated plan and policy in pursuit of particular ends. The Cult web functions in the same way. Each country has its own web as a subsidiary of the global one. They consist of networks of secret societies, semi-secret groups and bloodline families and their job is to impose the will of the spider and the global web in their particular country. Subsidiary networks control and manipulate the national political system, finance, corporations, media, medicine, etc. to

ensure that they follow the globally-dictated Cult agenda. These networks were the means through which the ‘Covid’ hoax could be played out with almost every country responding in the same way.

The ‘Yessir’ pyramid

Compartmentalisation is the key to understanding how a tiny few can dictate the lives of billions when combined with a top-down sequence of imposition and acquiescence. The inner core of the Cult sits at the peak of the pyramidal hierarchy of human society (Fig 2 overleaf). It imposes its will – its agenda for the world – on the level immediately below which acquiesces to that imposition. This level then imposes the Cult will on the level below them which acquiesces and imposes on the next level. Very quickly we meet levels in the hierarchy that have no idea there even is a Cult, but the sequence of imposition and acquiescence continues down the pyramid in just the same way. ‘I don’t know why we are doing this but the order came from “on-high” and so we be er just do it.’ Alfred Lord Tennyson said of the cannon fodder levels in his poem The Charge of the Light Brigade: ‘Theirs not to reason why; theirs but to do and die.’ The next line says that ‘into the valley of death rode the six hundred’ and they died because they obeyed without question what their perceived ‘superiors’ told them to do. In the same way the population capitulated to ‘Covid’. The whole hierarchical pyramid functions like this to allow the very few to direct the enormous many. Eventually imposition-acquiescence-imposition-acquiescence comes down to the mass of the population at the foot of the pyramid. If they acquiesce to those levels of the hierarchy imposing on them (governments/law enforcement/doctors/media) a circuit is completed between the population and the handful of superpsychopaths in the Cult inner core at the top of the pyramid. Without a circuit-breaking refusal to obey, the sequence of imposition and acquiescence allows a staggeringly few people to impose their will upon the entirety of humankind. We are looking at the very sequence that has subjugated billions since the start of 2020. Our freedom has not been taken from us. Humanity has given it

away. Fascists do not impose fascism because there are not enough of them. Fascism is imposed by the population acquiescing to fascism. Put another way allowing their perceptions to be programmed to the extent that leads to the population giving their freedom away by giving their perceptions – their mind – away. If this circuit is not broken by humanity ceasing to cooperate with their own enslavement then nothing can change. For that to happen people have to critically think and see through the lies and window dressing and then summon the backbone to act upon what they see. The Cult spends its days working to stop either happening and its methodology is systematic and highly detailed, but it can be overcome and that is what this book is all about.

Figure 2: The simple sequence of imposition and compliance that allows a handful of people at the peak of the pyramid to dictate the lives of billions.

The Life Program

Okay, back to world number 1 or the world of the ‘masses’. Observe the process of what we call ‘life’ and it is a perceptual download from cradle to grave. The Cult has created a global structure in which perception can be programmed and the program continually topped-up with what appears to be constant confirmation that the program is indeed true reality. The important word here is ‘appears’.

This is the structure, the fly-trap, the Postage Stamp Consensus or Perceptual Mainframe, which represents that incredibly narrow band of perceived possibility delivered by the ‘education’ system, mainstream media, science and medicine. From the earliest age the download begins with parents who have themselves succumbed to the very programming their children are about to go through. Most parents don’t do this out of malevolence and mostly it is quite the opposite. They do what they believe is best for their children and that is what the program has told them is best. Within three or four years comes the major transition from parental programming to fullblown state (Cult) programming in school, college and university where perceptually-programmed teachers and academics pass on their programming to the next generations. Teachers who resist are soon marginalised and their careers ended while children who resist are called a problem child for whom Ritalin may need to be prescribed. A few years a er entering the ‘world’ children are under the control of authority figures representing the state telling them when they have to be there, when they can leave and when they can speak, eat, even go to the toilet. This is calculated preparation for a lifetime of obeying authority in all its forms. Reflex-action fear of authority is instilled by authority from the start. Children soon learn the carrot and stick consequences of obeying or defying authority which is underpinned daily for the rest of their life. Fortunately I daydreamed through this crap and never obeyed authority simply because it told me to. This approach to my alleged ‘be ers’ continues to this day. There can be consequences of pursuing open-minded freedom in a world of closed-minded conformity. I spent a lot of time in school corridors a er being ejected from the classroom for not taking some of it seriously and now I spend a lot of time being ejected from Facebook, YouTube and Twi er. But I can tell you that being true to yourself and not compromising your self-respect is far more exhilarating than bowing to authority for authority’s sake. You don’t have to be a sheep to the shepherd (authority) and the sheep dog (fear of not obeying authority).

The perceptual download continues throughout the formative years in school, college and university while script-reading ‘teachers’, ‘academics’ ‘scientists’, ‘doctors’ and ‘journalists’ insist that ongoing generations must be as programmed as they are. Accept the program or you will not pass your ‘exams’ which confirm your ‘degree’ of programming. It is tragic to think that many parents pressure their offspring to work hard at school to download the program and qualify for the next stage at college and university. The late, great, American comedian George Carlin said: ‘Here’s a bumper sticker I’d like to see: We are proud parents of a child who has resisted his teachers’ a empts to break his spirit and bend him to the will of his corporate masters.’ Well, the best of luck finding many of those, George. Then comes the moment to leave the formal programming years in academia and enter the ‘adult’ world of work. There you meet others in your chosen or prescribed arena who went through the same Postage Stamp Consensus program before you did. There is therefore overwhelming agreement between almost everyone on the basic foundations of Postage Stamp reality and the rejection, even contempt, of the few who have a mind of their own and are prepared to use it. This has two major effects. Firstly, the consensus confirms to the programmed that their download is really how things are. I mean, everyone knows that, right? Secondly, the arrogance and ignorance of Postage Stamp adherents ensure that anyone questioning the program will have unpleasant consequences for seeking their own truth and not picking their perceptions from the shelf marked: ‘Things you must believe without question and if you don’t you’re a dangerous lunatic conspiracy theorist and a harebrained nu er’. Every government, agency and corporation is founded on the same Postage Stamp prison cell and you can see why so many people believe the same thing while calling it their own ‘opinion’. Fusion of governments and corporations in pursuit of the same agenda was the definition of fascism described by Italian dictator Benito Mussolini. The pressure to conform to perceptual norms downloaded for a lifetime is incessant and infiltrates society right

down to family groups that become censors and condemners of their own ‘black sheep’ for not, ironically, being sheep. We have seen an explosion of that in the ‘Covid’ era. Cult-owned global media unleashes its propaganda all day every day in support of the Postage Stamp and targets with abuse and ridicule anyone in the public eye who won’t bend their mind to the will of the tyranny. Any response to this is denied (certainly in my case). They don’t want to give a platform to expose official lies. Cult-owned-and-created Internet giants like Facebook, Google, YouTube and Twi er delete you for having an unapproved opinion. Facebook boasts that its AI censors delete 97-percent of ‘hate speech’ before anyone even reports it. Much of that ‘hate speech’ will simply be an opinion that Facebook and its masters don’t want people to see. Such perceptual oppression is widely known as fascism. Even Facebook executive Benny Thomas, a ‘CEO Global Planning Lead’, said in comments secretly recorded by investigative journalism operation Project Veritas that Facebook is ‘too powerful’ and should be broken up: I mean, no king in history has been the ruler of two billion people, but Mark Zuckerberg is … And he’s 36. That’s too much for a 36-year-old ... You should not have power over two billion people. I just think that’s wrong.

Thomas said Facebook-owned platforms like Instagram, Oculus, and WhatsApp needed to be separate companies. ‘It’s too much power when they’re all one together’. That’s the way the Cult likes it, however. We have an executive of a Cult organisation in Benny Thomas that doesn’t know there is a Cult such is the compartmentalisation. Thomas said that Facebook and Google ‘are no longer companies, they’re countries’. Actually they are more powerful than countries on the basis that if you control information you control perception and control human society.

I love my oppressor

Another expression of this psychological trickery is for those who realise they are being pressured into compliance to eventually

convince themselves to believe the official narratives to protect their self-respect from accepting the truth that they have succumbed to meek and subservient compliance. Such people become some of the most vehement defenders of the system. You can see them everywhere screaming abuse at those who prefer to think for themselves and by doing so reminding the compliers of their own capitulation to conformity. ‘You are talking dangerous nonsense you Covidiot!!’ Are you trying to convince me or yourself? It is a potent form of Stockholm syndrome which is defined as: ‘A psychological condition that occurs when a victim of abuse identifies and a aches, or bonds, positively with their abuser.’ An example is hostages bonding and even ‘falling in love’ with their kidnappers. The syndrome has been observed in domestic violence, abused children, concentration camp inmates, prisoners of war and many and various Satanic cults. These are some traits of Stockholm syndrome listed at goodtherapy.org: • Positive regard towards perpetrators of abuse or captor [see ‘Covid’]. • Failure to cooperate with police and other government authorities when it comes to holding perpetrators of abuse or kidnapping accountable [or in the case of ‘Covid’ cooperating with the police to enforce and defend their captors’ demands]. • Li le or no effort to escape [see ‘Covid’]. • Belief in the goodness of the perpetrators or kidnappers [see ‘Covid’]. • Appeasement of captors. This is a manipulative strategy for maintaining one’s safety. As victims get rewarded – perhaps with less abuse or even with life itself – their appeasing behaviours are reinforced [see ‘Covid’]. • Learned helplessness. This can be akin to ‘if you can’t beat ‘em, join ‘em’. As the victims fail to escape the abuse or captivity, they may start giving up and soon realize it’s just easier for everyone if they acquiesce all their power to their captors [see ‘Covid’].

Feelings of pity toward the abusers, believing they are actually • victims themselves. Because of this, victims may go on a crusade or mission to ‘save’ [protect] their abuser [see the venom unleashed on those challenging the official ‘Covid’ narrative]. • Unwillingness to learn to detach from their perpetrators and heal. In essence, victims may tend to be less loyal to themselves than to their abuser [ definitely see ‘Covid’]. Ponder on those traits and compare them with the behaviour of great swathes of the global population who have defended governments and authorities which have spent every minute destroying their lives and livelihoods and those of their children and grandchildren since early 2020 with fascistic lockdowns, house arrest and employment deletion to ‘protect’ them from a ‘deadly virus’ that their abusers’ perceptually created to bring about this very outcome. We are looking at mass Stockholm syndrome. All those that agree to concede their freedom will believe those perceptions are originating in their own independent ‘mind’ when in fact by conceding their reality to Stockholm syndrome they have by definition conceded any independence of mind. Listen to the ‘opinions’ of the acquiescing masses in this ‘Covid’ era and what gushes forth is the repetition of the official version of everything delivered unprocessed, unfiltered and unquestioned. The whole programming dynamic works this way. I must be free because I’m told that I am and so I think that I am. You can see what I mean with the chapter theme of ‘I’m thinking – Oh, but are you?’ The great majority are not thinking, let alone for themselves. They are repeating what authority has told them to believe which allows them to be controlled. Weaving through this mentality is the fear that the ‘conspiracy theorists’ are right and this again explains the o en hysterical abuse that ensues when you dare to contest the official narrative of anything. Denial is the mechanism of hiding from yourself what you don’t want to be true. Telling people what they want to hear is easy, but it’s an infinitely greater challenge to tell them what they would rather not be happening.

One is akin to pushing against an open door while the other is met with vehement resistance no ma er what the scale of evidence. I don’t want it to be true so I’ll convince myself that it’s not. Examples are everywhere from the denial that a partner is cheating despite all the signs to the reflex-action rejection of any idea that world events in which country a er country act in exactly the same way are centrally coordinated. To accept the la er is to accept that a force of unspeakable evil is working to destroy your life and the lives of your children with nothing too horrific to achieve that end. Who the heck wants that to be true? But if we don’t face reality the end is duly achieved and the consequences are far worse and ongoing than breaking through the walls of denial today with the courage to make a stand against tyranny.

Connect the dots – but how?

A crucial aspect of perceptual programming is to portray a world in which everything is random and almost nothing is connected to anything else. Randomness cannot be coordinated by its very nature and once you perceive events as random the idea they could be connected is waved away as the rantings of the tinfoil-hat brigade. You can’t plan and coordinate random you idiot! No, you can’t, but you can hide the coldly-calculated and long-planned behind the illusion of randomness. A foundation manifestation of the Renegade Mind is to scan reality for pa erns that connect the apparently random and turn pixels and dots into pictures. This is the way I work and have done so for more than 30 years. You look for similarities in people, modus operandi and desired outcomes and slowly, then ever quicker, the picture forms. For instance: There would seem to be no connection between the ‘Covid pandemic’ hoax and the human-caused global-warming hoax and yet they are masks (appropriately) on the same face seeking the same outcome. Those pushing the global warming myth through the Club of Rome and other Cult agencies are driving the lies about ‘Covid’ – Bill Gates is an obvious one, but they are endless. Why would the same people be involved in both when they are clearly not connected? Oh, but they

are. Common themes with personnel are matched by common goals. The ‘solutions’ to both ‘problems’ are centralisation of global power to impose the will of the few on the many to ‘save’ humanity from ‘Covid’ and save the planet from an ‘existential threat’ (we need ‘zero Covid’ and ‘zero carbon emissions’). These, in turn, connect with the ‘dot’ of globalisation which was coined to describe the centralisation of global power in every area of life through incessant political and corporate expansion, trading blocks and superstates like the European Union. If you are the few and you want to control the many you have to centralise power and decision-making. The more you centralise power the more power the few at the centre will have over the many; and the more that power is centralised the more power those at the centre have to centralise even quicker. The momentum of centralisation gets faster and faster which is exactly the process we have witnessed. In this way the hoaxed ‘pandemic’ and the fakery of human-caused global warming serve the interests of globalisation and the seizure of global power in the hands of the Cult inner-circle which is behind ‘Covid’, ‘climate change’ and globalisation. At this point random ‘dots’ become a clear and obvious picture or pa ern. Klaus Schwab, the classic Bond villain who founded the Cult’s Gates-funded World Economic Forum, published a book in 2020, The Great Reset, in which he used the ‘problem’ of ‘Covid’ to justify a total transformation of human society to ‘save’ humanity from ‘climate change’. Schwab said: ‘The pandemic represents a rare but narrow window of opportunity to reflect, reimagine, and reset our world.’ What he didn’t mention is that the Cult he serves is behind both hoaxes as I show in my book The Answer. He and the Cult don’t have to reimagine the world. They know precisely what they want and that’s why they destroyed human society with ‘Covid’ to ‘build back be er’ in their grand design. Their job is not to imagine, but to get humanity to imagine and agree with their plans while believing it’s all random. It must be pure coincidence that ‘The Great Reset’ has long been the Cult’s code name for the global imposition of fascism and replaced previous code-names of the ‘New World

Order’ used by Cult frontmen like Father George Bush and the ‘New Order of the Ages’ which emerged from Freemasonry and much older secret societies. New Order of the Ages appears on the reverse of the Great Seal of the United States as ‘Novus ordo seclorum’ underneath the Cult symbol used since way back of the pyramid and all seeing-eye (Fig 3). The pyramid is the hierarchy of human control headed by the illuminated eye that symbolises the force behind the Cult which I will expose in later chapters. The term ‘Annuit Coeptis’ translates as ‘He favours our undertaking’. We are told the ‘He’ is the Christian god, but ‘He’ is not as I will be explaining.

Figure 3: The all-seeing eye of the Cult ‘god’ on the Freemason-designed Great Seal of the United States and also on the dollar bill.

Having you on

Two major Cult techniques of perceptual manipulation that relate to all this are what I have called since the 1990s Problem-ReactionSolution (PRS) and the Totalitarian Tiptoe (TT). They can be uncovered by the inquiring mind with a simple question: Who benefits? The answer usually identifies the perpetrators of a given action or happening through the concept of ‘he who most benefits from a crime is the one most likely to have commi ed it’. The Latin ‘Cue bono?’ – Who benefits? – is widely a ributed to the Roman orator and statesman Marcus Tullius Cicero. No wonder it goes back so far when the concept has been relevant to human behaviour since

history was recorded. Problem-Reaction-Solution is the technique used to manipulate us every day by covertly creating a problem (or the illusion of one) and offering the solution to the problem (or the illusion of one). In the first phase you create the problem and blame someone or something else for why it has happened. This may relate to a financial collapse, terrorist a ack, war, global warming or pandemic, anything in fact that will allow you to impose the ‘solution’ to change society in the way you desire at that time. The ‘problem’ doesn’t have to be real. PRS is manipulation of perception and all you need is the population to believe the problem is real. Human-caused global warming and the ‘Covid pandemic’ only have to be perceived to be real for the population to accept the ‘solutions’ of authority. I refer to this technique as NO-Problem-Reaction-Solution. Billions did not meekly accept house arrest from early 2020 because there was a real deadly ‘Covid pandemic’ but because they perceived – believed – that to be the case. The antidote to ProblemReaction-Solution is to ask who benefits from the proposed solution. Invariably it will be anyone who wants to justify more control through deletion of freedom and centralisation of power and decision-making. The two world wars were Problem-Reaction-Solutions that transformed and realigned global society. Both were manipulated into being by the Cult as I have detailed in books since the mid1990s. They dramatically centralised global power, especially World War Two, which led to the United Nations and other global bodies thanks to the overt and covert manipulations of the Rockefeller family and other Cult bloodlines like the Rothschilds. The UN is a stalking horse for full-blown world government that I will come to shortly. The land on which the UN building stands in New York was donated by the Rockefellers and the same Cult family was behind Big Pharma scalpel and drug ‘medicine’ and the creation of the World Health Organization as part of the UN. They have been stalwarts of the eugenics movement and funded Hitler’s race-purity expert’ Ernst Rudin. The human-caused global warming hoax has been orchestrated by the Club of Rome through the UN which is

manufacturing both the ‘problem’ through its Intergovernmental Panel on Climate Change and imposing the ‘solution’ through its Agenda 21 and Agenda 2030 which demand the total centralisation of global power to ‘save the world’ from a climate hoax the United Nations is itself perpetrating. What a small world the Cult can be seen to be particularly among the inner circles. The bedfellow of Problem-Reaction-Solution is the Totalitarian Tiptoe which became the Totalitarian Sprint in 2020. The technique is fashioned to hide the carefully-coordinated behind the cover of apparently random events. You start the sequence at ‘A’ and you know you are heading for ‘Z’. You don’t want people to know that and each step on the journey is presented as a random happening while all the steps strung together lead in the same direction. The speed may have quickened dramatically in recent times, but you can still see the incremental approach of the Tiptoe in the case of ‘Covid’ as each new imposition takes us deeper into fascism. Tell people they have to do this or that to get back to ‘normal’, then this and this and this. With each new demand adding to the ones that went before the population’s freedom is deleted until it disappears. The spider wraps its web around the flies more comprehensively with each new diktat. I’ll highlight this in more detail when I get to the ‘Covid’ hoax and how it has been pulled off. Another prime example of the Totalitarian Tiptoe is how the Cult-created European Union went from a ‘freetrade zone’ to a centralised bureaucratic dictatorship through the Tiptoe of incremental centralisation of power until nations became mere administrative units for Cult-owned dark suits in Brussels. The antidote to ignorance is knowledge which the Cult seeks vehemently to deny us, but despite the systematic censorship to that end the Renegade Mind can overcome this by vociferously seeking out the facts no ma er the impediments put in the way. There is also a method of thinking and perceiving – knowing – that doesn’t even need names, dates, place-type facts to identify the pa erns that reveal the story. I’ll get to that in the final chapter. All you need to know about the manipulation of human society and to what end is still out there – at the time of writing – in the form of books, videos

and websites for those that really want to breach the walls of programmed perception. To access this knowledge requires the abandonment of the mainstream media as a source of information in the awareness that this is owned and controlled by the Cult and therefore promotes mass perceptions that suit the Cult. Mainstream media lies all day, every day. That is its function and very reason for being. Where it does tell the truth, here and there, is only because the truth and the Cult agenda very occasionally coincide. If you look for fact and insight to the BBC, CNN and virtually all the rest of them you are asking to be conned and perceptually programmed.

Know the outcome and you’ll see the journey

Events seem random when you have no idea where the world is being taken. Once you do the random becomes the carefully planned. Know the outcome and you’ll see the journey is a phrase I have been using for a long time to give context to daily happenings that appear unconnected. Does a problem, or illusion of a problem, trigger a proposed ‘solution’ that further drives society in the direction of the outcome? Invariably the answer will be yes and the random – abracadabra – becomes the clearly coordinated. So what is this outcome that unlocks the door to a massively expanded understanding of daily events? I will summarise its major aspects – the fine detail is in my other books – and those new to this information will see that the world they thought they were living in is a very different place. The foundation of the Cult agenda is the incessant centralisation of power and all such centralisation is ultimately in pursuit of Cult control on a global level. I have described for a long time the planned world structure of top-down dictatorship as the Hunger Games Society. The term obviously comes from the movie series which portrayed a world in which a few living in military-protected hi-tech luxury were the overlords of a population condemned to abject poverty in isolated ‘sectors’ that were not allowed to interact. ‘Covid’ lockdowns and travel bans anyone? The ‘Hunger Games’ pyramid of structural control has the inner circle of the Cult at the top with pre y much the entire

population at the bo om under their control through dependency for survival on the Cult. The whole structure is planned to be protected and enforced by a military-police state (Fig 4). Here you have the reason for the global lockdowns of the fake pandemic to coldly destroy independent incomes and livelihoods and make everyone dependent on the ‘state’ (the Cult that controls the ‘states’). I have warned in my books for many years about the plan to introduce a ‘guaranteed income’ – a barely survivable pi ance – designed to impose dependency when employment was destroyed by AI technology and now even more comprehensively at great speed by the ‘Covid’ scam. Once the pandemic was played and lockdown consequences began to delete independent income the authorities began to talk right on cue about the need for a guaranteed income and a ‘Great Reset’. Guaranteed income will be presented as benevolent governments seeking to help a desperate people – desperate as a direct result of actions of the same governments. The truth is that such payments are a trap. You will only get them if you do exactly what the authorities demand including mass vaccination (genetic manipulation). We have seen this theme already in Australia where those dependent on government benefits have them reduced if parents don’t agree to have their children vaccinated according to an insane healthdestroying government-dictated schedule. Calculated economic collapse applies to governments as well as people. The Cult wants rid of countries through the creation of a world state with countries broken up into regions ruled by a world government and super states like the European Union. Countries must be bankrupted, too, to this end and it’s being achieved by the trillions in ‘rescue packages’ and furlough payments, trillions in lost taxation, and money-no-object spending on ‘Covid’ including constant allmedium advertising (programming) which has made the media dependent on government for much of its income. The day of reckoning is coming – as planned – for government spending and given that it has been made possible by printing money and not by production/taxation there is inflation on the way that has the

potential to wipe out monetary value. In that case there will be no need for the Cult to steal your money. It just won’t be worth anything (see the German Weimar Republic before the Nazis took over). Many have been okay with lockdowns while ge ing a percentage of their income from so-called furlough payments without having to work. Those payments are dependent, however, on people having at least a theoretical job with a business considered non-essential and ordered to close. As these business go under because they are closed by lockdown a er lockdown the furlough stops and it will for everyone eventually. Then what? The ‘then what?’ is precisely the idea.

Figure 4: The Hunger Games Society structure I have long warned was planned and now the ‘Covid’ hoax has made it possible. This is the real reason for lockdowns.

Hired hands

Between the Hunger Games Cult elite and the dependent population is planned to be a vicious military-police state (a fusion of the two into one force). This has been in the making for a long time with police looking ever more like the military and carrying weapons to match. The pandemic scam has seen this process accelerate so fast as

lockdown house arrest is brutally enforced by carefully recruited fascist minds and gormless system-servers. The police and military are planned to merge into a centrally-directed world army in a global structure headed by a world government which wouldn’t be elected even by the election fixes now in place. The world army is not planned even to be human and instead wars would be fought, primarily against the population, using robot technology controlled by artificial intelligence. I have been warning about this for decades and now militaries around the world are being transformed by this very AI technology. The global regime that I describe is a particular form of fascism known as a technocracy in which decisions are not made by clueless and co-opted politicians but by unelected technocrats – scientists, engineers, technologists and bureaucrats. Cult-owned-and-controlled Silicon Valley giants are examples of technocracy and they already have far more power to direct world events than governments. They are with their censorship selecting governments. I know that some are calling the ‘Great Reset’ a Marxist communist takeover, but fascism and Marxism are different labels for the same tyranny. Tell those who lived in fascist Germany and Stalinist Russia that there was a difference in the way their freedom was deleted and their lives controlled. I could call it a fascist technocracy or a Marxist technocracy and they would be equally accurate. The Hunger Games society with its world government structure would oversee a world army, world central bank and single world cashless currency imposing its will on a microchipped population (Fig 5). Scan its different elements and see how the illusory pandemic is forcing society in this very direction at great speed. Leaders of 23 countries and the World Health Organization (WHO) backed the idea in March, 2021, of a global treaty for ‘international cooperation’ in ‘health emergencies’ and nations should ‘come together as a global community for peaceful cooperation that extends beyond this crisis’. Cut the Orwellian bullshit and this means another step towards global government. The plan includes a cashless digital money system that I first warned about in 1993. Right at the start of ‘Covid’ the deeply corrupt Tedros

Adhanom Ghebreyesus, the crooked and merely gofer ‘head’ of the World Health Organization, said it was possible to catch the ‘virus’ by touching cash and it was be er to use cashless means. The claim was ridiculous nonsense and like the whole ‘Covid’ mind-trick it was nothing to do with ‘health’ and everything to do with pushing every aspect of the Cult agenda. As a result of the Tedros lie the use of cash has plummeted. The Cult script involves a single world digital currency that would eventually be technologically embedded in the body. China is a massive global centre for the Cult and if you watch what is happening there you will know what is planned for everywhere. The Chinese government is developing a digital currency which would allow fines to be deducted immediately via AI for anyone caught on camera breaking its fantastic list of laws and the money is going to be programmable with an expiry date to ensure that no one can accrue wealth except the Cult and its operatives.

Figure 5: The structure of global control the Cult has been working towards for so long and this has been enormously advanced by the ‘Covid’ illusion.

Serfdom is so smart

The Cult plan is far wider, extreme, and more comprehensive than even most conspiracy researchers appreciate and I will come to the true depths of deceit and control in the chapters ‘Who controls the

Cult?’ and ‘Escaping Wetiko’. Even the world that we know is crazy enough. We are being deluged with ever more sophisticated and controlling technology under the heading of ‘smart’. We have smart televisions, smart meters, smart cards, smart cars, smart driving, smart roads, smart pills, smart patches, smart watches, smart skin, smart borders, smart pavements, smart streets, smart cities, smart communities, smart environments, smart growth, smart planet ... smart everything around us. Smart technologies and methods of operation are designed to interlock to create a global Smart Grid connecting the entirety of human society including human minds to create a centrally-dictated ‘hive’ mind. ‘Smart cities’ is code for densely-occupied megacities of total surveillance and control through AI. Ever more destructive frequency communication systems like 5G have been rolled out without any official testing for health and psychological effects (colossal). 5G/6G/7G systems are needed to run the Smart Grid and each one becomes more destructive of body and mind. Deleting independent income is crucial to forcing people into these AI-policed prisons by ending private property ownership (except for the Cult elite). The Cult’s Great Reset now openly foresees a global society in which no one will own any possessions and everything will be rented while the Cult would own literally everything under the guise of government and corporations. The aim has been to use the lockdowns to destroy sources of income on a mass scale and when the people are destitute and in unrepayable amounts of debt (problem) Cult assets come forward with the pledge to write-off debt in return for handing over all property and possessions (solution). Everything – literally everything including people – would be connected to the Internet via AI. I was warning years ago about the coming Internet of Things (IoT) in which all devices and technology from your car to your fridge would be plugged into the Internet and controlled by AI. Now we are already there with much more to come. The next stage is the Internet of Everything (IoE) which is planned to include the connection of AI to the human brain and body to replace the human mind with a centrally-controlled AI mind. Instead of perceptions

being manipulated through control of information and censorship those perceptions would come direct from the Cult through AI. What do you think? You think whatever AI decides that you think. In human terms there would be no individual ‘think’ any longer. Too incredible? The ravings of a lunatic? Not at all. Cult-owned crazies in Silicon Valley have been telling us the plan for years without explaining the real motivation and calculated implications. These include Google executive and ‘futurist’ Ray Kurzweil who highlights the year 2030 for when this would be underway. He said: Our thinking ... will be a hybrid of biological and non-biological thinking ... humans will be able to extend their limitations and ‘think in the cloud’ ... We’re going to put gateways to the cloud in our brains ... We’re going to gradually merge and enhance ourselves ... In my view, that’s the nature of being human – we transcend our limitations. As the technology becomes vastly superior to what we are then the small proportion that is still human gets smaller and smaller and smaller until it’s just utterly negligible.

The sales-pitch of Kurzweil and Cult-owned Silicon Valley is that this would make us ‘super-human’ when the real aim is to make us post-human and no longer ‘human’ in the sense that we have come to know. The entire global population would be connected to AI and become the centrally-controlled ‘hive-mind’ of externally-delivered perceptions. The Smart Grid being installed to impose the Cult’s will on the world is being constructed to allow particular locations – even one location – to control the whole global system. From these prime control centres, which absolutely include China and Israel, anything connected to the Internet would be switched on or off and manipulated at will. Energy systems could be cut, communication via the Internet taken down, computer-controlled driverless autonomous vehicles driven off the road, medical devices switched off, the potential is limitless given how much AI and Internet connections now run human society. We have seen nothing yet if we allow this to continue. Autonomous vehicle makers are working with law enforcement to produce cars designed to automatically pull over if they detect a police or emergency vehicle flashing from up to 100 feet away. At a police stop the car would be unlocked and the

window rolled down automatically. Vehicles would only take you where the computer (the state) allowed. The end of petrol vehicles and speed limiters on all new cars in the UK and EU from 2022 are steps leading to electric computerised transport over which ultimately you have no control. The picture is far bigger even than the Cult global network or web and that will become clear when I get to the nature of the ‘spider’. There is a connection between all these happenings and the instigation of DNA-manipulating ‘vaccines’ (which aren’t ‘vaccines’) justified by the ‘Covid’ hoax. That connection is the unfolding plan to transform the human body from a biological to a synthetic biological state and this is why synthetic biology is such a fast-emerging discipline of mainstream science. ‘Covid vaccines’ are infusing self-replicating synthetic genetic material into the cells to cumulatively take us on the Totalitarian Tiptoe from Human 1.0 to the synthetic biological Human 2.0 which will be physically and perceptually a ached to the Smart Grid to one hundred percent control every thought, perception and deed. Humanity needs to wake up and fast. This is the barest explanation of where the ‘outcome’ is planned to go but it’s enough to see the journey happening all around us. Those new to this information will already see ‘Covid’ in a whole new context. I will add much more detail as we go along, but for the minutiae evidence see my mega-works, The Answer, The Trigger and Everything You Need to Know But Have Never Been Told. Now – how does a Renegade Mind see the ‘world’?

CHAPTER TWO Renegade Perception It is one thing to be clever and another to be wise George R.R. Martin

A

simple definition of the difference between a programmed mind and a Renegade Mind would be that one sees only dots while the other connects them to see the picture. Reading reality with accuracy requires the observer to (a) know the planned outcome and (b) realise that everything, but everything, is connected. The entirety of infinite reality is connected – that’s its very nature – and with human society an expression of infinite reality the same must apply. Simple cause and effect is a connection. The effect is triggered by the cause and the effect then becomes the cause of another effect. Nothing happens in isolation because it can’t. Life in whatever reality is simple choice and consequence. We make choices and these lead to consequences. If we don’t like the consequences we can make different choices and get different consequences which lead to other choices and consequences. The choice and the consequence are not only connected they are indivisible. You can’t have one without the other as an old song goes. A few cannot control the world unless those being controlled allow that to happen – cause and effect, choice and consequence. Control – who has it and who doesn’t – is a two-way process, a symbiotic relationship, involving the controller and controlled. ‘They took my freedom away!!’ Well, yes, but you also gave it to them. Humanity is

subjected to mass control because humanity has acquiesced to that control. This is all cause and effect and literally a case of give and take. In the same way world events of every kind are connected and the Cult works incessantly to sell the illusion of the random and coincidental to maintain the essential (to them) perception of dots that hide the picture. Renegade Minds know this and constantly scan the world for pa erns of connection. This is absolutely pivotal in understanding the happenings in the world and without that perspective clarity is impossible. First you know the planned outcome and then you identify the steps on the journey – the day-byday apparently random which, when connected in relation to the outcome, no longer appear as individual events, but as the proverbial chain of events leading in the same direction. I’ll give you some examples:

Political puppet show

We are told to believe that politics is ‘adversarial’ in that different parties with different beliefs engage in an endless tussle for power. There may have been some truth in that up to a point – and only a point – but today divisions between ‘different’ parties are rhetorical not ideological. Even the rhetorical is fusing into one-speak as the parties eject any remaining free thinkers while others succumb to the ever-gathering intimidation of anyone with the ‘wrong’ opinion. The Cult is not a new phenomenon and can be traced back thousands of years as my books have documented. Its intergenerational initiates have been manipulating events with increasing effect the more that global power has been centralised. In ancient times the Cult secured control through the system of monarchy in which ‘special’ bloodlines (of which more later) demanded the right to rule as kings and queens simply by birthright and by vanquishing others who claimed the same birthright. There came a time, however, when people had matured enough to see the unfairness of such tyranny and demanded a say in who governed them. Note the word – governed them. Not served them – governed them, hence government defined as ‘the political direction and control exercised over the

actions of the members, citizens, or inhabitants of communities, societies, and states; direction of the affairs of a state, community, etc.’ Governments exercise control over rather than serve just like the monarchies before them. Bizarrely there are still countries like the United Kingdom which are ruled by a monarch and a government that officially answers to the monarch. The UK head of state and that of Commonwealth countries such as Canada, Australia and New Zealand is ‘selected’ by who in a single family had unprotected sex with whom and in what order. Pinch me it can’t be true. Ouch! Shit, it is. The demise of monarchies in most countries offered a potential vacuum in which some form of free and fair society could arise and the Cult had that base covered. Monarchies had served its interests but they couldn’t continue in the face of such widespread opposition and, anyway, replacing a ‘royal’ dictatorship that people could see with a dictatorship ‘of the people’ hiding behind the concept of ‘democracy’ presented far greater manipulative possibilities and ways of hiding coordinated tyranny behind the illusion of ‘freedom’. Democracy is quite wrongly defined as government selected by the population. This is not the case at all. It is government selected by some of the population (and then only in theory). This ‘some’ doesn’t even have to be the majority as we have seen so o en in firstpast-the-post elections in which the so-called majority party wins fewer votes than the ‘losing’ parties combined. Democracy can give total power to a party in government from a minority of the votes cast. It’s a sleight of hand to sell tyranny as freedom. Seventy-four million Trump-supporting Americans didn’t vote for the ‘Democratic’ Party of Joe Biden in the distinctly dodgy election in 2020 and yet far from acknowledging the wishes and feelings of that great percentage of American society the Cult-owned Biden government set out from day one to destroy them and their right to a voice and opinion. Empty shell Biden and his Cult handlers said they were doing this to ‘protect democracy’. Such is the level of lunacy and sickness to which politics has descended. Connect the dots and relate them to the desired outcome – a world government run by self-appointed technocrats and no longer even elected

politicians. While operating through its political agents in government the Cult is at the same time encouraging public distain for politicians by pu ing idiots and incompetents in theoretical power on the road to deleting them. The idea is to instil a public reaction that says of the technocrats: ‘Well, they couldn’t do any worse than the pathetic politicians.’ It’s all about controlling perception and Renegade Minds can see through that while programmed minds cannot when they are ignorant of both the planned outcome and the manipulation techniques employed to secure that end. This knowledge can be learned, however, and fast if people choose to get informed. Politics may at first sight appear very difficult to control from a central point. I mean look at the ‘different’ parties and how would you be able to oversee them all and their constituent parts? In truth, it’s very straightforward because of their structure. We are back to the pyramid of imposition and acquiescence. Organisations are structured in the same way as the system as a whole. Political parties are not open forums of free expression. They are hierarchies. I was a national spokesman for the British Green Party which claimed to be a different kind of politics in which influence and power was devolved; but I can tell you from direct experience – and it’s far worse now – that Green parties are run as hierarchies like all the others however much they may try to hide that fact or kid themselves that it’s not true. A very few at the top of all political parties are directing policy and personnel. They decide if you are elevated in the party or serve as a government minister and to do that you have to be a yes man or woman. Look at all the maverick political thinkers who never ascended the greasy pole. If you want to progress within the party or reach ‘high-office’ you need to fall into line and conform. Exceptions to this are rare indeed. Should you want to run for parliament or Congress you have to persuade the local or state level of the party to select you and for that you need to play the game as dictated by the hierarchy. If you secure election and wish to progress within the greater structure you need to go on conforming to what is acceptable to those running the hierarchy

from the peak of the pyramid. Political parties are perceptual gulags and the very fact that there are party ‘Whips’ appointed to ‘whip’ politicians into voting the way the hierarchy demands exposes the ridiculous idea that politicians are elected to serve the people they are supposed to represent. Cult operatives and manipulation has long seized control of major parties that have any chance of forming a government and at least most of those that haven’t. A new party forms and the Cult goes to work to infiltrate and direct. This has reached such a level today that you see video compilations of ‘leaders’ of all parties whether Democrats, Republicans, Conservative, Labour and Green parroting the same Cult mantra of ‘Build Back Be er’ and the ‘Great Reset’ which are straight off the Cult song-sheet to describe the transformation of global society in response to the Cult-instigated hoaxes of the ‘Covid pandemic’ and human-caused ‘climate change’. To see Caroline Lucas, the Green Party MP that I knew when I was in the party in the 1980s, speaking in support of plans proposed by Cult operative Klaus Schwab representing the billionaire global elite is a real head-shaker.

Many parties – one master

The party system is another mind-trick and was instigated to change the nature of the dictatorship by swapping ‘royalty’ for dark suits that people believed – though now ever less so – represented their interests. Understanding this trick is to realise that a single force (the Cult) controls all parties either directly in terms of the major ones or through manipulation of perception and ideology with others. You don’t need to manipulate Green parties to demand your transformation of society in the name of ‘climate change’ when they are obsessed with the lie that this is essential to ‘save the planet’. You just give them a platform and away they go serving your interests while believing they are being environmentally virtuous. America’s political structure is a perfect blueprint for how the two or multiparty system is really a one-party state. The Republican Party is controlled from one step back in the shadows by a group made up of billionaires and their gofers known as neoconservatives or Neocons.

I have exposed them in fine detail in my books and they were the driving force behind the policies of the imbecilic presidency of Boy George Bush which included 9/11 (see The Trigger for a comprehensive demolition of the official story), the subsequent ‘war on terror’ (war of terror) and the invasions of Afghanistan and Iraq. The la er was a No-Problem-Reaction-Solution based on claims by Cult operatives, including Bush and British Prime Minister Tony Blair, about Saddam Hussein’s ‘weapons of mass destruction’ which did not exist as war criminals Bush and Blair well knew.

Figure 6: Different front people, different parties – same control system.

The Democratic Party has its own ‘Neocon’ group controlling from the background which I call the ‘Democons’ and here’s the penny-drop – the Neocons and Democons answer to the same masters one step further back into the shadows (Fig 6). At that level of the Cult the Republican and Democrat parties are controlled by the same people and no ma er which is in power the Cult is in power. This is how it works in almost every country and certainly in Britain with Conservative, Labour, Liberal Democrat and Green parties now all on the same page whatever the rhetoric may be in their feeble a empts to appear different. Neocons operated at the time of Bush through a think tank called The Project for the New American Century which in September, 2000, published a document entitled Rebuilding America’s Defenses: Strategies, Forces, and Resources

For a New Century demanding that America fight ‘multiple, simultaneous major theatre wars’ as a ‘core mission’ to force regimechange in countries including Iraq, Libya and Syria. Neocons arranged for Bush (‘Republican’) and Blair (‘Labour Party’) to frontup the invasion of Iraq and when they departed the Democons orchestrated the targeting of Libya and Syria through Barack Obama (‘Democrat’) and British Prime Minister David Cameron (‘Conservative Party’). We have ‘different’ parties and ‘different’ people, but the same unfolding script. The more the Cult has seized the reigns of parties and personnel the more their policies have transparently pursued the same agenda to the point where the fascist ‘Covid’ impositions of the Conservative junta of Jackboot Johnson in Britain were opposed by the Labour Party because they were not fascist enough. The Labour Party is likened to the US Democrats while the Conservative Party is akin to a British version of the Republicans and on both sides of the Atlantic they all speak the same language and support the direction demanded by the Cult although some more enthusiastically than others. It’s a similar story in country a er country because it’s all centrally controlled. Oh, but what about Trump? I’ll come to him shortly. Political ‘choice’ in the ‘party’ system goes like this: You vote for Party A and they get into government. You don’t like what they do so next time you vote for Party B and they get into government. You don’t like what they do when it’s pre y much the same as Party A and why wouldn’t that be with both controlled by the same force? Given that only two, sometimes three, parties have any chance of forming a government to get rid of Party B that you don’t like you have to vote again for Party A which … you don’t like. This, ladies and gentlemen, is what they call ‘democracy’ which we are told – wrongly – is a term interchangeable with ‘freedom’.

The cult of cults

At this point I need to introduce a major expression of the Global Cult known as Sabbatian-Frankism. Sabbatian is also spelt as Sabbatean. I will summarise here. I have published major exposés

and detailed background in other works. Sabbatian-Frankism combines the names of two frauds posing as ‘Jewish’ men, Sabbatai Zevi (1626-1676), a rabbi, black magician and occultist who proclaimed he was the Jewish messiah; and Jacob Frank (1726-1791), the Polish ‘Jew’, black magician and occultist who said he was the reincarnation of ‘messiah’ Zevi and biblical patriarch Jacob. They worked across two centuries to establish the Sabbatian-Frankist cult that plays a major, indeed central, role in the manipulation of human society by the Global Cult which has its origins much further back in history than Sabbatai Zevi. I should emphasise two points here in response to the shrill voices that will scream ‘anti-Semitism’: (1) Sabbatian-Frankists are NOT Jewish and only pose as such to hide their cult behind a Jewish façade; and (2) my information about this cult has come from Jewish sources who have long realised that their society and community has been infiltrated and taken over by interloper Sabbatian-Frankists. Infiltration has been the foundation technique of Sabbatian-Frankism from its official origin in the 17th century. Zevi’s Sabbatian sect a racted a massive following described as the biggest messianic movement in Jewish history, spreading as far as Africa and Asia, and he promised a return for the Jews to the ‘Promised Land’ of Israel. Sabbatianism was not Judaism but an inversion of everything that mainstream Judaism stood for. So much so that this sinister cult would have a feast day when Judaism had a fast day and whatever was forbidden in Judaism the Sabbatians were encouraged and even commanded to do. This included incest and what would be today called Satanism. Members were forbidden to marry outside the sect and there was a system of keeping their children ignorant of what they were part of until they were old enough to be trusted not to unknowingly reveal anything to outsiders. The same system is employed to this day by the Global Cult in general which Sabbatian-Frankism has enormously influenced and now largely controls. Zevi and his Sabbatians suffered a setback with the intervention by the Sultan of the Islamic O oman Empire in the Middle East and what is now the Republic of Turkey where Zevi was located. The

Sultan gave him the choice of proving his ‘divinity’, converting to Islam or facing torture and death. Funnily enough Zevi chose to convert or at least appear to. Some of his supporters were disillusioned and dri ed away, but many did not with 300 families also converting – only in theory – to Islam. They continued behind this Islamic smokescreen to follow the goals, rules and rituals of Sabbatianism and became known as ‘crypto-Jews’ or the ‘Dönmeh’ which means ‘to turn’. This is rather ironic because they didn’t ‘turn’ and instead hid behind a fake Islamic persona. The process of appearing to be one thing while being very much another would become the calling card of Sabbatianism especially a er Zevi’s death and the arrival of the Satanist Jacob Frank in the 18th century when the cult became Sabbatian-Frankism and plumbed still new depths of depravity and infiltration which included – still includes – human sacrifice and sex with children. Wherever Sabbatians go paedophilia and Satanism follow and is it really a surprise that Hollywood is so infested with child abuse and Satanism when it was established by Sabbatian-Frankists and is still controlled by them? Hollywood has been one of the prime vehicles for global perceptual programming and manipulation. How many believe the version of ‘history’ portrayed in movies when it is a travesty and inversion (again) of the truth? Rabbi Marvin Antelman describes Frankism in his book, To Eliminate the Opiate, as ‘a movement of complete evil’ while Jewish professor Gershom Scholem said of Frank in The Messianic Idea in Judaism: ‘In all his actions [he was] a truly corrupt and degenerate individual ... one of the most frightening phenomena in the whole of Jewish history.’ Frank was excommunicated by traditional rabbis, as was Zevi, but Frank was undeterred and enjoyed vital support from the House of Rothschild, the infamous banking dynasty whose inner-core are Sabbatian-Frankists and not Jews. Infiltration of the Roman Church and Vatican was instigated by Frank with many Dönmeh ‘turning’ again to convert to Roman Catholicism with a view to hijacking the reins of power. This was the ever-repeating modus operandi and continues to be so. Pose as an advocate of the religion, culture or country that you want to control and then

manipulate your people into the positions of authority and influence largely as advisers, administrators and Svengalis for those that appear to be in power. They did this with Judaism, Christianity (Christian Zionism is part of this), Islam and other religions and nations until Sabbatian-Frankism spanned the world as it does today.

Sabbatian Saudis and the terror network

One expression of the Sabbatian-Frankist Dönmeh within Islam is the ruling family of Saudi Arabia, the House of Saud, through which came the vile distortion of Islam known as Wahhabism. This is the violent creed followed by terrorist groups like Al-Qaeda and ISIS or Islamic State. Wahhabism is the hand-chopping, head-chopping ‘religion’ of Saudi Arabia which is used to keep the people in a constant state of fear so the interloper House of Saud can continue to rule. Al-Qaeda and Islamic State were lavishly funded by the House of Saud while being created and directed by the Sabbatian-Frankist network in the United States that operates through the Pentagon, CIA and the government in general of whichever ‘party’. The front man for the establishment of Wahhabism in the middle of the 18th century was a Sabbatian-Frankist ‘crypto-Jew’ posing as Islamic called Muhammad ibn Abd al-Wahhab. His daughter would marry the son of Muhammad bin Saud who established the first Saudi state before his death in 1765 with support from the British Empire. Bin Saud’s successors would establish modern Saudi Arabia in league with the British and Americans in 1932 which allowed them to seize control of Islam’s major shrines in Mecca and Medina. They have dictated the direction of Sunni Islam ever since while Iran is the major centre of the Shiite version and here we have the source of at least the public conflict between them. The Sabbatian network has used its Wahhabi extremists to carry out Problem-Reaction-Solution terrorist a acks in the name of ‘Al-Qaeda’ and ‘Islamic State’ to justify a devastating ‘war on terror’, ever-increasing surveillance of the population and to terrify people into compliance. Another insight of the Renegade Mind is the streetwise understanding that

just because a country, location or people are a acked doesn’t mean that those apparently representing that country, location or people are not behind the a ackers. O en they are orchestrating the a acks because of the societal changes that can be then justified in the name of ‘saving the population from terrorists’. I show in great detail in The Trigger how Sabbatian-Frankists were the real perpetrators of 9/11 and not ‘19 Arab hijackers’ who were blamed for what happened. Observe what was justified in the name of 9/11 alone in terms of Middle East invasions, mass surveillance and control that fulfilled the demands of the Project for the New American Century document published by the Sabbatian Neocons. What appear to be enemies are on the deep inside players on the same Sabbatian team. Israel and Arab ‘royal’ dictatorships are all ruled by Sabbatians and the recent peace agreements between Israel and Saudi Arabia, the United Arab Emirates (UAE) and others are only making formal what has always been the case behind the scenes. Palestinians who have been subjected to grotesque tyranny since Israel was bombed and terrorised into existence in 1948 have never stood a chance. Sabbatian-Frankists have controlled Israel (so the constant theme of violence and war which Sabbatians love) and they have controlled the Arab countries that Palestinians have looked to for real support that never comes. ‘Royal families’ of the Arab world in Saudi Arabia, Bahrain, UAE, etc., are all Sabbatians with allegiance to the aims of the cult and not what is best for their Arabic populations. They have stolen the oil and financial resources from their people by false claims to be ‘royal dynasties’ with a genetic right to rule and by employing vicious militaries to impose their will.

Satanic ‘illumination’

The Satanist Jacob Frank formed an alliance in 1773 with two other Sabbatians, Mayer Amschel Rothschild (1744-1812), founder of the Rothschild banking dynasty, and Jesuit-educated fraudulent Jew, Adam Weishaupt, and this led to the formation of the Bavarian Illuminati, firstly under another name, in 1776. The Illuminati would

be the manipulating force behind the French Revolution (1789-1799) and was also involved in the American Revolution (1775-1783) before and a er the Illuminati’s official creation. Weishaupt would later become (in public) a Protestant Christian in archetypal Sabbatian style. I read that his name can be decoded as Adam-Weishaupt or ‘the first man to lead those who know’. He wasn’t a leader in the sense that he was a subordinate, but he did lead those below him in a crusade of transforming human society that still continues today. The theme was confirmed as early as 1785 when a horseman courier called Lanz was reported to be struck by lighting and extensive Illuminati documents were found in his saddlebags. They made the link to Weishaupt and detailed the plan for world takeover. Current events with ‘Covid’ fascism have been in the making for a very long time. Jacob Frank was jailed for 13 years by the Catholic Inquisition a er his arrest in 1760 and on his release he headed for Frankfurt, Germany, home city and headquarters of the House of Rothschild where the alliance was struck with Mayer Amschel Rothschild and Weishaupt. Rothschild arranged for Frank to be given the title of Baron and he became a wealthy nobleman with a big following of Jews in Germany, the Austro-Hungarian Empire and other European countries. Most of them would have believed he was on their side. The name ‘Illuminati’ came from the Zohar which is a body of works in the Jewish mystical ‘bible’ called the Kabbalah. ‘Zohar’ is the foundation of Sabbatian-Frankist belief and in Hebrew ‘Zohar’ means ‘splendour’, ‘radiance’, ‘illuminated’, and so we have ‘Illuminati’. They claim to be the ‘Illuminated Ones’ from their knowledge systematically hidden from the human population and passed on through generations of carefully-chosen initiates in the global secret society network or Cult. Hidden knowledge includes an awareness of the Cult agenda for the world and the nature of our collective reality that I will explore later. Cult ‘illumination’ is symbolised by the torch held by the Statue of Liberty which was gi ed to New York by French Freemasons in Paris who knew exactly what it represents. ‘Liberty’ symbolises the goddess worshipped in

Babylon as Queen Semiramis or Ishtar. The significance of this will become clear. Notice again the ubiquitous theme of inversion with the Statue of ‘Liberty’ really symbolising mass control (Fig 7). A mirror-image statute stands on an island in the River Seine in Paris from where New York Liberty originated (Fig 8). A large replica of the Liberty flame stands on top of the Pont de l’Alma tunnel in Paris where Princess Diana died in a Cult ritual described in The Biggest Secret. Lucifer ‘the light bringer’ is related to all this (and much more as we’ll see) and ‘Lucifer’ is a central figure in Sabbatian-Frankism and its associated Satanism. Sabbatians reject the Jewish Torah, or Pentateuch, the ‘five books of Moses’ in the Old Testament known as Genesis, Exodus, Leviticus, Numbers, and Deuteronomy which are claimed by Judaism and Christianity to have been dictated by ‘God’ to Moses on Mount Sinai. Sabbatians say these do not apply to them and they seek to replace them with the Zohar to absorb Judaism and its followers into their inversion which is an expression of a much greater global inversion. They want to delete all religions and force humanity to worship a one-world religion – Sabbatian Satanism that also includes worship of the Earth goddess. Satanic themes are being more and more introduced into mainstream society and while Christianity is currently the foremost target for destruction the others are planned to follow.

Figure 7: The Cult goddess of Babylon disguised as the Statue of Liberty holding the flame of Lucifer the ‘light bringer’.

Figure 8: Liberty’s mirror image in Paris where the New York version originated.

Marx brothers

Rabbi Marvin Antelman connects the Illuminati to the Jacobins in To Eliminate the Opiate and Jacobins were the force behind the French Revolution. He links both to the Bund der Gerechten, or League of the Just, which was the network that inflicted communism/Marxism on the world. Antelman wrote: The original inner circle of the Bund der Gerechten consisted of born Catholics, Protestants and Jews [Sabbatian-Frankist infiltrators], and those representatives of respective subdivisions formulated schemes for the ultimate destruction of their faiths. The heretical Catholics laid plans which they felt would take a century or more for the ultimate destruction of the church; the apostate Jews for the ultimate destruction of the Jewish religion.

Sabbatian-created communism connects into this anti-religion agenda in that communism does not allow for the free practice of religion. The Sabbatian ‘Bund’ became the International Communist Party and Communist League and in 1848 ‘Marxism’ was born with the Communist Manifesto of Sabbatian assets Karl Marx and Friedrich Engels. It is absolutely no coincidence that Marxism, just a different name for fascist and other centrally-controlled tyrannies, is being imposed worldwide as a result of the ‘Covid’ hoax and nor that Marxist/fascist China was the place where the hoax originated. The reason for this will become very clear in the chapter ‘Covid: The calculated catastrophe’. The so-called ‘Woke’ mentality has hijacked

traditional beliefs of the political le and replaced them with farright make-believe ‘social justice’ be er known as Marxism. Woke will, however, be swallowed by its own perceived ‘revolution’ which is really the work of billionaires and billionaire corporations feigning being ‘Woke’. Marxism is being touted by Wokers as a replacement for ‘capitalism’ when we don’t have ‘capitalism’. We have cartelism in which the market is stitched up by the very Cult billionaires and corporations bankrolling Woke. Billionaires love Marxism which keeps the people in servitude while they control from the top. Terminally naïve Wokers think they are ‘changing the world’ when it’s the Cult that is doing the changing and when they have played their vital part and become surplus to requirements they, too, will be targeted. The Illuminati-Jacobins were behind the period known as ‘The Terror’ in the French Revolution in 1793 and 1794 when Jacobin Maximillian de Robespierre and his Orwellian ‘Commi ee of Public Safety’ killed 17,000 ‘enemies of the Revolution’ who had once been ‘friends of the Revolution’. Karl Marx (1818-1883), whose Sabbatian creed of Marxism has cost the lives of at least 100 million people, is a hero once again to Wokers who have been systematically kept ignorant of real history by their ‘education’ programming. As a result they now promote a Sabbatian ‘Marxist’ abomination destined at some point to consume them. Rabbi Antelman, who spent decades researching the Sabbatian plot, said of the League of the Just and Karl Marx: Contrary to popular opinion Karl Marx did not originate the Communist Manifesto. He was paid for his services by the League of the Just, which was known in its country of origin, Germany, as the Bund der Geaechteten.

Antelman said the text a ributed to Marx was the work of other people and Marx ‘was only repeating what others already said’. Marx was ‘a hired hack – lackey of the wealthy Illuminists’. Marx famously said that religion was the ‘opium of the people’ (part of the Sabbatian plan to demonise religion) and Antelman called his books, To Eliminate the Opiate. Marx was born Jewish, but his family converted to Christianity (Sabbatian modus operandi) and he

a acked Jews, not least in his book, A World Without Jews. In doing so he supported the Sabbatian plan to destroy traditional Jewishness and Judaism which we are clearly seeing today with the vindictive targeting of orthodox Jews by the Sabbatian government of Israel over ‘Covid’ laws. I don’t follow any religion and it has done much damage to the world over centuries and acted as a perceptual straightjacket. Renegade Minds, however, are always asking why something is being done. It doesn’t ma er if they agree or disagree with what is happening – why is it happening is the question. The ‘why?’ can be answered with regard to religion in that religions create interacting communities of believers when the Cult wants to dismantle all discourse, unity and interaction (see ‘Covid’ lockdowns) and the ultimate goal is to delete all religions for a oneworld religion of Cult Satanism worshipping their ‘god’ of which more later. We see the same ‘why?’ with gun control in America. I don’t have guns and don’t want them, but why is the Cult seeking to disarm the population at the same time that law enforcement agencies are armed to their molars and why has every tyrant in history sought to disarm people before launching the final takeover? They include Hitler, Stalin, Pol Pot and Mao who followed confiscation with violent seizing of power. You know it’s a Cult agenda by the people who immediately race to the microphones to exploit dead people in multiple shootings. Ultra-Zionist Cult lackey Senator Chuck Schumer was straight on the case a er ten people were killed in Boulder, Colorado in March, 2121. Simple rule … if Schumer wants it the Cult wants it and the same with his ultraZionist mate the wild-eyed Senator Adam Schiff. At the same time they were calling for the disarmament of Americans, many of whom live a long way from a police response, Schumer, Schiff and the rest of these pampered clowns were si ing on Capitol Hill behind a razor-wired security fence protected by thousands of armed troops in addition to their own armed bodyguards. Mom and pop in an isolated home? They’re just potential mass shooters.

Zion Mainframe

Sabbatian-Frankists and most importantly the Rothschilds were behind the creation of ‘Zionism’, a political movement that demanded a Jewish homeland in Israel as promised by Sabbatai Zevi. The very symbol of Israel comes from the German meaning of the name Rothschild. Dynasty founder Mayer Amschel Rothschild changed the family name from Bauer to Rothschild, or ‘Red-Shield’ in German, in deference to the six-pointed ‘Star of David’ hexagram displayed on the family’s home in Frankfurt. The symbol later appeared on the flag of Israel a er the Rothschilds were centrally involved in its creation. Hexagrams are not a uniquely Jewish symbol and are widely used in occult (‘hidden’) networks o en as a symbol for Saturn (see my other books for why). Neither are Zionism and Jewishness interchangeable. Zionism is a political movement and philosophy and not a ‘race’ or a people. Many Jews oppose Zionism and many non-Jews, including US President Joe Biden, call themselves Zionists as does Israel-centric Donald Trump. America’s support for the Israel government is pre y much a gimme with ultra-Zionist billionaires and corporations providing fantastic and dominant funding for both political parties. Former Congresswoman Cynthia McKinney has told how she was approached immediately she ran for office to ‘sign the pledge’ to Israel and confirm that she would always vote in that country’s best interests. All American politicians are approached in this way. Anyone who refuses will get no support or funding from the enormous and all-powerful Zionist lobby that includes organisations like mega-lobby group AIPAC, the American Israel Public Affairs Commi ee. Trump’s biggest funder was ultra-Zionist casino and media billionaire Sheldon Adelson while major funders of the Democratic Party include ultra-Zionist George Soros and ultraZionist financial and media mogul, Haim Saban. Some may reel back at the suggestion that Soros is an Israel-firster (Sabbatian-controlled Israel-firster), but Renegade Minds watch the actions not the words and everywhere Soros donates his billions the Sabbatian agenda benefits. In the spirit of Sabbatian inversion Soros pledged $1 billion for a new university network to promote ‘liberal values and tackle intolerance’. He made the announcement during his annual speech

at the Cult-owned World Economic Forum in Davos, Switzerland, in January, 2020, a er his ‘harsh criticism’ of ‘authoritarian rulers’ around the world. You can only laugh at such brazen mendacity. How he doesn’t laugh is the mystery. Translated from the Orwellian ‘liberal values and tackle intolerance’ means teaching non-white people to hate white people and for white people to loathe themselves for being born white. The reason for that will become clear.

The ‘Anti-Semitism’ fraud

Zionists support the Jewish homeland in the land of Palestine which has been the Sabbatian-Rothschild goal for so long, but not for the benefit of Jews. Sabbatians and their global Anti-Semitism Industry have skewed public and political opinion to equate opposing the violent extremes of Zionism to be a blanket a ack and condemnation of all Jewish people. Sabbatians and their global Anti-Semitism Industry have skewed public and political opinion to equate opposing the violent extremes of Zionism to be a blanket a ack and condemnation of all Jewish people. This is nothing more than a Sabbatian protection racket to stop legitimate investigation and exposure of their agendas and activities. The official definition of ‘anti-Semitism’ has more recently been expanded to include criticism of Zionism – a political movement – and this was done to further stop exposure of Sabbatian infiltrators who created Zionism as we know it today in the 19th century. Renegade Minds will talk about these subjects when they know the shit that will come their way. People must decide if they want to know the truth or just cower in the corner in fear of what others will say. Sabbatians have been trying to label me as ‘anti-Semitic’ since the 1990s as I have uncovered more and more about their background and agendas. Useless, gutless, fraudulent ‘journalists’ then just repeat the smears without question and on the day I was writing this section a pair of unquestioning repeaters called Ben Quinn and Archie Bland (how appropriate) outright called me an ‘anti-Semite’ in the establishment propaganda sheet, the London Guardian, with no supporting evidence. The

Sabbatian Anti-Semitism Industry said so and who are they to question that? They wouldn’t dare. Ironically ‘Semitic’ refers to a group of languages in the Middle East that are almost entirely Arabic. ‘Anti-Semitism’ becomes ‘anti-Arab’ which if the consequences of this misunderstanding were not so grave would be hilarious. Don’t bother telling Quinn and Bland. I don’t want to confuse them, bless ‘em. One reason I am dubbed ‘anti-Semitic’ is that I wrote in the 1990s that Jewish operatives (Sabbatians) were heavily involved in the Russian Revolution when Sabbatians overthrew the Romanov dynasty. This apparently made me ‘antiSemitic’. Oh, really? Here is a section from The Trigger: British journalist Robert Wilton confirmed these themes in his 1920 book The Last Days of the Romanovs when he studied official documents from the Russian government to identify the members of the Bolshevik ruling elite between 1917 and 1919. The Central Committee included 41 Jews among 62 members; the Council of the People’s Commissars had 17 Jews out of 22 members; and 458 of the 556 most important Bolshevik positions between 1918 and 1919 were occupied by Jewish people. Only 17 were Russian. Then there were the 23 Jews among the 36 members of the vicious Cheka Soviet secret police established in 1917 who would soon appear all across the country. Professor Robert Service of Oxford University, an expert on 20th century Russian history, found evidence that [‘Jewish’] Leon Trotsky had sought to make sure that Jews were enrolled in the Red Army and were disproportionately represented in the Soviet civil bureaucracy that included the Cheka which performed mass arrests, imprisonment and executions of ‘enemies of the people’. A US State Department Decimal File (861.00/5339) dated November 13th, 1918, names [Rothschild banking agent in America] Jacob Schiff and a list of ultra-Zionists as funders of the Russian Revolution leading to claims of a ‘Jewish plot’, but the key point missed by all is they were not ‘Jews’ – they were Sabbatian-Frankists.

Britain’s Winston Churchill made the same error by mistake or otherwise. He wrote in a 1920 edition of the Illustrated Sunday Herald that those behind the Russian revolution were part of a ‘worldwide conspiracy for the overthrow of civilisation and for the reconstitution of society on the basis of arrested development, of envious malevolence, and impossible equality’ (see ‘Woke’ today because that has been created by the same network). Churchill said there was no need to exaggerate the part played in the creation of Bolshevism and in the actual bringing about of the Russian

Revolution ‘by these international and for the most part atheistical Jews’ [‘atheistical Jews’ = Sabbatians]. Churchill said it is certainly a very great one and probably outweighs all others: ‘With the notable exception of Lenin, the majority of the leading figures are Jews.’ He went on to describe, knowingly or not, the Sabbatian modus operandi of placing puppet leaders nominally in power while they control from the background: Moreover, the principal inspiration and driving power comes from the Jewish leaders. Thus Tchitcherin, a pure Russian, is eclipsed by his nominal subordinate, Litvinoff, and the influence of Russians like Bukharin or Lunacharski cannot be compared with the power of Trotsky, or of Zinovieff, the Dictator of the Red Citadel (Petrograd), or of Krassin or Radek – all Jews. In the Soviet institutions the predominance of Jews is even more astonishing. And the prominent, if not indeed the principal, part in the system of terrorism applied by the Extraordinary Commissions for Combatting Counter-Revolution has been taken by Jews, and in some notable cases by Jewesses.

What I said about seriously disproportionate involvement in the Russian Revolution by Jewish ‘revolutionaries’ (Sabbatians) is provable fact, but truth is no defence against the Sabbatian AntiSemitism Industry, its repeater parrots like Quinn and Bland, and the now breathtaking network of so-called ‘Woke’ ‘anti-hate’ groups with interlocking leaderships and funding which have the role of discrediting and silencing anyone who gets too close to exposing the Sabbatians. We have seen ‘truth is no defence’ confirmed in legal judgements with the Saskatchewan Human Rights Commission in Canada decreeing this: ‘Truthful statements can be presented in a manner that would meet the definition of hate speech, and not all truthful statements must be free from restriction.’ Most ‘anti-hate’ activists, who are themselves consumed by hatred, are too stupid and ignorant of the world to know how they are being used. They are far too far up their own virtue-signalling arses and it’s far too dark for them to see anything.

The ‘revolution’ game

The background and methods of the ‘Russian’ Revolution are straight from the Sabbatian playbook seen in the French Revolution

and endless others around the world that appear to start as a revolution of the people against tyrannical rule and end up with a regime change to more tyrannical rule overtly or covertly. Wars, terror a acks and regime overthrows follow the Sabbatian cult through history with its agents creating them as Problem-ReactionSolutions to remove opposition on the road to world domination. Sabbatian dots connect the Rothschilds with the Illuminati, Jacobins of the French Revolution, the ‘Bund’ or League of the Just, the International Communist Party, Communist League and the Communist Manifesto of Karl Marx and Friedrich Engels that would lead to the Rothschild-funded Russian Revolution. The sequence comes under the heading of ‘creative destruction’ when you advance to your global goal by continually destroying the status quo to install a new status quo which you then also destroy. The two world wars come to mind. With each new status quo you move closer to your planned outcome. Wars and mass murder are to Sabbatians a collective blood sacrifice ritual. They are obsessed with death for many reasons and one is that death is an inversion of life. Satanists and Sabbatians are obsessed with death and o en target churches and churchyards for their rituals. Inversion-obsessed Sabbatians explain the use of inverted symbolism including the inverted pentagram and inverted cross. The inversion of the cross has been related to targeting Christianity, but the cross was a religious symbol long before Christianity and its inversion is a statement about the Sabbatian mentality and goals more than any single religion. Sabbatians operating in Germany were behind the rise of the occult-obsessed Nazis and the subsequent Jewish exodus from Germany and Europe to Palestine and the United States a er World War Two. The Rothschild dynasty was at the forefront of this both as political manipulators and by funding the operation. Why would Sabbatians help to orchestrate the horrors inflicted on Jews by the Nazis and by Stalin a er they organised the Russian Revolution? Sabbatians hate Jews and their religion, that’s why. They pose as Jews and secure positions of control within Jewish society and play the ‘anti-Semitism’ card to protect themselves from exposure

through a global network of organisations answering to the Sabbatian-created-and-controlled globe-spanning intelligence network that involves a stunning web of military-intelligence operatives and operations for a tiny country of just nine million. Among them are Jewish assets who are not Sabbatians but have been convinced by them that what they are doing is for the good of Israel and the Jewish community to protect them from what they have been programmed since childhood to believe is a Jew-hating hostile world. The Jewish community is just a highly convenient cover to hide the true nature of Sabbatians. Anyone ge ing close to exposing their game is accused by Sabbatian place-people and gofers of ‘antiSemitism’ and claiming that all Jews are part of a plot to take over the world. I am not saying that. I am saying that Sabbatians – the real Jew-haters – have infiltrated the Jewish community to use them both as a cover and an ‘anti-Semitic’ defence against exposure. Thus we have the Anti-Semitism Industry targeted researchers in this way and most Jewish people think this is justified and genuine. They don’t know that their ‘Jewish’ leaders and institutions of state, intelligence and military are not controlled by Jews at all, but cultists and stooges of Sabbatian-Frankism. I once added my name to a proJewish freedom petition online and the next time I looked my name was gone and text had been added to the petition blurb to a ack me as an ‘anti-Semite’ such is the scale of perceptual programming.

Moving on America

I tell the story in The Trigger and a chapter called ‘Atlantic Crossing’ how particularly a er Israel was established the Sabbatians moved in on the United States and eventually grasped control of government administration, the political system via both Democrats and Republicans, the intelligence community like the CIA and National Security Agency (NSA), the Pentagon and mass media. Through this seriously compartmentalised network Sabbatians and their operatives in Mossad, Israeli Defense Forces (IDF) and US agencies pulled off 9/11 and blamed it on 19 ‘Al-Qaeda hijackers’ dominated by men from, or connected to, Sabbatian-ruled Saudi

Arabia. The ‘19’ were not even on the planes let alone flew those big passenger jets into buildings while being largely incompetent at piloting one-engine light aircra . ‘Hijacker’ Hani Hanjour who is said to have flown American Airlines Flight 77 into the Pentagon with a turn and manoeuvre most professional pilots said they would have struggled to do was banned from renting a small plane by instructors at the Freeway Airport in Bowie, Maryland, just six weeks earlier on the grounds that he was an incompetent pilot. The Jewish population of the world is just 0.2 percent with even that almost entirely concentrated in Israel (75 percent Jewish) and the United States (around two percent). This two percent and globally 0.2 percent refers to Jewish people and not Sabbatian interlopers who are a fraction of that fraction. What a sobering thought when you think of the fantastic influence on world affairs of tiny Israel and that the Project for the New America Century (PNAC) which laid out the blueprint in September, 2000, for America’s war on terror and regime change wars in Iraq, Libya and Syria was founded and dominated by Sabbatians known as ‘Neocons’. The document conceded that this plan would not be supported politically or publicly without a major a ack on American soil and a Problem-Reaction-Solution excuse to send troops to war across the Middle East. Sabbatian Neocons said: ... [The] process of transformation ... [war and regime change] ... is likely to be a long one, absent some catastrophic and catalysing event – like a new Pearl Harbor.

Four months later many of those who produced that document came to power with their inane puppet George Bush from the longtime Sabbatian Bush family. They included Sabbatian Dick Cheney who was officially vice-president, but really de-facto president for the entirety of the ‘Bush’ government. Nine months a er the ‘Bush’ inauguration came what Bush called at the time ‘the Pearl Harbor of the 21st century’ and with typical Sabbatian timing and symbolism 2001 was the 60th anniversary of the a ack in 1941 by the Japanese Air Force on Pearl Harbor, Hawaii, which allowed President Franklin Delano Roosevelt to take the United States into a Sabbatian-

instigated Second World War that he said in his election campaign that he never would. The evidence is overwhelming that Roosevelt and his military and intelligence networks knew the a ack was coming and did nothing to stop it, but they did make sure that America’s most essential naval ships were not in Hawaii at the time. Three thousand Americans died in the Pearl Harbor a acks as they did on September 11th. By the 9/11 year of 2001 Sabbatians had widely infiltrated the US government, military and intelligence operations and used their compartmentalised assets to pull off the ‘Al-Qaeda’ a acks. If you read The Trigger it will blow your mind to see the u erly staggering concentration of ‘Jewish’ operatives (Sabbatian infiltrators) in essential positions of political, security, legal, law enforcement, financial and business power before, during, and a er the a acks to make them happen, carry them out, and then cover their tracks – and I do mean staggering when you think of that 0.2 percent of the world population and two percent of Americans which are Jewish while Sabbatian infiltrators are a fraction of that. A central foundation of the 9/11 conspiracy was the hijacking of government, military, Air Force and intelligence computer systems in real time through ‘back-door’ access made possible by Israeli (Sabbatian) ‘cyber security’ so ware. Sabbatian-controlled Israel is on the way to rivalling Silicon Valley for domination of cyberspace and is becoming the dominant force in cyber-security which gives them access to entire computer systems and their passcodes across the world. Then add to this that Zionists head (officially) Silicon Valley giants like Google (Larry Page and Sergey Brin), Googleowned YouTube (Susan Wojcicki), Facebook (Mark Zuckerberg and Sheryl Sandberg), and Apple (Chairman Arthur D. Levinson), and that ultra-Zionist hedge fund billionaire Paul Singer has a $1 billion stake in Twi er which is only nominally headed by ‘CEO’ pothead Jack Dorsey. As cable news host Tucker Carlson said of Dorsey: ‘There used to be debate in the medical community whether dropping a ton of acid had permanent effects and I think that debate has now ended.’ Carlson made the comment a er Dorsey told a hearing on Capitol Hill (if you cut through his bullshit) that he

believed in free speech so long as he got to decide what you can hear and see. These ‘big names’ of Silicon Valley are only front men and women for the Global Cult, not least the Sabbatians, who are the true controllers of these corporations. Does anyone still wonder why these same people and companies have been ferociously censoring and banning people (like me) for exposing any aspect of the Cult agenda and especially the truth about the ‘Covid’ hoax which Sabbatians have orchestrated? The Jeffrey Epstein paedophile ring was a Sabbatian operation. He was officially ‘Jewish’ but he was a Sabbatian and women abused by the ring have told me about the high number of ‘Jewish’ people involved. The Epstein horror has Sabbatian wri en all over it and matches perfectly their modus operandi and obsession with sex and ritual. Epstein was running a Sabbatian blackmail ring in which famous people with political and other influence were provided with young girls for sex while everything was being filmed and recorded on hidden cameras and microphones at his New York house, Caribbean island and other properties. Epstein survivors have described this surveillance system to me and some have gone public. Once the famous politician or other figure knew he or she was on video they tended to do whatever they were told. Here we go again …when you’ve got them by the balls their hearts and minds will follow. Sabbatians use this blackmail technique on a wide scale across the world to entrap politicians and others they need to act as demanded. Epstein’s private plane, the infamous ‘Lolita Express’, had many well-known passengers including Bill Clinton while Bill Gates has flown on an Epstein plane and met with him four years a er Epstein had been jailed for paedophilia. They subsequently met many times at Epstein’s home in New York according to a witness who was there. Epstein’s infamous side-kick was Ghislaine Maxwell, daughter of Mossad agent and ultra-Zionist mega-crooked British businessman, Bob Maxwell, who at one time owned the Daily Mirror newspaper. Maxwell was murdered at sea on his boat in 1991 by Sabbatian-controlled Mossad when he became a liability with his

business empire collapsing as a former Mossad operative has confirmed (see The Trigger).

Money, money, money, funny money …

Before I come to the Sabbatian connection with the last three US presidents I will lay out the crucial importance to Sabbatians of controlling banking and finance. Sabbatian Mayer Amschel Rothschild set out to dominate this arena in his family’s quest for total global control. What is freedom? It is, in effect, choice. The more choices you have the freer you are and the fewer your choices the more you are enslaved. In the global structure created over centuries by Sabbatians the biggest decider and restrictor of choice is … money. Across the world if you ask people what they would like to do with their lives and why they are not doing that they will reply ‘I don’t have the money’. This is the idea. A global elite of multibillionaires are described as ‘greedy’ and that is true on one level; but control of money – who has it and who doesn’t – is not primarily about greed. It’s about control. Sabbatians have seized ever more control of finance and sucked the wealth of the world out of the hands of the population. We talk now, a er all, about the ‘Onepercent’ and even then the wealthiest are a lot fewer even than that. This has been made possible by a money scam so outrageous and so vast it could rightly be called the scam of scams founded on creating ‘money’ out of nothing and ‘loaning’ that with interest to the population. Money out of nothing is called ‘credit’. Sabbatians have asserted control over governments and banking ever more completely through the centuries and secured financial laws that allow banks to lend hugely more than they have on deposit in a confidence trick known as fractional reserve lending. Imagine if you could lend money that doesn’t exist and charge the recipient interest for doing so. You would end up in jail. Bankers by contrast end up in mansions, private jets, Malibu and Monaco. Banks are only required to keep a fraction of their deposits and wealth in their vaults and they are allowed to lend ‘money’ they don’t have called ‘credit. Go into a bank for a loan and if you succeed

the banker will not move any real wealth into your account. They will type into your account the amount of the agreed ‘loan’ – say £100,000. This is not wealth that really exists; it is non-existent, freshair, created-out-of-nothing ‘credit’ which has never, does not, and will never exist except in theory. Credit is backed by nothing except wind and only has buying power because people think that it has buying power and accept it in return for property, goods and services. I have described this situation as like those cartoon characters you see chasing each other and when they run over the edge of a cliff they keep running forward on fresh air until one of them looks down, realises what’s happened, and they all crash into the ravine. The whole foundation of the Sabbatian financial system is to stop people looking down except for periodic moments when they want to crash the system (as in 2008 and 2020 ongoing) and reap the rewards from all the property, businesses and wealth their borrowers had signed over as ‘collateral’ in return for a ‘loan’ of fresh air. Most people think that money is somehow created by governments when it comes into existence from the start as a debt through banks ‘lending’ illusory money called credit. Yes, the very currency of exchange is a debt from day one issued as an interest-bearing loan. Why don’t governments create money interest-free and lend it to their people interest-free? Governments are controlled by Sabbatians and the financial system is controlled by Sabbatians for whom interest-free money would be a nightmare come true. Sabbatians underpin their financial domination through their global network of central banks, including the privately-owned US Federal Reserve and Britain’s Bank of England, and this is orchestrated by a privately-owned central bank coordination body called the Bank for International Se lements in Basle, Switzerland, created by the usual suspects including the Rockefellers and Rothschilds. Central bank chiefs don’t answer to governments or the people. They answer to the Bank for International Se lements or, in other words, the Global Cult which is dominated today by Sabbatians.

Built-in disaster

There are so many constituent scams within the overall banking scam. When you take out a loan of thin-air credit only the amount of that loan is theoretically brought into circulation to add to the amount in circulation; but you are paying back the principle plus interest. The additional interest is not created and this means that with every ‘loan’ there is a shortfall in the money in circulation between what is borrowed and what has to be paid back. There is never even close to enough money in circulation to repay all outstanding public and private debt including interest. Coldly weaved in the very fabric of the system is the certainty that some will lose their homes, businesses and possessions to the banking ‘lender’. This is less obvious in times of ‘boom’ when the amount of money in circulation (and the debt) is expanding through more people wanting and ge ing loans. When a downturn comes and the money supply contracts it becomes painfully obvious that there is not enough money to service all debt and interest. This is less obvious in times of ‘boom’ when the amount of money in circulation (and the debt) is expanding through more people wanting and ge ing loans. When a downturn comes and the money supply contracts and it becomes painfully obvious – as in 2008 and currently – that there is not enough money to service all debt and interest. Sabbatian banksters have been leading the human population through a calculated series of booms (more debt incurred) and busts (when the debt can’t be repaid and the banks get the debtor’s tangible wealth in exchange for non-existent ‘credit’). With each ‘bust’ Sabbatian bankers have absorbed more of the world’s tangible wealth and we end up with the One-percent. Governments are in bankruptcy levels of debt to the same system and are therefore owned by a system they do not control. The Federal Reserve, ‘America’s central bank’, is privately-owned and American presidents only nominally appoint its chairman or woman to maintain the illusion that it’s an arm of government. It’s not. The ‘Fed’ is a cartel of private banks which handed billions to its associates and friends a er the crash of 2008 and has been Sabbatiancontrolled since it was manipulated into being in 1913 through the covert trickery of Rothschild banking agents Jacob Schiff and Paul

Warburg, and the Sabbatian Rockefeller family. Somehow from a Jewish population of two-percent and globally 0.2 percent (Sabbatian interlopers remember are far smaller) ultra-Zionists headed the Federal Reserve for 31 years between 1987 and 2018 in the form of Alan Greenspan, Bernard Bernanke and Janet Yellen (now Biden’s Treasury Secretary) with Yellen’s deputy chairman a IsraeliAmerican duel citizen and ultra-Zionist Stanley Fischer, a former governor of the Bank of Israel. Ultra-Zionist Fed chiefs spanned the presidencies of Ronald Reagan (‘Republican’), Father George Bush (‘Republican’), Bill Clinton (‘Democrat’), Boy George Bush (‘Republican’) and Barack Obama (‘Democrat’). We should really add the pre-Greenspan chairman, Paul Adolph Volcker, ‘appointed’ by Jimmy Carter (‘Democrat’) who ran the Fed between 1979 and 1987 during the Carter and Reagan administrations before Greenspan took over. Volcker was a long-time associate and business partner of the Rothschilds. No ma er what the ‘party’ officially in power the United States economy was directed by the same force. Here are members of the Obama, Trump and Biden administrations and see if you can make out a common theme.

Barack Obama (‘Democrat’)

Ultra-Zionists Robert Rubin, Larry Summers, and Timothy Geithner ran the US Treasury in the Clinton administration and two of them reappeared with Obama. Ultra-Zionist Fed chairman Alan Greenspan had manipulated the crash of 2008 through deregulation and jumped ship just before the disaster to make way for ultraZionist Bernard Bernanke to hand out trillions to Sabbatian ‘too big to fail’ banks and businesses, including the ubiquitous ultra-Zionist Goldman Sachs which has an ongoing staff revolving door operation between itself and major financial positions in government worldwide. Obama inherited the fallout of the crash when he took office in January, 2009, and fortunately he had the support of his ultra-Zionist White House Chief of Staff Rahm Emmanuel, son of a terrorist who helped to bomb Israel into being in 1948, and his ultraZionist senior adviser David Axelrod, chief strategist in Obama’s two

successful presidential campaigns. Emmanuel, later mayor of Chicago and former senior fundraiser and strategist for Bill Clinton, is an example of the Sabbatian policy a er Israel was established of migrating insider families to America so their children would be born American citizens. ‘Obama’ chose this financial team throughout his administration to respond to the Sabbatian-instigated crisis: Timothy Geithner (ultra-Zionist) Treasury Secretary; Jacob J. Lew, Treasury Secretary; Larry Summers (ultra-Zionist), director of the White House National Economic Council; Paul Adolph Volcker (Rothschild business partner), chairman of the Economic Recovery Advisory Board; Peter Orszag (ultra-Zionist), director of the Office of Management and Budget overseeing all government spending; Penny Pritzker (ultra-Zionist), Commerce Secretary; Jared Bernstein (ultra-Zionist), chief economist and economic policy adviser to Vice President Joe Biden; Mary Schapiro (ultra-Zionist), chair of the Securities and Exchange Commission (SEC); Gary Gensler (ultraZionist), chairman of the Commodity Futures Trading Commission (CFTC); Sheila Bair (ultra-Zionist), chair of the Federal Deposit Insurance Corporation (FDIC); Karen Mills (ultra-Zionist), head of the Small Business Administration (SBA); Kenneth Feinberg (ultraZionist), Special Master for Executive [bail-out] Compensation. Feinberg would be appointed to oversee compensation (with strings) to 9/11 victims and families in a campaign to stop them having their day in court to question the official story. At the same time ultraZionist Bernard Bernanke was chairman of the Federal Reserve and these are only some of the ultra-Zionists with allegiance to Sabbatian-controlled Israel in the Obama government. Obama’s biggest corporate donor was ultra-Zionist Goldman Sachs which had employed many in his administration.

Donald Trump (‘Republican’)

Trump claimed to be an outsider (he wasn’t) who had come to ‘drain the swamp’. He embarked on this goal by immediately appointing ultra-Zionist Steve Mnuchin, a Goldman Sachs employee for 17

years, as his Treasury Secretary. Others included Gary Cohn (ultraZionist), chief operating officer of Goldman Sachs, his first Director of the National Economic Council and chief economic adviser, who was later replaced by Larry Kudlow (ultra-Zionist). Trump’s senior adviser throughout his four years in the White House was his sinister son-in-law Jared Kushner, a life-long friend of Israel Prime Minister Benjamin Netanyahu. Kushner is the son of a convicted crook who was pardoned by Trump in his last days in office. Other ultra-Zionists in the Trump administration included: Stephen Miller, Senior Policy Adviser; Avrahm Berkowitz, Deputy Adviser to Trump and his Senior Adviser Jared Kushner; Ivanka Trump, Adviser to the President, who converted to Judaism when she married Jared Kushner; David Friedman, Trump lawyer and Ambassador to Israel; Jason Greenbla , Trump Organization executive vice president and chief legal officer, who was made Special Representative for International Negotiations and the Israeli-Palestinian Conflict; Rod Rosenstein, Deputy A orney General; Elliot Abrams, Special Representative for Venezuela, then Iran; John Eisenberg, National Security Council Legal Adviser and Deputy Council to the President for National Security Affairs; Anne Neuberger, Deputy National Manager, National Security Agency; Ezra Cohen-Watnick, Acting Under Secretary of Defense for Intelligence; Elan Carr, Special Envoy to monitor and combat anti-Semitism; Len Khodorkovsky, Deputy Special Envoy to monitor and combat anti-Semitism; Reed Cordish, Assistant to the President, Intragovernmental and Technology Initiatives. Trump Vice President Mike Pence and Secretary of State Mike Pompeo, both Christian Zionists, were also vehement supporters of Israel and its goals and ambitions. Donald ‘free-speech believer’ Trump pardoned a number of financial and violent criminals while ignoring calls to pardon Julian Assange and Edward Snowden whose crimes are revealing highly relevant information about government manipulation and corruption and the widespread illegal surveillance of the American people by US ‘security’ agencies. It’s so good to know that Trump is on the side of freedom and justice and not mega-criminals with

allegiance to Sabbatian-controlled Israel. These included a pardon for Israeli spy Jonathan Pollard who was jailed for life in 1987 under the Espionage Act. Aviem Sella, the Mossad agent who recruited Pollard, was also pardoned by Trump while Assange sat in jail and Snowden remained in exile in Russia. Sella had ‘fled’ (was helped to escape) to Israel in 1987 and was never extradited despite being charged under the Espionage Act. A Trump White House statement said that Sella’s clemency had been ‘supported by Benjamin Netanyahu, Ron Dermer, Israel’s US Ambassador, David Friedman, US Ambassador to Israel and Miriam Adelson, wife of leading Trump donor Sheldon Adelson who died shortly before. Other friends of Jared Kushner were pardoned along with Sholom Weiss who was believed to be serving the longest-ever white-collar prison sentence of more than 800 years in 2000. The sentence was commuted of Ponzi-schemer Eliyahu Weinstein who defrauded Jews and others out of $200 million. I did mention that Assange and Snowden were ignored, right? Trump gave Sabbatians almost everything they asked for in military and political support, moving the US Embassy from Tel Aviv to Jerusalem with its critical symbolic and literal implications for Palestinian statehood, and the ‘deal of the Century’ designed by Jared Kushner and David Friedman which gave the Sabbatian Israeli government the green light to substantially expand its already widespread program of building illegal Jewish-only se lements in the occupied land of the West Bank. This made a two-state ‘solution’ impossible by seizing all the land of a potential Palestinian homeland and that had been the plan since 1948 and then 1967 when the Arab-controlled Gaza Strip, West Bank, Sinai Peninsula and Syrian Golan Heights were occupied by Israel. All the talks about talks and road maps and delays have been buying time until the West Bank was physically occupied by Israeli real estate. Trump would have to be a monumentally ill-informed idiot not to see that this was the plan he was helping to complete. The Trump administration was in so many ways the Kushner administration which means the Netanyahu administration which means the Sabbatian administration. I understand why many opposing Cult fascism in all its forms gravitated to Trump, but he

was a crucial part of the Sabbatian plan and I will deal with this in the next chapter.

Joe Biden (‘Democrat’)

A barely cognitive Joe Biden took over the presidency in January, 2021, along with his fellow empty shell, Vice-President Kamala Harris, as the latest Sabbatian gofers to enter the White House. Names on the door may have changed and the ‘party’ – the force behind them remained the same as Zionists were appointed to a stream of pivotal areas relating to Sabbatian plans and policy. They included: Janet Yellen, Treasury Secretary, former head of the Federal Reserve, and still another ultra-Zionist running the US Treasury a er Mnuchin (Trump), Lew and Geithner (Obama), and Summers and Rubin (Clinton); Anthony Blinken, Secretary of State; Wendy Sherman, Deputy Secretary of State (so that’s ‘Biden’s’ Sabbatian foreign policy sorted); Jeff Zients, White House coronavirus coordinator; Rochelle Walensky, head of the Centers for Disease Control; Rachel Levine, transgender deputy health secretary (that’s ‘Covid’ hoax policy under control); Merrick Garland, A orney General; Alejandro Mayorkas, Secretary of Homeland Security; Cass Sunstein, Homeland Security with responsibility for new immigration laws; Avril Haines, Director of National Intelligence; Anne Neuberger, National Security Agency cybersecurity director (note, cybersecurity); David Cohen, CIA Deputy Director; Ronald Klain, Biden’s Chief of Staff (see Rahm Emanuel); Eric Lander, a ‘leading geneticist’, Office of Science and Technology Policy director (see Smart Grid, synthetic biology agenda); Jessica Rosenworcel, acting head of the Federal Communications Commission (FCC) which controls Smart Grid technology policy and electromagnetic communication systems including 5G. How can it be that so many pivotal positions are held by two-percent of the American population and 0.2 percent of the world population administration a er administration no ma er who is the president and what is the party? It’s a coincidence? Of course it’s not and this is why Sabbatians have built their colossal global web of interlocking ‘anti-

hate’ hate groups to condemn anyone who asks these glaring questions as an ‘anti-Semite’. The way that Jewish people horrifically abused in Sabbatian-backed Nazi Germany are exploited to this end is stomach-turning and disgusting beyond words.

Political fusion

Sabbatian manipulation has reversed the roles of Republicans and Democrats and the same has happened in Britain with the Conservative and Labour Parties. Republicans and Conservatives were always labelled the ‘right’ and Democrats and Labour the ‘le ’, but look at the policy positions now and the Democrat-Labour ‘le ’ has moved further to the ‘right’ than Republicans and Conservatives under the banner of ‘Woke’, the Cult-created far-right tyranny. Where once the Democrat-Labour ‘le ’ defended free speech and human rights they now seek to delete them and as I said earlier despite the ‘Covid’ fascism of the Jackboot Johnson Conservative government in the UK the Labour Party of leader Keir Starmer demanded even more extreme measures. The Labour Party has been very publicly absorbed by Sabbatians a er a political and media onslaught against the previous leader, the weak and inept Jeremy Corbyn, over made-up allegations of ‘anti-Semitism’ both by him and his party. The plan was clear with this ‘anti-Semite’ propaganda and what was required in response was a swi and decisive ‘fuck off’ from Corbyn and a statement to expose the Anti-Semitism Industry (Sabbatian) a empt to silence Labour criticism of the Israeli government (Sabbatians) and purge the party of all dissent against the extremes of ultra-Zionism (Sabbatians). Instead Corbyn and his party fell to their knees and appeased the abusers which, by definition, is impossible. Appeasing one demand leads only to a new demand to be appeased until takeover is complete. Like I say – ‘fuck off’ would have been a much more effective policy and I have used it myself with great effect over the years when Sabbatians are on my case which is most of the time. I consider that fact a great compliment, by the way. The outcome of the Labour Party capitulation is that we now have a Sabbatian-controlled

Conservative Party ‘opposed’ by a Sabbatian-controlled Labour Party in a one-party Sabbatian state that hurtles towards the extremes of tyranny (the Sabbatian cult agenda). In America the situation is the same. Labour’s Keir Starmer spends his days on his knees with his tongue out pointing to Tel Aviv, or I guess now Jerusalem, while Boris Johnson has an ‘anti-Semitism czar’ in the form of former Labour MP John Mann who keeps Starmer company on his prayer mat. Sabbatian influence can be seen in Jewish members of the Labour Party who have been ejected for criticism of Israel including those from families that suffered in Nazi Germany. Sabbatians despise real Jewish people and target them even more harshly because it is so much more difficult to dub them ‘anti-Semitic’ although in their desperation they do try.

CHAPTER THREE The Pushbacker sting Until you realize how easy it is for your mind to be manipulated, you remain the puppet of someone else’s game Evita Ochel

I

will use the presidencies of Trump and Biden to show how the manipulation of the one-party state plays out behind the illusion of political choice across the world. No two presidencies could – on the face of it – be more different and apparently at odds in terms of direction and policy. A Renegade Mind sees beyond the obvious and focuses on outcomes and consequences and not image, words and waffle. The Cult embarked on a campaign to divide America between those who blindly support its agenda (the mentality known as ‘Woke’) and those who are pushing back on where the Cult and its Sabbatians want to go. This presents infinite possibilities for dividing and ruling the population by se ing them at war with each other and allows a perceptual ring fence of demonisation to encircle the Pushbackers in a modern version of the Li le Big Horn in 1876 when American cavalry led by Lieutenant Colonel George Custer were drawn into a trap, surrounded and killed by Native American tribes defending their land of thousands of years from being seized by the government. In this modern version the roles are reversed and it’s those defending themselves from the Sabbatian government who are surrounded and the government that’s seeking to destroy them. This trap was set years ago and to explain how we must return to 2016

and the emergence of Donald Trump as a candidate to be President of the United States. He set out to overcome the best part of 20 other candidates in the Republican Party before and during the primaries and was not considered by many in those early stages to have a prayer of living in the White House. The Republican Party was said to have great reservations about Trump and yet somehow he won the nomination. When you know how American politics works – politics in general – there is no way that Trump could have become the party’s candidate unless the Sabbatian-controlled ‘Neocons’ that run the Republican Party wanted that to happen. We saw the proof in emails and documents made public by WikiLeaks that the Democratic Party hierarchy, or Democons, systematically undermined the campaign of Bernie Sanders to make sure that Sabbatian gofer Hillary Clinton won the nomination to be their presidential candidate. If the Democons could do that then the Neocons in the Republican Party could have derailed Trump in the same way. But they didn’t and at that stage I began to conclude that Trump could well be the one chosen to be president. If that was the case the ‘why’ was pre y clear to see – the goal of dividing America between Cult agenda-supporting Wokers and Pushbackers who gravitated to Trump because he was telling them what they wanted to hear. His constituency of support had been increasingly ignored and voiceless for decades and profoundly through the eight years of Sabbatian puppet Barack Obama. Now here was someone speaking their language of pulling back from the incessant globalisation of political and economic power, the exporting of American jobs to China and elsewhere by ‘American’ (Sabbatian) corporations, the deletion of free speech, and the mass immigration policies that had further devastated job opportunities for the urban working class of all races and the once American heartlands of the Midwest.

Beware the forked tongue

Those people collectively sighed with relief that at last a political leader was apparently on their side, but another trait of the Renegade Mind is that you look even harder at people telling you

what you want to hear than those who are telling you otherwise. Obviously as I said earlier people wish what they want to hear to be true and genuine and they are much more likely to believe that than someone saying what they don’t want to here and don’t want to be true. Sales people are taught to be skilled in eliciting by calculated questioning what their customers want to hear and repeating that back to them as their own opinion to get their targets to like and trust them. Assets of the Cult are also sales people in the sense of selling perception. To read Cult manipulation you have to play the long and expanded game and not fall for the Vaudeville show of party politics. Both American parties are vehicles for the Cult and they exploit them in different ways depending on what the agenda requires at that moment. Trump and the Republicans were used to be the focus of dividing America and isolating Pushbackers to open the way for a Biden presidency to become the most extreme in American history by advancing the full-blown Woke (Cult) agenda with the aim of destroying and silencing Pushbackers now labelled Nazi Trump supporters and white supremacists. Sabbatians wanted Trump in office for the reasons described by ultra-Zionist Saul Alinsky (1909-1972) who was promoting the Woke philosophy through ‘community organising’ long before anyone had heard of it. In those days it still went by its traditional name of Marxism. The reason for the manipulated Trump phenomenon was laid out in Alinsky’s 1971 book, Rules for Radicals, which was his blueprint for overthrowing democratic and other regimes and replacing them with Sabbatian Marxism. Not surprisingly his to-do list was evident in the Sabbatian French and Russian ‘Revolutions’ and that in China which will become very relevant in the next chapter about the ‘Covid’ hoax. Among Alinsky’s followers have been the deeply corrupt Barack Obama, House Speaker Nancy Pelosi and Hillary Clinton who described him as a ‘hero’. All three are Sabbatian stooges with Pelosi personifying the arrogant corrupt idiocy that so widely fronts up for the Cult inner core. Predictably as a Sabbatian advocate of the ‘light-bringer’ Alinsky features Lucifer on the dedication page of his book as the original radical who gained

his own kingdom (‘Earth’ as we shall see). One of Alinsky’s golden radical rules was to pick an individual and focus all a ention, hatred and blame on them and not to target faceless bureaucracies and corporations. Rules for Radicals is really a Sabbatian handbook with its contents repeatedly employed all over the world for centuries and why wouldn’t Sabbatians bring to power their designer-villain to be used as the individual on which all a ention, hatred and blame was bestowed? This is what they did and the only question for me is how much Trump knew that and how much he was manipulated. A bit of both, I suspect. This was Alinsky’s Trump technique from a man who died in 1972. The technique has spanned history: Pick the target, freeze it, personalize it, polarize it. Don’t try to attack abstract corporations or bureaucracies. Identify a responsible individual. Ignore attempts to shift or spread the blame.

From the moment Trump came to illusory power everything was about him. It wasn’t about Republican policy or opinion, but all about Trump. Everything he did was presented in negative, derogatory and abusive terms by the Sabbatian-dominated media led by Cult operations such as CNN, MSNBC, The New York Times and the Jeff Bezos-owned Washington Post – ‘Pick the target, freeze it, personalize it, polarize it.’ Trump was turned into a demon to be vilified by those who hated him and a demi-god loved by those who worshipped him. This, in turn, had his supporters, too, presented as equally demonic in preparation for the punchline later down the line when Biden was about to take office. It was here’s a Trump, there’s a Trump, everywhere a Trump, Trump. Virtually every news story or happening was filtered through the lens of ‘The Donald’. You loved him or hated him and which one you chose was said to define you as Satan’s spawn or a paragon of virtue. Even supporting some Trump policies or statements and not others was enough for an assault on your character. No shades of grey were or are allowed. Everything is black and white (literally and figuratively). A Californian I knew had her head u erly scrambled by her hatred for Trump while telling people they should love each other. She was so totally consumed by

Trump Derangement Syndrome as it became to be known that this glaring contradiction would never have occurred to her. By definition anyone who criticised Trump or praised his opponents was a hero and this lady described Joe Biden as ‘a kind, honest gentleman’ when he’s a provable liar, mega-crook and vicious piece of work to boot. Sabbatians had indeed divided America using Trump as the fall-guy and all along the clock was ticking on the consequences for his supporters.

In hock to his masters

Trump gave Sabbatians via Israel almost everything they wanted in his four years. Ask and you shall receive was the dynamic between himself and Benjamin Netanyahu orchestrated by Trump’s ultraZionist son-in-law Jared Kushner, his ultra-Zionist Ambassador to Israel, David Friedman, and ultra-Zionist ‘Israel adviser’, Jason Greenbla . The last two were central to the running and protecting from collapse of his business empire, the Trump Organisation, and colossal business failures made him forever beholding to Sabbatian networks that bailed him out. By the start of the 1990s Trump owed $4 billion to banks that he couldn’t pay and almost $1billion of that was down to him personally and not his companies. This megadisaster was the result of building two new casinos in Atlantic City and buying the enormous Taj Mahal operation which led to crippling debt payments. He had borrowed fantastic sums from 72 banks with major Sabbatian connections and although the scale of debt should have had him living in a tent alongside the highway they never foreclosed. A plan was devised to li Trump from the mire by BT Securities Corporation and Rothschild Inc. and the case was handled by Wilber Ross who had worked for the Rothschilds for 27 years. Ross would be named US Commerce Secretary a er Trump’s election. Another crucial figure in saving Trump was ultraZionist ‘investor’ Carl Icahn who bought the Taj Mahal casino. Icahn was made special economic adviser on financial regulation in the Trump administration. He didn’t stay long but still managed to find time to make a tidy sum of a reported $31.3 million when he sold his

holdings affected by the price of steel three days before Trump imposed a 235 percent tariff on steel imports. What amazing bits of luck these people have. Trump and Sabbatian operatives have long had a close association and his mentor and legal adviser from the early 1970s until 1986 was the dark and genetically corrupt ultraZionist Roy Cohn who was chief counsel to Senator Joseph McCarthy’s ‘communist’ witch-hunt in the 1950s. Esquire magazine published an article about Cohn with the headline ‘Don’t mess with Roy Cohn’. He was described as the most feared lawyer in New York and ‘a ruthless master of dirty tricks ... [with] ... more than one Mafia Don on speed dial’. Cohn’s influence, contacts, support and protection made Trump a front man for Sabbatians in New York with their connections to one of Cohn’s many criminal employers, the ‘Russian’ Sabbatian Mafia. Israel-centric media mogul Rupert Murdoch was introduced to Trump by Cohn and they started a long friendship. Cohn died in 1986 weeks a er being disbarred for unethical conduct by the Appellate Division of the New York State Supreme Court. The wheels of justice do indeed run slow given the length of Cohn’s crooked career.

QAnon-sense

We are asked to believe that Donald Trump with his fundamental connections to Sabbatian networks and operatives has been leading the fight to stop the Sabbatian agenda for the fascistic control of America and the world. Sure he has. A man entrapped during his years in the White House by Sabbatian operatives and whose biggest financial donor was casino billionaire Sheldon Adelson who was Sabbatian to his DNA?? Oh, do come on. Trump has been used to divide America and isolate Pushbackers on the Cult agenda under the heading of ‘Trump supporters’, ‘insurrectionists’ and ‘white supremacists’. The US Intelligence/Mossad Psyop or psychological operation known as QAnon emerged during the Trump years as a central pillar in the Sabbatian campaign to lead Pushbackers into the trap set by those that wished to destroy them. I knew from the start that QAnon was a scam because I had seen the same scenario many

times before over 30 years under different names and I had wri en about one in particular in the books. ‘Not again’ was my reaction when QAnon came to the fore. The same script is pulled out every few years and a new name added to the le erhead. The story always takes the same form: ‘Insiders’ or ‘the good guys’ in the governmentintelligence-military ‘Deep State’ apparatus were going to instigate mass arrests of the ‘bad guys’ which would include the Rockefellers, Rothschilds, Barack Obama, Hillary Clinton, George Soros, etc., etc. Dates are given for when the ‘good guys’ are going to move in, but the dates pass without incident and new dates are given which pass without incident. The central message to Pushbackers in each case is that they don’t have to do anything because there is ‘a plan’ and it is all going to be sorted by the ‘good guys’ on the inside. ‘Trust the plan’ was a QAnon mantra when the only plan was to misdirect Pushbackers into pu ing their trust in a Psyop they believed to be real. Beware, beware, those who tell you what you want to hear and always check it out. Right up to Biden’s inauguration QAnon was still claiming that ‘the Storm’ was coming and Trump would stay on as president when Biden and his cronies were arrested and jailed. It was never going to happen and of course it didn’t, but what did happen as a result provided that punchline to the Sabbatian Trump/QAnon Psyop. On January 6th, 2021, a very big crowd of Trump supporters gathered in the National Mall in Washington DC down from the Capitol Building to protest at what they believed to be widespread corruption and vote fraud that stopped Trump being re-elected for a second term as president in November, 2020. I say as someone that does not support Trump or Biden that the evidence is clear that major vote-fixing went on to favour Biden, a man with cognitive problems so advanced he can o en hardly string a sentence together without reading the words wri en for him on the Teleprompter. Glaring ballot discrepancies included serious questions about electronic voting machines that make vote rigging a comparative cinch and hundreds of thousands of paper votes that suddenly appeared during already advanced vote counts and virtually all of

them for Biden. Early Trump leads in crucial swing states suddenly began to close and disappear. The pandemic hoax was used as the excuse to issue almost limitless numbers of mail-in ballots with no checks to establish that the recipients were still alive or lived at that address. They were sent to streams of people who had not even asked for them. Private organisations were employed to gather these ballots and who knows what they did with them before they turned up at the counts. The American election system has been manipulated over decades to become a sick joke with more holes than a Swiss cheese for the express purpose of dictating the results. Then there was the criminal manipulation of information by Sabbatian tech giants like Facebook, Twi er and Google-owned YouTube which deleted pro-Trump, anti-Biden accounts and posts while everything in support of Biden was le alone. Sabbatians wanted Biden to win because a er the dividing of America it was time for full-on Woke and every aspect of the Cult agenda to be unleashed.

Hunter gatherer

Extreme Silicon Valley bias included blocking information by the New York Post exposing a Biden scandal that should have ended his bid for president in the final weeks of the campaign. Hunter Biden, his monumentally corrupt son, is reported to have sent a laptop to be repaired at a local store and failed to return for it. Time passed until the laptop became the property of the store for non-payment of the bill. When the owner saw what was on the hard drive he gave a copy to the FBI who did nothing even though it confirmed widespread corruption in which the Joe Biden family were using his political position, especially when he was vice president to Obama, to make multiple millions in countries around the world and most notably Ukraine and China. Hunter Biden’s one-time business partner Tony Bobulinski went public when the story broke in the New York Post to confirm the corruption he saw and that Joe Biden not only knew what was going on he also profited from the spoils. Millions were handed over by a Chinese company with close

connections – like all major businesses in China – to the Chinese communist party of President Xi Jinping. Joe Biden even boasted at a meeting of the Cult’s World Economic Forum that as vice president he had ordered the government of Ukraine to fire a prosecutor. What he didn’t mention was that the same man just happened to be investigating an energy company which was part of Hunter Biden’s corrupt portfolio. The company was paying him big bucks for no other reason than the influence his father had. Overnight Biden’s presidential campaign should have been over given that he had lied publicly about not knowing what his son was doing. Instead almost the entire Sabbatian-owned mainstream media and Sabbatianowned Silicon Valley suppressed circulation of the story. This alone went a mighty way to rigging the election of 2020. Cult assets like Mark Zuckerberg at Facebook also spent hundreds of millions to be used in support of Biden and vote ‘administration’. The Cult had used Trump as the focus to divide America and was now desperate to bring in moronic, pliable, corrupt Biden to complete the double-whammy. No way were they going to let li le things like the will of the people thwart their plan. Silicon Valley widely censored claims that the election was rigged because it was rigged. For the same reason anyone claiming it was rigged was denounced as a ‘white supremacist’ including the pathetically few Republican politicians willing to say so. Right across the media where the claim was mentioned it was described as a ‘false claim’ even though these excuses for ‘journalists’ would have done no research into the subject whatsoever. Trump won seven million more votes than any si ing president had ever achieved while somehow a cognitively-challenged soon to be 78-year-old who was hidden away from the public for most of the campaign managed to win more votes than any presidential candidate in history. It makes no sense. You only had to see election rallies for both candidates to witness the enthusiasm for Trump and the apathy for Biden. Tens of thousands would a end Trump events while Biden was speaking in empty car parks with o en only television crews a ending and framing their shots to hide the fact that no one was there. It was pathetic to see

footage come to light of Biden standing at a podium making speeches only to TV crews and party fixers while reading the words wri en for him on massive Teleprompter screens. So, yes, those protestors on January 6th had a point about election rigging, but some were about to walk into a trap laid for them in Washington by the Cult Deep State and its QAnon Psyop. This was the Capitol Hill riot ludicrously dubbed an ‘insurrection’.

The spider and the fly

Renegade Minds know there are not two ‘sides’ in politics, only one side, the Cult, working through all ‘sides’. It’s a stage show, a puppet show, to direct the perceptions of the population into focusing on diversions like parties and candidates while missing the puppeteers with their hands holding all the strings. The Capitol Hill ‘insurrection’ brings us back to the Li le Big Horn. Having created two distinct opposing groupings – Woke and Pushbackers – the trap was about to be sprung. Pushbackers were to be encircled and isolated by associating them all in the public mind with Trump and then labelling Trump as some sort of Confederate leader. I knew immediately that the Capitol riot was a set-up because of two things. One was how easy the rioters got into the building with virtually no credible resistance and secondly I could see – as with the ‘Covid’ hoax in the West at the start of 2020 – how the Cult could exploit the situation to move its agenda forward with great speed. My experience of Cult techniques and activities over more than 30 years has showed me that while they do exploit situations they haven’t themselves created this never happens with events of fundamental agenda significance. Every time major events giving cultists the excuse to rapidly advance their plan you find they are manipulated into being for the specific reason of providing that excuse – ProblemReaction-Solution. Only a tiny minority of the huge crowd of Washington protestors sought to gain entry to the Capitol by smashing windows and breaching doors. That didn’t ma er. The whole crowd and all Pushbackers, even if they did not support Trump, were going to be lumped together as dangerous

insurrectionists and conspiracy theorists. The la er term came into widespread use through a CIA memo in the 1960s aimed at discrediting those questioning the nonsensical official story of the Kennedy assassination and it subsequently became widely employed by the media. It’s still being used by inept ‘journalists’ with no idea of its origin to discredit anyone questioning anything that authority claims to be true. When you are perpetrating a conspiracy you need to discredit the very word itself even though the dictionary definition of conspiracy is merely ‘the activity of secretly planning with other people to do something bad or illegal‘ and ‘a general agreement to keep silent about a subject for the purpose of keeping it secret’. On that basis there are conspiracies almost wherever you look. For obvious reasons the Cult and its lapdog media have to claim there are no conspiracies even though the word appears in state laws as with conspiracy to defraud, to murder, and to corrupt public morals. Agent provocateurs are widely used by the Cult Deep State to manipulate genuine people into acting in ways that suit the desired outcome. By genuine in this case I mean protestors genuinely supporting Trump and claims that the election was stolen. In among them, however, were agents of the state wearing the garb of Trump supporters and QAnon to pump-prime the Capital riot which some genuine Trump supporters naively fell for. I described the situation as ‘Come into my parlour said the spider to the fly’. Leaflets appeared through the Woke paramilitary arm Antifa, the anti-fascist fascists, calling on supporters to turn up in Washington looking like Trump supporters even though they hated him. Some of those arrested for breaching the Capitol Building were sourced to Antifa and its stable mate Black Lives Ma er. Both organisations are funded by Cult billionaires and corporations. One man charged for the riot was according to his lawyer a former FBI agent who had held top secret security clearance for 40 years. A orney Thomas Plofchan said of his client, 66-year-old Thomas Edward Caldwell: He has held a Top Secret Security Clearance since 1979 and has undergone multiple Special Background Investigations in support of his clearances. After retiring from the Navy, he

worked as a section chief for the Federal Bureau of Investigation from 2009-2010 as a GS-12 [mid-level employee]. He also formed and operated a consulting firm performing work, often classified, for U.S government customers including the US. Drug Enforcement Agency, Department of Housing and Urban Development, the US Coast Guard, and the US Army Personnel Command.

A judge later released Caldwell pending trial in the absence of evidence about a conspiracy or that he tried to force his way into the building. The New York Post reported a ‘law enforcement source‘ as saying that ‘at least two known Antifa members were spo ed’ on camera among Trump supporters during the riot while one of the rioters arrested was John Earle Sullivan, a seriously extreme Black Lives Ma er Trump-hater from Utah who was previously arrested and charged in July, 2020, over a BLM-Antifa riot in which drivers were threatened and one was shot. Sullivan is the founder of Utahbased Insurgence USA which is an affiliate of the Cult-created-andfunded Black Lives Ma er movement. Footage appeared and was then deleted by Twi er of Trump supporters calling out Antifa infiltrators and a group was filmed changing into pro-Trump clothing before the riot. Security at the building was pathetic – as planned. Colonel Leroy Fletcher Prouty, a man with long experience in covert operations working with the US security apparatus, once described the tell-tale sign to identify who is involved in an assassination. He said: No one has to direct an assassination – it happens. The active role is played secretly by permitting it to happen. This is the greatest single clue. Who has the power to call off or reduce the usual security precautions?

This principle applies to many other situations and certainly to the Capitol riot of January 6th, 2021.

The sting

With such a big and potentially angry crowd known to be gathering near the Capitol the security apparatus would have had a major police detail to defend the building with National Guard troops on

standby given the strength of feeling among people arriving from all over America encouraged by the QAnon Psyop and statements by Donald Trump. Instead Capitol Police ‘security’ was flimsy, weak, and easily breached. The same number of officers was deployed as on a regular day and that is a blatant red flag. They were not staffed or equipped for a possible riot that had been an obvious possibility in the circumstances. No protective and effective fencing worth the name was put in place and there were no contingency plans. The whole thing was basically a case of standing aside and waving people in. Once inside police mostly backed off apart from one Capitol police officer who ridiculously shot dead unarmed Air Force veteran protestor Ashli Babbi without a warning as she climbed through a broken window. The ‘investigation’ refused to name or charge the officer a er what must surely be considered a murder in the circumstances. They just li ed a carpet and swept. The story was endlessly repeated about five people dying in the ‘armed insurrection’ when there was no report of rioters using weapons. Apart from Babbi the other four died from a heart a ack, strokes and apparently a drug overdose. Capitol police officer Brian Sicknick was reported to have died a er being bludgeoned with a fire extinguisher when he was alive a er the riot was over and died later of what the Washington Medical Examiner’s Office said was a stroke. Sicknick had no external injuries. The lies were delivered like rapid fire. There was a narrative to build with incessant repetition of the lie until the lie became the accepted ‘everybody knows that’ truth. The ‘Big Lie’ technique of Nazi Propaganda Minister Joseph Goebbels is constantly used by the Cult which was behind the Nazis and is today behind the ‘Covid’ and ‘climate change’ hoaxes. Goebbels said: If you tell a lie big enough and keep repeating it, people will eventually come to believe it. The lie can be maintained only for such time as the State can shield the people from the political, economic and/or military consequences of the lie. It thus becomes vitally important for the State to use all of its powers to repress dissent, for the truth is the mortal enemy of the lie, and thus by extension, the truth is the greatest enemy of the State.

Most protestors had a free run of the Capitol Building. This allowed pictures to be taken of rioters in iconic parts of the building including the Senate chamber which could be used as propaganda images against all Pushbackers. One Congresswoman described the scene as ‘the worst kind of non-security anybody could ever imagine’. Well, the first part was true, but someone obviously did imagine it and made sure it happened. Some photographs most widely circulated featured people wearing QAnon symbols and now the Psyop would be used to dub all QAnon followers with the ubiquitous fit-all label of ‘white supremacist’ and ‘insurrectionists’. When a Muslim extremist called Noah Green drove his car at two police officers at the Capitol Building killing one in April, 2021, there was no such political and media hysteria. They were just disappointed he wasn’t white.

The witch-hunt

Government prosecutor Michael Sherwin, an aggressive, dark-eyed, professional Ro weiler led the ‘investigation’ and to call it over the top would be to understate reality a thousand fold. Hundreds were tracked down and arrested for the crime of having the wrong political views and people were jailed who had done nothing more than walk in the building, commi ed no violence or damage to property, took a few pictures and le . They were labelled a ‘threat to the Republic’ while Biden sat in the White House signing executive orders wri en for him that were dismantling ‘the Republic’. Even when judges ruled that a mother and son should not be in jail the government kept them there. Some of those arrested have been badly beaten by prison guards in Washington and lawyers for one man said he suffered a fractured skull and was made blind in one eye. Meanwhile a woman is shot dead for no reason by a Capitol Police officer and we are not allowed to know who he is never mind what has happened to him although that will be nothing. The Cult’s QAnon/Trump sting to identify and isolate Pushbackers and then target them on the road to crushing and deleting them was a resounding success. You would have thought the Russians had

invaded the building at gunpoint and lined up senators for a firing squad to see the political and media reaction. Congresswoman Alexandria Ocasio-Cortez is a child in a woman’s body, a terribletwos, me, me, me, Woker narcissist of such proportions that words have no meaning. She said she thought she was going to die when ‘insurrectionists’ banged on her office door. It turned out she wasn’t even in the Capitol Building when the riot was happening and the ‘banging’ was a Capitol Police officer. She referred to herself as a ‘survivor’ which is an insult to all those true survivors of violent and sexual abuse while she lives her pampered and privileged life talking drivel for a living. Her Woke colleague and fellow meganarcissist Rashida Tlaib broke down describing the devastating effect on her, too, of not being in the building when the rioters were there. Ocasio-Cortez and Tlaib are members of a fully-Woke group of Congresswomen known as ‘The Squad’ along with Ilhan Omar and Ayanna Pressley. The Squad from what I can see can be identified by its vehement anti-white racism, anti-white men agenda, and, as always in these cases, the absence of brain cells on active duty. The usual suspects were on the riot case immediately in the form of Democrat ultra-Zionist senators and operatives Chuck Schumer and Adam Schiff demanding that Trump be impeached for ‘his part in the insurrection’. The same pair of prats had led the failed impeachment of Trump over the invented ‘Russia collusion’ nonsense which claimed Russia had helped Trump win the 2016 election. I didn’t realise that Tel Aviv had been relocated just outside Moscow. I must find an up-to-date map. The Russia hoax was a Sabbatian operation to keep Trump occupied and impotent and to stop any rapport with Russia which the Cult wants to retain as a perceptual enemy to be pulled out at will. Puppet Biden began a acking Russia when he came to office as the Cult seeks more upheaval, division and war across the world. A two-year stage show ‘Russia collusion inquiry’ headed by the not-very-bright former 9/11 FBI chief Robert Mueller, with support from 19 lawyers, 40 FBI agents plus intelligence analysts, forensic accountants and other

staff, devoured tens of millions of dollars and found no evidence of Russia collusion which a ten-year-old could have told them on day one. Now the same moronic Schumer and Schiff wanted a second impeachment of Trump over the Capitol ‘insurrection’ (riot) which the arrested development of Schumer called another ‘Pearl Harbor’ while others compared it with 9/11 in which 3,000 died and, in the case of CNN, with the Rwandan genocide in the 1990s in which an estimated 500,000 to 600,000 were murdered, between 250, 000 and 500,000 women were raped, and populations of whole towns were hacked to death with machetes. To make those comparisons purely for Cult political reasons is beyond insulting to those that suffered and lost their lives and confirms yet again the callous inhumanity that we are dealing with. Schumer is a monumental idiot and so is Schiff, but they serve the Cult agenda and do whatever they’re told so they get looked a er. Talking of idiots – another inane man who spanned the Russia and Capitol impeachment a empts was Senator Eric Swalwell who had the nerve to accuse Trump of collusion with the Russians while sleeping with a Chinese spy called Christine Fang or ‘Fang Fang’ which is straight out of a Bond film no doubt starring Klaus Schwab as the bloke living on a secret island and controlling laser weapons positioned in space and pointing at world capitals. Fang Fang plays the part of Bond’s infiltrator girlfriend which I’m sure she would enjoy rather more than sharing a bed with the brainless Swalwell, lying back and thinking of China. The FBI eventually warned Swalwell about Fang Fang which gave her time to escape back to the Chinese dictatorship. How very thoughtful of them. The second Trump impeachment also failed and hardly surprising when an impeachment is supposed to remove a si ing president and by the time it happened Trump was no longer president. These people are running your country America, well, officially anyway. Terrifying isn’t it?

Outcomes tell the story - always

The outcome of all this – and it’s the outcome on which Renegade Minds focus, not the words – was that a vicious, hysterical and

obviously pre-planned assault was launched on Pushbackers to censor, silence and discredit them and even targeted their right to earn a living. They have since been condemned as ‘domestic terrorists’ that need to be treated like Al-Qaeda and Islamic State. ‘Domestic terrorists’ is a label the Cult has been trying to make stick since the period of the Oklahoma bombing in 1995 which was blamed on ‘far-right domestic terrorists’. If you read The Trigger you will see that the bombing was clearly a Problem-Reaction-Solution carried out by the Deep State during a Bill Clinton administration so corrupt that no dictionary definition of the term would even nearly suffice. Nearly 30, 000 troops were deployed from all over America to the empty streets of Washington for Biden’s inauguration. Ten thousand of them stayed on with the pretext of protecting the capital from insurrectionists when it was more psychological programming to normalise the use of the military in domestic law enforcement in support of the Cult plan for a police-military state. Biden’s fascist administration began a purge of ‘wrong-thinkers’ in the military which means anyone that is not on board with Woke. The Capitol Building was surrounded by a fence with razor wire and the Land of the Free was further symbolically and literally dismantled. The circle was completed with the installation of Biden and the exploitation of the QAnon Psyop. America had never been so divided since the civil war of the 19th century, Pushbackers were isolated and dubbed terrorists and now, as was always going to happen, the Cult immediately set about deleting what li le was le of freedom and transforming American society through a swish of the hand of the most controlled ‘president’ in American history leading (officially at least) the most extreme regime since the country was declared an independent state on July 4th, 1776. Biden issued undebated, dictatorial executive orders almost by the hour in his opening days in office across the whole spectrum of the Cult wish-list including diluting controls on the border with Mexico allowing thousands of migrants to illegally enter the United States to transform the demographics of America and import an election-changing number of perceived Democrat

voters. Then there were Biden deportation amnesties for the already illegally resident (estimated to be as high as 20 or even 30 million). A bill before Congress awarded American citizenship to anyone who could prove they had worked in agriculture for just 180 days in the previous two years as ‘Big Ag’ secured its slave labour long-term. There were the plans to add new states to the union such as Puerto Rico and making Washington DC a state. They are all parts of a plan to ensure that the Cult-owned Woke Democrats would be permanently in power.

Border – what border?

I have exposed in detail in other books how mass immigration into the United States and Europe is the work of Cult networks fuelled by the tens of billions spent to this and other ends by George Soros and his global Open Society (open borders) Foundations. The impact can be seen in America alone where the population has increased by 100 million in li le more than 30 years mostly through immigration. I wrote in The Answer that the plan was to have so many people crossing the southern border that the numbers become unstoppable and we are now there under Cult-owned Biden. El Salvador in Central America puts the scale of what is happening into context. A third of the population now lives in the United States, much of it illegally, and many more are on the way. The methodology is to crush Central and South American countries economically and spread violence through machete-wielding psychopathic gangs like MS-13 based in El Salvador and now operating in many American cities. Biden-imposed lax security at the southern border means that it is all but open. He said before his ‘election’ that he wanted to see a surge towards the border if he became president and that was the green light for people to do just that a er election day to create the human disaster that followed for both America and the migrants. When that surge came the imbecilic Alexandria Ocasio-Cortez said it wasn’t a ‘surge’ because they are ‘children, not insurgents’ and the term ‘surge’ (used by Biden) was a claim of ‘white supremacists’.

This disingenuous lady may one day enter the realm of the most basic intelligence, but it won’t be any time soon. Sabbatians and the Cult are in the process of destroying America by importing violent people and gangs in among the genuine to terrorise American cities and by overwhelming services that cannot cope with the sheer volume of new arrivals. Something similar is happening in Europe as Western society in general is targeted for demographic and cultural transformation and upheaval. The plan demands violence and crime to create an environment of intimidation, fear and division and Soros has been funding the election of district a orneys across America who then stop prosecuting many crimes, reduce sentences for violent crimes and free as many violent criminals as they can. Sabbatians are creating the chaos from which order – their order – can respond in a classic Problem-Reaction-Solution. A Freemasonic moto says ‘Ordo Ab Chao’ (Order out of Chaos) and this is why the Cult is constantly creating chaos to impose a new ‘order’. Here you have the reason the Cult is constantly creating chaos. The ‘Covid’ hoax can be seen with those entering the United States by plane being forced to take a ‘Covid’ test while migrants flooding through southern border processing facilities do not. Nothing is put in the way of mass migration and if that means ignoring the government’s own ‘Covid’ rules then so be it. They know it’s all bullshit anyway. Any pushback on this is denounced as ‘racist’ by Wokers and Sabbatian fronts like the ultra-Zionist Anti-Defamation League headed by the appalling Jonathan Greenbla which at the same time argues that Israel should not give citizenship and voting rights to more Palestinian Arabs or the ‘Jewish population’ (in truth the Sabbatian network) will lose control of the country.

Society-changing numbers

Biden’s masters have declared that countries like El Salvador are so dangerous that their people must be allowed into the United States for humanitarian reasons when there are fewer murders in large parts of many Central American countries than in US cities like

Baltimore. That is not to say Central America cannot be a dangerous place and Cult-controlled American governments have been making it so since way back, along with the dismantling of economies, in a long-term plan to drive people north into the United States. Parts of Central America are very dangerous, but in other areas the story is being greatly exaggerated to justify relaxing immigration criteria. Migrants are being offered free healthcare and education in the United States as another incentive to head for the border and there is no requirement to be financially independent before you can enter to prevent the resources of America being drained. You can’t blame migrants for seeking what they believe will be a be er life, but they are being played by the Cult for dark and nefarious ends. The numbers since Biden took office are huge. In February, 2021, more than 100,000 people were known to have tried to enter the US illegally through the southern border (it was 34,000 in the same month in 2020) and in March it was 170,000 – a 418 percent increase on March, 2020. These numbers are only known people, not the ones who get in unseen. The true figure for migrants illegally crossing the border in a single month was estimated by one congressman at 250,000 and that number will only rise under Biden’s current policy. Gangs of murdering drug-running thugs that control the Mexican side of the border demand money – thousands of dollars – to let migrants cross the Rio Grande into America. At the same time gun ba les are breaking out on the border several times a week between rival Mexican drug gangs (which now operate globally) who are equipped with sophisticated military-grade weapons, grenades and armoured vehicles. While the Capitol Building was being ‘protected’ from a non-existent ‘threat’ by thousands of troops, and others were still deployed at the time in the Cult Neocon war in Afghanistan, the southern border of America was le to its fate. This is not incompetence, it is cold calculation. By March, 2021, there were 17,000 unaccompanied children held at border facilities and many of them are ensnared by people traffickers for paedophile rings and raped on their journey north to America. This is not conjecture – this is fact. Many of those designated

children are in reality teenage boys or older. Meanwhile Wokers posture their self-purity for encouraging poor and tragic people to come to America and face this nightmare both on the journey and at the border with the disgusting figure of House Speaker Nancy Pelosi giving disingenuous speeches about caring for migrants. The woman’s evil. Wokers condemned Trump for having children in cages at the border (so did Obama, Shhhh), but now they are sleeping on the floor without access to a shower with one border facility 729 percent over capacity. The Biden insanity even proposed flying migrants from the southern border to the northern border with Canada for ‘processing’. The whole shambles is being overseen by ultra-Zionist Secretary of Homeland Security, the moronic liar Alejandro Mayorkas, who banned news cameras at border facilities to stop Americans seeing what was happening. Mayorkas said there was not a ban on news crews; it was just that they were not allowed to film. Alongside him at Homeland Security is another ultra-Zionist Cass Sunstein appointed by Biden to oversee new immigration laws. Sunstein despises conspiracy researchers to the point where he suggests they should be banned or taxed for having such views. The man is not bonkers or anything. He’s perfectly well-adjusted, but adjusted to what is the question. Criticise what is happening and you are a ‘white supremacist’ when earlier non-white immigrants also oppose the numbers which effect their lives and opportunities. Black people in poor areas are particularly damaged by uncontrolled immigration and the increased competition for work opportunities with those who will work for less. They are also losing voting power as Hispanics become more dominant in former black areas. It’s a downward spiral for them while the billionaires behind the policy drone on about how much they care about black people and ‘racism’. None of this is about compassion for migrants or black people – that’s just wind and air. Migrants are instead being mercilessly exploited to transform America while the countries they leave are losing their future and the same is true in Europe. Mass immigration may now be the work of Woke Democrats, but it can be traced back to the 1986 Immigration Reform and Control Act (it

wasn’t) signed into law by Republican hero President Ronald Reagan which gave amnesty to millions living in the United States illegally and other incentives for people to head for the southern border. Here we have the one-party state at work again.

Save me syndrome

Almost every aspect of what I have been exposing as the Cult agenda was on display in even the first days of ‘Biden’ with silencing of Pushbackers at the forefront of everything. A Renegade Mind will view the Trump years and QAnon in a very different light to their supporters and advocates as the dots are connected. The QAnon/Trump Psyop has given the Cult all it was looking for. We may not know how much, or li le, that Trump realised he was being used, but that’s a side issue. This pincer movement produced the desired outcome of dividing America and having Pushbackers isolated. To turn this around we have to look at new routes to empowerment which do not include handing our power to other people and groups through what I will call the ‘Save Me Syndrome’ – ‘I want someone else to do it so that I don’t have to’. We have seen this at work throughout human history and the QAnon/Trump Psyop is only the latest incarnation alongside all the others. Religion is an obvious expression of this when people look to a ‘god’ or priest to save them or tell them how to be saved and then there are ‘save me’ politicians like Trump. Politics is a diversion and not a ‘saviour’. It is a means to block positive change, not make it possible. Save Me Syndrome always comes with the same repeating theme of handing your power to whom or what you believe will save you while your real ‘saviour’ stares back from the mirror every morning. Renegade Minds are constantly vigilant in this regard and always asking the question ‘What can I do?’ rather than ‘What can someone else do for me?’ Gandhi was right when he said: ‘You must be the change you want to see in the world.’ We are indeed the people we have been waiting for. We are presented with a constant ra of reasons to concede that power to others and forget where the real power is. Humanity has the numbers and the Cult does not. It has to

use diversion and division to target the unstoppable power that comes from unity. Religions, governments, politicians, corporations, media, QAnon, are all different manifestations of this powerdiversion and dilution. Refusing to give your power to governments and instead handing it to Trump and QAnon is not to take a new direction, but merely to recycle the old one with new names on the posters. I will explore this phenomenon as we proceed and how to break the cycles and recycles that got us here through the mists of repeating perception and so repeating history. For now we shall turn to the most potent example in the entire human story of the consequences that follow when you give your power away. I am talking, of course, of the ‘Covid’ hoax.

CHAPTER FOUR ‘Covid’: Calculated catastrophe Facts are threatening to those invested in fraud DaShanne Stokes

W

e can easily unravel the real reason for the ‘Covid pandemic’ hoax by employing the Renegade Mind methodology that I have outlined this far. We’ll start by comparing the long-planned Cult outcome with the ‘Covid pandemic’ outcome. Know the outcome and you’ll see the journey. I have highlighted the plan for the Hunger Games Society which has been in my books for so many years with the very few controlling the very many through ongoing dependency. To create this dependency it is essential to destroy independent livelihoods, businesses and employment to make the population reliant on the state (the Cult) for even the basics of life through a guaranteed pi ance income. While independence of income remained these Cult ambitions would be thwarted. With this knowledge it was easy to see where the ‘pandemic’ hoax was going once talk of ‘lockdowns’ began and the closing of all but perceived ‘essential’ businesses to ‘save’ us from an alleged ‘deadly virus’. Cult corporations like Amazon and Walmart were naturally considered ‘essential’ while mom and pop shops and stores had their doors closed by fascist decree. As a result with every new lockdown and new regulation more small and medium, even large businesses not owned by the Cult, went to the wall while Cult giants and their frontmen and women grew financially fa er by the second. Mom and pop were

denied an income and the right to earn a living and the wealth of people like Jeff Bezos (Amazon), Mark Zuckerberg (Facebook) and Sergei Brin and Larry Page (Google/Alphabet) have reached record levels. The Cult was increasing its own power through further dramatic concentrations of wealth while the competition was being destroyed and brought into a state of dependency. Lockdowns have been instigated to secure that very end and were never anything to do with health. My brother Paul spent 45 years building up a bus repair business, but lockdowns meant buses were running at a fraction of normal levels for months on end. Similar stories can told in their hundreds of millions worldwide. Efforts of a lifetime coldly destroyed by Cult multi-billionaires and their lackeys in government and law enforcement who continued to earn their living from the taxation of the people while denying the right of the same people to earn theirs. How different it would have been if those making and enforcing these decisions had to face the same financial hardships of those they affected, but they never do.

Gates of Hell

Behind it all in the full knowledge of what he is doing and why is the psychopathic figure of Cult operative Bill Gates. His puppet Tedros at the World Health Organization declared ‘Covid’ a pandemic in March, 2020. The WHO had changed the definition of a ‘pandemic’ in 2009 just a month before declaring the ‘swine flu pandemic’ which would not have been so under the previous definition. The same applies to ‘Covid’. The definition had included… ‘an infection by an infectious agent, occurring simultaneously in different countries, with a significant mortality rate relative to the proportion of the population infected’. The new definition removed the need for ‘significant mortality’. The ‘pandemic’ has been fraudulent even down to the definition, but Gates demanded economy-destroying lockdowns, school closures, social distancing, mandatory masks, a ‘vaccination’ for every man, woman and child on the planet and severe consequences and restrictions for those that refused. Who gave him this power? The

Cult did which he serves like a li le boy in short trousers doing what his daddy tells him. He and his psychopathic missus even smiled when they said that much worse was to come (what they knew was planned to come). Gates responded in the ma er-of-fact way of all psychopaths to a question about the effect on the world economy of what he was doing: Well, it won’t go to zero but it will shrink. Global GDP is probably going to take the biggest hit ever [Gates was smiling as he said this] … in my lifetime this will be the greatest economic hit. But you don’t have a choice. People act as if you have a choice. People don’t feel like going to the stadium when they might get infected … People are deeply affected by seeing these stats, by knowing they could be part of the transmission chain, old people, their parents and grandparents, could be affected by this, and so you don’t get to say ignore what is going on here. There will be the ability to open up, particularly in rich countries, if things are done well over the next few months, but for the world at large normalcy only returns when we have largely vaccinated the entire population.

The man has no compassion or empathy. How could he when he’s a psychopath like all Cult players? My own view is that even beyond that he is very seriously mentally ill. Look in his eyes and you can see this along with his crazy flailing arms. You don’t do what he has done to the world population since the start of 2020 unless you are mentally ill and at the most extreme end of psychopathic. You especially don’t do it when to you know, as we shall see, that cases and deaths from ‘Covid’ are fakery and a product of monumental figure massaging. ‘These stats’ that Gates referred to are based on a ‘test’ that’s not testing for the ‘virus’ as he has known all along. He made his fortune with big Cult support as an infamously ruthless so ware salesman and now buys global control of ‘health’ (death) policy without the population he affects having any say. It’s a breathtaking outrage. Gates talked about people being deeply affected by fear of ‘Covid’ when that was because of him and his global network lying to them minute-by-minute supported by a lying media that he seriously influences and funds to the tune of hundreds of millions. He’s handed big sums to media operations including the BBC, NBC, Al Jazeera, Univision, PBS NewsHour,

ProPublica, National Journal, The Guardian, The Financial Times, The Atlantic, Texas Tribune, USA Today publisher Ganne , Washington Monthly, Le Monde, Center for Investigative Reporting, Pulitzer Center on Crisis Reporting, National Press Foundation, International Center for Journalists, Solutions Journalism Network, the Poynter Institute for Media Studies, and many more. Gates is everywhere in the ‘Covid’ hoax and the man must go to prison – or a mental facility – for the rest of his life and his money distributed to those he has taken such enormous psychopathic pleasure in crushing.

The Muscle

The Hunger Games global structure demands a police-military state – a fusion of the two into one force – which viciously imposes the will of the Cult on the population and protects the Cult from public rebellion. In that regard, too, the ‘Covid’ hoax just keeps on giving. O en unlawful, ridiculous and contradictory ‘Covid’ rules and regulations have been policed across the world by moronic automatons and psychopaths made faceless by face-nappy masks and acting like the Nazi SS and fascist blackshirts and brownshirts of Hitler and Mussolini. The smallest departure from the rules decreed by the psychos in government and their clueless gofers were jumped upon by the face-nappy fascists. Brutality against public protestors soon became commonplace even on girls, women and old people as the brave men with the batons – the Face-Nappies as I call them – broke up peaceful protests and handed out fines like confe i to people who couldn’t earn a living let alone pay hundreds of pounds for what was once an accepted human right. Robot Face-Nappies of No ingham police in the English East Midlands fined one group £11,000 for a ending a child’s birthday party. For decades I charted the transformation of law enforcement as genuine, decent officers were replaced with psychopaths and the brain dead who would happily and brutally do whatever their masters told them. Now they were let loose on the public and I would emphasise the point that none of this just happened. The step-by-step change in the dynamic between police and public was orchestrated from the shadows by

those who knew where this was all going and the same with the perceptual reframing of those in all levels of authority and official administration through ‘training courses’ by organisations such as Common Purpose which was created in the late 1980s and given a massive boost in Blair era Britain until it became a global phenomenon. Supposed public ‘servants’ began to view the population as the enemy and the same was true of the police. This was the start of the explosion of behaviour manipulation organisations and networks preparing for the all-war on the human psyche unleashed with the dawn of 2020. I will go into more detail about this later in the book because it is a core part of what is happening. Police desecrated beauty spots to deter people gathering and arrested women for walking in the countryside alone ‘too far’ from their homes. We had arrogant, clueless sergeants in the Isle of Wight police where I live posting on Facebook what they insisted the population must do or else. A schoolmaster sergeant called Radford looked young enough for me to ask if his mother knew he was out, but he was posting what he expected people to do while a Sergeant Wilkinson boasted about fining lads for meeting in a McDonald’s car park where they went to get a lockdown takeaway. Wilkinson added that he had even cancelled their order. What a pair of prats these people are and yet they have increasingly become the norm among Jackboot Johnson’s Yellowshirts once known as the British police. This was the theme all over the world with police savagery common during lockdown protests in the United States, the Netherlands, and the fascist state of Victoria in Australia under its tyrannical and again moronic premier Daniel Andrews. Amazing how tyrannical and moronic tend to work as a team and the same combination could be seen across America as arrogant, narcissistic Woke governors and mayors such as Gavin Newsom (California), Andrew Cuomo (New York), Gretchen Whitmer (Michigan), Lori Lightfoot (Chicago) and Eric Garce i (Los Angeles) did their Nazi and Stalin impressions with the full support of the compliant brutality of their enforcers in uniform as they arrested small business owners defying

fascist shutdown orders and took them to jail in ankle shackles and handcuffs. This happened to bistro owner Marlena Pavlos-Hackney in Gretchen Whitmer’s fascist state of Michigan when police arrived to enforce an order by a state-owned judge for ‘pu ing the community at risk’ at a time when other states like Texas were dropping restrictions and migrants were pouring across the southern border without any ‘Covid’ questions at all. I’m sure there are many officers appalled by what they are ordered to do, but not nearly enough of them. If they were truly appalled they would not do it. As the months passed every opportunity was taken to have the military involved to make their presence on the streets ever more familiar and ‘normal’ for the longer-term goal of police-military fusion. Another crucial element to the Hunger Games enforcement network has been encouraging the public to report neighbours and others for ‘breaking the lockdown rules’. The group faced with £11,000 in fines at the child’s birthday party would have been dobbed-in by a neighbour with a brain the size of a pea. The technique was most famously employed by the Stasi secret police in communist East Germany who had public informants placed throughout the population. A police chief in the UK says his force doesn’t need to carry out ‘Covid’ patrols when they are flooded with so many calls from the public reporting other people for visiting the beach. Dorset police chief James Vaughan said people were so enthusiastic about snitching on their fellow humans they were now operating as an auxiliary arm of the police: ‘We are still ge ing around 400 reports a week from the public, so we will respond to reports …We won’t need to be doing hotspot patrols because people are very quick to pick the phone up and tell us.’ Vaughan didn’t say that this is a pillar of all tyrannies of whatever complexion and the means to hugely extend the reach of enforcement while spreading distrust among the people and making them wary of doing anything that might get them reported. Those narcissistic Isle of Wight sergeants Radford and Wilkinson never fail to add a link to their Facebook posts where the public can inform on their fellow slaves.

Neither would be self-aware enough to realise they were imitating the Stasi which they might well never have heard of. Government psychologists that I will expose later laid out a policy to turn communities against each other in the same way.

A coincidence? Yep, and I can knit fog

I knew from the start of the alleged pandemic that this was a Cult operation. It presented limitless potential to rapidly advance the Cult agenda and exploit manipulated fear to demand that every man, woman and child on the planet was ‘vaccinated’ in a process never used on humans before which infuses self-replicating synthetic material into human cells. Remember the plan to transform the human body from a biological to a synthetic biological state. I’ll deal with the ‘vaccine’ (that’s not actually a vaccine) when I focus on the genetic agenda. Enough to say here that mass global ‘vaccination’ justified by this ‘new virus’ set alarms ringing a er 30 years of tracking these people and their methods. The ‘Covid’ hoax officially beginning in China was also a big red flag for reasons I will be explaining. The agenda potential was so enormous that I could dismiss any idea that the ‘virus’ appeared naturally. Major happenings with major agenda implications never occur without Cult involvement in making them happen. My questions were twofold in early 2020 as the media began its campaign to induce global fear and hysteria: Was this alleged infectious agent released on purpose by the Cult or did it even exist at all? I then did what I always do in these situations. I sat, observed and waited to see where the evidence and information would take me. By March and early April synchronicity was strongly – and ever more so since then – pointing me in the direction of there is no ‘virus’. I went public on that with derision even from swathes of the alternative media that voiced a scenario that the Chinese government released the ‘virus’ in league with Deep State elements in the United States from a toplevel bio-lab in Wuhan where the ‘virus’ is said to have first appeared. I looked at that possibility, but I didn’t buy it for several reasons. Deaths from the ‘virus’ did not in any way match what they

would have been with a ‘deadly bioweapon’ and it is much more effective if you sell the illusion of an infectious agent rather than having a real one unless you can control through injection who has it and who doesn’t. Otherwise you lose control of events. A made-up ‘virus’ gives you a blank sheet of paper on which you can make it do whatever you like and have any symptoms or mutant ‘variants’ you choose to add while a real infectious agent would limit you to what it actually does. A phantom disease allows you to have endless ludicrous ‘studies’ on the ‘Covid’ dollar to widen the perceived impact by inventing ever more ‘at risk’ groups including one study which said those who walk slowly may be almost four times more likely to die from the ‘virus’. People are in psychiatric wards for less. A real ‘deadly bioweapon’ can take out people in the hierarchy that are not part of the Cult, but essential to its operation. Obviously they don’t want that. Releasing a real disease means you immediately lose control of it. Releasing an illusory one means you don’t. Again it’s vital that people are extra careful when dealing with what they want to hear. A bioweapon unleashed from a Chinese laboratory in collusion with the American Deep State may fit a conspiracy narrative, but is it true? Would it not be far more effective to use the excuse of a ‘virus’ to justify the real bioweapon – the ‘vaccine’? That way your disease agent does not have to be transmi ed and arrives directly through a syringe. I saw a French virologist Luc Montagnier quoted in the alternative media as saying he had discovered that the alleged ‘new’ severe acute respiratory syndrome coronavirus , or SARS-CoV-2, was made artificially and included elements of the human immunodeficiency ‘virus’ (HIV) and a parasite that causes malaria. SARS-CoV-2 is alleged to trigger an alleged illness called Covid-19. I remembered Montagnier’s name from my research years before into claims that an HIV ‘retrovirus’ causes AIDs – claims that were demolished by Berkeley virologist Peter Duesberg who showed that no one had ever proved that HIV causes acquired immunodeficiency syndrome or AIDS. Claims that become accepted as fact, publicly and medically, with no proof whatsoever are an ever-recurring story that profoundly applies to

‘Covid’. Nevertheless, despite the lack of proof, Montagnier’s team at the Pasteur Institute in Paris had a long dispute with American researcher Robert Gallo over which of them discovered and isolated the HIV ‘virus’ and with no evidence found it to cause AIDS. You will see later that there is also no evidence that any ‘virus’ causes any disease or that there is even such a thing as a ‘virus’ in the way it is said to exist. The claim to have ‘isolated’ the HIV ‘virus’ will be presented in its real context as we come to the shocking story – and it is a story – of SARS-CoV-2 and so will Montagnier’s assertion that he identified the full SARS-CoV-2 genome.

Hoax in the making

We can pick up the ‘Covid’ story in 2010 and the publication by the Rockefeller Foundation of a document called ‘Scenarios for the Future of Technology and International Development’. The inner circle of the Rockefeller family has been serving the Cult since John D. Rockefeller (1839-1937) made his fortune with Standard Oil. It is less well known that the same Rockefeller – the Bill Gates of his day – was responsible for establishing what is now referred to as ‘Big Pharma’, the global network of pharmaceutical companies that make outrageous profits dispensing scalpel and drug ‘medicine’ and are obsessed with pumping vaccines in ever-increasing number into as many human arms and backsides as possible. John D. Rockefeller was the driving force behind the creation of the ‘education’ system in the United States and elsewhere specifically designed to program the perceptions of generations therea er. The Rockefeller family donated exceptionally valuable land in New York for the United Nations building and were central in establishing the World Health Organization in 1948 as an agency of the UN which was created from the start as a Trojan horse and stalking horse for world government. Now enter Bill Gates. His family and the Rockefellers have long been extremely close and I have seen genealogy which claims that if you go back far enough the two families fuse into the same bloodline. Gates has said that the Bill and Melinda Gates Foundation was inspired by the Rockefeller Foundation and why not

when both are serving the same Cult? Major tax-exempt foundations are overwhelmingly criminal enterprises in which Cult assets fund the Cult agenda in the guise of ‘philanthropy’ while avoiding tax in the process. Cult operatives can become mega-rich in their role of front men and women for the psychopaths at the inner core and they, too, have to be psychopaths to knowingly serve such evil. Part of the deal is that a big percentage of the wealth gleaned from representing the Cult has to be spent advancing the ambitions of the Cult and hence you have the Rockefeller Foundation, Bill and Melinda Gates Foundation (and so many more) and people like George Soros with his global Open Society Foundations spending their billions in pursuit of global Cult control. Gates is a global public face of the Cult with his interventions in world affairs including Big Tech influence; a central role in the ‘Covid’ and ‘vaccine’ scam; promotion of the climate change shakedown; manipulation of education; geoengineering of the skies; and his food-control agenda as the biggest owner of farmland in America, his GMO promotion and through other means. As one writer said: ‘Gates monopolizes or wields disproportionate influence over the tech industry, global health and vaccines, agriculture and food policy (including biopiracy and fake food), weather modification and other climate technologies, surveillance, education and media.’ The almost limitless wealth secured through Microso and other not-allowedto-fail ventures (including vaccines) has been ploughed into a long, long list of Cult projects designed to enslave the entire human race. Gates and the Rockefellers have been working as one unit with the Rockefeller-established World Health Organization leading global ‘Covid’ policy controlled by Gates through his mouth-piece Tedros. Gates became the WHO’s biggest funder when Trump announced that the American government would cease its donations, but Biden immediately said he would restore the money when he took office in January, 2021. The Gates Foundation (the Cult) owns through limitless funding the world health system and the major players across the globe in the ‘Covid’ hoax.

Okay, with that background we return to that Rockefeller Foundation document of 2010 headed ‘Scenarios for the Future of Technology and International Development’ and its ‘imaginary’ epidemic of a virulent and deadly influenza strain which infected 20 percent of the global population and killed eight million in seven months. The Rockefeller scenario was that the epidemic destroyed economies, closed shops, offices and other businesses and led to governments imposing fierce rules and restrictions that included mandatory wearing of face masks and body-temperature checks to enter communal spaces like railway stations and supermarkets. The document predicted that even a er the height of the Rockefellerenvisaged epidemic the authoritarian rule would continue to deal with further pandemics, transnational terrorism, environmental crises and rising poverty. Now you may think that the Rockefellers are our modern-day seers or alternatively, and rather more likely, that they well knew what was planned a few years further on. Fascism had to be imposed, you see, to ‘protect citizens from risk and exposure’. The Rockefeller scenario document said: During the pandemic, national leaders around the world flexed their authority and imposed airtight rules and restrictions, from the mandatory wearing of face masks to body-temperature checks at the entries to communal spaces like train stations and supermarkets. Even after the pandemic faded, this more authoritarian control and oversight of citizens and their activities stuck and even intensified. In order to protect themselves from the spread of increasingly global problems – from pandemics and transnational terrorism to environmental crises and rising poverty – leaders around the world took a firmer grip on power. At first, the notion of a more controlled world gained wide acceptance and approval. Citizens willingly gave up some of their sovereignty – and their privacy – to more paternalistic states in exchange for greater safety and stability. Citizens were more tolerant, and even eager, for topdown direction and oversight, and national leaders had more latitude to impose order in the ways they saw fit. In developed countries, this heightened oversight took many forms: biometric IDs for all citizens, for example, and tighter regulation of key industries whose stability was deemed vital to national interests. In many developed countries, enforced cooperation with a suite of new regulations and agreements slowly but steadily restored both order and, importantly, economic growth.

There we have the prophetic Rockefellers in 2010 and three years later came their paper for the Global Health Summit in Beijing, China, when government representatives, the private sector, international organisations and groups met to discuss the next 100 years of ‘global health’. The Rockefeller Foundation-funded paper was called ‘Dreaming the Future of Health for the Next 100 Years and more prophecy ensued as it described a dystopian future: ‘The abundance of data, digitally tracking and linking people may mean the ‘death of privacy’ and may replace physical interaction with transient, virtual connection, generating isolation and raising questions of how values are shaped in virtual networks.’ Next in the ‘Covid’ hoax preparation sequence came a ‘table top’ simulation in 2018 for another ‘imaginary’ pandemic of a disease called Clade X which was said to kill 900 million people. The exercise was organised by the Gates-funded Johns Hopkins University’s Center for Health Security in the United States and this is the very same university that has been compiling the disgustingly and systematically erroneous global figures for ‘Covid’ cases and deaths. Similar Johns Hopkins health crisis scenarios have included the Dark Winter exercise in 2001 and Atlantic Storm in 2005.

Nostradamus 201

For sheer predictive genius look no further prophecy-watchers than the Bill Gates-funded Event 201 held only six weeks before the ‘coronavirus pandemic’ is supposed to have broken out in China and Event 201 was based on a scenario of a global ‘coronavirus pandemic’. Melinda Gates, the great man’s missus, told the BBC that he had ‘prepared for years’ for a coronavirus pandemic which told us what we already knew. Nostradamugates had predicted in a TED talk in 2015 that a pandemic was coming that would kill a lot of people and demolish the world economy. My god, the man is a machine – possibly even literally. Now here he was only weeks before the real thing funding just such a simulated scenario and involving his friends and associates at Johns Hopkins, the World Economic Forum Cult-front of Klaus Schwab, the United Nations,

Johnson & Johnson, major banks, and officials from China and the Centers for Disease Control in the United States. What synchronicity – Johns Hopkins would go on to compile the fraudulent ‘Covid’ figures, the World Economic Forum and Schwab would push the ‘Great Reset’ in response to ‘Covid’, the Centers for Disease Control would be at the forefront of ‘Covid’ policy in the United States, Johnson & Johnson would produce a ‘Covid vaccine’, and everything would officially start just weeks later in China. Spooky, eh? They were even accurate in creating a simulation of a ‘virus’ pandemic because the ‘real thing’ would also be a simulation. Event 201 was not an exercise preparing for something that might happen; it was a rehearsal for what those in control knew was going to happen and very shortly. Hours of this simulation were posted on the Internet and the various themes and responses mirrored what would soon be imposed to transform human society. News stories were inserted and what they said would be commonplace a few weeks later with still more prophecy perfection. Much discussion focused on the need to deal with misinformation and the ‘anti-vax movement’ which is exactly what happened when the ‘virus’ arrived – was said to have arrived – in the West. Cult-owned social media banned criticism and exposure of the official ‘virus’ narrative and when I said there was no ‘virus’ in early April, 2020, I was banned by one platform a er another including YouTube, Facebook and later Twi er. The mainstream broadcast media in Britain was in effect banned from interviewing me by the Tony-Blair-created government broadcasting censor Ofcom headed by career government bureaucrat Melanie Dawes who was appointed just as the ‘virus’ hoax was about to play out in January, 2020. At the same time the Ickonic media platform was using Vimeo, another ultra-Zionist-owned operation, while our own player was being created and they deleted in an instant hundreds of videos, documentaries, series and shows to confirm their unbelievable vindictiveness. We had copies, of course, and they had to be restored one by one when our player was ready. These people have no class. Sabbatian Facebook promised free advertisements for the Gates-

controlled World Health Organization narrative while deleting ‘false claims and conspiracy theories’ to stop ‘misinformation’ about the alleged coronavirus. All these responses could be seen just a short while earlier in the scenarios of Event 201. Extreme censorship was absolutely crucial for the Cult because the official story was so ridiculous and unsupportable by the evidence that it could never survive open debate and the free-flow of information and opinion. If you can’t win a debate then don’t have one is the Cult’s approach throughout history. Facebook’s li le boy front man – front boy – Mark Zuckerberg equated ‘credible and accurate information’ with official sources and exposing their lies with ‘misinformation’.

Silencing those that can see

The censorship dynamic of Event 201 is now the norm with an army of narrative-supporting ‘fact-checker’ organisations whose entire reason for being is to tell the public that official narratives are true and those exposing them are lying. One of the most appalling of these ‘fact-checkers’ is called NewsGuard founded by ultra-Zionist Americans Gordon Crovitz and Steven Brill. Crovitz is a former publisher of The Wall Street Journal, former Executive Vice President of Dow Jones, a member of the Council on Foreign Relations (CFR), and on the board of the American Association of Rhodes Scholars. The CFR and Rhodes Scholarships, named a er Rothschild agent Cecil Rhodes who plundered the gold and diamonds of South Africa for his masters and the Cult, have featured widely in my books. NewsGuard don’t seem to like me for some reason – I really can’t think why – and they have done all they can to have me censored and discredited which is, to quote an old British politician, like being savaged by a dead sheep. They are, however, like all in the censorship network, very well connected and funded by organisations themselves funded by, or connected to, Bill Gates. As you would expect with anything associated with Gates NewsGuard has an offshoot called HealthGuard which ‘fights online health care hoaxes’. How very kind. Somehow the NewsGuard European Managing Director Anna-Sophie Harling, a remarkably young-

looking woman with no broadcasting experience and li le hands-on work in journalism, has somehow secured a position on the ‘Content Board’ of UK government broadcast censor Ofcom. An executive of an organisation seeking to discredit dissidents of the official narratives is making decisions for the government broadcast ‘regulator’ about content?? Another appalling ‘fact-checker’ is Full Fact funded by George Soros and global censors Google and Facebook. It’s amazing how many activists in the ‘fact-checking’, ‘anti-hate’, arena turn up in government-related positions – people like UK Labour Party activist Imran Ahmed who heads the Center for Countering Digital Hate founded by people like Morgan McSweeney, now chief of staff to the Labour Party’s hapless and useless ‘leader’ Keir Starmer. Digital Hate – which is what it really is – uses the American spelling of Center to betray its connection to a transatlantic network of similar organisations which in 2020 shapeshi ed from a acking people for ‘hate’ to a acking them for questioning the ‘Covid’ hoax and the dangers of the ‘Covid vaccine’. It’s just a coincidence, you understand. This is one of Imran Ahmed’s hysterical statements: ‘I would go beyond calling anti-vaxxers conspiracy theorists to say they are an extremist group that pose a national security risk.’ No one could ever accuse this prat of understatement and he’s including in that those parents who are now against vaccines a er their children were damaged for life or killed by them. He’s such a nice man. Ahmed does the rounds of the Woke media ge ing so -ball questions from spineless ‘journalists’ who never ask what right he has to campaign to destroy the freedom of speech of others while he demands it for himself. There also seems to be an overrepresentation in Ofcom of people connected to the narrative-worshipping BBC. This incredible global network of narrative-support was super-vital when the ‘Covid’ hoax was played in the light of the mega-whopper lies that have to be defended from the spotlight cast by the most basic intelligence.

Setting the scene

The Cult plays the long game and proceeds step-by-step ensuring that everything is in place before major cards are played and they don’t come any bigger than the ‘Covid’ hoax. The psychopaths can’t handle events where the outcome isn’t certain and as li le as possible – preferably nothing – is le to chance. Politicians, government and medical officials who would follow direction were brought to illusory power in advance by the Cult web whether on the national stage or others like state governors and mayors of America. For decades the dynamic between officialdom, law enforcement and the public was changed from one of service to one of control and dictatorship. Behaviour manipulation networks established within government were waiting to impose the coming ‘Covid’ rules and regulations specifically designed to subdue and rewire the psyche of the people in the guise of protecting health. These included in the UK the Behavioural Insights Team part-owned by the British government Cabinet Office; the Scientific Pandemic Insights Group on Behaviours (SPI-B); and a whole web of intelligence and military groups seeking to direct the conversation on social media and control the narrative. Among them are the cyberwarfare (on the people) 77th Brigade of the British military which is also coordinated through the Cabinet Office as civilian and military leadership continues to combine in what they call the Fusion Doctrine. The 77th Brigade is a British equivalent of the infamous Israeli (Sabbatian) military cyberwarfare and Internet manipulation operation Unit 8200 which I expose at length in The Trigger. Also carefully in place were the medical and science advisers to government – many on the payroll past or present of Bill Gates – and a whole alternative structure of unelected government stood by to take control when elected parliaments were effectively closed down once the ‘Covid’ card was slammed on the table. The structure I have described here and so much more was installed in every major country through the Cult networks. The top-down control hierarchy looks like this: The Cult – Cult-owned Gates – the World Health Organization and Tedros – Gates-funded or controlled chief medical officers and science ‘advisers’ (dictators) in each country –

political ‘leaders’– law enforcement – The People. Through this simple global communication and enforcement structure the policy of the Cult could be imposed on virtually the entire human population so long as they acquiesced to the fascism. With everything in place it was time for the bu on to be pressed in late 2019/early 2020. These were the prime goals the Cult had to secure for its will to prevail: 1) Locking down economies, closing all but designated ‘essential’ businesses (Cult-owned corporations were ‘essential’), and pu ing the population under house arrest was an imperative to destroy independent income and employment and ensure dependency on the Cult-controlled state in the Hunger Games Society. Lockdowns had to be established as the global blueprint from the start to respond to the ‘virus’ and followed by pre y much the entire world. 2) The global population had to be terrified into believing in a deadly ‘virus’ that didn’t actually exist so they would unquestioningly obey authority in the belief that authority must know how best to protect them and their families. So ware salesman Gates would suddenly morph into the world’s health expert and be promoted as such by the Cult-owned media. 3) A method of testing that wasn’t testing for the ‘virus’, but was only claimed to be, had to be in place to provide the illusion of ‘cases’ and subsequent ‘deaths’ that had a very different cause to the ‘Covid-19’ that would be scribbled on the death certificate. 4) Because there was no ‘virus’ and the great majority testing positive with a test not testing for the ‘virus’ would have no symptoms of anything the lie had to be sold that people without symptoms (without the ‘virus’) could still pass it on to others. This was crucial to justify for the first time quarantining – house arresting – healthy people. Without this the economy-destroying lockdown of everybody could not have been credibly sold. 5) The ‘saviour’ had to be seen as a vaccine which beyond evil drug companies were working like angels of mercy to develop as quickly as possible, with all corners cut, to save the day. The public must absolutely not know that the ‘vaccine’ had nothing to do with a ‘virus’ or that the contents were ready and waiting with a very different motive long before the ‘Covid’ card was even li ed from the pack.

I said in March, 2020, that the ‘vaccine’ would have been created way ahead of the ‘Covid’ hoax which justified its use and the following December an article in the New York Intelligencer magazine said the Moderna ‘vaccine’ had been ‘designed’ by

January, 2020. This was ‘before China had even acknowledged that the disease could be transmi ed from human to human, more than a week before the first confirmed coronavirus case in the United States’. The article said that by the time the first American death was announced a month later ‘the vaccine had already been manufactured and shipped to the National Institutes of Health for the beginning of its Phase I clinical trial’. The ‘vaccine’ was actually ‘designed’ long before that although even with this timescale you would expect the article to ask how on earth it could have been done that quickly. Instead it asked why the ‘vaccine’ had not been rolled out then and not months later. Journalism in the mainstream is truly dead. I am going to detail in the next chapter why the ‘virus’ has never existed and how a hoax on that scale was possible, but first the foundation on which the Big Lie of ‘Covid’ was built.

The test that doesn’t test

Fraudulent ‘testing’ is the bo om line of the whole ‘Covid’ hoax and was the means by which a ‘virus’ that did not exist appeared to exist. They could only achieve this magic trick by using a test not testing for the ‘virus’. To use a test that was testing for the ‘virus’ would mean that every test would come back negative given there was no ‘virus’. They chose to exploit something called the RT-PCR test invented by American biochemist Kary Mullis in the 1980s who said publicly that his PCR test … cannot detect infectious disease. Yes, the ‘test’ used worldwide to detect infectious ‘Covid’ to produce all the illusory ‘cases’ and ‘deaths’ compiled by Johns Hopkins and others cannot detect infectious disease. This fact came from the mouth of the man who invented PCR and was awarded the Nobel Prize in Chemistry in 1993 for doing so. Sadly, and incredibly conveniently for the Cult, Mullis died in August, 2019, at the age of 74 just before his test would be fraudulently used to unleash fascism on the world. He was said to have died from pneumonia which was an irony in itself. A few months later he would have had ‘Covid-19’ on his death certificate. I say the timing of his death was convenient because had he lived Mullis, a brilliant, honest and decent man, would have been

vociferously speaking out against the use of his test to detect ‘Covid’ when it was never designed, or able, to do that. I know that to be true given that Mullis made the same point when his test was used to ‘detect’ – not detect – HIV. He had been seriously critical of the Gallo/Montagnier claim to have isolated the HIV ‘virus’ and shown it to cause AIDS for which Mullis said there was no evidence. AIDS is actually not a disease but a series of diseases from which people die all the time. When they die from those same diseases a er a positive ‘test’ for HIV then AIDS goes on their death certificate. I think I’ve heard that before somewhere. Countries instigated a policy with ‘Covid’ that anyone who tested positive with a test not testing for the ‘virus’ and died of any other cause within 28 days and even longer ‘Covid-19’ had to go on the death certificate. Cases have come from the test that can’t test for infectious disease and the deaths are those who have died of anything a er testing positive with a test not testing for the ‘virus’. I’ll have much more later about the death certificate scandal. Mullis was deeply dismissive of the now US ‘Covid’ star Anthony Fauci who he said was a liar who didn’t know anything about anything – ‘and I would say that to his face – nothing.’ He said of Fauci: ‘The man thinks he can take a blood sample, put it in an electron microscope and if it’s got a virus in there you’ll know it – he doesn’t understand electron microscopy and he doesn’t understand medicine and shouldn’t be in a position like he’s in.’ That position, terrifyingly, has made him the decider of ‘Covid’ fascism policy on behalf of the Cult in his role as director since 1984 of the National Institute of Allergy and Infectious Diseases (NIAID) while his record of being wrong is laughable; but being wrong, so long as it’s the right kind of wrong, is why the Cult loves him. He’ll say anything the Cult tells him to say. Fauci was made Chief Medical Adviser to the President immediately Biden took office. Biden was installed in the White House by Cult manipulation and one of his first decisions was to elevate Fauci to a position of even more control. This is a coincidence? Yes, and I identify as a flamenco dancer called Lola. How does such an incompetent criminal like Fauci remain in that

pivotal position in American health since the 1980s? When you serve the Cult it looks a er you until you are surplus to requirements. Kary Mullis said prophetically of Fauci and his like: ‘Those guys have an agenda and it’s not an agenda we would like them to have … they make their own rules, they change them when they want to, and Tony Fauci does not mind going on television in front of the people who pay his salary and lie directly into the camera.’ Fauci has done that almost daily since the ‘Covid’ hoax began. Lying is in Fauci’s DNA. To make the situation crystal clear about the PCR test this is a direct quote from its inventor Kary Mullis: It [the PCR test] doesn’t tell you that you’re sick and doesn’t tell you that the thing you ended up with was really going to hurt you ...’

Ask yourself why governments and medical systems the world over have been using this very test to decide who is ‘infected’ with the SARS-CoV-2 ‘virus’ and the alleged disease it allegedly causes, ‘Covid-19’. The answer to that question will tell you what has been going on. By the way, here’s a li le show-stopper – the ‘new’ SARSCoV-2 ‘virus’ was ‘identified’ as such right from the start using … the PCR test not testing for the ‘virus’. If you are new to this and find that shocking then stick around. I have hardly started yet. Even worse, other ‘tests’, like the ‘Lateral Flow Device’ (LFD), are considered so useless that they have to be confirmed by the PCR test! Leaked emails wri en by Ben Dyson, adviser to UK ‘Health’ Secretary Ma Hancock, said they were ‘dangerously unreliable’. Dyson, executive director of strategy at the Department of Health, wrote: ‘As of today, someone who gets a positive LFD result in (say) London has at best a 25 per cent chance of it being a true positive, but if it is a selfreported test potentially as low as 10 per cent (on an optimistic assumption about specificity) or as low as 2 per cent (on a more pessimistic assumption).’ These are the ‘tests’ that schoolchildren and the public are being urged to have twice a week or more and have to isolate if they get a positive. Each fake positive goes in the statistics as a ‘case’ no ma er how ludicrously inaccurate and the

‘cases’ drive lockdown, masks and the pressure to ‘vaccinate’. The government said in response to the email leak that the ‘tests’ were accurate which confirmed yet again what shocking bloody liars they are. The real false positive rate is 100 percent as we’ll see. In another ‘you couldn’t make it up’ the UK government agreed to pay £2.8 billion to California’s Innova Medical Group to supply the irrelevant lateral flow tests. The company’s primary test-making centre is in China. Innova Medical Group, established in March, 2020, is owned by Pasaca Capital Inc, chaired by Chinese-American millionaire Charles Huang who was born in Wuhan.

How it works – and how it doesn’t

The RT-PCR test, known by its full title of Polymerase chain reaction, is used across the world to make millions, even billions, of copies of a DNA/RNA genetic information sample. The process is called ‘amplification’ and means that a tiny sample of genetic material is amplified to bring out the detailed content. I stress that it is not testing for an infectious disease. It is simply amplifying a sample of genetic material. In the words of Kary Mullis: ‘PCR is … just a process that’s used to make a whole lot of something out of something.’ To emphasise the point companies that make the PCR tests circulated around the world to ‘test’ for ‘Covid’ warn on the box that it can’t be used to detect ‘Covid’ or infectious disease and is for research purposes only. It’s okay, rest for a minute and you’ll be fine. This is the test that produces the ‘cases’ and ‘deaths’ that have been used to destroy human society. All those global and national medical and scientific ‘experts’ demanding this destruction to ‘save us’ KNOW that the test is not testing for the ‘virus’ and the cases and deaths they claim to be real are an almost unimaginable fraud. Every one of them and so many others including politicians and psychopaths like Gates and Tedros must be brought before Nuremburg-type trials and jailed for the rest of their lives. The more the genetic sample is amplified by PCR the more elements of that material become sensitive to the test and by that I don’t mean sensitive for a ‘virus’ but for elements of the genetic material which

is naturally in the body or relates to remnants of old conditions of various kinds lying dormant and causing no disease. Once the amplification of the PCR reaches a certain level everyone will test positive. So much of the material has been made sensitive to the test that everyone will have some part of it in their body. Even lying criminals like Fauci have said that once PCR amplifications pass 35 cycles everything will be a false positive that cannot be trusted for the reasons I have described. I say, like many proper doctors and scientists, that 100 percent of the ‘positives’ are false, but let’s just go with Fauci for a moment. He says that any amplification over 35 cycles will produce false positives and yet the US Centers for Disease Control (CDC) and Food and Drug Administration (FDA) have recommended up to 40 cycles and the National Health Service (NHS) in Britain admi ed in an internal document for staff that it was using 45 cycles of amplification. A long list of other countries has been doing the same and at least one ‘testing’ laboratory has been using 50 cycles. Have you ever heard a doctor, medical ‘expert’ or the media ask what level of amplification has been used to claim a ‘positive’. The ‘test’ comes back ‘positive’ and so you have the ‘virus’, end of story. Now we can see how the government in Tanzania could send off samples from a goat and a pawpaw fruit under human names and both came back positive for ‘Covid-19’. Tanzania president John Magufuli mocked the ‘Covid’ hysteria, the PCR test and masks and refused to import the DNA-manipulating ‘vaccine’. The Cult hated him and an article sponsored by the Bill Gates Foundation appeared in the London Guardian in February, 2021, headed ‘It’s time for Africa to rein in Tanzania’s anti-vaxxer president’. Well, ‘reined in’ he shortly was. Magufuli appeared in good health, but then, in March, 2021, he was dead at 61 from ‘heart failure’. He was replaced by Samia Hassan Suhulu who is connected to Klaus Schwab’s World Economic Forum and she immediately reversed Magufuli’s ‘Covid’ policy. A sample of cola tested positive for ‘Covid’ with the PCR test in Germany while American actress and singer-songwriter Erykah Badu tested positive in one nostril and negative in the other. Footballer Ronaldo called

the PCR test ‘bullshit’ a er testing positive three times and being forced to quarantine and miss matches when there was nothing wrong with him. The mantra from Tedros at the World Health Organization and national governments (same thing) has been test, test, test. They know that the more tests they can generate the more fake ‘cases’ they have which go on to become ‘deaths’ in ways I am coming to. The UK government has its Operation Moonshot planned to test multiple millions every day in workplaces and schools with free tests for everyone to use twice a week at home in line with the Cult plan from the start to make testing part of life. A government advertisement for an ‘Interim Head of Asymptomatic Testing Communication’ said the job included responsibility for delivering a ‘communications strategy’ (propaganda) ‘to support the expansion of asymptomatic testing that ‘normalises testing as part of everyday life’. More tests means more fake ‘cases’, ‘deaths’ and fascism. I have heard of, and from, many people who booked a test, couldn’t turn up, and yet got a positive result through the post for a test they’d never even had. The whole thing is crazy, but for the Cult there’s method in the madness. Controlling and manipulating the level of amplification of the test means the authorities can control whenever they want the number of apparent ‘cases’ and ‘deaths’. If they want to justify more fascist lockdown and destruction of livelihoods they keep the amplification high. If they want to give the illusion that lockdowns and the ‘vaccine’ are working then they lower the amplification and ‘cases’ and ‘deaths’ will appear to fall. In January, 2021, the Cult-owned World Health Organization suddenly warned laboratories about over-amplification of the test and to lower the threshold. Suddenly headlines began appearing such as: ‘Why ARE “Covid” cases plummeting?’ This was just when the vaccine rollout was underway and I had predicted months before they would make cases appear to fall through amplification tampering when the ‘vaccine’ came. These people are so predictable.

Cow vaccines?

The question must be asked of what is on the test swabs being poked far up the nose of the population to the base of the brain? A nasal swab punctured one woman’s brain and caused it to leak fluid. Most of these procedures are being done by people with li le training or medical knowledge. Dr Lorraine Day, former orthopaedic trauma surgeon and Chief of Orthopaedic Surgery at San Francisco General Hospital, says the tests are really a ‘vaccine’. Cows have long been vaccinated this way. She points out that masks have to cover the nose and the mouth where it is claimed the ‘virus’ exists in saliva. Why then don’t they take saliva from the mouth as they do with a DNA test instead of pushing a long swab up the nose towards the brain? The ethmoid bone separates the nasal cavity from the brain and within that bone is the cribriform plate. Dr Day says that when the swab is pushed up against this plate and twisted the procedure is ‘depositing things back there’. She claims that among these ‘things’ are nanoparticles that can enter the brain. Researchers have noted that a team at the Gates-funded Johns Hopkins have designed tiny, star-shaped micro-devices that can latch onto intestinal mucosa and release drugs into the body. Mucosa is the thin skin that covers the inside surface of parts of the body such as the nose and mouth and produces mucus to protect them. The Johns Hopkins micro-devices are called ‘theragrippers’ and were ‘inspired’ by a parasitic worm that digs its sharp teeth into a host’s intestines. Nasal swabs are also coated in the sterilisation agent ethylene oxide. The US National Cancer Institute posts this explanation on its website: At room temperature, ethylene oxide is a flammable colorless gas with a sweet odor. It is used primarily to produce other chemicals, including antifreeze. In smaller amounts, ethylene oxide is used as a pesticide and a sterilizing agent. The ability of ethylene oxide to damage DNA makes it an effective sterilizing agent but also accounts for its cancer-causing activity.

The Institute mentions lymphoma and leukaemia as cancers most frequently reported to be associated with occupational exposure to ethylene oxide along with stomach and breast cancers. How does anyone think this is going to work out with the constant testing

regime being inflicted on adults and children at home and at school that will accumulate in the body anything that’s on the swab?

Doctors know best

It is vital for people to realise that ‘hero’ doctors ‘know’ only what the Big Pharma-dominated medical authorities tell them to ‘know’ and if they refuse to ‘know’ what they are told to ‘know’ they are out the door. They are mostly not physicians or healers, but repeaters of the official narrative – or else. I have seen alleged professional doctors on British television make shocking statements that we are supposed to take seriously. One called ‘Dr’ Amir Khan, who is actually telling patients how to respond to illness, said that men could take the birth pill to ‘help slow down the effects of Covid-19’. In March, 2021, another ridiculous ‘Covid study’ by an American doctor proposed injecting men with the female sex hormone progesterone as a ‘Covid’ treatment. British doctor Nighat Arif told the BBC that face coverings were now going to be part of ongoing normal. Yes, the vaccine protects you, she said (evidence?) … but the way to deal with viruses in the community was always going to come down to hand washing, face covering and keeping a physical distance. That’s not what we were told before the ‘vaccine’ was circulating. Arif said she couldn’t imagine ever again going on the underground or in a li without a mask. I was just thanking my good luck that she was not my doctor when she said – in March, 2021 – that if ‘we are behaving and we are doing all the right things’ she thought we could ‘have our nearest and dearest around us at home … around Christmas and New Year! Her patronising delivery was the usual school teacher talking to six-year-olds as she repeated every government talking point and probably believed them all. If we have learned anything from the ‘Covid’ experience surely it must be that humanity’s perception of doctors needs a fundamental rethink. NHS ‘doctor’ Sara Kayat told her television audience that the ‘Covid vaccine’ would ‘100 percent prevent hospitalisation and death’. Not even Big Pharma claimed that. We have to stop taking ‘experts’ at their word without question when so many of them are

clueless and only repeating the party line on which their careers depend. That is not to say there are not brilliants doctors – there are and I have spoken to many of them since all this began – but you won’t see them in the mainstream media or quoted by the psychopaths and yes-people in government.

Remember the name – Christian Drosten

German virologist Christian Drosten, Director of Charité Institute of Virology in Berlin, became a national star a er the pandemic hoax began. He was feted on television and advised the German government on ‘Covid’ policy. Most importantly to the wider world Drosten led a group that produced the ‘Covid’ testing protocol for the PCR test. What a remarkable feat given the PCR cannot test for infectious disease and even more so when you think that Drosten said that his method of testing for SARS-CoV-2 was developed ‘without having virus material available’. He developed a test for a ‘virus’ that he didn’t have and had never seen. Let that sink in as you survey the global devastation that came from what he did. The whole catastrophe of Drosten’s ‘test’ was based on the alleged genetic sequence published by Chinese scientists on the Internet. We will see in the next chapter that this alleged ‘genetic sequence’ has never been produced by China or anyone and cannot be when there is no SARS-CoV-2. Drosten, however, doesn’t seem to let li le details like that get in the way. He was the lead author with Victor Corman from the same Charité Hospital of the paper ‘Detection of 2019 novel coronavirus (2019-nCoV) by real-time PCR‘ published in a magazine called Eurosurveillance. This became known as the Corman-Drosten paper. In November, 2020, with human society devastated by the effects of the Corman-Drosten test baloney, the protocol was publicly challenged by 22 international scientists and independent researchers from Europe, the United States, and Japan. Among them were senior molecular geneticists, biochemists, immunologists, and microbiologists. They produced a document headed ‘External peer review of the RTPCR test to detect SARS-Cov-2 Reveals 10 Major Flaws At The Molecular and Methodological Level: Consequences

For False-Positive Results’. The flaws in the Corman-Drosten test included the following: • The test is non-specific because of erroneous design • Results are enormously variable • The test is unable to discriminate between the whole ‘virus’ and viral fragments • It doesn’t have positive or negative controls • The test lacks a standard operating procedure • It is unsupported by proper peer view The scientists said the PCR ‘Covid’ testing protocol was not founded on science and they demanded the Corman-Drosten paper be retracted by Eurosurveillance. They said all present and previous Covid deaths, cases, and ‘infection rates’ should be subject to a massive retroactive inquiry. Lockdowns and travel restrictions should be reviewed and relaxed and those diagnosed through PCR to have ‘Covid-19’ should not be forced to isolate. Dr Kevin Corbe , a health researcher and nurse educator with a long academic career producing a stream of peer-reviewed publications at many UK universities, made the same point about the PCR test debacle. He said of the scientists’ conclusions: ‘Every scientific rationale for the development of that test has been totally destroyed by this paper. It’s like Hiroshima/Nagasaki to the Covid test.’ He said that China hadn’t given them an isolated ‘virus’ when Drosten developed the test. Instead they had developed the test from a sequence in a gene bank.’ Put another way … they made it up! The scientists were supported in this contention by a Portuguese appeals court which ruled in November, 2020, that PCR tests are unreliable and it is unlawful to quarantine people based solely on a PCR test. The point about China not providing an isolated virus must be true when the ‘virus’ has never been isolated to this day and the consequences of that will become clear. Drosten and company produced this useless ‘protocol’ right on cue in January, 2020, just as the ‘virus’ was said to

be moving westward and it somehow managed to successfully pass a peer-review in 24 hours. In other words there was no peer-review for a test that would be used to decide who had ‘Covid’ and who didn’t across the world. The Cult-created, Gates-controlled World Health Organization immediately recommended all its nearly 200 member countries to use the Drosten PCR protocol to detect ‘cases’ and ‘deaths’. The sting was underway and it continues to this day. So who is this Christian Drosten that produced the means through which death, destruction and economic catastrophe would be justified? His education background, including his doctoral thesis, would appear to be somewhat shrouded in mystery and his track record is dire as with another essential player in the ‘Covid’ hoax, the Gates-funded Professor Neil Ferguson at the Gates-funded Imperial College in London of whom more shortly. Drosten predicted in 2003 that the alleged original SARS ‘virus’ (SARS-1’) was an epidemic that could have serious effects on economies and an effective vaccine would take at least two years to produce. Drosten’s answer to every alleged ‘outbreak’ is a vaccine which you won’t be shocked to know. What followed were just 774 official deaths worldwide and none in Germany where there were only nine cases. That is even if you believe there ever was a SARS ‘virus’ when the evidence is zilch and I will expand on this in the next chapter. Drosten claims to be co-discoverer of ‘SARS-1’ and developed a test for it in 2003. He was screaming warnings about ‘swine flu’ in 2009 and how it was a widespread infection far more severe than any dangers from a vaccine could be and people should get vaccinated. It would be helpful for Drosten’s vocal chords if he simply recorded the words ‘the virus is deadly and you need to get vaccinated’ and copies could be handed out whenever the latest made-up threat comes along. Drosten’s swine flu epidemic never happened, but Big Pharma didn’t mind with governments spending hundreds of millions on vaccines that hardly anyone bothered to use and many who did wished they hadn’t. A study in 2010 revealed that the risk of dying from swine flu, or H1N1, was no higher than that of the annual seasonal flu which is what at least most of ‘it’ really was as in

the case of ‘Covid-19’. A media investigation into Drosten asked how with such a record of inaccuracy he could be the government adviser on these issues. The answer to that question is the same with Drosten, Ferguson and Fauci – they keep on giving the authorities the ‘conclusions’ and ‘advice’ they want to hear. Drosten certainly produced the goods for them in January, 2020, with his PCR protocol garbage and provided the foundation of what German internal medicine specialist Dr Claus Köhnlein, co-author of Virus Mania, called the ‘test pandemic’. The 22 scientists in the Eurosurveillance challenge called out conflicts of interest within the Drosten ‘protocol’ group and with good reason. Olfert Landt, a regular co-author of Drosten ‘studies’, owns the biotech company TIB Molbiol Syntheselabor GmbH in Berlin which manufactures and sells the tests that Drosten and his mates come up with. They have done this with SARS, Enterotoxigenic E. coli (ETEC), MERS, Zika ‘virus’, yellow fever, and now ‘Covid’. Landt told the Berliner Zeitung newspaper: The testing, design and development came from the Charité [Drosten and Corman]. We simply implemented it immediately in the form of a kit. And if we don’t have the virus, which originally only existed in Wuhan, we can make a synthetic gene to simulate the genome of the virus. That’s what we did very quickly.

This is more confirmation that the Drosten test was designed without access to the ‘virus’ and only a synthetic simulation which is what SARS-CoV-2 really is – a computer-generated synthetic fiction. It’s quite an enterprise they have going here. A Drosten team decides what the test for something should be and Landt’s biotech company flogs it to governments and medical systems across the world. His company must have made an absolute fortune since the ‘Covid’ hoax began. Dr Reiner Fuellmich, a prominent German consumer protection trial lawyer in Germany and California, is on Drosten’s case and that of Tedros at the World Health Organization for crimes against humanity with a class-action lawsuit being prepared in the United States and other legal action in Germany.

Why China?

Scamming the world with a ‘virus’ that doesn’t exist would seem impossible on the face of it, but not if you have control of the relatively few people that make policy decisions and the great majority of the global media. Remember it’s not about changing ‘real’ reality it’s about controlling perception of reality. You don’t have to make something happen you only have make people believe that it’s happening. Renegade Minds understand this and are therefore much harder to swindle. ‘Covid-19’ is not a ‘real’ ‘virus’. It’s a mind virus, like a computer virus, which has infected the minds, not the bodies, of billions. It all started, publically at least, in China and that alone is of central significance. The Cult was behind the revolution led by its asset Mao Zedong, or Chairman Mao, which established the People’s Republic of China on October 1st, 1949. It should have been called The Cult’s Republic of China, but the name had to reflect the recurring illusion that vicious dictatorships are run by and for the people (see all the ‘Democratic Republics’ controlled by tyrants). In the same way we have the ‘Biden’ Democratic Republic of America officially ruled by a puppet tyrant (at least temporarily) on behalf of Cult tyrants. The creation of Mao’s merciless communist/fascist dictatorship was part of a frenzy of activity by the Cult at the conclusion of World War Two which, like the First World War, it had instigated through its assets in Germany, Britain, France, the United States and elsewhere. Israel was formed in 1948; the Soviet Union expanded its ‘Iron Curtain’ control, influence and military power with the Warsaw Pact communist alliance in 1955; the United Nations was formed in 1945 as a Cult precursor to world government; and a long list of world bodies would be established including the World Health Organization (1948), World Trade Organization (1948 under another name until 1995), International Monetary Fund (1945) and World Bank (1944). Human society was redrawn and hugely centralised in the global Problem-ReactionSolution that was World War Two. All these changes were significant. Israel would become the headquarters of the Sabbatians

and the revolution in China would prepare the ground and control system for the events of 2019/2020. Renegade Minds know there are no borders except for public consumption. The Cult is a seamless, borderless global entity and to understand the game we need to put aside labels like borders, nations, countries, communism, fascism and democracy. These delude the population into believing that countries are ruled within their borders by a government of whatever shade when these are mere agencies of a global power. America’s illusion of democracy and China’s communism/fascism are subsidiaries – vehicles – for the same agenda. We may hear about conflict and competition between America and China and on the lower levels that will be true; but at the Cult level they are branches of the same company in the way of the McDonald’s example I gave earlier. I have tracked in the books over the years support by US governments of both parties for Chinese Communist Party infiltration of American society through allowing the sale of land, even military facilities, and the acquisition of American business and university influence. All this is underpinned by the infamous stealing of intellectual property and technological know-how. Cult-owned Silicon Valley corporations waive their fraudulent ‘morality’ to do business with human-rightsfree China; Cult-controlled Disney has become China’s PR department; and China in effect owns ‘American’ sports such as basketball which depends for much of its income on Chinese audiences. As a result any sports player, coach or official speaking out against China’s horrific human rights record is immediately condemned or fired by the China-worshipping National Basketball Association. One of the first acts of China-controlled Biden was to issue an executive order telling federal agencies to stop making references to the ‘virus’ by the ‘geographic location of its origin’. Long-time Congressman Jerry Nadler warned that criticising China, America’s biggest rival, leads to hate crimes against Asian people in the United States. So shut up you bigot. China is fast closing in on Israel as a country that must not be criticised which is apt, really, given that Sabbatians control them both. The two countries have

developed close economic, military, technological and strategic ties which include involvement in China’s ‘Silk Road’ transport and economic initiative to connect China with Europe. Israel was the first country in the Middle East to recognise the establishment of Mao’s tyranny in 1950 months a er it was established.

Project Wuhan – the ‘Covid’ Psyop

I emphasise again that the Cult plays the long game and what is happening to the world today is the result of centuries of calculated manipulation following a script to take control step-by-step of every aspect of human society. I will discuss later the common force behind all this that has spanned those centuries and thousands of years if the truth be told. Instigating the Mao revolution in China in 1949 with a 2020 ‘pandemic’ in mind is not only how they work – the 71 years between them is really quite short by the Cult’s standards of manipulation preparation. The reason for the Cult’s Chinese revolution was to create a fiercely-controlled environment within which an extreme structure for human control could be incubated to eventually be unleashed across the world. We have seen this happen since the ‘pandemic’ emerged from China with the Chinese controlstructure founded on AI technology and tyrannical enforcement sweep across the West. Until the moment when the Cult went for broke in the West and put its fascism on public display Western governments had to pay some lip-service to freedom and democracy to not alert too many people to the tyranny-in-the-making. Freedoms were more subtly eroded and power centralised with covert government structures put in place waiting for the arrival of 2020 when that smokescreen of ‘freedom’ could be dispensed with. The West was not able to move towards tyranny before 2020 anything like as fast as China which was created as a tyranny and had no limits on how fast it could construct the Cult’s blueprint for global control. When the time came to impose that structure on the world it was the same Cult-owned Chinese communist/fascist government that provided the excuse – the ‘Covid pandemic’. It was absolutely crucial to the Cult plan for the Chinese response to the ‘pandemic’ –

draconian lockdowns of the entire population – to become the blueprint that Western countries would follow to destroy the livelihoods and freedom of their people. This is why the Cultowned, Gates-owned, WHO Director-General Tedros said early on: The Chinese government is to be congratulated for the extraordinary measures it has taken to contain the outbreak. China is actually setting a new standard for outbreak response and it is not an exaggeration.

Forbes magazine said of China: ‘… those measures protected untold millions from ge ing the disease’. The Rockefeller Foundation ‘epidemic scenario’ document in 2010 said ‘prophetically’: However, a few countries did fare better – China in particular. The Chinese government’s quick imposition and enforcement of mandatory quarantine for all citizens, as well as its instant and near-hermetic sealing off of all borders, saved millions of lives, stopping the spread of the virus far earlier than in other countries and enabling a swifter post-pandemic recovery.

Once again – spooky. The first official story was the ‘bat theory’ or rather the bat diversion. The source of the ‘virus outbreak’ we were told was a ‘‘wet market’ in Wuhan where bats and other animals are bought and eaten in horrifically unhygienic conditions. Then another story emerged through the alternative media that the ‘virus’ had been released on purpose or by accident from a BSL-4 (biosafety level 4) laboratory in Wuhan not far from the wet market. The lab was reported to create and work with lethal concoctions and bioweapons. Biosafety level 4 is the highest in the World Health Organization system of safety and containment. Renegade Minds are aware of what I call designer manipulation. The ideal for the Cult is for people to buy its prime narrative which in the opening salvoes of the ‘pandemic’ was the wet market story. It knows, however, that there is now a considerable worldwide alternative media of researchers sceptical of anything governments say and they are o en given a version of events in a form they can perceive as credible while misdirecting them from the real truth. In this case let them

think that the conspiracy involved is a ‘bioweapon virus’ released from the Wuhan lab to keep them from the real conspiracy – there is no ‘virus’. The WHO’s current position on the source of the outbreak at the time of writing appears to be: ‘We haven’t got a clue, mate.’ This is a good position to maintain mystery and bewilderment. The inner circle will know where the ‘virus’ came from – nowhere. The bo om line was to ensure the public believed there was a ‘virus’ and it didn’t much ma er if they thought it was natural or had been released from a lab. The belief that there was a ‘deadly virus’ was all that was needed to trigger global panic and fear. The population was terrified into handing their power to authority and doing what they were told. They had to or they were ‘all gonna die’. In March, 2020, information began to come my way from real doctors and scientists and my own additional research which had my intuition screaming: ‘Yes, that’s it! There is no virus.’ The ‘bioweapon’ was not the ‘virus’; it was the ‘vaccine’ already being talked about that would be the bioweapon. My conclusion was further enhanced by happenings in Wuhan. The ‘virus’ was said to be sweeping the city and news footage circulated of people collapsing in the street (which they’ve never done in the West with the same ‘virus’). The Chinese government was building ‘new hospitals’ in a ma er of ten days to ‘cope with demand’ such was the virulent nature of the ‘virus’. Yet in what seemed like no time the ‘new hospitals’ closed – even if they even opened – and China declared itself ‘virus-free’. It was back to business as usual. This was more propaganda to promote the Chinese draconian lockdowns in the West as the way to ‘beat the virus’. Trouble was that we subsequently had lockdown a er lockdown, but never business as usual. As the people of the West and most of the rest of the world were caught in an ever-worsening spiral of lockdown, social distancing, masks, isolated old people, families forced apart, and livelihood destruction, it was party-time in Wuhan. Pictures emerged of thousands of people enjoying pool parties and concerts. It made no sense until you realised there never was a ‘virus’ and the

whole thing was a Cult set-up to transform human society out of one its major global strongholds – China. How is it possible to deceive virtually the entire world population into believing there is a deadly virus when there is not even a ‘virus’ let alone a deadly one? It’s nothing like as difficult as you would think and that’s clearly true because it happened. See end of book Postscript for more on the ‘Wuhan lab virus release’ story which the authorities and media were pushing heavily in the summer of 2021 to divert a ention from the truth that the ‘Covid virus’ is pure invention. Postscript:

CHAPTER FIVE There

is no

‘virus’

You can fool some of the people all of the time, and all of the people some of the time, but you cannot fool all of the people all of the time Abraham Lincoln

T

he greatest form of mind control is repetition. The more you repeat the same mantra of alleged ‘facts’ the more will accept them to be true. It becomes an ‘everyone knows that, mate’. If you can also censor any other version or alternative to your alleged ‘facts’ you are pre y much home and cooking. By the start of 2020 the Cult owned the global mainstream media almost in its entirety to spew out its ‘Covid’ propaganda and ignore or discredit any other information and view. Cult-owned social media platforms in Cult-owned Silicon Valley were poised and ready to unleash a campaign of ferocious censorship to obliterate all but the official narrative. To complete the circle many demands for censorship by Silicon Valley were led by the mainstream media as ‘journalists’ became full-out enforcers for the Cult both as propagandists and censors. Part of this has been the influx of young people straight out of university who have become ‘journalists’ in significant positions. They have no experience and a headful of programmed perceptions from their years at school and university at a time when today’s young are the most perceptually-targeted generations in known human history given the insidious impact of technology. They enter the media perceptually prepared and ready to repeat the narratives of the system that programmed them to

repeat its narratives. The BBC has a truly pathetic ‘specialist disinformation reporter’ called Marianna Spring who fits this bill perfectly. She is clueless about the world, how it works and what is really going on. Her role is to discredit anyone doing the job that a proper journalist would do and system-serving hacks like Spring wouldn’t dare to do or even see the need to do. They are too busy licking the arse of authority which can never be wrong and, in the case of the BBC propaganda programme, Panorama, contacting payments systems such as PayPal to have a donations page taken down for a film company making documentaries questioning vaccines. Even the BBC soap opera EastEnders included a disgracefully biased scene in which an inarticulate white working class woman was made to look foolish for questioning the ‘vaccine’ while a well-spoken black man and Asian woman promoted the government narrative. It ticked every BBC box and the fact that the black and minority community was resisting the ‘vaccine’ had nothing to do with the way the scene was wri en. The BBC has become a disgusting tyrannical propaganda and censorship operation that should be defunded and disbanded and a free media take its place with a brief to stop censorship instead of demanding it. A BBC ‘interview’ with Gates goes something like: ‘Mr Gates, sir, if I can call you sir, would you like to tell our audience why you are such a great man, a wonderful humanitarian philanthropist, and why you should absolutely be allowed as a so ware salesman to decide health policy for approaching eight billion people? Thank you, sir, please sir.’ Propaganda programming has been incessant and merciless and when all you hear is the same story from the media, repeated by those around you who have only heard the same story, is it any wonder that people on a grand scale believe absolute mendacious garbage to be true? You are about to see, too, why this level of information control is necessary when the official ‘Covid’ narrative is so nonsensical and unsupportable by the evidence.

Structure of Deceit

The pyramid structure through which the ‘Covid’ hoax has been manifested is very simple and has to be to work. As few people as possible have to be involved with full knowledge of what they are doing – and why – or the real story would get out. At the top of the pyramid are the inner core of the Cult which controls Bill Gates who, in turn, controls the World Health Organization through his pivotal funding and his puppet Director-General mouthpiece, Tedros. Before he was appointed Tedros was chair of the Gates-founded Global Fund to ‘fight against AIDS, tuberculosis and malaria’, a board member of the Gates-funded ‘vaccine alliance’ GAVI, and on the board of another Gates-funded organisation. Gates owns him and picked him for a specific reason – Tedros is a crook and worse. ‘Dr’ Tedros (he’s not a medical doctor, the first WHO chief not to be) was a member of the tyrannical Marxist government of Ethiopia for decades with all its human rights abuses. He has faced allegations of corruption and misappropriation of funds and was exposed three times for covering up cholera epidemics while Ethiopia’s health minister. Tedros appointed the mass-murdering genocidal Zimbabwe dictator Robert Mugabe as a WHO goodwill ambassador for public health which, as with Tedros, is like appointing a psychopath to run a peace and love campaign. The move was so ridiculous that he had to drop Mugabe in the face of widespread condemnation. American economist David Steinman, a Nobel peace prize nominee, lodged a complaint with the International Criminal Court in The Hague over alleged genocide by Tedros when he was Ethiopia’s foreign minister. Steinman says Tedros was a ‘crucial decision maker’ who directed the actions of Ethiopia’s security forces from 2013 to 2015 and one of three officials in charge when those security services embarked on the ‘killing’ and ‘torturing’ of Ethiopians. You can see where Tedros is coming from and it’s sobering to think that he has been the vehicle for Gates and the Cult to direct the global response to ‘Covid’. Think about that. A psychopathic Cult dictates to psychopath Gates who dictates to psychopath Tedros who dictates how countries of the world must respond to a ‘Covid virus’ never scientifically shown to exist. At the same time psychopathic Cult-owned Silicon Valley information

giants like Google, YouTube, Facebook and Twi er announced very early on that they would give the Cult/Gates/Tedros/WHO version of the narrative free advertising and censor those who challenged their intelligence-insulting, mendacious story. The next layer in the global ‘medical’ structure below the Cult, Gates and Tedros are the chief medical officers and science ‘advisers’ in each of the WHO member countries which means virtually all of them. Medical officers and arbiters of science (they’re not) then take the WHO policy and recommended responses and impose them on their country’s population while the political ‘leaders’ say they are deciding policy (they’re clearly not) by ‘following the science’ on the advice of the ‘experts’ – the same medical officers and science ‘advisers’ (dictators). In this way with the rarest of exceptions the entire world followed the same policy of lockdown, people distancing, masks and ‘vaccines’ dictated by the psychopathic Cult, psychopathic Gates and psychopathic Tedros who we are supposed to believe give a damn about the health of the world population they are seeking to enslave. That, amazingly, is all there is to it in terms of crucial decision-making. Medical staff in each country then follow like sheep the dictates of the shepherds at the top of the national medical hierarchies – chief medical officers and science ‘advisers’ who themselves follow like sheep the shepherds of the World Health Organization and the Cult. Shepherds at the national level o en have major funding and other connections to Gates and his Bill and Melinda Gates Foundation which carefully hands out money like confe i at a wedding to control the entire global medical system from the WHO down.

Follow the money

Christopher Whi y, Chief Medical Adviser to the UK Government at the centre of ‘virus’ policy, a senior adviser to the government’s Scientific Advisory Group for Emergencies (SAGE), and Executive Board member of the World Health Organization, was gi ed a grant of $40 million by the Bill and Melinda Gates Foundation for malaria research in Africa. The BBC described the unelected Whi y as ‘the

official who will probably have the greatest impact on our everyday lives of any individual policymaker in modern times’ and so it turned out. What Gates and Tedros have said Whi y has done like his equivalents around the world. Patrick Vallance, co-chair of SAGE and the government’s Chief Scientific Adviser, is a former executive of Big Pharma giant GlaxoSmithKline with its fundamental financial and business connections to Bill Gates. In September, 2020, it was revealed that Vallance owned a deferred bonus of shares in GlaxoSmithKline worth £600,000 while the company was ‘developing’ a ‘Covid vaccine’. Move along now – nothing to see here – what could possibly be wrong with that? Imperial College in London, a major player in ‘Covid’ policy in Britain and elsewhere with its ‘Covid-19’ Response Team, is funded by Gates and has big connections to China while the now infamous Professor Neil Ferguson, the useless ‘computer modeller’ at Imperial College is also funded by Gates. Ferguson delivered the dramatically inaccurate excuse for the first lockdowns (much more in the next chapter). The Institute for Health Metrics and Evaluation (IHME) in the United States, another source of outrageously false ‘Covid’ computer models to justify lockdowns, is bankrolled by Gates who is a vehement promotor of lockdowns. America’s version of Whi y and Vallance, the again now infamous Anthony Fauci, has connections to ‘Covid vaccine’ maker Moderna as does Bill Gates through funding from the Bill and Melinda Gates Foundation. Fauci is director of the National Institute of Allergy and Infectious Diseases (NIAID), a major recipient of Gates money, and they are very close. Deborah Birx who was appointed White House Coronavirus Response Coordinator in February, 2020, is yet another with ties to Gates. Everywhere you look at the different elements around the world behind the coordination and decision making of the ‘Covid’ hoax there is Bill Gates and his money. They include the World Health Organization; Centers for Disease Control (CDC) in the United States; National Institutes of Health (NIH) of Anthony Fauci; Imperial College and Neil Ferguson; the London School of Hygiene where Chris Whi y worked; Regulatory agencies like the UK Medicines & Healthcare products Regulatory Agency (MHRA)

which gave emergency approval for ‘Covid vaccines’; Wellcome Trust; GAVI, the Vaccine Alliance; the Coalition for Epidemic Preparedness Innovations (CEPI); Johns Hopkins University which has compiled the false ‘Covid’ figures; and the World Economic Forum. A Nationalfile.com article said: Gates has a lot of pull in the medical world, he has a multi-million dollar relationship with Dr. Fauci, and Fauci originally took the Gates line supporting vaccines and casting doubt on [the drug hydroxychloroquine]. Coronavirus response team member Dr. Deborah Birx, appointed by former president Obama to serve as United States Global AIDS Coordinator, also sits on the board of a group that has received billions from Gates’ foundation, and Birx reportedly used a disputed Bill Gates-funded model for the White House’s Coronavirus effort. Gates is a big proponent for a population lockdown scenario for the Coronavirus outbreak.

Another funder of Moderna is the Defense Advanced Research Projects Agency (DARPA), the technology-development arm of the Pentagon and one of the most sinister organisations on earth. DARPA had a major role with the CIA covert technology-funding operation In-Q-Tel in the development of Google and social media which is now at the centre of global censorship. Fauci and Gates are extremely close and openly admit to talking regularly about ‘Covid’ policy, but then why wouldn’t Gates have a seat at every national ‘Covid’ table a er his Foundation commi ed $1.75 billion to the ‘fight against Covid-19’. When passed through our Orwellian Translation Unit this means that he has bought and paid for the Cultdriven ‘Covid’ response worldwide. Research the major ‘Covid’ response personnel in your own country and you will find the same Gates funding and other connections again and again. Medical and science chiefs following World Health Organization ‘policy’ sit atop a medical hierarchy in their country of administrators, doctors and nursing staff. These ‘subordinates’ are told they must work and behave in accordance with the policy delivered from the ‘top’ of the national ‘health’ pyramid which is largely the policy delivered by the WHO which is the policy delivered by Gates and the Cult. The whole ‘Covid’ narrative has been imposed on medical staff by a climate of fear although great numbers don’t even need that to comply. They do so through breathtaking levels of ignorance and

include doctors who go through life simply repeating what Big Pharma and their hierarchical masters tell them to say and believe. No wonder Big Pharma ‘medicine’ is one of the biggest killers on Planet Earth. The same top-down system of intimidation operates with regard to the Cult Big Pharma cartel which also dictates policy through national and global medical systems in this way. The Cult and Big Pharma agendas are the same because the former controls and owns the la er. ‘Health’ administrators, doctors, and nursing staff are told to support and parrot the dictated policy or they will face consequences which can include being fired. How sad it’s been to see medical staff meekly repeating and imposing Cult policy without question and most of those who can see through the deceit are only willing to speak anonymously off the record. They know what will happen if their identity is known. This has le the courageous few to expose the lies about the ‘virus’, face masks, overwhelmed hospitals that aren’t, and the dangers of the ‘vaccine’ that isn’t a vaccine. When these medical professionals and scientists, some renowned in their field, have taken to the Internet to expose the truth their articles, comments and videos have been deleted by Cult-owned Facebook, Twi er and YouTube. What a real head-shaker to see YouTube videos with leading world scientists and highly qualified medical specialists with an added link underneath to the notorious Cult propaganda website Wikipedia to find the ‘facts’ about the same subject.

HIV – the ‘Covid’ trial-run

I’ll give you an example of the consequences for health and truth that come from censorship and unquestioning belief in official narratives. The story was told by PCR inventor Kary Mullis in his book Dancing Naked in the Mind Field. He said that in 1984 he accepted as just another scientific fact that Luc Montagnier of France’s Pasteur Institute and Robert Gallo of America’s National Institutes of Health had independently discovered that a ‘retrovirus’ dubbed HIV (human immunodeficiency virus) caused AIDS. They

were, a er all, Mullis writes, specialists in retroviruses. This is how the medical and science pyramids work. Something is announced or assumed and then becomes an everybody-knows-that purely through repetition of the assumption as if it is fact. Complete crap becomes accepted truth with no supporting evidence and only repetition of the crap. This is how a ‘virus’ that doesn’t exist became the ‘virus’ that changed the world. The HIV-AIDS fairy story became a multibillion pound industry and the media poured out propaganda terrifying the world about the deadly HIV ‘virus’ that caused the lethal AIDS. By then Mullis was working at a lab in Santa Monica, California, to detect retroviruses with his PCR test in blood donations received by the Red Cross. In doing so he asked a virologist where he could find a reference for HIV being the cause of AIDS. ‘You don’t need a reference,’ the virologist said … ‘Everybody knows it.’ Mullis said he wanted to quote a reference in the report he was doing and he said he felt a li le funny about not knowing the source of such an important discovery when everyone else seemed to. The virologist suggested he cite a report by the Centers for Disease Control and Prevention (CDC) on morbidity and mortality. Mullis read the report, but it only said that an organism had been identified and did not say how. The report did not identify the original scientific work. Physicians, however, assumed (key recurring theme) that if the CDC was convinced that HIV caused AIDS then proof must exist. Mullis continues: I did computer searches. Neither Montagnier, Gallo, nor anyone else had published papers describing experiments which led to the conclusion that HIV probably caused AIDS. I read the papers in Science for which they had become well known as AIDS doctors, but all they had said there was that they had found evidence of a past infection by something which was probably HIV in some AIDS patients. They found antibodies. Antibodies to viruses had always been considered evidence of past disease, not present disease. Antibodies signaled that the virus had been defeated. The patient had saved himself. There was no indication in these papers that this virus caused a disease. They didn’t show that everybody with the antibodies had the disease. In fact they found some healthy people with antibodies.

Mullis asked why their work had been published if Montagnier and Gallo hadn’t really found this evidence, and why had they been fighting so hard to get credit for the discovery? He says he was hesitant to write ‘HIV is the probable cause of AIDS’ until he found published evidence to support that. ‘Tens of thousands of scientists and researchers were spending billions of dollars a year doing research based on this idea,’ Mullis writes. ‘The reason had to be there somewhere; otherwise these people would not have allowed their research to se le into one narrow channel of investigation.’ He said he lectured about PCR at numerous meetings where people were always talking about HIV and he asked them how they knew that HIV was the cause of AIDS: Everyone said something. Everyone had the answer at home, in the office, in some drawer. They all knew, and they would send me the papers as soon as they got back. But I never got any papers. Nobody ever sent me the news about how AIDS was caused by HIV.

Eventually Mullis was able to ask Montagnier himself about the reference proof when he lectured in San Diego at the grand opening of the University of California AIDS Research Center. Mullis says this was the last time he would ask his question without showing anger. Montagnier said he should reference the CDC report. ‘I read it’, Mullis said, and it didn’t answer the question. ‘If Montagnier didn’t know the answer who the hell did?’ Then one night Mullis was driving when an interview came on National Public Radio with Peter Duesberg, a prominent virologist at Berkeley and a California Scientist of the Year. Mullis says he finally understood why he could not find references that connected HIV to AIDS – there weren’t any! No one had ever proved that HIV causes AIDS even though it had spawned a multi-billion pound global industry and the media was repeating this as fact every day in their articles and broadcasts terrifying the shit out of people about AIDS and giving the impression that a positive test for HIV (see ‘Covid’) was a death sentence. Duesberg was a threat to the AIDS gravy train and the agenda that underpinned it. He was therefore abused and castigated a er he told the Proceedings of the National Academy of Sciences

there was no good evidence implicating the new ‘virus’. Editors rejected his manuscripts and his research funds were deleted. Mullis points out that the CDC has defined AIDS as one of more than 30 diseases if accompanied by a positive result on a test that detects antibodies to HIV; but those same diseases are not defined as AIDS cases when antibodies are not detected: If an HIV-positive woman develops uterine cancer, for example, she is considered to have AIDS. If she is not HIV positive, she simply has uterine cancer. An HIV-positive man with tuberculosis has AIDS; if he tests negative he simply has tuberculosis. If he lives in Kenya or Colombia, where the test for HIV antibodies is too expensive, he is simply presumed to have the antibodies and therefore AIDS, and therefore he can be treated in the World Health Organization’s clinic. It’s the only medical help available in some places. And it’s free, because the countries that support WHO are worried about AIDS.

Mullis accuses the CDC of continually adding new diseases (see ever more ‘Covid symptoms’) to the grand AIDS definition and of virtually doctoring the books to make it appear as if the disease continued to spread. He cites how in 1993 the CDC enormously broadened its AIDS definition and county health authorities were delighted because they received $2,500 per year from the Federal government for every reported AIDS case. Ladies and gentlemen, I have just described, via Kary Mullis, the ‘Covid pandemic’ of 2020 and beyond. Every element is the same and it’s been pulled off in the same way by the same networks.

The ‘Covid virus’ exists? Okay – prove it. Er … still waiting

What Kary Mullis described with regard to ‘HIV’ has been repeated with ‘Covid’. A claim is made that a new, or ‘novel’, infection has been found and the entire medical system of the world repeats that as fact exactly as they did with HIV and AIDS. No one in the mainstream asks rather relevant questions such as ‘How do you know?’ and ‘Where is your proof?’ The SARS-Cov-2 ‘virus’ and the ‘Covid-19 disease’ became an overnight ‘everybody-knows-that’. The origin could be debated and mulled over, but what you could not suggest was that ‘SARS-Cov-2’ didn’t exist. That would be

ridiculous. ‘Everybody knows’ the ‘virus’ exists. Well, I didn’t for one along with American proper doctors like Andrew Kaufman and Tom Cowan and long-time American proper journalist Jon Rappaport. We dared to pursue the obvious and simple question: ‘Where’s the evidence?’ The overwhelming majority in medicine, journalism and the general public did not think to ask that. A er all, everyone knew there was a new ‘virus’. Everyone was saying so and I heard it on the BBC. Some would eventually argue that the ‘deadly virus’ was nothing like as deadly as claimed, but few would venture into the realms of its very existence. Had they done so they would have found that the evidence for that claim had gone AWOL as with HIV causes AIDS. In fact, not even that. For something to go AWOL it has to exist in the first place and scientific proof for a ‘SARS-Cov-2’ can be filed under nothing, nowhere and zilch. Dr Andrew Kaufman is a board-certified forensic psychiatrist in New York State, a Doctor of Medicine and former Assistant Professor and Medical Director of Psychiatry at SUNY Upstate Medical University, and Medical Instructor of Hematology and Oncology at the Medical School of South Carolina. He also studied biology at the Massachuse s Institute of Technology (MIT) and trained in Psychiatry at Duke University. Kaufman is retired from allopathic medicine, but remains a consultant and educator on natural healing, I saw a video of his very early on in the ‘Covid’ hoax in which he questioned claims about the ‘virus’ in the absence of any supporting evidence and with plenty pointing the other way. I did everything I could to circulate his work which I felt was asking the pivotal questions that needed an answer. I can recommend an excellent pull-together interview he did with the website The Last Vagabond entitled Dr Andrew Kaufman: Virus Isolation, Terrain Theory and Covid-19 and his website is andrewkaufmanmd.com. Kaufman is not only a forensic psychiatrist; he is forensic in all that he does. He always reads original scientific papers, experiments and studies instead of second-third-fourth-hand reports about the ‘virus’ in the media which are repeating the repeated repetition of the narrative. When he did so with the original Chinese ‘virus’ papers Kaufman

realised that there was no evidence of a ‘SARS-Cov-2’. They had never – from the start – shown it to exist and every repeat of this claim worldwide was based on the accepted existence of proof that was nowhere to be found – see Kary Mullis and HIV. Here we go again.

Let’s postulate

Kaufman discovered that the Chinese authorities immediately concluded that the cause of an illness that broke out among about 200 initial patients in Wuhan was a ‘new virus’ when there were no grounds to make that conclusion. The alleged ‘virus’ was not isolated from other genetic material in their samples and then shown through a system known as Koch’s postulates to be the causative agent of the illness. The world was told that the SARS-Cov-2 ‘virus’ caused a disease they called ‘Covid-19’ which had ‘flu-like’ symptoms and could lead to respiratory problems and pneumonia. If it wasn’t so tragic it would almost be funny. ‘Flu-like’ symptoms’? Pneumonia? Respiratory disease? What in CHINA and particularly in Wuhan, one of the most polluted cities in the world with a resulting epidemic of respiratory disease?? Three hundred thousand people get pneumonia in China every year and there are nearly a billion cases worldwide of ‘flu-like symptoms’. These have a whole range of causes – including pollution in Wuhan – but no other possibility was credibly considered in late 2019 when the world was told there was a new and deadly ‘virus’. The global prevalence of pneumonia and ‘flu-like systems’ gave the Cult networks unlimited potential to rediagnose these other causes as the mythical ‘Covid-19’ and that is what they did from the very start. Kaufman revealed how Chinese medical and science authorities (all subordinates to the Cult-owned communist government) took genetic material from the lungs of only a few of the first patients. The material contained their own cells, bacteria, fungi and other microorganisms living in their bodies. The only way you could prove the existence of the ‘virus’ and its responsibility for the alleged ‘Covid-19’ was to isolate the virus from all the other material – a process also known as ‘purification’ – and

then follow the postulates sequence developed in the late 19th century by German physician and bacteriologist Robert Koch which became the ‘gold standard’ for connecting an alleged causation agent to a disease: 1. The microorganism (bacteria, fungus, virus, etc.) must be present in every case of the disease and all patients must have the same symptoms. It must also not be present in healthy individuals. 2. The microorganism must be isolated from the host with the disease. If the microorganism is a bacteria or fungus it must be grown in a pure culture. If it is a virus, it must be purified (i.e. containing no other material except the virus particles) from a clinical sample. 3. The specific disease, with all of its characteristics, must be reproduced when the infectious agent (the purified virus or a pure culture of bacteria or fungi) is inoculated into a healthy, susceptible host. 4. The microorganism must be recoverable from the experimentally infected host as in step 2.

Not one of these criteria has been met in the case of ‘SARS-Cov-2’ and ‘Covid-19’. Not ONE. EVER. Robert Koch refers to bacteria and not viruses. What are called ‘viral particles’ are so minute (hence masks are useless by any definition) that they could only be seen a er the invention of the electron microscope in the 1930s and can still only be observed through that means. American bacteriologist and virologist Thomas Milton Rivers, the so-called ‘Father of Modern Virology’ who was very significantly director of the Rockefeller Institute for Medical Research in the 1930s, developed a less stringent version of Koch’s postulates to identify ‘virus’ causation known as ‘Rivers criteria’. ‘Covid’ did not pass that process either. Some even doubt whether any ‘virus’ can be isolated from other particles containing genetic material in the Koch method. Freedom of Information requests in many countries asking for scientific proof that the ‘Covid virus’ has been purified and isolated and shown to exist have all come back with a ‘we don’t have that’ and when this happened with a request to the UK Department of Health they added this comment:

However, outside of the scope of the [Freedom of Information Act] and on a discretionary basis, the following information has been advised to us, which may be of interest. Most infectious diseases are caused by viruses, bacteria or fungi. Some bacteria or fungi have the capacity to grow on their own in isolation, for example in colonies on a petri dish. Viruses are different in that they are what we call ‘obligate pathogens’ – that is, they cannot survive or reproduce without infecting a host ... … For some diseases, it is possible to establish causation between a microorganism and a disease by isolating the pathogen from a patient, growing it in pure culture and reintroducing it to a healthy organism. These are known as ‘Koch’s postulates’ and were developed in 1882. However, as our understanding of disease and different disease-causing agents has advanced, these are no longer the method for determining causation [Andrew Kaufman asks why in that case are there two published articles falsely claiming to satisfy Koch’s postulates]. It has long been known that viral diseases cannot be identified in this way as viruses cannot be grown in ‘pure culture’. When a patient is tested for a viral illness, this is normally done by looking for the presence of antigens, or viral genetic code in a host with molecular biology techniques [Kaufman asks how you could know the origin of these chemicals without having a pure culture for comparison]. For the record ‘antigens’ are defined so: Invading microorganisms have antigens on their surface that the human body can recognise as being foreign – meaning not belonging to it. When the body recognises a foreign antigen, lymphocytes (white blood cells) produce antibodies, which are complementary in shape to the antigen.

Notwithstanding that this is open to question in relation to ‘SARSCov-2’ the presence of ‘antibodies’ can have many causes and they are found in people that are perfectly well. Kary Mullis said: ‘Antibodies … had always been considered evidence of past disease, not present disease.’

‘Covid’ really is a computer ‘virus’

Where the UK Department of Health statement says ‘viruses’ are now ‘diagnosed’ through a ‘viral genetic code in a host with molecular biology techniques’, they mean … the PCR test which its inventor said cannot test for infectious disease. They have no credible method of connecting a ‘virus’ to a disease and we will see that there is no scientific proof that any ‘virus’ causes any disease or there is any such thing as a ‘virus’ in the way that it is described. Tenacious Canadian researcher Christine Massey and her team made

some 40 Freedom of Information requests to national public health agencies in different countries asking for proof that SARS-CoV-2 has been isolated and not one of them could supply that information. Massey said of her request in Canada: ‘Freedom of Information reveals Public Health Agency of Canada has no record of ‘SARSCOV-2’ isolation performed by anyone, anywhere, ever.’ If you accept the comment from the UK Department of Health it’s because they can’t isolate a ‘virus’. Even so many ‘science’ papers claimed to have isolated the ‘Covid virus’ until they were questioned and had to admit they hadn’t. A reply from the Robert Koch Institute in Germany was typical: ‘I am not aware of a paper which purified isolated SARS-CoV-2.’ So what the hell was Christian Drosten and his gang using to design the ‘Covid’ testing protocol that has produced all the illusory Covid’ cases and ‘Covid’ deaths when the head of the Chinese version of the CDC admi ed there was a problem right from the start in that the ‘virus’ had never been isolated/purified? Breathe deeply: What they are calling ‘Covid’ is actually created by a computer program i.e. they made it up – er, that’s it. They took lung fluid, with many sources of genetic material, from one single person alleged to be infected with Covid-19 by a PCR test which they claimed, without clear evidence, contained a ‘virus’. They used several computer programs to create a model of a theoretical virus genome sequence from more than fi y-six million small sequences of RNA, each of an unknown source, assembling them like a puzzle with no known solution. The computer filled in the gaps with sequences from bits in the gene bank to make it look like a bat SARS-like coronavirus! A wave of the magic wand and poof, an in silico (computer-generated) genome, a scientific fantasy, was created. UK health researcher Dr Kevin Corbe made the same point with this analogy: … It’s like giving you a few bones and saying that’s your fish. It could be any fish. Not even a skeleton. Here’s a few fragments of bones. That’s your fish … It’s all from gene bank and the bits of the virus sequence that weren’t there they made up. They synthetically created them to fill in the blanks. That’s what genetics is; it’s a code. So it’s ABBBCCDDD and you’re missing some what you think is EEE so you put it in. It’s all

synthetic. You just manufacture the bits that are missing. This is the end result of the geneticization of virology. This is basically a computer virus.

Further confirmation came in an email exchange between British citizen journalist Frances Leader and the government’s Medicines & Healthcare Products Regulatory Agency (the Gates-funded MHRA) which gave emergency permission for untested ‘Covid vaccines’ to be used. The agency admi ed that the ‘vaccine’ is not based on an isolated ‘virus’, but comes from a computer-generated model. Frances Leader was naturally banned from Cult-owned fascist Twi er for making this exchange public. The process of creating computergenerated alleged ‘viruses’ is called ‘in silico’ or ‘in silicon’ – computer chips – and the term ‘in silico’ is believed to originate with biological experiments using only a computer in 1989. ‘Vaccines’ involved with ‘Covid’ are also produced ‘in silico’ or by computer not a natural process. If the original ‘virus’ is nothing more than a made-up computer model how can there be ‘new variants’ of something that never existed in the first place? They are not new ‘variants’; they are new computer models only minutely different to the original program and designed to further terrify the population into having the ‘vaccine’ and submi ing to fascism. You want a ‘new variant’? Click, click, enter – there you go. Tell the medical profession that you have discovered a ‘South African variant’, ‘UK variants’ or a ‘Brazilian variant’ and in the usual HIV-causes-AIDS manner they will unquestioningly repeat it with no evidence whatsoever to support these claims. They will go on television and warn about the dangers of ‘new variants’ while doing nothing more than repeating what they have been told to be true and knowing that any deviation from that would be career suicide. Big-time insiders will know it’s a hoax, but much of the medical community is clueless about the way they are being played and themselves play the public without even being aware they are doing so. What an interesting ‘coincidence’ that AstraZeneca and Oxford University were conducting ‘Covid vaccine trials’ in the three countries – the UK, South Africa and Brazil – where the first three ‘variants’ were claimed to have ‘broken out’.

Here’s your ‘virus’ – it’s a unicorn

Dr Andrew Kaufman presented a brilliant analysis describing how the ‘virus’ was imagined into fake existence when he dissected an article published by Nature and wri en by 19 authors detailing alleged ‘sequencing of a complete viral genome’ of the ‘new SARSCoV-2 virus’. This computer-modelled in silico genome was used as a template for all subsequent genome sequencing experiments that resulted in the so-called variants which he said now number more than 6,000. The fake genome was constructed from more than 56 million individual short strands of RNA. Those li le pieces were assembled into longer pieces by finding areas of overlapping sequences. The computer programs created over two million possible combinations from which the authors simply chose the longest one. They then compared this to a ‘bat virus’ and the computer ‘alignment’ rearranged the sequence and filled in the gaps! They called this computer-generated abomination the ‘complete genome’. Dr Tom Cowan, a fellow medical author and collaborator with Kaufman, said such computer-generation constitutes scientific fraud and he makes this superb analogy: Here is an equivalency: A group of researchers claim to have found a unicorn because they found a piece of a hoof, a hair from a tail, and a snippet of a horn. They then add that information into a computer and program it to re-create the unicorn, and they then claim this computer re-creation is the real unicorn. Of course, they had never actually seen a unicorn so could not possibly have examined its genetic makeup to compare their samples with the actual unicorn’s hair, hooves and horn. The researchers claim they decided which is the real genome of SARS-CoV-2 by ‘consensus’, sort of like a vote. Again, different computer programs will come up with different versions of the imaginary ‘unicorn’, so they come together as a group and decide which is the real imaginary unicorn.

This is how the ‘virus’ that has transformed the world was brought into fraudulent ‘existence’. Extraordinary, yes, but as the Nazis said the bigger the lie the more will believe it. Cowan, however, wasn’t finished and he went on to identify what he called the real blockbuster in the paper. He quotes this section from a paper wri en

by virologists and published by the CDC and then explains what it means: Therefore, we examined the capacity of SARS-CoV-2 to infect and replicate in several common primate and human cell lines, including human adenocarcinoma cells (A549), human liver cells (HUH 7.0), and human embryonic kidney cells (HEK-293T). In addition to Vero E6 and Vero CCL81 cells. ... Each cell line was inoculated at high multiplicity of infection and examined 24h post-infection. No CPE was observed in any of the cell lines except in Vero cells, which grew to greater than 10 to the 7th power at 24 h post-infection. In contrast, HUH 7.0 and 293T showed only modest viral replication, and A549 cells were incompatible with SARS CoV-2 infection.

Cowan explains that when virologists a empt to prove infection they have three possible ‘hosts’ or models on which they can test. The first was humans. Exposure to humans was generally not done for ethical reasons and has never been done with SARS-CoV-2 or any coronavirus. The second possible host was animals. Cowan said that forge ing for a moment that they never actually use purified virus when exposing animals they do use solutions that they claim contain the virus. Exposure to animals has been done with SARS-CoV-2 in an experiment involving mice and this is what they found: None of the wild (normal) mice got sick. In a group of genetically-modified mice, a statistically insignificant number lost weight and had slightly bristled fur, but they experienced nothing like the illness called ‘Covid-19’. Cowan said the third method – the one they mostly rely on – is to inoculate solutions they say contain the virus onto a variety of tissue cultures. This process had never been shown to kill tissue unless the sample material was starved of nutrients and poisoned as part of the process. Yes, incredibly, in tissue experiments designed to show the ‘virus’ is responsible for killing the tissue they starve the tissue of nutrients and add toxic drugs including antibiotics and they do not have control studies to see if it’s the starvation and poisoning that is degrading the tissue rather than the ‘virus’ they allege to be in there somewhere. You want me to pinch you? Yep, I understand. Tom Cowan said this about the whole nonsensical farce as he explains what that quote from the CDC paper really means:

The shocking thing about the above quote is that using their own methods, the virologists found that solutions containing SARS-CoV-2 – even in high amounts – were NOT, I repeat NOT, infective to any of the three human tissue cultures they tested. In plain English, this means they proved, on their terms, that this ‘new coronavirus’ is not infectious to human beings. It is ONLY infective to monkey kidney cells, and only then when you add two potent drugs (gentamicin and amphotericin), known to be toxic to kidneys, to the mix. My friends, read this again and again. These virologists, published by the CDC, performed a clear proof, on their terms, showing that the SARS-CoV-2 virus is harmless to human beings. That is the only possible conclusion, but, unfortunately, this result is not even mentioned in their conclusion. They simply say they can provide virus stocks cultured only on monkey Vero cells, thanks for coming.

Cowan concluded: ‘If people really understood how this “science” was done, I would hope they would storm the gates and demand honesty, transparency and truth.’ Dr Michael Yeadon, former Vice President and Chief Scientific Adviser at drug giant Pfizer has been a vocal critic of the ‘Covid vaccine’ and its potential for multiple harm. He said in an interview in April, 2021, that ‘not one [vaccine] has the virus. He was asked why vaccines normally using a ‘dead’ version of a disease to activate the immune system were not used for ‘Covid’ and instead we had the synthetic methods of the ‘mRNA Covid vaccine’. Yeadon said that to do the former ‘you’d have to have some of [the virus] wouldn’t you?’ He added: ‘No-one’s got any – seriously.’ Yeadon said that surely they couldn’t have fooled the whole world for a year without having a virus, ‘but oddly enough ask around – no one’s got it’. He didn’t know why with all the ‘great labs’ around the world that the virus had not been isolated – ‘Maybe they’ve been too busy running bad PCR tests and vaccines that people don’t need.’ What is today called ‘science’ is not ‘science’ at all. Science is no longer what is, but whatever people can be manipulated to believe that it is. Real science has been hijacked by the Cult to dispense and produce the ‘expert scientists’ and contentions that suit the agenda of the Cult. How big-time this has happened with the ‘Covid’ hoax which is entirely based on fake science delivered by fake ‘scientists’ and fake ‘doctors’. The human-caused climate change hoax is also entirely based on fake science delivered by fake ‘scientists’ and fake ‘climate experts’. In both cases real

scientists, climate experts and doctors have their views suppressed and deleted by the Cult-owned science establishment, media and Silicon Valley. This is the ‘science’ that politicians claim to be ‘following’ and a common denominator of ‘Covid’ and climate are Cult psychopaths Bill Gates and his mate Klaus Schwab at the Gatesfunded World Economic Forum. But, don’t worry, it’s all just a coincidence and absolutely nothing to worry about. Zzzzzzzz.

What is a ‘virus’ REALLY?

Dr Tom Cowan is one of many contesting the very existence of viruses let alone that they cause disease. This is understandable when there is no scientific evidence for a disease-causing ‘virus’. German virologist Dr Stefan Lanka won a landmark case in 2017 in the German Supreme Court over his contention that there is no such thing as a measles virus. He had offered a big prize for anyone who could prove there is and Lanka won his case when someone sought to claim the money. There is currently a prize of more than 225,000 euros on offer from an Isolate Truth Fund for anyone who can prove the isolation of SARS-CoV-2 and its genetic substance. Lanka wrote in an article headed ‘The Misconception Called Virus’ that scientists think a ‘virus’ is causing tissue to become diseased and degraded when in fact it is the processes they are using which do that – not a ‘virus’. Lanka has done an important job in making this point clear as Cowan did in his analysis of the CDC paper. Lanka says that all claims about viruses as disease-causing pathogens are wrong and based on ‘easily recognisable, understandable and verifiable misinterpretations.’ Scientists believed they were working with ‘viruses’ in their laboratories when they were really working with ‘typical particles of specific dying tissues or cells …’ Lanka said that the tissue decaying process claimed to be caused by a ‘virus’ still happens when no alleged ‘virus’ is involved. It’s the process that does the damage and not a ‘virus’. The genetic sample is deprived of nutrients, removed from its energy supply through removal from the body and then doused in toxic antibiotics to remove any bacteria. He confirms again that establishment scientists do not (pinch me)

conduct control experiments to see if this is the case and if they did they would see the claims that ‘viruses’ are doing the damage is nonsense. He adds that during the measles ‘virus’ court case he commissioned an independent laboratory to perform just such a control experiment and the result was that the tissues and cells died in the exact same way as with alleged ‘infected’ material. This is supported by a gathering number of scientists, doctors and researchers who reject what is called ‘germ theory’ or the belief in the body being infected by contagious sources emi ed by other people. Researchers Dawn Lester and David Parker take the same stance in their highly-detailed and sourced book What Really Makes You Ill – Why everything you thought you knew about disease is wrong which was recommended to me by a number of medical professionals genuinely seeking the truth. Lester and Parker say there is no provable scientific evidence to show that a ‘virus’ can be transmi ed between people or people and animals or animals and people: The definition also claims that viruses are the cause of many diseases, as if this has been definitively proven. But this is not the case; there is no original scientific evidence that definitively demonstrates that any virus is the cause of any disease. The burden of proof for any theory lies with those who proposed it; but none of the existing documents provides ‘proof’ that supports the claim that ‘viruses’ are pathogens.

Dr Tom Cowan employs one of his clever analogies to describe the process by which a ‘virus’ is named as the culprit for a disease when what is called a ‘virus’ is only material released by cells detoxing themselves from infiltration by chemical or radiation poisoning. The tidal wave of technologically-generated radiation in the ‘smart’ modern world plus all the toxic food and drink are causing this to happen more than ever. Deluded ‘scientists’ misread this as a gathering impact of what they wrongly label ‘viruses’.

Paper can infect houses

Cowan said in an article for davidicke.com – with his tongue only mildly in his cheek – that he believed he had made a tremendous

discovery that may revolutionise science. He had discovered that small bits of paper are alive, ‘well alive-ish’, can ‘infect’ houses, and then reproduce themselves inside the house. The result was that this explosion of growth in the paper inside the house causes the house to explode, blowing it to smithereens. His evidence for this new theory is that in the past months he had carefully examined many of the houses in his neighbourhood and found almost no scraps of paper on the lawns and surrounds of the house. There was an occasional stray label, but nothing more. Then he would return to these same houses a week or so later and with a few, not all of them, particularly the old and decrepit ones, he found to his shock and surprise they were li ered with stray bits of paper. He knew then that the paper had infected these houses, made copies of itself, and blew up the house. A young boy on a bicycle at one of the sites told him he had seen a demolition crew using dynamite to explode the house the previous week, but Cowan dismissed this as the idle thoughts of silly boys because ‘I was on to something big’. He was on to how ‘scientists’ mistake genetic material in the detoxifying process for something they call a ‘virus’. Cowan said of his house and paper story: If this sounds crazy to you, it’s because it should. This scenario is obviously nuts. But consider this admittedly embellished, for effect, current viral theory that all scientists, medical doctors and virologists currently believe.

He takes the example of the ‘novel SARS-Cov2’ virus to prove the point. First they take someone with an undefined illness called ‘Covid-19’ and don’t even a empt to find any virus in their sputum. Never mind the scientists still describe how this ‘virus’, which they have not located a aches to a cell receptor, injects its genetic material, in ‘Covid’s’ case, RNA, into the cell. The RNA once inserted exploits the cell to reproduce itself and makes ‘thousands, nay millions, of copies of itself … Then it emerges victorious to claim its next victim’:

If you were to look in the scientific literature for proof, actual scientific proof, that uniform SARS-CoV2 viruses have been properly isolated from the sputum of a sick person, that actual spike proteins could be seen protruding from the virus (which has not been found), you would find that such evidence doesn’t exist. If you go looking in the published scientific literature for actual pictures, proof, that these spike proteins or any viral proteins are ever attached to any receptor embedded in any cell membrane, you would also find that no such evidence exists. If you were to look for a video or documented evidence of the intact virus injecting its genetic material into the body of the cell, reproducing itself and then emerging victorious by budding off the cell membrane, you would find that no such evidence exists. The closest thing you would find is electron micrograph pictures of cellular particles, possibly attached to cell debris, both of which to be seen were stained by heavy metals, a process that completely distorts their architecture within the living organism. This is like finding bits of paper stuck to the blown-up bricks, thereby proving the paper emerged by taking pieces of the bricks on its way out.

The Enders baloney

Cowan describes the ‘Covid’ story as being just as make-believe as his paper story and he charts back this fantasy to a Nobel Prize winner called John Enders (1897-1985), an American biomedical scientist who has been dubbed ‘The Father of Modern Vaccines’. Enders is claimed to have ‘discovered’ the process of the viral culture which ‘proved’ that a ‘virus’ caused measles. Cowan explains how Enders did this ‘by using the EXACT same procedure that has been followed by every virologist to find and characterize every new virus since 1954’. Enders took throat swabs from children with measles and immersed them in 2ml of milk. Penicillin (100u/ml) and the antibiotic streptomycin (50,g/ml) were added and the whole mix was centrifuged – rotated at high speed to separate large cellular debris from small particles and molecules as with milk and cream, for example. Cowan says that if the aim is to find li le particles of genetic material (‘viruses’) in the snot from children with measles it would seem that the last thing you would do is mix the snot with other material – milk –that also has genetic material. ‘How are you ever going to know whether whatever you found came from the snot or the milk?’ He points out that streptomycin is a ‘nephrotoxic’ or poisonous-to-the-kidney drug. You will see the relevance of that

shortly. Cowan says that it gets worse, much worse, when Enders describes the culture medium upon which the virus ‘grows’: ‘The culture medium consisted of bovine amniotic fluid (90%), beef embryo extract (5%), horse serum (5%), antibiotics and phenol red as an indicator of cell metabolism.’ Cowan asks incredulously: ‘Did he just say that the culture medium also contained fluids and tissues that are themselves rich sources of genetic material?’ The genetic cocktail, or ‘medium’, is inoculated onto tissue and cells from rhesus monkey kidney tissue. This is where the importance of streptomycin comes in and currently-used antimicrobials and other drugs that are poisonous to kidneys and used in ALL modern viral cultures (e.g. gentamicin, streptomycin, and amphotericin). Cowan asks: ‘How are you ever going to know from this witch’s brew where any genetic material comes from as we now have five different sources of rich genetic material in our mix?’ Remember, he says, that all genetic material, whether from monkey kidney tissues, bovine serum, milk, etc., is made from the exact same components. The same central question returns: ‘How are you possibly going to know that it was the virus that killed the kidney tissue and not the toxic antibiotic and starvation rations on which you are growing the tissue?’ John Enders answered the question himself – you can’t: A second agent was obtained from an uninoculated culture of monkey kidney cells. The cytopathic changes [death of the cells] it induced in the unstained preparations could not be distinguished with confidence from the viruses isolated from measles.

The death of the cells (‘cytopathic changes’) happened in exactly the same manner, whether they inoculated the kidney tissue with the measles snot or not, Cowan says. ‘This is evidence that the destruction of the tissue, the very proof of viral causation of illness, was not caused by anything in the snot because they saw the same destructive effect when the snot was not even used … the cytopathic, i.e., cell-killing, changes come from the process of the culture itself, not from any virus in any snot, period.’ Enders quotes in his 1957 paper a virologist called Ruckle as reporting similar findings ‘and in addition has isolated an agent from monkey kidney tissue that is so

far indistinguishable from human measles virus’. In other words, Cowan says, these particles called ‘measles viruses’ are simply and clearly breakdown products of the starved and poisoned tissue. For measles ‘virus’ see all ‘viruses’ including the so-called ‘Covid virus’. Enders, the ‘Father of Modern Vaccines’, also said: There is a potential risk in employing cultures of primate cells for the production of vaccines composed of attenuated virus, since the presence of other agents possibly latent in primate tissues cannot be definitely excluded by any known method.

Cowan further quotes from a paper published in the journal Viruses in May, 2020, while the ‘Covid pandemic’ was well underway in the media if not in reality. ‘EVs’ here refers to particles of genetic debris from our own tissues, such as exosomes of which more in a moment: ‘The remarkable resemblance between EVs and viruses has caused quite a few problems in the studies focused on the analysis of EVs released during viral infections.’ Later the paper adds that to date a reliable method that can actually guarantee a complete separation (of EVs from viruses) DOES NOT EXIST. This was published at a time when a fairy tale ‘virus’ was claimed in total certainty to be causing a fairy tale ‘viral disease’ called ‘Covid-19’ – a fairy tale that was already well on the way to transforming human society in the image that the Cult has worked to achieve for so long. Cowan concludes his article: To summarize, there is no scientific evidence that pathogenic viruses exist. What we think of as ‘viruses’ are simply the normal breakdown products of dead and dying tissues and cells. When we are well, we make fewer of these particles; when we are starved, poisoned, suffocated by wearing masks, or afraid, we make more. There is no engineered virus circulating and making people sick. People in laboratories all over the world are making genetically modified products to make people sick. These are called vaccines. There is no virome, no ‘ecosystem’ of viruses, viruses are not 8%, 50% or 100 % of our genetic material. These are all simply erroneous ideas based on the misconception called a virus.

What is ‘Covid’? Load of bollocks

The background described here by Cowan and Lanka was emphasised in the first video presentation that I saw by Dr Andrew Kaufman when he asked whether the ‘Covid virus’ was in truth a natural defence mechanism of the body called ‘exosomes’. These are released by cells when in states of toxicity – see the same themes returning over and over. They are released ever more profusely as chemical and radiation toxicity increases and think of the potential effect therefore of 5G alone as its destructive frequencies infest the human energetic information field with a gathering pace (5G went online in Wuhan in 2019 as the ‘virus’ emerged). I’ll have more about this later. Exosomes transmit a warning to the rest of the body that ‘Houston, we have a problem’. Kaufman presented images of exosomes and compared them with ‘Covid’ under an electron microscope and the similarity was remarkable. They both a ach to the same cell receptors (claimed in the case of ‘Covid’), contain the same genetic material in the form of RNA or ribonucleic acid, and both are found in ‘viral cell cultures’ with damaged or dying cells. James Hildreth MD, President and Chief Executive Officer of the Meharry Medical College at Johns Hopkins, said: ‘The virus is fully an exosome in every sense of the word.’ Kaufman’s conclusion was that there is no ‘virus’: ‘This entire pandemic is a completely manufactured crisis … there is no evidence of anyone dying from [this] illness.’ Dr Tom Cowan and Sally Fallon Morell, authors of The Contagion Myth, published a statement with Dr Kaufman in February, 2021, explaining why the ‘virus’ does not exist and you can read it that in full in the Appendix. ‘Virus’ theory can be traced to the ‘cell theory’ in 1858 of German physician Rudolf Virchow (1821-1920) who contended that disease originates from a single cell infiltrated by a ‘virus’. Dr Stefan Lanka said that findings and insights with respect to the structure, function and central importance of tissues in the creation of life, which were already known in 1858, comprehensively refute the cell theory. Virchow ignored them. We have seen the part later played by John Enders in the 1950s and Lanka notes that infection theories were only established as a global dogma through the policies and

eugenics of the Third Reich in Nazi Germany (creation of the same Sabbatian cult behind the ‘Covid’ hoax). Lanka said: ‘Before 1933, scientists dared to contradict this theory; a er 1933, these critical scientists were silenced’. Dr Tom Cowan’s view is that ill-heath is caused by too much of something, too li le of something, or toxification from chemicals and radiation – not contagion. We must also highlight as a major source of the ‘virus’ theology a man still called the ‘Father of Modern Virology’ – Thomas Milton Rivers (1888-1962). There is no way given the Cult’s long game policy that it was a coincidence for the ‘Father of Modern Virology’ to be director of the Rockefeller Institute for Medical Research from 1937 to 1956 when he is credited with making the Rockefeller Institute a leader in ‘viral research’. Cult Rockefellers were the force behind the creation of Big Pharma ‘medicine’, established the World Health Organisation in 1948, and have long and close associations with the Gates family that now runs the WHO during the pandemic hoax through mega-rich Cult gofer and psychopath Bill Gates. Only a Renegade Mind can see through all this bullshit by asking the questions that need to be answered, not taking ‘no’ or prevarication for an answer, and certainly not hiding from the truth in fear of speaking it. Renegade Minds have always changed the world for the be er and they will change this one no ma er how bleak it may currently appear to be.

CHAPTER SIX Sequence of deceit If you tell the truth, you don’t have to remember anything Mark Twain

A

gainst the background that I have laid out this far the sequence that took us from an invented ‘virus’ in Cult-owned China in late 2019 to the fascist transformation of human society can be seen and understood in a whole new context. We were told that a deadly disease had broken out in Wuhan and the world media began its campaign (coordinated by behavioural psychologists as we shall see) to terrify the population into unquestioning compliance. We were shown images of Chinese people collapsing in the street which never happened in the West with what was supposed to be the same condition. In the earliest days when alleged cases and deaths were few the fear register was hysterical in many areas of the media and this would expand into the common media narrative across the world. The real story was rather different, but we were never told that. The Chinese government, one of the Cult’s biggest centres of global operation, said they had discovered a new illness with flu-like and pneumoniatype symptoms in a city with such toxic air that it is overwhelmed with flu-like symptoms, pneumonia and respiratory disease. Chinese scientists said it was a new – ‘novel’ – coronavirus which they called Sars-Cov-2 and that it caused a disease they labelled ‘Covid-19’. There was no evidence for this and the ‘virus’ has never to this day been isolated, purified and its genetic code established from that. It

was from the beginning a computer-generated fiction. Stories of Chinese whistleblowers saying the number of deaths was being supressed or that the ‘new disease’ was related to the Wuhan bio-lab misdirected mainstream and alternative media into cul-de-sacs to obscure the real truth – there was no ‘virus’. Chinese scientists took genetic material from the lung fluid of just a few people and said they had found a ‘new’ disease when this material had a wide range of content. There was no evidence for a ‘virus’ for the very reasons explained in the last two chapters. The ‘virus’ has never been shown to (a) exist and (b) cause any disease. People were diagnosed on symptoms that are so widespread in Wuhan and polluted China and with a PCR test that can’t detect infectious disease. On this farce the whole global scam was sold to the rest of the world which would also diagnose respiratory disease as ‘Covid-19’ from symptoms alone or with a PCR test not testing for a ‘virus’. Flu miraculously disappeared worldwide in 2020 and into 2021 as it was redesignated ‘Covid-19’. It was really the same old flu with its ‘flu-like’ symptoms a ributed to ‘flu-like’ ‘Covid-19’. At the same time with very few exceptions the Chinese response of draconian lockdown and fascism was the chosen weapon to respond across the West as recommended by the Cult-owned Tedros at the Cult-owned World Health Organization run by the Cult-owned Gates. All was going according to plan. Chinese scientists – everything in China is controlled by the Cult-owned government – compared their contaminated RNA lung-fluid material with other RNA sequences and said it appeared to be just under 80 percent identical to the SARS-CoV-1 ‘virus’ claimed to be the cause of the SARS (severe acute respiratory syndrome) ‘outbreak’ in 2003. They decreed that because of this the ‘new virus’ had to be related and they called it SARS-CoV-2. There are some serious problems with this assumption and assumption was all it was. Most ‘factual’ science turns out to be assumptions repeated into everyone-knows-that. A match of under 80-percent is meaningless. Dr Kaufman makes the point that there’s a 96 percent genetic correlation between humans and chimpanzees, but ‘no one would say our genetic material is part

of the chimpanzee family’. Yet the Chinese authorities were claiming that a much lower percentage, less than 80 percent, proved the existence of a new ‘coronavirus’. For goodness sake human DNA is 60 percent similar to a banana.

You are feeling sleepy

The entire ‘Covid’ hoax is a global Psyop, a psychological operation to program the human mind into believing and fearing a complete fantasy. A crucial aspect of this was what appeared to happen in Italy. It was all very well streaming out daily images of an alleged catastrophe in Wuhan, but to the Western mind it was still on the other side of the world in a very different culture and se ing. A reaction of ‘this could happen to me and my family’ was still nothing like as intense enough for the mind-doctors. The Cult needed a Western example to push people over that edge and it chose Italy, one of its major global locations going back to the Roman Empire. An Italian ‘Covid’ crisis was manufactured in a particular area called Lombardy which just happens to be notorious for its toxic air and therefore respiratory disease. Wuhan, China, déjà vu. An hysterical media told horror stories of Italians dying from ‘Covid’ in their droves and how Lombardy hospitals were being overrun by a tidal wave of desperately ill people needing treatment a er being struck down by the ‘deadly virus’. Here was the psychological turning point the Cult had planned. Wow, if this is happening in Italy, the Western mind concluded, this indeed could happen to me and my family. Another point is that Italian authorities responded by following the Chinese blueprint so vehemently recommended by the Cult-owned World Health Organization. They imposed fascistic lockdowns on the whole country viciously policed with the help of surveillance drones sweeping through the streets seeking out anyone who escaped from mass house arrest. Livelihoods were destroyed and psychology unravelled in the way we have witnessed since in all lockdown countries. Crucial to the plan was that Italy responded in this way to set the precedent of suspending freedom and imposing fascism in a ‘Western liberal democracy’. I emphasised in an

animated video explanation on davidicke.com posted in the summer of 2020 how important it was to the Cult to expand the Chinese lockdown model across the West. Without this, and the bare-faced lie that non-symptomatic people could still transmit a ‘disease’ they didn’t have, there was no way locking down the whole population, sick and not sick, could be pulled off. At just the right time and with no evidence Cult operatives and gofers claimed that people without symptoms could pass on the ‘disease’. In the name of protecting the ‘vulnerable’ like elderly people, who lockdowns would kill by the tens of thousands, we had for the first time healthy people told to isolate as well as the sick. The great majority of people who tested positive had no symptoms because there was nothing wrong with them. It was just a trick made possible by a test not testing for the ‘virus’. Months a er my animated video the Gates-funded Professor Neil Ferguson at the Gates-funded Imperial College confirmed that I was right. He didn’t say it in those terms, naturally, but he did say it. Ferguson will enter the story shortly for his outrageously crazy ‘computer models’ that led to Britain, the United States and many other countries following the Chinese and now Italian methods of response. Put another way, following the Cult script. Ferguson said that SAGE, the UK government’s scientific advisory group which has controlled ‘Covid’ policy from the start, wanted to follow the Chinese lockdown model (while they all continued to work and be paid), but they wondered if they could possibly, in Ferguson’s words, ‘get away with it in Europe’. ‘Get away with it’? Who the hell do these moronic, arrogant people think they are? This appalling man Ferguson said that once Italy went into national lockdown they realised they, too, could mimic China: It’s a communist one-party state, we said. We couldn’t get away with it in Europe, we thought … and then Italy did it. And we realised we could. Behind this garbage from Ferguson is a simple fact: Doing the same as China in every country was the plan from the start and Ferguson’s ‘models’ would play a central role in achieving that. It’s just a coincidence, of course, and absolutely nothing to worry your little head about.

Oops, sorry, our mistake

Once the Italian segment of the Psyop had done the job it was designed to do a very different story emerged. Italian authorities revealed that 99 percent of those who had ‘died from Covid-19’ in Italy had one, two, three, or more ‘co-morbidities’ or illnesses and health problems that could have ended their life. The US Centers for Disease Control and Prevention (CDC) published a figure of 94 percent for Americans dying of ‘Covid’ while having other serious medical conditions – on average two to three (some five or six) other potential causes of death. In terms of death from an unproven ‘virus’ I say it is 100 percent. The other one percent in Italy and six percent in the US would presumably have died from ‘Covid’s’ flu-like symptoms with a range of other possible causes in conjunction with a test not testing for the ‘virus’. Fox News reported that even more startling figures had emerged in one US county in which 410 of 422 deaths a ributed to ‘Covid-19’ had other potentially deadly health conditions. The Italian National Health Institute said later that the average age of people dying with a ‘Covid-19’ diagnosis in Italy was about 81. Ninety percent were over 70 with ten percent over 90. In terms of other reasons to die some 80 percent had two or more chronic diseases with half having three or more including cardiovascular problems, diabetes, respiratory problems and cancer. Why is the phantom ‘Covid-19’ said to kill overwhelmingly old people and hardly affect the young? Old people continually die of many causes and especially respiratory disease which you can rediagnose ‘Covid-19’ while young people die in tiny numbers by comparison and rarely of respiratory disease. Old people ‘die of Covid’ because they die of other things that can be redesignated ‘Covid’ and it really is that simple.

Flu has flown

The blueprint was in place. Get your illusory ‘cases’ from a test not testing for the ‘virus’ and redesignate other causes of death as ‘Covid-19’. You have an instant ‘pandemic’ from something that is nothing more than a computer-generated fiction. With near-on a

billion people having ‘flu-like’ symptoms every year the potential was limitless and we can see why flu quickly and apparently miraculously disappeared worldwide by being diagnosed ‘Covid-19’. The painfully bloody obvious was explained away by the childlike media in headlines like this in the UK ‘Independent’: ‘Not a single case of flu detected by Public Health England this year as Covid restrictions suppress virus’. I kid you not. The masking, social distancing and house arrest that did not make the ‘Covid virus’ disappear somehow did so with the ‘flu virus’. Even worse the article, by a bloke called Samuel Love , suggested that maybe the masking, sanitising and other ‘Covid’ measures should continue to keep the flu away. With a ridiculousness that disturbs your breathing (it’s ‘Covid-19’) the said Love wrote: ‘With widespread social distancing and mask-wearing measures in place throughout the UK, the usual routes of transmission for influenza have been blocked.’ He had absolutely no evidence to support that statement, but look at the consequences of him acknowledging the obvious. With flu not disappearing at all and only being relabelled ‘Covid-19’ he would have to contemplate that ‘Covid’ was a hoax on a scale that is hard to imagine. You need guts and commitment to truth to even go there and that’s clearly something Samuel Love does not have in abundance. He would never have got it through the editors anyway. Tens of thousands die in the United States alone every winter from flu including many with pneumonia complications. CDC figures record 45 million Americans diagnosed with flu in 2017-2018 of which 61,000 died and some reports claim 80,000. Where was the same hysteria then that we have seen with ‘Covid-19’? Some 250,000 Americans are admi ed to hospital with pneumonia every year with about 50,000 cases proving fatal. About 65 million suffer respiratory disease every year and three million deaths makes this the third biggest cause of death worldwide. You only have to redesignate a portion of all these people ‘Covid-19’ and you have an instant global pandemic or the appearance of one. Why would doctors do this? They are told to do this and all but a few dare not refuse those who must be obeyed. Doctors in general are not researching their own

knowledge and instead take it direct and unquestioned from the authorities that own them and their careers. The authorities say they must now diagnose these symptoms ‘Covid-19’ and not flu, or whatever, and they do it. Dark suits say put ‘Covid-19’ on death certificates no ma er what the cause of death and the doctors do it. Renegade Minds don’t fall for the illusion that doctors and medical staff are all highly-intelligent, highly-principled, seekers of medical truth. Some are, but not the majority. They are repeaters, gofers, and yes sir, no sir, purveyors of what the system demands they purvey. The ‘Covid’ con is not merely confined to diseases of the lungs. Instructions to doctors to put ‘Covid-19’ on death certificates for anyone dying of anything within 28 days (or much more) of a positive test not testing for the ‘virus’ opened the floodgates. The term dying with ‘Covid’ and not of ‘Covid’ was coined to cover the truth. Whether it was a with or an of they were all added to the death numbers a ributed to the ‘deadly virus’ compiled by national governments and globally by the Gates-funded Johns Hopkins operation in the United States that was so involved in those ‘pandemic’ simulations. Fraudulent deaths were added to the evergrowing list of fraudulent ‘cases’ from false positives from a false test. No wonder Professor Walter Ricciardi, scientific advisor to the Italian minister of health, said a er the Lombardy hysteria had done its job that ‘Covid’ death rates were due to Italy having the second oldest population in the world and to how hospitals record deaths: The way in which we code deaths in our country is very generous in the sense that all the people who die in hospitals with the coronavirus are deemed to be dying of the coronavirus. On re-evaluation by the National Institute of Health, only 12 per cent of death certificates have shown a direct causality from coronavirus, while 88 per cent of patients who have died have at least one pre-morbidity – many had two or three.

This is extraordinary enough when you consider the propaganda campaign to use Italy to terrify the world, but how can they even say twelve percent were genuine when the ‘virus’ has not been shown to exist, its ‘code’ is a computer program, and diagnosis comes from a test not testing for it? As in China, and soon the world, ‘Covid-19’ in

Italy was a redesignation of diagnosis. Lies and corruption were to become the real ‘pandemic’ fuelled by a pathetically-compliant medical system taking its orders from the tiny few at the top of their national hierarchy who answered to the World Health Organization which answers to Gates and the Cult. Doctors were told – ordered – to diagnose a particular set of symptoms ‘Covid-19’ and put that on the death certificate for any cause of death if the patient had tested positive with a test not testing for the virus or had ‘Covid’ symptoms like the flu. The United States even introduced big financial incentives to manipulate the figures with hospitals receiving £4,600 from the Medicare system for diagnosing someone with regular pneumonia, $13,000 if they made the diagnosis from the same symptoms ‘Covid-19’ pneumonia, and $39, 000 if they put a ‘Covid’ diagnosed patient on a ventilator that would almost certainly kill them. A few – painfully and pathetically few – medical whistleblowers revealed (before Cult-owned YouTube deleted their videos) that they had been instructed to ‘let the patient crash’ and put them straight on a ventilator instead of going through a series of far less intrusive and dangerous methods as they would have done before the pandemic hoax began and the financial incentives kicked in. We are talking cold-blooded murder given that ventilators are so damaging to respiratory systems they are usually the last step before heaven awaits. Renegade Minds never fall for the belief that people in white coats are all angels of mercy and cannot be full-on psychopaths. I have explained in detail in The Answer how what I am describing here played out across the world coordinated by the World Health Organization through the medical hierarchies in almost every country.

Medical scientist calls it

Information about the non-existence of the ‘virus’ began to emerge for me in late March, 2020, and mushroomed a er that. I was sent an email by Sir Julian Rose, a writer, researcher, and organic farming promotor, from a medical scientist friend of his in the United States. Even at that early stage in March the scientist was able to explain

how the ‘Covid’ hoax was being manipulated. He said there were no reliable tests for a specific ‘Covid-19 virus’ and nor were there any reliable agencies or media outlets for reporting numbers of actual ‘Covid-19’ cases. We have seen in the long period since then that he was absolutely right. ‘Every action and reaction to Covid-19 is based on totally flawed data and we simply cannot make accurate assessments,’ he said. Most people diagnosed with ‘Covid-19’ were showing nothing more than cold and flu-like symptoms ‘because most coronavirus strains are nothing more than cold/flu-like symptoms’. We had farcical situations like an 84-year-old German man testing positive for ‘Covid-19’ and his nursing home ordered to quarantine only for him to be found to have a common cold. The scientist described back then why PCR tests and what he called the ‘Mickey Mouse test kits’ were useless for what they were claimed to be identifying. ‘The idea these kits can isolate a specific virus like Covid-19 is nonsense,’ he said. Significantly, he pointed out that ‘if you want to create a totally false panic about a totally false pandemic – pick a coronavirus’. This is exactly what the Cult-owned Gates, World Economic Forum and Johns Hopkins University did with their Event 201 ‘simulation’ followed by their real-life simulation called the ‘pandemic’. The scientist said that all you had to do was select the sickest of people with respiratory-type diseases in a single location – ‘say Wuhan’ – and administer PCR tests to them. You can then claim that anyone showing ‘viral sequences’ similar to a coronavirus ‘which will inevitably be quite a few’ is suffering from a ‘new’ disease: Since you already selected the sickest flu cases a fairly high proportion of your sample will go on to die. You can then say this ‘new’ virus has a CFR [case fatality rate] higher than the flu and use this to infuse more concern and do more tests which will of course produce more ‘cases’, which expands the testing, which produces yet more ‘cases’ and so on and so on. Before long you have your ‘pandemic’, and all you have done is use a simple test kit trick to convert the worst flu and pneumonia cases into something new that doesn’t ACTUALLY EXIST [my emphasis].

He said that you then ‘just run the same scam in other countries’ and make sure to keep the fear message running high ‘so that people

will feel panicky and less able to think critically’. The only problem to overcome was the fact there is no actual new deadly pathogen and only regular sick people. This meant that deaths from the ‘new deadly pathogen’ were going to be way too low for a real new deadly virus pandemic, but he said this could be overcome in the following ways – all of which would go on to happen: 1. You can claim this is just the beginning and more deaths are imminent [you underpin this with fantasy ‘computer projections’]. Use this as an excuse to quarantine everyone and then claim the quarantine prevented the expected millions of dead. 2. You can [say that people] ‘minimizing’ the dangers are irresponsible and bully them into not talking about numbers. 3. You can talk crap about made up numbers hoping to blind people with pseudoscience. 4. You can start testing well people (who, of course, will also likely have shreds of coronavirus [RNA] in them) and thus inflate your ‘case figures’ with ‘asymptomatic carriers’ (you will of course have to spin that to sound deadly even though any virologist knows the more symptom-less cases you have the less deadly is your pathogen).

The scientist said that if you take these simple steps ‘you can have your own entirely manufactured pandemic up and running in weeks’. His analysis made so early in the hoax was brilliantly prophetic of what would actually unfold. Pulling all the information together in these recent chapters we have this is simple 1, 2, 3, of how you can delude virtually the entire human population into believing in a ‘virus’ that doesn’t exist: • A ‘Covid case’ is someone who tests positive with a test not testing for the ‘virus’. • A ‘Covid death’ is someone who dies of any cause within 28 days (or much longer) of testing positive with a test not testing for the ‘virus. • Asymptomatic means there is nothing wrong with you, but they claim you can pass on what you don’t have to justify locking

down (quarantining) healthy people in totality. The foundations of the hoax are that simple. A study involving ten million people in Wuhan, published in November, 2020, demolished the whole lie about those without symptoms passing on the ‘virus’. They found ‘300 asymptomatic cases’ and traced their contacts to find that not one of them was detected with the ‘virus’. ‘Asymptomatic’ patients and their contacts were isolated for no less than two weeks and nothing changed. I know it’s all crap, but if you are going to claim that those without symptoms can transmit ‘the virus’ then you must produce evidence for that and they never have. Even World Health Organization official Dr Maria Van Kerkhove, head of the emerging diseases and zoonosis unit, said as early as June, 2020, that she doubted the validity of asymptomatic transmission. She said that ‘from the data we have, it still seems to be rare that an asymptomatic person actually transmits onward to a secondary individual’ and by ‘rare’ she meant that she couldn’t cite any case of asymptomatic transmission.

The Ferguson factor

The problem for the Cult as it headed into March, 2020, when the script had lockdown due to start, was that despite all the manipulation of the case and death figures they still did not have enough people alleged to have died from ‘Covid’ to justify mass house arrest. This was overcome in the way the scientist described: ‘You can claim this is just the beginning and more deaths are imminent … Use this as an excuse to quarantine everyone and then claim the quarantine prevented the expected millions of dead.’ Enter one Professor Neil Ferguson, the Gates-funded ‘epidemiologist’ at the Gates-funded Imperial College in London. Ferguson is Britain’s Christian Drosten in that he has a dire record of predicting health outcomes, but is still called upon to advise government on the next health outcome when another ‘crisis’ comes along. This may seem to be a strange and ridiculous thing to do. Why would you keep turning for policy guidance to people who have a history of being

monumentally wrong? Ah, but it makes sense from the Cult point of view. These ‘experts’ keep on producing predictions that suit the Cult agenda for societal transformation and so it was with Neil Ferguson as he revealed his horrific (and clearly insane) computer model predictions that allowed lockdowns to be imposed in Britain, the United States and many other countries. Ferguson does not have even an A-level in biology and would appear to have no formal training in computer modelling, medicine or epidemiology, according to Derek Winton, an MSc in Computational Intelligence. He wrote an article somewhat aghast at what Ferguson did which included taking no account of respiratory disease ‘seasonality’ which means it is far worse in the winter months. Who would have thought that respiratory disease could be worse in the winter? Well, certainly not Ferguson. The massively China-connected Imperial College and its bizarre professor provided the excuse for the long-incubated Chinese model of human control to travel westward at lightning speed. Imperial College confirms on its website that it collaborates with the Chinese Research Institute; publishes more than 600 research papers every year with Chinese research institutions; has 225 Chinese staff; 2,600 Chinese students – the biggest international group; 7,000 former students living in China which is the largest group outside the UK; and was selected for a tour by China’s President Xi Jinping during his state visit to the UK in 2015. The college takes major donations from China and describes itself as the UK’s number one university collaborator with Chinese research institutions. The China communist/fascist government did not appear phased by the woeful predictions of Ferguson and Imperial when during the lockdown that Ferguson induced the college signed a five-year collaboration deal with China tech giant Huawei that will have Huawei’s indoor 5G network equipment installed at the college’s West London tech campus along with an ‘AI cloud platform’. The deal includes Chinese sponsorship of Imperial’s Venture Catalyst entrepreneurship competition. Imperial is an example of the enormous influence the Chinese government has within British and North American

universities and research centres – and further afield. Up to 200 academics from more than a dozen UK universities are being investigated on suspicion of ‘unintentionally’ helping the Chinese government build weapons of mass destruction by ‘transferring world-leading research in advanced military technology such as aircra , missile designs and cyberweapons’. Similar scandals have broken in the United States, but it’s all a coincidence. Imperial College serves the agenda in many other ways including the promotion of every aspect of the United Nations Agenda 21/2030 (the Great Reset) and produced computer models to show that human-caused ‘climate change’ is happening when in the real world it isn’t. Imperial College is driving the climate agenda as it drives the ‘Covid’ agenda (both Cult hoaxes) while Patrick Vallance, the UK government’s Chief Scientific Adviser on ‘Covid’, was named Chief Scientific Adviser to the UN ‘climate change’ conference known as COP26 hosted by the government in Glasgow, Scotland. ‘Covid’ and ‘climate’ are fundamentally connected.

Professor Woeful

From Imperial’s bosom came Neil Ferguson still advising government despite his previous disasters and it was announced early on that he and other key people like UK Chief Medical Adviser Chris Whi y had caught the ‘virus’ as the propaganda story was being sold. Somehow they managed to survive and we had Prime Minister Boris Johnson admi ed to hospital with what was said to be a severe version of the ‘virus’ in this same period. His whole policy and demeanour changed when he returned to Downing Street. It’s a small world with these government advisors – especially in their communal connections to Gates – and Ferguson had partnered with Whi y to write a paper called ‘Infectious disease: Tough choices to reduce Ebola transmission’ which involved another scare-story that didn’t happen. Ferguson’s ‘models’ predicted that up to150, 000 could die from ‘mad cow disease’, or BSE, and its version in sheep if it was transmi ed to humans. BSE was not transmi ed and instead triggered by an organophosphate pesticide used to treat a pest on

cows. Fewer than 200 deaths followed from the human form. Models by Ferguson and his fellow incompetents led to the unnecessary culling of millions of pigs, ca le and sheep in the foot and mouth outbreak in 2001 which destroyed the lives and livelihoods of farmers and their families who had o en spent decades building their herds and flocks. Vast numbers of these animals did not have foot and mouth and had no contact with the infection. Another ‘expert’ behind the cull was Professor Roy Anderson, a computer modeller at Imperial College specialising in the epidemiology of human, not animal, disease. Anderson has served on the Bill and Melinda Gates Grand Challenges in Global Health advisory board and chairs another Gates-funded organisation. Gates is everywhere. In a precursor to the ‘Covid’ script Ferguson backed closing schools ‘for prolonged periods’ over the swine flu ‘pandemic’ in 2009 and said it would affect a third of the world population if it continued to spread at the speed he claimed to be happening. His mates at Imperial College said much the same and a news report said: ‘One of the authors, the epidemiologist and disease modeller Neil Ferguson, who sits on the World Health Organisation’s emergency commi ee for the outbreak, said the virus had “full pandemic potential”.’ Professor Liam Donaldson, the Chris Whi y of his day as Chief Medical Officer, said the worst case could see 30 percent of the British people infected by swine flu with 65,000 dying. Ferguson and Donaldson were indeed proved correct when at the end of the year the number of deaths a ributed to swine flu was 392. The term ‘expert’ is rather liberally applied unfortunately, not least to complete idiots. Swine flu ‘projections’ were great for GlaxoSmithKline (GSK) as millions rolled in for its Pandemrix influenza vaccine which led to brain damage with children most affected. The British government (taxpayers) paid out more than £60 million in compensation a er GSK was given immunity from prosecution. Yet another ‘Covid’ déjà vu. Swine flu was supposed to have broken out in Mexico, but Dr Wolfgang Wodarg, a German doctor, former member of parliament and critic of the ‘Covid’ hoax, observed ‘the spread of swine flu’ in Mexico City at the time. He

said: ‘What we experienced in Mexico City was a very mild flu which did not kill more than usual – which killed even fewer people than usual.’ Hyping the fear against all the facts is not unique to ‘Covid’ and has happened many times before. Ferguson is reported to have over-estimated the projected death toll of bird flu (H5N1) by some three million-fold, but bird flu vaccine makers again made a killing from the scare. This is some of the background to the Neil Ferguson who produced the perfectly-timed computer models in early 2020 predicting that half a million people would die in Britain without draconian lockdown and 2.2 million in the United States. Politicians panicked, people panicked, and lockdowns of alleged short duration were instigated to ‘fla en the curve’ of cases gleaned from a test not testing for the ‘virus’. I said at the time that the public could forget the ‘short duration’ bit. This was an agenda to destroy the livelihoods of the population and force them into mass control through dependency and there was going to be nothing ‘short’ about it. American researcher Daniel Horowitz described the consequences of the ‘models’ spewed out by Gates-funded Ferguson and Imperial College: What led our government and the governments of many other countries into panic was a single Imperial College of UK study, funded by global warming activists, that predicted 2.2 million deaths if we didn’t lock down the country. In addition, the reported 8-9% death rate in Italy scared us into thinking there was some other mutation of this virus that they got, which might have come here. Together with the fact that we were finally testing and had the ability to actually report new cases, we thought we were headed for a death spiral. But again … we can’t flatten a curve if we don’t know when the curve started.

How about it never started?

Giving them what they want

An investigation by German news outlet Welt Am Sonntag (World on Sunday) revealed how in March, 2020, the German government gathered together ‘leading scientists from several research institutes and universities’ and ‘together, they were to produce a [modelling]

paper that would serve as legitimization for further tough political measures’. The Cult agenda was justified by computer modelling not based on evidence or reality; it was specifically constructed to justify the Cult demand for lockdowns all over the world to destroy the independent livelihoods of the global population. All these modellers and everyone responsible for the ‘Covid’ hoax have a date with a trial like those in Nuremberg a er World War Two when Nazis faced the consequences of their war crimes. These corruptbeyond-belief ‘modellers’ wrote the paper according to government instructions and it said that that if lockdown measures were li ed then up to one million Germans would die from ‘Covid-19’ adding that some would die ‘agonizingly at home, gasping for breath’ unable to be treated by hospitals that couldn’t cope. All lies. No ma er – it gave the Cult all that it wanted. What did long-time government ‘modeller’ Neil Ferguson say? If the UK and the United States didn’t lockdown half a million would die in Britain and 2.2 million Americans. Anyone see a theme here? ‘Modellers’ are such a crucial part of the lockdown strategy that we should look into their background and follow the money. Researcher Rosemary Frei produced an excellent article headlined ‘The Modelling-paper Mafiosi’. She highlights a guy called John Edmunds, a British epidemiologist, and professor in the Faculty of Epidemiology and Population Health at the London School of Hygiene & Tropical Medicine. He studied at Imperial College. Edmunds is a member of government ‘Covid’ advisory bodies which have been dictating policy, the New and Emerging Respiratory Virus Threats Advisory Group (NERVTAG) and the Scientific Advisory Group for Emergencies (SAGE). Ferguson, another member of NERVTAG and SAGE, led the way with the original ‘virus’ and Edmunds has followed in the ‘variant’ stage and especially the so-called UK or Kent variant known as the ‘Variant of Concern’ (VOC) B.1.1.7. He said in a co-wri en report for the Centre for Mathematical modelling of Infectious Diseases at the London School of Hygiene and Tropical Medicine, with input from the Centre’s ‘Covid-19’ Working Group, that there was ‘a realistic

possibility that VOC B.1.1.7 is associated with an increased risk of death compared to non-VOC viruses’. Fear, fear, fear, get the vaccine, fear, fear, fear, get the vaccine. Rosemary Frei reveals that almost all the paper’s authors and members of the modelling centre’s ‘Covid-19’ Working Group receive funding from the Bill and Melinda Gates Foundation and/or the associated Gates-funded Wellcome Trust. The paper was published by e-journal Medr χiv which only publishes papers not peer-reviewed and the journal was established by an organisation headed by Facebook’s Mark Zuckerberg and his missus. What a small world it is. Frei discovered that Edmunds is on the Scientific Advisory Board of the Coalition for Epidemic Preparedness Innovations (CEPI) which was established by the Bill and Melinda Gates Foundation, Klaus Schwab’s Davos World Economic Forum and Big Pharma giant Wellcome. CEPI was ‘launched in Davos [in 2017] to develop vaccines to stop future epidemics’, according to its website. ‘Our mission is to accelerate the development of vaccines against emerging infectious diseases and enable equitable access to these vaccines for people during outbreaks.’ What kind people they are. Rosemary Frei reveals that Public Health England (PHE) director Susan Hopkins is an author of her organisation’s non-peer-reviewed reports on ‘new variants’. Hopkins is a professor of infectious diseases at London’s Imperial College which is gi ed tens of millions of dollars a year by the Bill and Melinda Gates Foundation. Gates-funded modelling disaster Neil Ferguson also co-authors Public Health England reports and he spoke in December, 2020, about the potential danger of the B.1.1.7. ‘UK variant’ promoted by Gates-funded modeller John Edmunds. When I come to the ‘Covid vaccines’ the ‘new variants’ will be shown for what they are – bollocks.

Connections, connections

All these people and modellers are lockdown-obsessed or, put another way, they demand what the Cult demands. Edmunds said in January, 2021, that to ease lockdowns too soon would be a disaster and they had to ‘vaccinate much, much, much more widely than the

elderly’. Rosemary Frei highlights that Edmunds is married to Jeanne Pimenta who is described in a LinkedIn profile as director of epidemiology at GlaxoSmithKline (GSK) and she held shares in the company. Patrick Vallance, co-chair of SAGE and the government’s Chief Scientific Adviser, is a former executive of GSK and has a deferred bonus of shares in the company worth £600,000. GSK has serious business connections with Bill Gates and is collaborating with mRNA-’vaccine’ company CureVac to make ‘vaccines’ for the new variants that Edmunds is talking about. GSK is planning a ‘Covid vaccine’ with drug giant Sanofi. Puppet Prime Minister Boris Johnson announced in the spring of 2021 that up to 60 million vaccine doses were to be made at the GSK facility at Barnard Castle in the English North East. Barnard Castle, with a population of just 6,000, was famously visited in breach of lockdown rules in April, 2020, by Johnson aide Dominic Cummings who said that he drove there ‘to test his eyesight’ before driving back to London. Cummings would be be er advised to test his integrity – not that it would take long. The GSK facility had nothing to do with his visit then although I’m sure Patrick Vallance would have been happy to arrange an introduction and some tea and biscuits. Ruthless psychopath Gates has made yet another fortune from vaccines in collaboration with Big Pharma companies and gushes at the phenomenal profits to be made from vaccines – more than a 20-to-1 return as he told one interviewer. Gates also tweeted in December, 2019, with the foreknowledge of what was coming: ‘What’s next for our foundation? I’m particularly excited about what the next year could mean for one of the best buys in global health: vaccines.’ Modeller John Edmunds is a big promotor of vaccines as all these people appear to be. He’s the dean of the London School of Hygiene & Tropical Medicine’s Faculty of Epidemiology and Population Health which is primarily funded by the Bill and Melinda Gates Foundation and the Gates-established and funded GAVI vaccine alliance which is the Gates vehicle to vaccinate the world. The organisation Doctors Without Borders has described GAVI as being ‘aimed more at supporting drug-industry desires to promote new

products than at finding the most efficient and sustainable means for fighting the diseases of poverty’. But then that’s why the psychopath Gates created it. John Edmunds said in a video that the London School of Hygiene & Tropical Medicine is involved in every aspect of vaccine development including large-scale clinical trials. He contends that mathematical modelling can show that vaccines protect individuals and society. That’s on the basis of shit in and shit out, I take it. Edmunds serves on the UK Vaccine Network as does Ferguson and the government’s foremost ‘Covid’ adviser, the grimfaced, dark-eyed Chris Whi y. The Vaccine Network says it works ‘to support the government to identify and shortlist targeted investment opportunities for the most promising vaccines and vaccine technologies that will help combat infectious diseases with epidemic potential, and to address structural issues related to the UK’s broader vaccine infrastructure’. Ferguson is acting Director of the Imperial College Vaccine Impact Modelling Consortium which has funding from the Bill and Melina Gates Foundation and the Gates-created GAVI ‘vaccine alliance’. Anyone wonder why these characters see vaccines as the answer to every problem? Ferguson is wildly enthusiastic in his support for GAVI’s campaign to vaccine children en masse in poor countries. You would expect someone like Gates who has constantly talked about the need to reduce the population to want to fund vaccines to keep more people alive. I’m sure that’s why he does it. The John Edmunds London School of Hygiene & Tropical Medicine (LSHTM) has a Vaccines Manufacturing Innovation Centre which develops, tests and commercialises vaccines. Rosemary Frei writes: The vaccines centre also performs affiliated activities like combating ‘vaccine hesitancy’. The latter includes the Vaccine Confidence Project. The project’s stated purpose is, among other things, ‘to provide analysis and guidance for early response and engagement with the public to ensure sustained confidence in vaccines and immunisation’. The Vaccine Confidence Project’s director is LSHTM professor Heidi Larson. For more than a decade she’s been researching how to combat vaccine hesitancy.

How the bloody hell can blokes like John Edmunds and Neil Ferguson with those connections and financial ties model ‘virus’ case

and death projections for the government and especially in a way that gives their paymasters like Gates exactly what they want? It’s insane, but this is what you find throughout the world.

‘Covid’ is not dangerous, oops, wait, yes it is

Only days before Ferguson’s nightmare scenario made Jackboot Johnson take Britain into a China-style lockdown to save us from a deadly ‘virus’ the UK government website gov.uk was reporting something very different to Ferguson on a page of official government guidance for ‘high consequence infectious diseases (HCID)’. It said this about ‘Covid-19’: As of 19 March 2020, COVID-19 is no longer considered to be a high consequence infectious diseases (HCID) in the UK [my emphasis]. The 4 nations public health HCID group made an interim recommendation in January 2020 to classify COVID-19 as an HCID. This was based on consideration of the UK HCID criteria about the virus and the disease with information available during the early stages of the outbreak. Now that more is known about COVID-19, the public health bodies in the UK have reviewed the most up to date information about COVID-19 against the UK HCID criteria. They have determined that several features have now changed; in particular, more information is available about mortality rates (low overall), and there is now greater clinical awareness and a specific and sensitive laboratory test, the availability of which continues to increase. The Advisory Committee on Dangerous Pathogens (ACDP) is also of the opinion that COVID-19 should no longer be classified as an HCID.

Soon a er the government had been exposed for downgrading the risk they upgraded it again and everyone was back to singing from the same Cult hymn book. Ferguson and his fellow Gates clones indicated that lockdowns and restrictions would have to continue until a Gates-funded vaccine was developed. Gates said the same because Ferguson and his like were repeating the Gates script which is the Cult script. ‘Fla en the curve’ became an ongoing nightmare of continuing lockdowns with periods in between of severe restrictions in pursuit of destroying independent incomes and had nothing to do with protecting health about which the Cult gives not a shit. Why wouldn’t Ferguson be pushing a vaccine ‘solution’ when he’s owned by vaccine-obsessive Gates who makes a fortune from them and

when Ferguson heads the Vaccine Impact Modelling Consortium at Imperial College funded by the Gates Foundation and GAVI, the ‘vaccine alliance’, created by Gates as his personal vaccine promotion operation? To compound the human catastrophe that Ferguson’s ‘models’ did so much to create he was later exposed for breaking his own lockdown rules by having sexual liaisons with his married girlfriend Antonia Staats at his home while she was living at another location with her husband and children. Staats was a ‘climate’ activist and senior campaigner at the Soros-funded Avaaz which I wouldn’t trust to tell me that grass is green. Ferguson had to resign as a government advisor over this hypocrisy in May, 2020, but a er a period of quiet he was back being quoted by the ridiculous media on the need for more lockdowns and a vaccine rollout. Other government-advising ‘scientists’ from Imperial College’ held the fort in his absence and said lockdown could be indefinite until a vaccine was found. The Cult script was being sung by the payrolled choir. I said there was no intention of going back to ‘normal’ when the ‘vaccine’ came because the ‘vaccine’ is part of a very different agenda that I will discuss in Human 2.0. Why would the Cult want to let the world go back to normal when destroying that normal forever was the whole point of what was happening? House arrest, closing businesses and schools through lockdown, (un)social distancing and masks all followed the Ferguson fantasy models. Again as I predicted (these people are so predictable) when the ‘vaccine’ arrived we were told that house arrest, lockdown, (un)social distancing and masks would still have to continue. I will deal with the masks in the next chapter because they are of fundamental importance.

Where’s the ‘pandemic’?

Any mildly in-depth assessment of the figures revealed what was really going on. Cult-funded and controlled organisations still have genuine people working within them such is the number involved. So it is with Genevieve Briand, assistant program director of the Applied Economics master’s degree program at Johns Hopkins

University. She analysed the impact that ‘Covid-19’ had on deaths from all causes in the United States using official data from the CDC for the period from early February to early September, 2020. She found that allegedly ‘Covid’ related-deaths exceeded those from heart disease which she found strange with heart disease always the biggest cause of fatalities. Her research became even more significant when she noted the sudden decline in 2020 of all non-’Covid’ deaths: ‘This trend is completely contrary to the pa ern observed in all previous years … the total decrease in deaths by other causes almost exactly equals the increase in deaths by Covid-19.’ This was such a game, set and match in terms of what was happening that Johns Hopkins University deleted the article on the grounds that it ‘was being used to support false and dangerous inaccuracies about the impact of the pandemic’. No – because it exposed the scam from official CDC figures and this was confirmed when those figures were published in January, 2021. Here we can see the effect of people dying from heart a acks, cancer, road accidents and gunshot wounds – anything – having ‘Covid-19’ on the death certificate along with those diagnosed from ‘symptoms’ who had even not tested positive with a test not testing for the ‘virus’. I am not kidding with the gunshot wounds, by the way. Brenda Bock, coroner in Grand County, Colorado, revealed that two gunshot victims tested positive for the ‘virus’ within the previous 30 days and were therefore classified as ‘Covid deaths’. Bock said: ‘These two people had tested positive for Covid, but that’s not what killed them. A gunshot wound is what killed them.’ She said she had not even finished her investigation when the state listed the gunshot victims as deaths due to the ‘virus’. The death and case figures for ‘Covid-19’ are an absolute joke and yet they are repeated like parrots by the media, politicians and alleged medical ‘experts’. The official Cult narrative is the only show in town. Genevieve Briand found that deaths from all causes were not exceptional in 2020 compared with previous years and a Spanish magazine published figures that said the same about Spain which was a ‘Covid’ propaganda hotspot at one point. Discovery Salud, a

health and medicine magazine, quoted government figures which showed how 17,000 fewer people died in Spain in 2020 than in 2019 and more than 26,000 fewer than in 2018. The age-standardised mortality rate for England and Wales when age distribution is taken into account was significantly lower in 2020 than the 1970s, 80s and 90s, and was only the ninth highest since 2000. Where is the ‘pandemic’? Post mortems and autopsies virtually disappeared for ‘Covid’ deaths amid claims that ‘virus-infected’ bodily fluids posed a risk to those carrying out the autopsy. This was rejected by renowned German pathologist and forensic doctor Klaus Püschel who said that he and his staff had by then done 150 autopsies on ‘Covid’ patients with no problems at all. He said they were needed to know why some ‘Covid’ patients suffered blood clots and not severe respiratory infections. The ‘virus’ is, a er all, called SARS or ‘severe acute respiratory syndrome’. I highlighted in the spring of 2020 this phenomenon and quoted New York intensive care doctor Cameron Kyle-Sidell who posted a soon deleted YouTube video to say that they had been told to prepare to treat an infectious disease called ‘Covid-19’, but that was not what they were dealing with. Instead he likened the lung condition of the most severely ill patients to what you would expect with cabin depressurisation in a plane at 30,000 feet or someone dropped on the top of Everest without oxygen or acclimatisation. I have never said this is not happening to a small minority of alleged ‘Covid’ patients – I am saying this is not caused by a phantom ‘contagious virus’. Indeed Kyle-Sidell said that ‘Covid-19’ was not the disease they were told was coming their way. ‘We are operating under a medical paradigm that is untrue,’ he said, and he believed they were treating the wrong disease: ‘These people are being slowly starved of oxygen.’ Patients would take off their oxygen masks in a state of fear and stress and while they were blue in the face on the brink of death. They did not look like patients dying of pneumonia. You can see why they don’t want autopsies when their virus doesn’t exist and there is another condition in some people that they don’t wish to be uncovered. I should add here that

the 5G system of millimetre waves was being rapidly introduced around the world in 2020 and even more so now as they fire 5G at the Earth from satellites. At 60 gigahertz within the 5G range that frequency interacts with the oxygen molecule and stops people breathing in sufficient oxygen to be absorbed into the bloodstream. They are installing 5G in schools and hospitals. The world is not mad or anything. 5G can cause major changes to the lungs and blood as I detail in The Answer and these consequences are labelled ‘Covid19’, the alleged symptoms of which can be caused by 5G and other electromagnetic frequencies as cells respond to radiation poisoning.

The ‘Covid death’ scam

Dr Sco Jensen, a Minnesota state senator and medical doctor, exposed ‘Covid’ Medicare payment incentives to hospitals and death certificate manipulation. He said he was sent a seven-page document by the US Department of Health ‘coaching’ him on how to fill out death certificates which had never happened before. The document said that he didn’t need to have a laboratory test for ‘Covid-19’ to put that on the death certificate and that shocked him when death certificates are supposed to be about facts. Jensen described how doctors had been ‘encouraged, if not pressured’ to make a diagnosis of ‘Covid-19’ if they thought it was probable or ‘presumed’. No positive test was necessary – not that this would have ma ered anyway. He said doctors were told to diagnose ‘Covid’ by symptoms when these were the same as colds, allergies, other respiratory problems, and certainly with influenza which ‘disappeared’ in the ‘Covid’ era. A common sniffle was enough to get the dreaded verdict. Ontario authorities decreed that a single care home resident with one symptom from a long list must lead to the isolation of the entire home. Other courageous doctors like Jensen made the same point about death figure manipulation and how deaths by other causes were falling while ‘Covid-19 deaths’ were rising at the same rate due to re-diagnosis. Their videos rarely survive long on YouTube with its Cult-supporting algorithms courtesy of CEO Susan Wojcicki and her bosses at Google. Figure-tampering was so glaring

and ubiquitous that even officials were le ing it slip or outright saying it. UK chief scientific adviser Patrick Vallance said on one occasion that ‘Covid’ on the death certificate doesn’t mean ‘Covid’ was the cause of death (so why the hell is it there?) and we had the rare sight of a BBC reporter telling the truth when she said: ‘Someone could be successfully treated for Covid, in say April, discharged, and then in June, get run over by a bus and die … That person would still be counted as a Covid death in England.’ Yet the BBC and the rest of the world media went on repeating the case and death figures as if they were real. Illinois Public Health Director Dr Ngozi Ezike revealed the deceit while her bosses must have been clenching their bu ocks: If you were in a hospice and given a few weeks to live and you were then found to have Covid that would be counted as a Covid death. [There might be] a clear alternate cause, but it is still listed as a Covid death. So everyone listed as a Covid death doesn’t mean that was the cause of the death, but that they had Covid at the time of death.

Yes, a ‘Covid virus’ never shown to exist and tested for with a test not testing for the ‘virus’. In the first period of the pandemic hoax through the spring of 2020 the process began of designating almost everything a ‘Covid’ death and this has continued ever since. I sat in a restaurant one night listening to a loud conversation on the next table where a family was discussing in bewilderment how a relative who had no symptoms of ‘Covid’, and had died of a long-term problem, could have been diagnosed a death by the ‘virus’. I could understand their bewilderment. If they read this book they will know why this medical fraud has been perpetrated the world over.

Some media truth shock

The media ignored the evidence of death certificate fraud until eventually one columnist did speak out when she saw it first-hand. Bel Mooney is a long-time national newspaper journalist in Britain currently working for the Daily Mail. Her article on February 19th, 2021, carried this headline: ‘My dad Ted passed three Covid tests

and died of a chronic illness yet he’s officially one of Britain’s 120,000 victims of the virus and is far from alone ... so how many more are there?’ She told how her 99-year-old father was in a care home with a long-standing chronic obstructive pulmonary disease and vascular dementia. Maybe, but he was still aware enough to tell her from the start that there was no ‘virus’ and he refused the ‘vaccine’ for that reason. His death was not unexpected given his chronic health problems and Mooney said she was shocked to find that ‘Covid-19’ was declared the cause of death on his death certificate. She said this was a ‘bizarre and unacceptable untruth’ for a man with long-time health problems who had tested negative twice at the home for the ‘virus’. I was also shocked by this story although not by what she said. I had been highlighting the death certificate manipulation for ten months. It was the confirmation that a professional full-time journalist only realised this was going on when it affected her directly and neither did she know that whether her dad tested positive or negative was irrelevant with the test not testing for the ‘virus’. Where had she been? She said she did not believe in ‘conspiracy theories’ without knowing I’m sure that this and ‘conspiracy theorists’ were terms put into widespread circulation by the CIA in the 1960s to discredit those who did not accept the ridiculous official story of the Kennedy assassination. A blanket statement of ‘I don’t believe in conspiracy theories’ is always bizarre. The dictionary definition of the term alone means the world is drowning in conspiracies. What she said was even more da when her dad had just been affected by the ‘Covid’ conspiracy. Why else does she think that ‘Covid-19’ was going on the death certificates of people who died of something else? To be fair once she saw from personal experience what was happening she didn’t mince words. Mooney was called by the care home on the morning of February 9th to be told her father had died in his sleep. When she asked for the official cause of death what came back was ‘Covid-19’. Mooney challenged this and was told there had been deaths from Covid on the dementia floor (confirmed by a test not testing for the ‘virus’) so they considered it ‘reasonable

to assume’. ‘But doctor,’ Mooney rightly protested, ‘an assumption isn’t a diagnosis.’ She said she didn’t blame the perfectly decent and sympathetic doctor – ‘he was just doing his job’. Sorry, but that’s bullshit. He wasn’t doing his job at all. He was pu ing a false cause of death on the death certificate and that is a criminal offence for which he should be brought to account and the same with the millions of doctors worldwide who have done the same. They were not doing their job they were following orders and that must not wash at new Nuremberg trials any more than it did at the first ones. Mooney’s doctor was ‘assuming’ (presuming) as he was told to, but ‘just following orders’ makes no difference to his actions. A doctor’s job is to serve the patient and the truth, not follow orders, but that’s what they have done all over the world and played a central part in making the ‘Covid’ hoax possible with all its catastrophic consequences for humanity. Shame on them and they must answer for their actions. Mooney said her disquiet worsened when she registered her father’s death by telephone and was told by the registrar there had been very many other cases like hers where ‘the deceased’ had not tested positive for ‘Covid’ yet it was recorded as the cause of death. The test may not ma er, but those involved at their level think it ma ers and it shows a callous disregard for accurate diagnosis. The pressure to do this is coming from the top of the national ‘health’ pyramids which in turn obey the World Health Organization which obeys Gates and the Cult. Mooney said the registrar agreed that this must distort the national figures adding that ‘the strangest thing is that every winter we record countless deaths from flu, and this winter there have been none. Not one!’ She asked if the registrar thought deaths from flu were being misdiagnosed and lumped together with ‘Covid’ deaths. The answer was a ‘puzzled yes’. Mooney said that the funeral director said the same about ‘Covid’ deaths which had nothing to do with ‘Covid’. They had lost count of the number of families upset by this and other funeral companies in different countries have had the same experience. Mooney wrote:

The nightly shroud-waving and shocking close-ups of pain imposed on us by the TV news bewildered and terrified the population into eager compliance with lockdowns. We were invited to ‘save the NHS’ and to grieve for strangers – the real-life loved ones behind those shocking death counts. Why would the public imagine what I now fear, namely that the way Covid-19 death statistics are compiled might make the numbers seem greater than they are?

Oh, just a li le bit – like 100 percent.

Do the maths

Mooney asked why a country would wish to skew its mortality figures by wrongly certifying deaths? What had been going on? Well, if you don’t believe in conspiracies you will never find the answer which is that it’s a conspiracy. She did, however, describe what she had discovered as a ‘national scandal’. In reality it’s a global scandal and happening everywhere. Pillars of this conspiracy were all put into place before the bu on was pressed with the Drosten PCR protocol and high amplifications to produce the cases and death certificate changes to secure illusory ‘Covid’ deaths. Mooney notes that normally two doctors were needed to certify a death, with one having to know the patient, and how the rules were changed in the spring of 2020 to allow one doctor to do this. In the same period ‘Covid deaths’ were decreed to be all cases where Covid-19 was put on the death certificate even without a positive test or any symptoms. Mooney asked: ‘How many of the 30,851 (as of January 15) care home resident deaths with Covid-19 on the certificate (32.4 per cent of all deaths so far) were based on an assumption, like that of my father? And what has that done to our national psyche?’All of them is the answer to the first question and it has devastated and dismantled the national psyche, actually the global psyche, on a colossal scale. In the UK case and death data is compiled by organisations like Public Health England (PHE) and the Office for National Statistics (ONS). Mooney highlights the insane policy of counting a death from any cause as ‘Covid-19’ if this happens within 28 days of a positive test (with a test not testing for the ‘virus’) and she points out that ONS statistics reflect deaths ‘involving Covid’ ‘or due to Covid’ which meant in practice any

death where ‘Covid-19’ was mentioned on the death certificate. She described the consequences of this fraud: Most people will accept the narrative they are fed, so panicky governments here and in Europe witnessed the harsh measures enacted in totalitarian China and jumped into lockdown. Headlines about Covid deaths tolled like the knell that would bring doomsday to us all. Fear stalked our empty streets. Politicians parroted the frankly ridiculous aim of ‘zero Covid’ and shut down the economy, while most British people agreed that lockdown was essential and (astonishingly to me, as a patriotic Brit) even wanted more restrictions. For what? Lies on death certificates? Never mind the grim toll of lives ruined, suicides, schools closed, rising inequality, depression, cancelled hospital treatments, cancer patients in a torture of waiting, poverty, economic devastation, loneliness, families kept apart, and so on. How many lives have been lost as a direct result of lockdown?

She said that we could join in a national chorus of shock and horror at reaching the 120,000 death toll which was surely certain to have been totally skewed all along, but what about the human cost of lockdown justified by these ‘death figures’? The British Medical Journal had reported a 1,493 percent increase in cases of children taken to Great Ormond Street Hospital with abusive head injuries alone and then there was the effect on families: Perhaps the most shocking thing about all this is that families have been kept apart – and obeyed the most irrational, changing rules at the whim of government – because they believed in the statistics. They succumbed to fear, which his generation rejected in that war fought for freedom. Dad (God rest his soul) would be angry. And so am I.

Another theme to watch is that in the winter months when there are more deaths from all causes they focus on ‘Covid’ deaths and in the summer when the British Lung Foundation says respiratory disease plummets by 80 percent they rage on about ‘cases’. Either way fascism on population is always the answer.

Nazi eugenics in the 21st century

Elderly people in care homes have been isolated from their families month a er lonely month with no contact with relatives and grandchildren who were banned from seeing them. We were told

that lockdown fascism was to ‘protect the vulnerable’ like elderly people. At the same time Do Not Resuscitate (DNR) orders were placed on their medical files so that if they needed resuscitation it wasn’t done and ‘Covid-19’ went on their death certificates. Old people were not being ‘protected’ they were being culled – murdered in truth. DNR orders were being decreed for disabled and young people with learning difficulties or psychological problems. The UK Care Quality Commission, a non-departmental body of the Department of Health and Social Care, found that 34 percent of those working in health and social care were pressured into placing ‘do not a empt cardiopulmonary resuscitation’ orders on ‘Covid’ patients who suffered from disabilities and learning difficulties without involving the patient or their families in the decision. UK judges ruled that an elderly woman with dementia should have the DNA-manipulating ‘Covid vaccine’ against her son’s wishes and that a man with severe learning difficulties should have the jab despite his family’s objections. Never mind that many had already died. The judiciary always supports doctors and government in fascist dictatorships. They wouldn’t dare do otherwise. A horrific video was posted showing fascist officers from Los Angeles police forcibly giving the ‘Covid’ shot to women with special needs who were screaming that they didn’t want it. The same fascists are seen giving the jab to a sleeping elderly woman in a care home. This is straight out of the Nazi playbook. Hitler’s Nazis commi ed mass murder of the mentally ill and physically disabled throughout Germany and occupied territories in the programme that became known as Aktion T4, or just T4. Sabbatian-controlled Hitler and his grotesque crazies set out to kill those they considered useless and unnecessary. The Reich Commi ee for the Scientific Registering of Hereditary and Congenital Illnesses registered the births of babies identified by physicians to have ‘defects’. By 1941 alone more than 5,000 children were murdered by the state and it is estimated that in total the number of innocent people killed in Aktion T4 was between 275,000 and 300,000. Parents were told their children had been sent away for ‘special treatment’ never to return. It is rather pathetic to see claims about plans for new extermination camps being dismissed today

when the same force behind current events did precisely that 80 years ago. Margaret Sanger was a Cult operative who used ‘birth control’ to sanitise her programme of eugenics. Organisations she founded became what is now Planned Parenthood. Sanger proposed that ‘the whole dysgenic population would have its choice of segregation or sterilization’. These included epileptics, ‘feebleminded’, and prostitutes. Sanger opposed charity because it perpetuated ‘human waste‘. She reveals the Cult mentality and if anyone thinks that extermination camps are a ‘conspiracy theory’ their naivety is touching if breathtakingly stupid. If you don’t believe that doctors can act with callous disregard for their patients it is worth considering that doctors and medical staff agreed to put government-decreed DNR orders on medical files and do nothing when resuscitation is called for. I don’t know what you call such people in your house. In mine they are Nazis from the Josef Mengele School of Medicine. Phenomenal numbers of old people have died worldwide from the effects of lockdown, depression, lack of treatment, the ‘vaccine’ (more later) and losing the will to live. A common response at the start of the manufactured pandemic was to remove old people from hospital beds and transfer them to nursing homes. The decision would result in a mass cull of elderly people in those homes through lack of treatment – not ‘Covid’. Care home whistleblowers have told how once the ‘Covid’ era began doctors would not come to their homes to treat patients and they were begging for drugs like antibiotics that o en never came. The most infamous example was ordered by New York governor Andrew Cuomo, brother of a moronic CNN host, who amazingly was given an Emmy Award for his handling of the ‘Covid crisis’ by the ridiculous Wokers that hand them out. Just how ridiculous could be seen in February, 2021, when a Department of Justice and FBI investigation began into how thousands of old people in New York died in nursing homes a er being discharged from hospital to make way for ‘Covid’ patients on Cuomo’s say-so – and how he and his staff covered up these facts. This couldn’t have happened to a nicer psychopath. Even then there was a ‘Covid’ spin. Reports said that

thousands of old people who tested positive for ‘Covid’ in hospital were transferred to nursing homes to both die of ‘Covid’ and transmit it to others. No – they were in hospital because they were ill and the fact that they tested positive with a test not testing for the ‘virus’ is irrelevant. They were ill o en with respiratory diseases ubiquitous in old people near the end of their lives. Their transfer out of hospital meant that their treatment stopped and many would go on to die.

They’re old. Who gives a damn?

I have exposed in the books for decades the Cult plan to cull the world’s old people and even to introduce at some point what they call a ‘demise pill’ which at a certain age everyone would take and be out of here by law. In March, 2021, Spain legalised euthanasia and assisted suicide following the Netherlands, Belgium, Luxembourg and Canada on the Tiptoe to the demise pill. Treatment of old people by many ‘care’ homes has been a disgrace in the ‘Covid’ era. There are many, many, caring staff – I know some. There have, however, been legions of stories about callous treatment of old people and their families. Police were called when families came to take their loved ones home in the light of isolation that was killing them. They became prisoners of the state. Care home residents in insane, fascist Ontario, Canada, were not allowed to leave their room once the ‘Covid’ hoax began. UK staff have even wheeled elderly people away from windows where family members were talking with them. Oriana Criscuolo from Stockport in the English North West dropped off some things for her 80-year-old father who has Parkinson’s disease and dementia and she wanted to wave to him through a ground-floor window. She was told that was ‘illegal’. When she went anyway they closed the curtains in the middle of the day. Oriana said: It’s just unbelievable. I cannot understand how care home staff – people who are being paid to care – have become so uncaring. Their behaviour is inhumane and cruel. It’s beyond belief.

She was right and this was not a one-off. What a way to end your life in such loveless circumstances. UK registered nurse Nicky Millen, a proper old school nurse for 40 years, said that when she started her career care was based on dignity, choice, compassion and empathy. Now she said ‘the things that are important to me have gone out of the window.’ She was appalled that people were dying without their loved ones and saying goodbye on iPads. Nicky described how a distressed 89-year-old lady stroked her face and asked her ‘how many paracetamol would it take to finish me off’. Life was no longer worth living while not seeing her family. Nicky said she was humiliated in front of the ward staff and patients for le ing the lady stroke her face and giving her a cuddle. Such is the dehumanisation that the ‘Covid’ hoax has brought to the surface. Nicky worked in care homes where patients told her they were being held prisoner. ‘I want to live until I die’, one said to her. ‘I had a lady in tears because she hadn’t seen her great-grandson.’ Nicky was compassionate old school meeting psychopathic New Normal. She also said she had worked on a ‘Covid’ ward with no ‘Covid’ patients. Jewish writer Shai Held wrote an article in March, 2020, which was headlined ‘The Staggering, Heartless Cruelty Toward the Elderly’. What he described was happening from the earliest days of lockdown. He said ‘the elderly’ were considered a group and not unique individuals (the way of the Woke). Shai Held said: Notice how the all-too-familiar rhetoric of dehumanization works: ‘The elderly’ are bunched together as a faceless mass, all of them considered culprits and thus effectively deserving of the suffering the pandemic will inflict upon them. Lost entirely is the fact that the elderly are individual human beings, each with a distinctive face and voice, each with hopes and dreams, memories and regrets, friendships and marriages, loves lost and loves sustained.

‘The elderly’ have become another dehumanised group for which anything goes and for many that has resulted in cold disregard for their rights and their life. The distinctive face that Held talks about is designed to be deleted by masks until everyone is part of a faceless mass.

‘War-zone’ hospitals myth

Again and again medical professionals have told me what was really going on and how hospitals ‘overrun like war zones’ according to the media were virtually empty. The mantra from medical whistleblowers was please don’t use my name or my career is over. Citizen journalists around the world sneaked into hospitals to film evidence exposing the ‘war-zone’ lie. They really were largely empty with closed wards and operating theatres. I met a hospital worker in my town on the Isle of Wight during the first lockdown in 2020 who said the only island hospital had never been so quiet. Lockdown was justified by the psychopaths to stop hospitals being overrun. At the same time that the island hospital was near-empty the military arrived here to provide extra beds. It was all propaganda to ramp up the fear to ensure compliance with fascism as were never-used temporary hospitals with thousands of beds known as Nightingales and never-used make-shi mortuaries opened by the criminal UK government. A man who helped to install those extra island beds a ributed to the army said they were never used and the hospital was empty. Doctors and nurses ‘stood around talking or on their phones, wandering down to us to see what we were doing’. There were no masks or social distancing. He accused the useless local island paper, the County Press, of ‘pumping the fear as if our hospital was overrun and we only have one so it should have been’. He described ambulances parked up with crews outside in deck chairs. When his brother called an ambulance he was told there was a twohour backlog which he called ‘bullshit’. An old lady on the island fell ‘and was in a bad way’, but a caller who rang for an ambulance was told the situation wasn’t urgent enough. Ambulance stations were working under capacity while people would hear ambulances with sirens blaring driving through the streets. When those living near the stations realised what was going on they would follow them as they le , circulated around an urban area with the sirens going, and then came back without stopping. All this was to increase levels of fear and the same goes for the ‘ventilator shortage crisis’ that cost tens of millions for hastily produced ventilators never to be used.

Ambulance crews that agreed to be exploited in this way for fear propaganda might find themselves a mirror. I wish them well with that. Empty hospitals were the obvious consequence of treatment and diagnoses of non-’Covid’ conditions cancelled and those involved handed a death sentence. People have been dying at home from undiagnosed and untreated cancer, heart disease and other lifethreatening conditions to allow empty hospitals to deal with a ‘pandemic’ that wasn’t happening.

Death of the innocent

‘War-zones’ have been laying off nursing staff, even doctors where they can. There was no work for them. Lockdown was justified by saving lives and protecting the vulnerable they were actually killing with DNR orders and preventing empty hospitals being ‘overrun’. In Britain the mantra of stay at home to ‘save the NHS’ was everywhere and across the world the same story was being sold when it was all lies. Two California doctors, Dan Erickson and Artin Massihi at Accelerated Urgent Care in Bakersfield, held a news conference in April, 2020, to say that intensive care units in California were ‘empty, essentially’, with hospitals shu ing floors, not treating patients and laying off doctors. The California health system was working at minimum capacity ‘ge ing rid of doctors because we just don’t have the volume’. They said that people with conditions such as heart disease and cancer were not coming to hospital out of fear of ‘Covid19’. Their video was deleted by Susan Wojcicki’s Cult-owned YouTube a er reaching five million views. Florida governor Ron Desantis, who rejected the severe lockdowns of other states and is being targeted for doing so, said that in March, 2020, every US governor was given models claiming they would run out of hospital beds in days. That was never going to happen and the ‘modellers’ knew it. Deceit can be found at every level of the system. Urgent children’s operations were cancelled including fracture repairs and biopsies to spot cancer. Eric Nicholls, a consultant paediatrician, said ‘this is obviously concerning and we need to return to normal operating and to increase capacity as soon as possible’. Psychopaths

in power were rather less concerned because they are psychopaths. Deletion of urgent care and diagnosis has been happening all over the world and how many kids and others have died as a result of the actions of these cold and heartless lunatics dictating ‘health’ policy? The number must be stratospheric. Richard Sullivan, professor of cancer and global health at King’s College London, said people feared ‘Covid’ more than cancer such was the campaign of fear. ‘Years of lost life will be quite dramatic’, Sullivan said, with ‘a huge amount of avoidable mortality’. Sarah Woolnough, executive director for policy at Cancer Research UK, said there had been a 75 percent drop in urgent referrals to hospitals by family doctors of people with suspected cancer. Sullivan said that ‘a lot of services have had to scale back – we’ve seen a dramatic decrease in the amount of elective cancer surgery’. Lockdown deaths worldwide has been absolutely fantastic with the New York Post reporting how data confirmed that ‘lockdowns end more lives than they save’: There was a sharp decline in visits to emergency rooms and an increase in fatal heart attacks because patients didn’t receive prompt treatment. Many fewer people were screened for cancer. Social isolation contributed to excess deaths from dementia and Alzheimer’s. Researchers predicted that the social and economic upheaval would lead to tens of thousands of “deaths of despair” from drug overdoses, alcoholism and suicide. As unemployment surged and mental-health and substance-abuse treatment programs were interrupted, the reported levels of anxiety, depression and suicidal thoughts increased dramatically, as did alcohol sales and fatal drug overdoses.

This has been happening while nurses and other staff had so much time on their hands in the ‘war-zones’ that Tic-Tok dancing videos began appearing across the Internet with medical staff dancing around in empty wards and corridors as people died at home from causes that would normally have been treated in hospital.

Mentions in dispatches

One brave and truth-commi ed whistleblower was Louise Hampton, a call handler with the UK NHS who made a viral Internet video saying she had done ‘fuck all’ during the ‘pandemic’

which was ‘a load of bollocks’. She said that ‘Covid-19’ was rebranded flu and of course she lost her job. This is what happens in the medical and endless other professions now when you tell the truth. Louise filmed inside ‘war-zone’ accident and emergency departments to show they were empty and I mean empty as in no one there. The mainstream media could have done the same and blown the gaff on the whole conspiracy. They haven’t to their eternal shame. Not that most ‘journalists’ seem capable of manifesting shame as with the psychopaths they slavishly repeat without question. The relative few who were admi ed with serious health problems were le to die alone with no loved ones allowed to see them because of ‘Covid’ rules and they included kids dying without the comfort of mum and dad at their bedside while the evil behind this couldn’t give a damn. It was all good fun to them. A Sco ish NHS staff nurse publicly quit in the spring of 2021 saying: ‘I can no longer be part of the lies and the corruption by the government.’ She said hospitals ‘aren’t full, the beds aren’t full, beds have been shut, wards have been shut’. Hospitals were never busy throughout ‘Covid’. The staff nurse said that Nicola Sturgeon, tragically the leader of the Sco ish government, was on television saying save the hospitals and the NHS – ‘but the beds are empty’ and ‘we’ve not seen flu, we always see flu every year’. She wrote to government and spoke with her union Unison (the unions are Cult-compromised and useless, but nothing changed. Many of her colleagues were scared of losing their jobs if they spoke out as they wanted to. She said nursing staff were being affected by wearing masks all day and ‘my head is spli ing every shi from wearing a mask’. The NHS is part of the fascist tyranny and must be dismantled so we can start again with human beings in charge. (Ironically, hospitals were reported to be busier again when official ‘Covid’ cases fell in spring/summer of 2021 and many other conditions required treatment at the same time as the fake vaccine rollout.) I will cover the ‘Covid vaccine’ scam in detail later, but it is another indicator of the sickening disregard for human life that I am highlighting here. The DNA-manipulating concoctions do not fulfil

the definition of a ‘vaccine’, have never been used on humans before and were given only emergency approval because trials were not completed and they continued using the unknowing public. The result was what a NHS senior nurse with responsibility for ‘vaccine’ procedure said was ‘genocide’. She said the ‘vaccines’ were not ‘vaccines’. They had not been shown to be safe and claims about their effectiveness by drug companies were ‘poetic licence’. She described what was happening as a ‘horrid act of human annihilation’. The nurse said that management had instigated a policy of not providing a Patient Information Leaflet (PIL) before people were ‘vaccinated’ even though health care professionals are supposed to do this according to protocol. Patients should also be told that they are taking part in an ongoing clinical trial. Her challenges to what is happening had seen her excluded from meetings and ridiculed in others. She said she was told to ‘watch my step … or I would find myself surplus to requirements’. The nurse, who spoke anonymously in fear of her career, said she asked her NHS manager why he/she was content with taking part in genocide against those having the ‘vaccines’. The reply was that everyone had to play their part and to ‘put up, shut up, and get it done’. Government was ‘leaning heavily’ on NHS management which was clearly leaning heavily on staff. This is how the global ‘medical’ hierarchy operates and it starts with the Cult and its World Health Organization. She told the story of a doctor who had the Pfizer jab and when questioned had no idea what was in it. The doctor had never read the literature. We have to stop treating doctors as intellectual giants when so many are moral and medical pygmies. The doctor did not even know that the ‘vaccines’ were not fully approved or that their trials were ongoing. They were, however, asking their patients if they minded taking part in follow-ups for research purposes – yes, the ongoing clinical trial. The nurse said the doctor’s ignorance was not rare and she had spoken to a hospital consultant who had the jab without any idea of the background or that the ‘trials’ had not been completed. Nurses and pharmacists had shown the same ignorance.

‘My NHS colleagues have forsaken their duty of care, broken their code of conduct – Hippocratic Oath – and have been brainwashed just the same as the majority of the UK public through propaganda …’ She said she had not been able to recruit a single NHS colleague, doctor, nurse or pharmacist to stand with her and speak out. Her union had refused to help. She said that if the genocide came to light she would not hesitate to give evidence at a Nuremberg-type trial against those in power who could have affected the outcomes but didn’t.

And all for what?

To put the nonsense into perspective let’s say the ‘virus’ does exist and let’s go completely crazy and accept that the official manipulated figures for cases and deaths are accurate. Even then a study by Stanford University epidemiologist Dr John Ioannidis published on the World Health Organization website produced an average infection to fatality rate of … 0.23 percent! Ioannidis said: ‘If one could sample equally from all locations globally, the median infection fatality rate might even be substantially lower than the 0.23% observed in my analysis.’ For healthy people under 70 it was … 0.05 percent! This compares with the 3.4 percent claimed by the Cult-owned World Health Organization when the hoax was first played and maximum fear needed to be generated. An updated Stanford study in April, 2021, put the ‘infection’ to ‘fatality’ rate at just 0.15 percent. Another team of scientists led by Megan O’Driscoll and Henrik Salje studied data from 45 countries and published their findings on the Nature website. For children and young people the figure is so small it virtually does not register although authorities will be hyping dangers to the young when they introduce DNAmanipulating ‘vaccines’ for children. The O’Driscoll study produced an average infection-fatality figure of 0.003 for children from birth to four; 0.001 for 5 to 14; 0.003 for 15 to 19; and it was still only 0.456 up to 64. To claim that children must be ‘vaccinated’ to protect them from ‘Covid’ is an obvious lie and so there must be another reason and there is. What’s more the average age of a ‘Covid’ death is akin

to the average age that people die in general. The average age of death in England is about 80 for men and 83 for women. The average age of death from alleged ‘Covid’ is between 82 and 83. California doctors, Dan Erickson and Artin Massihi, said at their April media conference that projection models of millions of deaths had been ‘woefully inaccurate’. They produced detailed figures showing that Californians had a 0.03 chance of dying from ‘Covid’ based on the number of people who tested positive (with a test not testing for the ‘virus’). Erickson said there was a 0.1 percent chance of dying from ‘Covid’ in the state of New York, not just the city, and a 0.05 percent chance in Spain, a centre of ‘Covid-19’ hysteria at one stage. The Stanford studies supported the doctors’ data with fatality rate estimates of 0.23 and 0.15 percent. How close are these figures to my estimate of zero? Death-rate figures claimed by the World Health Organization at the start of the hoax were some 15 times higher. The California doctors said there was no justification for lockdowns and the economic devastation they caused. Everything they had ever learned about quarantine was that you quarantine the sick and not the healthy. They had never seen this before and it made no medical sense. Why in the in the light of all this would governments and medical systems the world over say that billions must go under house arrest; lose their livelihood; in many cases lose their mind, their health and their life; force people to wear masks dangerous to health and psychology; make human interaction and even family interaction a criminal offence; ban travel; close restaurants, bars, watching live sport, concerts, theatre, and any activity involving human togetherness and discourse; and closing schools to isolate children from their friends and cause many to commit suicide in acts of hopelessness and despair? The California doctors said lockdown consequences included increased child abuse, partner abuse, alcoholism, depression, and other impacts they were seeing every day. Who would do that to the entire human race if not mentally-ill psychopaths of almost unimaginable extremes like Bill Gates? We must face the reality of what we are dealing with and come out of

denial. Fascism and tyranny are made possible only by the target population submi ing and acquiescing to fascism and tyranny. The whole of human history shows that to be true. Most people naively and unquestioning believed what they were told about a ‘deadly virus’ and meekly and weakly submi ed to house arrest. Those who didn’t believe it – at least in total – still submi ed in fear of the consequences of not doing so. For the rest who wouldn’t submit draconian fines have been imposed, brutal policing by psychopaths for psychopaths, and condemnation from the meek and weak who condemn the Pushbackers on behalf of the very force that has them, too, in its gunsights. ‘Pathetic’ does not even begin to suffice. Britain’s brainless ‘Health’ Secretary Ma Hancock warned anyone lying to border officials about returning from a list of ‘hotspot’ countries could face a jail sentence of up to ten years which is more than for racially-aggravated assault, incest and a empting to have sex with a child under 13. Hancock is a lunatic, but he has the state apparatus behind him in a Cult-led chain reaction and the same with UK ‘Vaccine Minister’ Nadhim Zahawi, a prominent member of the mega-Cult secret society, Le Cercle, which featured in my earlier books. The Cult enforces its will on governments and medical systems; government and medical systems enforce their will on business and police; business enforces its will on staff who enforce it on customers; police enforce the will of the Cult on the population and play their essential part in creating a world of fascist control that their own children and grandchildren will have to live in their entire lives. It is a hierarchical pyramid of imposition and acquiescence and, yes indeedy, of clinical insanity. Does anyone bright enough to read this book have to ask what the answer is? I think not, but I will reveal it anyway in the fewest of syllables: Tell the psychos and their moronic lackeys to fuck off and let’s get on with our lives. We are many – They are few.

CHAPTER SEVEN War on your mind One believes things because one has been conditioned to believe them Aldous Huxley, Brave New World

I

have described the ‘Covid’ hoax as a ‘Psyop’ and that is true in every sense and on every level in accordance with the definition of that term which is psychological warfare. Break down the ‘Covid pandemic’ to the foundation themes and it is psychological warfare on the human individual and collective mind. The same can be said for the entire human belief system involving every subject you can imagine. Huxley was right in his contention that people believe what they are conditioned to believe and this comes from the repetition throughout their lives of the same falsehoods. They spew from government, corporations, media and endless streams of ‘experts’ telling you what the Cult wants you to believe and o en believing it themselves (although far from always). ‘Experts’ are rewarded with ‘prestigious’ jobs and titles and as agents of perceptual programming with regular access to the media. The Cult has to control the narrative – control information – or they lose control of the vital, crucial, without-which-they-cannot-prevail public perception of reality. The foundation of that control today is the Internet made possible by the Defense Advanced Research Projects Agency (DARPA), the incredibly sinister technological arm of the Pentagon. The Internet is the result of military technology.

DARPA openly brags about establishing the Internet which has been a long-term project to lasso the minds of the global population. I have said for decades the plan is to control information to such an extreme that eventually no one would see or hear anything that the Cult does not approve. We are closing in on that end with ferocious censorship since the ‘Covid’ hoax began and in my case it started back in the 1990s in terms of books and speaking venues. I had to create my own publishing company in 1995 precisely because no one else would publish my books even then. I think they’re all still running.

Cult Internet

To secure total control of information they needed the Internet in which pre-programmed algorithms can seek out ‘unclean’ content for deletion and even stop it being posted in the first place. The Cult had to dismantle print and non-Internet broadcast media to ensure the transfer of information to the appropriate-named ‘Web’ – a critical expression of the Cult web. We’ve seen the ever-quickening demise of traditional media and control of what is le by a tiny number of corporations operating worldwide. Independent journalism in the mainstream is already dead and never was that more obvious than since the turn of 2020. The Cult wants all information communicated via the Internet to globally censor and allow the plug to be pulled any time. Lockdowns and forced isolation has meant that communication between people has been through electronic means and no longer through face-to-face discourse and discussion. Cult psychopaths have targeted the bars, restaurants, sport, venues and meeting places in general for this reason. None of this is by chance and it’s to stop people gathering in any kind of privacy or number while being able to track and monitor all Internet communications and block them as necessary. Even private messages between individuals have been censored by these fascists that control Cult fronts like Facebook, Twi er, Google and YouTube which are all officially run by Sabbatian place-people and from the background by higher-level Sabbatian place people.

Facebook, Google, Amazon and their like were seed-funded and supported into existence with money-no-object infusions of funds either directly or indirectly from DARPA and CIA technology arm In-Q-Tel. The Cult plays the long game and prepares very carefully for big plays like ‘Covid’. Amazon is another front in the psychological war and pre y much controls the global market in book sales and increasingly publishing. Amazon’s limitless funds have deleted fantastic numbers of independent publishers to seize global domination on the way to deciding which books can be sold and circulated and which cannot. Moves in that direction are already happening. Amazon’s leading light Jeff Bezos is the grandson of Lawrence Preston Gise who worked with DARPA predecessor ARPA. Amazon has big connections to the CIA and the Pentagon. The plan I have long described went like this: 1. Employ military technology to establish the Internet. 2. Sell the Internet as a place where people can freely communicate without censorship and allow that to happen until the Net becomes the central and irreversible pillar of human society. If the Internet had been highly censored from the start many would have rejected it. 3. Fund and manipulate major corporations into being to control the circulation of information on your Internet using cover stories about geeks in garages to explain how they came about. Give them unlimited funds to expand rapidly with no need to make a profit for years while non-Cult companies who need to balance the books cannot compete. You know that in these circumstances your Googles, YouTubes, Facebooks and Amazons are going to secure near monopolies by either crushing or buying up the opposition. 4. Allow freedom of expression on both the Internet and communication platforms to draw people in until the Internet is the central and irreversible pillar of human society and your communication corporations have reached a stage of near monopoly domination. 5. Then unleash your always-planned frenzy of censorship on the basis of ‘where else are you going to go?’ and continue to expand that until nothing remains that the Cult does not want its human targets to see.

The process was timed to hit the ‘Covid’ hoax to ensure the best chance possible of controlling the narrative which they knew they had to do at all costs. They were, a er all, about to unleash a ‘deadly virus’ that didn’t really exist. If you do that in an environment of free-flowing information and opinion you would be dead in the

water before you could say Gates is a psychopath. The network was in place through which the Cult-created-and-owned World Health Organization could dictate the ‘Covid’ narrative and response policy slavishly supported by Cult-owned Internet communication giants and mainstream media while those telling a different story were censored. Google, YouTube, Facebook and Twi er openly announced that they would do this. What else would we expect from Cult-owned operations like Facebook which former executives have confirmed set out to make the platform more addictive than cigare es and coldly manipulates emotions of its users to sow division between people and groups and scramble the minds of the young? If Zuckerberg lives out the rest of his life without going to jail for crimes against humanity, and most emphatically against the young, it will be a travesty of justice. Still, no ma er, cause and effect will catch up with him eventually and the same with Sergey Brin and Larry Page at Google with its CEO Sundar Pichai who fix the Google search results to promote Cult narratives and hide the opposition. Put the same key words into Google and other search engines like DuckDuckGo and you will see how different results can be. Wikipedia is another intensely biased ‘encyclopaedia’ which skews its content to the Cult agenda. YouTube links to Wikipedia’s version of ‘Covid’ and ‘climate change’ on video pages in which experts in their field offer a different opinion (even that is increasingly rare with Wojcicki censorship). Into this ‘Covid’ silencethem network must be added government media censors, sorry ‘regulators’, such as Ofcom in the UK which imposed tyrannical restrictions on British broadcasters that had the effect of banning me from ever appearing. Just to debate with me about my evidence and views on ‘Covid’ would mean breaking the fascistic impositions of Ofcom and its CEO career government bureaucrat Melanie Dawes. Gutless British broadcasters tremble at the very thought of fascist Ofcom.

Psychos behind ‘Covid’

The reason for the ‘Covid’ catastrophe in all its facets and forms can be seen by whom and what is driving the policies worldwide in such a coordinated way. Decisions are not being made to protect health, but to target psychology. The dominant group guiding and ‘advising’ government policy are not medical professionals. They are psychologists and behavioural scientists. Every major country has its own version of this phenomenon and I’ll use the British example to show how it works. In many ways the British version has been affecting the wider world in the form of the huge behaviour manipulation network in the UK which operates in other countries. The network involves private companies, government, intelligence and military. The Cabinet Office is at the centre of the government ‘Covid’ Psyop and part-owns, with ‘innovation charity’ Nesta, the Behavioural Insights Team (BIT) which claims to be independent of government but patently isn’t. The BIT was established in 2010 and its job is to manipulate the psyche of the population to acquiesce to government demands and so much more. It is also known as the ‘Nudge Unit’, a name inspired by the 2009 book by two ultraZionists, Cass Sunstein and Richard Thaler, called Nudge: Improving Decisions About Health, Wealth, and Happiness. The book, as with the Behavioural Insights Team, seeks to ‘nudge’ behaviour (manipulate it) to make the public follow pa erns of action and perception that suit those in authority (the Cult). Sunstein is so skilled at this that he advises the World Health Organization and the UK Behavioural Insights Team and was Administrator of the White House Office of Information and Regulatory Affairs in the Obama administration. Biden appointed him to the Department of Homeland Security – another ultra-Zionist in the fold to oversee new immigration laws which is another policy the Cult wants to control. Sunstein is desperate to silence anyone exposing conspiracies and co-authored a 2008 report on the subject in which suggestions were offered to ban ‘conspiracy theorizing’ or impose ‘some kind of tax, financial or otherwise, on those who disseminate such theories’. I guess a psychiatrist’s chair is out of the question?

Sunstein’s mate Richard Thaler, an ‘academic affiliate’ of the UK Behavioural Insights Team, is a proponent of ‘behavioural economics’ which is defined as the study of ‘the effects of psychological, cognitive, emotional, cultural and social factors on the decisions of individuals and institutions’. Study the effects so they can be manipulated to be what you want them to be. Other leading names in the development of behavioural economics are ultraZionists Daniel Kahneman and Robert J. Shiller and they, with Thaler, won the Nobel Memorial Prize in Economic Sciences for their work in this field. The Behavioural Insights Team is operating at the heart of the UK government and has expanded globally through partnerships with several universities including Harvard, Oxford, Cambridge, University College London (UCL) and Pennsylvania. They claim to have ‘trained’ (reframed) 20,000 civil servants and run more than 750 projects involving 400 randomised controlled trials in dozens of countries’ as another version of mind reframers Common Purpose. BIT works from its office in New York with cities and their agencies, as well as other partners, across the United States and Canada – this is a company part-owned by the British government Cabinet Office. An executive order by President Cult-servant Obama established a US Social and Behavioral Sciences Team in 2015. They all have the same reason for being and that’s to brainwash the population directly and by brainwashing those in positions of authority.

‘Covid’ mind game

Another prime aspect of the UK mind-control network is the ‘independent’ [joke] Scientific Pandemic Insights Group on Behaviours (SPI-B) which ‘provides behavioural science advice aimed at anticipating and helping people adhere to interventions that are recommended by medical or epidemiological experts’. That means manipulating public perception and behaviour to do whatever government tells them to do. It’s disgusting and if they really want the public to be ‘safe’ this lot should all be under lock and key. According to the government website SPI-B consists of

‘behavioural scientists, health and social psychologists, anthropologists and historians’ and advises the Whi y-Vallance-led Scientific Advisory Group for Emergencies (SAGE) which in turn advises the government on ‘the science’ (it doesn’t) and ‘Covid’ policy. When politicians say they are being guided by ‘the science’ this is the rabble in each country they are talking about and that ‘science’ is dominated by behaviour manipulators to enforce government fascism through public compliance. The Behaviour Insight Team is headed by psychologist David Solomon Halpern, a visiting professor at King’s College London, and connects with a national and global web of other civilian and military organisations as the Cult moves towards its goal of fusing them into one fascistic whole in every country through its ‘Fusion Doctrine’. The behaviour manipulation network involves, but is not confined to, the Foreign Office; National Security Council; government communications headquarters (GCHQ); MI5; MI6; the Cabinet Office-based Media Monitoring Unit; and the Rapid Response Unit which ‘monitors digital trends to spot emerging issues; including misinformation and disinformation; and identifies the best way to respond’. There is also the 77th Brigade of the UK military which operates like the notorious Israeli military’s Unit 8200 in manipulating information and discussion on the Internet by posing as members of the public to promote the narrative and discredit those who challenge it. Here we have the military seeking to manipulate domestic public opinion while the Nazis in government are fine with that. Conservative Member of Parliament Tobias Ellwood, an advocate of lockdown and control through ‘vaccine passports’, is a Lieutenant Colonel reservist in the 77th Brigade which connects with the military operation jHub, the ‘innovation centre’ for the Ministry of Defence and Strategic Command. jHub has also been involved with the civilian National Health Service (NHS) in ‘symptom tracing’ the population. The NHS is a key part of this mind control network and produced a document in December, 2020, explaining to staff how to use psychological manipulation with different groups and ages to get them to have the DNA-manipulating ‘Covid vaccine’

that’s designed to cumulatively rewrite human genetics. The document, called ‘Optimising Vaccination Roll Out – Do’s and Dont’s for all messaging, documents and “communications” in the widest sense’, was published by NHS England and the NHS Improvement Behaviour Change Unit in partnership with Public Health England and Warwick Business School. I hear the mantra about ‘save the NHS’ and ‘protect the NHS’ when we need to scrap the NHS and start again. The current version is far too corrupt, far too anti-human and totally compromised by Cult operatives and their assets. UK government broadcast media censor Ofcom will connect into this web – as will the BBC with its tremendous Ofcom influence – to control what the public see and hear and dictate mass perception. Nuremberg trials must include personnel from all these organisations.

The fear factor

The ‘Covid’ hoax has led to the creation of the UK Cabinet Officeconnected Joint Biosecurity Centre (JBC) which is officially described as providing ‘expert advice on pandemics’ using its independent [all Cult operations are ‘independent’] analytical function to provide real-time analysis about infection outbreaks to identify and respond to outbreaks of Covid-19’. Another role is to advise the government on a response to spikes in infections – ‘for example by closing schools or workplaces in local areas where infection levels have risen’. Put another way, promoting the Cult agenda. The Joint Biosecurity Centre is modelled on the Joint Terrorism Analysis Centre which analyses intelligence to set ‘terrorism threat levels’ and here again you see the fusion of civilian and military operations and intelligence that has led to military intelligence producing documents about ‘vaccine hesitancy’ and how it can be combated. Domestic civilian ma ers and opinions should not be the business of the military. The Joint Biosecurity Centre is headed by Tom Hurd, director general of the Office for Security and Counter-Terrorism from the establishment-to-its-fingertips Hurd family. His father is former Foreign Secretary Douglas Hurd. How coincidental that Tom

Hurd went to the elite Eton College and Oxford University with Boris Johnson. Imperial College with its ridiculous computer modeller Neil Ferguson will connect with this gigantic web that will itself interconnect with similar set-ups in other major and not so major countries. Compared with this Cult network the politicians, be they Boris Johnson, Donald Trump or Joe Biden, are bit-part players ‘following the science’. The network of psychologists was on the ‘Covid’ case from the start with the aim of generating maximum fear of the ‘virus’ to ensure compliance by the population. A government behavioural science group known as SPI-B produced a paper in March, 2020, for discussion by the main government science advisory group known as SAGE. It was headed ‘Options for increasing adherence to social distancing measures’ and it said the following in a section headed ‘Persuasion’: • A substantial number of people still do not feel sufficiently personally threatened; it could be that they are reassured by the low death rate in their demographic group, although levels of concern may be rising. Having a good understanding of the risk has been found to be positively associated with adoption of COVID-19 social distancing measures in Hong Kong. • The perceived level of personal threat needs to be increased among those who are complacent, using hard-hi ing evaluation of options for increasing social distancing emotional messaging. To be effective this must also empower people by making clear the actions they can take to reduce the threat. • Responsibility to others: There seems to be insufficient understanding of, or feelings of responsibility about, people’s role in transmi ing the infection to others … Messaging about actions need to be framed positively in terms of protecting oneself and the community, and increase confidence that they will be effective. • Some people will be more persuaded by appeals to play by the rules, some by duty to the community, and some to personal risk.

All these different approaches are needed. The messaging also needs to take account of the realities of different people’s lives. Messaging needs to take account of the different motivational levers and circumstances of different people. All this could be achieved the SPI-B psychologists said by using the media to increase the sense of personal threat which translates as terrify the shit out of the population, including children, so they all do what we want. That’s not happened has it? Those excuses for ‘journalists’ who wouldn’t know journalism if it bit them on the arse (the great majority) have played their crucial part in serving this Cultgovernment Psyop to enslave their own kids and grandkids. How they live with themselves I have no idea. The psychological war has been underpinned by constant government ‘Covid’ propaganda in almost every television and radio ad break, plus the Internet and print media, which has pounded out the fear with taxpayers footing the bill for their own programming. The result has been people terrified of a ‘virus’ that doesn’t exist or one with a tiny fatality rate even if you believe it does. People walk down the street and around the shops wearing face-nappies damaging their health and psychology while others report those who refuse to be that naïve to the police who turn up in their own face-nappies. I had a cameraman come to my flat and he was so frightened of ‘Covid’ he came in wearing a mask and refused to shake my hand in case he caught something. He had – naïveitis – and the thought that he worked in the mainstream media was both depressing and made his behaviour perfectly explainable. The fear which has gripped the minds of so many and frozen them into compliance has been carefully cultivated by these psychologists who are really psychopaths. If lives get destroyed and a lot of young people commit suicide it shows our plan is working. SPI-B then turned to compulsion on the public to comply. ‘With adequate preparation, rapid change can be achieved’, it said. Some countries had introduced mandatory self-isolation on a wide scale without evidence of major public unrest and a large majority of the UK’s population appeared to be supportive of more coercive measures with 64 percent of adults saying they would

support pu ing London under a lockdown (watch the ‘polls’ which are designed to make people believe that public opinion is in favour or against whatever the subject in hand). For ‘aggressive protective measures’ to be effective, the SPI-B paper said, special a ention should be devoted to those population groups that are more at risk. Translated from the Orwellian this means making the rest of population feel guilty for not protecting the ‘vulnerable’ such as old people which the Cult and its agencies were about to kill on an industrial scale with lockdown, lack of treatment and the Gates ‘vaccine’. Psychopath psychologists sold their guilt-trip so comprehensively that Los Angeles County Supervisor Hilda Solis reported that children were apologising (from a distance) to their parents and grandparents for bringing ‘Covid’ into their homes and ge ing them sick. ‘… These apologies are just some of the last words that loved ones will ever hear as they die alone,’ she said. Gut-wrenchingly Solis then used this childhood tragedy to tell children to stay at home and ‘keep your loved ones alive’. Imagine heaping such potentially life-long guilt on a kid when it has absolutely nothing to do with them. These people are deeply disturbed and the psychologists behind this even more so.

Uncivil war – divide and rule

Professional mind-controllers at SPI-B wanted the media to increase a sense of responsibility to others (do as you’re told) and promote ‘positive messaging’ for those actions while in contrast to invoke ‘social disapproval’ by the unquestioning, obedient, community of anyone with a mind of their own. Again the compliant Goebbels-like media obliged. This is an old, old, trick employed by tyrannies the world over throughout human history. You get the target population to keep the target population in line – your line. SPI-B said this could ‘play an important role in preventing anti-social behaviour or discouraging failure to enact pro-social behaviour’. For ‘anti-social’ in the Orwellian parlance of SPI-B see any behaviour that government doesn’t approve. SPI-B recommendations said that ‘social disapproval’ should be accompanied by clear messaging and

promotion of strong collective identity – hence the government and celebrity mantra of ‘we’re all in this together’. Sure we are. The mind doctors have such contempt for their targets that they think some clueless comedian, actor or singer telling them to do what the government wants will be enough to win them over. We have had UK comedian Lenny Henry, actor Michael Caine and singer Elton John wheeled out to serve the propagandists by urging people to have the DNA-manipulating ‘Covid’ non-’vaccine’. The role of Henry and fellow black celebrities in seeking to coax a ‘vaccine’ reluctant black community into doing the government’s will was especially stomach-turning. An emotion-manipulating script and carefully edited video featuring these black ‘celebs’ was such an insult to the intelligence of black people and where’s the self-respect of those involved selling their souls to a fascist government agenda? Henry said he heard black people’s ‘legitimate worries and concerns’, but people must ‘trust the facts’ when they were doing exactly that by not having the ‘vaccine’. They had to include the obligatory reference to Black Lives Ma er with the line … ‘Don’t let coronavirus cost even more black lives – because we ma er’. My god, it was pathetic. ‘I know the vaccine is safe and what it does.’ How? ‘I’m a comedian and it says so in my script.’ SPI-B said social disapproval needed to be carefully managed to avoid victimisation, scapegoating and misdirected criticism, but they knew that their ‘recommendations’ would lead to exactly that and the media were specifically used to stir-up the divide-and-conquer hostility. Those who conform like good li le baa, baas, are praised while those who have seen through the tidal wave of lies are ‘Covidiots’. The awake have been abused by the fast asleep for not conforming to fascism and impositions that the awake know are designed to endanger their health, dehumanise them, and tear asunder the very fabric of human society. We have had the curtaintwitchers and morons reporting neighbours and others to the facenappied police for breaking ‘Covid rules’ with fascist police delighting in posting links and phone numbers where this could be done. The Cult cannot impose its will without a compliant police

and military or a compliant population willing to play their part in enslaving themselves and their kids. The words of a pastor in Nazi Germany are so appropriate today: First they came for the socialists and I did not speak out because I was not a socialist. Then they came for the trade unionists and I did not speak out because I was not a trade unionist. Then they came for the Jews and I did not speak out because I was not a Jew. Then they came for me and there was no one left to speak for me.

Those who don’t learn from history are destined to repeat it and so many are.

‘Covid’ rules: Rewiring the mind

With the background laid out to this gigantic national and global web of psychological manipulation we can put ‘Covid’ rules into a clear and sinister perspective. Forget the claims about protecting health. ‘Covid’ rules are about dismantling the human mind, breaking the human spirit, destroying self-respect, and then pu ing Humpty Dumpty together again as a servile, submissive slave. Social isolation through lockdown and distancing have devastating effects on the human psyche as the psychological psychopaths well know and that’s the real reason for them. Humans need contact with each other, discourse, closeness and touch, or they eventually, and literarily, go crazy. Masks, which I will address at some length, fundamentally add to the effects of isolation and the Cult agenda to dehumanise and de-individualise the population. To do this while knowing – in fact seeking – this outcome is the very epitome of evil and psychologists involved in this are the epitome of evil. They must like all the rest of the Cult demons and their assets stand trial for crimes against humanity on a scale that defies the imagination. Psychopaths in uniform use isolation to break enemy troops and agents and make them subservient and submissive to tell what they know. The technique is rightly considered a form of torture and

torture is most certainly what has been imposed on the human population. Clinically-insane American psychologist Harry Harlow became famous for his isolation experiments in the 1950s in which he separated baby monkeys from their mothers and imprisoned them for months on end in a metal container or ‘pit of despair’. They soon began to show mental distress and depression as any idiot could have predicted. Harlow put other monkeys in steel chambers for three, six or twelve months while denying them any contact with animals or humans. He said that the effects of total social isolation for six months were ‘so devastating and debilitating that we had assumed initially that twelve months of isolation would not produce any additional decrement’; but twelve months of isolation ‘almost obliterated the animals socially’. This is what the Cult and its psychopaths are doing to you and your children. Even monkeys in partial isolation in which they were not allowed to form relationships with other monkeys became ‘aggressive and hostile, not only to others, but also towards their own bodies’. We have seen this in the young as a consequence of lockdown. UK government psychopaths launched a public relations campaign telling people not to hug each other even a er they received the ‘Covid-19 vaccine’ which we were told with more lies would allow a return to ‘normal life’. A government source told The Telegraph: ‘It will be along the lines that it is great that you have been vaccinated, but if you are going to visit your family and hug your grandchildren there is a chance you are going to infect people you love.’ The source was apparently speaking from a secure psychiatric facility. Janet Lord, director of Birmingham University’s Institute of Inflammation and Ageing, said that parents and grandparents should avoid hugging their children. Well, how can I put it, Ms Lord? Fuck off. Yep, that’ll do.

Destroying the kids – where are the parents?

Observe what has happened to people enslaved and isolated by lockdown as suicide and self-harm has soared worldwide,

particularly among the young denied the freedom to associate with their friends. A study of 49,000 people in English-speaking countries concluded that almost half of young adults are at clinical risk of mental health disorders. A national survey in America of 1,000 currently enrolled high school and college students found that 5 percent reported a empting suicide during the pandemic. Data from the US CDC’s National Syndromic Surveillance Program from January 1st to October 17th, 2020, revealed a 31 percent increase in mental health issues among adolescents aged 12 to 17 compared with 2019. The CDC reported that America in general suffered the biggest drop in life expectancy since World War Two as it fell by a year in the first half of 2020 as a result of ‘deaths of despair’ – overdoses and suicides. Deaths of despair have leapt by more than 20 percent during lockdown and include the highest number of fatal overdoses ever recorded in a single year – 81,000. Internet addiction is another consequence of being isolated at home which lowers interest in physical activities as kids fall into inertia and what’s the point? Children and young people are losing hope and giving up on life, sometimes literally. A 14-year-old boy killed himself in Maryland because he had ‘given up’ when his school district didn’t reopen; an 11-year-old boy shot himself during a zoom class; a teenager in Maine succumbed to the isolation of the ‘pandemic’ when he ended his life a er experiencing a disrupted senior year at school. Children as young as nine have taken their life and all these stories can be repeated around the world. Careers are being destroyed before they start and that includes those in sport in which promising youngsters have not been able to take part. The plan of the psycho-psychologists is working all right. Researchers at Cambridge University found that lockdowns cause significant harm to children’s mental health. Their study was published in the Archives of Disease in Childhood, and followed 168 children aged between 7 and 11. The researchers concluded: During the UK lockdown, children’s depression symptoms have increased substantially, relative to before lockdown. The scale of this effect has direct relevance for the continuation of different elements of lockdown policy, such as complete or partial school closures …

… Specifically, we observed a statistically significant increase in ratings of depression, with a medium-to-large effect size. Our findings emphasise the need to incorporate the potential impact of lockdown on child mental health in planning the ongoing response to the global pandemic and the recovery from it.

Not a chance when the Cult’s psycho-psychologists were ge ing exactly what they wanted. The UK’s Royal College of Paediatrics and Child Health has urged parents to look for signs of eating disorders in children and young people a er a three to four fold increase. Specialists say the ‘pandemic’ is a major reason behind the rise. You don’t say. The College said isolation from friends during school closures, exam cancellations, loss of extra-curricular activities like sport, and an increased use of social media were all contributory factors along with fears about the virus (psycho-psychologists again), family finances, and students being forced to quarantine. Doctors said young people were becoming severely ill by the time they were seen with ‘Covid’ regulations reducing face-to-face consultations. Nor is it only the young that have been devastated by the psychopaths. Like all bullies and cowards the Cult is targeting the young, elderly, weak and infirm. A typical story was told by a British lady called Lynn Parker who was not allowed to visit her husband in 2020 for the last ten and half months of his life ‘when he needed me most’ between March 20th and when he died on December 19th. This vacates the criminal and enters the territory of evil. The emotional impact on the immune system alone is immense as are the number of people of all ages worldwide who have died as a result of Cult-demanded, Gates-demanded, lockdowns.

Isolation is torture

The experience of imposing solitary confinement on millions of prisoners around the world has shown how a large percentage become ‘actively psychotic and/or acutely suicidal’. Social isolation has been found to trigger ‘a specific psychiatric syndrome, characterized by hallucinations; panic a acks; overt paranoia; diminished impulse control; hypersensitivity to external stimuli; and difficulties with thinking, concentration and memory’. Juan Mendez,

a United Nations rapporteur (investigator), said that isolation is a form of torture. Research has shown that even a er isolation prisoners find it far more difficult to make social connections and I remember cha ing to a shop assistant a er one lockdown who told me that when her young son met another child again he had no idea how to act or what to do. Hannah Flanagan, Director of Emergency Services at Journey Mental Health Center in Dane County, Wisconsin, said: ‘The specificity about Covid social distancing and isolation that we’ve come across as contributing factors to the suicides are really new to us this year.’ But they are not new to those that devised them. They are ge ing the effect they want as the population is psychologically dismantled to be rebuilt in a totally different way. Children and the young are particularly targeted. They will be the adults when the full-on fascist AI-controlled technocracy is planned to be imposed and they are being prepared to meekly submit. At the same time older people who still have a memory of what life was like before – and how fascist the new normal really is – are being deleted. You are going to see efforts to turn the young against the old to support this geriatric genocide. Hannah Flanagan said the big increase in suicide in her county proved that social isolation is not only harmful, but deadly. Studies have shown that isolation from others is one of the main risk factors in suicide and even more so with women. Warnings that lockdown could create a ‘perfect storm’ for suicide were ignored. A er all this was one of the reasons for lockdown. Suicide, however, is only the most extreme of isolation consequences. There are many others. Dr Dhruv Khullar, assistant professor of healthcare policy at Weill Cornell Medical College, said in a New York Times article in 2016 long before the fake ‘pandemic’: A wave of new research suggests social separation is bad for us. Individuals with less social connection have disrupted sleep patterns, altered immune systems, more inflammation and higher levels of stress hormones. One recent study found that isolation increases the risk of heart disease by 29 percent and stroke by 32 percent. Another analysis that pooled data from 70 studies and 3.4 million people found that socially isolated individuals had a 30 percent higher risk of dying in the next seven years, and that this effect was largest in middle age.

Loneliness can accelerate cognitive decline in older adults, and isolated individuals are twice as likely to die prematurely as those with more robust social interactions. These effects start early: Socially isolated children have significantly poorer health 20 years later, even after controlling for other factors. All told, loneliness is as important a risk factor for early death as obesity and smoking.

There you have proof from that one article alone four years before 2020 that those who have enforced lockdown, social distancing and isolation knew what the effect would be and that is even more so with professional psychologists that have been driving the policy across the globe. We can go back even further to the years 2000 and 2003 and the start of a major study on the effects of isolation on health by Dr Janine Gronewold and Professor Dirk M. Hermann at the University Hospital in Essen, Germany, who analysed data on 4,316 people with an average age of 59 who were recruited for the long-term research project. They found that socially isolated people are more than 40 percent more likely to have a heart a ack, stroke, or other major cardiovascular event and nearly 50 percent more likely to die from any cause. Given the financial Armageddon unleashed by lockdown we should note that the study found a relationship between increased cardiovascular risk and lack of financial support. A er excluding other factors social isolation was still connected to a 44 percent increased risk of cardiovascular problems and a 47 percent increased risk of death by any cause. Lack of financial support was associated with a 30 percent increase in the risk of cardiovascular health events. Dr Gronewold said it had been known for some time that feeling lonely or lacking contact with close friends and family can have an impact on physical health and the study had shown that having strong social relationships is of high importance for heart health. Gronewold said they didn’t understand yet why people who are socially isolated have such poor health outcomes, but this was obviously a worrying finding, particularly during these times of prolonged social distancing. Well, it can be explained on many levels. You only have to identify the point in the body where people feel loneliness and missing people they are parted from – it’s in the centre of the chest where they feel the ache of loneliness and the ache of missing people. ‘My heart aches for

you’ … ‘My heart aches for some company.’ I will explain this more in the chapter Escaping Wetiko, but when you realise that the body is the mind – they are expressions of each other – the reason why state of the mind dictates state of the body becomes clear. American psychologist Ranjit Powar was highlighting the effects of lockdown isolation as early as April, 2020. She said humans have evolved to be social creatures and are wired to live in interactive groups. Being isolated from family, friends and colleagues could be unbalancing and traumatic for most people and could result in short or even long-term psychological and physical health problems. An increase in levels of anxiety, aggression, depression, forgetfulness and hallucinations were possible psychological effects of isolation. ‘Mental conditions may be precipitated for those with underlying pre-existing susceptibilities and show up in many others without any pre-condition.’ Powar said personal relationships helped us cope with stress and if we lost this outlet for le ing off steam the result can be a big emotional void which, for an average person, was difficult to deal with. ‘Just a few days of isolation can cause increased levels of anxiety and depression’ – so what the hell has been the effect on the global population of 18 months of this at the time of writing? Powar said: ‘Add to it the looming threat of a dreadful disease being repeatedly hammered in through the media and you have a recipe for many shades of mental and physical distress.’ For those with a house and a garden it is easy to forget that billions have had to endure lockdown isolation in tiny overcrowded flats and apartments with nowhere to go outside. The psychological and physical consequences of this are unimaginable and with lunatic and abusive partners and parents the consequences have led to tremendous increases in domestic and child abuse and alcoholism as people seek to shut out the horror. Ranjit Powar said: Staying in a confined space with family is not all a rosy picture for everyone. It can be extremely oppressive and claustrophobic for large low-income families huddled together in small single-room houses. Children here are not lucky enough to have many board/electronic games or books to keep them occupied.

Add to it the deep insecurity of running out of funds for food and basic necessities. On the other hand, there are people with dysfunctional family dynamics, such as domineering, abusive or alcoholic partners, siblings or parents which makes staying home a period of trial. Incidence of suicide and physical abuse against women has shown a worldwide increase. Heightened anxiety and depression also affect a person’s immune system, making them more susceptible to illness.

To think that Powar’s article was published on April 11th, 2020.

Six-feet fantasy

Social (unsocial) distancing demanded that people stay six feet or two metres apart. UK government advisor Robert Dingwall from the New and Emerging Respiratory Virus Threats Advisory Group said in a radio interview that the two-metre rule was ‘conjured up out of nowhere’ and was not based on science. No, it was not based on medical science, but it didn’t come out of nowhere. The distance related to psychological science. Six feet/two metres was adopted in many countries and we were told by people like the criminal Anthony Fauci and his ilk that it was founded on science. Many schools could not reopen because they did not have the space for sixfeet distancing. Then in March, 2021, a er a year of six-feet ‘science’, a study published in the Journal of Infectious Diseases involving more than 500,000 students and almost 100,000 staff over 16 weeks revealed no significant difference in ‘Covid’ cases between six feet and three feet and Fauci changed his tune. Now three feet was okay. There is no difference between six feet and three inches when there is no ‘virus’ and they got away with six feet for psychological reasons for as long as they could. I hear journalists and others talk about ‘unintended consequences’ of lockdown. They are not unintended at all; they have been coldly-calculated for a specific outcome of human control and that’s why super-psychopaths like Gates have called for them so vehemently. Super-psychopath psychologists have demanded them and psychopathic or clueless, spineless, politicians have gone along with them by ‘following the science’. But it’s not science at all. ‘Science’ is not what is; it’s only what people can be manipulated to believe it is. The whole ‘Covid’ catastrophe is

founded on mind control. Three word or three statement mantras issued by the UK government are a well-known mind control technique and so we’ve had ‘Stay home/protect the NHS/save lives’, ‘Stay alert/control the virus/save lives’ and ‘hands/face/space’. One of the most vocal proponents of extreme ‘Covid’ rules in the UK has been Professor Susan Michie, a member of the British Communist Party, who is not a medical professional. Michie is the director of the Centre for Behaviour Change at University College London. She is a behavioural psychologist and another filthy rich ‘Marxist’ who praised China’s draconian lockdown. She was known by fellow students at Oxford University as ‘Stalin’s nanny’ for her extreme Marxism. Michie is an influential member of the UK government’s Scientific Advisory Group for Emergencies (SAGE) and behavioural manipulation groups which have dominated ‘Covid’ policy. She is a consultant adviser to the World Health Organization on ‘Covid-19’ and behaviour. Why the hell are lockdowns anything to do with her when they are claimed to be about health? Why does a behavioural psychologist from a group charged with changing the behaviour of the public want lockdown, human isolation and mandatory masks? Does that question really need an answer? Michie absolutely has to explain herself before a Nuremberg court when humanity takes back its world again and even more so when you see the consequences of masks that she demands are compulsory. This is a Michie classic: The benefits of getting primary school children to wear masks is that regardless of what little degree of transmission is occurring in those age groups it could help normalise the practice. Young children wearing masks may be more likely to get their families to accept masks.

Those words alone should carry a prison sentence when you ponder on the callous disregard for children involved and what a statement it makes about the mind and motivations of Susan Michie. What a lovely lady and what she said there encapsulates the mentality of the psychopaths behind the ‘Covid’ horror. Let us compare what Michie said with a countrywide study in Germany published at researchsquare.com involving 25,000 school children and 17,854 health complaints submi ed by parents. Researchers

found that masks are harming children physically, psychologically, and behaviourally with 24 health issues associated with mask wearing. They include: shortness of breath (29.7%); dizziness (26.4%); increased headaches (53%); difficulty concentrating (50%); drowsiness or fatigue (37%); and malaise (42%). Nearly a third of children experienced more sleep issues than before and a quarter developed new fears. Researchers found health issues and other impairments in 68 percent of masked children covering their faces for an average of 4.5 hours a day. Hundreds of those taking part experienced accelerated respiration, tightness in the chest, weakness, and short-term impairment of consciousness. A reminder of what Michie said again: The benefits of getting primary school children to wear masks is that regardless of what little degree of transmission is occurring in those age groups it could help normalise the practice. Young children wearing masks may be more likely to get their families to accept masks.

Psychopaths in government and psychology now have children and young people – plus all the adults – wearing masks for hours on end while clueless teachers impose the will of the psychopaths on the young they should be protecting. What the hell are parents doing?

Cult lab rats

We have some schools already imposing on students microchipped buzzers that activate when they get ‘too close’ to their pals in the way they do with lab rats. How apt. To the Cult and its brain-dead servants our children are lab rats being conditioned to be unquestioning, dehumanised slaves for the rest of their lives. Children and young people are being weaned and frightened away from the most natural human instincts including closeness and touch. I have tracked in the books over the years how schools were banning pupils from greeting each other with a hug and the whole Cult-induced Me Too movement has terrified men and boys from a relaxed and natural interaction with female friends and work colleagues to the point where many men try never to be in a room

alone with a woman that’s not their partner. Airhead celebrities have as always played their virtue-signalling part in making this happen with their gross exaggeration. For every monster like Harvey Weinstein there are at least tens of thousands of men that don’t treat women like that; but everyone must be branded the same and policy changed for them as well as the monster. I am going to be using the word ‘dehumanise’ many times in this chapter because that is what the Cult is seeking to do and it goes very deep as we shall see. Don’t let them kid you that social distancing is planned to end one day. That’s not the idea. We are seeing more governments and companies funding and producing wearable gadgets to keep people apart and they would not be doing that if this was meant to be short-term. A tech start-up company backed by GCHQ, the British Intelligence and military surveillance headquarters, has created a social distancing wrist sensor that alerts people when they get too close to others. The CIA has also supported tech companies developing similar devices. The wearable sensor was developed by Tended, one of a number of start-up companies supported by GCHQ (see the CIA and DARPA). The device can be worn on the wrist or as a tag on the waistband and will vibrate whenever someone wearing the device breaches social distancing and gets anywhere near natural human contact. The company had a lucky break in that it was developing a distancing sensor when the ‘Covid’ hoax arrived which immediately provided a potentially enormous market. How fortunate. The government in big-time Cult-controlled Ontario in Canada is investing $2.5 million in wearable contact tracing technology that ‘will alert users if they may have been exposed to the Covid-19 in the workplace and will beep or vibrate if they are within six feet of another person’. Facedrive Inc., the technology company behind this, was founded in 2016 with funding from the Ontario Together Fund and obviously they, too, had a prophet on the board of directors. The human surveillance and control technology is called TraceSCAN and would be worn by the human cyborgs in places such as airports, workplaces, construction sites, care homes and … schools.

I emphasise schools with children and young people the prime targets. You know what is planned for society as a whole if you keep your eyes on the schools. They have always been places where the state program the next generation of slaves to be its compliant worker-ants – or Woker-ants these days; but in the mist of the ‘Covid’ madness they have been transformed into mind laboratories on a scale never seen before. Teachers and head teachers are just as programmed as the kids – o en more so. Children are kept apart from human interaction by walk lanes, classroom distancing, staggered meal times, masks, and the rolling-out of buzzer systems. Schools are now physically laid out as a laboratory maze for lab-rats. Lunatics at a school in Anchorage, Alaska, who should be prosecuted for child abuse, took away desks and forced children to kneel (know your place) on a mat for five hours a day while wearing a mask and using their chairs as a desk. How this was supposed to impact on a ‘virus’ only these clinically insane people can tell you and even then it would be clap-trap. The school banned recess (interaction), art classes (creativity), and physical exercise (ge ing body and mind moving out of inertia). Everyone behind this outrage should be in jail or be er still a mental institution. The behavioural manipulators are all for this dystopian approach to schools. Professor Susan Michie, the mind-doctor and British Communist Party member, said it was wrong to say that schools were safe. They had to be made so by ‘distancing’, masks and ventilation (si ing all day in the cold). I must ask this lady round for dinner on a night I know I am going to be out and not back for weeks. She probably wouldn’t be able to make it, anyway, with all the visits to her own psychologist she must have block-booked.

Masking identity

I know how shocking it must be for you that a behaviour manipulator like Michie wants everyone to wear masks which have long been a feature of mind-control programs like the infamous MKUltra in the United States, but, there we are. We live and learn. I spent many years from 1996 to right across the millennium

researching mind control in detail on both sides of the Atlantic and elsewhere. I met a large number of mind-control survivors and many had been held captive in body and mind by MKUltra. MK stands for mind-control, but employs the German spelling in deference to the Nazis spirited out of Germany at the end of World War Two by Operation Paperclip in which the US authorities, with help from the Vatican, transported Nazi mind-controllers and engineers to America to continue their work. Many of them were behind the creation of NASA and they included Nazi scientist and SS officer Wernher von Braun who swapped designing V-2 rockets to bombard London with designing the Saturn V rockets that powered the NASA moon programme’s Apollo cra . I think I may have mentioned that the Cult has no borders. Among Paperclip escapees was Josef Mengele, the Angel of Death in the Nazi concentration camps where he conducted mind and genetic experiments on children o en using twins to provide a control twin to measure the impact of his ‘work’ on the other. If you want to observe the Cult mentality in all its extremes of evil then look into the life of Mengele. I have met many people who suffered mercilessly under Mengele in the United States where he operated under the name Dr Greene and became a stalwart of MKUltra programming and torture. Among his locations was the underground facility in the Mojave Desert in California called the China Lake Naval Weapons Station which is almost entirely below the surface. My books The Biggest Secret, Children of the Matrix and The Perception Deception have the detailed background to MKUltra. The best-known MKUltra survivor is American Cathy O’Brien. I first met her and her late partner Mark Phillips at a conference in Colorado in 1996. Mark helped her escape and deprogram from decades of captivity in an offshoot of MKUltra known as Project Monarch in which ‘sex slaves’ were provided for the rich and famous including Father George Bush, Dick Cheney and the Clintons. Read Cathy and Mark’s book Trance-Formation of America and if you are new to this you will be shocked to the core. I read it in 1996 shortly before, with the usual synchronicity of my life, I found

myself given a book table at the conference right next to hers. MKUltra never ended despite being very publicly exposed (only a small part of it) in the 1970s and continues in other guises. I am still in touch with Cathy. She contacted me during 2020 a er masks became compulsory in many countries to tell me how they were used as part of MKUltra programming. I had been observing ‘Covid regulations’ and the relationship between authority and public for months. I saw techniques that I knew were employed on individuals in MKUltra being used on the global population. I had read many books and manuals on mind control including one called Silent Weapons for Quiet Wars which came to light in the 1980s and was a guide on how to perceptually program on a mass scale. ‘Silent Weapons’ refers to mind-control. I remembered a line from the manual as governments, medical authorities and law enforcement agencies have so obviously talked to – or rather at – the adult population since the ‘Covid’ hoax began as if they are children. The document said: If a person is spoken to by a T.V. advertiser as if he were a twelve-year-old, then, due to suggestibility, he will, with a certain probability, respond or react to that suggestion with the uncritical response of a twelve-year-old and will reach in to his economic reservoir and deliver its energy to buy that product on impulse when he passes it in the store.

That’s why authority has spoken to adults like children since all this began.

Why did Michael Jackson wear masks?

Every aspect of the ‘Covid’ narrative has mind-control as its central theme. Cathy O’Brien wrote an article for davidicke.com about the connection between masks and mind control. Her daughter Kelly who I first met in the 1990s was born while Cathy was still held captive in MKUltra. Kelly was forced to wear a mask as part of her programming from the age of two to dehumanise her, target her sense of individuality and reduce the amount of oxygen her brain and body received. Bingo. This is the real reason for compulsory

masks, why they have been enforced en masse, and why they seek to increase the number they demand you wear. First one, then two, with one disgraceful alleged ‘doctor’ recommending four which is nothing less than a death sentence. Where and how o en they must be worn is being expanded for the purpose of mass mind control and damaging respiratory health which they can call ‘Covid-19’. Canada’s government headed by the man-child Justin Trudeau, says it’s fine for children of two and older to wear masks. An insane ‘study’ in Italy involving just 47 children concluded there was no problem for babies as young as four months wearing them. Even a er people were ‘vaccinated’ they were still told to wear masks by the criminal that is Anthony Fauci. Cathy wrote that mandating masks is allowing the authorities literally to control the air we breathe which is what was done in MKUltra. You might recall how the singer Michael Jackson wore masks and there is a reason for that. He was subjected to MKUltra mind control through Project Monarch and his psyche was scrambled by these simpletons. Cathy wrote: In MKUltra Project Monarch mind control, Michael Jackson had to wear a mask to silence his voice so he could not reach out for help. Remember how he developed that whisper voice when he wasn’t singing? Masks control the mind from the outside in, like the redefining of words is doing. By controlling what we can and cannot say for fear of being labeled racist or beaten, for example, it ultimately controls thought that drives our words and ultimately actions (or lack thereof). Likewise, a mask muffles our speech so that we are not heard, which controls voice … words … mind. This is Mind Control. Masks are an obvious mind control device, and I am disturbed so many people are complying on a global scale. Masks depersonalize while making a person feel as though they have no voice. It is a barrier to others. People who would never choose to comply but are forced to wear a mask in order to keep their job, and ultimately their family fed, are compromised. They often feel shame and are subdued. People have stopped talking with each other while media controls the narrative.

The ‘no voice’ theme has o en become literal with train passengers told not to speak to each other in case they pass on the ‘virus’, singing banned for the same reason and bonkers California officials telling people riding roller coasters that they cannot shout and scream. Cathy said she heard every day from healed MKUltra survivors who cannot wear a mask without flashing back on ways

their breathing was controlled – ‘from ball gags and penises to water boarding’. She said that through the years when she saw images of people in China wearing masks ‘due to pollution’ that it was really to control their oxygen levels. ‘I knew it was as much of a population control mechanism of depersonalisation as are burkas’, she said. Masks are another Chinese communist/fascist method of control that has been swept across the West as the West becomes China at lightning speed since we entered 2020.

Mask-19

There are other reasons for mandatory masks and these include destroying respiratory health to call it ‘Covid-19’ and stunting brain development of children and the young. Dr Margarite GrieszBrisson MD, PhD, is a Consultant Neurologist and Neurophysiologist and the Founder and Medical Director of the London Neurology and Pain Clinic. Her CV goes down the street and round the corner. She is clearly someone who cares about people and won’t parrot the propaganda. Griesz-Brisson has a PhD in pharmacology, with special interest in neurotoxicology, environmental medicine, neuroregeneration and neuroplasticity (the way the brain can change in the light of information received). She went public in October, 2020, with a passionate warning about the effects of mask-wearing laws: The reinhalation of our exhaled air will without a doubt create oxygen deficiency and a flooding of carbon dioxide. We know that the human brain is very sensitive to oxygen deprivation. There are nerve cells for example in the hippocampus that can’t be longer than 3 minutes without oxygen – they cannot survive. The acute warning symptoms are headaches, drowsiness, dizziness, issues in concentration, slowing down of reaction time – reactions of the cognitive system.

Oh, I know, let’s tell bus, truck and taxi drivers to wear them and people working machinery. How about pilots, doctors and police? Griesz-Brisson makes the important point that while the symptoms she mentions may fade as the body readjusts this does not alter the fact that people continue to operate in oxygen deficit with long list of

potential consequences. She said it was well known that neurodegenerative diseases take years or decades to develop. ‘If today you forget your phone number, the breakdown in your brain would have already started 20 or 30 years ago.’ She said degenerative processes in your brain are ge ing amplified as your oxygen deprivation continues through wearing a mask. Nerve cells in the brain are unable to divide themselves normally in these circumstances and lost nerve cells will no longer be regenerated. ‘What is gone is gone.’ Now consider that people like shop workers and schoolchildren are wearing masks for hours every day. What in the name of sanity is going to be happening to them? ‘I do not wear a mask, I need my brain to think’, Griesz-Brisson said, ‘I want to have a clear head when I deal with my patients and not be in a carbon dioxide-induced anaesthesia’. If you are told to wear a mask anywhere ask the organisation, police, store, whatever, for their risk assessment on the dangers and negative effects on mind and body of enforcing mask-wearing. They won’t have one because it has never been done not even by government. All of them must be subject to class-action lawsuits as the consequences come to light. They don’t do mask risk assessments for an obvious reason. They know what the conclusions would be and independent scientific studies that have been done tell a horror story of consequences.

‘Masks are criminal’

Dr Griesz-Brisson said that for children and adolescents, masks are an absolute no-no. They had an extremely active and adaptive immune system and their brain was incredibly active with so much to learn. ‘The child’s brain, or the youth’s brain, is thirsting for oxygen.’ The more metabolically active an organ was, the more oxygen it required; and in children and adolescents every organ was metabolically active. Griesz-Brisson said that to deprive a child’s or adolescent’s brain of oxygen, or to restrict it in any way, was not only dangerous to their health, it was absolutely criminal. ‘Oxygen deficiency inhibits the development of the brain, and the damage that has taken place as a result CANNOT be reversed.’ Mind

manipulators of MKUltra put masks on two-year-olds they wanted to neurologically rewire and you can see why. Griesz-Brisson said a child needs the brain to learn and the brain needs oxygen to function. ‘We don’t need a clinical study for that. This is simple, indisputable physiology.’ Consciously and purposely induced oxygen deficiency was an absolutely deliberate health hazard, and an absolute medical contraindication which means that ‘this drug, this therapy, this method or measure should not be used, and is not allowed to be used’. To coerce an entire population to use an absolute medical contraindication by force, she said, there had to be definite and serious reasons and the reasons must be presented to competent interdisciplinary and independent bodies to be verified and authorised. She had this warning of the consequences that were coming if mask wearing continued: When, in ten years, dementia is going to increase exponentially, and the younger generations couldn’t reach their god-given potential, it won’t help to say ‘we didn’t need the masks’. I know how damaging oxygen deprivation is for the brain, cardiologists know how damaging it is for the heart, pulmonologists know how damaging it is for the lungs. Oxygen deprivation damages every single organ. Where are our health departments, our health insurance, our medical associations? It would have been their duty to be vehemently against the lockdown and to stop it and stop it from the very beginning. Why do the medical boards issue punishments to doctors who give people exemptions? Does the person or the doctor seriously have to prove that oxygen deprivation harms people? What kind of medicine are our doctors and medical associations representing? Who is responsible for this crime? The ones who want to enforce it? The ones who let it happen and play along, or the ones who don’t prevent it?

All of the organisations and people she mentions there either answer directly to the Cult or do whatever hierarchical levels above them tell them to do. The outcome of both is the same. ‘It’s not about masks, it’s not about viruses, it’s certainly not about your health’, Griesz-Brisson said. ‘It is about much, much more. I am not participating. I am not afraid.’ They were taking our air to breathe and there was no unfounded medical exemption from face masks. Oxygen deprivation was dangerous for every single brain. It had to be the free decision of every human being whether they want to

wear a mask that was absolutely ineffective to protect themselves from a virus. She ended by rightly identifying where the responsibility lies for all this: The imperative of the hour is personal responsibility. We are responsible for what we think, not the media. We are responsible for what we do, not our superiors. We are responsible for our health, not the World Health Organization. And we are responsible for what happens in our country, not the government.

Halle-bloody-lujah.

But surgeons wear masks, right?

Independent studies of mask-wearing have produced a long list of reports detailing mental, emotional and physical dangers. What a definition of insanity to see police officers imposing mask-wearing on the public which will cumulatively damage their health while the police themselves wear masks that will cumulatively damage their health. It’s u er madness and both public and police do this because ‘the government says so’ – yes a government of brain-donor idiots like UK Health Secretary Ma Hancock reading the ‘follow the science’ scripts of psychopathic, lunatic psychologists. The response you get from Stockholm syndrome sufferers defending the very authorities that are destroying them and their families is that ‘surgeons wear masks’. This is considered the game, set and match that they must work and don’t cause oxygen deficit. Well, actually, scientific studies have shown that they do and oxygen levels are monitored in operating theatres to compensate. Surgeons wear masks to stop spi le and such like dropping into open wounds – not to stop ‘viral particles’ which are so miniscule they can only be seen through an electron microscope. Holes in the masks are significantly bigger than ‘viral particles’ and if you sneeze or cough they will breach the mask. I watched an incredibly disingenuous ‘experiment’ that claimed to prove that masks work in catching ‘virus’ material from the mouth and nose. They did this with a slow motion camera and the mask did block big stuff which stayed inside the mask and

against the face to be breathed in or cause infections on the face as we have seen with many children. ‘Viral particles’, however, would never have been picked up by the camera as they came through the mask when they are far too small to be seen. The ‘experiment’ was therefore disingenuous and useless. Studies have concluded that wearing masks in operating theatres (and thus elsewhere) make no difference to preventing infection while the opposite is true with toxic shite building up in the mask and this had led to an explosion in tooth decay and gum disease dubbed by dentists ‘mask mouth’. You might have seen the Internet video of a furious American doctor urging people to take off their masks a er a four-year-old patient had been rushed to hospital the night before and nearly died with a lung infection that doctors sourced to mask wearing. A study in the journal Cancer Discovery found that inhalation of harmful microbes can contribute to advanced stage lung cancer in adults and long-term use of masks can help breed dangerous pathogens. Microbiologists have said frequent mask wearing creates a moist environment in which microbes can grow and proliferate before entering the lungs. The Canadian Agency for Drugs and Technologies in Health, or CADTH, a Canadian national organisation that provides research and analysis to healthcare decision-makers, said this as long ago as 2013 in a report entitled ‘Use of Surgical Masks in the Operating Room: A Review of the Clinical Effectiveness and Guidelines’. It said: • No evidence was found to support the use of surgical face masks to reduce the frequency of surgical site infections • No evidence was found on the effectiveness of wearing surgical face masks to protect staff from infectious material in the operating room. • Guidelines recommend the use of surgical face masks by staff in the operating room to protect both operating room staff and patients (despite the lack of evidence).

We were told that the world could go back to ‘normal’ with the arrival of the ‘vaccines’. When they came, fraudulent as they are, the story changed as I knew that it would. We are in the midst of transforming ‘normal’, not going back to it. Mary Ramsay, head of immunisation at Public Health England, echoed the words of US criminal Anthony Fauci who said masks and other regulations must stay no ma er if people are vaccinated. The Fauci idiot continued to wear two masks – different colours so both could be clearly seen – a er he claimed to have been vaccinated. Senator Rand Paul told Fauci in one exchange that his double-masks were ‘theatre’ and he was right. It’s all theatre. Mary Ramsay back-tracked on the vaccinereturn-to-normal theme when she said the public may need to wear masks and social-distance for years despite the jabs. ‘People have got used to those lower-level restrictions now, and [they] can live with them’, she said telling us what the idea has been all along. ‘The vaccine does not give you a pass, even if you have had it, you must continue to follow all the guidelines’ said a Public Health England statement which reneged on what we had been told before and made having the ‘vaccine’ irrelevant to ‘normality’ even by the official story. Spain’s fascist government trumped everyone by passing a law mandating the wearing of masks on the beach and even when swimming in the sea. The move would have devastated what’s le of the Spanish tourist industry, posed potential breathing dangers to swimmers and had Northern European sunbathers walking around with their forehead brown and the rest of their face white as a sheet. The ruling was so crazy that it had to be retracted a er pressure from public and tourist industry, but it confirmed where the Cult wants to go with masks and how clinically insane authority has become. The determination to make masks permanent and hide the serious dangers to body and mind can be seen in the censorship of scientist Professor Denis Rancourt by Bill Gatesfunded academic publishing website ResearchGate over his papers exposing the dangers and uselessness of masks. Rancourt said: ResearchGate today has permanently locked my account, which I have had since 2015. Their reasons graphically show the nature of their attack against democracy, and their corruption of

science … By their obscene non-logic, a scientific review of science articles reporting on harms caused by face masks has a ‘potential to cause harm’. No criticism of the psychological device (face masks) is tolerated, if the said criticism shows potential to influence public policy.

This is what happens in a fascist world.

Where are the ‘greens’ (again)?

Other dangers of wearing masks especially regularly relate to the inhalation of minute plastic fibres into the lungs and the deluge of discarded masks in the environment and oceans. Estimates predicted that more than 1.5 billion disposable masks will end up in the world’s oceans every year polluting the water with tons of plastic and endangering marine wildlife. Studies project that humans are using 129 billion face masks each month worldwide – about three million a minute. Most are disposable and made from plastic, nonbiodegradable microfibers that break down into smaller plastic particles that become widespread in ecosystems. They are li ering cities, clogging sewage channels and turning up in bodies of water. I have wri en in other books about the immense amounts of microplastics from endless sources now being absorbed into the body. Rolf Halden, director of the Arizona State University (ASU) Biodesign Center for Environmental Health Engineering, was the senior researcher in a 2020 study that analysed 47 human tissue samples and found microplastics in all of them. ‘We have detected these chemicals of plastics in every single organ that we have investigated’, he said. I wrote in The Answer about the world being deluged with microplastics. A study by the Worldwide Fund for Nature (WWF) found that people are consuming on average every week some 2,000 tiny pieces of plastic mostly through water and also through marine life and the air. Every year humans are ingesting enough microplastics to fill a heaped dinner plate and in a life-time of 79 years it is enough to fill two large waste bins. Marco Lambertini, WWF International director general said: ‘Not only are plastics polluting our oceans and waterways and killing marine life – it’s in all of us and we can’t escape consuming plastics,’ American

geologists found tiny plastic fibres, beads and shards in rainwater samples collected from the remote slopes of the Rocky Mountain National Park near Denver, Colorado. Their report was headed: ‘It is raining plastic.’ Rachel Adams, senior lecturer in Biomedical Science at Cardiff Metropolitan University, said that among health consequences are internal inflammation and immune responses to a ‘foreign body’. She further pointed out that microplastics become carriers of toxins including mercury, pesticides and dioxins (a known cause of cancer and reproductive and developmental problems). These toxins accumulate in the fa y tissues once they enter the body through microplastics. Now this is being compounded massively by people pu ing plastic on their face and throwing it away. Workers exposed to polypropylene plastic fibres known as ‘flock’ have developed ‘flock worker’s lung’ from inhaling small pieces of the flock fibres which can damage lung tissue, reduce breathing capacity and exacerbate other respiratory problems. Now … commonly used surgical masks have three layers of melt-blown textiles made of … polypropylene. We have billions of people pu ing these microplastics against their mouth, nose and face for hours at a time day a er day in the form of masks. How does anyone think that will work out? I mean – what could possibly go wrong? We posted a number of scientific studies on this at davidicke.com, but when I went back to them as I was writing this book the links to the science research website where they were hosted were dead. Anything that challenges the official narrative in any way is either censored or vilified. The official narrative is so unsupportable by the evidence that only deleting the truth can protect it. A study by Chinese scientists still survived – with the usual twist which it why it was still active, I guess. Yes, they found that virtually all the masks they tested increased the daily intake of microplastic fibres, but people should still wear them because the danger from the ‘virus’ was worse said the crazy ‘team’ from the Institute of Hydrobiology in Wuhan. Scientists first discovered microplastics in lung tissue of some patients who died of lung cancer

in the 1990s. Subsequent studies have confirmed the potential health damage with the plastic degrading slowly and remaining in the lungs to accumulate in volume. Wuhan researchers used a machine simulating human breathing to establish that masks shed up to nearly 4,000 microplastic fibres in a month with reused masks producing more. Scientists said some masks are laced with toxic chemicals and a variety of compounds seriously restricted for both health and environmental reasons. They include cobalt (used in blue dye) and formaldehyde known to cause watery eyes, burning sensations in the eyes, nose, and throat, plus coughing, wheezing and nausea. No – that must be ‘Covid-19’.

Mask ‘worms’

There is another and potentially even more sinister content of masks. Mostly new masks of different makes filmed under a microscope around the world have been found to contain strange black fibres or ‘worms’ that appear to move or ‘crawl’ by themselves and react to heat and water. The nearest I have seen to them are the selfreplicating fibres that are pulled out through the skin of those suffering from Morgellons disease which has been connected to the phenomena of ‘chemtrails’ which I will bring into the story later on. Morgellons fibres continue to grow outside the body and have a form of artificial intelligence. Black ‘worm’ fibres in masks have that kind of feel to them and there is a nanotechnology technique called ‘worm micelles’ which carry and release drugs or anything else you want to deliver to the body. For sure the suppression of humanity by mind altering drugs is the Cult agenda big time and the more excuses they can find to gain access to the body the more opportunities there are to make that happen whether through ‘vaccines’ or masks pushed against the mouth and nose for hours on end. So let us summarise the pros and cons of masks:

Against masks: Breathing in your own carbon dioxide; depriving the body and brain of sufficient oxygen; build-up of toxins in the mask that can be breathed into the lungs and cause rashes on the face and ‘mask-mouth’; breathing microplastic fibres and toxic chemicals into the lungs; dehumanisation and deleting individualisation by literally making people faceless; destroying human emotional interaction through facial expression and deleting parental connection with their babies which look for guidance to their facial expression. For masks: They don’t protect you from a ‘virus’ that doesn’t exist and even if it did ‘viral’ particles are so minute they are smaller than the holes in the mask. Governments, police, supermarkets, businesses, transport companies, and all the rest who seek to impose masks have done no risk assessment on their consequences for health and psychology and are now open to group lawsuits when the impact becomes clear with a cumulative epidemic of respiratory and other disease. Authorities will try to exploit these effects and hide the real cause by dubbing them ‘Covid-19’. Can you imagine se ing out to force the population to wear health-destroying masks without doing any assessment of the risks? It is criminal and it is evil, but then how many people targeted in this way, who see their children told to wear them all day at school, have asked for a risk assessment? Billions can’t be imposed upon by the few unless the billions allow it. Oh, yes, with just a tinge of irony, 85 percent of all masks made worldwide come from China.

Wash your hands in toxic shite

‘Covid’ rules include the use of toxic sanitisers and again the health consequences of constantly applying toxins to be absorbed through the skin is obvious to any level of Renegade Mind. America’s Food and Drug Administration (FDA) said that sanitisers are drugs and issued a warning about 75 dangerous brands which contain

methanol used in antifreeze and can cause death, kidney damage and blindness. The FDA circulated the following warning even for those brands that it claims to be safe: Store hand sanitizer out of the reach of pets and children, and children should use it only with adult supervision. Do not drink hand sanitizer. This is particularly important for young children, especially toddlers, who may be attracted by the pleasant smell or brightly colored bottles of hand sanitizer. Drinking even a small amount of hand sanitizer can cause alcohol poisoning in children. (However, there is no need to be concerned if your children eat with or lick their hands after using hand sanitizer.) During this coronavirus pandemic, poison control centers have had an increase in calls about accidental ingestion of hand sanitizer, so it is important that adults monitor young children’s use. Do not allow pets to swallow hand sanitizer. If you think your pet has eaten something potentially dangerous, call your veterinarian or a pet poison control center right away. Hand sanitizer is flammable and should be stored away from heat and flames. When using hand sanitizer, rub your hands until they feel completely dry before performing activities that may involve heat, sparks, static electricity, or open flames.

There you go, perfectly safe, then, and that’s without even a mention of the toxins absorbed through the skin. Come on kids – sanitise your hands everywhere you go. It will save you from the ‘virus’. Put all these elements together of the ‘Covid’ normal and see how much health and psychology is being cumulatively damaged, even devastated, to ‘protect your health’. Makes sense, right? They are only imposing these things because they care, right? Right?

Submitting to insanity

Psychological reframing of the population goes very deep and is done in many less obvious ways. I hear people say how contradictory and crazy ‘Covid’ rules are and how they are ever changing. This is explained away by dismissing those involved as idiots. It is a big mistake. The Cult is delighted if its cold calculation is perceived as incompetence and idiocy when it is anything but. Oh, yes, there are idiots within the system – lots of them – but they are administering the Cult agenda, mostly unknowingly. They are not deciding and dictating it. The bulwark against tyranny is self-

respect, always has been, always will be. It is self-respect that has broken every tyranny in history. By its very nature self-respect will not bow to oppression and its perpetrators. There is so li le selfrespect that it’s always the few that overturn dictators. Many may eventually follow, but the few with the iron spines (self-respect) kick it off and generate the momentum. The Cult targets self-respect in the knowledge that once this has gone only submission remains. Crazy, contradictory, ever-changing ‘Covid’ rules are systematically applied by psychologists to delete self-respect. They want you to see that the rules make no sense. It is one thing to decide to do something when you have made the choice based on evidence and logic. You still retain your self-respect. It is quite another when you can see what you are being told to do is insane, ridiculous and makes no sense, and yet you still do it. Your self-respect is extinguished and this has been happening as ever more obviously stupid and nonsensical things have been demanded and the great majority have complied even when they can see they are stupid and nonsensical. People walk around in face-nappies knowing they are damaging their health and make no difference to a ‘virus’. They do it in fear of not doing it. I know it’s da , but I’ll do it anyway. When that happens something dies inside of you and submissive reframing has begun. Next there’s a need to hide from yourself that you have conceded your self-respect and you convince yourself that you have not really submi ed to fear and intimidation. You begin to believe that you are complying with craziness because it’s the right thing to do. When first you concede your self-respect of 2+2 = 4 to 2+2 = 5 you know you are compromising your self-respect. Gradually to avoid facing that fact you begin to believe that 2+2=5. You have been reframed and I have been watching this process happening in the human psyche on an industrial scale. The Cult is working to break your spirit and one of its major tools in that war is humiliation. I read how former American soldier Bradley Manning (later Chelsea Manning a er a sex-change) was treated a er being jailed for supplying WikiLeaks with documents exposing the enormity of

government and elite mendacity. Manning was isolated in solitary confinement for eight months, put under 24-hour surveillance, forced to hand over clothing before going to bed, and stand naked for every roll call. This is systematic humiliation. The introduction of anal swab ‘Covid’ tests in China has been done for the same reason to delete self-respect and induce compliant submission. Anal swabs are mandatory for incoming passengers in parts of China and American diplomats have said they were forced to undergo the indignity which would have been calculated humiliation by the Cult-owned Chinese government that has America in its sights.

Government-people: An abusive relationship

Spirit-breaking psychological techniques include giving people hope and apparent respite from tyranny only to take it away again. This happened in the UK during Christmas, 2020, when the psychopsychologists and their political lackeys announced an easing of restrictions over the holiday only to reimpose them almost immediately on the basis of yet another lie. There is a big psychological difference between ge ing used to oppression and being given hope of relief only to have that dashed. Psychologists know this and we have seen the technique used repeatedly. Then there is traumatising people before you introduce more extreme regulations that require compliance. A perfect case was the announcement by the dark and sinister Whi y and Vallance in the UK that ‘new data’ predicted that 4,000 could die every day over the winter of 2020/2021 if we did not lockdown again. I think they call it lying and a er traumatising people with that claim out came Jackboot Johnson the next day with new curbs on human freedom. Psychologists know that a frightened and traumatised mind becomes suggestable to submission and behaviour reframing. Underpinning all this has been to make people fearful and suspicious of each other and see themselves as a potential danger to others. In league with deleted self-respect you have the perfect psychological recipe for self-loathing. The relationship between authority and public is now demonstrably the same as that of

subservience to an abusive partner. These are signs of an abusive relationship explained by psychologist Leslie Becker-Phelps: Undermining a partner’s self-worth with verbal a acks, name-calling, and beli ling. Humiliating the partner in public, unjustly accusing them of having an affair, or interrogating them about their every behavior. Keeping partner confused or off balance by saying they were just kidding or blaming the partner for ‘making’ them act this way … Feigning in public that they care while turning against them in private. This leads to victims frequently feeling confused, incompetent, unworthy, hopeless, and chronically self-doubting. [Apply these techniques to how governments have treated the population since New Year, 2020, and the parallels are obvious.] Physical abuse: The abuser might physically harm their partner in a range of ways, such as grabbing, hi ing, punching, or shoving them. They might throw objects at them or harm them with a weapon. [Observe the physical harm imposed by masks, lockdown, and so on.] Threats and intimidation: One way abusers keep their partners in line is by instilling fear. They might be verbally threatening, or give threatening looks or gestures. Abusers o en make it known that they are tracking their partner’s every move. They might destroy their partner’s possessions, threaten to harm them, or threaten to harm their family members. Not surprisingly, victims of this abuse o en feel anxiety, fear, and panic. [No words necessary.] Isolation: Abusers o en limit their partner’s activities, forbidding them to talk or interact with friends or family. They might limit access to a car or even turn off their phone. All of this might be done by physically holding them against their will, but is o en accomplished through psychological abuse and intimidation. The more isolated a person feels, the fewer resources they have to help gain perspective on their situation and to escape from it. [No words necessary.] Psychological and emotional abuse:

Abusers o en make their partners beholden to them for money by controlling access to funds of any kind. They might prevent their partner from ge ing a job or withhold access to money they earn from a job. This creates financial dependency that makes leaving the relationship very difficult. [See destruction of livelihoods and the proposed meagre ‘guaranteed income’ so long as you do whatever you are told.] Using children: An abuser might disparage their partner’s parenting skills, tell their children lies about their partner, threaten to take custody of their children, or threaten to harm their children. These tactics instil fear and o en elicit compliance. [See reframed social service mafia and how children are being mercilessly abused by the state over ‘Covid’ while their parents look on too frightened to do anything.] A further recurring trait in an abusive relationship is the abused blaming themselves for their abuse and making excuses for the abuser. We have the public blaming each other for lockdown abuse by government and many making excuses for the government while a acking those who challenge the government. How o en we have heard authorities say that rules are being imposed or reimposed only because people have refused to ‘behave’ and follow the rules. We don’t want to do it – it’s you. Renegade Minds are an antidote to all of these things. They will never concede their self-respect no ma er what the circumstances. Even when apparent humiliation is heaped upon them they laugh in its face and reflect back the humiliation on the abuser where it belongs. Renegade Minds will never wear masks they know are only imposed to humiliate, suppress and damage both physically and psychologically. Consequences will take care of themselves and they will never break their spirit or cause them to concede to tyranny. UK newspaper columnist Peter Hitchens was one of the few in the mainstream media to speak out against lockdowns and forced vaccinations. He then announced he had taken the jab. He wanted to see family members abroad and he believed vaccine passports were inevitable even though they had not yet been introduced. Hitchens Economic abuse:

has a questioning and critical mind, but not a Renegade one. If he had no amount of pressure would have made him concede. Hitchens excused his action by saying that the ba le has been lost. Renegade Minds never accept defeat when freedom is at stake and even if they are the last one standing the self-respect of not submi ing to tyranny is more important than any outcome or any consequence. That’s why Renegade Minds are the only minds that ever changed anything worth changing.

CHAPTER EIGHT ‘Reframing’ insanity Insanity is relative. It depends on who has who locked in what cage Ray Bradbury

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eframing’ a mind means simply to change its perception and behaviour. This can be done subconsciously to such an extent that subjects have no idea they have been ‘reframed’ while to any observer changes in behaviour and a itudes are obvious. Human society is being reframed on a ginormous scale since the start of 2020 and here we have the reason why psychologists rather than doctors have been calling the shots. Ask most people who have succumbed to ‘Covid’ reframing if they have changed and most will say ‘no’; but they have and fundamentally. The Cult’s long-game has been preparing for these times since way back and crucial to that has been to prepare both population and officialdom mentally and emotionally. To use the mind-control parlance they had to reframe the population with a mentality that would submit to fascism and reframe those in government and law enforcement to impose fascism or at least go along with it. The result has been the factdeleted mindlessness of ‘Wokeness’ and officialdom that has either enthusiastically or unquestioningly imposed global tyranny demanded by reframed politicians on behalf of psychopathic and deeply evil cultists. ‘Cognitive reframing’ identifies and challenges the way someone sees the world in the form of situations, experiences and emotions and then restructures those perceptions to view the same set of circumstances in a different way. This can have

benefits if the a itudes are personally destructive while on the other side it has the potential for individual and collective mind control which the subject has no idea has even happened. Cognitive therapy was developed in the 1960s by Aaron T. Beck who was born in Rhode Island in 1921 as the son of Jewish immigrants from the Ukraine. He became interested in the techniques as a treatment for depression. Beck’s daughter Judith S. Beck is prominent in the same field and they founded the Beck Institute for Cognitive Behavior Therapy in Philadelphia in 1994. Cognitive reframing, however, began to be used worldwide by those with a very dark agenda. The Cult reframes politicians to change their a itudes and actions until they are completely at odds with what they once appeared to stand for. The same has been happening to government administrators at all levels, law enforcement, military and the human population. Cultists love mind control for two main reasons: It allows them to control what people think, do and say to secure agenda advancement and, by definition, it calms their legendary insecurity and fear of the unexpected. I have studied mind control since the time I travelled America in 1996. I may have been talking to next to no one in terms of an audience in those years, but my goodness did I gather a phenomenal amount of information and knowledge about so many things including the techniques of mind control. I have described this in detail in other books going back to The Biggest Secret in 1998. I met a very large number of people recovering from MKUltra and its offshoots and successors and I began to see how these same techniques were being used on the population in general. This was never more obvious than since the ‘Covid’ hoax began.

Reframing the enforcers

I have observed over the last two decades and more the very clear transformation in the dynamic between the police, officialdom and the public. I tracked this in the books as the relationship mutated from one of serving the public to seeing them as almost the enemy and certainly a lower caste. There has always been a class divide

based on income and always been some psychopathic, corrupt, and big-I-am police officers. This was different. Wholesale change was unfolding in the collective dynamic; it was less about money and far more about position and perceived power. An us-and-them was emerging. Noses were li ed skyward by government administration and law enforcement and their a itude to the public they were supposed to be serving changed to one of increasing contempt, superiority and control. The transformation was so clear and widespread that it had to be planned. Collective a itudes and dynamics do not change naturally and organically that quickly on that scale. I then came across an organisation in Britain called Common Purpose created in the late 1980s by Julia Middleton who would work in the office of Deputy Prime Minister John Presco during the long and disastrous premiership of war criminal Tony Blair. When Blair speaks the Cult is speaking and the man should have been in jail a long time ago. Common Purpose proclaims itself to be one of the biggest ‘leadership development’ organisations in the world while functioning as a charity with all the financial benefits which come from that. It hosts ‘leadership development’ courses and programmes all over the world and claims to have ‘brought together’ what it calls ‘leaders’ from more than 100 countries on six continents. The modus operandi of Common Purpose can be compared with the work of the UK government’s reframing network that includes the Behavioural Insights Team ‘nudge unit’ and ‘Covid’ reframing specialists at SPI-B. WikiLeaks described Common Purpose long ago as ‘a hidden virus in our government and schools’ which is unknown to the general public: ‘It recruits and trains “leaders” to be loyal to the directives of Common Purpose and the EU, instead of to their own departments, which they then undermine or subvert, the NHS [National Health Service] being an example.’ This is a vital point to understand the ‘Covid’ hoax. The NHS, and its equivalent around the world, has been u erly reframed in terms of administrators and much of the medical personnel with the transformation underpinned by recruitment policies. The outcome has been the criminal and psychopathic behaviour of the

NHS over ‘Covid’ and we have seen the same in every other major country. WikiLeaks said Common Purpose trainees are ‘learning to rule without regard to democracy’ and to usher in a police state (current events explained). Common Purpose operated like a ‘glue’ and had members in the NHS, BBC, police, legal profession, church, many of Britain’s 7,000 quangos, local councils, the Civil Service, government ministries and Parliament, and controlled many RDA’s (Regional Development Agencies). Here we have one answer for how and why British institutions and their like in other countries have changed so negatively in relation to the public. This further explains how and why the beyond-disgraceful reframed BBC has become a propaganda arm of ‘Covid’ fascism. They are all part of a network pursuing the same goal. By 2019 Common Purpose was quoting a figure of 85,000 ‘leaders’ that had a ended its programmes. These ‘students’ of all ages are known as Common Purpose ‘graduates’ and they consist of government, state and local government officials and administrators, police chiefs and officers, and a whole range of others operating within the national, local and global establishment. Cressida Dick, Commissioner of the London Metropolitan Police, is the Common Purpose graduate who was the ‘Gold Commander’ that oversaw what can only be described as the murder of Brazilian electrician Jean Charles de Menezes in 2005. He was held down by psychopathic police and shot seven times in the head by a psychopathic lunatic a er being mistaken for a terrorist when he was just a bloke going about his day. Dick authorised officers to pursue and keep surveillance on de Menezes and ordered that he be stopped from entering the underground train system. Police psychopaths took her at her word clearly. She was ‘disciplined’ for this outrage by being promoted – eventually to the top of the ‘Met’ police where she has been a disaster. Many Chief Constables controlling the police in different parts of the UK are and have been Common Purpose graduates. I have heard the ‘graduate’ network described as a sort of Mafia or secret society operating within the fabric of government at all levels pursuing a collective policy

ingrained at Common Purpose training events. Founder Julia Middleton herself has said: Locally and internationally, Common Purpose graduates will be ‘lighting small fires’ to create change in their organisations and communities … The Common Purpose effect is best illustrated by the many stories of small changes brought about by leaders, who themselves have changed.

A Common Purpose mission statement declared: Common Purpose aims to improve the way society works by expanding the vision, decisionmaking ability and influence of all kinds of leaders. The organisation runs a variety of educational programmes for leaders of all ages, backgrounds and sectors, in order to provide them with the inspirational, information and opportunities they need to change the world.

Yes, but into what? Since 2020 the answer has become clear.

NLP and the Delphi technique

Common Purpose would seem to be a perfect name or would common programming be be er? One of the foundation methods of reaching ‘consensus’ (group think) is by se ing the agenda theme and then encouraging, cajoling or pressuring everyone to agree a ‘consensus’ in line with the core theme promoted by Common Purpose. The methodology involves the ‘Delphi technique’, or an adaption of it, in which opinions are expressed that are summarised by a ‘facilitator or change agent’ at each stage. Participants are ‘encouraged’ to modify their views in the light of what others have said. Stage by stage the former individual opinions are merged into group consensus which just happens to be what Common Purpose wants them to believe. A key part of this is to marginalise anyone refusing to concede to group think and turn the group against them to apply pressure to conform. We are seeing this very technique used on the general population to make ‘Covid’ group-thinkers hostile to those who have seen through the bullshit. People can be reframed by using perception manipulation methods such as Neuro-Linguistic Programming (NLP) in which you change perception with the use of

carefully constructed language. An NLP website described the technique this way: … A method of influencing brain behaviour (the ‘neuro’ part of the phrase) through the use of language (the ‘linguistic’ part) and other types of communication to enable a person to ‘recode’ the way the brain responds to stimuli (that’s the ‘programming’) and manifest new and better behaviours. Neuro-Linguistic Programming often incorporates hypnosis and selfhypnosis to help achieve the change (or ‘programming’) that is wanted.

British alternative media operation UKColumn has done very detailed research into Common Purpose over a long period. I quoted co-founder and former naval officer Brian Gerrish in my book Remember Who You Are, published in 2011, as saying the following years before current times: It is interesting that many of the mothers who have had children taken by the State speak of the Social Services people being icily cool, emotionless and, as two ladies said in slightly different words, ‘… like little robots’. We know that NLP is cumulative, so people can be given small imperceptible doses of NLP in a course here, another in a few months, next year etc. In this way, major changes are accrued in their personality, but the day by day change is almost unnoticeable.

In these and other ways ‘graduates’ have had their perceptions uniformly reframed and they return to their roles in the institutions of government, law enforcement, legal profession, military, ‘education’, the UK National Health Service and the whole swathe of the establishment structure to pursue a common agenda preparing for the ‘post-industrial’, ‘post-democratic’ society. I say ‘preparing’ but we are now there. ‘Post-industrial’ is code for the Great Reset and ‘post-democratic’ is ‘Covid’ fascism. UKColumn has spoken to partners of those who have a ended Common Purpose ‘training’. They have described how personalities and a itudes of ‘graduates’ changed very noticeably for the worse by the time they had completed the course. They had been ‘reframed’ and told they are the ‘leaders’ – the special ones – who know be er than the population. There has also been the very demonstrable recruitment of psychopaths and narcissists into government administration at all

levels and law enforcement. If you want psychopathy hire psychopaths and you get a simple cause and effect. If you want administrators, police officers and ‘leaders’ to perceive the public as lesser beings who don’t ma er then employ narcissists. These personalities are identified using ‘psychometrics’ that identifies knowledge, abilities, a itudes and personality traits, mostly through carefully-designed questionnaires and tests. As this policy has passed through the decades we have had power-crazy, powertrippers appointed into law enforcement, security and government administration in preparation for current times and the dynamic between public and law enforcement/officialdom has been transformed. UKColumn’s Brian Gerrish said of the narcissistic personality: Their love of themselves and power automatically means that they will crush others who get in their way. I received a major piece of the puzzle when a friend pointed out that when they made public officials re-apply for their own jobs several years ago they were also required to do psychometric tests. This was undoubtedly the start of the screening process to get ‘their’ sort of people in post.

How obvious that has been since 2020 although it was clear what was happening long before if people paid a ention to the changing public-establishment dynamic.

Change agents

At the centre of events in ‘Covid’ Britain is the National Health Service (NHS) which has behaved disgracefully in slavishly following the Cult agenda. The NHS management structure is awash with Common Purpose graduates or ‘change agents’ working to a common cause. Helen Bevan, a Chief of Service Transformation at the NHS Institute for Innovation and Improvement, co-authored a document called ‘Towards a million change agents, a review of the social movements literature: implications for large scale change in the NHS‘. The document compared a project management approach to that of change and social movements where ‘people change

themselves and each other – peer to peer’. Two definitions given for a ‘social movement’ were: A group of people who consciously attempt to build a radically new social order; involves people of a broad range of social backgrounds; and deploys politically confrontational and socially disruptive tactics – Cyrus Zirakzadeh 1997 Collective challenges, based on common purposes and social solidarities, in sustained interaction with elites, opponents, and authorities – Sidney Tarrow 1994 Helen Bevan wrote another NHS document in which she defined ‘framing’ as ‘the process by which leaders construct, articulate and put across their message in a powerful and compelling way in order to win people to their cause and call them to action’. I think I could come up with another definition that would be rather more accurate. The National Health Service and institutions of Britain and the wider world have been taken over by reframed ‘change agents’ and that includes everything from the United Nations to national governments, local councils and social services which have been kidnapping children from loving parents on an extraordinary and gathering scale on the road to the end of parenthood altogether. Children from loving homes are stolen and kidnapped by the state and put into the ‘care’ (inversion) of the local authority through council homes, foster parents and forced adoption. At the same time children are allowed to be abused without response while many are under council ‘care’. UKColumn highlighted the Common Purpose connection between South Yorkshire Police and Rotherham council officers in the case of the scandal in that area of the sexual exploitation of children to which the authorities turned not one blind eye, but both:

We were alarmed to discover that the Chief Executive, the Strategic Director of Children and Young People’s Services, the Manager for the Local Strategic Partnership, the Community Cohesion Manager, the Cabinet Member for Cohesion, the Chief Constable and his predecessor had all attended Leadership training courses provided by the pseudo-charity Common Purpose.

Once ‘change agents’ have secured positions of hire and fire within any organisation things start to move very quickly. Personnel are then hired and fired on the basis of whether they will work towards the agenda the change agent represents. If they do they are rapidly promoted even though they may be incompetent. Those more qualified and skilled who are pre-Common Purpose ‘old school’ see their careers stall and even disappear. This has been happening for decades in every institution of state, police, ‘health’ and social services and all of them have been transformed as a result in their a itudes to their jobs and the public. Medical professions, including nursing, which were once vocations for the caring now employ many cold, callous and couldn’t give a shit personality types. The UKColumn investigation concluded: By blurring the boundaries between people, professions, public and private sectors, responsibility and accountability, Common Purpose encourages ‘graduates’ to believe that as new selected leaders, they can work together, outside of the established political and social structures, to achieve a paradigm shift or CHANGE – so called ‘Leading Beyond Authority’. In doing so, the allegiance of the individual becomes ‘reframed’ on CP colleagues and their NETWORK.

Reframing the Face-Nappies

Nowhere has this process been more obvious than in the police where recruitment of psychopaths and development of unquestioning mind-controlled group-thinkers have transformed law enforcement into a politically-correct ‘Woke’ joke and a travesty of what should be public service. Today they wear their face-nappies like good li le gofers and enforce ‘Covid’ rules which are fascism under another name. Alongside the specifically-recruited psychopaths we have so ware minds incapable of free thought. Brian Gerrish again:

An example is the policeman who would not get on a bike for a press photo because he had not done the cycling proficiency course. Normal people say this is political correctness gone mad. Nothing could be further from the truth. The policeman has been reframed, and in his reality it is perfect common sense not to get on the bike ‘because he hasn’t done the cycling course’. Another example of this is where the police would not rescue a boy from a pond until they had taken advice from above on the ‘risk assessment’. A normal person would have arrived, perhaps thought of the risk for a moment, and dived in. To the police now ‘reframed’, they followed ‘normal’ procedure.

There are shocking cases of reframed ambulance crews doing the same. Sheer unthinking stupidity of London Face-Nappies headed by Common Purpose graduate Cressida Dick can be seen in their behaviour at a vigil in March, 2021, for a murdered woman, Sarah Everard. A police officer had been charged with the crime. Anyone with a brain would have le the vigil alone in the circumstances. Instead they ‘manhandled’ women to stop them breaking ‘Covid rules’ to betray classic reframing. Minds in the thrall of perception control have no capacity for seeing a situation on its merits and acting accordingly. ‘Rules is rules’ is their only mind-set. My father used to say that rules and regulations are for the guidance of the intelligent and the blind obedience of the idiot. Most of the intelligent, decent, coppers have gone leaving only the other kind and a few old school for whom the job must be a daily nightmare. The combination of psychopaths and rule-book so ware minds has been clearly on public display in the ‘Covid’ era with automaton robots in uniform imposing fascistic ‘Covid’ regulations on the population without any personal initiative or judging situations on their merits. There are thousands of examples around the world, but I’ll make my point with the infamous Derbyshire police in the English East Midlands – the ones who think pouring dye into beauty spots and using drones to track people walking in the countryside away from anyone is called ‘policing’. To them there are rules decreed by the government which they have to enforce and in their bewildered state a group gathering in a closed space and someone walking alone in the countryside are the same thing. It is beyond idiocy and enters the realm of clinical insanity.

Police officers in Derbyshire said they were ‘horrified’ – horrified – to find 15 to 20 ‘irresponsible’ kids playing a football match at a closed leisure centre ‘in breach of coronavirus restrictions’. When they saw the police the kids ran away leaving their belongings behind and the reframed men and women of Derbyshire police were seeking to establish their identities with a view to fining their parents. The most natural thing for youngsters to do – kicking a ball about – is turned into a criminal activity and enforced by the moronic so ware programs of Derbyshire police. You find the same mentality in every country. These barely conscious ‘horrified’ officers said they had to take action because ‘we need to ensure these rules are being followed’ and ‘it is of the utmost importance that you ensure your children are following the rules and regulations for Covid-19’. Had any of them done ten seconds of research to see if this parroting of their masters’ script could be supported by any evidence? Nope. Reframed people don’t think – others think for them and that’s the whole idea of reframing. I have seen police officers one a er the other repeating without question word for word what officialdom tells them just as I have seen great swathes of the public doing the same. Ask either for ‘their’ opinion and out spews what they have been told to think by the official narrative. Police and public may seem to be in different groups, but their mentality is the same. Most people do whatever they are told in fear not doing so or because they believe what officialdom tells them; almost the entirety of the police do what they are told for the same reason. Ultimately it’s the tiny inner core of the global Cult that’s telling both what to do. So Derbyshire police were ‘horrified’. Oh, really? Why did they think those kids were playing football? It was to relieve the psychological consequences of lockdown and being denied human contact with their friends and interaction, touch and discourse vital to human psychological health. Being denied this month a er month has dismantled the psyche of many children and young people as depression and suicide have exploded. Were Derbyshire police horrified by that? Are you kidding? Reframed people don’t have those

mental and emotional processes that can see how the impact on the psychological health of youngsters is far more dangerous than any ‘virus’ even if you take the mendacious official figures to be true. The reframed are told (programmed) how to act and so they do. The Derbyshire Chief Constable in the first period of lockdown when the black dye and drones nonsense was going on was Peter Goodman. He was the man who severed the connection between his force and the Derbyshire Constabulary Male Voice Choir when he decided that it was not inclusive enough to allow women to join. The fact it was a male voice choir making a particular sound produced by male voices seemed to elude a guy who terrifyingly ran policing in Derbyshire. He retired weeks a er his force was condemned as disgraceful by former Supreme Court Justice Jonathan Sumption for their behaviour over extreme lockdown impositions. Goodman was replaced by his deputy Rachel Swann who was in charge when her officers were ‘horrified’. The police statement over the boys commi ing the hanging-offence of playing football included the line about the youngsters being ‘irresponsible in the times we are all living through’ missing the point that the real relevance of the ‘times we are all living through’ is the imposition of fascism enforced by psychopaths and reframed minds of police officers playing such a vital part in establishing the fascist tyranny that their own children and grandchildren will have to live in their entire lives. As a definition of insanity that is hard to beat although it might be run close by imposing masks on people that can have a serious effect on their health while wearing a face nappy all day themselves. Once again public and police do it for the same reason – the authorities tell them to and who are they to have the self-respect to say no?

Wokers in uniform

How reframed do you have to be to arrest a six-year-old and take him to court for picking a flower while waiting for a bus? Brain dead police and officialdom did just that in North Carolina where criminal proceedings happen regularly for children under nine. A orney Julie Boyer gave the six-year-old crayons and a colouring book

during the ‘flower’ hearing while the ‘adults’ decided his fate. County Chief District Court Judge Jay Corpening asked: ‘Should a child that believes in Santa Claus, the Easter Bunny and the tooth fairy be making life-altering decisions?’ Well, of course not, but common sense has no meaning when you have a common purpose and a reframed mind. Treating children in this way, and police operating in American schools, is all part of the psychological preparation for children to accept a police state as normal all their adult lives. The same goes for all the cameras and biometric tracking technology in schools. Police training is focused on reframing them as snowflake Wokers and this is happening in the military. Pentagon top brass said that ‘training sessions on extremism’ were needed for troops who asked why they were so focused on the Capitol Building riot when Black Lives Ma er riots were ignored. What’s the difference between them some apparently and rightly asked. Actually, there is a difference. Five people died in the Capitol riot, only one through violence, and that was a police officer shooting an unarmed protestor. BLM riots killed at least 25 people and cost billions. Asking the question prompted the psychopaths and reframed minds that run the Pentagon to say that more ‘education’ (programming) was needed. Troop training is all based on psychological programming to make them fodder for the Cult – ‘Military men are just dumb, stupid animals to be used as pawns in foreign policy’ as Cult-to-his-DNA former Secretary of State Henry Kissinger famously said. Governments see the police in similar terms and it’s time for those among them who can see this to defend the people and stop being enforcers of the Cult agenda upon the people. The US military, like the country itself, is being targeted for destruction through a long list of Woke impositions. Cult-owned gaga ‘President’ Biden signed an executive order when he took office to allow taxpayer money to pay for transgender surgery for active military personnel and veterans. Are you a man soldier? No, I’m a LGBTQIA+ with a hint of Skoliosexual and Spectrasexual. Oh, good man. Bad choice of words you bigot. The Pentagon announced in March, 2021, the appointment of the first ‘diversity and inclusion

officer’ for US Special Forces. Richard Torres-Estrada arrived with the publication of a ‘D&I Strategic Plan which will guide the enterprise-wide effort to institutionalize and sustain D&I’. If you think a Special Forces ‘Strategic Plan’ should have something to do with defending America you haven’t been paying a ention. Defending Woke is now the military’s new role. Torres-Estrada has posted images comparing Donald Trump with Adolf Hitler and we can expect no bias from him as a representative of the supposedly non-political Pentagon. Cable news host Tucker Carlson said: ‘The Pentagon is now the Yale faculty lounge but with cruise missiles.’ Meanwhile Secretary of Defense Lloyd Austin, a board member of weapons-maker Raytheon with stock and compensation interests in October, 2020, worth $1.4 million, said he was purging the military of the ‘enemy within’ – anyone who isn’t Woke and supports Donald Trump. Austin refers to his targets as ‘racist extremists’ while in true Woke fashion being himself a racist extremist. Pentagon documents pledge to ‘eradicate, eliminate and conquer all forms of racism, sexism and homophobia’. The definitions of these are decided by ‘diversity and inclusion commi ees’ peopled by those who see racism, sexism and homophobia in every situation and opinion. Woke (the Cult) is dismantling the US military and purging testosterone as China expands its military and gives its troops ‘masculinity training’. How do we think that is going to end when this is all Cult coordinated? The US military, like the British military, is controlled by Woke and spineless top brass who just go along with it out of personal career interests.

‘Woke’ means fast asleep

Mind control and perception manipulation techniques used on individuals to create group-think have been unleashed on the global population in general. As a result many have no capacity to see the obvious fascist agenda being installed all around them or what ‘Covid’ is really all about. Their brains are firewalled like a computer system not to process certain concepts, thoughts and realisations that are bad for the Cult. The young are most targeted as the adults they

will be when the whole fascist global state is planned to be fully implemented. They need to be prepared for total compliance to eliminate all pushback from entire generations. The Cult has been pouring billions into taking complete control of ‘education’ from schools to universities via its operatives and corporations and not least Bill Gates as always. The plan has been to transform ‘education’ institutions into programming centres for the mentality of ‘Woke’. James McConnell, professor of psychology at the University of Michigan, wrote in Psychology Today in 1970: The day has come when we can combine sensory deprivation with drugs, hypnosis, and astute manipulation of reward and punishment, to gain almost absolute control over an individual’s behaviour. It should then be possible to achieve a very rapid and highly effective type of brainwashing that would allow us to make dramatic changes in a person’s behaviour and personality ... … We should reshape society so that we all would be trained from birth to want to do what society wants us to do. We have the techniques to do it... no-one owns his own personality you acquired, and there’s no reason to believe you should have the right to refuse to acquire a new personality if your old one is anti-social.

This was the potential for mass brainwashing in 1970 and the mentality there displayed captures the arrogant psychopathy that drives it forward. I emphasise that not all young people have succumbed to Woke programming and those that haven’t are incredibly impressive people given that today’s young are the most perceptually-targeted generations in history with all the technology now involved. Vast swathes of the young generations, however, have fallen into the spell – and that’s what it is – of Woke. The Woke mentality and perceptual program is founded on inversion and you will appreciate later why that is so significant. Everything with Woke is inverted and the opposite of what it is claimed to be. Woke was a term used in African-American culture from the 1900s and referred to an awareness of social and racial justice. This is not the meaning of the modern version or ‘New Woke’ as I call it in The Answer. Oh, no, Woke today means something very different no ma er how much Wokers may seek to hide that and insist Old Woke and New

Woke are the same. See if you find any ‘awareness of social justice’ here in the modern variety: • Woke demands ‘inclusivity’ while excluding anyone with a different opinion and calls for mass censorship to silence other views. • Woke claims to stand against oppression when imposing oppression is the foundation of all that it does. It is the driver of political correctness which is nothing more than a Cult invention to manipulate the population to silence itself. • Woke believes itself to be ‘liberal’ while pursuing a global society that can only be described as fascist (see ‘anti-fascist’ fascist Antifa). • Woke calls for ‘social justice’ while spreading injustice wherever it goes against the common ‘enemy’ which can be easily identified as a differing view. • Woke is supposed to be a metaphor for ‘awake’ when it is solidgold asleep and deep in a Cult-induced coma that meets the criteria for ‘off with the fairies’. I state these points as obvious facts if people only care to look. I don’t do this with a sense of condemnation. We need to appreciate that the onslaught of perceptual programming on the young has been incessant and merciless. I can understand why so many have been reframed, or, given their youth, framed from the start to see the world as the Cult demands. The Cult has had access to their minds day a er day in its ‘education’ system for their entire formative years. Perception is formed from information received and the Cultcreated system is a life-long download of information delivered to elicit a particular perception, thus behaviour. The more this has expanded into still new extremes in recent decades and everincreasing censorship has deleted other opinions and information why wouldn’t that lead to a perceptual reframing on a mass scale? I

have described already cradle-to-grave programming and in more recent times the targeting of young minds from birth to adulthood has entered the stratosphere. This has taken the form of skewing what is ‘taught’ to fit the Cult agenda and the omnipresent techniques of group-think to isolate non-believers and pressure them into line. There has always been a tendency to follow the herd, but we really are in a new world now in relation to that. We have parents who can see the ‘Covid’ hoax told by their children not to stop them wearing masks at school, being ‘Covid’ tested or having the ‘vaccine’ in fear of the peer-pressure consequences of being different. What is ‘peer-pressure’ if not pressure to conform to group-think? Renegade Minds never group-think and always retain a set of perceptions that are unique to them. Group-think is always underpinned by consequences for not group-thinking. Abuse now aimed at those refusing DNA-manipulating ‘Covid vaccines’ are a potent example of this. The biggest pressure to conform comes from the very group which is itself being manipulated. ‘I am programmed to be part of a hive mind and so you must be.’ Woke control structures in ‘education’ now apply to every mainstream organisation. Those at the top of the ‘education’ hierarchy (the Cult) decide the policy. This is imposed on governments through the Cult network; governments impose it on schools, colleges and universities; their leadership impose the policy on teachers and academics and they impose it on children and students. At any level where there is resistance, perhaps from a teacher or university lecturer, they are targeted by the authorities and o en fired. Students themselves regularly demand the dismissal of academics (increasingly few) at odds with the narrative that the students have been programmed to believe in. It is quite a thought that students who are being targeted by the Cult become so consumed by programmed group-think that they launch protests and demand the removal of those who are trying to push back against those targeting the students. Such is the scale of perceptual inversion. We see this with ‘Covid’ programming as the Cult imposes the rules via psycho-psychologists and governments on

shops, transport companies and businesses which impose them on their staff who impose them on their customers who pressure Pushbackers to conform to the will of the Cult which is in the process of destroying them and their families. Scan all aspects of society and you will see the same sequence every time.

Fact free Woke and hijacking the ‘left’

There is no more potent example of this than ‘Woke’, a mentality only made possible by the deletion of factual evidence by an ‘education’ system seeking to produce an ever more uniform society. Why would you bother with facts when you don’t know any? Deletion of credible history both in volume and type is highly relevant. Orwell said: ‘Who controls the past controls the future: who controls the present controls the past.’ They who control the perception of the past control the perception of the future and they who control the present control the perception of the past through the writing and deleting of history. Why would you oppose the imposition of Marxism in the name of Wokeism when you don’t know that Marxism cost at least 100 million lives in the 20th century alone? Watch videos and read reports in which Woker generations are asked basic historical questions – it’s mind-blowing. A survey of 2,000 people found that six percent of millennials (born approximately early1980s to early 2000s) believed the Second World War (1939-1945) broke out with the assassination of President Kennedy (in 1963) and one in ten thought Margaret Thatcher was British Prime Minister at the time. She was in office between 1979 and 1990. We are in a post-fact society. Provable facts are no defence against the fascism of political correctness or Silicon Valley censorship. Facts don’t ma er anymore as we have witnessed with the ‘Covid’ hoax. Sacrificing uniqueness to the Woke group-think religion is all you are required to do and that means thinking for yourself is the biggest Woke no, no. All religions are an expression of group-think and censorship and Woke is just another religion with an orthodoxy defended by group-think and censorship. Burned at

the stake becomes burned on Twi er which leads back eventually to burned at the stake as Woke humanity regresses to ages past. The biggest Woke inversion of all is its creators and funders. I grew up in a traditional le of centre political household on a council estate in Leicester in the 1950s and 60s – you know, the le that challenged the power of wealth-hoarding elites and threats to freedom of speech and opinion. In those days students went on marches defending freedom of speech while today’s Wokers march for its deletion. What on earth could have happened? Those very elites (collectively the Cult) that we opposed in my youth and early life have funded into existence the antithesis of that former le and hijacked the ‘brand’ while inverting everything it ever stood for. We have a mentality that calls itself ‘liberal’ and ‘progressive’ while acting like fascists. Cult billionaires and their corporations have funded themselves into control of ‘education’ to ensure that Woke programming is unceasing throughout the formative years of children and young people and that non-Wokers are isolated (that word again) whether they be students, teachers or college professors. The Cult has funded into existence the now colossal global network of Woke organisations that have spawned and promoted all the ‘causes’ on the Cult wish-list for global transformation and turned Wokers into demanders of them. Does anyone really think it’s a coincidence that the Cult agenda for humanity is a carbon (sorry) copy of the societal transformations desired by Woke?? These are only some of them: The means by which the Cult deletes all public debates that it knows it cannot win if we had the free-flow of information and evidence. Political correctness:

The means by which the Cult seeks to transform society into a globally-controlled dictatorship imposing its will over the fine detail of everyone’s lives ‘to save the planet’ which doesn’t actually need saving. Human-caused ‘climate change’:

Preparing collective perception to accept the ‘new human’ which would not have genders because it would be created technologically and not through procreation. I’ll have much more on this in Human 2.0. Transgender obsession:

The means by which the Cult seeks to divide and rule the population by triggering racial division through the perception that society is more racist than ever when the opposite is the case. Is it perfect in that regard? No. But to compare today with the racism of apartheid and segregation brought to an end by the civil rights movement in the 1960s is to insult the memory of that movement and inspirations like Martin Luther King. Why is the ‘anti-racism’ industry (which it is) so dominated by privileged white people? Race obsession:

This is a label used by privileged white people to demonise poor and deprived white people pushing back on tyranny to marginalise and destroy them. White people are being especially targeted as the dominant race by number within Western society which the Cult seeks to transform in its image. If you want to change a society you must weaken and undermine its biggest group and once you have done that by using the other groups you next turn on them to do the same … ‘Then they came for the Jews and I was not a Jew so I did nothing.’ White supremacy:

The mass movement of people from the Middle East, Africa and Asia into Europe, from the south into the United States and from Asia into Australia are another way the Cult seeks to dilute the racial, cultural and political influence of white people on Western society. White people ask why their governments appear to be working against them while being politically and culturally biased towards incoming cultures. Well, here’s your answer. In the same way sexually ‘straight’ people, men and women, ask why the Mass migration:

authorities are biased against them in favour of other sexualities. The answer is the same – that’s the way the Cult wants it to be for very sinister motives. These are all central parts of the Cult agenda and central parts of the Woke agenda and Woke was created and continues to be funded to an immense degree by Cult billionaires and corporations. If anyone begins to say ‘coincidence’ the syllables should stick in their throat.

Billionaire ‘social justice warriors’

Joe Biden is a 100 percent-owned asset of the Cult and the Wokers’ man in the White House whenever he can remember his name and for however long he lasts with his rapidly diminishing cognitive function. Even walking up the steps of an aircra without falling on his arse would appear to be a challenge. He’s not an empty-shell puppet or anything. From the minute Biden took office (or the Cult did) he began his executive orders promoting the Woke wish-list. You will see the Woke agenda imposed ever more severely because it’s really the Cult agenda. Woke organisations and activist networks spawned by the Cult are funded to the extreme so long as they promote what the Cult wants to happen. Woke is funded to promote ‘social justice’ by billionaires who become billionaires by destroying social justice. The social justice mantra is only a cover for dismantling social justice and funded by billionaires that couldn’t give a damn about social justice. Everything makes sense when you see that. One of Woke’s premier funders is Cult billionaire financier George Soros who said: ‘I am basically there to make money, I cannot and do not look at the social consequences of what I do.’ This is the same Soros who has given more than $32 billion to his Open Society Foundations global Woke network and funded Black Lives Ma er, mass immigration into Europe and the United States, transgender activism, climate change activism, political correctness and groups targeting ‘white supremacy’ in the form of privileged white thugs that dominate Antifa. What a scam it all is and when

you are dealing with the unquestioning fact-free zone of Woke scamming them is child’s play. All you need to pull it off in all these organisations are a few in-the-know agents of the Cult and an army of naïve, reframed, uninformed, narcissistic, know-nothings convinced of their own self-righteousness, self-purity and virtue. Soros and fellow billionaires and billionaire corporations have poured hundreds of millions into Black Lives Ma er and connected groups and promoted them to a global audience. None of this is motivated by caring about black people. These are the billionaires that have controlled and exploited a system that leaves millions of black people in abject poverty and deprivation which they do absolutely nothing to address. The same Cult networks funding BLM were behind the slave trade! Black Lives Ma er hijacked a phrase that few would challenge and they have turned this laudable concept into a political weapon to divide society. You know that BLM is a fraud when it claims that All Lives Ma er, the most inclusive statement of all, is ‘racist’. BLM and its Cult masters don’t want to end racism. To them it’s a means to an end to control all of humanity never mind the colour, creed, culture or background. What has destroying the nuclear family got to do with ending racism? Nothing – but that is one of the goals of BLM and also happens to be a goal of the Cult as I have been exposing in my books for decades. Stealing children from loving parents and giving schools ever more power to override parents is part of that same agenda. BLM is a Marxist organisation and why would that not be the case when the Cult created Marxism and BLM? Patrisse Cullors, a BLM co-founder, said in a 2015 video that she and her fellow organisers, including co-founder Alicia Garza, are ‘trained Marxists’. The lady known a er marriage as Patrisse Khan-Cullors bought a $1.4 million home in 2021 in one of the whitest areas of California with a black population of just 1.6 per cent and has so far bought four high-end homes for a total of $3.2 million. How very Marxist. There must be a bit of spare in the BLM coffers, however, when Cult corporations and billionaires have handed over the best part of $100 million. Many black people can see that Black Lives Ma er is not

working for them, but against them, and this is still more confirmation. Black journalist Jason Whitlock, who had his account suspended by Twi er for simply linking to the story about the ‘Marxist’s’ home buying spree, said that BLM leaders are ‘making millions of dollars off the backs of these dead black men who they wouldn’t spit on if they were on fire and alive’.

Black Lies Matter

Cult assets and agencies came together to promote BLM in the wake of the death of career criminal George Floyd who had been jailed a number of times including for forcing his way into the home of a black woman with others in a raid in which a gun was pointed at her stomach. Floyd was filmed being held in a Minneapolis street in 2020 with the knee of a police officer on his neck and he subsequently died. It was an appalling thing for the officer to do, but the same technique has been used by police on peaceful protestors of lockdown without any outcry from the Woke brigade. As unquestioning supporters of the Cult agenda Wokers have supported lockdown and all the ‘Covid’ claptrap while a acking anyone standing up to the tyranny imposed in its name. Court documents would later include details of an autopsy on Floyd by County Medical Examiner Dr Andrew Baker who concluded that Floyd had taken a fatal level of the drug fentanyl. None of this ma ered to fact-free, question-free, Woke. Floyd’s death was followed by worldwide protests against police brutality amid calls to defund the police. Throwing babies out with the bathwater is a Woke speciality. In the wake of the murder of British woman Sarah Everard a Green Party member of the House of Lords, Baroness Jones of Moulescoomb (Nincompoopia would have been be er), called for a 6pm curfew for all men. This would be in breach of the Geneva Conventions on war crimes which ban collective punishment, but that would never have crossed the black and white Woke mind of Baroness Nincompoopia who would have been far too convinced of her own self-righteousness to compute such details. Many American cities did defund the police in the face of Floyd riots

and a er $15 million was deleted from the police budget in Washington DC under useless Woke mayor Muriel Bowser carjacking alone rose by 300 percent and within six months the US capital recorded its highest murder rate in 15 years. The same happened in Chicago and other cities in line with the Cult/Soros plan to bring fear to streets and neighbourhoods by reducing the police, releasing violent criminals and not prosecuting crime. This is the mob-rule agenda that I have warned in the books was coming for so long. Shootings in the area of Minneapolis where Floyd was arrested increased by 2,500 percent compared with the year before. Defunding the police over George Floyd has led to a big increase in dead people with many of them black. Police protection for politicians making these decisions stayed the same or increased as you would expect from professional hypocrites. The Cult doesn’t actually want to abolish the police. It wants to abolish local control over the police and hand it to federal government as the psychopaths advance the Hunger Games Society. Many George Floyd protests turned into violent riots with black stores and businesses destroyed by fire and looting across America fuelled by Black Lives Ma er. Woke doesn’t do irony. If you want civil rights you must loot the liquor store and the supermarket and make off with a smart TV. It’s the only way.

It’s not a race war – it’s a class war

Black people are patronised by privileged blacks and whites alike and told they are victims of white supremacy. I find it extraordinary to watch privileged blacks supporting the very system and bloodline networks behind the slave trade and parroting the same Cult-serving manipulative crap of their privileged white, o en billionaire, associates. It is indeed not a race war but a class war and colour is just a diversion. Black Senator Cory Booker and black Congresswoman Maxine Waters, more residents of Nincompoopia, personify this. Once you tell people they are victims of someone else you devalue both their own responsibility for their plight and the power they have to impact on their reality and experience. Instead

we have: ‘You are only in your situation because of whitey – turn on them and everything will change.’ It won’t change. Nothing changes in our lives unless we change it. Crucial to that is never seeing yourself as a victim and always as the creator of your reality. Life is a simple sequence of choice and consequence. Make different choices and you create different consequences. You have to make those choices – not Black Lives Ma er, the Woke Mafia and anyone else that seeks to dictate your life. Who are they these Wokers, an emotional and psychological road traffic accident, to tell you what to do? Personal empowerment is the last thing the Cult and its Black Lives Ma er want black people or anyone else to have. They claim to be defending the underdog while creating and perpetuating the underdog. The Cult’s worst nightmare is human unity and if they are going to keep blacks, whites and every other race under economic servitude and control then the focus must be diverted from what they have in common to what they can be manipulated to believe divides them. Blacks have to be told that their poverty and plight is the fault of the white bloke living on the street in the same poverty and with the same plight they are experiencing. The difference is that your plight black people is due to him, a white supremacist with ‘white privilege’ living on the street. Don’t unite as one human family against your mutual oppressors and suppressors – fight the oppressor with the white face who is as financially deprived as you are. The Cult knows that as its ‘Covid’ agenda moves into still new levels of extremism people are going to respond and it has been spreading the seeds of disunity everywhere to stop a united response to the evil that targets all of us. Racist a acks on ‘whiteness’ are ge ing ever more outrageous and especially through the American Democratic Party which has an appalling history for anti-black racism. Barack Obama, Joe Biden, Hillary Clinton and Nancy Pelosi all eulogised about Senator Robert Byrd at his funeral in 2010 a er a nearly 60-year career in Congress. Byrd was a brutal Ku Klux Klan racist and a violent abuser of Cathy O’Brien in MKUltra. He said he would never fight in the military ‘with a negro by my side’ and ‘rather I should die a thousand times,

and see Old Glory trampled in the dirt never to rise again, than to see this beloved land of ours become degraded by race mongrels, a throwback to the blackest specimen from the wilds’. Biden called Byrd a ‘very close friend and mentor’. These ‘Woke’ hypocrites are not anti-racist they are anti-poor and anti-people not of their perceived class. Here is an illustration of the scale of anti-white racism to which we have now descended. Seriously Woke and moronic New York Times contributor Damon Young described whiteness as a ‘virus’ that ‘like other viruses will not die until there are no bodies le for it to infect’. He went on: ‘… the only way to stop it is to locate it, isolate it, extract it, and kill it.’ Young can say that as a black man with no consequences when a white man saying the same in reverse would be facing a jail sentence. That’s racism. We had super-Woke numbskull senators Tammy Duckworth and Mazie Hirono saying they would object to future Biden Cabinet appointments if he did not nominate more Asian Americans and Pacific Islanders. Never mind the ability of the candidate what do they look like? Duckworth said: ‘I will vote for racial minorities and I will vote for LGBTQ, but anyone else I’m not voting for.’ Appointing people on the grounds of race is illegal, but that was not a problem for this ludicrous pair. They were on-message and that’s a free pass in any situation.

Critical race racism

White children are told at school they are intrinsically racist as they are taught the divisive ‘critical race theory’. This claims that the law and legal institutions are inherently racist and that race is a socially constructed concept used by white people to further their economic and political interests at the expense of people of colour. White is a ‘virus’ as we’ve seen. Racial inequality results from ‘social, economic, and legal differences that white people create between races to maintain white interests which leads to poverty and criminality in minority communities‘. I must tell that to the white guy sleeping on the street. The principal of East Side Community School in New York sent white parents a manifesto that called on

them to become ‘white traitors’ and advocate for full ‘white abolition’. These people are teaching your kids when they urgently need a psychiatrist. The ‘school’ included a chart with ‘eight white identities’ that ranged from ‘white supremacist’ to ‘white abolition’ and defined the behaviour white people must follow to end ‘the regime of whiteness’. Woke blacks and their privileged white associates are acting exactly like the slave owners of old and Ku Klux Klan racists like Robert Byrd. They are too full of their own selfpurity to see that, but it’s true. Racism is not a body type; it’s a state of mind that can manifest through any colour, creed or culture. Another racial fraud is ‘equity’. Not equality of treatment and opportunity – equity. It’s a term spun as equality when it means something very different. Equality in its true sense is a raising up while ‘equity’ is a race to the bo om. Everyone in the same level of poverty is ‘equity’. Keep everyone down – that’s equity. The Cult doesn’t want anyone in the human family to be empowered and BLM leaders, like all these ‘anti-racist’ organisations, continue their privileged, pampered existence by perpetuating the perception of gathering racism. When is the last time you heard an ‘anti-racist’ or ‘anti-Semitism’ organisation say that acts of racism and discrimination have fallen? It’s not in the interests of their fundraising and power to influence and the same goes for the professional soccer anti-racism operation, Kick It Out. Two things confirmed that the Black Lives Ma er riots in the summer of 2020 were Cult creations. One was that while anti-lockdown protests were condemned in this same period for ‘transmi ing ‘Covid’ the authorities supported mass gatherings of Black Lives Ma er supporters. I even saw self-deluding people claiming to be doctors say the two types of protest were not the same. No – the non-existent ‘Covid’ was in favour of lockdowns and a acked those that protested against them while ‘Covid’ supported Black Lives Ma er and kept well away from its protests. The whole thing was a joke and as lockdown protestors were arrested, o en brutally, by reframed Face-Nappies we had the grotesque sight of police officers taking the knee to Black Lives Ma er, a Cult-funded Marxist

organisation that supports violent riots and wants to destroy the nuclear family and white people.

He’s not white? Shucks!

Woke obsession with race was on display again when ten people were shot dead in Boulder, Colorado, in March, 2021. Cult-owned Woke TV channels like CNN said the shooter appeared to be a white man and Wokers were on Twi er condemning ‘violent white men’ with the usual mantras. Then the shooter’s name was released as Ahmad Al Aliwi Alissa, an anti-Trump Arab-American, and the sigh of disappointment could be heard five miles away. Never mind that ten people were dead and what that meant for their families. Race baiting was all that ma ered to these sick Cult-serving people like Barack Obama who exploited the deaths to further divide America on racial grounds which is his job for the Cult. This is the man that ‘racist’ white Americans made the first black president of the United States and then gave him a second term. Not-very-bright Obama has become filthy rich on the back of that and today appears to have a big influence on the Biden administration. Even so he’s still a downtrodden black man and a victim of white supremacy. This disingenuous fraud reveals the contempt he has for black people when he puts on a Deep South Alabama accent whenever he talks to them, no, at them. Another BLM red flag was how the now fully-Woke (fully-Cult) and fully-virtue-signalled professional soccer authorities had their teams taking the knee before every match in support of Marxist Black Lives Ma er. Soccer authorities and clubs displayed ‘Black Lives Ma er’ on the players’ shirts and flashed the name on electronic billboards around the pitch. Any fans that condemned what is a Freemasonic taking-the-knee ritual were widely condemned as you would expect from the Woke virtue-signallers of professional sport and the now fully-Woke media. We have reverse racism in which you are banned from criticising any race or culture except for white people for whom anything goes – say what you like, no problem. What has this got to do with racial harmony and

equality? We’ve had black supremacists from Black Lives Ma er telling white people to fall to their knees in the street and apologise for their white supremacy. Black supremacists acting like white supremacist slave owners of the past couldn’t breach their selfobsessed, race-obsessed sense of self-purity. Joe Biden appointed a race-obsessed black supremacist Kristen Clarke to head the Justice Department Civil Rights Division. Clarke claimed that blacks are endowed with ‘greater mental, physical and spiritual abilities’ than whites. If anyone reversed that statement they would be vilified. Clarke is on-message so no problem. She’s never seen a black-white situation in which the black figure is anything but a virtuous victim and she heads the Civil Rights Division which should treat everyone the same or it isn’t civil rights. Another perception of the Renegade Mind: If something or someone is part of the Cult agenda they will be supported by Woke governments and media no ma er what. If they’re not, they will be condemned and censored. It really is that simple and so racist Clarke prospers despite (make that because of) her racism.

The end of culture

Biden’s administration is full of such racial, cultural and economic bias as the Cult requires the human family to be divided into warring factions. We are now seeing racially-segregated graduations and everything, but everything, is defined through the lens of perceived ‘racism. We have ‘racist’ mathematics, ‘racist’ food and even ‘racist’ plants. World famous Kew Gardens in London said it was changing labels on plants and flowers to tell its pre-‘Covid’ more than two million visitors a year how racist they are. Kew director Richard Deverell said this was part of an effort to ‘move quickly to decolonise collections’ a er they were approached by one Ajay Chhabra ‘an actor with an insight into how sugar cane was linked to slavery’. They are plants you idiots. ‘Decolonisation’ in the Woke manual really means colonisation of society with its mentality and by extension colonisation by the Cult. We are witnessing a new Chinese-style ‘Cultural Revolution’ so essential to the success of all

Marxist takeovers. Our cultural past and traditions have to be swept away to allow a new culture to be built-back-be er. Woke targeting of long-standing Western cultural pillars including historical monuments and cancelling of historical figures is what happened in the Mao revolution in China which ‘purged remnants of capitalist and traditional elements from Chinese society‘ and installed Maoism as the dominant ideology‘. For China see the Western world today and for ‘dominant ideology’ see Woke. Be er still see Marxism or Maoism. The ‘Covid’ hoax has specifically sought to destroy the arts and all elements of Western culture from people meeting in a pub or restaurant to closing theatres, music venues, sports stadiums, places of worship and even banning singing. Destruction of Western society is also why criticism of any religion is banned except for Christianity which again is the dominant religion as white is the numericallydominant race. Christianity may be fading rapidly, but its history and traditions are weaved through the fabric of Western society. Delete the pillars and other structures will follow until the whole thing collapses. I am not a Christian defending that religion when I say that. I have no religion. It’s just a fact. To this end Christianity has itself been turned Woke to usher its own downfall and its ranks are awash with ‘change agents’ – knowing and unknowing – at every level including Pope Francis (definitely knowing) and the clueless Archbishop of Canterbury Justin Welby (possibly not, but who can be sure?). Woke seeks to coordinate a acks on Western culture, traditions, and ways of life through ‘intersectionality’ defined as ‘the complex, cumulative way in which the effects of multiple forms of discrimination (such as racism, sexism, and classism) combine, overlap, or intersect especially in the experiences of marginalised individuals or groups’. Wade through the Orwellian Woke-speak and this means coordinating disparate groups in a common cause to overthrow freedom and liberal values. The entire structure of public institutions has been infested with Woke – government at all levels, political parties, police, military, schools, universities, advertising, media and trade unions. This abomination has been achieved through the Cult web by appointing

Wokers to positions of power and ba ering non-Wokers into line through intimidation, isolation and threats to their job. Many have been fired in the wake of the empathy-deleted, vicious hostility of ‘social justice’ Wokers and the desire of gutless, spineless employers to virtue-signal their Wokeness. Corporations are filled with Wokers today, most notably those in Silicon Valley. Ironically at the top they are not Woke at all. They are only exploiting the mentality their Cult masters have created and funded to censor and enslave while the Wokers cheer them on until it’s their turn. Thus the Woke ‘liberal le ’ is an inversion of the traditional liberal le . Campaigning for justice on the grounds of power and wealth distribution has been replaced by campaigning for identity politics. The genuine traditional le would never have taken money from today’s billionaire abusers of fairness and justice and nor would the billionaires have wanted to fund that genuine le . It would not have been in their interests to do so. The division of opinion in those days was between the haves and have nots. This all changed with Cult manipulated and funded identity politics. The division of opinion today is between Wokers and non-Wokers and not income brackets. Cult corporations and their billionaires may have taken wealth disparity to cataclysmic levels of injustice, but as long as they speak the language of Woke, hand out the dosh to the Woke network and censor the enemy they are ‘one of us’. Billionaires who don’t give a damn about injustice are laughing at them till their bellies hurt. Wokers are not even close to self-aware enough to see that. The transformed ‘le ’ dynamic means that Wokers who drone on about ‘social justice’ are funded by billionaires that have destroyed social justice the world over. It’s why they are billionaires.

The climate con

Nothing encapsulates what I have said more comprehensively than the hoax of human-caused global warming. I have detailed in my books over the years how Cult operatives and organisations were the pump-primers from the start of the climate con. A purpose-built vehicle for this is the Club of Rome established by the Cult in 1968

with the Rockefellers and Rothschilds centrally involved all along. Their gofer frontman Maurice Strong, a Canadian oil millionaire, hosted the Earth Summit in Rio de Janeiro, Brazil, in 1992 where the global ‘green movement’ really expanded in earnest under the guiding hand of the Cult. The Earth Summit established Agenda 21 through the Cult-created-and-owned United Nations to use the illusion of human-caused climate change to justify the transformation of global society to save the world from climate disaster. It is a No-Problem-Reaction-Solution sold through governments, media, schools and universities as whole generations have been terrified into believing that the world was going to end in their lifetimes unless what old people had inflicted upon them was stopped by a complete restructuring of how everything is done. Chill, kids, it’s all a hoax. Such restructuring is precisely what the Cult agenda demands (purely by coincidence of course). Today this has been given the codename of the Great Reset which is only an updated term for Agenda 21 and its associated Agenda 2030. The la er, too, is administered through the UN and was voted into being by the General Assembly in 2015. Both 21 and 2030 seek centralised control of all resources and food right down to the raindrops falling on your own land. These are some of the demands of Agenda 21 established in 1992. See if you recognise this society emerging today: • End national sovereignty • State planning and management of all land resources, ecosystems, deserts, forests, mountains, oceans and fresh water; agriculture; rural development; biotechnology; and ensuring ‘equity’ • The state to ‘define the role’ of business and financial resources • Abolition of private property • ‘Restructuring’ the family unit (see BLM) • Children raised by the state • People told what their job will be • Major restrictions on movement • Creation of ‘human se lement zones’

• Mass rese lement as people are forced to vacate land where they live • Dumbing down education • Mass global depopulation in pursuit of all the above The United Nations was created as a Trojan horse for world government. With the climate con of critical importance to promoting that outcome you would expect the UN to be involved. Oh, it’s involved all right. The UN is promoting Agenda 21 and Agenda 2030 justified by ‘climate change’ while also driving the climate hoax through its Intergovernmental Panel on Climate Change (IPCC), one of the world’s most corrupt organisations. The IPCC has been lying ferociously and constantly since the day it opened its doors with the global media hanging unquestioningly on its every mendacious word. The Green movement is entirely Woke and has long lost its original environmental focus since it was coopted by the Cult. An obsession with ‘global warming’ has deleted its values and scrambled its head. I experienced a small example of what I mean on a beautiful country walk that I have enjoyed several times a week for many years. The path merged into the fields and forests and you felt at one with the natural world. Then a ‘Green’ organisation, the Hampshire and Isle of Wight Wildlife Trust, took over part of the land and proceeded to cut down a large number of trees, including mature ones, to install a horrible big, bright steel ‘this-is-ours-stay-out’ fence that destroyed the whole atmosphere of this beautiful place. No one with a feel for nature would do that. Day a er day I walked to the sound of chainsaws and a magnificent mature weeping willow tree that I so admired was cut down at the base of the trunk. When I challenged a Woke young girl in a green shirt (of course) about this vandalism she replied: ‘It’s a weeping willow – it will grow back.’ This is what people are paying for when they donate to the Hampshire and Isle of Wight Wildlife Trust and many other ‘green’ organisations today. It is not the environmental movement that I knew and instead has become a support-system – as with Extinction Rebellion – for a very dark agenda.

Private jets for climate justice

The Cult-owned, Gates-funded, World Economic Forum and its founder Klaus Schwab were behind the emergence of Greta Thunberg to harness the young behind the climate agenda and she was invited to speak to the world at … the UN. Schwab published a book, Covid-19: The Great Reset in 2020 in which he used the ‘Covid’ hoax and the climate hoax to lay out a new society straight out of Agenda 21 and Agenda 2030. Bill Gates followed in early 2021 when he took time out from destroying the world to produce a book in his name about the way to save it. Gates flies across the world in private jets and admi ed that ‘I probably have one of the highest greenhouse gas footprints of anyone on the planet … my personal flying alone is gigantic.’ He has also bid for the planet’s biggest private jet operator. Other climate change saviours who fly in private jets include John Kerry, the US Special Presidential Envoy for Climate, and actor Leonardo DiCaprio, a ‘UN Messenger of Peace with special focus on climate change’. These people are so full of bullshit they could corner the market in manure. We mustn’t be sceptical, though, because the Gates book, How to Avoid a Climate Disaster: The Solutions We Have and the Breakthroughs We Need, is a genuine a empt to protect the world and not an obvious pile of excrement a ributed to a mega-psychopath aimed at selling his masters’ plans for humanity. The Gates book and the other shite-pile by Klaus Schwab could have been wri en by the same person and may well have been. Both use ‘climate change’ and ‘Covid’ as the excuses for their new society and by coincidence the Cult’s World Economic Forum and Bill and Melinda Gates Foundation promote the climate hoax and hosted Event 201 which pre-empted with a ‘simulation’ the very ‘coronavirus’ hoax that would be simulated for real on humanity within weeks. The British ‘royal’ family is promoting the ‘Reset’ as you would expect through Prince ‘climate change caused the war in Syria’ Charles and his hapless son Prince William who said that we must ‘reset our relationship with nature and our trajectory as a species’ to avoid a climate disaster. Amazing how many promotors of the ‘Covid’ and ‘climate change’ control

systems are connected to Gates and the World Economic Forum. A ‘study’ in early 2021 claimed that carbon dioxide emissions must fall by the equivalent of a global lockdown roughly every two years for the next decade to save the planet. The ‘study’ appeared in the same period that the Schwab mob claimed in a video that lockdowns destroying the lives of billions are good because they make the earth ‘quieter’ with less ‘ambient noise’. They took down the video amid a public backlash for such arrogant, empathy-deleted stupidity You see, however, where they are going with this. Corinne Le Quéré, a professor at the Tyndall Centre for Climate Change Research, University of East Anglia, was lead author of the climate lockdown study, and she writes for … the World Economic Forum. Gates calls in ‘his’ book for changing ‘every aspect of the economy’ (long-time Cult agenda) and for humans to eat synthetic ‘meat’ (predicted in my books) while cows and other farm animals are eliminated. Australian TV host and commentator Alan Jones described what carbon emission targets would mean for farm animals in Australia alone if emissions were reduced as demanded by 35 percent by 2030 and zero by 2050: Well, let’s take agriculture, the total emissions from agriculture are about 75 million tonnes of carbon dioxide, equivalent. Now reduce that by 35 percent and you have to come down to 50 million tonnes, I’ve done the maths. So if you take for example 1.5 million cows, you’re going to have to reduce the herd by 525,000 [by] 2030, nine years, that’s 58,000 cows a year. The beef herd’s 30 million, reduce that by 35 percent, that’s 10.5 million, which means 1.2 million cattle have to go every year between now and 2030. This is insanity! There are 75 million sheep. Reduce that by 35 percent, that’s 26 million sheep, that’s almost 3 million a year. So under the Paris Agreement over 30 million beasts. dairy cows, cattle, pigs and sheep would go. More than 8,000 every minute of every hour for the next decade, do these people know what they’re talking about?

Clearly they don’t at the level of campaigners, politicians and administrators. The Cult does know; that’s the outcome it wants. We are faced with not just a war on humanity. Animals and the natural world are being targeted and I have been saying since the ‘Covid’ hoax began that the plan eventually was to claim that the ‘deadly virus’ is able to jump from animals, including farm animals and

domestic pets, to humans. Just before this book went into production came this story: ‘Russia registers world’s first Covid-19 vaccine for cats & dogs as makers of Sputnik V warn pets & farm animals could spread virus’. The report said ‘top scientists warned that the deadly pathogen could soon begin spreading through homes and farms’ and ‘the next stage is the infection of farm and domestic animals’. Know the outcome and you’ll see the journey. Think what that would mean for animals and keep your eye on a term called zoonosis or zoonotic diseases which transmit between animals and humans. The Cult wants to break the connection between animals and people as it does between people and people. Farm animals fit with the Cult agenda to transform food from natural to synthetic.

The gas of life is killing us

There can be few greater examples of Cult inversion than the condemnation of carbon dioxide as a dangerous pollutant when it is the gas of life. Without it the natural world would be dead and so we would all be dead. We breathe in oxygen and breathe out carbon dioxide while plants produce oxygen and absorb carbon dioxide. It is a perfect symbiotic relationship that the Cult wants to dismantle for reasons I will come to in the final two chapters. Gates, Schwab, other Cult operatives and mindless repeaters, want the world to be ‘carbon neutral’ by at least 2050 and the earlier the be er. ‘Zero carbon’ is the cry echoed by lunatics calling for ‘Zero Covid’ when we already have it. These carbon emission targets will deindustrialise the world in accordance with Cult plans – the postindustrial, post-democratic society – and with so-called renewables like solar and wind not coming even close to meeting human energy needs blackouts and cold are inevitable. Texans got the picture in the winter of 2021 when a snow storm stopped wind turbines and solar panels from working and the lights went down along with water which relies on electricity for its supply system. Gates wants everything to be powered by electricity to ensure that his masters have the kill switch to stop all human activity, movement, cooking, water and warmth any time they like. The climate lie is so

stupendously inverted that it claims we must urgently reduce carbon dioxide when we don’t have enough. Co2 in the atmosphere is a li le above 400 parts per million when the optimum for plant growth is 2,000 ppm and when it falls anywhere near 150 ppm the natural world starts to die and so do we. It fell to as low as 280 ppm in an 1880 measurement in Hawaii and rose to 413 ppm in 2019 with industrialisation which is why the planet has become greener in the industrial period. How insane then that psychopathic madman Gates is not satisfied only with blocking the rise of Co2. He’s funding technology to suck it out of the atmosphere. The reason why will become clear. The industrial era is not destroying the world through Co2 and has instead turned around a potentially disastrous ongoing fall in Co2. Greenpeace cofounder and scientist Patrick Moore walked away from Greenpeace in 1986 and has exposed the green movement for fear-mongering and lies. He said that 500 million years ago there was 17 times more Co2 in the atmosphere than we have today and levels have been falling for hundreds of millions of years. In the last 150 million years Co2 levels in Earth’s atmosphere had reduced by 90 percent. Moore said that by the time humanity began to unlock carbon dioxide from fossil fuels we were at ‘38 seconds to midnight’ and in that sense: ‘Humans are [the Earth’s] salvation.’ Moore made the point that only half the Co2 emi ed by fossil fuels stays in the atmosphere and we should remember that all pollution pouring from chimneys that we are told is carbon dioxide is in fact nothing of the kind. It’s pollution. Carbon dioxide is an invisible gas. William Happer, Professor of Physics at Princeton University and long-time government adviser on climate, has emphasised the Co2 deficiency for maximum growth and food production. Greenhouse growers don’t add carbon dioxide for a bit of fun. He said that most of the warming in the last 100 years, a er the earth emerged from the super-cold period of the ‘Li le Ice Age’ into a natural warming cycle, was over by 1940. Happer said that a peak year for warming in 1988 can be explained by a ‘monster El Nino’ which is a natural and cyclical warming of the Pacific that has nothing to do with ‘climate

change’. He said the effect of Co2 could be compared to painting a wall with red paint in that once two or three coats have been applied it didn’t ma er how much more you slapped on because the wall will not get much redder. Almost all the effect of the rise in Co2 has already happened, he said, and the volume in the atmosphere would now have to double to increase temperature by a single degree. Climate hoaxers know this and they have invented the most ridiculously complicated series of ‘feedback’ loops to try to overcome this rather devastating fact. You hear puppet Greta going on cluelessly about feedback loops and this is why.

The Sun affects temperature? No you

climate denier

Some other nonsense to contemplate: Climate graphs show that rises in temperature do not follow rises in Co2 – it’s the other way round with a lag between the two of some 800 years. If we go back 800 years from present time we hit the Medieval Warm Period when temperatures were higher than now without any industrialisation and this was followed by the Li le Ice Age when temperatures plummeted. The world was still emerging from these centuries of serious cold when many climate records began which makes the ever-repeated line of the ‘ho est year since records began’ meaningless when you are not comparing like with like. The coldest period of the Li le Ice Age corresponded with the lowest period of sunspot activity when the Sun was at its least active. Proper scientists will not be at all surprised by this when it confirms the obvious fact that earth temperature is affected by the scale of Sun activity and the energetic power that it subsequently emits; but when is the last time you heard a climate hoaxer talking about the Sun as a source of earth temperature?? Everything has to be focussed on Co2 which makes up just 0.117 percent of so-called greenhouse gases and only a fraction of even that is generated by human activity. The rest is natural. More than 90 percent of those greenhouse gases are water vapour and clouds (Fig 9). Ban moisture I say. Have you noticed that the climate hoaxers no longer use the polar bear as their promotion image? That’s because far from becoming extinct polar

bear communities are stable or thriving. Joe Bastardi, American meteorologist, weather forecaster and outspoken critic of the climate lie, documents in his book The Climate Chronicles how weather pa erns and events claimed to be evidence of climate change have been happening since long before industrialisation: ‘What happened before naturally is happening again, as is to be expected given the cyclical nature of the climate due to the design of the planet.’ If you read the detailed background to the climate hoax in my other books you will shake your head and wonder how anyone could believe the crap which has spawned a multi-trillion dollar industry based on absolute garbage (see HIV causes AIDs and Sars-Cov-2 causes ‘Covid-19’). Climate and ‘Covid’ have much in common given they have the same source. They both have the contradictory everything factor in which everything is explained by reference to them. It’s hot – ‘it’s climate change’. It’s cold – ‘it’s climate change’. I got a sniffle – ‘it’s Covid’. I haven’t got a sniffle – ‘it’s Covid’. Not having a sniffle has to be a symptom of ‘Covid’. Everything is and not having a sniffle is especially dangerous if you are a slow walker. For sheer audacity I offer you a Cambridge University ‘study’ that actually linked ‘Covid’ to ‘climate change’. It had to happen eventually. They concluded that climate change played a role in ‘Covid-19’ spreading from animals to humans because … wait for it … I kid you not … the two groups were forced closer together as populations grow. Er, that’s it. The whole foundation on which this depended was that ‘Bats are the likely zoonotic origin of SARS-CoV-1 and SARS-CoV-2’. Well, they are not. They are nothing to do with it. Apart from bats not being the origin and therefore ‘climate change’ effects on bats being irrelevant I am in awe of their academic insight. Where would we be without them? Not where we are that’s for sure.

Figure 9: The idea that the gas of life is disastrously changing the climate is an insult to brain cell activity.

One other point about the weather is that climate modification is now well advanced and not every major weather event is natural – or earthquake come to that. I cover this subject at some length in other books. China is openly planning a rapid expansion of its weather modification programme which includes changing the climate in an area more than one and a half times the size of India. China used weather manipulation to ensure clear skies during the 2008 Olympics in Beijing. I have quoted from US military documents detailing how to employ weather manipulation as a weapon of war and they did that in the 1960s and 70s during the conflict in Vietnam with Operation Popeye manipulating monsoon rains for military purposes. Why would there be international treaties on weather modification if it wasn’t possible? Of course it is. Weather is energetic information and it can be changed.

How was the climate hoax pulled off? See ‘Covid’

If you can get billions to believe in a ‘virus’ that doesn’t exist you can get them to believe in human-caused climate change that doesn’t exist. Both are being used by the Cult to transform global society in the way it has long planned. Both hoaxes have been achieved in pre y much the same way. First you declare a lie is a fact. There’s a

‘virus’ you call SARS-Cov-2 or humans are warming the planet with their behaviour. Next this becomes, via Cult networks, the foundation of government, academic and science policy and belief. Those who parrot the mantra are given big grants to produce research that confirms the narrative is true and ever more ‘symptoms’ are added to make the ‘virus’/’climate change’ sound even more scary. Scientists and researchers who challenge the narrative have their grants withdrawn and their careers destroyed. The media promote the lie as the unquestionable truth and censor those with an alternative view or evidence. A great percentage of the population believe what they are told as the lie becomes an everybody-knows-that and the believing-masses turn on those with a mind of their own. The technique has been used endlessly throughout human history. Wokers are the biggest promotors of the climate lie and ‘Covid’ fascism because their minds are owned by the Cult; their sense of self-righteous self-purity knows no bounds; and they exist in a bubble of reality in which facts are irrelevant and only get in the way of looking without seeing. Running through all of this like veins in a blue cheese is control of information, which means control of perception, which means control of behaviour, which collectively means control of human society. The Cult owns the global media and Silicon Valley fascists for the simple reason that it has to. Without control of information it can’t control perception and through that human society. Examine every facet of the Cult agenda and you will see that anything supporting its introduction is never censored while anything pushing back is always censored. I say again: Psychopaths that know why they are doing this must go before Nuremberg trials and those that follow their orders must trot along behind them into the same dock. ‘I was just following orders’ didn’t work the first time and it must not work now. Nuremberg trials must be held all over the world before public juries for politicians, government officials, police, compliant doctors, scientists and virologists, and all Cult operatives such as Gates, Tedros, Fauci, Vallance, Whi y, Ferguson, Zuckerberg, Wojcicki, Brin, Page, Dorsey, the whole damn lot of

them – including, no especially, the psychopath psychologists. Without them and the brainless, gutless excuses for journalists that have repeated their lies, none of this could be happening. Nobody can be allowed to escape justice for the psychological and economic Armageddon they are all responsible for visiting upon the human race. As for the compliant, unquestioning, swathes of humanity, and the self-obsessed, all-knowing ignorance of the Wokers … don’t start me. God help their kids. God help their grandkids. God help them.

CHAPTER NINE We must have it? So what is it? Well I won’t back down. No, I won’t back down. You can stand me up at the Gates of Hell. But I won’t back down Tom Petty

I

will now focus on the genetically-manipulating ‘Covid vaccines’ which do not meet this official definition of a vaccine by the US Centers for Disease Control (CDC): ‘A product that stimulates a person’s immune system to produce immunity to a specific disease, protecting the person from that disease.’ On that basis ‘Covid vaccines’ are not a vaccine in that the makers don’t even claim they stop infection or transmission. They are instead part of a multi-levelled conspiracy to change the nature of the human body and what it means to be ‘human’ and to depopulate an enormous swathe of humanity. What I shall call Human 1.0 is on the cusp of becoming Human 2.0 and for very sinister reasons. Before I get to the ‘Covid vaccine’ in detail here’s some background to vaccines in general. Government regulators do not test vaccines – the makers do – and the makers control which data is revealed and which isn’t. Children in America are given 50 vaccine doses by age six and 69 by age 19 and the effect of the whole combined schedule has never been tested. Autoimmune diseases when the immune system a acks its own body have soared in the mass vaccine era and so has disease in general in children and the young. Why wouldn’t this be the case when vaccines target the immune system? The US government gave Big Pharma drug

companies immunity from prosecution for vaccine death and injury in the 1986 National Childhood Vaccine Injury Act (NCVIA) and since then the government (taxpayer) has been funding compensation for the consequences of Big Pharma vaccines. The criminal and satanic drug giants can’t lose and the vaccine schedule has increased dramatically since 1986 for this reason. There is no incentive to make vaccines safe and a big incentive to make money by introducing ever more. Even against a ridiculously high bar to prove vaccine liability, and with the government controlling the hearing in which it is being challenged for compensation, the vaccine court has so far paid out more than $4 billion. These are the vaccines we are told are safe and psychopaths like Zuckerberg censor posts saying otherwise. The immunity law was even justified by a ruling that vaccines by their nature were ‘unavoidably unsafe’. Check out the ingredients of vaccines and you will be shocked if you are new to this. They put that in children’s bodies?? What?? Try aluminium, a brain toxin connected to dementia, aborted foetal tissue and formaldehyde which is used to embalm corpses. Worldrenowned aluminium expert Christopher Exley had his research into the health effect of aluminium in vaccines shut down by Keele University in the UK when it began taking funding from the Bill and Melinda Gates Foundation. Research when diseases ‘eradicated’ by vaccines began to decline and you will find the fall began long before the vaccine was introduced. Sometimes the fall even plateaued a er the vaccine. Diseases like scarlet fever for which there was no vaccine declined in the same way because of environmental and other factors. A perfect case in point is the polio vaccine. Polio began when lead arsenate was first sprayed as an insecticide and residues remained in food products. Spraying started in 1892 and the first US polio epidemic came in Vermont in 1894. The simple answer was to stop spraying, but Rockefeller-created Big Pharma had a be er idea. Polio was decreed to be caused by the poliovirus which ‘spreads from person to person and can infect a person’s spinal cord’. Lead arsenate was replaced by the lethal DDT which had the same effect of causing paralysis by damaging the brain and central nervous

system. Polio plummeted when DDT was reduced and then banned, but the vaccine is still given the credit for something it didn’t do. Today by far the biggest cause of polio is the vaccines promoted by Bill Gates. Vaccine justice campaigner Robert Kennedy Jr, son of assassinated (by the Cult) US A orney General Robert Kennedy, wrote: In 2017, the World Health Organization (WHO) reluctantly admitted that the global explosion in polio is predominantly vaccine strain. The most frightening epidemics in Congo, Afghanistan, and the Philippines, are all linked to vaccines. In fact, by 2018, 70% of global polio cases were vaccine strain.

Vaccines make fortunes for Cult-owned Gates and Big Pharma while undermining the health and immune systems of the population. We had a glimpse of the mentality behind the Big Pharma cartel with a report on WION (World is One News), an international English language TV station based in India, which exposed the extraordinary behaviour of US drug company Pfizer over its ‘Covid vaccine’. The WION report told how Pfizer had made fantastic demands of Argentina, Brazil and other countries in return for its ‘vaccine’. These included immunity from prosecution, even for Pfizer negligence, government insurance to protect Pfizer from law suits and handing over as collateral sovereign assets of the country to include Argentina’s bank reserves, military bases and embassy buildings. Pfizer demanded the same of Brazil in the form of waiving sovereignty of its assets abroad; exempting Pfizer from Brazilian laws; and giving Pfizer immunity from all civil liability. This is a ‘vaccine’ developed with government funding. Big Pharma is evil incarnate as a creation of the Cult and all must be handed tickets to Nuremberg.

Phantom ‘vaccine’ for a phantom ‘disease’

I’ll expose the ‘Covid vaccine’ fraud and then go on to the wider background of why the Cult has set out to ‘vaccinate’ every man, woman and child on the planet for an alleged ‘new disease’ with a survival rate of 99.77 percent (or more) even by the grotesquely-

manipulated figures of the World Health Organization and Johns Hopkins University. The ‘infection’ to ‘death’ ratio is 0.23 to 0.15 percent according to Stanford epidemiologist Dr John Ioannidis and while estimates vary the danger remains tiny. I say that if the truth be told the fake infection to fake death ratio is zero. Never mind all the evidence I have presented here and in The Answer that there is no ‘virus’ let us just focus for a moment on that death-rate figure of say 0.23 percent. The figure includes all those worldwide who have tested positive with a test not testing for the ‘virus’ and then died within 28 days or even longer of any other cause – any other cause. Now subtract all those illusory ‘Covid’ deaths on the global data sheets from the 0.23 percent. What do you think you would be le with? Zero. A vaccination has never been successfully developed for a so-called coronavirus. They have all failed at the animal testing stage when they caused hypersensitivity to what they were claiming to protect against and made the impact of a disease far worse. Cultowned vaccine corporations got around that problem this time by bypassing animal trials, going straight to humans and making the length of the ‘trials’ before the public rollout as short as they could get away with. Normally it takes five to ten years or more to develop vaccines that still cause demonstrable harm to many people and that’s without including the long-term effects that are never officially connected to the vaccination. ‘Covid’ non-vaccines have been officially produced and approved in a ma er of months from a standing start and part of the reason is that (a) they were developed before the ‘Covid’ hoax began and (b) they are based on computer programs and not natural sources. Official non-trials were so short that government agencies gave emergency, not full, approval. ‘Trials’ were not even completed and full approval cannot be secured until they are. Public ‘Covid vaccination’ is actually a continuation of the trial. Drug company ‘trials’ are not scheduled to end until 2023 by which time a lot of people are going to be dead. Data on which government agencies gave this emergency approval was supplied by the Big Pharma corporations themselves in the form of Pfizer/BioNTech, AstraZeneca, Moderna, Johnson & Johnson, and

others, and this is the case with all vaccines. By its very nature emergency approval means drug companies do not have to prove that the ‘vaccine’ is ‘safe and effective’. How could they with trials way short of complete? Government regulators only have to believe that they could be safe and effective. It is criminal manipulation to get products in circulation with no testing worth the name. Agencies giving that approval are infested with Big Pharma-connected placepeople and they act in the interests of Big Pharma (the Cult) and not the public about whom they do not give a damn.

More human lab rats

‘Covid vaccines’ produced in record time by Pfizer/BioNTech and Moderna employ a technique never approved before for use on humans. They are known as mRNA ‘vaccines’ and inject a synthetic version of ‘viral’ mRNA or ‘messenger RNA’. The key is in the term ‘messenger’. The body works, or doesn’t, on the basis of information messaging. Communications are constantly passing between and within the genetic system and the brain. Change those messages and you change the state of the body and even its very nature and you can change psychology and behaviour by the way the brain processes information. I think you are going to see significant changes in personality and perception of many people who have had the ‘Covid vaccine’ synthetic potions. Insider Aldous Huxley predicted the following in 1961 and mRNA ‘vaccines’ can be included in the term ‘pharmacological methods’: There will be, in the next generation or so, a pharmacological method of making people love their servitude, and producing dictatorship without tears, so to speak, producing a kind of painless concentration camp for entire societies, so that people will in fact have their own liberties taken away from them, but rather enjoy it, because they will be distracted from any desire to rebel by propaganda or brainwashing, or brainwashing enhanced by pharmacological methods. And this seems to be the final revolution.

Apologists claim that mRNA synthetic ‘vaccines’ don’t change the DNA genetic blueprint because RNA does not affect DNA only the other way round. This is so disingenuous. A process called ‘reverse

transcription’ can convert RNA into DNA and be integrated into DNA in the cell nucleus. This was highlighted in December, 2020, by scientists at Harvard and Massachuse s Institute of Technology (MIT). Geneticists report that more than 40 percent of mammalian genomes results from reverse transcription. On the most basic level if messaging changes then that sequence must lead to changes in DNA which is receiving and transmi ing those communications. How can introducing synthetic material into cells not change the cells where DNA is located? The process is known as transfection which is defined as ‘a technique to insert foreign nucleic acid (DNA or RNA) into a cell, typically with the intention of altering the properties of the cell’. Researchers at the Sloan Ke ering Institute in New York found that changes in messenger RNA can deactivate tumour-suppressing proteins and thereby promote cancer. This is what happens when you mess with messaging. ‘Covid vaccine’ maker Moderna was founded in 2010 by Canadian stem cell biologist Derrick J. Rossi a er his breakthrough discovery in the field of transforming and reprogramming stem cells. These are neutral cells that can be programmed to become any cell including sperm cells. Moderna was therefore founded on the principle of genetic manipulation and has never produced any vaccine or drug before its genetically-manipulating synthetic ‘Covid’ shite. Look at the name – Mode-RNA or Modify-RNA. Another important point is that the US Supreme Court has ruled that genetically-modified DNA, or complementary DNA (cDNA) synthesized in the laboratory from messenger RNA, can be patented and owned. These psychopaths are doing this to the human body. Cells replicate synthetic mRNA in the ‘Covid vaccines’ and in theory the body is tricked into making antigens which trigger antibodies to target the ‘virus spike proteins’ which as Dr Tom Cowan said have never been seen. Cut the crap and these ‘vaccines’ deliver self-replicating synthetic material to the cells with the effect of changing human DNA. The more of them you have the more that process is compounded while synthetic material is all the time selfreplicating. ‘Vaccine’-maker Moderna describes mRNA as ‘like

so ware for the cell’ and so they are messing with the body’s so ware. What happens when you change the so ware in a computer? Everything changes. For this reason the Cult is preparing a production line of mRNA ‘Covid vaccines’ and a long list of excuses to use them as with all the ‘variants’ of a ‘virus’ never shown to exist. The plan is further to transfer the mRNA technique to other vaccines mostly given to children and young people. The cumulative consequences will be a transformation of human DNA through a constant infusion of synthetic genetic material which will kill many and change the rest. Now consider that governments that have given emergency approval for a vaccine that’s not a vaccine; never been approved for humans before; had no testing worth the name; and the makers have been given immunity from prosecution for any deaths or adverse effects suffered by the public. The UK government awarded permanent legal indemnity to itself and its employees for harm done when a patient is being treated for ‘Covid-19’ or ‘suspected Covid-19’. That is quite a thought when these are possible ‘side-effects’ from the ‘vaccine’ (they are not ‘side’, they are effects) listed by the US Food and Drug Administration: Guillain-Barre syndrome; acute disseminated encephalomyelitis; transverse myelitis; encephalitis; myelitis; encephalomyelitis; meningoencephalitis; meningitis; encephalopathy; convulsions; seizures; stroke; narcolepsy; cataplexy; anaphylaxis; acute myocardial infarction (heart a ack); myocarditis; pericarditis; autoimmune disease; death; implications for pregnancy, and birth outcomes; other acute demyelinating diseases; non anaphylactic allergy reactions; thrombocytopenia ; disseminated intravascular coagulation; venous thromboembolism; arthritis; arthralgia; joint pain; Kawasaki disease; multisystem inflammatory syndrome in children; vaccine enhanced disease. The la er is the way the ‘vaccine’ has the potential to make diseases far worse than they would otherwise be.

UK doctor and freedom campaigner Vernon Coleman described the conditions in this list as ‘all unpleasant, most of them very serious, and you can’t get more serious than death’. The thought that anyone at all has had the ‘vaccine’ in these circumstances is testament to the potential that humanity has for clueless, unquestioning, stupidity and for many that programmed stupidity has already been terminal.

An insider speaks

Dr Michael Yeadon is a former Vice President, head of research and Chief Scientific Adviser at vaccine giant Pfizer. Yeadon worked on the inside of Big Pharma, but that did not stop him becoming a vocal critic of ‘Covid vaccines’ and their potential for multiple harms, including infertility in women. By the spring of 2021 he went much further and even used the no, no, term ‘conspiracy’. When you begin to see what is going on it is impossible not to do so. Yeadon spoke out in an interview with freedom campaigner James Delingpole and I mentioned earlier how he said that no one had samples of ‘the virus’. He explained that the mRNA technique originated in the anticancer field and ways to turn on and off certain genes which could be advantageous if you wanted to stop cancer growing out of control. ‘That’s the origin of them. They are a very unusual application, really.’ Yeadon said that treating a cancer patient with an aggressive procedure might be understandable if the alternative was dying, but it was quite another thing to use the same technique as a public health measure. Most people involved wouldn’t catch the infectious agent you were vaccinating against and if they did they probably wouldn’t die: If you are really using it as a public health measure you really want to as close as you can get to zero sides-effects … I find it odd that they chose techniques that were really cutting their teeth in the field of oncology and I’m worried that in using gene-based vaccines that have to be injected in the body and spread around the body, get taken up into some cells, and the regulators haven’t quite told us which cells they get taken up into … you are going to be generating a wide range of responses … with multiple steps each of which could go well or badly.

I doubt the Cult intends it to go well. Yeadon said that you can put any gene you like into the body through the ‘vaccine’. ‘You can certainly give them a gene that would do them some harm if you wanted.’ I was intrigued when he said that when used in the cancer field the technique could turn genes on and off. I explore this process in The Answer and with different genes having different functions you could create mayhem – physically and psychologically – if you turned the wrong ones on and the right ones off. I read reports of an experiment by researchers at the University of Washington’s school of computer science and engineering in which they encoded DNA to infect computers. The body is itself a biological computer and if human DNA can inflict damage on a computer why can’t the computer via synthetic material mess with the human body? It can. The Washington research team said it was possible to insert malicious malware into ‘physical DNA strands’ and corrupt the computer system of a gene sequencing machine as it ‘reads gene le ers and stores them as binary digits 0 and 1’. They concluded that hackers could one day use blood or spit samples to access computer systems and obtain sensitive data from police forensics labs or infect genome files. It is at this level of digital interaction that synthetic ‘vaccines’ need to be seen to get the full picture and that will become very clear later on. Michael Yeadon said it made no sense to give the ‘vaccine’ to younger people who were in no danger from the ‘virus’. What was the benefit? It was all downside with potential effects: The fact that my government in what I thought was a civilised, rational country, is raining [the ‘vaccine’] on people in their 30s and 40s, even my children in their 20s, they’re getting letters and phone calls, I know this is not right and any of you doctors who are vaccinating you know it’s not right, too. They are not at risk. They are not at risk from the disease, so you are now hoping that the side-effects are so rare that you get away with it. You don’t give new technology … that you don’t understand to 100 percent of the population.

Blood clot problems with the AstraZeneca ‘vaccine’ have been affecting younger people to emphasise the downside risks with no benefit. AstraZeneca’s version, produced with Oxford University, does not use mRNA, but still gets its toxic cocktail inside cells where

it targets DNA. The Johnson & Johnson ‘vaccine’ which uses a similar technique has also produced blood clot effects to such an extent that the United States paused its use at one point. They are all ‘gene therapy’ (cell modification) procedures and not ‘vaccines’. The truth is that once the content of these injections enter cells we have no idea what the effect will be. People can speculate and some can give very educated opinions and that’s good. In the end, though, only the makers know what their potions are designed to do and even they won’t know every last consequence. Michael Yeadon was scathing about doctors doing what they knew to be wrong. ‘Everyone’s mute’, he said. Doctors in the NHS must know this was not right, coming into work and injecting people. ‘I don’t know how they sleep at night. I know I couldn’t do it. I know that if I were in that position I’d have to quit.’ He said he knew enough about toxicology to know this was not a good risk-benefit. Yeadon had spoken to seven or eight university professors and all except two would not speak out publicly. Their universities had a policy that no one said anything that countered the government and its medical advisors. They were afraid of losing their government grants. This is how intimidation has been used to silence the truth at every level of the system. I say silence, but these people could still speak out if they made that choice. Yeadon called them ‘moral cowards’ – ‘This is about your children and grandchildren’s lives and you have just buggered off and le it.’

‘Variant’ nonsense

Some of his most powerful comments related to the alleged ‘variants’ being used to instil more fear, justify more lockdowns, and introduce more ‘vaccines’. He said government claims about ‘variants’ were nonsense. He had checked the alleged variant ‘codes’ and they were 99.7 percent identical to the ‘original’. This was the human identity difference equivalent to pu ing a baseball cap on and off or wearing it the other way round. A 0.3 percent difference would make it impossible for that ‘variant’ to escape immunity from the ‘original’. This made no sense of having new ‘vaccines’ for

‘variants’. He said there would have to be at least a 30 percent difference for that to be justified and even then he believed the immune system would still recognise what it was. Gates-funded ‘variant modeller’ and ‘vaccine’-pusher John Edmunds might care to comment. Yeadon said drug companies were making new versions of the ‘vaccine’ as a ‘top up’ for ‘variants’. Worse than that, he said, the ‘regulators’ around the world like the MHRA in the UK had got together and agreed that because ‘vaccines’ for ‘variants’ were so similar to the first ‘vaccines’ they did not have to do safety studies. How transparently sinister that is. This is when Yeadon said: ‘There is a conspiracy here.’ There was no need for another vaccine for ‘variants’ and yet we were told that there was and the country had shut its borders because of them. ‘They are going into hundreds of millions of arms without passing ‘go’ or any regulator. Why did they do that? Why did they pick this method of making the vaccine?’ The reason had to be something bigger than that it seemed and ‘it’s not protection against the virus’. It’s was a far bigger project that meant politicians and advisers were willing to do things and not do things that knowingly resulted in avoidable deaths – ‘that’s already happened when you think about lockdown and deprivation of health care for a year.’ He spoke of people prepared to do something that results in the avoidable death of their fellow human beings and it not bother them. This is the penny-drop I have been working to get across for more than 30 years – the level of pure evil we are dealing with. Yeadon said his friends and associates could not believe there could be that much evil, but he reminded them of Stalin, Pol Pot and Hitler and of what Stalin had said: ‘One death is a tragedy. A million? A statistic.’ He could not think of a benign explanation for why you need top-up vaccines ‘which I’m sure you don’t’ and for the regulators ‘to just get out of the way and wave them through’. Why would the regulators do that when they were still wrestling with the dangers of the ‘parent’ vaccine? He was clearly shocked by what he had seen since the ‘Covid’ hoax began and now he was thinking the previously unthinkable:

If you wanted to depopulate a significant proportion of the world and to do it in a way that doesn’t involve destruction of the environment with nuclear weapons, poisoning everyone with anthrax or something like that, and you wanted plausible deniability while you had a multi-year infectious disease crisis, I actually don’t think you could come up with a better plan of work than seems to be in front of me. I can’t say that’s what they are going to do, but I can’t think of a benign explanation why they are doing it.

He said he never thought that they would get rid of 99 percent of humans, but now he wondered. ‘If you wanted to that this would be a hell of a way to do it – it would be unstoppable folks.’ Yeadon had concluded that those who submi ed to the ‘vaccine’ would be allowed to have some kind of normal life (but for how long?) while screws were tightened to coerce and mandate the last few percent. ‘I think they’ll put the rest of them in a prison camp. I wish I was wrong, but I don’t think I am.’ Other points he made included: There were no coronavirus vaccines then suddenly they all come along at the same time; we have no idea of the long term affect with trials so short; coercing or forcing people to have medical procedures is against the Nuremberg Code instigated when the Nazis did just that; people should at least delay having the ‘vaccine’; a quick Internet search confirms that masks don’t reduce respiratory viral transmission and ‘the government knows that’; they have smashed civil society and they know that, too; two dozen peer-reviewed studies show no connection between lockdown and reducing deaths; he knew from personal friends the elite were still flying around and going on holiday while the public were locked down; the elite were not having the ‘vaccines’. He was also asked if ‘vaccines’ could be made to target difference races. He said he didn’t know, but the document by the Project for the New American Century in September, 2000, said developing ‘advanced forms of biological warfare that can target specific genotypes may transform biological warfare from the realm of terror to a politically useful tool.’ Oh, they’re evil all right. Of that we can be absolutely sure.

Another cull of old people

We have seen from the CDC definition that the mRNA ‘Covid vaccine’ is not a vaccine and nor are the others that claim to reduce ‘severity of symptoms’ in some people, but not protect from infection or transmission. What about all the lies about returning to ‘normal’ if people were ‘vaccinated’? If they are not claimed to stop infection and transmission of the alleged ‘virus’, how does anything change? This was all lies to manipulate people to take the jabs and we are seeing that now with masks and distancing still required for the ‘vaccinated’. How did they think that elderly people with fragile health and immune responses were going to be affected by infusing their cells with synthetic material and other toxic substances? They knew that in the short and long term it would be devastating and fatal as the culling of the old that began with the first lockdowns was continued with the ‘vaccine’. Death rates in care homes soared immediately residents began to be ‘vaccinated’ – infused with synthetic material. Brave and commi ed whistleblower nurses put their careers at risk by exposing this truth while the rest kept their heads down and their mouths shut to put their careers before those they are supposed to care for. A long-time American Certified Nursing Assistant who gave his name as James posted a video in which he described emotionally what happened in his care home when vaccination began. He said that during 2020 very few residents were sick with ‘Covid’ and no one died during the entire year; but shortly a er the Pfizer mRNA injections 14 people died within two weeks and many others were near death. ‘They’re dropping like flies’, he said. Residents who walked on their own before the shot could no longer and they had lost their ability to conduct an intelligent conversation. The home’s management said the sudden deaths were caused by a ‘super-spreader’ of ‘Covid-19’. Then how come, James asked, that residents who refused to take the injections were not sick? It was a case of inject the elderly with mRNA synthetic potions and blame their illness and death that followed on the ‘virus’. James described what was happening in care homes as ‘the greatest crime of genocide this country has ever seen’. Remember the NHS staff nurse from earlier who used the same

word ‘genocide’ for what was happening with the ‘vaccines’ and that it was an ‘act of human annihilation’. A UK care home whistleblower told a similar story to James about the effect of the ‘vaccine’ in deaths and ‘outbreaks’ of illness dubbed ‘Covid’ a er ge ing the jab. She told how her care home management and staff had zealously imposed government regulations and no one was allowed to even question the official narrative let alone speak out against it. She said the NHS was even worse. Again we see the results of reframing. A worker at a local care home where I live said they had not had a single case of ‘Covid’ there for almost a year and when the residents were ‘vaccinated’ they had 19 positive cases in two weeks with eight dying.

It’s not the ‘vaccine’ – honest

The obvious cause and effect was being ignored by the media and most of the public. Australia’s health minister Greg Hunt (a former head of strategy at the World Economic Forum) was admi ed to hospital a er he had the ‘vaccine’. He was suffering according to reports from the skin infection ‘cellulitis’ and it must have been a severe case to have warranted days in hospital. Immediately the authorities said this was nothing to do with the ‘vaccine’ when an effect of some vaccines is a ‘cellulitis-like reaction’. We had families of perfectly healthy old people who died a er the ‘vaccine’ saying that if only they had been given the ‘vaccine’ earlier they would still be alive. As a numbskull rating that is off the chart. A father of four ‘died of Covid’ at aged 48 when he was taken ill two days a er having the ‘vaccine’. The man, a health administrator, had been ‘shielding during the pandemic’ and had ‘not really le the house’ until he went for the ‘vaccine’. Having the ‘vaccine’ and then falling ill and dying does not seem to have qualified as a possible cause and effect and ‘Covid-19’ went on his death certificate. His family said they had no idea how he ‘caught the virus’. A family member said: ‘Tragically, it could be that going for a vaccination ultimately led to him catching Covid …The sad truth is that they are never going to know where it came from.’ The family warned people to remember

that the virus still existed and was ‘very real’. So was their stupidity. Nurses and doctors who had the first round of the ‘vaccine’ were collapsing, dying and ending up in a hospital bed while they or their grieving relatives were saying they’d still have the ‘vaccine’ again despite what happened. I kid you not. You mean if your husband returned from the dead he’d have the same ‘vaccine’ again that killed him?? Doctors at the VCU Medical Center in Richmond, Virginia, said the Johnson & Johnson ‘vaccine’ was to blame for a man’s skin peeling off. Patient Richard Terrell said: ‘It all just happened so fast. My skin peeled off. It’s still coming off on my hands now.’ He said it was stinging, burning and itching and when he bent his arms and legs it was very painful with ‘the skin swollen and rubbing against itself’. Pfizer/BioNTech and Moderna vaccines use mRNA to change the cell while the Johnson & Johnson version uses DNA in a process similar to AstraZeneca’s technique. Johnson & Johnson and AstraZeneca have both had their ‘vaccines’ paused by many countries a er causing serious blood problems. Terrell’s doctor Fnu Nutan said he could have died if he hadn’t got medical a ention. It sounds terrible so what did Nutan and Terrell say about the ‘vaccine’ now? Oh, they still recommend that people have it. A nurse in a hospital bed 40 minutes a er the vaccination and unable to swallow due to throat swelling was told by a doctor that he lost mobility in his arm for 36 hours following the vaccination. What did he say to the ailing nurse? ‘Good for you for ge ing the vaccination.’ We are dealing with a serious form of cognitive dissonance madness in both public and medical staff. There is a remarkable correlation between those having the ‘vaccine’ and trumpeting the fact and suffering bad happenings shortly a erwards. Witold Rogiewicz, a Polish doctor, made a video of his ‘vaccination’ and ridiculed those who were questioning its safety and the intentions of Bill Gates: ‘Vaccinate yourself to protect yourself, your loved ones, friends and also patients. And to mention quickly I have info for anti-vaxxers and anti-Coviders if you want to contact Bill Gates you can do this through me.’ He further ridiculed the dangers of 5G. Days later he

was dead, but naturally the vaccination wasn’t mentioned in the verdict of ‘heart a ack’.

Lies, lies and more lies

So many members of the human race have slipped into extreme states of insanity and unfortunately they include reframed doctors and nursing staff. Having a ‘vaccine’ and dying within minutes or hours is not considered a valid connection while death from any cause within 28 days or longer of a positive test with a test not testing for the ‘virus’ means ‘Covid-19’ goes on the death certificate. How could that ‘vaccine’-death connection not have been made except by calculated deceit? US figures in the initial rollout period to February 12th, 2020, revealed that a third of the deaths reported to the CDC a er ‘Covid vaccines’ happened within 48 hours. Five men in the UK suffered an ‘extremely rare’ blood clot problem a er having the AstraZeneca ‘vaccine’, but no causal link was established said the Gates-funded Medicines and Healthcare products Regulatory Agency (MHRA) which had given the ‘vaccine’ emergency approval to be used. Former Pfizer executive Dr Michael Yeadon explained in his interview how the procedures could cause blood coagulation and clots. People who should have been at no risk were dying from blood clots in the brain and he said he had heard from medical doctor friends that people were suffering from skin bleeding and massive headaches. The AstraZeneca ‘shot’ was stopped by some 20 countries over the blood clo ing issue and still the corrupt MHRA, the European Medicines Agency (EMA) and the World Health Organization said that it should continue to be given even though the EMA admi ed that it ‘still cannot rule out definitively’ a link between blood clo ing and the ‘vaccine’. Later Marco Cavaleri, head of EMA vaccine strategy, said there was indeed a clear link between the ‘vaccine’ and thrombosis, but they didn’t know why. So much for the trials showing the ‘vaccine’ is safe. Blood clots were affecting younger people who would be under virtually no danger from ‘Covid’ even if it existed which makes it all the more stupid and sinister.

The British government responded to public alarm by wheeling out June Raine, the terrifyingly weak infant school headmistress sound-alike who heads the UK MHRA drug ‘regulator’. The idea that she would stand up to Big Pharma and government pressure is laughable and she told us that all was well in the same way that she did when allowing untested, never-used-on-humans-before, genetically-manipulating ‘vaccines’ to be exposed to the public in the first place. Mass lying is the new normal of the ‘Covid’ era. The MHRA later said 30 cases of rare blood clots had by then been connected with the AstraZeneca ‘vaccine’ (that means a lot more in reality) while stressing that the benefits of the jab in preventing ‘Covid-19’ outweighed any risks. A more ridiculous and disingenuous statement with callous disregard for human health it is hard to contemplate. Immediately a er the mendacious ‘all-clears’ two hospital workers in Denmark experienced blood clots and cerebral haemorrhaging following the AstraZeneca jab and one died. Top Norwegian health official Pål Andre Holme said the ‘vaccine’ was the only common factor: ‘There is nothing in the patient history of these individuals that can give such a powerful immune response … I am confident that the antibodies that we have found are the cause, and I see no other explanation than it being the vaccine which triggers it.’ Strokes, a clot or bleed in the brain, were clearly associated with the ‘vaccine’ from word of mouth and whistleblower reports. Similar consequences followed with all these ‘vaccines’ that we were told were so safe and as the numbers grew by the day it was clear we were witnessing human carnage.

Learning the hard way

A woman interviewed by UKColumn told how her husband suffered dramatic health effects a er the vaccine when he’d been in good health all his life. He went from being a li le unwell to losing all feeling in his legs and experiencing ‘excruciating pain’. Misdiagnosis followed twice at Accident and Emergency (an ‘allergy’ and ‘sciatica’) before he was admi ed to a neurology ward where doctors said his serious condition had been caused by the

‘vaccine’. Another seven ‘vaccinated’ people were apparently being treated on the same ward for similar symptoms. The woman said he had the ‘vaccine’ because they believed media claims that it was safe. ‘I didn’t think the government would give out a vaccine that does this to somebody; I believed they would be bringing out a vaccination that would be safe.’ What a tragic way to learn that lesson. Another woman posted that her husband was transporting stroke patients to hospital on almost every shi and when he asked them if they had been ‘vaccinated’ for ‘Covid’ they all replied ‘yes’. One had a ‘massive brain bleed’ the day a er his second dose. She said her husband reported the ‘just been vaccinated’ information every time to doctors in A and E only for them to ignore it, make no notes and appear annoyed that it was even mentioned. This particular report cannot be verified, but it expresses a common theme that confirms the monumental underreporting of ‘vaccine’ consequences. Interestingly as the ‘vaccines’ and their brain blood clot/stroke consequences began to emerge the UK National Health Service began a publicity campaign telling the public what to do in the event of a stroke. A Sco ish NHS staff nurse who quit in disgust in March, 2021, said: I have seen traumatic injuries from the vaccine, they’re not getting reported to the yellow card [adverse reaction] scheme, they’re treating the symptoms, not asking why, why it’s happening. It’s just treating the symptoms and when you speak about it you’re dismissed like you’re crazy, I’m not crazy, I’m not crazy because every other colleague I’ve spoken to is terrified to speak out, they’ve had enough.

Videos appeared on the Internet of people uncontrollably shaking a er the ‘vaccine’ with no control over muscles, limbs and even their face. A Sco ish mother broke out in a severe rash all over her body almost immediately a er she was given the AstraZeneca ‘vaccine’. The pictures were horrific. Leigh King, a 41-year-old hairdresser from Lanarkshire said: ‘Never in my life was I prepared for what I was about to experience … My skin was so sore and constantly hot … I have never felt pain like this …’ But don’t you worry, the ‘vaccine’ is perfectly safe. Then there has been the effect on medical

staff who have been pressured to have the ‘vaccine’ by psychopathic ‘health’ authorities and government. A London hospital consultant who gave the name K. Polyakova wrote this to the British Medical Journal or BMJ: I am currently struggling with … the failure to report the reality of the morbidity caused by our current vaccination program within the health service and staff population. The levels of sickness after vaccination is unprecedented and staff are getting very sick and some with neurological symptoms which is having a huge impact on the health service function. Even the young and healthy are off for days, some for weeks, and some requiring medical treatment. Whole teams are being taken out as they went to get vaccinated together. Mandatory vaccination in this instance is stupid, unethical and irresponsible when it comes to protecting our staff and public health. We are in the voluntary phase of vaccination, and encouraging staff to take an unlicensed product that is impacting on their immediate health … it is clearly stated that these vaccine products do not offer immunity or stop transmission. In which case why are we doing it?

Not to protect health that’s for sure. Medical workers are lauded by governments for agenda reasons when they couldn’t give a toss about them any more than they can for the population in general. Schools across America faced the same situation as they closed due to the high number of teachers and other staff with bad reactions to the Pfizer/BioNTech, Moderna, and Johnson & Johnson ‘Covid vaccines’ all of which were linked to death and serious adverse effects. The BMJ took down the consultant’s comments pre y quickly on the grounds that they were being used to spread ‘disinformation’. They were exposing the truth about the ‘vaccine’ was the real reason. The cover-up is breathtaking.

Hiding the evidence

The scale of the ‘vaccine’ death cover-up worldwide can be confirmed by comparing official figures with the personal experience of the public. I heard of many people in my community who died immediately or soon a er the vaccine that would never appear in the media or even likely on the official totals of ‘vaccine’ fatalities and adverse reactions when only about ten percent are estimated to be

reported and I have seen some estimates as low as one percent in a Harvard study. In the UK alone by April 29th, 2021, some 757,654 adverse reactions had been officially reported from the Pfizer/BioNTech, Oxford/AstraZeneca and Moderna ‘vaccines’ with more than a thousand deaths linked to jabs and that means an estimated ten times this number in reality from a ten percent reporting rate percentage. That’s seven million adverse reactions and 10,000 potential deaths and a one percent reporting rate would be ten times those figures. In 1976 the US government pulled the swine flu vaccine a er 53 deaths. The UK data included a combined 10,000 eye disorders from the ‘Covid vaccines’ with more than 750 suffering visual impairment or blindness and again multiply by the estimated reporting percentages. As ‘Covid cases’ officially fell hospitals virtually empty during the ‘Covid crisis’ began to fill up with a range of other problems in the wake of the ‘vaccine’ rollout. The numbers across America have also been catastrophic. Deaths linked to all types of vaccine increased by 6,000 percent in the first quarter of 2021 compared with 2020. A 39-year-old woman from Ogden, Utah, died four days a er receiving a second dose of Moderna’s ‘Covid vaccine’ when her liver, heart and kidneys all failed despite the fact that she had no known medical issues or conditions. Her family sought an autopsy, but Dr Erik Christensen, Utah’s chief medical examiner, said proving vaccine injury as a cause of death almost never happened. He could think of only one instance where an autopsy would name a vaccine as the official cause of death and that would be anaphylaxis where someone received a vaccine and died almost instantaneously. ‘Short of that, it would be difficult for us to definitively say this is the vaccine,’ Christensen said. If that is true this must be added to the estimated ten percent (or far less) reporting rate of vaccine deaths and serious reactions and the conclusion can only be that vaccine deaths and serious reactions – including these ‘Covid’ potions’ – are phenomenally understated in official figures. The same story can be found everywhere. Endless accounts of deaths and serious reactions among the public, medical

and care home staff while official figures did not even begin to reflect this. Professional script-reader Dr David Williams, a ‘top public-health official’ in Ontario, Canada, insulted our intelligence by claiming only four serious adverse reactions and no deaths from the more than 380,000 vaccine doses then given. This bore no resemblance to what people knew had happened in their owns circles and we had Dirk Huyer in charge of ge ing millions vaccinated in Ontario while at the same time he was Chief Coroner for the province investigating causes of death including possible death from the vaccine. An aide said he had stepped back from investigating deaths, but evidence indicated otherwise. Rosemary Frei, who secured a Master of Science degree in molecular biology at the Faculty of Medicine at Canada’s University of Calgary before turning to investigative journalism, was one who could see that official figures for ‘vaccine’ deaths and reactions made no sense. She said that doctors seldom reported adverse events and when people got really sick or died a er ge ing a vaccination they would a ribute that to anything except the vaccines. It had been that way for years and anyone who wondered aloud whether the ‘Covid vaccines’ or other shots cause harm is immediately branded as ‘anti-vax’ and ‘anti-science’. This was ‘career-threatening’ for health professionals. Then there was the huge pressure to support the push to ‘vaccinate’ billions in the quickest time possible. Frei said: So that’s where we’re at today. More than half a million vaccine doses have been given to people in Ontario alone. The rush is on to vaccinate all 15 million of us in the province by September. And the mainstream media are screaming for this to be sped up even more. That all adds up to only a very slim likelihood that we’re going to be told the truth by officials about how many people are getting sick or dying from the vaccines.

What is true of Ontario is true of everywhere.

They KNEW – and still did it

The authorities knew what was going to happen with multiple deaths and adverse reactions. The UK government’s Gates-funded

and Big Pharma-dominated Medicines and Healthcare products Regulatory Agency (MHRA) hired a company to employ AI in compiling the projected reactions to the ‘vaccine’ that would otherwise be uncountable. The request for applications said: ‘The MHRA urgently seeks an Artificial Intelligence (AI) so ware tool to process the expected high volume of Covid-19 vaccine Adverse Drug Reaction …’ This was from the agency, headed by the disingenuous June Raine, that gave the ‘vaccines’ emergency approval and the company was hired before the first shot was given. ‘We are going to kill and maim you – is that okay?’ ‘Oh, yes, perfectly fine – I’m very grateful, thank you, doctor.’ The range of ‘Covid vaccine’ adverse reactions goes on for page a er page in the MHRA criminally underreported ‘Yellow Card’ system and includes affects to eyes, ears, skin, digestion, blood and so on. Raine’s MHRA amazingly claimed that the ‘overall safety experience … is so far as expected from the clinical trials’. The death, serious adverse effects, deafness and blindness were expected? When did they ever mention that? If these human tragedies were expected then those that gave approval for the use of these ‘vaccines’ must be guilty of crimes against humanity including murder – a definition of which is ‘killing a person with malice aforethought or with recklessness manifesting extreme indifference to the value of human life.’ People involved at the MHRA, the CDC in America and their equivalent around the world must go before Nuremberg trials to answer for their callous inhumanity. We are only talking here about the immediate effects of the ‘vaccine’. The longer-term impact of the DNA synthetic manipulation is the main reason they are so hysterically desperate to inoculate the entire global population in the shortest possible time. Africa and the developing world are a major focus for the ‘vaccine’ depopulation agenda and a mass vaccination sales-pitch is underway thanks to caring people like the Rockefellers and other Cult assets. The Rockefeller Foundation, which pre-empted the ‘Covid pandemic’ in a document published in 2010 that ‘predicted’ what happened a decade later, announced an initial $34.95 million grant in February, 2021, ‘to ensure more equitable access to Covid-19

testing and vaccines’ among other things in Africa in collaboration with ‘24 organizations, businesses, and government agencies’. The pan-Africa initiative would focus on 10 countries: Burkina Faso, Ethiopia, Ghana, Kenya, Nigeria, Rwanda, South Africa, Tanzania, Uganda, and Zambia’. Rajiv Shah, President of the Rockefeller Foundation and former administrator of CIA-controlled USAID, said that if Africa was not mass-vaccinated (to change the DNA of its people) it was a ‘threat to all of humanity’ and not fair on Africans. When someone from the Rockefeller Foundation says they want to do something to help poor and deprived people and countries it is time for a belly-laugh. They are doing this out of the goodness of their ‘heart’ because ‘vaccinating’ the entire global population is what the ‘Covid’ hoax set out to achieve. Official ‘decolonisation’ of Africa by the Cult was merely a prelude to financial colonisation on the road to a return to physical colonisation. The ‘vaccine’ is vital to that and the sudden and convenient death of the ‘Covid’ sceptic president of Tanzania can be seen in its true light. A lot of people in Africa are aware that this is another form of colonisation and exploitation and they need to stand their ground.

The ‘vaccine is working’ scam

A potential problem for the Cult was that the ‘vaccine’ is meant to change human DNA and body messaging and not to protect anyone from a ‘virus’ never shown to exist. The vaccine couldn’t work because it was not designed to work and how could they make it appear to be working so that more people would have it? This was overcome by lowering the amplification rate of the PCR test to produce fewer ‘cases’ and therefore fewer ‘deaths’. Some of us had been pointing out since March, 2020, that the amplification rate of the test not testing for the ‘virus’ had been made artificially high to generate positive tests which they could call ‘cases’ to justify lockdowns. The World Health Organization recommended an absurdly high 45 amplification cycles to ensure the high positives required by the Cult and then remained silent on the issue until January 20th, 2021 – Biden’s Inauguration Day. This was when the

‘vaccinations’ were seriously underway and on that day the WHO recommended a er discussions with America’s CDC that laboratories lowered their testing amplification. Dr David Samadi, a certified urologist and health writer, said the WHO was encouraging all labs to reduce their cycle count for PCR tests. He said the current cycle was much too high and was ‘resulting in any particle being declared a positive case’. Even one mainstream news report I saw said this meant the number of ‘Covid’ infections may have been ‘dramatically inflated’. Oh, just a li le bit. The CDC in America issued new guidance to laboratories in April, 2021, to use 28 cycles but only for ‘vaccinated’ people. The timing of the CDC/WHO interventions were cynically designed to make it appear the ‘vaccines’ were responsible for falling cases and deaths when the real reason can be seen in the following examples. New York’s state lab, the Wadsworth Center, identified 872 positive tests in July, 2020, based on a threshold of 40 cycles. When the figure was lowered to 35 cycles 43 percent of the 872 were no longer ‘positives’. At 30 cycles the figure was 63 percent. A Massachuse s lab found that between 85 to 90 percent of people who tested positive in July with a cycle threshold of 40 would be negative at 30 cycles, Ashish Jha, MD, director of the Harvard Global Health Institute, said: ‘I’m really shocked that it could be that high … Boy, does it really change the way we need to be thinking about testing.’ I’m shocked that I could see the obvious in the spring of 2020, with no medical background, and most medical professionals still haven’t worked it out. No, that’s not shocking – it’s terrifying. Three weeks a er the WHO directive to lower PCR cycles the London Daily Mail ran this headline: ‘Why ARE Covid cases plummeting? New infections have fallen 45% in the US and 30% globally in the past 3 weeks but experts say vaccine is NOT the main driver because only 8% of Americans and 13% of people worldwide have received their first dose.’ They acknowledged that the drop could not be a ributed to the ‘vaccine’, but soon this morphed throughout the media into the ‘vaccine’ has caused cases and deaths to fall when it was the PCR threshold. In December, 2020, there was

chaos at English Channel ports with truck drivers needing negative ‘Covid’ tests before they could board a ferry home for Christmas. The government wanted to remove the backlog as fast as possible and they brought in troops to do the ‘testing’. Out of 1,600 drivers just 36 tested positive and the rest were given the all clear to cross the Channel. I guess the authorities thought that 36 was the least they could get away with without the unquestioning catching on. The amplification trick which most people believed in the absence of information in the mainstream applied more pressure on those refusing the ‘vaccine’ to succumb when it ‘obviously worked’. The truth was the exact opposite with deaths in care homes soaring with the ‘vaccine’ and in Israel the term used was ‘skyrocket’. A reanalysis of published data from the Israeli Health Ministry led by Dr Hervé Seligmann at the Medicine Emerging Infectious and Tropical Diseases at Aix-Marseille University found that Pfizer’s ‘Covid vaccine’ killed ‘about 40 times more [elderly] people than the disease itself would have killed’ during a five-week vaccination period and 260 times more younger people than would have died from the ‘virus’ even according to the manipulated ‘virus’ figures. Dr Seligmann and his co-study author, Haim Yativ, declared a er reviewing the Israeli ‘vaccine’ death data: ‘This is a new Holocaust.’ Then, in mid-April, 2021, a er vast numbers of people worldwide had been ‘vaccinated’, the story changed with clear coordination. The UK government began to prepare the ground for more future lockdowns when Nuremberg-destined Boris Johnson told yet another whopper. He said that cases had fallen because of lockdowns not ‘vaccines’. Lockdowns are irrelevant when there is no ‘virus’ and the test and fraudulent death certificates are deciding the number of ‘cases’ and ‘deaths’. Study a er study has shown that lockdowns don’t work and instead kill and psychologically destroy people. Meanwhile in the United States Anthony Fauci and Rochelle Walensky, the ultra-Zionist head of the CDC, peddled the same line. More lockdown was the answer and not the ‘vaccine’, a line repeated on cue by the moron that is Canadian Prime Minister Justin Trudeau. Why all the hysteria to get everyone ‘vaccinated’ if lockdowns and

not ‘vaccines’ made the difference? None of it makes sense on the face of it. Oh, but it does. The Cult wants lockdowns and the ‘vaccine’ and if the ‘vaccine’ is allowed to be seen as the total answer lockdowns would no longer be justified when there are still livelihoods to destroy. ‘Variants’ and renewed upward manipulation of PCR amplification are planned to instigate never-ending lockdown and more ‘vaccines’.

You must have it – we’re desperate

Israel, where the Jewish and Arab population are ruled by the Sabbatian Cult, was the front-runner in imposing the DNAmanipulating ‘vaccine’ on its people to such an extent that Jewish refusers began to liken what was happening to the early years of Nazi Germany. This would seem to be a fantastic claim. Why would a government of Jewish people be acting like the Nazis did? If you realise that the Sabbatian Cult was behind the Nazis and that Sabbatians hate Jews the pieces start to fit and the question of why a ‘Jewish’ government would treat Jews with such callous disregard for their lives and freedom finds an answer. Those controlling the government of Israel aren’t Jewish – they’re Sabbatian. Israeli lawyer Tamir Turgal was one who made the Nazi comparison in comments to German lawyer Reiner Fuellmich who is leading a class action lawsuit against the psychopaths for crimes against humanity. Turgal described how the Israeli government was vaccinating children and pregnant women on the basis that there was no evidence that this was dangerous when they had no evidence that it wasn’t dangerous either. They just had no evidence. This was medical experimentation and Turgal said this breached the Nuremberg Code about medical experimentation and procedures requiring informed consent and choice. Think about that. A Nuremberg Code developed because of Nazi experimentation on Jews and others in concentration camps by people like the evil-beyond-belief Josef Mengele is being breached by the Israeli government; but when you know that it’s a Sabbatian government along with its intelligence and military agencies like Mossad, Shin Bet and the Israeli Defense Forces, and that Sabbatians

were the force behind the Nazis, the kaleidoscope comes into focus. What have we come to when Israeli Jews are suing their government for violating the Nuremberg Code by essentially making Israelis subject to a medical experiment using the controversial ‘vaccines’? It’s a shocker that this has to be done in the light of what happened in Nazi Germany. The Anshe Ha-Emet, or ‘People of the Truth’, made up of Israeli doctors, lawyers, campaigners and public, have launched a lawsuit with the International Criminal Court. It says: When the heads of the Ministry of Health as well as the prime minister presented the vaccine in Israel and began the vaccination of Israeli residents, the vaccinated were not advised, that, in practice, they are taking part in a medical experiment and that their consent is required for this under the Nuremberg Code.

The irony is unbelievable, but easily explained in one word: Sabbatians. The foundation of Israeli ‘Covid’ apartheid is the ‘green pass’ or ‘green passport’ which allows Jews and Arabs who have had the DNA-manipulating ‘vaccine’ to go about their lives – to work, fly, travel in general, go to shopping malls, bars, restaurants, hotels, concerts, gyms, swimming pools, theatres and sports venues, while non-’vaccinated’ are banned from all those places and activities. Israelis have likened the ‘green pass’ to the yellow stars that Jews in Nazi Germany were forced to wear – the same as the yellow stickers that a branch of UK supermarket chain Morrisons told exempt mask-wears they had to display when shopping. How very sensitive. The Israeli system is blatant South African-style apartheid on the basis of compliance or non-compliance to fascism rather than colour of the skin. How appropriate that the Sabbatian Israeli government was so close to the pre-Mandela apartheid regime in Pretoria. The Sabbatian-instigated ‘vaccine passport’ in Israel is planned for everywhere. Sabbatians struck a deal with Pfizer that allowed them to lead the way in the percentage of a national population infused with synthetic material and the result was catastrophic. Israeli freedom activist Shai Dannon told me how chairs were appearing on beaches that said ‘vaccinated only’. Health Minister Yuli Edelstein said that anyone unwilling or unable to get

the jabs that ‘confer immunity’ will be ‘le behind’. The man’s a liar. Not even the makers claim the ‘vaccines’ confer immunity. When you see those figures of ‘vaccine’ deaths these psychopaths were saying that you must take the chance the ‘vaccine’ will kill you or maim you while knowing it will change your DNA or lockdown for you will be permanent. That’s fascism. The Israeli parliament passed a law to allow personal information of the non-vaccinated to be shared with local and national authorities for three months. This was claimed by its supporters to be a way to ‘encourage’ people to be vaccinated. Hadas Ziv from Physicians for Human Rights described this as a ‘draconian law which crushed medical ethics and the patient rights’. But that’s the idea, the Sabbatians would reply.

Your papers, please

Sabbatian Israel was leading what has been planned all along to be a global ‘vaccine pass’ called a ‘green passport’ without which you would remain in permanent lockdown restriction and unable to do anything. This is how badly – desperately – the Cult is to get everyone ‘vaccinated’. The term and colour ‘green’ was not by chance and related to the psychology of fusing the perception of the green climate hoax with the ‘Covid’ hoax and how the ‘solution’ to both is the same Great Reset. Lying politicians, health officials and psychologists denied there were any plans for mandatory vaccinations or restrictions based on vaccinations, but they knew that was exactly what was meant to happen with governments of all countries reaching agreements to enforce a global system. ‘Free’ Denmark and ‘free’ Sweden unveiled digital vaccine certification. Cyprus, Czech Republic, Estonia, Greece, Hungary, Iceland, Italy, Poland, Portugal, Slovakia, and Spain have all commi ed to a vaccine passport system and the rest including the whole of the EU would follow. The satanic UK government will certainly go this way despite mendacious denials and at the time of writing it is trying to manipulate the public into having the ‘vaccine’ so they could go abroad on a summer holiday. How would that work without something to prove you had the synthetic toxicity injected into you?

Documents show that the EU’s European Commission was moving towards ‘vaccine certificates’ in 2018 and 2019 before the ‘Covid’ hoax began. They knew what was coming. Abracadabra – Ursula von der Leyen, the German President of the Commission, announced in March, 2021, an EU ‘Digital Green Certificate’ – green again – to track the public’s ‘Covid status’. The passport sting is worldwide and the Far East followed the same pa ern with South Korea ruling that only those with ‘vaccination’ passports – again the green pass – would be able to ‘return to their daily lives’. Bill Gates has been preparing for this ‘passport’ with other Cult operatives for years and beyond the paper version is a Gates-funded ‘digital ta oo’ to identify who has been vaccinated and who hasn’t. The ‘ta oo’ is reported to include a substance which is externally readable to confirm who has been vaccinated. This is a bio-luminous light-generating enzyme (think fireflies) called … Luciferase. Yes, named a er the Cult ‘god’ Lucifer the ‘light bringer’ of whom more to come. Gates said he funded the readable ta oo to ensure children in the developing world were vaccinated and no one was missed out. He cares so much about poor kids as we know. This was just the cover story to develop a vaccine tagging system for everyone on the planet. Gates has been funding the ID2020 ‘alliance’ to do just that in league with other lovely people at Microso , GAVI, the Rockefeller Foundation, Accenture and IDEO.org. He said in interviews in March, 2020, before any ‘vaccine’ publicly existed, that the world must have a globalised digital certificate to track the ‘virus’ and who had been vaccinated. Gates knew from the start that the mRNA vaccines were coming and when they would come and that the plan was to tag the ‘vaccinated’ to marginalise the intelligent and stop them doing anything including travel. Evil just doesn’t suffice. Gates was exposed for offering a $10 million bribe to the Nigerian House of Representatives to invoke compulsory ‘Covid’ vaccination of all Nigerians. Sara Cunial, a member of the Italian Parliament, called Gates a ‘vaccine criminal’. She urged the Italian President to hand him over to the International Criminal Court for crimes against

humanity and condemned his plans to ‘chip the human race’ through ID2020. You know it’s a long-planned agenda when war criminal and Cult gofer Tony Blair is on the case. With the scale of arrogance only someone as dark as Blair can muster he said: ‘Vaccination in the end is going to be your route to liberty.’ Blair is a disgusting piece of work and he confirms that again. The media has given a lot of coverage to a bloke called Charlie Mullins, founder of London’s biggest independent plumbing company, Pimlico Plumbers, who has said he won’t employ anyone who has not been vaccinated or have them go to any home where people are not vaccinated. He said that if he had his way no one would be allowed to walk the streets if they have not been vaccinated. Gates was cheering at the time while I was alerting the white coats. The plan is that people will qualify for ‘passports’ for having the first two doses and then to keep it they will have to have all the follow ups and new ones for invented ‘variants’ until human genetics is transformed and many are dead who can’t adjust to the changes. Hollywood celebrities – the usual propaganda stunt – are promoting something called the WELL Health-Safety Rating to verify that a building or space has ‘taken the necessary steps to prioritize the health and safety of their staff, visitors and other stakeholders’. They included Lady Gaga, Jennifer Lopez, Michael B. Jordan, Robert DeNiro, Venus Williams, Wolfgang Puck, Deepak Chopra and 17th Surgeon General Richard Carmona. Yawn. WELL Health-Safety has big connections with China. Parent company Delos is headed by former Goldman Sachs partner Paul Scialla. This is another example – and we will see so many others – of using the excuse of ‘health’ to dictate the lives and activities of the population. I guess one confirmation of the ‘safety’ of buildings is that only ‘vaccinated’ people can go in, right?

Electronic concentration camps

I wrote decades ago about the plans to restrict travel and here we are for those who refuse to bow to tyranny. This can be achieved in one go with air travel if the aviation industry makes a blanket decree.

The ‘vaccine’ and guaranteed income are designed to be part of a global version of China’s social credit system which tracks behaviour 24/7 and awards or deletes ‘credits’ based on whether your behaviour is supported by the state or not. I mean your entire lifestyle – what you do, eat, say, everything. Once your credit score falls below a certain level consequences kick in. In China tens of millions have been denied travel by air and train because of this. All the locations and activities denied to refusers by the ‘vaccine’ passports will be included in one big mass ban on doing almost anything for those that don’t bow their head to government. It’s beyond fascist and a new term is required to describe its extremes – I guess fascist technocracy will have to do. The way the Chinese system of technological – technocratic – control is sweeping the West can be seen in the Los Angeles school system and is planned to be expanded worldwide. Every child is required to have a ‘Covid’tracking app scanned daily before they can enter the classroom. The so-called Daily Pass tracking system is produced by Gates’ Microso which I’m sure will shock you rigid. The pass will be scanned using a barcode (one step from an inside-the-body barcode) and the information will include health checks, ‘Covid’ tests and vaccinations. Entry codes are for one specific building only and access will only be allowed if a student or teacher has a negative test with a test not testing for the ‘virus’, has no symptoms of anything alleged to be related to ‘Covid’ (symptoms from a range of other illness), and has a temperature under 100 degrees. No barcode, no entry, is planned to be the case for everywhere and not only schools. Kids are being psychologically prepared to accept this as ‘normal’ their whole life which is why what they can impose in schools is so important to the Cult and its gofers. Long-time American freedom campaigner John Whitehead of the Rutherford Institute was not exaggerating when he said: ‘Databit by databit, we are building our own electronic concentration camps.’ Canada under its Cult gofer prime minister Justin Trudeau has taken a major step towards the real thing with people interned against their will if they test positive with a test not testing for the ‘virus’ when they arrive at a Canadian

airport. They are jailed in internment hotels o en without food or water for long periods and with many doors failing to lock there have been sexual assaults. The interned are being charged sometimes $2,000 for the privilege of being abused in this way. Trudeau is fully on board with the Cult and says the ‘Covid pandemic’ has provided an opportunity for a global ‘reset’ to permanently change Western civilisation. His number two, Deputy Prime Minister Chrystia Freeland, is a trustee of the World Economic Forum and a Rhodes Scholar. The Trudeau family have long been servants of the Cult. See The Biggest Secret and Cathy O’Brien’s book Trance-Formation of America for the horrific background to Trudeau’s father Pierre Trudeau another Canadian prime minister. Hide your fascism behind the façade of a heart-on-the-sleeve liberal. It’s a wellhoned Cult technique.

What can the ‘vaccine’ really do?

We have a ‘virus’ never shown to exist and ‘variants’ of the ‘virus’ that have also never been shown to exist except, like the ‘original’, as computer-generated fictions. Even if you believe there’s a ‘virus’ the ‘case’ to ‘death’ rate is in the region of 0.23 to 0.15 percent and those ‘deaths’ are concentrated among the very old around the same average age that people die anyway. In response to this lack of threat (in truth none) psychopaths and idiots, knowingly and unknowingly answering to Gates and the Cult, are seeking to ‘vaccinate’ every man, woman and child on Planet Earth. Clearly the ‘vaccine’ is not about ‘Covid’ – none of this ever has been. So what is it all about really? Why the desperation to infuse genetically-manipulating synthetic material into everyone through mRNA fraudulent ‘vaccines’ with the intent of doing this over and over with the excuses of ‘variants’ and other ‘virus’ inventions? Dr Sherri Tenpenny, an osteopathic medical doctor in the United States, has made herself an expert on vaccines and their effects as a vehement campaigner against their use. Tenpenny was board certified in emergency medicine, the director of a level two trauma centre for 12 years, and moved to Cleveland in 1996 to start an integrative

medicine practice which has treated patients from all 50 states and some 17 other countries. Weaning people off pharmaceutical drugs is a speciality. She became interested in the consequences of vaccines a er a ending a meeting at the National Vaccine Information Center in Washington DC in 2000 where she ‘sat through four days of listening to medical doctors and scientists and lawyers and parents of vaccine injured kids’ and asked: ‘What’s going on?’ She had never been vaccinated and never got ill while her father was given a list of vaccines to be in the military and was ‘sick his entire life’. The experience added to her questions and she began to examine vaccine documents from the Centers for Disease Control (CDC). A er reading the first one, the 1998 version of The General Recommendations of Vaccination, she thought: ‘This is it?’ The document was poorly wri en and bad science and Tenpenny began 20 years of research into vaccines that continues to this day. She began her research into ‘Covid vaccines’ in March, 2020, and she describes them as ‘deadly’. For many, as we have seen, they already have been. Tenpenny said that in the first 30 days of the ‘vaccine’ rollout in the United States there had been more than 40,000 adverse events reported to the vaccine adverse event database. A document had been delivered to her the day before that was 172 pages long. ‘We have over 40,000 adverse events; we have over 3,100 cases of [potentially deadly] anaphylactic shock; we have over 5,000 neurological reactions.’ Effects ranged from headaches to numbness, dizziness and vertigo, to losing feeling in hands or feet and paraesthesia which is when limbs ‘fall asleep’ and people have the sensation of insects crawling underneath their skin. All this happened in the first 30 days and remember that only about ten percent (or far less) of adverse reactions and vaccine-related deaths are estimated to be officially reported. Tenpenny said: So can you think of one single product in any industry, any industry, for as long as products have been made on the planet that within 30 days we have 40,000 people complaining of side effects that not only is still on the market but … we’ve got paid actors telling us how great

they are for getting their vaccine. We’re offering people $500 if they will just get their vaccine and we’ve got nurses and doctors going; ‘I got the vaccine, I got the vaccine’.

Tenpenny said they were not going to be ‘happy dancing folks’ when they began to suffer Bell’s palsy (facial paralysis), neuropathies, cardiac arrhythmias and autoimmune reactions that kill through a blood disorder. ‘They’re not going to be so happy, happy then, but we’re never going to see pictures of those people’ she said. Tenpenny described the ‘vaccine’ as ‘a well-designed killing tool’.

No off-switch

Bad as the initial consequences had been Tenpenny said it would be maybe 14 months before we began to see the ‘full ravage’ of what is going to happen to the ‘Covid vaccinated’ with full-out consequences taking anything between two years and 20 years to show. You can understand why when you consider that variations of the ‘Covid vaccine’ use mRNA (messenger RNA) to in theory activate the immune system to produce protective antibodies without using the actual ‘virus’. How can they when it’s a computer program and they’ve never isolated what they claim is the ‘real thing’? Instead they use synthetic mRNA. They are inoculating synthetic material into the body which through a technique known as the Trojan horse is absorbed into cells to change the nature of DNA. Human DNA is changed by an infusion of messenger RNA and with each new ‘vaccine’ of this type it is changed even more. Say so and you are banned by Cult Internet platforms. The contempt the contemptuous Mark Zuckerberg has for the truth and human health can be seen in an internal Facebook video leaked to the Project Veritas investigative team in which he said of the ‘Covid vaccines’: ‘… I share some caution on this because we just don’t know the long term side-effects of basically modifying people’s DNA and RNA.’ At the same time this disgusting man’s Facebook was censoring and banning anyone saying exactly the same. He must go before a Nuremberg trial for crimes against humanity when he knows that he

is censoring legitimate concerns and denying the right of informed consent on behalf of the Cult that owns him. People have been killed and damaged by the very ‘vaccination’ technique he cast doubt on himself when they may not have had the ‘vaccine’ with access to information that he denied them. The plan is to have at least annual ‘Covid vaccinations’, add others to deal with invented ‘variants’, and change all other vaccines into the mRNA system. Pfizer executives told shareholders at a virtual Barclays Global Healthcare Conference in March, 2021, that the public may need a third dose of ‘Covid vaccine’, plus regular yearly boosters and the company planned to hike prices to milk the profits in a ‘significant opportunity for our vaccine’. These are the professional liars, cheats and opportunists who are telling you their ‘vaccine’ is safe. Given this volume of mRNA planned to be infused into the human body and its ability to then replicate we will have a transformation of human genetics from biological to synthetic biological – exactly the long-time Cult plan for reasons we’ll see – and many will die. Sherri Tenpenny said of this replication: It’s like having an on-button but no off-button and that whole mechanism … they actually give it a name and they call it the Trojan horse mechanism, because it allows that [synthetic] virus and that piece of that [synthetic] virus to get inside of your cells, start to replicate and even get inserted into other parts of your DNA as a Trojan-horse.

Ask the overwhelming majority of people who have the ‘vaccine’ what they know about the contents and what they do and they would reply: ‘The government says it will stop me ge ing the virus.’ Governments give that false impression on purpose to increase takeup. You can read Sherri Tenpenny’s detailed analysis of the health consequences in her blog at Vaxxter.com, but in summary these are some of them. She highlights the statement by Bill Gates about how human beings can become their own ‘vaccine manufacturing machine’. The man is insane. [‘Vaccine’-generated] ‘antibodies’ carry synthetic messenger RNA into the cells and the damage starts, Tenpenny contends, and she says that lungs can be adversely affected through varying degrees of pus and bleeding which

obviously affects breathing and would be dubbed ‘Covid-19’. Even more sinister was the impact of ‘antibodies’ on macrophages, a white blood cell of the immune system. They consist of Type 1 and Type 2 which have very different functions. She said Type 1 are ‘hypervigilant’ white blood cells which ‘gobble up’ bacteria etc. However, in doing so, this could cause inflammation and in extreme circumstances be fatal. She says these affects are mitigated by Type 2 macrophages which kick in to calm down the system and stop it going rogue. They clear up dead tissue debris and reduce inflammation that the Type 1 ‘fire crews’ have caused. Type 1 kills the infection and Type 2 heals the damage, she says. This is her punchline with regard to ‘Covid vaccinations’: She says that mRNA ‘antibodies’ block Type 2 macrophages by a aching to them and deactivating them. This meant that when the Type 1 response was triggered by infection there was nothing to stop that ge ing out of hand by calming everything down. There’s an on-switch, but no offswitch, she says. What follows can be ‘over and out, see you when I see you’.

Genetic suicide

Tenpenny also highlights the potential for autoimmune disease – the body a acking itself – which has been associated with vaccines since they first appeared. Infusing a synthetic foreign substance into cells could cause the immune system to react in a panic believing that the body is being overwhelmed by an invader (it is) and the consequences can again be fatal. There is an autoimmune response known as a ‘cytokine storm’ which I have likened to a homeowner panicked by an intruder and picking up a gun to shoot randomly in all directions before turning the fire on himself. The immune system unleashes a storm of inflammatory response called cytokines to a threat and the body commits hara-kiri. The lesson is that you mess with the body’s immune response at your peril and these ‘vaccines’ seriously – fundamentally – mess with immune response. Tenpenny refers to a consequence called anaphylactic shock which is a severe and highly dangerous allergic reaction when the immune system

floods the body with chemicals. She gives the example of having a bee sting which primes the immune system and makes it sensitive to those chemicals. When people are stung again maybe years later the immune response can be so powerful that it leads to anaphylactic shock. Tenpenny relates this ‘shock’ with regard to the ‘Covid vaccine’ to something called polyethylene glycol or PEG. Enormous numbers of people have become sensitive to this over decades of use in a whole range of products and processes including food, drink, skin creams and ‘medicine’. Studies have claimed that some 72 percent of people have antibodies triggered by PEG compared with two percent in the 1960s and allergic hypersensitive reactions to this become a gathering cause for concern. Tenpenny points out that the ‘mRNA vaccine’ is coated in a ‘bubble’ of polyethylene glycol which has the potential to cause anaphylactic shock through immune sensitivity. Many reports have appeared of people reacting this way a er having the ‘Covid vaccine’. What do we think is going to happen as humanity has more and more of these ‘vaccines’? Tenpenny said: ‘All these pictures we have seen with people with these rashes … these weepy rashes, big reactions on their arms and things like that – it’s an acute allergic reaction most likely to the polyethylene glycol that you’ve been previously primed and sensitised to.’ Those who have not studied the conspiracy and its perpetrators at length might think that making the population sensitive to PEG and then pu ing it in these ‘vaccines’ is just a coincidence. It is not. It is instead testament to how carefully and coldly-planned current events have been and the scale of the conspiracy we are dealing with. Tenpenny further explains that the ‘vaccine’ mRNA procedure can breach the blood-brain barrier which protects the brain from toxins and other crap that will cause malfunction. In this case they could make two proteins corrupt brain function to cause Amyotrophic lateral sclerosis (ALS) , a progressive nervous system disease leading to loss of muscle control, and frontal lobe degeneration – Alzheimer’s and dementia. Immunologist J. Bart Classon published a paper connecting mRNA ‘vaccines’ to prion

disease which can lead to Alzheimer’s and other forms of neurogenerative disease while others have pointed out the potential to affect the placenta in ways that make women infertile. This will become highly significant in the next chapter when I will discuss other aspects of this non-vaccine that relate to its nanotechnology and transmission from the injected to the uninjected.

Qualified in idiocy

Tenpenny describes how research has confirmed that these ‘vaccine’generated antibodies can interact with a range of other tissues in the body and a ack many other organs including the lungs. ‘This means that if you have a hundred people standing in front of you that all got this shot they could have a hundred different symptoms.’ Anyone really think that Cult gofers like the Queen, Tony Blair, Christopher Whi y, Anthony Fauci, and all the other psychopaths have really had this ‘vaccine’ in the pictures we’ve seen? Not a bloody chance. Why don’t doctors all tell us about all these dangers and consequences of the ‘Covid vaccine’? Why instead do they encourage and pressure patients to have the shot? Don’t let’s think for a moment that doctors and medical staff can’t be stupid, lazy, and psychopathic and that’s without the financial incentives to give the jab. Tenpenny again: Some people are going to die from the vaccine directly but a large number of people are going to start to get horribly sick and get all kinds of autoimmune diseases 42 days to maybe a year out. What are they going to do, these stupid doctors who say; ‘Good for you for getting that vaccine.’ What are they going to say; ‘Oh, it must be a mutant, we need to give an extra dose of that vaccine.’ Because now the vaccine, instead of one dose or two doses we need three or four because the stupid physicians aren’t taking the time to learn anything about it. If I can learn this sitting in my living room reading a 19 page paper and several others so can they. There’s nothing special about me, I just take the time to do it.

Remember how Sara Kayat, the NHS and TV doctor, said that the ‘Covid vaccine’ would ‘100 percent prevent hospitalisation and death’. Doctors can be idiots like every other profession and they

should not be worshipped as infallible. They are not and far from it. Behind many medical and scientific ‘experts’ lies an uninformed prat trying to hide themselves from you although in the ‘Covid’ era many have failed to do so as with UK narrative-repeating ‘TV doctor’ Hilary Jones. Pushing back against the minority of proper doctors and scientists speaking out against the ‘vaccine’ has been the entire edifice of the Cult global state in the form of governments, medical systems, corporations, mainstream media, Silicon Valley, and an army of compliant doctors, medical staff and scientists willing to say anything for money and to enhance their careers by promoting the party line. If you do that you are an ‘expert’ and if you won’t you are an ‘anti-vaxxer’ and ‘Covidiot’. The pressure to be ‘vaccinated’ is incessant. We have even had reports claiming that the ‘vaccine’ can help cure cancer and Alzheimer’s and make the lame walk. I am waiting for the announcement that it can bring you coffee in the morning and cook your tea. Just as the symptoms of ‘Covid’ seem to increase by the week so have the miracles of the ‘vaccine’. American supermarket giant Kroger Co. offered nearly 500,000 employees in 35 states a $100 bonus for having the ‘vaccine’ while donut chain Krispy Kreme promised ‘vaccinated’ customers a free glazed donut every day for the rest of 2021. Have your DNA changed and you will get a doughnut although we might not have to give you them for long. Such offers and incentives confirm the desperation. Perhaps the worse vaccine-stunt of them all was UK ‘Health’ Secretary Ma -the-prat Hancock on live TV a er watching a clip of someone being ‘vaccinated’ when the roll-out began. Hancock faked tears so badly it was embarrassing. Brain-of-Britain Piers Morgan, the lockdown-supporting, ‘vaccine’ supporting, ‘vaccine’ passportsupporting, TV host played along with Hancock – ‘You’re quite emotional about that’ he said in response to acting so atrocious it would have been called out at a school nativity which will presumably today include Mary and Jesus in masks, wise men keeping their camels six feet apart, and shepherds under tent arrest. System-serving Morgan tweeted this: ‘Love the idea of covid vaccine passports for everywhere: flights, restaurants, clubs, football, gyms,

shops etc. It’s time covid-denying, anti-vaxxer loonies had their bullsh*t bluff called & bar themselves from going anywhere that responsible citizens go.’ If only I could aspire to his genius. To think that Morgan, who specialises in shouting over anyone he disagrees with, was lauded as a free speech hero when he lost his job a er storming off the set of his live show like a child throwing his dolly out of the pram. If he is a free speech hero we are in real trouble. I have no idea what ‘bullsh*t’ means, by the way, the * throws me completely. The Cult is desperate to infuse its synthetic DNA-changing concoction into everyone and has been using every lie, trick and intimidation to do so. The question of ‘Why?’ we shall now address.

CHAPTER TEN Human 2.0 I believe that at the end of the century the use of words and general educated opinion will have altered so much that one will be able to speak of machines thinking without expecting to be contradicted – Alan Turing (1912-1954), the ‘Father of artificial intelligence‘

I

have been exposing for decades the plan to transform the human body from a biological to a synthetic-biological state. The new human that I will call Human 2.0 is planned to be connected to artificial intelligence and a global AI ‘Smart Grid’ that would operate as one global system in which AI would control everything from your fridge to your heating system to your car to your mind. Humans would no longer be ‘human’, but post-human and subhuman, with their thinking and emotional processes replaced by AI. What I said sounded crazy and beyond science fiction and I could understand that. To any balanced, rational, mind it is crazy. Today, however, that world is becoming reality and it puts the ‘Covid vaccine’ into its true context. Ray Kurzweil is the ultra-Zionist ‘computer scientist, inventor and futurist’ and co-founder of the Singularity University. Singularity refers to the merging of humans with machines or ‘transhumanism’. Kurzweil has said humanity would be connected to the cyber ‘cloud’ in the period of the everrecurring year of 2030: Our thinking … will be a hybrid of biological and non-biological thinking … humans will be able to extend their limitations and ‘think in the cloud’ … We’re going to put gateways to the

cloud in our brains ... We’re going to gradually merge and enhance ourselves ... In my view, that’s the nature of being human – we transcend our limitations. As the technology becomes vastly superior to what we are then the small proportion that is still human gets smaller and smaller and smaller until it’s just utterly negligible.

They are trying to sell this end-of-humanity-as-we-know-it as the next stage of ‘evolution’ when we become super-human and ‘like the gods’. They are lying to you. Shocked, eh? The population, and again especially the young, have been manipulated into addiction to technologies designed to enslave them for life. First they induced an addiction to smartphones (holdables); next they moved to technology on the body (wearables); and then began the invasion of the body (implantables). I warned way back about the plan for microchipped people and we are now entering that era. We should not be diverted into thinking that this refers only to chips we can see. Most important are the nanochips known as smart dust, neural dust and nanobots which are far too small to be seen by the human eye. Nanotechnology is everywhere, increasingly in food products, and released into the atmosphere by the geoengineering of the skies funded by Bill Gates to ‘shut out the Sun’ and ‘save the planet from global warming’. Gates has been funding a project to spray millions of tonnes of chalk (calcium carbonate) into the stratosphere over Sweden to ‘dim the Sun’ and cool the Earth. Scientists warned the move could be disastrous for weather systems in ways no one can predict and opposition led to the Swedish space agency announcing that the ‘experiment’ would not be happening as planned in the summer of 2021; but it shows where the Cult is going with dimming the impact of the Sun and there’s an associated plan to change the planet’s atmosphere. Who gives psychopath Gates the right to dictate to the entire human race and dismantle planetary systems? The world will not be safe while this man is at large. The global warming hoax has made the Sun, like the gas of life, something to fear when both are essential to good health and human survival (more inversion). The body transforms sunlight into vital vitamin D through a process involving … cholesterol. This is the cholesterol we are also told to fear. We are urged to take Big Pharma

statin drugs to reduce cholesterol and it’s all systematic. Reducing cholesterol means reducing vitamin D uptake with all the multiple health problems that will cause. At least if you take statins long term it saves the government from having to pay you a pension. The delivery system to block sunlight is widely referred to as chemtrails although these have a much deeper agenda, too. They appear at first to be contrails or condensation trails streaming from aircra into cold air at high altitudes. Contrails disperse very quickly while chemtrails do not and spread out across the sky before eventually their content falls to earth. Many times I have watched aircra crosscross a clear blue sky releasing chemtrails until it looks like a cloudy day. Chemtrails contain many things harmful to humans and the natural world including toxic heavy metals, aluminium (see Alzheimer’s) and nanotechnology. Ray Kurzweil reveals the reason without actually saying so: ‘Nanobots will infuse all the ma er around us with information. Rocks, trees, everything will become these intelligent creatures.’ How do you deliver that? From the sky. Self-replicating nanobots would connect everything to the Smart Grid. The phenomenon of Morgellons disease began in the chemtrail era and the correlation has led to it being dubbed the ‘chemtrail disease’. Self-replicating fibres appear in the body that can be pulled out through the skin. Morgellons fibres continue to grow outside the body and have a form of artificial intelligence. I cover this at greater length in Phantom Self.

‘Vaccine’ operating system

‘Covid vaccines’ with their self-replicating synthetic material are also designed to make the connection between humanity and Kurzweil’s ‘cloud’. American doctor and dedicated campaigner for truth, Carrie Madej, an Internal Medicine Specialist in Georgia with more than 20 years medical experience, has highlighted the nanotechnology aspect of the fake ‘vaccines’. She explains how one of the components in at least the Moderna and Pfizer synthetic potions are ‘lipid nanoparticles’ which are ‘like li le tiny computer bits’ – a ‘sci-fi substance’ known as nanobots and hydrogel which can be ‘triggered

at any moment to deliver its payload’ and act as ‘biosensors’. The synthetic substance had ‘the ability to accumulate data from your body like your breathing, your respiration, thoughts and emotions, all kind of things’ and each syringe could carry a million nanobots: This substance because it’s like little bits of computers in your body, crazy, but it’s true, it can do that, [and] obviously has the ability to act through Wi-Fi. It can receive and transmit energy, messages, frequencies or impulses. That issue has never been addressed by these companies. What does that do to the human? Just imagine getting this substance in you and it can react to things all around you, the 5G, your smart device, your phones, what is happening with that? What if something is triggering it, too, like an impulse, a frequency? We have something completely foreign in the human body.

Madej said her research revealed that electromagnetic (EMF) frequencies emi ed by phones and other devices had increased dramatically in the same period of the ‘vaccine’ rollout and she was seeing more people with radiation problems as 5G and other electromagnetic technology was expanded and introduced to schools and hospitals. She said she was ‘floored with the EMF coming off’ the devices she checked. All this makes total sense and syncs with my own work of decades when you think that Moderna refers in documents to its mRNA ‘vaccine’ as an ‘operating system’: Recognizing the broad potential of mRNA science, we set out to create an mRNA technology platform that functions very much like an operating system on a computer. It is designed so that it can plug and play interchangeably with different programs. In our case, the ‘program’ or ‘app’ is our mRNA drug – the unique mRNA sequence that codes for a protein … … Our MRNA Medicines – ‘The ‘Software Of Life’: When we have a concept for a new mRNA medicine and begin research, fundamental components are already in place. Generally, the only thing that changes from one potential mRNA medicine to another is the coding region – the actual genetic code that instructs ribosomes to make protein. Utilizing these instruction sets gives our investigational mRNA medicines a software-like quality. We also have the ability to combine different mRNA sequences encoding for different proteins in a single mRNA investigational medicine.

Who needs a real ‘virus’ when you can create a computer version to justify infusing your operating system into the entire human race on the road to making living, breathing people into cyborgs? What is missed with the ‘vaccines’ is the digital connection between synthetic material and the body that I highlighted earlier with the study that hacked a computer with human DNA. On one level the body is digital, based on mathematical codes, and I’ll have more about that in the next chapter. Those who ridiculously claim that mRNA ‘vaccines’ are not designed to change human genetics should explain the words of Dr Tal Zaks, chief medical officer at Moderna, in a 2017 TED talk. He said that over the last 30 years ‘we’ve been living this phenomenal digital scientific revolution, and I’m here today to tell you, that we are actually hacking the software of life, and that it’s changing the way we think about prevention and treatment of disease’: In every cell there’s this thing called messenger RNA, or mRNA for short, that transmits the critical information from the DNA in our genes to the protein, which is really the stuff we’re all made out of. This is the critical information that determines what the cell will do. So we think about it as an operating system. So if you could change that, if you could introduce a line of code, or change a line of code, it turns out, that has profound implications for everything, from the flu to cancer.

Zaks should more accurately have said that this has profound implications for the human genetic code and the nature of DNA. Communications within the body go both ways and not only one. But, hey, no, the ‘Covid vaccine’ will not affect your genetics. Cult fact-checkers say so even though the man who helped to develop the mRNA technique says that it does. Zaks said in 2017: If you think about what it is we’re trying to do. We’ve taken information and our understanding of that information and how that information is transmitted in a cell, and we’ve taken our understanding of medicine and how to make drugs, and we’re fusing the two. We think of it as information therapy.

I have been writing for decades that the body is an information field communicating with itself and the wider world. This is why

radiation which is information can change the information field of body and mind through phenomena like 5G and change their nature and function. ‘Information therapy’ means to change the body’s information field and change the way it operates. DNA is a receivertransmi er of information and can be mutated by information like mRNA synthetic messaging. Technology to do this has been ready and waiting in the underground bases and other secret projects to be rolled out when the ‘Covid’ hoax was played. ‘Trials’ of such short and irrelevant duration were only for public consumption. When they say the ‘vaccine’ is ‘experimental’ that is not true. It may appear to be ‘experimental’ to those who don’t know what’s going on, but the trials have already been done to ensure the Cult gets the result it desires. Zaks said that it took decades to sequence the human genome, completed in 2003, but now they could do it in a week. By ‘they’ he means scientists operating in the public domain. In the secret projects they were sequencing the genome in a week long before even 2003.

Deluge of mRNA

Highly significantly the Moderna document says the guiding premise is that if using mRNA as a medicine works for one disease then it should work for many diseases. They were leveraging the flexibility afforded by their platform and the fundamental role mRNA plays in protein synthesis to pursue mRNA medicines for a broad spectrum of diseases. Moderna is confirming what I was saying through 2020 that multiple ‘vaccines’ were planned for ‘Covid’ (and later invented ‘variants’) and that previous vaccines would be converted to the mRNA system to infuse the body with massive amounts of genetically-manipulating synthetic material to secure a transformation to a synthetic-biological state. The ‘vaccines’ are designed to kill stunning numbers as part of the long-exposed Cult depopulation agenda and transform the rest. Given this is the goal you can appreciate why there is such hysterical demand for every human to be ‘vaccinated’ for an alleged ‘disease’ that has an estimated ‘infection’ to ‘death’ ratio of 0.23-0.15 percent. As I write

children are being given the ‘vaccine’ in trials (their parents are a disgrace) and ever-younger people are being offered the vaccine for a ‘virus’ that even if you believe it exists has virtually zero chance of harming them. Horrific effects of the ‘trials’ on a 12-year-old girl were revealed by a family member to be serious brain and gastric problems that included a bowel obstruction and the inability to swallow liquids or solids. She was unable to eat or drink without throwing up, had extreme pain in her back, neck and abdomen, and was paralysed from the waist down which stopped her urinating unaided. When the girl was first taken to hospital doctors said it was all in her mind. She was signed up for the ‘trial’ by her parents for whom no words suffice. None of this ‘Covid vaccine’ insanity makes any sense unless you see what the ‘vaccine’ really is – a bodychanger. Synthetic biology or ‘SynBio’ is a fast-emerging and expanding scientific discipline which includes everything from genetic and molecular engineering to electrical and computer engineering. Synthetic biology is defined in these ways: • A multidisciplinary area of research that seeks to create new biological parts, devices, and systems, or to redesign systems that are already found in nature. • The use of a mixture of physical engineering and genetic engineering to create new (and therefore synthetic) life forms. • An emerging field of research that aims to combine the knowledge and methods of biology, engineering and related disciplines in the design of chemically-synthesized DNA to create organisms with novel or enhanced characteristics and traits (synthetic organisms including humans). We now have synthetic blood, skin, organs and limbs being developed along with synthetic body parts produced by 3D printers. These are all elements of the synthetic human programme and this comment by Kurzweil’s co-founder of the Singularity University,

Peter Diamandis, can be seen in a whole new light with the ‘Covid’ hoax and the sanctions against those that refuse the ‘vaccine’: Anybody who is going to be resisting the progress forward [to transhumanism] is going to be resisting evolution and, fundamentally, they will die out. It’s not a matter of whether it’s good or bad. It’s going to happen.

‘Resisting evolution’? What absolute bollocks. The arrogance of these people is without limit. His ‘it’s going to happen’ mantra is another way of saying ‘resistance is futile’ to break the spirit of those pushing back and we must not fall for it. Ge ing this geneticallytransforming ‘vaccine’ into everyone is crucial to the Cult plan for total control and the desperation to achieve that is clear for anyone to see. Vaccine passports are a major factor in this and they, too, are a form of resistance is futile. It’s NOT. The paper funded by the Rockefeller Foundation for the 2013 ‘health conference’ in China said: We will interact more with artificial intelligence. The use of robotics, bio-engineering to augment human functioning is already well underway and will advance. Re-engineering of humans into potentially separate and unequal forms through genetic engineering or mixed human-robots raises debates on ethics and equality. A new demography is projected to emerge after 2030 [that year again] of technologies (robotics, genetic engineering, nanotechnology) producing robots, engineered organisms, ‘nanobots’ and artificial intelligence (AI) that can self-replicate. Debates will grow on the implications of an impending reality of human designed life.

What is happening today is so long planned. The world army enforcing the will of the world government is intended to be a robot army, not a human one. Today’s military and its technologically ‘enhanced’ troops, pilotless planes and driverless vehicles are just stepping stones to that end. Human soldiers are used as Cult fodder and its time they woke up to that and worked for the freedom of the population instead of their own destruction and their family’s destruction – the same with the police. Join us and let’s sort this out. The phenomenon of enforce my own destruction is widespread in the ‘Covid’ era with Woker ‘luvvies’ in the acting and entertainment

industries supporting ‘Covid’ rules which have destroyed their profession and the same with those among the public who put signs on the doors of their businesses ‘closed due to Covid – stay safe’ when many will never reopen. It’s a form of masochism and most certainly insanity.

Transgender = transhumanism

When something explodes out of nowhere and is suddenly everywhere it is always the Cult agenda and so it is with the tidal wave of claims and demands that have infiltrated every aspect of society under the heading of ‘transgenderism’. The term ‘trans’ is so ‘in’ and this is the dictionary definition: A prefix meaning ‘across’, ’through’, occurring … in loanwords from Latin, used in particular for denoting movement or conveyance from place to place (transfer; transmit; transplant) or complete change (transform; transmute), or to form adjectives meaning ’crossing’, ‘on the other side of’, or ‘going beyond’ the place named (transmontane; transnational; transSiberian).

Transgender means to go beyond gender and transhuman means to go beyond human. Both are aspects of the Cult plan to transform the human body to a synthetic state with no gender. Human 2.0 is not designed to procreate and would be produced technologically with no need for parents. The new human would mean the end of parents and so men, and increasingly women, are being targeted for the deletion of their rights and status. Parental rights are disappearing at an ever-quickening speed for the same reason. The new human would have no need for men or women when there is no procreation and no gender. Perhaps the transgender movement that appears to be in a permanent state of frenzy might now contemplate on how it is being used. This was never about transgender rights which are only the interim excuse for confusing gender, particularly in the young, on the road to fusing gender. Transgender activism is not an end; it is a means to an end. We see again the technique of creative destruction in which you destroy the status quo to ‘build back be er’ in the form that you want. The gender status quo had to be

destroyed by persuading the Cult-created Woke mentality to believe that you can have 100 genders or more. A programme for 9 to 12 year olds produced by the Cult-owned BBC promoted the 100 genders narrative. The very idea may be the most monumental nonsense, but it is not what is true that counts, only what you can make people believe is true. Once the gender of 2 + 2 = 4 has been dismantled through indoctrination, intimidation and 2 + 2 = 5 then the new no-gender normal can take its place with Human 2.0. Aldous Huxley revealed the plan in his prophetic Brave New World in 1932: Natural reproduction has been done away with and children are created, decanted’, and raised in ‘hatcheries and conditioning centres’. From birth, people are genetically designed to fit into one of five castes, which are further split into ‘Plus’ and ‘Minus’ members and designed to fulfil predetermined positions within the social and economic strata of the World State.

How could Huxley know this in 1932? For the same reason George Orwell knew about the Big Brother state in 1948, Cult insiders I have quoted knew about it in 1969, and I have known about it since the early 1990s. If you are connected to the Cult or you work your balls off to uncover the plan you can predict the future. The process is simple. If there is a plan for the world and nothing intervenes to stop it then it will happen. Thus if you communicate the plan ahead of time you are perceived to have predicted the future, but you haven’t. You have revealed the plan which without intervention will become the human future. The whole reason I have done what I have is to alert enough people to inspire an intervention and maybe at last that time has come with the Cult and its intentions now so obvious to anyone with a brain in working order.

The future is here

Technological wombs that Huxley described to replace parent procreation are already being developed and they are only the projects we know about in the public arena. Israeli scientists told The Times of Israel in March, 2021, that they have grown 250-cell embryos

into mouse foetuses with fully formed organs using artificial wombs in a development they say could pave the way for gestating humans outside the womb. Professor Jacob Hanna of the Weizmann Institute of Science said: We took mouse embryos from the mother at day five of development, when they are just of 250 cells, and had them in the incubator from day five until day 11, by which point they had grown all their organs. By day 11 they make their own blood and have a beating heart, a fully developed brain. Anybody would look at them and say, ‘this is clearly a mouse foetus with all the characteristics of a mouse.’ It’s gone from being a ball of cells to being an advanced foetus.

A special liquid is used to nourish embryo cells in a laboratory dish and they float on the liquid to duplicate the first stage of embryonic development. The incubator creates all the right conditions for its development, Hanna said. The liquid gives the embryo ‘all the nutrients, hormones and sugars they need’ along with a custom-made electronic incubator which controls gas concentration, pressure and temperature. The cu ing-edge in the underground bases and other secret locations will be light years ahead of that, however, and this was reported by the London Guardian in 2017: We are approaching a biotechnological breakthrough. Ectogenesis, the invention of a complete external womb, could completely change the nature of human reproduction. In April this year, researchers at the Children’s Hospital of Philadelphia announced their development of an artificial womb.

The article was headed ‘Artificial wombs could soon be a reality. What will this mean for women?’ What would it mean for children is an even bigger question. No mother to bond with only a machine in preparation for a life of soulless interaction and control in a world governed by machines (see the Matrix movies). Now observe the calculated manipulations of the ‘Covid’ hoax as human interaction and warmth has been curtailed by distancing, isolation and fear with people communicating via machines on a scale never seen before.

These are all dots in the same picture as are all the personal assistants, gadgets and children’s toys through which kids and adults communicate with AI as if it is human. The AI ‘voice’ on SatNav should be included. All these things are psychological preparation for the Cult endgame. Before you can make a physical connection with AI you have to make a psychological connection and that is what people are being conditioned to do with this ever gathering human-AI interaction. Movies and TV programmes depicting the transhuman, robot dystopia relate to a phenomenon known as ‘pre-emptive programming’ in which the world that is planned is portrayed everywhere in movies, TV and advertising. This is conditioning the conscious and subconscious mind to become familiar with the planned reality to dilute resistance when it happens for real. What would have been a shock such is the change is made less so. We have young children put on the road to transgender transition surgery with puberty blocking drugs at an age when they could never be able to make those life-changing decisions. Rachel Levine, a professor of paediatrics and psychiatry who believes in treating children this way, became America’s highestranked openly-transgender official when she was confirmed as US Assistant Secretary at the Department of Health and Human Services a er being nominated by Joe Biden (the Cult). Activists and governments press for laws to deny parents a say in their children’s transition process so the kids can be isolated and manipulated into agreeing to irreversible medical procedures. A Canadian father Robert Hoogland was denied bail by the Vancouver Supreme Court in 2021 and remained in jail for breaching a court order that he stay silent over his young teenage daughter, a minor, who was being offered life-changing hormone therapy without parental consent. At the age of 12 the girl’s ‘school counsellor’ said she may be transgender, referred her to a doctor and told the school to treat her like a boy. This is another example of state-serving schools imposing ever more control over children’s lives while parents have ever less.

Contemptible and extreme child abuse is happening all over the world as the Cult gender-fusion operation goes into warp-speed.

Why the war on men – and now women?

The question about what artificial wombs mean for women should rightly be asked. The answer can be seen in the deletion of women’s rights involving sport, changing rooms, toilets and status in favour of people in male bodies claiming to identify as women. I can identify as a mountain climber, but it doesn’t mean I can climb a mountain any more than a biological man can be a biological woman. To believe so is a triumph of belief over factual reality which is the very perceptual basis of everything Woke. Women’s sport is being destroyed by allowing those with male bodies who say they identify as female to ‘compete’ with girls and women. Male body ‘women’ dominate ‘women’s’ competition with their greater muscle mass, bone density, strength and speed. With that disadvantage sport for women loses all meaning. To put this in perspective nearly 300 American high school boys can run faster than the quickest woman sprinter in the world. Women are seeing their previously protected spaces invaded by male bodies simply because they claim to identify as women. That’s all they need to do to access all women’s spaces and activities under the Biden ‘Equality Act’ that destroys equality for women with the usual Orwellian Woke inversion. Male sex offenders have already commi ed rapes in women’s prisons a er claiming to identify as women to get them transferred. Does this not ma er to the Woke ‘equality’ hypocrites? Not in the least. What ma ers to Cult manipulators and funders behind transgender activists is to advance gender fusion on the way to the no-gender ‘human’. When you are seeking to impose transparent nonsense like this, or the ‘Covid’ hoax, the only way the nonsense can prevail is through censorship and intimidation of dissenters, deletion of factual information, and programming of the unquestioning, bewildered and naive. You don’t have to scan the world for long to see that all these things are happening.

Many women’s rights organisations have realised that rights and status which took such a long time to secure are being eroded and that it is systematic. Kara Dansky of the global Women’s Human Rights Campaign said that Biden’s transgender executive order immediately he took office, subsequent orders, and Equality Act legislation that followed ‘seek to erase women and girls in the law as a category’. Exactly. I said during the long ago-started war on men (in which many women play a crucial part) that this was going to turn into a war on them. The Cult is phasing out both male and female genders. To get away with that they are brought into conflict so they are busy fighting each other while the Cult completes the job with no unity of response. Unity, people, unity. We need unity everywhere. Transgender is the only show in town as the big step towards the no-gender human. It’s not about rights for transgender people and never has been. Woke political correctness is deleting words relating to genders to the same end. Wokers believe this is to be ‘inclusive’ when the opposite is true. They are deleting words describing gender because gender itself is being deleted by Human 2.0. Terms like ‘man’, ‘woman’, ‘mother’ and ‘father’ are being deleted in the universities and other institutions to be replaced by the no-gender, not trans-gender, ‘individuals’ and ‘guardians’. Women’s rights campaigner Maria Keffler of Partners for Ethical Care said: ‘Children are being taught from kindergarten upward that some boys have a vagina, some girls have a penis, and that kids can be any gender they want to be.’ Do we really believe that suddenly countries all over the world at the same time had the idea of having drag queens go into schools or read transgender stories to very young children in the local library? It’s coldly-calculated confusion of gender on the way to the fusion of gender. Suzanne Vierling, a psychologist from Southern California, made another important point: Yesterday’s slave woman who endured gynecological medical experiments is today’s girlchild being butchered in a booming gender-transitioning sector. Ovaries removed, pushing her into menopause and osteoporosis, uncharted territory, and parents’ rights and authority decimated.

The erosion of parental rights is a common theme in line with the Cult plans to erase the very concept of parents and ‘ovaries removed, pushing her into menopause’ means what? Those born female lose the ability to have children – another way to discontinue humanity as we know it.

Eliminating Human 1.0 (before our very eyes)

To pave the way for Human 2.0 you must phase out Human 1.0. This is happening through plummeting sperm counts and making women infertile through an onslaught of chemicals, radiation (including smartphones in pockets of men) and mRNA ‘vaccines’. Common agriculture pesticides are also having a devastating impact on human fertility. I have been tracking collapsing sperm counts in the books for a long time and in 2021 came a book by fertility scientist and reproductive epidemiologist Shanna Swan, Count Down: How Our Modern World Is Threatening Sperm Counts, Altering Male and Female Reproductive Development and Imperiling the Future of the Human Race. She reports how the global fertility rate dropped by half between 1960 and 2016 with America’s birth rate 16 percent below where it needs to be to sustain the population. Women are experiencing declining egg quality, more miscarriages, and more couples suffer from infertility. Other findings were an increase in erectile dysfunction, infant boys developing more genital abnormalities, male problems with conception, and plunging levels of the male hormone testosterone which would explain why so many men have lost their backbone and masculinity. This has been very evident during the ‘Covid’ hoax when women have been prominent among the Pushbackers and big strapping blokes have bowed their heads, covered their faces with a nappy and quietly submi ed. Mind control expert Cathy O’Brien also points to how global education introduced the concept of ‘we’re all winners’ in sport and classrooms: ‘Competition was defused, and it in turn defused a sense of fighting back.’ This is another version of the ‘equity’ doctrine in which you drive down rather than raise up. What a contrast in Cult-controlled China with its global ambitions

where the government published plans in January, 2021, to ‘cultivate masculinity’ in boys from kindergarten through to high school in the face of a ‘masculinity crisis’. A government adviser said boys would be soon become ‘delicate, timid and effeminate’ unless action was taken. Don’t expect any similar policy in the targeted West. A 2006 study showed that a 65-year-old man in 2002 had testosterone levels 15 percent lower than a 65-year-old man in 1987 while a 2020 study found a similar story with young adults and adolescents. Men are ge ing prescriptions for testosterone replacement therapy which causes an even greater drop in sperm count with up to 99 percent seeing sperm counts drop to zero during the treatment. More sperm is defective and malfunctioning with some having two heads or not pursuing an egg. A class of synthetic chemicals known as phthalates are being blamed for the decline. These are found everywhere in plastics, shampoos, cosmetics, furniture, flame retardants, personal care products, pesticides, canned foods and even receipts. Why till receipts? Everyone touches them. Let no one delude themselves that all this is not systematic to advance the long-time agenda for human body transformation. Phthalates mimic hormones and disrupt the hormone balance causing testosterone to fall and genital birth defects in male infants. Animals and fish have been affected in the same way due to phthalates and other toxins in rivers. When fish turn gay or change sex through chemicals in rivers and streams it is a pointer to why there has been such an increase in gay people and the sexually confused. It doesn’t ma er to me what sexuality people choose to be, but if it’s being affected by chemical pollution and consumption then we need to know. Does anyone really think that this is not connected to the transgender agenda, the war on men and the condemnation of male ‘toxic masculinity’? You watch this being followed by ‘toxic femininity’. It’s already happening. When breastfeeding becomes ‘chest-feeding’, pregnant women become pregnant people along with all the other Woke claptrap you know that the world is going insane and there’s a Cult scam in progress. Transgender activists are promoting the Cult agenda while Cult

billionaires support and fund the insanity as they laugh themselves to sleep at the sheer stupidity for which humans must be infamous in galaxies far, far away.

‘Covid vaccines’ and female infertility

We can now see why the ‘vaccine’ has been connected to potential infertility in women. Dr Michael Yeadon, former Vice President and Chief Scientific Advisor at Pfizer, and Dr Wolfgang Wodarg in Germany, filed a petition with the European Medicines Agency in December, 2020, urging them to stop trials for the Pfizer/BioNTech shot and all other mRNA trials until further studies had been done. They were particularly concerned about possible effects on fertility with ‘vaccine’-produced antibodies a acking the protein Syncytin-1 which is responsible for developing the placenta. The result would be infertility ‘of indefinite duration’ in women who have the ‘vaccine’ with the placenta failing to form. Section 10.4.2 of the Pfizer/BioNTech trial protocol says that pregnant women or those who might become so should not have mRNA shots. Section 10.4 warns men taking mRNA shots to ‘be abstinent from heterosexual intercourse’ and not to donate sperm. The UK government said that it did not know if the mRNA procedure had an effect on fertility. Did not know? These people have to go to jail. UK government advice did not recommend at the start that pregnant women had the shot and said they should avoid pregnancy for at least two months a er ‘vaccination’. The ‘advice’ was later updated to pregnant women should only have the ‘vaccine’ if the benefits outweighed the risks to mother and foetus. What the hell is that supposed to mean? Then ‘spontaneous abortions’ began to appear and rapidly increase on the adverse reaction reporting schemes which include only a fraction of adverse reactions. Thousands and ever-growing numbers of ‘vaccinated’ women are describing changes to their menstrual cycle with heavier blood flow, irregular periods and menstruating again a er going through the menopause – all links to reproduction effects. Women are passing blood clots and the lining of their uterus while men report erectile dysfunction and blood effects. Most

significantly of all unvaccinated women began to report similar menstrual changes a er interaction with ‘vaccinated’ people and men and children were also affected with bleeding noses, blood clots and other conditions. ‘Shedding’ is when vaccinated people can emit the content of a vaccine to affect the unvaccinated, but this is different. ‘Vaccinated’ people were not shedding a ‘live virus’ allegedly in ‘vaccines’ as before because the fake ‘Covid vaccines’ involve synthetic material and other toxicity. Doctors exposing what is happening prefer the term ‘transmission’ to shedding. Somehow those that have had the shots are transmi ing effects to those that haven’t. Dr Carrie Madej said the nano-content of the ‘vaccines’ can ‘act like an antenna’ to others around them which fits perfectly with my own conclusions. This ‘vaccine’ transmission phenomenon was becoming known as the book went into production and I deal with this further in the Postscript. Vaccine effects on sterility are well known. The World Health Organization was accused in 2014 of sterilising millions of women in Kenya with the evidence confirmed by the content of the vaccines involved. The same WHO behind the ‘Covid’ hoax admi ed its involvement for more than ten years with the vaccine programme. Other countries made similar claims. Charges were lodged by Tanzania, Nicaragua, Mexico, and the Philippines. The Gardasil vaccine claimed to protect against a genital ‘virus’ known as HPV has also been linked to infertility. Big Pharma and the WHO (same thing) are criminal and satanic entities. Then there’s the Bill Gates Foundation which is connected through funding and shared interests with 20 pharmaceutical giants and laboratories. He stands accused of directing the policy of United Nations Children’s Fund (UNICEF), vaccine alliance GAVI, and other groupings, to advance the vaccine agenda and silence opposition at great cost to women and children. At the same time Gates wants to reduce the global population. Coincidence?

Great Reset = Smart Grid = new human

The Cult agenda I have been exposing for 30 years is now being openly promoted by Cult assets like Gates and Klaus Schwab of the World Economic Forum under code-terms like the ‘Great Reset’, ‘Build Back Be er’ and ‘a rare but narrow window of opportunity to reflect, reimagine, and reset our world’. What provided this ‘rare but narrow window of opportunity’? The ‘Covid’ hoax did. Who created that? They did. My books from not that long ago warned about the planned ‘Internet of Things’ (IoT) and its implications for human freedom. This was the plan to connect all technology to the Internet and artificial intelligence and today we are way down that road with an estimated 36 billion devices connected to the World Wide Web and that figure is projected to be 76 billion by 2025. I further warned that the Cult planned to go beyond that to the Internet of Everything when the human brain was connected via AI to the Internet and Kurzweil’s ‘cloud’. Now we have Cult operatives like Schwab calling for precisely that under the term ‘Internet of Bodies’, a fusion of the physical, digital and biological into one centrally-controlled Smart Grid system which the Cult refers to as the ‘Fourth Industrial Revolution’. They talk about the ‘biological’, but they really mean the synthetic-biological which is required to fully integrate the human body and brain into the Smart Grid and artificial intelligence planned to replace the human mind. We have everything being synthetically manipulated including the natural world through GMO and smart dust, the food we eat and the human body itself with synthetic ‘vaccines’. I said in The Answer that we would see the Cult push for synthetic meat to replace animals and in February, 2021, the so predictable psychopath Bill Gates called for the introduction of synthetic meat to save us all from ‘climate change’. The climate hoax just keeps on giving like the ‘Covid’ hoax. The war on meat by vegan activists is a carbon (oops, sorry) copy of the manipulation of transgender activists. They have no idea (except their inner core) that they are being used to promote and impose the agenda of the Cult or that they are only the vehicle and not the reason. This is not to say those who choose not to eat meat shouldn’t be respected and supported in that right, but there are ulterior motives

for those in power. A Forbes article in December, 2019, highlighted the plan so beloved of Schwab and the Cult under the heading: ‘What Is The Internet of Bodies? And How Is It Changing Our World?’ The article said the human body is the latest data platform (remember ‘our vaccine is an operating system’). Forbes described the plan very accurately and the words could have come straight out of my books from long before: The Internet of Bodies (IoB) is an extension of the IoT and basically connects the human body to a network through devices that are ingested, implanted, or connected to the body in some way. Once connected, data can be exchanged, and the body and device can be remotely monitored and controlled.

They were really describing a human hive mind with human perception centrally-dictated via an AI connection as well as allowing people to be ‘remotely monitored and controlled’. Everything from a fridge to a human mind could be directed from a central point by these insane psychopaths and ‘Covid vaccines’ are crucial to this. Forbes explained the process I mentioned earlier of holdable and wearable technology followed by implantable. The article said there were three generations of the Internet of Bodies that include: • Body external: These are wearable devices such as Apple Watches or Fitbits that can monitor our health. • Body internal: These include pacemakers, cochlear implants, and digital pills that go inside our bodies to monitor or control various aspects of health. • Body embedded: The third generation of the Internet of Bodies is embedded technology where technology and the human body are melded together and have a real-time connection to a remote machine.

Forbes noted the development of the Brain Computer Interface (BCI) which merges the brain with an external device for monitoring and controlling in real-time. ‘The ultimate goal is to help restore function to individuals with disabilities by using brain signals rather than conventional neuromuscular pathways.’ Oh, do fuck off. The goal of brain interface technology is controlling human thought and emotion from the central point in a hive mind serving its masters wishes. Many people are now agreeing to be chipped to open doors without a key. You can recognise them because they’ll be wearing a mask, social distancing and lining up for the ‘vaccine’. The Cult plans a Great Reset money system a er they have completed the demolition of the global economy in which ‘money’ will be exchanged through communication with body operating systems. Rand Corporation, a Cult-owned think tank, said of the Internet of Bodies or IoB: Internet of Bodies technologies fall under the broader IoT umbrella. But as the name suggests, IoB devices introduce an even more intimate interplay between humans and gadgets. IoB devices monitor the human body, collect health metrics and other personal information, and transmit those data over the Internet. Many devices, such as fitness trackers, are already in use … IoB devices … and those in development can track, record, and store users’ whereabouts, bodily functions, and what they see, hear, and even think.

Schwab’s World Economic Forum, a long-winded way of saying ‘fascism’ or ‘the Cult’, has gone full-on with the Internet of Bodies in the ‘Covid’ era. ‘We’re entering the era of the Internet of Bodies’, it declared, ‘collecting our physical data via a range of devices that can be implanted, swallowed or worn’. The result would be a huge amount of health-related data that could improve human wellbeing around the world, and prove crucial in fighting the ‘Covid-19 pandemic’. Does anyone think these clowns care about ‘human wellbeing’ a er the death and devastation their pandemic hoax has purposely caused? Schwab and co say we should move forward with the Internet of Bodies because ‘Keeping track of symptoms could help us stop the spread of infection, and quickly detect new cases’. How wonderful, but keeping track’ is all they are really bothered

about. Researchers were investigating if data gathered from smartwatches and similar devices could be used as viral infection alerts by tracking the user’s heart rate and breathing. Schwab said in his 2018 book Shaping the Future of the Fourth Industrial Revolution: The lines between technologies and beings are becoming blurred and not just by the ability to create lifelike robots or synthetics. Instead it is about the ability of new technologies to literally become part of us. Technologies already influence how we understand ourselves, how we think about each other, and how we determine our realities. As the technologies … give us deeper access to parts of ourselves, we may begin to integrate digital technologies into our bodies.

You can see what the game is. Twenty-four hour control and people – if you could still call them that – would never know when something would go ping and take them out of circulation. It’s the most obvious rush to a global fascist dictatorship and the complete submission of humanity and yet still so many are locked away in their Cult-induced perceptual coma and can’t see it.

Smart Grid control centres

The human body is being transformed by the ‘vaccines’ and in other ways into a synthetic cyborg that can be a ached to the global Smart Grid which would be controlled from a central point and other sublocations of Grid manipulation. Where are these planned to be? Well, China for a start which is one of the Cult’s biggest centres of operation. The technological control system and technocratic rule was incubated here to be unleashed across the world a er the ‘Covid’ hoax came out of China in 2020. Another Smart Grid location that will surprise people new to this is Israel. I have exposed in The Trigger how Sabbatian technocrats, intelligence and military operatives were behind the horrors of 9/11 and not `19 Arab hijackers’ who somehow manifested the ability to pilot big passenger airliners when instructors at puddle-jumping flying schools described some of them as a joke. The 9/11 a acks were made possible through control of civilian and military air computer systems and those of the White House, Pentagon and connected agencies. See The Trigger – it

will blow your mind. The controlling and coordinating force were the Sabbatian networks in Israel and the United States which by then had infiltrated the entire US government, military and intelligence system. The real name of the American Deep State is ‘Sabbatian State’. Israel is a tiny country of only nine million people, but it is one of the global centres of cyber operations and fast catching Silicon Valley in importance to the Cult. Israel is known as the ‘start-up nation’ for all the cyber companies spawned there with the Sabbatian specialisation of ‘cyber security’ that I mentioned earlier which gives those companies access to computer systems of their clients in real time through ‘backdoors’ wri en into the coding when security so ware is downloaded. The Sabbatian centre of cyber operations outside Silicon Valley is the Israeli military Cyber Intelligence Unit, the biggest infrastructure project in Israel’s history, headquartered in the desert-city of Beersheba and involving some 20,000 ‘cyber soldiers’. Here are located a literal army of Internet trolls scanning social media, forums and comment lists for anyone challenging the Cult agenda. The UK military has something similar with its 77th Brigade and associated operations. The Beersheba complex includes research and development centres for other Cult operations such as Intel, Microso , IBM, Google, Apple, Hewle Packard, Cisco Systems, Facebook and Motorola. Techcrunch.com ran an article about the Beersheba global Internet technology centre headlined ‘Israel’s desert city of Beersheba is turning into a cybertech oasis’: The military’s massive relocation of its prestigious technology units, the presence of multinational and local companies, a close proximity to Ben Gurion University and generous government subsidies are turning Beersheba into a major global cybertech hub. Beersheba has all of the ingredients of a vibrant security technology ecosystem, including Ben Gurion University with its graduate program in cybersecurity and Cyber Security Research Center, and the presence of companies such as EMC, Deutsche Telekom, PayPal, Oracle, IBM, and Lockheed Martin. It’s also the future home of the INCB (Israeli National Cyber Bureau); offers a special income tax incentive for cyber security companies, and was the site for the relocation of the army’s intelligence corps units.

Sabbatians have taken over the cyber world through the following process: They scan the schools for likely cyber talent and develop them at Ben Gurion University and their period of conscription in the Israeli Defense Forces when they are stationed at the Beersheba complex. When the cyber talented officially leave the army they are funded to start cyber companies with technology developed by themselves or given to them by the state. Much of this is stolen through backdoors of computer systems around the world with America top of the list. Others are sent off to Silicon Valley to start companies or join the major ones and so we have many major positions filled by apparently ‘Jewish’ but really Sabbatian operatives. Google, YouTube and Facebook are all run by ‘Jewish’ CEOs while Twi er is all but run by ultra-Zionist hedge-fund shark Paul Singer. At the centre of the Sabbatian global cyber web is the Israeli army’s Unit 8200 which specialises in hacking into computer systems of other countries, inserting viruses, gathering information, instigating malfunction, and even taking control of them from a distance. A long list of Sabbatians involved with 9/11, Silicon Valley and Israeli cyber security companies are operatives of Unit 8200. This is not about Israel. It’s about the Cult. Israel is planned to be a Smart Grid hub as with China and what is happening at Beersheba is not for the benefit of Jewish people who are treated disgustingly by the Sabbatian elite that control the country. A glance at the Nuremberg Codes will tell you that. The story is much bigger than ‘Covid’, important as that is to where we are being taken. Now, though, it’s time to really strap in. There’s more … much more …

CHAPTER ELEVEN Who controls the Cult? Awake, arise or be forever fall’n John Milton, Paradise Lost

I

have exposed this far the level of the Cult conspiracy that operates in the world of the seen and within the global secret society and satanic network which operates in the shadows one step back from the seen. The story, however, goes much deeper than that. The ‘Covid’ hoax is major part of the Cult agenda, but only part, and to grasp the biggest picture we have to expand our a ention beyond the realm of human sight and into the infinity of possibility that we cannot see. It is from here, ultimately, that humanity is being manipulated into a state of total control by the force which dictates the actions of the Cult. How much of reality can we see? Next to damn all is the answer. We may appear to see all there is to see in the ‘space’ our eyes survey and observe, but li le could be further from the truth. The human ‘world’ is only a tiny band of frequency that the body’s visual and perceptual systems can decode into perception of a ‘world’. According to mainstream science the electromagnetic spectrum is 0.005 percent of what exists in the Universe (Fig 10). The maximum estimate I have seen is 0.5 percent and either way it’s miniscule. I say it is far, far, smaller even than 0.005 percent when you compare reality we see with the totality of reality that we don’t. Now get this if you are new to such information: Visible light, the only band of frequency that we can see, is a fraction of the 0.005

percent (Fig 11 overleaf). Take this further and realise that our universe is one of infinite universes and that universes are only a fragment of overall reality – infinite reality. Then compare that with the almost infinitesimal frequency band of visible light or human sight. You see that humans are as near blind as it is possible to be without actually being so. Artist and filmmaker, Sergio Toporek, said:

Figure 10: Humans can perceive such a tiny band of visual reality it’s laughable.

Figure 11: We can see a smear of the 0.005 percent electromagnetic spectrum, but we still know it all. Yep, makes sense. Consider that you can see less than 1% of the electromagnetic spectrum and hear less than 1% of the acoustic spectrum. 90% of the cells in your body carry their own microbial DNA and are not ‘you’. The atoms in your body are 99.9999999999999999% empty space and none of them are the ones you were born with ... Human beings have 46 chromosomes, two less than a potato.

The existence of the rainbow depends on the conical photoreceptors in your eyes; to animals without cones, the rainbow does not exist. So you don’t just look at a rainbow, you create it. This is pretty amazing, especially considering that all the beautiful colours you see represent less than 1% of the electromagnetic spectrum.

Suddenly the ‘world’ of humans looks a very different place. Take into account, too, that Planet Earth when compared with the projected size of this single universe is the equivalent of a billionth of a pinhead. Imagine the ratio that would be when compared to infinite reality. To think that Christianity once insisted that Earth and humanity were the centre of everything. This background is vital if we are going to appreciate the nature of ‘human’ and how we can be manipulated by an unseen force. To human visual reality virtually everything is unseen and yet the prevailing perception within the institutions and so much of the public is that if we can’t see it, touch it, hear it, taste it and smell it then it cannot exist. Such perception is indoctrinated and encouraged by the Cult and its agents because it isolates believers in the strictly limited, village-idiot, realm of the five senses where perceptions can be firewalled and information controlled. Most of those perpetuating the ‘this-world-is-all-there-is’ insanity are themselves indoctrinated into believing the same delusion. While major players and influencers know that official reality is laughable most of those in science, academia and medicine really believe the nonsense they peddle and teach succeeding generations. Those who challenge the orthodoxy are dismissed as nu ers and freaks to protect the manufactured illusion from exposure. Observe the dynamic of the ‘Covid’ hoax and you will see how that takes the same form. The inner-circle psychopaths knows it’s a gigantic scam, but almost the entirety of those imposing their fascist rules believe that ‘Covid’ is all that they’re told it is.

Stolen identity

Ask people who they are and they will give you their name, place of birth, location, job, family background and life story. Yet that is not who they are – it is what they are experiencing. The difference is absolutely crucial. The true ‘I’, the eternal, infinite ‘I’, is consciousness,

a state of being aware. Forget ‘form’. That is a vehicle for a brief experience. Consciousness does not come from the brain, but through the brain and even that is more symbolic than literal. We are awareness, pure awareness, and this is what withdraws from the body at what we call ‘death’ to continue our eternal beingness, isness, in other realms of reality within the limitlessness of infinity or the Biblical ‘many mansions in my father’s house’. Labels of a human life, man, woman, transgender, black, white, brown, nationality, circumstances and income are not who we are. They are what we are – awareness – is experiencing in a brief connection with a band of frequency we call ‘human’. The labels are not the self; they are, to use the title of one of my books, a Phantom Self. I am not David Icke born in Leicester, England, on April 29th, 1952. I am the consciousness having that experience. The Cult and its non-human masters seek to convince us through the institutions of ‘education’, science, medicine, media and government that what we are experiencing is who we are. It’s so easy to control and direct perception locked away in the bewildered illusions of the five senses with no expanded radar. Try, by contrast, doing the same with a humanity aware of its true self and its true power to consciously create its reality and experience. How is it possible to do this? We do it all day every day. If you perceive yourself as ‘li le me’ with no power to impact upon your life and the world then your life experience will reflect that. You will hand the power you don’t think you have to authority in all its forms which will use it to control your experience. This, in turn, will appear to confirm your perception of ‘li le me’ in a self-fulfilling feedback loop. But that is what ‘li le me’ really is – a perception. We are all ‘big-me’, infinite me, and the Cult has to make us forget that if its will is to prevail. We are therefore manipulated and pressured into self-identifying with human labels and not the consciousness/awareness experiencing those human labels. The phenomenon of identity politics is a Cult-instigated manipulation technique to sub-divide previous labels into even smaller ones. A United States university employs this list of le ers to

describe student identity: LGBTTQQFAGPBDSM or lesbian, gay, bisexual, transgender, transsexual, queer, questioning, flexual, asexual, gender-fuck, polyamorous, bondage/discipline, dominance/submission and sadism/masochism. I’m sure other lists are even longer by now as people feel the need to self-identity the ‘I’ with the minutiae of race and sexual preference. Wokers programmed by the Cult for generations believe this is about ‘inclusivity’ when it’s really the Cult locking them away into smaller and smaller versions of Phantom Self while firewalling them from the influence of their true self, the infinite, eternal ‘I’. You may notice that my philosophy which contends that we are all unique points of a ention/awareness within the same infinite whole or Oneness is the ultimate non-racism. The very sense of Oneness makes the judgement of people by their body-type, colour or sexuality u erly ridiculous and confirms that racism has no understanding of reality (including anti-white racism). Yet despite my perception of life Cult agents and fast-asleep Wokers label me racist to discredit my information while they are themselves phenomenally racist and sexist. All they see is race and sexuality and they judge people as good or bad, demons or untouchables, by their race and sexuality. All they see is Phantom Self and perceive themselves in terms of Phantom Self. They are pawns and puppets of the Cult agenda to focus a ention and self-identity in the five senses and play those identities against each other to divide and rule. Columbia University has introduced segregated graduations in another version of social distancing designed to drive people apart and teach them that different racial and cultural groups have nothing in common with each other. The last thing the Cult wants is unity. Again the pumpprimers of this will be Cult operatives in the knowledge of what they are doing, but the rest are just the Phantom Self blind leading the Phantom Self blind. We do have something in common – we are all the same consciousness having different temporary experiences.

What is this ‘human’?

Yes, what is ‘human’? That is what we are supposed to be, right? I mean ‘human’? True, but ‘human’ is the experience not the ‘I’. Break it down to basics and ‘human’ is the way that information is processed. If we are to experience and interact with this band of frequency we call the ‘world’ we must have a vehicle that operates within that band of frequency. Our consciousness in its prime form cannot do that; it is way beyond the frequency of the human realm. My consciousness or awareness could not tap these keys and pick up the cup in front of me in the same way that radio station A cannot interact with radio station B when they are on different frequencies. The human body is the means through which we have that interaction. I have long described the body as a biological computer which processes information in a way that allows consciousness to experience this reality. The body is a receiver, transmi er and processor of information in a particular way that we call human. We visually perceive only the world of the five senses in a wakened state – that is the limit of the body’s visual decoding system. In truth it’s not even visual in the way we experience ‘visual reality’ as I will come to in a moment. We are ‘human’ because the body processes the information sources of human into a reality and behaviour system that we perceive as human. Why does an elephant act like an elephant and not like a human or a duck? The elephant’s biological computer is a different information field and processes information according to that program into a visual and behaviour type we call an elephant. The same applies to everything in our reality. These body information fields are perpetuated through procreation (like making a copy of a so ware program). The Cult wants to break that cycle and intervene technologically to transform the human information field into one that will change what we call humanity. If it can change the human information field it will change the way that field processes information and change humanity both ‘physically’ and psychologically. Hence the messenger (information) RNA ‘vaccines’ and so much more that is targeting human genetics by changing the body’s information – messaging – construct through food, drink, radiation, toxicity and other means.

Reality that we experience is nothing like reality as it really is in the same way that the reality people experience in virtual reality games is not the reality they are really living in. The game is only a decoded source of information that appears to be a reality. Our world is also an information construct – a simulation (more later). In its base form our reality is a wavefield of information much the same in theme as Wi-Fi. The five senses decode wavefield information into electrical information which they communicate to the brain to decode into holographic (illusory ‘physical’) information. Different parts of the brain specialise in decoding different senses and the information is fused into a reality that appears to be outside of us but is really inside the brain and the genetic structure in general (Fig 12 overleaf). DNA is a receiver-transmi er of information and a vital part of this decoding process and the body’s connection to other realities. Change DNA and you change the way we decode and connect with reality – see ‘Covid vaccines’. Think of computers decoding Wi-Fi. You have information encoded in a radiation field and the computer decodes that information into a very different form on the screen. You can’t see the Wi-Fi until its information is made manifest on the screen and the information on the screen is inside the computer and not outside. I have just described how we decode the ‘human world’. All five senses decode the waveform ‘WiFi’ field into electrical signals and the brain (computer) constructs reality inside the brain and not outside – ‘You don’t just look at a rainbow, you create it’. Sound is a simple example. We don’t hear sound until the brain decodes it. Waveform sound waves are picked up by the hearing sense and communicated to the brain in an electrical form to be decoded into the sounds that we hear. Everything we hear is inside the brain along with everything we see, feel, smell and taste. Words and language are waveform fields generated by our vocal chords which pass through this process until they are decoded by the brain into words that we hear. Different languages are different frequency fields or sound waves generated by vocal chords. Late British philosopher Alan Wa s said:

Figure 12: The brain receives information from the five senses and constructs from that our perceived reality. [Without the brain] the world is devoid of light, heat, weight, solidity, motion, space, time or any other imaginable feature. All these phenomena are interactions, or transactions, of vibrations with a certain arrangement of neurons.

That’s exactly what they are and scientist Robert Lanza describes in his book, Biocentrism, how we decode electromagnetic waves and energy into visual and ‘physical’ experience. He uses the example of a flame emi ing photons, electromagnetic energy, each pulsing electrically and magnetically: … these … invisible electromagnetic waves strike a human retina, and if (and only if) the waves happen to measure between 400 and 700 nano meters in length from crest to crest, then their energy is just right to deliver a stimulus to the 8 million cone-shaped cells in the retina. Each in turn send an electrical pulse to a neighbour neuron, and on up the line this goes, at 250 mph, until it reaches the … occipital lobe of the brain, in the back of the head. There, a cascading complex of neurons fire from the incoming stimuli, and we subjectively perceive this experience as a yellow brightness occurring in a place we have been conditioned to call the ‘external world’.

You hear what you decode

If a tree falls or a building collapses they make no noise unless someone is there to decode the energetic waves generated by the disturbance into what we call sound. Does a falling tree make a noise? Only if you hear it – decode it. Everything in our reality is a frequency field of information operating within the overall ‘Wi-Fi’ field that I call The Field. A vibrational disturbance is generated in The Field by the fields of the falling tree or building. These disturbance waves are what we decode into the sound of them falling. If no one is there to do that then neither will make any noise. Reality is created by the observer – decoder – and the perceptions of the observer affect the decoding process. For this reason different people – different perceptions – will perceive the same reality or situation in a different way. What one may perceive as a nightmare another will see as an opportunity. The question of why the Cult is so focused on controlling human perception now answers itself. All experienced reality is the act of decoding and we don’t experience Wi-Fi until it is decoded on the computer screen. The sight and sound of an Internet video is encoded in the Wi-Fi all around us, but we don’t see or hear it until the computer decodes that information. Taste, smell and touch are all phenomena of the brain as a result of the same process. We don’t taste, smell or feel anything except in the brain and there are pain relief techniques that seek to block the signal from the site of discomfort to the brain because if the brain doesn’t decode that signal we don’t feel pain. Pain is in the brain and only appears to be at the point of impact thanks to the feedback loop between them. We don’t see anything until electrical information from the sight senses is decoded in an area at the back of the brain. If that area is damaged we can go blind when our eyes are perfectly okay. So why do we go blind if we damage an eye? We damage the information processing between the waveform visual information and the visual decoding area of the brain. If information doesn’t reach the brain in a form it can decode then we can’t see the visual reality that it represents. What’s more the brain is decoding only a fraction of the information it receives and the rest is absorbed by the

sub-conscious mind. This explanation is from the science magazine, Wonderpedia: Every second, 11 million sensations crackle along these [brain] pathways ... The brain is confronted with an alarming array of images, sounds and smells which it rigorously filters down until it is left with a manageable list of around 40. Thus 40 sensations per second make up what we perceive as reality.

The ‘world’ is not what people are told to believe that is it and the inner circles of the Cult know that.

Illusory ‘physical’ reality

We can only see a smear of 0.005 percent of the Universe which is only one of a vast array of universes – ‘mansions’ – within infinite reality. Even then the brain decodes only 40 pieces of information (‘sensations’) from a potential 11 million that we receive every second. Two points strike you from this immediately: The sheer breathtaking stupidity of believing we know anything so rigidly that there’s nothing more to know; and the potential for these processes to be manipulated by a malevolent force to control the reality of the population. One thing I can say for sure with no risk of contradiction is that when you can perceive an almost indescribable fraction of infinite reality there is always more to know as in tidal waves of it. Ancient Greek philosopher Socrates was so right when he said that wisdom is to know how li le we know. How obviously true that is when you think that we are experiencing a physical world of solidity that is neither physical nor solid and a world of apartness when everything is connected. Cult-controlled ‘science’ dismisses the socalled ‘paranormal’ and all phenomena related to that when the ‘para’-normal is perfectly normal and explains the alleged ‘great mysteries’ which dumbfound scientific minds. There is a reason for this. A ‘scientific mind’ in terms of the mainstream is a material mind, a five-sense mind imprisoned in see it, touch it, hear it, smell it and taste it. Phenomena and happenings that can’t be explained that way leave the ‘scientific mind’ bewildered and the rule is that if they

can’t account for why something is happening then it can’t, by definition, be happening. I beg to differ. Telepathy is thought waves passing through The Field (think wave disturbance again) to be decoded by someone able to connect with that wavelength (information). For example: You can pick up the thought waves of a friend at any distance and at the very least that will bring them to mind. A few minutes later the friend calls you. ‘My god’, you say, ‘that’s incredible – I was just thinking of you.’ Ah, but they were thinking of you before they made the call and that’s what you decoded. Native peoples not entrapped in five-sense reality do this so well it became known as the ‘bush telegraph’. Those known as psychics and mediums (genuine ones) are doing the same only across dimensions of reality. ‘Mind over ma er’ comes from the fact that ma er and mind are the same. The state of one influences the state of the other. Indeed one and the other are illusions. They are aspects of the same field. Paranormal phenomena are all explainable so why are they still considered ‘mysteries’ or not happening? Once you go down this road of understanding you begin to expand awareness beyond the five senses and that’s the nightmare for the Cult.

Figure 13: Holograms are not solid, but the best ones appear to be.

Figure 14: How holograms are created by capturing a waveform version of the subject image.

Holographic ‘solidity’

Our reality is not solid, it is holographic. We are now well aware of holograms which are widely used today. Two-dimensional information is decoded into a three-dimensional reality that is not solid although can very much appear to be (Fig 13). Holograms are created with a laser divided into two parts. One goes directly onto a holographic photographic print (‘reference beam’) and the other takes a waveform image of the subject (‘working beam’) before being directed onto the print where it ‘collides’ with the other half of the laser (Fig 14). This creates a waveform interference pa ern which contains the wavefield information of whatever is being photographed (Fig 15 overleaf). The process can be likened to dropping pebbles in a pond. Waves generated by each one spread out across the water to collide with the others and create a wave representation of where the stones fell and at what speed, weight and distance. A waveform interference pa ern of a hologram is akin to the waveform information in The Field which the five senses decode into electrical signals to be decoded by the brain into a holographic illusory ‘physical’ reality. In the same way when a laser (think human a ention) is directed at the waveform interference pa ern a three-dimensional version of the subject is projected into apparently ‘solid’ reality (Fig 16). An amazing trait of holograms reveals more ‘paranormal mysteries’. Information of the whole

hologram is encoded in waveform in every part of the interference pa ern by the way they are created. This means that every part of a hologram is a smaller version of the whole. Cut the interference wave-pa ern into four and you won’t get four parts of the image. You get quarter-sized versions of the whole image. The body is a hologram and the same applies. Here we have the basis of acupuncture, reflexology and other forms of healing which identify representations of the whole body in all of the parts, hands, feet, ears, everywhere. Skilled palm readers can do what they do because the information of whole body is encoded in the hand. The concept of as above, so below, comes from this.

Figure 15: A waveform interference pattern that holds the information that transforms into a hologram.

Figure 16: Holographic people including ‘Elvis’ holographically inserted to sing a duet with Celine Dion.

The question will be asked of why, if solidity is illusory, we can’t just walk through walls and each other. The resistance is not solid against solid; it is electromagnetic field against electromagnetic field and we decode this into the experience of solid against solid. We should also not underestimate the power of belief to dictate reality. What you believe is impossible will be. Your belief impacts on your decoding processes and they won’t decode what you think is impossible. What we believe we perceive and what we perceive we experience. ‘Can’t dos’ and ‘impossibles’ are like a firewall in a computer system that won’t put on the screen what the firewall blocks. How vital that is to understanding how human experience has been hijacked. I explain in The Answer, Everything You Need To Know But Have Never Been Told and other books a long list of ‘mysteries’ and ‘paranormal’ phenomena that are not mysterious and perfectly normal once you realise what reality is and how it works. ‘Ghosts’ can be seen to pass through ‘solid’ walls because the walls are not solid and the ghost is a discarnate entity operating on a frequency so different to that of the wall that it’s like two radio stations sharing the same space while never interfering with each other. I have seen ghosts do this myself. The apartness of people and objects is also an illusion. Everything is connected by the Field like all sea life is connected by the sea. It’s just that within the limits of our visual reality we only ‘see’ holographic information and not the field of information that connects everything and from which the holographic world is made manifest. If you can only see holographic ‘objects’ and not the field that connects them they will appear to you as unconnected to each other in the same way that we see the computer while not seeing the Wi-Fi.

What you don’t know can hurt you

Okay, we return to those ‘two worlds’ of human society and the Cult with its global network of interconnecting secret societies and satanic groups which manipulate through governments, corporations, media, religions, etc. The fundamental difference between them is knowledge. The idea has been to keep humanity

ignorant of the plan for its total enslavement underpinned by a crucial ignorance of reality – who we are and where we are – and how we interact with it. ‘Human’ should be the interaction between our expanded eternal consciousness and the five-sense body experience. We are meant to be in this world in terms of the five senses but not of this world in relation to our greater consciousness and perspective. In that state we experience the small picture of the five senses within the wider context of the big picture of awareness beyond the five senses. Put another way the five senses see the dots and expanded awareness connects them into pictures and pa erns that give context to the apparently random and unconnected. Without the context of expanded awareness the five senses see only apartness and randomness with apparently no meaning. The Cult and its other-dimensional controllers seek to intervene in the frequency realm where five-sense reality is supposed to connect with expanded reality and to keep the two apart (more on this in the final chapter). When that happens five-sense mental and emotional processes are no longer influenced by expanded awareness, or the True ‘I’, and instead are driven by the isolated perceptions of the body’s decoding systems. They are in the world and of it. Here we have the human plight and why humanity with its potential for infinite awareness can be so easily manipulatable and descend into such extremes of stupidity. Once the Cult isolates five-sense mind from expanded awareness it can then program the mind with perceptions and beliefs by controlling information that the mind receives through the ‘education’ system of the formative years and the media perceptual bombardment and censorship of an entire lifetime. Limit perception and a sense of the possible through limiting knowledge by limiting and skewing information while censoring and discrediting that which could set people free. As the title of another of my books says … And The Truth Shall Set You Free. For this reason the last thing the Cult wants in circulation is the truth about anything – especially the reality of the eternal ‘I’ – and that’s why it is desperate to control information. The Cult knows that information becomes perception

which becomes behaviour which, collectively, becomes human society. Cult-controlled and funded mainstream ‘science’ denies the existence of an eternal ‘I’ and seeks to dismiss and trash all evidence to the contrary. Cult-controlled mainstream religion has a version of ‘God’ that is li le more than a system of control and dictatorship that employs threats of damnation in an a erlife to control perceptions and behaviour in the here and now through fear and guilt. Neither is true and it’s the ‘neither’ that the Cult wishes to suppress. This ‘neither’ is that everything is an expression, a point of a ention, within an infinite state of consciousness which is the real meaning of the term ‘God’. Perceptual obsession with the ‘physical body’ and five-senses means that ‘God’ becomes personified as a bearded bloke si ing among the clouds or a raging bully who loves us if we do what ‘he’ wants and condemns us to the fires of hell if we don’t. These are no more than a ‘spiritual’ fairy tales to control and dictate events and behaviour through fear of this ‘God’ which has bizarrely made ‘Godfearing’ in religious circles a state to be desired. I would suggest that fearing anything is not to be encouraged and celebrated, but rather deleted. You can see why ‘God fearing’ is so beneficial to the Cult and its religions when they decide what ‘God’ wants and what ‘God’ demands (the Cult demands) that everyone do. As the great American comedian Bill Hicks said satirising a Christian zealot: ‘I think what God meant to say.’ How much of this infinite awareness (‘God’) that we access is decided by how far we choose to expand our perceptions, self-identity and sense of the possible. The scale of self-identity reflects itself in the scale of awareness that we can connect with and are influenced by – how much knowing and insight we have instead of programmed perception. You cannot expand your awareness into the infinity of possibility when you believe that you are li le me Peter the postman or Mary in marketing and nothing more. I’ll deal with this in the concluding chapter because it’s crucial to how we turnaround current events.

Where the Cult came from

When I realised in the early 1990s there was a Cult network behind global events I asked the obvious question: When did it start? I took it back to ancient Rome and Egypt and on to Babylon and Sumer in Mesopotamia, the ‘Land Between Two Rivers’, in what we now call Iraq. The two rivers are the Tigris and Euphrates and this region is of immense historical and other importance to the Cult, as is the land called Israel only 550 miles away by air. There is much more going with deep esoteric meaning across this whole region. It’s not only about ‘wars for oil’. Priceless artefacts from Mesopotamia were stolen or destroyed a er the American and British invasion of Iraq in 2003 justified by the lies of Boy Bush and Tony Blair (their Cult masters) about non-existent ‘weapons of mass destruction’. Mesopotamia was the location of Sumer (about 5,400BC to 1,750BC), and Babylon (about 2,350BC to 539BC). Sabbatians may have become immensely influential in the Cult in modern times but they are part of a network that goes back into the mists of history. Sumer is said by historians to be the ‘cradle of civilisation’. I disagree. I say it was the re-start of what we call human civilisation a er cataclysmic events symbolised in part as the ‘Great Flood’ destroyed the world that existed before. These fantastic upheavals that I have been describing in detail in the books since the early1990s appear in accounts and legends of ancient cultures across the world and they are supported by geological and biological evidence. Stone tablets found in Iraq detailing the Sumer period say the cataclysms were caused by nonhuman ‘gods’ they call the Anunnaki. These are described in terms of extraterrestrial visitations in which knowledge supplied by the Anunnaki is said to have been the source of at least one of the world’s oldest writing systems and developments in astronomy, mathematics and architecture that were way ahead of their time. I have covered this subject at length in The Biggest Secret and Children of the Matrix and the same basic ‘Anunnaki’ story can be found in Zulu accounts in South Africa where the late and very great Zulu high shaman Credo Mutwa told me that the Sumerian Anunnaki were known by Zulus as the Chitauri or ‘children of the serpent’. See my six-hour video interview with Credo on this subject entitled The

Reptilian Agenda recorded at his then home near Johannesburg in 1999 which you can watch on the Ickonic media platform. The Cult emerged out of Sumer, Babylon and Egypt (and elsewhere) and established the Roman Empire before expanding with the Romans into northern Europe from where many empires were savagely imposed in the form of Cult-controlled societies all over the world. Mass death and destruction was their calling card. The Cult established its centre of operations in Europe and European Empires were Cult empires which allowed it to expand into a global force. Spanish and Portuguese colonialists headed for Central and South America while the British and French targeted North America. Africa was colonised by Britain, France, Belgium, the Netherlands, Portugal, Spain, Italy, and Germany. Some like Britain and France moved in on the Middle East. The British Empire was by far the biggest for a simple reason. By now Britain was the headquarters of the Cult from which it expanded to form Canada, the United States, Australia and New Zealand. The Sun never set on the British Empire such was the scale of its occupation. London remains a global centre for the Cult along with Rome and the Vatican although others have emerged in Israel and China. It is no accident that the ‘virus’ is alleged to have come out of China while Italy was chosen as the means to terrify the Western population into compliance with ‘Covid’ fascism. Nor that Israel has led the world in ‘Covid’ fascism and mass ‘vaccination’. You would think that I would mention the United States here, but while it has been an important means of imposing the Cult’s will it is less significant than would appear and is currently in the process of having what power it does have deleted. The Cult in Europe has mostly loaded the guns for the US to fire. America has been controlled from Europe from the start through Cult operatives in Britain and Europe. The American Revolution was an illusion to make it appear that America was governing itself while very different forces were pulling the strings in the form of Cult families such as the Rothschilds through the Rockefellers and other subordinates. The Rockefellers are extremely close to Bill Gates and

established both scalpel and drug ‘medicine’ and the World Health Organization. They play a major role in the development and circulation of vaccines through the Rockefeller Foundation on which Bill Gates said his Foundation is based. Why wouldn’t this be the case when the Rockefellers and Gates are on the same team? Cult infiltration of human society goes way back into what we call history and has been constantly expanding and centralising power with the goal of establishing a global structure to dictate everything. Look how this has been advanced in great leaps with the ‘Covid’ hoax.

The non-human dimension

I researched and observed the comings and goings of Cult operatives through the centuries and even thousands of years as they were born, worked to promote the agenda within the secret society and satanic networks, and then died for others to replace them. Clearly there had to be a coordinating force that spanned this entire period while operatives who would not have seen the end goal in their lifetimes came and went advancing the plan over millennia. I went in search of that coordinating force with the usual support from the extraordinary synchronicity of my life which has been an almost daily experience since 1990. I saw common themes in religious texts and ancient cultures about a non-human force manipulating human society from the hidden. Christianity calls this force Satan, the Devil and demons; Islam refers to the Jinn or Djinn; Zulus have their Chitauri (spelt in other ways in different parts of Africa); and the Gnostic people in Egypt in the period around and before 400AD referred to this phenomena as the ‘Archons’, a word meaning rulers in Greek. Central American cultures speak of the ‘Predators’ among other names and the same theme is everywhere. I will use ‘Archons’ as a collective name for all of them. When you see how their nature and behaviour is described all these different sources are clearly talking about the same force. Gnostics described the Archons in terms of ‘luminous fire’ while Islam relates the Jinn to ‘smokeless fire’. Some refer to beings in form that could occasionally be seen, but the most common of common theme is that they operate from

unseen realms which means almost all existence to the visual processes of humans. I had concluded that this was indeed the foundation of human control and that the Cult was operating within the human frequency band on behalf of this hidden force when I came across the writings of Gnostics which supported my conclusions in the most extraordinary way. A sealed earthen jar was found in 1945 near the town of Nag Hammadi about 75-80 miles north of Luxor on the banks of the River Nile in Egypt. Inside was a treasure trove of manuscripts and texts le by the Gnostic people some 1,600 years earlier. They included 13 leather-bound papyrus codices (manuscripts) and more than 50 texts wri en in Coptic Egyptian estimated to have been hidden in the jar in the period of 400AD although the source of the information goes back much further. Gnostics oversaw the Great or Royal Library of Alexandria, the fantastic depository of ancient texts detailing advanced knowledge and accounts of human history. The Library was dismantled and destroyed in stages over a long period with the death-blow delivered by the Cult-established Roman Church in the period around 415AD. The Church of Rome was the Church of Babylon relocated as I said earlier. Gnostics were not a race. They were a way of perceiving reality. Whenever they established themselves and their information circulated the terrorists of the Church of Rome would target them for destruction. This happened with the Great Library and with the Gnostic Cathars who were burned to death by the psychopaths a er a long period of oppression at the siege of the Castle of Monségur in southern France in 1244. The Church has always been terrified of Gnostic information which demolishes the official Christian narrative although there is much in the Bible that supports the Gnostic view if you read it in another way. To anyone studying the texts of what became known as the Nag Hammadi Library it is clear that great swathes of Christian and Biblical belief has its origin with Gnostics sources going back to Sumer. Gnostic themes have been twisted to manipulate the perceived reality of Bible believers. Biblical texts have been in the open for centuries where they could be changed while Gnostic

documents found at Nag Hammadi were sealed away and untouched for 1,600 years. What you see is what they wrote.

Use your pneuma not your nous

Gnosticism and Gnostic come from ‘gnosis’ which means knowledge, or rather secret knowledge, in the sense of spiritual awareness – knowledge about reality and life itself. The desperation of the Cult’s Church of Rome to destroy the Gnostics can be understood when the knowledge they were circulating was the last thing the Cult wanted the population to know. Sixteen hundred years later the same Cult is working hard to undermine and silence me for the same reason. The dynamic between knowledge and ignorance is a constant. ‘Time’ appears to move on, but essential themes remain the same. We are told to ‘use your nous’, a Gnostic word for head/brain/intelligence. They said, however, that spiritual awakening or ‘salvation’ could only be secured by expanding awareness beyond what they called nous and into pneuma or Infinite Self. Obviously as I read these texts the parallels with what I have been saying since 1990 were fascinating to me. There is a universal truth that spans human history and in that case why wouldn’t we be talking the same language 16 centuries apart? When you free yourself from the perception program of the five senses and explore expanded realms of consciousness you are going to connect with the same information no ma er what the perceived ‘era’ within a manufactured timeline of a single and tiny range of manipulated frequency. Humans working with ‘smart’ technology or knocking rocks together in caves is only a timeline appearing to operate within the human frequency band. Expanded awareness and the knowledge it holds have always been there whether the era be Stone Age or computer age. We can only access that knowledge by opening ourselves to its frequency which the five-sense prison cell is designed to stop us doing. Gates, Fauci, Whi y, Vallance, Zuckerberg, Brin, Page, Wojcicki, Bezos, and all the others behind the ‘Covid’ hoax clearly have a long wait before their range of frequency can make that connection given that an open heart is

crucial to that as we shall see. Instead of accessing knowledge directly through expanded awareness it is given to Cult operatives by the secret society networks of the Cult where it has been passed on over thousands of years outside the public arena. Expanded realms of consciousness is where great artists, composers and writers find their inspiration and where truth awaits anyone open enough to connect with it. We need to go there fast.

Archon hijack

A fi h of the Nag Hammadi texts describe the existence and manipulation of the Archons led by a ‘Chief Archon’ they call ‘Yaldabaoth’, or the ‘Demiurge’, and this is the Christian ‘Devil’, ‘Satan’, ‘Lucifer’, and his demons. Archons in Biblical symbolism are the ‘fallen ones’ which are also referred to as fallen angels a er the angels expelled from heaven according to the Abrahamic religions of Judaism, Christianity and Islam. These angels are claimed to tempt humans to ‘sin’ ongoing and you will see how accurate that symbolism is during the rest of the book. The theme of ‘original sin’ is related to the ‘Fall’ when Adam and Eve were ‘tempted by the serpent’ and fell from a state of innocence and ‘obedience’ (connection) with God into a state of disobedience (disconnection). The Fall is said to have brought sin into the world and corrupted everything including human nature. Yaldabaoth, the ‘Lord Archon’, is described by Gnostics as a ‘counterfeit spirit’, ‘The Blind One’, ‘The Blind God’, and ‘The Foolish One’. The Jewish name for Yaldabaoth in Talmudic writings is Samael which translates as ‘Poison of God’, or ‘Blindness of God’. You see the parallels. Yaldabaoth in Islamic belief is the Muslim Jinn devil known as Shaytan – Shaytan is Satan as the same themes are found all over the world in every religion and culture. The ‘Lord God’ of the Old Testament is the ‘Lord Archon’ of Gnostic manuscripts and that’s why he’s such a bloodthirsty bastard. Satan is known by Christians as ‘the Demon of Demons’ and Gnostics called Yaldabaoth the ‘Archon of Archons’. Both are known as ‘The Deceiver’. We are talking about the same ‘bloke’ for sure and these common themes

using different names, storylines and symbolism tell a common tale of the human plight. Archons are referred to in Nag Hammadi documents as mind parasites, inverters, guards, gatekeepers, detainers, judges, pitiless ones and deceivers. The ‘Covid’ hoax alone is a glaring example of all these things. The Biblical ‘God’ is so different in the Old and New Testaments because they are not describing the same phenomenon. The vindictive, angry, hate-filled, ‘God’ of the Old Testament, known as Yahweh, is Yaldabaoth who is depicted in Cult-dictated popular culture as the ‘Dark Lord’, ‘Lord of Time’, Lord (Darth) Vader and Dormammu, the evil ruler of the ‘Dark Dimension’ trying to take over the ‘Earth Dimension’ in the Marvel comic movie, Dr Strange. Yaldabaoth is both the Old Testament ‘god’ and the Biblical ‘Satan’. Gnostics referred to Yaldabaoth as the ‘Great Architect of the Universe’and the Cult-controlled Freemason network calls their god ‘the ‘Great Architect of the Universe’ (also Grand Architect). The ‘Great Architect’ Yaldabaoth is symbolised by the Cult as the allseeing eye at the top of the pyramid on the Great Seal of the United States and the dollar bill. Archon is encoded in arch-itect as it is in arch-angels and arch-bishops. All religions have the theme of a force for good and force for evil in some sort of spiritual war and there is a reason for that – the theme is true. The Cult and its non-human masters are quite happy for this to circulate. They present themselves as the force for good fighting evil when they are really the force of evil (absence of love). The whole foundation of Cult modus operandi is inversion. They promote themselves as a force for good and anyone challenging them in pursuit of peace, love, fairness, truth and justice is condemned as a satanic force for evil. This has been the game plan throughout history whether the Church of Rome inquisitions of non-believers or ‘conspiracy theorists’ and ‘anti-vaxxers’ of today. The technique is the same whatever the timeline era.

Yaldabaoth is revolting (true)

Yaldabaoth and the Archons are said to have revolted against God with Yaldabaoth claiming to be God – the All That Is. The Old Testament ‘God’ (Yaldabaoth) demanded to be worshipped as such: ‘ I am the LORD, and there is none else, there is no God beside me’ (Isaiah 45:5). I have quoted in other books a man who said he was the unofficial son of the late Baron Philippe de Rothschild of the Mouton-Rothschild wine producing estates in France who died in 1988 and he told me about the Rothschild ‘revolt from God’. The man said he was given the name Phillip Eugene de Rothschild and we shared long correspondence many years ago while he was living under another identity. He said that he was conceived through ‘occult incest’ which (within the Cult) was ‘normal and to be admired’. ‘Phillip’ told me about his experience a ending satanic rituals with rich and famous people whom he names and you can see them and the wider background to Cult Satanism in my other books starting with The Biggest Secret. Cult rituals are interactions with Archontic ‘gods’. ‘Phillip’ described Baron Philippe de Rothschild as ‘a master Satanist and hater of God’ and he used the same term ‘revolt from God’ associated with Yaldabaoth/Satan/Lucifer/the Devil in describing the Sabbatian Rothschild dynasty. ‘I played a key role in my family’s revolt from God’, he said. That role was to infiltrate in classic Sabbatian style the Christian Church, but eventually he escaped the mind-prison to live another life. The Cult has been targeting religion in a plan to make worship of the Archons the global one-world religion. Infiltration of Satanism into modern ‘culture’, especially among the young, through music videos, stage shows and other means, is all part of this. Nag Hammadi texts describe Yaldabaoth and the Archons in their prime form as energy – consciousness – and say they can take form if they choose in the same way that consciousness takes form as a human. Yaldabaoth is called ‘formless’ and represents a deeply inverted, distorted and chaotic state of consciousness which seeks to a ached to humans and turn them into a likeness of itself in an a empt at assimilation. For that to happen it has to manipulate

humans into low frequency mental and emotional states that match its own. Archons can certainly appear in human form and this is the origin of the psychopathic personality. The energetic distortion Gnostics called Yaldabaoth is psychopathy. When psychopathic Archons take human form that human will be a psychopath as an expression of Yaldabaoth consciousness. Cult psychopaths are Archons in human form. The principle is the same as that portrayed in the 2009 Avatar movie when the American military travelled to a fictional Earth-like moon called Pandora in the Alpha Centauri star system to infiltrate a society of blue people, or Na’vi, by hiding within bodies that looked like the Na’vi. Archons posing as humans have a particular hybrid information field, part human, part Archon, (the ancient ‘demigods’) which processes information in a way that manifests behaviour to match their psychopathic evil, lack of empathy and compassion, and stops them being influenced by the empathy, compassion and love that a fully-human information field is capable of expressing. Cult bloodlines interbreed, be they royalty or dark suits, for this reason and you have their obsession with incest. Interbreeding with full-blown humans would dilute the Archontic energy field that guarantees psychopathy in its representatives in the human realm. Gnostic writings say the main non-human forms that Archons take are serpentine (what I have called for decades ‘reptilian’ amid unbounded ridicule from the Archontically-programmed) and what Gnostics describe as ‘an unborn baby or foetus with grey skin and dark, unmoving eyes’. This is an excellent representation of the ET ‘Greys’ of UFO folklore which large numbers of people claim to have seen and been abducted by – Zulu shaman Credo Mutwa among them. I agree with those that believe in extraterrestrial or interdimensional visitations today and for thousands of years past. No wonder with their advanced knowledge and technological capability they were perceived and worshipped as gods for technological and other ‘miracles’ they appeared to perform. Imagine someone arriving in a culture disconnected from the modern world with a smartphone and computer. They would be

seen as a ‘god’ capable of ‘miracles’. The Renegade Mind, however, wants to know the source of everything and not only the way that source manifests as human or non-human. In the same way that a Renegade Mind seeks the original source material for the ‘Covid virus’ to see if what is claimed is true. The original source of Archons in form is consciousness – the distorted state of consciousness known to Gnostics as Yaldabaoth.

‘Revolt from God’ is energetic disconnection

Where I am going next will make a lot of sense of religious texts and ancient legends relating to ‘Satan’, Lucifer’ and the ‘gods’. Gnostic descriptions sync perfectly with the themes of my own research over the years in how they describe a consciousness distortion seeking to impose itself on human consciousness. I’ve referred to the core of infinite awareness in previous books as Infinite Awareness in Awareness of Itself. By that I mean a level of awareness that knows that it is all awareness and is aware of all awareness. From here comes the frequency of love in its true sense and balance which is what love is on one level – the balance of all forces into a single whole called Oneness and Isness. The more we disconnect from this state of love that many call ‘God’ the constituent parts of that Oneness start to unravel and express themselves as a part and not a whole. They become individualised as intellect, mind, selfishness, hatred, envy, desire for power over others, and such like. This is not a problem in the greater scheme in that ‘God’, the All That Is, can experience all these possibilities through different expressions of itself including humans. What we as expressions of the whole experience the All That Is experiences. We are the All That Is experiencing itself. As we withdraw from that state of Oneness we disconnect from its influence and things can get very unpleasant and very stupid. Archontic consciousness is at the extreme end of that. It has so disconnected from the influence of Oneness that it has become an inversion of unity and love, an inversion of everything, an inversion of life itself. Evil is appropriately live wri en backwards. Archontic consciousness is obsessed with death, an inversion of life,

and so its manifestations in Satanism are obsessed with death. They use inverted symbols in their rituals such as the inverted pentagram and cross. Sabbatians as Archontic consciousness incarnate invert Judaism and every other religion and culture they infiltrate. They seek disunity and chaos and they fear unity and harmony as they fear love like garlic to a vampire. As a result the Cult, Archons incarnate, act with such evil, psychopathy and lack of empathy and compassion disconnected as they are from the source of love. How could Bill Gates and the rest of the Archontic psychopaths do what they have to human society in the ‘Covid’ era with all the death, suffering and destruction involved and have no emotional consequence for the impact on others? Now you know. Why have Zuckerberg, Brin, Page, Wojcicki and company callously censored information warning about the dangers of the ‘vaccine’ while thousands have been dying and having severe, sometimes lifechanging reactions? Now you know. Why have Tedros, Fauci, Whi y, Vallance and their like around the world been using case and death figures they’re aware are fraudulent to justify lockdowns and all the deaths and destroyed lives that have come from that? Now you know. Why did Christian Drosten produce and promote a ‘testing’ protocol that he knew couldn’t test for infectious disease which led to a global human catastrophe. Now you know. The Archontic mind doesn’t give a shit (Fig 17). I personally think that Gates and major Cult insiders are a form of AI cyborg that the Archons want humans to become.

Figure 17: Artist Neil Hague’s version of the ‘Covid’ hierarchy.

Human batteries

A state of such inversion does have its consequences, however. The level of disconnection from the Source of All means that you withdraw from that source of energetic sustenance and creativity. This means that you have to find your own supply of energetic power and it has – us. When the Morpheus character in the first Matrix movie held up a ba ery he spoke a profound truth when he said: ‘The Matrix is a computer-generated dream world built to keep us under control in order to change the human being into one of

these.’ The statement was true in all respects. We do live in a technologically-generated virtual reality simulation (more very shortly) and we have been manipulated to be an energy source for Archontic consciousness. The Disney-Pixar animated movie Monsters, Inc. in 2001 symbolised the dynamic when monsters in their world had no energy source and they would enter the human world to terrify children in their beds, catch the child’s scream, terror (low-vibrational frequencies), and take that energy back to power the monster world. The lead character you might remember was a single giant eye and the symbolism of the Cult’s all-seeing eye was obvious. Every thought and emotion is broadcast as a frequency unique to that thought and emotion. Feelings of love and joy, empathy and compassion, are high, quick, frequencies while fear, depression, anxiety, suffering and hate are low, slow, dense frequencies. Which kind do you think Archontic consciousness can connect with and absorb? In such a low and dense frequency state there’s no way it can connect with the energy of love and joy. Archons can only feed off energy compatible with their own frequency and they and their Cult agents want to delete the human world of love and joy and manipulate the transmission of low vibrational frequencies through low-vibrational human mental and emotional states. We are their energy source. Wars are energetic banquets to the Archons – a world war even more so – and think how much low-frequency mental and emotional energy has been generated from the consequences for humanity of the ‘Covid’ hoax orchestrated by Archons incarnate like Gates. The ancient practice of human sacrifice ‘to the gods’, continued in secret today by the Cult, is based on the same principle. ‘The gods’ are Archontic consciousness in different forms and the sacrifice is induced into a state of intense terror to generate the energy the Archontic frequency can absorb. Incarnate Archons in the ritual drink the blood which contains an adrenaline they crave which floods into the bloodstream when people are terrorised. Most of the sacrifices, ancient and modern, are children and the theme of ‘sacrificing young virgins to the gods’ is just code for children. They

have a particular pre-puberty energy that Archons want more than anything and the energy of the young in general is their target. The California Department of Education wants students to chant the names of Aztec gods (Archontic gods) once worshipped in human sacrifice rituals in a curriculum designed to encourage them to ‘challenge racist, bigoted, discriminatory, imperialist/colonial beliefs’, join ‘social movements that struggle for social justice’, and ‘build new possibilities for a post-racist, post-systemic racism society’. It’s the usual Woke crap that inverts racism and calls it antiracism. In this case solidarity with ‘indigenous tribes’ is being used as an excuse to chant the names of ‘gods’ to which people were sacrificed (and still are in secret). What an example of Woke’s inability to see beyond black and white, us and them, They condemn the colonisation of these tribal cultures by Europeans (quite right), but those cultures sacrificing people including children to their ‘gods’, and mass murdering untold numbers as the Aztecs did, is just fine. One chant is to the Aztec god Tezcatlipoca who had a man sacrificed to him in the 5th month of the Aztec calendar. His heart was cut out and he was eaten. Oh, that’s okay then. Come on children … a er three … Other sacrificial ‘gods’ for the young to chant their allegiance include Quetzalcoatl, Huitzilopochtli and Xipe Totec. The curriculum says that ‘chants, affirmations, and energizers can be used to bring the class together, build unity around ethnic studies principles and values, and to reinvigorate the class following a lesson that may be emotionally taxing or even when student engagement may appear to be low’. Well, that’s the cover story, anyway. Chanting and mantras are the repetition of a particular frequency generated from the vocal cords and chanting the names of these Archontic ‘gods’ tunes you into their frequency. That is the last thing you want when it allows for energetic synchronisation, a achment and perceptual influence. Initiates chant the names of their ‘Gods’ in their rituals for this very reason.

Vampires of the Woke

Paedophilia is another way that Archons absorb the energy of children. Paedophiles possessed by Archontic consciousness are used as the conduit during sexual abuse for discarnate Archons to vampire the energy of the young they desire so much. Stupendous numbers of children disappear every year never to be seen again although you would never know from the media. Imagine how much low-vibrational energy has been generated by children during the ‘Covid’ hoax when so many have become depressed and psychologically destroyed to the point of killing themselves. Shocking numbers of children are now taken by the state from loving parents to be handed to others. I can tell you from long experience of researching this since 1996 that many end up with paedophiles and assets of the Cult through corrupt and Cult-owned social services which in the reframing era has hired many psychopaths and emotionless automatons to do the job. Children are even stolen to order using spurious reasons to take them by the corrupt and secret (because they’re corrupt) ‘family courts’. I have wri en in detail in other books, starting with The Biggest Secret in 1997, about the ubiquitous connections between the political, corporate, government, intelligence and military elites (Cult operatives) and Satanism and paedophilia. If you go deep enough both networks have an interlocking leadership. The Woke mentality has been developed by the Cult for many reasons: To promote almost every aspect of its agenda; to hijack the traditional political le and turn it fascist; to divide and rule; and to target agenda pushbackers. But there are other reasons which relate to what I am describing here. How many happy and joyful Wokers do you ever see especially at the extreme end? They are a mental and psychological mess consumed by emotional stress and constantly emotionally cocked for the next explosion of indignation at someone referring to a female as a female. They are walking, talking, ba eries as Morpheus might say emi ing frequencies which both enslave them in low-vibrational bubbles of perceptual limitation and feed the Archons. Add to this the hatred claimed to be love; fascism claimed to ‘anti-fascism’, racism claimed to be ‘anti-racism’;

exclusion claimed to inclusion; and the abuse-filled Internet trolling. You have a purpose-built Archontic energy system with not a wind turbine in sight and all founded on Archontic inversion. We have whole generations now manipulated to serve the Archons with their actions and energy. They will be doing so their entire adult lives unless they snap out of their Archon-induced trance. Is it really a surprise that Cult billionaires and corporations put so much money their way? Where is the energy of joy and laughter, including laughing at yourself which is confirmation of your own emotional security? Mark Twain said: ‘The human race has one really effective weapon, and that is laughter.‘ We must use it all the time. Woke has destroyed comedy because it has no humour, no joy, sense of irony, or self-deprecation. Its energy is dense and intense. Mmmmm, lunch says the Archontic frequency. Rudolf Steiner (1861-1925) was the Austrian philosopher and famous esoteric thinker who established Waldorf education or Steiner schools to treat children like unique expressions of consciousness and not minds to be programmed with the perceptions determined by authority. I’d been writing about this energy vampiring for decades when I was sent in 2016 a quote by Steiner. He was spot on: There are beings in the spiritual realms for whom anxiety and fear emanating from human beings offer welcome food. When humans have no anxiety and fear, then these creatures starve. If fear and anxiety radiates from people and they break out in panic, then these creatures find welcome nutrition and they become more and more powerful. These beings are hostile towards humanity. Everything that feeds on negative feelings, on anxiety, fear and superstition, despair or doubt, are in reality hostile forces in super-sensible worlds, launching cruel attacks on human beings, while they are being fed ... These are exactly the feelings that belong to contemporary culture and materialism; because it estranges people from the spiritual world, it is especially suited to evoke hopelessness and fear of the unknown in people, thereby calling up the above mentioned hostile forces against them.

Pause for a moment from this perspective and reflect on what has happened in the world since the start of 2020. Not only will pennies drop, but billion dollar bills. We see the same theme from Don Juan Matus, a Yaqui Indian shaman in Mexico and the information source for Peruvian-born writer, Carlos Castaneda, who wrote a series of

books from the 1960s to 1990s. Don Juan described the force manipulating human society and his name for the Archons was the predator: We have a predator that came from the depths of the cosmos and took over the rule of our lives. Human beings are its prisoners. The predator is our lord and master. It has rendered us docile, helpless. If we want to protest, it suppresses our protest. If we want to act independently, it demands that we don’t do so ... indeed we are held prisoner! They took us over because we are food to them, and they squeeze us mercilessly because we are their sustenance. Just as we rear chickens in coops, the predators rear us in human coops, humaneros. Therefore, their food is always available to them. Different cultures, different eras, same recurring theme.

The ‘ennoia’ dilemma

Nag Hammadi Gnostic manuscripts say that Archon consciousness has no ‘ennoia’. This is directly translated as ‘intentionality’, but I’ll use the term ‘creative imagination’. The All That Is in awareness of itself is the source of all creativity – all possibility – and the more disconnected you are from that source the more you are subsequently denied ‘creative imagination’. Given that Archon consciousness is almost entirely disconnected it severely lacks creativity and has to rely on far more mechanical processes of thought and exploit the creative potential of those that do have ‘ennoia’. You can see cases of this throughout human society. Archon consciousness almost entirely dominates the global banking system and if we study how that system works you will appreciate what I mean. Banks manifest ‘money’ out of nothing by issuing lines of ‘credit’ which is ‘money’ that has never, does not, and will never exist except in theory. It’s a confidence trick. If you think ‘credit’ figures-on-a-screen ‘money’ is worth anything you accept it as payment. If you don’t then the whole system collapses through lack of confidence in the value of that ‘money’. Archontic bankers with no ‘ennoia’ are ‘lending’ ‘money’ that doesn’t exist to humans that do have creativity – those that have the inspired ideas and create businesses and products. Archon banking feeds off human creativity

which it controls through ‘money’ creation and debt. Humans have the creativity and Archons exploit that for their own benefit and control while having none themselves. Archon Internet platforms like Facebook claim joint copyright of everything that creative users post and while Archontic minds like Zuckerberg may officially head that company it will be human creatives on the staff that provide the creative inspiration. When you have limitless ‘money’ you can then buy other companies established by creative humans. Witness the acquisition record of Facebook, Google and their like. Survey the Archon-controlled music industry and you see non-creative dark suit executives making their fortune from the human creativity of their artists. The cases are endless. Research the history of people like Gates and Zuckerberg and how their empires were built on exploiting the creativity of others. Archon minds cannot create out of nothing, but they are skilled (because they have to be) in what Gnostic texts call ‘countermimicry’. They can imitate, but not innovate. Sabbatians trawl the creativity of others through backdoors they install in computer systems through their cybersecurity systems. Archon-controlled China is globally infamous for stealing intellectual property and I remember how Hong Kong, now part of China, became notorious for making counterfeit copies of the creativity of others – ‘countermimicry’. With the now pervasive and all-seeing surveillance systems able to infiltrate any computer you can appreciate the potential for Archons to vampire the creativity of humans. Author John Lamb Lash wrote in his book about the Nag Hammadi texts, Not In His Image: Although they cannot originate anything, because they lack the divine factor of ennoia (intentionality), Archons can imitate with a vengeance. Their expertise is simulation (HAL, virtual reality). The Demiurge [Yaldabaoth] fashions a heaven world copied from the fractal patterns [of the original] ... His construction is celestial kitsch, like the fake Italianate villa of a Mafia don complete with militant angels to guard every portal.

This brings us to something that I have been speaking about since the turn of the millennium. Our reality is a simulation; a virtual reality that we think is real. No, I’m not kidding.

Human reality? Well, virtually

I had pondered for years about whether our reality is ‘real’ or some kind of construct. I remembered being immensely affected on a visit as a small child in the late 1950s to the then newly-opened Planetarium on the Marylebone Road in London which is now closed and part of the adjacent Madame Tussauds wax museum. It was in the middle of the day, but when the lights went out there was the night sky projected in the Planetarium’s domed ceiling and it appeared to be so real. The experience never le me and I didn’t know why until around the turn of the millennium when I became certain that our ‘night sky’ and entire reality is a projection, a virtual reality, akin to the illusory world portrayed in the Matrix movies. I looked at the sky one day in this period and it appeared to me like the domed roof of the Planetarium. The release of the first Matrix movie in 1999 also provided a synchronistic and perfect visual representation of where my mind had been going for a long time. I hadn’t come across the Gnostic Nag Hammadi texts then. When I did years later the correlation was once again astounding. As I read Gnostic accounts from 1,600 years and more earlier it was clear that they were describing the same simulation phenomenon. They tell how the Yaldabaoth ‘Demiurge’ and Archons created a ‘bad copy’ of original reality to rule over all that were captured by its illusions and the body was a prison to trap consciousness in the ‘bad copy’ fake reality. Read how Gnostics describe the ‘bad copy’ and update that to current times and they are referring to what we would call today a virtual reality simulation. Author John Lamb Lash said ‘the Demiurge fashions a heaven world copied from the fractal pa erns’ of the original through expertise in ‘HAL’ or virtual reality simulation. Fractal pa erns are part of the energetic information construct of our reality, a sort of blueprint. If these pa erns were copied in computer terms it would indeed give you a copy of a ‘natural’ reality in a non-natural frequency and digital form. The principle is the same as making a copy of a website. The original website still exists, but now you can change the copy version to make it whatever you like and it can

become very different to the original website. Archons have done this with our reality, a synthetic copy of prime reality that still exists beyond the frequency walls of the simulation. Trapped within the illusions of this synthetic Matrix, however, were and are human consciousness and other expressions of prime reality and this is why the Archons via the Cult are seeking to make the human body synthetic and give us synthetic AI minds to complete the job of turning the entire reality synthetic including what we perceive to be the natural world. To quote Kurzweil: ‘Nanobots will infuse all the ma er around us with information. Rocks, trees, everything will become these intelligent creatures.’ Yes, synthetic ‘creatures’ just as ‘Covid’ and other genetically-manipulating ‘vaccines’ are designed to make the human body synthetic. From this perspective it is obvious why Archons and their Cult are so desperate to infuse synthetic material into every human with their ‘Covid’ scam.

Let there be (electromagnetic) light

Yaldabaoth, the force that created the simulation, or Matrix, makes sense of the Gnostic reference to ‘The Great Architect’ and its use by Cult Freemasonry as the name of its deity. The designer of the Matrix in the movies is called ‘The Architect’ and that trilogy is jam-packed with symbolism relating to these subjects. I have contended for years that the angry Old Testament God (Yaldabaoth) is the ‘God’ being symbolically ‘quoted’ in the opening of Genesis as ‘creating the world’. This is not the creation of prime reality – it’s the creation of the simulation. The Genesis ‘God’ says: ‘Let there be Light: and there was light.’ But what is this ‘Light’? I have said for decades that the speed of light (186,000 miles per second) is not the fastest speed possible as claimed by mainstream science and is in fact the frequency walls or outer limits of the Matrix. You can’t have a fastest or slowest anything within all possibility when everything is possible. The human body is encoded to operate within the speed of light or within the simulation and thus we see only the tiny frequency band of visible light. Near-death experiencers who perceive reality outside the body during temporary ‘death’ describe a very different

form of light and this is supported by the Nag Hammadi texts. Prime reality beyond the simulation (‘Upper Aeons’ to the Gnostics) is described as a realm of incredible beauty, bliss, love and harmony – a realm of ‘watery light’ that is so powerful ‘there are no shadows’. Our false reality of Archon control, which Gnostics call the ‘Lower Aeons’, is depicted as a realm with a different kind of ‘light’ and described in terms of chaos, ‘Hell’, ‘the Abyss’ and ‘Outer Darkness’, where trapped souls are tormented and manipulated by demons (relate that to the ‘Covid’ hoax alone). The watery light theme can be found in near-death accounts and it is not the same as simulation ‘light’ which is electromagnetic or radiation light within the speed of light – the ‘Lower Aeons’. Simulation ‘light’ is the ‘luminous fire’ associated by Gnostics with the Archons. The Bible refers to Yaldabaoth as ‘that old serpent, called the Devil, and Satan, which deceiveth the whole world’ (Revelation 12:9). I think that making a simulated copy of prime reality (‘countermimicry’) and changing it dramatically while all the time manipulating humanity to believe it to be real could probably meet the criteria of deceiving the whole world. Then we come to the Cult god Lucifer – the Light Bringer. Lucifer is symbolic of Yaldabaoth, the bringer of radiation light that forms the bad copy simulation within the speed of light. ‘He’ is symbolised by the lighted torch held by the Statue of Liberty and in the name ‘Illuminati’. Sabbatian-Frankism declares that Lucifer is the true god and Lucifer is the real god of Freemasonry honoured as their ‘Great or Grand Architect of the Universe’ (simulation). I would emphasise, too, the way Archontic technologicallygenerated luminous fire of radiation has deluged our environment since I was a kid in the 1950s and changed the nature of The Field with which we constantly interact. Through that interaction technological radiation is changing us. The Smart Grid is designed to operate with immense levels of communication power with 5G expanding across the world and 6G, 7G, in the process of development. Radiation is the simulation and the Archontic manipulation system. Why wouldn’t the Archon Cult wish to unleash radiation upon us to an ever-greater extreme to form

Kurzweil’s ‘cloud’? The plan for a synthetic human is related to the need to cope with levels of radiation beyond even anything we’ve seen so far. Biological humans would not survive the scale of radiation they have in their script. The Smart Grid is a technological sub-reality within the technological simulation to further disconnect five-sense perception from expanded consciousness. It’s a technological prison of the mind.

Infusing the ‘spirit of darkness’

A recurring theme in religion and native cultures is the manipulation of human genetics by a non-human force and most famously recorded as the biblical ‘sons of god’ (the gods plural in the original) who interbred with the daughters of men. The Nag Hammadi Apocryphon of John tells the same story this way: He [Yaldabaoth] sent his angels [Archons/demons] to the daughters of men, that they might take some of them for themselves and raise offspring for their enjoyment. And at first they did not succeed. When they had no success, they gathered together again and they made a plan together ... And the angels changed themselves in their likeness into the likeness of their mates, filling them with the spirit of darkness, which they had mixed for them, and with evil ... And they took women and begot children out of the darkness according to the likeness of their spirit.

Possession when a discarnate entity takes over a human body is an age-old theme and continues today. It’s very real and I’ve seen it. Satanic and secret society rituals can create an energetic environment in which entities can a ach to initiates and I’ve heard many stories of how people have changed their personality a er being initiated even into lower levels of the Freemasons. I have been inside three Freemasonic temples, one at a public open day and two by just walking in when there was no one around to stop me. They were in Ryde, the town where I live, Birmingham, England, when I was with a group, and Boston, Massachuse s. They all felt the same energetically – dark, dense, low-vibrational and sinister. Demonic a achment can happen while the initiate has no idea what is going on. To them it’s just a ritual to get in the Masons and do a bit of good

business. In the far more extreme rituals of Satanism human possession is even more powerful and they are designed to make possession possible. The hierarchy of the Cult is dictated by the power and perceived status of the possessing Archon. In this way the Archon hierarchy becomes the Cult hierarchy. Once the entity has a ached it can influence perception and behaviour and if it a aches to the extreme then so much of its energy (information) infuses into the body information field that the hologram starts to reflect the nature of the possessing entity. This is the Exorcist movie type of possession when facial features change and it’s known as shapeshi ing. Islam’s Jinn are said to be invisible tricksters who change shape, ‘whisper’, confuse and take human form. These are all traits of the Archons and other versions of the same phenomenon. Extreme possession could certainty infuse the ‘spirit of darkness’ into a partner during sex as the Nag Hammadi texts appear to describe. Such an infusion can change genetics which is also energetic information. Human genetics is information and the ‘spirit of darkness’ is information. Mix one with the other and change must happen. Islam has the concept of a ‘Jinn baby’ through possession of the mother and by Jinn taking human form. There are many ways that human genetics can be changed and remember that Archons have been aware all along of advanced techniques to do this. What is being done in human society today – and far more – was known about by Archons at the time of the ‘fallen ones’ and their other versions described in religions and cultures. Archons and their human-world Cult are obsessed with genetics as we see today and they know this dictates how information is processed into perceived reality during a human life. They needed to produce a human form that would decode the simulation and this is symbolically known as ‘Adam and Eve’ who le the ‘garden’ (prime reality) and ‘fell’ into Matrix reality. The simulation is not a ‘physical’ construct (there is no ‘physical’); it is a source of information. Think Wi-Fi again. The simulation is an energetic field encoded with information and body-brain systems are designed to decode that information encoded in wave or frequency form which

is transmi ed to the brain as electrical signals. These are decoded by the brain to construct our sense of reality – an illusory ‘physical’ world that only exists in the brain or the mind. Virtual reality games mimic this process using the same sensory decoding system. Information is fed to the senses to decode a virtual reality that can appear so real, but isn’t (Figs 18 and 19). Some scientists believe – and I agree with them – that what we perceive as ‘physical’ reality only exists when we are looking or observing. The act of perception or focus triggers the decoding systems which turn waveform information into holographic reality. When we are not observing something our reality reverts from a holographic state to a waveform state. This relates to the same principle as a falling tree not making a noise unless someone is there to hear it or decode it. The concept makes sense from the simulation perspective. A computer is not decoding all the information in a Wi-Fi field all the time and only decodes or brings into reality on the screen that part of Wi-Fi that it’s decoding – focusing upon – at that moment.

Figure 18: Virtual reality technology ‘hacks’ into the body’s five-sense decoding system.

Figure 19: The result can be experienced as very ‘real’.

Interestingly, Professor Donald Hoffman at the Department of Cognitive Sciences at the University of California, Irvine, says that our experienced reality is like a computer interface that shows us only the level with which we interact while hiding all that exists beyond it: ‘Evolution shaped us with a user interface that hides the truth. Nothing that we see is the truth – the very language of space and time and objects is the wrong language to describe reality.’ He is correct in what he says on so many levels. Space and time are not a universal reality. They are a phenomenon of decoded simulation reality as part of the process of enslaving our sense of reality. Neardeath experiencers report again and again how space and time did not exist as we perceive them once they were free of the body – body decoding systems. You can appreciate from this why Archons and their Cult are so desperate to entrap human a ention in the five senses where we are in the Matrix and of the Matrix. Opening your mind to expanded states of awareness takes you beyond the information confines of the simulation and you become aware of knowledge and insights denied to you before. This is what we call ‘awakening’ – awakening from the Matrix – and in the final chapter I will relate this to current events.

Where are the ‘aliens’?

A simulation would explain the so-called ‘Fermi Paradox’ named a er Italian physicist Enrico Fermi (1901-1954) who created the first nuclear reactor. He considered the question of why there is such a lack of extraterrestrial activity when there are so many stars and planets in an apparently vast universe; but what if the night sky that we see, or think we do, is a simulated projection as I say? If you control the simulation and your aim is to hold humanity fast in essential ignorance would you want other forms of life including advanced life coming and going sharing information with humanity? Or would you want them to believe they were isolated and apparently alone? Themes of human isolation and apartness are common whether they be the perception of a lifeless universe or the fascist isolation laws of the ‘Covid’ era. Paradoxically the very

existence of a simulation means that we are not alone when some force had to construct it. My view is that experiences that people have reported all over the world for centuries with Reptilians and Grey entities are Archon phenomena as Nag Hammadi texts describe; and that benevolent ‘alien’ interactions are non-human groups that come in and out of the simulation by overcoming Archon a empts to keep them out. It should be highlighted, too, that Reptilians and Greys are obsessed with genetics and technology as related by cultural accounts and those who say they have been abducted by them. Technology is their way of overcoming some of the limitations in their creative potential and our technology-driven and controlled human society of today is archetypical ArchonReptilian-Grey modus operandi. Technocracy is really Archontocracy. The Universe does not have to be as big as it appears with a simulation. There is no space or distance only information decoded into holographic reality. What we call ‘space’ is only the absence of holographic ‘objects’ and that ‘space’ is The Field of energetic information which connects everything into a single whole. The same applies with the artificially-generated information field of the simulation. The Universe is not big or small as a physical reality. It is decoded information, that’s all, and its perceived size is decided by the way the simulation is encoded to make it appear. The entire night sky as we perceive it only exists in our brain and so where are those ‘millions of light years’? The ‘stars’ on the ceiling of the Planetarium looked a vast distance away. There’s another point to mention about ‘aliens’. I have been highlighting since the 1990s the plan to stage a fake ‘alien invasion’ to justify the centralisation of global power and a world military. Nazi scientist Werner von Braun, who was taken to America by Operation Paperclip a er World War Two to help found NASA, told his American assistant Dr Carol Rosin about the Cult agenda when he knew he was dying in 1977. Rosin said that he told her about a sequence that would lead to total human control by a one-world government. This included threats from terrorism, rogue nations, meteors and asteroids before finally an ‘alien invasion’. All of these

things, von Braun said, would be bogus and what I would refer to as a No-Problem-Reaction-Solution. Keep this in mind when ‘the aliens are coming’ is the new mantra. The aliens are not coming – they are already here and they have infiltrated human society while looking human. French-Canadian investigative journalist Serge Monast said in 1994 that he had uncovered a NASA/military operation called Project Blue Beam which fits with what Werner von Braun predicted. Monast died of a ‘heart a ack’ in 1996 the day a er he was arrested and spent a night in prison. He was 51. He said Blue Beam was a plan to stage an alien invasion that would include religious figures beamed holographically into the sky as part of a global manipulation to usher in a ‘new age’ of worshipping what I would say is the Cult ‘god’ Yaldabaoth in a one-world religion. Fake holographic asteroids are also said to be part of the plan which again syncs with von Braun. How could you stage an illusory threat from asteroids unless they were holographic inserts? This is pre y straightforward given the advanced technology outside the public arena and the fact that our ‘physical’ reality is holographic anyway. Information fields would be projected and we would decode them into the illusion of a ‘physical’ asteroid. If they can sell a global ‘pandemic’ with a ‘virus’ that doesn’t exist what will humans not believe if government and media tell them? All this is particularly relevant as I write with the Pentagon planning to release in June, 2021, information about ‘UFO sightings’. I have been following the UFO story since the early 1990s and the common theme throughout has been government and military denials and cover up. More recently, however, the Pentagon has suddenly become more talkative and apparently open with Air Force pilot radar images released of unexplained cra moving and changing direction at speeds well beyond anything believed possible with human technology. Then, in March, 2021, former Director of National Intelligence John Ratcliffe said a Pentagon report months later in June would reveal a great deal of information about UFO sightings unknown to the public. He said the report would have ‘massive implications’. The order to do this was included bizarrely

in a $2.3 trillion ‘coronavirus’ relief and government funding bill passed by the Trump administration at the end of 2020. I would add some serious notes of caution here. I have been pointing out since the 1990s that the US military and intelligence networks have long had cra – ‘flying saucers’ or anti-gravity cra – which any observer would take to be extraterrestrial in origin. Keeping this knowledge from the public allows cra flown by humans to be perceived as alien visitations. I am not saying that ‘aliens’ do not exist. I would be the last one to say that, but we have to be streetwise here. President Ronald Reagan told the UN General Assembly in 1987: ‘I occasionally think how quickly our differences worldwide would vanish if we were facing an alien threat from outside this world.’ That’s the idea. Unite against a common ‘enemy’ with a common purpose behind your ‘saviour force’ (the Cult) as this age-old technique of mass manipulation goes global.

Science moves this way …

I could find only one other person who was discussing the simulation hypothesis publicly when I concluded it was real. This was Nick Bostrom, a Swedish-born philosopher at the University of Oxford, who has explored for many years the possibility that human reality is a computer simulation although his version and mine are not the same. Today the simulation and holographic reality hypothesis have increasingly entered the scientific mainstream. Well, the more open-minded mainstream, that is. Here are a few of the ever-gathering examples. American nuclear physicist Silas Beane led a team of physicists at the University of Bonn in Germany pursuing the question of whether we live in a simulation. They concluded that we probably do and it was likely based on a la ice of cubes. They found that cosmic rays align with that specific pa ern. The team highlighted the Greisen–Zatsepin–Kuzmin (GZK) limit which refers to cosmic ray particle interaction with cosmic background radiation that creates an apparent boundary for cosmic ray particles. They say in a paper entitled ‘Constraints on the Universe as a Numerical Simulation’ that this ‘pa ern of constraint’ is exactly what you

would find with a computer simulation. They also made the point that a simulation would create its own ‘laws of physics’ that would limit possibility. I’ve been making the same point for decades that the perceived laws of physics relate only to this reality, or what I would later call the simulation. When designers write codes to create computer and virtual reality games they are the equivalent of the laws of physics for that game. Players interact within the limitations laid out by the coding. In the same way those who wrote the codes for the simulation decided the laws of physics that would apply. These can be overridden by expanded states of consciousness, but not by those enslaved in only five-sense awareness where simulation codes rule. Overriding the codes is what people call ‘miracles’. They are not. They are bypassing the encoded limits of the simulation. A population caught in simulation perception would have no idea that this was their plight. As the Bonn paper said: ‘Like a prisoner in a pitch-black cell we would not be able to see the “walls” of our prison,’ That’s true if people remain mesmerised by the five senses. Open to expanded awareness and those walls become very clear. The main one is the speed of light. American theoretical physicist James Gates is another who has explored the simulation question and found considerable evidence to support the idea. Gates was Professor of Physics at the University of Maryland, Director of The Center for String and Particle Theory, and on Barack Obama’s Council of Advisors on Science and Technology. He and his team found computer codes of digital data embedded in the fabric of our reality. They relate to on-off electrical charges of 1 and 0 in the binary system used by computers. ‘We have no idea what they are doing there’, Gates said. They found within the energetic fabric mathematical sequences known as errorcorrecting codes or block codes that ‘reboot’ data to its original state or ‘default se ings’ when something knocks it out of sync. Gates was asked if he had found a set of equations embedded in our reality indistinguishable from those that drive search engines and browsers and he said: ‘That is correct.’ Rich Terrile, director of the Centre for Evolutionary Computation and Automated Design at NASA’s Jet

Propulsion Laboratory, has said publicly that he believes the Universe is a digital hologram that must have been created by a form of intelligence. I agree with that in every way. Waveform information is delivered electrically by the senses to the brain which constructs a digital holographic reality that we call the ‘world’. This digital level of reality can be read by the esoteric art of numerology. Digital holograms are at the cu ing edge of holographics today. We have digital technology everywhere designed to access and manipulate our digital level of perceived reality. Synthetic mRNA in ‘Covid vaccines’ has a digital component to manipulate the body’s digital ‘operating system’.

Reality is numbers

How many know that our reality can be broken down to numbers and codes that are the same as computer games? Max Tegmark, a physicist at the Massachuse s Institute of Technology (MIT), is the author of Our Mathematical Universe in which he lays out how reality can be entirely described by numbers and maths in the way that a video game is encoded with the ‘physics’ of computer games. Our world and computer virtual reality are essentially the same. Tegmark imagines the perceptions of characters in an advanced computer game when the graphics are so good they don’t know they are in a game. They think they can bump into real objects (electromagnetic resistance in our reality), fall in love and feel emotions like excitement. When they began to study the apparently ‘physical world’ of the video game they would realise that everything was made of pixels (which have been found in our energetic reality as must be the case when on one level our world is digital). What computer game characters thought was physical ‘stuff’, Tegmark said, could actually be broken down into numbers: And we’re exactly in this situation in our world. We look around and it doesn’t seem that mathematical at all, but everything we see is made out of elementary particles like quarks and electrons. And what properties does an electron have? Does it have a smell or a colour or a texture? No! ... We physicists have come up with geeky names for [Electron] properties, like

electric charge, or spin, or lepton number, but the electron doesn’t care what we call it, the properties are just numbers.

This is the illusory reality Gnostics were describing. This is the simulation. The A, C, G, and T codes of DNA have a binary value – A and C = 0 while G and T = 1. This has to be when the simulation is digital and the body must be digital to interact with it. Recurring mathematical sequences are encoded throughout reality and the body. They include the Fibonacci sequence in which the two previous numbers are added to get the next one, as in ... 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, etc. The sequence is encoded in the human face and body, proportions of animals, DNA, seed heads, pine cones, trees, shells, spiral galaxies, hurricanes and the number of petals in a flower. The list goes on and on. There are fractal pa erns – a ‘neverending pa ern that is infinitely complex and self-similar across all scales in the as above, so below, principle of holograms. These and other famous recurring geometrical and mathematical sequences such as Phi, Pi, Golden Mean, Golden Ratio and Golden Section are computer codes of the simulation. I had to laugh and give my head a shake the day I finished this book and it went into the production stage. I was sent an article in Scientific American published in April, 2021, with the headline ‘Confirmed! We Live in a Simulation’. Two decades a er I first said our reality is a simulation and the speed of light is it’s outer limit the article suggested that we do live in a simulation and that the speed of light is its outer limit. I le school at 15 and never passed a major exam in my life while the writer was up to his eyes in qualifications. As I will explain in the final chapter knowing is far be er than thinking and they come from very different sources. The article rightly connected the speed of light to the processing speed of the ‘Matrix’ and said what has been in my books all this time … ‘If we are in a simulation, as it appears, then space is an abstract property wri en in code. It is not real’. No it’s not and if we live in a simulation something created it and it wasn’t us. ‘That David Icke says we are manipulated by aliens’ – he’s crackers.’

Wow …

The reality that humanity thinks is so real is an illusion. Politicians, governments, scientists, doctors, academics, law enforcement, media, school and university curriculums, on and on, are all founded on a world that does not exist except as a simulated prison cell. Is it such a stretch to accept that ‘Covid’ doesn’t exist when our entire ‘physical’ reality doesn’t exist? Revealed here is the knowledge kept under raps in the Cult networks of compartmentalised secrecy to control humanity’s sense of reality by inducing the population to believe in a reality that’s not real. If it wasn’t so tragic in its experiential consequences the whole thing would be hysterically funny. None of this is new to Renegade Minds. Ancient Greek philosopher Plato (about 428 to about 347BC) was a major influence on Gnostic belief and he described the human plight thousands of years ago with his Allegory of the Cave. He told the symbolic story of prisoners living in a cave who had never been outside. They were chained and could only see one wall of the cave while behind them was a fire that they could not see. Figures walked past the fire casting shadows on the prisoners’ wall and those moving shadows became their sense of reality. Some prisoners began to study the shadows and were considered experts on them (today’s academics and scientists), but what they studied was only an illusion (today’s academics and scientists). A prisoner escaped from the cave and saw reality as it really is. When he returned to report this revelation they didn’t believe him, called him mad and threatened to kill him if he tried to set them free. Plato’s tale is not only a brilliant analogy of the human plight and our illusory reality. It describes, too, the dynamics of the ‘Covid’ hoax. I have only skimmed the surface of these subjects here. The aim of this book is to crisply connect all essential dots to put what is happening today into its true context. All subject areas and their connections in this chapter are covered in great evidential detail in Everything You Need To Know, But Have Never Been Told and The Answer. They say that bewildered people ‘can’t see the forest for the trees’. Humanity, however, can’t see the forest for the twigs. The five senses

see only twigs while Renegade Minds can see the forest and it’s the forest where the answers lie with the connections that reveals. Breaking free of perceptual programming so the forest can be seen is the way we turn all this around. Not breaking free is how humanity got into this mess. The situation may seem hopeless, but I promise you it’s not. We are a perceptual heartbeat from paradise if only we knew.

CHAPTER TWELVE Escaping Wetiko Life is simply a vacation from the infinite Dean Cavanagh

R

enegade Minds weave the web of life and events and see common themes in the apparently random. They are always there if you look for them and their pursuit is aided by incredible synchronicity that comes when your mind is open rather than mesmerised by what it thinks it can see. Infinite awareness is infinite possibility and the more of infinite possibility that we access the more becomes infinitely possible. That may be stating the apparently obvious, but it is a devastatinglypowerful fact that can set us free. We are a point of a ention within an infinity of consciousness. The question is how much of that infinity do we choose to access? How much knowledge, insight, awareness, wisdom, do we want to connect with and explore? If your focus is only in the five senses you will be influenced by a fraction of infinite awareness. I mean a range so tiny that it gives new meaning to infinitesimal. Limitation of self-identity and a sense of the possible limit accordingly your range of consciousness. We are what we think we are. Life is what we think it is. The dream is the dreamer and the dreamer is the dream. Buddhist philosophy puts it this way: ‘As a thing is viewed, so it appears.’ Most humans live in the realm of touch, taste, see, hear, and smell and that’s the limit of their sense of the possible and sense of self. Many will follow a religion and speak of a God in his heaven, but their lives are still

dominated by the five senses in their perceptions and actions. The five senses become the arbiter of everything. When that happens all except a smear of infinity is sealed away from influence by the rigid, unyielding, reality bubbles that are the five-sense human or Phantom Self. Archon Cult methodology is to isolate consciousness within five-sense reality – the simulation – and then program that consciousness with a sense of self and the world through a deluge of life-long information designed to instil the desired perception that allows global control. Efforts to do this have increased dramatically with identity politics as identity bubbles are squeezed into the minutiae of five-sense detail which disconnect people even more profoundly from the infinite ‘I’. Five-sense focus and self-identity are like a firewall that limits access to the infinite realms. You only perceive one radio or television station and no other. We’ll take that literally for a moment. Imagine a vast array of stations giving different information and angles on reality, but you only ever listen to one. Here we have the human plight in which the population is overwhelmingly confined to CultFM. This relates only to the frequency range of CultFM and limits perception and insight to that band – limits possibility to that band. It means you are connecting with an almost imperceptibly minuscule range of possibility and creative potential within the infinite Field. It’s a world where everything seems apart from everything else and where synchronicity is rare. Synchronicity is defined in the dictionary as ‘the happening by chance of two or more related or similar events at the same time‘. Use of ‘by chance’ betrays a complete misunderstanding of reality. Synchronicity is not ‘by chance’. As people open their minds, or ‘awaken’ to use the term, they notice more and more coincidences in their lives, bits of ‘luck’, apparently miraculous happenings that put them in the right place at the right time with the right people. Days become peppered with ‘fancy meeting you here’ and ‘what are the chances of that?’ My entire life has been lived like this and ever more so since my own colossal awakening in 1990 and 91 which transformed my sense of reality. Synchronicity is not ‘by chance’; it is by accessing expanded

realms of possibility which allow expanded potential for manifestation. People broadcasting the same vibe from the same openness of mind tend to be drawn ‘by chance’ to each other through what I call frequency magnetism and it’s not only people. In the last more than 30 years incredible synchronicity has also led me through the Cult maze to information in so many forms and to crucial personal experiences. These ‘coincidences’ have allowed me to put the puzzle pieces together across an enormous array of subjects and situations. Those who have breached the bubble of fivesense reality will know exactly what I mean and this escape from the perceptual prison cell is open to everyone whenever they make that choice. This may appear super-human when compared with the limitations of ‘human’, but it’s really our natural state. ‘Human’ as currently experienced is consciousness in an unnatural state of induced separation from the infinity of the whole. I’ll come to how this transformation into unity can be made when I have described in more detail the force that holds humanity in servitude by denying this access to infinite self.

The Wetiko factor

I have been talking and writing for decades about the way five-sense mind is systematically barricaded from expanded awareness. I have used the analogy of a computer (five-sense mind) and someone at the keyboard (expanded awareness). Interaction between the computer and the operator is symbolic of the interaction between five-sense mind and expanded awareness. The computer directly experiences the Internet and the operator experiences the Internet via the computer which is how it’s supposed to be – the two working as one. Archons seek to control that point where the operator connects with the computer to stop that interaction (Fig 20). Now the operator is banging the keyboard and clicking the mouse, but the computer is not responding and this happens when the computer is taken over – possessed – by an appropriately-named computer ‘virus’. The operator has lost all influence over the computer which goes its own way making decisions under the control of the ‘virus’. I have

just described the dynamic through which the force known to Gnostics as Yaldabaoth and Archons disconnects five-sense mind from expanded awareness to imprison humanity in perceptual servitude.

Figure 20: The mind ‘virus’ I have been writing about for decades seeks to isolate five-sense mind (the computer) from the true ‘I’. (Image by Neil Hague).

About a year ago I came across a Native American concept of Wetiko which describes precisely the same phenomenon. Wetiko is the spelling used by the Cree and there are other versions including wintiko and windigo used by other tribal groups. They spell the name with lower case, but I see Wetiko as a proper noun as with Archons and prefer a capital. I first saw an article about Wetiko by writer and researcher Paul Levy which so synced with what I had been writing about the computer/operator disconnection and later the Archons. I then read his book, the fascinating Dispelling Wetiko, Breaking the Spell of Evil. The parallels between what I had concluded long before and the Native American concept of Wetiko were so clear and obvious that it was almost funny. For Wetiko see the Gnostic Archons for sure and the Jinn, the Predators, and every other name for a force of evil, inversion and chaos. Wetiko is the Native American name for the force that divides the computer from

the operator (Fig 21). Indigenous author Jack D. Forbes, a founder of the Native American movement in the 1960s, wrote another book about Wetiko entitled Columbus And Other Cannibals – The Wetiko Disease of Exploitation, Imperialism, and Terrorism which I also read. Forbes says that Wetiko refers to an evil person or spirit ‘who terrorizes other creatures by means of terrible acts, including cannibalism’. Zulu shaman Credo Mutwa told me that African accounts tell how cannibalism was brought into the world by the Chitauri ‘gods’ – another manifestation of Wetiko. The distinction between ‘evil person or spirit’ relates to Archons/Wetiko possessing a human or acting as pure consciousness. Wetiko is said to be a sickness of the soul or spirit and a state of being that takes but gives nothing back – the Cult and its operatives perfectly described. Black Hawk, a Native American war leader defending their lands from confiscation, said European invaders had ‘poisoned hearts’ – Wetiko hearts – and that this would spread to native societies. Mention of the heart is very significant as we shall shortly see. Forbes writes: ‘Tragically, the history of the world for the past 2,000 years is, in great part, the story of the epidemiology of the wetiko disease.’ Yes, and much longer. Forbes is correct when he says: ‘The wetikos destroyed Egypt and Babylon and Athens and Rome and Tenochtitlan [capital of the Aztec empire] and perhaps now they will destroy the entire earth.’ Evil, he said, is the number one export of a Wetiko culture – see its globalisation with ‘Covid’. Constant war, mass murder, suffering of all kinds, child abuse, Satanism, torture and human sacrifice are all expressions of Wetiko and the Wetiko possessed. The world is Wetiko made manifest, but it doesn’t have to be. There is a way out of this even now.

Figure 21: The mind ‘virus’ is known to Native Americans as ‘Wetiko’. (Image by Neil Hague).

Cult of Wetiko

Wetiko is the Yaldabaoth frequency distortion that seeks to a ach to human consciousness and absorb it into its own. Once this connection is made Wetiko can drive the perceptions of the target which they believe to be coming from their own mind. All the horrors of history and today from mass killers to Satanists, paedophiles like Jeffrey Epstein and other psychopaths, are the embodiment of Wetiko and express its state of being in all its grotesqueness. The Cult is Wetiko incarnate, Yaldabaoth incarnate, and it seeks to facilitate Wetiko assimilation of humanity in totality into its distortion by manipulating the population into low frequency states that match its own. Paul Levy writes: ‘Holographically enforced within the psyche of every human being the wetiko virus pervades and underlies the entire field of consciousness, and can therefore potentially manifest through any one of us at any moment if we are not mindful.’ The ‘Covid’ hoax has achieved this with many people, but others have not fallen into Wetiko’s frequency lair. Players in the ‘Covid’ human catastrophe including Gates, Schwab, Tedros, Fauci, Whi y, Vallance, Johnson, Hancock, Ferguson, Drosten, and all the rest, including the psychopath psychologists, are expressions of Wetiko. This is why

they have no compassion or empathy and no emotional consequence for what they do that would make them stop doing it. Observe all the people who support the psychopaths in authority against the Pushbackers despite the damaging impact the psychopaths have on their own lives and their family’s lives. You are again looking at Wetiko possession which prevents them seeing through the lies to the obvious scam going on. Why can’t they see it? Wetiko won’t let them see it. The perceptual divide that has now become a chasm is between the Wetikoed and the non-Wetikoed. Paul Levy describes Wetiko in the same way that I have long described the Archontic force. They are the same distorted consciousness operating across dimensions of reality: ‘… the subtle body of wetiko is not located in the third dimension of space and time, literally existing in another dimension … it is able to affect ordinary lives by mysteriously interpenetrating into our threedimensional world.’ Wetiko does this through its incarnate representatives in the Cult and by weaving itself into The Field which on our level of reality is the electromagnetic information field of the simulation or Matrix. More than that, the simulation is Wetiko / Yaldabaoth. Caleb Scharf, Director of Astrobiology at Columbia University, has speculated that ‘alien life’ could be so advanced that it has transcribed itself into the quantum realm to become what we call physics. He said intelligence indistinguishable from the fabric of the Universe would solve many of its greatest mysteries: Perhaps hyper-advanced life isn’t just external. Perhaps it’s already all around. It is embedded in what we perceive to be physics itself, from the root behaviour of particles and fields to the phenomena of complexity and emergence ... In other words, life might not just be in the equations. It might BE the equations [My emphasis].

Scharf said it is possible that ‘we don’t recognise advanced life because it forms an integral and unsuspicious part of what we’ve considered to be the natural world’. I agree. Wetiko/Yaldabaoth is the simulation. We are literally in the body of the beast. But that doesn’t mean it has to control us. We all have the power to overcome Wetiko

influence and the Cult knows that. I doubt it sleeps too well because it knows that.

Which Field?

This, I suggest, is how it all works. There are two Fields. One is the fierce electromagnetic light of the Matrix within the speed of light; the other is the ‘watery light’ of The Field beyond the walls of the Matrix that connects with the Great Infinity. Five-sense mind and the decoding systems of the body a ach us to the Field of Matrix light. They have to or we could not experience this reality. Five-sense mind sees only the Matrix Field of information while our expanded consciousness is part of the Infinity Field. When we open our minds, and most importantly our hearts, to the Infinity Field we have a mission control which gives us an expanded perspective, a road map, to understand the nature of the five-sense world. If we are isolated only in five-sense mind there is no mission control. We’re on our own trying to understand a world that’s constantly feeding us information to ensure we do not understand. People in this state can feel ‘lost’ and bewildered with no direction or radar. You can see ever more clearly those who are influenced by the Fields of Big Infinity or li le five-sense mind simply by their views and behaviour with regard to the ‘Covid’ hoax. We have had this division throughout known human history with the mass of the people on one side and individuals who could see and intuit beyond the walls of the simulation – Plato’s prisoner who broke out of the cave and saw reality for what it is. Such people have always been targeted by Wetiko/Archon-possessed authority, burned at the stake or demonised as mad, bad and dangerous. The Cult today and its global network of ‘anti-hate’, ‘anti-fascist’ Woke groups are all expressions of Wetiko a acking those exposing the conspiracy, ‘Covid’ lies and the ‘vaccine’ agenda. Woke as a whole is Wetiko which explains its black and white mentality and how at one it is with the Wetiko-possessed Cult. Paul Levy said: ‘To be in this paradigm is to still be under the thrall of a two-valued logic – where things are either true or false – of a

wetikoized mind.’ Wetiko consciousness is in a permanent rage, therefore so is Woke, and then there is Woke inversion and contradiction. ‘Anti-fascists’ act like fascists because fascists and ‘antifascists’ are both Wetiko at work. Political parties act the same while claiming to be different for the same reason. Secret society and satanic rituals are a aching initiates to Wetiko and the cold, ruthless, psychopathic mentality that secures the positions of power all over the world is Wetiko. Reframing ‘training programmes’ have the same cumulative effect of a aching Wetiko and we have their graduates described as automatons and robots with a cold, psychopathic, uncaring demeanour. They are all traits of Wetiko possession and look how many times they have been described in this book and elsewhere with regard to personnel behind ‘Covid’ including the police and medical profession. Climbing the greasy pole in any profession in a Wetiko society requires traits of Wetiko to get there and that is particularly true of politics which is not about fair competition and pre-eminence of ideas. It is founded on how many backs you can stab and arses you can lick. This culminated in the global ‘Covid’ coordination between the Wetiko possessed who pulled it off in all the different countries without a trace of empathy and compassion for their impact on humans. Our sight sense can see only holographic form and not the Field which connects holographic form. Therefore we perceive ‘physical’ objects with ‘space’ in between. In fact that ‘space’ is energy/consciousness operating on multiple frequencies. One of them is Wetiko and that connects the Cult psychopaths, those who submit to the psychopaths, and those who serve the psychopaths in the media operations of the world. Wetiko is Gates. Wetiko is the mask-wearing submissive. Wetiko is the fake journalist and ‘fact-checker’. The Wetiko Field is coordinating the whole thing. Psychopaths, gofers, media operatives, ‘anti-hate’ hate groups, ‘fact-checkers’ and submissive people work as one unit even without human coordination because they are a ached to the same Field which is organising it all (Fig 22). Paul Levy is here describing how Wetiko-possessed people are drawn together and refuse to let any information breach their rigid

perceptions. He was writing long before ‘Covid’, but I think you will recognise followers of the ‘Covid’ religion oh just a little bit: People who are channelling the vibratory frequency of wetiko align with each other through psychic resonance to reinforce their unspoken shared agreement so as to uphold their deranged view of reality. Once an unconscious content takes possession of certain individuals, it irresistibly draws them together by mutual attraction and knits them into groups tied together by their shared madness that can easily swell into an avalanche of insanity. A psychic epidemic is a closed system, which is to say that it is insular and not open to any new information or informing influences from the outside world which contradict its fixed, limited, and limiting perspective.

There we have the Woke mind and the ‘Covid’ mind. Compatible resonance draws the awakening together, too, which is clearly happening today.

Figure 22: The Wetiko Field from which the Cult pyramid and its personnel are made manifest. (Image by Neil Hague).

Spiritual servitude

Wetiko doesn’t care about humans. It’s not human; it just possesses humans for its own ends and the effect (depending on the scale of

possession) can be anything from extreme psychopathy to unquestioning obedience. Wetiko’s worst nightmare is for human consciousness to expand beyond the simulation. Everything is focussed on stopping that happening through control of information, thus perception, thus frequency. The ‘education system’, media, science, medicine, academia, are all geared to maintaining humanity in five-sense servitude as is the constant stimulation of low-vibrational mental and emotional states (see ‘Covid’). Wetiko seeks to dominate those subconscious spaces between five-sense perception and expanded consciousness where the computer meets the operator. From these subconscious hiding places Wetiko speaks to us to trigger urges and desires that we take to be our own and manipulate us into anything from low-vibrational to psychopathic states. Remember how Islam describes the Jinn as invisible tricksters that ‘whisper’ and confuse. Wetiko is the origin of the ‘trickster god’ theme that you find in cultures all over the world. Jinn, like the Archons, are Wetiko which is terrified of humans awakening and reconnecting with our true self for then its energy source has gone. With that the feedback loop breaks between Wetiko and human perception that provides the energetic momentum on which its very existence depends as a force of evil. Humans are both its target and its source of survival, but only if we are operating in low-vibrational states of fear, hate, depression and the background anxiety that most people suffer. We are Wetiko’s target because we are its key to survival. It needs us, not the other way round. Paul Levy writes: A vampire has no intrinsic, independent, substantial existence in its own right; it only exists in relation to us. The pathogenic, vampiric mind-parasite called wetiko is nothing in itself – not being able to exist from its own side – yet it has a ‘virtual reality’ such that it can potentially destroy our species … …The fact that a vampire is not reflected by a mirror can also mean that what we need to see is that there’s nothing, no-thing to see, other than ourselves. The fact that wetiko is the expression of something inside of us means that the cure for wetiko is with us as well. The critical issue is finding this cure within us and then putting it into effect.

Evil begets evil because if evil does not constantly expand and find new sources of energetic sustenance its evil, its distortion, dies with the assimilation into balance and harmony. Love is the garlic to Wetiko’s vampire. Evil, the absence of love, cannot exist in the presence of love. I think I see a way out of here. I have emphasised so many times over the decades that the Archons/Wetiko and their Cult are not all powerful. They are not. I don’t care how it looks even now they are not. I have not called them li le boys in short trousers for effect. I have said it because it is true. Wetiko’s insatiable desire for power over others is not a sign of its omnipotence, but its insecurity. Paul Levy writes: ‘Due to the primal fear which ultimately drives it and which it is driven to cultivate, wetiko’s body politic has an intrinsic and insistent need for centralising power and control so as to create imagined safety for itself.’ Yeeeeeees! Exactly! Why does Wetiko want humans in an ongoing state of fear? Wetiko itself is fear and it is petrified of love. As evil is an absence of love, so love is an absence of fear. Love conquers all and especially Wetiko which is fear. Wetiko brought fear into the world when it wasn’t here before. Fear was the ‘fall’, the fall into low-frequency ignorance and illusion – fear is False Emotion Appearing Real. The simulation is driven and energised by fear because Wetiko/Yaldabaoth (fear) are the simulation. Fear is the absence of love and Wetiko is the absence of love.

Wetiko today

We can now view current events from this level of perspective. The ‘Covid’ hoax has generated momentous amounts of ongoing fear, anxiety, depression and despair which have empowered Wetiko. No wonder people like Gates have been the instigators when they are Wetiko incarnate and exhibit every trait of Wetiko in the extreme. See how cold and unemotional these people are like Gates and his cronies, how dead of eye they are. That’s Wetiko. Sabbatians are Wetiko and everything they control including the World Health Organization, Big Pharma and the ‘vaccine’ makers, national ‘health’

hierarchies, corporate media, Silicon Valley, the banking system, and the United Nations with its planned transformation into world government. All are controlled and possessed by the Wetiko distortion into distorting human society in its image. We are with this knowledge at the gateway to understanding the world. Divisions of race, culture, creed and sexuality are diversions to hide the real division between those possessed and influenced by Wetiko and those that are not. The ‘Covid’ hoax has brought both clearly into view. Human behaviour is not about race. Tyrants and dictatorships come in all colours and creeds. What unites the US president bombing the innocent and an African tribe commi ing genocide against another as in Rwanda? What unites them? Wetiko. All wars are Wetiko, all genocide is Wetiko, all hunger over centuries in a world of plenty is Wetiko. Children going to bed hungry, including in the West, is Wetiko. Cult-generated Woke racial divisions that focus on the body are designed to obscure the reality that divisions in behaviour are manifestations of mind, not body. Obsession with body identity and group judgement is a means to divert a ention from the real source of behaviour – mind and perception. Conflict sown by the Woke both within themselves and with their target groups are Wetiko providing lunch for itself through still more agents of the division, chaos, and fear on which it feeds. The Cult is seeking to assimilate the entirety of humanity and all children and young people into the Wetiko frequency by manipulating them into states of fear and despair. Witness all the suicide and psychological unravelling since the spring of 2020. Wetiko psychopaths want to impose a state of unquestioning obedience to authority which is no more than a conduit for Wetiko to enforce its will and assimilate humanity into itself. It needs us to believe that resistance is futile when it fears resistance and even more so the game-changing non-cooperation with its impositions. It can use violent resistance for its benefit. Violent impositions and violent resistance are both Wetiko. The Power of Love with its Power of No will sweep Wetiko from our world. Wetiko and its Cult know that. They just don’t want us to know.

AI Wetiko

This brings me to AI or artificial intelligence and something else Wetikos don’t want us to know. What is AI really? I know about computer code algorithms and AI that learns from data input. These, however, are more diversions, the expeditionary force, for the real AI that they want to connect to the human brain as promoted by Silicon Valley Wetikos like Kurzweil. What is this AI? It is the frequency of Wetiko, the frequency of the Archons. The connection of AI to the human brain is the connection of the Wetiko frequency to create a Wetiko hive mind and complete the job of assimilation. The hive mind is planned to be controlled from Israel and China which are both 100 percent owned by Wetiko Sabbatians. The assimilation process has been going on minute by minute in the ‘smart’ era which fused with the ‘Covid’ era. We are told that social media is scrambling the minds of the young and changing their personality. This is true, but what is social media? Look more deeply at how it works, how it creates divisions and conflict, the hostility and cruelty, the targeting of people until they are destroyed. That’s Wetiko. Social media is manipulated to tune people to the Wetiko frequency with all the emotional exploitation tricks employed by platforms like Facebook and its Wetiko front man, Zuckerberg. Facebook’s Instagram announced a new platform for children to overcome a legal bar on them using the main site. This is more Wetiko exploitation and manipulation of kids. Amnesty International likened the plan to foxes offering to guard the henhouse and said it was incompatible with human rights. Since when did Wetiko or Zuckerberg (I repeat myself) care about that? Would Brin and Page at Google, Wojcicki at YouTube, Bezos at Amazon and whoever the hell runs Twi er act as they do if they were not channelling Wetiko? Would those who are developing technologies for no other reason than human control? How about those designing and selling technologies to kill people and Big Pharma drug and ‘vaccine’ producers who know they will end or devastate lives? Quite a thought for these people to consider is that if you are Wetiko in a human life you are Wetiko on the ‘other side’ unless your frequency

changes and that can only change by a change of perception which becomes a change of behaviour. Where Gates is going does not bear thinking about although perhaps that’s exactly where he wants to go. Either way, that’s where he’s going. His frequency will make it so.

The frequency lair

I have been saying for a long time that a big part of the addiction to smartphones and devices is that a frequency is coming off them that entraps the mind. People spend ages on their phones and sometimes even a minute or so a er they put them down they pick them up again and it all repeats. ‘Covid’ lockdowns will have increased this addiction a million times for obvious reasons. Addictions to alcohol overindulgence and drugs are another way that Wetiko entraps consciousness to a ach to its own. Both are symptoms of lowvibrational psychological distress which alcoholism and drug addiction further compound. Do we think it’s really a coincidence that access to them is made so easy while potions that can take people into realms beyond the simulation are banned and illegal? I have explored smartphone addiction in other books, the scale is mind-blowing, and that level of addiction does not come without help. Tech companies that make these phones are Wetiko and they will have no qualms about destroying the minds of children. We are seeing again with these companies the Wetiko perceptual combination of psychopathic enforcers and weak and meek unquestioning compliance by the rank and file. The global Smart Grid is the Wetiko Grid and it is crucial to complete the Cult endgame. The simulation is radiation and we are being deluged with technological radiation on a devastating scale. Wetiko frauds like Elon Musk serve Cult interests while occasionally criticising them to maintain his street-cred. 5G and other forms of Wi-Fi are being directed at the earth from space on a volume and scale that goes on increasing by the day. Elon Musk’s (officially) SpaceX Starlink project is in the process of pu ing tens of thousands of satellites in low orbit to cover every inch of the planet with 5G and other Wi-Fi to create Kurzweil’s global ‘cloud’ to which the

human mind is planned to be a ached very soon. SpaceX has approval to operate 12,000 satellites with more than 1,300 launched at the time of writing and applications filed for 30,000 more. Other operators in the Wi-Fi, 5G, low-orbit satellite market include OneWeb (UK), Telesat (Canada), and AST & Science (US). Musk tells us that AI could be the end of humanity and then launches a company called Neuralink to connect the human brain to computers. Musk’s (in theory) Tesla company is building electric cars and the driverless vehicles of the smart control grid. As frauds and bullshi ers go Elon Musk in my opinion is Major League. 5G and technological radiation in general are destructive to human health, genetics and psychology and increasing the strength of artificial radiation underpins the five-sense perceptual bubbles which are themselves expressions of radiation or electromagnetism. Freedom activist John Whitehead was so right with his ‘databit by databit, we are building our own electronic concentration camps’. The Smart Grid and 5G is a means to control the human mind and infuse perceptual information into The Field to influence anyone in sync with its frequency. You can change perception and behaviour en masse if you can manipulate the population into those levels of frequency and this is happening all around us today. The arrogance of Musk and his fellow Cult operatives knows no bounds in the way that we see with Gates. Musk’s satellites are so many in number already they are changing the night sky when viewed from Earth. The astronomy community has complained about this and they have seen nothing yet. Some consequences of Musk’s Wetiko hubris include: Radiation; visible pollution of the night sky; interference with astronomy and meteorology; ground and water pollution from intensive use of increasingly many spaceports; accumulating space debris; continual deorbiting and burning up of aging satellites, polluting the atmosphere with toxic dust and smoke; and everincreasing likelihood of collisions. A collective public open le er of complaint to Musk said: We are writing to you … because SpaceX is in process of surrounding the Earth with a network of thousands of satellites whose very purpose is to irradiate every square inch of the

Earth. SpaceX, like everyone else, is treating the radiation as if it were not there. As if the mitochondria in our cells do not depend on electrons moving undisturbed from the food we digest to the oxygen we breathe. As if our nervous systems and our hearts are not subject to radio frequency interference like any piece of electronic equipment. As if the cancer, diabetes, and heart disease that now afflict a majority of the Earth’s population are not metabolic diseases that result from interference with our cellular machinery. As if insects everywhere, and the birds and animals that eat them, are not starving to death as a result.

People like Musk and Gates believe in their limitless Wetiko arrogance that they can do whatever they like to the world because they own it. Consequences for humanity are irrelevant. It’s absolutely time that we stopped taking this shit from these selfstyled masters of the Earth when you consider where this is going.

Why is the Cult so anti-human?

I hear this question o en: Why would they do this when it will affect them, too? Ah, but will it? Who is this them? Forget their bodies. They are just vehicles for Wetiko consciousness. When you break it all down to the foundations we are looking at a state of severely distorted consciousness targeting another state of consciousness for assimilation. The rest is detail. The simulation is the fly-trap in which unique sensations of the five senses create a cycle of addiction called reincarnation. Renegade Minds see that everything which happens in our reality is a smaller version of the whole picture in line with the holographic principle. Addiction to the radiation of smart technology is a smaller version of addiction to the whole simulation. Connecting the body/brain to AI is taking that addiction on a giant step further to total ongoing control by assimilating human incarnate consciousness into Wetiko. I have watched during the ‘Covid’ hoax how many are becoming ever more profoundly a ached to Wetiko’s perceptual calling cards of aggressive response to any other point of view (‘There is no other god but me’), psychopathic lack of compassion and empathy, and servile submission to the narrative and will of authority. Wetiko is the psychopaths and subservience to psychopaths. The Cult of Wetiko is

so anti-human because it is not human. It embarked on a mission to destroy human by targeting everything that it means to be human and to survive as human. ‘Covid’ is not the end, just a means to an end. The Cult with its Wetiko consciousness is seeking to change Earth systems, including the atmosphere, to suit them, not humans. The gathering bombardment of 5G alone from ground and space is dramatically changing The Field with which the five senses interact. There is so much more to come if we sit on our hands and hope it will all go away. It is not meant to go away. It is meant to get ever more extreme and we need to face that while we still can – just. Carbon dioxide is the gas of life. Without that human is over. Kaput, gone, history. No natural world, no human. The Cult has created a cock and bull story about carbon dioxide and climate change to justify its reduction to the point where Gates and the ignoramus Biden ‘climate chief’ John Kerry want to suck it out of the atmosphere. Kerry wants to do this because his master Gates does. Wetikos have made the gas of life a demon with the usual support from the Wokers of Extinction Rebellion and similar organisations and the bewildered puppet-child that is Greta Thunberg who was put on the world stage by Klaus Schwab and the World Economic Forum. The name Extinction Rebellion is both ironic and as always Wetiko inversion. The gas that we need to survive must be reduced to save us from extinction. The most basic need of human is oxygen and we now have billions walking around in face nappies depriving body and brain of this essential requirement of human existence. More than that 5G at 60 gigahertz interacts with the oxygen molecule to reduce the amount of oxygen the body can absorb into the bloodstream. The obvious knock-on consequences of that for respiratory and cognitive problems and life itself need no further explanation. Psychopaths like Musk are assembling a global system of satellites to deluge the human atmosphere with this insanity. The man should be in jail. Here we have two most basic of human needs, oxygen and carbon dioxide, being dismantled. Two others, water and food, are ge ing similar treatment with the United Nations Agendas 21 and 2030 – the Great Reset – planning to

centrally control all water and food supplies. People will not even own rain water that falls on their land. Food is affected at the most basic level by reducing carbon dioxide. We have genetic modification or GMO infiltrating the food chain on a mass scale, pesticides and herbicides polluting the air and destroying the soil. Freshwater fish that provide livelihoods for 60 million people and feed hundreds of millions worldwide are being ‘pushed to the brink’ according the conservationists while climate change is the only focus. Now we have Gates and Schwab wanting to dispense with current food sources all together and replace them with a synthetic version which the Wetiko Cult would control in terms of production and who eats and who doesn’t. We have been on the Totalitarian Tiptoe to this for more than 60 years as food has become ever more processed and full of chemical shite to the point today when it’s not natural food at all. As Dr Tom Cowan says: ‘If it has a label don’t eat it.’ Bill Gates is now the biggest owner of farmland in the United States and he does nothing without an ulterior motive involving the Cult. Klaus Schwab wrote: ‘To feed the world in the next 50 years we will need to produce as much food as was produced in the last 10,000 years … food security will only be achieved, however, if regulations on genetically modified foods are adapted to reflect the reality that gene editing offers a precise, efficient and safe method of improving crops.’ Liar. People and the world are being targeted with aluminium through vaccines, chemtrails, food, drink cans, and endless other sources when aluminium has been linked to many health issues including dementia which is increasing year a er year. Insects, bees and wildlife essential to the food chain are being deleted by pesticides, herbicides and radiation which 5G is dramatically increasing with 6G and 7G to come. The pollinating bee population is being devastated while wildlife including birds, dolphins and whales are having their natural radar blocked by the effects of ever-increasing radiation. In the summer windscreens used to be spla ered with insects so numerous were they. It doesn’t happen now. Where have they gone?

Synthetic everything

The Cult is introducing genetically-modified versions of trees, plants and insects including a Gates-funded project to unleash hundreds of millions of genetically-modified, lab-altered and patented male mosquitoes to mate with wild mosquitoes and induce genetic flaws that cause them to die out. Clinically-insane Gates-funded Japanese researchers have developed mosquitos that spread vaccine and are dubbed ‘flying vaccinators’. Gates is funding the modification of weather pa erns in part to sell the myth that this is caused by carbon dioxide and he’s funding geoengineering of the skies to change the atmosphere. Some of this came to light with the Gates-backed plan to release tonnes of chalk into the atmosphere to ‘deflect the Sun and cool the planet’. Funny how they do this while the heating effect of the Sun is not factored into climate projections focussed on carbon dioxide. The reason is that they want to reduce carbon dioxide (so don’t mention the Sun), but at the same time they do want to reduce the impact of the Sun which is so essential to human life and health. I have mentioned the sun-cholesterol-vitamin D connection as they demonise the Sun with warnings about skin cancer (caused by the chemicals in sun cream they tell you to splash on). They come from the other end of the process with statin drugs to reduce cholesterol that turns sunlight into vitamin D. A lack of vitamin D leads to a long list of health effects and how vitamin D levels must have fallen with people confined to their homes over ‘Covid’. Gates is funding other forms of geoengineering and most importantly chemtrails which are dropping heavy metals, aluminium and self-replicating nanotechnology onto the Earth which is killing the natural world. See Everything You Need To Know, But Have Never Been Told for the detailed background to this. Every human system is being targeted for deletion by a force that’s not human. The Wetiko Cult has embarked on the process of transforming the human body from biological to synthetic biological as I have explained. Biological is being replaced by the artificial and synthetic – Archontic ‘countermimicry’ – right across human society. The plan eventually is to dispense with the human body altogether

and absorb human consciousness – which it wouldn’t really be by then – into cyberspace (the simulation which is Wetiko/Yaldabaoth). Preparations for that are already happening if people would care to look. The alternative media rightly warns about globalism and ‘the globalists’, but this is far bigger than that and represents the end of the human race as we know it. The ‘bad copy’ of prime reality that Gnostics describe was a bad copy of harmony, wonder and beauty to start with before Wetiko/Yaldabaoth set out to change the simulated ‘copy’ into something very different. The process was slow to start with. Entrapped humans in the simulation timeline were not technologically aware and they had to be brought up to intellectual speed while being suppressed spiritually to the point where they could build their own prison while having no idea they were doing so. We have now reached that stage where technological intellect has the potential to destroy us and that’s why events are moving so fast. Central American shaman Don Juan Matus said: Think for a moment, and tell me how you would explain the contradictions between the intelligence of man the engineer and the stupidity of his systems of belief, or the stupidity of his contradictory behaviour. Sorcerers believe that the predators have given us our systems of beliefs, our ideas of good and evil; our social mores. They are the ones who set up our dreams of success or failure. They have given us covetousness, greed, and cowardice. It is the predator who makes us complacent, routinary, and egomaniacal. In order to keep us obedient and meek and weak, the predators engaged themselves in a stupendous manoeuvre – stupendous, of course, from the point of view of a fighting strategist; a horrendous manoeuvre from the point of those who suffer it. They gave us their mind. The predators’ mind is baroque, contradictory, morose, filled with the fear of being discovered any minute now.

For ‘predators’ see Wetiko, Archons, Yaldabaoth, Jinn, and all the other versions of the same phenomenon in cultures and religions all over the world. The theme is always the same because it’s true and it’s real. We have reached the point where we have to deal with it. The question is – how?

Don’t fight – walk away

I thought I’d use a controversial subheading to get things moving in terms of our response to global fascism. What do you mean ‘don’t fight’? What do you mean ‘walk away’? We’ve got to fight. We can’t walk away. Well, it depends what we mean by fight and walk away. If fighting means physical combat we are playing Wetiko’s game and falling for its trap. It wants us to get angry, aggressive, and direct hate and hostility at the enemy we think we must fight. Every war, every ba le, every conflict, has been fought with Wetiko leading both sides. It’s what it does. Wetiko wants a fight, anywhere, any place. Just hit me, son, so I can hit you back. Wetiko hits Wetiko and Wetiko hits Wetiko in return. I am very forthright as you can see in exposing Wetikos of the Cult, but I don’t hate them. I refuse to hate them. It’s what they want. What you hate you become. What you fight you become. Wokers, ‘anti-haters’ and ‘anti-fascists’ prove this every time they reach for their keyboards or don their balaclavas. By walk away I mean to disengage from Wetiko which includes ceasing to cooperate with its tyranny. Paul Levy says of Wetiko: The way to ‘defeat’ evil is not to try to destroy it (for then, in playing evil’s game, we have already lost), but rather, to find the invulnerable place within ourselves where evil is unable to vanquish us – this is to truly ‘win’ our battle with evil.

Wetiko is everywhere in human society and it’s been on steroids since the ‘Covid’ hoax. Every shouting match over wearing masks has Wetiko wearing a mask and Wetiko not wearing one. It’s an electrical circuit of push and resist, push and resist, with Wetiko pushing and resisting. Each polarity is Wetiko empowering itself. Dictionary definitions of ‘resist’ include ‘opposing, refusing to accept or comply with’ and the word to focus on is ‘opposing’. What form does this take – se ing police cars alight or ‘refusing to accept or comply with’? The former is Wetiko opposing Wetiko while the other points the way forward. This is the difference between those aggressively demanding that government fascism must be obeyed who stand in stark contrast to the great majority of Pushbackers. We saw this clearly with a march by thousands of Pushbackers against lockdown in London followed days later by a Woker-hijacked

protest in Bristol in which police cars were set on fire. Masks were virtually absent in London and widespread in Bristol. Wetiko wants lockdown on every level of society and infuses its aggression to police it through its unknowing stooges. Lockdown protesters are the ones with the smiling faces and the hugs, The two blatantly obvious states of being – ge ing more obvious by the day – are the result of Wokers and their like becoming ever more influenced by the simulation Field of Wetiko and Pushbackers ever more influenced by The Field of a far higher vibration beyond the simulation. Wetiko can’t invade the heart which is where most lockdown opponents are coming from. It’s the heart that allows them to see through the lies to the truth in ways I will be highlighting. Renegade Minds know that calmness is the place from which wisdom comes. You won’t find wisdom in a hissing fit and wisdom is what we need in abundance right now. Calmness is not weakness – you don’t have to scream at the top of your voice to be strong. Calmness is indeed a sign of strength. ‘No’ means I’m not doing it. NOOOO!!! doesn’t mean you’re not doing it even more. Volume does not advance ‘No – I’m not doing it’. You are just not doing it. Wetiko possessed and influenced don’t know how to deal with that. Wetiko wants a fight and we should not give it one. What it needs more than anything is our cooperation and we should not give that either. Mass rallies and marches are great in that they are a visual representation of feeling, but if it ends there they are irrelevant. You demand that Wetikos act differently? Well, they’re not going to are they? They are Wetikos. We don’t need to waste our time demanding that something doesn’t happen when that will make no difference. We need to delete the means that allows it to happen. This, invariably, is our cooperation. You can demand a child stop firing a peashooter at the dog or you can refuse to buy the peashooter. If you provide the means you are cooperating with the dog being smacked on the nose with a pea. How can the authorities enforce mask-wearing if millions in a country refuse? What if the 74 million Pushbackers that voted for Trump in 2020 refused to wear masks, close their businesses or stay in their homes. It would be unenforceable. The

few control the many through the compliance of the many and that’s always been the dynamic be it ‘Covid’ regulations or the Roman Empire. I know people can find it intimidating to say no to authority or stand out in a crowd for being the only one with a face on display; but it has to be done or it’s over. I hope I’ve made clear in this book that where this is going will be far more intimidating than standing up now and saying ‘No’ – I will not cooperate with my own enslavement and that of my children. There might be consequences for some initially, although not so if enough do the same. The question that must be addressed is what is going to happen if we don’t? It is time to be strong and unyieldingly so. No means no. Not here and there, but everywhere and always. I have refused to wear a mask and obey all the other nonsense. I will not comply with tyranny. I repeat: Fascism is not imposed by fascists – there are never enough of them. Fascism is imposed by the population acquiescing to fascism. I will not do it. I will die first, or my body will. Living meekly under fascism is a form of death anyway, the death of the spirit that Martin Luther King described.

Making things happen

We must not despair. This is not over till it’s over and it’s far from that. The ‘fat lady’ must refuse to sing. The longer the ‘Covid’ hoax has dragged on and impacted on more lives we have seen an awakening of phenomenal numbers of people worldwide to the realisation that what they have believed all their lives is not how the world really is. Research published by the system-serving University of Bristol and King’s College London in February, 2021, concluded: ‘One in every 11 people in Britain say they trust David Icke’s take on the coronavirus pandemic.’ It will be more by now and we have gathering numbers to build on. We must urgently progress from seeing the scam to ceasing to cooperate with it. Prominent German lawyer Reiner Fuellmich, also licenced to practice law in America, is doing a magnificent job taking the legal route to bring the psychopaths to justice through a second Nuremberg tribunal for crimes against humanity. Fuellmich has an impressive record of

beating the elite in court and he formed the German Corona Investigative Commi ee to pursue civil charges against the main perpetrators with a view to triggering criminal charges. Most importantly he has grasped the foundation of the hoax – the PCR test not testing for the ‘virus’ – and Christian Drosten is therefore on his charge sheet along with Gates frontman Tedros at the World Health Organization. Major players must be not be allowed to inflict their horrors on the human race without being brought to book. A life sentence must follow for Bill Gates and the rest of them. A group of researchers has also indicted the government of Norway for crimes against humanity with copies sent to the police and the International Criminal Court. The lawsuit cites participation in an internationally-planned false pandemic and violation of international law and human rights, the European Commission’s definition of human rights by coercive rules, Nuremberg and Hague rules on fundamental human rights, and the Norwegian constitution. We must take the initiative from hereon and not just complain, protest and react. There are practical ways to support vital mass non-cooperation. Organising in numbers is one. Lockdown marches in London in the spring in 2021 were mass non-cooperation that the authorities could not stop. There were too many people. Hundreds of thousands walked the London streets in the centre of the road for mile a er mile while the Face-Nappies could only look on. They were determined, but calm, and just did it with no histrionics and lots of smiles. The police were impotent. Others are organising group shopping without masks for mutual support and imagine if that was happening all over. Policing it would be impossible. If the store refuses to serve people in these circumstances they would be faced with a long line of trolleys full of goods standing on their own and everything would have to be returned to the shelves. How would they cope with that if it kept happening? I am talking here about moving on from complaining to being pro-active; from watching things happen to making things happen. I include in this our relationship with the police. The behaviour of many Face-Nappies

has been disgraceful and anyone who thinks they would never find concentration camp guards in the ‘enlightened’ modern era have had that myth busted big-time. The period and se ing may change – Wetikos never do. I watched film footage from a London march in which a police thug viciously kicked a protestor on the floor who had done nothing. His fellow Face-Nappies stood in a ring protecting him. What he did was a criminal assault and with a crowd far outnumbering the police this can no longer be allowed to happen unchallenged. I get it when people chant ‘shame on you’ in these circumstances, but that is no longer enough. They have no shame those who do this. Crowds needs to start making a citizen’s arrest of the police who commit criminal offences and brutally a ack innocent people and defenceless women. A citizen’s arrest can be made under section 24A of the UK Police and Criminal Evidence (PACE) Act of 1984 and you will find something similar in other countries. I prefer to call it a Common Law arrest rather than citizen’s for reasons I will come to shortly. Anyone can arrest a person commi ing an indictable offence or if they have reasonable grounds to suspect they are commi ing an indictable offence. On both counts the a ack by the police thug would have fallen into this category. A citizen’s arrest can be made to stop someone: • • • •

Causing physical injury to himself or any other person Suffering physical injury Causing loss of or damage to property Making off before a constable can assume responsibility for him

A citizen’s arrest may also be made to prevent a breach of the peace under Common Law and if they believe a breach of the peace will happen or anything related to harm likely to be done or already done in their presence. This is the way to go I think – the Common Law version. If police know that the crowd and members of the public will no longer be standing and watching while they commit

their thuggery and crimes they will think twice about acting like Brownshirts and Blackshirts.

Common Law – common sense

Mention of Common Law is very important. Most people think the law is the law as in one law. This is not the case. There are two bodies of law, Common Law and Statute Law, and they are not the same. Common Law is founded on the simple premise of do no harm. It does not recognise victimless crimes in which no harm is done while Statute Law does. There is a Statute Law against almost everything. So what is Statute Law? Amazingly it’s the law of the sea that was brought ashore by the Cult to override the law of the land which is Common Law. They had no right to do this and as always they did it anyway. They had to. They could not impose their will on the people through Common Law which only applies to do no harm. How could you stitch up the fine detail of people’s lives with that? Instead they took the law of the sea, or Admiralty Law, and applied it to the population. Statute Law refers to all the laws spewing out of governments and their agencies including all the fascist laws and regulations relating to ‘Covid’. The key point to make is that Statute Law is contract law. It only applies between contracting corporations. Most police officers don’t even know this. They have to be kept in the dark, too. Long ago when merchants and their sailing ships began to trade with different countries a contractual law was developed called Admiralty Law and other names. Again it only applied to contracts agreed between corporate entities. If there is no agreed contract the law of the sea had no jurisdiction and that still applies to its new alias of Statute Law. The problem for the Cult when the law of the sea was brought ashore was an obvious one. People were not corporations and neither were government entities. To overcome the la er they made governments and all associated organisations corporations. All the institutions are private corporations and I mean governments and their agencies, local councils, police, courts, military, US states, the whole lot. Go to the

Dun and Bradstreet corporate listings website for confirmation that they are all corporations. You are arrested by a private corporation called the police by someone who is really a private security guard and they take you to court which is another private corporation. Neither have jurisdiction over you unless you consent and contract with them. This is why you hear the mantra about law enforcement policing by consent of the people. In truth the people ‘consent’ only in theory through monumental trickery. Okay, the Cult overcame the corporate law problem by making governments and institutions corporate entities; but what about people? They are not corporations are they? Ah ... well in a sense, and only a sense, they are. Not people exactly – the illusion of people. The Cult creates a corporation in the name of everyone at the time that their birth certificate is issued. Note birth/ berth certificate and when you go to court under the law of the sea on land you stand in a dock. These are throwbacks to the origin. My Common Law name is David Vaughan Icke. The name of the corporation created by the government when I was born is called Mr David Vaughan Icke usually wri en in capitals as MR DAVID VAUGHAN ICKE. That is not me, the living, breathing man. It is a fictitious corporate entity. The trick is to make you think that David Vaughan Icke and MR DAVID VAUGHAN ICKE are the same thing. They are not. When police charge you and take you to court they are prosecuting the corporate entity and not the living, breathing, man or woman. They have to trick you into identifying as the corporate entity and contracting with them. Otherwise they have no jurisdiction. They do this through a language known as legalese. Lawful and legal are not the same either. Lawful relates to Common Law and legal relates to Statute Law. Legalese is the language of Statue Law which uses terms that mean one thing to the public and another in legalese. Notice that when a police officer tells someone why they are being charged he or she will say at the end: ‘Do you understand?’ To the public that means ‘Do you comprehend?’ In legalese it means ‘Do you stand under me?’ Do you stand under my authority? If you say

yes to the question you are unknowingly agreeing to give them jurisdiction over you in a contract between two corporate entities. This is a confidence trick in every way. Contracts have to be agreed between informed parties and if you don’t know that David Vaughan Icke is agreeing to be the corporation MR DAVID VAUGHAN ICKE you cannot knowingly agree to contract. They are deceiving you and another way they do this is to ask for proof of identity. You usually show them a driving licence or other document on which your corporate name is wri en. In doing so you are accepting that you are that corporate entity when you are not. Referring to yourself as a ‘person’ or ‘citizen’ is also identifying with your corporate fiction which is why I made the Common Law point about the citizen’s arrest. If you are approached by a police officer you identify yourself immediately as a living, breathing, man or woman and say ‘I do not consent, I do not contract with you and I do not understand’ or stand under their authority. I have a Common Law birth certificate as a living man and these are available at no charge from commonlawcourt.com. Businesses registered under the Statute Law system means that its laws apply. There are, however, ways to run a business under Common Law. Remember all ‘Covid’ laws and regulations are Statute Law – the law of contracts and you do not have to contract. This doesn’t mean that you can kill someone and get away with it. Common Law says do no harm and that applies to physical harm, financial harm etc. Police are employees of private corporations and there needs to be a new system of noncorporate Common Law constables operating outside the Statute Law system. If you go to davidicke.com and put Common Law into the search engine you will find videos that explain Common Law in much greater detail. It is definitely a road we should walk.

With all my heart

I have heard people say that we are in a spiritual war. I don’t like the term ‘war’ with its Wetiko dynamic, but I know what they mean. Sweep aside all the bodily forms and we are in a situation in which two states of consciousness are seeking very different realities.

Wetiko wants upheaval, chaos, fear, suffering, conflict and control. The other wants love, peace, harmony, fairness and freedom. That’s where we are. We should not fall for the idea that Wetiko is allpowerful and there’s nothing we can do. Wetiko is not all-powerful. It’s a joke, pathetic. It doesn’t have to be, but it has made that choice for now. A handful of times over the years when I have felt the presence of its frequency I have allowed it to a ach briefly so I could consciously observe its nature. The experience is not pleasant, the energy is heavy and dark, but the ease with which you can kick it back out the door shows that its real power is in persuading us that it has power. It’s all a con. Wetiko is a con. It’s a trickster and not a power that can control us if we unleash our own. The con is founded on manipulating humanity to give its power to Wetiko which recycles it back to present the illusion that it has power when its power is ours that we gave away. This happens on an energetic level and plays out in the world of the seen as humanity giving its power to Wetiko authority which uses that power to control the population when the power is only the power the population has handed over. How could it be any other way for billions to be controlled by a relative few? I have had experiences with people possessed by Wetiko and again you can kick its arse if you do it with an open heart. Oh yes – the heart which can transform the world of perceived ‘ma er’. We are receiver-transmi ers and processors of information, but what information and where from? Information is processed into perception in three main areas – the brain, the heart and the belly. These relate to thinking, knowing, and emotion. Wetiko wants us to be head and belly people which means we think within the confines of the Matrix simulation and low-vibrational emotional reaction scrambles balance and perception. A few minutes on social media and you see how emotion is the dominant force. Woke is all emotion and is therefore thought-free and fact-free. Our heart is something different. It knows while the head thinks and has to try to work it out because it doesn’t know. The human energy field has seven prime vortexes which connect us with wider reality (Fig 23). Chakra means

‘wheels of light’ in the Sanskrit language of ancient India. The main ones are: The crown chakra on top of the head; brow (or ‘third eye’) chakra in the centre of the forehead; throat chakra; heart chakra in the centre of the chest; solar plexus chakra below the sternum; sacral chakra beneath the navel; and base chakra at the bo om of the spine. Each one has a particular function or functions. We feel anxiety and nervousness in the belly where the sacral chakra is located and this processes emotion that can affect the colon to give people ‘the shits’ or make them ‘shit scared’ when they are nervous. Chakras all play an important role, but the Mr and Mrs Big is the heart chakra which sits at the centre of the seven, above the chakras that connect us to the ‘physical’ and below those that connect with higher realms (or at least should). Here in the heart chakra we feel love, empathy and compassion – ‘My heart goes out to you’. Those with closed hearts become literally ‘heart-less’ in their a itudes and behaviour (see Bill Gates). Native Americans portrayed Wetiko with what Paul Levy calls a ‘frigid, icy heart, devoid of mercy’ (see Bill Gates).

Figure 23: The chakra system which interpenetrates the human energy field. The heart chakra is the governor – or should be.

Wetiko trembles at the thought of heart energy which it cannot infiltrate. The frequency is too high. What it seeks to do instead is close the heart chakra vortex to block its perceptual and energetic influence. Psychopaths have ‘hearts of stone’ and emotionallydamaged people have ‘heartache’ and ‘broken hearts’. The astonishing amount of heart disease is related to heart chakra

disruption with its fundamental connection to the ‘physical’ heart. Dr Tom Cowan has wri en an outstanding book challenging the belief that the heart is a pump and making the connection between the ‘physical’ and spiritual heart. Rudolph Steiner who was way ahead of his time said the same about the fallacy that the heart is a pump. What? The heart is not a pump? That’s crazy, right? Everybody knows that. Read Cowan’s Human Heart, Cosmic Heart and you will realise that the very idea of the heart as a pump is ridiculous when you see the evidence. How does blood in the feet so far from the heart get pumped horizontally up the body by the heart?? Cowan explains in the book the real reason why blood moves as it does. Our ‘physical’ heart is used to symbolise love when the source is really the heart vortex or spiritual heart which is our most powerful energetic connection to ‘out there’ expanded consciousness. That’s why we feel knowing – intuitive knowing – in the centre of the chest. Knowing doesn’t come from a process of thoughts leading to a conclusion. It is there in an instant all in one go. Our heart knows because of its connection to levels of awareness that do know. This is the meaning and source of intuition – intuitive knowing. For the last more than 30 years of uncovering the global game and the nature of reality my heart has been my constant antenna for truth and accuracy. An American intelligence insider once said that I had quoted a disinformer in one of my books and yet I had only quoted the part that was true. He asked: ‘How do you do that?’ By using my heart antenna was the answer and anyone can do it. Heartcentred is how we are meant to be. With a closed heart chakra we withdraw into a closed mind and the bubble of five-sense reality. If you take a moment to focus your a ention on the centre of your chest, picture a spinning wheel of light and see it opening and expanding. You will feel it happening, too, and perceptions of the heart like joy and love as the heart impacts on the mind as they interact. The more the chakra opens the more you will feel expressions of heart consciousness and as the process continues, and becomes part of you, insights and knowings will follow. An open

heart is connected to that level of awareness that knows all is One. You will see from its perspective that the fault-lines that divide us are only illusions to control us. An open heart does not process the illusions of race, creed and sexuality except as brief experiences for a consciousness that is all. Our heart does not see division, only unity (Figs 24 and 25). There’s something else, too. Our hearts love to laugh. Mark Twain’s quote that says ‘The human race has one really effective weapon, and that is laughter’ is really a reference to the heart which loves to laugh with the joy of knowing the true nature of infinite reality and that all the madness of human society is an illusion of the mind. Twain also said: ‘Against the assault of laughter nothing can stand.’ This is so true of Wetiko and the Cult. Their insecurity demands that they be taken seriously and their power and authority acknowledged and feared. We should do nothing of the sort. We should not get aggressive or fearful which their insecurity so desires. We should laugh in their face. Even in their no-face as police come over in their face-nappies and expect to be taken seriously. They don’t take themselves seriously looking like that so why should we? Laugh in the face of intimidation. Laugh in the face of tyranny. You will see by its reaction that you have pressed all of its bu ons. Wetiko does not know what to do in the face of laughter or when its targets refuse to concede their joy to fear. We have seen many examples during the ‘Covid’ hoax when people have expressed their energetic power and the string puppets of Wetiko retreat with their tail limp between their knees. Laugh – the world is bloody mad a er all and if it’s a choice between laughter and tears I know which way I’m going.

Figure 24: Head consciousness without the heart sees division and everything apart from everything else.

Figure 25: Heart consciousness sees everything as One.

‘Vaccines’ and the soul

The foundation of Wetiko/Archon control of humans is the separation of incarnate five-sense mind from the infinite ‘I’ and closing the heart chakra where the True ‘I’ lives during a human life. The goal has been to achieve complete separation in both cases. I was interested therefore to read an account by a French energetic healer of what she said she experienced with a patient who had been given the ‘Covid’ vaccine. Genuine energy healers can sense information and consciousness fields at different levels of being which are referred to as ‘subtle bodies’. She described treating the patient who later returned a er having, without the healer’s knowledge, two doses of the ‘Covid vaccine’. The healer said: I noticed immediately the change, very heavy energy emanating from [the] subtle bodies. The scariest thing was when I was working on the heart chakra, I connected with her soul: it was detached from the physical body, it had no contact and it was, as if it was floating in a state of total confusion: a damage to the consciousness that loses contact with the physical body, i.e. with our biological machine, there is no longer any communication between them. I continued the treatment by sending light to the heart chakra, the soul of the person, but it seemed that the soul could no longer receive any light, frequency or energy. It was a very powerful experience for me. Then I understood that this substance is indeed used to detach consciousness so that this consciousness can no longer interact through this body that it possesses in life, where there is no longer any contact, no frequency, no light, no more energetic balance or mind.

This would create a human that is rudderless and at the extreme almost zombie-like operating with a fractional state of consciousness at the mercy of Wetiko. I was especially intrigued by what the healer said in the light of the prediction by the highly-informed Rudolf Steiner more than a hundred years ago. He said: In the future, we will eliminate the soul with medicine. Under the pretext of a ‘healthy point of view’, there will be a vaccine by which the human body will be treated as soon as possible directly at birth, so that the human being cannot develop the thought of the existence of soul and Spirit. To materialistic doctors will be entrusted the task of removing the soul of humanity. As today, people are vaccinated against this disease or that disease, so in the future, children will be vaccinated with a substance that can be produced precisely in such a way that people, thanks to this vaccination, will be immune to being subjected to the ‘madness’ of spiritual life. He would be extremely smart, but he would not develop a conscience, and that is the true goal of some materialistic circles.

Steiner said the vaccine would detach the physical body from the etheric body (subtle bodies) and ‘once the etheric body is detached the relationship between the universe and the etheric body would become extremely unstable, and man would become an automaton’. He said ‘the physical body of man must be polished on this Earth by spiritual will – so the vaccine becomes a kind of arymanique (Wetiko) force’ and ‘man can no longer get rid of a given materialistic feeling’. Humans would then, he said, become ‘materialistic of constitution and can no longer rise to the spiritual’. I have been writing for years about DNA being a receiver-transmi er of information that connects us to other levels of reality and these ‘vaccines’ changing DNA can be likened to changing an antenna and what it can transmit and receive. Such a disconnection would clearly lead to changes in personality and perception. Steiner further predicted the arrival of AI. Big Pharma ‘Covid vaccine’ makers, expressions of Wetiko, are testing their DNA-manipulating evil on children as I write with a view to giving the ‘vaccine’ to babies. If it’s a soul-body disconnector – and I say that it is or can be – every child would be disconnected from ‘soul’ at birth and the ‘vaccine’ would create a closed system in which spiritual guidance from the greater self would play no part. This has been the ambition of Wetiko all

along. A Pentagon video from 2005 was leaked of a presentation explaining the development of vaccines to change behaviour by their effect on the brain. Those that believe this is not happening with the ‘Covid’ genetically-modifying procedure masquerading as a ‘vaccine’ should make an urgent appointment with Naivety Anonymous. Klaus Schwab wrote in 2018: Neurotechnologies enable us to better influence consciousness and thought and to understand many activities of the brain. They include decoding what we are thinking in fine levels of detail through new chemicals and interventions that can influence our brains to correct for errors or enhance functionality.

The plan is clear and only the heart can stop it. With every heart that opens, every mind that awakens, Wetiko is weakened. Heart and love are far more powerful than head and hate and so nothing like a majority is needed to turn this around.

Beyond the Phantom

Our heart is the prime target of Wetiko and so it must be the answer to Wetiko. We are our heart which is part of one heart, the infinite heart. Our heart is where the true self lives in a human life behind firewalls of five-sense illusion when an imposter takes its place – Phantom Self; but our heart waits patiently to be set free any time we choose to see beyond the Phantom, beyond Wetiko. A Wetikoed Phantom Self can wreak mass death and destruction while the love of forever is locked away in its heart. The time is here to unleash its power and let it sweep away the fear and despair that is Wetiko. Heart consciousness does not seek manipulated, censored, advantage for its belief or religion, its activism and desires. As an expression of the One it treats all as One with the same rights to freedom and opinion. Our heart demands fairness for itself no more than for others. From this unity of heart we can come together in mutual support and transform this Wetikoed world into what reality is meant to be – a place of love, joy, happiness, fairness, justice and freedom. Wetiko has another agenda and that’s why the world is as

it is, but enough of this nonsense. Wetiko can’t stay where hearts are open and it works so hard to keep them closed. Fear is its currency and its food source and love in its true sense has no fear. Why would love have fear when it knows it is All That Is, Has Been, And Ever Can Be on an eternal exploration of all possibility? Love in this true sense is not the physical a raction that passes for love. This can be an expression of it, yes, but Infinite Love, a love without condition, goes far deeper to the core of all being. It is the core of all being. Infinite realty was born from love beyond the illusions of the simulation. Love infinitely expressed is the knowing that all is One and the swi ly-passing experience of separation is a temporary hallucination. You cannot disconnect from Oneness; you can only perceive that you have and withdraw from its influence. This is the most important of all perception trickery by the mind parasite that is Wetiko and the foundation of all its potential for manipulation. If we open our hearts, open the sluice gates of the mind, and redefine self-identity amazing things start to happen. Consciousness expands or contracts in accordance with self-identity. When true self is recognised as infinite awareness and label self – Phantom Self – is seen as only a series of brief experiences life is transformed. Consciousness expands to the extent that self-identity expands and everything changes. You see unity, not division, the picture, not the pixels. From this we can play the long game. No more is an experience something in and of itself, but a fleeting moment in the eternity of forever. Suddenly people in uniform and dark suits are no longer intimidating. Doing what your heart knows to be right is no longer intimidating and consequences for those actions take on the same nature of a brief experience that passes in the blink of an infinite eye. Intimidation is all in the mind. Beyond the mind there is no intimidation. An open heart does not consider consequences for what it knows to be right. To do so would be to consider not doing what it knows to be right and for a heart in its power that is never an option. The Renegade Mind is really the Renegade Heart. Consideration of consequences will always provide a getaway car for the mind and

the heart doesn’t want one. What is right in the light of what we face today is to stop cooperating with Wetiko in all its forms and to do it without fear or compromise. You cannot compromise with tyranny when tyranny always demands more until it has everything. Life is your perception and you are your destiny. Change your perception and you change your life. Change collective perception and we change the world. Come on people … One human family, One heart, One goal … FREEEEEEDOM! We must se le for nothing less.

Postscript

T

he big scare story as the book goes to press is the ‘Indian’ variant and the world is being deluged with propaganda about the ‘Covid catastrophe’ in India which mirrors in its lies and misrepresentations what happened in Italy before the first lockdown in 2020. The New York Post published a picture of someone who had ‘collapsed in the street from Covid’ in India in April, 2021, which was actually taken during a gas leak in May, 2020. Same old, same old. Media articles in mid-February were asking why India had been so untouched by ‘Covid’ and then as their vaccine rollout gathered pace the alleged ‘cases’ began to rapidly increase. Indian ‘Covid vaccine’ maker Bharat Biotech was funded into existence by the Bill and Melinda Gates Foundation (the pair announced their divorce in May, 2021, which is a pity because they so deserve each other). The Indian ‘Covid crisis’ was ramped up by the media to terrify the world and prepare people for submission to still more restrictions. The scam that worked the first time was being repeated only with far more people seeing through the deceit. Davidicke.com and Ickonic.com have sought to tell the true story of what is happening by talking to people living through the Indian nightmare which has nothing to do with ‘Covid’. We posted a le er from ‘Alisha’ in Pune who told a very different story to government and media mendacity. She said scenes of dying people and overwhelmed hospitals were designed to hide what was really happening – genocide and starvation. Alisha said that millions had already died of starvation during the ongoing lockdowns while government and media were lying and making it look like the ‘virus’:

Restaurants, shops, gyms, theatres, basically everything is shut. The cities are ghost towns. Even so-called ‘essential’ businesses are only open till 11am in the morning. You basically have just an hour to buy food and then your time is up. Inter-state travel and even inter-district travel is banned. The cops wait at all major crossroads to question why you are traveling outdoors or to fine you if you are not wearing a mask. The medical community here is also complicit in genocide, lying about hospitals being full and turning away people with genuine illnesses, who need immediate care. They have even created a shortage of oxygen cylinders.

This is the classic Cult modus operandi played out in every country. Alisha said that people who would not have a PCR test not testing for the ‘virus’ were being denied hospital treatment. She said the people hit hardest were migrant workers and those in rural areas. Most businesses employed migrant workers and with everything closed there were no jobs, no income and no food. As a result millions were dying of starvation or malnutrition. All this was happening under Prime Minister Narendra Modi, a 100-percent asset of the Cult, and it emphasises yet again the scale of pure antihuman evil we are dealing with. Australia banned its people from returning home from India with penalties for trying to do so of up to five years in jail and a fine of £37,000. The manufactured ‘Covid’ crisis in India was being prepared to justify further fascism in the West. Obvious connections could be seen between the Indian ‘vaccine’ programme and increased ‘cases’ and this became a common theme. The Seychelles, the most per capita ‘Covid vaccinated’ population in the world, went back into lockdown a er a ‘surge of cases’. Long ago the truly evil Monsanto agricultural biotechnology corporation with its big connections to Bill Gates devastated Indian farming with genetically-modified crops. Human rights activist Gurcharan Singh highlighted the efforts by the Indian government to complete the job by destroying the food supply to hundreds of millions with ‘Covid’ lockdowns. He said that 415 million people at the bo om of the disgusting caste system (still going whatever they say) were below the poverty line and struggled to feed themselves every year. Now the government was imposing lockdown at just the

time to destroy the harvest. This deliberate policy was leading to mass starvation. People may reel back at the suggestion that a government would do that, but Wetiko-controlled ‘leaders’ are capable of any level of evil. In fact what is described in India is in the process of being instigated worldwide. The food chain and food supply are being targeted at every level to cause world hunger and thus control. Bill Gates is not the biggest owner of farmland in America for no reason and destroying access to food aids both the depopulation agenda and the plan for synthetic ‘food’ already being funded into existence by Gates. Add to this the coming hyperinflation from the suicidal creation of fake ‘money’ in response to ‘Covid’ and the breakdown of container shipping systems and you have a cocktail that can only lead one way and is meant to. The Cult plan is to crash the entire system to ‘build back be er’ with the Great Reset.

‘Vaccine’ transmission

Reports from all over the world continue to emerge of women suffering menstrual and fertility problems a er having the fake ‘vaccine’ and of the non-’vaccinated’ having similar problems when interacting with the ‘vaccinated’. There are far too many for ‘coincidence’ to be credible. We’ve had menopausal women ge ing periods, others having periods stop or not stopping for weeks, passing clots, sometimes the lining of the uterus, breast irregularities, and miscarriages (which increased by 400 percent in parts of the United States). Non-‘vaccinated’ men and children have suffered blood clots and nose bleeding a er interaction with the ‘vaccinated’. Babies have died from the effects of breast milk from a ‘vaccinated’ mother. Awake doctors – the small minority – speculated on the cause of non-’vaccinated’ suffering the same effects as the ‘vaccinated’. Was it nanotechnology in the synthetic substance transmi ing frequencies or was it a straight chemical bioweapon that was being transmi ed between people? I am not saying that some kind of chemical transmission is not one possible answer, but the foundation of all that the Cult does is frequency and

this is fertile ground for understanding how transmission can happen. American doctor Carrie Madej, an internal medicine physician and osteopath, has been practicing for the last 20 years, teaching medical students, and she says a ending different meetings where the agenda for humanity was discussed. Madej, who operates out of Georgia, did not dismiss other possible forms of transmission, but she focused on frequency in search of an explanation for transmission. She said the Moderna and Pfizer ‘vaccines’ contained nano-lipid particles as a key component. This was a brand new technology never before used on humanity. ‘They’re using a nanotechnology which is pre y much li le tiny computer bits … nanobots or hydrogel.’ Inside the ‘vaccines’ was ‘this sci-fi kind of substance’ which suppressed immune checkpoints to get into the cell. I referred to this earlier as the ‘Trojan horse’ technique that tricks the cell into opening a gateway for the self-replicating synthetic material and while the immune system is artificially suppressed the body has no defences. Madej said the substance served many purposes including an on-demand ability to ‘deliver the payload’ and using the nano ‘computer bits’ as biosensors in the body. ‘It actually has the ability to accumulate data from your body, like your breathing, your respiration, thoughts, emotions, all kinds of things.’ She said the technology obviously has the ability to operate through Wi-Fi and transmit and receive energy, messages, frequencies or impulses. ‘Just imagine you’re ge ing this new substance in you and it can react to things all around you, the 5G, your smart device, your phones.’ We had something completely foreign in the human body that had never been launched large scale at a time when we were seeing 5G going into schools and hospitals (plus the Musk satellites) and she believed the ‘vaccine’ transmission had something to do with this: ‘… if these people have this inside of them … it can act like an antenna and actually transmit it outwardly as well.’ The synthetic substance produced its own voltage and so it could have that kind of effect. This fits with my own contention that the nano receiver-transmi ers are designed to connect people to the

Smart Grid and break the receiver-transmi er connection to expanded consciousness. That would explain the French energy healer’s experience of the disconnection of body from ‘soul’ with those who have had the ‘vaccine’. The nanobots, self-replicating inside the body, would also transmit the synthetic frequency which could be picked up through close interaction by those who have not been ‘vaccinated’. Madej speculated that perhaps it was 5G and increased levels of other radiation that was causing the symptoms directly although interestingly she said that non-‘vaccinated’ patients had shown improvement when they were away from the ‘vaccinated’ person they had interacted with. It must be remembered that you can control frequency and energy with your mind and you can consciously create energetic barriers or bubbles with the mind to stop damaging frequencies from penetrating your field. American paediatrician Dr Larry Palevsky said the ‘vaccine’ was not a ‘vaccine’ and was never designed to protect from a ‘viral’ infection. He called it ‘a massive, brilliant propaganda of genocide’ because they didn’t have to inject everyone to get the result they wanted. He said the content of the jabs was able to infuse any material into the brain, heart, lungs, kidneys, liver, sperm and female productive system. ‘This is genocide; this is a weapon of mass destruction.’ At the same time American colleges were banning students from a ending if they didn’t have this life-changing and potentially life-ending ‘vaccine’. Class action lawsuits must follow when the consequences of this college fascism come to light. As the book was going to press came reports about fertility effects on sperm in ‘vaccinated’ men which would absolutely fit with what I have been saying and hospitals continued to fill with ‘vaccine’ reactions. Another question is what about transmission via blood transfusions? The NHS has extended blood donation restrictions from seven days a er a ‘Covid vaccination’ to 28 days a er even a sore arm reaction. I said in the spring of 2020 that the then touted ‘Covid vaccine’ would be ongoing each year like the flu jab. A year later Pfizer CEO, the appalling Albert Bourla, said people would ‘likely’ need a ‘booster dose’ of the ‘vaccine’ within 12 months of ge ing ‘fully

vaccinated’ and then a yearly shot. ‘Variants will play a key role’, he said confirming the point. Johnson & Johnson CEO Alex Gorsky also took time out from his ‘vaccine’ disaster to say that people may need to be vaccinated against ‘Covid-19’ each year. UK Health Secretary, the psychopath Ma Hancock, said additional ‘boosters’ would be available in the autumn of 2021. This is the trap of the ‘vaccine passport’. The public will have to accept every last ‘vaccine’ they introduce, including for the fake ‘variants’, or it would cease to be valid. The only other way in some cases would be continuous testing with a test not testing for the ‘virus’ and what is on the swabs constantly pushed up your noise towards the brain every time?

‘Vaccines’ changing behaviour

I mentioned in the body of the book how I believed we would see gathering behaviour changes in the ‘vaccinated’ and I am already hearing such comments from the non-‘vaccinated’ describing behaviour changes in friends, loved ones and work colleagues. This will only increase as the self-replicating synthetic material and nanoparticles expand in body and brain. An article in the Guardian in 2016 detailed research at the University of Virginia in Charlo esville which developed a new method for controlling brain circuits associated with complex animal behaviour. The method, dubbed ‘magnetogenetics’, involves genetically-engineering a protein called ferritin, which stores and releases iron, to create a magnetised substance – ‘Magneto’ – that can activate specific groups of nerve cells from a distance. This is claimed to be an advance on other methods of brain activity manipulation known as optogenetics and chemogenetics (the Cult has been developing methods of brain control for a long time). The ferritin technique is said to be noninvasive and able to activate neurons ‘rapidly and reversibly’. In other words, human thought and perception. The article said that earlier studies revealed how nerve cell proteins ‘activated by heat and mechanical pressure can be genetically engineered so that they become sensitive to radio waves and magnetic fields, by a aching them to an iron-storing protein called ferritin, or to inorganic

paramagnetic particles’. Sensitive to radio waves and magnetic fields? You mean like 5G, 6G and 7G? This is the human-AI Smart Grid hive mind we are talking about. The Guardian article said: … the researchers injected Magneto into the striatum of freely behaving mice, a deep brain structure containing dopamine-producing neurons that are involved in reward and motivation, and then placed the animals into an apparatus split into magnetised and non-magnetised sections. Mice expressing Magneto spent far more time in the magnetised areas than mice that did not, because activation of the protein caused the striatal neurons expressing it to release dopamine, so that the mice found being in those areas rewarding. This shows that Magneto can remotely control the firing of neurons deep within the brain, and also control complex behaviours.

Make no mistake this basic methodology will be part of the ‘Covid vaccine’ cocktail and using magnetics to change brain function through electromagnetic field frequency activation. The Pentagon is developing a ‘Covid vaccine’ using ferritin. Magnetics would explain changes in behaviour and why videos are appearing across the Internet as I write showing how magnets stick to the skin at the point of the ‘vaccine’ shot. Once people take these ‘vaccines’ anything becomes possible in terms of brain function and illness which will be blamed on ‘Covid-19’ and ‘variants’. Magnetic field manipulation would further explain why the non-‘vaccinated’ are reporting the same symptoms as the ‘vaccinated’ they interact with and why those symptoms are reported to decrease when not in their company. Interestingly ‘Magneto’, a ‘mutant’, is a character in the Marvel Comic X-Men stories with the ability to manipulate magnetic fields and he believes that mutants should fight back against their human oppressors by any means necessary. The character was born Erik Lehnsherr to a Jewish family in Germany.

Cult-controlled courts

The European Court of Human Rights opened the door for mandatory ‘Covid-19 vaccines’ across the continent when it ruled in a Czech Republic dispute over childhood immunisation that legally

enforced vaccination could be ‘necessary in a democratic society’. The 17 judges decided that compulsory vaccinations did not breach human rights law. On the face of it the judgement was so inverted you gasp for air. If not having a vaccine infused into your body is not a human right then what is? Ah, but they said human rights law which has been specifically wri en to delete all human rights at the behest of the state (the Cult). Article 8 of the European Convention on Human Rights relates to the right to a private life. The crucial word here is ‘except’: There shall be no interference by a public authority with the exercise of this right EXCEPT such as is in accordance with the law and is necessary in a democratic society in the interests of national security, public safety or the economic wellbeing of the country, for the prevention of disorder or crime, for the protection of health or morals, or for the protection of the rights and freedoms of others [My emphasis].

No interference except in accordance with the law means there are no ‘human rights’ except what EU governments decide you can have at their behest. ‘As is necessary in a democratic society’ explains that reference in the judgement and ‘in the interests of national security, public safety or the economic well-being of the country, for the prevention of disorder or crime, for the protection of health or morals, or for the protection of the rights and freedoms of others’ gives the EU a coach and horses to ride through ‘human rights’ and sca er them in all directions. The judiciary is not a check and balance on government extremism; it is a vehicle to enforce it. This judgement was almost laughably predictable when the last thing the Cult wanted was a decision that went against mandatory vaccination. Judges rule over and over again to benefit the system of which they are a part. Vaccination disputes that come before them are invariably delivered in favour of doctors and authorities representing the view of the state which owns the judiciary. Oh, yes, and we have even had calls to stop pu ing ‘Covid-19’ on death certificates within 28 days of a ‘positive test’ because it is claimed the practice makes the ‘vaccine’ appear not to work. They are laughing at you.

The scale of madness, inhumanity and things to come was highlighted when those not ‘vaccinated’ for ‘Covid’ were refused evacuation from the Caribbean island of St Vincent during massive volcanic eruptions. Cruise ships taking residents to the safety of another island allowed only the ‘vaccinated’ to board and the rest were le to their fate. Even in life and death situations like this we see ‘Covid’ stripping people of their most basic human instincts and the insanity is even more extreme when you think that fake ‘vaccine’-makers are not even claiming their body-manipulating concoctions stop ‘infection’ and ‘transmission’ of a ‘virus’ that doesn’t exist. St Vincent Prime Minister Ralph Gonsalves said: ‘The chief medical officer will be identifying the persons already vaccinated so that we can get them on the ship.’ Note again the power of the chief medical officer who, like Whi y in the UK, will be answering to the World Health Organization. This is the Cult network structure that has overridden politicians who ‘follow the science’ which means doing what WHO-controlled ‘medical officers’ and ‘science advisers’ tell them. Gonsalves even said that residents who were ‘vaccinated’ a er the order so they could board the ships would still be refused entry due to possible side effects such as ‘wooziness in the head’. The good news is that if they were woozy enough in the head they could qualify to be prime minister of St Vincent.

Microchipping freedom

The European judgement will be used at some point to justify moves to enforce the ‘Covid’ DNA-manipulating procedure. Sandra Ro, CEO of the Global Blockchain Business Council, told a World Economic Forum event that she hoped ‘vaccine passports’ would help to ‘drive forced consent and standardisation’ of global digital identity schemes: ‘I’m hoping with the desire and global demand for some sort of vaccine passport – so that people can get travelling and working again – [it] will drive forced consent, standardisation, and frankly, cooperation across the world.’ The lady is either not very bright, or thoroughly mendacious, to use the term ‘forced consent’.

You do not ‘consent’ if you are forced – you submit. She was describing what the plan has been all along and that’s to enforce a digital identity on every human without which they could not function. ‘Vaccine passports’ are opening the door and are far from the end goal. A digital identity would allow you to be tracked in everything you do in cyberspace and this is the same technique used by Cult-owned China to enforce its social credit system of total control. The ultimate ‘passport’ is planned to be a microchip as my books have warned for nearly 30 years. Those nice people at the Pentagon working for the Cult-controlled Defense Advanced Research Projects Agency (DARPA) claimed in April, 2021, they have developed a microchip inserted under the skin to detect ‘asymptomatic Covid-19 infection’ before it becomes an outbreak and a ‘revolutionary filter’ that can remove the ‘virus’ from the blood when a ached to a dialysis machine. The only problems with this are that the ‘virus’ does not exist and people transmi ing the ‘virus’ with no symptoms is brain-numbing bullshit. This is, of course, not a ruse to get people to be microchipped for very different reasons. DARPA also said it was producing a one-stop ‘vaccine’ for the ‘virus’ and all ‘variants’. One of the most sinister organisations on Planet Earth is doing this? Be er have it then. These people are insane because Wetiko that possesses them is insane. Researchers from the Salk Institute in California announced they have created an embryo that is part human and part monkey. My books going back to the 1990s have exposed experiments in top secret underground facilities in the United States where humans are being crossed with animal and non-human ‘extraterrestrial’ species. They are now easing that long-developed capability into the public arena and there is much more to come given we are dealing with psychiatric basket cases. Talking of which – Elon Musk’s scientists at Neuralink trained a monkey to play Pong and other puzzles on a computer screen using a joystick and when the monkey made the correct move a metal tube squirted banana smoothie into his mouth which is the basic technique for training humans into unquestioning compliance. Two Neuralink chips were in the monkey’s skull and

more than 2,000 wires ‘fanned out’ into its brain. Eventually the monkey played a video game purely with its brain waves. Psychopathic narcissist Musk said the ‘breakthrough’ was a step towards pu ing Neuralink chips into human skulls and merging minds with artificial intelligence. Exactly. This man is so dark and Cult to his DNA.

World Economic Fascism (WEF)

The World Economic Forum is telling you the plan by the statements made at its many and various events. Cult-owned fascist YouTube CEO Susan Wojcicki spoke at the 2021 WEF Global Technology Governance Summit (see the name) in which 40 governments and 150 companies met to ensure ‘the responsible design and deployment of emerging technologies’. Orwellian translation: ‘Ensuring the design and deployment of long-planned technologies will advance the Cult agenda for control and censorship.’ Freedomdestroyer and Nuremberg-bound Wojcicki expressed support for tech platforms like hers to censor content that is ‘technically legal but could be harmful’. Who decides what is ‘harmful’? She does and they do. ‘Harmful’ will be whatever the Cult doesn’t want people to see and we have legislation proposed by the UK government that would censor content on the basis of ‘harm’ no ma er if the information is fair, legal and provably true. Make that especially if it is fair, legal and provably true. Wojcicki called for a global coalition to be formed to enforce content moderation standards through automated censorship. This is a woman and mega-censor so selfdeluded that she shamelessly accepted a ‘free expression’ award – Wojcicki – in an event sponsored by her own YouTube. They have no shame and no self-awareness. You know that ‘Covid’ is a scam and Wojcicki a Cult operative when YouTube is censoring medical and scientific opinion purely on the grounds of whether it supports or opposes the Cult ‘Covid’ narrative. Florida governor Ron DeSantis compiled an expert panel with four professors of medicine from Harvard, Oxford, and Stanford Universities who spoke against forcing children and

vaccinated people to wear masks. They also said there was no proof that lockdowns reduced spread or death rates of ‘Covid-19’. Cultgofer Wojcicki and her YouTube deleted the panel video ‘because it included content that contradicts the consensus of local and global health authorities regarding the efficacy of masks to prevent the spread of Covid-19’. This ‘consensus’ refers to what the Cult tells the World Health Organization to say and the WHO tells ‘local health authorities’ to do. Wojcicki knows this, of course. The panellists pointed out that censorship of scientific debate was responsible for deaths from many causes, but Wojcicki couldn’t care less. She would not dare go against what she is told and as a disgrace to humanity she wouldn’t want to anyway. The UK government is seeking to pass a fascist ‘Online Safety Bill’ to specifically target with massive fines and other means non-censored video and social media platforms to make them censor ‘lawful but harmful’ content like the Cult-owned Facebook, Twi er, Google and YouTube. What is ‘lawful but harmful’ would be decided by the fascist Blair-created Ofcom. Another WEF obsession is a cyber-a ack on the financial system and this is clearly what the Cult has planned to take down the bank accounts of everyone – except theirs. Those that think they have enough money for the Cult agenda not to ma er to them have got a big lesson coming if they continue to ignore what is staring them in the face. The World Economic Forum, funded by Gates and fronted by Klaus Schwab, announced it would be running a ‘simulation’ with the Russian government and global banks of just such an a ack called Cyber Polygon 2021. What they simulate – as with the ‘Covid’ Event 201 – they plan to instigate. The WEF is involved in a project with the Cult-owned Carnegie Endowment for International Peace called the WEF-Carnegie Cyber Policy Initiative which seeks to merge Wall Street banks, ‘regulators’ (I love it) and intelligence agencies to ‘prevent’ (arrange and allow) a cyber-a ack that would bring down the global financial system as long planned by those that control the WEF and the Carnegie operation. The Carnegie Endowment for International Peace sent an instruction to First World

War US President Woodrow Wilson not to let the war end before society had been irreversibly transformed.

The Wuhan lab diversion

As I close, the Cult-controlled authorities and lapdog media are systematically pushing ‘the virus was released from the Wuhan lab’ narrative. There are two versions – it happened by accident and it happened on purpose. Both are nonsense. The perceived existence of the never-shown-to-exist ‘virus’ is vital to sell the impression that there is actually an infective agent to deal with and to allow the endless potential for terrifying the population with ‘variants’ of a ‘virus’ that does not exist. The authorities at the time of writing are going with the ‘by accident’ while the alternative media is promoting the ‘on purpose’. Cable news host Tucker Carlson who has questioned aspects of lockdown and ‘vaccine’ compulsion has bought the Wuhan lab story. ‘Everyone now agrees’ he said. Well, I don’t and many others don’t and the question is why does the system and its media suddenly ‘agree’? When the media moves as one unit with a narrative it is always a lie – witness the hour by hour mendacity of the ‘Covid’ era. Why would this Cult-owned combination which has unleashed lies like machine gun fire suddenly ‘agree’ to tell the truth?? Much of the alternative media is buying the lie because it fits the conspiracy narrative, but it’s the wrong conspiracy. The real conspiracy is that there is no virus and that is what the Cult is desperate to hide. The idea that the ‘virus’ was released by accident is ludicrous when the whole ‘Covid’ hoax was clearly long-planned and waiting to be played out as it was so fast in accordance with the Rockefeller document and Event 201. So they prepared everything in detail over decades and then sat around strumming their fingers waiting for an ‘accidental’ release from a bio-lab? What?? It’s crazy. Then there’s the ‘on purpose’ claim. You want to circulate a ‘deadly virus’ and hide the fact that you’ve done so and you release it down the street from the highest-level bio-lab in China? I repeat – What??

You would release it far from that lab to stop any association being made. But, no, we’ll do it in a place where the connection was certain to be made. Why would you need to scam ‘cases’ and ‘deaths’ and pay hospitals to diagnose ‘Covid-19’ if you had a real ‘virus’? What are sections of the alternative media doing believing this crap? Where were all the mass deaths in Wuhan from a ‘deadly pathogen’ when the recovery to normal life a er the initial propaganda was dramatic in speed? Why isn’t the ‘deadly pathogen’ now circulating all over China with bodies in the street? Once again we have the technique of tell them what they want to hear and they will likely believe it. The alternative media has its ‘conspiracy’ and with Carlson it fits with his ‘China is the danger’ narrative over years. China is a danger as a global Cult operations centre, but not for this reason. The Wuhan lab story also has the potential to instigate conflict with China when at some stage the plan is to trigger a Problem-Reaction-Solution confrontation with the West. Question everything – everything – and especially when the media agrees on a common party line.

Third wave … fourth wave … fifth wave …

As the book went into production the world was being set up for more lockdowns and a ‘third wave’ supported by invented ‘variants’ that were increasing all the time and will continue to do so in public statements and computer programs, but not in reality. India became the new Italy in the ‘Covid’ propaganda campaign and we were told to be frightened of the new ‘Indian strain’. Somehow I couldn’t find it within myself to do so. A document produced for the UK government entitled ‘Summary of further modelling of easing of restrictions – Roadmap Step 2’ declared that a third wave was inevitable (of course when it’s in the script) and it would be the fault of children and those who refuse the health-destroying fake ‘Covid vaccine’. One of the computer models involved came from the Cultowned Imperial College and the other from Warwick University which I wouldn’t trust to tell me the date in a calendar factory. The document states that both models presumed extremely high uptake

of the ‘Covid vaccines’ and didn’t allow for ‘variants’. The document states: ‘The resurgence is a result of some people (mostly children) being ineligible for vaccination; others choosing not to receive the vaccine; and others being vaccinated but not perfectly protected.’ The mendacity takes the breath away. Okay, blame those with a brain who won’t take the DNA-modifying shots and put more pressure on children to have it as ‘trials’ were underway involving children as young as six months with parents who give insanity a bad name. Massive pressure is being put on the young to have the fake ‘vaccine’ and child age consent limits have been systematically lowered around the world to stop parents intervening. Most extraordinary about the document was its claim that the ‘third wave’ would be driven by ‘the resurgence in both hospitalisations and deaths … dominated by those that have received two doses of the vaccine, comprising around 60-70% of the wave respectively’. The predicted peak of the ‘third wave’ suggested 300 deaths per day with 250 of them fully ‘vaccinated’ people. How many more lies do acquiescers need to be told before they see the obvious? Those who took the jab to ‘protect themselves’ are projected to be those who mostly get sick and die? So what’s in the ‘vaccine’? The document went on: It is possible that a summer of low prevalence could be followed by substantial increases in incidence over the following autumn and winter. Low prevalence in late summer should not be taken as an indication that SARS-CoV-2 has retreated or that the population has high enough levels of immunity to prevent another wave.

They are telling you the script and while many British people believed ‘Covid’ restrictions would end in the summer of 2021 the government was preparing for them to be ongoing. Authorities were awarding contracts for ‘Covid marshals’ to police the restrictions with contracts starting in July, 2021, and going through to January 31st, 2022, and the government was advertising for ‘Media Buying Services’ to secure media propaganda slots worth a potential £320 million for ‘Covid-19 campaigns’ with a contract not ending until March, 2022. The recipient – via a list of other front companies – was reported to be American media marketing giant Omnicom Group

Inc. While money is no object for ‘Covid’ the UK waiting list for all other treatment – including life-threatening conditions – passed 4.5 million. Meantime the Cult is seeking to control all official ‘inquiries’ to block revelations about what has really been happening and why. It must not be allowed to – we need Nuremberg jury trials in every country. The cover-up doesn’t get more obvious than appointing ultra-Zionist professor Philip Zelikow to oversee two dozen US virologists, public health officials, clinicians, former government officials and four American ‘charitable foundations’ to ‘learn the lessons’ of the ‘Covid’ debacle. The personnel will be those that created and perpetuated the ‘Covid’ lies while Zelikow is the former executive director of the 9/11 Commission who ensured that the truth about those a acks never came out and produced a report that must be among the most mendacious and manipulative documents ever wri en – see The Trigger for the detailed exposure of the almost unimaginable 9/11 story in which Sabbatians can be found at every level.

Passive no more

People are increasingly challenging the authorities with amazing numbers of people taking to the streets in London well beyond the ability of the Face-Nappies to stop them. Instead the Nappies choose situations away from the mass crowds to target, intimidate, and seek to promote the impression of ‘violent protestors’. One such incident happened in London’s Hyde Park. Hundreds of thousands walking through the streets in protest against ‘Covid’ fascism were ignored by the Cult-owned BBC and most of the rest of the mainstream media, but they delighted in reporting how police were injured in ‘clashes with protestors’. The truth was that a group of people gathered in Hyde Park at the end of one march when most had gone home and they were peacefully having a good time with music and chat. Face-Nappies who couldn’t deal with the full-march crowd then waded in with their batons and got more than they bargained for. Instead of just standing for this criminal brutality the crowd used their numerical superiority to push the Face-Nappies out of the

park. Eventually the Nappies turned and ran. Unfortunately two or three idiots in the crowd threw drink cans striking two officers which gave the media and the government the image they wanted to discredit the 99.9999 percent who were peaceful. The idiots walked straight into the trap and we must always be aware of potential agent provocateurs used by the authorities to discredit their targets. This response from the crowd – the can people apart – must be a turning point when the public no longer stand by while the innocent are arrested and brutally a acked by the Face-Nappies. That doesn’t mean to be violent, that’s the last thing we need. We’ll leave the violence to the Face-Nappies and government. But it does mean that when the Face-Nappies use violence against peaceful people the numerical superiority is employed to stop them and make citizen’s arrests or Common Law arrests for a breach of the peace. The time for being passive in the face of fascism is over. We are the many, they are the few, and we need to make that count before there is no freedom le and our children and grandchildren face an ongoing fascist nightmare. COME ON PEOPLE – IT’S TIME.

One final thought …

The power of love A force from above Cleaning my soul Flame on burn desire Love with tongues of fire Purge the soul Make love your goal

I’ll protect you from the hooded claw Keep the vampires from your door When the chips are down I’ll be around With my undying, death-defying Love for you Envy will hurt itself Let yourself be beautiful Sparkling love, flowers And pearls and pre y girls Love is like an energy Rushin’ rushin’ inside of me This time we go sublime Lovers entwine, divine, divine, Love is danger, love is pleasure Love is pure – the only treasure I’m so in love with you Purge the soul Make love your goal The power of love A force from above Cleaning my soul The power of love A force from above A sky-scraping dove

Flame on burn desire Love with tongues of fire Purge the soul Make love your goal

Frankie Goes To Hollywood

Appendix Cowan-Kaufman-Morell Statement on Virus Isolation (SOVI) Isolation: The action of isolating; the fact or condition of being isolated or standing alone; separation from other things or persons; solitariness Oxford English Dictionary

T

he controversy over whether the SARS-CoV-2 virus has ever been isolated or purified continues. However, using the above definition, common sense, the laws of logic and the dictates of science, any unbiased person must come to the conclusion that the SARS-CoV-2 virus has never been isolated or purified. As a result, no confirmation of the virus’ existence can be found. The logical, common sense, and scientific consequences of this fact are: • the structure and composition of something not shown to exist can’t be known, including the presence, structure, and function of any hypothetical spike or other proteins; • the genetic sequence of something that has never been found can’t be known; • “variants” of something that hasn’t been shown to exist can’t be known; • it’s impossible to demonstrate that SARS-CoV-2 causes a disease called Covid-19.

In as concise terms as possible, here’s the proper way to isolate, characterize and demonstrate a new virus. First, one takes samples (blood, sputum, secretions) from many people (e.g. 500) with symptoms which are unique and specific enough to characterize an illness. Without mixing these samples with ANY tissue or products that also contain genetic material, the virologist macerates, filters and ultracentrifuges i.e. purifies the specimen. This common virology technique, done for decades to isolate bacteriophages1 and so-called giant viruses in every virology lab, then allows the virologist to demonstrate with electron microscopy thousands of identically sized and shaped particles. These particles are the isolated and purified virus. These identical particles are then checked for uniformity by physical and/or microscopic techniques. Once the purity is determined, the particles may be further characterized. This would include examining the structure, morphology, and chemical composition of the particles. Next, their genetic makeup is characterized by extracting the genetic material directly from the purified particles and using genetic-sequencing techniques, such as Sanger sequencing, that have also been around for decades. Then one does an analysis to confirm that these uniform particles are exogenous (outside) in origin as a virus is conceptualized to be, and not the normal breakdown products of dead and dying tissues.2 (As of May 2020, we know that virologists have no way to determine whether the particles they’re seeing are viruses or just normal breakdown products of dead and dying tissues.)3 1

Isolation, characterization and analysis of bacteriophages from the haloalkaline lake Elmenteita, KenyaJuliah Khayeli Akhwale et al, PLOS One, Published: April 25, 2019. https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0215734 – accessed 2/15/21

2 “Extracellular Vesicles Derived From Apoptotic Cells: An Essential Link Between Death and Regeneration,” Maojiao Li1 et al, Frontiers in Cell and Developmental Biology, 2020 October 2. https://www.frontiersin.org/articles/10.3389/fcell.2020.573511/full – accessed 2/15/21

3 “The Role of Extraellular Vesicles as Allies of HIV, HCV and SARS Viruses,” Flavia Giannessi, et al, Viruses, 2020 May

If we have come this far then we have fully isolated, characterized, and genetically sequenced an exogenous virus particle. However, we still have to show it is causally related to a disease. This is carried out by exposing a group of healthy subjects (animals are usually used) to this isolated, purified virus in the manner in which the disease is thought to be transmi ed. If the animals get sick with the same disease, as confirmed by clinical and autopsy findings, one has now shown that the virus actually causes a disease. This demonstrates infectivity and transmission of an infectious agent. None of these steps has even been a empted with the SARS-CoV-2 virus, nor have all these steps been successfully performed for any so-called pathogenic virus. Our research indicates that a single study showing these steps does not exist in the medical literature. Instead, since 1954, virologists have taken unpurified samples from a relatively few people, o en less than ten, with a similar disease. They then minimally process this sample and inoculate this unpurified sample onto tissue culture containing usually four to six other types of material – all of which contain identical genetic material as to what is called a “virus.” The tissue culture is starved and poisoned and naturally disintegrates into many types of particles, some of which contain genetic material. Against all common sense, logic, use of the English language and scientific integrity, this process is called “virus isolation.” This brew containing fragments of genetic material from many sources is then subjected to genetic analysis, which then creates in a computersimulation process the alleged sequence of the alleged virus, a so called in silico genome. At no time is an actual virus confirmed by electron microscopy. At no time is a genome extracted and sequenced from an actual virus. This is scientific fraud.

The observation that the unpurified specimen — inoculated onto tissue culture along with toxic antibiotics, bovine fetal tissue, amniotic fluid and other tissues — destroys the kidney tissue onto which it is inoculated is given as evidence of the virus’ existence and pathogenicity. This is scientific fraud. From now on, when anyone gives you a paper that suggests the SARS-CoV-2 virus has been isolated, please check the methods sections. If the researchers used Vero cells or any other culture method, you know that their process was not isolation. You will hear the following excuses for why actual isolation isn’t done: 1. There were not enough virus particles found in samples from patients to analyze. 2. Viruses are intracellular parasites; they can’t be found outside the cell in this manner.

If No. 1 is correct, and we can’t find the virus in the sputum of sick people, then on what evidence do we think the virus is dangerous or even lethal? If No. 2 is correct, then how is the virus spread from person to person? We are told it emerges from the cell to infect others. Then why isn’t it possible to find it? Finally, questioning these virology techniques and conclusions is not some distraction or divisive issue. Shining the light on this truth is essential to stop this terrible fraud that humanity is confronting. For, as we now know, if the virus has never been isolated, sequenced or shown to cause illness, if the virus is imaginary, then why are we wearing masks, social distancing and pu ing the whole world into prison? Finally, if pathogenic viruses don’t exist, then what is going into those injectable devices erroneously called “vaccines,” and what is their purpose? This scientific question is the most urgent and relevant one of our time.

We are correct. The SARS-CoV2 virus does not exist. Sally Fallon Morell, MA Dr. Thomas Cowan, MD Dr. Andrew Kaufman, MD

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Saul: Rules for Radicals (Vintage, 1989)

Antelman,

Rabbi Marvin: To Eliminate the Opiate (Zahavia, 1974)

, Joe: The Climate Chronicles (Relentless Thunder Press, 2018)

Bastardi

Tom: Human Heart, Cosmic Heart (Chelsea Green Publishing, 2016)

Cowan,

, Tom, and Fallon Morell, Sally: The Contagion Myth (Skyhorse Publishing, 2020)

Cowan

Jack D: Columbus And Other Cannibals – The Wetiko Disease of Exploitation, Imperialism, and Terrorism (Seven Stories Press, 2008 – originally published in 1979)

Forbes,

Bill: How to Avoid a Climate Disaster: The Solutions We Have and the Breakthroughs We Need (Allen Lane, 2021)

Gates,

Aldous: Brave New World (Cha o & Windus, 1932)

Huxley,

Köhnlein, Lanza,

Dr Claus, and Engelbrecht, Torsten: Virus Mania (emu-Vertag, Lahnstein, 2020)

Robert, and Berman, Bob: Biocentrism (BenBella Books, 2010)

John Lamb: Not In His Image (Chelsea Green Publishing, 2006)

Lash,

Dawn, and Parker, David: What Really Makes You Ill – Why everything you thought you knew about disease is wrong (Independently Published, 2019)

Lester,

Levy,

Paul: Dispelling Wetiko, Breaking the Spell of Evil (North Atlantic Books, 2013)

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Karl: A World Without Jews (Philosophical Library, first edition, 1959)

Mullis,

Kary: Dancing Naked in the Mine Field (Bloomsbury, 1999)

O’Brien,

Cathy: Trance-Formation of America (Reality Marketing, 1995)

Scholem,

Gershon: The Messianic Idea in Judaism (Schocken Books, 1994)

Klaus, and Davis, Nicholas: Shaping the Future of the Fourth Industrial Revolution: A guide to building a better world (Penguin Books, 2018)

Schwab,

, Klaus: The Great Reset (Agentur Schweiz, 2020)

Schwab

Cass and Thaler, Richard: Nudge: Improving Decisions About Health, Wealth, and Happiness (Penguin, 2009)

Sunstein,

Shanna: Count Down: How Our Modern World Is Threatening Sperm Counts, Altering Male and Female Reproductive Development and Imperiling the Future of the Human Race (Scribner, 2021)

Swan,

Max: Our Mathematical Universe: My Quest for the Ultimate Nature of Reality (Penguin, 2015)

Tegmark,

Velikovsky,

Immanuel: Worlds in Collision (Paradigma, 2009)

Wilton,

Robert: The Last Days of the Romanovs (Blurb, 2018, first published 1920)

Index A abusive relationships

blaming themselves, abused as ref1 children ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10 conspiracy theories ref1 domestic abuse ref1, ref2 economic abuse and dependency ref1 isolation ref1 physical abuse ref1 psychological abuse ref1 signs of abuse ref1

addiction

alcoholism ref1 frequencies ref1 substance abuse ref1, ref2 technology ref1, ref2, ref3 Adelson, Sheldon ref1, ref2, ref3 Agenda 21/Agenda 2030 (UN) ref1, ref2, ref3, ref4 AIDs/HIV ref1 causal link between HIV and AIDs ref1, ref2 retroviruses ref1 testing ref1, ref2 trial-run for Covid-19, as ref1, ref2 aliens/extraterrestrials ref1, ref2 aluminium ref1 Amazon ref1, ref2, ref3

ref1, ref2 anaphylactic shock ref1, ref2, ref3, ref4 animals ref1, ref2, ref3 antibodies ref1, ref2, ref3, ref4, ref5 Antifa ref1, ref2, ref3, ref4 antigens ref1, ref2 anti-Semitism ref1, ref2, ref3 Archons ref1, ref2 consciousness ref1, ref2, ref3 energy ref1, ref2, ref3 ennoia ref1 genetic manipulation ref1, ref2 inversion ref1, ref2, ref3 lockdowns ref1 money ref1 radiation ref1 religion ref1, ref2 technology ref1, ref2, ref3 Wetiko factor ref1, ref2, ref3, ref4 artificial intelligence (AI) ref1 army made up of robots ref1, ref2 Human 2.0 ref1, ref2 Internet ref1 MHRA ref1 Morgellons fibres ref1, ref2 Smart Grid ref1 Wetiko factor ref1 asymptomatic, Covid-19 as ref1, ref2, ref3 aviation industry ref1 amplification cycles

B

banking, finance and money

ref1, ref2, ref3

2008 crisis ref1, ref2 boom and bust ref1 cashless digital money systems ref1 central banks ref1 credit ref1 digital currency ref1 fractional reserve lending ref1 Great Reset ref1 guaranteed income ref1, ref2, ref3 Human 2.0 ref1 incomes, destruction of ref1, ref2 interest ref1 one per cent ref1, ref2 scams ref1 BBC ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 Becker-Phelps, Leslie ref1 ref1, ref2, ref3 behavioural scientists and psychologists, advice from ref1, ref2 Bezos, Jeff ref1, ref2, ref3, ref4 Biden, Hunter ref1 Biden, Joe ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12, ref13, ref14, ref15, ref16, ref17 Behavioural Insights Team (BIT) (Nudge Unit)

Big Pharma

cholesterol ref1 health professionals ref1, ref2 immunity from prosecution in US ref1 vaccines ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 Wetiko factor ref1, ref2 WHO ref1, ref2, ref3 Bill and Melinda Gates Foundation ref1, ref2, ref3, ref4, ref5, ref6, ref7

ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 ref10, ref11 bird flu (H5N1) ref1 Black Lives Matter (BLM) ref1, ref2, ref3, ref4, ref5 Blair, Tony ref1, ref2, ref3, ref4, ref5, ref6, ref7 Brin, Sergei ref1, ref2, ref3, ref4, ref5, ref6, ref7 British Empire ref1 Bush, George HW ref1, ref2 Bush, George W ref1, ref2, ref3, ref4 Byrd, Robert ref1 billionaires

C Canada

Global Cult ref1 hate speech ref1 internment ref1 masks ref1 old people ref1 SARS-COV-2 ref1 satellites ref1 vaccines ref1 wearable technology ref1 Capitol Hill riot ref1, ref2 agents provocateur ref1 Antifa ref1 Black Lives Ma er (BLM) ref1, ref2 QAnon ref1 security precautions, lack of ref1, ref2, ref3 carbon dioxide ref1, ref2 care homes, deaths in ref1, ref2 cashless digital money systems ref1 censorship ref1, ref2, ref3, ref4, ref5

fact-checkers ref1 masks ref1 media ref1, ref2 private messages ref1 social media ref1, ref2, ref3, ref4, ref5, ref6 transgender persons ref1 vaccines ref1, ref2, ref3 Wokeness ref1 Centers for Disease Control (CDC) (United States) ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12, ref13 centralisation ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 chakras ref1 change agents ref1, ref2, ref3 chemtrails ref1, ref2, ref3 chief medical officers and scientific advisers ref1, ref2, ref3, ref4, ref5, ref6 children see also young people abuse ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10 care, taken into ref1, ref2, ref3 education ref1, ref2, ref3, ref4 energy ref1 family courts ref1 hand sanitisers ref1 human sacrifice ref1 lockdowns ref1, ref2, ref3 masks ref1, ref2, ref3, ref4, ref5 mental health ref1 old people ref1 parents, replacement of ref1, ref2 Psyop (psychological operation), Covid as a ref1, ref2 reframing ref1 smartphone addiction ref1

social distancing and isolation ref1 social media ref1 transgender persons ref1, ref2 United States ref1 vaccines ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10 Wetiko factor ref1 China ref1, ref2, ref3, ref4 anal swab tests ref1 Chinese Revolution ref1, ref2, ref3 digital currency ref1 Global Cult ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 guaranteed income ref1 Imperial College ref1 Israel ref1 lockdown ref1, ref2 masculinity crisis ref1 masks ref1 media ref1 origins of virus in China ref1, ref2, ref3, ref4, ref5 pollution causing respiratory diseases ref1 Sabbatians ref1, ref2 Smart Grid ref1, ref2 social credit system ref1 testing ref1, ref2 United States ref1, ref2 vaccines ref1, ref2 Wetiko factor ref1 wet market conspiracy ref1 Wuhan ref1, ref2, ref3, ref4, ref5, ref6, ref7 cholesterol ref1, ref2 Christianity ref1, ref2, ref3, ref4, ref5 criticism ref1 cross, inversion of the ref1

Nag Hammadi texts ref1, ref2, ref3 Roman Catholic Church ref1, ref2 Sabbatians ref1, ref2 Satan ref1, ref2, ref3, ref4 Wokeness ref1 class ref1, ref2 climate change hoax ref1, ref2, ref3, ref4, ref5 Agenda 21/Agenda 2030 ref1, ref2, ref3 carbon dioxide ref1, ref2 Club of Rome ref1, ref2, ref3, ref4, ref5 fear ref1 funding ref1 Global Cult ref1 green new deals ref1 green parties ref1 inversion ref1 perception, control of ref1 PICC ref1 reframing ref1 temperature, increases in ref1 United Nations ref1, ref2 Wikipedia ref1 Wokeness ref1, ref2 Clinton, Bill ref1, ref2, ref3, ref4, ref5, ref6 Clinton, Hillary ref1, ref2, ref3 the cloud ref1, ref2, ref3, ref4, ref5, ref6, ref7 Club of Rome and climate change hoax ref1, ref2, ref3, ref4, ref5 cognitive therapy ref1 Cohn, Roy ref1 Common Law ref1 Admiralty Law ref1 arrests ref1, ref2

contractual law, Statute Law as ref1 corporate entities, people as ref1 legalese ref1 sea, law of the ref1 Statute Law ref1 Common Purpose leadership programme

ref1, ref2

ref1, ref2 co-morbidities ref1 communism

computer-generated virus,

as ref1, ref2, ref3 computer models ref1, ref2, ref3, ref4, ref5 connections ref1, ref2, ref3, ref4 consciousness ref1, ref2, ref3, ref4 Archons ref1, ref2, ref3 expanded ref1, ref2, ref3, ref4, ref5, ref6, ref7 experience ref1 heart ref1 infinity ref1, ref2 religion ref1, ref2 self-identity ref1 simulation thesis ref1 vaccines ref1 Wetiko factor ref1, ref2 conspiracy theorists ref1, ref2, ref3, ref4, ref5 contradictory rules ref1 contrails ref1 Corman-Drosten test ref1, ref2, ref3, ref4 countermimicry ref1, ref2, ref3 Covid-19 vaccines see vaccines Covidiots ref1, ref2 Cowan, Tom ref1, ref2, ref3, ref4 crimes against humanity ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 Covid-19

ref1 cyberwarfare ref1 cyber-operations

D DARPA (Defense Advanced Research Projects Agency)

ref1

deaths

care homes ref1 certificates ref1, ref2, ref3, ref4 mortality rate ref1 post-mortems/autopsies ref1 recording ref1, ref2, ref3, ref4, ref5, ref6, ref7 vaccines ref1, ref2, ref3, ref4, ref5

deceit

pyramid of deceit ref1, ref2 sequence of deceit ref1 decoding ref1, ref2, ref3 dehumanisation ref1, ref2, ref3 Delphi technique ref1 democracy ref1 dependency ref1, ref2, ref3, ref4, ref5 Descartes, René ref1 DNA

numbers ref1 vaccines ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10 DNR (do not resuscitate) orders ref1 domestic abuse ref1, ref2 downgrading of Covid-19 ref1 Drosten, Christian ref1, ref2, ref3, ref4, ref5, ref6, ref7 Duesberg, Peter ref1, ref2

E

ref1 Edmunds, John ref1, ref2 education ref1, ref2, ref3, ref4 electromagnetic spectrum ref1, ref2 Enders, John ref1 economic abuse

energy

Archons ref1, ref2, ref3 children and young people ref1 consciousness ref1 decoding ref1 frequencies ref1, ref2, ref3, ref4 heart ref1 human energy field ref1 source, humans as an energy ref1, ref2 vaccines ref1 viruses ref1 ennoia ref1 Epstein, Jeffrey ref1, ref2 eternal ‘I’ ref1, ref2 ethylene oxide ref1 European Union ref1, ref2, ref3, ref4 Event ref1 and Bill Gates ref2 exosomes, Covid-19 as natural defence mechanism called ref1 experience ref1, ref2 Extinction Rebellion ref1, ref2

F Facebook

addiction ref1, 448–50 Facebook

Archons ref1 censorship ref1, ref2, ref3 hate speech ref1 monopoly, as ref1 private messages, censorship of ref1 Sabbatians ref1 United States election fraud ref1 vaccines ref1 Wetiko factor ref1 fact-checkers ref1 Fauci, Anthony ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12 fear ref1, ref2, ref3, ref4 climate change ref1 computer models ref1 conspiracy theories ref1 empty hospitals ref1 Italy ref1, ref2, ref3 lockdowns ref1, ref2, ref3, ref4 masks ref1, ref2 media ref1, ref2 medical staff ref1 Psyop (psychological operation), Covid as a ref1 Wetiko factor ref1, ref2 female infertility ref1 Fermi Paradox ref1 Ferguson, Neil ref1, ref2, ref3, ref4, ref5, ref6, ref7 fertility, decline in ref1 The Field ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 finance see banking, finance and money five-senses ref1, ref2 Archons ref1, ref2, ref3

censorship ref1 consciousness, expansion of ref1, ref2, ref3, ref4, ref5, ref6 decoding ref1 education ref1, ref2 the Field ref1, ref2 God, personification of ref1 infinity ref1, ref2 media ref1 paranormal ref1 perceptual programming ref1, ref2 Phantom Self ref1 pneuma not nous, using ref1 reincarnation ref1 self-identity ref1 Wetiko factor ref1, ref2, ref3, ref4, ref5, ref6 5G ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 Floyd, George and protests, killing of ref1 flu, re-labelling of ref1, ref2, ref3 food and water, control of ref1, ref2 Freemasons ref1, ref2, ref3, ref4, ref5, ref6 Frei, Rosemary ref1 frequencies

addictions ref1 Archons ref1, ref2, ref3 awareness ref1 chanting and mantras ref1 consciousness ref1 decoding ref1, ref2 education ref1 electromagnetic (EMF) frequencies ref1 energy ref1, ref2, ref3, ref4 fear ref1

the Field ref1, ref2 5G ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10 five-senses ref1, ref2 ghosts ref1 Gnostics ref1 hive-minds ref1 human, meaning of ref1 light ref1, ref2 love ref1, ref2 magnetism ref1 perception ref1 reality ref1, ref2, ref3 simulation ref1 terror ref1 vaccines ref1 Wetiko ref1, ref2, ref3 Fuellmich, Reiner ref1, ref2, ref3 furlough/rescue payments ref1

G Gallo, Robert Gates, Bill

ref1, ref2, ref3

Archons ref1, ref2, ref3 climate change ref1, ref2, ref3, ref4 Daily Pass tracking system ref1 Epstein ref1 fascism ref1 five senses ref1 GAVI ref1 Great Reset ref1 GSK ref1 Imperial College ref1, ref2 Johns Hopkins University ref1, ref2, ref3

lockdowns ref1, ref2 masks ref1 Nuremberg trial, proposal for ref1, ref2 Rockefellers ref1, ref2 social distancing and isolation ref1 Sun, dimming the ref1 synthetic meat ref1, ref2 vaccines ref1, ref2, ref3, ref4, ref5, ref6, ref7 Wellcome Trust ref1 Wetiko factor ref1, ref2, ref3 WHO ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10 Wokeness ref1 World Economic Forum ref1, ref2, ref3, ref4 Gates, Melinda ref1, ref2, ref3 GAVI vaccine alliance ref1 genetics, manipulation of ref1, ref2, ref3 Germany ref1, ref2, ref3, ref4, ref5, ref6 see also Nazi Germany Global Cult ref1, ref2, ref3, ref4, ref5 anti-human, why Global Cult is ref1 Black Lives Ma er (BLM) ref1, ref2, ref3, ref4 China ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 climate change hoax ref1 contradictory rules ref1 Covid-19 ref1, ref2, ref3 fascism ref1 geographical origins ref1 immigration ref1 Internet ref1 mainstream media ref1, ref2 masks ref1, ref2 monarchy ref1 non-human dimension ref1

perception ref1 political parties ref1, ref2 pyramidal hierarchy ref1, ref2, ref3 reframing ref1 Sabbantian-Frankism ref1, ref2 science, manipulation of ref1 spider and the web ref1 transgender persons ref1 vaccines ref1 who controls the Cult ref1 Wokeness ref1, ref2, ref3, ref4 globalisation ref1, ref2 Gnostics ref1, ref2, ref3, ref4, ref5 Google ref1, ref2, ref3, ref4 government

behavioural scientists and psychologists, advice from ref1, ref2 definition ref1 Joint Biosecurity Centre (JBC) ref1 people, abusive relationship with ref1 Great Reset ref1, ref2, ref3, ref4, ref5, ref6 fascism ref1, ref2, ref3 financial system ref1 Human 2.0 ref1 water and food, control of ref1 green parties ref1 Griesz-Brisson, Margarite ref1 guaranteed income ref1, ref2, ref3

H

ref1, ref2, ref3, ref4, ref5 hand sanitisers ref1 heart ref1, ref2 Hancock, Matt

ref1, ref2, ref3 holographs ref1, ref2, ref3, ref4 hospitals, empty ref1 human, meaning of ref1 Human 2.0 ref1 addiction to technology ref1 artificial intelligence (AI) ref1, ref2 elimination of Human 1.0 ref1 fertility, decline in ref1 Great Reset ref1 implantables ref1 money ref1 mRNA ref1 nanotechnology ref1 parents, replacement of ref1, ref2 Smart Grid, connection to ref1, ref2 synthetic biology ref1, ref2, ref3, ref4 testosterone levels, decrease in ref1 transgender = transhumanism ref1, ref2, ref3 vaccines ref1, ref2, ref3, ref4 human sacrifice ref1, ref2, ref3 Hunger Games Society ref1, ref2, ref3, ref4, ref5, ref6, ref7 Huxley, Aldous ref1, ref2, ref3 hive-minds/groupthink

I

ref1, ref2, ref3 Illuminati ref1, ref2 illusory physical reality ref1 immigration ref1, ref2, ref3, ref4 Imperial College ref1, ref2, ref3, ref4, ref5, ref6 implantables ref1, ref2 identity politics

ref1, ref2 Infinite Awareness ref1, ref2, ref3, ref4 Internet ref1, ref2 see also social media artificial intelligence (AI) ref1 independent journalism, lack of ref1 Internet of Bodies (IoB) ref1 Internet of Everything (IoE) ref1, ref2 Internet of Things (IoT) ref1, ref2 lockdowns ref1 Psyop (psychological operation), Covid as a ref1 trolls ref1 intersectionality ref1 incomes, destruction of

inversion

Archons ref1, ref2, ref3 climate change hoax ref1 energy ref1 Judaism ref1, ref2, ref3 symbolism ref1 Wetiko factor ref1 Wokeness ref1, ref2, ref3

Islam

Archons ref1 crypto-Jews ref1 Islamic State ref1, ref2 Jinn and Djinn ref1, ref2, ref3 O oman Empire ref1 Wahhabism ref1 isolation see social distancing and isolation Israel

China ref1 Cyber Intelligence Unit Beersheba complex ref1 expansion of illegal se lements ref1

formation ref1 Global Cult ref1 Judaism ref1, ref2, ref3, ref4, ref5 medical experiments, consent for ref1 Mossad ref1, ref2, ref3, ref4 Palestine-Israel conflict ref1, ref2, ref3 parents, replacement of ref1 Sabbatians ref1, ref2, ref3, ref4, ref5 September 11, 2001, terrorist a acks on United States ref1 Silicon Valley ref1 Smart Grid ref1, ref2 United States ref1, ref2 vaccines ref1 Wetiko factor ref1 Italy

fear ref1, ref2, ref3 Lombardy ref1, ref2, ref3 vaccines ref1

J

ref1, ref2, ref3, ref4, ref5, ref6, ref7 Johnson, Boris ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 Joint Biosecurity Centre (JBC) ref1 Johns Hopkins University

Judaism

anti-Semitism ref1, ref2, ref3 Archons ref1, ref2 crypto-Jews ref1 inversion ref1, ref2, ref3 Israel ref1, ref2, ref3, ref4, ref5 Labour Party ref1 Nazi Germany ref1, ref2, ref3, ref4 Sabbatians ref1, ref2, ref3, ref4, ref5

Silicon Valley ref1 Torah ref1 United States ref1, ref2 Zionists ref1, ref2, ref3

K

ref1, ref2, ref3, ref4 knowledge ref1, ref2, ref3, ref4, ref5, ref6 Koch’s postulates ref1 Kurzweil, Ray ref1, ref2, ref3, ref4, ref5, ref6, ref7 Kaufman, Andrew

Kushner, Jared ref1, ref2

L

ref1, ref2 Lanka, Stefan ref1, ref2 Labour Party

Lateral Flow Device (LFD)

ref1

ref1, ref2, ref3 Life Program ref1 lockdowns ref1, ref2, ref3 amplification tampering ref1 Archons ref1 Behavioural Insights Team ref1 Black Lives Ma er (BLM) ref1 care homes, deaths in ref1 children abuse ref1, ref2 mental health ref1 China ref1, ref2 computer models ref1 consequences ref1, ref2 dependency ref1, ref2, ref3 Levy, Paul

domestic abuse ref1 fall in cases ref1 fear ref1, ref2, ref3, ref4 guaranteed income ref1 Hunger Games Society ref1, ref2, ref3 interaction, destroying ref1 Internet ref1, ref2 overdoses ref1 perception ref1 police-military state ref1, ref2 protests ref1, ref2, ref3, ref4, ref5 psychopathic personality ref1, ref2, ref3 reporting/snitching, encouragement of ref1, ref2 testing ref1 vaccines ref1 Wetiko factor ref1 WHO ref1 love ref1, ref2, ref3 Lucifer ref1, ref2, ref3

M

ref1, ref2 Magufuli, John ref1, ref2 mainstream media ref1 BBC ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 censorship ref1, ref2 China ref1 climate change hoax ref1 fear ref1, ref2 Global Cult ref1, ref2 independent journalism, lack of ref1 Ofcom ref1, ref2, ref3 Madej, Carrie

perception ref1, ref2 Psyop (psychological operation), Covid as a ref1 Sabbatians ref1, ref2 social disapproval ref1 social distancing and isolation ref1 United States ref1, ref2 vaccines ref1, ref2, ref3, ref4, ref5 Mao Zedong ref1, ref2, ref3 Marx and Marxism ref1, ref2, ref3, ref4, ref5, ref6 masculinity ref1 masks/face coverings ref1, ref2, ref3 censorship ref1 children ref1, ref2, ref3, ref4, ref5 China, made in ref1 dehumanisation ref1, ref2, ref3 fear ref1, ref2 flu ref1 health professionals ref1, ref2, ref3, ref4 isolation ref1 laughter ref1 mass non-cooperation ref1 microplastics, risk of ref1 mind control ref1 multiple masks ref1 oxygen deficiency ref1, ref2, ref3 police ref1, ref2, ref3, ref4, ref5 pollution, as cause of plastic ref1 Psyop (psychological operation), Covid as a ref1 reframing ref1, ref2 risk assessments, lack of ref1, ref2 self-respect ref1 surgeons ref1

United States ref1 vaccines ref1, ref2, ref3, ref4, ref5 Wetiko factor ref1 ‘worms’ ref1 The Matrix movies ref1, ref2, ref3 measles ref1, ref2 media see mainstream media Medicines and Healthcare products Regulatory Agency (MHRA) ref1, ref2, ref3, ref4 Mesopotamia ref1 messaging ref1 military-police state ref1, ref2, ref3 mind control ref1, ref2, ref3, ref4, ref5, ref6 see also MKUltra MKUltra ref1, ref2, ref3 monarchy ref1

see banking, finance and money Montagnier, Luc ref1, ref2, ref3 Mooney, Bel ref1 Morgellons disease ref1, ref2 mortality rate ref1 Mullis, Kary ref1, ref2, ref3 Musk, Elon ref1 money

N

ref1, ref2, ref3 nanotechnology ref1, ref2, ref3 narcissism ref1 Nazi Germany ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 near-death experiences ref1, ref2 Neocons ref1, ref2, ref3 Nag Hammadi texts

Neuro-Linguistic Programming (NLP) and the Delphi technique ref1 NHS (National Health Service)

amplification cycles ref1 Common Purpose ref1, ref2 mind control ref1 NHS England ref1 saving the NHS ref1, ref2 vaccines ref1, ref2, ref3, ref4, ref5 whistle-blowers ref1, ref2, ref3 No-Problem-Reaction-Solution ref1, ref2, ref3, ref4 non-human dimension of Global Cult ref1 nous ref1 numbers, reality as ref1 Nuremberg Codes ref1, ref2, ref3 Nuremberg-like tribunal, proposal for ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12

O

ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10 O’Brien, Cathy ref1, ref2, ref3, ref4 Ochel, Evita ref1 Ofcom ref1, ref2, ref3 old people ref1, ref2, ref3, ref4, ref5 Oneness ref1, ref2, ref3 Open Society Foundations (Soros) ref1, ref2, ref3 oxygen 406, 528–34 Obama, Barack

P

ref1, ref2 Page, Larry ref1, ref2, ref3, ref4, ref5, ref6, ref7 paedophilia

ref1, ref2, ref3 pandemic, definition of ref1 Palestine-Israel conflict

pandemic and health crisis scenarios/simulations

ref4

ref1 PCR tests ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 Pearl Harbor attacks, prior knowledge of ref1 Pelosi, Nancy ref1, ref2, ref3 perception ref1, ref2, ref3, ref4 climate change hoax ref1 control ref1, ref2, ref3 decoding ref1, ref2 enslavement ref1 externally-delivered perceptions ref1 five senses ref1 human labels ref1 media ref1, ref2 political parties ref1, ref2 Psyop (psychological operation), Covid as a ref1 sale of perception ref1 self-identity ref1, ref2 Wokeness ref1 Phantom Self ref1, ref2, ref3 pharmaceutical industry see Big Pharma phthalates ref1 Plato’s Allegory of the Cave ref1, ref2 pneuma ref1 paranormal

police

Black Lives Ma er (BLM) ref1 brutality ref1 citizen’s arrests ref1, ref2 common law arrests ref1, ref2

ref1, ref2, ref3,

Common Purpose ref1 defunding ref1 lockdowns ref1, ref2 masks ref1, ref2, ref3, ref4 police-military state ref1, ref2, ref3 psychopathic personality ref1, ref2, ref3, ref4 reframing ref1 United States ref1, ref2, ref3, ref4 Wokeness ref1 polio ref1 political correctness ref1, ref2, ref3, ref4 political parties ref1, ref2, ref3, ref4 political puppets ref1 pollution ref1, ref2, ref3 post-mortems/autopsies ref1 Postage Stamp Consensus ref1, ref2 pre-emptive programming ref1 Problem-Reaction-Solution ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 Project for the New American Century ref1, ref2, ref3, ref4 psychopathic personality ref1 Archons ref1 heart energy ref1 lockdowns ref1, ref2, ref3 police ref1, ref2, ref3, ref4 recruitment ref1, ref2 vaccines ref1 wealth ref1 Wetiko ref1, ref2 Psyop (psychological operation), Covid as a ref1, ref2, ref3, ref4, ref5 Pushbackers ref1, ref2, ref3, ref4 pyramid structure ref1, ref2, ref3, ref4

Q QAnon Psyop

R

ref1, ref2, ref3

see also Black Lives Ma er (BLM) anti-racism industry ref1 class ref1 critical race theory ref1 culture ref1 intersectionality ref1 reverse racism ref1 white privilege ref1, ref2 white supremacy ref1, ref2, ref3, ref4, ref5 Wokeness ref1, ref2, ref3 radiation ref1, ref2 randomness, illusion of ref1, ref2, ref3 reality ref1, ref2, ref3 reframing ref1, ref2 change agents ref1, ref2 children ref1 climate change ref1 Common Purpose leadership programme ref1, ref2 contradictory rules ref1 enforcers ref1 masks ref1, ref2 NLP and the Delphi technique ref1 police ref1 Wetiko factor ref1 Wokeness ref1, ref2 religion see also particular religions alien invasions ref1 racism

Archons ref1, ref2 consciousness ref1, ref2 control, system of ref1, ref2, ref3 criticism, prohibition on ref1 five senses ref1 good and evil, war between ref1 hidden non-human forces ref1, ref2 Sabbatians ref1 save me syndrome ref1 Wetiko ref1 Wokeness ref1 repetition and mind control ref1, ref2, ref3 reporting/snitching, encouragement of ref1, ref2 Reptilians/Grey entities ref1 rewiring the mind ref1 Rivers, Thomas Milton ref1, ref2 Rockefeller family ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 Rockefeller Foundation documents ref1, ref2, ref3, ref4 Roman Empire ref1 Rothschild family ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 RT-PCR tests ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 Russia

collusion inquiry in US ref1 Russian Revolution ref1, ref2 Sabbatians ref1

S

ref1, ref2 anti-Semitism ref1, ref2 banking and finance ref1, ref2, ref3 China ref1, ref2 Israel ref1, ref2, ref3, ref4, ref5

Sabbantian-Frankism

Judaism ref1, ref2, ref3, ref4, ref5 Lucifer ref1 media ref1, ref2 Nazis ref1, ref2 QAnon ref1 Rothschilds ref1, ref2, ref3, ref4, ref5, ref6 Russia ref1 Saudi Arabia ref1 Silicon Valley ref1 Sumer ref1 United States ref1, ref2, ref3 Wetiko factor ref1 Wokeness ref1, ref2, ref3 SAGE (Scientific Advisory Group for Emergencies)

ref4

SARS-1

ref1, ref2, ref3,

ref1

ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 Satan/Satanism ref1, ref2, ref3, ref4, ref5, ref6, ref7 satellites in low-orbit ref1 Saudi Arabia ref1 Save Me Syndrome ref1 scapegoating ref1 Schwab, Klaus ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12 science, manipulation of ref1 self-identity ref1, ref2, ref3, ref4 self-respect, attacks on ref1 September 11, 2001, terrorist attacks on United States ref1, ref2, ref3, ref4 77th Brigade of UK military ref1, ref2, ref3 Silicon Valley/tech giants ref1, ref2, ref3, ref4, ref5, ref6 see also SARs-CoV-2

Facebook

Israel ref1 Sabbatians ref1 technocracy ref1 Wetiko factor ref1 Wokeness ref1 ref1, ref2, ref3, ref4, ref5 Smart Grid ref1, ref2, ref3 artificial intelligence (AI) ref1 China ref1, ref2 control centres ref1 the Field ref1 Great Reset ref1 Human 2.0 ref1, ref2 Israel ref1, ref2 vaccines ref1 Wetiko factor ref1 social disapproval ref1 social distancing and isolation ref1, ref2, ref3 abusive relationships ref1, ref2 children ref1 flats and apartments ref1 heart issues ref1 hugs ref1 Internet ref1 masks ref1 media ref1 older people ref1, ref2 one-metre (three feet) rule ref1 rewiring the mind ref1 simulation, universe as a ref1 SPI-B ref1 substance abuse ref1 simulation hypothesis

suicide and self-harm ref1, ref2, ref3, ref4, ref5 technology ref1 torture, as ref1, ref2 two-metre (six feet) rule ref1 women ref1 social justice ref1, ref2, ref3, ref4 social media see also Facebook bans on alternative views ref1 censorship ref1, ref2, ref3, ref4, ref5, ref6 children ref1 emotion ref1 perception ref1 private messages ref1 Twi er ref1, ref2, ref3, ref4, ref5, ref6, ref7 Wetiko factor ref1 YouTube ref1, ref2, ref3, ref4, ref5 Soros, George ref1, ref2, ref3, ref4, ref5, ref6 Spain ref1 SPI-B (Scientific Pandemic Insights Group on Behaviours) ref1, ref2, ref3, ref4 spider and the web ref1, ref2, ref3, ref4 Starmer, Keir ref1 Statute Law ref1 Steiner, Rudolf ref1, ref2, ref3 Stockholm syndrome ref1 streptomycin ref1 suicide and self-harm ref1, ref2, ref3, ref4, ref5 Sumer ref1, ref2 Sunstein, Cass ref1, ref2, ref3 swine flu (H1N1) ref1, ref2, ref3 synchronicity ref1 synthetic biology ref1, ref2, ref3, ref4 synthetic meat ref1, ref2

T

see also artificial intelligence (AI); Internet; social media addiction ref1, ref2, ref3, ref4 Archons ref1, ref2 the cloud ref1, ref2, ref3, ref4, ref5, ref6, ref7 cyber-operations ref1 cyberwarfare ref1 radiation ref1, ref2 social distancing and isolation ref1 technocracy ref1 Tedros Adhanom Ghebreyesus ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12, ref13 telepathy ref1 Tenpenny, Sherri ref1 Tesla, Nikola ref1 testosterone levels, decrease in ref1 testing for Covid-19 ref1, ref2 anal swab tests ref1 cancer ref1 China ref1, ref2, ref3 Corman-Drosten test ref1, ref2, ref3, ref4 death certificates ref1, ref2 fraudulent testing ref1 genetic material, amplification of ref1 Lateral Flow Device (LFD) ref1 PCR tests ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8 vaccines ref1, ref2, ref3 Thunberg, Greta ref1, ref2, ref3 Totalitarian Tiptoe ref1, ref2, ref3, ref4 technology

transgender persons

activism ref1 artificial wombs ref1

censorship ref1 child abuse ref1, ref2 Human 2.0 ref1, ref2, ref3 Wokeness ref1, ref2, ref3, ref4, ref5 women, deletion of rights and status of ref1, ref2 young persons ref1 travel restrictions ref1 Trudeau, Justin ref1, ref2, ref3 Trump, Donald ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11 Twitter ref1, ref2, ref3, ref4, ref5, ref6, ref7

U UKColumn

ref1, ref2

United Nations (UN)

ref1, ref2, ref3, ref4, ref5 see also Agenda

21/Agenda 2030 (UN)

ref1, ref2 American Revolution ref1 borders ref1, ref2 Capitol Hill riot ref1, ref2 children ref1 China ref1, ref2 CIA ref1, ref2 Daily Pass tracking system ref1 demographics by immigration, changes in ref1 Democrats ref1, ref2, ref3, ref4, ref5, ref6, ref7 election fraud ref1 far-right domestic terrorists, pushbackers as ref1 Federal Reserve ref1 flu/respiratory diseases statistics ref1 Global Cult ref1, ref2 hand sanitisers, FDA warnings on ref1

United States

immigration, effects of illegal ref1 impeachment ref1 Israel ref1, ref2 Judaism ref1, ref2, ref3 lockdown ref1 masks ref1 mass media ref1, ref2 nursing homes ref1 Pentagon ref1, ref2, ref3, ref4 police ref1, ref2, ref3, ref4 pushbackers ref1 Republicans ref1, ref2 borders ref1, ref2 Democrats ref1, ref2, ref3, ref4, ref5 Russia, inquiry into collusion with ref1 Sabbatians ref1, ref2, ref3 September 11, 2001, terrorist a acks ref1, ref2, ref3, ref4 UFO sightings, release of information on ref1 vaccines ref1 white supremacy ref1, ref2, ref3, ref4 Woke Democrats ref1, ref2

V

vaccines ref1, ref2, ref3 adverse reactions ref1, ref2, ref3, ref4, ref5 Africa ref1 anaphylactic shock ref1, ref2, ref3, ref4 animals ref1, ref2 anti-vax movement ref1, ref2, ref3, ref4, ref5 AstraZeneca/Oxford ref1, ref2, ref3, ref4 autoimmune diseases, rise in ref1, ref2 Big Pharma ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8

bioweapon, as real ref1, ref2 black and ethnic minority communities ref1 blood clots ref1, ref2 Brain Computer Interface (BCI) ref1 care homes, deaths in ref1 censorship ref1, ref2, ref3 chief medical officers and scientific advisers, financial interests of ref1, ref2 children ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10 China ref1, ref2 clinical trials ref1, ref2, ref3, ref4, ref5, ref6 compensation ref1 compulsory vaccinations ref1, ref2, ref3 computer programs ref1 consciousness ref1 cover-ups ref1 creation before Covid ref1 cytokine storm ref1 deaths and illnesses caused by vaccines ref1, ref2, ref3, ref4, ref5 definition ref1 developing countries ref1 digital ta oos ref1 DNA-manipulation ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10 emergency approval ref1, ref2, ref3, ref4, ref5 female infertility ref1 funding ref1 genetic suicide ref1 Global Cult ref1 heart chakras ref1 hesitancy ref1 Human 2.0 ref1, ref2, ref3, ref4 immunity from prosecution ref1, ref2, ref3

implantable technology ref1 Israel ref1 Johnson & Johnson ref1, ref2, ref3, ref4 lockdowns ref1 long-term effects ref1 mainstream media ref1, ref2, ref3, ref4, ref5 masks ref1, ref2, ref3, ref4, ref5 Medicines and Healthcare products Regulatory Agency (MHRA) ref1, ref2 messaging ref1 Moderna ref1, ref2, ref3, ref4, ref5, ref6 mRNA vaccines ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 nanotechnology ref1, ref2 NHS ref1, ref2, ref3, ref4, ref5 older people ref1, ref2 operating system ref1 passports ref1, ref2, ref3, ref4 Pfizer/BioNTech ref1, ref2, ref3, ref4, ref5, ref6, ref7 polyethylene glycol ref1 pregnant women ref1 psychopathic personality ref1 races, targeting different ref1 reverse transcription ref1 Smart Grid ref1 social distancing ref1 social media ref1 sterility ref1 synthetic material, introduction of ref1 tests ref1, ref2, ref3 travel restrictions ref1 variants ref1, ref2 viruses, existence of ref1 whistle-blowing ref1

WHO ref1, ref2, ref3, ref4 Wokeness ref1 working, vaccine as ref1 young people ref1 Vallance, Patrick ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 variants ref1, ref2, ref3 vegans ref1 ventilators ref1, ref2 virology ref1, ref2 virtual reality ref1, ref2, ref3 viruses, existence of ref1 visual reality ref1, ref2 vitamin D ref1, ref2 von Braun, Wernher ref1, ref2

W war-zone hospital myths

ref1

ref1, ref2 wealth ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 ref10, ref11 wet market conspiracy ref1 Wetiko factor ref1 alcoholism and drug addiction ref1 anti-human, why Global Cult is ref1 Archons ref1, ref2, ref3, ref4 artificial intelligence (AI) ref1 Big Pharma ref1, ref2 children ref1 China ref1 consciousness ref1, ref2 education ref1 Facebook ref1 waveforms

fear ref1, ref2 frequency ref1, ref2 Gates ref1, ref2 Global Cult ref1, ref2 heart ref1, ref2 lockdowns ref1 masks ref1 Native American concept ref1 psychopathic personality ref1, ref2 reframing/retraining programmes ref1 religion ref1 Silicon Valley ref1 Smart Grid ref1 smartphone addiction ref1, ref2 social media ref1 war ref1, ref2 WHO ref1 Wokeness ref1, ref2, ref3 Yaldabaoth ref1, ref2, ref3, ref4 whistle-blowing ref1, ref2, ref3, ref4, ref5, ref6, ref7 white privilege ref1, ref2 white supremacy ref1, ref2, ref3, ref4, ref5 Whitty, Christopher ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10 ‘who benefits’ ref1 Wi-Fi ref1, ref2, ref3, ref4 Wikipedia ref1, ref2 Wojcicki, Susan ref1, ref2, ref3, ref4, ref5, ref6, ref7 Wokeness

Antifa ref1, ref2, ref3, ref4 anti-Semitism ref1 billionaire social justice warriors ref1, ref2, ref3

Capitol Hill riot ref1, ref2 censorship ref1 Christianity ref1 climate change hoax ref1, ref2 culture ref1 education, control of ref1 emotion ref1 facts ref1 fascism ref1, ref2, ref3 Global Cult ref1, ref2, ref3, ref4 group-think ref1 immigration ref1 indigenous people, solidarity with ref1 inversion ref1, ref2, ref3 le , hijacking the ref1, ref2 Marxism ref1, ref2, ref3 mind control ref1 New Woke ref1 Old Woke ref1 Oneness ref1 perceptual programming ref1 Phantom Self ref1 police ref1 defunding the ref1 reframing ref1 public institutions ref1 Pushbackers ref1, ref2, ref3 racism ref1, ref2, ref3 reframing ref1, ref2 religion, as ref1 Sabbatians ref1, ref2, ref3 Silicon Valley ref1 social justice ref1, ref2, ref3, ref4

transgender ref1, ref2, ref3, ref4, ref5 United States ref1, ref2 vaccines ref1 Wetiko factor ref1, ref2, ref3 young people ref1, ref2, ref3 ref1, ref2 World Economic Forum (WEF) ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 World Health Organization (WHO) ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9 AIDs/HIV ref1 amplification cycles ref1 Big Pharma ref1, ref2, ref3 cooperation in health emergencies ref1 creation ref1, ref2 fatality rate ref1 funding ref1, ref2, ref3 Gates ref1 Internet ref1 lockdown ref1 vaccines ref1, ref2, ref3, ref4 Wetiko factor ref1 world number 1 (masses) ref1, ref2 world number 2 ref1 Wuhan ref1, ref2, ref3, ref4, ref5, ref6, ref7 ref8 women, deletion of rights and status of

Y

ref1, ref2, ref3, ref4, ref5, ref6 Yeadon, Michael ref1, ref2, ref3, ref4 young people see also children addiction to technology ref1 Human 2.0 ref1 vaccines ref1, ref2 Yaldabaoth

Wokeness ref1, ref2, ref3 YouTube ref1, ref2, ref3, ref4, ref5 WHO 548

Z

ref1 Zionism ref1, ref2, ref3 Zuckerberg, Mark ref1, ref2, ref3, ref4, ref5, ref6, ref7, ref8, ref9, ref10, ref11, ref12 Zulus ref1 Zaks, Tal

Before you go … For more detail, background and evidence about the subjects in Perceptions of a Renegade Mind – and so much more – see my others books including And The Truth Shall Set You Free; The Biggest Secret; Children of the Matrix; The David Icke Guide to the Global Conspiracy; Tales from the Time Loop; The Perception Deception; Remember Who You Are; Human Race Get Off Your Knees; Phantom Self; Everything You Need To Know But Have Never Been Told, The Trigger and The Answer. You can subscribe to the fantastic new Ickonic media platform where there are many hundreds of hours of cu ing-edge information in videos, documentaries and series across a whole range of subjects which are added to every week. This includes my 90 minute breakdown of the week’s news every Friday to explain why events are happening and to what end.