Fundamentals of Fluid Mechanics [3 ed.]
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TM

Fundamentals of Fluid Mechanics

Fundamentals of Fundamentals of

G.S. Sawhney has served as Professor and Head, Department of Mechanical Engineering, GNIT Greater Noida. Earlier he was at Lord Krishna College of Engineering, Ghaziabad in the same capacity. Prior to joining teaching, he had served for 28 years in the Corps of Engineers and for 10 years in industry. His other books published by I.K. International Pvt. Ltd. are: ? Biomedical Electronics and Instrumentation ? Fluid Machinery Made Easy ? Fundamentals of Computer Aided Manufacturing ? Fundamentals of Biomedical Engineering Made Easy ? Heat and Mass Transfer ? Manufacturing Science (Volume-I, II) ? Materials Science and Engineering ? Mechanical Experiments and Workshop Practice ? Project Management Made Easy

Fluid Mechanics

Written for the second-year engineering students of undergraduate level, this well set out textbook explains the fundamentals of Fluid Mechanics. Written in questionanswer form, the book is precise and easy to understand. The book presents an exhaustive coverage of the theory, definitions, formulae and examples which are well supported by diagrams (wherever necessary) and problems in order to make the underlying principles more comprehensive. In the present third edition, the book has been thoroughly revised and enlarged. Modification to every chapter has been carried out on the basis of suggestions received and new developments in this area. Additional typical problems based on the examination papers of various technical universities have been included with solutions for easy understanding by the students. Objective questions of competitive examinations of GATE, IES and IAS have also been included with solutions and explanations at the end of each chapter.

Third Edition

Fluid Mechanics Third Edition

G.S. Sawhney

G.S. Sawhney

978-93-89795-22-6

Distributed by: 9 789389 795226

TM

Fundamentals of Fluid Mechanics

Fundamentals of

Fluid Mechanics Third Edition

G.S. SAWHNEY B.Tech. (IIT Madras), ME, Ptsc GNIT, Greater Noida AIMT, Greater Noida Formerly, Professor and HoD Lord Krishna College of Engineering Ghaziabad

Fundamentals of Fluid Mechanics, 3/E Authors: G.S. Sawhney Published by I.K. International Pvt. Ltd. 4435, 36/7, Ansari Rd, Daryaganj, New Delhi, Delhi 110002 ISBN: 978-93-90455-20-1 EISBN: 978-93-90455-25-6 ©Copyright 2020 I.K. International Pvt. Ltd., New Delhi-110002. This book may not be duplicated in any way without the express written consent of the publisher, except in the form of brief excerpts or quotations for the purposes of review. The information contained herein is for the personal use of the reader and may not be incorporated in any commercial programs, other books, databases, or any kind of software without written consent of the publisher. Making copies of this book or any portion for any purpose other than your own is a violation of copyright laws. Limits of Liability/disclaimer of Warranty: The author and publisher have used their best efforts in preparing this book. The author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness of any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. The accuracy and completeness of the information provided herein and the opinions stated herein are not guaranteed or warranted to produce any particulars results, and the advice and strategies contained herein may not be suitable for every individual. Neither Dreamtech Press nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Trademarks: All brand names and product names used in this book are trademarks, registered trademarks, or trade names of their respective holders. Dreamtech Press is not associated with any product or vendor mentioned in this book. Edition: 2020 Printed at: Rekha Printers

PREFACE TO THE THIRD EDITION

In the third edition, the book has been thoroughly revised and enlarged. Additional typical problems based on the examination papers of various technical universities have been included with solutions for easy understanding by the students. Objective questions of competitive examinations of GATE, IES and IAS have been included with solutions and explanations at the end of each chapter. I once again request students and teachers to send constructive suggestions and criticism by emailing at [email protected]

G.S. Sawhney

PREFACE TO THE SECOND EDITION

In the second edition, the book has been thoroughly revised and enlarged. Additional typical problems based on the examination papers of various technical universities have been included with solutions for easy understanding by the students. I thank all the faculty members of Mechanical Department of GNIT, Greater Noida for their assistance in completing the second edition of the book. I once again request students and teachers to send constructive suggestions and criticism by emailing at [email protected] G.S. Sawhney

PREFACE TO THE FIRST EDITION

Fluid Mechanics is an important subject which has been given equal weightage in Mechanical, &LYLO&KHPLFDODQG$HURQDXWLFDOXQGHUJUDGXDWHHQJLQHHULQJFXUULFXOXP,WGHDOVZLWKWKHÀRZ RIÀXLGV7KLVERRNLVGHVLJQHGWRH[SODLQWKHIXQGDPHQWDOVRIÀXLGPHFKDQLFVLQWKHDUHDVRI SURSHUWLHV RI ÀXLGV SUHVVXUH DQG LWV PHDVXUHPHQWV K\GURVWDWLF IRUFHV EXR\DQF\ GLPHQVLRQDO DQDO\VLVK\GUDXOLFVLPLOLWXGHDQGPRGHOVWXGLHVÀXLGNLQHPDWLFVÀXLGG\QDPLFVODPLQDUDQG WXUEXOHQW ÀRZV ERXQGDU\ OD\HU DQDO\VLV ÀRZ WKURXJK SLSHV LGHDO ÀXLG ÀRZ DQG ÀRZ SDVW VXEPHUJHGERGLHV(൵RUWVKDYHEHHQPDGHWRFRYHUWKHV\OODELRIVHYHUDOXQLYHUVLWLHV Based on my teaching experience, I have tried to explain principles and concepts of Fluid Mechanics in simple and clear terms. The endeavour is to present the subject matter in the most comprehensive and useable form. The derivation of fundamental relations has been kept as VLPSOHDVSRVVLEOH'LDJUDPVKDYHEHHQXVHGDEXQGDQWO\WRHOXFLGDWHWKHGL൶FXOWFRQFHSWVZKLFK FDQQRW EH H[SODLQHG H൵HFWLYHO\7KH WKHRU\ LV IXUWKHU VXSSRUWHG ZLWK LOOXVWUDWLRQV DQG VXLWDEO\ worked out examples. The book has an easy-to-read style, as it is written in question-answer IRUPWKDWLVJRLQJWREHQH¿WWKHUHDGHUVLPPHQVHO\ I express my gratitude to K.K Aggarwal, Chairman, and Dr. A.M. Chandra, Director, Lord Krishna College of Engineering, Ghaziabad for their support and encouragement extended to me during the compilation of this book. I am grateful to my doctoral guide, Dr. Prasad, Galgotia Engineering College, Greater Noida for moral support extended to me. Above all, I wish to record my sincere thanks to my wife, Jasbeer Kaur for her patience shown throughout the preparation of the book. I would appreciate constructive suggestions and objective criticism from students and teachers alike with a view to enhance further usefulness of the book. They may mail me their views at [email protected] G.S. Sawhney

CONTENTS

Preface to the Third Edition ............................................................................................................ v Preface to the Second Edition ........................................................................................................ vi Preface to the First Edition ........................................................................................................... vii 

 )OXLGV'H¿QLWLRQV 3URSHUWLHV 1.1 Introduction 1.2 States of Matter 1.3 Fluids 1.4 Applications of Fluid Mechanics 1.5 Types of Fluids 1.6 Properties of Fluids 1.7 Newton’s Law of Viscosity 1.8 Change of Phase 1.9 Cohesion and Adhesion 1.10 Surface Tension 1.11 Pressure Inside a Droplet 1.12 Capillary Rise or Fall 1.13 Questions from Competitive Examinations

 1 2 4 6 13 14 18 43 45 46 49 51 65



 3UHVVXUHDQG+HDG 2.1 Introduction 2.2 Hydrostatic Law and Aerostatic Law 2.3 Absolute and Gauge Pressures 2.4 Pressure Measuring Devices 2.5 Manometers 2.6 Questions from Competitive Examinations

 79 79 83 85 87 119

x

Contents



 +\GURVWDWLF)RUFHV 3.1 Introduction 3.2 Total Pressure and Centre of Pressure 3.3 Pressure on Plane Surface 3.4 Pressure on Inclined Surface 3.5 Centre of Pressure: Vertical and Inclined Surfaces 3.6 Hydrostatic Forces on Plane and Curved Surfaces 3.7 Pressure Diagram 3.8 Stability of Dam 3.9 Gate and Lock Gates 3.10 Questions from Competitive Examinations

 133 133 135 136 138 166 192 196 203 221



 %XR\DQF\DQG)ORDWDWLRQ 4.1 Introduction 4.2 Buoyancy 4.3 Floatation 4.4 Stability of Floating and Submerged Body 4.5 Metacentre 4.6 Rolling of Floating Body 4.7 Questions from Competitive Examinations

 233 233 235 237 240 244 285



 )OXLG0DVVHV6XEMHFWHGWR$FFHOHUDWLRQ 5.1 Introduction 5.2 Dynamic Equilibrium 5.3 Translational Acceleration 5.4 Rotational Acceleration 5.5 D’Alembert’s Principle 5.6 Horizontal Acceleration 5.7 Vertical Acceleration 5.8 Acceleration on Inclined Plane 5.9 Radial Acceleration 5.10 Questions from Competitive Examinations

 300 300 301 302 302 303 305 307 309 330



 'LPHQVLRQDO$QDO\VLV 6.1 Introduction 6.2 Dimensional Analysis 6.3 Fundamental and Secondary Dimensions 6.4 Dimensional Homogeneity

 332 332 334 335

Contents

xi

6.5 Methods of Dimensional Analysis 6.6 Dimensionless Numbers 6.7 Questions from Competitive Examinations

337 342 352



 6LPLOLWXGHDQG0RGHO$QDO\VLV 7.1 Introduction 7.2 Model and Prototype 7.3 Hydraulic Similitude 7.4 Geometric Similarity 7.5 Kinematic Similarity 7.6 Dynamic Similarity 7.7 Reynolds Model Law 7.8 Froude’s Model Law 7.9 Euler’s Model Law 7.10 Weber’s Model Law 7.11 Mach’s Model Law 7.12 Questions from Competitive Examinations

 358 358 360 361 362 362 366 367 369 370 370 386



 )OXLG.LQHPDWLFV 8.1 Introduction 8.2 Methods of Describing Fluid Motion 8.3 Lines of Flow 8.4 Types of Displacements of Fluid Particles 8.5 Types of Fluid Flows 8.6 One-, Two- and Three-Dimensional Flows 8.7 Equation of Continuity 8.8 Rotation and Vorticity 8.9 Circulation 8.10 Stream Function 8.11 Flow Net 8.12 Velocity and Acceleration Vector 8.13 Forced and Free Vortex Flow 8.14 Questions from Competitive Examinations

 395 396 398 400 402 404 406 411 414 415 421 425 428 461



 )OXLG'\QDPLFV, 9.1 Introduction 9.2 Forces and Energies in Fluid Flow 9.3 Concept of Control Volume

 485 486 487

xii

 

Contents

9.4 9.5 9.6 9.7 9.8  9.10  9.12 9.13 9.14 9.15

Reynolds Transport Theory Navier-Stokes Equations Euler’s Equations Bernoulli’s Equations Pitot Tube +\GUDXOLF&RH¿FLHQWV Vena Contracta 2UL¿FH Mouthpiece Free Liquid Jet Weir and Notch Questions from Competitive Examinations

488 489 490 492 493  497  505 517 520 567

  )OXLG'\QDPLFV,, 10.1 Introduction 10.2 Momentum Principle 10.3 Force on Vane 10.4 Moment of Momentum 10.5 Vortex Motion 10.6 Force Exerted on a Bend 10.7 Torque Acting on a Sprinkler   9HQWXULPHWHU)ORZPHWHUDQG2UL¿FH0HWHU 10.9 Rotometer 10.10 Questions from Competitive Examinations

 579 579 581 589 591 594 603  615 625

  /DPLQDU)ORZ 11.1 Introduction 11.2 Laminar Flow 11.3 Turbulent Flow 11.4 Reynolds Number 11.5 Flow in Horizontal Pipes 11.6 Average Velocity 11.7 Hagen Poiseuille Formula 11.8 Friction Factor 11.9 Flow in Parallel Plates 11.10 Kinetic Energy Correction Factor 11.11 Flow in Inclined Pipe 11.12 Flow in Inclined Plates

 636 637 638 638 640 642 644 645 646 659 663 671

Contents

11.13 11.14 11.15 11.16 11.17

xiii

Darcy’s Equation Oil Bearings Dashpot Measurement of Viscosity Questions from Competitive Examinations

673 674 680 682 691

  7XUEXOHQW)ORZ 12.1 Introduction 12.2 Turbulent Flow 12.3 Instantaneous Velocity 12.4 Eddy Viscosity 12.5 Turbulence Shear Stress 12.6 Mixing Length Concept 12.7 Similarity Concept 12.8 Velocity Distribution 12.9 Smooth and Rough Boundary 12.10 Average Velocity 12.11 Friction Factor 12.12 Moody’s Diagram 12.13 Hot Wire Anemometer 12.14 Questions from Competitive Examinations

 700 701 704 706 707 707 709 712 713 720 724 725 726 740

  %RXQGDU\/D\HU$QDO\VLV 13.1 Introduction 13.2 Concept of Boundary Layer 13.3 Laminar Sublayer 13.4 Boundary Layer Thickness 13.5 Von Karman Momentum Equation 13.6 Separation of Boundary Layer 13.7 Questions from Competitive Examinations

 746 746 750 751 757 764 795

  )ORZ7KURXJK3LSHVDQG&RPSUHVVLELOLW\(൵HFWV 14.1 Introduction 14.2 Major and Minor Energy Losses 14.3 Friction Head Loss 14.4 Friction Factor 14.5 Chenzy’s Formula 14.6 Sudden Enlargement and Head Loss 14.7 Sudden Contraction and Energy Loss

 808 809 812 813 815 818 820

xiv

Contents

14.8 14.9 14.10 14.11 14.12 14.13 14.14 14.15 14.16 14.17

Bend and Energy Loss Total Energy and Hydraulic Gradient Line Equivalent Pipe Parallel Pipes Pipe Networking Syphon Power Transmitted by a Flow Water Hammer Time to Empty a Tank Questions from Competitive Examinations

821 822 826 826 841 844 850 852 854 875

  ,GHDO)OXLG)ORZ 15.1 Introduction 15.2 Ideal Fluid Flow 15.3 Uniform Flow 15.4 Source Flow 15.5 Sink Flow 15.6 Vortex Flow 15.7 Doublet Flow 15.8 Superposition of Flow 15.9 Plotting Streamlines 15.10 Questions from Competitive Examinations

 893 893 895 896 899 900 903 903 906 908

  )ORZ3DVW6XEPHUJHG%RGLHV 16.1 Introduction 16.2 Submerged Bodies 16.3 Drag and Lift Force 16.4 Friction Pressure and Shear Drag   &RH൶FLHQWRI'UDJDQG/LIW 16.6 Stokes’ Law Concerning Skin and Pressure Drag 16.7 Circulation in Free Vortex   0DJQXV(൵HFW 16.9 Karman Vortex Trail 16.10 Aerofoil 16.11 Aeroplane in Equilibrium 16.12 Questions from Competitive Examinations

 910 910 911 912  916 918  922 923 926 942

,QGH[



Chapter

1

FLUIDS: DEFINITIONS & PROPERTIES

KEYWORDS AND TOPICS        

CONCEPT OF CONTINUUM IDEAL FLUID REAL FLUID VISCOSITY KINEMATIC VISCOSITY NEWTON’S LAW OF VISCOSITY NEWTONIAN FLUIDS NON-NEWTONIAN FLUIDS

       

COMPRESSIBILITY VAPOUR PRESSURE COHESION ADHESION SURFACE TENSION CAVITATION DROPLET & PRESSURE CAPILLARITY

1.1 INTRODUCTION )OXLGPHFKDQLFVLVWKDWEUDQFKRIVFLHQFHZKLFKGHDOVZLWKWKHEHKDYLRXURIÀXLGVDWUHVWDQG LQPRWLRQ7KHVWXG\RIÀXLGVDWUHVWLVFDOOHGfluid static7KHVWXG\RIÀXLGVLQPRWLRQGXHWR pressure forces is called fluid dynamics7KHVWXG\RIÀXLGVLQPRWLRQZLWKRXWSUHVVXUHIRUFHV is called fluid kinematics. $ VROLG FDQ UHVLVW WHQVLOH FRPSUHVVLYH DQG VKHDU IRUFHV XSWR D FHUWDLQ OLPLW ZKLOH D ÀXLG KDVQRRUQHJOLJLEOHWHQVLOHVWUHQJWK$ÀXLGFDQUHVLVWFRPSUHVVLYHIRUFHVRQO\ZKHQLWLVNHSW LQDYHVVHO$ÀXLGGHIRUPVFRQWLQXRXVO\ZKHQVXEMHFWHGWRDVKHDULQJIRUFH7KHÀXLGWKHUHIRUH FKDQJHVVKDSHRULWÀRZVDVLWR൵HUVOLWWOHUHVLVWDQFHWRVKHDULQJVWUHVVHV7KHUHH[LVWKRZHYHU VKHDULQJVWUHVVHVEHWZHHQWKHDGMDFHQWÀXLGOD\HUVZKLFKWU\WRRSSRVHWKHPRYHPHQWRIRQH OD\HURYHUWKHDQRWKHU7KHPDJQLWXGHRIWKLVVKHDULQJVWUHVVLQDÀXLGGHSHQGVRQWKHUDWHRI GHIRUPDWLRQRIWKHÀXLGHOHPHQWV$VGXULQJUHVWWKHÀXLGKDVQRGHIRUPDWLRQUHVXOWLQJLQQRQ SUHVHQFHRIDQ\VKHDULQJIRUFHZKHQÀXLGLVVWDWLRQDU\ ,Q RUGLQDU\ FRQGLWLRQV D JDV FDQ EH FRPSUHVVHG HDVLO\ ZKLOH D OLTXLG LV GL൶FXOW WR EH compressed. Hence, liquids are regarded as incompressible.

2

Fundamentals of Fluid Mechanics

1.2

STATES OF MATTER

 ‡ What do you understand by the composition and intermolecular forces in matter? How does intermolecular force vary? Every matter in nature is made up of molecules. Each molecule has positively charged nucleus and negatively charged electrons. The molecule is electrically neutral as negative charge in electrons is equal to the positive charge of its nucleus. When two molecules come close to each other, the distribution of charge in them becomes such that the force of attraction between opposite charges (positive charge of nucleus of one molecule and negative charge of electron of other molecule) becomes greater than the force of repulsion between similar charges (nucleus – nucleus and electron – electron of one and other molecule). This net force of attraction between the molecules is called “intermolecular force”. As the distance between the molecules increases, the net attractive force between them decreases and vice versa. However, at mean distance r0, no net force acts between them. In case distance is decreased below r0, then a repulsive force is generated.  ‡ 7KHGLDJUDPJLYHQEHORZVKRZVDJUDSKRIGLVWDQFH r EHWZHHQWZRPROHFXOHVDJDLQVW WKH LQWHUPROHFXODU SRWHQWLDO HQHUJ\ U  6WDWH ZLWK SURSHU DUJXPHQW L  ZKLFK SRLQWRIWKHJUDSKLQGLFDWHVWKHHTXLOLEULXPVWDWHRIWKHPROHFXOHV" LL :KLFKSDUW of the intermolecular force is attractive?

Intermolecular Potential Energy

+

A

r0 O

B

r E

F

D

The equilibrium state of the molecules is at point D because it is a point of minimum potential energy. In portion DBA of the curve, the potential energy increases as r decreases, i.e., intermolecular force is repulsive. In portion DEF of the curve, U increases as r increases.  ‡ :KDWDUHWKHGLIIHUHQWVWDWHVRIPDWWHU"   0DWWHU H[LVWV LQ WKUHH VWDWHV QDPHO\   VROLG   OLTXLG DQG   JDV7KH OLTXLG DQG JDV VWDWHV FDQ EH JURXSHG WRJHWKHU ZKLFK LV FDOOHG ÀXLG VWDWH 6SDFLQJ RI PROHFXOHV r0) is ODUJHLQJDVVPDOOLQOLTXLGDQGH[WUHPHO\VPDOOLQDVROLG+HQFHLQWHUPROHFXODUERQGLQJ IRUFHVDUHYHU\ZHDNLQJDVVX൶FLHQWLQOLTXLGDQGYHU\VWURQJLQDVROLG'XHWRWKHVWURQJ LQWHUPROHFXODU IRUFHV D VROLG LV YHU\ FRPSDFW DQG ULJLG LQ IRUP $ OLTXLG KDV VX൶FLHQW intermolecular cohesive bonding forces to hold it as a continuous mass but without a shape. 7KLV LV WKH UHDVRQ ZK\ D OLTXLG DGMXVWV WR WKH VKDSH RI WKH FRQWDLQHU7KH LQWHUPROHFXODU ERQGLQJ IRUFHV DUH YHU\ ZHDN LQ JDVHV DQG D JDV FDQ ¿OO XS WKH ZKROH RI D FRQWDLQHU DV LW GRHV QRW KDYH DQ\ GH¿QLWH YROXPH )RU D JLYHQ PDVV D VROLG KDV GH¿QLWH YROXPH DQG

(NWKFU&GſPKVKQPU2TQRGTVKGU

3

VKDSH D OLTXLG KDV GH¿QLWH YROXPH ZKLOH D JDV KDV QHLWKHU VKDSH QRU YROXPH 6ROLGV DUH incompressible, liquids are slightly compressible and gases are compressible.

Gas Solid

Vessel

Definite Volume and Shape

Liquid

Definite Volume

No Volume, No Shape

Solid, Liquid & Gas

 ‡ ([SODLQWKHUHVSRQVHVRIVROLGVDQGIOXLGVWRH[WHUQDOIRUFHV   )OXLGVGL൵HUIURPVROLGVLQWKHLUUHVSRQVHVWRH[WHUQDODSSOLHGIRUFHV:LWKLQHODVWLFOLPLW deformation occurs in solids on the application of shear force and deformation disappears when intermolecular force between the molecules restores the molecules to their original position on the removal of the shear force. T = Torque A

B

B A

f

B¢ comes back to same positive

B¢ T is removed



 )RUH[DPSOHDVROLGWZLVWVZKHQDWRUTXH T) is applied by shear angle f, which disappears on the removal of the torque (T +RZHYHUWZLVWEHKDYLRXURIDÀXLGLVFRPSOHWHO\GL൵HUHQW 7KH ÀXLG FDQ VXVWDLQ RQO\ QRUPDO IRUFHV DQG GHIRUPV FRQWLQXRXVO\ ZKHQ VKHDU IRUFH LV DSSOLHG)OXLGVGRQRWDFTXLUHHTXLOLEULXPRQWKHDSSOLFDWLRQRIVKHDUIRUFHOLNHVROLGVEXWLW FRQWLQXHVWRGHIRUPDVORQJDVWKHVKHDUIRUFHLVDFWLQJHYHQZKHQLWLVVPDOO)RUH[DPSOH Fixed cylinder Torque (T )

Fluid

Outer cylinder

4

Fundamentals of Fluid Mechanics

if a torque (T LVDSSOLHGRQDQH[WHUQDOF\OLQGHUZKLFKLVURWDWDEOHDQGVHSDUDWHGIURPWKH ¿[HG F\OLQGHU E\ D WKLFNQHVV RI ÀXLG DV VKRZQ LQ WKH ¿JXUH WKH RXWHU F\OLQGHU GRHV QRW have equilibrium position with the applied torque (T DQGWKHRXWHUF\OLQGHUVWDUWVSLFNLQJ up velocity from the stationary state. However, the outer cylinder will attain an equilibrium YHORFLW\ GHSHQGLQJ XSRQ WKH PDJQLWXGH RI WKH WRUTXH DQG WKH SURSHUWLHV RI WKH ÀXLG VSHFLDOO\YLVFRVLW\ VHSDUDWLQJLWIURPWKH¿[HGF\OLQGHU,IT is large then the equilibrium velocity will be high and equilibrium velocity will be small, when T is made smaller. It is HYLGHQWIURPWKLVWKDWWKHÀXLGGRHVUHVLVWWKHDSSOLHGWRUTXHDVH[WHUQDOF\OLQGHUDFKLHYHV Ê df ˆ the equilibrium rate of deformation Á ˜ or equilibrium velocity. The deformation (f) in Ë dt ¯ ÀXLGVXQOLNHVROLGVGRHVQRWGLVDSSHDUZKHQWRUTXH T) is removed.

1.3 FLUIDS  ‡ :KDWGR\RXXQGHUVWDQGIURPIOXLGV"   $ÀXLGLVDVXEVWDQFHZKLFKGHIRUPVFRQWLQXRXVO\ZKHQVXEMHFWHGWRH[WHUQDOVKHDULQJIRUFH   $ ÀXLG LV D VXEVWDQFH ZKLFK FDQQRW DFTXLUH DQ\ VWDWLF HTXLOLEULXP XQGHU WKH DFWLRQ RI DQ\ VKHDU IRUFH RI HYHQ D VPDOO PDJQLWXGH ,Q RWKHU ZRUGV ÀXLGV FDQQRW DFTXLUH D VWDWLF equivalent deformation (f  LQ RUGHU WR DFKLHYH D HTXLOLEULXP ZLWK WKH H[WHUQDO DSSOLHG WRUTXH+RZHYHUWKHÀXLGUHVLVWVWKHDSSOLHGWRUTXHE\DWWDLQLQJWKHFRQVWDQWYDOXHRIWKH Ê df ˆ rate of change of deformation Á ˜ DQGWKHQWKHDFFHOHUDWLRQRIÀXLGEHFRPHV]HUR+HQFH Ë dt ¯

 ‡      ‡  

DÀXLGLVDVXEVWDQFHZKLFKFDQ  DWWDLQWKHVKDSHRIWKHFRQWDLQHUDVLWKDVQRGH¿QLWH VKDSH RI LWV RZQ   GHIRUP FRQWLQXRXVO\ XQGHU WKH DFWLRQ RI VKHDU IRUFH   DWWDLQ WKH equilibrium rate of deformation and in which (4) the deformation does not disappear on the removal of the torque. :KDWDUHWKHFKDUDFWHULVWLFVRIDIOXLG" 7KHFKDUDFWHULVWLFVRIÀXLGDUH D  ,IKDVQRGH¿QLWHVKDSHRILWVRZQEXWLWFRQIRUPVWRWKHVKDSHRIWKHFRQWDLQLQJYHVVHO E  (YHQDVPDOODPRXQWRIVKHDUIRUFHDSSOLHGRQDÀXLGZLOOFDXVHLWWRXQGHUJRGHIRU mation which is continuous as the long as the force is applied. F  $VROLGVX൵HUVVWUDLQZKHQVXEMHFWWRVKHDUIRUFHZKLOHDÀXLGVX൵HUVDUDWHRIVWUDLQ :KDWLVYLVFRVLW\RIDIOXLG" 7KH SURSHUW\ ZKLFK FKDUDFWHUL]HV WKH UHVLVWDQFH ZKLFK D ÀXLG R൵HUV WR WKH DSSOLHG VKHDU force is called viscosity. The resistance does not depend upon the deformation (f) but on Ê df ˆ the rate of deformation Á ˜ :KHQWKHUHLVDUHODWLYHPRWLRQEHWZHHQGL൵HUHQWOD\HUVRI Ë dt ¯ ÀXLGVWKHQWKHUHLVDWDQJHQtial friction force in between the layers. Viscosity is less in gases but larger in liquids.

(NWKFU&GſPKVKQPU2TQRGTVKGU

5

 ‡ &RPSDULVRQEHWZHHQDVROLGDQGDIOXLG" Features

Solid

Fluid

1.

Spacing of molecules

Closely spaced

Spacing is large

2.

Intermolecular force

Large

Low

3.

External shape

Does not change

6KDSHFKDQJHVDQGLWFDQÀRZ

4.

Shear force

Can withstand elastic limit

&DQQRWZLWKVWDQGDQGLWÀRZVLQWKH direction of the shear force

5.

Extent of deformation

Deformation is constant and force dependent

Continuous with force

6.

Deformation

Disappears on the removal of shear force

Deformation does not disappear

7.

Rate of deformation

Deformation does not change with time, i.e., rate of change of deformation is zero

The rate of deformation becomes constant

8.

Motion on shear force Solid deforms without motion It is in motion in the direction of shear force. It comes to rest on the removal of shear force.

 ‡ :KDWLVIOXLGPHFKDQLFV" Fluid mechanics is the science in which we study the behaviour of fluids, which are either at rest or in motion. The study of the fluids under static conditions (fluids at rest) is FDOOHGIOXLGVWDWLFV)RUH[DPSOHWKHVWXG\RIZDWHULQDGDPRURWKHUVWDWLFFRQGLWLRQV are done in fluid statics. Aerostatics is the term used for the study of incompressible gases under static conditions. The study of mechanics of fluids in motion is called fluid dynamics. The flow of fluids through pipes and channels due to shear force is studied in fluid dynamics.

 ‡ :KDWLVIOXLGNLQHPDWLFV"   )OXLGNLQHPDWLFVGHDOVZLWKWKHPRWLRQ WUDQVODWLRQDQGURWDWLRQ DQGGHIRUPDWLRQRIWKHÀXLG elements without any consideration to the force or energy which is causing such motion or

6

Fundamentals of Fluid Mechanics

 ‡  

 ‡    ‡  

GHIRUPDWLRQ +HQFH NLQHPDWLFV GHDOV ZLWK WKH YHORFLWLHV DFFHOHUDWLRQV DQG ÀRZ SDWWHUQV RIDÀXLG :KDWLVIOXLGNLQHWLFV" ,QÀXLGNLQHWLFVZHVWXG\WKHUHODWLRQVKLSRIWKHYHORFLWLHVDFFHOHUDWLRQVGHYHORSHGE\WKH ÀXLGZKHQDQH[WHUQDOIRUFHLVDSSOLHGRQWKHÀXLGRUWKHIRUFHVH[HUWHGE\WKHPRYLQJÀXLG ZLWKFHUWDLQYHORFLWLHVDFFHOHUDWLRQV :KDWLVK\GUDXOLFV" +\GUDXOLFV LV D EUDQFK RI ÀXLG PHFKDQLFV LQ ZKLFK ZH VWXG\ WKH EHKDYLRXU RI ZDWHU (incompressible) in motion or rest. :KDWLVSQHXPDWLFV" 3QHXPDWLFVLVDEUDQFKRIÀXLGPHFKDQLFVLQZKLFKZHVWXG\WKHEHKDYLRXURIFRPSUHVVLEOH ÀXLGVZKLFKDUHLQPRWLRQRUUHVW

1.4 APPLICATIONS OF FLUID MECHANICS  ‡ 'HVFULEHVRPHDSSOLFDWLRQVRIIOXLGPHFKDQLFV   7KHDSSOLFDWLRQVRIÀXLGPHFKDQLFVDUH   7KHGHVLJQRIGDPVFDQDOVDQGZHLUV$GDPLVDPDVRQU\VWUXFWXUHWRVWRUHZDWHUDQG the main forces acting on a dam is static water pressure (F) against upstream face and the weight of the masonry (W). The resultant of these two forces must pass through WKHPLGGOHWKLUGRIWKHEDVHRIWKHGDPIRULWVVWDELOLW\&DQDOVDUHPHDQWWRWDNHZDWHU from one place to another by gravitational action. Weir is any angular obstruction in an RSHQVWUHDPRYHUZKLFKWKHÀRZWDNHVSODFH Dam Weir Height of F water

W R

Flow

Middle third Dam



Weir

 'HVLJQRISXPSVRYHUKHDGUHVHUYRLUVDQGSLSHOLQHVIRUWUDQVSRUWLQJZDWHUWRGRPHVWLF VHUYLFHOLQHV7KHUHLVSUHVVXUHKHDGORVVGXHWR  IULFWLRQLQSLSHOLQHVDQG  OLIWLQJ of water against gravity.

(NWKFU&GſPKVKQPU2TQRGTVKGU Overhead reservoir

7

Service tank

Pump

Underground water supply line Water Supply Installation and Distribution



 'HVLJQRIÀXLGFRQWUROGHYLFHVVXFKDVK\GUDXOLFJDWHVYDOYHVFRFNVDQGWDSV7KHVH GHYLFHVDUHXVHGIRUFRQWUROOLQJWKHÀRZRIWKHÀXLGV$SOXJFRFNLVDVLPSOHGHYLFH Hole in plug

Plug cock Disc & washer

Seat Close

Open Stop Cock

Disc & washer

Water

Pin at the centre of curvature of the gate

Seat Angle Valve

Hydraulic Radial Gate

Head race Dam Penstock

Tailrace Dam

8



Fundamentals of Fluid Mechanics

LQZKLFKWKHÀXLGSDVVDJHLVDKROHLQDURWDWDEOHSOXJ¿WWHGLQWKHERG\RIWKHFRFN ,Q VFUHZ GRZQ VWRS YDOYH WKH VWRS KDV GLVF ZLWK ZDVKHU WR PDNH WKH DFWXDO FRQWDFW ZLWKVHDWWRVWRSWKHÀRZRIZDWHU$QJOHYDOYHDOVRZRUNVRQWKHVDPHDUUDQJHPHQW Hydraulic gates are used in dams to allow the water out of the dam by rotating the gate.  ,QSUHVVXUHPHDVXUHPHQWXVLQJPDQRPHWHUV7KHGL൵HUHQFHRIWKHOHYHORIWKHÀXLGLQ the limbs of U-tube is an indicator of the pressure of the liquid in the vessel or pipe.

Pipe with filled liquid

h

Fluid



 ,QPHDVXULQJÀRZRIWKHÀXLGVLQWKHSLSHVE\YHQWXUHRURUL¿FHPHWHU

Orifice Metering

Venturi Metering

6. In designing an airplane. As an airplane is a mechanically driven machine and is heavier than air, it is supported by the dynamic reaction of the air over its wings, which are in the shape of aerofoils to provide the required lift. The airstream moves at higher speed over the upper surface of aerofoil as compared to airstream moving at the lower surface, thereby providing lift. Lift

V1 V1>V2 Propulsive thrust

Drag

V2 Weight

Aerofoil Flying of Aircraft

Lift

(NWKFU&GſPKVKQPU2TQRGTVKGU

9

7. In using wind power for moving sail boat. It is possible to move the boat in the desired GLUHFWLRQE\XVLQJVDLOVSURSHUO\ZKHQWKHZLQGLVDSSURDFKLQJIURPGL൵HUHQWGLUHFWLRQV



 6ZLQJLQJRIDFULFNHWEDOOLVDFKLHYHGE\PDNLQJRQHVLGHVPRRWKDQGRWKHUVLGHURXJK The air stream moves with lesser velocity on the rough side as compared to smooth side, thereby providing side force.



 6SLQQLQJRIWDEOHWHQQLVEDOODQGFULFNHWEDOOLVSRVVLEOHZLWKFKDQJHRIÀRZRIDLUVWUHDP on rotating ball.

10 

Fundamentals of Fluid Mechanics

 8VLQJ WKH SURSHUW\ WKDW WKH ÀXLGV H[HUW VDPH SUHVVXUH LQ DOO GLUHFWLRQV LQ GHVLJQLQJ hydraulic ram or press. It is possible to lift a heavy load with the application of small H൵RUWE\WKHPHWKRGVKRZQEHORZ Load = 1 ton Effort = 50 kg Lift



 8VLQJVWUHDPOLQHGERG\WRUHGXFHGUDJZKLOHGHVLJQLQJWKHKXOORIDVKLSRULQFUHDVLQJ WKHGUDJWRDFWDVDLUEUDNHVLQDQDLUFUDIWDVZHOODVLQWKHUHPRYDORIFKD൵IURPWKH grain in a separator. Steam lined hull

Hull

Higher drag

Lower drag Stream Lining to Reduce Drag

Part of wing is raised

Grains with chaff

Wing

Gains

Chaff Separator

Airbrake of an aircraft Application of Drag

(NWKFU&GſPKVKQPU2TQRGTVKGU



11

 7RXQGHUVWDQGWKHFLUFXODWLRQRIEORRGLQWKHDUWHULHVYHLQVKHDUWDQGOXQJV7KHEORRG KDVWREHR[\JHQDWHGDQGVXSSOLHGWRDOOWLVVXHVRIWKHERG\VRWKDWWKH\FDQSHUIRUP WKHLUIXQFWLRQV+XPDQKHDUWLVDWZRVWDJHIRXUFKDPEHUSXPS$UWL¿FLDOKHDUWPDFKLQH is used as a cardio-pulmonary bypass to enable the operation of the heart.

Heart Lungs

Oxygenation O2

Artery Deoxygenated Blood Vein

Pump Oxygenated blood

Artificial Blood Oxygenation



 7RGHVLJQSXPSVIRUSXPSLQJOLTXLGWKURXJKSLSHV7KHSULQFLSOHLVWRSURGXFHVXFWLRQ DWRQHVLGHWRGUDZWKHÀXLGDQGWRFUHDWHH[FHVVSUHVVXUHRQWKHRWKHUVLGHWRIRUFHWKH OLTXLGRXW3XPSVFDQEH  SLVWRQW\SH  FHQWULIXJDOW\SHDQG  MHWSXPSZKLFK XWLOL]HVWKHHQHUJ\RIÀRZLQJÀXLGV7KHFHQWULIXJDOSXPSSURGXFHVGHOLYHU\KHDGE\ FHQWULIXJDODFFHOHUDWLRQRIWKHÀXLGLQWKHURWDWLQJLPSHOOHU

12

Fundamentals of Fluid Mechanics

Impeller

Diffiuser

Inlet

Centrifugal Pump



 7R GHVLJQ ZDWHU WXUELQHV WR REWDLQ PHFKDQLFDO ZRUN IURP WKH NLQHWLF HQHUJ\ RI WKH falling water. The oldest type of water turbine is the water wheel in which wheel is URWDWHGE\WKHGL൵HUHQFHRIZDWHUOHYHOLQDVWUHDP7KHSHOWRQZKHHOLVDOVREDVHGRQ WKHSULQFLSOHRIWKHROGZDWHUZKHHO)UDQFLV .DSODQWXUELQHVDUHXVHGZKHUHÀRZV and heads are small. The principle of these turbines is that the water is diverted to obtain FKDQJHRIPRPHQWXPLQVLGHWKHWXUELQH7KHGLYHUVLRQWDNHVSODFHDWULJKWDQJOHVWRWKH direction of entry, causing turbine rotor to rotate. Guide vanes are provided to ensure PD[LPXPURWDU\PRWLRQRIWKHWXUELQHURWRU

(NWKFU&GſPKVKQPU2TQRGTVKGU

13

1.5 TYPES OF FLUIDS  ‡ :KDWDUHWKHGLIIHUHQFHVEHWZHHQOLTXLGV JDVHV" Liquids

Gases

1.

/LTXLGVKDYHGH¿QLWHYROXPH

1.

*DVHVGRQRWKDYHGH¿QLWHYROXPH

2.

/LTXLGVKDYHVLJQL¿FDQWFRKHVLYHIRUFHV due to closely packed molecules.

2.

Gases have weak cohesive forces due to widely spaced molecules.

3.

Liquids cannot expand.

3.

*DVHVFDQH[SDQGWR¿OODQ\YROXPH

4.

Liquids are almost incompressible.

4.

Gases are compressible.

5.

The change of temperature and pressure does not alter the volume of liquid.

5.

The volume of gas changes with temperature & pressure.

6.

Liquids form a free surface in gravitational ¿HOG

6.

Gases do not form free surface.

Free surface

‡ :KDWDUHWKHGLIIHUHQFHVEHWZHHQLGHDOIOXLG UHDOIOXLG" Ideal Fluids

Real Fluids

1.

Have no viscosity.

1.

Have viscosity

2.

Have no surface tension.

2.

Have surface tension

3.

Incompressible.

3.

Slightly compressible

4.

6KHDUIRUFHLV]HURGXULQJÀRZDVYLVFRVLW\LV zero.

4.

6KHDUIRUFHLVDFWLQJGXULQJÀRZ

5.

,GHDOÀXLGGRHVQRWH[LVW

5.

$OOÀXLGVDUHUHDO

 ‡ :KDWLVFRQWLQXXP",VDLUDFRQWLQXXP"'RHVLWDOZD\VUHPDLQVR" 8378 A substance is composed of a vast number of molecules. In the concept of continuum, the VXEVWDQFHLVFRQVLGHUHGIUHHIURPDQ\NLQGRIGLVFRQWLQXLW\7KHFRQFHSWLVYDOLGDVPRVW HQJLQHHULQJV\VWHPVDUHFRQFHUQHGZLWKWKHPDFURVFRSLFRUEXONEHKDYLRXURIDVXEVWDQFH rather than the microscopic or molecular behaviour. Hence in most cases, it is convenient to WKLQNRIDVXEVWDQFHDVDFRQWLQXRXVGLVWULEXWLRQRIPHGLXPRUDFRQWLQXXP$VWKHVFDOHRI analysis is large, the discontinuity of the order of intermolecular spacing or the free mean path is negligible The properties such as pressure, temperature and velocity, which are PHDVXUHGDWLQ¿QLWHO\VPDOOSRLQWVDWGL൵HUHQWSODFHVDUHFRQVLGHUHGWRYDU\FRQWLQXRXVO\ from one point to another if they are not the same. However, there are certain instances ZKHUH SUHVVXUHV DUH ORZ DQG KLJK DFFXUDF\ LV UHTXLUHG VXFK DV URFNHW SURSXOVLRQ  WKH concept of continuum cannot be applied.

14

Fundamentals of Fluid Mechanics

Concept of continuum

Molecules with spacing

Continuous matter

Air is considered a continuum in most engineering applications. The atmospheric air has mean free path of about 5 to 7.5 ¥–6 cm. The molecular spacing of air increases under low pressures and this large spacing between the molecules cannot be neglected. In that case, the concept of continuum cannot be applied on air while determining its properties.

1.6 PROPERTIES OF FLUIDS  ‡ :KDW GR \RX XQGHUVWDQG IURP WKH SURSHUWLHV" :KDW DUH WKH GLIIHUHQW W\SHV RI properties?   (YHU\ ÀXLG KDV FHUWDLQ FKDUDFWHULVWLFV ZKLFK GHVFULEH WKH SK\VLFDO FRQGLWLRQ RI WKH ÀXLG 6XFK FKDUDFWHULVWLFV DUH FDOOHG SURSHUWLHV RI WKH ÀXLG 7KH SURSHUWLHV RI DQ\ ÀXLG DUH pressure, temperature, density, mass, volume, etc.    7KHSURSHUWLHVFDQEHLQWHQVLYHRUH[WHQVLYH,QWHQVLYHSURSHUWLHVGRQRWGHSHQGXSRQWKH PDVVRIWKHÀXLG3UHVVXUHWHPSHUDWXUHDQGGHQVLW\DUHH[WHQVLYHSURSHUWLHV7KHH[WHQVLYH SURSHUWLHVGHSHQGRQWKHPDVVRIWKHÀXLG9ROXPHDQGZHLJKWDUHH[WHQVLYHSURSHUWLHV    :H FDQ FRQYHUW WKH H[WHQVLYH SURSHUWLHV LQWR LQWHQVLYH SURSHUWLHV E\ GLYLGLQJ WKHP E\ WKHPDVVRIWKHÀXLG+HQFHYROXPHDQGZHLJKWDUHH[WHQVLYHSURSHUWLHVEXWWKH\FDQEH converted into intensive properties by dividing them by mass, which gives us the properties DVVSHFL¿FYROXPHDQGVSHFL¿FZHLJKW  ‡ :KDWGR\RXXQGHUVWDQGIURPWKHIROORZLQJIOXLGSURSHUWLHV  GHQVLW\  VSHFL¿F ZHLJKW  VSHFL¿FYROXPHDQG  UHODWLYHGHQVLW\"  'HQVLW\,WLVGH¿QHGDVWKHPDVVRIWKHÀXLGSHUXQLWYROXPH7KHGHQVLW\RIZDWHULV NJP. Density (r) = 

0DVV NJ = Volume m

 6SHFL¿F:HLJKW IWLVGH¿QHGDVWKHZHLJKWRIWKHÀXLGSHUXQLWYROXPH SpHFL¿FZHLJKW  =

Weight Volume Mass ¥ g Volume

(NWKFU&GſPKVKQPU2TQRGTVKGU

   

15

=r¥g =g   7KHVSHFL¿FZHLJKWRIZDWHU ¥ 9.8    QHZWRQP    N1P  6SHFL¿F9ROXPH,WLVGH¿QHGDVWKHYROXPHRIWKHÀXLGSHUXQLWYROXPH v=

9ROXPH  = = mNJ Mass r

 5HODWLYH'HQVLW\ 6* ,WLVWKHUDWLRRIVSHFL¿FZHLJKWRIWKHÀXLGWRWKHVSHFL¿FZHLJKW RIWKHVWDQGDUGÀXLG:DWHULVWKHVWDQGDUGÀXLGIRUOLTXLGZKLOHDLURUK\GURJHQLVWKH VWDQGDUGÀXLGIRUWKHJDVHV SG of liquid =

=

Specific weight of the liquid Specific weight of water

g g = g w 9800

‡ OLWUHVRISHWUROZHLJKV1&DOFXODWH  VSZHLJKW  GHQVLW\  VSYROXPHDQG  6* Guidance: All dimensions are to be in MKS. Litre has to be changed into m.

 

   

Volume (V  OLWUHV  ¥± m :HLJKW 1 Sp. weight (g) =

:HLJKW  = 9ROXPH  ¥ -

= 7 ¥1P Sp. weight = r ¥ g r=

\  

g  ¥  = J 

 NJP Sp. volume = =

1 r 1 713.6 ¥± mNJ

16

Fundamentals of Fluid Mechanics

SG =

rPetrol 713.6 = rWater 1000

   V  ‡ $OLTXLGKDV6* )LQGLWV  GHQVLW\  VSYROXPHDQG  VSZHLJKW SG =

rliquid rwater

= 0.9

r liquid = 0.9 ¥ NJP

\

Sp. volume =

1 rliquid

=

1 900

 ¥± mNJ Sp. weight = rliquid ¥ g = 900 ¥1P     1P  ‡ $EORZHUGHOLYHUVPDLUSHUVHFRQGDWƒ&DQGRQHDWPRVSKHULFSUHVVXUH EDU  )LQGPDVVRIWKHDLUGHOLYHUHGLIPROHFXODUZHLJKWRIWKHDLULV$OVR¿QG  GHQVLW\  VSYROXPHDQG  VSZHLJKWRIWKHDLUEHLQJVXSSOLHG Guidance The mass of air can be calculated by the gas equation PV = mRT where P = Pressure, V = Volume, m = Mass, R = Gas constant & T 7HPSHUDWXUHLQNHOYLQXQLYHUVDO  

gas constant R = MR where M = molecular weight and R = gas constant. The value of R 1PNJPRO.

 

Gas constant (R) for air =

R Mair

=

8314 30

  T  N P EDU ¥51P V = 8 m Mass of air, m =

PV RT

1 ¥ 10 5 ¥ 8 = 9.62 kg = 277.13 ¥ 300 Density (r) =

Mass Volume

(NWKFU&GſPKVKQPU2TQRGTVKGU

=

17

9.62 8

 NJP

 

Sp. volume (v) = =

1 r 1 1.2

 PNJ

 

Sp. weight (g ) = r ¥ g     ¥     1P  ‡ :KDWLVWKHLPSRUWDQFHRIIOXLGPHFKDQLFV"   )OXLGV OLNH ZDWHU DQG DLU DUH UHDGLO\ DYDLODEOH LQ QDWXUH:H HPSOR\ WKHVH ÀXLGV LQ PDQ\ HQJLQHHULQJDSSOLFDWLRQV0DQ\VXFKDSSOLFDWLRQVLQYROYHWKHÀRZRIWKHVHÀXLGV7KHÀRZ through pipes; energy conversion by pump and turbine, the storage capacity and stability RI GDPV WKH PRYHPHQW RI DLUFUDIW VXEPDULQHV DQG URFNHWV FDQ EH DQDO\VHG E\ WKH ÀXLGV PHFKDQLFV7KUHHEDVLFSULQFLSOHVZKLFKDUHXVHGLQWKHÀXLGPHFKDQLFVIRUDQDO\VLQJÀXLG ÀRZSUREOHPVDUH  WKHSULQFLSOHRIWKHFRQVHUYDWLRQRIPDVVÀRZ  WKHSULQFLSOHRI WKHFRQVHUYDWLRQRIHQHUJ\DQG  WKHSULQFLSOHRIWKHFRQVHUYDWLRQRIPRPHQWXP  ‡ :KDWGR\RXXQGHUVWDQGIURPWKHYLVFRVLW\RIWKHIOXLGV" When a solid mass slides over a surface, a friction force is developed to oppose the motion. 6LPLODUO\ZKHQDOD\HURIDÀXLGVOLGHVRYHUDQRWKHUOD\HURIWKHVDPHÀXLGDIULFWLRQIRUFH is developed between them opposing the relative motion. This tangential force is called WKHYLVFRXVIRUFH6XSSRVHDÀXLGLVPRYLQJLQVWUHDPOLQHGPDQQHURQD¿[HGKRUL]RQWDO surface ABDVVKRZQLQWKH¿JXUHEHORZ7KHOD\HURIWKHÀXLGLQWKHFRQWDFWRIWKHVXUIDFH remains stationary (at rest), while the velocity of other layers increases with distance from WKH ¿[HG VXUIDFH ,Q WKH ¿JXUH WKH OHQJWK RI DUURZV UHSUHVHQWV WKH LQFUHDVLQJ YHORFLW\ RI WKHOD\HUV,WFDQEHVHHQWKDWWKHUHLVDUHODWLYHPRWLRQEHWZHHQWKHGL൵HUHQWOD\HUV,IZH WDNHWKUHHSDUDOOHOOD\HUVa, b, and cIURPWKH¿[HGVXUIDFHWKHQWKHLUYHORFLWLHVDUHLQWKH

c

uc

Velocity a O A

uc > ub > ua

ub

b ua

B

Velocity Gradient Due to Viscosity

18

Fundamentals of Fluid Mechanics

increasing order. The layer a tends to retard the layer b and layer b tends to retard the layer c. Therefore, each layer tries to decrease the velocity of the layer above it. It also means that each layer tries to increase the velocity of the layer below it. There is therefore an internal tangential resistive force acting between two layers which is opposing their relative motion. 7KLV IRUFH LV FDOOHG YLVFRXV IRUFH ,Q RUGHU WR PDLQWDLQ WKH ÀRZ RI WKH ÀXLG ZH KDYH WR DSSO\DQH[WHUQDOIRUFHWRRYHUFRPHWKLVWDQJHQWLDOUHVLVWLYHRUYLVFRXVIRUFH,QWKHDEVHQFH RIH[WHUQDOIRUFHWKHYLVFRXVIRUFHZLOOUHVLVWWKHÀRZDQGEULQJWKHÀXLGWRUHVW  ‡ 'H¿QHYLVFRVLW\*LYHH[DPSOHVWRVKRZWKHH[LVWHQFHRIYLVFRVLW\   7KH SURSHUW\ RI WKH ÀXLG E\ YLUWXH RI ZKLFK LW RSSRVHV WKH UHODWLYH PRWLRQ EHWZHHQ LWV DGMDFHQWOD\HUVLVNQRZQDVYLVFRVLW\   7KHSURSHUW\RIYLVFRVLW\FDQEHVHHQLQWKHIROORZLQJQDWXUDOSKHQRPHQRQ   :H FDQQRW ZDON IDVW LQ ZDWHU DV FRPSDUHG WR DLU DV ZDWHU KDV FRPSDUDWLYHO\ ODUJHU viscosity.   $VWLUUHGOLTXLGFRPHVWRUHVWRQDFFRXQWRIYLVFRVLW\   ,IZHSRXUKRQH\DQGZDWHURQDWDEOHWKHQWKHKRQH\ZLOOVWRSÀRZLQJVRRQZKLOHWKH ZDWHUZLOOÀRZIRUDODUJHGLVWDQFHDVLWKDVVPDOOYLVFRVLW\DVFRPSDUHGWRKRQH\

1.7 NEWTON’S LAW OF VISCOSITY  ‡ 'HULYHWKH1HZWRQ¶VHTXDWLRQRIYLVFRVLW\ 8378   $WKLQOD\HURIÀXLGLVFRQWDLQHGEHWZHHQWZRSDUDOOHOSODWHVDVVKRZQEHORZ7KHXSSHU plate is moved by tangential force PZKLOHWKHORZHUSODWHLVNHSWVWDWLRQDU\7KHPRYHPHQW of upper plate is resisted by the liquid through its viscosity. Therefore, the magnitude of YHORFLW\RIXSSHUSODWHZLOOGHSHQGXSRQWKHYLVFRVLW\RIWKHÀXLG,QQRVOLSFRQGLWLRQWKH ÀXLGOD\HURQWKHORZHUVXUIDFHUHPDLQVVWDWLRQDU\ZKHUHDVWKHWRSÀXLGOD\HUPRYHVZLWK WKHVSHHGRIWKHXSSHUSODWH+HQFHDYHORFLW\JUDGLHQWZLOODFWUHVXOWLQJDYHUWLFDOÀXLG line OA deforms to the position OA. Point B on line OA at a height dy from O, will move to B due to its velocity du.

(NWKFU&GſPKVKQPU2TQRGTVKGU

BB = =  BB = du ¥ dt =

Now, Also, \

19

velocity ¥ time du ¥ dt dy ¥ dq dy ¥ dq

du dq = dy dt The angle of deformation dq or strain increases with time but the time rate of shear strain per

or

Ê dq ˆ time Á ˜ remains constant as long force P is unchanged. As shear stress (t = PA) depends Ë dt ¯ upon the applied force (P) and area of contact (A), it remains constant and it depends upon the time rate of shear strain

dq dt du a dy

ta

t=m

du dy

This is called the Newton’s equation of viscosity. The constant of proportionality m is called YLVFRVLW\$ÀXLGLQZKLFKWKHVKHDUVWUHVV t ) is linearly proportional to time rate of strain LVFDOOHG1HZWRQLDQÀXLG  ‡ (QXQFLDWH 1HZWRQ¶V ODZ RI YLVFRVLW\ 'LVWLQJXLVK EHWZHHQ 1HZWRQLDQ DQG QRQ Newtonian fluids. 8378   1HZWRQ¶V /DZ RI 9LVFRVLW\ 7KH WDQJHQWLDO VKHDU VWUHVV H[LVWLQJ EHWZHHQ WZR DGMDFHQW ÀXLGOD\HUVLVGLUHFWO\SURSRUWLRQDOWRWKHYHORFLW\JUDGLHQWH[LVWLQJLQWKHÀXLGLQDGLUHFWLRQ SHUSHQGLFXODUWRWKHÀXLGOD\HUV Shear stress (t ) a

\

or

t=m where,

du dy

du dy

du Ê dq ˆ = velocity gradient which is equal to the time rate of shear strain Á ˜ Ë dt ¯ dy m = viscosity

20

Fundamentals of Fluid Mechanics

Newtonian Fluids 1.

1.

1HZWRQLDQÀXLGVIROORZ1HZWRQ¶VODZRI viscosity, t = m

2.

Non-Newtonian Fluids

du dy

1RQ1HZWRQLDQÀXLGVGRQRWIROORZ 1HZWRQ¶VODZRIGHIRUPLW\ tπm

The value of m is constant and it does not changes with the rate of deformation.

2.

du ndy

The value of m is not constant and it changes with the rate of deformation. Non-Newtonian fluids

t

Newtonian fluids

t

tan q = μ = Viscosity

Ideal fluid

q du dy

3.

7KHFRPPRQÀXLGVVXFKDVZDWHUNHURVHQH DQGDLUDUH1HZWRQLDQÀXLGV

3.

7KHFRPPRQQRQ1HZWRQLDQÀXLGVDUH slurries, gels, blood, lubricants oils, polymer solutions and paints.

 ‡ :KDWDUHSVHXGRSODVWLFIOXLGV" Fluids in which the apparent viscosity decreases with increasing rate of deformation are FDOOHG SVHXGRSODVWLF ÀXLGV 6OXUULHV JXPV EORRG DQG PLON DUH SVHXGRSODVWLF ÀXLGV7KH\ DUH DOVR FDOOHG VKHDU WKLQQLQJ ÀXLGV DV WKH UDWH RI LQFUHDVH RI VKHDU VWUHVV GHFUHDVHV ZLWK WKHUDWHRIGHIRUPDWLRQRIWKHÀXLGV

Ê du ˆ t = mÁ ˜ Ë dy ¯ 1

2

n

and

n

3 tan q1 > tan q2 > tan q3

t

μ1 > μ2 > μ3

q 1 q2 q3

q

(NWKFU&GſPKVKQPU2TQRGTVKGU

21

 ‡ :KDWDUHGLODWDQWRUVKHDUWKLFNHQLQJIOXLGV" Fluids in which the apparent viscosity increases with increasing rate of deformation are FDOOHG GLODWDQW RU VKHDU WKLFNHQLQJ ÀXLGV 6XVSHQVLRQ RI VDQG EXWWHU VWDUFK DQG VXJDU VROXWLRQVDUHVKHDUWKLFNHQLQJÀXLGV

Ê du ˆ t = mÁ ˜ Ë dy ¯

n

and

n

3 tan q3 > tan q2 > tan q1

2

t

μ3 > μ2 > μ1 1 q3 q2 q1 du dy

 ‡ :KDWDUH%LQJKDPSODVWLFVRULGHDOSODVWLFIOXLGV" 7KHÀXLGVZKLFKDFWOLNHVROLGVE\ZLWKVWDQGLQJDGH¿QLWHVKHDUVWUHVVZLWKRXWVWUDLQFKDQJH DQGODWHUUDWHRIVWUDLQLQFUHDVHZLWKVKHDUVWUHVVDIWHU\LHOGLQJ2LOSDLQWVMHOOLHVGULOOLQJ VDQGDQGVHZDJHVOXGJHDUHLGHDOSODVWLFÀXLGV

t

du dy

 ‡ :KDWDUHWKL[RWURSKLFRUSODVWLFIOXLGV"   7KH ÀXLGV ZKLFK SRVVHVV D GH¿QLWH \LHOG VKHDU VWUHVV DQG ODWHU WKHLU VKHDU VWUHVV YDULHV QRQOLQHDUO\ZLWKWKHUDWHRIVWUDLQDUHFDOOHGWKL[RWURSLFRUSODVWLFÀXLGV/LSVWLFNDQGSULQWHU LQNDUHSODVWLFÀXLGV

t

du dy

22

Fundamentals of Fluid Mechanics

Ê du ˆ t = tym Á ˜ Ë dy ¯

n

and

n!

‡ 6KRZWKHFXUYHRIVROLGDQGLQYLVFLGRULGHDOIOXLGRQVKHDUVWUHVVYHUVXVVKHDUVWUDLQ UDWHGLDJUDP The solids deform or undergo strain on application of shear strain but their strains do not df du = = 0. Hence,VROLGVDUHVKRZQDORQJVWUHVVD[LV vary with time, i.e., dt dy

t Solid

du dy

Behaviour of Solids

  7KHLGHDORULQYLVFLGÀXLGKDV]HURYLVFRVLW\+HQFHLWLVVKRZQDORQJ]HURVKHDUVWUHVVD[LV  ‡ )ORZRIDIOXLGLVJLYHQE\ t m

Ê du ˆ ÁË dy ˜¯



 7KHYHORFLW\GLVWULEXWLRQLVJLYHQE\ u 

uPD[ 

È y y ˘ Í -  ˙ Î h h ˚

where uPD[ PD[YHORFLW\DQGh ¿OPWKLFNQHVV,IYLVFRVLW\ 1VPWKHQ¿QG   shear stress at solid surface for uPD[ PVDQGh PPDQG  ZKDW1HZWRQLDQ IOXLGFDQLQGXFHVDPHVKHDUVWUHVVIRUVDPHYHORFLW\JUDGLHQWDQGPD[LPXPYHORFLW\"

Ê du ˆ Guidance: The velocity gradient Á ˜ can be founG RXW E\ GL൵HUHQWLDWLQJ WKH JLYHQ Ë dy ¯ relationship of velocity distribution. Shear stress can be found out by using shear stress and velocity gradient relationship. u=

umax 3

È y y3 ˘ Í2 - 3 ˙ Î h h ˚

(NWKFU&GſPKVKQPU2TQRGTVKGU

du umax È 2 3 y 2 ˘ = Í ˙ dy 3 Î h h3 ˚

t Ideal or non-viscous fluid du dy

Behaviour of Ideal Fluids At solid surface y = 0,

Ê du ˆ ÁË dy ˜¯ Ê du ˆ ÁË dy ˜¯

=

umax 2 ¥ 3 h

=

0.3 2 ¥ 3 0.01

y =0

y =0

 V±

 

Ê du ˆ t0 = m Á ˜ Ë dy ¯

1.5

= 0.5 ¥    1P

   )RU1HZWRQLDQÀXLG

Ê du ˆ t0 = m Á ˜ Ë dy ¯  m ¥

  m=

Ns 24.56   2 m 20

  1HZWRQLDQÀXLGRIYLVFRVLW\

Ns can induce same shear stress. m2

23

24

Fundamentals of Fluid Mechanics

 ‡ :KDW GR \RX XQGHUVWDQG IURP   SRLVH   FHQWLSRLVH   VSHFL¿F YLVFRVLW\ DQG   kinematic viscosity? Or Differentiate between dynamic and kinematic viscosity.    8378 Poise is the unit of viscosity in CGS system. One SRLVH  



Ns . Poise is a relatively m2

large unit. Hence, centipoise is used which is equal WR  SRLVH:DWHU KDV YLVFRVLW\ DW ƒ& FHQWLSRLVH  .LQHPDWLFYLVFRVLW\LVWKHUDWLRRIWKHYLVFRVLW\RIÀXLGWRLWVGHQVLW\ Kinematic viscosity =

Viscosity m Ê m2 ˆ = Density r ÁË s ˜¯

 7KHNLQHPDWLFYLVFRVLW\LQ&*6V\VWHPLVVWRNH –4 mV 6WRNHLVDUHODWLYHODUJHXQLW DQGFHQWLVWRNH cst ZKLFKLVHTXDOWRVWRNHLVXVHG7KHNLQHPDWLFYLVFRVLW\RIZDWHU DWƒ&DQGDLUDW673DUHcstDQGcstUHVSHFWLYHO\$LUKDVKLJKHUNLQHPDWLFYLVFRVLW\ than water. Dynamic Viscosity 1.

It is the ratio of shear stress produced by unit rate of shear strain, t =

2.

Kinematic Viscosity

1 Ê du ˆ ÁË dy ˜¯

1.

.

It has unit of poise in CGS system and

It is the ratio of dynamic viscosity by unit density u =

2.

Ns in SI units m2

m r

It has unit of stoke in CGS system or m2/s in SI units

104 stokes = 1 m2/s

Ns 10 poise = 1 m2 4

Poise = 100 centipoises

1 stoke = 100 centistokes.

3.

It is an indicator of only viscous force of the liquid.

3.

It is the ratio of viscous force to the inertia IRUFHRIWKHÀXLG,IYLVFRXVIRUFHLVODUJH kinematic viscosity is large and vice versa.

4.

It is used where viscosity forces are predominant.

4.

It is used where both viscous and inertia forces are predominant.

(NWKFU&GſPKVKQPU2TQRGTVKGU

25

 ‡ :KDWDUHWKHHIIHFWVRIWHPSHUDWXUHDQGSUHVVXUHRQYLVFRVLW\" Increase of temperature causes a decrease in the viscosity of liquid while it increases the YLVFRVLW\RIJDVHV7KHYLVFRXVIRUFHVLQDÀXLGDUHSURGXFHGE\WKHLQWHUPROHFXODUFRKHVLRQ and molecular moment transfer. The molecules of a liquid are comparatively more closely SDFNHGDQGPROHFXODUDFWLYLW\LVUDWKHUVPDOO+HQFHYLVFRVLW\RIDOLTXLGLVSULPDULO\GXH to molecular cohesion. The molecular cohesion decreases with elevated temperature which results into drop in viscosity of liquid. The molecular cohesive forces are very small in a gas and the viscosity is primarily due to the molecular momentum transfer. The rise of temperatures increases the molecular activity thereby increasing the viscosity of gas.    +LJKSUHVVXUHVDOVRD൵HFWWKHYLVFRVLW\RIDOLTXLG7KHYLVFRVLW\LVIRXQGWREHLQFUHDVLQJ with rising pressure. Viscosity increases, as more energy is required for the relative movement of liquid molecules at elevated pressure.  ‡ $SODWHPPGLVWDQFHIURPZLWKD¿[HGSODWHIOXLGLQEHWZHHQPRYHVRIPV,W UHTXLUHVDIRUFHRI1P to maintain the speed. Determine the viscosity. du = 1 m/s t = 2 N/m

2

0.05

Guidance: We have to apply the Newton’s law of viscosity, which relates velocity gradient to the applied shear stress. Shear force and velocity gradient are given. t=m

 

du dy

 m ¥

1 0.05 ¥ 10 -3

m ¥ 0.05 ¥–61VP  

 ¥± poise

 ‡ $SODWHKDYLQJVL]H¥PPLVSXOOHGZLWKYHORFLW\RIPVRYHUD¿[HGSODWH DWGLVWDQFHRIPP)LQG  IRUFHDQG  SRZHUWRPDLQWDLQYHORFLW\LIIOXLGKDV m SRLVH Guidance 7KHVKHDUVWUHVVLVIRUFHGLYLGHGE\DUHDRIWKHSODWH1HZWRQ¶VODZRIYLVFRVLW\ LVXVHGWR¿QGWKHVKHDUVWUHVVIURPWKHYHORFLW\JUDGLHQW du and dy are given)

26

Fundamentals of Fluid Mechanics

t=m   

du dy

 ¥

0.5  1P  ¥ -

F =t A F=A¥t  ¥¥–6 ¥  N1 Power = Force × Velocity

    

du = 0.05 m/s

0.25

    ¥ ¥ 0.5     N:  ‡ $ VNDWHU ZHLJKLQJ  1 LV VNDWLQJ ZLWK VSHHG RI  PV7KH VNDWHV KDYH DUHD RI  cm)LQGWKHWKLFNQHVVRIZDWHUOD\HUEHWZHHQVNDWHVDQGLFHLIm ¥± poise and G\QDPLFFRHIILFLHQWRIIULFWLRQRIWKHLFH  Guidance 7KHVKHDUIRUFHHQDEOLQJVNDWHUWRPRYHLVIULFWLRQIRUFH7KHWKLFNQHVVRIZDWHU layer can be found out from Newton’s law of viscosity. Friction force

dy

     

Ice

Friction force = md ¥ mg  ¥ 900  1

du = 20 m/s

(NWKFU&GſPKVKQPU2TQRGTVKGU

27

t = mwater du dy dy =

\  

 ¥ -5 ¥  m 

 ¥–5 m

   ¥± mm  ‡ $ SODWH RI VL]H  FP ¥  FP DQG ZHLJKW  1 VOLGHV GRZQ RQ LQFOLQHG VXUIDFH LQFOLQLQJƒWRWKHKRUL]RQWDOZKLFKKDVFHUWDLQWKLFNQHVVRIOXEULFDWLRQZLWKm  SRLVH 7KLV DWWDLQV YHORFLW\ RI  PV RYHU WKH OXEULFDWHG VXUIDFH )LQG WKLFNQHVV RI lubrication. Guidance 7KHFRPSRQHQWRIWKHZHLJKWDFWLQJDORQJWKHLQFOLQHGVXUIDFHZLOODFWDVVKHDU force. Shear force = wVLQ t ¥ t=

1 = 500 N 2

F 500 = = 0.8 ¥41P  ¥  ¥ -4 A

Plate W cos q

W sin q W = 100 30°

t=m

8 ¥ =

dy =

du dy

 ¥  dy  ¥ -  ¥ 

 ¥± mm

28

Fundamentals of Fluid Mechanics

 ‡ $SLVWRQKDYLQJPPGLDPHWHUDQGPPOHQJWKPRYHVLQVLGHDF\OLQGHURI PPGLDPHWHU7KHZHLJKWRIWKHSLVWRQLV1DQGDQQXODUVSDFHEHWZHHQSLVWRQDQG cylinder has lubrication oil of m 1VP)LQGWKHYHORFLW\ZLWKZKLFKWKHSLVWRQ will slide inside the cylinder. 100.4 Piston Cylinder 100

L = 150

w = 50 N

Guidance ,WLVWKHZHLJKWRIWKHSLVWRQZKLFKLVSURYLGLQJVKHDUIRUFH7KHVXUIDFHRI shear force is the surface area of the piston. Shear force = Weight of piston = 50 N Surface area = p DL = p ¥¥¥–6 m        

 ¥± m 2LO¿OPWKLFNQHVVdy =

 -  ¥± 

 ¥± t=

50  ¥ N  ¥ -

t=m

du dy

  ¥

\

du =

du  ¥ -

 ¥  ¥  ¥ - 0.5

   PV  ‡ 'HWHUPLQHWKHYLVFRVLW\RIDIOXLGKDYLQJNLQHPDWLFYLVFRVLW\RIVWRNHVDQG6*RI m Guidance .LQHPDWLFYLVcosity = , m FDQEHIRXQGRXWIURPNQRZQNLQHPDWLFviscosity r and density.

(NWKFU&GſPKVKQPU2TQRGTVKGU

SG =

29

r rwater

r = rwater ¥ SG ¥ NJP g= or

m r

m=r¥g  

 ¥¥–4

 

 1VP

VWRNH –4 mV

 ‡ 7KURXJK D YHU\ QDUURZ JDS RI KHLJKW h D WKLQ SODWH RI ODUJH H[WHQW LV SXOOHG DW D velocity v. On one side of the plate is oil of viscosity m and on other side oil of viscosity m &DOFXODWH WKH SRVLWLRQ RI WKH SODWH VR WKDW WKH SXOO UHTXLUHG WR GUDJ WKH SODWH LV minimum. 8378 

μ1

y h

V μ2

Guidance :KHQSODWHLVSXOOHGGUDJIRUFHVF & FZLOOEHH[HUWHGDWXSSHUDQGlower VXUIDFHRIWKHSODWHE\ÀXLGV7KHWRWDOGUDJ FF) will be minimum, when = 0, where y = distance of the plate from the upper surface of the gap. F GUDJIRUFHH[HUWHGE\XSSHUÀXLG = m1 F = m 2

V ¥A y V ¥A h-y

Êm m ˆ F = VA Á 1 + 2 ˜ Ë y b - y¯

d (F1 + F2 ) dy

30

Fundamentals of Fluid Mechanics

dF =0 dy

For minimum F,

dF Ê m m ˆ = VA Á - 1 - 2 ˜ = 0 dy Ë y b - y¯ 2

(h - y ) y

=

2

m2 m1

m h 2 + y 2 - 2hy = 2 m1 y2 2

Ê hˆ Ê hˆ Ê m2 ˆ ÁË y ˜¯ - 2 ÁË y ˜¯ + ÁË 1 - m ˜¯ = 0 1

h = y h y y=

Ê m ˆ 2 ± 4 - 4 Á1 - 2 ˜ m1 ¯ Ë 2 

m2 m1

Êh ˆ ÁË y KDVWREH > ˜¯

h 1 + m 2 m1

‡ 7ZRODUJHSODQHVXUIDFHVDUHPPDSDUW7KHVSDFHLQEHWZHHQ¿OOHGZLWKJO\FHULQ :KDWIRUFHLVUHTXLUHGWRPRYHDYHU\WKLQSODWHKDYLQJVXUIDFHDUHDP between WKH WZR SODQH VXUIDFHV ZLWK YHORFLW\ RI  PV ZKHQ   WKLQ SODWH LV LQ WKH PLGGOH RI WZRDQG  WKLQSODWHLVDWPPIURPRQHVXUIDFH7DNHm 16P. Guidance 7KHVKHDUIRUFHLVWKHVXPRIWKHVKHDUIRUFHRIXSSHUDQGORZHUÀXLG

dy1 = 16

dy1 = 12

du = 1 m/s dy2 = 12

du = 1 m/s

dy2 = 8

Case 1

Case 2

(NWKFU&GſPKVKQPU2TQRGTVKGU

Case dy dy  \

31

t = t or F = F

  ¥ - =  1P  F = A ¥ t =  ¥  1

t = 0.5 ¥

=  1 F = F + F =  ¥  =  1 Case  dy = 8, dy    ¥ - =  1P  F =  ¥ 

  ¥ - =  1P  F =  ¥ 

t = 0.5 ¥

t  = 0.5 ¥

=  1 F = F + F

=  1

=  +  = 1

du = 0.5

Drag force

Drag force

‡ $VTXDUHSODWHPVLGHDQGPPWKLFNZHLJKLQJ1LVWREHOLIWHGWKURXJKDYHUWLFDO JDSRIPPRILQ¿QLWHH[WHQW7KHRLO¿OOLQJWKHJDVKDVm 1VPDQG6*  7KHSODWHLVPRYHGZLWKPV¿QGIRUFHDQGSRZHUUHTXLUHGIRUPRYLQJWKHSODWH

Plate Vertical gap

w Fb = w1

Ê du ˆ Guidance 7KHSODWHKDVWREHPRYHGDJDLQVWGUag force Á ª Am ˜ acting both surfaces dy ¯ Ë and weight (w) but the buoyancy force wl (weight of liquid displaced) is acting upwards.

32

Fundamentals of Fluid Mechanics

Weight of liquid displaced (w) = volume of the plate ¥ SG of liquid ¥ density of water  ¥¥¥± ¥ 0.9 ¥  1  1 = t ¥A

        %XR\DQF\IRUFH Drag force

= ¥ 4 ¥ m

du dy

 ¥ 4 ¥¥

  

0.5 10 ¥ 10 -3

21 - 1 ˆ Ê ÁË dy = 2 ˜¯

 1

  )RUFHUHTXLUHG 'UDJIRUFHZHLJKW±Fb      

 ±  1

Power required = force required ¥ velocity     ¥ 0.5     ZDWWV  ‡ 7KH YHORFLW\ GLVWULEXWLRQ LQ D YLVFRXV IORZ RYHU D SODWH LV JLYHQ DV u  uy – y for y £PZKHUHu YHORFLW\ PV DWGLVWDQFHy from the plate. If m 1VP)LQG shear stress at y DQGy  Guidance %\GL൵HUHQWLDWLQJWKHYHORFLW\HTXDWLRQZHFDQ¿QGYHORFLW\JUDGLHQWDQGVKHDU stress. u = 4y – y

du  ±y dy

2

y

u = 4y – y

Plate

Ê du ˆ ÁË dy ˜¯

y =0

Ê du ˆ =4& Á ˜ =0 Ë dy ¯ y = 2

(NWKFU&GſPKVKQPU2TQRGTVKGU

33

Ê du ˆ Ê du ˆ t0 = m Á ˜ t0 = m Á ˜ Ë dy ¯ y = 0 Ë dy ¯ y = 2  ¥ 4

 

   1P t = 0  ‡ $ IOXLG NLQHPDWLF YLVFRVLW\  VWRNHV IORZV RYHU D KRUL]RQWDO SODWH ZLWK DUHD  P. 7KHYHORFLW\LVJLYHQE\u y±y where y is distance from the plate. If shear force PHDVXUHGRQSODWHLV1¿QGVSZHLJKWDQG6*RIWKHIOXLG Guidance7KHYLVFRVLW\JUDGLHQWLVWREHIRXQGRXWE\GL൵HUHQWLDWLQJYHORFLW\HTXDWLRQ y

u = 2y – y

3

t = 0.4 N (due to flow) Plate F0 = t0 ´ A = t0 ´ 1

u y±y

du =±y dy Ê du ˆ ÁË dy ˜¯

  y =0

Ê du ˆ t0 = m Á ˜ and F0 = t0 ¥ A = t 0 ¥ Ë dy ¯ y = 0 F0 = m ¥

 

m=

Ns 0.4    m 2

g=

m r

¥– 4 = r=

\  

0.2 r 0.2 NJP 2.5 ¥ 10 -4

 NJP SG =

r rwater

=

800 = 0.8 1000

34

Fundamentals of Fluid Mechanics

‡ 7KH YHORFLW\ RI D IOXLG RYHU D KRUL]RQWDO SODWH LV YDU\LQJ DV SDUDEROD ZLWK YHUWH[ DW  FP IURP WKH SODWH ZKHUH WKH YHORFLW\ LV  PV ,I m   16P ¿QG YHORFLW\ JUDGLHQWDQGVKHDUVWUHVVDWy DQGy PIURPWKHSODWH Guidance 7KH YHORFLW\ SUR¿OH FDQ EH DVVXPHG DV u = ay  by  c as it is varying SDUDEROLF7KHUHDUHWKUHHXQNQRZQVDQGZHUHTXLUHWKUHHFRQGLWLRQV$WSODWHZKHQy = 0,

Ê du ˆ then u DWYHUWH[y YHORFLW\u PV JLYHQ $OVRYHORFLW\JUDGLHQW Á ˜ Ë dy ¯ y = 0.1 DWWKHYHUWH[ U = 0.1 m/s

Velocity profit of the fluid y = 0.1 M Plate

u = ay 2 + by + c

u = aybyc



0=c u = ayby



 ¥ ab ab 



Putting y = 0, u = 0 \ Putting y u     or

Ê du ˆ =0 Puttting Á ˜ Ë dy ¯ y = 0.1 Ê du ˆ ÁË dy ˜¯ or

= ¥ a ¥b y = 0.1

a = – 5b

 )URPHTXDWLRQV  DQG 

and \

– 5bb  b  a = – u ±yy

(4)

(NWKFU&GſPKVKQPU2TQRGTVKGU

35

Ê du ˆ ÁË dy ˜¯ = – y Ê du ˆ ÁË dy ˜¯

Ê du ˆ   Á ˜  ±  Ë dy ¯ y = 0.05 y =0 Ê du ˆ Ê du ˆ t0 = m Á ˜ t0.05 = m Á ˜ Ë dy ¯ y = 0 Ë dy ¯ y = 0.05 = 0.9 ¥  

 

 1P  

¥ 1P

Ns LVPRYLQJRYHUD¿[HGVXUIDFHZLWKKHLJKW PA m2 IORDWPRYHVPLQVHFRQGV)LQGWKH YHORFLW\JUDGLHQWDQGVKHDUVWUHVVDFWLQJRQ the float.

 ‡ $IOXLGRIYLVFRVLW\P 

Guidance 7KHYHORFLW\JUDGLHQWLVHTXDOWRWLPHUDWHRIVWUDLQ

x - x d q tan q ª = h Rate of strain =  dt dt =

 = 0.05 per sec.  ¥ 

du dq = = 0.05 per sec. dy dt But,

t=m

du   ¥ 1 dy

‡ )LQGWKHWRUTXHDQGSRZHUUHTXLUHGWRWXUQFPORQJFPGLDPHWHUVKDIWDW USP LQ D  FP GLDPHWHU FRQFHQWULF EHDULQJ IORRGHG ZLWK OXEULFDWLQJ RLO RI YLVFRVLW\ 

Ns . m2

36

Fundamentals of Fluid Mechanics

Oil

Shaft Bearing

dy

Guidance 7KHSUREOHPLVVLPLODUWRDWKLQSODWHPRYLQJEHWZHHQWZR¿[HGSODWHV7KH YHORFLW\ FDQ EH FDOFXODWHG IURP USP DQG ZKHQ LW LV GLYLGHG E\ WKH WKLFNQHVV RI ÀXLG WKH velocity gradient is found out. d bearing - dshaft dy =  0.07 - 0.05 =  =  P

p dN p ¥ 0.05 ¥ 600 = 60 60 =  PV du t=m dy  = ¥ =  1P   ¥ -

du =

Area = p dL = p ¥  ¥  ¥ -4 =  ¥ - P  \ F = p¥ A =  ¥  ¥ - =  N Torque = F ¥

d 

5 ¥ -  =  ¥ - 1P =  ¥

(NWKFU&GſPKVKQPU2TQRGTVKGU

37

p NT 60 p ¥  ¥  ¥ - = 60 =  ZDWWV

Power =

 ‡ $YHUWLFDOF\OLQGHURIGLDPHWHUPPURWDWHVFRQFHQWULFDOO\LQVLGHDELJJHUF\OLQGHU RIGLDPHWHUPPERWKKDYLQJKHLJKW PP7KHJDSEHWZHHQWKHF\OLQGHUV FRQWDLQVDIOXLGZLWKYLVFRVLW\XQNQRZQ)LQGYLVFRVLW\LIWRUTXHRI1PLVDSSOLHG WRURWDWHWKHLQQHUF\OLQGHUZLWKUSP  T = 30 Nm Fluid

200 200.4

Guidance 9HORFLW\ JUDGLHQW FDQ EH IRXQG IURP WKH USP DQG ÀXLG JDS %\ 1HZWRQ¶V equation of viscosity, viscosity can be calculated. du =

=  

p ¥ 0.2 ¥ 200 60

 PV dy =

 

p dN 60

do - di 200.4 - 200 = ¥± 2 2

 ¥± m Force = t ¥ A Torque =

t¥ A¥d .2

A = pdL = p ¥¥¥– 6   

 ¥± m  t ¥¥± ¥

200 ¥± 2

38

Fundamentals of Fluid Mechanics

t=

30 ¥ 2 18.84 ¥ 200 ¥ 10 -5

 ¥1P

 

t=m

  

du dy

¥ = m ¥

m=

2.093 0.2 ¥ 10 -3

1.592 ¥ 10 3 ¥ 0.2 ¥ 10 -3 2.093

  Ns2 m  ‡ A circular disc haviQJUDGLXVRLVNHSWDWVPDOOKHLJKWDERYHD¿[HGVXUIDFHE\DOD\HU RIRLOKDYLQJYLVFRVLW\m'HWHUPLQHWKHH[SUHVVLRQIRUWKHYLVFRXVWRUTXHRQWKHGLVF Disc

w

R

Oil

h

r



dr

 ,QRUGHUWR¿QGYLVFRXVWRUTXHZHWDNHDQHOHPHQWDU\ULQJDWUDGLXVrDQGWKLFNQHVVdr. If w is angular velocity of the disc, the tangential velocity at ring is u = r.w and area of ring pr.dr t=m

=

du dy

m ¥ rw h

dFr = tr ¥ area of ring

(NWKFU&GſPKVKQPU2TQRGTVKGU

=

m ¥ rw ¥p rdr h

=

2pmw ¥ r dr h

dTr =

2pmw ¥ r dr h

39

Torque = Force ¥ radius

Total torque =

Ú

R

0

dTr =

2pmw R 3 r dr h Ú0 T=

2pmw R 4 ¥ 4 h

but

=

2pmw D4 ¥ 4 ¥ 16 h

=

pmwD4 32h

R=

D 2

 ‡ $WKUXVWEHDULQJKDVFPGLDPHWHULVVHSDUDWHGE\DQRLO¿OPRIKHLJKWPPIURP ¿[HGEDVH)LQGSRZHUGLVVLSDWHGLQWKHEHDULQJLQFDVHLWURWDWHVDWUSP*LYHQm 

Ns m2

N = 400 rpm

h = 2 ´ 10

3

0.1 m 4 Guidance 9LVFRXVWRUTXHT = pmwD can be calculated. Power is equal T ¥ w 32h

w = 2p N / 60 2p ¥ 0.1 ¥ 400 = 60 = 41.86 rad / sec.

40

Fundamentals of Fluid Mechanics

pmwD4 32h p ¥ 0.9 ¥ 41.86 ¥ (0.1)4 = 32 ¥ 2 ¥ 10 -3 = 0.1851 ¥ 10 -3 Nm

T=

Power = T ¥ w = 0.1851 ¥ 10 -3 ¥ 41.86 = 7.74 ¥ 10 -3 watts  ‡ $VROLGFRne of radius RDQGYHUWH[DQJOHqLVPDGHWRURWDWHDWDQJXODUYHORFLW\w in WKHFRQLFDOFDYLW\FRQWDLQLQJRLOZLWKYLVFRVLW\m. If hLVWKHJDSEHWZHHQWKHFRQHDQG FDYLW\¿QGWRUTXHT to rotate the cone. R R q

r + dr 20

r

h

ds q

r r + dr

Oil

  7DNHDVODQWKHLJKWds between a radius r and rdr sin q = or

ds =

dr ds dr sin q

Velocity at radius, r = u = wr  9HORFLW\GL൵HUHQFHdu at r from stationary outer cone surface = wr

(NWKFU&GſPKVKQPU2TQRGTVKGU

Shear stress at strip, tr = m

41

du dy =m

wr h

Force at strip, dF = tr ¥ dA

wr dr ¥pr ¥ h sin q

=m =

as

ds =

dr sin q

2pmw  r dr h sin q

Torque at strip = dF ¥ r dTr = \

Torque = =

2pmw  r dr h sin q

Ú

R

0

dTr =

2pmw h sin q

Ú

R

0

r 3 dr

2pmw R 4 pmwR 4 ¥ = h sin q 4 2h sin q

‡ :KDWLVWKHFRPSUHVVLELOLW\DQGFRHIILFLHQWRIFRPSUHVVLELOLW\RIDIOXLG" Compressibility. )OXLGV FDQ EH FRPSUHVVLEOH RU LQFRPSUHVVLEOH ,QFRPSUHVVLEOH ÀXLGV KDYH FRQVWDQW GHQVLW\ ZKLOH LQFRPSUHVVLEOH ÀXLGV KDYH YDULDEOH GHQVLW\ ,Q FRPSUHVVLEOH ÀXLGVZKHQSUHVVXUHLVDSSOLHGÀXLGVFRQWUDFWDQGZKHQSUHVVXUHLVUHGXFHGWKH\H[SDQG &RPSUHVVLELOLW\LVWKHSURSHUW\RIDÀXLGZKLFKFKDUDFWHUL]HVLWVDELOLW\WRFKDQJHLWVYROXPH under pressure.

dp

dv V

Coefficient of compressibility. It is the ratio of the change of volume per unit volume, i.e., volumetric strain to change of pressure. If there is change of volume – dv from the original volume V and the change of pressure is dp, then

42

Fundamentals of Fluid Mechanics

dv V dp

  &RH൶FLHQWRIFRPSUHVVLbility (b) =

‡ :KDWLVEXONPRGXOXVRIHODVWLFLW\")LQGWKHYDOXHRIEXONPRGXOXVIRULVRWKHUPDODQG adiabable processes.   7KHEXONPRGXOXVRIHODVWLFLW\ k) is the ratio of compressive stress to the volumetric strain

dp 1 = dV b V  7KHEXONPRGXOXVFDQEHDOVRH[SUHVVHGLQUHODWLYHFKDQJHLQGHQVLW\ k=

m = rV dm = dr ¥ Vr ¥ dV \ \

\



dV dr = V r K=

dP dr r

 ,VRWKHUPDOSURFHVVFDQEHH[SUHVVHGDV PV = constant \

pdVVdp = 0 p=

\

dp =k dV V

%XONPRGXOXVRILVRWKHUPDOSURFHVVLVHTXDOWRWKHSUHVVXUHRIWKHÀXLG

 $GLDEDWLFSURFHVVFDQEHH[SUHVVHGDVPV g constant Pg Vg ±  dVVg dp = 0 gP = –

dp =k dV V

\ %XONPRGXOXVLQDGLDEDWLFSURFHVVLVHTXDOWRg time the pressure. ‡ $QLQFUHDVHLQSUHVVXUHRIDOLTXLGIURP03DWR03DUHVXOWVLQWRGHFUHDVH LQLWVYROXPH)LQG  EXONPRGXOXVDQG  FRHIILFLHQWRIFRPSUHVVLELOLW\ P ± ¥61P

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43

dV    V dp 7 ¥ 106 = dV .002 V

k=–

 ¥1P

 

b=

1 1 = k 3.5 ¥ 109

 ¥– 9 m1

 

 ‡ $F\OLQGHUFRQWDLQVPDLUDWN1PZKLFKLVFRPSUHVVHGWRP)LQGEXON PRGXOXVLIFRPSUHVVLRQLVGRQH  LVRWKHUPDOO\DQG  DGLDEDWLFDOO\*LYHQg  Guidance %XONPRGXOXVLQLVRWKHUPDOSURFHVVLVHTXDOWRP which it is g P for adiabatic process PV = PV for isothermal process P =

\

PV 100 ¥ 10 3 ¥ 0.4 1 1 = V2 0.08

= 500 ¥1P k = P

\%XONPRGXOXV

= 500 ¥1P PV g = PV g for adiabatic process

Ê V1 ˆ P = P ÁË V ˜¯

g

2

    

Ê 0.4 ˆ  ¥ Á Ë 0.08 ˜¯

1.4

¥

 ¥51P K = g P



 ¥¥51P

1.8 CHANGE OF PHASE  ‡ ([SODLQ  (YDSRUDWLRQ  YDSRXUSUHVVXUHDQG  ERLOLQJ Evaporation (YDSRUDWLRQ PHDQV D FKDQJH RI SKDVH IURP OLTXLG WR JDVHRXV 7KH UDWH RI evaporation depends upon the pressure and temperature. Evaporation increases with the rise of temperature or lowering of pressure.

44

Fundamentals of Fluid Mechanics

Vapour Pressure&RQVLGHUDOLTXLGZKLFKLVHQFORVHGLQDFORVHGVSDFH7KHPROHFXOHVRI the liquid which attain high energy leave the liquid phase to vapour phase. However, some of the molecules in vapour phase have a tendency to rebound and get absorbed in the liquid. Hence, there is continuous interchange of molecules between the liquid and the vapour above it. The vapour pressure will have a constant value when the molecules leaving and HQWHULQJLQWRYDSRXUVWDWHLVVDPH7KHFRQVWDQWYDSRXUSUHVVXUHH[HUWHGE\WKHYDSRXURQ liquid is called the saturated vapour pressure. Higher the vapour pressure, more volatile is WKH OLTXLG 3HWURO KDV YDSRXU SUHVVXUH RI  N1P while water has vapour pressure of N1PDWƒ&+HQFHSHWUROYDSRUL]HVIDVWHUWKDQZDWHU0HUFXU\KDVYHU\ORZYDSRXU SUHVVXUHDQGKLJKGHQVLW\'XHWRWKLVLWLVYHU\VXLWDEOHDVÀXLGIRUPDQRPHWHU Boiling (YDSRUDWLRQRIOLTXLGFHDVHVZKHQVDWXUDWHGYDSRXUSUHVVXUHKDVEHHQDFKLHYHGRQ the liquid. In case vapour pressure decreases below the saturated vapuor pressure, evaporation of liquid starts again and it continues until new equilibrium condition is attained. If the vapour pressure falls considerably, then the molecules leave the liquid surface very rapidly to vapour state and this is called boiling. The boiling point of a liquid is the temperature at which its YDSRXUSUHVVXUHHTXDOVWKHH[WHUQDOSUHVVXUH)RUZDWHUWKHYDSRXUSUHVVXUHEHFRPHV EDUDWƒ&ZKLFKLVHTXDOWRWKHDWPRVSKHULFSUHVVXUHUHVXOWLQJERLOLQJRIWKHZDWHU$W ORZHUDWPRVSKHULFSUHVVXUHOLNHLQKLOOVWDWLRQVZDWHUDWWDLQVYDSRXUSUHVVXUHHTXDOWRORZHU atmospheric pressure at lower temperature and water boils at lower temperature.  ‡ :KDWLVFDYLWDWLRQ":KDWLVLWVHIIHFW"+RZLVLWDYRLGHG"   7KHÀXLGVWDUWVERLOLQJHYHQDWORZWHPSHUDWXUHLISUHVVXUHRQWKHÀXLGIDOOVWRLWVYDSRXU SUHVVXUHDWWKDWWHPSHUDWXUH6XFKERLOLQJRIWHQRFFXUVLQÀRZLQJÀXLGVVSHFLDOO\RQVXFWLRQ VLGH RI WKH SXPS :KHQ VXFK ERLOLQJ RFFXUV LQ WKH ÀRZLQJ ÀXLGV YDSRXU EXEEOHV EHJLQ to grow in local regions of very low pressure. The formation of bubbles of vapour in the ÀRZLQJÀXLGLVFDOOHGFDYLWDWLRQ Air at Patm

Vapour

Vapour at Pvapour Patm > Pvapour

Pvapour > Patm

Evaporation

Bubbles

Boiling

The bubbles of vapuor are formed due to cavitation in the region of low pressure and EXEEOHV DUH FDUULHG E\ WKH ÀXLG ÀRZ WR RWKHU UHJLRQV VXFK DV GHOLYHU\ VLGH RI WKH SXPS

(NWKFU&GſPKVKQPU2TQRGTVKGU



45

where the pressure is high. In the region of high pressure, bubbles suddenly collapse and vapour condenses, resulting in drop in volume occupied. The space occupied by the bubbles LVQRZDYDLODEOHIRUWKHÀXLGWRUXVKLQWR¿OOLW  7KHUXVKLQJRIÀXLGJHQHUDWHVQRLVHDQGYLEUDWLRQV,IEXEEOHVFROODSVHRQWKHLPSHOOHU YDQHV WKH UXVKLQJ ÀXLG FRUURGHV WKH YDQHV DQG RWKHU VXUIDFHV RI WKH SXPS &DYLWDWLRQ can also occur if a liquid contains dissolved air or other gases in liquid. The solubility of gases in a liquid decreases as the pressure is reduced. Gas or air bubbles are released in WKHVDPHZD\DVYDSRXUEXEEOHVZLWKWKHVDPHGDPDJLQJH൵HFWV7KHFDYLWDWLRQRFFXUVLQ K\GUDXOLFPDFKLQHVKDYLQJÀRZRIKLJKVSHHGÀXLGVVXFKDVWXUELQHVSXPSVSURSHOOHUVRI VKLSDQGSLSHOLQH7KHFDYLWDWLRQKDVWREHDYRLGHGE\HQVXULQJWKDWSUHVVXUHRQWKHÀXLG VKRXOGQRWIDOOEHORZLWVYDSRXUSUHVVXUHLQDQ\UHJLRQRIWKHÀRZE\SURSHUGHVLJQLQJ WKHÀXLGÀRZ

Pfluid < Pvapour

Bubble

Vapour

Pvapour > Pfluid

Fluid Collapse of Bubble

Cavitation

1.9 COHESION AND ADHESION  ‡ :KDWDUHGLIIHUHQFHVEHWZHHQFRKHVLRQDQGDGKHVLRQ" Cohesion

Adhesion

1.

Cohesion is the intermolecular attraction EHWZHHQWKHPROHFXOHVRIVDPHÀXLG OLNH molecules).

1.

Adhesion is the force of attraction between alike molecules, i.e., the PROHFXOHVRIÀXLGDQGWKHPROHFXOHVRI solid boundary surface in contact with WKHÀXLG

2.

7KHÀXLGWHQGVWRUHVLVWWKHVKHDUVWUHVV due to cohesion.

2.

The property of adhesion enables a liquid to stick to another surface.

3.

If cohesive force is more than adhesive force, liquids form droplets without wetting the contact surface. Example: mercury.

3.

If adhesive force is more than cohesive force, liquids spread on the contact surface and wet the surface. Example: water

4.

For higher cohesion, contact angle is more than 90° for most surfaces.

4.

For higher adhesion contact angle is less than 90°.

46

Fundamentals of Fluid Mechanics

 ‡ :KDWDUHZHWWLQJDQGQRQZHWWLQJOLTXLGV" Adhesion < Cohesion

Contact surface Cohesion < Adhesion q = Contact angle q Liquid

Liquid q

Contact surface

Wetting Liquid (Spread Out)



Contact surface

Non-Wetting Liquid (forms Droplet)

 :KHQ D OLTXLG OLNH PHUFXU\ LV VSLOOHG RQ D VPRRWK VXUIDFH LW WHQGV WR IRUP GURSOHWV7KH cohesive molecular forces between the molecules of mercury are more than the adhesive forces between the molecules of mercury and the contact surface. Mercury tends to stay away from the contact surface. Hence, mercury is said to be a non-wetting liquid. In the case of water, cohesive molecular forces between molecules of water are smaller than the adhesive molecular forces between the molecules of water and the contact surface. Hence, when water is poured on the contact surface, it spreads and wets the contact surface. The wetting and non-wetting liquids are decided by the angle of contact between the liquids and the surface material. Liquids wet the contact surface if contact angle q < p7KHGHJUHH RIZHWWLQJLQFUHDVHVDVFRQWDFWDQJOHGHFUHDVHVWR]HUR,QWKHFDVHRIQRQZHWWLQJOLTXLGV the contact angle > p7KHFRQWDFWDQJOHIRUZDWHULVDOPRVW]HURZKLOHLWLVƒWRƒ in case of mercury.

1.10 SURFACE TENSION  ‡ 'H¿QHVXUIDFHWHQVLRQDQGH[SODLQLWVFDXVHV 8378   7KHVXUIDFHWHQVLRQRIDOLTXLGLVGH¿QHGDVWKHIRUFHSHUXQLWOHQJWKLQWKHSODQHRIOLTXLG surface which is acting at right angles on either side of an imaginary line drawn in that surface.     /HWDOLTXLGVXUIDFHEHLQGLFDWHGE\SRLQWVDQGDQLPDJLQDU\OLQHAB is drawn LQDQ\GLUHFWLRQLQWKHOLTXLGVXUIDFHDVVKRZQLQWKH¿JXUH7KHVXUIDFHRQHLWKHUVLGHRI this line (AB  H[HUWV D SXOOLQJ IRUFH RQ WKH OLQH ZKLFK LV LQ WKH SODQH RI WKH VXUIDFH DQG at right angle to the line (AB). The magnitude of this force per unit length of the line is a measure of the surface tension of the liquid. In case length of line = l and surface tension = sWKHQIRUFHDFWLQJRQRQHVLGHRIWKHOLQHLV F=s¥l or

surface tension

s=

F . The XQLWLV1P l

(NWKFU&GſPKVKQPU2TQRGTVKGU

47

2 F B 3

1 A F

Liquid surface

4 Surface Tension on Liquid Surface

Cause of Surface Tension:5DLQGURSVDQGVRDSEXEEOHVH[LVWLQSHUIHFWO\VSKHULFDOVKDSH Hence there is some other force besides gravity which is controlling the behaviour of the liquids. For a given volume, the surface area of a sphere is the least. Hence, we can say that free surface of the liquid has a tendency to contract to a minimum possible area under the action of this force, which is called surface tension. The free surface of the liquid tends to KDYHDPLQLPXPSRVVLEOHDUHDDQGLWEHKDYHVOLNHDVWUHWFKHGHODVWLFPHPEUDQH$OLTXLG VXUIDFHLVDOZD\VLQDVWDWHRIWHQVLRQ7KLVWHQVLRQLQWKHVXUIDFHLVNQRZQDVVXUIDFHWHQVLRQ The tension in a liquid surface always remains constant. The surface tension is caused by intermolecular cohesive forces between the molecules of the liquid. In order to understand how is surface tension caused on the free surface of a liquid, we FRQVLGHUIRXUOLTXLGPROHFXOHVOLNHA, B, C, and D with their sphere of molecular activity DVVKRZQLQWKH¿JXUHDERYH$VPROHFXOHA is well inside the liquid, it is attracted equally by other molecules in all directions, resulting no force acting on it. Similarly, force on the molecule BLVDOVR]HUR7KHVSKHUHRIDFWLYLW\RIPROHFXOHC is partly outside the liquid and more molecules are pulling it down as compared to molecules, which are pulling it up. Hence, there is a net force acting on it to pull it down. Similarly, molecule DZLOOH[SHULHQFH D QHW IRUFH SXOOLQJ LW GRZQ ZKLFK LV LQ IDFW PD[LPXP$OO PROHFXOHV RQ WKH OLTXLG IUHH VXUIDFH RU QHDU DERXW KDYH WR ZRUN DJDLQVW WKH GRZQZDUG FRKHVLYH IRUFH WR UHDFK WKHUH +HQFH ZRUN LV VWRUHG LQ WKHVH PROHFXOHV LQ WKH IRUP RI SRWHQWLDO HQHUJ\ 7KH SRWHQWLDO energy of the molecules lying on the free surface is greater than the molecules inside the OLTXLG:HNQRZWKDWDV\VWHPDWWDLQVHTXLOLEULXPZKHQLWKDVPLQLPXPSRWHQWLDOHQHUJ\,Q order to have minimum potential energy, the liquid surface tries to have minimum number of molecules in the surface, which is possible by contracting its surface to a minimum area. The contraction produces tension, which is called surface tension.  ‡ (VWDEOLVKUHODWLRQEHWZHHQVXUIDFHWHQVLRQRIDOLTXLGDQGWKHZRUNGRQHLQLQFUHDVLQJ its surface area. The molecules in the surface have some additional energy due to their position. This additional energy per unit area of the liquid free surface is called surface energy.    /HWDOLTXLG¿OPEHIRUPHGEHWZHHQDEHQWZLUHABCD in which the portion of the wire CD FDQ VOLGH RQ WKH UHVW RI WKH ZLUH$V WKH ¿OP VXUIDFH WHQGV WR FRQWUDFW WR UHGXFH WKH surface area, the wire CD tends to move upwards which is opposed by applying a force F.

48

Fundamentals of Fluid Mechanics

,WFDQEHVHHQWKDWWKHUHDUHWZRIUHHVXUIDFHVRIWKH¿OP IURQWDQGEDFN /HWWKHZLUHCD move a distance dx.    or

F  ¥ l ¥ s (l = length of wire CD & s = surface tension) :RUN F ¥ dx ¥ l ¥ s ¥ dx W = DA ¥ s (DA LQFUHDVHRIVXUIDFHDUHDRI¿OPIRUERWKVLGHV l ¥ dx) s=

W DA Molecules Liquid surface

Central molecule

B

Downward cohesive force

C

D

A Sphere of molecular activity Cohesive Force – Downward on Surface

Molecule and Sphere of Molecular Activity



If DA  WKHQ s = W  +HQFHWKHVXUIDFHWHQVLRQRIDOLTXLGLVHTXDOWRWKHZRUNUHTXLUHGWRWKHLQFUHDVHWKHVXUIDFH DUHDRIWKHOLTXLG¿OPE\XQLW\DWFRQVWDQWWHPSHUDWXUH7KHUHIRUHVXUIDFHWHQVLRQFDQDOVR EHH[SUHVVHGLQMRXOHPHWUH A

B

Liquid film

Surface tension

C

D

dx F Surface Tension & Work Done

 ‡ ([SODLQ  IORDWDWLRQRIQHHGOHRQZDWHU  ELJJHUEXEEOHVFDQEHIRUPHGIURPVRDS VROXWLRQDQG  VRDSKHOSVLQFOHDQLQJWKHFORWKHV   7KHQHHGOHÀRDWVRQWKHZDWHUGXHWRWKHVXUIDFHWHQVLRQ7KHUHDUHWKUHHIRUFHVDFWLQJRQ WKHQHHGOHDVVKRZQLQWKH¿JXUHLHVXUIDFHWHQVLRQIRUFHVT & T and the weight of the QHHGOH7KH KRUL]RQWDO FRPSRQHQWV RI WKH VXUIDFH WHQVLRQ T & T cancel each other while vertical components add up to balance the weight W.

(NWKFU&GſPKVKQPU2TQRGTVKGU T

49

T Needle

W Floatation of Needle



 %XEEOHV IRUPHG IURP SXUH ZDWHU FROODSVH LQ FDVH WKHLU VL]H JURZ 7KH UHDVRQ LV WKDW WKH surface tension of the water is large and pressure density inside the bubble varies inversely

4s ˆ , resulting in the formation of smaller bubbles. However, the with the diameter ÊÁ P = Ë d ˜¯ soap solution has a comparatively much lower surface tension (s is less) and it has two surfaces, resulting in the formation of bigger bubbles. The soap reduces the surface tension of the solution. Therefore a drop of soap solution wets a larger surface area of the cloth in comparison to a drop of pure water. Hence, the soap solution can enter and reach the restricted points of the cloth surface where pure water cannot reach and bring out the dirt particles with it. Also the cohesive forces between the molecules of soap solution are smaller than cohesive force between the molecule of soap solution and the dirt, resulting removal of the dirt.

1.11 PRESSURE INSIDE A DROPLET  ‡ :K\LVWKHUHDQLQWHUQDOSUHVVXUHLQVLGHDGURSOHW" 8378   7KH H൵HFW RI VXUIDFH WHQVLRQ RI D OLTXLG LV WR UHGXFH WKH VXUIDFH DUHD RI LWV IUHH VXUIDFH UHVXOWLQJGURSVRIOLTXLGWHQGWRWDNHDVSKHULFDOVKDSHLQRUGHUWRPLQLPL]HLWVVXUIDFHDUHD Formation of such droplet due to surface tension will cause an increase of pressure (P) of the liquid inside the droplet, which balances the surface tension. Consider a spherical droplet of diameter dDQGH[FHVVSUHVVXUHGHYHORSHGLQVLGHWKHGURSOHWLVP. If the surface tension is s on the spherical surface of the droplet, then applying the equilibrium on the half droplet DVVKRZQLQWKH¿JXUHZHJHW   

6XUIDFHWHQVLRQIRUFH )RUFHGXHWRH[FHVVSUHVVXUH 2 s ¥ p ¥ d = P ¥ pd 4

or

P=

4s d

Since d is very small, the value of P becomes very large. Also if the pressure P is greater than the pressure of vapour or gas in a bubble, the bubble will collapse.

50

Fundamentals of Fluid Mechanics s s

s P

d

s s

s

Spherical Droplet

 ‡ :KDWLVWKHSUHVVXUHLQWHQVLW\LQVLGHDVRDSEXEEOH" Soap bubble has two surfaces in contact with air i.,e., inside and outside spherical surfaces. )RUHTXLOLEULXP   

)RUFHGXHWRVXUIDFHWHQVLRQ )RUFHGXHWRH[FHVVSUHVVXUH spd = P ¥

  

P=

p  d 4

8s d

The intensity of pressure inside the soap bubble is twice as that of water droplet of the same diameter.  ‡ :KDWLVWKHSUHVVXUHLQWHQVLW\LQVLGHDOLTXLGMHW"   6XUIDFHWHQVLRQDOVRLQFUHDVHVWKHLQWHUQDOSUHVVXUHLQDF\OLQGULFDOMHWRIÀXLG&RQVLGHUD OLTXLGMHWRIF\OLQGULFDOVKDSHZLWKGLDPHWHUd and length lLVDVVKRZQLQWKH¿JXUH)RU HTXLOLEULXP    )RUFHGXHWRH[FHVVSUHVVXUH )RUFHGXHWRVXUIDFHWHQVLRQ P(d ¥ l) = s l) P=

2s d

 3UHVVXUHLVKLJKDVGLDPHWHURIWKHMet is small. d

l

l

Jet of Water

s s s

s s s s s

s

Forces on Jet

s s s

P

(NWKFU&GſPKVKQPU2TQRGTVKGU

51

 ‡ $LULVVHQWLQWRDZDWHUWDQNWKURXJKDQR]]OHWRIRUPDVWUHDPRIEXEEOHVRIGLDPHWHU  PP +RZ PXFK SUHVVXUH RI WKH DLU DW WKH QX]]OH PXVW H[FHHG IURP WKH ZDWHU IRU EXEEOHVIRUPDWLRQLIVXUIDFHWHQVLRQRIZDWHULV¥±1P   )RUEXEEOHVWKHH[FHVVSUHVVXUHLV

4s d 4 ¥ 72 ¥ 10 -3 = 3 ¥ 10 -3 = 96 N/m 2

P=

1.12 CAPILLARY RISE OR FALL  ‡ ([SODLQWKHSKHQRPHQRQRIFDSLOODU\2EWDLQDQH[SUHVVLRQIRUFDSLOODU\ULVHRUIDOORI DOLTXLGLQDYHU\VPDOOGLDPHWHUWXEH 8378 When a small diameter tube (diameter < 6 mm) open at both ends is lowered vertically into a liquid, which wets the tube, the liquid rises in the tube. However, if the liquid is such that it does not wet the tube, then the level of liquid in the tube depresses below the level of the liquid outside the tube. This phenomenon of the rise or fall of liquid level in the capillary tube is called capillarity. The cause of capillarity is the molecular forces of cohesion and adhesion. If the adhesion force between the molecules of liquid and the tube is more than the cohesion force between the liquid molecules, the liquid has the property to wet the glass as well as it rises in the capillary. The free surface of the liquid attains concave shape when liquid rises in the capillary. The surface tension forms angle q with the vertical, which is DOVROHVVWKDQƒ,QFDVHWKHOLTXLGLVVXFKWKDWLWKDVJUHDWHUFRKHVLYHIRUFHEHWZHHQLWV molecules than the adhesive force between its molecules and the molecules of the tube material, then liquid level falls below its level outside the tube. The free surface of the liquid LQ WKH WXEH DWWDLQV FRQYH[ VKDSH DQG WKH VXUIDFH WHQVLRQ IRUPV DQJOH q with the vertical ZKLFKLVJUHDWHUWKDQƒ

52



Fundamentals of Fluid Mechanics

Expression for capillary rise and fall: Let h be the capillary rise or fall of the liquid w.r.t. RXWVLGHÀXLGOHYHOd = diameter of tube, s = surface tension, and r GHQVLW\RIWKHÀXLG  )RUFHGXHWRVXUIDFHWHQVLRQDFWLQJRQÀXLGLQYHUWLFDOGLUHFWLRQ ZHLJKWRIWKHZDWHURI height hZUWRXWVLGHÀXLGOHYHO s cos q ¥ pd = h=

pd 2 ¥h¥r¥g 4 4s cos q rgd

‡ )LQGWKHFDSLOODU\ULVHRUIDOOIRUZDWHUDQGPHUFXU\ For capillary rise or fall h= For water,

q=0 h=

For mercury,

4s cos q rgd 4s rgd

q ƒ h=

=

+ 4s cos 140 rgd - 4s cos 50 rgd

=–

 ‡

       

2.57 s rgd

Minus sign indicates fall of mercury level inside the tube. 7KHVXUIDFHWHQVLRQRIVRDSVROXWLRQLV¥± 1P+RZPXFKZRUNZLOOEHGRQHLQ PDNLQJDEXEEOHRIGLDPHWHUFPE\EORZLQJ" Guidance 7KHZRUNGRQHLQPDNLQJDEXEEOHE\EORZLQJLVVWRUHGLQWKHIRUPRIHQHUJ\ LQWKHVXUIDFHRIWKHEXEEOH7KHVRDSEXEEOHKDVWZRVXUIDFHVLHLQWHUQDODQGH[WHUQDO  6XUIDFHDUHDRIEXEEOH ¥ 4pr = 8 ¥ p ¥     ¥± m  :RUNGRQH s ¥ area   ¥± ¥¥± = 5 ¥–5MRXOHV

(NWKFU&GſPKVKQPU2TQRGTVKGU

53

 ‡ $PHUFXU\GURSRIUDGLXVFPLVVSUD\HGLQWRGURSOHWVRIHTXDOVL]H&DOFXODWHWKH HQHUJ\H[SDQGHG7DNHs ¥± 1P Guidance :H KDYH WR ¿QG WKH FKDQJH RI VXUIDFH DUHD 7KH FKDQJH RI VXUIDFH DUHD LV multiplied by surface tension for obtaining energy. Let smaller droplets have radius = r

4  4 pR  6 ¥ ¥ pr (Volume remains same) 3 3 R r

\

r ¥– 4 m Let DAEHWKHGL൵HUHQFHRIVXUIDFHDUHDRIVPDOOHUGURSOHWVDQGELJJHUGURSOHWV DA 6 ¥ 4pr – 4pR = 4p >6r± r)] = 4p ¥ 99 ¥– 4 m :RUNGRQH s = DA

  

= ¥± ¥ 4p ¥ 99 ¥– 4     .98 ¥±MRXOHV  ‡ 7ZRFDSLOODU\WXEHVRIGLDPHWHUPPDQGPPDUHKHOGYHUWLFDOO\LQVLGHZDWHURQH E\RQH+RZPXFKKLJKWKHZDWHUZLOOULVHLQWRHDFKWXEH7DNHs ¥±1P   ,QFDSLOODU\WKHULVHRIOLTXLGLV h=

4s cos q rgd

For water, q = 0 for tube having d = 5 mm h = =

Now for

4s rgd 4 ¥ 7 ¥ 10 -2 5 ¥ 10 -3 ¥ 1 ¥ 10 3 ¥ 9.8

= 9.7 mm d = 4 mm h d  = h d  h =

9.7 ¥ 5 4

    PP   1RWHRise and fall increases with the decrease of tube diameter.

54

Fundamentals of Fluid Mechanics

 ‡ 'ULYH DQ H[SUHVVLRQ IRU WKH FDSLOODU\ ULVH RI D OLTXLG EHWZHHQ SDUDOOHO SODWHV DW D distance t apart. Let the length of parallel plates be l Let the liquid rise to height be h Weight of liquid of height h = The force of surface tension at surface acting vertically h ¥ r ¥ g ¥ t ¥ l = s ¥¥ l cos q or

h=

2s cos q t ◊r◊ g

‡ 7ZRSDUDOOHOSODWHVRIJODVVDUHKHOGZLWKDJDSRIPP)LQGWKHULVHRIZDWHULQWKH JDSZKHQSODWHVDUHGLSSHGLQZDWHU7DNHs 1P The rise of water due to capillary action h= =  

2s cos q t ¥r¥ g 2 ¥ 0.075 ¥ cos 0 0.001 ¥ 9.8 ¥ 10 3

 PP l s

s

h

t Capillary Rise in Parallel Plates

 ‡ $JODVVWXEHZLWKGLDPHWHUPPLVLPPHUVHGLQPHUFXU\:KDWLVWKHIDOORIPHUFXU\ in tube w.r.t. outside mercury? Take s 1PDQGq ƒ Capillary fall of mercury (h) is 4s cos q h= rgd =  

4 ¥ 0.5 ¥ cos 130 13.6 ¥ 10 3 ¥ 9.81 ¥ 3 ¥ 10 -3

 ±PP

(NWKFU&GſPKVKQPU2TQRGTVKGU

55

Minus sign shows that the mercury in the tube falls below the mercury level outside the tube.  ‡ )LQGPLQLPXPGLDPHWHURIJODVVWXEHLQZKLFKFDSLOODU\HIIHFWLVWREHUHVWULFWHGWR mm when immersed in water. Take s 1P Capillary rise (h) is h= ¥± =

 

d=

4s cos q rgd 4 ¥ 0.075 ¥ cos 0 10 3 ¥ 9.81 ¥ d 4 ¥ 0.075 3 ¥ 10 -3 ¥ 10 3 ¥ 9.81

   PP  ‡ $ 8 WXEH KDV WZR OLPEV RI LQWHUQDO GLDPHWHU RI  DQG  PP UHVSHFWLYHO\7KH WXEH FRQWDLQVZDWHU)LQGWKHGLIIHUHQFHRIOHYHOLQWKHOLPEVLIswater 1P   7KH ZDWHU ZLOO ULVH PRUH LQ WKH VPDOO GLDPHWHU OLPE7KH GL൵HUHQFH LQ WKH OHYHO ZDWHU LQ the limbs is 6

16

H

U Tube: 2 Capillaries

Ê 4s cos q ˆ Ê 4s cos q ˆ H = h1 - h2 = Á Ë rgd1 ˜¯ ÁË rgd2 ˜¯ q=0 H= =

\

cos q = 1

4s Ê 1 1ˆ - ˜ Á rg Ë d1 d2 ¯ 4 ¥ 0.075 È 1 1 ˘ -3 -3 ˙ 3 Í 10 ¥ 9.81 Î 6 ¥ 10 16 ¥ 10 ˚

56

Fundamentals of Fluid Mechanics

0.3 È 1 1 ˘ 9.81 ÍÎ 6 16 ˙˚ 0.3 = [0.166 - 0.0625] 9.81 0.3 = ¥ 0.1035 9.81 = 3.16 mm =

 ‡ +RZGRHVWKHULVHRIOLTXLGOHYHOLQFDSLOODU\WXEHJHWDIIHFWHG L ZKHQWKHWXEHKDV LQVXIILFLHQWOHQJWKWRWKHSRVVLEOHULVHRIWKHOLTXLGOHYHODQG LL ZKHQWRSRIWXEHLV closed. The radius r of capillary tube and the radius of curvature of liquid free surface (R) are related DV

r = R cos q

r =R cos q

or

2s cos q rgr 2s = rgR

h=

R

s

q

R

s

s

1

R q

q

s q1

1

R >R q1 > q 1 h R

\

h1 =

Rh = R1 h1 h1 < h

(NWKFU&GſPKVKQPU2TQRGTVKGU



  

57

 1RWHWater will never rise beyond the height of capillary and spill or fountain out of it.   ,IWRSRIFDSLOODU\LVFORVHGDLUHQWUDSSHGDWWRSRIWKHFDSLOODU\H[HUWVSUHVVXUHVWRRSSRVH WKHULVHRIÀXLGLQWKHFDSLOODU\GXHWRFDSLOODULW\+HQFHWKHULVHRIOLTXLGOHYHOZRXOGEH less than the normal rise. ‡ :KDWDUHWKHGLIIHUHQWPHWKRGVRIGHWHUPLQDWLRQRIYLVFRVLW\DQGNLQHPDWLFYLVFRVLW\"  7KH PHWKRGV RI PHDVXULQJ YLVFRVLW\ DQG NLQHPDWLF YLVFRVLW\ DUH   FDSLOODU\ WXEH   VSKHUHUHVLVWDQFH  URWDWLRQDOF\OLQGHUDQG  YLVFRPHWHU   Capillary Tube: The method is based on the fact that the volume (V ) per second of DÀRZWKURXJKDKRUL]RQWDOFDSLOODU\WXEHRIDJLYHQUDGLXV r) and length (l) under a FRQVWDQW GL൵HUHQFH RI SUHVVXUH dp) between its ends is inversely proportional to the YHORFLW\RIWKHÀXLG Flow Q =

V pr 4 dP = 8lm t Linear drop of pressure

dp Distance 2r

l Flow through Capillary



  Sphere Resistance /LNH IULFWLRQ H[HUWLQJ UHVLVWDQFH WR PRWLRQ RQ GU\ VXUIDFH WKH viscosity of a liquid causes a frictional resistance to the motion of any body moving WKURXJKWKHOLTXLG$PHWDOEDOOZKHQGURSSHGLQWRDMDURIDOLTXLGPRYHVGRZQDQG DWWDLQVDFRQVWDQWWHUPLQDOYHORFLW\ZKLFKGHSHQGVXSRQWKHYLVFRVLW\RIWKHÀXLG7KH terminal velocity also varies inversely with the radius of sphere. The viscous force acting on the sphere is given by F = 6pmru where u = velocity, m = viscosity, and r = radius At the time of terminal velocity, the weight of sphere is equal to the drag force and buoyancy force. No more variation of velocity is possible after achieving terminal velocity. 4 3 4 pr grsolid = 6 pmruterminal + pr 3 grfluid 3 3 2r 2 or uterminal = g (rsolid - rfluid ) gm

58

Fundamentals of Fluid Mechanics

Sphere Achieving Terminal Velocity



  Rotational Cylinder$F\OLQGHURIUDGLXVriLVURWDWHGFRD[LDOO\LQVLGHD¿[HGF\OLQGHURI radius ro7KHDQQXODUVSDFHEHWZHHQWKHF\OLQGHUVLV¿OOHGZLWKDOLTXLGKDYLQJYLVFRVLW\ m. The cylinders have equal length of l. Now a torque T is applied on the inner cylinder to move with angular velocity w the inner cylinder Shear stress on the ÀXLG  m

du dy T

Fluid

ri

r0

ro - ri , V = ri w  dv t=m ro - ri  F = t¥ A r +r A = DUHD = p o i ¥ l  r +r Torque, T = F ¥ o i  dy =

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The time measured in seconds to pass 60 cc of any liquid is called “second saybolt” ZKLFKKHOSVLQ¿QGLQJYLVFRVLW\E\

v=

 m = t t r  = t t

IRU t £  IRU t > 

 ‡ $ EDOO RI  PP GLDPHWHU LV GURSSHG LQWR D OLTXLG RI VS ZHLJKW  N1P. If the sp ZHLJKWRIPDWHULDORIEDOOLVN1PDQGYHORFLW\RIEDOOGRZQZDUGLVPPLQ WKHQ¿QGYLVFRVLW\RIWKHOLTXLG u terminal =

r  (rsolid ¥ g – rÀXLG ¥ g) 9m

 ¥  ¥ - 0.08 = [gsolid – gÀXLG] 9¥m 60 ¥– =

 

m=

 ¥  ¥ -6 [78 – 8] ¥ 9¥m  ¥  ¥ -  ¥  ¥ -

 

Ns m

 ‡ $Q RLO RI PHDQ GHQVLW\  NJIP IORZV XQGHU D KHDG RI  P WKURXJK  P RI SLSHRIFPGLDPHWHU2ZLQJWRFRROLQJYLVFRVLW\FKDQJHVDORQJWKHOHQJWKDQGPD\ EH WDNHQ DV  SHU VHFRQG RYHU WKH ¿UVW  P DQG  RYHU WKH VHFRQG  P Determine the flow in mVHFQHJOHFWLQJHQWUDQFHDQGH[LWORVVHV 3000 m

m2

m1

1500

0.3

1500

dP1 h = 30 d2P2 Head Variation

60

Fundamentals of Fluid Mechanics

Flow through a pipe per second,

Q=

pr 4 d r 8l m

dr =

or

Q ¥ 8l m pr 4

where dp SUHVVXUHGL൵HUHQFHDWERWKHQGV r = radius l = length m = viscosity Total head =

dP dP +  0 rg rg

Q ¥ 8 ¥ lm Q ¥ 8 ¥ l ¥ m  + =  pr 4 ¥ rg pr 4 ¥ rg Q= =

s ¥ P ¥ g ¥ pr 4 (lm + l m  )8

 ¥  ¥  ¥ p ¥  4   +  ¥ 

l = l = 

=  P V  ‡ ,QRUGHUWRFODPSWKHYLEUDWLRQVDGDVKSRWKDYLQJDSLVWRQDQGF\OLQGHUGLDPHWHUVRI DQGFPUHVSHFWLYHO\LVXVHG,IWKHSLVWRQZKLFKKDVOHQJWKRIFPVOLGHVLQRLO RIYLVFRVLW\SRLVHZLWKDYHORFLW\RIFPVHF¿QGWKHUHVLVWDQFHWRPRYHPHQW Cylinder

9.8

Piston

Oil

12 cm

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10 cm

du dy

 -  

 P

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t ¥

61

 

 1P

  Area, pdL = p ¥ 0.098 ¥

 P

   Resistance, F = t ¥ A      

 ¥  1

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Ideal plastic

Newtonian fluids Shear stress

Real plastic Ideal fluids

du dy

Rheology Diagram

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Fundamentals of Fluid Mechanics

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4cos q ¥ s rgd

s water =  1P \

 ¥  ¥   ¥  ¥  ¥ d  ¥  ¥  d=  ¥  - ¥  ¥  ¥  d =  ¥  -6 P h=

=  ¥  - PP  ‡ Discuss effect of temperature and pressure on the physical properties of the fluid.  8378   7KHH൵HFWRIWHPSHUDWXUHDQGSUHVVXUHRQWKHSK\VLFDOSURSHUWLHVDUH    Viscosity: Viscosity of liquids drops with increase of the temperature of the liquids. On the other hand, viscosity of gases rises with increase in the temperature of the gases. ,QOLTXLGVFRKHVLYHIRUFHVDUHSUHGRPLQDQWDVOLTXLGVKDYHFORVHO\SDFNHGPROHFXOHV The cohesive forces of the molecules in liquids decrease with increase in temperature. In gases, cohesive forces are small and the molecular momentum transfer predominates which increases with increase in temperature. Hence, viscosity of gases increases with WKHULVHRIWHPSHUDWXUH7KHYLVFRVLW\RIOLTXLGVDWDQ\WHPSHUDWXUHFDQEHJLYHQDV m=



m0 where m0, a & b are constants. On the other hand viscosity of gases  + at  + b t 

foUDQ\WHPSHUDWXUHFDQEHJLYHQDV m = m0a tbt   Density: Density of liquids may be considered constant. However, the density of gases varies as per the variation of pressure and temperature. The relationship is

PV = m RT V P = RT m P = RT r



The density varies inverse proportionally to temperature and proportionally to pressure.   Surface Tension: Surface tension of the liquids decreases with the increase of temperature.

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 ‡ )RUm SRLVHr JPFPNLQHPDWLFYLVFRVLW\LQVWRNHVLV  D   E    F   G   &LYLO6HUYLFHV u=

m 0.06 = = 0.067 r 0.9

Option (c) is correct.  ‡ $WURRPWHPSHUDWXUHWKHG\QDPLFDQGNLQHPDWLFYLVFRVLW\RIZDWHU  D  DUHERWKJUHDWHUWKDQDLU  E  DUHERWKOHVVWKDQDLU  F  DUHUHVSHFWLYHO\JUHDWHUWKDQDQGOHVVWKDQDLU  G  DUHUHVSHFWLYHO\OHVVWKDQDQGJUHDWHUWKDQDLU ,(6   7KHG\QDPLFYLVFRVLW\RIZDWHULVJUHDWHUWKDQDLUEXWLWVNLQHPDWLFYLVFRVLW\LVOHVVWKDQ air. Option (c) is correct.  ‡ 7KH VXUIDFH WHQVLRQ RI ZDWHU DW ƒ& LV  ¥ ± 1P7KH GLIIHUHQFH LQ WKH ZDWHU VXUIDFH ZLWKLQ DQG RXWVLGH DQ RSHQHQGHG FDSLOODU\ WXEH RI  PP LQWHUQDO ERUH LQVHUWHGDWWKHZDWHUVXUIDFHZRXOGEH  D  PP E  PP  F  PP G  PP ,(6 s

s

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s  ¥  ¥ - = rg ¥ d  ¥  ¥  ¥  ¥ -

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64

Fundamentals of Fluid Mechanics W sinq q

tan q = \

 

q ƒ Force = W sin q  ¥VLQ  1

     

du dy

Now shear stress = m ¥

=m¥  

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    m \ Force due to gravity = shear resistance  m

 Ns =    m ‡ $FPGLDVKDIWUXQVDWUSPLQDVOHHYHZLWKDUDGLDOGLVWDQFHRIPP,IWKH VOHHYH OHQJWK LV  FP ORQJ DQG WKH VSDFH LV ¿OOHG ZLWK DQ RLO RI G\QDPLF YLVFRVLW\ m NJPVGHWHUPLQHWKHWRUTXHUHVLVWDQFH 8378 \

m=

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1 mm

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 PV dy ¥± m t=m◊

du dy

¥

  ¥ -

= 4¥1P Force = t ¥ A ¥ ¥ pdl   

 ¥ ¥ p ¥ ¥±) ¥ ¥±)

  

 1

Torque = Force ¥5DGLXV 8 ¥

 ¥± 

 

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4s D  ¥   ¥ -

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E  1RQYLVFRXVDQGLQFRPSUHVVLEOH



F  1RQYLVFRXVDQGFRPSUHVVLEOH G  9LVFRXVDQGLQFRPSUHVVLEOH



 ,GHDOÀXLGKDVQRYLVFRVLW\DQGFRPSUHVVLELOLW\ Option (b) is correct.

,$6

66

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 D  μ d u  dy 

E  μ



⎛ ⎞ F  μ ⎜ du ⎟  ⎝ dy ⎠

du dy

G  μ ⎛ du ⎞ ⎜⎝ dy ⎟⎠

 



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⎛ du ⎞ ⎛ du ⎞ a) W  , t = μ ⎜ ⎟ , t = μ ⎜ ⎟ ⎝ dy ⎠ ⎝ dy ⎠

⎛ du ⎞ ⎛ du ⎞ b) W  , t = μ ⎜ ⎟ , t = μ ⎜ ⎟ ⎝ dy ⎠ ⎝ dy ⎠









F  W

⎛ du ⎞ ⎛ du ⎞ ⎛ du ⎞ μ⎜ ⎟,t = μ⎜ ⎟ ,t = μ⎜ ⎟ ⎝ dy ⎠ ⎝ dy ⎠ ⎝ dy ⎠





⎛ du ⎞ ⎛ du ⎞ G  W  μ ⎜ ⎟ , t = μ ⎜ ⎟ , t = 0  ⎝ dy ⎠ ⎝ dy ⎠



,$6

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du whLOHQRQ1HZWRQLDQÀXLGVKDYHVRPHSRZHURIYHORFLW\JUDGLHQWIRUVKHDUVtress dy

Option (b) is correct.  ‡ )OXLGVWKDWUHTXLUHDJUDGXDOO\LQFUHDVLQJVKHDUVWUHVVWRPDLQWDLQDFRQVWDQWVWUDLQ rate are known as  D  5KHGRSHWLFIOXLGV E  7KL[RWURSLFIOXLGV  F  3VHXGRSODVWLFIOXLGV G  1HZWRQLDQIOXLGV ,$6 

⎛ du ⎞  1HZWRQLDQÀXLGVKDYHDFRQVWDQWVWrain rate ⎜ ⎟ for a given shear stress (W) ⎝ dy ⎠ Option (d) is correct.

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 ‡ $WWKHLQWHUIDFHRIDOLTXLGDQGJDVDWUHVWWKHSUHVVXUHLV  D  +LJKHURQWKHFRQFDYHVLGHWRWKDWRQWKHFRQYH[VLGH  E  +LJKHURQWKHFRQYH[VLGHFRPSRUHGWRWKDWRQWKHFRQFDYHVLGH  F  (TXDOWRERWKVLGHV  G  (TXDOWRVXUIDFHWHQVLRQGLYLGHGE\UDGLXVRIFXUYDWXUHRQERWKHQGV ,$6 The pressure inside the air bubble is higher, i.e., concave side. Option (a) is correct.  ‡ :KHQ D ÀDW SODWH RI  P DUHD LV SXOOHG DW D FRQVWDQW YHORFLW\ RI  FPV SDUDOOHO WR DQRWKHUVWDWLRQDU\SODWHORFDWHGDWDGLVWDQFHFPIURPLWDQGWKHVSDFHLQEHWZHHQLV ¿OOHGZLWKDÀXLGRIG\QDPLFYLVFRVLW\ 1VPWKHIRUFHUHTXLUHGWREHDSSOLHGLV  D  1 E  1  F  1 G  1 ,$6 ⎛ du ⎞ W = m⋅ ⎜ ⎟ ⎝ dy ⎠ =  ×

 

 1P Force = W × A î  1

  

   Option (a) is correct.  ‡ &RQVLGHUWKHIROORZLQJVWDWHPHQWV   *DVHVDUHFRQVLGHUHGLQFRPSUHVVLEOHZKHQ0DFKQXPEHULVOHVVWKDQ   $1HZWRQLDQIOXLGLVLQFRPSUHVVLEOHDQGQRQYLVFRXV   $QLGHDOIOXLGKDVQHJOLJLEOHVXUIDFHWHQVLRQ   :KLFKRIWKHVHVWDWHPHQWVLVDUHFRUUHFW"   D  DQG           E  DORQH   F  DORQH           G  DQG ,$6 Option (d) is correct.  ‡ 0DWFK /LVW, SK\VLFDO SURSHUWLHV RI IOXLG  ZLWK /LVW,, GLPHQVLRQVGH¿QLWLRQV  DQG select the correct answer.        /LVW,        /LVW,, du is constant   D  $EVROXWHYHORFLW\   dy    

 E  .LQHPDWLFYHORFLW\  F  1HZWRQLDQÀXLG  G  6XUIDFHWHQVLRQ   

   

 1HZWRQSHUPHWUH  3RLVH  6WUHVVVWUDLQLVFRQVWDQW  6WRNHV

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Fundamentals of Fluid Mechanics

    

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or

dyne  dyne − s = cm = du FP  V cm  dy cm

τ

m

Option (c) is correct.  ‡ 7KHVKHDUVWUHVVGHYHORSHGLQOXEULFDWLQJRLORIYLVFRVLW\SRLVH¿OOHGEHWZHHQWZR SDUDOOHOSODWHVFPDSDUWDQGPRYLQJZLWKUHODWLYHYHORFLW\RIPVLV  D  1P E  1P 

F  

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70

Fundamentals of Fluid Mechanics

 7KHNLQHPDWLFYLVFosity v =



μ ρ

m = m × r  î–4 î î)    1VP Option (c) is correct.  ‡ 'HFUHDVHLQWHPSHUDWXUHLQJHQHUDOUHVXOWVLQ  D  DQLQFUHDVHLQYLVFRVLWLHVRIERWKJDVHVDQGOLTXLGV  E  DGHFUHDVHLQWKHYLVFRVLWLHVRIERWKOLTXLGVDQGJDVHV  F  DQLQFUHDVHLQWKHYLVFRVLWLHVRIOLTXLGVDQGGHFUHDVHLQWKDWRIJDVHV  G  DQGHFUHDVHLQWKHYLVFRVLWLHVRIOLTXLGVDQGDQLQFUHDVHLQWKDWRIJDVHV ,(6 The viscosity of water will increase with decrease of temperature as cohesion increases while the viscosity of air will decrease with decrease of temperature as transfer of molecular momentum decreases. Option (c) is correct.  ‡ Assertion (A): ,Q JHQHUDO YLVFRVLW\ LQ OLTXLGV LQFUHDVHV DQG JDVHV ZKLOH LW GHFUHDVHV with rise in temperature. Reason (R): 9LVFRVLW\LVFDXVHGE\LQWHUPROHFXODU IRUFHVRI FRKHVLRQDQGWUDQVIHURI PROHFXODUPRPHQWXPEHWZHHQIOXLGOD\HUVRIZKLFKLQOLTXLGVWKHIRUPHUDQGLQJDVHV WKHODWHUFRQWULEXWHWKHPDMRUSDUWWRZDUGVYLVFRVLW\  D  %RWK$DQG5DUHLQGLYLGXDOO\WUXHDQG5LVFRUUHFWH[SODQDWLRQRI$  E  %RWK$DQG5DUHLQGLYLGXDOO\WUXHEXW5LVQRWWKHFRUUHFWH[SODQDWLRQRI$  F  $LVWUXHEXW5LVIDOVH  G  $LVIDOVHEXW5LVWUXH    ,(6 The assertion (A) is false but reason (R) is true. Option (d) is correct. ‡ $VVHUWLRQ $  %ORRGLVD1HZWRQLDQIOXLG Reason (R): 7KHUDWHRIVWUDLQYDULHVQRQOLQHDUO\ZLWKVKHDUVWUHVVIRUEORRG  D  %RWK$DQG5DUHLQGLYLGXDOO\WUXHEXW5LVQRWWKHFRUUHFWH[SODQDWLRQRI$  E  %RWK$DQG5DUHLQGLYLGXDOO\WUXHEXW5LVQRWWKHFRUUHFWH[SODQDWLRQRI$  F  $LVWUXHEXW5LVIDOVH  G  $LVIDOVHEXW5LVWUXH ,(6   $VVHUWLRQ $ LVIDOVHDVEORRGLVDQRQ1HZWRQLDQÀXLG+RZHYHUUHDVRQ 5 LVWUXH Option (d) is correct.  ‡ ,IWKHUHODWLRQVKLSEHWZHHQWKHVKHDUVWUHVV W DQGWKHUDWHRIVKHDUVWUDLQLVH[SUHVVHG n

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⎛ du ⎞ F  W  μ ⎜ ⎟ , n !   ⎝ dy ⎠ n

⎛ du ⎞ G  W  τ  + μ ⎜ ⎟ , n   ⎝ dy ⎠     

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Chapter

3

HYDROSTATIC FORCES

KEYWORDS AND TOPICS        

HYDROSTATIC FORCE TOTAL PRESSURE PRESSURE INTENSITY CENTRE OF PRESSURE FREE SURFACE IMAGINARY FREE SURFACE PIEZOMETRIC SURFACE HORIZONTAL SURFACE

       

VERTICAL SURFACE INCLINED SURFACE CURVED SURFACE PRESSURE DIAGRAM STABILITY OF DAM SLUICE GATE LOCK GATE GRAVTIY & ARCHED DAM

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x

b

Expression for Centre of Pressure

x

h

Hydrostatic Forces

 

139

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142 4.

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\

bd 3 12 ¥ b ¥ d ¥ P

b2 12 P ‡ $UHFWDQJXODUJDWHRIPZLGWKDQGPKHLJKWLVORFDWHGLQDYHUWLFDOSODQHLQZDWHU DQGWKHRWKHUVLGHLVDLU)LQGWKHWRWDOSUHVVXUHDQGGHSWKRIFHQWUHRISUHVVXUHIURP WKHIUHHVXUIDFHLI  WRSHGJHRIWKHJDWHFRQFRLGZLWKIUHHVXUIDFHDQG  WRSHGJH LVDWGHSWKRIPIURPWKHIUHHVXUIDFH = P+

4 G C

G C

4 G

G C

C

2

2

Case-1

Case-2

Case  F = x ¥ A¥r¥ g F = x ¥ A¥r¥ g HHUH

x =

4 =  P A =  ¥  =  P 2 2

F =  ¥  ¥  ¥ 3 ¥   N1

4

146

Fundamentals of Fluid Mechanics

  &HQWUHRISUHVVXUHIURPWKHIUHHVXUIDFH I h=x+ G Ax bd 3 2 ¥ 43 IG = = =  12 12  h = 2+ ¥ =  +  =  P Case  F = x ¥ A¥r¥ g

x    \

F =  ¥  ¥  ¥ 3 ¥   N1 1RWH7RWDOSUHVVXUHKDVLQFUHDVHGZLWKGHSWK

h = x+ = 6+

IG Ax  ¥

=  +  =  P  ‡ 3URYHWKHGLVWDQFHEHWZHHQWKHFHQWUHRISUHVVXUH h DQGFHQWURLGRIDUHFWDQJXODU d SODWHORFDWHGYHUWLFDOO\KDYLQJZDWHUDWRQHVLGHDQGDLUDWRWKHUVLGHLV ZKHUHdLV  KHLJKWRIWKHSODWHDQGbLVZLGWKRIWKHSODWH

G

Air

G

C

d

C b

x =

d bd 3  IG =  DQGA = b ¥ d 2 12 h=x+

IG Ax

Hydrostatic Forces

=

 

bd 3 d bd ¥ ¥ 12 2 d + 6 2 ¥  +  = d 3

d + 2

d 2 d = 6 =

147

'LVWDQFH  h - x =

2 1 d d - d= 3 2 6

 ‡ )ORZLQDSLSHOLQHRIPLQGLDPHWHULVFRQWUROOHGE\DFLUFXODUJDWH7KH6*RIWKH IOXLGLV7KHSUHVVXUHDWWKHFHQWUHRIWKHJDWHLVN1P)LQG  WRWDOSUHVVXUH IRUFHRQWKHJDWHDQG  WKHGHSWKRIFHQWUHRISUHVVXUHIURPWKHIUHHVXUIDFH

_ x

_ h

4 Gate

Pipeline

Side view

PreVVXUHDW x  N1P2 \ or

P = x ¥r¥ g

 ¥ 3 = x ¥  ¥ 3 ¥  x =

  ¥ 

 P 

3UHVVXUHforce = x ¥ r ¥ g ¥ A =  ¥  ¥ 3 ¥  ¥ = 2261 kN

p  2 4

148

Fundamentals of Fluid Mechanics

  7KHGHSWKRIFHQWUHRISUHVVXUH

h = x+ IG =

IG Ax

p d 4 p ¥  4 = = p =  P 4 64 64

h =  +

  ¥ 

=  +  =  P  ‡ $WULDQJXODUSODWHLVXVHGDVJDWHDQGLWLVORFDWHGYHUWLFDOO\WRVWRSWKHIORZRIZDWHU DVVKRZQEHORZ)LQG  WRWDOSUHVVXUHDQG  GHSWKRIWKHFHQWUHRISUHVVXUH

x = +

2 ¥= += P 3

A=

1 ¥  ¥  =  P2 2

IG =

bh3 6 ¥ 33 = =  P 36 36

h = x+ = 4+

IG Ax  9¥4

    F = xArg =  ¥  ¥  ¥ 3 ¥   N1

Hydrostatic Forces

149

 ‡ $VTXDUHSODWHRIPVLGHLVXVHGDVJDWHWRVWRSWKHIORZRIZDWHUDVVKRZQEHORZ )LQG  K\GURVWDWLFIRUFHDQG  GHSWKRIWKHFHQWUHRISUHVVXUHIURPWKHIUHHVXUIDFH 3

_ x

_ h

G C

Side a P 'LDJRQDO d



2

\



Êdˆ Êdˆ ÁË 2 ˜¯ + ÁË 2 ˜¯

2

=4

or

2d2 = 16

or

d=

\

x = 3+

P



 2

  a 4 24   IG = = =  12 12 A = 2 ¥ P2

h = x+ = 4+

IG Ax   ¥ 

   P F = xArg =  ¥  ¥  ¥ 3 ¥   N1  ‡ $VTXDUHSODWHZLWKVLGHPDQGDKROHRIPGLDPHWHULVORFDWHGLQZDWHUYHUWLFDOO\ DVVKRZQEHORZ)LQG  WRWDOSUHVVXUHDQG  GHSWKRIFHQWUHRISUHVVXUH

a =  d = a VLQ  =

2¥3 2

=  P

150

Fundamentals of Fluid Mechanics

If A1 DUHDRIVTXDUHDQGA2 DLURIFLUFOHWKHQQHWDUHDA A = A1 ±A2 =  -

p  2 =  -  =  P 2 4

3

_ x 3

2

  1HWPRPHQWRILQHUWLD , I = I VTXDUH - I FLUFOH =

a 4 pd 4 12 64

=

 p ¥  12 64

 ±  



x = +

d =  +  =  P 2

7RWDOSUHVVXUHF = xArg =  ¥  ¥  ¥ 3 ¥  = 646 kN   7KHGHSWKRIFHQWUHRISUHVVXUH

h = x+

IG Ax

  ¥  =  +  =  P =  +

_ h

Hydrostatic Forces

151

 ‡ A vertical square having area 1 m ¥ 1 m submerged in water with layer edge 50 cm below the free surface. Find the distance of horizontal line from the upper edge of the square so that the forces on the upper and lower portions are equal on the square.

0.5 m F1

y

F2

Upper portion

(1) (2)

Horizontal line Lower portion

Guidance: We can solve the problem considering two rectangular surfaces at depth of 0.5 m and (0.5 + y) which have same total pressure force depending upon their areas. F1 on area 1 = x1rA1 g

yˆ Ê 3 = Á 0.5 + ˜ ¥ 1 ¥ 10 ¥ ( y ¥ 1) ¥ 9.81 Ë 2¯ F2 on area 2 = x2rA2 g

1- yˆ Ê ¥ 1 ¥ 103 ¥ [(1 - y ) ¥ 1] ¥ 9.81 = Á 0.5 + y + Ë 2 ˜¯ Now,

F1 = F2

yˆ Ê y + 1ˆ Ê 3 ¥ 1 ¥ 103 ¥ (1 - y )9.81 ÁË 0.5 + 2 ˜¯ ¥ 1 ¥ 10 ¥ ( y ¥ 1) ¥ 9.81 = Á 0.5 + Ë 2 ˜¯ 0.5 y +

y2 Ê y + 2ˆ ¥ (1 - y ) = Á Ë 2 ˜¯ 2

2 y2 + y = - y - y + 2

2 y2 + 2 y - 2 = 0 y2 + y - 1 = 0 y=

-1 ± 1 + 4 = 0.62 m 2

 ‡ A circular opening of 4 m diameter in a vertical wall of a water tank is controlled by a circular disc of 4 m diameter and which can rotate about its horizontal diameter. Find (1) total pressure on the disc, and (2) torque required to maintain the disc in vertical position, and when the free surface is 6 m from its centre.

152

Fundamentals of Fluid Mechanics

hinges

Thydrostatic

B

Tapplied 4

Guidance:7RWDOSUHVVXUHIRUFH F DFWVEHORZWKH&*DWFHQWUHRISUHVVXUH7KHWRUTXHWR VWRSWKHGLVFIURPURWDWLQJGXHWRFLVHTXDOWRK\GURVWDWLFWRUTXH = F ¥ h - x F = x ¥ A ¥ rg = ¥

p  2 ¥  ¥ 3 ¥  4

 N1

h = x+ = 6+

IG Ax

IG =

pd 4 p ¥ 44 = = 4p 64 64

4p 4p ¥ 6

   P \ \

h - x =  -  =  +\GURVWDWLFWRUTXH  h - x ¥ F =¥  1P DQWLFORFNZLVH

 +RZHYHUWRUTXHWREHDSSOLHGLQFORFNZLVHGLUHFWLRQ 1P  ‡ $VTXDUHGRRUZLWKVLGHPLVSURYLGHGLQWKHVLGHZDOORIDWDQNZKLFKLV¿OOHGZLWK ZDWHU:KDWIRUFHPXVWEHDSSOLHGDWWKHORZHUHQGRIWKHJDWHVRDVWRNHHSWKHKLQJHG GRRUFORVHG"7KHXSSHUHGJHRIWKHGRRULVDWPIURPWKHIUHHVXUIDFH hinge hinge 4

_ _ x h

A

A

G C

G C

P = applied force

a=1

a

_ h–4 _ 5–h

F P

Hydrostatic Forces

x = +

153

1 =  2

A=1¥1=1

F = x ¥ A¥r¥ g =  ¥  ¥  ¥ 3 ¥  =  N1 1 IG 12 =  + h = x+ Ax  ¥        1RZWDNLQJPRPHQWRIIRUFHVDFWLQJDWWKHJDWHIURPKLQJHGSRLQW A

F ¥ h -  = P ¥   -   P P=

 ¥  

 N1  ‡ $ SODWH RI VL]H  P ¥  P LV KLQJHG DW WKH ERWWRP DQG VXSSRUWHG DW PLGSRLQW E\ D SURSVRWKDWLWUHPDLQVLQFOLQHGLQZDWHUDWƒWRKRUL]RQWDODVVKRZQEHORZDVIXOO\ LPPHUVHG)LQGWKHFRPSUHVVLYHIRUFHLQWKHSURSLILWLVLQFOLQHGDWƒKRUL]RQWDOO\  -DGDYSXU8QLYHUVLW\ 4

A _ h

_ x

C 2 sin 60

G C

45° 60°

F y2 y1

B

x =

2 3 VLQ  = =  2 2

F = x ¥ A¥r¥ g

45° 60°

Fc sin 60

154

Fundamentals of Fluid Mechanics

=  ¥ ( ¥ ) ¥  ¥ 3 ¥   N1

 ¥ 3 ¥  2 I G VLQ q 12 =  + h = x+  ¥  ¥  Ax 2

=  +

 ¥  ¥  =  +   ¥  ¥ 

  y1 GLVWDQFHIURPWKHKLQJHWRWKHFHQWUHRISUHVVXUHDORQJWKHJDWHVXUIDFH = 2-

h VLQ 

= 2-

 ¥  13

 ±   y2 =

2 =1 2

 7DNLQJPRPHQWDERXWKLQJH B

F ¥ y1 = Fc VLQ ¥ y2 ZKHUHFc FRPSUHVVLYHIRUFHLQWKHSURS  ¥  = Fc ¥  ¥  \

Fc =

 ¥  

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Hydrostatic Forces

155

1 A _ h

h

3 3 sin 60

hinge

x

strut

1 60º 30º

Guidance 7KHÀDVKERDUGZLOOWULSZKHQWKHWRWDOSUHVVXUHIRUFHVWDUWVDFWLQJDERYHWKH KLQJH+HQFH h ≥ x -  ¥ VLQ  LVWKHFRQGLWLRQIRUWULSSLQJ  6XSSRVHWKHKHLJKWRIZDWHULVxWKHQ

h = CHQWUHRISUHVVXUH  x + x =

I G VLQ 2 q A¥ x

x x ¥1 =  x  A=  VLQ  3

Ê x ˆ 1¥ Á Ë VLQ  ˜¯ =  x 3 IG = 12 h =  x +

x 3 ¥ VLQ 2   x ¥  x

=  x +  x =  x   1RZIRUWULSSLQJRIÀDVKERDUG

h = x -  ¥ VLQ  

x = x± x  x=

 =  P 

F = x ¥ A¥r¥ g =

x ¥  x ¥  ¥ 3 ¥ g 2

=

 ¥  ¥  ¥  ¥ 3 ¥  2

 N1

156

Fundamentals of Fluid Mechanics

 ‡ A rectangular gate of size 6 ¥ 2 is hinged at the base at inclination angle of 60° with the horizontal. The gate is kept in position by a weight of 6 tonnes by an arrangement as shown below. Find the water level (x) when the gate is about to open. Neglect the weight of the gate. P 6 tonnes

2 90º

6

_ F h

6

x

F

sin 60 x0

y2

60º

h = x+

I G sin 2 q Ax 3

2 Ê x ˆ 2¥Á ¥ (sin 60) ˜ Ë sin 60 ¯ x = + x x 2 12 ¥ 2 ¥ ¥ sin 60 2

= 0.5 x +

2x 12

= 0.5 x + 0.167 = 0.667x Force (P) acting on top edge to hold the gate, P = 6 ¥ 1000 ¥ g = 58.86 kN y = x-

h 0.667 x = xsin 60 sin 60

= x - 0.767 x = 0.233 x F=

x x ¥2¥ ¥ 1 ¥ 103 ¥ g = 0.233 x 2 sin 60

During equilibrium, the moment from hinge is zero, i.e., SM = 0

F ¥ y = P¥6 11.28 x 2 ¥ 0.767 x = 58.86 ¥ 6

6

Hydrostatic Forces

x3 =

157

 ¥  =   ¥ 

x P  7KHJDWHZLOOWULSZKHQKHLJKWRIZDWHU x ≥  P   ‡ $ YHUWLFDO UHFWDQJXODU JDWH  P KLJK DQG  P ZLGH KDV ZDWHU WR D GHSWK RI  P DW XSVWUHDP DQG  P DW GRZQVWUHDP &DOFXODWH   WRWDO SUHVVXUH H[HUWHG DW XSVWUHDP DQGGRZQVWUHDPDQG  WKHUHVXOWDQWWRWDOSUHVVXUHDQGLWVORFDWLRQZLWKUHVSHFWWR WKHERWWRPHGJH 4

_ h1

x1 6 5

5 2

x2

2

F1 = x1 ¥ A1 ¥ r ¥ g =

 ¥  ¥  ¥ 3 ¥  2

 N1

h = x+

 = + 2

IG Ax 3

 ¥ ()   ¥  ¥ ¥  2

=  +

 6

 RUy1 ± P F2 = x2 ¥ A2 ¥ r ¥ g =  ¥  ¥  ¥  ¥ 3 ¥   N1

F1 y1

F2 y2

158

Fundamentals of Fluid Mechanics

h2 = x2 + = 1+

IG A2 x2

4 ¥1 12 ¥ 4 ¥ 1 ¥ 1

 RUy2 ±  5HVXOWDQWR = F1±F2  ± N1 Let RDFWDWGLVWDQFHyIURPWKHEDVH

R ¥ y = F1 y1 - F2 y2  ¥ y =  ¥  -  ¥  y= =

 -    

 P 

 +HQFHUHVXOWDQWIRUFHLVN1ZKLFKLVDFWLQJWRZDUGVGRZQVWUHDPDWPIURPWKH ERWWRPRIWKHJDWH

 ‡ 7KHSUHVVXUHRIDLURYHUDWDQNLVNJFP N1 7KHUHLVDQRSHQLQJLQWKH WDQNZLWKIODSRIPVTXDUHZLWKFHQWUHRIVTXDUHDWPIURPWKHIUHHVXUIDFH7KH IODSLVKLQJHGDWWKHFHQWUH)LQGWKHIRUFHPWREHDSSOLHGYHUWLFDOO\DWERWWRPHGJH RIWKHIODSDWSRLQWBWRKROGWKHZDWHU 1 Air

4

A

_ x

B

Flap

_A h

P

B

A

F

hinged P B

Hinged at centre



 6LQFHDLUKDVSUHVVXUHRINJIFP2 PRIZDWHUKHQFHLPDJLQDU\IUHHVXUIDFHLVP DERYHWKHDFWXDOIUHHVXUIDFH

Hydrostatic Forces

\

x   P

Force

F = x1 ¥ A ¥ r ¥ g

159

=  ¥ ( ¥ ) ¥  ¥ 3 ¥ g  N1   )RUWKHVTXDUHSODWHIURPIUHHVXUIDFH DFWXDO

I G VLQ 2 q Ax  ¥  ¥ VLQ 2  = 2+ 12 ¥ 1 ¥ 1 ¥ 2  = 2+  = 

h2 = x +

  1RZIRUFH F LVEDODQFLQJDJDLQVWWKHDSSOLHGIRUFH P DERXWWKHKLQJH \ SM IURPWKHKLQJH



 ¥

(h - x ) VLQ 

( - ) ¥  3

= P ¥ VLQ  ¥  = P¥

P=

1 ¥  2

 ¥  ¥  ¥   ¥ 

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F1

F2

h3 F3

_ h1

_ h2

_ h3

160

Fundamentals of Fluid Mechanics

Guidance 7KHJDWHGRFNLVGLYLGHGLQWKUHHSRUWLRQV h1h2 & h3 VRWKDWWKHWRWDOSUHVVXUH LQ HDFK SRUWLRQ LV HTXDO F1 = F2 = F3  DQG EHDPV DUH ORFDWHG DW WKH FHQWUHV RI SUHVVXUH JLYHQby h1  h2 h3 F = xArg =  ¥  ¥ b ¥ r ¥ g F1 =

F2 =

F3 =

NRZ

h12 ¥ b ¥ r ¥ g ZKHUHb ZLGWK 2

(h

2

- h12

2

2

(h

2 3

± h22 2

) brg

) b rg

F1 = F2 = F3 =

F 3

(

)

(

(

) (

)

h22 ± h12 h32 ± h22  h12 b rg = b rg = b rg b rg = 2 2 3 2

)

100 3

or

h 12 = h22 ± h12 = h32 ± h22 =

NRZSXW

h3 ZHJHWh1 Ph2 P

h1 =

h1 + 2

h = 1+ 2

IG A¥

h1 2 1 ¥ h13

12 ¥ h1 ¥ 1 ¥

=

h1 h1 + 2 6

=

2 h1 3

=

2 ¥  = P 3

h1 2

Hydrostatic Forces

161

  7KHSRVLWLRQRIUHVXOWDQWIRUFHF2FDQEHIRXQGRXWE\WDNLQJPRPHQWIURPWKHIUHHVXUIDFH

2

2 h1 + F2 ¥ h2 3 F F1 = F2 = 3

( F1 + F2 ) ¥ 3 ¥ h2

= F1 ¥

2F 2 F 2 F ¥ ¥ h2 = ¥ h1 + ¥ h 3 3 3 3 3 =

 ¥  ±  ¥  3

=

 ±  3  F1

h3 10 = h3

F2 F3

  1RZDJDLQWDNLQJPRPHQWIURPIUHHVXUIDFH

(F + F 1

2

+ F3 ) ¥

2 2 ¥ 10 = ( F1 + F2 ) ¥ ¥ h2 + F3 ¥ h3 3 3 3F 2 2F 2 F ¥ ¥  ¥ ¥  ¥ h3 = 3 3 3 3 3

or

h3 =

 -  3

 P

162

Fundamentals of Fluid Mechanics

 ‡ $FLUFXODUSODWHRIPGLDPHWHULVLPPHUVHGLQZDWHULQVXFKDZD\WKDWLWVJUHDWHVW DQG OHDVW GHSWK EHORZ WKH IUHH VXUIDFH DUH  P DQG  P UHVSHFWLYHO\ 'HWHUPLQH WKH WRWDOSUHVVXUHRQRQHIDFHRIWKHSODWHDQGSRVLWLRQRIWKHFHQWUHRISUHVVXUH 8378 q

2 q

4

G C G C 4

  7KHDQJOHRILQFOLQDWLRQ 

VLQq =

4-2 1 = 4 2

\ q = 30°

x VLQ  A=

pd 2 p ¥ 42 = 4 4

IG =

pd 4 64

h = x+

IG VLQ 2 q A+ x

pd 4 ¥ VLQ 2  pd 2 64 ¥ ¥3 4 16 ¥ 1 = 3+  ¥  = 3+

  F = xArg

p ¥ 42 ¥  ¥ 3 ¥  4  N1 = ¥

Hydrostatic Forces

163

 ‡ $QDQJXODUJDWH$%&LVGHVLJQHGWRWLSDXWRPDWLFDOO\ZKHQZDWHUULVHVDERYHFHUWDLQ OHYHOLQWKHWDQN7KHJDWHLVKLQJHGDWSRLQWBDQGABLVYHUWLFDOZKLOHBC P LV KRUL]RQWDO )LQG WKH KHLJKW h IURP WKH KLQJH B  ZKHQ ZDWHU ULVHV WR WLS WKH JDWH WR DOORZWKHZDWHUWRIORZRXW A F2 h

y1

C

B 1

F1 x1

Guidance :KHQ OHYHO h LQFUHDVHV WKH IRUFH F2 RQ YHUWLFDO SODWH LQFUHDVHV :KHQ

F2 y1 ≥ F1 x1 WKHJDWHZLOOWLS7DNHZLGWKRIJDWH P F1 = h ¥ AKRUL]RQWDO ¥ r ¥ g = h ¥  ¥  ¥  ¥ 3 ¥   K F2 = =

h ¥ AYHUWLFDO ¥ r ¥ g 2 h ¥ h ¥  ¥  ¥ 3 ¥  2

 h2 x1 = GLVWDQFH of CBIURPKLQJH  y1 = h - h = h -

1 =  2

2 1 h= h 3 3

F1 ¥ x1 = F2 ¥ y1 1 h ¥  = h 2 ¥ h 3 or

h2 = 3 h=

3  P

  7KHJDWHZLOODXWRPDWLFDOO\WLSZKHQZDWHUULVHVDERYHPIURPWKHKLQJHB

164

Fundamentals of Fluid Mechanics

 ‡ $ WDQN LV ¿OOHG XQGHU SUHVVXUH  N3D  ZLWK DQ RLO KDYLQJ 6*   ,W KDV EHHQ D FRYHURIVL]HP¥PKLQJHGDWWRSHGJH ƒWRKRUL]RQWDODQGKHOGLQSRVLWLRQ E\DIRUFHPDVVKRZQEHORZ)LQG  IRUFHPDQG  UHDFWLRQDWWKHKLQJHGSRLQW



*DXJHSUHVVXUHRIWKHRLO N3D 17 = h ¥ SG ¥ g or

h=

17  PRIRLO  ¥ 

x = +

 ¥ VLQ  2

  P F = x ¥ A ¥ r ¥ g =  ¥  ¥  ¥  ¥  N1  N1

h = x+

I G VLQ 2 q A¥ x

1 4 =  +  ¥  ¥  ¥   ¥  3 ¥



   P

  7DNLQJPRPHQWIURPWKHKLQJH SRLQWA P ¥ AB = F ¥ AC

AC =

P ¥ ¥

h -   = =  1 VLQ  2

Hydrostatic Forces

P=

165

 ¥  

 N1   7KHUHDFWLRQDWSRLQWA is R

R=F-P =  -  =  N1

 ‡ $WDQNZLWKVTXDUHYHUWLFDOVLGHRIPDQGGHSWKRIPFRQWDLQVZDWHUXSWRDKHLJKW RIPDQGRLORI6*XSWRPDVVKRZQLQWKH¿JXUH¿QG  WRWDOSUHVVXUHRQ WKHYHUWLFDOVLGHDQG  FHQWUHRISUHVVXUHIURPWKHIUHHVXUIDFH

Pressure Intensity

3UHVVXUHLQWHQVLW\DWA = 0 3UHVVXUHLQWHQVLW\DWB = hRLO ¥¥N1  ¥¥N1  N1 3UHVVXUHLQWHQVLW\DWC hZDWHU ¥¥ 1  ¥  N1

      

  ,QWKHSUHVVXUHLQWHQVLW\GLDJUDPDVVKRZQDERYHWKHYDOXHVDUH BD = CE N1 EF = CF±CE ± N1 1RZ SUHVVXUHIRUFHF1 DUHDRIDABD ¥ZLGWKRIWDQN \

F1 =

1 ¥  ¥  ¥  2

 N1  7KHIRUFHDFWVDWGHSWKh1IURPWKHIUHHVXUIDFH 

2 ¥  =  P 3

F2 DUHDRIWULDQJOHDEF ¥ZLGWKRIWDQN

166

Fundamentals of Fluid Mechanics

=

1 ¥  ¥  ¥  2

 N1 7KLVIRUFHDFWVDWGHSWKh3 =  +

1 ¥  =  P 2

7RWDOIRUFHF = F1F2F3 

   N1 7KHFHQWUHRISUHVVXUH h =

 ¥  +  ¥  +  ¥   P

3.6 HYDROSTATIC FORCES ON PLANE AND CURVED SURFACES  ‡ :KDWDUHWKHGL൵HUHQFHVRIK\GURVWDWLFIRUFHVDFWLQJRQDSODQHDQGFXUYHGVXUIDFH"   ,QFDVHRISODQHVXUIDFHVWKHGL൵HUHQWLDOIRUFHV dF DFWLQJRQHYHU\HOHPHQWDU\DUHDRIWKH VXUIDFHKDYHWKHVDPHGLUHFWLRQ+HQFHWKRVHIRUFHVIRUPDV\VWHPRISDUDOOHOIRUFHV'XH WRWKLVUHDVRQWKHPDJQLWXGHRIWRWDOIRUFHFDQGLWVSRLQWRIDSSOLFDWLRQFDQEHGHWHUPLQHG E\LQWHJUDWLRQPHWKRG

x

P

dF = dP ´ dA & dP = xrg F = ò dPdA = ò xrgdbx Area

P + dp



= rgxA ` (Note: Integration is possible)

 ,QFDVHRIFXUYHGVXUIDFHVDOOWKHHOHPHQWVRIWKHDUHDGRQRWOLHLQWKHVDPHSODQH$OWKRXJK WKHGL൵HUHQWLDOIRUFHVUHPDLQSHUSHQGLFXODUWRWKHLUUHVSHFWLYHHOHPHQWVRIDUHDEXWWKH\GRQRW IRUPDV\VWHPRISDUDOOHOIRUFHV'XHWRWKLVUHDVRQWKHPDJQLWXGHRIWRWDOIRUFH F DQGLWV ORFDWLRQ FHQWUHRISUHVVXUH FDQQRWEHGHWHUPLQHGE\WKHPHWKRGRILQWHJUDWLRQ+RZHYHU LQWHJUDWLRQFDQEHGRQHE\UHVROYLQJGL൵HUHQWLDOIRUFHVLQKRUL]RQWDODQGYHUWLFDOGLUHFWLRQV dFv dF

dF = normal force dfH = resolved horizontal force dfv = resolved vertical force, It can be upward or downward

dF H dF

dF H dFv Inside Curved

Outside Curved

Hydrostatic Forces

167

 ‡ +RZLVWRWDOIRUFH F DQGLWVORFDWLRQIRXQGRXWRQWKHFXUYHGVXUIDFH" 

8378±± C

D

dFv E

x

dF

B q

dFH A dA



 /HWDVPDOOHOHPHQWRIWKHFXUYHGVXUIDFHRIDUHDdAEHLQFOLQHGDWDQJOHqWRWKHKRUL]RQWDO 7KHGL൵HUHQWLDOIRUFHdFLVDFWLQJQRUPDOWRdA:HFDQUHVROYHdFDVXQGHU dFH +RUL]RQWDOFRPSRQHQW dFVLQq = PdFVLQq = xrgdAVLQq dFV 9HUWLFDOFRPSRQHQW dF cos q = PdA cos q = xrg dA cos q FH = rg Ú xdAVLQq FH = rg Ú xdA cos q  1RZdAVLQqDQGdA cos qUHSUHVHQWWKHSURMHFWLRQVRIWKHHOHPHQWDU\DUHDLQYHUWLFDODQG KRUL]RQWDOSODQHV+HQFHLWIROORZV



D  FH= rg Ú xdAVLQq = rgxAEA  3URMHFWHGDUHDRIFXUYHGVXUIDFHRQYHUWLFDOSODQH /HQJWK EA ¥ZLGWKRIWKHVXUIDFH

h= x+

IG ZKHUH h  &HQWUHRISUHVVXUHRISURMHFWHGDUHD EA ¥ZLGWKRIWKHVXUIDFH Ax

ThHSRLQWRIDSSOLFDWLRQRIWKLVIRUFHLVWKHFHQWUHRISUHVVXUHRIWKHSURMHFWHGDUHDRI WKHFXUYHGVXUIDFHRQWKHYHUWLFDOSODQH 

E  Fv = rg Ú xdA cos q = rg ¥ V

ZKHUH V YROXPH

168

Fundamentals of Fluid Mechanics



ZHLJKWRIOLTXLGLQSUR¿OHAEDCBA, i.e., over the curved surface up to free surface. The point of this force is vertically downward of the centroid of the liquid volume above the curved surface. Ê FV ˆ The resultant force F is then equal to FH 2 + FV 2 which acts at angle q = tan–1 Á Ë F ˜¯ H

 ‡ Find the resultant force due to water pressure acting on the curved surface AB (a quardant of a circle of radius 4 m) if width of the gate is unity.

The projected area of the curved surface on the vertical plane is a rectangle AA¢ B¢B as shown above. 4 x = 2+ = 2+2= 4m \ 2

h = x+

1 ¥ (4)3 IG IG = 12 Ax

A=4¥1=4

43 4 ¥ 4 ¥ 12 = 4.33 m = 4+

FH = xArg = 4 ¥ 4 ¥ 1 ¥ 103 ¥ 9.81 = 157 kN FH is acting at centre of pressure, h = 4.33 m FV = Weight of water supported by curved surface AB up to free surface = Weight in rectangular part (1) + Weight in curved surface (2) = 9.81 ¥ 2 ¥ 4 ¥ 1 ¥ 1 ¥ 103 + 9.81 ¥ 1 ¥

p 42 ¥ 1 ¥ 103 4

Hydrostatic Forces

  

169

   N1  N1  7KHSRLQWRIDSSOLFDWLRQRIFVFDQEHIRXQGE\9HULJQRQ¶VWKHRUHPE\WDNLQJPRPHQWRI ZHLJKWRIZDWHULQHDFKSRUWLRQ SRUWLRQ  IURPWKHSRLQWBZUWWKHLUFHQWURLGV7KLV PXVWEHHTXDOWRWKHZHLJKWRIWRWDOZDWHUZUWLWVFHQWURLGIURPWKHSRLQWB Y1

W1 B W2

Y2

W y = W1 y1 + W2 y2 4 ¥ 4ˆ Ê  ¥ y =  ¥  +  ¥ Á  Ë 3p ˜¯  ¥ y =  +   

y = F=

 =  IURP%  FH2 + FV2

=

()2 + ()2

=

 ¥ 4 +  ¥ 4

 N1 q WDQ±

FV   WDQ±  ƒ FH 

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170

Fundamentals of Fluid Mechanics 6

A

D

C

3

Fv

FH

Projected area on vertical

B

FH = xA ¥ rg where A is the projected area on the vertical plane.

x =

3 = 1.5 2

A = 3 ¥ 6 = 18 FH = 1.5 ¥ 18 ¥ 1 ¥ 103 ¥ 9.81 = 264.87 kN FV = Weight supported by the curved surface ACB = Volume ¥ r ¥ g =

1 Ê pd 2 ˆ ¥ ¥"¥r¥ g 2 ÁË 4 ˜¯

=

1 p ¥ 32 ¥ ¥ 6 ¥ 1 ¥ 103 ¥ 9.81 2 4

= 207.9 kN F= =

FH 2 + FV 2

(264.87)2 + (207.9)2

= 336.8 kN -1 q = tan

FV 207.9 = tan -1 = 38.2∞ FH 264.87

 ‡ A cylindrical roller gate of diameter 2 m and length 3 m is positioned as shown in the ¿JXUH,IWKHZHLJKWRIWKHJDWHLVN1¿QG  KRUL]RQWDOUHDFWLRQDWSRLQWADQG  YHUWLFDOUHDFWLRQDWSRLQWB D

C

FH

3 Projected area of

RA W

A RB

Fv B

2

roller

_ x

Hydrostatic Forces

171

FH = xArg =



2 ¥  ¥  ¥  ¥ 3 ¥  2

 N1 FV :HLJKWRIZDWHUVXSSRUWHGE\WKHFXUYHGVXUIDFHBCDLHKDOIDUHD   RIFLUFOH¥ZLGWK =

1 Ê p ¥ 22 ˆ ¥ ¥3¥r¥ g 2 ÁË 4 ˜¯

N1   $SSO\LQJWKHFRQGLWLRQVIRUHTXLOLEULXPRQWKHUROOHUJDWH SPx RA = FH N1 SPy W±FV = RB \ RB ±   N1  ‡ $ WDQN DV VKRZQ EHORZ FRQWDLQV D OLTXLG RI 6*  XS WR D GHSWK RI  P )LQG WKH PDJQLWXGHDQGGLUHFWLRQRIK\GURVWDWLFIRUFHSHUXQLWOHQJWKRIWKHWDQNRQWKHYHUWLFDO IDFHABFXUYHGVXUIDFHBCDQGKRUL]RQWDOERWWRPCD

9HUWLFDO)DFHAB

x =

1 =  2

$UHD AA1 B1 B = 1 ¥ 1 = 1 FAB = xArg =  ¥  ¥  ¥ 3 ¥   N1   &XUYHG6XUIDFHBC FH = xArg

172

Fundamentals of Fluid Mechanics

x = +

1 =    &HQWURLGRIBB1CC ZUWIUHHVXUIDFH 2

A = 1 ¥   $UHDRIBB1C1C FH =  ¥  ¥  ¥ 3 ¥   N1 FV ZHLJKWRIOLTXLGVXSSRUWHGRQBC  ZHLJKWRIOLTXLGLQYROXPH  YROXPH  



1 p ¥ 12 ¥ rg 4 3 =  ¥  ¥ ( + )

= (1 ¥ 1 ¥ 1) rg +

N1 F= =

2

(F ) + (F )

2

V

H2

()2 + ()2

=  +   N1 q = WDQ -1

FV  = WDQ -1 = ∞ FH 

 %RWWRP6XUIDFHCD

x =2 A=1¥1 FCD = xArg =  ¥  ¥  ¥ 3 ¥   N1  ‡ $F\OLQGULFDOUROOHUJDWHKDVGLDPHWHUPDQGOHQJWKP7KHZDWHUOHYHODWXSVWUHDP LVDERXWWRIORZRYHUWKHJDWHZKLOHDWGRZQVWUHDPLWLVP)LQG D PDJQLWXGHDQG GLUHFWLRQRIWKHUHVXOWDQWK\GURVWDWLFIRUFHDFWLQJRQWKHJDWH E WKHOHDVWZHLJKWRI WKHJDWHWRDYRLGWKHJDWHWREHOLIWHGE\K\GURVWDWLFIRUFHDQG F FKHFNZKHWKHUWKH UHVXOWDQWIRUFHSDVVHVWKURXJKWKHFHQWUHRIWKHF\OLQGHU 8SVWUHDP FH1 = xArg

Hydrostatic Forces

x =

2 =  2

A =  ¥  = 

FH1 =  ¥  ¥  ¥ 3 ¥   

h1 = 2 ¥  =  P IURPIUHHVXUIDFH 3 FV1 = rg ¥

FV1LVDFWLQJXSZDUGVDQGLVORFDWHG

1 p ¥ 2 ¥  =  N1 2

4r 4 ¥ 1 = =  P IURPSRLQWO 3p 3p

'RZQVWUHDP

FH 2 = xArg x =

1 =  P 2

A =  ¥  =  P 2

FH 2 =  ¥  ¥  ¥ 3 ¥   N1

h2 = 2 ¥  +  =  IURPIUHHVXUIDFH 3

FV2  :HLJKWRIWKHZDWHURQFXUYHGVXUIDFH

173

174

Fundamentals of Fluid Mechanics

=

1 ¥ p2 ¥  ¥  ¥ 3 ¥  4

 

FV2 ZLOODOVRDFWDWIURPSRLQWO 7RWDO FV = FV1 + FV2    N1



FH = FH1 - FH 2    1RZUHVXOWDQWIRUFH

±  N1 F= =

FH 2 + FV 2

()2 + ()2

 N1 q = WDQ -1

FV  = WDQ -1 = ∞ FH 

  7KHF\OLQGULFDOJDWHLVLQHTXLOLEULXP If or

SPy = 0

FV1 + FV2 = W W = FV N1

  7KHF\OLQGULFDOJDWHKDVWREHN1IRUWKHVWDELOLW\   1RZWDNHPRPHQWIURPSRLQWO

Hydrostatic Forces

175

SM = -  ¥  +  ¥  +  ¥  -  ¥  =   +HQFHWKHUHVXOWDQWRIDOIRUFHVSDVVHVWKURXJKWKHSRLQWO  ‡ ƒVHFWRUJDWHABCRIPUDGLXVRQWKHVSLOOZD\RIDGDPLVSODFHGDVVKRZQLQWKH ¿JXUH7KHJDWHLVKLQJHGDWCDQGACLVKRUL]RQWDODWWKHOHYHORIIUHHVXUIDFH,IWKH OHQJWKRIJDWHLVP¿QGWKHPDJQLWXGHDQGGLUHFWLRQRIWKHUHVXOWDQWSUHVVXUHRQWKH JDWH A

hinge

4m 60° 4

B

C

5

A

4 sin 60

B

Side View

Guidance 7KHFXUYHGVXUIDFHABZLOOKDYHK\GURVWDWLFIRUFHQRUPDOWRLWVVXUIDFHZKLFK FDQEHIRXQGRXWE\FDOFXODWLQJKRUL]RQWDODQGYHUWLFDOFRPSRQHQW FH = xArg =

VLQ  ¥ VLQ  ¥  ¥  ¥  ¥  2

= ¥¥



3 ¥  ¥  4

 N1 FV :HLJKWRILPDJLQDU\YROXPHRIZDWHURQWKHFXUYHGVXUIDFHAB   $UHDRIVHFWRUABC±$UHDRIDBDC ¥OHQJWK¥ r ¥ g

60∞ 1 Ê ˆ - CD ¥ DC ˜ ¥  ¥ r ¥ g = Á pr ¥ Ë ¯ 30∞ 2 1 Ê ˆ = Á  - ¥  ¥ ˜ ¥  ¥  ¥ 3 ¥  Ë ¯ 2 =  ¥  ¥  N1  N1 F= =

FH 2 + FV 2

()2 + ()2

176

Fundamentals of Fluid Mechanics

=

 ¥ 4 +  ¥ 4

=

 ¥ 4

 N1 -1 q = WDQ

FV  = WDQ -1 = ∞ FH 

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E

10

F

B

FH

4 mf

1

FH

2

A

C

Fv

Fv 12 m

D Side View

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x = + A=

pd 2 p ¥ 42 = = p =  P3 4 4

IG =

pd 2 p ¥ 42 = =  32 32

h = x+ = +



4 =P 2

IG A¥ x   ¥ 

 P FH +RUL]RQWDOIRUFHDFWLQJRQHDFKÀDWIDFH  RIWKHVXUIDFHDQGEDODQFHHDFKRWKHU

Hydrostatic Forces

177

= xArg =  ¥  ¥  ¥ 3 ¥   N1 FV 9HUWLFDOIRUFH FADCXSZDUGV±FABCGRZQZDUGV  :HLJKWRIZDWHULQDUHDADLFE ¥P  ±ZHLJKWRIZDWHULQDUHDABCFE ¥P  :HLJKWRIZDWHULQFLUFOHABCD ¥ 12

  

=

pd 2 ¥ r ¥ g ¥ 12 4

=

p ¥ 2 ¥  ¥ 2 ¥  ¥  4

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pd 2 p ¥ 22 = =  P 2 4 4

FH = xArg =  ¥  ¥  ¥ 3 ¥   N1



E

D Water

4m

C B

O A

4

C B

O B1

2

A

FV = FAB±FBC = :HLJKWRIZDWHURQFXUYHGVXUIDFHAB ±:HLJKWRIZDWHURQFXUYHGVXUIDFHBC 

 :HLJKWRIZDWHULQKHPLVSKHUH =

1 Ê 4 3ˆ ¥ pr ˜ ¥ r ¥ g ¯ 2 ÁË 3

178

Fundamentals of Fluid Mechanics

=

1 4 ¥ p ¥ 3 ¥  ¥ 3 ¥  2 3

 N1 

 6LQFH ZHLJKW RI ZDWHU RQ FXUYHG VXUIDFH ABED  LV PRUH WKDQ WKH ZHLJKW RI ZDWHU RQ FXUYHGVXUIDFH BCDE KHQFHQHWYHUWLFDOIRUFHRIN1ZLOOEHDFWLQJXSRQWRZDUGV IUHHVXUIDFH  ‡ $F\OLQGHURIVL]HPGLDPHWHUDQGRQHPHWUHOHQJWKVWD\VLQHTXLOLEULXPZKHQLWLV SODFHGLQDWDQNFRQWDLQLQJOLJKWRLO 6*  DQGKHDY\RLO 6*  DVVKRZQLQWKH ¿JXUH)LQGWKHGHQVLW\RIWKHPDWHULDORIWKHF\OLQGHU E

Light oil

D

A

FV

B

O

1

FV

W

2

C Heavy oil

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1 Ê ˆ = rgl Á DUHD AEDO - FLUFOH DUHD ˜ Ë ¯ 4 Ê 1 p ¥ 22 ˆ =  ¥ 3 ¥  ¥ Á ¥  - ¥ 4 4 ˜¯ Ë =  ¥  ¥ 3 ( - )  N1 GRZQZDUGV FV2 :HLJKWRIKHDY\RLOLQDUHDAEDC

1 Ê ˆ = rgl Á DUHD AEDO + FLUFOH DUHD ˜ Ë ¯ 4

Hydrostatic Forces

179

=  ¥ 3 ¥  ¥ ( + )  N1 XSZDUGV   1HWK\GURVWDWLFIRUFHDFWLQJXSZDUGV FV = FV2±FV2   ±   N1 XSZDUGV   'XULQJHTXLOLEULXPRIF\OLQGHUSPy = 0  :HLJKWRIF\OLQGHU ¥ 103

4 3 rr ¥ r PHWDO ¥ g ¥ 103 3 rPHWDO =

 ¥   ¥ p ¥ 3 ¥ 

 ¥ 103NJP3  ‡ $JDWHRIPVTXDUHOLHVLQDSODQHPDNLQJDQDQJOHRIƒ ZLWKWKHYHUWLFDO,WVXSSHU HGJHLVKRUL]RQWDODQGLVPEHORZWKHOLTXLGRI6* )LQGWKHUHVXOWDQWSUHVVXUH RQWKHJDWHDQGWKHORFDWLRQRIWKHFHQWUHRISUHVVXUH $0,(

x =  +  VLQ  

    F = xArg =  ¥  ¥  ¥  ¥ 3 ¥ 



 NN

1.6 SG = 1.5

F 30°

1 1

180

Fundamentals of Fluid Mechanics

h = x+

IG 1 ¥ 13 VLQ 2 q I G = 12 Ax

=  +

1 ¥  ¥  ¥  ¥ 

 PIURPIUHHVXUIDFH  ‡ $ VTXDUH SODWH RI VLGH  P LV LPPHUVHG YHUWLFDOO\ LQ ZDWHU VR WKDW DQ HGJH RI WKH VTXDUHPDNHVDQDQJOHRIƒZLWKWKHKRUL]RQWDO,IWKHKLJKHVWFRUQHURIWKHSODWHLV DWDGHSWKRIPEHORZWKHIUHHVXUIDFH¿QGWKHWRWDOSUHVVXUHRQWKHODPLQDDQG WKHGHSWKRIWKHFHQWUHRISUHVVXUH 3XQMDE8QLYHUVLW\

0.9

A 45°

B

D 0.6

0.6 C

x =  + AB VLQ  =  +  ¥  =  +  =  P A î  F = xArg =  ¥  ¥  ¥ 3 ¥   N1 IG =

a 4 4 = =  12 12

h = x+

IG Ax

  ¥    =  +

 P

Hydrostatic Forces

181

 ‡ $WDQNLV¿OOHGZLWKZDWHUXQGHUSUHVVXUH7KHSUHVVXUHJDXJHLQGLFDWHVSUHVVXUHRI  N3D 7DQN KDV OHQJWK RI  P 7KH WDQN KDV GHSUHVVLRQ RI D TXDUWHU RI FLUFXODU F\OLQGHU RI UDGLXV  P )LQG WKH KRUL]RQWDO DQG YHUWLFDO FRPSRQHQWV RI K\GURVWDWLF IRUFHRQWKLVFXUYHGVXUIDFH 15 kPa

B

Imaginary free surface

2m C

Projected area

D

1.53 A

2.5 B

The gauge reading = 15 kPa =

A

2

15 15 = = 1.53 m of water rg 1 ¥ 103 ¥ 9.81

2 = 2.53 m 2 A = 2 ¥ 2.5 = 5 m

x = 1.53 +

FH = xArg = 2.53 ¥ 5 ¥ 1 ¥ 103 ¥ 9.81 = 124.1 kN FV = Imaginary weight of water in ABCD

1 Ê ˆ = rg Á1.53 ¥ 2.0 + ¥ p ¥ 12 ˜ " Ë ¯ 4 = 1 ¥ 103 ¥ 9.81 ¥ (2.06 + 0.785) ¥ 2.5 = 51.31 kN F=

FH 2 + FV 2

=

(124.1)2 + (51.31)2

=

154 ¥ 104 + 0.26 ¥ 104

=

1.80 ¥ 104

= 134.16 kN -1 q = tan

FV 51.31 = tan -1 = 23.4 FH 124.1

182

Fundamentals of Fluid Mechanics

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A

1

A1

C 2

y= dy

x 16

Projected area on vertical plane

16

16 m

B

x

B

B

1

x 2 = 16y or

x=

16 y = 4 y

 &RQVLGHUWKHSURMHFWHGDUHDRIWKHGDPRQWKHYHUWLFDOSODQHDVVKRZQE\AA1 BB1 y=

16 = 2

FH = yArg 3 =  ¥  ¥  ¥  ¥  ¥ 

 N1 FV :HLJKWRIZDWHURQWKHFXUYHGVXUIDFHAB 16

= rg ¥ 1Ú xdy o

16

=  ¥ 3 ¥  ¥ Ú  ¥ ydy o

16

Ê y 32 ˆ =  ¥  ¥  Á 3 ˜ ÁË 2 ˜¯ o 3

=

3 2  ¥  16 ( ) ¥ 103 3

 N1 F= =

FH 2 + FV 2

()2 + ()2

Hydrostatic Forces

=

183

 +  4

 N1 F  =  = ∞ q = WDQ -1 V = WDQ -1 FH  

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1

A

B

9

y0 = 9

Projected area on vertical plane

dy æx æ y = y0 æx 0

2

x

O x0 = 6

æ

2

2

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x2 = 4y x= 2 y

y =

9 =  2

A=9¥1=9 FH = yArg =  ¥  ¥  ¥ 3 ¥   N1 FV :HLJKWRIZDWHULQOABIRUPHWHUOHQJWK = rg ¥ DUHD OAB ¥  9

= rg Ú xdy o

184

Fundamentals of Fluid Mechanics 9

= rg Úo 2 y dy 9

È 2 ¥ y 32 ˘ ˙ = rg ¥ Í Í 3 ˙ 2 ˚0 Î =  ¥ 3 ¥  ¥ =

3 4 ¥ 2 3

3 ¥  ¥  ¥  3

 N1 F= =

FH 2 + FV 2 2 + 2

 N1 WDQ q =



FV  = =  FH 

q WDQ± ƒ  ‡ $WDQNFRQWDLQVRLODQGZDWHUXSWRWKHKHLJKWRIPDQGPUHVSHFWLYHO\DVVKRZQ 7KH6*RIWKHRLOLV)LQGWKHUHVXOWDQWIRUFHRQVLGHABCLI6*RIRLOLVDQGWKH OHQJWKRIWDQNLVP A Oil

Imaginary water surface

10 B

8 C

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10 ¥  ¥  ¥  ¥ 3 ¥  2

 N1

Hydrostatic Forces

h1 = x1 +



185

IG 2 ¥ 103 =+ =  P  ¥  ¥  ¥  A1 x1

 1RZPRIRLOLVHTXLYDOHQWWRhPRIZDWHU \ h = 10 ¥6*RIRLO = 10 ¥ P  1RZIRUZDWHUOD\HUIURPLPDJLQDU\ZDWHUVXUIDFHZHFDQ¿QGRXW

x2 = h +  =  +  =  2 A 2 ¥ 2 = 16 FBC = x2 A2 ¥ rg =  ¥  ¥  ¥ 3 ¥   N1

h2 = x2 +

IG  ¥ 3 =  + =  +  =  12 ¥ 16 ¥ 12 A2 x2

F = FAB + FBC =  +   N1   7DNLQJPRPHQWIURPWKHKRUL]RQWDOVXUIDFHSDVVLQJSRLQWAZHKDYH 

3 hcp =  ¥  +  ¥  hcp =

 +  =  P IURP A 

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h1 = x1 +

IG 2 ¥ 63 = + =P 3 ¥ 12 ¥ 12 A1 x1

F2 = x2 A2 ¥ rg

186

Fundamentals of Fluid Mechanics 4m 3m 6m

CG

2m 6m

CG

1 ˆ Ê1 Ê ˆ 3 = Á  + ¥ ˜ ¥ Á ¥  ¥ ˜ ¥  ¥  ¥  Ë ¯ Ë ¯ 3 2 =  ¥  ¥  = N1

IG 2 ¥ 63 =+ =  P 36 ¥ 6 A2 x2

h2 = x2 +

F = F1 + F2 =  +   N1  7DNLQJPRPHQWIURPIUHHVXUIDFH

 ¥ hcp =  ¥  +  ¥  hcp =

 +  =  P 

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Hydrostatic Forces

187

O A B 0.5 m

q D

FH

FH

1m

C Water

SG = 0.75

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1 ¥  ¥  ¥  ¥ 3 ¥  2

 N1 FH2 = x2 ¥ A2 ¥ r ¥ g  GXHWRZDWHU =

 ¥  ¥  ¥  ¥ 3 ¥  2 N1

\ 



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=

1 ¥ p ¥ 2 ¥  ¥  ¥ 3 ¥  4

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QRWH WKDW VHFWRUKDV∞ DVFRV q = LV

60∞ 1 = RIWKHFLUFOH  360∞ 6

1 RUq = ∞ DQGDUHD 2

188

Fundamentals of Fluid Mechanics

1 Ê1 ˆ = rJ ¥ " Á pr 2 - ¥  ¥ VLQ ∞˜ Ë6 ¯ 2 Ê  ˆ 3 2 ¥ =  ¥  ¥  ¥  Á p ¥  2 2 ˜¯ Ë6 =  ¥ 3  -   N1 \

FV   N1 W =  ¥  =  N1

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6m F AB 60°

B

C

F BC

3m

Guidance: FABLVK\GURVWDWLFIRUFHDFWLQJRQABSDUWRIWKHJDWHZKLFKLVQRUPDOWRAB DQGWHQGVWRPRYHWKHJDWHFORFNZLVHZKLOHFBC LVK\GURVWDWLFIRUFHDFWLQJQRUPDOWRBC SDUWRIWKHJDWHWU\LQJWRPRYHWKHJDWHDQWLFORFNZLVH FAB = x1 A1rg

x =

6 = 3M 2

Hydrostatic Forces

AB =

189

6 =  VLQ 

Ê 6 ˆ ¥ 2 ¥ 1 ¥ 103 ¥ g FAB = 3 ¥ Á Ë VLQ  ˜¯ =

3¥ 6¥ 2 ¥  ¥  

 N1 3

h1 = x1 +

 ¥ () IG = + =  +  =  P x¥A  ¥ ( ¥ ) ¥ 

FBC =  ¥  ¥  ¥  ¥ 3 ¥   N1   ,WZLOODFWXSZDUGVDWPIURPB   7DNLQJPRPHQWIURPSRLQWB 

8QEDODQFHGPRPHQW

(

)

FAB AB - h1 - FBC ¥  =  ¥  -  -  ¥  =   -  ¥  =  -  =  N1P FORFNZLVH

 ‡ $ F\OLQGHU SOXJ RI  P GLDPHWHU LV XVHG LQ D UHFWDQJXODU WDQN  P ORQJ:LWK ZKDW IRUFHWKHF\OLQGHULVSUHVVHGDJDLQVWWKHERWWRPRIWKHWDQNGXHWRPRIZDWHU"

Guidance 7KHYHUWLFDOFRPSRQHQWRIWKHK\GURVWDWLFIRUFHRQWKHFXUYHGVXUIDFHZLOOSUHVV WKHF\OLQGHUDJDLQVWWKHERWWRPRIWDQN

190

Fundamentals of Fluid Mechanics

VLQq =



  = =  3 2

q = 30° FV BCD   :HLJKW RI WKH ZDWHU LQ BB1 D1 D CB 1

= :HLJKW RI WKH ZDWHU LQ UHFWDQJOH BB1 D1 D - KDOI FLUFOH

1 2ˆ Ê = rg " Á ¥  - p ˜ Ë ¯ 2 3 =  ¥  ¥  ¥   - 

= 9N1 GRZQZDUGV

FV1 AB DE = :HLJKW RI WKH ZDWHU RQ AB DE = :HLJKW RI WKH ZDWHU LQ DUHD ABB1 A1 A + DUHD EE1 D1 DE =

 ¥ r ¥ g ¥ " UHFWDQOJHBB1 A1 F + DUHD VHFWRURIFLUFOHIRU∞ - DOAF

=  ¥  ¥ 3 ¥  ¥ >  -  r - r FRV  +

pr 2 ¥

300 1 - ¥ r - r FRV  r VLQ @ 360 2

=  ¥  ¥ > ¥  -  ¥ FRV  + =

p ¥ 32 1 3 -  - FRV  ¥ @ 12 2 2

=  ¥   ¥  +  -  =N1 XSZDUGV F =± =N1 GRZQZDUGV  ‡ $ F\OLQGULFDO UROOHU RI  P GLDPHWHU LV SODFHG LQ ZDWHU DV VKRZQ LQ WKH ¿JXUH )LQG WKH KRUL]RQWDO DQG YHUWLFDO FRPSRQHQWV RI K\GURVWDWLF IRUFH GXH WR ZDWHU DFWLQJ RQ F\OLQGULFDOUROOHUSHUZDWHUOHQJWK

Hydrostatic Forces K

J

L

H 2m E

F

C O

45°

G

D H A

B 45°

FH )RUFHRQFXUYHGVXUIDFHCDA±)RUFHRQFXUYHGVXUIDFHAB   )RUWKLVZHKDYHWR¿QGWKHIRUFHVRQWKHYHUWLFDOSURMHFWLRQRIWKHDUHDRICDA FH = xAr g x = JE +

1 EA 2

1 ¥  + FRV  2 1 =  +  ¥ 2 =  P A = EA ¥ 1 = +

=  ¥  =  P 2 FH CDA    ¥  ¥  ¥ 3 ¥   N1 WRZDUGVOHIW 1RZIRUFXUYHGDUHDABWKHSURMHFWLRQRQWKHYHUWLFDOSODQHJLYHVXV

x = 4 + EH +

1 HA 2

=  +  ¥  +

1  -  2

=  +  +   P A = HA ¥ 1 =  ¥  =  P 2

191

192

Fundamentals of Fluid Mechanics

\

FH AB   xArG =  ¥  ¥  ¥ 3 ¥  =  N1 WRZDUGV ULJKW FH =  -  =  N1 WRZDUGV OHIW F V   

   

9HUWLFDOIRUFHDFWLQJRQWKHF\OLQGULFDOUROOHU XSZDUGIRUFHRQDAB±GRZQZDUGIRUFHRQDC ZHLJKWRI YROXPHDABHKD ±YROXPHDCLK ZHLJKWRI YROXPHDCLK YROXPHRIUHFWDQJOHCFHL  1  YROXPHWULDQJOHCFB FLUFOHYROXPHCDAB ±YROXPHDCLK 2  1RWH9ROXPHDLCKFDQFHOVRXW

1 1 Ê 2ˆ = rg ¥ l Á  ¥  + ¥  ¥  + p ¥  ˜ Ë ¯ 2 2 =  ¥ 3 ¥  ¥   +  +  =  ¥  N1 =  N1 XSZDUGV

3.7 PRESSURE DIAGRAM  ‡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

x

h

x=h

Horizontal

Vertical

Inclined

Hydrostatic Forces



193

 7KHSUHVVXUHGLDJUDPVRIDKRUL]RQWDOYHUWLFDODQGLQFOLQHGVXUIDFHKDYHEHHQVKRZQLQWKH ¿JXUH,WLVSRVVLEOHWRXVHWKHSUHVVXUHGLDJUDPWR¿QGWKHWRWDOSUHVVXUH K\GURVWDWLFIRUFH  DQGFHQWUHRISUHVVXUHIRUWKHVXEPHUJHGVXUIDFH,WLVVSHFLDOO\XVHIXOLQFDVHWKHVXUIDFH LVVXEPHUJHGLQWZRRUPRUHGL൵HUHQWOLTXLGV)ROORZLQJVWHSVDUHWREHWDNHQ



  'HWHUPLQHWKHSUHVVXUHDWERWWRPRIHDFKOLTXLGDVVKRZQLQ¿JXUHP1 = rgh1 P1 SUHVVXUHLQWHQVLW\h1LVKHLJKWRIOLTXLG



  'UDZWKHSUHVVXUHGLDJUDPZKLFKLVWKHKHLJKWRIOLTXLGRQxD[LVDQGSUHVVXUHLQWHQVLW\ RQyD[LVRUYLFHYHUVD   )LQG WRWDO SUHVVXUH F1  RI HDFK ÀXLG ZKLFK LV DUHD RI SUHVVXUH GLDJUDP ¥ OHQJWK RU ZLGWKRIWDQN F1 = A1 lZKHUHA1 DUHDRISUHVVXUHGLDJUDPDQGl OHQJWKRIWDQN





  7RWDOK\GURVWDWLFIRUFHLVWKHVXPRIWRWDOSUHVVXUHVRIHDFKOLTXLG F3 = F1F2F3F4



  &HQWUHRISUHVVXUHFDQEHIRXQGRXWE\XVLQJODZRIPRPHQW±

Fh = F1h1 + F2 h2 + F3 h3  O h1

S1

1

F1 = wt of vol (1) F2 = wt of vol (2) + (3)

P1 h2

2

S2

F3 = wt of vol (4) + (5) + (6)

P2 h3

4

S3 P3 O

5

6

h1 (h1 + h2) (h1 + h2 + h3) Pressure Diagram

Different Liquids

 ‡ $WDQNRIOHQJWKPFRQWDLQVDQLPPLVFLEOHOLTXLGDQGZDWHUIRUWKHGHSWKRIPDQG PDVVKRZQLQWKH¿JXUH)LQG    7RWDOSUHVVXUHRQWKHYHUWLFDOVLGHRIWKHWDQN    7KHORFDWLRQRIWKHFHQWUHRISUHVVXUH Liquid (SG = 0.8)

A height

1m

1 F

B 2

0.5 m C

P1 = 7.85

3 P2 = 12.76 D

E

Pressure

Water Fluids

Pressure Diagram

194

Fundamentals of Fluid Mechanics

3UHVVXUH,QWHQVLW\    $WERWWRPRIOLTXLG

P1 = r1 gh1 =  ¥ 3 ¥  ¥  =  N1P 2

   $WERWWRPRIZDWHU

P2 = P1 + rgh2 =  +  ¥ 3 ¥  ¥  =  +  =    +RUL]RQWDOIRUFHRQWKHVLGHRIWKHWDQN FH $UHDRISUHVVXUHGLDJUDP¥OHQJWKRIWKHWDQN   )URPWKHSUHVVXUHGLDJUDP

Ê1 ˆ F1 = Á ¥ P1 ¥ h1 ˜ ¥ l Ë2 ¯ 1 ¥  ¥  ¥  =  N1 2 F2 = P1 ¥ h2 ¥ l =  ¥  ¥  =

=  N1 1 F3 = ( P2 - P1 ) ¥ h2 ¥ 2 2 1 =  -  ¥  ¥  =  N1 2   +HQFHWRWDOK\GURVWDWLFIRUFHRQWKHYHUWLFDOVLGH

F = F1 + F2 + F3 =  +  +  =  N1 

 1RZWKHFHQWUHRISUHVVXUHFDQEHIRXQGRXWE\DSSO\LQJWKH ODZRIPRPHQW/HWWKHORFDWLRQ RIWKHFHQWUHRISUHVVXUHEH h IURPWKHIUHHVXUIDFH7DNLQJPRPHQWIURPIUHHVXUIDFH

F ¥ h = F1 ¥ h1 + F2 ¥ h2 + F3 ¥ h3 h1 LVFHQWUoid of DABFh2LVFHQWURLGRIUHFWDQJOHBCDFDQGh3LVFHQWURLGRIDDEF

Hydrostatic Forces

195

Ê ˆ Ê  ˆ Ê ˆ  h =  ¥ Á ¥ ˜ +  ¥ Á + ˜ +  ¥ Á ¥  + ˜ Ë3 ¯ Ë 2 ¯ Ë3 ¯ =  +  +   h= =  P IURP IUHH VXUIDFH   ‡ $ YHUWLFDO WDQN RI VTXDUH FURVV VHFWLRQ ZLWK VLGH ZLGWK   P DQG KHLJKW  P ,W LV FORVHG RQ DOO VLGHV ,W FRQWDLQV RLO 6*    WR GHSWK  P ZDWHU XS WR GHSWK  P DQGUHPDLQLQJKHLJKWLV¿OOHGZLWKDLUXQGHUSUHVVXUHRIPRIZDWHU)LQG  WRWDO K\GURVWDWLFIRUFHRQWKHYHUWLFDOVLGHDQG  ORFDWLRQRIWKHFHQWUHRISUHVVXUH)URP WKHERWWRPRIWDQN Air

Oil

P1 P1 1

2

2.5

2

height

0.5

2.5

2

P1 3 P1

4

5 P1

Water

P2 6 P2

Pressure Pressure Distribution

P1 PRIZDWHU SUHVVXUHGXHWRDLU =  ¥ r ¥ g =  ¥  ¥ 3 ¥  =  N1P 2 P2 3UHVVXUHLQWHQVLW\DWWKHERWWRPRIRLO = P1 +  ¥  ¥ 3 ¥ g =  +  =  N1P 2 P3 3UHVVXUHLQWHQVLW\DWWKHERWWRP = P2 +  ¥ r ¥ g =  +  ¥  ¥ 3 ¥  =  N1P 2  +RUL]RQWDOIRUFHRQWKHYHUWLFDOVLGHRIWDQNLV FH DUHDRISUHVVXUHGLDJUDP¥OHQJWKRIWKHWDQN

P3

196

Fundamentals of Fluid Mechanics

  )URPSUHVVXUHGLDJUDP

F1 = P1 ¥ A1 =  ¥  ¥  =  N1 F2 = P1 ¥ h1 ¥ l =  ¥  ¥  =  N1 1 1 P2 - P1 ) ¥ h1 ¥ l = ¥  ¥  ¥  =  N1 ( 2 2 F4 = P1 ¥ h2 ¥ l =  ¥  ¥  =  N1 F3 =

F = ( P - P ) ¥ h ¥ l =  ¥  ¥  =  N1 1 1 P3 - P2 ) ¥ h2 ¥ l = ¥  ¥  ¥  =  N1 ( 2 2 F = F1 + F2 + F3 + F + F + F F6 =

=  +  +  +  +  +  =  N1 

 &HQWUHRISUHVVXUHFDQEHIRXQGRXWE\DSSO\LQJODZRIPRPHQWIURPWKHERWWRPRIWKHWDQN

F ¥ h = Fh + F h + F h + F h + F h + F h = F + F + F ¥ hC + F h + F h + F h = ( +  + ) ¥

 Ê ˆ +  ¥ Á + ˜ Ë3 ¯ 2

  +  ¥ ¥  2 3  +  +  +  h=   =  =  P +  ¥

3.8 STABILITY OF DAM  ‡ :KDWGR\RXXQGHUVWDQGIURPWKHVWDELOLW\RIDGDP"

Hydrostatic Forces

197

The forces acting on a dam are: (1) horizontal component of hydrostatic force (FH DFWLQJWRUZDUGVOHIWDVVKRZQLQWKH¿JXUH FH = xArg

 

  



where x is centroid of water and ALVZHWWHGDUHDRIYHUWLFDOVLGH (2) The weight of masonry of the dam acting down (W)  7KHGDPKDVWREHGHVLJQHGVXFKWKDWLWPXVWKDYH   7KHIDFWRUVDIHW\DJDLQVWVOLGLQJGXHWRK\GURVWDWLFIRUFH FH FH KDVWREHFRXQWHUDFWHG E\WKHZHLJKWRIWKHGDPDQGFRH൶FLHQWRIIULFWLRQEHWZHHQWKHEDVHRIWKHGDPDQG WKHIRXQGDWLRQRIVRLO+HQFHZHKDYH  D 6OLGLQJUHVLVWDQFH m ◊W  E 6OLGLQJIRUFH FH  F )RUVWDELOLW\m ◊W > FH where m FRH൶FLHQWRIIULFWLRQ$VVHHQIURPWKHGLDJUDP m = tan aLHDQJOHRILQFOLQDWLRQRIWKHUHVXOWDQW)RUVWDELOLW\WKHDQJOHRIIULFWLRQ PXVWEHJUHDWHUWKDQWKHDQJOHRIIULFWLRQLHl > a  G )DFWRURIVDIHW\DJDLQVWVOLGLQJLV

FSsliding = 

6OLGLQJ UHVLVWDQFH m◊W 6OLGLQJ IRUFH FH

  7RWDOULJKWLQJPRPHQWPXVWEHJUHDWHUWKDQWKHWRWDORYHUWXUQLQJPRPHQW7KHULJKWLQJ PRPHQWLVJHQHUDWHGE\WKHZHLJKWRIWKHGDPZKLFKVKRXOGDFWDVQHDUWRKHHOVRWKDW PRPHQWDUPZUWWKHWRHLVODUJH7KHIDFWRURIVDIHW\DJDLQVWRYHUWXUQLQJ FSRYHUWXUQLQJ =

Total righting moment 7RWDO RYHUWXUQLQJ PRPHQW =



W ¥x FH ¥ y

  7KHPD[LPXPDQGPLQLPXPWHQVLRQGHYHORSHGDWWKHEDVHGXHWRHFFHQWULFLW\VKRXOG EHVDIH7RDYRLGWHQVLRQDWWKHEDVHWKHUHVXOWDQW R VKRXOGEHZLWKLQPLGGOHWKLUGRI WKHEDVH  ‡ $ FRQFUHWH GDP LV UHWDLQLQJ  P RI ZDWHU DV VKRZQ LQ WKH ¿JXUH 7KH ZHLJKW RI WKH masonry work of the dam is 30 kN/m2 )LQG   IDFWRU RI VDIHW\ DJDLQVW VOLGLQJ   IDFWRURIVDIHW\DJDLQVWRYHUWXUQLQJDQG  WKHSUHVVXUHLQWHQVLW\RQWKHEDVH7DNH m EHWZHHQWKHEDVHDQGWKHIRXQGDWLRQVRLO

198

Fundamentals of Fluid Mechanics 3 C

6m

(1)

D

8m

(2)

W1E W2 3

A

B

3 x2 x1



:HLJKWRIWKHGDP :HLJKWRIUHFWDQJOHAECD:HLJKWRIDEBD = w1w2

1 Ê ˆ = Á ¥  ¥  + ¥  ¥ ˜ ¥ 30 Ë ¯ 2    N1 If x1DQGx2DUHGLVWDQFHVRIw1DQGw2IURPWKHWRHSRLQWB7KHQ&*RIWKHFRPSOHWHGDP x LVJLYHQE\ w x + w2 x2 x = 1 1 w 2 x1 =  +  =  P x2 = ¥  =  P 3

x =

 ¥  +  ¥   +  = =  P  

FH = yArg =

6 ¥  ¥  ¥  ¥ 3 ¥  2

N1 I = y+ G Ay = 3+

1 ¥ 63 12 ¥ 6 ¥ 1 ¥ 3

 P 

2YHUWXUQLQJPRPHQW  FH ¥ y  ¥ 4  N1P FORFNZLVH

Hydrostatic Forces



5LJKWLQJPRPHQW  w ¥ x  ¥  N1P DQWLFORFNZLVH

  7KHGDPLVVWDEOHDVULJKWLQJPRPHQW!RYHUWXUQLQJPRPHQW

5LJKWLQJ PRPHQW 2YHUWXUQLQJ PRPHQW  =  =  6OLGLQJ UHVLVWDQFH = 6OLGLQJ IRUFH

FSRYHUWXUQLQJ =

FSVOLGLQJ

=

mw FH

=

 ¥  

=  F= =

2

(w)2 + ( FH )

()2 + ()2

=  +  ¥ 4 =  N1 If xRLVWKHGLVWDQFHIURPWKHWRHIURPZKHUHRSDVVHVWKURXJKWKHEDVH

ÂM

5LJKWLQJ PRPHQW - 2YHUWXUQLQJ PRPHQW R R  -  =  3263 = =  P 

xR =

A

  (FFHQWULFLW\ froPFHQWUH 

=

6 6 -  =  P < =  2 6

 +HQFHWKHUHVXOWDQWOLHVZLWKLQPLGGOHWKLUGRIWKHEDVH

199

200

Fundamentals of Fluid Mechanics W

F

A

B 3

3

3.67

 7KHSUHVVXUHLQWHQVLW\YDULHVGXHWRHFFHQWULFLW\ P=

w w ¥  -  ¥  ¥  ± A 63

=

  ¥  ¥  ¥  ± 6 ¥1 63

=  ±  N1P 2 3UHVVXUHLQWHQVLW\DWKHHO SRLQWA    N1P2  3UHVVXUHLQWHQVLW\DWWRH SRLQWB  ±   N1  ‡ $FRQWDLQHURIPOHQJWKLV¿OOHGZLWKOLTXLGKDYLQJ6* XSWRWRSRILWDVVKRZQLQ WKH¿JXUH)LQG  IRUFHVDFWLQJRQABBCCD & AD  FHQWUHRIJUDYLW\IRUVXUIDFH  GUDZSUHVVXUHGLVWULEXWLRQGLDJUDPIRUVXUIDFHABBCCDDQGAD

0.2 1.5 m A

B

1m 60º D

C

E 1.5 m

PAB 3UHVVXUHLQWHQVLW\DWOHYHOAB = x rg  ¥¥ 103 ¥  N1P2

  )RUFHRQVXUIDFHAB

FAB = PAB ¥ ab ¥ 1 AB = CD±DE 

Hydrostatic Forces

   \

 ±¥FRW±  ±±  P FAB = ¥ N1

)RUFHRQVXUIDFHBC

x =  +

1 = 2

PBC = xArg =  ¥  ¥  ¥  ¥ 3 ¥   N1 )RUFHRQVXUIDFHCD

x =  +  =  P PCD = xArg =  ¥  ¥  ¥  ¥ 3 ¥   N1 )RUFHRQVXUIDFHAD

x =  + A = ¥

1 =P 2

1 =  P 2 VLQ 

F = xArg =  ¥  ¥  ¥ 3 ¥   N1   3UHVVXUHGLDJUDPRQAB A

≠ P ≠ P ≠ P ≠ P ≠ P ≠ P ≠B P N1P2

3UHVVXUHGLDJUDPRQBC 11.77

B

C 7.85

11.77

201

202

Fundamentals of Fluid Mechanics

PB N1P2 PC =  + =  +

1¥ r ¥ g 103  ¥  ¥ 3 ¥  103

=  +  =  3UHVVXUHGLDJUDPRQCD D

Ø Ø Ø Ø Ø Ø Ø PCD

3 PCD = x rg =  ¥  ¥  ¥ 

 N1 3UHVVXUHGLDJUDPRQAD P  PD = PA +

x rg 103

=  +

 ¥  ¥ 3 ¥  103

=  +  =  N1 A 11.77 PA

D 11.77 7.85

PD

 ‡ $ZRRGHQF\OLQGULFDOVWRUDJHYDWKDVPGLDPHWHUDWRXWVLGH,WLV¿OOHGZLWKDOFRKRO XS WR  P KHLJKW$OFRKRO KDV 6*   7KH ZRRGHQ SODQNV RI WKH YDW DUH EUDFHG ZLWKIODWVWHHOEDQGVPPZLGHDQGPPZLGHZLWKDOORZDEOHVWUHVV¥N1 P)LQGWKHVSDFLQJRIWKHEDQGVQHDUWKHERWWRPRIWKHYDW

FH = xArg =  ¥  ¥  ¥  ¥ 3 ¥  =  N1

Hydrostatic Forces

203

1

8

FH

FH

8

T 6

Band Planks of vat

  1RZLQFDVHWKHUHDUHnEDQGVWKHQ n ¥ 2T = FH Where TLVIRUFHLQHDFKEDQG s ¥$UHD =  ¥ 4 ¥  ¥  ¥ -6 = 16 kN n ¥ 2T = FH n ¥ 2 ¥ N1

 32  

n=

6SDFLQJ 

  = n +  

 FP

3.9 GATE, AND LOCK GATES  ‡ :KDWDUHWKHHQJLQHHULQJVWUXFWXUHVZKLFKDUHGHVLJQHGWRZLWKVWDQGK\GURVWDWLFIRUFHV"   7KHHQJLQHHULQJVWUXFWXUHVZKLFKDUHPHDQWWRKROGZDWHURUFRQWUROWKHÀRZRIZDWHU FDQEHJURXSHGDVXQGHU    *DWHV *DWHV DUH XVHG LQ ULYHUV GDPV SLSHV DQG FDQDO ORFNV HLWKHU WR FRQWURO WKH ÀRZ RI WKH ZDWHU RU WR PDLQWDLQ WKH OHYHO RI WKH ZDWHU7KH K\GURVWDWLF IRUFHV DFWLQJ RQWKHVHJDWHVFDQEHRQRQHVLGHRUERWKVLGHVGHSHQGLQJXSRQWKHOHYHORIZDWHUDW XSVWUHDP DQG GRZQVWUHDP ,I WKHUH LV QR ZDWHU OHYHO DW GRZQVWUHDP WKHQ K\GURVWDWLF IRUFHVZLOODFWRQRQHVLGHLHVWDWLFVLGHRQO\7KHJDWHVFDQEH  VLPSOHJDWH   VOXLFHJDWH  WDLQWHUJDWHRU  UROOHUJDWH$VOXLFHJDWHLVXVHGWRFRQWUROWKHÀRZ RIZDWHUE\PRYLQJLWLQDYHUWLFDOSODQHWKURXJKWKHJURRYHVRUUDLOVSURYLGHGRQWKH PDVRQU\RXWOHWV$VOXLFHJDWHFRQVLVWVRIWZRYHUWLFDOSODWHVFDOOHGVNLQSODWHVLQZKLFK KRUL]RQWDOIEHDPVDUHSURYLGHG7KHIEHDPVDUHORFDWHGFORVHO\DWERWWRP WRVWDQG PRUHK\GURVWDWLFIRUFH DVFRPSDUHGWRWKHWRSRIWKHJDWH7KHJDWHPRYHVXSDQGGRZQ ZLWKWKHKHOSRIUROOHUV¿[HGRQWKHVNLQSODWHV7KHUROOHUVWUDYHORQWKHYHUWLFDOUDLOV JXLGHV ZKLFKDUH¿[HGRQWKHSLHUVRUYHUWLFDOZDOOVRIWKHRXWOHWV

204

Fundamentals of Fluid Mechanics Cable to move sluice gate up & down

Rail Vertical wall of the outlet

Upstream

Downstream

Sluice gate

Sluice Gate (to Control Flow)



 $ORFNJDWHLVXVHGWRFKDQJHWKHZDWHUOHYHOLQWKHFDQDOE\FRQVWUXFWLQJDORFNIRUWKH SXUSRVHRIQDYLJDWLRQ:KHQDGDPLVFRQVWUXFWHGRQDFDQDORUULYHUWKHZDWHUOHYHO RQWKHWZRVLGHVRIWKHGDPZLOOEHGL൵HUHQW,QRUGHUWRSHUPLWERDWVWRPRYHGRZQ IURPWKHXSSHUOHYHORIZDWHUGXHWRGDPWRWKHORZHUOHYHORIZDWHURUYLFHYHUVDD FKDPEHUFDOOHGµORFN¶LVFRQVWUXFWHGEHWZHHQWKHVHWZROHYHOVRIZDWHUDVVKRZQLQWKH ¿JXUH7ZRVHWVRIORFNJDWHVDUHSURYLGHGWRFRQWUROWKHZDWHUOHYHOLQWKHORFN Lock gate (closed)

Lock gate (open)

Downstream lock gate

Length of lock

Lock gate (closed)

Boat Transfer to Upstream

q

Width of lock

Lock and Lock Gates



 ,QRUGHUWRSHUPLWDERDWIURPWKHORZHUOHYHOWRKLJKHUOHYHOLILV¿UVWDGPLWWHGLQWRWKH ORFNE\RSHQLQJWKHGRZQVWUHDPORFNJDWHDQGWKHQFORVLQJLW1RZWKHXSVWUHDPORFN

Hydrostatic Forces





205

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0.5 5m 6m

5m

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Width of lock

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206

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h=x+

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 2 =  P

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(F H)up (F H)down

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5

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208

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15 m

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Hydrostatic Forces

209

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210

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211

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=

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212

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215

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p × 12 ¥1 2

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m=

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 î3 î î  

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h F

P Piston

\

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P = g◊h F =P×a = g◊h × a

Force

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A

B 4r CP s = 3p

=h

 × π kN 2

r  ×     = = P 3π 3π 3π

 0RPHQWRIWKUXVWDERXWAB m = F ×h =

 π  × 2 3π

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E  r

D  r F 

225

  π

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π  

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 +RUL]RQWDOFRPSRQHQWRIWKUXVW PH = ρ⋅ g ⋅ A ⋅ x ⎛ πr 2 ⎞ 4r  rg  ⎜ × ⎝ 2 ⎟⎠ 3π =

2 ρg ⋅ r 3 3

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π ρg × r 3 3

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  +\GURVWDWLFIRUFHDWODUJHUYHUWLFDOVXUIDFH FV = ρg A ⋅ x 

 



 rg  x◊2x ◊

2x 2

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 FBFV =

 rg x3  ρg x3 =1  ρg x3

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N SDVVLQJWKURXJKSRLQWEHWZHHQ0DQG& P

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Hydrostatic Forces

227

Fy 9HUWLFDOFRPSRQHQWPfK\GURVWDWLFIRUFHRQBC = ρ⋅ g ⋅ A ⋅ h ⎛ AB + BD ⎞ 3 =  ×  × DC ×  × ⎜ ⎟⎠ ⎝ 2 ⎛   ⎞ ⎛  +  ⎞ 3 × ⎟ × ⎜ =  ×  × ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ îJî31P

 

FV 3URMHFWHGDUHDRIBCî GHSWKRIFHQWUHRIBC îr × g  ⎞ ⎛ 3 =  ×  × ⎜  + ⎟ ×  × g ⎝ 2⎠ îg × 1031P

 

FV2 + FH2

F =

3 2 2 = g ×   + 

  

J1P

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2 rd of BDLHDSRLQWEHWZHHQ 3

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F = ρ⋅ g ⋅ A ⋅ x = 3 ×  × = 10 kN

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π 2 − 2 ×  4

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N P

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Hydrostatic Forces

     ‡

229

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h  

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h  

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IG −h Ah

E 

IG + Ah  Ah

F 

Ah +h IC

G 

IG + Ah  h

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IG +h Ah

E 

IG A h

F 

IG ⋅ h +h A

G 

Ah +h IG

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230

Fundamentals of Fluid Mechanics

The second moment of area, i.e., IG does not depend on angle q. Reason (R) is false. Option (c) is correct.  ‡ A triangular dam of height hLV¿OOHGWRLWVWRSZLWKZDWHUDVVKRZQLQWKH¿JXUH7KH condition of stability

Fy

Dam in masonry (SG = 2.56)

h

0

b



D  E K (c) b =

E  E K 3h 

G  E K

,(6

The hydrostatic force FH is acting horizontal to topple the dam while weight W is acting to prevent this toppling about point O. Taking moment about point O, We get FH ×

(ρ⋅ g ⋅ h) ×

or

h 2h =W× 3 3 h ⎛ bh ⎞ 2b × 1⎟ × = (2.56 × ρ) × g × ⎜ ⎝ 2 ⎠ 3 3

or b ª 0.425 h Option (d) is correct.  ‡ :KDWDFFHOHUDWLRQZRXOGFDXVHWKHIUHHVXUIDFHRIDOLTXLGFRQWDLQHGLQDQRSHQWDQN PRYLQJLQDKRUL]RQWDOWDQNWRGLSE\ƒ" (a)

g 2

(b) 2g

(c)

3 ⋅g 2

(d)

3 ⋅g  2

,(6

Option (c) is correct.  ‡ 7KHFHQWUHRISUHVVXUHRIDSODQHVXEPHUJHGVXUIDFH  D  LV D SRLQW RQ WKH VXEPHUJHG DUHD DW ZKLFK WKH UHVXOWDQW K\GURVWDWLF IRUFH LV VXSSRVHGWRDFW  E  VKRXOGDOZD\VFRLQFLGHZLWKLQWKHFHQWUHRIVXEPHUJHGDUHD

Hydrostatic Forces

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h = x+

I G ⋅ VLQ 2 θ A⋅ x

231

,(6

or h > x

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232

Fundamentals of Fluid Mechanics 2D

D

W2



D  W =

W  

F  W W

W1

E  W = W G  W W

  3UHVVXUHLVVDPHLQERWKF\OLQGHUV P=

P=

\



W1 4W = IRUWKH¿UVWF\OLQGHU 2 πD πD 2 4 W2 W = 22 for the VHFRQGF\OLQGHU 2   D πD 4 4W1 W2 2 = πD πD 2

or W2 = 4W1  2SWLRQ G LVFRUUHFW

Chapter

4

BUOYANCY AND FLOATATION

KEYWORDS AND TOPICS         

BUOYANCY BUOYANCY FORCE CENTRE OF BUOYANCY FLOATATION PRINCIPLE OF FLOATATION CONDITIONS OF STABILITY STABLE EQUILIBRIUM UNSTABLE EQUILIBRIUM NEUTRAL EQUILIBRIUM

        

STABILITY FOR SUBMERGED BODY STABILITY FOR FLOATING BODY METACENTRE METACENTRIC HEIGHT ROLLING PITCHING TIME PERIOD RADIUS OF GYRATION BILGE WATER

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234

Fundamentals of Fluid Mechanics

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dV rgV

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Buoyancy and Floatation

235

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h

h1

h dA

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Horizontal Prism

Forces Acting on Submerged Body

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V1 V2

r3 V3

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236

Fundamentals of Fluid Mechanics

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Chapter

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KEYWORDS AND TOPICS       

STATIC BODY RELATIVE EQUILIBRIUM TRANSLATION ACCELERATION HORIZONTAL ACCELERATION VERTICAL ACCELERATION INCLINED FREE SURFACE CONSTANT PRESSURE LINE

      

D’ALEMBERT’S PRINCIPLE ROTATIONAL ACCELERATION INCLINED ACCELERATION PARABOLOID FREE SURFACE FORCED VORTEX VERTEX AND RISE OF LIQUID IN VORTEX ROTATION OF CLOSED VESSEL

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5.3 TRANSLATIONAL ACCELERATION  ‡ How can a fluid be subjected to horizontal translational acceleration?   7KH ÀXLG FDQ EH VXEMHFWHG WR WUDQVODWLRQDO DFFHOHUDWLRQ ZLWKRXW UHODWLYH PRWLRQ EHWZHHQ WKHÀXLGSDUWLFOHVE\PRYLQJWKHYHVVHOFRQWDLQLQJWKHÀXLGZLWKKRUL]RQWDODFFHOHUDWLRQ$ ÀXLGLQYHVVHOWUDQVSRUWHGE\DYHKLFOHLVVXEMHFWHGWRKRUL]RQWDOWUDQVODWLRQDODFFHOHUDWLRQ DVVKRZQLQWKH¿JXUH Fluid is sloped

Acceleration =a

Vehicle of Rest

Vehicle While Accelerating

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302

Fundamentals of Fluid Mechanics

Vessel with fluid Fluid remains horizontal

Acceleration = a

Rocket Before Launch

Rocket After Launch

5.4 ROTATIONAL ACCELERATION  ‡ How is a fluid subjected to rotational acceleration?   7KH ÀXLG FDQ EH VXEMHFWHG WR URWDWLRQDO DFFHOHUDWLRQ ZLWKRXW UHODWLYH PRWLRQ EHWZHHQ WKH ÀXLGSDUWLFOHVE\URWDWLQJWKHYHVVHOFRQWDLQLQJWKHÀXLGE\H[WHUQDODJHQF\

Fluid at Rest

     

Fluid Subjected Tp Rotational Acceleration

‡ 'HSHQGLQJXSRQWKHW\SHRIDFFHOHUDWLRQZKDWDUHWKHGL൵HUHQWHTXLOLEULXPVZKLFKD fluid may have?  7KHÀXLGFDQKDYHWKHIROORZLQJUHODWLYHHTXLOLEULXPV D  +RUL]RQWDOUHODWLYHHTXLOLEULXP E  9HUWLFDOUHODWLYHHTXLOLEULXP F  5RWDWLRQDOUHODWLYHHTXLOLEULXP ‡ What are the conditions for the relative equilibrium?  7KH FRQGLWLRQV IRU WKH UHODWLYH HTXLOLEULXP DUH   WKH ÀXLG LV IUHH IURP VKHDU IRUFH   QRPRWLRQEHWZHHQWKHÀXLGSDUWLFOHVDQG  QRPRWLRQEHWZHHQWKHÀXLGDQGWKHYHVVHO FRQWDLQLQJÀXLG

5.5 D’ALEMBERT’S PRINCIPLE  ‡ What is D’Alembert’s principle?   '¶$OHPEHUW¶VSULQFLSOHVWDWHVWKDWDPRYLQJÀXLGPDVVPD\EHEURXJKWWRDVWDWLFHTXLOLEULXP SRVLWLRQ E\ DSSO\LQJ DQ LPDJLQDU\ LQHUWLD IRUFH RI WKH VDPH PDJQLWXGH DV WKDW RI WKH DFFHOHUDWLQJIRUFHEXWLQWKHRSSRVLWHGLUHFWLRQ

Fluid Masses Subjected to Acceleration

303

 ‡ What is achieved by applying D’Alembert’s principle?   '¶$OHPEHUW¶VSULQFLSOHKHOSVLQFKDQJLQJWKHG\QDPLFHTXLOLEULXPRIÀXLGPDVVLQWRDVWDWLF HTXLOLEULXP$QLPDJLQDU\LQHUWLDIRUFHRIVDPHPDJQLWXGHEXWRSSRVLWHLQGLUHFWLRQLQSODFH RIDFWXDODFFHOHUDWLRQIRUFHLVDSSOLHGWRWKHPRYLQJÀXLGVRWKDWLWFDQEHEURXJKWWRVWDWLF HTXLOLEULXP    &RQVLGHUDWDQN¿OOHGZLWKOLTXLGZLWKPDVV m LVPRYLQJZLWKXQLIRUPDFFHOHUDWLRQ a  IURPULJKWWROHIWDVVKRZQLQ¿JXUH7KHDFFHOHUDWLQJIRUFHZLOOEHHTXDOWRM ¥ aDFWLQJ IURP ULJKW WR OHIW $FFRUGLQJ WR '¶$OHPEHUW¶V SULQFLSOH WKLV DFFHOHUDWLQJ IRUFH LV WR EH UHSODFHGE\WKHLQHUWLDIRUFHRIPDJQLWXGH M ¥ aEXWLWZLOODFWIURPOHIWWRULJKW

D’ Alembert’s F = m.a

Inertia force = m.a

Principle

Application of D’ Alembert’s Principle

5.6 HORIZONTAL ACCELERATION  ‡ What happens to a liquid when it is subjected to a horizontal acceleration? Determine the slope of the free liquid surface when subjected to the constant horizontal acceleration.   :KHQDOLTXLGLVVXEMHFWHGWRDFRQVWDQWKRUL]RQWDODFFHOHUDWLRQWKHOHYHORIWKHOLTXLGRQ WKHIURQWVLGHRIWDQNIDOOVZKLOHWKHOHYHORIWKHOLTXLGRQWKHEDFNVLGHULVHVWKLVPDNHVWKH IUHH OLTXLG VXUIDFH WR VORSH XSZDUGV LQ WKH GLUHFWLRQ RSSRVLWH WR WKH GLUHFWLRQ RI FRQVWDQW DFFHOHUDWLRQDVVKRZQLQWKH¿JXUH Free surface of liquid

Free surface of liquid q

Lines of constant pressure

Acceleration =a

h1

F = m.a F1 F2

h2

Free Surface on Horizontal Acceleration from Right to Left



Free Body Diagram of Liquid Mass

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304

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a g

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P1 P2 P3 P4

Lines 1 – 1, 2 – 2, 3 – 3, & 4 – 4 are pressure lines

1 2 3 4

P1 < P2 < P3 < P4 ---

1 2 3 Constant pressure lines

4

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5.7 VERTICAL ACCELERATION  ‡ What happens when a liquid is subjected to constant vertical acceleration?   :KHQDOLTXLGLVVXEMHFWHGWRFRQVWDQWYHUWLFDODFFHOHUDWLRQWKHIUHHVXUIDFHRIWKHOLTXLG UHPDLQV KRUL]RQWDO DV LW ZDV HDUOLHU ZLWKRXW VXEMHFWLQJ WKH OLTXLG WR WKH DFFHOHUDWLRQ +RZHYHU WKH SUHVVXUH LQWHQVLW\ QRZ DW DQ\ SRLQW LQ WKH OLTXLG LV JUHDWHU WKDQ WKH VWDWLF

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rgh

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a g

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H

a mg F = P, dA

Liquid Subjected to Vertical Acceleration



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306



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ma h h mg

a

F

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a  g

 ‡ What will happen if a liquid is subjected to downward acceleration equal to the FRH൶FLHQWRIDFFHOHUDWLRQ g " 8378   7KHSUHVVXUHDWDQ\SRLQWZLWKGRZQZDUGDFFHOHUDWLRQ

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Fluid Masses Subjected to Acceleration

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5.8 ACCELERATION ON INCLINED PLANE  ‡ What happens when a liquid is subjected to constant acceleration while moving up along an inclined plane? Find the angle of slope of the free surface of the liquid. 8378   :KHQDOLTXLGFRQWDLQHURQDWUDQVSRUWLVVWDQGLQJRQDLQFOLQHGSODQHWKHIUHHVXUIDFHRI WKH OLTXLG UHPDLQV KRUL]RQWDO LQ VSLWH RI WKH QRUPDO D[LV RI WKH OLTXLG VORSLQJ ZLWK DQJOH LQFOLQDWLRQDQJOH ZLWKWKHYHUWLFDO Vertical Normal axis f Liquid surface horizontal

Inclined plane or road

f

Liquid at Rest on Inclined Plane

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m.az

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308 

Fundamentals of Fluid Mechanics

SPx FVLQq max SPy mg + maz FFRVq

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\

tan q 

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‡ What happens when a liquid is subjected to constant acceleration while moving down along an inclined plane?

A q Mg

f

tion era l e c Ac ‘a’

Max

A

f

F Maz



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 ‡ The transport tanks for carrying liquid in mountainous region are specially designed to avoid any spillover of the liquid. Elucidate.   7KHWUDQVSRUWWDQNVIRUWUDQVSRUWLQJOLTXLGDUHVSHFLDOO\GHVLJQHGLQVXFKDZD\WKDWOLTXLG GHSWK DW ERWK HQGV UHPDLQV WKH VDPH ZKHQ WKH WUDQVSRUW LV DW UHVW RQ DQ LQFOLQHG URDG DV VKRZQLQWKH¿JXUH7KHWDQNKDVUHDUHQGUDLVHGXSWRHQVXUHWKDWEDVHRIWKHWDQNLVSDUDOOHO WRWKHKRUL]RQWDO Free surface horizontal

Base is almost horizontal f Tanks for Inclined Plane

Fluid Masses Subjected to Acceleration

309

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Railway Wagons for Petroleum



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Forced Vortex in Open Vessel



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310

Fundamentals of Fluid Mechanics

 ‡ What are constant pressure surfaces when a liquid is subjected to a forced vortex?   :KHQDOLTXLGLVURWDWHGLQDQRSHQYHVVHOLWVIUHHVXUIDFHDWWDLQVWKHIRUPRISDUDERORLG $Q\VXUIDFHLQWKHOLTXLGKDYLQJDOOLWVSRLQWVDWHTXDOGLVWDQFHIURPWKHIUHHVXUIDFHZLOO KDYH FRQVWDQW SUHVVXUH +HQFH DOO VXFK VXUIDFHV DUH FDOOHG FRQVWDQW SUHVVXUH VXUIDFHV DV VKRZQLQWKH¿JXUH

Free surface Constant pressure surfaces

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311

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B Original level

D

Rise of water

O Axial depth of water A

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312

Fundamentals of Fluid Mechanics

 ‡ What is the volume of empty space in the paraboloid?

B¢¢

A¢¢

B

r2

A

r1





Z2 Z1

O (vertex)



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 ‡ What is spill out? When does it happen?

r

r

R

R z

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z z¢

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r¢ r



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314

Fundamentals of Fluid Mechanics q

B

D

1m

Acceleration = 3 m/s

2

C

A

6

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Fluid Masses Subjected to Acceleration

315

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a g a 

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O

B

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O

2 Acceleration 2

a = 0.8 m/s 3

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(d) None of these

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Chapter

6

DIMENSIONAL ANALYSIS

KEYWORDS AND TOPICS        

FUNDAMENTAL DIMENSIONS SECONDARY DIMENSIONS HOMOGENEOUS EQUATION DIMENSIONAL FORM NON-DIMENSIONAL FORM RAYLEIGH’S METHOD p THEOREM VARIABLES

       

DEPENDENT VARIABLES REPEATING VARIABLES DIMENSIONLESS NUMBERS REYNOLDS NUMBER FORDE’S NUMBER WEBER’S NUMBER MACH’S NUMBER EULER’S NUMBER

6.1 INTRODUCTION Dimensional analysis is a mathematical technique which uses the dimensions of quantities in solving the engineering problems. Dimensional analysis helps in determining a possible arrangement of variables in a physical relationship. Physical relationship is rational in case the dimensions of the left hand of the relation are equal to the dimensions of the right hand. All physical relationships are dimensionally homogeneous. ,QÀXLGPHFKDQLFVDÀXLGSDUWLFOHLVVXEMHFWHGWRDQXPEHURIIRUFHVVXFKDV L LQHUWLDIRUFH LL YLVFRXVIRUFH LLL JUDYLWDWLRQDOIRUFH LY SUHVVXUHIRUFH Y VXUIDFHWHQVLRQIRUFHDQG YL  HODVWLFIRUFH'LPHQVLRQOHVVSDUDPHWHUVDUHXVHIXOLQÀXLGPHFKDQLFVZKHQDQ\WZRIRUFHVDUH SUHGRPLQDQW 7KH GLPHQVLRQOHVV SDUDPHWHUV DUH L  5H\QROGV QXPEHU LL  (XOHU¶V QXPEHU LLL  and Weber’s number.

6.2 DIMENSIONAL ANALYSIS  ‡ What do you understand from dimensions and units?   $Q\ SK\VLFDO VLWXDWLRQ FDQ EH GHVFULEHG E\ FHUWDLQ IDPLOLDU SURSHUWLHV VXFK DV OHQJWK YHORFLW\ DUHD YROXPH RU DFFHOHUDWLRQ 7KHVH DUH DOO NQRZQ DV GLPHQVLRQV DQG DUH RI QR use without a magnitude being attached. Dimensions are properties which can be measured

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Dimensions of common physical quantities

Qunatity

SI unit

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m/s

LT –1

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m/s2

LT –2 2

MLT –2

3.

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Newton (N) or kg m/s

4.

Energy or work

Nm or kg m2/s2

ML2T –2

5.

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Nm/s or kg m2s–3

ML2T –3

2

6.

Pressure or stress

N/m kg/ms

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kg/m3

2

2

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8.

Viscosity

Ns/m or kg/ms

ML–1T –1

9.

Surface tension

N/m or kg/s2

MT –2

 ‡ What do you understand from dimensional analysis? Dimensional analysis is a branch of mathematics which deals with the dimensions of quantities. Dimensional analysis helps in determining a possible arrangement of variables in a physical relationship. This is achieved by forming a number of non-dimensional group free from units from a given number of dimensional quantities in such a way that variables can be reduced and the physical relationship can be expressed in a possible arrangement of remaining variables.  ‡ What are the uses of dimensional analysis? The uses of dimensional analysis are:    7R DVFHUWDLQ ZKHWKHU DQ\ HTXDWLRQ RI SK\VLFDO SKHQRPHQRQ LV UDWLRQDO RU QRW ,I WKH HTXDWLRQLVGLPHQVLRQDOO\KRPRJHQHRXV GLPHQVLRQDOXQLWVDWERWKVLGHVRIWKHHTXDWLRQ DUHDVDPH WKHSK\VLFDOSKHQRPHQRQLVUDWLRQDO2WKHUZLVHLWLVLPSRVVLEOH    7RDVFHUWDLQZKHWKHUWKHUHODWLRQVKLSEHWZHHQYDULRXVTXDQWLWLHVLQDQHTXDWLRQVLJQLI\ a physical phenomenon.    7RUHGXFHWKHQXPEHURIYDULDEOHVLQYROYHGLQDSK\VLFDOSKHQRPHQRQ7KHSHUIRUPDQFH of experiment becomes easy due to this reduction of variables.    7RFRQYHUWWKHWKHRUHWLFDOHTXDWLRQLQWRDVLPSOHGLPHQVLRQOHVVIRUP    7RVROYHDQ\FRPSOH[SK\VLFDOSKHQRPHQRQ    7RFRQYHUWWKHXQLWVRITXDQWLWLHVIURPRQHV\VWHPWRDQRWKHUV\VWHP    7R IDFLOLWDWH PRGHO WHVWLQJ E\ UHGXFLQJ WKH QXPEHU RI YDULDEOHV LQWR IRXU SULPDU\ quantities.

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6.3 FUNDAMENTAL AND SECONDARY DIMENSIONS  ‡ What are the fundamental dimensions or primary quantities?   0DVV OHQJWK WLPH DQG WHPSHUDWXUH DUH IXQGDPHQWDO RU SULPDU\ GLPHQVLRQV 7KHVH IRXU physical quantities are called fundamental dimensions or primary quantities as the units of these quantities do not depend on the units of any other physical quantities.  ‡ What are the derived or secondary dimensions? The unit of physical quantity which depends on the unit of other physical quantities is called derived or secondary dimension.  ‡ Find the dimensions of (1) force, (2) angular velocity, (3) discharge, (4) torque, (5) power, (6) density, and (7) viscosity  )RUFH 0DVV¥ Acceleration F M ¥ Angular velRFLW\  w 

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6.4 DIMENSIONAL HOMOGENEITY  ‡ What is dimensional homogeneity? What are variables and dependent variables? Or What do you understand by a dimensionally homogenous equation? Write three dimensionless parameters. A physical phenomenon is given in variables in the form of an equation. The physical phenomenon is rational in case the dimensions of left hand of the relation is equal to the GLPHQVLRQV RI ULJKW KDQG ,I LW LV VR ZH VD\ WKDW HTXDWLRQ LV GLPHQVLRQDOO\ KRPRJHQRXV )RUH[DPSOHWDNHWKHH[SUHVVLRQRIVWDWLFSUHVVXUH P rgh Mass ¥ acceleration  3UHVVXUH )RUFH$UHD  Area M ¥ LT±L ML–1 T± r GHQVLW\  

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ML–1 T± ML± ¥ LT± ¥ L   ML–1 T±  GLPHQVLRQVRIOHIWVLGH GLPHQVLRQVRIULJKWVLGH   ,QWKHDERYHHTXDWLRQP is a dependent variable and rg and h are variables.  ‡ What are the applications of dimensional homogeneity? The applications of dimensional homogeneity are:    7R DVFHUWDLQ ZKHWKHU WKH HTXDWLRQ LV KRPRJHQHRXV WKHUHE\ ¿QGLQJ RXW ZKHWKHU WKH phenomenon is rational.    7RDVFHUWDLQWKHGLPHQVLRQVRIDYDULDEOH    7RIDFLOLWDWHHDV\FRQYHUVLRQRIXQLWVIURPRQHV\VWHPWRDQRWKHU    7RKHOSLQGLPHQVLRQDODQDO\VLVDQGPRGHOWHVWLQJ   ,QÀXLGPHFKDQLFVDÀXLGSDUWLFOHLVVXEMHFWHGWRDQXPEHURIIRUFHVVXFKDVLQHUWLDIRUFH YLVFRXV IRUFH JUDYLWDWLRQDO IRUFH SUHVVXUH IRUFH VXUIDFH WHQVLRQ IRUFH DQG HODVWLF IRUFH 'LPHQVLRQOHVV SDUDPHWHUV DUH XVHIXO LQ ÀXLG PHFKDQLFV ZKHQ DQ\ RI WKH WZR IRUFHV DUH SUHGRPLQDQW +HQFH GLPHQVLRQOHVV SDUDPHWHUV DUH WKH UDWLRV RI WZR IRUFHV ZKLFK DUH predominant. The dimensionless parameters are:  D  Reynolds number. ,WLVWKHUDWLRRILQHUWLDIRUFHWRYLVFRXVIRUFHLQÀXLG R e 

r l2V 2 mVL

rVL m b  Euler’s number. ,WLVWKHUDWLRRILQHUWLDIRUFHWRSUHVVXUHIRUFHLQWKHÀXLG  



E 

r l 2V 2 DPL2

rV 2 DP F  Weber’s number. It is the ratio of inertia force to surface tension force.  

W 

r l 2V 2 r V 2L   sl s

‡ What are dimensional form and non-dimensional form of homogenous equation?   ,QGLPHQVLRQDOIRUPRIKRPRJHQRXVHTXDWLRQVWKHGLPHQVLRQVRIWKHOHIWVLGHDUHHTXDOWR the dimension of the right side. For example: Q 

8 Cd 2 g tan q H 15

On reduction L T –1 L T –1

Dimensional Analysis



337

 ,Q QRQGLPHQVLRQDO IRUP RI KRPRJHQRXV HTXDWLRQV WKH OHIW VLGH DQG ULJKW VLGH KDYH QR GLPHQVLRQ )RU H[DPSOH WKH DERYH KRPRJHQHRXV HTXDWLRQ FDQ EH ZULWWHQ LQ QRQ dimensional form as:

Q g ◊H

5 /2

 

8 2 Cd tan q 15

6.5 METHODS OF DIMENSIONAL ANALYSIS  ‡ What are different methods of dimensional analysis?   7KHUH DUH WZR PHWKRGV RI GLPHQVLRQDO DQDO\VLV YL]   5D\OHLJK¶V PHWKRG DQG   %XFNLQJKDP¶VPHWKRGRU%XFNLQJKDP¶Vp–theorem.  ‡ Explain dimensional analysis by Rayleigh’s method. The Rayleigh’s method is used for determining the expression for a dependent variable ZKLFK GHSHQGV XSRQ PD[LPXP WKUHH RU IRXU YDULDEOHV RU RWKHUZLVH LW EHFRPHV GL൶FXOW WR¿QGWKHH[SUHVVLRQIRUWKHGHSHQGHQWYDULDEOH,IX is a variable which is dependent on three variables X1X and XWKHQWKHH[SUHVVLRQFDQEHZULWWHQDVX f X1XX 7KLV H[SUHVVLRQFDQEHVLPSOL¿HGE\5D\OHLJK¶VPHWKRGE\ZULWLQJX m◊X1a Xb Xc where m  constant and ab and c are powers. The values of ab and c can be found out by comparing fundamental dimensions of both sides. Hence the expression for dependent variable can be obtained.  ‡ Prove by the method of dimensional analysis that R, the resistance to be motion of a sphere of radius r falling with a velocity V through a fluid of viscosity h is given by R = kh rV where k is a dimensional constant. Guidance: 7KHGHSHQGHQWYDULDEOHGHSHQGVXSRQWKUHHYDULDEOHVZKLFKPDNHVLWVXLWDEOH for Rayleigh’s method of dimensional analysis R μ ra ¥ rb ¥ Vc ¥ hd Putting fundamental dimensions MLT± μ ML± a ¥ L b ¥ LT1 c ¥ ML–1 T–1 d MLT± μ Ma+d ¥ L±a+b+c–d ¥ T   1RZHTXDWHSRZHUVRIERWKVLGHVZHJHWWKUHHHTXDWLRQVDVXQGHU   

 a + d  ±a + b + c – d ± ±c – d

  

  )URPHTXDWLRQV    DQG  ZHFDQ¿QGWKHYDOXHVRIab and c in term of d as under a ±d b ±d c ±d

338

Fundamentals of Fluid Mechanics

Putting the values to the equation R μ r1–d r±d V±d hd

Ê n ˆ  k rr V Á Ë r rV ˜¯



d

PXWWLQJ

d  R kh rV  ‡ State Buckingham’s theorem. What are repeating variables? How are these selected in dimensional analysis? (UPTU 2004-5) Or Describe Buckingham’s theorem? Why this theorem is considered superior to Rayleigh’s method for dimensional analysis? (UPTU 2005-6) BUCKINGHAM’S THEOREM   %XFNLQJKDP¶V WKHRUHP VWDWHV WKDW LI WKHUH DUH n number of variables and m number of IXQGDPHQWDOGLPHQVLRQVLQDGLPHQVLRQDOO\KRPRJHQHRXVHTXDWLRQWKHQWKHYDULDEOHVFDQ EHDUUDQJHGLQ n – m QXPEHURIGLPHQVLRQOHVVWHUPV7KHVHGLPHQVLRQOHVVWHUPVDUHFDOOHG p terms. Let X1 f XX … Xn or f X1X … Xn   If there are nYDULDEOHVZLWKIXQGDPHQWDOGLPHQVLRQVWKHQWKHHTXDWLRQFDQEHZULWWHQDV f p1p … pn±   Each pWHUPLVIRUPHGZLWKYDULDEOHVLQZKLFKWKUHHYDULDEOHVDUHUHSHDWLQJYDULDEOHV VD\ XX and X   +HQFHp terms can be written as p1 Xa1 Xb1 Xc1 X1 p Xa Xb Xc X and so on but up to pn± Xan± Xbn± Xcn± Xn± The above equations are solved by the principle of dimensional homogeneity. When the values of p1p … pn±DUHDNQRZQWKHUHTXLUHGH[SUHVVLRQFDQEHZULWWHQDV f p1p … pn±  

 

REPEATING VARIABLES These are variables which are selected to appear in each p term. The number of repeating variables is equal to the number of fundamental dimensions appearing in the variables of the equation. The choice of repeating variables is made on basis of the following:   7KHGHSHQGHQWYDULDEOHVVKRXOGQRWEHVHOHFWHG   9DULDEOHVKDYLQJGL൵HUHQWFDWHJRULHVRISURSHUW\LQVWHDGRIRQHVKRXOGEHVHOHFWHG)RU H[DPSOH YDULDEOHV VKRXOG EH RQH HDFK IURP JHRPHWULF SURSHUW\ OHQJWK KHLJKW DQG GLDPHWHU ÀRZSURSHUW\ YHORFLW\DFFHOHUDWLRQ DQGÀXLGSURSHUW\ YLVFRVLW\GHQVLW\ 

Dimensional Analysis

339

  

  6KRXOGQRWEHGLPHQVLRQOHVV   6KRXOGKDYHGL൵HUHQWGLPHQVLRQV   6KRXOGKDYHVDPHQXPEHURIIXQGDPHQWDOGLPHQVLRQV Superiority of p theorem over Rayleigh’s method: The Rayleigh’s method of dimensional analysis becomes more laborious as the number of variables becomes more than fundamental dimensions. Such problem is not faced when p theorem is used.  ‡ The drag force F on a partially submerged body depends on the relative velocity V between the body and the fluid, characteristics linear dimension l, height of surface roughness k, fluid density r, the viscosity m and the acceleration due to gravity (g). Obtain an expression for the drag force using the method of dimensional analysis. (UPTU 2001-2) Guidance: 7KHUH DUH VL[ YDULDEOHV ZKLFK VXJJHVW WR XVH %XFNLQJKDP¶V p theorem for solving the expression. Let the expression be f FlkVmrg   

7RWDOQXPEHURIYDULDEOHV m   7RWDOQXPEHURIIXQGDPHQWDOGLPHQVLRQ n   ML & T



Number of pWHUPV ±  Each p term consists of MYDULDEOHVLH  We select repeating variables as LV and r. p1 La1 Vb1 rc1 F

\

p La Vb rc m p La Vb rc g p La Vb rc k 1RZ



p1 La1 LT–1 b1 ML± c1 MLT± MLT La1+b±c1+1 Mc1+1 T–b± –b1± RUb1 ± + c1 RUc1 ± a1 + b1±c1 RUa1 ± p1 L± V±± rF

\   



 

F r L2V 2

p La Vb rcm

1RZ 

 La LT–1 b ML± c2 ML–1 T–1

340

Fundamentals of Fluid Mechanics

MLT Lab±c± Mc T –b± c RUc ±

\

–b± RUb ± a + b±c± RUa ± p L–1 V–1 r–1 m 

\ 1RZ



m LV r

p La Vb rc g MLT La LT–1 b ML± c LT±  Lab±c Mc T–b± c 

\

–b± RUb ± a + b±c RUa  p L1 V± g  1RZ

L g V2

p La Vb rc k p La LT–1 b ML± c L



 Lab±c Mc T–b

\

c b a ±

\

p L–1 k 

k L

f p1ppp  

Ê F m L kˆ , 2 g , ˜   fÁ 2 2, L¯ Ë r L V LV r V F Ê m Lg k ˆ  f Á , , r L2 V 2 Ë LV r V 2 L ˜¯ or

Ê m Lg k ˆ , , F rLV ¥ f Á Ë LV r V 2 L ˜¯

‡ Show by p theorem that a general equation for discharge Q over a weir of any shape is given by ⎛ 5 1 n s ⎞ Q = H 2 g 2 f ⎜ 2/ 3 1/ 2  2 ⎟ H gr ⎠ ⎝H g

Dimensional Analysis

341

where H = head over weir, n = kinematic viscosity, r = density of the fluid g = acceleration due to gravity and s = surface tension. (UPTU 2006-7) F QHrngs   Select Hr and g as repeating variables.   +HQFHZHZLOOKDYHM  YDULDEOHVLQHDFKp term and there will n – m  ± p terms p1 Ha1gb1rc1Q 

 La1 LT± b1 ML± c1 L T–1



 La+b±c Mc T±b–1 c1 b1 

\



-1 and a1 ± 2

\

p1 H± g± Q 

1RZ

p Ha gb rc n

Q H g1/2 5 /2

 La LT± b ML± c LT–1



 Lab±c Mc T±b± \

c b ±DQGa ±

\

p

H± g± n

p Ha gb rc s 



 La LT± b ML± c MT±





 Lab±c T±b± Mc

\

c ±b ±DQGa ± p H± g–1 r–1 s

Ê n s ˆ Q F Á 5/2 1/2 , 3/2 1/2 , 2 ˜   H g H gr ¯ ËH g Q Ê n s ˆ  f Á 3/2 1/2 , 2 ˜ H 5/2 g1/2 H gr ¯ ËH g Ê n s ˆ Q H g f Á 3/2 1/2 , 2 ˜ H gr ¯ ËH g

342

Fundamentals of Fluid Mechanics

6.6 DIMENSIONLESS NUMBERS  ‡ What do you mean by dimensionless numbers? Derive an expression for any two dimensionless numbers. (UPTU 2002-3, 2007-08) Or What is the importance of dimensionless numbers? Derive their expressions and write their applications.   $ÀXLGLVJHQHUDOO\VXEMHFWHGWRYDULRXVIRUFHVVXFKDV  LQHUWLDIRUFH  YLVFRXVIRUFH  JUDYLWDWLRQDOIRUFH  ඉUHVVXUHIRUFH  VXUIDFHWHQVLRQIRUFHDQG  HODVWLFIRUFH7KH ratio of any two forces is a dimensionless parameter. These parameters have great physical VLJQL¿FDQFHDVÀXLGVKDYLQJVDPHSDUDPHWHUKDYHG\QDPLFDOVLPLODULW\7KHPRVWLPSRUWDQW GLPHQVLRQOHVVQXPEHUVDUH  5H\QROGVQXPEHU  )URXGH¶VQXPEHU  (XOHU¶VQXPEHU  0DFK¶VQXPEHUDQG  :HEHU¶VQXPEHU    Reynolds Number: ,I LQHUWLD IRUFH DQG YLVFRXV IRUFH DUH SUHGRPLQDQW LQ DQ\ ÀRZ SKHQRPHQRQWKHQ5H\QROGVQXPEHULVXVHGWRFRPSDUHPRGHODQGSURWRW\SH5H\QROGV number is the ratio of inertia force to the viscous force. R e 

Inertia force Viscous force

,QHUWLDIRUFH PDVV¥ acceleration  YROXPH¥ density ¥ acceleration



L ¥ r ¥ LT± 2





Ê Lˆ  ÁË T ˜¯ ¥ L ¥ r but



VLr 9LVFRXVIRUFH 7HQVLRQ¥ area



 m

dV ¥ area dy



 m

LT -1 ¥ L L

mL T–1 mVL \

R e 

 

L2V 2r mVL LV r m

L  V 9HORFLW\ T

Dimensional Analysis

     

5H\QROGVQXPEHULVXVHGIRUVWXG\RI  PRWLRQRIFRPSOHWHO\VXEPHUJHGERGLHVOLNH VXEPDULQHV DHURSODQHV DQG DXWRPRWLYHV DQG   LQFRPSUHVVLEOH ÀRZ WKURXJK SLSHV bends and turbines in which inertia and viscous forces are predominant.   Froude’s Number: Froude’s number is used where inertia and gravitational forces are predominant. It is the square root of the ratio of inertia force to the gravitational force. ,QHUWLDIRUFH PDVV¥ acceleration  VLr *UDYLWDWLRQDOIRUFH PDVV¥ g  YROXPH¥ r ¥ g  Lrg F 

Inertia force Gravitational force



V 2 L2r L3r g

 

V2 Lg





343

V L◊ g

)URXGH¶V QXPEHU LV XVHG IRU WKH VWXG\ RI ÀXLG ÀRZ IURP RSHQ K\GUDXOLF VWUXFWXUHV VXFK DV VSLOOZD\V ZHLUV VOXLFHV DQG RSHQ FKDQQHOV ZKHUH JUDYLWDWLRQDO IRUFHV DUH predominant.   Weber’s number: It is the square root of the ratio of the inertial force to the surface WHQVLRQIRUFHLQWKHÀXLG W 

Inertia force Surface tension force

IQHUWLDIRUFH VLr 6XUIDFHWHQVLRQIRUFH s ¥ L W   

V 2 L2r sL V s rL

344



Fundamentals of Fluid Mechanics

7KH:HEHU¶VQXPEHULVXVHGIRUWKHVWXG\RI  FDSLOODU\PRYHPHQWRIZDWHULQVRLODQG  ÀRZRIEORRGLQDUWHULHVDQGYHLQVZKHQWKHVXUIDFHWHQVLRQIRUFHVDUHSUHGRPLQDQW LQWKHÀRZ   Euler’s number: It is the square root of the ratio of inertia force to the pressure force LQWKHÀXLG E 

Inertia force Pressure force

,QHUWLDIRUFH VLr 3UHVVXUHIRUFH P◊A PL E  



V P r

7KH(XOHU¶VQXPEHULVXVHGIRUWKHVWXG\RIÀXLGZKHUHSUHVVXUHIRUFHLVSUHGRPLQDQW VXFKDV  ZDWHUKDPPHUH൵HFWVLQSHQVWRFNVRIK\GURSRZHUSODQWVDQG  GLVFKDUJH FRH൶FLHQWVRIRUL¿FHVVOXLFHJDWHVDQGPRXWKSLHFHV   Mach’s Number: (ODVWLFIRUFHEHFRPHVSUHGRPLQDQWLQFRPSUHVVLEOHÀXLGV0DFK¶V QXPEHULVVLJQL¿FDQWLQ¿QGLQJG\QDPLFDOVLPLODULW\RIWKHÀXLGVZKHUHHODVWLFIRUFHV are predominant. Mach’s number is the square root of the ratio of inertia force to the HODVWLFIRUFHLQWKHÀXLG M 



V 2 L2r PL2

Inertia force Elastic force

 

V 2 L2r  ZKHUHE 9ROXPHWULFPRGXOXVRIHODVWLFLW\ E ¥ L2

 

rV 2 E

M 

V E r

0DFK¶VQXPEHULVXVHGIRUWKHVWXG\RIÀRZDWKLJKVSHHGVXFKDV  ÀRZLQSLSHV  ZDWHUKDPPHUH൵HFWDQG  KLJKVSHHGPRWLRQREMHFWVOLNHDHURSODQHVSURMHFWLYH and missiles in atmosphere where the elastic forces are predominant.

Dimensional Analysis

345

 ‡ Develop an expression for the distance travelled by a freely falling body in time (T). Take it to be dependent on mass of the body, the acceleration of gravity and time. Guidance: 7KHUHDUHYDULDEOHVYL]GLVWDQFH S PDVV M DFFHOHUDWLRQ g DQGWLPH T  +HQFHZHFDQXVH5D\OHLJK¶VPHWKRGRIGLPHQVLRQDODQDO\VLV S f MgT  1 

M L T  k ¥ Ma ¥ LT± b ¥ T c  1 

a b

M L T  M L T \

where k FRQVWDQW

±b+c

a b c 

\ S k g T  ‡ Derive an expression for an ideal fluid flow (Q  ZKLOH IORZLQJ WKURXJK DQ RUL¿FH ZKLFKLVGHSHQGHQWRQWKHGHQVLW\RIIOXLGWKHGLDPHWHURIWKHRUL¿FHDQGWKHSUHVVXUH difference. Q f rDP  k ra Db Pc



MLT–1 k ML± a L b ML–1T± c  k Ma+c L±a+b–c T±c



1 a + b RUa ±±a + b – c ± 2

\

c 

or

b  Q k r± D P kD

P r

 ‡ Derive an expression for the power delivered to a pump which is dependent on the sp. wt. of the fluid, the flow and the head delivered. 

3RZHUGHOLYHUHG P f vQH MLT± k ML± T± a L T–1 b L c kMa L±ab+c T±a–b \ \

a b c  P k VQH

 ‡ 8VLQJ GLPHQVLRQDO DQDO\VLV ¿QG WKH H[SUHVVLRQ RI G\QDPLF SUHVVXUH H[HUWHG E\ D flowing incompressible fluid on an immersed object if it is dependent on density and velocity.  3UHVVXUH P f rV or

P k ra Vb

346

Fundamentals of Fluid Mechanics a

Ê Mˆ Ê Lˆ ML–1 T ± k Á 3 ˜ Á ˜ Ë L ¯ ËT¯

b

 k M a L±a+b T –b \ b  a  \ P krV  ‡ Using p theorem show that the shear stress of the pipe wall is given by

Ê r Vd ˆ t = rV2 f Á Ë m ˜¯ where,

r = fluid density, m = viscosity, V = average velocity, d = diameter of pipe

The relationship can be expressed as f trmVd   

 +HUHm DQGn +HQFHWKHUHZLOOEHWZRp terms with each p term having four variables. Select dV & r as repeated variables \ \

p1 da Vb rc ¥ t MLT La LT–1 b ML± c MLT± Mc+1 La+b±c+1 T–b±

\

c ±b ±DQGa  p1 d  V± r–1 t  



1RZ

t V 2r

p da Vb rc m La LT–1 b ML± c ML–1 T–1 MLT Mc+1 La+b±c–1 T–b–1 b ±c ±DQGa ±

\

p d–1 V–1 r–1 m 

Ê t m ˆ f Á 2 ,   Ë V r dV r ˜¯

m dV r

Dimensional Analysis

347

Ê m ˆ t  f Á V 2r Ë dV r ˜¯

or

Ê m ˆ t Vr f Á Ë dV r ˜¯

or

‡ Assuming that thrust T of a screw propeller is dependent upon the diameter d, speed of advance V, fluid density r, revolution per second n and coefficient of viscosity m, show using the principle of dimensional homogeneity that it can be represented by

Ê m d ◊nˆ , T = rd2V2 f Á Ë rVd V ˜¯ (Allahabad University) Let the expression be given by F TrdVmn     +HUHm DQGn +HQFHWKHUHZLOOEHWKUHHp terms and each p term will have four variables. Select rV & d as repeated variables p1 ra Vb dcT    ML± a LT–1 b L c MLT± MLT Ma+1 L±a+b+c+1 T–b± a ±b ±DQGc ±

\

p1 r–1 V– d± T  p ra Vb dc m

NRZ 

  ML± a LT–1 b L c ML–1 T–1 MLT Ma+1 L±a+b+c–1 T–b–1 a ±b ±DQGc ±

\

p r–1 V–1 d–1 m 

\

m rVd

p ra Vb dc r

1RZ

\

T rV 2 d 2



  ML± a LT–1 b L c T–1



 Ma L±a+b+c T–b–1 a b ±c  p V–1 d1 n  

d ◊n V

348

Fundamentals of Fluid Mechanics

m d ◊nˆ Ê T , , FÁ   2 2 Ë rV d rVd V ˜¯ or

T rV 2 d 2

Ê m dn ˆ , f Á Ë rVd V ¯˜

Ê m dn ˆ , T rV d f Á Ë rVd V ¯˜ ‡ Show by the method of dimensional analysis that the volume rate of flow of a gas WKURXJKDVKDUSHGJHRUL¿FHLVJLYHQE\ Q = d2

P ÊV fÁ ¥ r Ëd

Pˆ r ˜¯

where dLVWKHGLDPHWHURIWKHRUL¿FHr is the pressure difference between the two sides RIWKHRUL¿FHr and V are the density and kinematic viscosity of the gas respectively. Let the expression be given by F QdPrV   Here m DQGn +HQFHWKHUHZLOOEHWZRp terms and each will have four variables. Select dP and r as repeated variables. p1 da pb rc Q   L a ML–1 T± b ML± c LT–1 MLT Mb+c La–b±c T±b–1 \

b 

1 -1 c  and a ± 2 2

\

p1 d± P± r Q

NRZ

p da Pb rc v   L a ML–1 T± b ML± c LT–1 MLT Mb+c La–b±c T±b–1

\

b ±c 

1 and a ± 2

p d–1 P± r v 

V r ◊ d P

Ê Q V , FÁ Á d2 P d ÁË r

Pˆ   r ˜˜ ˜¯

Dimensional Analysis

P ÊV f r ÁË d

Q d

or

349

Pˆ r ˜¯

 ‡ Prove by the method of dimensions that in the rotation of similar discs in fluid with turbulent motion, the frictional torque T of a disc diameter D rotating at speed N, in a fluid of viscosity m and density r.

Ê m ˆ T = D 2N 2r f Á 2 Ë D N r ˜¯ Let the expression be F TDNrm     +HUHm DQGn +HQFHWKHUHDUHWZRp terms and each p term has four variables p1 Da Nb rc T   L a T–1 b ML± c ML T±



MLT Mc+1 La±c T–b± c ±b ±DQGa ±

\

p 1 

T D N 2r 2

p Da Nb rcm

NRZ 

  L a T–1 b ML± c Mc–1 T–1 MLT Mc+1 La±c–1 T–b–1

\

c ±b ±a  p 

m D Nr 2

\

m ˆ Ê T F Á 2 2 , 2 ˜   Ë D N r D Nr ¯

or

Ê m ˆ T DNr f Á 2 ˜ Ë D Nr ¯

‡ In turbomachinery, the relevant parameters are volume flow rate, density, viscosity, bulk modulus, pressure difference, power consumption, rotation speed and a characteristic dimension. According to Buckingham’s pi(p) theorem, the number of independent non-dimensional groups for this case is (a) 3 (b) 4 (c) 5 (d) 6 (IES 93)

350

Fundamentals of Fluid Mechanics

 +HUHWRWDOQXPEHUVRIYDULDEOHV m    1XPEHURISULPDU\GLPHQVLRQ n    +HQFHQXPEHURILQGHSHQGHQWQRQGLPHQVLRQDOJURXSV m – n ±    2SWLRQ F LVFRUUHFW  ‡ 8VLQJ %XFNLQJKDP¶V pWKHRUHP VKRZ WKDW WKH YHORFLW\ WKURXJK D FLUFXODU RUL¿FH LV given by ⎛D m ⎞ V  2 g H fn ⎜  ⎝ H rVH ⎟⎠ where,

H +HDGFDXVLQJÀRZ D 'LDPHWHURIWKHRUL¿FH m = Coefficient of viscosity r = Mass density g = Acceleration due to gravity

  :HKDYH V f HDmrg or F VHDmrg   Select Hg and r as repeating variables \ m   ZKLOH n  \ Variables in each p term will be n – m ±  \

p1 Ha1 ◊ gb1 ◊ rc1 ◊ V p Ha ◊ gb ◊ rc ◊ D

p H a1 ◊ g b1 ◊ r c ◊ m Now putting dimensions in p1ZHKDYH MLT  L a1 LT± b1 ML c1 LT–1 Equating the power of ML and r on boWKVLGHVZHKDYH c1  a1 ± b1 ± \

p1 H

or

p1 

-

1 2

. g± . r . V

V H◊g

Putting now dimensions in pZHKDYH MLT  L a LT± b ML± c L

(UPTU 2006-7)

Dimensional Analysis

Equating the powers of c  a b p 

ML and rRQERWKVLGHVZHKDYH  ±  H–1 g r D 

D H

Putting now dimensions in pZHKDYH M L T La ◊ gb ◊ rc ◊ m Equating the power of ML and tRQERWKVLGHVZHKDYH c ± a   -

3 2

b ± \ 

p H± ◊ g± ◊ P–1 ◊ m m r ⋅ H  ⋅ g m ¥ V ⋅H ⋅r

  







  





 p1 ¥

V gH

m V ⋅H ⋅r

\

F p1pp  

or

⎛ V D m ⎞   p1 × F⎜ ⎟   VH r ⎠ ⎝ gH H

or

⎛ V D m ⎞   F⎜ ⎟   ⎝ gH H VH r ⎠

or

or

V

⎛D m ⎞  f ⎜  gH ⎝ H VH r ⎟⎠

⎛D m ⎞ V  gH f ⎜  ⎝ H rVH ⎟⎠ ⎛D m ⎞  2 gH fn ⎜  ⎝ H rVH ⎟⎠

351

352

Fundamentals of Fluid Mechanics

6.7 QUESTIONS FROM COMPETITIVE EXAMINATIONS  ‡ The drag force D on a certain object in a certain flow is a function of the coefficient of viscosity m, the flow speed n and the body dimension L (for geometrical similar objectives), then D is proportion to 2 2 (a) L × m × v (b) μ ⋅V L2 (d) μ⋅ L V

(c) m2V2L2

(IAS 2001)

D is function of m n and L   2SWLRQ D LVFRUUHFW  ‡ If the number of fundamental dimensions is equal to m, then the repeating variables shall be equal to (a) m and none of the repeating variables shall represent the dependent variable (b) m + 1 and one of the repeating variables shall represent the dependent (c) m + 1 and none of the repeating variables shall represent the dependent variable (d) m and one of the repeated variables shall represent the dependent variable (GATE 2002, IES 1998, 1999)   2SWLRQ F LVFRUUHFW  ‡ The dimensionless group formed by wavelength O, density of fluid r, acceleration due to gravity (g) and surface tension s, (a)

σ λ 2 gρ

(b)

σ λg 2 ρ

(c)

σ⋅ g λ 2ρ

(d)

σ λ ⋅ g ⋅σ

salbgcrd 0LT  

MT ± a L b LT± c ML± d MLT

 M a+d L b+c±d T ±a±c MLT \ a + d RUa ±d7DNHa d ± b + c±d RUb + c RUb + c ± ±a±c RU±C ±DQGb ± \

salbgcrd s × l± × g–1 × r–1 



 2SWLRQ D LVFRUUHFW

σ λ  ⋅ g ⋅σ

(IES 2000)

Dimensional Analysis

353

 ‡ Match List-I (Fluid parameters) with List-II (Basic dimensions) and select the correct answer List-I (a) Dynamic viscosity (b) Chezy’s roughness coefficient (c) Bulk modulus of elasticity

List-II 1. M/T2 2. M/LT2 3. M/LT

(d) Surface tension Code: A B (a) 3 2 (b) 1 4 (c) 3 4 (d) 1 2

4. C 4 ? 2 4

L/T

D 1 3 1 3

(IES 2002)

OSWLRQ F LVFRUUHFW  ‡ In M-L-T system, what is the dimension of sp speed for a rotodynamic pump? 1

1

(a) L–3/4 T 3/2

(b) M 2 L4 T – 5/ 2

(c) L3/4 T–3/2

(b) L3/4 T3/2 N 

(IES 2006)

n Q T ± L T = H  L 

 L



 

×T

−1 −

 

 Lu T±   2SWLRQ F LVFRUUHFW  ‡ A dimensionless group formed with the variables r (density), Z (angular velocity), m (dynamic viscosity) and D (characteristic diameter) is (a)

ρωμ D

(b)

(c) rZmD2

(d) r◊Z◊m◊D

Let F  ρa ⋅ D b ⋅μ c ⋅ω = M  L T  or  ML± a ◊ L b ◊ ML–1T –1 c ◊ T –1  MLT On solving a b DQGc ± \ OSWLRQ E LVFRUUHFW

ρ⋅ω⋅ D  μ

F 

ρ⋅ω⋅ D  μ

(IES 1995)

354

Fundamentals of Fluid Mechanics

 ‡ :KLFKRIWKHIROORZLQJLVQRWDGLPHQVLRQOHVVJURXS" (a)

ΔP ρN 2 D 2

(b)

gH N 2 D2

(c)

ρ⋅ wD 2 μ

(d)

ΔP ρV 3

(IES 1992)

OSWLRQ G LVFRUUHFW  ‡ Which is the correct dimensionless group formed with the variable r (density), N (rotational speed), d (diameter) and p (coefficient of viscosity)? (a)

ρNd 2 π

(b) ρ⋅ N ⋅ d π

(c)

Nd ρ⋅π

(d)

Nd 2 ρ⋅π

(IES 2009)

OSWLRQ D LVFRUUHFW  ‡ In a steady flow through a nozzle, the flow velocity on the nozzle axis is given by V = μ o ⎛⎜ 1 + 3 x ⎞⎟ i , where x is distance along the axis of the nozzle from its inlet plane ⎝ I L⎠

and L is the length of the nozzle. The time required for a fluid particle on the axis to travel from the inlet to the exit plane of the nozzle is (a)

L μo

(b)

1 log e 4 3μ o

(c)

L 4μ o

(d)

L 2.5 μ o

(GATE 2007)

dx  μ ⎛1 +  x ⎞ ⎟ o ⎜ ⎝ L⎠ dt

or

dx  x 1+ L

mo dt

 2QLQWHJUDWLRQZHJHW



L



t dx   ∫ μ o dt  x L+ L L

or

L⎡ x ⎤ mo × t  ⎢log e ⎛⎜1 + ⎞⎟ ⎥ ⎝ ⎣  ⎠ ⎦

Dimensional Analysis



 

or



L × ORJ e  

t 

L ORJ e  μ o

355

OSWLRQ E LVFRUUHFW  ‡ 7KH5H\QROGVQXPEHUIRUIORZRIDFHUWDLQIOXLGLQDFLUFXODUWXEHLVVSHFL¿HGDV What will be the Reynolds number when the tube diameter is increased by 25% and the fluid viscosity is decreased by 40% keeping the fluid same (a) 1200

(b) 1800

(c) 3000

(d) 200

Reynolds numEHURe  or

(GATE 1997)

ρ ⋅V ⋅ D μ R′e  

ρ ×  V  D μ

îî   ‡ The square root of the ratio of inertia force to gravity force is called (a) Reynolds number

(b) Froude’s number

(c) Mach’s number

(d) Euler’s number (IAS 2003, GATE 1994)



 2SWLRQ E LVFRUUHFW

 ‡ Given power P of a pump, the head H and the discharge Q and the sp wt. W of the liquid, dimensionless analysis would lead to the result that P is proportional to (a) H1/2 Q2W

(b) H1/2 W

(c) HQ1/2W

(d) H◊Q◊W P μ H a Qb W MLT ± μ  L a LT –1 b ML±T ± μ Lab± ◊ M ◊ T–b±

ab± DQG± ±b± or 

or  2SWLRQ G LVFRUUHFW

a RUb  P μ HQW

(IES 1998)

356

Fundamentals of Fluid Mechanics

 ‡ Volumetric flow rate Q, acceleration due to gravity g and H form a dimensionless group, which is given by Q g⋅H (b) (a) Q gH 5 (c)

Q gH

(d)

3

Q g2 H

(IES 2002)

F MLT Q ◊ ga ◊ Hb  LT –1 1 ◊ L ◊ T ± a ◊ Lb 0 ◊ La+b±  a 

−1 b 

− 

F Qg± H± 

Q gH 

  2SWLRQ E LVFRUUHFW  ‡ The time period of a simple pendulum depends on its effective length l and the local acceleration due to gravity g. What is the number of dimensionless parameter involved? (a) two (b) one (c) three (d) zero (IES 2009) Here n LQFOXGHV L WLPHSHULRG LL OHQJWKDQG LLL g m LQFOXGHV L OHQJWKDQG LL WLPH Dimensionless terms or p terms are   n – m ±    2SWLRQ E LVFRUUHFW  ‡ In fluid mechanics, the relevant parameters are volume, flow rate, density, viscosity, bulk modulus, pressure difference, power consumption, rotational speed and characteristic dimension. Using Buckingham S theorem, what would be the number of independent non-dimensional group? (a) 3 (b) 4 (c) 5 (d) None of the above (IES 1993, 2007)   1RRIYDULDEOH n     1RRILQGHSHQGHQWGLPHQVLRQ m   No. of pWHUPV n – m ±    2SWLRQ F LVFRUUHFW

Dimensional Analysis

357

 ‡ Consider the following statements: 1. Dimensional analysis is used to determine the number of variables involved in a certain phenomenon. 2. The group of repeating variables in dimensional analysis should include all the fundamental units. 3. Buckingham’s p theorem stipulates the number of dimensionless group for a given phenomenon. 4. The coefficient in Chezy’s equation has no dimension. Which of these are correct? (a) 1, 2, 3, and 4 (b) 2, 3, and 4 (c) 1 and 4 (d) 2 and 3 (IES 2003)   2SWLRQ G LVFRUUHFW  ‡ The variables controlling the motion of a floating vessel through water are drag force F, the speed V, the length l, the density U, dynamic viscosity m of water and gravitational constant g. If non-dimensional groups are Reynolds number (Re), Weber’s number (We), Prandtl’s number (Pr) and Froude’s number (Fr), the expression F is given by (a)

F = f ρV 2 l 2

(b)

F = f R e , Pr ρV 2 l 2

(c)

F = f (R e  We ) ρV 2 l 2

(d)

F = f (R e , Fr ) ρV 2 l 2

(IES 1997)

OSWLRQ G LVFRUUHFW  ‡ (XOHUQXPEHULVGH¿QHGDVWKHUDWLRRILQHUWLDIRUFHWR

   ‡

   ‡



(a) Viscous force (b) Elastic force (c) Pressure force (d) Gravity force (IES 1997) 2SWLRQ F LVFRUUHFW :KLFKRQHRIWKHGLPHQVLRQOHVVQXPEHULGHQWL¿HVWKHFRPSUHVVLELOLW\HIIHFWRIDIOXLG" (a) Euler’s number (b) Froude’s number (c) Mach’s number (d) Weber’s number (IES 2005) 2SWLRQ F LVFRUUHFW A phenomenon is modeled using n dimensional variables with k primary dimensions. The number of non-dimensional variable is

(a) k (c) n – k  2SWLRQ F LVFRUUHFW

(b) n (d) n + k

(GATE 2012)

Chapter

7

SIMILITUDE AND MODEL ANALYSIS

KEYWORDS AND TOPICS  PROTOTYPE

 SIMILARITY LAW

 MODEL

 MODEL LAW

 MODEL TESTING

 REYNOLDS MODEL LAW

 MODEL ANALYSIS

 FROUDE’S MODEL LAW

 HYDRAULIC SIMILITUDE

 EULER’S MODEL LAW

 GEOMETRIC SIMILARITY

 WEBER’S MODEL LAW

 KINEMATIC SIMILARITY

 MACH’S MODEL LAW

 DYNAMIC SIMILARITY

 DISTORTED MODEL

7.1 INTRODUCTION The behaviour of the hydraulic structures and machines can be predicted by performing tests on their models. Since prototypes are costly as compared to the models, any failure of model does not therefore involve much loss of material and human labour. Geometric similarity implies similarity of shape between the model and prototype. Kinematic similarity means the similarity of motion. Dynamic similarity implies that the forces acting on the matching points of the prototype and its model are equal in magnitude and direction. Model testing is used in design of (i) aeroplanes, rockets and missiles, (ii) harbours, (iii) ships and submarines, (iv) skyscrapers, Y  WXUELQHV SXPSV DQG FRPSUHVVRUV YL  GDPV ZHLUV VSLOOZD\V DQG FDQDOV DQG YLL  ÀRRG control measures.

7.2 MODEL AND PROTOTYPE  ‡ What is a model? A model is small-scale replica of the actual machine or structure.

Similitude and Model Analysis

359

 ‡ What is a prototype? The actual structure or machine is called prototype.  ‡ Can a model be bigger than its prototype? It is not necessary that the models should be smaller than the prototypes. However, in most of cases, models are smaller than prototypes.  ‡ What is the model analysis? Or :KDWLVWKHGL൵HUHQFHEHWZHHQDPRGHODQGSURWRW\SH" (UPTU 2006-7) It can be appreciated that when hydraulic structure or machine has been built, it becomes GL൶FXOWWRPRGLI\LWWRPHHWWKHGHVLUHGUHTXLUHPHQWV,QRUGHUWRSUHGLFWWKHSHUIRUPDQFH of hydraulic structures or machines before they are built, it is advantageous to make their models and perform experiments on them to see that the performance is as required. In case any changes are necessary to modify the performance, it is very convenient to incorporate LQ WKH PRGHO +HQFH PRGHO PDNLQJ DQG PRGHO DQDO\VLV DQ XVHIXO WHFKQLTXH LQ WKH ÀXLG machines.  ‡ ([SODLQH[SHULPHQWDOPRGHOWHVWLQJRUZLQGWXQQHOWHVWLQJ A prototype of tractor-trailer unit is to be manufactured and there is a need to measure the drag fore before it is manufactured. A model of smaller size (say 1 : 16 scale) with length about 0.901 metre which is geometrically similar to the prototype is made and tested in a ZLQGWXQQHOZLWKPD[LPXPVSHHGRIPVDVVKRZQLQ¿JXUH7KHDLULQZLQGWXQQHOLV DWWKHVDPHWHPSHUDWXUHDQGSUHVVXUHDVWKHDLUÀRZLQJDURXQGSURWRW\SH%\HQVXULQJDLU velocity of 70 m/s in testing, it has been ensured that air velocity of 60 kmph or 26.8 m/s with resultant drag force would be acting on the prototype.

Model Testing in a Wind Tunnel

 ‡ What are the advantages of model testing? The advantages of model testing are: (1) The behaviour and performance of the structures or machines can be predicted by performing tests on their models. Since prototypes are costly as compared to the models, any failure of model does not involve much loss of material and human labour.

360

Fundamentals of Fluid Mechanics

(2) It is possible to make models of a prototype based on alternative designs which facilitates selecting most economical, safe and sound design of the prototype. (3) Safety and reliability of a structure or machine can be ascertained by model testing before actually constructing or putting into the use of the prototype. (4) Model testing also helps in identifying the defects in existing structure or machines.  ‡ :KLFKDUHWKH¿HOGVZKHUHPRGHOWHVWLQJFDQEHXVHG" 

 0RGHOWHVWLQJLVXVHGLQWKHIROORZLQJ¿HOGV (1) Design of aeroplane, rockets and missiles. The models are tested in wind tunnels by VXEMHFWLQJWKHPWRVLPLODUDLUÀRZ (2) Design of harbour (3) Design of ships and submarines by testing their models (4) Design of skyscrapers by subjecting their models to wind loads (5) Design of irrigation channels (6) Design of turbines, pumps and compressors (7) Design of civil engineering structures such as dams, weirs, spillways, and canals



  'HVLJQRIÀRRGFRQWUROPHDVXUHV

7.3 HYDRAULIC SIMILITUDE  ‡ :KDWLVK\GUDXOLFVLPLOLWXGH" For ascertaining the soundness and performance of the hydraulic structure or machine, it is most essential that model should represent its prototype completely in all aspects. This similarity between the prototype and its model is known as hydraulic similitude.  ‡ :KDWLVWKHRXWFRPHRIWKHK\GUDXOLFVLPLOLWXGH" The outcome of hydraulic similitude is that the results of model tests can be successfully applied to the prototype of the hydraulic structure or machine. Similitude helps us to assume WKDWWKHÀRZVRIWKHÀXLGDUHPHFKDQLFDOO\VLPLODUIRUERWKSURWRW\SHDQGLWVPRGHO  ‡ :KDWDUHWKHGL൵HUHQWVLPLODULWLHVZKLFKVKRXOGH[LVWEHWZHHQPRGHOVDQGSURWRW\SHV" Or 'LVFXVV JHRPHWULF NLQHPDWLF DQG G\QDPLF VLPLODULWLHV $UH WKHVH VLPLODULWLHV WUXO\ (UPTU 2005-6) DWWDLQDEOH",IQRWZK\" Models must reproduce the behaviour of the prototype. Hence, it must possess the following similarities: (1) Geometric similarity (2) Kinematic similarity (3) Dynamic similarity   $PRGHOZKLFKVDWLV¿HVDOOWKHDERYHVLPLODULWLHVZLWKLWVSURWRW\SHLVNQRZQDVFRPSOHWHO\ similar and true model.

Similitude and Model Analysis

361

In practice, it is not possible to achieve complete similarity in models. It is, therefore, common to consider only those forces which are predominant in a phenomenon and design WKHPRGHOVXFKWKDWWKHVDPHIRUFHVLQÀXHQFHWKHÀRZSKHQRPHQRQLQWKHPRGHODOVR7KH H൵HFWV RI DOO RWKHU IRUFHV ZKLFK DUH LQVLJQL¿FDQW DUH HLWKHU QHJOHFWHG RU FRQVLGHUHG E\ D correction factor based on experimentation.

7.4 GEOMETRIC SIMILARITY ‡ :KDWLVJHRPHWULFVLPLODULW\" Geometric similarity implies similarity of shape between the model and prototype. The model is an exact replica of the prototype having identical shape but smaller in size. The model and prototype have same ratio for all corresponding linear dimensions but all included angles are same. For geometric similarity,

hp hm

=

Wp Wm

=

Lp Lm

= Ls = length scale

And angle aP = aM = 60° in this case

Similarity leads to: (1) Area scale ratio: As =

Ap Am

(2) Volume scale ratio: Vs =

=

Vp Vm

Lp ¥ Wp Lm ¥ Wm

=

= Ls2

Ls ¥ W p ¥ hp Lm ¥ Wm ¥ hm

= Ls3

362

Fundamentals of Fluid Mechanics

7.5 KINEMATIC SIMILARITY  ‡ :KDWLVNLQHPDWLFVLPLODULW\" Kinematic similarity means the similarity of motion. For obtaining similarity of motion, both the model and its prototype must produce identical time rates of change of motion. In RWKHUZRUGVWKHUDWLRRIYHORFLWLHVDQGDFFHOHUDWLRQVRIDÀXLGSDUWLFOHDWFHUWDLQSRLQWVLQ the model and at the matching point of prototype must be same in magnitude and direction. For kinematic similarity, we must have: (1) Same velocity ratio, Vs =

Vp Vm

(2) Same acceleration ratio, as =

ap am

(3) Same direction of velocity VP Vm

Prototype

Model

7.6 DYNAMIC SIMILARITY  ‡ :KDWLVG\QDPLFVLPLODULW\" $0,( The dynamic similarity implies that the forces acting on the matching points of the prototype and its model are equal in magnitude and direction. Dynamic similarity can be said to exist EHWZHHQWKHPRGHODQGLWVSURWRW\SHLIWKHUDWLRVRIDOOIRUFHVDFWLQJRQFRUUHVSRQGLQJÀXLG particles or corresponding boundary surfaces of the model and prototype are identical. Both geometric and kinematic similarities are prerequisites for dynamic similarity. For dynamic similarity: (a) Same force ratio, Fs =

Fp Fm

(b) Same direction of forces.

Similitude and Model Analysis

363

WP FP RP

Fm

Prototype

Wm Rm Model

Fs =

Fp Fm

=

Wp Wm

=

Rp Pm

‡ '\QDPLFVLPLODULW\LVWKHVLPLODULW\RIIRUFHV([SODLQYDULRXVIRUFHVDFWLQJRQDIOXLG HOHPHQW In dynamically similar systems, the magnitude of forces at the corresponding similar points in HDFKV\VWHP SURWRW\SHDQGPRGHO DUHLQD¿[HGUDWLR,QDV\VWHPLQYROYLQJÀXLGÀRZGL൵HUHQW IRUFHVGXHWRGL൵HUHQWFDXVHVPD\DFWRQDÀXLGHOHPHQW7KHVHIRUFHVDUHDVSHU7DEOH Table 7.1 'LႇHUHQWIRUFHVDFWLQJRQÀXLGHOHPHQW S. No.

Force

Cause of Force

Force Parameters

1.

9LVFRXVIRUFH

9LVFRVLW\

FV μ m ◊v ◊l

2.

3UHVVXUHIRUFH

3UHVVXUHGLႇHUHQFH

FP μ DP ◊l 2

3.

*UDYLW\IRUFH

*UDYLWDWLRQDODWWUDFWLRQ

Fg μ r◊ l 3 ◊ g

4.

&DSLOODU\IRUFHRUVXUIDFH WHQVLRQIRUFH

6XUIDFHWHQVLRQ

Fs μ s ◊l

5.

&RPSUHVVLELOLW\RU HODVWLFIRUFH

(ODVWLFLW\

Fe μ E ◊l 2

6.

,QHUWLDIRUFH

$FFHOHUDWLRQ

FL μ r◊ l 2 ◊V 2

1. ,QHUWLD )RUFH 7KH LQHUWLD IRUFH DFWLQJ RQ D ÀXLG HOHPHQW LV HTXDO WR WKH PDVV RI WKH element multiplied by its acceleration & 3 Fi = m ¥ a = rl ¥ a a=

V V V2 = = t l /V l

& V2 = r◊ l 2 ◊V 2 Fi = (r◊ l 3 ) ¥ l

364

Fundamentals of Fluid Mechanics Inertia force (man-falling backward)

Accerlation

&  9LVFRXV)RUFH ( FV ) 7KHYLVFRXVIRUFHDULVHVIURPVKHDUVWUHVVLQDÀRZRIÀXLG9LVFRXV force can be written, & ( FV ) = Shear stress ¥ Surface area The shear stress, t = viscosity ¥ rate of shear strain ÊV ˆ μ m ¥ velocity gradient Á ˜ Ël¯  7KHYLVFRXVIRUFHFDQEHVLPSOL¿HG FV μ m◊

V 2 ◊l l

μ m◊V ◊ l Moving plate Viscous force (FV)

V Applied force

l

Viscous fluid (m)

Stationary plate

&   3UHVVXUH)RUFH ( Fp )  7KHSUHVVXUHIRUFHDULVHVGXHWRGL൵HUHQFHRISUHVVXUHLQDÀRZ ¿HOG & Pressure force FP = DP ¥ area = DP ◊ l 2 F2

F1

Piston (A2) F1 = P × A A2 F 2 = P × A 2 A5 = A2 >> A1 Hence, F2 >> F1

A1 Piston (A1)

Fluid

Similitude and Model Analysis

365

&   *UDYLW\)RUFH ( Fg )  7KHJUDYLW\IRUFHRQDÀXLGHOHPHQWLVLWVZHLJKW Fg = r (volume) ¥ g 3 μ = r◊l ◊ g

Fruit falling (gravity force)

  6XUIDFH7HQVLRQ)RUFH Fs  The surface tension force acts tangentially to a surface. Fs μ s ◊ 2pr μ r◊l

T

T

r

  (ODVWLF)RUFH Fe  (ODVWLFIRUFHDULVHVGXHWRWKHFRPSUHVVLELOLW\RIWKHÀXLGLQWKHFRXUVH ofÀRZ)RUDJLYHQFRPSUHVVLRQWKHLQFUHDVHLQSUHVVXUHLVSURSRUWLRQDOWRWKHEXONPRGXOXV of elasticity (E), i.e., DP μ E Force Fe μ DP ¥ area μ E ◊ l2

V

V 2

366

Fundamentals of Fluid Mechanics

 ‡ :KDWDUHVLPLODULW\ODZVRUPRGHOODZV" $0,(   7KHVLPLODULW\LVGL൶FXOWWREHDFKLHYHGLQDOOUHVSHFWVDV (1) For the dynamic similarity, the ratio of corresponding forces acting at the matching points of the model and the prototype must be equal. (2) The dimensionless numbers should be have same magnitude for the model and prototype. %XW LW LV YHU\ GL൶FXOW WR KDYH DOO GLPHQVLRQOHVV QXPEHUV VXFK DV 5H\QROGV QXPEHU Froude’s number, Euler’s number, Weber’s number and Mach’s number same for both model and prototype. (3) The models are therefore designed on the basis of the few forces out of all forces such as inertia force, viscous force, elastic force, pressure force and gravitational force which DUHSUHGRPLQDQWLQWKHÀRZVLWXDWLRQ The laws on which models are designed for dynamic similarity are called model laws or similarity laws. The model laws are: (1) Reynolds model law, (2) Froude’s model law, (3) Euler’s model law, (4) Weber’s model law, and (5) Mach’s model law.

7.7 REYNOLDS MODEL LAW  ‡ :KDWLV5H\QROGVPRGHOODZ" If the model is designed on the basis of Reynolds number then the model is said to be based on Reynolds model law. Reynolds model law is used whenever inertia and viscous forces DUHSUHGRPLQDQWLQWKHÀXLGDVFRPSDUHGWRRWKHUIRUFHV (Re)prototype = (Re)model

r p Vp Lp mp

=

rm Vm Lm mm

r p Vp Lp ◊ ◊ Pm Vm Lm =1 mp mm rr Vr Lr = 1 r = scale ratio mr

\

 ‡ )LQGWKHYDOXHVRIIROORZLQJE\WKH5H\QROGVPRGHOODZ  DFFHOHUDWLRQVFDOHUDWLR   IRUFHVFDOHUDWLRDQG  GLVFKDUJHVXFKUDWLR (1) Acceleration scale ratio, ar = But,

a =

ap am Velocity V = Time T

Similitude and Model Analysis

367

Vp \

ar =

=

(2) Force scale ratio,

Fr = =

=

Tp Vp T V = ◊ m = r Vm T Tr Vm p Tm Vr Tr Fp Fm mp a p mm am r p L3p a p rm L3m am

= rr L r 3 ar (3) Discharge scale ratio,

Qr =

=

Qp Qm r p L3p V p rm L3m Vm

= rr Lr3 Vr3  ‡ :KHUHFDQZHXVH5H\QROGVPRGHOODZ"   7KH5H\QROGVPRGHOODZFDQEHXVHGIRUWKHIROORZLQJÀXLGÀRZVLWXDWLRQV   )ORZRIÀXLGLQSLSHVÀRZPHWHUVDQGIDQV (2) Flow around the submerged bodies such as aeroplanes and submarines.

7.8 FROUDE’S MODEL LAW  ‡ :KLFKLV)URXGH¶VPRGHOODZ"   :KHQHYHULQHUWLDDQGJUDYLWDWLRQDOIRUFHVDUHSUHGRPLQDQWLQWKHÀXLGWKH)URXGH¶VQXPEHU for the model and the prototype must be equal for dynamic stability. The Froude’s model law is based on the Froude’s number for establishing similarity between prototype and its model. (F)protype = (F)model

Vp gL p

=

Vm gLm

368

Fundamentals of Fluid Mechanics

Vp Vm

or

Lp

=1

Lm Vr

or

Lr

or

=1

Lr

Vr =

 ‡ )LQG WKH IROORZLQJ E\ WKH )URXGH¶V PRGHO ODZ   DFFHOHUDWLRQ VFDOH UDWLR  GLVFKDUJHVFDOHUDWLR  IRUFHVFDOHUDWLRDQG  SUHVVXUHVFDOHUDWLR ap 1. Acceleration scale ratio, ar = am

or

ar =

=

ÊV ˆ ÁË T ˜¯ p ÊV ˆ ÁË T ˜¯ m Lr

=

1 Lr

Vp Vm

1 ◊ Tp Tm

=1

Lp Vp

As,

Vm Vp

But,

Vm

Hence,

=

=

Tr =

Tp 1 = Lr ¥ , Lm Tr Tm Lr and Lr

2. Discharge scale ratio, 3 Ê L3p ˆ 1 Ê Tm ˆ Ê Lp ˆ Qr = = Á ˜ ¥ Á 3 ˜ = Á ˜ ◊ T Ê pˆ Qm Ë Lm ¯ Ë Lm ¯ Ë Tp ¯ ÁË T ˜¯ m 1 = Lr3 ¥ = L r5/2 Lr

Qp

Similitude and Model Analysis

369

3. Force scale ratio, Fr =

Fp Fm

=

r p L2p V p2 rm L2m Vm2

Generally, r p = rm GHQVLW\RIWKHÀXLG 2

ÊL ˆ ÊV ˆ Fr = Á p ˜ ¥ Á p ˜ Ë Lm ¯ Ë Vm ¯

\

2

= L r2 ◊ L r = L r3 4. Pressure scale ratio, pr =

Pp

=

Pm

r p V p2 rm Vm2

Generally, r p = r m = dHQVLW\RIÀXLG 2

Ê Vp ˆ = Lr pr = ÁË V ˜¯

\

m

7.9 EULER’S MODEL LAW  ‡ :KDWLV(XOHU¶VPRGHOODZ" Euler’s model law is applicable to the models which are designed on the basis of the Euler’s QXPEHU :KHUHYHU SUHVVXUH DQG LQHUWLDO IRUFHV DUH SUHGRPLQDQW LQ D ÀXLG IRU G\QDPLF stability the Euler’s number of the model and its prototype must be equal. (E)prototype = (E)model

Vp Pp

=

rp

Vm Pm rm

Generally, rp = rm GHQVLW\RIWKHÀXLG

Vp Vm

¥

1 Pp

=1

Pm ‡ :KHUHFDQ(XOHU¶VPRGHOODZEHDSSOLHG"   (XOHU¶VPRGHOODZLVDSSOLFDEOHLQWKHIROORZLQJÀXLGPHFKDQLFVVLWXDWLRQV (1) To avoid water hammer phenomenon (2) Ascertaining pressure distribution on ship

370



Fundamentals of Fluid Mechanics

(3) To avoid cavitation phenomenon (4) To ascertain pressure forces on aircraft wings and fan blades   7RDVFHUWDLQIXOO\WXUEXOHQWÀRZLQFDVHRIFORVHGSLSHV

7.10 WEBER’S MODEL LAW  ‡ :KDWLV:HEHU¶VPRGHOODZ" The models which are based on the Weber’s number, they obey Weber’s model law. :KHQHYHU LQHUWLD DQG VXUIDFH WHQVLRQ IRUFHV DUH SUHGRPLQDQW LQ D ÀXLG IRU G\QDPLF similarity Weber’s number for the model and its prototype must be equal. (W)protoype = (W)model

Vp

=

sp r p Lp

Vm sm rm Lm

‡ :KHUHFDQ:HEHU¶VPRGHOODZEHDSSOLHG" Weber’s model law can be applied for: (1) Study of capillary movement of water in soil    6WXG\RIÀRZRYHUZHLUVIRUVPDOOKHDGV (3) Study of droplets and very small jet (4) Study of capillary waves in channels

7.11 MACH’S MODEL LAW  ‡ :KDWLV0DFK¶VPRGHOODZ" Mach’s model law is applicable for the models which are based on Mach’s number. :KHQHYHULQHUWLDDQGHODVWLFIRUFHVDUHSUHGRPLQDQWLQWKHÀXLG0DFK¶VQXPEHUIRUPRGHO and its prototype must be equal for dynamic stability. (M)prototype = (M)model

Vp Ep rp

=

Vm Em rm

where E = Elastic modulus

‡ :KHUHFDQ0DFK¶VPRGHOODZEHDSSOLHG" The Mach’s model law can be applied for:   6WXG\RIÀRZRIDLURQWKHDHURSODQHVZLWKVXSHUVRQLFVSHHG (2) Study of movement of torpedoes underwater (3) Overcoming the problem of water hammer phenomenon

Similitude and Model Analysis

371

(4) Study of movement of rockets and missiles (5) aerodynamic testing.  ‡ 'H¿QH VLJQL¿FDQFH DQG DUHD RI DSSOLFDWLRQ RI 5H\QROGV QXPEHU )URXGH¶V QXPEHU 0DFK¶VQXPEHU:HEHU¶VQXPEHUDQG(XOHU¶VQXPEHU ,(6   7KHV\PEROJURXSRIYDULDEOHVVLJQL¿FDQFHDQG¿HOGRIDSSOLFDWLRQRIYDULRXVGLPHQVLRQDO numbers are as per Table 7.2 Table 7.2 'LPHQVLRQOHVVQXPEHUV Dimensionless Number 5H\QROGV QXPEHU

Symbol

Group of Variables

RH

r◊V ◊ L m

)URXGH¶VQXPEHU

FU

V L ◊g

(XOHU¶VQXPEHU

E

V P r

:HEHU¶VQXPEHU

W

rV 2L s

0DFK¶VQXPEHU

M

V k /r

6LJQL¿FDQFH ,QHUWLDO IRUFH 9LVFRXV IRUFH ,QHUWLDO IRUFH *UDYLW\ IRUFH ,QHUWLDO IRUFH 3UHVVXUH IRUFH

,QHUWLD IRUFH 6XUIDFH WHQVLRQ IRUFH ,QHUWLD IRUFH (ODVWLF IRUFH

Field of Application :KHUHYLVFRXVHႇRUWV DUHVLJQL¿FDQW :KHUHJUDYLW\HႇRUWV DUHVLJQL¿FDQW &RQGXLWÀRZ

:KHUHVXUIDFHWHQVLRQ LVVLJQL¿FDQW :KHUHFRPSUHVVLELOLW\ HႇRUWLVVLJQL¿FDQW

 ‡ :KDWLVPHDQWE\JHRPHWULFDONLQHPDWLFDQGG\QDPLFVLPLODULWLHV"$UHWKHVHVLPLODULWHV WUXHO\DWWDLQDEOH",IQRWZK\" $0,) For geometrical similarity, the ratio of corresponding length in the model and prototype must be same and included angles between two corresponding sides must be same. For kinematic similarity, there should be similarity of motion, i.e., the direction of velocity DQGDFFHOHUDWLRQDWFRUUHVSRQGLQJSRLQWVLQWKHWZRÀRZVVKRXOGEHWKHVDPH (a ) (a ) (V1 )m (V2 )m and 1 m = 2 m = (a1 ) P (a1 ) P (V1 ) P (V2 ) P  '\QDPLFVLPLODULW\LVWKHVLPLODULW\RIIRUFHVDWWKHFRUUHVSRQGLQJSRLQWVLQWKHÀRZV (Fgravity ) m ( Finertia )m (F ) = Force ratio = viscous m = ( Finertia ) P ( FViscous ) P ( Fgravity ) P 

These similarities are not truely attainable for the following reasons: D  7KHJHRPHWULFVLPLODULW\FDQEHIXOO\DWWDLQDEOHZKHQVXUIDFHURXJKQHVVSUR¿OHVDUHDOVRLQ WKHVDPHUDWLR$VLWLVGL൶FXOWWRSUHSDUHDPRGHOWRKDYHWLPHVEHWWHUVXUIDFH¿QLVKLQ 1:20 scale, hence complete geometrical similarity cannot be achieved.

372

Fundamentals of Fluid Mechanics



E  7KHNLQHPDWLFVLPLODULW\LVPRUHGL൶FXOWWRDFKLHYHDVWKHÀRZSDWWHUQDURXQGWKHVPDOO PRGHOWHQGVWREHGL൵HUHQWIURPWKRVHDURXQGODUJHSURWRW\SH (c) Dynamic similarity is almost impossible as Reynolds number and Froude’s number cannot be equated simultaneously.  ‡ 2LORINLQHPDWLFYLVFRVLW\¥ –5 m2VLVWREHXVHGLQDSURWRW\SHLQZKLFKERWKYLVFRXV DQG JUDYLW\ IRUFHV GRPLQDQW$ PRGHO VFDOH RI  LV DOVR GHVLUHG :KDW YLVFRVLW\ RI PRGHOOLTXLGLVQHFHVVDU\WRPDNHWKHERWKWKH)URXGH¶VQXPEHUDQG5H\QROGVQXPEHU the same in model and prototype? If Reynolds number is same, then

rr Vr Lr = 1 or mr

Vr Lr =1 vr

Also Froude’s number is same. Vr =

Lr

Given,

Lr = 4

\

Vr =

Now,

\ \ or

Lr =

4 =2

Vr Lr =1 vr 2¥4 =1 vr 1 vr = 8 vp vm

=

1 8

vm = 8vp = 8 ¥ 5 ¥ 10–5 m2/s = 40 ¥ 10 –5 m2/s  ‡ $PRGHORIVXEPDULQHLVWREHWHVWHGLQDWRZLQJWDQNFRQWDLQLQJVDOWZDWHU,IWKH VXEPDULQHPRYHVDWNPKDWZKDWYHORFLW\VKRXOGWKHPRGHOEHWRZHGIRUG\QDPLF VWDELOLW\ Since the submarine is fully submerged, hence viscous force is predominant and Reynolds model rule is applicable. Given,

L r = 20 rr = 1 mr = 1

Similitude and Model Analysis

Now,

373

rr Vr Lr =1 mr 1 ¥ Vr ¥ 20 =1 1

or

Vr =

Vp

or

=

Vm

1 20 1 20

or

Vm = 20 ¥ Vp = 20 ¥ 30 = 600 km/h  ‡ $PRGHORIUHVHUYRLULVGUDLQHGLQPLQE\RSHQLQJDVOXLFHJDWH7KHPRGHOVFDOHLV +RZPXFKWLPHZRXOGLWWDNHWRHPSW\WKHSURWRW\SH" Here gravitational force is predominant and we have to apply Froude’s model law.

Vr Lr

= 1 and given L r = 256

Vr = But,

256 = 16

Lr = Vr Tr Lr 256 = = 16 Vr 16

\

Tr =

\

Tr =

\

Tp = 5 ¥ 16 = 80 min

Tp Tm

=

Tp 5

= 16

 ‡ $ UHFWDQJXODU SLHU LQ ULYHU LV  P ZLGH DQG  P ORQJ 7KH DYHUDJH GHSWK RI ZDWHU LVP$PRGHOLVEXLOWWRDVFDOHRI7KHYHORFLW\RIIORZLQWKHPRGHOLVPV DQGWKHIRUFHDFWLQJRQWKHPRGHOLV1)LQG D WKHYDOXHVRIYHORFLW\DQGIRUFHRQ SURWRW\SHDQG E WKHKHLJKWRIWKHVWDQGLQJZDYHDWWKHSLHULILWLVPLQPRGHO DQG F FRH൶FLHQWRIGUDJUHVLVWDQFH As gravitational force is predominant, Froude’s model law is applicable Vr =

Lr

374

Fundamentals of Fluid Mechanics

L r = 16 (given) \ \ \

Vr = 16 = 4 Vp =4 Vm Vp = 4 ¥ Vm = 4 ¥ 1 = 4 m/s

Force acting is gravitational force which is

\ Now,

F = mg = r L3g Fr = L r3 = 16 3 = 4096 Fp = 4096 ¥ 5 N = 20480 N Vr =

Lr =

\

hr = 4

\

hp = 4 ¥ =4¥

or

hr

hm 0.09

= 4 ¥ 0.3 = 1.2 m hp = 1.44 m = standing wave height 2

Drag force = CD r A V 2

3ˆ Ê 2 12 Drag force = 5 = CD ¥ 1 ¥ 103 ¥ Á ¥ ˜ ¥ Ë 16 16 ¯ 2 \

CD =

5 ¥ 2 ¥ 16 ¥ 16 103 ¥ 6

= 42.67 ¥ 10 –2 = 0.427  ‡ $PRGHORIUHVHUYRLUHPSWLHGLQPLQ,IWKHPRGHOVFDOHLVWKHWLPHWDNHQE\ WKHSURWRW\SHWRHPSW\LWVHOIFRXOGEH (a) 250 min (b) 50 min  F  PLQ G  PLQ ,(6 

Similitude and Model Analysis

Tr =

Tp Tm

=

375

Lr

L r = 25 (given) Tr = Tr =

25 = 5 Tp Tm

=5

TP = 5 ¥ Tm = 5 ¥ 10 = 50 min Option (b) is correct.  ‡ $RFHDQOLQHUPORQJKDVPD[LPXPVSHHGRIPV7KHWRZLQJVSHHGRIDPRGHO PORQJWRVLPXODWHWKHZDYHUHVLVWDQFHVKRXOGEH  D  PV E  PV  F  PV G  PV 173& Guidance: The wave resistance is a function of Froude’s number. In order to determine the model speed, Froude model law is used

Vr Lr

=1

Lr =

250 = 25 10

Vr =

25 = 5

Vp Vm

=5

Vm =

15 = 3 m/s 5

Option (b) is correct.  ‡ $ K\GUDXOLF PRGHO RI D FDSLOODU\ LV FRQVWUXFWHG ZLWK D VFDOH  ,I WKH SURWRW\SH GLVFKDUJHLVP3VWKHQWKHFRUUHVSRQGLQJGLVFKDUJHIRUZKLFKWKHPRGHOVKRXOG be tested is  D  P3/s (b) 2 m3/s G  P3V &LYLO6HUYLFHV  F  P3V Here Froude’s model law is applicable

Vr Lr

=1

376

Fundamentals of Fluid Mechanics

Vr =

16 = 4

L3r Qr = Tr L r = 16, Tr = Qr =

\

Qp Qm

163 = 1024 4

= 1024

Qm =

\

16 Lr = =4 4 Vr

2048 = 2 m3/s 1024

Option (b) is correct.  ‡ :KDW IORZ UDWH LQ P2V LV QHHGHG XVLQJ D  VFDOH PRGHO RI D GDP RYHU ZKLFK  m3VRIZDWHUIORZV"  D   E    F   G   ,(6 Froude’s model law is to be used.

Vr Lr

=1

Vr = Tr = Qr =

Lr =

20 = 4.472

Lr 20 = = 4.472 Vr 4.472 (Lr )3 203 = Tr 4.472

= 1.789 ¥ 10 3

Qp Qm

= 1.789 ¥ 10 3

Qm =

4 = 2.2 ¥ 10 –3 1.789 ¥ 103

= .0022 m3/s Option (d) is correct.

Similitude and Model Analysis

377

 ‡ ,Q  PRGHO RI D VWLOOLQJ EDVLQ WKH KHLJKW RI WKH K\GUDXOLF MXPS LQ WKH PRGHO LV REVHUYHGWREHP7KHKHLJKWRIWKHK\GUDXOLFMXPSLQWKHSURWRW\SHZLOOEH  D  P E  P  F  P G  P ,(6 Given, L r = 20 Now,

hp hm

= L r = 20

\ hp = 20 ¥ 0.2 = 4 m Option (c) is correct.  ‡ 7KHPRGHORIDSURSHOOHUPLQGLDPHWHUFUXLVLQJPVLQDLULVWHVWHGLQDZLQG WXQQHORQDVFDOHPRGHO,IWKHWKUXVWRI1LVPHDVXUHGRQWKHPRGHODWPV ZLQGVSHHGWKHQWKHWKUXVWRQWKHSURWRW\SHZLOOEH  D  1 E  1  F  1 G  QRQH ,(6 Fr = r r Vr2 L r2 2

10 10 = 1 ¥ ÊÁ ˆ˜ ÊÁ ˆ˜ Ë 5¯ Ë 5¯

2

4 = 10 5

Fr =

\

Fp =

Fp Fm

=

104 52

Fm ¥ 104 50 ¥ 10000 = 25 25

= 20,000 N Option (a) is correct.  ‡ :KDWGR\RXXQGHUVWDQGE\GLVWRUWHGPRGHO" A distorted model is one which has its one or more characteristics not similar to the FRUUHVSRQGLQJ FKDUDFWHULVWLFV RI WKH SURWRW\SH $ PRGHO KDYLQJ GL൵HUHQW KRUL]RQWDO DQG vertical scale ratio is a distorted model. In order to predict the performance of a prototype, the law of distortion has to be applied to the results obtained from the model test. The GLVWRUWLRQFDQEH  JHRPHWULFGLVWRUWLRQLHKRUL]RQWDODQGYHUWLFDOVFDOHVDUHGL൵HUHQW   K\GUDXOLFGLVWRUWLRQLHÀXLGVLQPRGHODQGSURWRW\SHVDUHGL൵HUHQW  PDWHULDOGLVWRUWLRQ LH PDWHULDOV LQ PRGHO DQG SURWRW\SH DUH GL൵HUHQW DQG   FRQ¿JXUDWLRQ GLVWRUWLRQ LH VORSHLQPRGHODQGSURWRW\SHLVGL൵HUHQW

378

Fundamentals of Fluid Mechanics

 ‡ +RZDUHVFDOHUDWLRVIRUGLVWRUWHGPRGHOVXWLOL]HGIRU¿QGLQJ  YHORFLW\  YHUWLFDO FURVVVHFWLRQ  GLVFKDUJHDQG  WLPHRIIORZ In case the horizontal scale for non-tidal model be 1: m, then

Lp Lm

=

Bp Bm

= m (where L = length & B = breath)

In case the vertical scale is 1: n, then

hp

= n (where h = height)

hm

Generally, the horizontal scale is greater than the vertical scale, i.e., m > n (1) Velocity ratio: As per Froude’s law

Vp Vm

Ê hp ˆ = Vr = Á ˜ = Ë hm ¯

n

(2) If A is vertical cross section, then

Ê Ap ˆ B p ¥ hp = =m¥n ÁË A ˜¯ Bm ¥ hm m vertical (3) If Q is the discharge, then

Qp Qm

=

V p ¥ Ap Vm ¥ Am

=

n ¥m¥n

= m n3/2 (4) If T is the timeRIÀRZWKHQ

Lp Tp Tm

=

Tr =

Vm Lp Vp = ¥ Vp Lm Lm Vm m n

‡ $KRUL]RQWDOPRGHOKDVKRUL]RQWDOUDWLRRIDQGDYHUWLFDOVFDOHUDWLRRI ,IWKHIORRGSHDNUHTXLUHVKRXUVWRWUDYHODGLVWDQFHNPLQWKHPRGHO¿QGWLPH WKHIORRGSHDNZLOOWDNHLQWKHDFWXDOULYHU 3DWQD8QLYHUVLW\ Here,

m = 2500 & n = 225

Similitude and Model Analysis

Tr =

Tp Tm \

=

Tp =

2500 225

=

379

2500 15

500 3 500 ¥ 10 = 1666.67 hours 3

= 69 days 11 hours 40 min ‡ A model of spillway is built to a scale of 1:36. If the model velocity and discharge are 1.25 m/s and 2.5 m3/s respectively, what are the corresponding values for the prototype? (Punjab University) Applying Froude’s model law Vr = =

Lr 36 = 6

Q r = Vr (L r)2 = 6 ¥ 362 = 7776 Vr = \

Vp Vm

=6

V p = 6 ¥ 1.25 = 7.50 m/s Qr =

Qp Qm

= 7776

Q p = 7776 ¥ 2.5 m3/s = 19440 m3/s  ‡ A model of a reservoir is drained in 6 min by operating the sluice gate. How long should it take to empty the prototype if the scale ratio is 1:256? (Poona University) Applying Froude’s model law Vr = =

Lr 256

= 16 Tr =

Lr 256 = = 16 Vr 16

380

Fundamentals of Fluid Mechanics

Tr = or

Tp Tm

= 16

Tp = 16 ¥ 6 = 96 min

 ‡ $JHRPHWULFDOPRGHORIDVXUIDFHYHVVHOLVWHVWHGLQDODERUDWRU\7KHOLQHDUVFDOHRIWKH PRGHOLV,WLVREVHUYHGWKDWZLWKDVSHHGRIPVWKHUHVLVWDQFHRIWKHPRGHO LV17KHOLTXLGXVHGIRUWKHWHVWLVWKHVDPHDVWKDWRQHZKLFKWKHVXUIDFHYHVVHOLV WRVDLO&DOFXODWHWKHFRUUHVSRQGLQJVSHHGDQGWKHUHVLVWDQFHRIPRWLRQRIWKHVXUIDFH YHVVHO&RQVLGHUWKHH൵HFWRIJUDYLW\RQO\ 3XQMDE8QLYHUVLW\   $VH൵HFWRIJUDYLW\LVWREHFRQVLGHUHGWKH)URXGH¶VPRGHOODZLVDSSOLFDEOH Vr =

Vp Vm

Lr =

49 = 7

= 7, or Vp = 7 ¥ 10 = 70 m/s

Fr = r r Vr2 L r2

Fr = resistance force scale

= 72 ¥ 492 Fp = 49 ¥ 492 ¥ 3

Now,

as

Fm = 3 N (given)

= 353 kN  ‡ (VWLPDWH D WKHVSHHGRIURWDWLRQ E WKHWKUXVWSURGXFHG F WKHWRUTXHGHYHORSHG DQG G WKHH൶FLHQF\RISURSXOVLRQE\DPGLDPHWHUSURSHOOHUWRFUXLVHDWPVLI DVFDOHPRGHOSURGXFHGWKHIROORZLQJUHVXOWV

For dynamic similitude,

Vm PV Nm = 750 rpm Fm = 50 N and Tm 1P

Nr dr =1 Vr d r = 10 (given) Vr =

Vp Vm

=

10 =2 5

N r ¥ 10 =1 2 \

Nr =

1 5

Np =

1 ¥ 750 = 150 rpm 5

Similitude and Model Analysis

Now,

Fr = Pr Vr2 N r2 2

381

where Fr = thrust ratio scale.

= 1 ¥ 2 ¥ 10

2

= 400 F p = 400 ¥ 50 = 20 kN where Fp = thrust produced Tr = Torque ratio = r r N r2 dr5 2

Ê 150 ˆ =1¥ Á ¥ (10)5 Ë 750 ˜¯ =

100 ¥ 103 25

= 40 kN 3URWRW\SHH൶FLHQF\

=

Fp ¥ V p T ¥ 2 pN /60

=

20 ¥ 10 ¥ 60 40 ¥ 2p ¥ 150

= 31.8% 0RGHOH൶FLHQF\

=

50 ¥ 5 ¥ 60 Fm ¥ Vm = T ¥ 2 pN 10 ¥ 2p ¥ 750 60

= 31.8%  ‡ 7KHGUDJRIDVPDOOVXEPHUJHGKXOOLVGHVLUHGZKHQLWLVPRYLQJIDUEHORZWKHVXUIDFH RI ZDWHU $  VFDOH PRGHO LV WR EH WHVWHG :KDW GLPHQVLRQOHVV JURXS VKRXOG EH GXSOLFDWHG EHWZHHQ WKH PRGHO DQG SURWRW\SH DQG ZK\" ,I WKH GUDJ RI WKH SURWRW\SH DWNQRWLVGHVLUHGDWZKDWVSHHGVKRXOGWKHPRGHOEHPRYHGWRJLYHWKHGUDJWREH H[SHFWHGE\WKHSURWRW\SH":RXOGWKLVUHVXOWVWLOOEHWUXHLIWKLVSURWRW\SHZHUHWREH FORVHGWRWKHVXUIDFH"([SODLQ When the body is moving fully submerged, the viscous forces are predominant and Reynolds model law is applicable. If the prototype is moving close the surface, then gravitational forces are also acting due to wave resistance, thereby drag force also contributes to the resistance Using Reynolds model law rr Vr Lr =1 mr Now, L r = 10 and r r = 1 and m r = 1 Vr L r = 1 or

Vr =

1 10

382

Fundamentals of Fluid Mechanics

Vp Vm Now,

=

1 10

Vp = 1 Knot Vm = 10 Vp Knot = 10 Knot F r = Drag ratio = rr Vr2 Lr2 =1¥

1 ¥ 100 100

=1 Hence, drag force experienced by the prototype at 1 knot and model at 10 knot is same.  ‡ &DOFXODWH WKH VSHHG RI URWDWLRQ WKH WRUTXH SURGXFHG DQG WKH SRZHU RI D ZLQGPLOO RI  P GLDPHWHU LQ D ZLQG VSHHG RI  NPKU IURP WKH SHUIRUPDQFH RI D  VFDOH JHRPHWULFDOPRGHOLQDZLQGWXQQHOZLWKPVIUHHVWUHDP7KHPRGHOURWDWHGDW USPSURGXFHGDWRUTXHRI1P Apply speed parameter for the model & the prototype.

N r Dr =1 Vr Dr =

Dp Dm

= 10

Vp = 30 kMh =

30 ¥ 103 = 8.33 m/s 3600

Vm = 10 m/s Vr =

Now,

\

Vp Vm

=

8.33 = 0.833 10

N r Dr =1 Vr Nr =

=

Vr Dr 0.833 = 0.0833 10

Similitude and Model Analysis

Np Nm

383

= 0.0833

N p = 1200 ¥ 0.0833 = 100 rpm Tr = Torque ratio = r r N r2 D r5 2

Ê 100 ˆ =1¥ Á (10)5 Ë 1200 ˜¯ Tp = Tm ¥ =

1 ¥ 105 144

2.5 ¥ 100 kNm 144

= 1.736 kNm Power of wind mill = Tp ¥ w = Tp ¥ = 1736 ¥

2p NP 60

100 ¥ 2p 60

= 18.2 kW  ‡ $QR൵VKRUHRLOGULOOLQJSODWIRUPLVH[SHFWHGWRHQFRXQWHUZDYHVRIPKHLJKWDW +]IUHTXHQF\DQGDVWHDG\FXUUHQWRIPV'HWHUPLQHWKHSDUDPHWHUVIRUWKHPRGHO ZDYHFKDQQHOZKHUHDRQHVL[WHHQWKPRGHORIWKHSODWIRUPFDQEHWHVWHG Froude model scale is to be applied as gravitational force is predominant

Vr Lr

= 1 or Vr = Vr =

Now, \

Lr 16 = 4

Vp = 1 m/s (given) Vm =

Vp

= 0.25 m/s

4

h r = Lr = 16 hm = Frequency ratio

hp 16

=

4 = 0.25 m 16

= fr =

1 V = r Tr Lr

384

Fundamentals of Fluid Mechanics

fr =

4 1 = 16 4

= fr =

1 4

fm = 4 ¥ fp = 4 ¥ 0.1 = 0.4 Hz  ‡ $QDXWRPRELOHPRYLQJDWDYHORFLW\RINPKULVH[SHULHQFLQJDZLQGUHVLVWDQFHRI  N1 ,I WKH DXWRPRELOH LV PRYLQJ DW D YHORFLW\ RI  NPKU WKH SRZHU UHTXLUHG WR RYHUFRPHWKHZLQGUHVLVWDQFHLV  D  N: E  N:  F  N: G  N: ,(6 F = r ◊ V 2 ◊ d2 F2 ÊV ˆ r V22 d 2 = = Á 2˜ 2 2 F1 r V1 d Ë V1 ¯

\

Ê 50 ˆ F2 = 2 Á ˜ Ë 40 ¯

or

2

2

= 3.125 kN Ê 50 ¥ 103 ˆ Power = 3.125 ¥ Á ˜ kW Ë 3600 ¯ = 43.4 kW Option (a) is correct.  ‡ $VFDOHPRGHORIDUHVHUYRLULVGUDLQHGLQPLQE\RSHQLQJWKHVOXLFHJDWH7KH WLPHUHTXLUHGWRHPSW\WKHSURWRW\SHZLOOEH  D  PLQ E  PLQ  F  PLQ G  PLQ ,(6 tμ

tr tm

=

=

h hr hm 256 1

Similitude and Model Analysis

\

tr =

385

256 ¥ tm

tr = 16 ¥ 4 = 64 min Option (b) is correct.  ‡ $VKLSPRGHOVFDOHZLWKQHJOLJLEOHIULFWLRQLVWHVWHGLQDWRZLQJWDQNDWDVSHHGRI PLQV,IDIRUFHRINJLVUHTXLUHGWRWRZWKHPRGHOWKHSURSXOVLYHIRUFHUHTXLUHG WRWRZSURWRW\SHVKLSZLOOEH  D  01 E  01  F  01 G  01 ,(6 As per Froude’s law for dynamic similarity, we have Ê V ˆ Ê V ˆ Á ˜ = Á ˜ Ë gl ¯ m Ë gl ¯ r or Now, or

\

Vm = Vr

lm = lr

1 60

F = r V2 L2 Ê Fm ˆ Ê Vm ˆ Á ˜ = Á ˜ Ë Fr ¯ Ë Vr ¯

2

Ê Lm ˆ Á ˜ Ë Lr ¯

Ê Vr ˆ Fr = Fm ¥ Á ˜ ËV ¯

2

m

= 0.5 ¥ 9.81 ¥

2

Ê Lr ˆ ÁË L ˜¯

2

m

2

( 60 )

¥ (60)2

= 0.5 ¥ 9.81 ¥ 63 ¥ 103 = 1.06 ¥ 106 N = 1.06 MN ª 1 MN Option (c) is correct.  ‡ $PRGHOLVWREHFRQGXFWHGLQDZDWHUWXQQHOXVLQJDPRGHORIDVXEPDULQHZKLFK LVWRWUDYHODWDVSHHGRINPKGHHSXQGHUVHDVXUIDFH7KHZDWHUWHPSHUDWXUHLQ WKHWXQQHOLVPDLQWDLQHGVRWKDWLWVNLQHPDWLFYLVFRVLW\LVWKHKDOIWKDWRIVHDZDWHU$W ZKDWVSHHGLVWKHPRGHOWHVWWREHFRQGXFWHGWRSURGXFHXVHIXOGDWDIRUWKHSURWRW\SH"  D  NPK E  NPK  F  NPK G  NPK ,(6

386

Fundamentals of Fluid Mechanics

(Re)model = (Re)prototype Ê r ◊V ◊ d ˆ Ê rVd ˆ ÁË m ˜¯ = ÁË m ˜¯ r M Ê d r ˆ Ê m m ˆ Ê rr ˆ Vm = Vp Á Á ˜ Ë d M ˜¯ Ë m r ¯ ÁË rM ˜¯ Ê 20 ˆ Ê 1 ˆ = 12 Á ˜ Á ˜ Ë 1 ¯ Ë 2¯ = 120 km/h Option (d) is correct.

7.12 QUESTIONS FROM COMPETITIVE EXAMINATIONS  ‡ ,W LV REVHUYHG LQ D IORZ SUREOHP WKDW WRWDO SUHVVXUH LQHUWLD DQG JUDYLW\ IRUFHV DUH LPSRUWDQW7KHQVLPLODULW\UHTXLUHVWKDW  D  5H\QROGVDQG:HEHU¶VQXPEHUVEHHTXDO  E  0DFK¶VDQG)URXGH¶VQXPEHUVEHHTXDO  F  (XOHU¶VDQG)URXGH¶VQXPEHUVEHHTXDO  G  5H\QROGVDQG0DFK¶VQXPEHUVEHHTXDO ,(6 Option (c) is correct.  ‡ $ VSKHUH LV PRYLQJ LQ ZDWHU ZLWK D YHORFLW\ RI  PV$QRWKHU VSKHUH RI WZLFH WKH GLDPHWHULVSODFHGLQDZLQGWXQQHODQGWHVWHGZLWKDLUWLPHVOHVVYLVFRXVWKDQZDWHU 7KHYHORFLW\RIDLUWKDWZLOOJLYHG\QDPLFDOO\VLPLODUFRQGLWLRQLV  D  PV E  PV  F  PV G  PV ,(6 Re is to be maintained r ¥ V ¥ 2d r ¥ 1.6 ¥ d 750 = m m 60 = 10 m/s Option (b) is correct.  ‡ 7KHPRGHORIDSURSHOOHUPLQGLDPHWHUFUXLVLQJDWPVLQDLULVWHVWHGLQDZLQG RQDVFDOHPRGHO,IDWKUXVWRI1LVPHDVXUHGRQWKHPRGHODWPVZLQGWKHQ WKUXVWRQWKHSURWRW\SHZLOOEH

Similitude and Model Analysis

 

D  1 F  1

E  1 G  1

387

,(6

Fm Pm ◊ L2m ◊Vm2 = FP PP ◊ L2P ◊VP2 2

Ê 1ˆ Ê 5ˆ = (1) ¥ Á ˜ Á ˜ Ë 10 ¯ Ë 10 ¯

2

50 1 25 1 ¥ = = FP 100 100 400 FP = 50 × 400 = 20,000 N Option (a) is correct.  ‡ $PRGHOWHVWLVWREHFRQGXFWHGLQDZLQGWXQQHOXVLQJDPRGHORIVXEPDULQHZKLFK LVWRWUDYHODWDVSHHGRINPSKGHHSXQGHUVHDVXUIDFH7KHZDWHUWHPSHUDWXUHLQ WKHWXQQHOLVPDLQWDLQHGVRWKDWLWVNLQHPDWLFYLVFRVLW\LVKDOIWKDWRIVHDZDWHU$WZKDW VSHHGLVWKHPRGHOWHVWWREHFRQGXFWHGWRSURGXFHXVHIXOGDWDIRUWKHSURWRW\SH"  D  NPSK E  NPSK  F  NPSK G  NPSK (IES 2002) rm ◊Vm ◊ Lm Ê r ˆ ÊV ˆ Ê L ˆ Ê m ˆ (Re) m mm =Á m˜ Á m˜ Á m˜ Á P˜ = rP ¥ VP ¥ LP Ë rP ¯ Ë VP ¯ Ë LP ¯ Ë m m ¯ (Re) P mP

ÊV ˆ 1 = (1) Á m ˜ Ë 12 ¯

Ê 1ˆ ÁË 20 ˜¯ (2)

Vp = 12 ¥ 20 = 120 kmph 2

or

Option (d) is correct.  ‡ )RUDPVFDOHPRGHORIK\GUDXOLFWXUELQHWKHVSHFL¿FVSHHGRIWKHPRGHONm is related to the prototype sp speed Np as (a) Nm =  

NP m

(b) NM = m  NP

F  Nm = (NP)m (d) Nm = NP  7KHVSHFL¿FVSHHGRIPRGHODQGSURWRW\SHUHPDLQVVDPH Option (d) is correct.

,$6

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 ‡ A

 

 model RIDVKLSLVWREHWHVWHGIRUHVWLPDWLQJWKHZDYHGUDJ,IVSHHGRIVKLSLV 25

PVWKHQVSHed aWZKLFKWKHPRGHOPXVWEHWHVWHGLV D  PV E  PV F  PV G  PV Froude model law: (Fr)m = (Fr)p

VP g ◊ LP

Vm = g ◊ Lm

or

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or

=

,$6,(6

Lm =1¥ LP

1 25

1 = 0.2 m/s 5

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=

=

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(Froude Model Law)

Similitude and Model Analysis

389

4 ¥ 10 = 2 m/s 100

Vm = Option (a) is correct.

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(i)

( Fr )m V = 1, m = ( Fr ) P VP

(ii)

Lm LP

From Eqs. (i) and (ii), we have Ê Lm ˆ ÁË L ˜¯ P



 ‫׵‬

3/ 2

=

nP = 0.0894 nM

Lm = 1 : 5 = model scale LP

Option (d) is correct  ‡ $PRGHORIVSLOOZD\GLVVLSDWHVKS7KHFRUUHVSRQGLQJKRUVHSRZHUGLVVLSDWHG ZLOOEH 

D  

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F  

G  

Prandtl’s number (Pr) remains same ÊL ˆ PP = Á P˜ Pm Ë Lm ¯

3.5

= (20)3.5

PP = 0.25 × (20)35 = 8944 hp.

,(6

390

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E  PLQ



F  PLQ

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Froude’s number (Fr) remains constant (Fr)p = (Fr)m L1/P 2 L1/2 = m TP Tm ÊL ˆ TP = Tm ¥ Á P ˜ ËL ¯

or

1/ 2

= 4 ¥ 256

m

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391

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 Pr /(Urlr)

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1 25

1 = 2 m/s 5

Option (c) is correct.  ‡ $PRGHORIWRUSHGRLVWHVWHGLQDWRZLQJWDQNDWDYHORFLW\RIPV7KHSURWRW\SHLV H[SHFWHGWRDWWDLQDYHORFLW\RIPV:KDWPRGHOVFDOHKDVEHHQXVHG"  D   E    F   G  

394

Fundamentals of Fluid Mechanics

As per Froude’s model law

or

Vm = VP

Lm LP

5 = 25

Lm 1 = Lp 5

Lm:LP = 1:52 = 1:25

Option (b) is correct.  ‡ $PRGHORIDUHVHUYRLULVGUDLQHGLQPLQE\RSHQLQJWKHVOXLFHJDWH7KHPRGHOVFDOH LV+RZORQJVKRXOGLWWDNHWRHPSW\WKHSURWRW\SH"  D  PLQ E  PLQ  F  PLQ G  PLQ The gravitational force is predominant and we have to apply Froude’s model law Vm = Vp Now,

  ‫׵‬

1 1 = 225 15

L =V T Lm Vm Tm L T = m ¥ P = LP VP LP Tm TP =

  ‫׵‬

Lm = LP

T 1 1 ¥ P = 225 Tm 15

Tp = Tm ¥ 225 ¥

1 15

= 4 × 15 = 60 min Option (b) is correct.

Chapter

8

FLUID KINEMATICS

KEYWORDS AND TOPICS            

FLUID KINEMATICS STREAMLINE STREAM TUBE STREAK LINE PATH LINE EQUIPOTENTIAL LINE ROTATION DISTORTION VORTICITY CIRCULATION ROTATION & IRROTATIONAL FLOW STEADY & UNSTEADY FLOW

           

UNIFORM & NON-UNIFORM FLOW LAMINAR & TURBULENT FLOW CONTINUITY EQUATION ONE-DIMENSIONAL FLOW TWO-DIMENSIONAL FLOW THREE-DIMENSIONAL FLOW STREAM FUNCTION POTENTIAL FUNCTION LAPLACE EQUATIONS FLOW NETS FREE VORTEX FORCED VORTEX

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396

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∂y ∂x ∂x v = &w= ∂t ∂t ∂t

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∂ x ∂ y ∂ x  a y =  & az = ∂t ∂t ∂ t

Fluid Kinematics

397

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Stream Line Around a Solid Cylinder

Stream Lines within a Closed Conduct

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Stream Tube

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Streak lines

Path lines

1.

Imaginary lines showing YDULRXVÀXLGSDUWLFOHV positions of

1.

Real line showing instantaneous positions of various particles

1.

Real line showing successive position of one particle.

2.

Particle may change VWUHDPOLQHGHSHQGLQJ RQW\SHRIÀRZ

2.

May change from instant to instant

2.

Particle may cross its path line

3.

Streamlines cannot LQWHUVHFWHDFKRWKHU they are always parallel

3.

Streak line changes with time. Two streak lines may intersect each other

3.

Two path lines for two particles may intersect each other

4.

1RÀRZDFURVV streamline

4.

Flow across streak line is possible

4.

Flow across a path line is possible by other particles.

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Fluid Kinematics Element at original position D

C

Original position D

Element at new position C¢



C





401

New position

B A

B







A A¢

Translation

Translation

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dq - dq = 0 and shear stresV LV ]HUo as strain rate 

 Ê ∂v ∂ u ˆ =0  ÁË ∂ x ∂ y ˜¯ D¢ C



D

dq

B¢ dq A

B

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D

C



dq2



dq1 A

D D¢

B¢ B

Angular Distortion

A

B Volume Distortion

402

Fundamentals of Fluid Mechanics

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∂r ∂v dP ∂ v π  π π  π0 ∂t ∂t ∂t ∂t  ,QVWHDG\ÀRZWKHSDWKOLQHDQGVWUHDPOLQHZLOOFRQFLGH,QXQVWHDG\ÀXLGÀRZWKHSDWKOLQH RIVXFFHVVLYHSDUWLFOHZLOOEHGL൵HUHQWDQGVWUHDPOLQHSDWWHUQRIWKHÀRZZLOOEHFKDQJLQJ DWHYHU\LQVWDQW  ‡ :KDWGR\RXXQGHUVWDQGE\XQLIRUPDQGQRQXQLIRUPIORZ"   8QLIRUPÀRZLVWKHÀRZLQZKLFKYHORFLW\RIWKHÀRZGRHVQRWFKDQJHDORQJLWVGLUHFWLRQ RI ÀRZ DW DQ\ SRLQW RI WLPH$ ÀRZ WKURXJK D FRQVWDQW GLDPHWHU SLSH OLQH LV DQ XQLIRUP ÀRZ)RUXQLIRUPÀRZZHKDYH

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Uniform Flow

Non-Uniform Flow

Fluid Kinematics

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Laminar Flow



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Turbulent Flow

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r VL Inertia force = m Viscous force

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B

A

A B

B

B A

A

A

B

A B

A Irrotational Motion

A Element AB circling but not rotating

B

Element AB circling and rotating

Rotational Motion

8.6 ONE-, TWO- AND THREE-DIMENSIONAL FLOWS  ‡ 'L൵HUHQWLDWHRQHWZRDQGWKUHHGLPHQVLRQDOIORZV  2QHGLPHQVLRQDOÀRZ2QHGLPHQVLRQDOÀRZLVWKHÀRZLQZKLFKWKHYHORFLW\RIWKH ÀRZLVDIXQFWLRQRIWLPHDQGRQHVSDFHFRRUGLQDWH xy or z 7KHÀRZWKURXJKDSLSH LVRQHGLPHQVLRQDOÀRZ)RURQHGLPHQVLRQDOÀRZZHKDYHIRU D  6WHDG\ÀRZV = f x  E  8QVWHDG\ÀRZV = f xt V = f (x)

x axis

One-Dimensional Flow

Fluid Kinematics







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Volume Time Area ¥ Length Time

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Fundamentals of Fluid Mechanics

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A

B¢ E¢

B

E

C¢ D¢

C

2-2 section D 1-1 section



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∂ rAV ds ∂s

Ê ˆ d \ 5DWHRILQFUHDVHRIPDVVZLWKLQWKHVWUHDPWXEH rAV± Á rAV  rAV ds˜ ds Ë ¯ ±

∂  r A◊V ds ∂s

Fluid Kinematics

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∂  r ¥ Ads ∂t

407

∂  GHQVLW\¥YROXPH ∂t

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∂ ∂ rA Kence rAV   ∂t ∂s

E  )RUVWHDG\DQGLQFRPSUHVVLEOHÀRZr is constant As

or or or

∂ rAV = 0 ∂s ∂ AV   r FRQVWDQW ∂s AV = constaQW RQLQWHJUDWLRQ AV = AV

 ‡ 'HULYHWKHFRQWLQXLW\HTXDWLRQLQWKUHHGLPHQVLRQDO&DUWHVLDQFRRUGLQDWHV    8378 Or 'HULYH WKH H[SUHVVLRQ IRU WKH FRQWLQXLW\ HTXDWLRQ IRU WKH VWHDG\VWDWH ' IORZ RI D FRPSUHVVLEOHIOXLG 8378   'HULYDWLRQLVEDVHGRQWKHSULQFLSOHRIFRQVHUYDWLRQRIPDVVZKLFKVWDWHVWKDWWKHTXDQWLW\ RIÀXLGSHUVHFRQGUHPDLQVFRQVWDQWZKHQDÀXLGÀRZVWKURXJKDQ\VHFWLRQRIWKHSLSH

408

Fundamentals of Fluid Mechanics Z E G F

w H

dz

A B Y

v

D dy

u

dx

X

C

Fluid Element

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∂ ru ◊ dy dz dx ∂x

∂ È ˘ ru ◊ d y dz dx ˙ ± ru dy ◊ dz Rate of mass increase in x-direction = Í ru ◊ d y dz + ∂ x Î ˚ =

∂ ru dx dy dz ∂x

 6LPLODUO\ZHFDQ¿QGIRUy and z direction as under Rate in mass increase in y-direction =

∂ rv ◊dx dydz ∂y

Rate in mass increase in z-direction =

∂ rw dx dy dz ∂z

È∂ ˘ ∂ ∂ Total rate in mass increase = dx ◊ dy ◊ dz Í ru + rv + rw ˙ ∂y ∂z Î ∂x ˚  $V WKHUH LV QR DFFXPXODWLRQ RI ÀXLG PDVV WKHUH LV QR PDVV LQFUHDVH DV SHU WKH ODZ RI FRQVHUYDWLRQRIPDVV HencH

È ∂ ru ∂ rv ∂ r w ˘ + + Í ˙ dx ◊ dy ◊ dz = 0 ∂y ∂z ˚ Î ∂x

Fluid Kinematics

409

∂ rv ∂ ru ∂ rw + + =0 ∂y ∂x ∂z

or

If r IRULQFRPSUHVVLEOHÀXLGV

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‡ 'HULYHDFRQWLQXLW\HTXDWLRQLQWZRGLPHQVLRQVLQSRODUFRRUGLQDWHVIRULQFRPSUHVVLEOH IOXLGV Vq +

∂Vq ⋅ dq ∂q

D

Vr +

∂Vr dr ∂r

A C dq Vr

B Vq

q O r r + dr



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410

Fundamentals of Fluid Mechanics

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∂Vq ¥ dq ¥ dr ¥ dt ∂q

± r ¥ ∂Vq Ê rd q ¥ dr ¥ dt ˆ ˜¯ ∂q ÁË r  ±

r ∂Vq (rdq ¥ dr ¥ dt) r ∂q

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∂ PDVV ¥ dt ∂t





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∂ (r ¥ r dq ¥ dr) ¥ dt ∂t

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∂ (r ¥ r dq ¥ dr) ◊ dt ± r ∂t

ÈVr ∂Vr ˘ dr (rdq)dt± r ∂Vq (rdq ¥ dr ¥ d t) Í r + ∂r ˙ r ∂q Î ˚

Fluid Kinematics

If dt 

411

∂  IRUVWHDG\ÀRZDQGr = constant ∂t Vr +

∂ Vr ¥ r + ∂ Vq   ∂r ∂q ∂ ∂ r ¥ Vr  Vq   ∂r ∂q

or

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D



C

dq B¢ dq A

dx

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Ê ˆ ∂u   7KHYHORFLW\FRPSRQHQWVLQx-direction at A and D = u and Á u + ◊ d y˜  ∂y Ë ¯ ∂v Ê ˆ   7KHYHORFLW\FRPSRQHQWVLQy-direction at A and B = v and Á v + ◊ dx˜ Ë ¯ ∂x

412

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ÈÊ ˘ ∂v ˆ Distance BB¢ = ÍÁ v + ◊ dx˜ - v ˙ dt Ë ¯ ∂ x Î ˚ =

∂v dx ◊ dt ∂x

È Ê ˆ˘ ∂u ◊ d y˜ ˙ ¥ dt Distance DD¢ = Íu - Á u + ∂y Ë ¯˚ Î ±

dq =

∂u ◊dy ◊dt ∂y

BB ¢ DD ¢ = dx dy

∂u ∂v ◊ d y ◊ dt ◊ dx ◊ dt ∂y ∂x = dy dx ∂u ∂v ◊ dt ± ◊ dt ∂y ∂x  1RZDQJXODUYHORFLW\Z =

\

dq dt

ZAB

∂v ◊ dt ∂v ∂x = = ∂x dt -

ZAD =

∂u ◊ dt ∂u ∂y  ± dt ∂y

 7KHDYHUDJHRIZAB and ZADZLOOJLYHWKHURWDWLRQRIÀXLGÀRZDERXWzD[LV \

Zz =

=

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Fluid Kinematics

413

 :HFDQDOVR¿QGRXWWKHFRPSRQHQWRIURWDWLRQDERXWx- and yD[LV

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Zx =

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Zy =

 È ∂u ∂w ˘  ÍÎ ∂ z ∂ x ˙˚

Z=

w x + w y + w z

5RWDWLRQDOYHFWRUZ = Zx i + Zy j + Zzk 9RUWLFLW\ ¥ rotation W Z   7KUHHFRPSRQHQWVRIYRUWLFLW\DUH W x Zx =

∂w ∂v ± ∂y ∂z

W y Zy =

∂u ∂w ± ∂z ∂x

W z Zz =

∂u ∂v ± ∂y ∂x

Note: 9RUWLFLW\LQSRODUFRRUGLQDWHVLVW =

∂Vq V  ∂Vr + q ± ∂r r ∂q r

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 ∂Vr V ∂Vq + q + =0 r ∂q r ∂r

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G=

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Ê Ê ∂v ˆ ∂u ˆ = udx + Á v + dx˜ dy± Á u + d y dx± ∂y ¯ ∂ y ˜¯ Ë Ë Ê ∂v ∂ uˆ = Á dx ◊dy Ë ∂ x ∂ y ˜¯ = W z ¥ dx ¥ dy = W z ¥DUHDHQFORVHGE\FORVHGFXUYHLQSODQHDWULJKWDQJOHWRzD[LV   +HQFHYRUWLFLW\LVWKHFLUFXODWLRQSHUXQLWVXUIDFHDUHD \

W=

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8.10

415

STREAM FUNCTION

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Streamline A¢

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V

w

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dx dy dz dx  d y  dz

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= = = =

u dt vdt wdt u v w

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or

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y A

u

v

dy

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∂y ∂y ◊ dy + ◊ dx = 0 ∂y ∂x

Fluid Kinematics

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417

dy = 0

or

y = constant

+HQFHSURYHG

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  1RZWRSURYHWKDWVWUHDPIXQFWLRQUHSUHVHQWVVWUHDPOLQHZHNQRZWKDWVWUHDPOLQHHTXDWLRQ FDQEHJLYHQDV u dy±v dx = 0 

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∂y ∂y and v = ∂y ∂x

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∂y ∂y ◊ dy + ◊ dx = 0 ∂x ∂y or

dy = 0

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∂u ∂v + =  ∂x ∂y



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±

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∂ y ∂ y + =0 ∂x ∂y ∂ y ∂x

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418

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 ‡ +RZy and fLQSRODUFRRUGLQDWHVDUHXVHGWR¿QGYHORFLW\FRPSRQHQWVLQUDGLDODQG WDQJHQWLDOGLUHFWLRQV y = f rq Radial YHORFLW\ur = 7DQJHQWLDOYHORFLW\uq =

 ∂y r ∂q ∂y ∂r

f = f rq 5DGLDOYHORFLW\ur =

∂f ∂r

 ∂f r ∂q Note: )ROORZLQJPD\EHXVHGDVDVLPSOHZD\WRUHPHPEHU 7DQJHQWLDOYHORFLW\uq =

  u ∫ uqv ∫ ur     dx ∫ r dq dy ∫ d r  ‡ 'HULYH/DSODFHHTXDWLRQRIVWUHDPIXQFWLRQIRULUURWDWLRQIORZ   )RUWZRGLPHQVLRQDOÀRZLQ xy SODQHWKHYRUWLFLW\WLVJLYHQE\ W=

∂v ∂u ±  ∂x ∂y

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∂y ∂y

v=+

∂y ∂x

 3XWWKHYDOXHVRIu and vLQHTQ   W=

= FRULUURWDWLRQÀRZW = 0

∂ Ê ∂y ˆ ∂ Ê ∂y ˆ ± Á ˜ ∂ y ÁË ∂ y ˜¯ ∂x Ë ∂x ¯ ∂y  ∂ y + ∂y  ∂ x



Fluid Kinematics

419

∂y  ∂ y + =0 ∂y  ∂ x

\

 7KLVLV/DSODFHHTXDWLRQ  ‡ Enumerate the properties of stream function y   7KHSURSHUWLHVRIVWUHDPIXQFWLRQDUH   yLVFRQVWDQWDWDOOSODFHVLQDVWUHDPOLQH    7KHÀRZDURXQGDQ\SDWKLQWKHÀXLGLV]HUR    7KHYHORFLW\YHFWRUDWDQ\SRLQWRQWKHVWUHDPOLQHFDQEHIRXQGRXWE\GL൵HUHQWLDWLQJ its stream functiRQLHu = ±

∂y ∂y and v = ∂x ∂y

  7KHGLVFKDUJHEHWZHHQWKHVWUHDPOLQHVLVHTXDOWRGL൵HUHQFHRIWKHLUVWUHDPIXQFWLRQV LH q = y±y 

  7KHUDWHRIFKDQJHRIVWUHDPIXQFWLRQZLWKGLVWDQFHLVSURSRUWLRQDOWRWKHFRPSRQHQW RIYHORFLW\QRUPDOWRWKDWGLUHFWLRQ



  ,IWZRVWUHDPOLQHVDUHVXSHULPSRVHGWKHQ

∂ y + y  ∂y ∂y  + = ∂s ∂s ∂s

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∂f ∂f v = ± ∂y ∂x

and

w=±

∂f ∂z

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∂u ∂v ∂w + + =0 ∂x ∂y ∂z

420

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1RZ u = ±

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∂ ∂ Ê ∂f ˆ Ê ∂f ˆ ÁË - ∂ x ˜¯ + ∂ y Á - ˜ + ∂ z Ë ∂y ¯

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\

Ê ∂f ˆ ÁË - ∂ z ˜¯ = 0

∂ f ∂ f ∂ f + + =0 ∂y  ∂x  ∂ z

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u=±

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\

u=±

∂y ∂f =± ∂y ∂x

and

v=

∂y ∂f =± ∂x ∂y

 +HQFHZHFDQZULWH

∂y ∂f = ∂y ∂x and

∂f ∂y =± ∂y ∂x

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421

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\

∂f ∂f ◊ dx + dy = 0 ∂y ∂x

∂f ∂f  ± u and  ±vZHJHW ∂y ∂x ± udx±v d y = 0

\

u dy =± v dx

or

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\ or As

dy =

∂y ∂y ◊ dx + ◊ dy = 0 ∂y ∂x

∂y ∂y = v and  ± uZHJet ∂y ∂x v dx±u dy = 0

\

dy v = dx u

 +HQFHWKHDERYHLVVORSHm of streamline   

1RZm ¥ m = ±

u v ¥  ± v u

 +HQFHHTXLSRWHQWLDOOLQHDQGVWUHDPOLQHDUHRUWKRJRQDO

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If qLVWKHÀRZDQGV & VDUHYHORFLWLHVDORQJVWUHDPOLQHVy and y q = V Dn = V Dn = constant

\

Where Dn and DnDUHWKHGLVWDQFHVEHWZHHQWZRVWUHDPOLQHV

V D n = V D n

\ f1

f3

f2

f4 y1

ds

q y2

dn

q y3 q y4

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But Df = VDs = V Ds = constant \

V D s = V D s D s D n = D s D n

or

or In case

D n D n = D s D s Dn = constant Ds Dn  WKHQDn = Ds:HJHWDVHWRIVTXDUHLQWKHÀRZQHW Ds

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Fluid Kinematics

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Flow

Flow Pipes or parallel plates

Converging pipe Flow

Flow Bend pipe

Outlet from tank

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 ‡ 7KHUH LV VLPLODULW\ EHWZHHQ IORZ RI IOXLGV DQG IORZ RI HOHFWULFLW\ :KDW DUH WKHVH similarities? Flow of electricity

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Velocity potential ϕLVHVVHQWLDOIRUÀXLGÀRZ +HQFHϕ is analogous to voltage (V)

2.

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3.

Phenomenon takes place in homogeneous ÀXLGV

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4.

Darcy-Weisbach equation is applicable

Ohm’s law is applicable

5.

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,WLVSRVVLEOHWRGLYLGHFXUUHQWE\HOHFWULFDO network

Q1

(OHFWULFLW\SRWHQWLDO Y LVHVVHQWLDOIRUÀRZ of current

i1 Q

i

i i2

Q2

Q = Q1 + Q2 i = i1 + i 2

424

Fundamentals of Fluid Mechanics

 ‡ 'HVFULEHWKHHOHFWULFDODQDORJPHWKRGRIGUDZLQJIORZQHW Conducting boundary

Vessel

Water with electrolyte

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433

components,

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u = 5x 2 v = – 15x 2y ∂y =

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\

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= 10 x3y (y)1,2 = 10 ¥ 1 ¥ 2 = 20 m2/s Option (c) is correct.  ‡ Of the possible irrotation flow functions given below, the incorrect relation is: (a) y = xy (b) y = A (x 2 – y2) (c) f = u r cos q –

4 sin q r

2ˆ Ê (d) f = Á r - ˜ sin q Ë r¯

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1 ∂f 1 ∂2 f ∂2 f + + =0 r ∂r r 2 ∂q ∂ r2

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(IES 1995)

434

Fundamentals of Fluid Mechanics

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\

  PV 

‡ A pipe AB EUDQFKHV LQWR WZR SLSHV C and D DV VKRZQ LQ WKH ¿JXUH7KH SLSH KDV D GLDPHWHURIFPDWAFPDWBFPDWC FPDWD'HWHUPLQHWKHGLVFKDUJH at A if the velocity at ALVPV$OVRGHWHUPLQHWKHYHORFLWLHVDWB and D if velocity at CLVPV C dc = 0.5 m A da = 1.2 m

B db = 0.9 m

D dc = 0.2 m

Va = 2 m/s

QA = = 1RZ

Vc = 4 m/s

p d a ¥ Va  p ¥  ¥ P3V 

Aa Va = Ab Vb

 =  PV p ¥  Vb =  Q B = Q c + Q d

QB = QA

Fluid Kinematics

441

 Ac Vc + Ad Vd =

p ¥  p ¥  ¥  ¥ Vd  

 Vd Vd PV

\

 ‡ A stream function is give by: y x  y t y Find the velocity vector and its value at position vector r

ij±kZKHQt VHF

y x y t y 

∂y  ± v = 8xy ∂x ∂y = + u x  t y ∂y V = ui + vj  >x t y] i±xyj 

 1RZSXWx y z ±DQGt = 3

V   -   >@i±j  i±j V=

 + 

=  +   PV  ‡ ,IIRUDWZRGLPHQVLRQDOSRWHQWLDOIORZWKHYHORFLW\LVJLYHQE\f = x y± 'HWHUPLQH the velocity and stream function at point P   $0,( f = x y±

\ 

u=±

∂f  ± y±  ±y ∂x

v=±

∂f  ±x ∂y

V = ui + vj   ±y i±xj

442

Fundamentals of Fluid Mechanics

Put

x  y  V  ± i±j

\

V=

9 + 8  PV

y = f xy dy =

∂y ∂y dx + dy ∂x ∂y

= vdx±udy  ±xdx± ±y dy

 \

y=

-x  y ±y +  

 ± x  + y±y Put

x y  y ±±  



 ‡ 6NHWFK WKH VWUHDPOLQH UHSUHVHQWHG E\ y = x  + y $OVR ¿QG RXW WKH YHORFLW\ DQG GLUHFWLRQRISRLQW   $0,(  y = x + y u ±

v=

∂y ∂y

±y

∂y  x ∂x

V = ui + vj±yixj V  ±¥ij 

 ± ij \

V= tan q =

 +  =

  PV

v  =   u 

q ƒZLWKxD[LV y = x + y = r If r «VRRQ then y = r «DQGVRRQ

Fluid Kinematics 5 4 3

443

y=4 y=9

2

y = 25

1

2

3 4

5

Streamlines

 ‡ Determine the circulation GDURXQGDUHFWDQJOHGH¿QHGE\x y x DQGy = IRUWKHYHORFLW\RIIOXLGu xy and v ±y. C (1, 4)

y

y=4 B (5, 4)

x=1

x=5 A (5, 1)

y=1

D (1, 1) O

x

6SHFL¿HG5HFWDQJOH

 &LUFXODWLRQLVDOLQHLQWHJUDORIYHORFLW\DURXQGDFORVHGFXUYH u ds =

G=

Ú

=

Ú

ABCD

=

Ú

AB

=

FXUYH

Ú





Ú

ABCD

u dx + vdy

x + 3y dx±ydy

±y dy + ±ydy + 

Ú Ú

Bc





x + 3y dx +

CD

± x dx + 

Ê x ˆ Ê x ˆ = Á +  x˜ + Á + 3 x˜ Ë  ¯ Ë  ¯

 ±±±±  ± ±

±y dy +

Ú

Ú



Ú

DA

±y dy +

x + 3y dx

Ú





x± dx

444

Fundamentals of Fluid Mechanics

Another method W YRUWLFLW\ w =

∂u ∂v ± ∂y ∂x

 ± ± $UHD ¥  Circulation G = W ¥ A   ±¥ ±  ‡ $WZRGLPHQVLRQDOLQFRPSUHVVLEOHÀRZLQSRODUFRRUGLQDWHVLVJLYHQE\ Vr r sin q cos q and Vq ± r sin q )LQGZKHWKHUWKHVHFRPSRQHQWVUHSUHVHQWDSRVVLEOHIORZ Guidance7KHYHORFLW\FRPSRQHQWVPXVWVDWLVI\WKHFRQWLQXLW\HTXDWLRQ,QSRODUFRRUGLQDWHV WKHFRQWLQXLW\HTXDWLRQLV

∂Vr V  ∂Vq + + r =0 ∂r r ∂q r Vr r sin q cos q

∂Vr = sin q cos q ∂r Vq ±r sin q

∂Vq  ±r ¥VLQq cos q ∂q  /+6RIFRQWLQXLW\HTXDWLRQLV

  r VLQ q FRV q ±r ¥VLQ FRVq  VLn q cos q = r r  VLQq cos q ± = 0 = RHS   +HQFHWKHÀRZLVDSRVVLEOHÀRZ  ‡ ,QDWZRGLPHQVLRQDOIORZWKHWDQJHQWLDOFRPSRQHQWRIWKHYHORFLW\LV Vq = –

Asin q ZKHUHA = constant r

 )LQG D UDGLDOYHORFLW\Vr E PDJQLWXGHDQGGLUHFWLRQRIWKHUHVXOWDQW Guidance 8VHFRQWLQXLW\HTXDWLRQWR¿QGUDGLDOYHORFLW\

 ∂Vq V ∂Vr + + r =0 r ∂q r ∂r

Fluid Kinematics

or

445

∂Vq ∂ + rVr   ∂q ∂r

NoZ

q Vq = ± A sin  r ∂Vq A cos q  ± ∂q r

\

A cos q ∂rVr = r ∂r rVr = ±

A cos q r

A cos q r

\

Vr =

1RZ

)& V = Vq + Vr

\

V=

Vq + Vr A sin  q A cos  q + r r

=

=

A r

 ‡ $IOXLGIORZLVJLYHQE\

aˆ Ê aˆ Ê Vq = Á 1 +  ˜ sin q, Vr = Á 1 -  ˜ cos q Ë Ë r ¯ r ¯  6KRZWKDWLWUHSUHVHQWVDSRVVLEOHIORZDQG¿QGZKHWKHULWLVLUURWDWLRQIORZ Guidance 8VHFRQWLQXLW\HTXDWLRQLQSRODUFRRUGLQDWHVWRVKRZWKHÀRZLVSRVVLEOH)RU URWDWLRQDOÀRZYRUWLFLW\ZLOOEH]HUR

aˆ Ê Vr = Á -  ˜ cos q Ë r ¯ Ê r Vr = Á r Ë

aˆ cos q r ˜¯

∂ aˆ Ê rVr   Á +  ˜ cos q Ë ∂r r ¯

446

Fundamentals of Fluid Mechanics

aˆ Ê Vq ± Á +  ˜ sin q Ë r ¯ aˆ ∂Vq Ê  ± Á +  ˜ cos q Ë r ¯ ∂q \&RQWLQXLW\HTXDWLRQLV

∂Vq ∂rVr + =0 ∂q ∂r aˆ Ê aˆ Ê /+6 ± Á +  ˜ cos q + Á +  ˜ cos q Ë Ë r ¯ r ¯ = 0 = RHS 

 +HQFHÀRZLVSRVVLEOH W=

1RZ

∂ ∂V rVq ± r ∂r ∂q

aˆ Ê Vq ± Á +  ˜ sin q Ë r ¯ aˆ Ê rVq ± Á r + ˜ sin q Ë r¯ aˆ Ê ∂ rVq  ± ÁË -  ˜¯ sin q r ∂r aˆ Ê Vr = Á -  ˜ cos q Ë r ¯ ∂Vr  ± Ê - a ˆ sin q ÁË r  ˜¯ ∂q 1RZ

W=

∂V ∂ r Vq ± r ∂r ∂q

aˆ Ê aˆ Ê  ± Á -  ˜ sin q + Á -  ˜ sin q Ë Ë r ¯ r ¯ =0 

 +HQFHÀRZLVLUURWDWLRQDO

Fluid Kinematics

447

 ‡ 'R WKH IROORZLQJ YHORFLW\ SRWHQWLDOV UHSUHVHQW SRVVLEOH IORZV ,I VR GHWHUPLQH WKH VWUHDPIXQFWLRQV   D f = y + x – y   E f = ur cos q + u cos q r  Ê ˆ q a   F f = u Á r + ˜ cos q + p r ¯ Ë Case f = y + x ±y

∂f  x ∂x ∂ f   ∂ x ∂f  ±y ∂y ∂ f  ± ∂y  ∂ f ∂ f +  ±  ∂ x ∂y    +HQFH/DSODFHHTXDWLRQLVVDWLV¿HGDQGWKHÀRZLVSRVVLEOH ∂y ∂f u=+ =+  x ∂y ∂x v=+ \

dy =

∂f ∂y  ±   ±y ∂y ∂x

∂y ∂y ◊ dx + ◊dy ∂y ∂x

 ± ±y dxx dy

  Case 

y ± xxyxy  ± xxy xy± x f FRVq +

 &RQWLQXLW\HTXDWLRQ

 FRVq r

 ∂ f  ∂f ∂ f + + =0 r  ∂ r r ∂r ∂ r

448

Fundamentals of Fluid Mechanics

u ∂f  FRVq±  cos q r ∂r ∂ f = + u cos q r3 ∂ r ∂f u  ± ur sin q± sin q ∂q r ∂ f  ± ur cos q± u cos q r ∂q  L+6RIFRQWLQXLW\HTXDWLon =

u ˆ  Ê  Ê u u ˆ ÁË u cos q - r  cos q˜¯ + 3 cos q +  ÁË - ur cos q - cos q˜¯ r r r r   ˘ È  = u cos q Í - 3 + 3 - - 3 ˙ r r r ˚ Îr r = 0 = RHS

  +HQFHÀRZLVSRVVLEOHDQGLVLUURWDWLRQDO ur =

∂f  ∂y u = = u cos q±  cos q ∂r r ∂q r

uq =

∂f Ê u ∂y ˆ = = Á - ur sin q -  sin q˜ r ∂q Ë ¯ r r ∂r

dy = ur ◊r◊dq + uq dr

u ˆ Ê ˆ Ê dy = Á ur cos q - cos q˜ dq±u sin q Á + 3 ˜ ◊ dr Ë ¯ Ë r r ¯  2QLQWHJUDWLRQZHJHW

Ê y = u sin q Á r Ë

ˆ  ˆ Ê ± u sin q Á r -  ˜ ˜ ¯ r Ë r ¯

 ˆ Ê = u sin q Á r - - r +  ˜ Ë r r ¯ = ±

u Ê  ˆ VLQ q Á - Ë r ˜¯ r

Fluid Kinematics

Case

Ê q a ˆ f = u Á r + ˜ cos q + p r ¯ Ë Ê ∂f a ˆ = u Á -  ˜ cos q ∂r r ¯ Ë ∂ f ∂ r

 ± a u cos q r3

∂f Ê a  ˆ sin q +   ±u r + ÁË ∂q p r ˜¯ Ê ∂ f a ˆ  ±u Á r + ˜ cos q  ∂q r ¯ Ë CRQWLQXLW\HTXDWLRQ

LHS =

 ∂f  ∂ f ∂ f + + =0 r  ∂q  r ∂r ∂ r

˘ È Ê a ˆ u Ê a ˆ a  Íu Á -  ˜ FRV q ˙ ± 3 u cos q±  Á r + ˜ cos q r ÎÍ Ë r ¯ r Ë r ¯ r ˙˚

È  a  a   a  ˘ = u cos q Í - 3 - 3 - - 3 ˙ r r r ˚ Îr r =0 uq =

∂f -u ∂y = = r ∂q r ∂r

ur =

∂y ∂f Ê a ˆ = = u Á -  ˜ cos q r ∂q ∂r r ¯ Ë

dy =

 Ê a ˆ sin q + r + ÁË ˜  pr r ¯

∂y ∂y dr + dq ∂r ∂q

 ±

u r

 Ê  a ˆ sin q x dr + dr + ur Ê - a ˆ cos q x dq r + ÁË ˜ ÁË pr ¯ r  ˜¯

Ê Ê a ˆ a ˆ   ±u sin q Á +  ˜ dr + dr + u cos q Á r -  ˜ dq r ¯ r ¯ Ë Ë pr

449

450

Fundamentals of Fluid Mechanics

On integration, we get

Ê Ê log r a2 ˆ a2 ˆ y = – u sin q Á r - 2 ˜ + + u sin q Á r - 2 ˜ 2p r ¯ r ¯ Ë Ë Ê a2 a2 ˆ log r = – u sin q Á r +r- ˜ + r r ¯ 2p Ë Ê a2 ˆ log r = 2 u sin q Á - r˜ + Ë r ¯ 2p ‡ )RUJLYHQf = 4(x2 – y2 ¿QGy

(UPTU 2006-7) u=

∂y ∂f = 8x = ∂y ∂x

v=

∂f - ∂y = – 8y = ∂y ∂x

dy =

∂y - ∂y ◊dx + ◊dy ∂y ∂x

= 8y dx + 8x dy y = 8xy + 8xy = 16xy  ‡ $VWUHDPIXQFWLRQLVJLYHQE\ y = 3 x2y + (3 + t) y2   )LQGWKHIORZUDWHVDFURVVWKHIDFHVRIWKHWULDQJXODUSULVPDWt VHFLISULVPWKLFNQHVV LVPLQzGLUHFWLRQ A

y

2m

O B

O

3m Prism: 3 m thick

Prism: 3 m thick y = 3x2y + (3+ t) y2 At point A, x = 0, y = 2, t = 5

z x

Fluid Kinematics

451

yA = 3 ¥ 0 ¥  ¥ At point Bx y t  yB = 3 ¥ 9 ¥  ¥ 0 = 0 At point Ox y t   \      

yO = 0 )ORZUDWH  y±y ¥ thickness )ORZIURPAO  yA±y0 ¥ 3   ± ¥ 3  P3V )ORZIURPBO  yB±y0 =0 )ORZIURPAB  yA±yB ¥ 3   ± ¥ 3  

 ‡ ,IWKHYHORFLW\¿HOGLVJLYHQE\u y±x, v y±x¿QGWKHFLUFXODWLRQDURXQG WKHFORVHGFXUYHGH¿QHGE\x y x y  W=

∂v ∂u ± ∂x ∂y

u y±x

∂u   ∂y v = 8y±x

∂v  ± ∂x \     

W ±± ± $UHD  y±y x±x   ±  ±   G=W¥A  ±¥  ±

452

Fundamentals of Fluid Mechanics y 8

2 4

8

 ‡ $SLSHOLQHFPGLDPHWHUELIXUFDWHVDWD\MXQFWLRQLQWRWZREUDQFKHVFPDQG FPGLDPHWHU,IWKHUDWHRIIORZLQWKHPDLQSLSHLVFXPVDQGPDVVYHORFLW\RIIORZ LQFPGLDPHWHUSLSHLVPVGHWHUPLQHWKHUDWHRIIORZLQWKHFPGLDPHWHU SLSH 8378 dC = 0.4 m C dB = 0.6 m A

3

B

QB = 1.5 m /s D

dD = 0.3 m Vd = 7.7 m/s

qB = qC + qD \

qC = qB±qD  = ± p ¥  ¥ 

 ±  P3V  ‡ 7KHVWUHDPOLQHVSDFLQJLVFPDWDVHFWLRQKDYLQJDYHORFLW\RIPVLQWZRGLPHQVLRQDO IORZ,IDWDQRWKHUVHFWLRQWKHVSDFLQJLVFPZKDWLVYHORFLW\" D  PV E  PV  F  PV G  PV  H  PV VD n = VDn ¥ V ¥

 \ OSWLRQ D LVFRUUHFW

V =

 ¥   PV 

Fluid Kinematics

453

 ‡ If velocity along a streamline is given by V tDWs    7KHQWKHDFFHOHUDWLRQDIWHUVHFLVJLYHQE\ E  as at   D  as at  G  QRQHRIa, b and c  F  as at  V s + t + 3

∂V = ∂s ∂V   ∂t as = V ¥

∂V   t ¥ ∂s

  ¥   at =

∂V   ∂t

 2SWLRQ D LVFRUUHFW  ‡ $IOXLGSDUWLFOHPRYHVDORQJDFXUYHGSDWKRIUDGLXVPZLWKDYHORFLW\RIPV,WV  QRUPDODFFHOHUDWLRQLQPVLV D ]HUR E y F  G   F QRQHRIDERYH   v ar = r =

   

OSWLRQ E LVFRUUHFW  ‡ :KLFKRIIROORZLQJVWUHDPIXQFWLRQVVDWLV¿HV/DSODFHHTXDWLRQ" E  xy D  x + y G  x y  F  x + y y = x + y

\

∂y ∂ y  x    ∂x ∂x

∂y ∂ y  y    ∂y ∂y ∂ y ∂ y +   π'RHVQRWVDWLVI\/DSODFHHTXDWLRQ ∂y  ∂x 

454

Fundamentals of Fluid Mechanics

y xy

1RZ

∂y  y and ∂x

∂ y =0 ∂x 

∂y ∂ y  x  = 0 ∂y ∂y ∂ y ∂ y +  6DWLV¿HV/DSODFHHTXDWLRQ  ∂y  ∂x y = x3 + y3

∂y ∂ y = 3x   x ∂x ∂x

∂y ∂ y = 3y   y ∂y ∂y ∂ y ∂ y   x + y π'RHVQRWVDWLVI\/DSODFHHTXDWLRQ  + ∂x ∂y  y = xy

1RZ

 ∂y  xy ∂ y  y ∂x ∂x

∂y ∂ y  xy   x ∂y ∂y ∂ y ∂ y +   x + y π'RHVQRWVDWLVI\/DSODFHHTXDWLRQ ∂y  ∂x  OSWLRQ E LVFRUUHFW

& )& Æ  ‡ $ IOXLG IORZ LV UHSUHVHQWHG E\ WKH YHORFLW\ ¿HOG V = a ◊ x ◊ i + a ◊ y ◊ j  ZKHUH a is a FRQVWDQW7KHHTXDWLRQRIVWUeamlLQHSDVVLQJWKURXJKDSRLQW  LV  

D  x±y  F  x – y 

+HQFH and   (TXDWLRQRIDVWUHDPOLQHLV

E  x + y  G  xy  )& & Æ V = a◊x◊ i + a◊y◊ j 

ux = a ◊ x uy = a ◊ y dx dy = ux uy

*$7(

Fluid Kinematics

or

dy dx = a◊ y a◊x

or

dy dx = y x

455

2QLQWHJUDWLRQZHKDYH   

log x = logy + logc ZKHUH c = constant or x = c◊y  6LQFHWKHVWUHDPOLQHSDVVHGWKURXJKSRLQW  KHQFH  ◊ C or C 

  7KHHTXDWLRQRIVWUHDPOLQHLV x=

 y 

x±y = 0

RU OSWLRQ F LVFRUUHFW

 ‡ 7KHYHORFLW\FRPSRQHQWVLQx and yGLUHFWLRQVRIDWZRGLPHQVLRQDOSRWHQWLDOIORZDUH u and v respectively7KHn



∂u LVHTXDOWR ∂x

D 

∂v  ∂x

E  −

∂v ∂x

F 

∂v ∂y

G  −

∂v ∂y

*$7(

CRQWLQXLW\HTXDWLRQLVVDWLV¿HGLQDÀRZ+HQFH ∂v ∂u + =0 ∂y ∂x or

∂v ∂u = ∂y ∂x

\ 2SWLRQ G LVFRUUHFW  ‡ 6KRZWKDWWKHYHORFLW\YHFWRUVLVWDQJHQWLDOWRWKHVWUHDPOLQHVHYHU\ZKHUHLQWKHxy SODQH The stream function in x-y plane can be represented as y xy  FRQVWDQW

456

Fundamentals of Fluid Mechanics

  7KHVWUHDPIXQFWLRQFDQEHH[SUHVVHGDV dy =

∂y ∂y ◊ dx + ◊ dy ∂y ∂x

1RZ

y xy  FRQVWDQW+HQFH∂y = 0

\

∂y ∂y ◊ dx + ◊ dy = 0 ∂y ∂x

%XW

Vx =

\

Vx dy±Vy dx = 0

or

Vy dy = dx Vx

∂y - ∂y and Vy = ∂y ∂x

7KHDERYHLQdiFDWHVWKDWYHORFLW\YHFWRUV is tangent to lines of y xy  FRQVWDQW  ‡ $YHORFLW\¿HOGFDQEHJLYHQDVu = Vcosq, v = Vsinq and w 'HWHUPLQHWKHH[SUHVVLRQ RIWKHVWUHDPOLQHVRIWKLVIORZ   7KHVWUHDPOLQHVFDQEHH[SUHVVHGDV dx dz dy = = u w v dy dz dx = = V sin q 0 V cos q

or

dy V sin q = = tanq dx V cos q

or or

dy = tanq ◊ dx

 2QLQWHJUDWLRQZHJHW y = tanq ◊ x + c   +HQFHVWUHDPOLQHVDUHJURXSRIOLQHVLQFOLQHGDWDQJOHq to the xD[LV  ‡ $WZRGLPHQVLRQDOIORZ¿HOGLVVSHFL¿HGE\V yixj)LQGZKHWKHUWKHIORZLV VWHDG\LUURWDWLRQDODVZHOODVIHDVLEOH'HWHUPLQHWKHVWUHDPIXQFWLRQDQGYROXPHIORZ UDWHSDVVLQJEHWZHHQVWUHDPOLQHVWKURXJKWKHSRLQWV  DQG   *LYHQ





Vx = 3y and Vy = 3x and

1RZ





W=

 

∂V y ∂x

±

∂Vx ∂y

± 

∂V =0 dt

Fluid Kinematics



457

 7KHUHIRUHWKH¿HOGLVLUURWDWLRQDO7KH¿HOGLVIHDVLEOHLILWREH\VFRQWLQXLW\HTXDWLRQZKLFK LVJLYHQE\ ∂V y ∂Vx +   L ∂y ∂x BXW

and

∂  y ∂Vx = =0 ∂y ∂x ∂V y

=

∂y

∂ x   ∂y

3XWWLQJ WKHVH YDOXHV LQ HTXDWLRQ L  ZH JHW /+6  5+6 +HQFH WKH ¿HOG VDWLV¿HV WKH FRQWLQXLW\HTXDWLRQDQGLWLVIHDVLEOH1RZZHKDYH dy =

 

∂y ∂y ◊ dx + ◊ dy ∂y ∂x

= ±Vy dx + Vx dy ±xdx + 3ydy  2QLQWHJUDWLRQZHJHW y=

3  3  y ± x + c  

TKHGLVFKDUJHEHWZHHQVWUHDPOLQHV  DQG  LV = y  ±y  =

3  3  ± ± ±  

 XQLWV  ‡ 'HWHUPLQHZKHWKHUWKHWZRGLPHQVLRQDOIORZ¿HOGDVJLYHQEHORZLVURWDWLRQDO V = xyi + ∂V y

:HNQRZ

y=

+HUH

Vx x3y

∂x

\

∂Vx  x3 ∂y

and

Vy =

x 

±

x j  ∂Vx ∂y

458

Fundamentals of Fluid Mechanics

∂V y

\

∂x

 x3

y x3±x3 = 0

\

  +HQFHWKHJLYHQÀRZLVLUURWDWLRQDO  ‡ 2IWKHSRVVLEOHLUURWDWLRQDOIORZIXQFWLRQVJLYHQEHORZWKHLQFRUUHFWUHODWLRQLV ZKHUH y = stream function and f YHORFLW\SRWHQWLDO  D  y = xy E  y = A x – y  

F  f = u ◊ r cosq +

,(6

∂y =0 ∂y 

D  1RZy = xy ∂y =y ∂x SLPLODUO\



ˆ Ê G  f = Á r - ˜ sinq Ë r¯

 ,IVWUHDPIXQFWLRQ y VDWLV¿HVWKH/DSODFHHTXDWLRQWKHQWKHÀRZLVLUUotatiRQDOLH +



u cosq r

and

∂y =0 ∂x 

∂y ∂ y = x and =0 ∂y ∂y 

HHQFHy = xyVDWLV¿HVWKH/DSODFHHTXDWLRQDQGWKHÀRZLVLUURWDWLRQDO E  1RZy = A x±y \

∂y ∂y =Ax and  A ∂x ∂x 

$OVR

∂y ∂ y  ±Ay and  ±A ∂y ∂y 

HHQFHy = A x±y VDWLV¿HVWKH/DSODFHHTXDWLRQDQGWKHÀRZLVLUURWDWLRQDO 

F  1RZ

\ and

f = ur cosq +

Vr = -

u cosq r

∂f u  ±u cosq +  cosq ∂r r

Vq ±

u  ∂f = u sinq +  sinq r r ∂q

∂y ∂x 

Fluid Kinematics

Now,

2WZ =

=

1 ∂ 1 ∂2 [Vq] – [Vr] r ∂r r ∂q

1 ∂ È u 1 ˘ u sin q + 2 cos q ˙ – 2 r ∂r ÍÎ r r ˚

= -

459

∂ È- u cos q + u cos q ˘ Í ˙ r2 ˚ ∂q Î

u 2 u sin q u sin q - 4 sinq π 0 - 2 4 r r r

HeQFHÀRZLVQRWLUURWDWLRQDO 2ˆ Ê (d) Now, f = Á r - ˜ sinq Ë r¯ Vr = -

∂f 2˘ È = – Í1 + 2 ˙ sinq ∂r Î r ˚

Vq = –

1 È 1 ∂f 2˘ =– r - ˙ cosq r ÍÎ r ∂q r˚ = – È1 - 2 ˘ cosq Í r2 ˙ Î ˚

Now,

1 ∂ 1 ∂ (rVr) + (Vv) is r ∂r r ∂q =

˘ 2ˆ 1 ∂ ÈÊ - r - 2 ˆ sin q ˘ 1 ∂ È Ê + - Á - 1 - ˜ cos q ˙ ÍÁË ˙ ˜ Í r¯ r¯ r ∂r Î r ∂r Î Ë ˚ ˚

=

1 Ê 1 Ê 2ˆ 2ˆ - 1 - 2 ˜ sinq + ÁË1 - r ˜¯ sinq ¯ r ÁË r r

     DQGÀRZLVURWDWLRQDO   7KHUHODWLRQLQ F LVLQFRUUHFW  ‡ )RUVWHDG\LQFRPSUHVVLEOHIORZLIWKHuFRPSRQHQWRIYHORFLW\LVu = AexWKHQZKDWLV WKHvFRPSRQHQWRIYHORFLW\"  D  Ae E  Aexy G  ±Aex ,(6  F  ±Aexy   &RQWLQXLW\HTXDWLRQLV dv du + =0 dy dx

460

Fundamentals of Fluid Mechanics

dv d Ae x + =0 dy dx

or

dv =0 dy

or

Aex +

or

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ax = u \

ax  ±x  ±   x

AOVR

ay = u

dv dv dv dv +v +w + dy dx dt dz

 y ◊   y

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Vx + Vy =

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8.14 QUESTIONS FROM COMPETITIVE EXAMINATIONS ‡ :KLFKRQHRIWKHIROORZLQJLVWKHFRQWLQXLW\HTXDWLRQLQGL൵HUHQWLDOIRUP" 

D 

dA dV d r  &RQVW + + A V r

E 

dA dV d r   + + A V r



F 

r A V  &RQVW + + dA dV d r

G  AdA + Vdv + rdr 

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dA dV d r =0 + + A V r   2QLQWHJUDWLRQZHJHW logA + logV + logr = logC   RUORJ AÂVÂr  ORJC or A VÂr = C or AV = Constant   7KHDERYHLVDFRQWLQXLW\HTXDWLRQ   2SWLRQ E LVFRUUHFW  ‡ :KLFK RQH RI WKH IROORZLQJ HTXDWLRQV UHSUHVHQWV WKH FRQWLQXLW\ HTXDWLRQ IRU VWHDG\ FRPSUHVVLEOHIOXLGIORZ" D  Dr ◊ V +

∂r ∂t

E  Dr ◊ V +



F  DV 

∂r = Const ∂t

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,$6 

  *HQHUDOFRQWLQXLW\HTXDWLRQLV Dr ◊ V +

∂r =0 ∂t

L

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 ‡ 7KHFRQWLQXLW\HTXDWLRQIRUGLPHQVWLRQDOIORZ  

D 6WHDG\IORZ F ,GHDOIOXLGIORZ

∂u ∂v ∂w + +  LVDSSOLFDEOHWR ∂x ∂y ∂z

E  8QLIRUPIORZ G  ,GHDODVZHOODVYLVFRXVIORZ ,(6

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D  V = – c x + y 

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or

or

d dx

dv dy

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or 

= -

cx dv =  y dy

 2QLQWHJUDWLRQZHJHt V =

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  2SWLRQ E LVFRUUHFW  ‡ The components of velocity u and v along x and yGLUHFWLRQLQ'IORZRILQFRPSUHVVLEOH fluid are D  u = x cos y, v = ±x sin y E  u = x  v  – y F  u = xyt, v = x – y,

t 

G  u = log x + y, v = xy -

y x

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463

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  7KHHTXDWLRQLV du dv = =0 dx dy Checking for u and vYDOXHsRI D  d  d x FRV y + - x VLQ y dx dy   x cos y±x cos y = 0   &KHFNLQJLQJLYHQYDOXHVLQ E d d x +  +  - y  ±  dx dy Checking for u and vDVJLYHQLQ F d d Ê   tˆ xyt + ÁË x - y ◊ ˜¯ = yt±yt = 0 dx dy  Checking for u and vDVJLYHQLQ G d d Ê yˆ   ORJ x + y + -x- π0 ÁË xy - ˜¯ = dx dy x x x   +HQFHDQGDUHWUXH   2SWLRQ D LVFRUUHFW & &  ‡ If V is velocity vector of fluid, then DV  LVVWULFWO\WUXHIRUZKLFKRIWKHIROORZLQJ        ‡



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⎛ ∂v ∂u ⎞ ⎜⎝ ∂x − ∂y ⎟⎠

E wz =

 ⎛ ∂v ∂u ⎞ +  ⎜⎝ ∂x ∂y ⎟⎠

F wz =

 

⎛ ∂u ∂v ⎞ ⎜⎝ ∂x − ∂y ⎟⎠

G  wz =

 

 2SWLRQ D LVFRUUHFW

⎛ ∂u ∂v ⎞ ⎜⎝ ∂x + ∂y ⎟⎠

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∂v ∂v = ∂y ∂x

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F 

∂w ∂v = ∂y ∂z

G 

∂u ∂u =  ,$6,(6 ∂x ∂y

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∂u ∂v ∂w +v +w ∂x ∂y ∂z

E  ∂u + ∂v + ∂w ∂t ∂t ∂t

F  u

∂u ∂v ∂w +u +u ∂x ∂y ∂z

G  u

∂u ∂u ∂u +v +w ∂x ∂y ∂z

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du du du du +v +w + dx dy dz dt

du =0 dt

‫׵‬ԙax = – 2x(– 2) + 2y × 0 + 0 + 0 = 4x ay = u ◊

dv dv dv +v +w dx dy dz

= – 2x × 0 + 2y × 2 + 0 = 4y

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‫׵‬ԙ ax = 4x and ay = 4y   2SWLRQ G LVFRUUHFW  ‡ $ VWHDG\ LQFRPSUHVVLEOH IORZ LV JLYHQ E\ u  x + y and v = ± xy :KDW LV WKH convective acceleration along xGLUHFWLRQDWSRLQW  " E  ax XQLW G  ax ±XQLW

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 axDWSRLQW  LV  

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du  dQ ¥ = Ax dt dt =

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∂u ∂u ∂u ∂u +v +w + ∂t ∂y ∂z ∂t

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       ‡       

     

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wz =

 

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=

 

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Codes: A B C D (a) 1 4 3 2 (b) 3 2 2 4 (c) 1 2 3 4 (d) 3 4 1 2 (IES 2002) 2SWLRQ E LVFRUUHFW A streamline is a line D  :KLFKLVDORQJWKHSDWKRIWKHSDUWLFOH E  :KLFKLVDOZD\VSDUDOOHOWRWKHPDLQGLUHFWLRQRIIORZ F  $ORQJZKLFKWKHUHLVQRIORZ G  2QZKLFKWDQJHQWGUDZQDWDQ\SRLQWJLYHVWKHGLUHFWLRQRIYHORFLW\ ,(6  2SWLRQ G LVFRUUHFW Assertion (A): 6WUHDPOLQHVFDQFURVVRQHDQRWKHULIIOXLGKDVKLJKHUYHORFLW\ Reason (R): $WVX൶FLHQWO\KLJKYHORFLW\WKH5H\QROGVQXPEHULVKLJKDQGDWVX൶FLHQWO\ KLJK5H\QROGVQXPEHUVWKHVWUXFWXUHRIWKHIORZLVWXUEXOHQWW\SH D  %RWK$DQG5DUHLQGLYLGXDOO\WUXHDQG5LVWKHFRUUHFWH[SODQDWLRQRI$ E  %RWK$DQG5DUHLQGLYLGXDOO\WUXHEXW5LVQRWWKHFRUUHFWH[SODQDWLRQRI$ F  $LVWUXHEXW5LVIDOVH G  $LVIDOVHEXW5LVWUXH ,(6 2SWLRQ G LVFRUUHFW The streamlines and the lines of constant velocity potential in an inviscid irrotational IORZ¿HOGIRUP D  3DUDOOHOJULGOLQHVSODFHGLQDFFRUGDQFHZLWKWKHLUPDJQLWXGH E  ,QWHUVHFWLQJJULGQHWZLWKDUELWUDU\RULHQWDWLRQ F  $QRUWKRJRQDOJULGV\VWHP G  1RQHRIWKHDERYH ,(6 2SWLRQ F LVFRUUHFW  $ YHORFLW\ ¿HOG LV JLYHQ E\ u  xy and and v = x  - y   :KDW LV WKH UHOHYDQW  HTXDWLRQRIDVWUHDPOLQH"   xy E   D  dx = x - y  x - y dy xy F 

dx  xy =   dy x - y 

G 

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vdx = uÃdy

dx x  - y   = dy  xy

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u dx = v dy

or   3XWWLQJYDOXHVRIu and v

3 xy dx = 3 dy x - y   xy

= x - y  

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du dv + r◊ =  ZKHUHu and v are velocities in the x dx dy

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 ‡ 7KH JHQHUDO IRUP RI H[SUHVVLRQ IRU WKH FRQWLQXLW\ HTXDWLRQ LQ &DUWHVLDQ FRRUGLQDWH system for incompressible or compressible form is given by

 

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∂u ∂v ∂w + + = ∂x ∂y ∂z

E 

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F 

∂r ∂ru ∂rv ∂rw + + + = ∂t ∂x ∂y ∂z

G

∂r ∂ru ∂rv ∂rw + + + = ∂t ∂x ∂y ∂z

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S

∂ ∂t

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 AV = AV





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∂ ∂ x  y  x y + =  xy +  π  ∂x  ∂y 

E 

∂ ∂  xy +   xy =  +  =   ∂x ∂y

F 

∂ ∂ Ax  y +  Ax  y  =  Ay +  Ax π   ∂x ∂y

G 

∂ ∂ Ax + By  +  Ax + By  =  +  B π   ∂x ∂y

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dv ∂u D dy − ∂x =  

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∂u ∂w − = ∂z ∂x

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G 

∂v ∂u − = ∂x ∂y

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(a) ‫ × ׏‬Vr = 0

(b) ‫ × ׏‬Vr = 1

(c) ‫׏‬2 × Vr = 1

(d) |Vr × ‫ × |׏‬Vr =

∂V ∂r

(IES 1993)

 2SWLRQ D LVFRUUHFW

 ‡ The relation

∂ψ ∂ψ =  =   IRU DQ LUURWDWLRQDO IORZ LV NQRZQ E\ ZKLFK RQH RI WKH ∂x  ∂y



IROORZLQJ D  1DYLHU6WRNHVHTXDWLRQ

E  /DSODFHHTXDWLRQ



F  RH\QROGVHTXDWLRQ

G  (XOHU¶VHTXDWLRQ

Option E LVFRUUHFW  ‡ &RQVLGHUWKHIROORZLQJVWDWHPHQWV )RUDWZRGLPHQVLRQDOSRWHQWLDOIORZ   /DSODFHHTXDWLRQIRUVWUHDPIXQFWLRQPXVWEHVDWLV¿HG   /DSODFHHTXDWLRQIRUYHORFLW\SRWHQWLDOPXVWEHVDWLV¿HG

,(6

474

Fundamentals of Fluid Mechanics

 6WUHDPOLQHVDQGequipotential lines are mutually perpendicular  6WUHDPIXQFWLRQDQGSRWHQWLDOIXQFWLRQDUHQRWLQWHUFKDQJHDEOH D  DQGDUHFRUUHFW  E  DQGDUHFRUUHFW F  DQGDUHFRUUHFW  G  DQGDUHFRUUHFW  ,(6 2SWLRQ F LVFRUUHFW :KLFKRIWKHIROORZLQJIXQFWLRQVUHSUHVHQWWKHYHORFLW\SRWHQWLDOLQDWZRGLPHQVLRQDO IORZRIDQLGHDOIOXLG"

       ‡ 

 xy

 x±y

 FRV x – y 

 WDQ±

x y

6HOHFWWKHFRUUHFWDQVZHUXVLQJFRGHVJLYHQEHORZ D  DQG F  DQG

 

E  DQG G  DQG

,(6

 )RU'LGHDOÀXLGÀRZ ∂f ∂f + =0 ∂x  ∂y  If f = 2x + 2y, we have LHS =

Now,

= 0 + 0 = 0 = RHS f = cos (x – y) LHS =



∂  x +  y ∂  x +  y + ∂x  ∂x 

∂ ∂ FRV x - y +  FRV x ± y  ∂x ∂y

 ±FRV x±y FRV x±y = 0 = LHS

  2SWLRQ D LVFRUUHFW  ‡ ,IIRUDIORZDVWUHDPIXQFWLRQH[LVWVDQGVDWLV¿HVWKH/DSODFHHTXDWLRQWKHQZKLFK RQHRIWKHIROORZLQJLVWKHFRUUHFWVWDWHPHQW"  D  7KHFRQWLQXLW\HTXDWLRQVDWLV¿HGDQGWKHIORZLVLUURWDWLRQDO  E  7KHFRQWLQXLW\HTXDWLRQLVVDWLV¿HGDQGWKHIORZLVURWDWLRQDO  F  7KHIORZLVLUURWDWLRQDOEXWGRHVQRWVDWLVI\WKHFRQWLQXLW\HTXDWLRQ  G  7KHIORZLVURWDWLRQDO ,(6   2SWLRQ D LVFRUUHFW

Fluid Kinematics

475

 ‡ ,QDWZRGLPHQVLRQDOIORZWKHYHORFLW\FRPSRQHQWLQx- and yGLUHFWLRQVLQWHUPVRI stream function are ∂ψ ∂ψ  v = ∂x ∂y



D  u =



F  u = −



∂ψ ∂ψ v =  ∂y ∂x

E  u =

∂ψ ∂ψ v = ∂y ∂x

G  u =

∂ψ −∂ψ v =  ∂x ∂y

,(6

 2SWLRQ F LVFRUUHFW

 ‡ 2IWKHSRVVLEOHLUURWDWLRQDOIORZIXQFWLRQVJLYHQEHORZWKHLQFRUUHFWUHODWLRQLV  D  Y = xy E  Y = A x±y F  f = u × cos q +

u cos q r

ˆ Ê G  f = Á r - ˜ sin q Ë r¯

,(6

∂ y ∂ y +  LV/DSODFHHTXDWLRQIRUURWDWLRQDOÀRZ ∂x  ∂y    D Check ∂ ∂ xy + xy    \ LUURWDWLRQDO ∂x  ∂y    E Check ∂ ∂   A x y + A x  - y   ±  \ LUURWDWLRQDO ∂x  ∂y    F Check Vr = -

∂f  ∂f and VT = ∂r r ∂q

Vr = -

u ∂ Ê ˆ ur cos q + cos q˜ and Á Ë ¯ ∂r r

VT = -

 ∂ Ê u ˆ ur cos q + cos q˜ Á ¯ r ∂q Ë r

Vr = -u cos q + 

wz =

u cos q 

and Vq = u sin q +

u sin q r

u  ∂ Ê  ∂ Ê u ˆ ˆ ◊ u sin q +  sin q ˜ ±  u cos q +  cos q ˜ Á Á Ë ¯ ¯ r ∂r r r ∂q Ë r

476

Fundamentals of Fluid Mechanics

=

2u sin q u sin q u - 4 sin q π 0 r4 r2 r

(d) Check Vq =

1 ∂f 1Ê 2ˆ 2ˆ Ê = - Á r - ˜ cos q = - Á1 - 2 ˜ cos q , Ë r ∂q rË r¯ r ¯

Vr = 2wZ =

∂f 2ˆ Ê = - Á1 + 2 ˜ sin q Ë ∂r r ¯

˘ 1 ∂ 1 ∂ ÈÊ 2ˆ ÁË - r - ˜¯ sin q ˙ + ◊ Í r ∂r Î r ˚ r ∂r

ÈÊ ÍÁË -1 Î

˘ 2ˆ ˜¯ cos q ˙ = 0 r ˚

Option (d) is correct.  ‡ 7KHYHORFLW\SRWHQWLDORIYHORFLW\¿HOGLVJLYHQE\f = x2 – y2FRQVW,WVVWUHDPIXQFWLRQ ZLOOEHJLYHQE\ D  ±xyFRQVWDQW E  xyFRQVWDQW  F  ±xy + f(x  G  xy + f(y  ,(6 u =

-∂f -∂f and v = ∂x ∂y

u =

-∂ 2 -∂ 2 ( x - y 2 + c) ( x - y 2 + c) v = ∂y ∂x

Y =

∂y ∂y dx + ◊ dy ∂y dx

= (v) dx + (– u) dy = (2y) dx + (2x) dy = 2xy + 2xy = 2 ή 2xy  f (y) Option (d) is correct.  ‡ 7KH VWUHDP IXQFWLRQ LQ D GLPHQVLRQDO IORZ ¿HOG LV JLYHQ E\ Y = xy7KH SRWHQWLDO IXQFWLRQLV 2 2 x2 - y2 E   D  x + y  2 2 

G  x2y + y2x

F  xy Y = xy ∂y =–u=x ∂y

∂y =v=y ∂x

,(6

Fluid Kinematics

477

Ê ∂f ˆ Ê ∂f ˆ f = Á ˜ ◊ dx + Á ˜ ◊ dy Ë ∂x ¯ Ë ∂y ¯   ±u dx ±v dy   x dx ±y dy



=

x y  x - y =   

  2SWLRQ E LVFRUUHFW  ‡ 7KH YHORFLW\ FRPSRQHQWV IRU D ' LQFRPSUHVVLEOH IORZ RI IOXLG DUH u = x ± y and v = – y±x It can be concluded that  D  7KHIORZGRHVQRWVDWLVI\WKHFRQWLQXLW\HTXDWLRQ  E  7KHIORZLVURWDWLRQDO  F  7KHIORZLVLUURWDWLRQDO  G  1RQHRIWKHDERYH ,(6   )RUFRQWLQXLW\HTXDWLRQ ∂u ∂u + =0 ∂x ∂y 

/HWXVYHULI\LW ∂ ∂u x -  y = +  = ∂ x ∂x ∂ ∂u - y -  x = -  = ∂y ∂y

∂u ∂v + =1–1=0 ∂x ∂y

‫׵‬ 

 7KHVWDWHPHQWWKDWÀRZLVQRWVDWLVI\LQJFRQWLQXLW\HTXDWLRQLVIDOVH



 1RZFKHFNIRUURWDWLRQRULUURWDWLRQDO ∂v ∂u ∂x ∂y



  7KHÀRZLVLUURWDWLQDO



 2SWLRQ F LVFRUUHFW

=

∂ ∂ - y -  x x -  y ∂x ∂y

 ± 

478

Fundamentals of Fluid Mechanics

 ‡ The stream function Y = x – yLVREVHUYHGIRUD'IOXLGIORZ:KDWLVWKHPDJQLWXGH RIYHORFLW\DWSRLQW ± "  D   E    F   G   ,(6 u =

∂y = -  y  = -  ¥ -  = -  ∂y

v = 

5HVXOWDQWYHORFLW\ =

∂y = -  x  = -  ¥   = -  ∂x u  + v  = -  + - 

  2SWLRQ D LVForrHFW  ‡ 7KHVWUHDPIXQFWLRQLQDIORZ¿HOGLVJLYHQE\Y xy,QWKHVDPHIORZ¿HOGZKDW LVWKHYHORFLW\DWDSRLQW   D  XQLW E  XQLW  F  XQLW G  XQLW ,(6

-∂y -∂ =  xy = -  x = -  ¥  =  ∂y ∂y

v =

∂y ∂ =  xy =  y =  ∂x ∂y

5HVXOWDQWYHORFLW\  -  +  = 

 

u =

 2SWLRQ G LVFRUUHFW

 ‡ )RU LUURWDWLRQDO DQG LQFRPSUHVVLEOH IORZ WKH YHORFLW\ SRWHQWLDO DQG VWUHDP IXQFWLRQ are given by f and YUHVSHFWLYHO\:KLFKRQHRIWKHIROORZLQJVHWVLVFRUUHFW" 

D  ∇  φ =  ∇  ψ =  

E  ∇  φ ≠  ∇  ψ = 



F  ∇  φ =  ∇  ψ ≠  

E  ∇  φ ≠  ∇  ψ ≠  

,(6

  2SWLRQ D LVFRUUHFW  ‡ :KLFKRQHRIWKHIROORZLQJVWDWHPHQWVLVWUXHLQD'IORZRILGHDOIOXLG"  D  3RWHQWLDOIXQFWLRQH[LVWVLIVWUHDPIXQFWLRQH[LVWV  E  6WUHDPIXQFWLRQPD\RUPD\QRWH[LVW  F  %RWKSRWHQWLDOIXQFWLRQDQGVWUHDPIXQFWLRQPXVWH[LVWIRUHYHU\IORZ  G  6WUHDPIXQFWLRQZLOOH[LVWEXWSRWHQWLDOIXQFWLRQPD\RUPD\QRWH[LVW  ,(6   2SWLRQ G LVFRUUHFW

Fluid Kinematics

 ‡ 7KHUHDOL]DWLRQRIYHORFLW\SRWHQWLDOLQIOXLGIORZLQGLFDWHVWKDWWKH  D  )ORZPXVWEHLUURWDWLRQDO  E  &LUFXODWLRQDURXQGDQ\FORVHGFXUYHPXVWKDYHD¿QLWHYDOXH  F  )ORZLVURWDWLRQDODQGVDWLV¿HVWKHFRQ¿QXLW\HTXDWLRQ  G  9RUWLFLW\PXVWEHQRQ]HUR   2SWLRQ D LVFRUUHFW

479

,(6

 ‡ 7KHYHORFLW\SRWHQWLDOIXQFWLRQLQD'IORZ¿HOGLVJLYHQE\f = x – y7KHPDJQLWXGH RIYHORFLW\DWWKHSRLQW  LV  D   E   F    

 

,$6

 2SWLRQ D LVFRUUHFW u =

- ∂f - ∂  x - y = - x = -  = ∂x ∂x

v =

± ∂f - ∂  = x - y =  y = +  ∂y ∂y

    5HVXOWDQWYHORFLW\  u + v = -  +  =  

 

G    

 2SWLRQ F LVFRUUHFW

 ‡ ,QDWZRGLPHQVLRQDOYHORFLW\¿HOGZLWKYHORFLWLHVu and v along the x and y directions respectively, the convection acceleration along the x direction is given by 

D  u ◊



F  u



du du + v◊  dx dy

dv du + v◊  dx dy

E  u ◊

du dv +v dx dy

G  v

du ∂u +u  dx ∂y

*$7(

 2SWLRQ D LVFRUUHFW

 ‡ )RUDIOXLGIORZWKURXJKDGLYHUJHQWSLSHRIOHQJWKl having inlet and outlet radii of R and RUHVSHFWLYHO\DQGDFRQVWDQWIORZUDWHRIQDVVXPLQJWKLVYHORFLW\WREHD[LDO DQGXQLIRUPDWDQ\FURVVVHFWLRQWKHDFFHOHUDWLRQDWWKHH[LWLV D 



F 

Q R - R p ◊ L ◊ R



Q  Rc - R  p  ◊ L ◊ R

E 

G 

Q  R - R p LR

Q R - R  p  ◊ L ◊ R

*$7(

480

Fundamentals of Fluid Mechanics

Rx = R +

R - R ◊x L

 A = p Rx

u =

Q Q =  Ax R - R ˆ Ê p Á R + ◊ x˜ Ë ¯ L

ax = u ◊

du du du =0 but + dt dx dt

R - R L ¥ =  R R ˆ Ê R R  Ê ˆ  p Á R + ◊ x˜ p Á R +  ◊ x˜ Ë ¯ L Ë ¯ L - ◊Q ◊

Q

At x = LZHKDYH aH[LW =

Q  R - R p  ◊ L ◊ R

  2SWLRQ F LVFRUUHFW  ‡ $ WZRGLPHQVLRQDO IORZ ¿HOG KDV YHORFLWLHV DORQJ WKH x and y directions given by u = xÂt and v ±xyÂtUHVSHFWLYHO\ZKHUHtLVWLPH7KHHTXDWLRQRIVWUHDPOLQHLV 

D xy FRQVWDQW

E  xy = constant

F  xy FRQVWDQW

G  1RWSRVVLEOHWRGHWHUPLQH

dy dx = u v



dx x ◊t or





dy -  xy ◊ t

= -

dy y

log x + log y = 0

or RU

dx x

=



ORJ xy  

or xy = constant  2SWLRQ D LVFRUUHFW

*$7(

Fluid Kinematics

481

 ‡ The velocity components in the x- and yGLUHFWLRQVDUHJLYHQE\ u = lxy – xy

 v = xy -

and

 ¥ y 

The value of lIRUDSRVVLEOHIORZ¿HOGLQYROYLQJDQLQFRPSUHVVLEOHIOXLGLV 

D 

- 

E 



F 

  

G  



-  *$7( 

 &RQWLQXLW\HTXDWLRQLV

∂u ∂v + =0 ∂x ∂y ∂ ∂ Ê  3 ˆ lxy  - x  y + xy - ¥ y  ˜ = 0 Á ¯ ∂x ∂y Ë  or ly3±xyxy±y3 = 0 

or l = 3  2SWLRQ G LVFRUUHFW

 ‡ 7KH'IORZZLWKYHORFLW\n  xy i ±y j is  D  &RPSUHVVLEOHDQGLUURWDWLRQDO  E  &RPSUHVVLEOHDQGQRWLUURWDWLRQDO  F  ,QFRPSUHVVLEOHDQGLUURWDWLRQDO  G  ,QFRPSUHVVLEOHDQGQRWLUURWDWLRQDO u = xy v ±y ∂u ∂v + ∂x ∂y

=

∂ ∂ x +  y +  +  - y ∂x ∂y



  ±   &RQWLQXLW\HTXDWLRQLVVDWLV¿HG



 )RUURWDWLRQDOÀRZ 

∂v ∂u + =0 ∂x ∂y

∂ ∂  - y + x +  y +  ∂x ∂y =0+2=2്0

(GATE 2001)

482

Fundamentals of Fluid Mechanics

                     

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F    

G   u =

- ∂f - ∂  = x - y = - x = ±  ∂x ∂x

v =

- ∂f - ∂  x - y =  y =  = ∂y ∂y

,(6

Fluid Kinematics

483

  5HVXOWDQWYHORFLW\  -  +  = 



=   

 2SWLRQ F LVFRUUHFW

&

 ‡ $ ' IORZ ¿HOG LV GH¿QHG DV v = ix - j ◊ y  7KH HTXDWLRQ RI VWUHDPOLQH SDVVLQJ WKURXJKSRLQW  LV  D  xy±  E  xy   F  xy + 2  G  xy±  ,(6   7KHHTXDWLRQRIVWUHDPOLQHSDVVLQJ  LV y±y = m x±x y± = -

-  x -  LIm = x x

±xy + x = x



or xy    2SWLRQ E LVFRUUHFW  ‡ BWKHGLUHFWLRQRIÀRZLVIURPAWRB  /RVVRIKHDG KHDGDWA±KHDGDWB   ±   PRIZDWHU  ‡ $ SXPS RI  +3 ZLWK  H൶FLHQF\ LV GLVFKDUJLQJ FUXGH RLO 6*    WR WKH RYHUKHDGWDQN EL P IURPWKHVWRUDJHWDQN EL P DVVKRZQLQWKH¿JXUH ,IWKHORVVLQWKHZKROHV\VWHPLVPRIIORZLQJIOXLG¿QGWKHGLVFKDUJH3UHVVXUHLQ VWRUDJHWDQNLVN1P B Pressure = 37.67 kN/m

EL = 9 m

EL = 30

2

A

Pump

3XPSLQJ6WRUDJHWR2YHUKHDG7DQN

  $SSO\LQJ%HUQRXOOL¶VHTXDWLRQEHWZHHQSRLQWVA B PA P V V + A + zA + Hp = B + B + zB + HLOSS rg rg g g 1RZ

zA zB PA N1 VA ªVB ªPB  DWPRVSKHULF Hp SXPSHe  ¥  Hp   ¥  ¥ 

   

RU

Hp  Hp ±  P 3XPSKRUVHSRZHU HP h  (QHUJ\VXSSOLHGWRWKHV\VWHPE\WKHSXPS ¥ +3 HP =

r g ◊Q ◊ H p 

534

Fundamentals of Fluid Mechanics

1.8 ¥ 0.746 ¥ 103 Q= 0.8 ¥ 103 ¥ 9.81 ¥ 18.2

or

= 0.094 m3/s  ‡ Brine of SG 1.15 is draining from the bottom of a large open tank through a 80 mm pipe. The drain pipe ends at a point 10 m below the surface of the brine in the tank. &RQVLGHULQJ D VWUHDPOLQH VWDUWLQJ DW WKH RUL¿FH RI WKH EULQH LQ WKH WDQN DQG SDVVLQJ through the centre of the drain line to the point of discharge and assuming the friction is negligible, calculate the velocity of flow along the streamline at the point of discharge from the pipe. (AMIE 2000) 1

10

2

Draining from Pipe

Applying Bernoulli’s equation between points 1 and 2, we get P1 P V2 V2 + 1 + z 1 = 2 + 2 + z2 rg rg 2g 2g Now,

we get

or

z1 – z2 = 10,

P1 P P = 2 = atm and V1 = 0, rg rg rg 10 =

V2 =

V22 2g 2 g ¥ 10

= 14 m/s  ‡ 7KH¿JXUHVKRZVDWXUELQHZLWKLQOHWSLSHDQGDGUDIWWXEH,IWKHH൶FLHQF\RIWXUELQH LVDQGGLVFKDUJHLVOLWUHVV¿QG  WKHSRZHUGHYHORSHGE\WKHWXUELQH   the reading of gauge G.

Fluid Dynamics-I 1

535

2

350 kN/m

Dia = 0.4 m of inlet pipe

2m

Turbine 1m

Dia - 0.5 m

3m G

2

Datum Draft tube

Q OLWUHVV PV A DUHDRILQOHWSLSH 

p ¥  p d =  

 P 9HORFLW\DWLQOHW V =

Q  = A 

 PV /HWHT WXUELQHKHDGGHYHORSHGSHUXQLWZHLJKWRIWKHZDWHU  DQG HL KHDGORVVHV    $SSO\LQJ%HUQRXOOL¶VWKHRUHPEHWZHHQSRLQW  DWLQOHWDQGSRLQW  DWIUHHVXUIDFHDQG WDNLQJIUHHVXUIDFHDVGDWXP P P V V +  + z =  +  + z + HT + HL rg rg g g  ¥   +  HT   ¥   ¥  \

HT   P

PRZHUGHYHORSHG  

P = h m g HT = h ¥ Q ¥ r ¥ g ¥ HT  ¥¥¥¥¥  N:

536

Fundamentals of Fluid Mechanics

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M 6m 4m L 4m

Dia = 0.1 m

1

Pump sump

/LIWLQJIURPD6XPS

Q OLWUHVV PV A DUHDRISLSH 

p p ¥ d = ¥    

 P Q = AVZKHUHV YHORFLW\DWRXWOHW RU

YHORFLW\ 

Q  =  PV A 

  $SSO\LQJ %HUQRXOOL¶V HTXDWLRQ EHWZHHQ SRLQW  DW WKH VXUIDFH RI ZDWHU DQG SRLQW  DW WKH RXWOHWZHJHW P V P V +  + z + Hp =  +  + z + HL rg g rg g HP +HDGGHYHORSHGE\WKHSXPS HL +HDGORVVLQWKHV\VWHP    1RZ

P P =  =DWPRVSKHUHVV  V PV rg rg z  z  0 + 0 + 0 + HP = 0 +

    ¥ 

Fluid Dynamics-I

HP PRIZDWHU

RU PRZHUWRUXQSXPS

P = m g HP h =

r ¥ Q ¥ g ¥  

=

 ¥  ¥  ¥  ¥  

 N:   $SSO\LQJ%HQRXOOL¶VHTXDWLRQEHWZHHQSRLQWDQGSRLQWLZHJHW P P V V +  + z = L + L + zL rg rg g g 0+0+0=

PL   +  rg g

PL  ±  rg RU

PL ± ¥¥¥  ± N1P



  $SSO\LQJ%HUQRXOOL¶VHTXDWLRQEHWZHHQSRLQWVDQGMZHJHW P P V V +  + z + HP = M + M + zm rg rg g g NRZ

P  V  Z  HP P rg Zm P



 

PM   +  rg g

PM    ± ± rg  ¥   ±± P PM ¥ ¥¥ N1P

537

538

Fundamentals of Fluid Mechanics

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'\QDPLFKHDGHd =

=

      

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V = Cv g hd    ¥  ¥ 

hw FP

Fluid Dynamics-I

539

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PG\QDPLF r¥ g

=

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Chapter

10

FLUID DYNAMICS-II

KEYWORDS AND TOPICS         

PRINCIPLE OF DYNAMICS MOMENTUM PRINCIPLE MOMENT OF MOMENTUM LINEAR MOMENTUM ANGULAR MOMENTUM VANES FLOW EXERTED BY JET FORCES ON BENDS RECTILINEAR FLOW

        

RADIAL FLOW ROTARY FLOW FORCED VORTEX FREE VORTEX ROTATION OF SPRINKLER VENTURIMETER ORIFICE METER FLOW NOZZLE ROTO METER

10.1 INTRODUCTION The fundamental principle of dynamics is Newton’s second law of motion. It states that time rate FKDQJHRIPRPHQWXPFRQWDLQHGLQDYROXPHRIÀXLGLVSURSRUWLRQDOWRWKHDSSOLHGIRUFHDQGWKLV change of momentum takes place in the direction of force. This is called impulse momentum principle which is very useful in addition to the continuity and the energy principles. The impulse PRPHQWXPSLUQFLSOH¿QGVYDVWDSSOLFDWLRQVLQWKHVROXWLRQRIVHYHUDOÀXLGÀRZSUREOHPV7KH LPSXOVH PRPHQWXP HTXDWLRQ LV XVHG IRU WKH ÀRZ DQDO\VLV RI L  SLSH EHQGV DQG UHGXFHUV LL  MHW SURSXOVLRQV LLL  ¿[HG DQG PRYLQJ YDQHV LY  SURSHOOHUV RI VKLSV DLUFUDIW DQG KHOLFRSWHUV Y KHDGORVVGXHWRVXGGHQHQODUJHPHQWRUFRQWUDFWLRQLQSLSHV\VWHPDQG YL K\GUDXOLFMXPS GXULQJÀRZLQRSHQFKDQQHO

10.2 MOMENTUM PRINCIPLE ‡ :KDWLVWKHIXQGDPHQWDOSULQFLSOHRIG\QDPLFV" The fundamental principle of dynamics is Newton’s second law of motion. The second law states that the time rate of change of momentum contained in a volume is proportional

580

Fundamentals of Fluid Mechanics

to the applied force and this change of momentum takes place in the direction of force. $FFRUGLQJ WR WKLV ODZ WKH UHVXOWDQW H[WHUQDO IRUFH DFWLQJ RQ D PDVV SDUWLFOH DORQJ DQ\ DUELWUDULO\ FKRVHQ GLUHFWLRQ LV HTXDO WR WLPH UDWH FKDQJH RI LWV OLQHDU PRPHQWXP LQ WKH same direction. Dynamic force = Rate of change of momentum in the same direction or

F=

d mv ZKHUHm PDVVv = velocity and T = time dT

‡ :KDWLVLPSXOVHPRPHQWXPSULQFLSOH" 2U   'HVFULEHWKHPRPHQWXPHTXDWLRQ 8378 The impulse momentum principle states that the force FDFWLQJRQDÀXLGPDVVm in short Ê d ˆ interval of time dT is equal to the change of momentum Á mv ˜ in the direction of force. Ë dT ¯ 7KHLPSXOVHPRPHQWXPSULQFLSOHLVEDVHGRQWKHODZRIFRQVHUYDWLRQRIPRPHQWXPZKLFK states WKDWWKHQHWIRUFHDFWLQJRQDÀXLGPDVVLVHTXDOWRWKHFKDQJHLQWKHPRPHQWXPRI ÀRZSHUXQLWWLPHLQWKHGLUHFWLRQRIQHWIRUFH Force = change of momentum F= or

d mv dT

F ◊dT = d mv

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Fluid Dynamics-II

581

10.3 FORCE ON VANE  ‡ :KDWLVDYDQH"   $YDQHLVDÀDWRUFXUYHGSODWH$QXPEHURIYDQHVDUH¿[HGRQWKHULPRIDZKHHOWRIRUP DZDWHUZKHHORUWXUELQH Curved vanes Vane Guide vanes

Flow

Turbine

Water Wheel

 ‡ 'HULYHDQH[SUHVVLRQIRUWKHIRUFHH[HUWHGRQDVWDWLRQDU\YDQHE\DMHWRIZDWHUZKLFK LVVWULNLQJQRUPDOO\WRWKHYDQH

V Nozzel

Vane Water jet

Stationary Vane: Jet Normal



 7KH¿JXUHDERYHVKRZVDMHWRIZDWHUVWULNLQJQRUPDOO\WRDÀDWVWDWLRQDU\YDQH$VSHUWKH PRPHQWXPHTXDWLRQWKHIRUFHH[HUWHGE\WKHMHWRIZDWHULVHTXDOWRWKHUDWHRIFKDQJHRI momentum of the water. Mass of water striking the vane = m = r ◊a◊v

  ZKHUHr = density a = area of jet and v = velocity of jet The velocity of water is reduced to zero from velocity v after striking the vane. \   

Force on the vane = rate of change of momentum of water on vane surface = m ¥ dv   rAv v± F = rav

582

Fundamentals of Fluid Mechanics

 ‡ 'HULYH DQ H[SUHVVLRQ IRU IRUFH H[HUWHG E\ D MHW RI ZDWHU VWULNLQJ LQFOLQHG VWDWLRQDU\ YDQH Vane V Fx = F sin q

Nozzle q

F

Jet

Fy = F cos q

q

Jet Inclined to Vane

When a jet of water strikes inclined qWRWKHVWDWLRQDU\YDQHWKHQIRUFHH[HUWHGQRUPDOWR WKHYDQHLV F = m ¥ vsin q = mass ¥ normal velocity = ra v sin q   ZKHUHa = area of jet and v = velocity of jet 

 7KH DERYH IRUFH FDQ EH UHVROYHG LQWR WZR FRPSRQHQWV YL]   IRUFH Fx  LQ GLUHFWLRQ RI ÀRZDQG  IRUFH Fy DFWLQJQRUPDOWRWKHGLUHFWLRQRIÀRZ+HQFHZHKDYH Fx = = Fy = =

and

F sin q rav sin q F cos q ravsin q cos q

 ‡ $ MHW RI ZDWHU VWULNHV D SODWH KLQJHG DW WKH WRS HQG 7KH SODWH KDV ZHLJKW  W DQG OHQJWK L)LQG D WKHDQJOHqWRZKLFKWKHSODWHZLOOVZLQJRQDFWLRQRIWKHMHW E  IRUFHpWRH[HUWHGDWWKHERWWRPRIWKHSODWHWRVWRSWKHSODWHIURPVZLQJLQJZKHQ D  L  MHWLVDFWLQJDW  E MHWLVDFWLQJDW L  4 Hinged Jet

L 2

q F

W Swinging of Plate

3 L 4 L

F L 2

F P

Force P to Hold the Plate

P

Fluid Dynamics-II



583

 'XULQJHTXLOLEULXPWKHSODWHKDVVZXQJE\DQDQJOHq from vertical due to force FH[HUWHG E\WKHMHWRIOLTXLG7DNLQJPRPHQWZLWKUHVSHFWWRKLQJHGSRLQWZHKDYH Â Mhinged ± F ¥

L L – W ¥ sin q = 0  

 ZKHUHF IRUFHH[HUWHGE\MHWDQG W = weight of the plate or

sin q =

F W

Now we apply force p to stop the plate from swinging. Taking moment with respect to KLQJHGSRLQWZHKDYH Â MELQJHG ± F ¥

L +P¥L=0 

F 

or

p=

In case the jet is acting at a point

 LIURPWKHKLQJHGSRLQWWKHQRQWDNLQJPRPHQWZLWK 

reVSHFWWRKLQJHGSRLQWZHKDYH Â MELQJHG =± F ¥

 L+p¥L=0 

 F  ‡ 'HULYHDQH[SUHVVLRQIRUIRUFHH[HUWHGE\DMHWRIZDWHUZKLOH VWULNLQJRQDPRYLQJ YHUWLFDOSODWH)LQGDOVR  ZRUNGRQH  H൶FLHQF\DQG  FRQGLWLRQIRUPD[LPXP H൶FLHQF\ or

p=

Plate

V V1

Nozzle Jet of water Normal Jet on Moving Plate

Consider a jet of water striking a plate with a velocity of V. The plate is moving with a velocity of VLQWKHVDPHGLUHFWLRQRIWKHMHW+HQFHUHODWLYHYHORFLW\RIWKHMHWZLWKUHVSHFW to the moving plate is V – V. The velocity of the water of the jet is zero after it strikes.

584

Fundamentals of Fluid Mechanics

   or

)RUFHH[HUWHGE\WKHMHW UDWHRIFKDQJHRIPRPHQWXPRIWKHZDWHU F = mass ¥ change of velocity = r ¥ a ¥ V – V ◊ V – V   ZKHUHr = density & a = area of jet or F = r ◊ a ◊ V – V  

 1RZWKHIRUFHDFWLQJRQWKHSODWHLVJLYHQE\WKHDERYHHTXDWLRQ7KHZRUNGRQHE\WKH IRUFHLVHTXDOWRWKHIRUFHPXOWLSOLHGE\GLVWDQFH

  

:RUNGRQH W = force ¥ distance = r ◊a◊ V – V  ¥ V

  7KHMHWKDVEHHQVXSSOLHGNLQHWLFHQHrJ\E\WKHQR]]OH KE =



=

 ◊ r ◊ a ◊ V◊V  

=

 r aV  

(൶FLHQFy =

=

or

 mV 

h=

:RUNGRQHE\MHW Energy supplied to jet r ◊ a ◊ V - V  ◊ V  ◊ r ◊ a ◊ V   V - V  ◊ V V

 ,QRUGHUWR¿QGPD[LPXPH൶FLHQF\ ∂h KDVWREHHTXDWHGWR]HUR ∂V ∂h V ◊  V - V V 3 - V  ◊  V - V  = =0 ∂V V   RU or

V± ◊ V – V   V ◊V

Fluid Dynamics-II

585

  7KHDERYHLVWKHFRQGLWLRQIRUPD[LPXPH൶FLHQF\7KHPD[LPXPH൶FLHQF\  V - V  ◊ V

h ma[ =

V 

=

=

 ◊  ◊ V ◊ V  ◊ V 8 

‡ 'HULYH DQ H[SUHVVLRQ IRU ZRUN GRQH E\ WKH MHW RI ZDWHU RQ PRYLQJ LQFOLQHG SODWH RU IODWYDQH Nozzle

V

F

Fy = F cos q

y x q

Fx = F sin q

Jet on Moving Inclined Plate



 7KH MHW LV VWULNLQJ LQFOLQHG WR WKH SODWH DV VKRZQ LQ WKH ¿JXUH 7KH UHODWLYH YHORFLW\ LV V – V Force = F = m V – V VLQq = r ◊a◊ V – V sin q Force along xGLUHFWLRQFx = Fsin q \

Fx = r ◊ a ◊ V – V sin q

 6LPLODUO\ZHKDYH Fy = r ◊a◊ V – V  sin q cos q   :RUNGRQHE\WKHMHW W = Fx◊ displacement = r ◊ a◊ V – V  sin q ◊ V = r ◊a ◊V V – V  sin q  ‡ $PPGLDPHWHUZDWHUMHWVWULNHVDKLQJHGYHUWLFDOSODWHRI1ZHLJKWQRUPDOO\ DWWKHFHQWUHZLWKPVYHORFLW\'HWHUPLQHWKHDQJOHRIGHIOHFWLRQqRIWKHKLQJHG SODWHIURPYHUWLFDO$OVRGHWHUPLQHWKHPDJQLWXGHRIIRUFHFWKDWPXVWEHDSSOLHGDW LWVORZHUHGJHWRNHHSWKHSODWHYHUWLFDO 8378

586

Fundamentals of Fluid Mechanics

Jet with dia = 40 cm q Nozzle

F V = 1.5 m/s P 800 N

Area of jet

       

p ¥   pd =  

¥  ± m  )RUFHH[HUWHGE\MHWF = raV   ¥  ¥¥± ¥  1

Force PWRKROGWKHSODWHLQYHUWLFDO F  =  

P=  

 1

  1RZ

sin q =



or

F  =   W 800

q ƒ

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     or

$UHDRIWKHMHW a =

p ¥   pd =  ¥ ± m  

9HORFLW\RIMHW v = cv g h  ¥  ¥  ¥  v PV

Fluid Dynamics-II

587

  )RUFHH[HUWHGE\WKHMHWLV F = r ◊a◊v     ¥ ¥¥± ¥       N1  ‡ $MHWRIZDWHULVHPHUJLQJIURPDQR]]OHXQGHUDFRQVWDQWKHDGRIP&DOFXODWHWKH GLDPHWHURIWKHMHWLIWKHIRUFHH[HUWHGE\WKHMHWVWULNLQJDSODWHQRUPDOO\LVN1 $VVXPHcv    

9HORFLW\RIWKHMHW cv =

g h

v =¥     )RUFHH[HUWHGE\WKHMHWLV

 ¥  ¥ 

 PV F = r ◊a ◊ v

or

a=

=

 ¥   ¥  ¥  ◊  

 ¥± m

  or

F r ◊ v

pd  ¥ ± 

or

d=

 ¥  ¥ - p

= 8 ¥± m = .08 m = 80 mm  ‡ $MHWRIZDWHURIPPGLDPHWHUHPHUJHVZLWKDYHORFLW\RIPVDQGVWULNHVDSODWH )LQGWKHIRUFHH[HUWHGRQWKHSODWHLI  WKHMHWVWULNHVQRUPDOO\WRWKHSODWHDQG   WKHMHWLVLQFOLQHGDWƒWRWKHSODWH Area of jeWa =

p ¥   pd =  ¥ ± m  

588

Fundamentals of Fluid Mechanics

 &DVH,:KHQWKHMHWLVVWULNLQJQRUPDOO\WRWKHSODWHWKHIRUFH F = r ◊ a ◊v  ¥ ¥¥ ± ¥    N1

     

  &DVH,,7KHMHWLVLQFOLQHGƒWRWKHSODWHWKHIRUFHH[HUWHG F = r ◊a◊vsin  =¥  

 

 N1

 ‡ $MHWRIZDWHUZLWKPPGLDPHWHUH[HUWVDIRUFHRIN1LQWKHGLUHFWLRQRIWKHIORZ RQDIODWSODWH7KHMHWLVLQFOLQHGDWƒWRWKHSODWH)LQGWKHUDWHRIGLVFKDUJH The area ofMHWa =

p ¥    ¥ ±  m 

 7KHIRUFHH[HUWHGRQWKHSODWH F = r ◊a◊vsin     

¥  ¥ ¥¥– ¥ v ¥ v =

or

   

   ‡

 

 ¥  ¥   ¥ 

v PV 1RZ 'LVFKDUJHQ = a ◊v   ¥ ± ¥   PV $MHWRIZDWHUZLWKGLDPHWHURIPPVWULNHVQRUPDOO\DSODWHZLWKYHORFLW\RIPV 7KHSODWHKDVOHQJWKRIPDQGWKHMHWVWULNHVDWWKHFHQWUHRIWKHSODWH:KDWIRUFH LVWREHH[HUWHGDWPPEHORZWKHKLQJHGWRSWRVWRSSODWHWRPRYH"

200 V Nozzle 100 P

Fluid Dynamics-II

Area of the jet =

589

pd p ¥   =  ¥ ± ¥ m  

F = r ◊a◊v     ¥ ¥¥± ¥       1   7DNLQJPRPHQWIURPKLQJHGSRLQW Â m hinged ± F ¥P ¥  or

P= =

 

F ¥    ¥  

 1

 ‡ $MHWRIZDWHUKDYLQJPPGLDPHWHULVPRYLQJZLWKYHORFLW\RIPVDQGLPSLQJHV QRUPDOO\RQDSODWH)LQGWKHIRUFHRQWKHSODWHZKHQ  SODWHLV¿[HGDQG  SODWH LVPRYLQJZLWKYHORFLW\RIPV The area of the jeta =

p ¥    ¥ ± m 

Case3ODWHLV¿[HG Force = F = r◊ a◊v  or   

F ¥ ◊ ¥ ± ¥    1

Case 3ODWHLVPRYLQJZLWKPV F = r ◊a ◊ V – V      ¥ ¥¥ ± ¥       1

10.4 MOMENT OF MOMENTUM  ‡ 'HVFULEHGL൵HUHQFHVEHWZHHQOLQHDUPRPHQWXPDQGPRPHQWRIPRPHQWXP   :KHQDÀXLGSDUWLFOHLVPRYLQJLQDVWUDLJKWGLUHFWLRQLWKDVOLQHDUPRPHQWXP+RZHYHU ZKHQ D ÀXLG SDUWLFOH PRYHV DORQJ D FXUYHG SDWK VXFK WKDW LWV GLVWDQFH IURP WKH D[LV RI URWDWLRQFKDQJHVZLWKWLPHVWKHQWKHGLVWDQFHIURPWKHD[LVRIURWDWLRQRIWKHSDUWLFOHZLOO EHGL൵HUHQWDWGL൵HUHQWSRVLWLRQVRIWKHSDUWLFOH+HQFHWKHPRPHQWRIPRPHQWXPRIWKH SDUWLFOHZLOOFKDQJHIURPWKHD[LVRIURWDWLRQ7KHPRPHQWRIPRPHQWXPLVDOVRNQRZQDV DQJXODUPRPHQWXP$VSHUWKHHTXDWLRQRIPRPHQWRIPRPHQWXPRUDQJXODUPRPHQWXP

590

Fundamentals of Fluid Mechanics

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V2 r2

Vr

r Vr

r1 Vr

1

V1

2

2

2

V

A Vq

Vq

Vq

1

1

Consider a particle ALVPRYLQJRQDFXUYHGSDWKZLWKUHVSHFWWRWKHD[LVRIURWDWLRQSDVVLQJ through point ODVVKRZQLQWKH¿JXUH7KHYHORFLW\VRIWKHSDUWLFOHFDQEHUHVROYHGLQWR WZR FRPSRQHQWV YL]   Vr is normal velocity acting along the radius r and towards the centre of rotation O DQG   Vq is tangential velocity acting normal to the radius r. The DQJXODUPRPHQWXPVRIWKHSDUWLFOHDWSRLQWDQGDUH    $QJXODUPRPHQWXPDWSRLQW PRPHQWRIWKHPRPHQWXP = r mVq1    $QJXODUPRPHQWXPDWSRLQW r  m◊ Vq Change of angular momentum = m rVq – r Vq If the change of angular momentum takes place in time t Rate of change of angular momentum =

 m r Vq – rVq t

 1RZWKHUDWHRIFKDQJHPRPHQWXPLVHTXDOWRWRUTXH T  T=  %XW \

 m r Vq – rVq t

m = rgQ PDVVRIÀXLGÀRZLQJSHUXQLWWLPH t T = rgQ r Vq – r Vq

The toque T LV WKH WRTXH H[HUWHG E\ VRPH H[WHUQDO DJHQF\ RQ WKH ÀXLG DV WKH DQJXODU PRPHQWXP LV LQFUHDVLQJ ,Q FHQWULIXJDO SXPS WKH WRUTXH LV H[HUWHG RQ WKH ÀXLG E\ WKH FHQWULIXJDO SXPS ZKHQ LW PDNHV WKH ZDWHU WR PRYH RQ WKH FXUYHG SDWK 6LPLODUO\ D FRPSUHVVRUH[HUWVWRUTXHRQWKHDLUDQGDSURSHOOHUH[HUWVWRUTXHRQWKHDLU,QFDVHWKHÀXLG LVPDGHWRPRYHRQDFLUFXODUSDWKWKHQr = r = rDQGWRUTXH

Fluid Dynamics-II

591

T = rgQ r Vq – Vq 

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  7KHHTXDWLRQ T = rg ◊Q ◊r Vq – Vq is called the equation of moment of momentum or angular momentum.

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Flow Inwardly (as in Pump)

Flow Outwardly (as in Turbine)

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z

Original level

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592

Fundamentals of Fluid Mechanics

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Z U  ◊ zLVDOVRFDOOHGGHSWKRISDUDEROD g

‡ ([SODLQIUHHYRUWH[IORZ   ,QIUHHYHUWH[ÀRZZHKDYH  D  )ORZSDUWLFOHVPRYLQJLQFLUFOHVDERXWDSRLQW  E  7DQJHQWLDOYHORFLW\FRPSRQHQW Vq H[LVWVZKLOHUDGLDOYHORFLW\FRPSRQHQWLV]HURLH Vr = 0 and VqH[LVWV  F  7DQJHQWLDOYHORFLW\YDULHVUDGLXVr so that same circulation is maintained.  G  $OO VWUHDPOLQHV DUH FRQFHQWULF FLUFOHV DERXW D JLYHQ SRLQW ZKHUH WKH YHORFLW\ DORQJ HDFK VWUHDPOLQH LV LQYHUVHO\ SURSRUWLRQDO WR WKH GLVWDQFH IURP WKH FHQWUH7KH ÀRZ LV QHFHVVDULO\LUURWDWLRQDO9HORFLW\FRPSRQHQWVDUH y3

C r1 C y2 = r2 C y3 = r3

y1 =

y2 y1

r1 r2

r3



D  7DQJHQWLDOYHOocity Vq = = =

Circulation constant r G  2p    ZKHUHG = circulation r c r

Fluid Dynamics-II



593

E  5DGLDOYHORFLW\Vr = 0 F  6WUHDP funFWLRQY = -

G log e r + c1 p

‡ ([SODLQIRUFHGYRUWH[IORZ Flows where streamlines are concentric circles and tangential velocity is directly proportional WR WKH UDGLXV RI FXUYDWXUH DUH FDOOHG SODQH FLUFXODU IRUFHG YRUWH[ ÀRZV7KH ÀRZ ¿HOG LV GHVFULEHGLQDSRODUFRRUGLQDWHV\VWHPDV Vq = ω ◊ r and 

Vr = 0

 $OOÀXLGSDUWLFOHVURWDWHZLWKWKHVDPHDQJXODUYHORFLW\wOLNHDVROLGERG\7KLVLVWKHUHDVRQ ZK\DIRUFHGYRUWH[ÀRZLVDOVRFDOOHGDVROLGERG\URWDWLRQ7KHYRUWLFLW\WIRUWKHÀRZ W =

∂Vq 1 ∂Vr Vq + r ∂q r ∂r

= w – 0 + w w 

 7KHUHIRUHDIRUFHGYRUWH[PRWLRQLVQRWLUURWDWLRQDOEXWDURWDWLRQDOÀRZZLWKDFRQVWDQW YRUWLFLW\w.

  7KHGL൵HUHQFHRIWRWDOKHDGEHWZHHQSRLQWµ¶DQGµ¶RQVDPHSODQHLV È P2 P ˘ w2 2 r2 - r12 - 1˙+ H – H = Í 2 g g g r r Î ˚ But H – H =

w2 2 r2 - r12 DQGZHFDQZULWHSUHVVXUHGL൵HUHQFHDV g P2 - P1 w2 2 r2 - r12 =  r

 ‡ $ RSHQ FLUFXODU WDQN RI GLDPHWHU   FP DQG  FP ORQJ FRQWDLQV ZDWHU XS WR D KHLJKWRIFP7KHWDQNLVURWDWHGDERXWLWVYHUWLFDOD[LVDWUSP)LQGWKHGHSWK RISDUDERODIRUPHGDWWKHIUHHVXUIDFHRIZDWHU

z 60 cm

594

Fundamentals of Fluid Mechanics

  

 1RZ

Z=

p N  ¥ p ¥  =  

z=

Z U  g

=  

UDGV

  ¥    ¥ 

 P

10.6 FORCE EXERTED ON A BEND  ‡ 'HULYHDQH[SUHVVLRQIRUIRUFHH[HUWHGE\IOXLGIORZRQDEHQGZKHQWKH  EHQGLVLQ KRUL]RQWDOSODQHDQG  EHQGLVLQYHUWLFDOSODQH 2U 6WDWHWKHPRPHQWXPHTXDWLRQ+RZLVLWXVHGLQGHWHUPLQLQJWKHIRUFHH[HUWHGE\WKH IORZLQJOLTXLGRQDSLSHEHQG 8378 P2A2 sin q 2

V2 cos q

1

P1 A 1

P2A2 cos q

V2

Fy

2

V2 sin q

q

V1

P2 A 2

Fx

1

Forces on Bend (In Horizontal Plane)

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  3UHVVXUHIRUFH P◊A



  )RUFHRIÀXLGRQWKHEHQG F



  :HLJKWRIÀXLGÀRZLQJ WWREHFRQVLGHUHGLQEHQGLQYHUWLFDOSODQH

Fluid Dynamics-II

595

Applying momentum equation in xGLUHFWLRQZHJHW P A – PA cos q + Fx = rQ v cos q – v Fx = rQ vcos q – v P Acos q – PA

or

  6LPLODUO\HTXDWLQJIRUFHVLQyGLUHFWLRQZHJHW Fy = rQ v sin q PAsin q Resultant forFHF =

\

Fx + F y

 7KHDQJOHRIUHVXOWDQWZLWKKRUL]RQWDO q = tan±

Fy Fx

Case)RUFHVRQEHQGZKLFKKDVLQOHWLQKRUL]RQWDOSODQHDQGRXWOHWLQYHUWLFDOSODQH1HW force in xGLUHFWLRQ ZLOO UHPDLQ VDPH EXW WKH QHW IRUFH LQ y-direction will change as the ZHLJKWRIÀXLGSDUWLFOHLQyGLUHFWLRQKDVWREHFRQVLGHUHG 2 P2A2 Fy 1 V2 P1A1

V1

2

q Fx W

1

Fx = rQ vcos q – v P A cos q – P A Fy = rQ vsin q P A sin q + W  ‡ $EHQGLQSLSHOLQHFRQYH\LQJZDWHUJUDGXDOO\UHGXFHVIURPFPWRFPLQGLDPHWHU DQGWKHEHQGGHIOHFWVWKHIORZWKURXJKƒ7KHJDXJHSUHVVXUHDWLQOHWLVN3D )LQGWKHPDJQLWXGHDQGGLUHFWLRQRIWKHIRUFHRQWKHEHQGZKHQ  WKHUHLVQRIORZ DQG  IORZLVPV A = area at inlet = A = area at outlet =

p ¥   p d =  P   p d  p ¥   =  P  

596

Fundamentals of Fluid Mechanics

Case  :KHQ QR ÀRZ LV WDNLQJ SODFH WKHQ WKH QHW IRUFH LQ x-direction is the change of momentum in x-direction. P A – PAcos q + Fx = rQ v cos q – v   +HUH

q = 0 and P = P = P

\

Fx = P A cos q – P A = P A cos q – A

  

 ¥  FRV±

  

 ¥ ± 

  

 ±¥ ¥ 

  

 ± N1

  6LPLODUO\ZH¿QGRXWFy. Fy = P A sin q + rq vcos q   %XW \  

q=0 Fy ¥ ¥ 0.¥  N1 Fx + F y =

R= =

-   +  

 +  +   N1

q = tan± Case:KHQÀRZRI

 

 or q ƒ 

m LVWDNLQJSODFHZHFDQDSSO\WKHFRQWLQXLW\HTXDWLRQ s q = a  v = a v

\

v =

0.8 

and

v =

0.8  PV 

PV

Fluid Dynamics-II

Apply Bernoulli’VHTXDWLRQDWVHFWLRQDQG 2 P2 A2

1

V1

P1

60°

2

A1 V1 1

p p v v +  =  +  rg rg g g p  ¥      + = +  ¥  ¥   ¥   ¥   ¥  ¥   

 

p  ¥ 



p ¥¥

or   Now net force in xGLUHFWLRQLV

 N1P

P A – PA + Fx = rq v cos q – v Fx ±¥ ¥¥ ¥¥

or

  ¥ ¥FRV±

  

 ±¥¥ Fx ± N1

  

Now net force in yGLUHFWLRQLV Fy – P Asin q = rQ v sin q Fy ¥ ¥ VLQ¥ ¥ 0.8 ¥VLQ   

 ¥¥

  

 N1 R=

Fx + F y

=

-   +  

=

 + 

597

598

Fundamentals of Fluid Mechanics

 

  q = tan±

= tan–

Fy Fx  

   ƒ  ‡ $  PP GLDPHWHU SLSH FDUULHV ZDWHU XQGHU D KHDG RI  P ZLWK D YHORFLW\ RI  PV,IWKHD[LVRIWKHSLSHWXUQVWKURXJKƒ¿QGWKHPDJQLWXGHDQGGLUHFWLRQRIWKH UHVXOWDQWIRUFHDWWKHEHQG 8378 2

V2

1 2 H1

45°

V1 1

A = A =  1RZ  6LQFH

p ¥    P 

A V = AV A = AKHQFHV = V PV

  $SSO\LQJ%HUQRXOOL¶VHTXDWLRQEHWZHHQDQG p p v v +  + z =  +  + z  rg rg g g  1RZ  +HQFH

z = z

p p = H  = Hv = v r g rg

H = H  p = p ¥ rg ¥¥ ¥

 

 N1P

  1RZ P A = PA cos q + Fx = rQ vcos q – v

Fluid Dynamics-II

     

599

 ¥±¥¥FRVFx  ¥ ¥ ¥  FRV±  ¥ ±FRV Fx     ¥ FRV±

  \

Fx ± ± Fx ± N1

  6LPLODUO\ \ or

Fy – P A sin q = r◊Q vsin q Fy ¥¥VLQ¥  VLQ¥

 

 

 

 

\

 

R=

Fx + F y

=

  +  

=

 + 

 N1 q = tan±

 

ƒ  ‡ $SLSHKDVGLDPHWHURIPPDQGULJKWDQJOHEHQGLQDKRUL]RQWDOSODQH,IPV ZDWHULVIORZLQJ¿QGIRUFHVDFWLQJRQWKHEHQG7KHSUHVVXUHDWWKHLQOHWDQGRXWOHW DUHDQGN1PUHVSHFWLYHO\ P2, A2, V2 2

1 P1 A1 V1 1 Right Hand Bend (in Horizontal Plane)

2

600

Fundamentals of Fluid Mechanics

Area A = A = A =

p ¥   P 

A V = A V = Q

 &RQWLQXLW\HTXDWLRQ or

V =

Q Q =  PV A 

V = V PV Applying momentum equation in x-direction P A + Fx = rQv or Fx ¥ ¥¥±¥¥     ± N1 Applying momentum equation in y-direction Fy – P A = rQv or Fy ¥ ¥¥q ¥ ¥ Fy N1 R= =

f x + f y   +  ◊  

   N1 Angle of resultant from horizontal q = tan±

fy  = tan± fx 

q ∞  ‡ $ SLSH KDV GLDPHWHU RI  PP DQG EHQG RI ƒ LQ KRUL]RQWDO SODQH ,I WKH IORZ LV PV)LQGWKHPDJQLWXGHDQGGLUHFWLRQRIWKHUHVXOWDQWRIIRUFHV7KHSUHVVXUHLQ SLSHLVN1P 2 P2, A2, V2 2

Fy

1 135° P1 A1

Fx

V1 1 A Bend in Horizontal Plane

Fluid Dynamics-II

A = A = A =   &RQWLQXLW\HTXDWLRQ

  1RZ

A V = AV = Q V = V =

or

p ¥    P 

 

 PV 

P = P N1P

Applying momentum equation in x-direction P A + Fx + P Acos q = rQ ± vcos q – v or

Fx ¥ ¥ ±  FRV  ± ¥ ¥±¥ ¥FRV

   or  

Fx ± ±  ± N1

Applying momentum equation in y-direction – P A sin q + Fy = rQ –v sin q or

Fy = r ◊Q◊v VLQP AVLQ

  

 ¥ ¥¥VLQ¥¥VLQ

  

 

  

 N1 R=

 

Fx + F y

=

  +  

=

 + 

 N1 q = tan±

 

 ƒ

 

601

602

Fundamentals of Fluid Mechanics

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P2, A2, V2

A1 V1 2 1

Reducer

A =

p d  p ¥ 0.8 =  P  

A =

p d  p ¥  =  P  

Continuity equation = A V = A V   RU or



¥ ¥ v V =

 

 ¥  

 PV Q = A V ¥ PV

  $SSO\LQJ%HUQRXOOL¶VHTXDWLRQEHWZHHQVHFWLRQ  p p v v +  + z =  +  + z + hloss rg rg g g z = z p    ¥    + = +   ¥  ¥   ¥   ¥   ¥  ¥  or  

p 

 ¥  ¥ 

 ±±  

Fluid Dynamics-II

or

603

p N1

Applying momentum equation in x direction P A – P A + Fx = rQ v – v Fx = rQ v – v P A – PA Fx ¥ ¥¥ ± ¥   ¥±¥ ¥  ¥¥±¥  ± N1

or         

10.7 TORQUE ACTING ON A SPRINKLER  ‡ 'HULYH DQ H[SUHVVLRQ IRU WRUTXH DFWLQJ RQ D VSULQNOHU XVLQJ PRPHQW RI PRPHQWXP HTXDWLRQ r1

r2

1

2

A V1

V2

Sprinkler

  &RQVLGHUDVSULQNOHUZLWKQR]]OHDWDQGDQGKLQJHGDWSRLQWA. 0RPHQWXPDWSRLQW PDVV¥ velocity = rQ ¥ v

  

0RPHQWRIPRPHQWXPDWSRLQW r ◊Q◊v ¥ r

  

0RPHQWXPRISRLQW r ¥ Q ¥ v

  

0RPHQWRIPRPHQWXPDWSRLQW r ◊ Q◊v ¥ r Rate of change of moment of momentum = rQ vr – v r  $FFRUGLQJWRWKHPRPHQWRIPRPHQWXPHTXDWLRQWKHUDWHRIFKDQJHRIPRPHQWRIPRPHQWXP LVHTXDOWRWKHWRUTXH T 

   

T = rQ v r – vr  ‡ $VSULQNOHUKDVWZRQR]]OHVHDFKORFDWHGDWPIURPWKHFHQWUHZKLFKLVKLQJHG7KH GLDPHWHURIQR]]OHVLVPDQGGLVFKDUJHLV¥ ±PV)LQGWKHVSHHGRIURWDWLRQ LIWKHUHLVQRIULFWLRQORVV$OVR¿QGWRUTXHWRKROGWKHVSULQNOHUVWDWLRQDU\

d2 = 0.01 m

d1 = 0.01 m V1

0.3 m

0.3 m Sprinkler

V2

604

Fundamentals of Fluid Mechanics

A = A = A =

 ¥ - Q =  ¥  ¥ - A

v = v = 

pd p ¥   =  ¥± m  

 PV v = v = ZU

or

  or

Z=

  UDGV 

Z=

p N 

N=

 ¥   USP p

  7RUTXH T = rQ vr – vr  %XW v = – v \ T ¥ ¥¥± ¥¥      1P  ‡ $Q XQV\PPHWULFDO VSULQNOHU KDV HTXDO IORZ WKURXJK HDFK QR]]OH ZLWK YHORFLW\ RI PV)LQGWKHVSHHGRIURWDWLRQLQUSP 0.5 1

1m w

8 m/s

2 8 m/s

Since r > r   +HQFHZU > ZU. Sprinkler will rotate anticlockwise. v = 8 + ZU Z v = 8 – ZU = 8 – Z As per moment of momentum equation T = rQ v r – vr   $VQRH[WHUQDOWRUTXHLVDSSOLHGT = 0 \ v  r  = v r     ±Z ¥  Z ¥ or 8 – Z Z   RU  Z 

Fluid Dynamics-II

 1RZ



or

Z=

  UDGV 

Z=

p N 

N=

 ¥   USP p

605

‡ $ODZQVSULQNOHUKDVFPGLDPHWHUQR]]OHDWWKHHQGRIDURWDWLQJDUPDQGGLVFKDUJHV ZDWHUDWWKHUDWHRIPVYHORFLW\'HWHUPLQHWKHWRUTXHUHTXLUHGWRKROGWKHURWDWLQJ DUPVWDWLRQDU\$OVRGHWHUPLQHWKHFRQVWDQWVSHHGRIURWDWLRQRIWKHDUPLILWLVIUHHWR URWDWH 10 m/s w wr2 20 cm

wr1

25 cm

10 m/s

A=

p ¥  ¥ -  pd =  ¥± m  

Q = AV ¥± ¥ ¥ ± mV T = rQ vr – vr   

 ¥ ¥¥± ¥¥

  

 ¥± 

    ¥¥±     1P If Zis the angular velocity of the sprinkler. Since r > rWKHVSULQNOHUZLOOURWDWHFORFNZLVHDVPRPHQWRIPRPHQWXPRIVHFWLRQLV more    v DEVROXWH Z    v DEVROXWH Z   1RZT DVQRH[WHUQDOWRUTXHLVDSSOLHG T = rQ r v DEVROXWH – r v DEVROXWH   \  

r v DEsolute = r v DEVROXWH  Z   Z

606

Fundamentals of Fluid Mechanics

    

Z Z Z  Z=

  UDGV 

Z=

p N 

N=

 ¥   USP p

‡ $VSULQNOHUZLWKHTXDODUPVRIPEXWQR]]OHVLQFOLQHGƒWRVSULQNOHUD[LVDVVKRZQ LQWKH¿JXUH7KHQR]]OHVKDYHGLDPHWHURIFPDQGVSHHGRIURWDWLRQLVUSP)LQG WKHIORZRXWRIWKHVSULQNOHULIWKHUHLVQRIULFWLRQORVV V2 60 60

1m

1m

V1

Area of nozzle =  

p ¥   

 ¥± m

Now velocity component v VLQDQGv VLQKDYHPRPHQWRIPRPHQWXPZUWKLQJHG point of sprinkler v VLQ¥ r = vVLQ¥ r = ZU   %XW  r = r = r P Z= =   \  

p N   ¥ p ¥  

 UDGs v = v =

 ¥  VLQ 

 PV

Fluid Dynamics-II

607

Q for each nozzle = A ¥ V = A V   

 ¥¥±

  

  ¥± mV QIRUERWKQR]]OHV ¥± mV

 ‡ $YHUWLFDOO\XSZDUGMHWRIZDWHUFPLQGLDPHWHULVVXLQJIURPDQR]]OHDWYHORFLW\RI PVVWULNHVQRUPDOWRDIODWFLUFXODUSODWHRIPDVVNJDQGGLDPHWHUFPDQGVXSSRUWV LW)LQGWKHYHUWLFDOGLVWDQFHDERYHWKHQR]]OHZKHUHWKHSODWHLVKHOGLQHTXLOLEULXP Plate

Jet

Nozzle

  

)RUFHH[HUWHGE\WKHMHW rAV A=

pd p ¥   =  ¥± m  

 1RZSODWHLVLQHTXLOLEULXP or

F = mg ra v = mg

or

v =

 ¥   ¥  ¥  ¥ -

     or v PV Initial velocity of jet u PV v = u ±gh     ±¥¥ h or  

h=

 -   ¥ 

 P

 ‡ 7ZRWDQNVADQGBRQIULFWLRQOHVVZKHHOVDUHSODFHGDVVKRZQAMHWHPHUJLQJIURP WDQN A VWULNHV WKH EODGH RQ WDQN B )LQG WKH IRUFHV UHTXLUHG WR KROG WKH WDQNV VWDWLRQDU\:LOOWKHIRUFHVUHTXLUHGZLOOEHGL൵HUHQWLIHLWKHURQHRUERWKRIWKHPDUH DOORZHGWRUHFHGHDWPV"

608

Fundamentals of Fluid Mechanics

Consider control volume of tank A and jet is emerging from it with force FA. Then FA = rva  ¥ ¥

(

gH

 

  ¥¥¥

 

 1



)

¥

pd 

p ¥   

Jet diameter = 2 cm 2m

A

B

FA FB

Consider control volume of tank BWKHQIURFHDFWLQJRQLWGXHWRMHWLV MHWLVGHÀHFWHGEDFN FB = – rvB ¥ a – r vB ¥ a z  ± rvB ¥ a

     1RZ  



vA = vB =

 ¥   

 ± ¥ ¥¥¥

p ¥   

TB ±1 The force FB is opposite to FA. If tank ALVPRYHGWROHIWRUULJKWWKHIRUFHFA remains the VDPHLHUHDFWLRQRIWKHMHW 1+RZHYHULIWDQNBLVPRYHGWRULJKWYHORFLW\vB  LVUHGXFHGE\PVDQGFBEHFRPHV ¥ ¥   ¥ p ¥   1,IWDQN 

is moved WRWKHOHIWWKHYHORFLW\vBLVLQFUHDVHGE\PVDQGFBEHFRPHV 

 ¥ ¥   ¥ p ¥   1 

Fluid Dynamics-II

609

‡ $MHWHQJLQHZKHQWHVWHGWDNHVLQDLUDWPVDQGGLVFKDUJHVWKHH[KDXVWJDVHVDW  PV7KH LQOHW DQG RXWOHW FURVV VHFWLRQV DUH  P DQG IXHOWRDLU UDWLR LV  )LQGWKHIRUFHH[SHFWHGWRDFWRQWKHHQJLQH7DNHGHQVLW\RIDLUDWHQWU\ NJP DQGp p . mf

V1 . ma

V2

Jet Engine

Applying momentum equation in x-direction P A – PA + Fx = m a + m f v – m a v m a = r ¥ A ¥ V ¥¥   

 NJV m f =

 6LQFH   



 = NJV 

P = P and A = A Fx  m a + m f ± m a ¥    ±¥

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 PV v RXWOHWH[KDXVWJDVYHORFLW\

  

  

 PV air intake = m = rAV   ¥

p ¥   ¥ 

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610

Fundamentals of Fluid Mechanics

F = m v – v   ± 

     

 ¥ 1 Thrust = F 1

\   

3RZHUORVV NLQHWLFHQHUJ\RIH[KDXVWJDVHV =

 

 ¥  

 N: Power output from engine = thrust ¥ velocity of jet

     

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Flow

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h



Main pipe

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Venturimeter

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Fluid Dynamics-II

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d

D

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h

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612

Fundamentals of Fluid Mechanics

 ‡ 'HULYHDQH[SUHVVLRQIRUWKHIORZIRUYHQWXULPHWHU 1

2

Flow

h

Venturimeter

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z = z

\

p - p v  - v =  rg g

p - p = hWKHSUHVVXUHKHDGGL൵HUHQFHPHDVXUHGE\uWXEHPDQRPHWHU rg \

h=

v - v g

 $VSHUFRQWLQXLW\HTXDWLRQ a v= av where a = pipe cross-sectional area and a = throat cross-sectional area or

v =

a v a  

v \

h=

or

h=

or

v =

or

v =

Êa ˆ - Á  ˜ v Ëa ¯ 

g   v Ê a - a ˆ Á ˜   g Ë a ¯

a ¥  gh a - a a  gh a - a

È Ê a ˆ˘ v Í - Á  ˜ ˙ Í Ë a ¯ ˙ Î ˚ = g

Fluid Dynamics-II

Q = v a

  )ORZWKURXJKSLSH or 

613

Q=

a ◊ a  gh a - a

 7KHDERYHLVWKHRretical discharge. The actual discharge is less than the theoretical discharge ZKLFKLV Qactual = Cd

aa  gh a - a

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0

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614

Fundamentals of Fluid Mechanics

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v =

a0 v a 0 

v0



h=

\

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v0 =

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a a  gh a - a

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a a  gh a - a

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1.

In venturimeter the losses are less. Hence, FRHႈFLHQWRIGLVFKDUJHLVKLJK

1.

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2.

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p - p Ês ˆ  z – z = x Á m - ˜ rg Ë s ¯

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 P 

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=

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 ¥  ¥   ¥  ¥ 

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 Area a = p ¥   P 

Area a =

p ¥    P 

Q = Cd

 =

 ¥  ¥ h =  

a ◊ a  gh a - a

 ¥  ¥   ¥  ¥ h   -    ¥  -  ¥ -  ¥  ¥ 

  h=

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618

Fundamentals of Fluid Mechanics

Case'L൵HUHQFHRIUHDGLQJRIWZRSUHVVXUHJDXJHVLV Ê p p ˆ h = Á  -  ˜  z – z Ë rg rg ¯ =

p - p  ¥  ¥ 

 ± 

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 ¥ 0.8 ¥¥

  

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or

x=

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Flow

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h=

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Fluid Dynamics-II

=

Q=

=

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=

=     

619

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 î±  PV

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\

p p = 0 &  =P r g rg

v v =  g g a =

p ¥ d p ¥   =  P  

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620

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  v

 v  a v = Q 

or   1RZ

a =

   

p d     d  =

 ¥  p

 

  d P   1RZPDQRPHWHUVKRZVGHÀHFWLRQRIFP Ê sin ˆ h=xÁ - ˜ Ë s ¯

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Fluid Dynamics-II

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Fundamentals of Fluid Mechanics

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or

 





P – Pt  

r  Vt – V 

r = 

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p Ds ¥ P – Pt 

p Ds r ¥ =  

ÈÊ D ˆ  ˘ ÍÁ ˜ - ˙ u ÍË Dt ¯ ˙ Î ˚

Now spring force = pressure force rx =

\

r u 8

r u x= 8k

ÈÊ D ˆ  ˘ ÍÁ ˜ - ˙ pDs ÍË Dt ¯ ˙ Î ˚ ÈÊ D ˆ  ˘ ÍÁ ˜ - ˙ pDs ÍË Dt ¯ ˙ Î ˚

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Fluid Dynamics-II

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or

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ÊS ˆ  g x Á m - ˜ Ë S ¯

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x PP   2SWLRQ D LVFRUUHFW

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Z = Z h +

V V = h +  g g

Fluid Dynamics-II

or

h = h – h =

625

V  - V V V –  =  g g g

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V =

 1RZ





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or   









or

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a =

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 1RZ

a =

p d  

or

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D 



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2 î οP)free = – (οP)forced ¥

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or

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=

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r r

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r r

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Chapter

11

LAMINAR FLOW

KEYWORDS AND TOPICS          

LAMINAR FLOW TURBULENT FLOW FLOW IN HORIZONTAL PIPES FLOW IN PARALLEL PLATES AVERAGE VELOCITY MAXIMUM VELOCITY HEAD LOSS FRICTION FACTOR FLOW IN INCLINED PIPES CORRECTION FACTOR FOR KINETIC ENERGY  CORRECTION FACTOR FOR MOMENTUM

         

FLOW IN POROUS MATERIAL DARCY’S EQUATION FLUIDIZATION JOURNAL BEARING FOOT STOP BEARING COLLAR BEARING DASHPOT FALLING SPHERE METHOD CAPILLARY METHOD ROTATING CYLINDER METHOD  FLOW IN OPEN CHANNEL

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Laminar Flow

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\   %XWZHKDYHIRXQGRXWt = –

r ∂p  ∂x r ∂p du =  m ∂x dr

\

Ú du =

  Rr

  RU

du du = –m dy dr



u=

∂p m∂ x

Ú

rdr

 ∂p r +C  m ∂x 

C FRQVWDQW

 :KHQr = Ru DWÀXLGLVVWDWLRQDU\DWERXQGDU\ \

C=–

\

u=

 ∂p ¥ R  m ∂x

 Ê - ∂p ˆ  r – R   m ÁË ∂x ˜¯

642

Fundamentals of Fluid Mechanics

KHUH

∂p mDQGRDUHFRQVWDQWV+HQFHYHORFLW\DWDQ\VHFWLRQLQSLSHYDULHVZLWKWKHVTXDUH ∂x

RIUDGLXV r :KHQr WKHYHORFLW\LVPD[LPXPDQGZKHQr = RWKHYHORFLW\LV]HUR 7KHYHORFLW\YDULHVSDUDEROLFIURP]HURDWWKHERXQGDU\VXUIDFHWRWKHPD[LPXPDWFHQWUH RIWKHSLSH7KHYHORFLW\GLVWULEXWLRQDWDQ\VHFWLRQLVDVVKRZQEHORZ r=R

r=0

Velocity Distribution

11.6 AVERAGE VELOCITY  ‡ 'HULYH DQ H[SUHVVLRQ IRU DYHUDJH YHORFLW\ IRU D IORZ WKURXJK SLSH )LQG WKH UDWLR RI PD[LPXPYHORFLW\WRDYHUDJHYHORFLW\ Or   )RUDVWHDG\ODPLQDUIORZWKURXJKDFLUFXODUSLSHSURYHWKDWYHORFLW\GLVWULEXWLRQDFURVV WKHVHFWLRQLVSDUDEROLFDQGWKHDYHUDJHYHORFLW\LVKDOIRIWKHPD[LPXPYHORFLW\

(UPTU 2006-07) r

r + dr

  &RQVLGHU WKH ÀRZ WKURXJK D FLUFXODU ÀXLG ULQJ HOHPHQW RI UDGLXV r DQG WKLFNQHVV dr7KH GLVFKDUJH WKURXJK WKLV ULQJ HOHPHQW LV HTXDO WR WKH YHORFLW\ RI WKH ÀRZ PXOWLSOLHG E\ WKH FURVVVHFWLRQDODUHDRIWKHULQJHOHPHQW dQ = u ¥ prdr   %XW

\

u=–

dQ = –

 ∂p  R – r  m ∂x  ∂p  R – r ¥prdr  m ∂x

Laminar Flow

 RU



Q=–

p ∂ p  m ∂x

643

R

Ú

R – r rdr



R

p ∂p È  r r ˘ =– ◊ - ˙ ÍR  m ∂x Î   ˚ =–

p ∂p  ◊ R 8m ∂x

  ,QFDVHDYHUDJHYHORFLW\ u WKHQ ∂p  ◊R ∂x Q = u = $UHDRISLSH 8m ¥ p ¥ R -p

=

 Ê - ∂p ˆ  R 8 m ÁË ∂x ˜¯

 7KHDERYHLVWKHH[SUHVVLRQIRUDYHUDJHYHORFLW\RIWKHÀRZWKURXJKDSLSH   7KHYHORFLW\RIÀRZDWDQ\FURVVVHFWLRQLVJLYHQE\ u=

 Ê - ∂p ˆ  R – r   m ËÁ ∂x ¯˜

 7KHPD[LPXPYHORFLW\RIÀRZLVDWWKHFHQWUHRIWKHSLSHLHr  uPD[ =

 Ê - ∂p ˆ  R  m ÁË ∂x ˜¯

 7KHUDWLRRIPD[LPXPYHORFLW\DQGDYHUDJHYHORFLW\RIWKHÀRZLV

uPD[ u

 

 Ê ∂p ˆ  R  m ÁË ∂x ˜¯ =  Ê ∂p ˆ  R 8 m ÁË ∂x ˜¯  

  +HQFHWKHPD[LPXPYHORFLW\RIWKHÀRZLVWZRWLPHVWKHDYHUDJHYHORFLW\RIWKHÀRZLQ DSLSH

644

Fundamentals of Fluid Mechanics

11.7 HAGEN POISEUILLE FORMULA  ‡ Derive an expression for drop of pressure for a given length of pipe when the fluid is flowing through a pipe. Or Derive Hagen Poiseuille formula. 1

Flow

2

P1

P2

x1

L x2

 

 &RQVLGHUDÀRZWKURXJKDSLSHWKHOHQJWKRISLSHLVLEHWZHHQVHFWLRQDQGVHFWLRQDV VKRZQLQWKH¿JXUH  7KHDYHUDJHYHORFLW\RIÀRZWKURXJKDSLSHLVJLYHQE\ u =

or



 Ê - ∂p ˆ  R 8 m ÁË ∂x ˜¯ ∂p 8u m = ∂x R

 1RZLQWHJUDWLQJIRUVHFWLRQWRVHFWLRQ –

Ú



∂p =



or

p – p = = p – p =

8u m R

Ú

x x

∂x

8u m (x – x) R D 8u m  ◊L EXW R =  R u m ◊L D

hf /RVVRISUHVVXUHKHDG =

p - p rg

Laminar Flow

hf =

\

645

u m p - p = ◊L r g D rg

 7KHDERYHHTXDWLRQLVNQRZQDV+DJHQ3RLVHXLOOHIRUPXOD  ‡ What is the power required to overcome the viscous resistance to make the fluid flow through length L and discharge Q?    3RZHUUHTXLUHG :HLJKWRIWKHÀXLGÀRZLQJSHUVHFRQG¥KHDGORVV = m ¥ g ¥ hf = Q◊r ¥ g ¥ =

u m L r g D

u m L Q D

11.8 FRICTION FACTOR  ‡ What is friction factor ( f)?   ,Q WKH SLSH ÀRZ WKH ORVV RI HQHUJ\ RU IULFWLRQ KHDG ORVV WDNHV SODFH EHWZHHQ DQ\ WZR VHFWLRQVRIWKHSLSH$UDWLRQDOH[SUHVVLRQIRUIULFWLRQORVVKHDGFDQGHULYHGE\GH¿QLQJD GLPHQVLRQOHVVFRH൶FLHQW f NQRZQDVIULFWLRQIDFWRU)ULFWLRQIDFWRULV f =

8t  rV 

 ZKHUH t VKHDUVWUHVVDWSLSHZDOODQGVLVDYHUDJHYHORFLW\

‡ What is Darcy’s equation?   'DUF\¶VHTXDWLRQLVIULFWLRQORVVLQSLSHÀRZ7KHIULFWLRQKHDGLVJLYHQLQWHUPVRIIULFWLRQ IDFWRU f DVXQGHU hf =

f LV   ZKHUH  gD

L OHQJWKV DYHUDJHYHORFLW\DQGD GLDPHWHU

f 'DUF\FRH൶FLHQWDQGf IULFWLRQDOIDFWRU f  ‡ 'HULYH DQ H[SUHVVLRQ IRU FRH൶FLHQW RI IULFWLRQ LQ YLVFRXV IORZ LQ WHUPV RI 5H\QROGV QXPEHU Head loss in viscous flows by Hagen Poiseuille formula is: hf =

u m L r g D

 /RVVRIKHDGGXHWRIULFWLRQE\'DUF\¶VHTXDWLRQ hf =

f LV   gD

646

Fundamentals of Fluid Mechanics

  ,QFDVHRISLSHÀRZV = u LHYHORFLW\LVDOZD\VDYHUDJHYHORFLW\LQSLSHÀRZ  m u L f Lu  = rg D  gD f=

RU

 m ru D

 ru D = m rvD m

5H\QROGV QXPEHU = Re =

  %XW

f=

\

 Re

11.9 FLOW IN PARALLEL PLATES  ‡ Derive an expression for local velocity of viscous flow between two stationary flat SODWHV6WDWHVWKHDVVXPSWLRQVPDGH 8378   7KHDVVXPSWLRQVPDGHDUH   )ORZLVWZRGLPHQVLRQDOVWHDG\ODPLQDUÀRZ   )OXLGLV1HZWRQLDQ   7KHUHLVQRVOLSRIÀXLGSDUWLFOHVDWWKHVROLGERXQGDU\ t+ x

∂t . dy ∂y P+

dy

t

P dx

x1

y

dy

∂P . dx ∂x

dx t

L x2 Viscous Flow through Two Stationary Parallel Plates

  &RQVLGHUWZRVWDWLRQDU\SDUDOOHOSODWHVNHSWDWDGLVWDQFHtDSDUWDQGDYLVFRXVÀRZLVWDNLQJ SODFH DV VKRZQ LQ WKH ¿JXUH 7DNH D VPDOO ÀXLG HOHPHQW RI WKLFNQHVV dy OHQJWK dx DQG ZLGWKXQLW\DWDGLVWDQFHyIURPWKHERWWRPSODWH,IZHGUDZDIUHHERG\GLDJUDPRIWKLV ∂p ◊dx DFWLQJ DORQJ x ÀXLG HOHPHQW WKHQ IRUFHV DFWLQJ DUH   SUHVVXUH IRUFHV p DQG p + ∂x GiUHFWLRQDWERWKHQGIDFHV  WKHVKHDULQJIRUFHVUHVLVWLQJÀRZDWORZHUDQGXSSHUVXUIDFH

Laminar Flow

647

Ê ∂t ˆ RIWKHHOHPHQW6KHDUVWUHVVLVtDWORZHUIDFHDQGVKHDUVWUHVVLV Á t + ˜ DWXSSHUIDFH ∂y ¯ Ë FRUSUDFWLFDOSXUSRVHLWLVDVVXPHGWKDWJUDYLWDWLRQDOIRUFHLV]HURIRUKRUL]RQWDOÀRZDQG ∂p ∂y

,QHTXLOLEULXPFRQGLWLRQQHWIRUFHDFWLQJRQWKHÀXLGHOHPHQWLV]HUR Ê ˆ ∂p ◊ dx˜ dy ¥ ±t dx ¥ p dy ¥ – Á p + ∂x Ë ¯ Ê ˆ ∂t ◊ d y˜ dx ¥  + Át + ∂y Ë ¯ ∂p ∂t = ∂x ∂y

 7KHDERYHHTXDWLRQVKRZVWKDWSUHVVXUHJUDGLHQWLQxGLUHFWLRQLVHTXDOWRWKHUDWHRIFKDQJH RIVKHDUVWUHVVLQyGLUHFWLRQ   2QLQWHJUDWLRQ dp dt = dy dx

Ú

Ú

t=

dp ◊y + c ZKHUH c FRQVWDQW dx

 1RZDVSHU1HZWRQ¶VODZRIYLVFRVLW\ZHKDYH t=m

 RU



du =

 RU



Ú du =

 RU



u=

du dp = y + c dy dx

c  dp y dy +  dy m m dx  dp m dx

Ú

ydy +

c m

Ú

dy

c  d p y ◊ + y + c m m dx 

 $VWKHUHLVVOLSDWVROLGERXQGDULHVWKHERXQGDU\FRQGLWLRQVDUH y u  KHQFHc 

648

Fundamentals of Fluid Mechanics

  DQG

Ê d pˆ y = tu  KHQFH c =± Á Ë dx ˜¯



u=

\

 dp  ◊ y – ty m dx

 6LQFHt◊y > yZHFDQZULWH u=

 m

Ê - d pˆ  ÁË dx ˜¯ t◊y – y

 7KH DERYH HTXDWLRQ LV D HTXDWLRQ RI SDUDEROD +HQFH ORFDO YHORFLW\ YDULHV LQ SDUDEROLF PDQQHUZKHQDYLVFRXVÀRZWDNHVSODFHEHWZHHQWZRSDUDOOHODQGVWDWLRQDU\SODWHV

t

Local Velocity Distribution

 ‡ Draw the shear stress distribution for the viscous flow through two parallel and VWDWLRQDU\SODWHV   7KHORFDOYHORFLW\ u=

 Ê - d pˆ ty – y m ÁË dx ˜¯

 2QGL൵HUHQWLDWLQJZUWy du  Ê - d pˆ = t±y  dy m ÁË dx ˜¯  %XW

\

t=m

t=

du dy

 Ê -d pˆ t±\  ÁË dx ˜¯

t  ZKHQ y =

t 

t PD[LPXP ZKHQ y RUy = t

Laminar Flow

649

  7KHGLVWULEXWLRQRIVKHDUVWUHVVLVDVVKRZQEHORZ

t

Shear Stress Distribution

 ‡ )LQGWKHH[SUHVVLRQIRUDYHUDJHYHORFLW\IRUWKHYLVFRXVIORZEHWZHHQWZRSDUDOOHODQG VWDWLRQDU\SODWHV)LQGWKHUDWLRRIPD[LPXPYHORFLW\RIWKHDYHUDJHYHORFLW\   7KHORFDOYHORFLW\DWGLVWDQFHyLVJLYHQE\ u=

 Ê - d pˆ t◊y – y  m ÁË dx ˜¯

 'LVFKDUJHWKURXJKDVPDOOHOHPHQWRIKHLJKWdy dQ = u ¥ dy ¥ =

 Ê - d pˆ t◊y – y dy  m ÁË dx ˜¯

 1RZWRWDOGLVFKDUJHQLVE\LQWHJUDWLQJ \

Q=

 Ê - d pˆ  m ÁË dx ˜¯

Ú

t



ty – y dy t

 Ê - d p ˆ È ty  y  ˘ - ˙ = Í  m ÁË dx ˜¯ Î   ˚ =

 Ê - d p ˆ È t t ˘ Í - ˙  m ÁË dx ˜¯ Î   ˚

=

 Ê - d pˆ  t  m ÁË dx ˜¯

  1RZDYHUDJHYHORFLW\ u  Ê -d pˆ  t  ÁË dx ˜¯ Q = u =  m t ¥ A

650

Fundamentals of Fluid Mechanics

u =

 Ê -d pˆ  t  m ÁË dx ˜¯

 7R¿QGPD[LPXPYHORFLW\ZKLFKLVDWWKHHTXDOGLVWDQFHIURPERWKWKHSODWHSXWy =

u=

uPD[ =

=

t 

 Ê -d pˆ t◊y – y  m ÁË dx ˜¯  Ê -d pˆ  m ÁË dx ˜¯

Ê t t ˆ Át ◊  -  ˜ Ë ¯

 Ê -d pˆ  t 8 m ÁË dx ˜¯

  1RZWKHUDWLRRIuPD[WR u LV uPD[ = u

=

 Ê -d pˆ  t 8 m ÁË dx ˜¯  Ê -d pˆ  t  m ÁË dx ˜¯  

‡ Derive an expression for drop of pressure head of a viscous flow through two parallel and stationary plates for a length of L

Flow

t

P2

P1

x1

L x2

Ê - d pˆ   7KHDYHUDJHYHORFLW\ u IRUDSUHVVXUHJUDGLHQW Á YLVFRVLW\RIÀXLGDQGGLVWDQFH Ë dx ˜¯ t EHWZHHQWKHSODWHVLVJLYHQE\

Laminar Flow

u =

 RU



 RU

±∂p =

±

651

 Ê - ∂p ˆ  ◊t  m ÁË ∂x ˜¯

u m ∂p = ∂x t

u m ◊ ∂x t

 1RZLQWHJUDWLQJWKHHTXDWLRQIURPxWRxZHJHW

Ú





- ∂p =

p – p  

     RU

p – p =



 u m t

Ú

x x

∂x

u m x – x t  m u ◊ L t

 7KHKHDGORVVGXULQJÀRZLV  m u L p - p = hf = rg rgt  Note: 7KH KHDG ORVV LQFUHDVHV ZLWK YHORFLW\ DQG OHQJWK RI ÀRZ EXW GHFUHDVHV ZLWK LQFUHDVLQJGLVWDQFHEHWZHHQWKHSODWHV  ‡ Derive an expression of local velocity for viscous flow between parallel plates when one SODWHLV¿[HGDQGWKHRWKHULVPRYLQJ:KDWLVWKHQDPHJLYHQWRWKLVNLQGRIIORZ" Moving plate

t+

dx dy

t

P+

P

y

∂t . dy ∂y ∂P . dx ∂x

t

  7KLVW\SHRIÀRZLVFDOOHG&RXHWWHÀRZ,QFDVHZHFRQVLGHUWKHHTXLOLEULXPRIWKHÀXLG HOHPHQWZHJHW Ê ∂p ˆ p◊dy◊± Á p + dx dy◊ – t ◊dx◊ ∂x ˜¯ Ë ∂t ∂p = ∂y ∂x

Ê ∂t ˆ ÁË t + ∂ y d y˜¯ dx◊ 

652

Fundamentals of Fluid Mechanics

 RU



 RU



∂p ◊ dy ∂x

Ú ∂t = Ú

Ê ∂p ˆ t = Á ˜ y + c Ë ∂x ¯ t=m

 %XW

du =

\

du dy

 Ê ∂p ˆ y dy + c ∂y m ÁË ∂x ˜¯

  ,QWHJUDWLQJWKHHTXDWLRQZHJHW

Ú ∂u = u=

 Ê ∂p ˆ m ÁË ∂x ˜¯

Ú y dy + Ú dy

 Ê ∂ p ˆ y + c  y + c  m ÁË ∂x ˜¯ 

L

 $SSO\LQJWKHERXQGDU\FRQGLWLRQVWR¿QGcDQGc y u  KHQFHc  y = tu = v KHQFHc =

 Ê ∂p ˆ t v – Á ˜ m Ë ∂x ¯  t

 3XWWLQJWKHYDOXHRIC & CLQHTQ L u=

=

 Ê ∂ p ˆ y Ê v  ∂p t ˆ + Á y ◊ Á ˜ m Ë ∂x ¯  Ë t m ∂x  ˜¯ v  Ê - ∂p ˆ [ty – y] + ◊y Á ˜ t  m Ë ∂x ¯

 7KHHTXDWLRQVKRZVWKDWORFDOYHORFLW\GHSHQGVRQ

∂p ∂p DQGYHORFLW\ v RIWKHSODWH ∂x ∂x

FaQEHQHJDWLYHRUSRVLWLYH,QFDVHLWLV]HURWKHQ u=

v ◊y t

  7KH HTXDWLRQ VKRZV WKDW YHORFLW\ u FKDQJHV WR ]HUR IURP VWDWLRQDU\ SODWH WR PD[LPXP YHORFLW\vOLQHDUO\ZKLFKLVWKHYHORFLW\RIWKHPRYLQJSODWH,QFDVH YHORFLW\ZLOOYDU\SDUDEROLFDOO\

∂p πWKHQWKHORFDO ∂x

Laminar Flow

 ‡ )LQGWKHVWUHVVGLVWULEXWLRQLQ&RXHWWHIORZ   ,Q&RXHWWHÀRZWKHORFDOYHORFLW\ v LVJLYHQE\ u=

 Ê - ∂p ˆ v ty – y  ◊y  m ÁË ∂x ˜¯ t

V

V V

V

31

31

∂P π0 ∂x

13

Velocity distribution

13

Velocity distribution

∂P =0 ∂x

  'L൵HUHQWLDWLQJWKHHTXDWLRQZHJHW  Ê - ∂p ˆ ∂u v = t±y  Á ˜  m Ë ∂x ¯ ∂y t t=m

 %XWVKHDUVWUHVV

 RU

t=



 

∂u ∂y Ê ∂p ˆ mv ÁË - ∂x ˜¯ t±y  t

 6KHDUVWUHVVDWVWDWLRQDU\SODWHLVt FDQEHIRXQGRXWE\SXWWLQJy  t =

Ê - ∂p ˆ t mv + Á ◊ t Ë ∂x ˜¯ 

 6KHDUVWUHVVDWPRYLQJSODWHFDQEHIRXQGRXWE\SXWWLQJy = t tm =

=   6KHDUVWUHVVDWy =

Ê - ∂p ˆ Ê t mv ˆ + Á ÁË - t ˜¯ ˜ t Ë ∂x ¯  Ê ∂p ˆ t mv + Á ˜◊ Ë ∂x ¯  t

t LHFHQWUHRIWZRSODWHV  mv tc = t

653

654

Fundamentals of Fluid Mechanics

 $SRLQWZKHUHVKHDUVWUHVVLV]HURLV O=m

 RU



y=



t Ê - ∂p ˆ Ê t ˆ + Á - y˜ Á ˜  ¯ Ë ∂x ¯ Ë 

 t v –m ¥ Ê  - ∂p ˆ t ÁË ∂x ˜¯

 7KHVKHDUVWUHVVGLVWULEXWLRQLVDVVKRZQEHORZ V

t y = t/2

+ mv ×

1

F I H K ∂P

13

t 2

31

y =

Shear Stress Distribution

∂x

 ‡ $ IOXLG RI YLVFRVLW\  1VP2 DQG 6*  LV IORZLQJ WKURXJK D FLUFXODU SLSH RI GLDPHWHU  PP7KH PD[LPXP VKHDU VWUHVV DW WKH SLSH ZDOO LV JLYHQ DV  1P2 )LQG L SUHVVXUHJUDGLHQW LL DYHUDJHYHORFLW\DQG LLL 5H\QROGVQXPEHURIWKHIORZ (UPTU 2005-6) Ê - ∂p ˆ r Guidance: )RUÀRZWKURXJKSLSHVKHDUVWUHVVt = Á +HQFHIRUPD[LPXPVKHDU Ë ∂x ˜¯  VWUHVVDWSLSHZDOOLVZKHQr = R$YHUDJHYHORFLW\iV u =

 Ê - ∂p ˆ ¥ R 8 m ÁË ∂x ˜¯

Ê - ∂p ˆ R t PD[ = Á Ë ∂x ˜¯   

\

Ê ˆ    - ∂ p ¥ ÁË ∂x ˜¯   ¥  - ∂p = = 8 ¥1P  ∂x

Laminar Flow

655

 $YHUDJHYHORFLW\ u  u =

=

 Ê - ∂p ˆ  R 8 m ÁË ∂x ˜¯  ¥ 8 ¥ ¥    ¥ 

=PV  

5H\QROGVQXPEHU =

=

u ◊r◊ D m  ¥  ¥  ¥  

     ‡ Show that for viscous flow through a circular pipe mean velocity of flow occurs at a UDGLDOGLVWDQFHRIR from the centre of pipe where R is the radius of pipe   7KHORFDOYHORFLW\ u=

 Ê - ∂p ˆ  R – r    m ÁË ∂x ˜¯

L

 DQGDYHUDJHYHORFLW\ u =

 Ê - ∂p ˆ  R  8 m ÁË ∂x ˜¯

LL

 (TXDWLQJHTQV L DQG LL  Ê - ∂p ˆ   Ê - ∂p ˆ  R – r   R Á ˜  m Ë ∂x ¯ 8 m ÁË ∂x ˜¯  RU  RU   RU



R±r = R  R r 

r=

 

R

   R  ‡ $ OXEULFDWLQJ RLO RI YLVFRVLW\  1VP2 DQG 6*  LV SXPSHG WKURXJK D  PP GLDPHWHUSLSH,IWKHSUHVVXUHGURSSHUPHWUHRISLSHLVN1P2¿QGWKHIORZ   

$YHUDJHYHORFLW\ u =

 Ê - ∂p ˆ  p 8 m ÁË ∂x ˜¯

656

Fundamentals of Fluid Mechanics

 ¥ ¥    ¥ 

u =

 RU

u  PV  

'LVFKDUJHQ = u ¥ A  ¥ p ¥  

 PV

 

 ‡ 2LOKDYLQJ6*LVSXPSHGWKURXJKDPPGLDPHWHUSLSH7KHGLVFKDUJHLV mPLQDQGSUHVVXUHGURSLVN3DIRUPOHQJWKRISLSH)LQGYLVFRVLW\ 

$YHUDJHYHORFLW\ u =

 RU

u =



Q A 0.9 ¥ 4 6 ¥ p ¥ (0.20) 2

   PV   7KHSUHVVXUHGURSLQDSLSHIRUOHQJWKL p – p =

 

¥ =

 RU



 u m L D  m ¥  ¥   

m=

 ¥  ¥   ¥  ¥ 

 1VP

 

 ‡ $ SUHVVXUH GURS RI  1P2 WDNHV SODFH LQ D FLUFXODU SLSH RI OHQJWK  P 7KH 5H\QROGVQXPEHURIWKLVIORZLV)LQGGURSLQSUHVVXUHLQFDVHIORZLVPDGHGRXEOH ZLWKRXWFKDQJLQJOLTXLGSURSHUWLHV   6LQFH5H\QROGVQXPEHUWKHÀRZLVODPLQDU p – p =

 u m L   D

 :KHQÀRZLVPDGHGRXEOHLH Q¢ Q

Laminar Flow

657

A u ¢ ◊A u

ZHKDYH  RU



u ¢  u p¢ – p¢ =

\

 u ¢ ¥ m ¥ L D

=¥  u ¥ m ¥ L D     

 ¥  N1P

 ‡ A viscous flow of oil is taking place in a pipe of 12 cm diameter with discharge of ¥ 10±  mV:KDWSRZHUSHUNLORPHWUHLVUHTXLUHGWRPDLQWDLQWKHIORZLI6*RI WKHIOXLGLVDQGYLVFRVLW\LV1VP2? u =  

Q  ¥ -  ¥  = A p ¥  

 PV

  3UHVVXUHGURS  u m L p - p = hf = r gD rg  RU    



hf =

 ¥  ¥  ¥   ¥  ¥  ¥  

 PRIRLO 3RZHU rg◊Qhf

   ¥ ¥¥¥±  ¥    ZDWWV  ‡ $PPGLDPHWHUSLSHNPORQJLVXVHGIRUWUDQVSRUWLQJRLOIURPDWDQNHULQVHD WRVKRUHDWIORZRIPV)LQGWKHSRZHUUHTXLUHGWRPDLQWDLQWKHIORZ$VVXPHm 1PV 6* IRUWKHRLO   

)ORZ  u ¥$UHD  Q = u ¥ pd 

u =

 ¥   PV p ¥ 

658

Fundamentals of Fluid Mechanics

 1RZWRFKHFN5H\QROGVQXPEHU ru d m

Re = =  

 ¥  ¥  ¥  

 

  6LQFHReKHQFHÀRZLVODPLQDU p – p =

  1RZ

=     

 u m L D  ¥  ¥  ¥  ¥   

 ¥1P 3RZHUORVV  p – p ¥ Q

  

 ¥¥

  

 ZDWWV

  

 N:

 ‡ $QRLOKDVYLVFRVLW\RI1PVDQG6*RIIORZVWKURXJKDKRUL]RQWDOSLSHRI FPGLDPHWHUZLWKDSUHVVXUHGURSRIN1P2SHUPHWUHOHQJWKRISLSH)LQG  UDWH RIIORZ  VKHDUVWUHVVDWSLSHZDOO  GUDJIRUPOHQJWKRISLSHDQG  SRZHU WRPDLQWDLQIORZLQPSLSHOHQJWK -dp -dp = = 6 ¥1P dx dL  1RZ



Q=

=     1RZVKHDUVWUHVV

 

p Ê - ∂p ˆ  R 8 m ÁË ∂x ˜¯ p ¥ ¥    ¥ 

 PV t=–

∂p r ◊ ∂x 

DWZDOOt PD[ = –

∂p R ◊ ∂x 

Laminar Flow

= 6 ¥ ¥

659

 

 1P

 

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    ¥ p ¥¥     N1   3RZHU UHTXLUHG WR PDLQWDLQ WKH ÀRZ FDQ EH IRXQG RXW E\ PXOWLSO\LQJ GLVFKDUJH WR WKH SUHVVXUHGURS 3RZHU Q ¥

  

∂P ◊L ∂x

 ¥ 6 ¥ ¥   ¥ZDWWV  N:

     

11.10 KINETIC ENERGY CORRECTION FACTOR  ‡ The velocity distribution for viscous flow in a circular pipe of radius R is given by: u umax

± ÊÁ r ˆ˜ Ë R¯

2

Where u is the local velocity at radius r and umaxLVWKHPD[LPXPYHORFLW\DWWKHFHQWUH )LQGWKHYDOXHVRIFRUUHFWLRQIDFWRUVIRUNLQHWLFHQHUJ\DQGPRPHQWXPLQFDVHDYHUDJH YHORFLW\LVXVHGLQWKHLUFDOFXODWLRQ Or 3URYHWKDWIRUYLVFRXVIORZWKURXJKDFLUFXODUSLSHWKHNLQHWLFHQHUJ\FRUUHFWLRQIDFWRU LVHTXDOWR 8378 Guidance: :HKDYHWRDSSO\FRUUHFWLRQIDFWRUVLQFDVHNLQHWLFHQHUJ\DQGPRPHQWXPDUH FDOFXODWHGRQWKHEDVLVRIDYHUDJHYHORFLW\ Local velocity (u)

r r + dr

R

Viscous Flow through Pipe

660

Fundamentals of Fluid Mechanics

  &RQVLGHUDQHOHPHQWDU\DUHDDWUDGLXVrDQGWKLFNQHVVdr7KHQDUHDRIWKHFLUFXODUVWULS dA prdr A=

Ú prdr

 7KHDYHUDJHYHORFLW\ u 

  %XW

u =

Q A

Q=

Ú

udA =

Ú

Ê r ˆ uPD[ Á -  ˜ p ◊ r ◊ dr R ¯ Ë

u =

Ú

u ◊prdr =

Ú

Ê r ˆ uPD[ Á -  ˜ ¥pr ¥ dr R ¯ Ë

Ú p rdr uPD[

=

Ú

Ê r ˆ r dr Á R  ˜¯ Ë

R



Ú

R



rdr R

uPD[ =

=

=

È r r ˘ Í ˙ Î   R ˚ R

Ê r ˆ Á ˜ Ë ¯ uPD[ ◊

 > R  - R  @  R 

uPD[ 

 .LQHWLFHQHUJ\FRUUHFWLRQIDFWRU a=

$FWXDONLQHWLFHQHUJ\ .LQHWLFHQHUJ\FDOFXODWHGE\DYHUDJHYHORFLW\

 a= A

Ú

Ê uˆ ÁË ˜¯ u



dA

Laminar Flow

 = p R  %XW



\

Ú

u ¥prdr u

R



u =

uPD[ 

a=

 p R

 =  R  R

=

661

Ú

R



Ú

R



Ú

R



◊ u prdr  uPD[

Ê r ˆ Á - R  ˜ Ë ¯



r ◊dr

Ê r r r ˆ Á - R  + R  - R  ˜ r ◊dr Ë ¯ R

 È r   r  r  r ˘ + =  Í ˙ R Î   R   R   R ˚ 

 

=

 >R±RR±R] R

=

  ◊ R >@ R

 

  7KHPRPHQWXPFRUUHFWLRQIDFWRU b=

=

 A

Ú

 p R

8 = R   

 u dA =  p R u

Ú

R



Ú

R



◊

Ú

R



u  ¥ p r dr u

u ◊ prdr  uPD[ 

Ê r ˆ ◊ rdr  Á R  ˜¯ Ë

 

 +HQFHDFWXDONLQHWLFHQHUJ\DQGPRPHQWXPFDQEHIRXQGRXWE\¿QGLQJWKHVHZLWKDYHUDJH YHORFLW\ DQG WKHQ PXOWLSO\LQJ WKH FDOFXODWHG YDOXHV ZLWK FRUUHFWLRQ IDFWRU RI  DQG  UHVSHFWLYHO\

662

Fundamentals of Fluid Mechanics

 ‡ 2LORIDEVROXWHYLVFRVLW\SRLVHDQGGHQVLW\NJPIORZVWKURXJKDFPSLSH ,IWKHKHDGORVVLQOHQJWKSLSHLVPDVVXPLQJDODPLQDUIORZGHWHUPLQH L WKH YHORFLW\ LL 5H\QROGVQXPEHUDQG LLL IULFWLRQIDFWRU )DQQLQJ¶V  $0,(  p – p = hf =

 u m L p - p =   D rL rg

u =

 ¥ D  r L  m L

\

=     

 u m L D

 ¥   ¥  ¥   ¥  ¥ 

 PV 5H\QROGVQXPEHr =

ru D m

=  ¥  ¥       )ULFWLRQIDFWRU

 

f= =

 5H\QROGVQXPEHU   ¥ 

‡ Crude oil of m SRLVHDQGUHODWLYHGHQVLW\ IORZVWKURXJKDPPGLDPHWHU YHUWLFDOSLSH7KHSUHVVXUHJDXJHV¿[HGPDSDUWUHDGN1P2DQGN1P2 as VKRZQLQ¿JXUH)LQGGLUHFWLRQDQGUDWHRIIORZWKURXJKWKHSLSH 2

20

P2 = 200 kN/m

2

P1 = 600 kN/m

20 f

Laminar Flow

663

Guidance: 7KHSUHVVXUHGURSLQFOXGHVWKHGL൵HUHQFHRIKHLJKWRIWKHJDXJHV+HQFHDGG hrgDWULJKWVLGHRIWKHSUHVVXUHGURSHTXDWLRQ 32 u m L + hrg D2

p1 – p2 =

32 ¥ u ¥¥ 20 + 20 ¥ 900 ¥  2

(600 – 200) ¥ 103 =

> - @ ¥ 3 ¥  ¥ - 4  ¥  ¥ 

u =

\     1RZ5H\QROGVQXPEHU

 PV

Re =

ru D  ¥  ¥  = m 

    Since ReÀRZLVODPLQDU   1RZ  ÀRZUDWHQ = u ¥ A =

 ¥ p ¥  2 4

 ¥ 104P3V

 

11.11 FLOW IN INCLINED PIPE  ‡ A liquid of viscosity 0.07 poise and RD of 0.86 flows through an inclined pipe of 20 m diameter. A discharge of 0.013 m3/min is to be maintained through the pipe in such a way that pressure along the length is constant. Find the required inclination of the pipe. Flo

L L sin q

Z2 q

Z1

w

664

Fundamentals of Fluid Mechanics

Guidance: 7KHÀRZKDVWRWDNHIURPKLJKHUOHYHOWRORZHUOHYHOVRWKDWWKHSUHVVXUHGURS KHDGGXHWRYLVFRVLW\LVHTXDOWRVWDWLFKHDGGXHWRGL൵HUHQFHRIOHYHOVIRUOHQJWKL,QRWKHU ZRUGVhf = (z1 – z2) = LVLQq +HDGORVV hf =

  

u = =

 

32 u m L D2 ¥ r ¥ g Q A 0.013 p 60 ¥ ¥ (0.02)2 4

 PV hf =

\

32 ¥ 0.69 ¥ 0.07 ¥ L (0.02) 2 ¥ .86 ¥ 103 ¥ 9.81

 6LQFHSUHVVXUHLVFRQVWDQWKHQFH hf = LVLQq   RU



LVLQq =

32 ¥ 0.69 ¥ 0.07 ¥ L 4 ¥ 10- 4 ¥ 0.86 ¥ 103 ¥ 9.81

 RU  VLQq = 0.45   RU  q = 27.2  ‡ A tank has 4 cm diameter and 100 cm long pipe attached at its bottom. The tank has oil of SG = 0.9 and viscosity = 0.15 Ns/m2. Find flow through pipe when the height of oil in tank is 0.6 m above the pipe.

1 0.6 m

1m

Dia = 4 cm 2

  $SSO\LQJ%HUQRXOOL¶VHTXDWLRQEHWZHHQVHFWLRQVDQG p1 p v2 v2 + 1 + z1 = 2 + 2 + z2 + hL rg rg 2g 2g

Laminar Flow

665

p p =   v ªz – z P rg rg V= u  DYHUDJHYHORFLW\DWRXWOHWRISLSH

 DQG

  %XWKHDGORVVLV

u g

±hL

hL =

u m ◊ L D  ◊ rg

L

 ◊ u ¥  ◊    ¥  ¥  ¥ 

=

  u

 

 3XWWLQJWKHYDOXHRIhLLQHTQ L u  ± u g  RU



u   u ±  u = u =

    1RZ5H\QROGVQXPEHU

-  ±  +   -  ±  

 PV

Re = =

r uD m  ¥  ¥  

      6LQFHReWKHÀRZLVODPLQDU   1RZ

Q= u ¥A =

 

 ¥ p ¥   

 ¥ mV

666

Fundamentals of Fluid Mechanics

 ‡ A laminar flow is taking place between two parallel and stationary plates 100 mm DSDUW0D[LPXPYHORFLW\RIWKHIORZLVPV¿QG  IORZSHUPHWUHZLGWK  VKHDU VWUHVVDWWKHSODWHV  WKHSUHVVXUHGURSIRUIORZRIP  YHORFLW\JUDGLHQWEHWZHHQ WKHSODWHVDWDVHFWLRQDQG  ORFDOYHORFLW\DWPPIURPWKHSODWH7DNHYLVFRVLW\ 1VP2

t = 100 mm

Guidance: 7KHORFDOYHORFLW\RIWKHÀRZLVJLYHQE\u =

YHORFLW\ u =

 Ê - ∂p ˆ ty – y DYHUDJH m ÁË ∂ x ˜¯

u mL  Ê - ∂p ˆ   t DQGuPD[  u 3UHVVXUHGURS p – p = DQG m ÁË ∂ x ˜¯ t 

Ê - ∂p ˆ t±y t= Á Ë ∂ x ˜¯ uPD[ PV \

u =

  PV 

Q = u ◊A

 1RZ  

 ¥ ¥ = PV

 1RZ

\

u =

 Ê - ∂p ˆ  ◊t m ÁË ∂ x ˜¯

 ¥  ¥ m - ∂p = ∂x   =

 ¥   1P  ¥ -

 1RZWR¿QG p – p EHWZHHQWZRSRLQWVDWPDSDUWZHNQRZ p – p =  

u mL  ¥  ¥  ¥  =  t  

 N1P

Laminar Flow

667

  6KHDUVWUHVVLV  Ê - ∂p ˆ t±y  ÁË ∂ x ˜¯

t=

 7KHVKHDUVWUHVVDWSODWHVLVVDPHZKLFKFDQEHIRXQGRXWE\SXWWLQJy RUy = t  Ê - ∂p ˆ t  ÁË ∂ x ˜¯

t ZDOO =

 ¥¥ 

=

 1P

 

  1RZWR¿QGRXWYHORFLW\JUDGLHQWEHWZHHQWKHSODWHVZHXVH1HZWRQ¶VHTXDWLRQ t=m

 RU



YHORFLW\JUDGLHQW 

du dy

du t = dy m

du  =  V± dy 

\  7R¿QGYHORFLW\DWy PP u = =  

 Ê - ∂p ˆ ty – y m ÁË ∂ x ˜¯   >¥±  ]  ¥ 

 PV

 ‡ 7ZR SDUDOOHO SODWHV DUH  PP DSDUW DQG D VWHDG\ YLVFRXV IORZ RI RLO LV WDNLQJ SODFH EHWZHHQWKHP,ISUHVVXUHGURSLV.1P2 per metre length of plates and m = 5 ¥ 10–2 1VP2 IRU RLO ¿QG   IORZ SHU PHWHU ZLGWK   PD[LPXP VKHDU DQG   PD[LPXP YHORFLW\RIIORZ P1

t = 4 mm

L

P2

668

Fundamentals of Fluid Mechanics

  'URSRISUHVVXUHRIWKHÀRZLQOHQJWKLRIWKHSODWHV p – p =

  

¥  =

u =

\

u mL  +HUHL P t  ¥ u ¥  ¥ - ¥     ¥  ¥  ¥ -  ¥  ¥ -

 ¥±  PV

      'LVFKDUJHSHUPHWUHZLGWK         1RZPD[LPXPVKHDUVWUHVV

Q = u ¥ A = u ¥ t ¥  ¥¥  ¥± mV

Ê - ∂p ˆ t ¥ tPD[ = Á ˜ Ë ∂x ¯  Ê -  ¥  ˆ  = Á ¥ ˜   Ë ¯  

 1P

  1RZPD[LPXPYHORFLW\      

uPD[  u  ¥  PV

 ‡ $PDVRQU\ZDOORIDZDWHUWDQNLVPWKLFN,WKDVGHYHORSHGDFUDFNRIPPDQG PZLGHDQGWKLVFUDFNH[WHQGVWRHQWLUHWKLFNQHVVRIWKHZDOO7KHKHLJKWRIZDWHULV PDERYHWKHFUDFN)LQGOHDNDJHYROXPHSHUGD\LIm 1VP2

1 4m

Leakage Through Tank

Laminar Flow

669

Guidance: 7KH SUREOHP LV QRWKLQJ EXW WKH ÀRZ WKURXJK WZR SDUDOOHO VWDWLRQDU\ SODWHV IRUPHGE\WKHFUDFN+HUHt PPL PDQGZLGWK P hf = 

p - p u mL =  rg t ◊ rg

hf  

u mL t  rg

u =

\

=

 ¥  ¥ -  ¥  ¥  ¥   ¥  ¥ - ¥   ¥  ¥ - ¥  ¥   ¥  -

 ¥±PV Q ÀRZ A u   ¥ ¥± ¥   ¥± mV  /HDNDJHSHUGD\ Q ¥¥   ¥± ¥¥   P 7ZRKRUL]RQWDOSODWHVPDSDUWFRQWDLQRLOZLWKYLVFRVLW\RI1VP2,IWKHXSSHU SODWHLVPRYHGZLWKPVDQGWKHSUHVVXUHGL൵HUHQFHRIWZRVHFWLRQVPDSDUWLV  N1P2 )LQG L  YHORFLW\ GLVWULEXWLRQ LL  GLVFKDUJH SHU XQLW ZLGWK DQG LLL  VKHDU VWUHVVRQWKHXSSHUSODWH Guidance: 7KH YHORFLW\ GLVWULEXWLRQ RI WKH ÀXLG LQ FDVH RQH SODWH LV PRYLQJ LV JLYHQ DV

 

    

     ‡

u=

 Ê - ∂p ˆ ◊y + Á m Ë ∂ x ˜¯

Ú

t



ty – y 'LVFKDUJHFDQEHIRXQGE\LQWHJUDWLQJDV Q =

VWUHVVE\GL൵HUHQWLDWLQJYHORFLW\DQGDSSO\LQJt = m 1

Ú

t



∂u ∂y

2 Moving plate with v = 1.2 m/s

L = 12 mm

Stationary L = 100 m

udyVKHDU

670

Fundamentals of Fluid Mechanics

  +HUH

dx P dP ¥ ¥1P



- ∂p  ¥  =  1P ∂x 

\

  )LQGYHORFLW\GLVWULEXWLRQ u=

=

  y+    y – y  ¥  

 y y – y  yyy  y± y

      

V  Ê - ∂p ˆ ◊y + ty – y t m ÁË ∂ x ˜¯

  'LVFKDUJHSHUXQLWZLGWKRIWKHSODWHV Q=



Ú

t

u ◊dy



 )URPHTQ L Q=

Ú

t



=

Ú

y±y dy





y±y dy 

È y y ˘ = Í -  ˙   ˚ Î  ¥   –

 

 ±  ¥± mV

   

   

 

  6KHDUVWUHVVDWXSSHUSODWH t=m

∂ ∂u =m  y±y ∂y ∂y

 8VLQJHTQ L RIYHORFLW\ t = m ±y   

8SSHUSODWHPHDQVy P

L

Laminar Flow

\         

t XSSHU   

671

 ± ¥¥±  ± ¥ 1P

11.12 FLOW IN INCLINED PLATES  ‡ Laminar flow of a fluid (m 1P2DQG6*  LVPDLQWDLQHGLQSDUDOOHOSODWHV RIH[WHQVLYHZLGWK7KHSODWHVDUHLQFOLQHGDWƒWRWKHKRUL]RQWDODQGDUHNHSWPP DSDUW8SSHUSODWHLVPRYHGZLWKPVLQGLUHFWLRQRSSRVLWHWRWKHIORZ7ZRSUHVVXUH JDXJHV¿WWHGRQXSSHUSODWHDWPYHUWLFDOO\DSDUWUHDGVSUHVVXUHDVN1P2 and N1P2UHVSHFWLYHO\)LQG  WKHYHORFLW\DQGVKHDUVWUHVVGLVWULEXWLRQEHWZHHQWKH SODWHV  PD[LPXPIORZYHORFLW\DQG  VKHDUVWUHVVRQWKHXSSHUSODWH Moving plate 1 250 kN/m2 2

z=

10

/s

.5 m

1 V=

mm

L 1m

30 kN/m2

45°

z2

z1

  /HQJWKRIWZRVHFWLRQ L L   VLQ 

\  RU

L=



 m

 $SSO\LQJ%HUQRXOOL¶VHTXDWLRQEHWZHHQVHFWLRQVDQG P P V V +  + z =  +  + z + hL rg rg g g  6LQFHSLSHKDVFRQVWDQWGLDPHWHUKHQFHV = V \

HL = =

P - P + z – z rg  -  ¥    ¥  ¥ 

672

Fundamentals of Fluid Mechanics

= 14.44 + 1 = 15.44 m H ¥ r¥ g - ∂P = L 2 ∂x

Now,

as dx = L

15.44 ¥ 1.2 ¥ 103 ¥ 9.81 - ∂P = ∂x 2 = 128.5 ¥ 103 N/m3 The velocity distribution, u=

=

V 1 Ê - ∂P ˆ 2 y+ Á ˜ (ty – y ) t 2 m Ë ∂x ¯ - 1.5 1 y+ (128.5 ¥ 103)(0.01 y – y2) 0.01 2 ¥ 0.8

= –150 y + 80.3 ¥ 103(0.01y – y2) The condition for maximuPYHORFLW\RIWKHÀRZLVZKHQ

∂u = 0, hence ∂y

∂u = –150 + 80.3 ¥ 103(0.01 – 2y) = 0 ∂y or

0.01 – 2y =

150 = 1.86 ¥ 10–3 80.3 ¥ 103

= 0.00186 2y = 0.01 – 0.00186 = 8.14 ¥ 10–3 or y = 4.07 ¥ 10–3 m Put y = 4.07 ¥ 10–3 in velocity equation to get maximum velocity. umax = – 150 ¥ 4.07 ¥ 10–3 + 80.3 ¥ 103 [0.01 ¥ 4.07 ¥ 10–3 – (4.07 ¥ 10–3)2] = – 0.610 + 3.268 – 1.330 = 1.327 m/s Shear stress distribution is: or

Ê - ∂P ˆ Ê t ˆ t = mV + Á ˜¯ Á - y˜ Ë Ë ¯ ∂ 2 x t = 0.8 ¥

Ê 0.01 1.5 + (128.5 ¥ 103) Á Ë 2 0.01

ˆ y˜ ¯

Laminar Flow

673

= –120 + 642.5 – 128.5 ¥ 103 ¥ y = 522.5 – 128.5 ¥ 103 ¥ y Shear stress at upper plate can be found out by putting y = 0.01 t upper = 522.5 – 128.5 ¥ 103 ¥ 0.01 = 522.5 – 128.5 = –762.2 N/m2

11.13 DARCY’S EQUATION  ‡ Derive the expression for Darcy’s equation for laminar flow through a porous material. Darcy proved by experiments that the velocity of a fluid through a porous material varies linearly with the loss of head (hf ) and the flow is therefore laminar/viscous. Since the loss of head through a pipe is given by hf = by hf =

32 u mL and through parallel plates rg D2

12 u mL , hence a general expression for laminar flow is: D2 ◊ rg k ¢u m L rg D2

hf =

where

k ¢ = Constant

Now porous material has a number of small pores say n of diameter d p hence, we can write: D = n ◊dp Hence, we say that the head loss,

or

Now, \

hf =

k ¢u m L r g ◊ n 2 d P2

u =

r g n 2 d P2 Ê h f ˆ Á ˜ k ¢m Ë L ¯

hf L

= hydraulic gradient = i

u = ki

k LV FDOOHG FRH൶FLHQW RI SHUPHDELOLW\ 7KH DERYH HTXDWLRQ LV FDOOHG 'DUF\¶V HTXDWLRQ IRU ÀRZRIZDWHUWKURXJKVRLO7KHHTXDWLRQLVDSSOLFDEOHLIR e < 1. 41. Water at a rate of 8 ¥ 10–7 m3/s is flowing through a soil 10 cm height and 50 cm2 crossVHFWLRQDODUHDXQGHUDFRQVWDQWKHDGRIFP)LQGWKHFRH൶FLHQWRISHUPHDELOLW\   $YHUDJHYHORFLW\RIÀRZRIZDWHUWKURXJKWKHVRLO 8 ¥ 10-7 Q = u = 50 ¥ (10-2 ) 2 A

674

Fundamentals of Fluid Mechanics

 ¥ - =  ¥±PV  ¥ -   $VSHU'DUF\¶VHTXDWLRQ Êh ˆ u = ki = k Á f ˜ Ë L¯ ¥± = k ¥

 

 ¥ -  ¥ -

 RU  k ¥± PV  ‡ :KDWLVIOXLGL]DWLRQ":KHUHLVLWXVHG"   :KHQDOLTXLGLVIRUFHGXSZDUGVDWLQFUHDVLQJYHORFLW\WKURXJKDFROXPQRIJUDQXODUPDWHULDO RIVL]HdDVWDWHLVUHDFKHGZKHQDOOSDUWLFOHVDUHHQWUDLQHGRUFDUULHGLQWKHÀRZLHWKH EHGLVLQÀXLGVWDWH7KLVVWDWHLVFDOOHGÀXLGL]DWLRQ7KHFRUUHVSRQGLQJYHORFLW\LVFDOOHG ÀXLGL]DWLRQYHORFLW\ZKLFKLVJLYHQE\ Vf =

rg - r g d  c   m

 ZKHUH c SRURVLW\RIWKHPDWHULDOrg & rDUHGHQVLWLHV RIJUDQXOHVDQGZDWHU

  )OXLGL]DWLRQ LV XVHG LQ WKH EDFN ZDVKLQJ RI JUDYLW\ ¿OWHU IRU ZDWHU SXUL¿FDWLRQ %\ EDFN ZDVKLQJLPSXULWLHVLQWKH¿OWHUDUHZDVKHGRXWUHPRYHG

11.14 OIL BEARINGS  ‡ :KDW LV D MRXUQDO EHDULQJ" )LQG WKH YLVFRXV UHVLVWDQFH DQG SRZHU WR RYHUFRPH WKH UHVLVWDQFH Bearing Shaft N

D

Thickness of oil film

L

Lubricating oil

Journal Bearing

  ,QMRXUQDOEHDULQJDYHU\WKLQ¿OPRIOXEULFDWLQJRLOLVPDLQWDLQHGEHWZHHQVWDWLRQDU\VXUIDFH RIWKHEHDULQJDQGWKHURWDWLQJVKDIW7KHVKDIWKDVGLDPHWHUDDQGVSHHG N

Laminar Flow

\



$QJXODUYHORFLW\ Z =

675

p N UDGV 

 7KHWDQJHQWLDOYHORFLW\RIRLODWWKHVKDIWLV D 

u = ZR = Z u=

p DN p N D ¥ =   

 7KHYHORFLW\JUDGLHQWEHWZHHQVWDWLRQDU\VXUIDFHRIWKHEHDULQJDQGPRYLQJVXUIDFHRIWKH VKDIWLV du u- =  ZKHUH t WKLFNQHVVRIRLO¿OP dy t  1RZDSSO\LQJ1HZWRQ¶VODZRIYLVFRVLW\  

6KHDUVWUHVVt = m

du dy

= m◊

u t

= m◊

p DN  ◊ t

\6KHDUIRUFHRUYLVFRXVUHVLVWDQFH F = t ¥6XUIDFHDUHD = m◊

p DN ¥ pDL  ◊ t

= m◊

p D N ◊ L  ◊ t

 7RUTXHUHTXLUHGWRRYHUFRPHYLVFRXVUHVLVWDQFH T=F¥

=

D D m p  D  NL = ¥    ◊ t

m p  D N ◊ L  ◊ t

 3RZHUDEVRUEHGE\WKHEHDULQJWRRYHUFRPHWKHYLVFRXVUHVLVWDQFH P = T ◊w =

p NT 

676

Fundamentals of Fluid Mechanics

 ZKHUHT =

mp  D N ◊ L  ◊ t

‡ $VKDIWRIPPGLDPHWHUURWDWHVDWUSPLQDPPORQJEHDULQJ7KHVKDIWDQG EHDULQJDUHVHSDUDWHGE\DGLVWDQFHRIPPRIRLOZLWKm 1VP)LQGWKHSRZHU DEVRUEHGE\WKHEHDULQJ T=

         

mp  D NL  ◊ t

=

 ¥ p  ¥   ¥  ¥   ¥  ¥ -

=

 ¥ - ¥ p  ¥  ¥ - ¥   ¥ -

 ¥± Nm 3RZHUDEVRUEHG =

p NT p ¥  ¥  ¥ - =  

 ¥±ZDWWV  ZDWWV

 ‡ 'HULYHDQH[SUHVVLRQIRUSRZHUDEVRUEHGLQIRRWVWRSEHDULQJ D = 2R

Shaft

Bearing

Oil

Foot Stop Bearing

  $IRRWVWRSEHDULQJLVWRSURYLGHVXSSRUWWRDURWDWLQJYHUWLFDOVKDIW7KHVSDFHEHWZHHQWKH VXUIDFHRIWKHVKDIWDQGWKHEHDULQJLV¿OOHGZLWKRLO7KHUSPRIWKHVKDIWLVNWKHQ

Laminar Flow

    

$QJXODUYHORFLW\Z =

p N 

7DQJHQWLDOYHORFLW\ u = Z◊

D 

=

p N D ◊  

=

p DN 

u- u du = = t t dy 6KHDUVWUHVV t = m =m

du dy u t

 6KHDUIRUFHRQWKHHOHPHQWDU\ULQJRIUDGLXVrDQGWKLFNQHVVdr dF = m◊

u ¥pr ◊dr t

= m◊

p DN ¥pr ◊dr  t

=

m◊ p rN ◊ pr ◊ dr  ◊ t

=

m p  Nr  dr ◊  t

dT = dF◊r = \

T=

Ú

R



m p Nr dr  ¥ t

m   ◊p Nr dr  t

=

m  R  p N  t 

=

m p N R   t

677

678

Fundamentals of Fluid Mechanics

=

m p NR   t

  3RZHUDEVRUEHGE\EHDULQJ P = Tw =  ZKHUH T =

p NT 

m ¥ p ◊N◊R   ◊ t

‡ )LQG SRZHU UHTXLUHG WR URWDWH D YHUWLFDO VKDIW RI GLDPHWHU  PP DW  USP7KH ORZHUHQGRIWKHVKDIWUHVWVLQDIRRWVWRSEHDULQJ7KHHQGRIWKHVKDIWDQGVXUIDFHRI WKHEHDULQJDUHERWKIODWDQGDUHVHSDUDWHGE\DQRLO¿OPRIWKLFNQHVVPP2LOKDV YLVFRVLW\ SRLVH (Punjab University)   

7RUTXH T  

=    

m ◊pN R  t   Ê ˆ  ¥ p ¥ Á ˜ Ë  ¯  ¥  ¥ -

 1P 3RZHU 

=  

p NT  p ¥  ¥  

 :

 ‡ 'HULYHDQH[SUHVVLRQIRUSRZHUFRQVXPHGLQFROODUEHDULQJ Bearing Collar Shaft

r r + dr

Collar Bearing

Laminar Flow

679

  7KHFROODUEHDULQJVXSSRUWVWKHD[LDOWKUXVWRIWKHURWDWLQJVKDIWE\VHSDUDWLQJWKHFROODURI WKHVKDIWIURPWKHVXUIDFHRILWWKHEHDULQJZLWKD¿OPRIRLORIXQLIRUPWKLFNQHVVZKLFKLV PDLQWDLQHGE\DIRUFHGOXEULFDWLQJV\VWHP,INLVUSPRIWKHVKDIWWKHQ   

 

$QJXODUYHORFLW\Z =

p N 

7DQJHQWLDOYHORFLW\= u = Zr =

p N ◊r 

u t = m du = m◊ t dy dF = t ¥ dA = m◊ =m¥ =

u ◊pr◊dr t

p N ¥ r ◊prdr  ◊ t

m   p Nr dr  ◊ t

dT = dF ¥ r =

\

T=

m   p Nr dr  t

Ú

R

R

m   ◊p N r dr  t

R & RDUHUDGLLRIRXWHU LQQHUVXUIDFHRIWKHFROODU

 

T=

R m Ê ˆ  p N Á r ˜  t Ë  ¯R 

T=

m ◊pN R  – R   t

3RZHUFRQVXPHG 

pN ◊T W 

‡ $FROODUEHDULQJKDYLQJLQWHUQDODQGLQWHUQDOGLDPHWHURIPPDQGPPRLO¿OP WKLFNQHVV PPDQGm SRLVHLVWDNLQJWKHWKUXVWRIWKHVKDIWDQGRYHUFRPLQJ YLVFRXVUHVLVWDQFHZKHQWKHVKDIWURWDWHVZLWKUSP)LQGWKHSRZHUFRQVXPHGE\ WKHFROODUEHDULQJ

680

Fundamentals of Fluid Mechanics

  7RUTXHUHTXLUHGWRRYHUFRPHYLVFRXVUHVLVWDQFH T=

    

m p N R – R   t

=

 ¥ p  ¥ ±  ¥  ¥ -

=

 ¥ - ¥ p  ¥ ± ¥±   ¥ -

 ¥± ¥  1P

  3RZHUFRQVXPHG P=  

 ¥ p ¥   ¥ p N =  

 :

11.15 DASHPOT  ‡ :KDWLVDGDVKSRW"([SODLQLWVZRUNLQJDQGGHULYHWKHHTXDWLRQRIORDG

W

Piston

Oil

Cylinder Dashpot Mechanism





 'DVKSRWLVDK\GUDXOLFGHYLFHIRUGDPSLQJYLEUDWLRQVRIDPDFKLQH,WFRQVLVWVRIDF\OLQGHU ¿OOHGXSZLWKYLVFRXVÀXLGDQGDSLVWRQZKLFKLQVLGHWKHF\OLQGHUGLVSODFLQJWKHOLTXLGLQ RSSRVLWHGLUHFWLRQWRLWVPRYHPHQW7KHSLVWRQLVFRQQHFWHGWRWKHPRYLQJORDGwXSZDUGV DQGGRZQZDUGV   7KH SULQFLSOH RI WKH GDVKSRW LV EDVHG RQ WKH GUDJ IRUFH H[HUWHG E\ WKH ÀXLG ZKHQ D PRYHDEOHSODWHLVPRYHGSDUDOOHOWRDVWDWLRQDU\DQGSDUDOOHOSODWHV

Laminar Flow

681

If WLVORDGRQWKHSLVWRQWKHQGL൵HUHQFHRISUHVVXUHEHWZHHQWZRHQGVRIWKHSLVWRQ dp =

W 4W = 2 pd pd 2 4

 7KHÀRZRIRLOEHWZHHQWKHJDSRIWKHSLVWRQDQGF\OLQGHUVXUIDFHLVVLPLODUWRWKHYLVFRXV ÀRZEHWZHHQSDUDOOHOSODWHVDQGSUHVVXUHGURSIRUOHQJWKL, dp =

12 u mL  ZKHUH t WKLFNQHVVRIJDp and u  DYHUDJHYHORFLW\ t2 12 u mL 4W = 2 t2 pd

\

 RU 

u =



Wt 2 3p m Ld 2

 1RZVXSSRVHSLVWRQPRYHVZLWKYHORFLW\ vpWKHQÀRZQ is Q = vp ¥

pd 2 4

 $OVRZLWKDYHUDJHYHORFLW\ u WKHGLVFKDUJHQWKURXJKWKHJDS Q = u ◊p ◊d ◊t \

2 vp ¥ pd = u ◊p◊d◊t 4

\

u =

Vp ◊ d 4t

2

\

 RU

V d Wt = r 3pm Ld 2 4t 

m=

4Wt 3 3p L d 3 V p

‡ A oil dashpot is used for damping vibrations. Piston falls 50 mm in 50 seconds in the oil of the cylinder If additional load of 1.2 N is applied, the piston falls through 50 mm in 40 seconds. The diameter of piston is 50 mm and its length is 100 mm. The gap thickness is 1 mm. Find viscosity of the oil. W W + 1.2 = V V1

682

Fundamentals of Fluid Mechanics

V=

5 ¥ -  ¥±PV 

V¢ =

5 ¥ -  ¥±PV 

W +  W = -  ¥ -  ¥  W = W

  Rr

 1RZ



W=

  1 

m=

Wt  p L d  V p

=  

 ¥  ¥  ¥ -  p ¥  ¥   ¥ -

 1VP

11.16 MEASUREMENT OF VISCOSITY  ‡ :KDWLV6WRNHV¶ODZ":KDWDUHWKHFRQGLWLRQVDQGDVVXPSWLRQVPDGHIRU6WRNHV¶ODZ" 8378   $VSHUWKH6WRNHV¶ODZZKHQDVPDOOVSKHUHPRYHVWKURXJKDYLVFRXVÀXLGZLWKFRQVWDQW YHORFLW\ u DGUDJIRUFHH[SHULHQFHGE\WKHVSKHUHLQDGLUHFWLRQRSSRVLWHWRWKHPRWLRQLV JLYHQE\Fd p muDZKHUHu WHUPLQDOYHORFLW\m YLVFRVLW\RIÀXLGDQGD GLDPHWHU RIVSKHUH   7KHFRQGLWLRQIRUDSSO\LQJ6WRNHVWZRLVWKDWWKH5H\QROGVQXPEHURIWKHÀXLGLVOHVVWKDQRQH   7KHDVVXPSWLRQVPDGHLQ6WRNHV¶ODZDUH    ,QHUWLDIRUFHDFWLQJRQWKHERG\LVVPDOODVFRPSDUHGWRYLVFRXVIRUFH    :DOOVRIYHVVHOFRQWDLQLQJWKHÀXLGGRQRWD൵HFWWKHÀRZRIÀXLGDURXQGWKHVSKHUH    )OXLGGRHVQRWVOLSRYHUWKHVSKHUH    7KHVSKHUHLVULJLG  ‡ What are the various methods of measuring viscosity?   )ROORZLQJDUHYDULRXVPHWKRGVIRUPHDVXUHPHQWRIYLVFRVLW\    )DOOLQJVSKHUHPHWKRG    &DSLOODU\WXEHPHWKRG

Laminar Flow

     ‡

683

  5RWDWLQJF\OLQGHUPHWKRG   2UL¿FHW\SHYLVFRPHWHU 7KHGHYLFHVXVHGIRUWKHPHDVXUHPHQWRIYLVFRVLW\DUHFDOOHGYLVFRPHWHUV 'HVFULEHIDOOLQJVSKHUHPHWKRGRIPHDVXULQJYLVFRVLW\ Sphere Transparent tube with fluid

Bath for constant temperature

Fd = drag force u

Sphere

u

w = weight of sphere

u

FB = Buoyaucy force

Falling Sphere Viscometer



Free Body DIagram of Sphere

 7KHIDOOLQJVSKHUHPHWKRGLVEDVHGRQ6WRNHV¶ODZ$YHUWLFDOWUDQVSDUHQWWXEHLVXVHGZKLFK LV¿OOHGZLWKOLTXLGZKRVHYLVFRVLW\LVWREHDVFHUWDLQHG$VSKHULFDOEDOOLVUHOHDVHGLQWR WKHOLTXLG7KHEDOODWWDLQVDFRQVWDQWYHORFLW\DWDPHDVXUDEOHLQWHUYDORIWLPH'XULQJWKH VWDWHRIFRQVWDQWYHORFLW\WKHIRUFHVDFWLQJGRZQZDUGVPXVWEHHTXDOWRWKHIRUFHVDFWLQJ XSZDUGV,Irs GHQVLW\RIVSKHUHDQGrL GHQVLW\RIOLTXLGWKHQZHKDYH W = Fd + FBZKHUHFd GUDJIRUFHFB ERX\DQF\IRUFHDQGW¢ :HLJKW p p ¥ d ¥ rs ¥ g pmd◊u + ◊d◊rL ¥ g 6 6 \

pm◊u◊d =

m=

p  d ¥ g rs – rL 6 gd  rs – rL  u

684

Fundamentals of Fluid Mechanics

‡ Describe capillary tube method for measuring viscosity.   &DSLOODU\ WXEH PHWKRG LV EDVHG RQ WKH SULQFLSOH RI KHDG ORVV RI WKH OLTXLG ZKLOH ÀRZLQJ WKURXJKFDSLOODU\WXEHRIGLDPHWHUdDQGOHQJWKL. The head loss depends on viscosity of WKHOLTXLGZKLFKFDQEHGHWHUPLQHGE\PHDVXULQJÀRZ Q). Piezometer Liquid under test

Capillary tube of diameter ‘d ’

h

L

Measurement tank

Capillary Tube Method



  $FDSLOODU\WXEHLVDWWDFKHGWRWKHWDQNDVVKRZQLQWKH¿JXUHDQGGLVFKDUJHWKURXJKWKH FDSLOODU\LVFROOHFWHGLQWKHPHDVXULQJWDQNIRUDJLYHQWLPH7KHÀRZQ is found out by PHDVXULQJYROXPHRIOLTXLGFROOHFWHGLQPHDVXULQJWXEHGLYLGHGE\WLPH3LH]RPHWHUJLYHV the loss of head of the liquid. Q=

9ROXPHRIOLTXLGLQPHDVXULQJWDQN time

u =

Q  DYHUDJHYHORFLW\ area of capillary

=

Q 4Q 2 = pd pd 2 4

The head loss,

or

h=

32 ¥ 4Q ¥ m ¥ L 32u mL = 2 pd 4 r g d ¥ r¥ g

m=

prg hd 4 128 ◊ Q ◊ L

Viscosity can be determined from the above equation.

Laminar Flow

685

 ‡ 'HVFULEHURWDWLQJF\OLQGHUPHWKRGIRUPHDVXULQJYLVFRVLW\ Dial to measure deflection of torsional spring

Torsional spring Inner stationary cylinder of radius R1

Gap ‘t’ of liquid

h

Outer rotating cylinder of radius R2

Rotating Cylinder Method





 5RWDWLQJF\OLQGULFDOPHWKRGLVEDVHGRQ1HZWRQ¶VODZRIYLVFRVLW\$OLTXLGZKRVHYLVFRVLW\ LVWREHGHWHUPLQHGLV¿OOHGLQWKHJDSEHWZHHQLQQHUVWDWLRQDU\F\OLQGHUDQGRXWHUURWDWDEOH F\OLQGHUVXVSHQGHGE\DWRUVLRQDOVSULQJ1RZWKHRXWHUF\OLQGHULVURWDWHGZLWKFRQVWDQW VSHHGDQGGHÀHFWLRQRIWRUVLRQVSULQJLVQRWHG  ,IWKHRXWHUF\OLQGHUURWDWHVDWNUSPWKHQDQJXODUYHORFLW\ w=  

p N 

7DQJHQWLDOYHORFLW\u = w ¥ R =

p N R 

\ 9HORFLW\JUDGLHQWLQWKHJDS p N R du =  ZKHUH t WKLFNQHVVRIJDS  ◊ t dy \

t=m

p N R du = m◊  ◊ t dy

686

Fundamentals of Fluid Mechanics

 

9LVFRXVIRUFHRUGUDJIRUFH t ¥VXUIDFHDUHD Fd = m

 

7RUTXH T  Fd ¥UDGLXV Fd ¥ R =

 RU

pR ◊ N ¥pR◊ hZKHUHh KHLJKWRIOLTXLG  ◊ t

mp  R ◊ R ◊ h ◊ N  ◊ t

m = T◊



 ◊ t p R R ◊ h N 

 ‡ $VPDOODLUEXEEOHRIGLDPHWHUPPULVHVZLWKVWHDG\VWDWHYHORFLW\RIPVWKURXJK DQRLO)LQGWKHYLVFRVLW\RIWKHRLOLI6*RIWKHRLOLV   ,IZHQHJOHFWWKHZHLJKWRIEXEEOHWKHEXEEOHKDVWZRIRUFHVLQHTXLOLEULXPZKHQYHORFLW\ LVVWHDG\   

'UDJIRUFH ERX\DQWIRUFH pmu◊d =

    RU

m=



=  RU  1RZ

 

    6LQFHRe < 

rg ◊ d   u  ¥  ¥  ¥  ¥ -   ¥ 

m ¥±1VP Re = =

 

pd  ◊r ¥ g 6

rud m  ¥  ¥  ¥  ¥ -  ¥ -

 ¥±  

  7KH6WRNHV¶ODZLVYDOLG  ‡ $VPDOOVSKHUHLVXVHGIRUPHDVXULQJWKHYLVFRVLW\RIDQRLO,IVSKHUH 6*  KDV FPGLDPHWHUDQGIDOOVZLWKFRQVWDQWYHORFLW\RIPVWKURXJKWKHRLO 6*   ¿QGWKHYLVFRVLW\RIWKHRLO

Laminar Flow

gd  r s – r L  u

m=

 ¥  ¥    ¥  -  ¥   ¥ 

=  RU  1RZ



687

m 1VP rud m

Re =

 ¥  ¥  ¥  ¥ - 

=  

 

ReKHQFH6WRNHV¶ODZLVDSSOLFDEOH  ‡ ,Q D URWDWLQJ F\OLQGHU YLVFRPHWHU WKH UDGLXV RI WKH RXWHU DQG LQQHU F\OLQGHUV DUH  DQGPP7KHUSPRIWKHRXWHUF\OLQGHULV7KHKHLJKWRIOLTXLGLVPP7KH WRUVLRQDOVSULQJGLDOLQGLFDWHVWRUTXHDV¥ 10±1P)LQGYLVFRVLW\ m = T◊

=



p ¥

R

 ◊ t ¥ R ¥ h ¥ N

 ¥ - ¥  ¥  ¥ - p  ¥   ¥  ¥  ¥ 

   ¥– 61VP  ‡ $RLORI6* LVWREHPHDVXUHGE\WKHFDSLOODU\WXEHPHWKRG7KHWXEHKDVGLDPHWHU RI  PP DQG SUHVVXUH GURS RI  P RI ZDWHU EHWZHHQ WZR SRLQWV  P DSDUW 7KH ZHLJKWRIZDWHUFROOHFWHGLQPHDVXULQJWDQNLV1LQVHFRQGV)LQGYLVFRVLW\ RIWKHRLO :HLJKWRIRLOFROOHFWHG    'LVFKDUJH Q = WLPH ¥ r ¥ g =    1RZYLVFRVLW\LVJLYHQE\ m=

  ¥  ¥  ¥ 

 ¥± mV pr gh d   ◊ Q ◊ L

688

Fundamentals of Fluid Mechanics

=

p ¥  ¥  ¥  ¥  ¥    ¥  ¥  - ¥ 

 ¥±1VP  1VP

    

 ‡ :KDWLVDQRUL¿FHW\SHYLVFRPHWHU"   ,Q WKLV PHWKRG WKH WLPH WDNHQ E\ D FHUWDLQ TXDQWLW\ RI OLTXLG ZKRVH YLVFRVLW\ LV WR EH GHWHUPLQHGLVPDGHWRÀRZWKURXJKDVKRUWFDSLOODU\WXEH7KHFRH൶FLHQWRIYLVFRVLW\LV WKHQFDOFXODWHGE\FRPSDULQJWKLVWLPHZLWKWKHWLPHRIRWKHUOLTXLGZKRVHYLVFRVLW\LVNQRZQ

Liquid

Bath at constant temp. Capillary tube Measuring cylinder Saybolt viscometer

  6D\EROW YLVFRPHWHU LV EDVHG RQ WKH RUL¿FH W\SH YLVFRPHWHU 7KH WLPH WDNHQ E\  FF RI OLTXLGWRÀRZWKURXJKFDSLOODU\WXEHLVPHDVXUHG)URPWKHWLPHPHDVXUHPHQWWKHNLQHPDWLF YLVFRVLW\LVIRXQGRXWIURPWKHUHODWLRQ v $W–

B  ZKHUH t WLPHDQGADQGBDUHFRQVWDQWV t

‡ 'HULYHDQH[SUHVVLRQIRUWKHYHORFLW\RIODPLQDUIORZWKURXJKDQRSHQFKDQQHO   ,QRSHQFKDQQHOWKHXSSHUVXUIDFHRIWKHOLTXLGLVH[SRVHGWRWKHDWPRVSKHUH+HQFHWKH Ê - ∂P ˆ SUHVVXUH DW HYHU\ VHFWLRQ RI FKDQQHO LV VDPH DQG QR SUHVVXUH JUDGLHQW Á  FDQ EH Ë ∂x ˜¯ ∂z maiQWDLQHGIRUÀRZWRWDNHSODFH7KHUHIRUHWKHEHGRIWKHFKDQQHOLVPDGHVORSLQJ ÊÁ ˆ˜ Ë ∂x ¯ VRWKDWWKHKHDGORVVGXHWRÀRZRQDFFRXQWRIYLVFRXVÀRZFDQEHJLYHQWRÀRZE\WKH JUDYLW\IRUFHE\VORSLQJWKHEHG

Laminar Flow

  7KHSUHVVXUHJUDGLHQWLQWKHGLUHFWLRQRIÀRZ ∂t ∂P = ∂y ∂x  DQG

t=m



∂u ∂y

 ∂P =m∂ u ∂x ∂y 

\

 $VZDWHULVÀRZLQJGXHWRWKHVORSHRIEHGLH ∂ z ∂x ∂z ∂P = rg ∂x ∂x

\

\

m

∂ u = rg ∂z ∂y  ∂x

  2QLQWHJUDWLRQZHJHW ∂u rg = ∂y m  %XW

Ê ∂z ˆ ÁË ˜¯ y + c ∂x

∂u  DWVXUIDFHRIZDWHULQFKDQQHOZKHUHy = D 'HSWKRIFKDQQHO ∂y c =

\

- rg m

Ê ∂z ˆ ÁË ˜¯ D ∂x

r g Ê - ∂z ˆ ∂u = Á ˜ D – y m Ë ∂x ¯ ∂y

\  ,QWHJUDWLQJDJDLQZHJHW u=

r g Ê - ∂z ˆ Á ˜ m Ë ∂x ¯

Ê y ˆ ∂ y ÁË ˜ + c z ¯

689

690

Fundamentals of Fluid Mechanics

 1RZu LIy DQGKHQFHc  r g Ê - ∂z ˆ Ê y ˆ ÁË Dy - ˜¯ Á ˜ m Ë ∂x ¯ 

u=

 7KHDERYHLVWKHYHORFLW\GLVWULEXWLRQHTXDWLRQRIWKHÀRZWKURXJKWKHRSHQFKDQQHO  ‡ 'HULYHDQH[SUHVVLRQIRUWKHGLVFKDUJHWKURXJKRSHQFKDQQHO   7KHGLVFKDUJHSHUXQLWZLGWKRIWKHFKDQQHO dQ YHORFLW\¥DUHD u ¥ dy ¥ Q=

\

=

Ú

D



Ú

D



udy rg Ê ∂zˆ Ê y ˆ ÁË - ˜¯ ÁË Dy - ˜¯  m ∂x

r g Ê - ∂zˆ Á ˜ = m Ë ∂x ¯

È Dy  y  ˘ - ˙ Í ˚ Î 

D



=

r g È - ∂ z ˘ D m ÍÎ ∂ x ˙˚ 

‡ 'HULYHDQH[SUHVVLRQIRUDYHUDJHYHORFLW\RIIORZLQRSHQFKDQQHO Q = u ◊ D ◊  +HUHZLGWKLVXQLWOHQJWK r g Ê ∂ z ˆ D  u = Q = ¥ Á- ˜ m Ë ∂ x¯  D D =

r g Ê ∂ z ˆ D Á- ˜ m Ë ∂ x¯ 

‡ 'HULYHDQH[SUHVVLRQIRUKHDGORVVLQRSHQFKDQQHO  +HDGORVVLQRSHQFKDQQHO z – z =

- ∂z ◊L ∂x

 ZKHUHL KHDGGURSEHWZHHQWZRVHFWLRQV HL =  %XW



u =

- ∂z ◊L ∂x r g Ê - ∂ zˆ D Á ˜ m Ë ∂x ¯ 

Laminar Flow

 DQG

\



691

u m - ∂z = r gD  ∂x

H L=

u m◊ L r gD 

11.17 QUESTIONS FROM COMPETITIVE EXAMINATIONS  ‡ )RUODPLQDUIORZWKURXJKDORQJSLSHWKHSUHVVXUHGURSSHUXQLWOHQJWKLQFUHDVHV (a) In linear proportion to the cross-sectional area (b) In proportion to the diameter of the pipe (c) In inverse proportion to the cross-sectional area  G  ,QLQYHUVHSURSRUWLRQWRWKHVTXDUHRIFURVVVHFWLRQDODUHD *$7(

 ◊m◊ Q DP = l pD  ‫ן‬

 D

‫ן‬

 A

  2SWLRQ G LVFRUUHFW  ‡ $IORZWKURXJKDSLSHWKHWUDQVLWLRQIURPODPLQDUWRWXUEXOHQWIORZGRHVQRWGHSHQG on (a) Velocity of the fluid (b) Density of the fluid  F  'LDPHWHURIWKHSLSH G  /HQJWKRIWKHSLSH *$7(  )OXLGÀRZGHSHQGVXSRQ5H\QROGVQXPEHU Re =

r◊VD m

  7KHÀRZGRHVQRWGHSHQGRQOHQJWK   2SWLRQ G LVFRUUHFW  ‡ The lower critical Reynolds number for a pipe flow is  D  'L൵HUHQWIRUGL൵HUHQWIOXLGV (b) The Reynolds number at which the laminar flow changes to turbulent flow  F  0RUHWKDQ  G  7KHOHDVW5H\QROGVQXPEHUHYHUREWDLQHGIRUODPLQDUIORZ ,$6   2SWLRQ E LVFRUUHFW

692

Fundamentals of Fluid Mechanics

 ‡ Which one of the following is the characteristic of a fully developed laminar flow? (a) The pressure drop in the flow direction is zero.  E  7KHYHORFLW\SUR¿OHFKDQJHVXQLIRUPO\LQWKHIORZGLUHFWLRQ  F  7KHYHORFLW\SUR¿OHGRHVQRWFKDQJHLQWKHIORZGLUHFWLRQ (d) The Reynolds number for the flow is critical (IAS 2004) Option (c) is correct  ‡ The velocity distribution in laminar flow through a circular pipe follows the (a) Linear law (b) Parabolic law (c) Cubic power law (d) Logarithmic law (IAS 1996)   ,QODPLQDUÀRZZHKDYH V= -

1 ∂P 2 ◊ (R - r 2 ) 4m ∂x

È Ê r ˆ2˘ m = max Í1 - Á ˜ ˙ ÍÎ Ë R ¯ ˙˚ It shows that variation of velocity u as per distance r from centre, i.e., u ‫ ן‬r2 which is a parabolic curve Option (b) is correct.

Ê dP ˆ ˜ in the flow direction is ∂x ¯

‡ )RUIORZWKURXJKKRUL]RQWDOSLSHWKHSUHVVXUHJUDGLHQW Á Ë (a) + ve (c) zero Option (d) is correct.

(b) 1 (d) – ve

(IAS 1995)

 ‡ :KDW LV WKH GLVFKDUJH IRU ODPLQDU IORZ WKURXJK D SLSH RI GLDPHWHU  PP KDYLQJ centre line velocity of 1.5 m/s? (a) (c)

3p 3 m /s 50 3p m3 /s 5000 Q $UHDî$YHUDJHYHORFLW\

(b)

3p m3 /s 2500

(d)

3p m3 /s  

2

Option (d) is correct.

=

p Ê 40 ˆ Ê 1.5 ˆ ¥ ¥Á ˜ Ë 2¯ 4 ÁË 1000 ˜¯

=

3p m3 /s  

(IAS 1995)

Laminar Flow

693

 ‡ The minimum value of friction factor f that can occur in laminar flow through a circular pipe is  D   E    F   G   ,$6 

 )ULFWLRQIDFWRUf iV 4ήf =

 

  = 5H\QROGV QXPEHU 

  2SWLRQ E LVFRUUHFW

 ‡ 7KHGUDJFRH൶FLHQWIRUODPLQDUIORZYDULHVZLWK5H\QROGVQXPEHU Re) is (b) Re (a) Re –1 (c) Re (d) Re± ,$6   7KHFRH൶FLHQWRIIULFWLRQLV f=

 or f ‫ ן‬Re–1 RH

  2SWLRQ F LVFRUUHFW  ‡ Which one of the following statements is correct?   +\GURG\QDPLFHQWUDQFHOHQJWKIRU (a) Laminar flow is greater than the turbulent flow (b) Turbulent flow is greater than that for laminar flow (c) Laminar flow is equal to that for turbulent flow  G  $JLYHQIORZFDQEHGHWHUPLQHGRQO\LIWKH3UDQGWOQXPEHULVNQRZQ ,(6   2SWLRQ D LVFRUUHFW  ‡ ,QIXOO\GHYHORSHGODPLQDUIORZLQFXUFXODUSLSHWKHKHDGORVVGXHWRIULFWLRQLVGLUHFWO\ proportion to mean velocity  D  7UXH E  )DOVH  F  ,QVX൶FLHQWGDWD G  1RQHRIWKHDERYH *$7(   7KHIULFWLRQKHDGhfLV  ◊m ◊ u ◊ L hf = r◊ gD 

or hf ‫ ן‬u   2SWLRQ D LVFRUUHFW  ‡ 7KHYHORFLW\SUR¿OHLQIXOO\GHYHORSHGODPLQDUIORZLQDSLSHRIGLDPHWHUD is given by Ê r 2 ˆ u = uo Á 1 - 2 ˜ D ¯ Ë

694

Fundamentals of Fluid Mechanics

where r isUDGLDOGLVWDQFHIURPWKHFHQWUH,IWKHYLVFRVLW\RIWKHIOXLGLVȝWKHSUHVVXUH drop across length L of the pipe is



(a)

m ◊ uo L D2

(b)

m ◊ uo L D2

(c)

m ◊ uo L D2

(d)

16m ◊ uo 2 D2

 7KHSUHVVXUHGURS P =

(GATE 2006)

-  ◊m ◊ L ◊ uo D

  2SWLRQ G LVFRUUHFW  ‡ 7KH YHORFLW\ SUR¿OH RI D IXOO\ GHYHORSHG ODPLQDU IORZ LQ D VWUDLJKW FLUFXODU SLSH DV VKRZQLQWKH¿JXUHLVJLYHQE\WKHH[SUHVVLRQ u(r)

r

R x

u r  



 ZKHUH

R  ◊m

Ê dP ˆ ÁË dx ˜¯

Ê r ˆ  ÁË R  ˜¯

dP LVDFRQVWDQW7KHDYHUDJHYHORFLW\RIÀXLGLQWKHSLSHLV ∂x

(a)

R 2 Ê dP ˆ m ÁË dx ˜¯

(c)

-

R 2 Ê dP ˆ m ÁË dx ˜¯

(b)

- R 2 Ê dP ˆ 2m ÁË dx ˜¯

(d)

- R 2 Ê dP ˆ m ÁË dx ˜¯

*$7(

  2SWLRQ D LVFRUUHFW  ‡ A fully developed laminar viscous flow through a circular tube has the ratio of maximum velocity to average velocity as  D   E    F   G   *$7(,(6   2SWLRQ F LVFRUUHFW

Laminar Flow

695

 ‡ Which one of the following statements is correct for a fully developed pipe flow? (a) Pressure gradient balances the wall shear stress only and has a constant value (b) Pressure gradient is greater than the wall shear stress  F  7KHYHORFLW\SUR¿OHLVFKDQJLQJFRQWLQXRXVO\  G  ,QHUWLDIRUFHEDODQFHVWKHZDOOVKHDUVWUHVV ,(6   2SWLRQ D LVFRUUHFW  ‡ Which one of the following is correct? In a fully developed region of the pipe flow  D  7KHYHORFLW\SUR¿OHFRQWLQXRXVO\FKDQJHIURPOLQHDUWRSDUDEROLFVKDSH (b) The pressure gradient remains constant in the downstream direction (c) The pressure gradient continuously changes exceeding the wall shear stress in the downstream direction  G  7KHSLSHLVQRWUXQQLQJIXOO ,(6   2SWLRQ E LVFRUUHFW  ‡ ,QDVWHDG\IORZRIDQRLOLQDIXOO\GHYHORSHGODPLQDUUHJLPHWKHVKHDUVWUHVVLV (a) Constant across the pipe  E  0D[LPXP DW WKH FHQWUH DQG GHFUHDVHV SDUDEROLFDOO\ IRUZDUGV WKH SLSH ZDOO ERXQGDU\  F  =HURDWWKHERXQGDU\DQGLQFUHDVHVOLQHDUO\WRZDUGVWKHFHQWUH  G  =HURDWWKHFHQWUHDQGLQFUHDVHVWRZDUGVWKHSLSHZDOO ,(6   7KHVKHDUVWUHVVLQWKHÀRZ ∂P r ◊ W = ∂x       

 ZKHUHrLVWKHUDGLDOGLVWDQFHIURPWKHFHQWUH6KHDUVWUHVVLV]HURZKHQr DQGPD[LPXP ZKHQr = R  2SWLRQ G LVFRUUHFW ‡ The pressure drop for a relatively low Reynolds number flow in a 600 mm diameter DQGPORQJSLSHOLQHLVN3D:KDWLVWKHZDOOVKHDUVWUHVV" D   E  3D F  3D G  3D ,(6  7KHSUHVVXUHGURSSHUXQLWOHQJWK ΔP L



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697

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Laminar Flow

699

 ‡ Which one of the following statements a not correct in the context of laminar flow through a pipeline?  D  6KHDUVWUHVVLV]HURDWWKHFHQWUHDQGYDULHVOLQHDUO\ZLWKWKHUDGLXVRIWKHSLSH (b) Head loss is proportional to the square of the average flow velocity (c) The friction factor varies inversely with Reynolds number  G  1RGLVSHUVLRQRIGLHLQMHFWHGLQWRDIORZVWUHDP ,(6   2SWLRQ E LVFRUUHFW  ‡ Laminar flow between closely spaced parallel plates is governed by the consideration of which one of the following pair of forces? (a) Pressure and inertial forces (b) Gravity and inertial forces (c) Viscous and inertial forces  G  3UHVVXUHDQGYLVFRXVIRUFHV ,(6   2SWLRQ F LVFRUUHFW  ‡ )ORZ WDNHV SODFH DQG LW KDV 5H\QROGV QXPEHU RI  LQ WZR GL൵HUHQW SLSHV ZLWK UHODWLYHURXJKQHVVRIDQG7KHIULFWLRQIDFWRU  D  :LOOEHKLJKHULQWKHFDVHRISLSHZLWKUHODWLYHURXJKQHVV  E  :LOOEHKLJKHULQWKHFDVHRISLSHKDYLQJUHODWLYHURXJKQHVVRI (c) Will be same in both the pipes (d) In the two pipes cannot be compared on the basis of data given (IES 2000) 

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Chapter

12

TURBULENT FLOW

KEYWORDS AND TOPICS            

CRITICAL VELOCITY INSTANTANEOUS VELOCITY MAGNITUDE OF TURBULENCE INTENSITY OF TURBULENCE SCALE OF TURBULENCE EDDY VISCOSITY TURBULENCE SHEAR STRESS MIXING LENGTH CONCEPT SIMILARITY CONCEPT VELOCITY DISTRIBUTION IN PIPE MAXIMUM VELOCITY OF FLOW AVERAGE VELOCITY OF FLOW

 SHEAR VELOCITY

           

SMOOTH BOUNDARY ROUGH BOUNDARY ROUGHNESS REYNOLDS NUMBER FRICTION FACTOR DARCY’S EQUATION HEAD LOSS RELATIVE ROUGHNESS EQUIVALENT SAND GRAIN ROUGHNESS POWER LAW LAMINAR SUBLAYER MOODY’S DIAGRAM HOTWIRE ANEMOMETER

 LASER DOPPLER ANEMOMETER

12.1 INTRODUCTION /DPLQDUÀRZRFFXUVDWORZYHORFLW\ZKHQÀXLGÀRZVLQDSLSH7KHODPLQDUÀRZWDNHVSODFH LQ SDUDOOHO OD\HUV RU ODPLQDH ZKLFK DUH FRQFHQWULF F\OLQGULFDO VKHHWV LQ SLSH ÀRZ ,I ÀRZ KDV 5H\QROGV QXPEHU JUHDWHU WKDQ  WKHQ WKH ÀRZ LQ SLSH LV QR ORQJHU LQ SDUDOOHO OD\HUV DQG LW LV VDLG WR EH D WXUEXOHQW ÀRZ7KH 5H\QROGV QXPEHU LV WKH UDWLR RI LQHUWLDO IRUFH WR YLVFRXV IRUFH7KHYLVFRXVIRUFHWHQGVWRPDNHWKHPRWLRQRIWKHÀXLGLQSDUDOOHOOD\HUVZKLOHLQHUWLDO IRUFHWHQGVWRGL൵XVHWKHÀXLGSDUWLFOHV7KHÀXLGSDUWLFOHVDUHLQWKHH[WUHPHVWDWHRIGLVRUGHU LQWXUEXOHQWÀRZDQGWKH\GRQRWPRYHLQOD\HUV7KH\PRYHLQKDSKD]DUGPDQQHUDQGSURGXFH ODUJHVFDOH HGGLHV UHVXOWLQJ LQ FRPSOHWH PL[LQJ RI WKH ÀXLG7KH SUHVVXUH DQG YHORFLW\ RI WKH ÀRZGRQRWUHPDLQFRQVWDQWZLWKWLPHDQGWKH\ÀXFWXDWHLQLUUHJXODUZD\'XHWRÀXFWXDWLRQVLQ YHORFLW\LWLVLPSRVVLEOHWRGHYHORSDQ\WKHRU\IRUWKHDQDO\VLVRIWXUEXOHQWÀRZ7KHDQDO\VLV KDV WR EH WKHUHIRUH EDVHG RQ IRUPXODH GHULYHG RQ WKH EDVLV RI H[SHULPHQWDO UHVXOWV 'XH WR KDSKD]DUGPRYHPHQWRIÀXLGSDUWLFOHVLQWXUEXOHQWÀRZWKHYHORFLW\GLVWULEXWLRQLVPRUHXQLIRUP

Turbulent Flow

701

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12.2 TURBULENT FLOW ‡ What is turbulent flow? Compare velocity distribution in laminar and turbulent flows in a pipe.   $ÀXLGÀRZDVLQSLSHVKDYLQJ5H\QROGVQXPEHUJUHDWHUWKDQLVVDLGWREHDWXUEXOHQW ÀRZ7KHÀXLGSDUWLFOHVDUHLQWKHH[WUHPHVWDWHRIGLVRUGHULQWXUEXOHQWÀRZDQGWKH\GR QRWÀRZLQOD\HUV7KH\PRYHLQKDSKD]DUGPDQQHUDQGSURGXFHODUJHVFDOHHGGLHVUHVXOWLQJ FRPSOHWHPL[LQJRIWKHÀXLG7KHSUHVVXUHDQGYHORFLW\RIWKHÀRZGRQRWUHPDLQFRQVWDQW ZLWKWLPHEXWWKH\ÀXFWXDWHLQLUUHJXODUZD\'XHWRÀXFWXDWLRQVLQYHORFLW\LWLVLPSRVVLEOH WRGHYHORSDQ\WKHRU\IRUWKHDQDO\VLVRIWXUEXOHQWÀRZ7KHDQDO\VLVLVWKHUHIRUHFRPSOHWHO\ EDVHGRQIRUPXODHGHULYHGRQWKHEDVLVRIH[SHULPHQWDOUHVXOWV    7KHYHORFLW\GLVWULEXWLRQLQODPLQDUDQGWXUEXOHQWÀRZVDUHDVVKRZQLQWKH¿JXUH7KH YHORFLW\ GLVWULEXWLRQ LV PRUH XQLIRUP LQ WXUEXOHQW ÀRZ DV FRPSDUHG WR D ODPLQDU ÀRZ +RZHYHUWKHYHORFLW\JUDGLHQWLQWXUEXOHQWÀRZQHDUWKHSLSHZDOOLVYHU\ODUJHUHVXOWLQJ KLJKVKHDUVWUHVVDWWKHZDOORIWKHSLSH7KHÀDWQHVVRIYHORFLW\GLVWULEXWLRQFXUYH VOLJKW YDULDWLRQLQYHORFLW\ DZD\IURPWKHZDOOKDVUHVXOWHGGXHWRWKHPL[LQJRIÀXLGOD\HUVDQG H[FKDQJHRIPRPHQWXPEHWZHHQWKHOD\HUVDWPDFURVFRSLFOHYHO,QODPLQDUÀRZWKHUDWLR RIWKHDYHUDJHYHORFLW\WRWKHPD[LPXPLVZKLOHLQWXUEXOHQWÀRZLWLVJUHDWHUWKDQ DQGLWPD\LQFUHDVHXSWR7KHYHORFLW\GLVWULEXWLRQLVSDUDERORLGLQODPLQDUÀRZZKLOH LWREH\VSRZHUODZDQGORJDULWKPLFODZLQWXUEXOHQWÀRZ Turbulent

Flow

Laminar

Velocity Distribution in Laminar and Turbulent Flow

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702

Fundamentals of Fluid Mechanics

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Turbulent Flow

703

 ‡ Describe the mechanism of formation of vortices.   ,IWZRVWUHDPVZLWKXQHTXDOYHORFLWLHVMRLQHDFKRWKHUWKLVSKHQRPHQRQPD\EHUHSUHVHQWHG E\FRQVLGHULQJDVXUIDFHRIGLVFRQWLQXLW\ZLWKVWUHDPVRIYHORFLW\VLQRSSRVLWHGLUHFWLRQVRQ HLWKHUVLGHRILW,IDVOLJKWGLVWXUEDQFHLVPDGHWRWKHVXUIDFHRIGLVFRQWLQXLW\WKHVWUHDPOLQHV DERYHDQGEHORZZLOOFRQYHUJHDWVRPHSRLQWVDQGGLYHUJHDWRWKHUSRLQWV7KHYHORFLWLHV DWFRQYHUJLQJSRLQWVDUHJUHDWHU SHDN DQGGLYHUJLQJSRLQWVDUHORZHU YDOOH\ 3HDNVDQG YDOOH\VGLVWXUEWKHVXUIDFHIXUWKHUUHVXOWLQJLQWKHIRUPDWLRQRIYRUWLFHV Streamline V V V V

S

D

Surface of discontinuity

Streamline Stream Lines and Surface of Discontinuity

D

S

Slight disturbance of surface of discontinuity

D

S

Higher disturbance of surface of discontinuity

Vortices Formation



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704

Fundamentals of Fluid Mechanics

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12.3 INSTANTANEOUS VELOCITY  ‡ Explain (1) instantaneous velocity, (2) magnitude of turbulence, (3) intensity of turbulence, and (4) scale of turbulence.

Velocity

u¢ ut u

ut Time (t)

Variation of Velocity at a Point



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  6LPLODUO\ vt = v ± v¢ in y-direction wt = w ± w¢ in z-direction and

pt = P ± P¢ for pressure

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ut dt

Turbulent Flow

 v′ = T

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w′ =

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Ú

p′ =

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T

 T



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705

T



vt dt wt dt pt dt

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v′ =

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w′ =  T

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 p′ = T

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 T

 T

 T



u¢d t  v¢dt  w¢dt  p¢dt 

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p¢  VRWKDWSRVLWLYH

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du dy

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du LVWKHYHORFLW\JUDGLHQWIURPWKH dy

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du du h dy dy

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12.5 TURBULENCE SHEAR STRESS  ‡ What is the Reynolds expression for turbulence shear stress?   ,WZDVVKRZQE\5H\QROGVWKDWWKHUHLVDQH[FKDQJHRIWUDQVYHUVHPRPHQWXPGXHWRYHORFLW\ ÀXFWXDWLRQVEHWZHHQDGMDFHQWOD\HUVZKLFKGHYHORSHDWDQJHQWLDOVKHDUIRUFHEHWZHHQWKH DGMDFHQW OD\HUV +H GHYHORSHG DQ H[SUHVVLRQ IRU WKH WXUEXOHQW VKHDU VWUHVV EHWZHHQ WKH DGMDFHQW OD\HUV ZKLFK LV H[SUHVVHG DV t = r u¢v¢ ZKHUH r  GHQVLW\ DQG u¢ & v¢ DUH WKH ÀXFWXDWLQJFRPSRQHQWVRIYHORFLW\LQxDQGyGLUHFWLRQVGXHWRWXUEXOHQFH y

u¢ = l

Layer 1 (u + u¢ ) l

Layer 2 (u)

du dy

v¢ Slope =

du dy

x Momentum Exchange between Layers of Fluid



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r ◊ v¢ ◊ D A ◊ u ¢ = r u¢ v¢ DA

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  6LPLODUO\

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u¢ = l

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du dy

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du dy 0L[LQJOHQJWK l = k  d u d y

  

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l =

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Fundamentals of Fluid Mechanics

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=

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du =

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u=

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∂ d¢







k Smooth boundary (d¢ > k)

Turbulent boundary layer

S

Turbulent boundary layer Eddies

Eddies Laminar sublayer

k d¢ Rough boundary (d¢ < k)

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Fundamentals of Fluid Mechanics

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D

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4000

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Flow Through Pipes and Compressibility Effects

829

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AV = AV V Ê ˆ = D ÁË D ˜¯ V 

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831

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\

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There,

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= 0.25 L  ‡ The pipes of the same material and of equal lengths are used for connecting an overhead tank which supplies 85 ¥ 10 –3 m3/s of water. If diameter of pipes are 30 and FPUHVSHFWLYHO\¿QGWKHUDWLRRIKHDGORVVHVLISLSHVDUHFRQQHFWHGLQVHULHVDQGDUH parallel. Neglect minor losses. Series. Length of each pipe line = L and Q remains same hf =

f ¥ L ¥ Q2 12 ¥ d 5

hf =

f ¥ L ¥ (85 ¥ 10-3 ) 2 f ¥ L ¥ (85 ¥ 10-3 ) 2 + 12 ¥ (0.3)5 12 ¥ (0.15)5

Flow Through Pipes and Compressibility Effects

=  

833

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834

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V =



   

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Flow Through Pipes and Compressibility Effects

835

Q ¥ ± Q = AV $UHD¥9HORFLW\  ¥ -

V=

\



  4

 PV

V f Lq  H = hL KHDGORVV   g  ¥ d  =     

 ¥    ¥  ¥  ¥ -  +  ¥   ¥  

   P

  :DWHUOHYHOWREHPDLQWDLQHG H P

10 m

 ‡ :DWHU OHYHO LQ D UHVHUYRLU LV  P DERYH WKH RXWOHW :DWHU LV GLVFKDUJHG WKURXJK D FRPSRXQGSLSHFRQVLVWLQJRIPORQJSLSHVHJPHQWRIPGLDPHWHUDQGPVHJPHQW RI  PP GLDPHWHU %RWK VHJPHQWV DUH KRUL]RQWDO DQG f YDOXHV DUH  DQG  UHVSHFWLYHO\)LQG  GLVFKDUJHDQG  GLVFKDUJHZKHQPLQRUORVVHVDUHQHJOHFWHG

L1 = 30 m

L2 = 15 m

  $VSHUFRQWLQXLW\HTXDWLRQZHKDYH AV = A V V =

4 ¥ p d  4 ¥ p d 

¥ V

Ê  ˆ V = Á V Ë  ˜¯  RU



V V H = hL + hL

836

Fundamentals of Fluid Mechanics

  0DMRUORVVLVIULFWLRQORVVLQERWKSLSHV hf =

=

f ¥ L V f ¥ L V +   ◊ g ◊ d  ◊ g ◊ d  ¥  ¥ V  ¥  ¥ V  +  ¥  ¥   ¥  ¥ 

=  VV  V 0LQRUORVV KHDGORVVDWHQWUDQFHKHDGORVVGXHWRVXGGHQH[SDQVLRQ

     

 

V V - V  +  g g

=

V V - V  +  g g

=

 +   V g

  V

    

7RWDOKHDGORVV KHDGDYDLODEOHLQZDWHUWDQN

  

7RWDOKHDGORVV PDMRUORVVPLQRUORVV  V  V  V

     

V =

  RU \ \



V = PV 'LVFKDUJHQ = AV =

 

   

p ¥   ¥ 4

 ¥±PV

  ,QFDVHPLQRUORVVHVDUHQHJOHFWHGWKHQ         RU   1RZ

+HDGDYDLODEOH KHDGORVVGXHWRIULFWLRQRQO\  V  V PV GLVFKDUJHQ = AV

Flow Through Pipes and Compressibility Effects

=

837

p ¥   ¥ 4

 ¥ ± PV

 

 ‡ :DWHU LQ D WDQN LV PDLQWDLQHG DW FRQVWDQW KHDG RI  P :DWHU LV GLVFKDUJHG WKURXJK D KRUL]RQWDO SLSH  P ORQJ DQG  PP LQ GLDPHWHU 7KHUH LV D YDOYH DW WKH HQG RI WKH SLSH WR FRQWURO WKH GLVFKDUJH :KHQ YDOYH LV KDOI RSHQ GLVFKDUJH LV ¥±PV)LQGWKHYDOXHRIORVVFRHIILFLHQWRIWKHYDOYHLIf = 0.026.

Dia = 100 mm f

4m

L = 100 m

V=

 

 ¥ - ¥  Q =  PV A p ¥  

7RWDOKHDGORVV PDMRUORVVHVPLQRUORVVHV  /RVVGXHWRIULFWLRQORVVDWHQWUDQFHORVVDWYDOYH  7RWDOKHDGDYDLODEOHDWWDQN WRWDOKHDGORVVLQSLSHOLQH 4=

=

Ê V Vˆ f ¥ L ¥V +k + Á   ZKHUHK ORVVFRH൶FLHQWRIYDOYH g  g ˜¯ Ë ◊ g ◊d Ê  ¥   k   ˆ  ¥  ¥   + + Á  ¥  ¥   ¥  ˜¯ Ë  ¥ 

 K   K    RU K   ‡ :DWHUIORZVWKURXJKDSLSHRIPPLQGLDPHWHUZLWKGLVFKDUJHHTXDOWR¥ ± PV7KH SLSHOLQH FRQVLVWV RI VWUDLJKW OHQJWK RI  P RQH YDOYH K    RQH HOERZ (K = 2) and one T outlet (K  :KDWLVWKHHTXLYDOHQWOHQJWKRIWKHSLSHOLQH")LQG also total loss of head if f = 0.02. Q ¥±PV AV = Q ZKHUH A DUHDV YHORFLW\ RU

V=

 ¥ - ¥   PV p ¥  

838

Fundamentals of Fluid Mechanics



 RU  

0DMRUORVV IULFWLRQORVV hf =

hf =



 ¥  ¥    P  ¥  ¥ 

0LQRUORVVHV ORVVHVGXHWR¿WWLQJV hm = K + K + K

=

   

   

f LV   gd

V g

 +  +  ¥    ¥ 

  P  7RWDOORVVRIKHDG hf + hm      P 1RZHTXLYDOHQWOHQJWKIRUPLQRUORVVHV le = =

Kd f  +  +   

   P   7RWDOHTXLYDOHQWOHQJWKRIWKHSLSHOLQH LHTY = L + le    P

     

 ‡ )LQGWKHGLDPHWHURIDXQLIRUPSLSHWRUHSODFHDFRPSRXQGSLSHKDYLQJWKUHHVHJPHQWV RIPPDQGPRIFPFPDQGFPGLDPHWHUVUHVSHFWLYHO\7RWDO OHQJWKRISLSHLVVDPHDVWKDWRIFRPSRXQGSLSHOLQH L + L + L d

=

L d

+

L d 

+

L d

   +  +  =  + +       d

Flow Through Pipes and Compressibility Effects

839

   d    RU

  Ê  ˆ d= Á Ë  ˜¯



 



 P

 ‡ )RUWKHGLVWULEXWLRQPDLQVRIDWRZQZDWHUVXSSO\DFPPDLQLVUHTXLUHG$VSLSHV DERYHFPGLDPHWHUDUHQRWDYDLODEOHLWLVGHFLGHGWROD\WZRSDUDOOHOPDLQVRIVDPH GLDPHWHU)LQGWKHGLDPHWHURIWKHSDUDOOHOPDLQV 3XQMDE8QLYHUVLW\ hf =  1RZ

Q = Q + Q QDVQ = Q



Q =

RU

f ¥  ¥ Q =  ¥ d 

\

 RU

d  =



=

 RU  

f ¥ L ¥ Q  ¥ d 



d =

Q  Ê Qˆ f ¥¥ Á ˜ Ë ¯



 ¥ d   ¥ d 4  ¥   4   =   

 FP

  +HQFHQHDUHVWVWDQGDUGGLDPHWHULVFP7ZRFPGLDPHWHUSLSHVDUHWREHODLG  ‡ $VSHFLDOSLSHRIGLDPHWHUd carrying discharge QDWXSVWUHDPLVDUUDQJHGWRGLVFKDUJH qSHUXQLWOHQJWKXQLIRUPO\DORQJLWVOHQJWK'HULYHDQH[SUHVVLRQIRUKHDGORVVRYHUD length L and head loss for entire length.

840

Fundamentals of Fluid Mechanics q

q

q

q

q

Q d x



dx

 7KHUHLVIDOORIGLVFKDUJHµq¶SHUXQLWOHQJWK,IQLVWKHLQLWLDOÀRZLQWKHSLSHWKHQUHPDLQLQJ GLVFKDUJHDIWHUaGLVWDQFHx Qx = Q – q ◊x

  6XSSRVHWKLVGLVFKDUJHUHPDLQVIRUGLVWDQFHdxWKHQKHDGORVVLQWKLVVPDOOOHQJWK dh f =

f ⋅ dx ⋅ Q − q x   d 

 %\LQWHJUDWLQJZHFDQ¿QGKHDGORVVIRUFRPSOHWHOHQJWKL

Ú dhf = Ú

L

0

hf =

=

f Q – qx  dx   d

f  ¥ d 

L

È  x q  x ˘ ◊Q + ÍQ ◊ x -  ◊ q ˙   ˚0 Î

f Q  L È q  L q  ˘ - L˙ Í +  ◊ d  Î Q Q ˚

  7KHÀRZNHHSVRQGLVFKDUJLQJqSHUXQLWOHQJWKWLOOLWEHFRPHV]HURDIWHUOHQJWKL PLQ Q = q ¥ lPLQRU

hf =

=

q  = Q LPLQ

f LPLQ ¥ Q  Ê  ˆ ÁË + - ˜¯   ◊ d  f ¥ Q  ◊ d  ◊ q

Flow Through Pipes and Compressibility Effects

841

14.12 PIPE NETWORKING  ‡ :KDWLVSLSHQHWZRUNLQJ":KDWDUHWKHFRQGLWLRQVWREHVDWLV¿HGE\WKHSLSHQHWZRUN" C

D H E

B

A

F

G

Networking



  



 $ JURXS RI SLSHV LQWHUFRQQHFWHG DQG IRUPLQJ VHYHUDO FLUFXLWV RU ORRSV LV FDOOHG D SLSH QHWZRUN 1HWZRUNLQJ LV D FRPPRQ PHWKRG LQ ZDWHU GLVWULEXWLRQ V\VWHP DGRSWHG E\ WKH PXQLFLSDOLWLHVRIWKHFLWLHV  ,QQHWZRUNLQJRISLSHVIROORZLQJFRQGLWLRQVDUHDOZD\VVDWLV¿HG   Continuity equation. $WHDFKMXQFWLRQRISLSHVWRWDOLQÀRZLVHTXDOWRRXWÀRZDVSHU WKHSULQFLSOHRIFRQWLQXLW\   Energy equation. 7KHORVVRIKHDGGXHWRÀRZLQHDFKORRSLQFORFNZLVHGLUHFWLRQPXVW EHHTXDOWRWKHORVVRIKHDGGXHWRÀRZLQDQWLFORFNZLVHGLUHFWLRQ,QWKH¿JXUHWKHKHDG ORVVLQWKHFORFNZLVHÀRZIURPAWREPXVWEHHTXDOWRWKHKHDGORVVLQDQWLFORFNZLVH IURPAFE,QRWKHUZRUGVWKHDOJHEUDLFVXPRIKHDGORVVHVLQDQ\FORVHGFLUFXLWZLWKLQ WKHQHWZRUNPXVWEH]HUR   Darcy’s equation. 7KH KHDG ORVV LQ HDFK SLSH RI WKH QHWZRUN IRU VLPSOLFLW\ FDQ EH H[SUHVVHG DV hf = r Q ZKHUH r =

fL 7KH ÀRZ LQ HDFK SLSH PXVW VDWLVI\ WKH  ¥ d 

UHODWLRQEHWZHHQWKHKHDGORVVDQGGLVFKDUJHDWDOOWLPHV  ‡ 'HVFULEHWKH+DUG\FURVVPHWKRGIRUVROYLQJSLSHQHWZRUNSUREOHPV   7KHSLSHQHWZRUNSUREOHPVDUHGL൶FXOWWRVROYHDQDO\WLFDOO\+HQFHWKHPHWKRGRIVXFFHVVLYH DSSUR[LPDWLRQLVXVHGZKLFKLVFRPPRQO\NQRZQDV+DUG\FURVVPHWKRG)ROORZLQJVWHSV DUHXVHGLQWKLVPHWKRG    D WULDO GLVWULEXWLRQ RI ÀRZ LV PDGH IRU WKH QHWZRUN LQ VXFK D ZD\ WKDW WKH FRQWLQXLW\ HTXDWLRQLVDOZD\VVDWLV¿HGDWHDFKMXQFWLRQRUQRGH    KHDGORVVZLWKWULDOGLVWULEXWLRQRIÀRZIRUHDFKSLSHLQQHWZRUNLVIRXQGRXW+HDGORVV LQHDFKSLSHLVhf = kQ ZKHUHk =

fL  ◊ d 

  DSSO\WKHFRQGLWLRQDQGVHHZKHWKHUWKHVXPRIKHDGORVVHVLQHDFKORRSLV]HURLH Â h f = Â kQ  

842 

Fundamentals of Fluid Mechanics

  ,QFDVHVXPRIKHDGORVVHVLQDORRSLVQRW]HURFKDQJHWKHYDOXHRIWKHÀRZVRWKDW WKHVXPRIKHDGORVVRIÀRZPD\EHFRPH]HUR  1HZÀRZLVQ = Q0 + DQDQGÂhf = ÂK Q0 + DQ  = 0DQGDQ =



 KQ    KQ

  7KHDERYHSURFHGXUHKDVWREHUHSHDWHGWLOOZHJHWÂhf   ‡ )LQG WKH GLVFKDUJH LQ HDFK SLSH RI WKH QHWZRUN DV VKRZQ LQ WKH ¿JXUH7KH YDOXH RI kIRUHDFKSLSHLVDOVRVKRZQ7KHGLVFKDUJHIURPWKHQRGHVLVDOVRVKRZQ QA = –20 ¥±, QD ¥ ±, Qc = – 40 ¥ ± and QB = 45 ¥±PV  A

20

k=7

D 15

k=8

k=6

k=6

k=4

45

B

40

C

Guidance.+DUG\FURVVPHWKRGLVWREHXVHG7DNHORRSABCDQWLFORFNZLVHDQGCDAORRS FORFNZLVH$VVXPHÀRZLQHDFKSLSHRIWKHORRS)LQGhf = KQ IRUHDFKSLSH7KHYDOXH RI hf LV SRVLWLYH LQ WKH GLUHFWLRQ RI DVVXPHG ÀRZ DQG QHJDWLYH LQ RSSRVLWH GLUHFWLRQ7KH YDOXHRIÂhfRIWKHORRSLVZRUNHGRXW,QFDVHÂhfLV]HURWKHDVVXPHGYDOXHVRIÀRZLQ HDFKSLSHLVFRUUHFWRWKHUZLVH¿QGFKDQJHLQÀRZDQZKLFKLVJLYHQE\= 

 KQ    KQ

 :H DVVXPH WKH ÀRZ LQ HDFK SLSH RI WKH ORRSV VXFK WKDW WKH ÂQ   DW HDFK QRGH 7KH DVVXPHGÀRZLQHDFKSLSHLVDVVKRZQLQWKH¿JXUH 10

A

20

5

15

45

D

B

15

5

C

30

40

First trial Loop ABC

Loop CDA

K

Q

hf = KQ2

2KQ

Pipe

K

AB

6

15

–1350

180

CD

6

5

150

60

BC

4

30

3600

240

DA

7

10

700

140

CA

8

5

–200

80

AC

7

5

 = 2050

 = 500

Pipe

DQ =

- 2050 500

= –4

Q

hf = KQ2

2KQ

200

80

 = 750

 = 280

DQ = –

750 280

ª –3

Flow Through Pipes and Compressibility Effects

843

Second Trial   7KHFRUUHFWHGÀRZLHQ ± DQLQHDFKSLSHRIWKHQHWZRUNLVDVVKRZQLQWKH¿JXUH 10 – 3 = 7

A

20

6

15 + 4 = 19

45

D

B

5+3=8

C

30 – 4 = 26

Loop ABC Pipe

K

15

40

Loop CDA

Q

KQ2

2KQ

Pipe

K

Q

KQ2

2KQ

AB

6

19

–2166

238

CD

6

8

–384

96

BC

4

25

2704

208

DA

7

7

343

98

CA

8

6

–288

96

AC

8

6

286

96

 = 250

 = 532

 = –1247

 = 290

= – 0.5

DQ =

250

DQ =

532

- 247 290

= –1

Third Trial. 7KHFRUUHFWHGÀRZLHQ ± DQLQHDFKSLSHRIWKHQHWZRUNLVVKRZQLQWKH ¿JXUHEHORZ 7+1=8

A

20

6 – 0.5 = 5.5

19 + 0.5 = 19.5 45

D

B

26 – 0.5 = 25.5

8+1=9

C

Loop ABC Pipe

K

15

40

Loop CDA

Q

KQ2

2KQ

Pipe

K

Q

KQ2

2KQ

AB

6

19.5

–2281.5

254

CD

6

9

–486

108

BC

4

25.5

2601

204

DA

7

8

202

84

CA

8

55

–242

88

AC

8

5.5

 = 77.5

 = 526

DQ =

- 72.5 526

= – 0.15

242

88

Â= 8

 = 280

DQ =

-8 280

= – 0.03

844

Fundamentals of Fluid Mechanics

 7KH¿QDOÀRZLQHDFKSLSHLQWKHQHWZRUN 20

5.97

A

5.35

19.65

45

D

B

25.35

15

9.03

40

C

14.13 SYPHON  ‡ :KDWLVDV\SKRQ":KHUHLVLWXVHG"([SODLQLWVDFWLRQ'HULYHDQH[SUHVVLRQIRUWKH length of its inlet leg. 8378   6\SKRQLVDQDUUDQJHPHQWRISLSHV\VWHPE\ZKLFKZDWHUFDQEHPDGHWRUXQXSWKHKLOO XWLOL]LQJWKHIRUFHRIDWPRVSKHULFSUHVVXUH,WLVDEHQWSLSHFRQQHFWLQJWZRZDWHUVXUIDFHV DWGL൵HUHQWOHYHOVDQGVHSDUDWHGE\KLJKJURXQGRYHUZKLFKWKHSLSHLVODLGDVVKRZQLQWKH ¿JXUH7KHZDWHUZLOOÀRZIURPKLJKZDWHUOHYHOVXUIDFH VXUIDFHA WRWKHORZHUZDWHUOHYHO VXUIDFH VXUIDFHB HYHQZKHQDQREVWUXFWLRQLQZD\RIVXPPLWCLVSUHVHQW7KHSUHVVXUH DWSRLQWCLQSLSHLVOHVVWKDQDWPRVSKHULFSUHVVXUHDVLWOLHVDWDOHYHOZKLFKLVDERYHWKH IUHHVXUIDFHRIWKHZDWHUDWSRLQWA6LQFHDWPRVSKHULFSUHVVXUHLVHTXDOWRPRIZDWHU WKHSUHVVXUHDWCFDQEHWKHRUHWLFDOO\UHGXFHGWR±PRIZDWHUEXWLWLVOLPLWHGWR± PRIZDWHUWRDYRLGDLUDQGGLVVROYHGJDVHVVHSDUDWLQJIURPWKHZDWHUDQGJHWWLQJFROOHFWHG DWSRLQWC VXPPLW RIWKHSLSHOLQHZKLFKDUHOLNHO\WRREVWUXFWOLQHÀRZRIWKHZDWHU7KH LQOHWOHJLVLQIDFWWKHSLSHOLQHIURPWKHUHVHUYRLUAWRVXPPLWCDQGWKHRXWOHWOHJLVWKH SLSHIURPVXPPLWCWRUHVHUYRLUB C

Summit

A TEL

HGL B

Obstruction like hill

Syphon

  6\SKRQFDQEHXVHGIRU  

  7RWDNHZDWHUIURPRQHUHVHUYRLUWRDQRWKHUUHVHUYRLUORFDWHGDWORZHUOHYHOEXWZKHQ WKH\DUHVHSDUDWHGE\KLJKREVWDFOHOLNHKLOORUULGJH   7R GUDZ RXW ZDWHU IURP D FKDQQHO ZLWKRXW DQ\ RXWOHW WR ORZHU JURXQG DV VKRZQ EHORZ

Flow Through Pipes and Compressibility Effects

845

Syphon Low ground Channel

Flow

Drawing Out Water from a Channel



  7RGUDZRXWZDWHUIURPDWDQNKDYLQJQRRXWOHWDVVKRZQEHORZ

Flow Drawing Out Water from a Tank



  7RFRQQHFWWZRRSHQFDQDOVE\LQYHUWHGV\SKRQDVVKRZQEHORZ HGL

Syphon

Obstacle like big ditch

Inverted Syphon Connecting Two Channels



Principle of working  1HJDWLYHSUHVVXUHLVFUHDWHGDWVXPPLWCRIWKHV\SKRQVRWKDWZDWHUFDQEHSXVKHGWRZDUGV WKHVXPPLWE\WKHDWPRVSKHULFSUHVVXUHDFWLQJRQWKHIUHHVXUIDFHRIWKHZDWHU7KHÀRZ LQWKHV\SKRQLVPDLQWDLQHGWLOOWKHQHJDWLYHSUHVVXUHUHPDLQVDWVXPPLWC7KHÀRZDQG YHORFLW\RIWKHÀRZZKHQVWDUWHGLQWKHV\SKRQGHSHQGVRQWKHGL൵HUHQFHRIZDWHUOHYHODW SRLQW¶VADQGB,WGRHVQRWGHSHQGRQWKHOHYHORICRIWKHV\SKRQWLOOQHJDWLYHSUHVVXUHLV PDLQWDLQHGDWSRLQWC2QDSSO\LQJ%HUQRXOOL¶VHTXDWLRQEHWZHHQSRLQWVADQGBZHJHW PA P V V + A + ZA = B + B + ZBORVVHV r◊ g r◊ g g g  $V



P PA = B   DQG VA = VBZHKDYH r◊ g r◊g ZA – ZB ORVVHV

846

Fundamentals of Fluid Mechanics

 

+HDGORVV ORVVRIKHDGDWHQWU\ORVVRIKHDG

 

 GXHWRIULFWLRQORVVRIKHDGDWWKHH[LW  

V V f LV  + + g g g ◊ d

 1RZDSSO\%HUQRXOOL¶VHTXDWLRQEHWZHHQSRLQW¶VADQGCZHKDYH P V PA V + A + ZA = C + C + ZCORVVHV g r◊ g r◊ g g PA VA VC = V \

ZC – ZA = –

 PC Ê V  f LACV  ˆ – VC – Á + rg  ◊ g ◊ d ˜¯ g Ë g

= – hC PC V Ê f LAC ˆ = hC +  + rg  g ÁË d ˜¯ Inlet leg and outlet leg   $SSO\%HUQRXOOL¶VHTXDWLRQEHWZHHQSRLQWVA DQGC  PA P V + A + ZA = C + VC + ZC + hf rg rg g g

PA = Pa DWPRVSKHULFSUHVVXUH PC  DEVROXWH]HURSUHVVXUH VA = 0 VC = V LQOHWYHORFLW\WRV\SKRQ hf =

f l V ZKHUHl LQOHWOHQJWK ◊ g ◊d

Pa f l V V + 0 + ZA = 0 +  + ZC + rg ◊ g ◊d g  RU



V =

gd È Pa ˘ - ( Z C - Z A )˙  f ◊ l ÍÎ rg ˚

L

Flow Through Pipes and Compressibility Effects

847

 1RZDSSO\%HUQRXOOL¶VHTXDWLRQEHWZHHQSRLQWCDQGB PC V V P + C + ZC = B + B + ZB + h f rg g g rg

PC  VB VC = VPB = Pa DWPRVSKHULFSUHVVXUHDQGhf = P V f V L + ZC = a + 0 + ZB + rg g ◊ g ◊d

0+

 RU

f ◊ V L g ◊ d



V =

◊ g ◊d f ◊ L

Pa ˘ È Í Z C - Z B - r g ˙  Î ˚

LL

 1RZ WKH YHORFLW\ V PXVW EH HTXDO WR RU JUHDWHU WKDQ V VR WKDW QHJDWLYH SUHVVXUH LV PDLQWDLQHGDWVXPPLWCZLWKRXWOHWOHJÀRZLQJDOZD\VIXOO V ≥ V È Pa ˘ Í r g - ZC - Z A ˙ L Î ˚ £ P È L a ˘ Í Z C - Z B - r g ˙ Î ˚

RU

‡ 7KHGL൵HUHQFHLQWKHZDWHUVXUIDFHOHYHOVRIWZRUHVHUYRLUVZKLFKDUHFRQQHFWHGE\D V\SKRQLVP7KHOHQJWKRIWKHV\SKRQLVPDQGLWVGLDPHWHUFP$VVXPLQJ f GHWHUPLQHWKHGLVFKDUJHZKHQWKHV\SKRQLVUXQQLQJIXOO,IWKHYHUWH[RIWKH SLSHOLQHLVPDERYHWKHVXUIDFHOHYHORIWKHXSSHUUHVHUYRLUGHWHUPLQHWKHPD[LPXP length of the inlet leg for the pipe to run full. Neglect all losses other than that of friction. 3RRQD8QLYHUVLW\ C A

5m

H=8m B

H = 

 RU



V=

V V + fL ◊  ZKHUH H = 8 g d g

 gH fl  + d

848

Fundamentals of Fluid Mechanics

=

2 ¥ 9.81 ¥ 8 0.02 ¥ 500 1.5 + 0.3

= 1.945 m/s Discharge, Q = A◊V =

p ◊ (0.3)2 ¥ 1.945 4

= 0.1374 m3/s Inlet leg

But,

Pa - (ZC - Z A) L1 g £ P L2 (ZC - Z B ) - a r◊g Pa = atmospheric pressure head r◊ g = 10.3 m

\

10.3 - 5 L1 £ (5 + 8) - 10.3 L2 5.3 L1 £ 2.7 L2

But,

L2 = L – L1

\

L1 5.3 £ 2.7 L - L1

or

L - L1 2.7 ≥ 5.3 L1

or

L £ 0.51 + 1 L1

or

L1 ≥

or

L1 £ 39.7 m

600 1.51

Flow Through Pipes and Compressibility Effects

849

 ‡ $V\SKRQRIGLDPHWHUPPFRQQHFWVWZRUHVHUYRLUVKDYLQJDGL൵HUHQFHRIHOHYDWLRQ RIZDWHUVXUIDFHDVP7KHWRWDOOHQJWKRIV\SKRQLVPDQGWKHVXPPLWLVP DERYHWKHZDWHUOHYHOLQWKHXSSHUUHVHUYRLU,IVHSDUDWLRQWDNHVSODFHDWPRIZDWHU DEVROXWH¿QGPD[LPXPYDOXHRILQOHWOHJ$VVXPHf DQGDWPRVSKHULFSUHVVXUH KHDG RIZDWHU   1RZGL൵HUHQFHRIZDWHUOHYHOEHWZHHQWZRUHVHUYRLUVLVP f LV  ◊ g ◊d

H =  hf =

 RU

V =



=     RU



 g ◊ d ◊  f ◊L  ¥  ¥  ¥   ¥ 

  V PV

  1RZDSSO\%HUQRXOOL¶VHTXDWLRQEHWZHHQWKHXSSHUUHVHUYRLU A DQGVXPPLWC PA P V V + A + Za = C + B + ZCORVVHV rg rg g g  

\  

 

hf  ±

  ±  ¥ 

 P

%XW

hf =

RU

L =

=  

V PDMRUORVVHV h f g

f ¥ L ¥ V  ◊ g ◊d  ◊ g ◊ d ◊ hf f ¥V  ¥  ¥  ¥   ¥  

 P

\+HQFHPD[LPXPLQOHWOHJRIWKHV\SKRQ P

850

Fundamentals of Fluid Mechanics

14.14 POWER TRANSMITTED BY A FLOW  ‡ )LQGDQH[SUHVVLRQIRUSRZHUWUDQVPLWWHGE\DIORZWKURXJKDSLSHOLQH$OVR¿QGWKH H൶FLHQF\RISRZHUWUDQVPLWWHG

H

L



 ,QWKHWKFHQWXU\WKHUHZDVQRNQRZOHGJHDERXWHOHFWULFSRZHURULWVWUDQVPLVVLRQDVLWLV XVHGQRZDGD\V+HQFHWKHÀRZRIZDWHUXQGHUSUHVVXUHLQDSLSHOLQHZDVXVHGIRUSRZHU WUDQVPLVVLRQ,IHLVWKHWRWDOKHDGRIVXSSO\DWWKHHQWUDQFHRIWKHSLSHDQGhf LVWKHPDMRU ORVVGXHWRIULFWLRQWKHQWKHKHDGDYDLODEOHDWWKHRXWOHWRISLSHLVH – hf QHJOHFWLQJPLQRU ORVVHV  hf =

f LV  g ◊ d

 1RZZHLJKWRIZDWHUÀRZLQJWKRXJKWWKHSLSHOLQH W = r ◊g◊

pd  ¥V 4

 +HQFHSRZHUWUDQVPLWWHGDWRXWOHWRIWKHSLSHOLQH   

p Ê ˆ Ê f LV  ˆ 3RZHUP = Á rg ¥ d  ¥ V ˜ Á H Ë ¯ Ë 4  gd ˜¯

(৽FLHQF\(൶FLHQF\LVWKHUDWLRRISRZHUDYDLODEOHDWWKHRXWOHWRIWKHSLSHOLQHWRWKHSRZHU VXSSOLHGDWWKHLQOHWRIWKHSLSHOLQH h=

=

:HLJKWRIZDWHU ◊ H - h f :HLJKWRIZDWHU ◊ H H - hf H

‡ 'HULYH WKH H[SUHVVLRQ IRU PD[LPXP SRZHU WUDQVPLWWHG DQG PD[LPXP H൶FLHQF\ RI SRZHUWUDQVPLWWHGE\DIORZLQDSLSHOLQH (UPTU 2004-5)

Flow Through Pipes and Compressibility Effects

851

2ˆ 2 Ê ˆ Ê Power transmitted, P = Á rg ¥ p d ◊ V ˜ H - f LV Ë ¯ ÁË 2 gd ˜¯ 4

Since power transmitted depends on the velocity, the maximum power transmitted will be when

dP =0 dV dP d È rg p d 4 = Í dV dV ÍÎ 4

or

H–

Ê f LV 3 ˆ ˘ HV ˙ =0 ÁË 2 ◊ g ◊ d ˜¯ ˙˚

3 f LV 2 =0 2◊ g ◊d

or

H=

3 f LV 2 2 gd

= 3 hf Hence, for maximum power transmitted, the friction loss (hf) is one-third of the head available. Now,

h=

H - hf H

For maximum power transmitted, hf = \



H 3

0D[LPXPH൶FLHQF\ h max =

H-

H 3

H =

2 = 66.67 % 3

‡ The pressure at the inlet of pipe is 7 ¥ 103 kN/m2 and pressure drop is 700 kN/m2. The SLSHOLQHLVNPORQJ,IN:LVWREHWUDQVPLWWHGRYHUWKLVOLQH¿QGWKHGLDPHWHU (UPSC) RIWKHSLSHDQGH൶FLHQF\RIWUDQVPLVVLRQ7DNHf = 0.024 hf =

700 ¥ 103 700 ¥ 103 = = 71.35 m 1 ¥ 103 ¥ 9.81 r¥ g

H=

7000 ¥ 103 7000 ¥ 103 = = 713.5 m 103 ¥ 9.81 r¥ g

852

Fundamentals of Fluid Mechanics

h=    

H - hf H

=

 -  

 RU 3RZHUWUDQVPLWWHG rgQ H – hf ¥  ¥¥ Q ±

 

Q=

  RU

 ¥   ¥  ¥ 

 PV

 

 RU



 



 RU



hf =

f LQ     ¥ d 

d =

 ¥  ¥  ¥    ¥ 

 ¥± d FP

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Flow Through Pipes and Compressibility Effects

853

$VLWLVXQHFRQRPLFDOWRGHVLJQSLSHRUSLSHOLQHWRVWDQGKLJKSUHVVXUHRIZDWHUKDPPHUVDIHW\ GHYLFHVDUHXVHGWRRYHUFRPHWKLVSUREOHP:DWHUKDPPHUUHVXOWVGXHWRXQVWHDG\ÀRZ 8QVWHDG\ ÀRZ :KHQ D YDOYH LV FORVHG WKH NLQHWLF HQHUJ\ RI WKH ÀXLG LV FRQYHUWHG LQWR D ULVH RI SUHVVXUH$ VXGGHQ SUHVVXUH SXOVH LV IRUPHG DW WKH YDOYH DQG WKLV SXOVH WUDYHOV EDFNZDUGVDVUHÀHFWHGSXOVHDWWKHVSHHGRIVRXQG Magnitude of pressure rise. ,Q D VORZ FORVLQJ RI WKH YDOYH WKH ZKROH PDVV RI ÀXLG LQ SLSHOLQHLVGHFHOHUDWHGDQGLQHUWLDSUHVVXUH dP =

r A ¥ L d u ◊ dT A

  ZKHUH A DUHDL OHQJWK = r ◊L◊ 

du  GHFHOHUDWLRQ dT

du dT

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K r

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 Q = p ¥    PV V= A 4  3UHVVXUHULVHZKHQYDOYHLVJUDGXDOO\FORVHG dP = r◊ L◊

h= \



L dV dV RUSUHVVXUHKHDG h = ◊ g dt dt

 dP   ¥  ¥  ¥ rg  

 P!P

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14.16 TIME TO EMPTY A TANK  ‡ 'HULYH DQ H[SUHVVLRQ IRU WLPH WR HPSW\ D WDQN IURP OHYHO H to H2 through a pipe discharging in open.   &RQVLGHUDWDQNZLWKZDWHUZKLFKLVWREHHPSWLHGE\PHDQVRIDSLSH¿WWHGDWLWVERWWRP 7KHSLSHGLVFKDUJHVIUHHLQRSHQ$VVXPHADQGaDUHFURVVVHFWLRQDODUHDVRIWDQNDQGSLSH UHVSHFWLYHO\&RQVLGHUZDWHUOD\HUDWDKHLJKWhDQGOHYHOIDOOVE\dhLQWLPHdT –A◊dh = a ◊V ◊dT

L

Flow Through Pipes and Compressibility Effects

855

dh H1

h

H2

l

Datum line Time of Emptying a Tank

  $OVRWRWDOKHDGh V g

h=

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f lˆ Ê ÁË + ˜¯ d

 gh f ◊l  + d

V=

 6XEVWLWXWLQJWKHYDOXHRIVLQHTQ L –A ◊dh = a

 RU

RU



 gh ◊dT fl  + d

A dT = – a

T=–

A a

A = a

f ◊l d ¥ h– dh g

 +

f ◊l d g

 +

Ú

H

H

h± dh

f ◊l d H  – H    g

 +

‡ 'HULYHDQH[SUHVVLRQIRUWLPHIRUORZHULQJRIOHYHORIWZRUHVHUYRLUVIURPOHYHOH to H2ZKHQIORZLVPDGHWKURXJKDSLSHEHWZHHQWKHWZRUHVHUYRLUV

856

Fundamentals of Fluid Mechanics dh

A1

h dh ¥

A1 A2

A2

l

Two Reservoirs having Flow through a Pipe



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 +RZHYHUZKHQOHYHOLQWKHXSSHUUHVHUYRLUIDOOVWKHOHYHORIORZHUUHVHUYRLUULVHV7KHQHW GL൵HUHQFHRIKHDG dH  A dH = dh – dH ◊ A 

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NRZ

RU



dh A dH = +  ±AdH = aV◊dT V=

– A dH = a

RU

 gh f ◊L  + d  gh ◊ dT f ◊L  + d

A dT = –  a

f ◊L d ◊ dH  gh

 +

Flow Through Pipes and Compressibility Effects

A dT = – 1 a

1.5 + 2g

fl d ◊

dh

Ê A1 ˆ ÁË1 + A ˜¯ 2 f ◊l d 2g

1.5 +

A1 A2 T= a ( A1 + A2 )

1.5 +

2 A1 A2 = a ( A1 + A2 ) where,

h -1/2

857

2g

Ú

H2

H1

h–1/2 dh

fl d ◊ (H 1/2 – H 1/2) 1 2

H1 LQLWLDOGL൵HUHQFHRIOHYHOVEHWZHHQWZRUHVHUYRLUV H2 GL൵HUHQFHRIOHYHOVDIWHUWLPHT.

 ‡ A reservoir 930 m2LV¿OOHGZLWKZDWHUWRDGHSWKRIP$FPGLDPHWHUKRUL]RQWDO SLSHPORQJLV¿WWHGDVLWVERWWRP+RZORQJZLOOLWWDNHWRORZHUWKHOHYHORIZDWHU LQWKHUHVHUYRLUIURPWRPWKURXJKWKHSLSH"$VVXPHf     'HOKL8QLYHUVLW\ 2A T= a

=

f ◊l d (H11/2 – H21/2) 2g

1.5 +

2 ¥ 930

0.03 ¥ 450 0.15 (151/2 – 121/2) 2 ¥ 9.81

1.5 +

p ¥ (0.15) 2 4

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858

Fundamentals of Fluid Mechanics

A A

¥

¥

  P 

 )LQDOOHYHOGL൵HUHQFHEHWZHHQWKHUHVHUYRLUV H  ± ± P   1RZWLPHUHTXLUHGWRORZHUWKHOHYHOIRUHWRH

T=

 A A ◊ A + A ◊ a

f ◊L d ◊ H – H g

 +

 ¥  ¥  = p  +  ¥ ¥   4     

 ¥   ¥ ±  ¥ 

 +

 VHFV  +UPLQVHF

 ‡ :KDW LV D EUDQFKHG SLSH" :KDW DUH HTXDWLRQV DYDLODEOH WR VROYH SUREOHPV DERXW D branched pipe?   :KHQWKUHHRUPRUHUHVHUYRLUVDUHFRQQHFWHGE\PHDQVRISLSHVZKLFKDUHPHHWLQJDWRQH MXQFWLRQWKHQWKHSLSHOLQHLVFDOOHGDEUDQFKHGSLSHDVVKRZQEHORZ

B

A

ZB

ZA D ZD

C

ZC Datum line



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859

 )ROORZLQJIRXUHTXDWLRQVFDQEHREWDLQHG   %HUQRXOOL¶VHTXDWLRQEHWZHHQADQGD ZA = ZD +

PD  hf AD rg

  %HUQRXOOL¶VHTXDWLRQEHWZHHQBDQGD ZB = ZD +

PD  hf BD rg

  %HUQRXOOL¶VHTXDWLRQEHWZHHQDDQGC ZD +

PD = ZC hf DC rg

  &RQWLQXLW\HTXDWLRQDWMXQFWLRQD QAD + QBD = QDC   ,WLVSRVVLEOHWRVROYHWKHSUREOHPVDERXWEUDQFKHGSLSHZLWKWKHDERYHIRXUHTXDWLRQV  ‡ 7KUHHUHVHUYRLUVA, B, and CDUHFRQQHFWHGE\DSLSHV\VWHPDVVKRZQEHORZ)LQGWKH GLVFKDUJH IURP RU LQWR UHVHUYRLU B and C LI WKH UDWLR RI IORZ IURP UHVHUYRLU A is 60 OLWUHVVHF)LQGWKHKHLJKWRIOHYHOLQWKHUHVHUYRLUC. Take f IRUSLSHV\VWHP l1 = 1200 m d1 = 30 cm

l2 = 600 m d2 = 20 cm

A QA = 60 litres/s B

40 m

D 38 m l3 = 800 m d3 = 30 cm

Datum line

  $SSO\%HUQRXOOL¶VHTXDWLRQEHWZHHQADQGDZHJHW ZA = ZD +

PD  hf AD rg

C

860

Fundamentals of Fluid Mechanics

40 = ZD +

ZD +  

f LQA PD +  ¥ d  rg

 ¥  ¥   PD = 40 – rg  ¥    P

  1RZDSSO\%HUQRXOOL¶VHTXDWLRQEHWZHHQBDQGDZHJHW Ê P ˆ f L QB ZB = Á Z D + D ˜ + rg ¯ Ë  ¥ d    

 

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QB =

 RU



QB ¥ ±PV

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   %XW       \

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QC = QA + QB    PV



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= ZC  RU



ZC ± P

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Flow Through Pipes and Compressibility Effects

l = 1170 m

d = 30 cm l = 2340

72 m

Case 2

Case 1

Case  

V f ◊ L˘ È ¥ Í + d ˙˚ g Î

 ¥  ¥    ¥  = V =  +   

 

  V ªPV  Q = AV = p ¥  ¥ PV 4



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hf =

 

 

=

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Q =

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Ê Q ˆ Á  ˜ Ë ¯



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¥

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861

862

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Q – Q ±  

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=  

Q - Q Q  ¥ 

 

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863

 'XULQJVXGGHQFORVLQJRIWKHYDOYHWKHSUHVVXUHULVH dP = rU

K r

= U Kr  ¥  ¥  ¥ 

 

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f ◊ L ◊ q  ¥ d 

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f ◊ L ◊ Q  ¥  

 /RVVRIKHDGIRUWZRSDUDOOHOSLSHVKDYLQJGLDPHWHU dDQGÀRZ  hf  =

f ◊ L ◊ Q  >7KHORVVLQRQHSLSHLVWREHWDNHQ@  ¥ d  ¥ 

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864

Fundamentals of Fluid Mechanics

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PA = 10.3 rg

A

15 m B

hf 

 RU



V =

f ◊ L ◊V   ¥  ¥ V  = g ◊ d  ¥  ¥ 

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V PV 

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    g

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865

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‡ 'HULYH H[SUHVVLRQV IRU WKH ULVH RI SUHVVXUH ZKHQ YDOYH LV   JUDGXDOO\ FORVHG DQG (2) suddenly closed in rigid pipe. What is pressure rise in elastic pipe during sudden FORVLQJRIYDOYH Case3UHVVXUHULVHVGXHWRGHFHOHUDWLRQRIÀXLG+HQFHLIdPLVSUHVVXUHULVHWKHQ dP ¥DUHD PDVV¥GHFHOHUDWLRQ dP =

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dP = r◊u

Case,QHODVWLFSLSHVWKHSUHVVXUHULVHGXULQJVXGGHQFORVLQJRIDYDOYH r

P=U

dˆ Ê ÁË + ˜¯ K Et

‡ 7KUHH LGHQWLFDO SLSHV RI OHQJWK l GLDPHWHU d and friction factor f are connected in SDUDOOHOEHWZHHQWZRUHVHUYRLUV:KDWLVWKHVL]HRIDSLSHRIOHQJWKlDQGRIWKHVDPH friction factor fHTXLYDOHQWWRWKHDERYHSLSHV"  D  d E  d  F  d G  d (IES 2009) Q/3 Q/3

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Q = AV

RU

Q V = pd  4

\

hf =

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=

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L ? L

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=

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\

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\

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1 h2

2

(a)

2 gh

(b)

2 gh2

(c)

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V =

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VP = V

\

VP =

 g h - h

 g h - h

2SWLRQ F LVFRUUHFW  ‡ )RUPD[LPXPWUDQVPLVVLRQRISRZHUWKURXJKDSLSHOLQHZLWKWRWDOKHDGH, the head lost due to friction hfLVJLYHQE\ 

D  ¥ H

(b)

H 

H 2

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(,(6 V hf H



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hf =

\

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Ê f LV  ˆ = r ◊ g ◊ A ◊ Á H ◊V  gd ˜¯ Ë )RUPD[LPXPSRZHU

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RU

∂P =0 ∂V ∂P = 0 = rgA ∂V H=

Ê  f LV  ˆ H Á  gd ˜¯ Ë

 f LV   gd

 hf RU

hf =

H 

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V μ

Q Q

f LV   FRQVWDQW  gd d

pd ¥ d AV =   = 4 A V pd  ¥ d 4 =

d d 

Ê  ˆ = Á ˜ Ë  ¯



 

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D  N3D (c) 0.50 kPa  7KHVKHDUVWUHVV

E  N3D (d) 25 kPa t= -

,(6

r Ê ∂P ˆ  ÁË ∂x ˜¯

r = R PP ¥ ±P

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 ∂P = -  ¥   ±¥± ∂x 

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t=

\

=

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2 5

4

3

Pipe Label 1

Velocity

Area

(1)

5 cm/s

4 sq. m

(2)

6 cm/s

5 sq. m

(3)

V3 cm/s

2 sq. m

(4)

4 cm/s

10 sq. m

(5)

V5 cm/s

8 sq. m

  7KHYHORFLW\V5 would be  D  FPV   F  FPV

E  FPV G  FPV Q + Q = Q4

RU

AV + AV = A4V4

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4 ¥¥ V = 4 ¥  V =  PV 8   2SWLRQ D LVFRUUHFW  ‡ $SLSHLVFRQQHFWHGLQVHULHVWRDQRWKHUSLSHZKRVHGLDPHWHULVWZLFHDQGOHQJWKLV WLPHVWKDWRIWKH¿UVWSLSH7KHUDWLRRIIULFWLRQDOKHDGORVVHVIRUWKH¿UVWSLSHWRWKRVH IRUWKHVHFRQGSLSH ERWKWKHSLSHVKDYHWKHVDPHIULFWLRQDOFRQVWDQW LV (a) 8 (b) 4  F   G   ,(6 f ◊ L ◊ Q hf =  d  f ◊ L ◊ Q  d 5 =   f ◊  L ◊ Q   ¥ d 

hf  hf 

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D  PV F  PV

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(a) 5.25L F  

(b) 9.5L

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G  

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hfe = hf  + hf  + hf  f ◊ Le ◊ Q   ¥ D Le

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=

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=

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+

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=

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Q PV

RU

Q ¥  PV  ‡ 7KHWRWDOIORZLQWKUHHSDUDOOHOSLSHV\VWHPVLVPV7KHSLSHGLDPHWHUVDQGOHQJWKV of three pipes are: d PL Pd2 PL2 Pd PL P)LQGWKHGLVFKDUJHLQWKHWKUHHSLSHV 8378 Q1 Q

Q2 Q3



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=

f ◊ L ◊ Q  ¥ d 

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=

=

f ◊ L ◊ Q  ¥ d

800 ¥ Q  

=

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RUQ Q Q RU

Q Q Q

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Q = Q + Q  + Q  4 = QQQ = Q  4  PV 

\

Q =

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)ORZ V =

\





p2 = 60 kPa

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 7RWDOKHDGDWVHFWLRQ =

r rg

+

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p  p d = ¥ P 4 4

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V=

Q  =  PV A 

hf =

f ◊ L ◊V   WDNHf  ◊ g ◊d =







 Z  – Z =

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(c)

H 4

H =

 ◊ f ◊ LQ   gp  D 

H =

 ◊ f ◊ L ◊ Q  H  = H=  g ◊ p  ◊  D  8 

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(b)

H 2

(d)

H  8

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(a)

f =

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f =

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t 0

(d) f =

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t0 rV   rV   t0

,$6

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r ◊V   ◊t0

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L

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(d)

2◊L

2 L  4

,$6

 )RUVLQJOHSLSH hs =

 ◊ f ◊ Ls Q   g p  D



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L

Q LQHDFKSLSH 



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hs L =  ¥ s = hP L   RU

= Ls =

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1 2 2

P1 = 3.5 kg/cm

2

P2 = 3.4 kg/cm

V1 = 0.6 m/s Water

d2 = 5 cm d1 = 10 cm

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V

PV

E – E =

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,$6,(6

4 ◊ f ◊ L ◊V  RUh μ V  D◊◊ g

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 ◊ f ◊ L ◊ Q   g p  D

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RU

  f

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- df dQ ¥ =  f Q

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= -

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= 

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A P

2 (b) P A

(d)

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A 4m

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sin 2q ˆ (b) R 2 ÊÁ q ˜ Ë 2 ¯

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Q Q

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Chapter

15

IDEAL FLUID FLOW

KEYWORDS AND TOPICS      

IDEAL FLUID FLOW UNIFORM FLOW SOURCE FLOW SINK FLOW VORTEX FLOW DOUBLET FLOW

 SUPERPOSITION OF FLOWS

     

STREAM FUNCTION EQUIPOTENTIAL FUNCTION SOURCE STRENGTH HALF BODY PROFILE DIVIDING STREAM LINE MAGNUS EFFECT

 TORNADO

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Ideal Fluid Flow

ur = RU

k=

897

q k = ZKHUHq GLVFKDUJHSHUXQLWOHQJWKDQGk VRXUFHVWUHQJWK p r r q p

Source strength (k) 7KHVRXUFHVWUHQJWKLVDPHDVXUHRIWKHUDGLDOYHORFLW\DWXQLWUDGLXV ,WLVGH¿QHGDVYROXPHÀRZSHUXQLWGHSWK ÊÁ k = q ˆ˜  Ë p ¯ ur =

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k r

ur =

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ur =

q p r

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 RU



 %XW



∂y =

q ∂q p

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\

 7KHVWUHDPIXQFWLRQLVDIXQFWLRQRIDQJOHq y Streamlines y1

y2

y3

y4

q

O

x

y5

y6

y7

y8

Streamlines in Source Flow

898

Fundamentals of Fluid Mechanics

(TXLSRWHQWLDOIXQFWLRQRIVRXUFHÀRZ

\  RU



ur =

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ur =

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O f=C

Equipotential Lines in the Source Flow

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P - P0 u =– r rg g \

P – P0 =

- rur 

Ideal Fluid Flow

 1RZ

ur =



q p r

P – P0 = –

\

899

=–

rÊ q ˆ Á ˜  Ë p r ¯



r q 8p  r 

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Radial

(P – P0)

velocity = ur P0 at r = 1 O Source

r

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 ZKHUHk =

-q = kr p r

-q  VLQNVWUHQJWK p

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- r q - rk = 8p  r  r 

900

Fundamentals of Fluid Mechanics

y=C

f=C

A Sink Flow

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uq =

G C = p r r

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G p

uq =

∂y  ∂f = ∂r r ∂q

ur =

 ∂y ∂f = =0 r ∂q ∂r

- ∂y C = uq = ∂r r

\

y = – CORJr =

 RU  DQG

C=



 ∂f C = uq = r ∂q r

-G ORJr p

Ideal Fluid Flow

 RU

f = Cq =



901

G q p

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Equipotential lines

Free Vortex Flow Pattern

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- r uq - r c =  r 

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Fundamentals of Fluid Mechanics Velocity

uq

P0

Pressure

(P – P0)

902

r

Centre

Velocity and Pressure Variation in a Free Vortex

 ‡ :KDWDUHVXSHULPSRVHGIORZSDWWHUQV":K\DUHWKH\UHTXLUHG" 

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6WUHDPIXQFWLRQy = k q – q (TXLSRWHQWLDOIXQFWLRQf = kORJr – kORJr = kORJ

r r

Ideal Fluid Flow

903

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q p

q q ¥ ds Æ m DV OLPLW ds Æ  ZKHUH k = = 6WUHQJWK RI VRXUFHVLQN ds  GLVWDQFH p p EeWZHHQVRXUFH VLQNDQGm VWUHQJWKRIGRXEOHW  

6WUHDPIXQFWLRQy =

 

(TXLSRWHQWLDOIXQFWLRQf =

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Doublet

15.8 SUPERPOSITION OF FLOW  ‡ :KDWLVWKHIORZSDWWHUQREWDLQHGRQWKHVXSHUSRVLWLRQRIDXQLIRUPÀRZDQGDVRXUFH ÀRZ"   :KHQDVRXUFHRIVWUHQJWKkLVSODFHGDWWKHRULJLQO DWxD[LVLQWKHXQLIRUPÀRZZKLFK H[WHQGVWRLQ¿QLWHGLVWDQFHRQERWKWKHVLGHVRIWKHxD[LV7KHVWDJQDWLRQRFFXUVDWSRLQW SZKLFKLVNQRZQDVVWDJQDWLRQSRLQWIRUWKHFHQWUDOVWUHDPOLQHRIWKHXQLIRUPÀRZDORQJ xD[LV7KHRWKHUVWUHDPOLQHVRIXQLIRUPÀRZDUHGHÀHFWHGIURPxD[LV7KHÀRZSDWWHUQLV VLPLODUWRWKHÀRZREWDLQHGZKHQXQLIRUPÀRZHQFRXQWHUVDVROLGERG\7KHÀRZSUR¿OH VRIRUPHGLVNQRZQDVKDOIERG\6XFKW\SHRIÀRZSDWWHUQLVREWDLQHGZKHQZDWHUÀRZV DURXQGDURXQGQRVHGEULGJHSLHU

904

Fundamentals of Fluid Mechanics

u

U =

+

Uniform

S

O

Source

Uniform Flow and a Source

  9HORFLW\FRPSRQHQWVDQ\ZKHUHLQWKHÀRZDUHJLYHQE\ ur = UFRVq +

q p r

uq = –UVLQq   7KHVWDJQDWLRQSRLQWSLVJLYHQE\

Ê -q ˆ  ˜ ÁË pU ¯  ‡ :KDWLVWKHIORZSDWWHUQREWDLQHGE\WKHVXSHUSRVLWLRQRIDVRXUFHDQGVLQNSDLULQD uniform flow? y

U

U

y=0

S1

S2

x

a Source and Sink Pair in Uniform Flow



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Ideal Fluid Flow

905

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U

U

Doublet in Uniform Flow

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+

= S1

Uniform Flow Around Cylinder

Circulation

S2

Magnus Effect

Flow Around a Cylinder

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906

Fundamentals of Fluid Mechanics

Vortex and Sink Flows

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y=3 y=2 y=1

0

y=0

x

‡ 3ORWWKHVWUHDPOLQHVIRUXQLIRUPIORZKDYLQJYHORFLW\RIPVSDUDOOHOWRSRVLWLYHyD[LV y = –Ux y = –x DVU PV x  y = 0 x  y ± x  y ± x  y ± y = –1 y = –2 y = –3

y

x 0

1

2

3

Ideal Fluid Flow

907

 ‡ ,IDVRXUFHKDVVRXUFHVWUHQJWKRIPV¿QGWKHYHORFLW\RIIORZDWr DQGP 6WUHQJWKRIVRXUFH q PV

  

ur =

q  = p r p r

 ZKHQ



r P ur =

  PV 4p

 ZKHQ



r P ur =

  PV p

‡ $ VRXUFH IORZ KDV VWUHQJWK RI  PV 7KH SUHVVXUH DW GLVWDQFH r   P LV  N1P)LQG  SUHVVXUHDWr PDQG  VWUHDPIXQFWLRQDWDQJOHƒ u r =

q p r

=

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=

  PV  ¥ p

ur =

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ur ur P P + = +  g rg rg g  ¥  P     + = + rg  ¥  ¥   ¥   ¥   RU



P =¥ +

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 N1

  

6WUHDPIXQFWLRQy =

q ¥q p

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Fundamentals of Fluid Mechanics

=

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=

 

 PV

15.10 QUESTIONS FROM COMPETITIVE EXAMINATIONS  ‡ (OHPHQWDU\IORZSDWWHUQLVSRVVLEOHLQLGHDOIOXLGIORZE\   8QLIRUPÀRZ  6RXUFHRUVLQNÀRZ   9RUWH[  'RXEOHW The correct answer is  D   E    F   G     2SWLRQ G LVFRUUHFW  ‡ 6XSHUSRVLWLRQLVSRVVLEOHLQLGHDOIOXLGIORZE\FRPELQLQJ   6RXUFHDQGVLQN   8QLIRUPÀRZDQGVLQN   'RXEOHWDQGXQLIRUPÀRZDQGYRUWH[ The correct answer is  D  DQG E  DQG  F  DQG G  DQG   2SWLRQ D LVFRUUHFW  ‡ $GRXEOHWLVREWDLQHGE\VXSHUSRVLWLRQRIIROORZLQJHOHPHQWDU\IORZV   6RXUFHDQGVLQNRIHTXDOVWUHQJWK   6RXUFHDQGYRUWH[   6LQNDQGYRUWH[   8QLIRUPIORZDQGYRUWH[ The correct answer is  D  RQO\ E  RQO\  F  RQO\ G  RQO\   2SWLRQ F LVFRUUHFW  ‡ 7KHIORZLQZKLFKWKHYHORFLW\YHFWRULVLGHQWLFDOLQPDJQLWXGHDQGGLUHFWLRQDWHYHU\ SRLQWIRUDQ\JLYHQLQVWDQWLVNQRZQDV  D  2QHGLPHQVLRQDOIORZ E  6WHDG\IORZ  F  8QLIRUPIORZ G  6WUHDPOLQHIORZ   2SWLRQ F LVFRUUHFW

Ideal Fluid Flow

                       

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909

Chapter

16

FLOW PAST SUBMERGED BODIES

KEYWORDS AND TOPICS          

SUBMERGED BODIES DRAG LIFT DRAG COEFFICIENT LIFT COEFFICIENT SKIN FRICTION DRAG PRESSURE DRAG WAVE DRAG INDUCED DRAG COMPRESSIBILITY DRAG

         

STREAMLINED BODY BLUFF BODY STOKES’ LAW MAGNUS EFFECT KUTTA JOUKOWSKY EQUATION KARMAM VORTEX TRAIL AEROFOIL CHORD ANGLE OF ATTACK SPAN

16.1 INTRODUCTION 7KHUHVLVWDQFHGXHWRDERG\PRYLQJWKURXJKDODUJHPDVVRIVWDWLRQDU\ÀXLGRUUHVLVWDQFHGXHWR DÀXLGÀRZLQJDURXQGDVWDWLRQDU\VXEPHUJHGERG\KDVJUHDWWHFKQLFDOLPSRUWDQFHLQWKH¿HOGRI K\GURG\QDPLFVDQGDHURG\QDPLFV$HURSODQHVDQGDXWRPRELOHVDUHERGLHVZKLFKPRYHLQDOPRVW VWDWLRQDU\DLU6LPLODUO\VXEPDULQHVDQGVKLSVDUHERGLHVZKLFKPRYHLQDOPRVWVWDWLRQDU\ZDWHU +RZHYHUEXLOGLQJVDQGFKLPQH\VDUHVWDWLRQDU\ERGLHVZKLFKDUHVXEMHFWHGWRZLQG7KHIRUFH H[HUWHGRQWKHERG\RUÀXLGUHVXOWVIURPWKHUHODWLYHPRWLRQEHWZHHQWKHLPPHUVHGERG\DQGLWV VXUURXQGLQJÀXLG7KHIRUFHH[HUWHGE\WKHÀXLGRQWKHPRYLQJERG\FDQEHPDGHWRLQFOLQHWR WKHGLUHFWLRQRIPRWLRQ+HQFHWKLVIRUFHKDVDFRPSRQHQWLQWKHGLUHFWLRQRIPRWLRQDVZHOODV DQRWKHUFRPSRQHQWSHUSHQGLFXODUWRWKHGLUHFWLRQRIPRWLRQ7KHFRPSRQHQWRIWKHIRUFHLQWKH GLUHFWLRQRIPRWLRQLVFDOOHGWKHGUDJIRUFHDQGWKHFRPSRQHQWSHUSHQGLFXODUWRWKHGLUHFWLRQRI motion is called lift force.

16.2 SUBMERGED BODIES  ‡ What happens when bodies move through static fluid or when static bodies are subjected to fluid flow?

Flow Past Submerged Bodies



911

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16.3 DRAG AND LIFT FORCE  ‡ 'H¿QHGUDJIRUFHDQGOLIWIRUFHDFWLQJRQDERG\VXEPHUJHGLQIORZLQJIOXLG   :KHQ D ERG\ LV SODFHG LQ WKH XQLIRUP ÀRZ DV VKRZQ LQ ¿JXUH EHORZ WKHQ WKHUH DUH WZR IRUFHVDFWLQJRQWKHERG\YL]  SUHVVXUHIRUFHDFWLQJQRUPDOWRWKHVXUIDFHRIWKHERG\ DQG  VKHDUIRUFHDFWLQJDORQJWKHVXUIDFHRIWKHERG\7KHSUHVVXUHIRUFHDQGVKHDUIRUFH FDQEHFRPELQHGWRJLYHWRWDOIRUFHH[HUWHGE\WKHÀXLGRQWKHERG\ZKLFKFDQEHUHVROYHG LQWKHGLUHFWLRQSDUDOOHOWRWKHPRWLRQRIWKHÀXLGDQGSHUSHQGLFXODUWRWKHGLUHFWLRQRIWKH PRWLRQRIWKHÀXLG Left

Pressure force Uniform flow (U)

Total force exerted by fluid (F)

(FL)

90°

q q

Drag (FD)

Shear force

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 rUA 

= F cos q 

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 rU  A = F sin q 

912

Fundamentals of Fluid Mechanics

16.4 FRICTION PRESSURE AND SHEAR DRAG  ‡ 'HULYHH[SUHVVLRQIRU  IULFWLRQGUDJ  SUHVVXUHGUDJ  WRWDOGUDJDQG  OLIW Or   'L൵HUHQWLDWHEHWZHHQSUHVVXUHGUDJDQGVKHDUGUDJ 8378 FL P. dA P. dA

P. dA. cos q

U q

to

P. dA. sin q

dA

FD to dA

to dA sin q q to dA cos q



 &RQVLGHU D ERG\ LV KHOG VWDWLRQDU\ LQ WKH XQLIRUP ÀXLG ÀRZ KDYLQJ YHORFLW\ RI U ,I ZH consider a small area dARQWKHVXUIDFHWKHQIRUFHVDFWLQJRQWKLVDUHDDUH  SUHVVXUHIRUFH P◊dA DFWLQJ LQ WKH GLUHFWLRQ SHUSHQGLFXODU WR WKH VXUIDFH DQG   VKHDU IRUFH t ◊dA acting along the surface. Friction Drag. 7KLV LV DOVR FDOOHG VNLQ GUDJ )ULFWLRQ GUDJ LV WKH VXP RI FRPSRQHQWV RI VKHDUIRUFHVLQWKHGLUHFWLRQRIÀXLGÀRZDFWLQJRQWKHFRPSOHWHVXUIDFHRIWKHERG\ FDF =

Ú

A

t ◊dA ◊cos q

Pressure Drag. 7KHVXPRIFRPSRQHQWVRISUHVVXUHIRUFHVLQWKHGLUHFWLRQRIWKHÀXLGÀRZ IRUWKHFRPSOHWHVXUIDFHRIWKHERG\LVFDOOHGSUHVVXUHGUDJ,WLVDOVRFDOOHGIRUPGUDJ FDP =

Ú

A

P◊dA ◊sin q

Total Drag. 7KHVXPRIIULFWLRQGUDJDQGSUHVVXUHGUDJRQDERG\LVFDOOHGWRWDOGUDJ FD = FDF + FDP Lift. 7KHOLIWRQDERG\SODFHGLQÀXLGÀRZLVJLYHQE\WKHVXPPDWLRQRIWKHFRPSRQHQWV RISUHVVXUHIRUFHVDQGVKHDUIRUFHVDFWLQJRQWKHERG\LQWKHGLUHFWLRQSHUSHQGLFXODUWRWKH ÀXLGÀRZ FL =

Ú

A

P ◊dA ◊cos q +

Ú t ◊ dA◊sin q

 ‡ +RZGRHVV\PPHWU\RIWKHERG\D൵HFWGUDJDQGOLIW"   :KHQDERG\KDVDQD[LVRIV\PPHWU\DQGIUHHVWHDPDSSURDFKHVWKHERG\DORQJWKLVD[LV WKHIRUFHWKDWDFWVRQWKHERG\LVZKROO\GUDJIRUFHDFWLQJDORQJWKHVWUHDP Total Force, F = FD + FL While When

FD = F cos q & FL = F sin q q = 0, FL = 0 and F = FD

Flow Past Submerged Bodies



 

 7KH SURGXFWLRQ RI OLIW UHTXLUHV DV\PPHWU\ RI ÀRZ DERXW WKH GLUHFWLRQ RI WKH IUHH VWUHDP +HQFH ZKHQ q π  WKHQ ERWK GUDJ FD = F cos q  DQG OLIW FL = F sin q  DUH SURGXFHG under all circumstances. It is clear from the above.   ,WLVSRVVLEOHWRFUHDWHGUDJZLWKRXWOLIWE\NHHSLQJq = 0.   ,W LV LPSRVVLEOH WR FUHDWH OLIW ZLWKRXW GUDJ ,I q π 0, both F cos q and F sin q have GH¿QLWHYDOXHV Total force (F)

Lift (L)

Drag

q

q

Drag (D)

Lift & drag (q π 0)

Drag only (q = 0)



913

 $SODWHNHSWLQWKHÀRZSHUSHQGLFXODUWRGLUHFWLRQRIÀRZZLOOKDYHGUDJRQO\DVVKRZQLQ WKH¿JXUH7RWDOIRUFHDFWLQJRQDQDHURSODQHLVFZKLFKKDVFRPSRQHQWVDORQJDQGQRUPDO WRWKHVXUIDFHRIWKHERG\ZKLFKDUHGUDJD and lift LUHVSHFWLYHO\

16.5 COEFFICIENT OF DRAG AND LIFT  ‡ 'H¿QHFRH൶FLHQWRIGUDJDQGFRH൶FLHQWRIOLIW2QZKDWIDFWRUVZLOOWKHVHFRH൶FLHQWV GHSHQG" 8378   7KHFRH൶FLHQWVDUHGH¿QHGDVWKHUDWLRVRIFRUUHVSRQGLQJIRUFHVWRG\QDPLFIRUFHVRQWKH SURMHFWHGDUHDV   7KHFRH൶FLHQWRIGUDJ CD LVGH¿QHGDVWKHUDWLRRIGUDJWRG\QDPLFIRUFHRQWKHSURMHFWHG area. CD =

FD  rU  A 

ZKHUH FD = drag, U = free stream velocity, and A SURMHFWHGDUHD 

 7KHFRH൶FLHQWRIOLIW CL LVGH¿QHGDVWKHUDWLRRIOLIWWRWKHG\QDPLFIRUFHRQWKHSURMHFWHG area. CL =

FL  rU  A 

 7KHUDWLRRIFRH൶FLHQWRIGUDJWRWKHFRH൶FLHQWRIOLIWLVVDPHDVWKDWRIWKHUDWLRRIWKHOLIW to the drag.

914

Fundamentals of Fluid Mechanics

CD FL = CL FD  7KHFRH൶FLHQWVDUHIXQFWLRQVRI5H\QROGVQXPEHUDQGFKDUDFWHULVWLFDUHDRIWKHERG\ A   ‡ )LQGWKHSURMHFWHGDUHDVRI  FLUFXODUF\OLQGHU  VSKHUH  DHURIRLODQG  IODW SODWH Case &LUFXODUF\OLQGHUZLWKGLDPHWHUd and length l. l U

d d

3URMHFWHGDUHD d ◊l

   Case6SKHUHZLWKGLDPHWHUd

3URMHFWed area =

  

pd  4

U

Case 3. Aerofoil of chord c VSDQs Spa

n=

U

S

c

3URMHFWHGDUHD s ¥ c

  

Case 4. Plate having length lDQGZLGWKb b

b U

  

l

l

3URMHFWHGDUH l ¥ b

Flow Past Submerged Bodies

915

 ‡ ([SODLQWKHGL൵HUHQWUHJLRQVRIDIOXLGIORZSDVWDERG\&ODVVLI\GUDJRQWKHEDVLVRI GL൵HUHQWRULJLQV

a

b

c

d

Flow Past a Bridge Pier: Different Regions of Fluid Flow

 

     

 &RQVLGHUZDWHUÀRZLQJSDVWDEULGJHSLHUDVVKRZQLQWKH¿JXUH7KHÀRZFDQEHGLYLGHG LQWRIRXUUHJLRQVWRXQGHUVWDQGWKHÀRZ   Region ‘a¶7KHUHLVGLYHUJHQFHRIVWUHDPOLQHVZKLFKOHDGVWRGHFUHDVHLQYHORFLW\RI WKHÀRZ2FFXUUHQFHRIDVWDJQDWLRQSRLQWXSVWUHDPRIWKHERG\ZKLFKGLYLGHVWKHÀRZ LQWRWZRVWUHDPV   Region ‘b¶%RXQGDU\ODPLQDUÀRZFRQVLVWLQJRIODPLQDUDQGWXUEXOHQWÀRZLVWDNLQJ SODFH   Region ‘c¶7KHUHLVDVHSDUDWLRQRIERXQGDU\OD\HUGXHWRSRVLWLYHSUHVVXUHJUDGLHQW resulting in surface discontinuity and formation of vortices.   Region ‘d¶7KHUHLVVKHGGLQJRIYRUWLFHVWRFRQVWLWXWHWKHZDNH  'UDJFDQEHFODVVL¿HGRQWKHEDVLVRIGL൵HUHQWRULJLQVDVGHVFULEHGEHORZ  Skin friction drag. This drag results from the formation of the boundary layer and VKHDUVWUHVVHVDWWKHVXUIDFHRIWKHERG\LQWKHÀXLGÀRZLQWKHUHJLRQµb’.  Form drag )RUPDWLRQ RI HGGLHV DQG ZDNH LQ WKH UHJLRQ d  UHVXOWV LQWR UHVLVWDQFH ZKLFKLVFDOOHGIRUPGUDJ7KHPDJQLWXGHRIIRUPGUDJPD\IDUH[FHHGWKHVNLQIULFWLRQ GUDJDQGLWGHSHQGVRQWKHIRUPRUVKDSHRIWKHERG\ 3. Wave drag. :KHQ D ERG\ LV SDUWO\ VXEPHUJHG LQ D OLTXLG WKH ÀRZ LQGXFHV WKH IRUPDWLRQRIJUDYLWDWLRQDOZDYHVUHVXOWLQJLQWRZDYHGUDJ:KHQDVXEPHUJHGERG\ WUDYHOV DW WUDQVRQLF RU VXSHUVRQLF VSHHGV WKH FRPSUHVVLELOLW\ H൵HFW DOVR JLYHV ULVH WR ZDYHGUDJ 4. Induced drag. 7KLVLVWKHGUDJUHVXOWLQJGXHWRWKUHHGLPHQVLRQDOQDWXUHRIÀXLGÀRZ DQG¿QLWHVL]HRIWKHERG\,WLVPDLQO\LQGXFHGE\WKHLQGXFHGFRPSRQHQWRIYHORFLW\ DQGYRUWLFLW\GLVWULEXWLRQDORQJWKHVSDQRIWKHERG\,WLVOLIWGHSHQGHQW 5. Compressibility drag. ,WDULVHVGXHWRFKDQJHVLQGHQVLWLHVDVWKHÀRZSURFHHGV 6. Gravitational wave drag. ,W LV H[SHULHQFHG E\ D ERG\ ZKHQ LW LV WRZHG LQ D IUHH VXUIDFHÀRZRUZKHQDERG\LVSODFHGDWLQWHUIDFHRIDLUDQGOLTXLG,QVXFKVLWXDWLRQV WKHJUDYLWDWLRQDOIRUFHVDOPRVWKDYHVDPHPDJQLWXGHDVLQHUWLDIRUFHVUHVXOWLQJLQÀRZ SDWWHUQFRQVLGHUDEO\LQÀXHQFHGE\WKHIRUPDWLRQRIJUDYLWDWLRQDOZDYHV

916

Fundamentals of Fluid Mechanics

 ‡ 'L൵HUHQWLDWHEHWZHHQVWUHDPOLQHERG\DQGEOX൵ERG\ Streamline body

8378 %OXႇERG\

1.

Its surface coincides with the streamlines RIWKHÀRZ

1.

Its surface does not coincide with the VWUHDPOLQHVRIWKHÀRZ

2.

7KHÀRZGRHVQRWVHSDUDWHXQWLOQHDU trailing edge of the body. Hence, wake formation zone at trailing edge is very small.

2.

$ YHU\ ZLGH ZDNH LV GHYHORSHG RQ WKH down-stream of the bluff body. The SUHVVXUHDWGRZQVWUHDPIDOOVUHVXOWLQJLQ ODUJH SUHVVXUH GLႇHUHQFH IURP XSVWUHDP to downstream.

3.

3.

Wake

Wake

Aerofoil: stream line body (small wake)

Circular disc : (large wake) 4.

6PDOOSUHVVXUHGUDJUHVXOWVDQGGUDJ is mainly due to friction only.

4.

3UHVVXUHGUDJLVODUJHDVFRPSDUHG to friction drag.

5.

$HURSODQHVVXEPDULQHVDQGVSDFHVKLS are streamline bodies.

5.

Tall buildings, chimneys and advertising ERDUGVDUHEOXႇERGLHV

16.6 STOKES’ LAW CONCERNING SKIN AND PRESSURE DRAG  ‡ ([SODLQ6WRNHV¶ODZFRQFHUQLQJVNLQDQGSUHVVXUHGUDJIRUFHVIRUDVSKHUHPRYLQJLQ IOXLG:KDWLVWKHYDOXHRIFRH൶FLHQWRIGUDJDFFRUGLQJWRWKH6WRNHV¶ODZ" 

8378  :KHQ WKH YHORFLW\ RI ÀRZ LV YHU\ VPDOO 5H\QROGV QXPEHU OHVV WKDQ   WKHQ YLVFRXV IRUFHVDUHPRUHSUHGRPLQDQWWKDQLQHUWLDIRUFHV6WRNHVDQDO\VHGWKHÀRZDURXQGDVSKHUH DQGJDYHDUHODWLRQIRUGUDJDFWLQJRQWKHVSKHUHZKLOHLWLVPRYLQJZLWKFRQVWDQWYHORFLW\ WHUPLQDOYHORFLW\ LQWKHÀXLG7KHWRWDOGUDJ ZKHUHm = viscosity,

FD = 3pmDU D GLDPHWHURIVSKHUHDQGU = terminal velocity.

Stokes’ Law: 6WRNHVIXUWKHUGLYLGHGWKHWRWDOGUDJLQWZRSDUWVGHFLGLQJWZRWKLUGVGUDJ LVFRQWULEXWHGE\VNLQIULFWLRQDQGRQHWKLUGE\SUHVVXUH7KLVGLYLVLRQRIWKHWRWDOGUDJLV NQRZQDV6WRNHV¶ODZ+HQFHDFFRUGLQJWR6WRNHV¶ODZZHKDYH 

  6NLQIULFWLRQGUDJ FDF =   3UHVVXUHGUDJFDP =

 FD p mDU 3

 FD = pmDU 3

Flow Past Submerged Bodies

917

&RH৽FLHQWRIGUDJ CD   $VSHU6WRNHV¶ODZWRWDOGUDJ FD = 3pmDU

L

  $VSHUGH¿QLWLRQRIFRH൶FLHQWRIGUDJCD, the total drag, FD = CD ¥

 rU  ¥ A 

LL

pd  = CD ¥  rU ¥ 4   (TXDWLQJHTQV L DQG LL 3pmDU = CD ¥

CD =

\

=

p  rU ¥ ◊ D  4 

 m r ◊U ◊ D  Re

‡ :KDW DUH WKH YDOXHV RI FRH൶FLHQW RI GUDJ IRU VSKHUH LQ WKH IOXLG IORZ ZLWK KLJKHU 5H\QROGVQXPEHUV"  

 Reynolds numbers between 0.2 and 5  7KH6WRNHV¶ODZZDVLPSURYHGRYHURFHDQE\LQFUHDVLQJLQHUWLDIRUFHVLQWKHÀXLGÀRZV ZKLFKKDYH5H\QROGVQXPEHUVEHWZHHQDQG CD = 

 £ Re £  CD = 0.4

3.

 £ Re £  CD = 0.5

4.

Re > 4 CD 

 Re

3 ˘ È Í +  R ˙ e˚ Î

‡ :KDWLVWHUPLQDOYHORFLW\"   :KHQDERG\IDOOVIURPUHVWLQWRDÀXLGWKHERG\VWDUWVDFFHOHUDWLQJLQWKHÀXLGGXHWRWKH gravitational force. The drag on the body increases as its velocity increases. At one stage, WKHVXPRIGUDJDQGEXR\DQWIRUFHDFWLQJXSZDUGVRQWKHERG\EHFRPHVHTXDOWRWKHZHLJKW

918

Fundamentals of Fluid Mechanics

RIWKHERG\DFWLQJGRZQZDUGVDQGWKHQHWIRUFHDFWLQJRQWKHERG\EHFRPHV]HUR7KHERG\ LQ WKLV FRQGLWLRQ GRHV QRW DFFHOHUDWH DQ\ PRUH ZKLOH PRYLQJ GRZQ LQ WKH ÀXLG ,W VWDUWV PRYLQJGRZQZLWKDFRQVWDQWYHORFLW\ZKLFKLVFDOOHGWHUPLQDOYHORFLW\  ‡ :KDWDUHWKHYDOXHVRIFRH൶FLHQWRIGUDJIRUDORQJF\OLQGHUSODFHGLQIOXLGIORZZLWK YDU\LQJ5H\QROGVQXPEHU"   7KHÀRZDURXQGDORQJF\OLQGHULVVLPLODUWRWKHÀRZDURXQGDVSKHUH7KHSORWRICD versus R e IRU WKH F\OLQGHU ORRNV VRPHZKDW VLPLODU WR WKDW IRU D VSKHUH H[FHSW IRU WKH QXPHULFDO values. The value of CDLV    Reynolds number 7KHGUDJIRUFHLVSURSRUWLRQDOWRYHORFLW\RIÀRZLQVXFKÀRZV KHQFHFRH൶FLHQWRIGUDJLVLQYHUVHO\SURSRUWLRQWR5H\QROGVQXPEHU FD μ u and CD μ

   

 Re

   < R e < 7KHFRH൶FLHQWRIGUDJGHFUHDVHVZLWKLQFUHDVLQJ5H\QROGVQXPEHUCD reaches to a minimum value of 0.95 at Re     < R e < 3 ¥ 47KHFRH൶FLHQWRIGUDJLQFUHDVHVDQGLWDWWDLQVWKHPD[LPXPYDOXH RIDWRe = 3 ¥4.    ¥ 4 < R e < 3 ¥ 105.7KHFRH൶FLHQWRIGUDJGHFUHDVHVDQGLWEHFRPHVHTXDOWR at Re = 3 ¥5.    ¥ 5 < R e < 3 ¥ 6. CD increases and it attains the value of 0.7.  7KHFKDQJHRIWKHFRH൶FLHQWRIGUDJ CD ZLWKLQFUHDVLQJ5H\QROGVQXPEHULVGXHWR   ÀRZVHSDUDWLRQDQG  HGGLHVIRUPDWLRQ Vortex S1

S2

Re < 1 No separation

2 < Re Separation at S1 & S2

Re > 90 Vortex shed off

16.7 CIRCULATION IN FREE VORTEX  ‡ 'H¿QHFLUFXODWLRQ+RZLVLWGHWHUPLQHG"2QZKDWIDFWRUVZLOOLWGHSHQG":KDWLVWKH YDOXHRIFLUFXODWLRQLQIUHHYRUWH["   &LUFXODWLRQLVGH¿QHGDVOLQHLQWHJUDORIYHORFLW\DURXQGDFORVHGSDWK   &RQVLGHUDWZRGLPHQVLRQDOVWHDG\ÀXLGÀRZZLWKYHORFLW\uPDNLQJDQJOHq to an element dL DVVKRZQLQWKH¿JXUH7KHu cos qLVWKHYHORFLW\FRPSRQHQWWDQJHQWLDOO\WRWKHHOHPHQW dL. The circulation around the closed curve around the element dL,

Flow Past Submerged Bodies

dl

U

919

q

G=

³ U cos q dL

The circulation depends on (1) tangential velocity or vorticity, and (2) contour area in x and y plane. r

uq =

G 2pr

Circulation for a Free Vortex



 ,QFDVHRIIUHHYRUWH[ÀRZZHKDYHuq ¥ r = constant. Hence, the circulation, G=

³ U ◊ dL q

or

= Uq ¥ 2p r = 2p (Uq ¥ r) G = Constant

and

uq =

G 2p r

 7KLV VKRZV WKDW WKH YDOXH RI FLUFXODWLRQ IRU IUHH YRUWH[ LV LQGHSHQGHQW RI WKH VL]H RI WKH enclosed path and also circulation is constant.

16.8 MAGNUS EFFECT  ‡ 6WDWHPDJQXVH൵HFW



Or 6KRZ WKH IORZ SDWWHUQ ZKHQ D FLUFXODU F\OLQGHU LV VXEMHFWHG WR D FRPELQDWLRQ RI XQLIRUPIORZDQGDFLUFXODWRU\IORZ  0DJQXVKDGVKRZQWKDWLIDFLUFXODUF\OLQGHULVSODFHGLQDXQLIRUPÀRZDQGLWLVURWDWHG then a lift force or a transverse force is produced on the cylinder. The reason for this is WKDW WKH URWDWLRQ RI WKH F\OLQGHU GLVWXUEV WKH V\PPHWULFDO ÀRZ SDWWHUQ H[LVWLQJ GXULQJ WKH SUHVHQFHRIRQO\XQLIRUPÀRZUHVXOWLQJLQDQDV\PPHWULFDOÀRZ SDWWHUQRQWKHF\OLQGHU 7KH YHORFLW\ DQG SUHVVXUH GLVWULEXWLRQ RQ WKH F\OLQGHU EHFRPH DV\PPHWULFDO DW WKH XSSHU

920

Fundamentals of Fluid Mechanics

DQGWKHORZHUVLGHRIWKHF\OLQGHUDVVKRZQ2QWRSVLGHVWUHDPOLQHVDUHFORVHULQGLFDWLQJ KLJKHUYHORFLWLHVDQGKHQFHORZHUSUHVVXUH2QWKHORZHUVLGHWKHVWUHDPOLQHVDUHIXUWKHU DSDUWLQGLFDWLQJORZHUYHORFLWLHVDQGKLJKHUSUHVVXUH7KHF\OLQGHUH[SHULHQFHVDQXSZDUG IRUFHRUOLIW,WLVSRVVLEOHWRREWDLQOLIWIRUFHZKHQZHVLPSOLI\WKHDQDO\VLVE\DVVXPLQJ WKHÀXLGÀRZDVLGHDOÀXLGÀRZ:HZLOOJHWWKHVDPHÀRZSDWWHUQZKHQDF\OLQGHULVNHSW VWDWLRQDU\DQGLWLVVXEMHFWHGWRDFRPELQDWLRQRIXQLIRUPDQGYRUWH[ÀRZLQVWHDGRIURWDWLQJ WKHF\OLQGHULQWKHXQLIRUPÀRZ FL US U +

=

q S1

Uniform Flow

Circulator Fow

S2

Magnus Effect (Lift Generation)

Magnus Effect

The velocity at the surface of the cylinder, us U sin q +

G  pR

 $WVWDJQDWLRQSRLQWus = 0 Hence,

or

U sin q = –

sin q =

G  pR

G 4 pU R

qFDQWDNHYDOXH LQEHWZHHQƒWRƒ In case q ƒWKHQ   or

VLQ ±

G  ± 4 pU R

G = 4pUR

 ‡ 'HULYH DQ H[SUHVVLRQ IRU OLIW IRUFH DFWLQJ RQ WKH URWDWLQJ F\OLQGHU SODFHG LQ XQLIRUP flowing fluid?   &RQVLGHUDF\OLQGHUURWDWLQJLQDXQLIRUPÀRZ¿HOG7DNHDVPDOOOHQJWKRIHOHPHQWRQWKH VXUIDFHRIWKHF\OLQGHU$SSO\%HUQRXOOL¶VHTXDWLRQDWWZRSRLQWVRQHDWWKHVXUIDFHRIWKH

Flow Past Submerged Bodies

921

R

PS

dq

q q

F\OLQGHUDQGRWKHURQHDWYHU\ODUJHGLVWDQFHDZD\IURPWKHF\OLQGHUPs and UsDUHSUHVVXUH DQGYHORFLW\DWVXUIDFHZKLFKP and UDUHSUHVVXUHDQGYHORFLW\DWIDUGLVWDQFHSRLQW

P U Ps U + s + ZS = + +Z rg g g rg Z = ZsDVÀRZLVWDNLQJSODFHLQDSODQH

P Ps  = + [U  – US] rg g rg But,

G = velocity at surface of the cylinder  pR

Us U sin q +

 P  È  Ê Ps G ˆ ˘ ÍU - Á U VLQ q + ˙ = + rg g ÍÎ rg pR ˜¯ ˙˚ Ë

or

=

P  + rg g

Ps = P +

È  4sin q G ˘ G   ÍU - 4U sin q -   ˙ pR ˚ 4p R Î

4sin q G ˘ G rU  È  Í - VLQ q -    ˙  pU R ˚ 4p R U  Î

 7KHOLIWDFWLQJRQWKHHOHPHQWDODUHDZKHQPsLVWKHSUHVVXUHDFWLQJRQWKHVXUIDFH FL = –

Ú

=–

Ú

=

p

0 p

0

Ú

p

0

Ps ¥ sin q ¥ dA PS ¥ sin q ¥ R ◊dq ◊L

È rU  ÍP +  ÍÎ

as dA = R ◊dq ◊L

Ê G 4sin q G ˆ ˘   VLQ q ˙ R ◊L◊sin q dq Á  pU R ˜¯ ˙˚ 4 pU  R  Ë

922

Fundamentals of Fluid Mechanics

=–

+

+

But \

Ú

p

0

Ú

p

Psin q ◊RLdq –

0

Ú

p

usin3 q RLdq ◊

0

Ú

0

rU  sin q ◊ R◊L◊dq g

rU  + g

Ú

p

0

r U G  RL sin q dq p U  R 

 rU  4sin q ◊ G ◊ ◊R◊L ◊dq pU R 

p

0

sin q dq =

Ú

p

Ú

p

0

sin3 qdq = 0 FL = R◊L◊

Ú

p

0

rgU sin  q ◊ ◊G◊dq pg UR p

 L rU G Ê q sin q ˆ =  ˜¯ 0 p ÁË 

=

L ◊ r ◊UG ¥p p

= L ◊r◊UG  7KHDERYHHTXDWLRQLVNQRZQDV.XWWD-RXNRZVN\HTXDWLRQ

16.9 KARMAN VORTEX TRAIL  ‡ :KDWLV.DUPDQYRUWH[WUDLO"   :KHQDF\OLQGHULVVXEMHFWHGWRDXQLIRUPÀRZDWORZ5H\QROGVQXPEHU R e WKHÀRZ GRHV QRW VHSDUDWH IURP WKH F\OLQGHU$ERYH Re   VHSDUDWLRQ DQG HGG\ IRUPDWLRQ VWDUWV WDNLQJ SODFH :KHQ 5H\QROGV QXPEHU H[FHHGV DERYH  WZR YRUWLFHV DUH IRUPHG DW WKH SRLQWV RI VHSDUDWLRQ7KH YRUWLFHV URWDWH LQ RSSRVLWH GLUHFWLRQV DQG WKH\ HORQJDWH WR VXFK DQH[WHQWWKDWWKH\EHFRPHXQVWDEOHWKHUHE\EUHDNLQJR൵IURPWKHVXUIDFHRIWKHF\OLQGHU 7KLVSURFHVVRIIRUPDWLRQRIYRUWLFHVDQGWKHLUVKHGGLQJDZD\IURPWKHF\OLQGHUFRQWLQXHV +HQFHWKHUHDUHWZRURZVRIYRUWLFHVVHSDUDWLQJIURPWKHF\OLQGHUDUHIRUPHGZKLFKVWDUW PRYLQJGRZQVWUHDPZLWKVPDOOYHORFLW\7KHVHWUDLOVRIYRUWLFHVDUHFRPPRQO\NQRZQDV “Karman vortex trails”.

Flow Past Submerged Bodies

923

Separation Starts at Re > 2

a l Karman Vortex Trails

  'XULQJVWDEOHFRQGLWLRQVRIVKHGGLQJYRUWLFHV.DUPDQIRXQGWKDW  

a   l

  9RUWH[WUDLOPRYHVGRZQZDUGZLWKDYHORFLW\RIutZKLFKLV ut =

0.355 G  ZKHUHG is the circulation around the vortex. a

  WKHIUHTXHQF\RIYRUWH[VKHGGLQJ f  or

U D

Ê  ˆ ÁË ˜  ZKHUHD = diameter of cylinder Re ¯

fD Ê  ˆ   Á ˜ Ë U Re ¯

fD LVNQRZQDV6WURXKDOQXPEHU$V5H\QROGVQXPEHUYDULHVIURPWR4, Strouhal U nuPEHUYDULHVIURPWRIRUDF\OLQGHU

16.10 AEROFOIL  ‡ What is an aerofoil? What is its purpose?   $QDHURIRLOLVDVWUHDPOLQHGERG\GHVLJQHGLQVXFKDZD\WKDWDVHSDUDWLRQLIRFFXUVLWVKRXOG EHQHDUWKHWUDLOLQJHGJHRIWKHDHURIRLO$VVPDOOZDNHLVDOORZHGWREHSURGXFHGZKLFK UHVXOWVLQWRDVPDOOSUHVVXUHGUDJDFWLQJRQWKHDHURIRLO(YHQZKHQÀRZKDVKLJK5H\QROGV QXPEHUSUHVVXUHGUDJLVYHU\VPDOOEXWWKHVNLQIULFWLRQGUDJLVODUJH7KHUHVXOWLQJIRUFHRQ WKHDHURIRLOGXHWRVPDOOSUHVVXUHGUDJDORQJWKHÀRZDQGODUJHIULFWLRQGUDJSHUSHQGLFXODU WRWKHGLUHFWLRQRIÀRZLVDOVRDFWLQJQHDUO\SHUSHQGLFXODUWRÀRZ$ODUJHOLIWLVREWDLQHG from an aerofoil.

924

Fundamentals of Fluid Mechanics Lift Resulting force

Drag

Force on an Aerofoil

 ‡ ([SODLQ  FKRUGOLQHDQGFKRUG  DQJOHRIDWWDFN   FKDPEHU   SUR¿OH FHQWUH OLQH  DVSHFWUDWLRIRUDQDHURIRLODQG  VWDOO Mean camber line Upper camber line

Leading edge Chord

Flow Angle of attack (a)

Trailing edge

an

Sp Lower camber line Line and Chord



 

 

  Chord line and chord : &KRUG OLQH LV DQ LPDJLQDU\ OLQH ZKLFK FDQ EH GUDZQ IURP leading edge to the trailing edge of a cross section of an aerofoil. The length of the line is called chord of the aerofoil.   Angle of Attack. 7KH DFXWH DQJOH EHWZHHQ WKH FKRUG OLQH RI WKH DHURIRLO DQG WKH GLUHFWLRQRIWKHÀRZ   Camber. Camber is the curvature of the aerofoil from the leading edge to trailing edge. 7KHFDPEHUFDQEH  XSSHUFDPEHU  ORZHUFDPEHUDQG  PHDQFDPEHUZKLFKLV WKHPHDQOLQHWKDWLVHTXLGLVWDQFHIURPWKHXSSHUDQGORZHUVXUIDFHV,QWKHFDVHRID V\PPHWULFDODHURIRLOWKHSUR¿OHFHQWUHOLQHFRQFLGHVZLWKWKHFKRUGOLQH   3UR¿OH&HQWUH/LQH 7KHLPDJLQDU\OLQHMRLQLQJWKHPLGSRLQWVRIWKHSUR¿OHVWDUWLQJ IURPOHDGLQJHGJHWRH[LWDWWKHWUDLOLQJHGJHLVFDOOHGDSUR¿OHFHQWUHOLQH   Aspect Ratio. ,WLVWKHUDWLRRIVSDQRIWKHDHURIRLO l WRWKHOHQJWKRIPHDQFKRUG c  AsSHFWUatio =



VSDQ l = mean chord c

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Flow Past Submerged Bodies

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S1

S2

Aerofoil in Uniform Flow

Aerofoil in Circulatory Flow

S1 S2

Aerofoil in Combined Uniform and Vortex Flow



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926

Fundamentals of Fluid Mechanics

trailing HGJHDVWKHYHORFLW\RIÀRZLQFUHDVHV2QIXUWKHULQFUHDVHRIYHORFLW\LWJLYHVULVH WR WKH IRUPDWLRQ RI YRUWLFHV 7KH YRUWLFHV FUHDWH FLUFXODWLRQ DURXQG WKH DHURIRLO DQG ÀRZ SDWWHUQDURXQGWKHDHURIRLOEHFRPHVVLPLODUWRZKDWLVREWDLQHGZKHQDHURIRLOLVVXEMHFWHG WRFRPELQHGÀRZRIXQLIRUPDQGYRUWH[ÀRZ7KHUHDUVWDJQDWLRQSRLQWS2LVQRZDWWUDLOLQJ HGJH7KHUHLVGLVWRUWLRQRIWKHÀRZSDWWHUQDERXWWKHDHURIRLOZLWKVWUHDPOLQHVDUHFORVHURQ WKHXSSHUVXUIDFHDQGVWUHDPOLQHVDUHZLGHVSDFHGDWORZHUVXUIDFH+HQFHWKHÀRZSDWWHUQ LVDV\PPHWULFDODERXWWKHFHQWUDOD[LVDQGDHURIRLOLVQRZVXEMHFWHGWRDOLIWIRUFH

16.11 AEROPLANE IN EQUILIBRIUM  ‡ 'H¿QH DHURSODQH :KHQ ZLOO LW EH VDLG WR EH LQ HTXLOLEULXP" 'HULYH H[SUHVVLRQ IRU YHORFLW\DQGSURSXOVLRQSRZHU Lift

Propulsion Drag

or thrust

Weight Equilibrium of aeroplane



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1 rU 2 ¥ A 2

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1 rU 2 ¥ A 2 Velocity of plane = Velocity of air = u W = CL ¥

U=

ÊW ˆ 2 ÁË ˜¯ A r ◊ CL

 3RZHUUHTXLUHGIRUSURSXOVLRQ P = FD ¥ A 3 = CD ¥ rU ¥ A 2

Flow Past Submerged Bodies

927

‡ ([SODLQWKHYDULDWLRQRIWKHFRH൶FLHQWRIOLIWDQGWKHFRH൶FLHQWRIGUDJZLWKWKHDQJOH RIDWWDFNGXULQJWKHIORZRYHUDQDHURIRLO Stall point CL CD

CD

CL

Angle of attack (a)



 $VWKHDQJOHRIDWWDFNLQFUHDVHVOLIWIRUFHVLQFUHDVHDQGOLIWFRH൶FLHQWUHDFKHVPD[LPXP $IWHUWKLVWKHOLIWFRH൶FLHQWGHFUHDVHVZLWKKLJKHUYDOXHVRIWKHDQJOHRIDWWDFN7KHDHURIRLO LVWKHQVDLGWRVWDOOZKHQCLVWDUWVGHFUHDVLQJ7KHPD[LPXPSRLQWRQOLIWFRH൶FLHQWFXUYH LVNQRZQDVWKHVWDOOLQJSRLQW7KHUHDVRQIRUGHFUHDVHLQWKHOLIWFRH൶FLHQWLVGXHWRWKH VHSDUDWLRQ RI WKH ÀRZ RQ WKH DHURIRLO 6LPLODUO\ GUDJ FRH൶FLHQW IULVWLO\ GHFUHDVHV ZLWK LQFUHDVH RI DQJOH RI DWWDFN DQG ODWHU LV LQFUHDVHV UDSLGO\ GXHWR VHSDUDWLRQ RI ÀRZ RQ WKH aerofoil.  ‡ $IODWSODWHRIVL]H¥PLVVXEPHUJHGLQZDWHUIORZLQJZLWKYHORFLW\RIPV)LQG drag and lift if CD DQGCL    

$UHDRISODWH¥ 3 = 6 m

 rU ¥ A 

FD = CD ¥

FD = 0.04 ¥

  

= 0.04 ¥ 75 ¥3  N1 FL = C L ¥

  

 ¥¥3 ¥   ¥ 6 

 ¥ rU ¥ A 

 ¥

 ¥¥3 ¥ 5 ¥ 6 

 N1

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928

Fundamentals of Fluid Mechanics

A ¥ P FL = CL ¥  rU  ¥ A  = 0.9 ¥  

 ¥¥ ¥ 4 

 1 FD = CD ¥

 rU  ¥ A   ¥¥ ¥ 4 

 

 ¥

 

 1 F= =

 

  +  

 ¥ 4 +  ¥ 4

 1 tan q =

or   3RZHUH[HUWHGE\DLU

FL + FD =

FL  =  FD

= 4.5 q ƒ

  

P = FD ¥ U  ¥

  

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 ¥ 3 =PV 3600

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Flow Past Submerged Bodies

929

  7KHSRZHUUHTXLUHGWRRYHUFRPHGUDJ 3RZHU FD ¥ U  ¥  N:

         Case   

CD = 0.3 3RZHUUHPDLQVVDPH N: P = FD ¥ U 3¥3 = CD ¥

U3 =

or

 ¥ 3 ¥    ¥ ¥  ¥  

 ¥3

 or

 rU  ¥ A ¥ U 

U PV

 ‡ $ PDQ ZHLJKLQJ  1 GHVFHQGV WR WKH JURXQG IURP DQ DHURSODQH XVLQJ SDUDFKXWH 7KHSDUDFKXWHLVKHPLVSKHULFDOZLWKGLDPHWHURIP)LQGWKHYHORFLW\RIGHVFHQGRI WKHSDUDFKXWH$VVXPHCD DQGrair NJP Diameter = 2 m FD

800 N

  

3URMHFWHGDUHDRISDUachute =

A=

p ¥ d 4 p ¥   P 4

FD 1

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Fundamentals of Fluid Mechanics

FD = CD ¥  ¥ r ¥ U  ¥ A   

 ¥ U =

or   or

 ¥¥ U ¥ 

 ¥   ¥  ¥ 

  U PV

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3URMHFWHGDUHDA =

p ¥ d p ¥ 3 = = 7.065 m 4 4

FD = CD ¥ = 0.5 ¥  

 rU ¥ A   ¥¥ ¥ 7.065 

 1 FD :HLJKWRIPDQ:HLJKWRISDUDFKXWH

\

:HLJKWRIPDQ ±

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or

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Flow Past Submerged Bodies

NRZ

931

p ¥ d   4 d =

or

 ¥   P p

d = 3.57 m  ‡ $ VSKHUH KDYLQJ GLDPHWHU RI  PP DQG UHODWLYH GHQVLW\ RI  LV VXVSHQGHG LQ D XQLIRUPVWUHDPRIDLUZLWKWKHKHOSRIDVWULQJ$LUIORZKDVYHORFLW\RIPV)LQG  LQFOLQDWLRQRIVWULQJZLWKYHUWLFDODQG   WHQVLRQLQWKH VWULQJ$VVXPHCD   DQGDLUGHQVLW\ NJP q T

FD

U

W

Weight, W = =  

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 1 FD = CD ¥ A=

 ¥ rU  ¥ A 

p p ¥ d = ¥   4 4

 ¥–3 m

 

FD = 0.5 ¥

\     $SSO\LQJ/DPL¶VWKHRUHP

 ¥¥ ¥¥–3 

 1

T W FD = = sin 90 VLQ  + q VLQ  - q

932

Fundamentals of Fluid Mechanics

T=

or

tan q =

  = cos q sin q  



q ƒ and

T=

  1 sin 67.79

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U 15°

FD

45°

T = 2.6 kg

A = P, U =

W = 0.25 kg

 ¥ 3  PV 3600

T ¥ 1 W ¥ 1 FD = T cos 45   

 FRV 1 FL = T sin 45 + W

  

 VLQ

  

 1 CD =

 

FD

 rU  A   

=

  ¥  ¥   ¥  

Flow Past Submerged Bodies

CL =

933

FL  =   rU  A ¥  ¥   ¥   

= 0.466 ‡ $NLWHLVZHLJKLQJ1VRDUVDWDQDQJOHƒWRWKHKRUL]RQWDO7KHVWULQJRIWKHNLWH PDNHV ƒ WR WKH KRUL]RQWDO DQG LW KDV WHQVLRQ RI  1 )LQG WKH UDWLR RI OLIW WR GUDJ DFWLQJRQWKHUDWH FD = T cos 30 = 5 cos 30 = 4.33 N FL = T sin 30 + W =5¥

\

FL FD

 + 0.5 

=3N 3 = = 0.69 4.33 FL

FD 30° T = 25 N W=6N

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FD 45° T = 25 N W=6N

934

Fundamentals of Fluid Mechanics

  'XULQJHTXLOLEULXPRIWKHNLWHZHKDYH FD = TFRV FRV     1 FL = T sin 45 + W          1   1RZ

CD =

=

 

FD  rU  A    ¥  ¥   ¥  

  CL =

=

FL  rU  A    ¥  ¥   ¥  

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rVHDZDWHU NJP3 CD = 0.7 and CL  FD = CD ¥ = 0.7 ¥

 

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Flow Past Submerged Bodies

 ¥¥   ¥ 

 

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 ¥3N1 F= =

935

FD + FL  ¥   +  ¥  

 ¥3N1

 

3RZHUUHTXLUHG F ¥ U  ¥6 ¥  ¥4N:

       

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pDN 60 p ¥  ¥   PV 60

Circulation G p Ruq  ¥ p ¥¥  PV

      Lift

FL = L◊r◊U ¥ G  ¥¥3 ¥¥ = 47.3 ¥6 N

     $WVWDJQDWLRQSRLQWZHKDYH VLQq +

or

sin q = – 

or   

uq =0 U   ¥ 

 ± ± VLQƒ q VLQ  DQGVLQ ±  ƒDQGƒ

936

Fundamentals of Fluid Mechanics

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or  

360 ¥3 PV 3600

FL = W  ¥3 = CL ¥

CL =

 ¥ rU  ¥ A 

 ¥ 3 ¥   ¥   ¥ 

 

For CL IURPFODUNyZLQJGLDJUDP CD ª 0.05 \

FD = CD ¥

 rU ◊A 

= 0.05 ¥       

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Circulation is given by FL = r◊U ◊G◊ L =  

 ¥ 3  ¥  ¥ 

 PV

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Flow Past Submerged Bodies

937

0.4

U

2m

A = h ¥ D ¥ P 3 U = 60NPSK   ¥   PV 3600

FD = CD ¥

 

 ¥

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Resistance due to rolling =

 ¥ 1 

Resistance due to surface friction =

 ¥ 1 

+HQFHGUDJ ±   1 FD = CD ¥

  or

 ¥ 3  PV 3600

 CD ¥ CD = 0.934

 rU ¥ A   ¥¥   ¥ 6.5 

938

Fundamentals of Fluid Mechanics

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U

U NPK = uq =

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pDN p ¥  ¥  =  PV 60 60

Circulation generated by rotation of each cylinder, G p Ruq p ¥  

3 ¥ 

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Flow Past Submerged Bodies

939

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 N1

Case3ODWHLQDLU FD ¥

   

 ¥¥ 4 ¥ 9 

 

 1

   

 ±  1  N1

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 ¥

 

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Drag due to shear, FDS = CDS ¥

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940

Fundamentals of Fluid Mechanics

  1RZZHNQRZ \   

FD = FDS + FDP FDP = FD – FDS  ¥ –4 ¥ –4 = 4.05 ¥ –4 N

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r ¥3JPFP3 & v FPV m = r ¥ v ¥3 ¥

  

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gm cm ◊ s

 

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NJ ms

Weight of rDLQGURS =

=    1RZ   or

p ¥ ¥ –5 ¥¥ 6

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p ¥ D3 ¥ r ¥ g 6

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Flow Past Submerged Bodies

941

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10

°

Projected Area

  3URMHFWHGDUHD

 $ ¥ = 0.64 m

  ,QHTXLOLEULXPWKHUHDUHIRXUIRUFHVZKLFKDUHDFWLQJYL]P, W, FL and FD . SFy = 0, or

 RU

CL ¥

¥

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P  r ◊ A ◊ u = +4  

 ¥¥ 0.64 u = 0.707 P + 4  0.3 u = 0.707 P

or

L

SFx = 0, FD = Pcos45 or

or

or

CD ¥

0.6 ¥

P  r ◊ A ◊ u =  

P  ¥¥ 0.64 ¥ u =   u =

     )URPHTQV L DQG LL ZHKDYH     P  or P and u u or u

P  ( ¥  ¥ ) P

P + 4 1 P ¥ PV

LL

942

Fundamentals of Fluid Mechanics

16.12 QUESTIONS FROM COMPETITIVE EXAMINATIONS  r When pressure drag over a body is large as compared to the friction drag, then the shape of the body is that of  D  $QDHURIRLO E  $VWUHDPOLQHGERG\  F  $WZRGLPHQVLRQDOERG\ G  $EOX൵ERG\ ,(6   2SWLRQ G LVFRUUHFW  ‡ Assertion (A): $ ERG\ ZLWK ODUJH FXUYDWXUH FDXVHV D ODUJHU SUHVVXUH GUDJ DQG WKHUHIRUH larger resistance to motion. Reason (R): /DUJHFXUYDWXUHGLYHUJHVWKHVWUHDPOLQHVGHFUHDVHVWKHYHORFLW\UHVXOWLQJ in the increase in pressure and development of adverse pressure gradient leading to UHYHUVHIORZQHDUWKHERXQGDU\  D  %RWK$DQG5DUHLQGLYLGXDOO\WUXHDQG5LVWKHFRUUHFWH[SODQDWLRQRI$  E  %RWK$DQG5DUHLQGLYLGXDOO\WUXHEXW5LVQRWWKHFRUUHFWH[SODQDWLRQRI$  F  $LVWUXHEXW5LVIDOVH  G  $LVIDOVHEXW5LVWUXH ,$6   2SWLRQ D LVFRUUHFW  ‡ :KHQHYHU D SODWH LV LPPHUVHG DW DQ DQJOH ZLWK WKH GLUHFWLRQ RI IORZ RI OLTXLG LW LV VXEMHFWHGWRVRPHSUHVVXUH:KDWLVWKHFRPSRQHQWRIWKLVSUHVVXUHLQWKHGLUHFWLRQRI IORZRIOLTXLGNQRZQDV  D  6WUHDPOLQHSUHVVXUH E  /LIW  F  'UDJ G  %XONPRGXOXV ,(6   2SWLRQ F LVFRUUHFW  ‡ 7KHGUDJIRUFHH[HUWHGE\DIOXLGRQDERG\LPPHUVHGLQDIOXLGLVGXHWR  D  3UHVVXUHDQGYLVFRXVIRUFHV E  3UHVVXUHDQGJUDYLW\IRUFHV  F  3UHVVXUHDQGVXUIDFHWHQVLRQIRUFHV G  9LVFRXVDQGJUDYLW\IRUFHV ,(6   2SWLRQ D LVFRUUHFW  ‡ :KLFKRQHRIWKHIROORZLQJLVWUXHRIIORZDURXQGDVXEPHUJHGERG\"  D  )RUVXEVRQLFQRQYLVFRXVIORZWKHGUDJLV]HUR  E  )RU VXSHUVRQLF IORZ WKH GUDJ FRH൶FLHQW LV GHSHQGHQW HTXDOO\ RQ 0DFK QXPEHU DQG5H\QROGVQXPEHU  F  7KHOLIWDQGGUDJFRH൶FLHQWVRIDQDHURIRLODUHLQGHSHQGHQWRI5H\QROGVQXPEHU  G  )RULQFRPSUHVVLEOHIORZDURXQGDQDHURIRLOWKHSUR¿OHGUDJLVWKHVXPRIIRUP GUDJDQGVNLQIULFWLRQGUDJ ,(6 Option (d) is correct.  ‡ Assertion (A) ,QIORZRYHULPPHUVHGERG\SUHVVXUHDQGVKHDUIRUFHVDFWRQWKHERG\ Reason (R): 'UDJFDQEHFUHDWHGZLWKRXWOLIW/LIWFDQQRWEHFUHDWHGZLWKRXWGUDJ

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 D  %RWK$DQG5DUHLQGLYLGXDOO\WUXHDQG5LVWKHFRUUHFWH[SODQDWLRQRI$  E  %RWK$DQG5DUHLQGLYLGXDOO\WUXHEXW5LVQRWWKHFRUUHFWH[SODQDWLRQRI$  F  $LVWUXHEXW5LVIDOVH  G  $LVIDOVHEXW5LVWUXH ,$6   2SWLRQ E LVWUXH  ‡ 0DJQXVH൵HFWLVGH¿QHGDV  D  7KHJHQHUDWLRQRIOLIWSHUXQLWGUDJIRUFH  E  7KHFLUFXODWLRQLQGXFHGLQDQDLUFUDIWZLQJ  F  7KHVHSDUDWLRQRIERXQGDU\OD\HUQHDUWKHWUDLOLQJHGJHRIDVOHQGHUERG\  G  7KHJHQHUDWLRQRIOLIWRQDURWDWLQJF\OLQGHULQDXQLIRUPÀRZ ,$6   2SWLRQ G LVFRUUHFW  ‡ $QDXWRPRELOHPRYLQJDWDYHORFLW\RINPSKLVH[SHULHQFLQJDZLQGUHVLVWDQFHRI  N1 ,I WKH DXWRPRELOH LV PRYLQJ DW D YHORFLW\ RI  NPSK WKH SRZHU UHTXLUHG WR overcome the resistance is  D  N: E  N:  F  N: G  N: ,(6 Power = Drag force × Speed = FD × V = CD ¥

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Fundamentals of Fluid Mechanics

 ‡ ,PSURYHG VWUHDPLQJ SURGXFHV  UHGXFWLRQ LQ WKH GUDJ FRH൶FLHQW RI D WRUSHGR :KHQ LW LV PRYLQJ IXOO\ VXEPHUJHG DQG DVVXPLQJ WKH GULYLQJ SRZHU WR UHPDLQ WKH VDPHWKHLQFUHDVHLQVSHHGLV  D   E    F   G   ,(6 Power = FD ¥ Velocity = CD ¥

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D  



F  

N1  m

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N1  m

N1 m

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= 10.05 m2/s =P∙L∙V∙G = 1000 × L ∙ 4 ∙ 10.05

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or

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Codes A B (a) 2 3 (b) 3 2 (c) 2 3 (d) 3 2  2SWLRQ F LVFRUUHFW

C 1 4 4 1

D 4 1 1 4

(IES 2001)

946

Fundamentals of Fluid Mechanics

 ‡ )RU VROLG VSKHUH IDOOLQJ YHUWLFDOO\ GRZQZDUGV XQGHU JUDYLW\ LQ D YLVFRXV IOXLG WKH WHUPLQDOYHORFLW\V varies with diameter D of the sphere as  D  V μ D for all diameters  E  V μ D for all diameters  F  V μ D for large D and V μ D for small D ,(6  G  V μ D for large D and V μ D for small D   2SWLRQ E LVFRUUHFW  ‡ 0DWFK/LVW, W\SHRIIORZ ZLWK/LVW,, EDVLFLGHDOIORZV DQGVHOHFWWKHFRUUHFWDQVZHU    

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2

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 ‡ Assertion (A): $LUFUDIW ZLQJV DUH VORWWHG WR FRQWURO VHSDUDWLRQ RI ERXQGDU\ OD\HU VSHFLDOO\DWODUJHDQJOHVRIDWWDFN Reason (R): 7KLVKHOSVWRLQFUHDVHWKHOLIWDQGWKHDLUFUDIWFDQWDNHR൵IURPDQGODQG RQVKRUWUXQ  D  %RWK$DQG5DUHLQGLYLGXDOO\WUXHDQG5LVFRUUHFWH[SODQDWLRQRI$  E  %RWK$DQG5DUHLQGLYLGXDOO\WUXHEXW5LVQRWFRUUHFWH[SODQDWLRQRI$  F  $LVWUXHEXW5LVIDOVH  G  $LVIDOVHEXW5LVFRUUHFW ,(6   2SWLRQ F LVFRUUHFW  ‡ &RQVLGHUWKHIROORZLQJVWDWHPHQWV   7KHFDXVHRIVWDOOLQJRIDQDHURIRLOLVWKHERXQGDU\OD\HUVHSDUDWLRQDQGIRUPDWLRQ RILQFUHDVHG]RQHRIZDNH   $QDHURIRLOVKRXOGKDYHDURXQGHGQRVHLQVXSHUVRQLFIORZWRSUHYHQWIRUPDWLRQ RIQHZVKRFN   :KHQDQDHURIRLORSHUDWHVDWDQJOHRILQFLGHQFHJUHDWHUWKDQWKDWRIVWDOOLQJWKH lift decreases and drag increases   $URXJKEDOOZKHQDWFHUWDLQVSHHGVFDQDWWDLQORQJHUUDQJHGXHWRUHGXFWLRQRI OLIWDVWKHURXJKQHVVLQGXFHVHDUO\VHSDUDWLRQ   :KLFKRIWKHIROORZLQJVWDWHPHQWVDUHFRUUHFW  D  DQG E  DQG  F  DQG G  DQG ,(6   2SWLRQ G LVFRUUHFW  ‡ &RQVLGHUWKHIROORZLQJVWDWHPHQWV   7KH SKHQRPHQRQ RI OLIW SURGXFHG E\ LPSRVLQJ FLUFXODWLRQ RYHU D GRXEOHW LQ D XQLIRUPIORZLVNQRZQDV0DJQXVH൵HFW   7KHSDWKGHYLDWLRQRIFULFNHWEDOOIURPLWVRULJLQDOWUDMHFWRU\LVGXHWRWKH0DJQXV H൵HFW   :KLFKRIWKHVWDWHPHQWVJLYHQDERYHLVDUHFRUUHFW"  D  RQO\ E  RQO\  F  %RWKDQG G  1HLWKHUQRU ,(6   2SWLRQ F LVFRUUHFW

INDEX

A Acceleration on inclined plane 307 Acceleration vector 425 Adhesion 45 Aerofoil 923 Aeroplane in equilibrium 926 Aerostatic law 79, 82 $൷X[ Angular momentum 590 Anicut 548 Aquaduct 548 Archimedes’ principle 234 Atmospheric pressure 83 $YHUDJHFRH൶FLHQWRIGUDJ Average velocity 642, 720 $[LDOGHSWKRIZDWHU

B Backwater curve 548 Bend, forces on 594 Bernoulli’s equation 490, 492, 637 %LQJKDPSODVWLFV %OX൵ERG\ Boundary layer concept of 746 growth of 749 separation of 764 %RXQGDU\OD\HUWKLFNQHVV Bourdon tube gauge 85 Broad crested weir 528 Buckingham’s theorem 338

Bulk modulus of elasticity 42 Buoyancy 233 applications of 236 Buoyant force 233

C &DSLOODU\IDOO &DSLOODU\ULVH Capillary tube method 684 Cavitation 44 Centre of buoyancy 235 &HQWUHRISUHVVXUH &KHQ]\¶VIRUPXOD &LUFXODWLRQ &RH൶FLHQWRIFRPSUHVVLELOLW\ &RH൶FLHQWRIFRQWUDFWLRQ &RH൶FLHQWRIGLVFKDUJH &RH൶FLHQWRIGUDJ &RH൶FLHQWRIOLIW &RH൶FLHQWRIYHORFLW\ Cohesion 45 &RPSUHVVLELOLW\ &RPSUHVVLEOHÀRZ Conservation of mass, principle of 406 Continuity equation in three-dimensional Cartesian coordinates 407 &RQWLQXLW\RIÀRZHTXDWLRQRI &RQWLQXXP Control volume, concept of 487 Convergent divergent mouthpiece 506 Convergent mouthpiece 506

950

Index

&RXHWWHÀRZ &ULWLFDOYHORFLW\ &\OLQGULFDOH[WHUQDOPRXWKSLHFH

D D’Alembert’s principle 302 'DPVWDELOLW\RI Darcy’s equation 645, 673 for friction loss 722 Dashpot 680 'L൵HUHQWLDOPDQRPHWHUV 'LODWDQWÀXLG Dimensional analysis 332 by Rayleigh’s method 337, 339 methods of 337 Dimensional homogeneity 335 applications of 336 'LPHQVLRQOHVVQXPEHUV Dimensions 332 'LVSODFHPHQWWKLFNQHVV Distorted model 377 'RXEOHWÀRZ 'UDJIRUFH Drowned weir 529 Dynamic equilibrium 300 '\QDPLFVLPLODULW\ Dynamic viscosity 24 Dynamics, fundamental principle of 579

E Eddy viscosity 706 Elastic force 365 Electrical analogy method 424 End contraction 527 Energy gradient 822 Energy thickness 754 Equation of continuity 406 (TXDWLRQRIVWUHDPOLQH Equivalent pipe 826 Equivalent sand grain roughness 725 Euler’s equation of motion 490 Euler’s model law 369 Euler’s number 336, 344

Evaporation 43 ([SHULPHQWDOPRGHOWHVWLQJ

F Falling sphere viscometer 683 Filament line 400 Floatation 235 principle of 236 Floating, stability of 237 Flow in inclined pipe 663 LQFOLQHGSODWHV )ORZQHW uses and limitations of 424 )ORZQR]]OH Flow of electricity 423 )ORZRIÀXLGV Flow power transmitted by 850 superposition of 903 )OXLGG\QDPLFV )OXLGÀRZ )OXLGNLQHPDWLFV Fluid kinetics 6 Fluid mechanics 5 applications of 6 LPSRUWDQFHRI Fluid motion 396 Eulerian method of 397 Langrangian method of 396 Fluid particles displacements of 400 LUURWDWLRQDOÀRZV YRUWLFLW\RI )OXLGVWDWLF Fluidization 674 Fluids 4, 5 behaviour of 23 characteristics of 4 physical properties 62 SURSHUWLHVRI W\SHVRI YLVFRVLW\RI

Index

FRUFHGYRUWH[ÀRZ )UHHOLTXLGMHW )UHHYRUWH[ÀRZ )UHHYRUWH[FLUFXODWLRQLQ )ULFWLRQIDFWRU )ULFWLRQKHDGORVV Froude’s model law 367 Froude’s number 343 Fundamental dimensions 334 )XQGDPHQWDOHTXDWLRQRIÀRZ

G *DVHV *HRPHWULFVLPLODULW\ Gravity force 365

H

Inertia force 363 Instantaneous velocity 704 Inverted U-tube manometer 90

J -HWWUDMHFWRU\ Journal bearing 674

K Karman similarity concept 709 .DUPDQYRUWH[WUDLO Kinematic eddy viscosity 706 .LQHPDWLFVLPLODULW\ Kinematic viscosity 24 Kinematics 486 Kinetic energy 486 Kinetic energy correction factor 659

Hagen poiseuille formula 644 +DUG\FURVVPHWKRG Horizontal acceleration 303 +RUL]RQWDOSLSHVÀRZLQ +RUL]RQWDOWUDQVODWLRQDODFFHOHUDWLRQ Hot wire anemometer 726 +\GUDXOLFFRH¿FLHQWV Hydraulic gradient line 822 +\GUDXOLFUDGLXV Hydraulic similitude 360 Hydraulics 6 +\GURG\QDPLFDOO\URXJKERXQGDU\ +\GURG\QDPLFDOO\VPRRWKERXQGDU\ Hydrometer 237 +\GURVWDWLFODZ +\GURVWDWLFSDUDGR[

L

I

Mach’s model law 370 Mach’s number 344 0DJQXVH൵HFW Major energy loss 809 Manometers 84, 87 Manometery 84 Masonry dam 205 Matter, states of 2

,GHDOÀXLGÀRZ ,GHDOÀXLGV ,GHDOSODVWLFÀXLGV ,PDJLQDU\IUHHVXUIDFH ,PPHUVHGERGLHVVWDELOLW\FRQGLWLRQIRU Inclined single column micrometer 89 IncompressibleÀRZV

Laminar boundary, characteristics of 749 /DPLQDUÀRZ Laminar sublayer 750 /DPLQDUVXEOD\HUWKLFNQHVV Laminar, parameters of 764 /DSODFHHTXDWLRQ IRUWKHYHORFLW\SRWHQWLDO Laser doppler anemometer 726 /LIWIRUFH Linear momentum 589 /LQHVRIÀRZ /LTXLGV /RFDOFRH൶FLHQWRIGUDJ Lock gate 204

M

951

952

Index

Mechanical gauges 85 Mercury barometer 84 Metacentre 240 Metacentric height 240 Micromanometer 92 Micrometers 88 Minor energy loss 809 0L[LQJOHQJWKFRQFHSW Model 358 Model analysis 359 Moment of momentum 589 Momentum equation, applications of 580 Momentum principle 579 Momentum thickness 753 Moody’s diagram 725 Mouthpiece 505, 507

N Navier-Stokes equations 489 Neutral equilibrium 239 1HZWRQ¶VHTXDWLRQRIYLVFRVLW\ 1HZWRQ¶VODZRIYLVFRVLW\ 1HZWRQLDQÀXLGV 1RQ1HZWRQLDQÀXLGV 1RQXQLIRUPÀRZ Notch 520

O Oil bearings 674 2QHGLPHQVLRQDOÀRZ 2SHQVSLOOZD\ 2UL¿FH 2UL¿FHPHWHU 2UL¿FHW\SHYLVFRPHWHU Overhead water tanks 205

P 3DUDOOHOSODWHVÀRZLQ Pascal’s law 80 Path line 398 S theorem 339 Piezometer 86

3LH]RPHWULFVXUIDFH Pipe 809 3LSHQHWZRUNLQJ Pipeline 809 Pitot static 495 Pitot tube 493 3ODVWLFÀXLGV Pneumatics 6 Poise 24 Potential energy 486 Potential function 420 Potential line 400 Prandtl’s concept 709 3UDQGWO¶VYHORFLW\GLVWULEXWLRQ 3UHVVXUHGLDJUDP 3UHVVXUHGUDJ Pressure energy 487 Pressure force 364 Pressure measuring devices 85 Pressure on LQFOLQHGVXUIDFH SODQHVXUIDFH Pressure tube 86 Primary quantities 334 Prototype 359 3VHXGRSODVWLFÀXLGV

R Radial acceleration 309 5DGLDOÀRZ Raised weirs 548 5DWHRIÀRZ 5HDOÀXLGV 5HFWLOLQHDUÀRZ Repeating variables 338 5H\QROGVH[SHULPHQW Reynolds model law 366 Reynolds number 336, 342, 638 Reynolds transport theorem 488 5RWDU\PRWLRQ Rotating cylinder method 685

Index

Rotation FRQGLWLRQVIRU VWUHDPIXQFWLRQIRU YRUWLFLW\RI Rotational acceleration 302 5RWDWLRQDOÀRZ 5RWRPHWHU 5RXJKERXQGDU\ Running free Borda mouthpiece 507 Running full Borda mouthpiece 507

Streamline 398 6WUHDPOLQHERG\ Submerged body, equilibrium for 238 Submerged weir 529 Surface tension 46 cause of 47 Surface tension force 365 Surge tank 852 Syphon 844 6\SKRQVSLOOZD\

S

T

Secondary dimensions 334 Sensitive manometers 88 Shape factor 754 6KHDUGUDJ 6KHDUWKLFNHQLQJÀXLGV Short tube 505 Short tube mouthpiece 506 Similarity laws 366 6LPSOHPDQRPHWHU 6LQNÀRZ 6NLQDQGSUHVVXUHGUDJ 6PRRWKERXQGDU\ Solid 5 6RXUFHÀRZ 6SLOORXW 6SLOOZD\ Sprinkler, torque acting on 603 Stability condition for equilibrium of 240 ÀRDWLQJERGLHV rolling of 244 Stable equilibrium 238 Static body 300 6WHDG\ÀRZ 6WRNHV¶ODZ 6WUHDPIXQFWLRQ PDWKHPDWLFDOFRQFHSWRI SURSHUWLHVRI Stream tube 399

953

7KL[RWURSKLFSODVWLF 7KUHHGLPHQVLRQDOÀRZ Total energy line 822 7RWDOSUHVVXUH 7UDQVODWLRQDODFFHOHUDWLRQ Trapezoidal weir 528 Triangular notch 526 7XUEXOHQFHVKHDUVWUHVV5H\QROGVH[SUHVVLRQIRU Turbulent boundary, parameters of 764 7XUEXOHQWÀRZ Two piezometer tubes method 89 7ZRGLPHQVLRQDOÀRZ

U 8QLIRUPÀRZ Unstable equilibrium 239 8QVWHDG\ÀRZ 8WXEHGL൵HUHQWLDOPDQRPHWHU U-tube manometer 87

V 9DQH Vapour pressure 44 9HORFLW\SRWHQWLDO Velocity vector 425 Vena contracta 497 9HQWXULPHWHU Vertical acceleration 305 Vertical single column micrometer 89

954

Index

Viscosity determination of 57 H൵HFWVRIWHPSHUDWXUHDQGSUHVVXUHRQ measurement of 682 9LVFRXVÀRZWKHRU\ Viscous force 364 Von Karman momentum equation 757 9RUWH[ÀRZ 9RUWH[PRWLRQ Vortices, formation of 703

W Water hammer 852 Weber’s model law 370 Weber’s number 336, 343 Weir 520 applications of 548 Wind tunnel testing 359

Fundamentals of Fluid Mechanics

Fundamentals of Fundamentals of

G.S. Sawhney has served as Professor and Head, Department of Mechanical Engineering, GNIT Greater Noida. Earlier, he was at Lord Krishna College of Engineering, Ghaziabad in the same capacity. Prior to joining teaching, he had served for 28 years in the Corps of Engineers and for 10 years in industry. ? Biomedical Electronics and Instrumentation ? Fluid Machinery Made Easy ? Fundamentals of Computer Aided Manufacturing ? Fundamentals of Biomedical Engineering Made Easy ? Heat and Mass Transfer ? Manufacturing Science (Volume-I, II) ? Materials Science and Engineering ? Mechanical Experiments and Workshop Practice ? Project Management Made Easy

Third Edition

Fluid Mechanics

Written for the second-year engineering students of undergraduate level, this well-set-up textbook explains the fundamentals of Fluid Mechanics. Written in the question-answer form, the book is precise and easy to understand. The book presents an exhaustive coverage of the theory, definitions, formulae and examples which are well supported by diagrams (wherever necessary) and problems in order to make the underlying principles more comprehensive. In the present third edition, the book has been thoroughly revised and enlarged. Modification to every chapter has been carried out on the basis of the suggestions received and new developments in this area. Additional typical problems based on the examination papers of various technical universities have been included with solutions for easy understanding by the students. Objective questions of competitive examinations of GATE, IES and IAS have also been included with solutions and explanations at the end of each chapter.

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Fluid Mechanics Third Edition

G.S. Sawhney

G.S. Sawhney

978-93-90455-20-1

Distributed by: 9 789390 455201

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