# Fluid Mechanics 8th Edition [8th ed.] 0073398276

##### The eighth edition of White's Fluid Mechanics offers students a clear and comprehensive presentation of the materia

414 17 76MB

English Pages 865 Year 2015

Cover......Page 2
Title Page......Page 5
Contents......Page 9
Preface......Page 13
1.1 Preliminary Remarks......Page 19
1.2 The Concept of a Fluid......Page 20
1.3 The Fluid as a Continuum......Page 22
1.4 Dimensions and Units......Page 23
1.6 Thermodynamic Properties of a Fluid......Page 31
1.7 Viscosity and Other Secondary Properties......Page 39
1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines......Page 55
1.11 The History of Fluid Mechanics......Page 59
Summary......Page 60
Problems......Page 61
Fundamentals of Engineering Exam Problems......Page 68
Comprehensive Problems......Page 69
References......Page 72
2.1 Pressure and Pressure Gradient......Page 75
2.2 Equilibrium of a Fluid Element......Page 77
2.3 Hydrostatic Pressure Distributions......Page 78
2.4 Application to Manometry......Page 85
2.5 Hydrostatic Forces on Plane Surfaces......Page 88
2.6 Hydrostatic Forces on Curved Surfaces......Page 96
2.7 Hydrostatic Forces in Layered Fluids......Page 99
2.8 Buoyancy and Stability......Page 101
2.9 Pressure Distribution in Rigid-Body Motion......Page 107
2.10 Pressure Measurement......Page 115
Problems......Page 119
Fundamentals of Engineering Exam Problems......Page 142
Comprehensive Problems......Page 143
Design Projects......Page 145
References......Page 146
3.1 Basic Physical Laws of Fluid Mechanics......Page 149
3.2 The Reynolds Transport Theorem......Page 153
3.3 Conservation of Mass......Page 160
3.4 The Linear Momentum Equation......Page 165
3.5 Frictionless Flow: The Bernoulli Equation......Page 179
3.6 The Angular Momentum Theorem......Page 188
3.7 The Energy Equation......Page 194
Problems......Page 205
Word Problems......Page 232
Fundamentals of Engineering Exam Problems......Page 233
Comprehensive Problems......Page 234
References......Page 235
Chapter 4 Differential Relations for Fluid Flow......Page 237
4.1 The Acceleration Field of a Fluid......Page 238
4.2 The Differential Equation of Mass Conservation......Page 240
4.3 The Differential Equation of Linear Momentum......Page 246
4.4 The Differential Equation of Angular Momentum......Page 253
4.5 The Differential Equation of Energy......Page 254
4.6 Boundary Conditions for the Basic Equations......Page 257
4.7 The Stream Function......Page 262
4.8 Vorticity and Irrotationality......Page 269
4.9 Frictionless Irrotational Flows......Page 271
4.10 Some Illustrative Incompressible Viscous Flows......Page 277
Problems......Page 285
Word Problems......Page 296
Comprehensive Problems......Page 297
References......Page 298
5.1 Introduction......Page 301
5.2 The Principle of Dimensional Homogeneity......Page 304
5.3 The Pi Theorem......Page 310
5.4 Nondimensionalization of the Basic Equations......Page 320
5.5 Modeling and Similarity......Page 329
Problems......Page 341
Word Problems......Page 349
Comprehensive Problems......Page 350
Design Projects......Page 351
References......Page 352
6.1 Reynolds Number Regimes......Page 355
6.2 Internal versus External Viscous Flows......Page 360
6.3 Head Loss-The Friction Factor......Page 363
6.4 Laminar Fully Developed Pipe Flow......Page 365
6.5 Turbulence Modeling......Page 367
6.6 Turbulent Pipe Flow......Page 374
6.7 Four Types of Pipe Flow Problems......Page 382
6.8 Flow in Noncircular Ducts......Page 387
6.9 Minor or Local Losses in Pipe Systems......Page 396
6.10 Multiple-Pipe Systems......Page 405
6.11 Experimental Duct Flows: Diffuser Performance......Page 411
6.12 Fluid Meters......Page 416
Summary......Page 437
Problems......Page 438
Word Problems......Page 456
Fundamentals of Engineering Exam Problems......Page 457
Comprehensive Problems......Page 458
References......Page 460
7.1 Reynolds Number and Geometry Effects......Page 465
7.2 Momentum Integral Estimates......Page 469
7.3 The Boundary Layer Equations......Page 472
7.4 The Flat-Plate Boundary Layer......Page 475
7.5 Boundary Layers with Pressure Gradient......Page 484
7.6 Experimental External Flows......Page 490
Summary......Page 517
Problems......Page 518
Fundamentals of Engineering Exam Problems......Page 531
Comprehensive Problems......Page 532
References......Page 533
8.1 Introduction and Review......Page 537
8.2 Elementary Plane Flow Solutions......Page 540
8.3 Superposition of Plane Flow Solutions......Page 547
8.4 Plane Flow Past Closed-Body Shapes......Page 553
8.5 Other Plane Potential Flows......Page 563
8.6 Images......Page 567
8.7 Airfoil Theory......Page 570
8.8 Axisymmetric Potential Flow......Page 578
8.9 Numerical Analysis......Page 584
Problems......Page 593
Comprehensive Problems......Page 604
Design Projects......Page 605
References......Page 606
9.1 Introduction: Review of Thermodynamics......Page 609
9.2 The Speed of Sound......Page 614
9.4 Isentropic Flow with Area Changes......Page 622
9.5 The Normal Shock Wave......Page 629
9.6 Operation of Converging and Diverging Nozzles......Page 637
9.7 Compressible Duct Flow with Friction......Page 642
9.8 Frictionless Duct Flow with Heat Transfer......Page 653
9.9 Mach Waves and Oblique Shock Waves......Page 658
9.10 Prandtl-Meyer Expansion Waves......Page 668
Summary......Page 680
Problems......Page 681
Fundamentals of Engineering Exam Problems......Page 694
Comprehensive Problems......Page 695
Design Projects......Page 696
References......Page 697
10.1 Introduction......Page 699
10.2 Uniform Flow; The Chézy Formula......Page 705
10.3 Efficient Uniform-Flow Channels......Page 711
10.4 Specific Energy; Critical Depth......Page 713
10.5 The Hydraulic Jump......Page 720
10.7 Flow Measurement and Control by Weirs......Page 732
Summary......Page 739
Problems......Page 740
Comprehensive Problems......Page 752
References......Page 754
11.1 Introduction and Classification......Page 757
11.2 The Centrifugal Pump......Page 760
11.3 Pump Performance Curves and Similarity Rules......Page 766
11.4 Mixed- and Axial-Flow Pumps: The Specific Speed......Page 776
11.5 Matching Pumps to System Characteristics......Page 783
11.6 Turbines......Page 791
Summary......Page 805
Problems......Page 807
Comprehensive Problems......Page 820
References......Page 822
Appendix A Physical Properties of Fluids......Page 824
Appendix B Compressible Flow Tables......Page 829
Appendix C Conversion Factors......Page 836
Appendix D Equations of Motion in Cylindrical Coordinates......Page 838
Appendix E Estimating Uncertainty in Experimental Data......Page 840
Index......Page 849
##### Citation preview

Eighth Edition

Fluid Mechanics Frank M. White

Eighth Edition

Fluid Mechanics Frank M. White

Fluid Mechanics

Fluid Mechanics Eighth Edition

Frank M. White University of Rhode Island

Frank M. White is Professor Emeritus of Mechanical and Ocean Engineering at the University of Rhode Island. He studied at Georgia Tech and M.I.T. In 1966 he helped found, at URI, the first department of ocean engineering in the country. Known primarily as a teacher and writer, he has received eight teaching awards and has written four textbooks on fluid mechanics and heat transfer. From 1979 to 1990, he was editor-in-chief of the ASME Journal of Fluids Engineering and then served from 1991 to 1997 as chairman of the ASME Board of Editors and of the Publications Committee. He is a Fellow of ASME and in 1991 received the ASME Fluids Engineering Award. He lives with his wife, Jeanne, in Narragansett, Rhode Island.

v

To Jeanne

Contents

Preface xi

2.9 2.10

Chapter 1 Introduction 3 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11

Preliminary Remarks 3 The Concept of a Fluid 4 The Fluid as a Continuum 6 Dimensions and Units 7 Properties of the Velocity Field 15 Thermodynamic Properties of a Fluid 15 Viscosity and Other Secondary Properties 23 Basic Flow Analysis Techniques 39 Flow Patterns: Streamlines, Streaklines, and Pathlines 39 The Fundamentals of Engineering (FE) Examination 43 The History of Fluid Mechanics 43 Summary 44 Problems 45 Fundamentals of Engineering Exam Problems 52 Comprehensive Problems 53 References 56

Chapter 3 Integral Relations for a Control Volume 133 3.1 3.2 3.3 3.4 3.5 3.6 3.7

Chapter 2 Pressure Distribution in a Fluid 59 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Pressure and Pressure Gradient 59 Equilibrium of a Fluid Element 61 Hydrostatic Pressure Distributions 62 Application to Manometry 69 Hydrostatic Forces on Plane Surfaces 72 Hydrostatic Forces on Curved Surfaces 80 Hydrostatic Forces in Layered Fluids 83 Buoyancy and Stability 85

Pressure Distribution in Rigid-Body Motion 91 Pressure Measurement 99 Summary 103 Problems 103 Word Problems 126 Fundamentals of Engineering Exam Problems 126 Comprehensive Problems 127 Design Projects 129 References 130

Basic Physical Laws of Fluid Mechanics 133 The Reynolds Transport Theorem 137 Conservation of Mass 144 The Linear Momentum Equation 149 Frictionless Flow: The Bernoulli Equation 163 The Angular Momentum Theorem 172 The Energy Equation 178 Summary 189 Problems 189 Word Problems 216 Fundamentals of Engineering Exam Problems 217 Comprehensive Problems 218 Design Project 219 References 219

Chapter 4 Differential Relations for Fluid Flow 221 4.1 4.2

The Acceleration Field of a Fluid 222 The Differential Equation of Mass Conservation

224 vii

viii Contents 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10

The Differential Equation of Linear Momentum 230 The Differential Equation of Angular Momentum 237 The Differential Equation of Energy 238 Boundary Conditions for the Basic Equations 241 The Stream Function 246 Vorticity and Irrotationality 253 Frictionless Irrotational Flows 255 Some Illustrative Incompressible Viscous Flows 261 Summary 269 Problems 269 Word Problems 280 Fundamentals of Engineering Exam Problems 281 Comprehensive Problems 281 References 282

Chapter 5 Dimensional Analysis and Similarity 285 5.1 5.2 5.3 5.4 5.5

Introduction 285 The Principle of Dimensional Homogeneity 288 The Pi Theorem 294 Nondimensionalization of the Basic Equations 304 Modeling and Similarity 313 Summary 325 Problems 325 Word Problems 333 Fundamentals of Engineering Exam Problems 334 Comprehensive Problems 334 Design Projects 335 References 336

Chapter 6 Viscous Flow in Ducts 339 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9

Reynolds Number Regimes 339 Internal versus External Viscous Flows 344 Head Loss—The Friction Factor 347 Laminar Fully Developed Pipe Flow 349 Turbulence Modeling 351 Turbulent Pipe Flow 358 Four Types of Pipe Flow Problems 366 Flow in Noncircular Ducts 371 Minor or Local Losses in Pipe Systems 380

6.10 6.11 6.12

Multiple-Pipe Systems 389 Experimental Duct Flows: Diffuser Performance 395 Fluid Meters 400 Summary 421 Problems 422 Word Problems 440 Fundamentals of Engineering Exam Problems 441 Comprehensive Problems 442 Design Projects 444 References 444

Chapter 7 Flow Past Immersed Bodies 449 7.1 7.2 7.3 7.4 7.5 7.6

Reynolds Number and Geometry Effects 449 Momentum Integral Estimates 453 The Boundary Layer Equations 456 The Flat-Plate Boundary Layer 459 Boundary Layers with Pressure Gradient 468 Experimental External Flows 474 Summary 501 Problems 502 Word Problems 515 Fundamentals of Engineering Exam Problems 515 Comprehensive Problems 516 Design Project 517 References 517

Chapter 8 Potential Flow and Computational Fluid Dynamics 521 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9

Introduction and Review 521 Elementary Plane Flow Solutions 524 Superposition of Plane Flow Solutions 531 Plane Flow Past Closed-Body Shapes 537 Other Plane Potential Flows 547 Images 551 Airfoil Theory 554 Axisymmetric Potential Flow 562 Numerical Analysis 568 Summary 577 Problems 577 Word Problems 588

Contents Comprehensive Problems Design Projects 589 References 590

Problems 724 Word Problems 736 Fundamentals of Engineering Exam Problems Comprehensive Problems 736 Design Projects 738 References 738

588

Chapter 9 Compressible Flow 593 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10

Introduction: Review of Thermodynamics 593 The Speed of Sound 598 Adiabatic and Isentropic Steady Flow 600 Isentropic Flow with Area Changes 606 The Normal Shock Wave 613 Operation of Converging and Diverging Nozzles 621 Compressible Duct Flow with Friction 626 Frictionless Duct Flow with Heat Transfer 637 Mach Waves and Oblique Shock Waves 642 Prandtl-Meyer Expansion Waves 652 Summary 664 Problems 665 Word Problems 678 Fundamentals of Engineering Exam Problems 678 Comprehensive Problems 679 Design Projects 680 References 681

Chapter 10 Open-Channel Flow 683 10.1 10.2 10.3 10.4 10.5 10.6 10.7

Introduction 683 Uniform Flow; The Chézy Formula 689 Efficient Uniform-Flow Channels 695 Specific Energy; Critical Depth 697 The Hydraulic Jump 704 Gradually Varied Flow 708 Flow Measurement and Control by Weirs 716 Summary 723

ix

736

Chapter 11 Turbomachinery 741 11.1 11.2 11.3 11.4 11.5 11.6

Introduction and Classification 741 The Centrifugal Pump 744 Pump Performance Curves and Similarity Rules 750 Mixed- and Axial-Flow Pumps: The Specific Speed 760 Matching Pumps to System Characteristics 767 Turbines 775 Summary 789 Problems 791 Word Problems 804 Comprehensive Problems 804 Design Project 806 References 806

Appendix A Physical Properties of Fluids 808 Appendix B Compressible Flow Tables 813 Appendix C Conversion Factors 820 Appendix D Equations of Motion in Cylindrical Coordinates 822 Appendix E Estimating Uncertainty in Experimental Data 824 Answers to Selected Problems 826 Index 833

Preface

General Approach

The eighth edition of Fluid Mechanics sees some additions and deletions but no philosophical change. The basic outline of eleven chapters, plus appendices, remains the same. The triad of integral, differential, and experimental approaches is retained. Many problem exercises, and some fully worked examples, have been changed. The informal, student-oriented style is retained. A number of new photographs and figures have been added. Many new references have been added, for a total of 445. The writer is a firm believer in “further reading,” especially in the postgraduate years.

Learning Tools

The total number of problem exercises continues to increase, from 1089 in the first edition, to 1683 in this eighth edition. There are approximately 20 new problems in each chapter. Most of these are basic end-of-chapter problems, classified according to topic. There are also Word Problems, multiple-choice Fundamentals of Engineering Problems, Comprehensive Problems, and Design Projects. The appendix lists approximately 700 Answers to Selected Problems. The example problems are structured in the text to follow the sequence of recommended steps outlined in Section 1.7. Most of the problems in this text can be solved with a hand calculator. Some can even be simply explained in words. A few problems, especially in Chapters 6, 9, and 10, involve solving complicated algebraic expressions, laborious for a hand calculator. Check to see if your institution has a license for equation-solving software. Here the writer solves complicated example problems by using the iterative power of Microsoft Office Excel, as illustrated, for example, in Example 6.5. For further use in your work, Excel also contains several hundred special mathematical functions for engineering and statistics. Another benefit: Excel is free.

Content Changes

There are some revisions in each chapter. Chapter 1 has been substantially revised. The pre-reviewers felt, correctly, that it was too long, too detailed, and at too high a level for an introduction. Former Section 1.2, History of Fluid Mechanics, has been shortened and moved to the end of the chapter. Former Section 1.3, Problem-Solving Techniques, has been moved to appear just before Example 1.7, where these techniques are first used. Eulerian and Lagrangian descriptions have been moved to Chapter 4. A temperature-entropy chart for steam xi

xii Preface

Preface

xiii

Appendices B and D are unchanged. Appendix A adds a list of liquid kinematic viscosities to Table A.4. A few more conversion factors are added to Appendix C. There is a new Appendix E, Estimating Uncertainty in Experimental Data, which was moved from its inappropriate position in Chapter 1. The writer believes that “uncertainty” is vital to reporting measurements and always insisted upon it when he was an engineering journal editor.

Online Supplements

A number of supplements are available to instructors at McGraw-Hill’s Connect Engineering®. Instructors may obtain the text images in PowerPoint format and the full Solutions Manual in PDF format. The solutions manual provides complete and detailed solutions, including problem statements and artwork, to the end-of-chapter problems. Instructors can also obtain access to the Complete Online Solutions Manual Organization System (C.O.S.M.O.S.) for Fluid Mechanics, 8th edition. Instructors can use C.O.S.M.O.S. to create exams and assignments, to create custom content, and to edit supplied problems and solutions.

Acknowledgments

xiv

As usual, so many people have helped me that I may fail to list them all. Material help, in the form of photos, articles, and problems, came from Scott Larwood of the University of the Pacific; Sukanta Dash of the Indian Institute of Technology at Kharagpur; Mark Coffey of the Colorado School of Mines; Mac Stevens of Oregon State University; Stephen Carrington of Malvern Instruments; Carla Cioffi of NASA; Lisa Lee and Robert Pacquette of the Rhode Island Department of Environmental Management; Vanessa Blakeley and Samuel Schweighart of Terrafugia Inc.; Beric Skews of the University of the Witwatersrand, South Africa; Kelly Irene Knorr and John Merrill of the School of Oceanography at the University of Rhode Island; Adam Rein of Altaeros Energies Inc.; Dasari Abhinav of Anna University, India; Kris Allen of Transcanada Corporation; Bruce Findlayson of the University of Washington; Wendy Koch of USA Today; Liz  Boardman of the South County Independent; Beth Darchi and Colin McAteer of the American Society of Mechanical Engineers; Catherine Hines of the William Beebe Web Site; Laura Garrison of York College of Pennsylvania. The following pre-reviewers gave many excellent suggestions for improving the manuscript: Steve Baker, Naval Postgraduate School; Suresh Aggarwal, University of Illinois at Chicago; Edgar Caraballo, Miami University; Chang-Hwan Choi, Stevens Institute of Technology; Drazen Fabris, Santa Clara University; James Liburdy, Oregon State University; Daniel Maynes, Brigham Young University; Santosh Sahu, Indian Institute of Technology Indore; Brian Savilonis, Worcester Polytechnic Institute; Eric Savory, University of Western Ontario; Rick Sellens, Queen’s University; Gordon Stubley, University of Waterloo. Many others have supported me, throughout my revision efforts, with comments and suggestions: Gordon Holloway of the University of New Brunswick; David Taggart, Donna Meyer, Arun Shukla, and Richard Lessmann of the University of Rhode Island; Debendra K. Das of the University of Alaska–Fairbanks; Elizabeth Kenyon of Mathworks; Deborah V. Pence of Oregon State University; Sheldon Green of the University of British Columbia; Elena Bingham of the DuPont Corporation; Jane Bates of Broad Rock School; Kim Mather of West Kingston School; Nancy Dreier of Curtiss Corner School; Richard Kline, co-inventor of the Kline-Fogelman airfoil. The McGraw-Hill staff was, as usual, very helpful. Thanks are due to Bill Stenquist, Katherine Neubauer, Lorraine Buczek, Samantha Donisi-Hamm, Raghu Srinivasan, Tammy Juran, Thomas Scaife, and Lisa Bruflodt. Finally, I am thankful for the continuing support of my family, especially Jeanne, who remains in my heart, and my sister Sally White GNSH, my dog Jack, and my cats Cole and Kerry.

Fluid Mechanics

Falls on the Nesowadnehunk Stream in Baxter State Park, Maine, which is the northern terminus of the Appalachian Trail. Such flows, open to the atmosphere, are driven simply by gravity and do not depend much upon fluid properties such as density and viscosity. They are discussed later in Chap. 10. To the writer, one of the joys of fluid mechanics is that visualization of a fluid-flow process is simple and beautiful [Photo Credit: Design Pics/Natural Selection Robert Cable].

2

Chapter 1 Introduction

1.1 Preliminary Remarks

Fluid mechanics is the study of fluids either in motion (fluid dynamics) or at rest (fluid statics). Both gases and liquids are classified as fluids, and the number of fluid engineering applications is enormous: breathing, blood flow, swimming, pumps, fans, turbines, airplanes, ships, rivers, windmills, pipes, missiles, icebergs, engines, filters, jets, and sprinklers, to name a few. When you think about it, almost everything on this planet either is a fluid or moves within or near a fluid. The essence of the subject of fluid flow is a judicious compromise between theory and experiment. Since fluid flow is a branch of mechanics, it satisfies a set of welldocumented basic laws, and thus a great deal of theoretical treatment is available. However, the theory is often frustrating because it applies mainly to idealized situations, which may  be invalid in practical problems. The two chief obstacles to a workable theory are geometry and viscosity. The basic equations of fluid motion (Chap. 4) are too difficult to enable the analyst to attack arbitrary geometric configurations. Thus most textbooks concentrate on flat plates, circular pipes, and other easy geometries. It is possible to apply numerical computer techniques to complex geometries, and specialized textbooks are now available to explain the new computational fluid dynamics (CFD) approximations and methods [1–4].1 This book will present many theoretical results while keeping their limitations in mind. The second obstacle to a workable theory is the action of viscosity, which can be neglected only in certain idealized flows (Chap.  8). First, viscosity increases the difficulty of the basic equations, although the boundary-layer approximation found by Ludwig Prandtl in 1904 (Chap. 7) has greatly simplified viscous-flow analyses. Second, viscosity has a destabilizing effect on all fluids, giving rise, at frustratingly small velocities, to a disorderly, random phenomenon called turbulence. The theory of turbulent flow is crude and heavily backed up by experiment (Chap. 6), yet it can be quite serviceable as an engineering estimate. This textbook only introduces the standard experimental correlations for turbulent time-mean flow. Meanwhile, there are advanced texts on both time-mean turbulence and turbulence modeling [5, 6] and on the newer, computer-intensive direct numerical simulation (DNS) of fluctuating turbulence [7, 8]. 1

Numbered references appear at the end of each chapter.

3

4

Chapter 1 Introduction

Thus there is theory available for fluid flow problems, but in all cases it should be backed up by experiment. Often the experimental data provide the main source of information about specific flows, such as the drag and lift of immersed bodies (Chap.  7). Fortunately, fluid mechanics is a highly visual subject, with good instrumentation [9–11], and the use of dimensional analysis and modeling concepts (Chap. 5) is widespread. Thus experimentation provides a natural and easy complement to the theory. You should keep in mind that theory and experiment should go hand in hand in all studies of fluid mechanics.

1.2 The Concept of a Fluid

From the point of view of fluid mechanics, all matter consists of only two states, fluid and solid. The difference between the two is perfectly obvious to the layperson, and it is an interesting exercise to ask a layperson to put this difference into words. The technical distinction lies with the reaction of the two to an applied shear or tangential stress. A solid can resist a shear stress by a static deflection; a fluid cannot. Any shear stress applied to a fluid, no matter how small, will result in motion of that fluid. The fluid moves and deforms continuously as long as the shear stress is applied. As a corollary, we can say that a fluid at rest must be in a state of zero shear stress, a state often called the hydrostatic stress condition in structural analysis. In this condition, Mohr’s circle for stress reduces to a point, and there is no shear stress on any plane cut through the element under stress. Given this definition of a fluid, every layperson also knows that there are two classes of fluids, liquids and gases. Again the distinction is a technical one concerning the effect of cohesive forces. A liquid, being composed of relatively close-packed molecules with strong cohesive forces, tends to retain its volume and will form a free surface in a gravitational field if unconfined from above. Free-surface flows are dominated by gravitational effects and are studied in Chaps. 5 and 10. Since gas molecules are widely spaced with negligible cohesive forces, a gas is free to expand until it encounters confining walls. A gas has no definite volume, and when left to itself without confinement, a gas forms an atmosphere that is essentially hydrostatic. The hydrostatic behavior of liquids and gases is taken up in Chap.  2. Gases cannot form a free surface, and thus gas flows are rarely concerned with gravitational effects other than buoyancy. Figure 1.1 illustrates a solid block resting on a rigid plane and stressed by its own weight. The solid sags into a static deflection, shown as a highly exaggerated dashed line, resisting shear without flow. A free-body diagram of element A on the side of the block shows that there is shear in the block along a plane cut at an angle θ through  A. Since the block sides are unsupported, element A has zero stress on the left and right sides and compression stress σ 5 2p on the top and bottom. Mohr’s circle does not reduce to a point, and there is nonzero shear stress in the block. By contrast, the liquid and gas at rest in Fig.  1.1 require the supporting walls in order to eliminate shear stress. The walls exert a compression stress of 2p and reduce Mohr’s circle to a point with zero shear everywhere—that is, the hydrostatic condition. The liquid retains its volume and forms a free surface in the container. If the walls are removed, shear develops in the liquid and a big splash results. If the container is  tilted, shear again develops, waves form, and the free surface seeks a horizontal configuration, pouring out over the lip if necessary. Meanwhile, the gas is unrestrained

1.2 The Concept of a Fluid Free surface

Static deflection

A

A Solid

A Liquid

Gas

(a)

(c) p

σ1 θ

θ τ1

0

τ= 0

p 0

A

Fig. 1.1 A solid at rest can resist shear. (a) Static deflection of the solid; (b) equilibrium and Mohr’s circle for solid element A. A fluid cannot resist shear. (c) Containing walls are needed; (d ) equilibrium and Mohr’s circle for fluid element A.

5

p

A

–σ = p

–σ = p

τ

τ

(1) Hydrostatic condition

σ

–p

(b)

σ

–p

(d )

and expands out of the container, filling all available space. Element A in the gas is also hydrostatic and exerts a compression stress 2p on the walls. In the previous discussion, clear decisions could be made about solids, liquids, and gases. Most engineering fluid mechanics problems deal with these clear cases—that is, the common liquids, such as water, oil, mercury, gasoline, and alcohol, and the common gases, such as air, helium, hydrogen, and steam, in their common temperature and pressure ranges. There are many borderline cases, however, of which you should be aware. Some apparently “solid” substances such as asphalt and lead resist shear stress for short periods but actually deform slowly and exhibit definite fluid behavior over long periods. Other substances, notably colloid and slurry mixtures, resist small shear stresses but “yield” at large stress and begin to flow as fluids do. Specialized textbooks are devoted to this study of more general deformation and flow, a field called rheology [16]. Also, liquids and gases can coexist in two-phase mixtures, such as steam–water mixtures or water with entrapped air bubbles. Specialized textbooks present the analysis of such multiphase flows [17]. Finally, in some situations the distinction between a liquid and a gas blurs. This is the case at temperatures and

6

Chapter 1 Introduction

pressures above the so-called critical point of a substance, where only a single phase exists, primarily resembling a gas. As pressure increases far above the critical point, the gaslike substance becomes so dense that there is some resemblance to a liquid, and the usual thermodynamic approximations like the perfect-gas law become inaccurate. The critical temperature and pressure of water are Tc 5 647 K and pc 5 219 atm (atmosphere)2 so that typical problems involving water and steam are below the critical point. Air, being a mixture of gases, has no distinct critical point, but its principal component, nitrogen, has Tc 5 126 K and pc 5 34 atm. Thus typical problems involving air are in the range of high temperature and low pressure where air is distinctly and definitely a gas. This text will be concerned solely with clearly identifiable liquids and gases, and the borderline cases just discussed will be beyond our scope.

1.3 The Fluid as a Continuum

We have already used technical terms such as fluid pressure and density without a rigorous discussion of their definition. As far as we know, fluids are aggregations of molecules, widely spaced for a gas, closely spaced for a liquid. The distance between molecules is very large compared with the molecular diameter. The molecules are not fixed in a lattice but move about freely relative to each other. Thus fluid density, or mass per unit volume, has no precise meaning because the number of molecules occupying a given volume continually changes. This effect becomes unimportant if the unit volume is large compared with, say, the cube of the molecular spacing, when the number of molecules within the volume will remain nearly constant in spite of the enormous interchange of particles across the boundaries. If, however, the chosen unit volume is too large, there could be a noticeable variation in the bulk aggregation of the particles. This situation is illustrated in Fig. 1.2, where the “density” as calculated from molecular mass δm within a given volume δ 9 is plotted versus the size of the unit volume. There is a limiting volume δ 9* below which molecular variations may be important and above which aggregate variations may be important. The density ρ of a fluid is best defined as ρ5

lim

δ 9Sδ 9*

ρ Elemental volume

ρ = 1000 kg/m3

δm δ9

(1.1)

Microscopic uncertainty Macroscopic uncertainty

ρ = 1100 δυ 1200 ρ = 1200

Fig. 1.2 The limit definition of continuum fluid density: (a) an elemental volume in a fluid region of variable continuum density; (b) calculated density versus size of the elemental volume.

ρ = 1300 0

δ9* ≈ 10–9 mm3

Region containing fluid (a) 2

One atmosphere equals 2116 lbf/ft2 5 101,300 Pa.

(b)

δ9

1.4 Dimensions and Units

7

The limiting volume δ 9* is about 1029 mm3 for all liquids and for gases at atmospheric pressure. For example, 1029 mm3 of air at standard conditions contains approximately 3 3 107 molecules, which is sufficient to define a nearly constant density according to Eq. (1.1). Most engineering problems are concerned with physical dimensions much larger than this limiting volume, so that density is essentially a point function and fluid properties can be thought of as varying continually in space, as sketched in Fig. 1.2a. Such a fluid is called a continuum, which simply means that its variation in properties is so smooth that differential calculus can be used to analyze the substance. We shall assume that continuum calculus is valid for all the analyses in this book. Again there are borderline cases for gases at such low pressures that molecular spacing and mean free path3 are comparable to, or larger than, the physical size of the system. This requires that the continuum approximation be dropped in favor of a molecular theory of rarefied gas flow [18]. In principle, all fluid mechanics problems can be attacked from the molecular viewpoint, but no such attempt will be made here. Note that the use of continuum calculus does not preclude the possibility of discontinuous jumps in fluid properties across a free surface or fluid interface or across a shock wave in a compressible fluid (Chap. 9). Our calculus in analyzing fluid flow must be flexible enough to handle discontinuous boundary conditions.

1.4 Dimensions and Units

A dimension is the measure by which a physical variable is expressed quantitatively. A unit is a particular way of attaching a number to the quantitative dimension. Thus length is a dimension associated with such variables as distance, displacement, width, deflection, and height, while centimeters and inches are both numerical units for expressing length. Dimension is a powerful concept about which a splendid tool called dimensional analysis has been developed (Chap.  5), while units are the numerical quantity that the customer wants as the final answer. In 1872 an international meeting in France proposed a treaty called the Metric Convention, which was signed in 1875 by 17 countries including the United States. It was an improvement over British systems because its use of base 10 is the foundation of our number system, learned from childhood by all. Problems still remained because even the metric countries differed in their use of kiloponds instead of dynes or newtons, kilograms instead of grams, or calories instead of joules. To standardize the metric system, a General Conference of Weights and Measures, attended in 1960 by 40 countries, proposed the International System of Units (SI). We are now undergoing a painful period of transition to SI, an adjustment that may take many more years to complete. The professional societies have led the way. Since July 1, 1974, SI units have been required by all papers published by the American Society of Mechanical Engineers, and there is a textbook explaining the SI [19]. The present text will use SI units together with British gravitational (BG) units.

Primary Dimensions

In fluid mechanics there are only four primary dimensions from which all other dimensions can be derived: mass, length, time, and temperature.4 These dimensions and 3

The mean distance traveled by molecules between collisions (see Prob. P1.5). If electromagnetic effects are important, a fifth primary dimension must be included, electric current {I}, whose SI unit is the ampere (A). 4

8

Chapter 1 Introduction

Table  1.1 Primary Dimensions in SI and BG Systems

Primary dimension Mass {M} Length {L} Time {T} Temperature {Θ}

SI unit

BG unit

Kilogram (kg) Meter (m) Second (s) Kelvin (K)

Slug Foot (ft) Second (s) Rankine (8R)

Conversion factor 1 1 1 1

slug 5 14.5939 kg ft 5 0.3048 m s51s K 5 1.88R

their units in both systems are given in Table  1.1. Note that the Kelvin unit uses no degree symbol. The braces around a symbol like {M} mean “the dimension” of mass. All other variables in fluid mechanics can be expressed in terms of {M}, {L}, {T}, and {Θ}. For example, acceleration has the dimensions {LT 22}. The most crucial of these secondary dimensions is force, which is directly related to mass, length, and time by Newton’s second law. Force equals the time rate of change of momentum or, for constant mass, F 5 ma From this we see that, dimensionally, {F} 5 {MLT

The International System (SI)

(1.2) 22

}.

The use of a constant of proportionality in Newton’s law, Eq.  (1.2), is avoided by defining the force unit exactly in terms of the other basic units. In the SI system, the basic units are newtons {F}, kilograms {M}, meters {L}, and seconds {T}. We define 1 newton of force 5 1 N 5 1 kg # 1 m/s2 The newton is a relatively small force, about the weight of an apple (0.225 lbf ). In addition, the basic unit of temperature {®} in the SI system is the degree Kelvin, K. Use of these SI units (N, kg, m, s, K) will require no conversion factors in our equations.

The British Gravitational (BG) System

In the BG system also, a constant of proportionality in Eq. (1.2) is avoided by defining the force unit exactly in terms of the other basic units. In the BG system, the basic units are pound-force {F}, slugs {M}, feet {L}, and seconds {T}. We define 1 pound of force 5 1 lbf 5 1 slug # 1 ft/s2 One lbf < 4.4482 N and approximates the weight of four apples. We will use the abbreviation lbf for pound-force and lbm for pound-mass. The slug is a rather hefty  mass, equal to 32.174 lbm. The basic unit of temperature {Θ} in the BG system is the degree Rankine, 8R. Recall that a temperature difference 1 K 5 1.88R. Use of these BG units (lbf, slug, ft, s, 8R) will require no conversion factors in our equations.

Other Unit Systems

There are other unit systems still in use. At least one needs no proportionality constant: the CGS system (dyne, gram, cm, s, K). However, CGS units are too small for most applications (1 dyne 5 1025 N) and will not be used here.

1.4 Dimensions and Units Table  1.2 Secondary Dimensions in Fluid Mechanics

Secondary dimension 2

Area {L } Volume {L3} Velocity {LT 21} Acceleration {LT 22} Pressure or stress {ML21T 22} Angular velocity {T 21} Energy, heat, work {ML2T 22} Power {ML2T 23} Density {ML23} Viscosity {ML21T 21} Specific heat {L2T 22Q21}

SI unit 2

m m3 m/s m/s2 Pa 5 N/m2 s21 J5N∙m W 5 J/s kg/m3 kg/(m ∙ s) m2/(s2 ∙ K)

BG unit 2

ft ft3 ft/s ft/s2 lbf/ft2 s21 ft ∙ lbf ft ∙ lbf/s slugs/ft3 slugs/(ft ∙ s) ft2/(s2 ∙ 8R)

9

Conversion factor 1 1 1 1 1 1 1 1 1 1 1

2

m 5 10.764 ft2 m3 5 35.315 ft3 ft/s 5 0.3048 m/s ft/s2 5 0.3048 m/s2 lbf/ft2 5 47.88 Pa s21 5 1 s21 ft ∙ lbf 5 1.3558 J ft ∙ lbf/s 5 1.3558 W slug/ft3 5 515.4 kg/m3 slug/(ft ∙ s) 5 47.88 kg/(m ∙ s) m2/(s2 ∙ K) 5 5.980 ft2/(s2 ∙ 8R)

In the USA, some still use the English Engineering system (lbf, lbm, ft, s, 8R), where the basic mass unit is the pound of mass. Newton’s law (1.2) must be rewritten: ma ft # lbm F5 , where gc 5 32.174 (1.3) gc lbf # s2 The constant of proportionality, gc, has both dimensions and a numerical value not equal to 1.0. The present text uses only the SI and BG systems and will not solve problems or examples in the English Engineering system. Because Americans still use them, a few problems in the text will be stated in truly awkward units: acres, gallons, ounces, or miles. Your assignment will be to convert these and solve in the SI or BG systems.

The Principle of Dimensional Homogeneity

In engineering and science, all equations must be dimensionally homogeneous, that is, each additive term in an equation must have the same dimensions. For example, take Bernoulli’s incompressible equation, to be studied and used throughout this text: p1

1 2 ρV 1 ρgZ 5 constant 2

Each and every term in this equation must have dimensions of pressure {ML21T 22}. We will examine the dimensional homogeneity of this equation in detail in Example 1.3. A list of some important secondary variables in fluid mechanics, with dimensions derived as combinations of the four primary dimensions, is given in Table 1.2. A more complete list of conversion factors is given in App. C.

EXAMPLE 1.1 A body weighs 1000 lbf when exposed to a standard earth gravity g 5 32.174 ft/s2. (a) What is its mass in kg? (b) What will the weight of this body be in N if it is exposed to the moon’s standard acceleration gmoon 5 1.62 m/s2? (c) How fast will the body accelerate if a net force of 400 lbf is applied to it on the moon or on the earth?

10

Chapter 1 Introduction

Solution We need to find the (a) mass; (b) weight on the moon; and (c) acceleration of this body. This is a fairly simple example of conversion factors for differing unit systems. No property data is needed. The example is too low-level for a sketch.

Part (a)

Newton’s law (1.2) holds with known weight and gravitational acceleration. Solve for m: F 5 W 5 1000 lbf 5 mg 5 (m) (32.174 ft/s2 ), or m 5

1000 lbf 5 31.08 slugs 32.174 ft/s2

Convert this to kilograms: m 5 31.08 slugs 5 (31.08 slugs) (14.5939 kg/slug) 5 454 kg

Part (b)

The mass of the body remains 454 kg regardless of its location. Equation (1.2) applies with a new gravitational acceleration and hence a new weight: F 5 Wmoon 5 mgmoon 5 (454 kg) (1.62 m/s2 ) 5 735 N 5 165 lbf

Part (c)

Ans. (a)

Ans. (b)

This part does not involve weight or gravity or location. It is simply an application of Newton’s law with a known mass and known force: F 5 400 lbf 5 ma 5 (31.08 slugs) a Solve for a5 Comment (c):

400 lbf ft m m 5 12.87 2 a0.3048 b 5 3.92 2 31.08 slugs ft s s

Ans. (c)

This acceleration would be the same on the earth or moon or anywhere.

Many data in the literature are reported in inconvenient or arcane units suitable only to some industry or specialty or country. The engineer should convert these data to the SI or BG system before using them. This requires the systematic application of conversion factors, as in the following example. EXAMPLE 1.2 Industries involved in viscosity measurement [27, 29] continue to use the CGS system of units, since centimeters and grams yield convenient numbers for many fluids. The absolute viscosity (μ) unit is the poise, named after J. L. M. Poiseuille, a French physician who in 1840 performed pioneering experiments on water flow in pipes; 1 poise 5 1 g/(cm-s). The kinematic viscosity (ν) unit is the stokes, named after G. G. Stokes, a British physicist who in 1845 helped develop the basic partial differential equations of fluid momentum; 1 stokes 5 1 cm2/s. Water at 208C has μ < 0.01 poise and also ν < 0.01 stokes. Express these results in (a) SI and (b) BG units.

Solution Part (a)

• Approach: Systematically change grams to kg or slugs and change centimeters to meters or feet.

1.4 Dimensions and Units

11

• Property values: Given μ 5 0.01 g/(cm-s) and ν 5 0.01 cm2/s. • Solution steps: (a) For conversion to SI units, g(1 kg/1000 g) g kg μ 5 0.01 cm # s 5 0.01 5 0.001 m # s cm(0.01 m/cm)s ν 5 0.01

Part (b)

cm2 (0.01 m/cm) 2 cm2 m2 5 0.01 5 0.000001 s s s

Ans. (a)

• For conversion to BG units μ 5 0.01

g slug g(1 kg/1000 g) (1 slug/14.5939 kg) 5 0.0000209 # 5 0.01 cm # s (0.01 m/cm) (1 ft/0.3048 m)s ft s

ν 5 0.01

cm2 (0.01 m/cm) 2 (1 ft/0.3048 m) 2 ft2 cm2 5 0.01 5 0.0000108 s s s

Ans. (b)

• Comments: This was a laborious conversion that could have been shortened by using the direct viscosity conversion factors in App. C or the inside front cover. For example, μBG 5 μSI/47.88.

We repeat our advice: Faced with data in unusual units, convert them immediately to either SI or BG units because (1) it is more professional and (2) theoretical equations in fluid mechanics are dimensionally consistent and require no further conversion factors when these two fundamental unit systems are used, as the following example shows. EXAMPLE 1.3 A useful theoretical equation for computing the relation between pressure, velocity, and altitude in a steady flow of a nearly inviscid, nearly incompressible fluid with negligible heat transfer and shaft work5 is the Bernoulli relation, named after Daniel Bernoulli, who published a hydrodynamics textbook in 1738: p0 5 p 1 12 ρV2 1 ρgZ where p0 p V ρ Z g

5 5 5 5 5 5

(1)

stagnation pressure pressure in moving fluid velocity density altitude gravitational acceleration

(a) Show that Eq. (1) satisfies the principle of dimensional homogeneity, which states that all additive terms in a physical equation must have the same dimensions. (b) Show that consistent units result without additional conversion factors in SI units. (c) Repeat (b) for BG units.

5

That’s an awful lot of assumptions, which need further study in Chap.  3.

12

Chapter 1 Introduction

Solution Part (a)

We can express Eq. (1) dimensionally, using braces, by entering the dimensions of each term from Table 1.2: 5ML 21T 22 6 5 5ML 21T 22 6 1 5ML 23 65L2T 22 6 1 5ML 23 65LT 22 65L6 5 5ML 21T 22 6 for all terms

Part (b)

Ans. (a)

Enter the SI units for each quantity from Table 1.2: 5N/m2 6 5 5N/m2 6 1 5kg/m3 65m2/s2 6 1 5kg/m3 65m/s2 65m6 5 5N/m2 6 1 5kg/(m # s2 ) 6 The right-hand side looks bad until we remember from Eq. (1.3) that 1 kg 5 1 N ∙ s2/m. 5kg/(m # s2 ) 6 5

5N # s2/m6 5 5N/m2 6 5m # s2 6

Ans. (b)

Thus all terms in Bernoulli’s equation will have units of pascals, or newtons per square meter, when SI units are used. No conversion factors are needed, which is true of all theoretical equations in fluid mechanics.

Part (c)

Introducing BG units for each term, we have 5lbf/ft2 6 5 5lbf/ft2 6 1 5slugs/ft3 65ft2/s2 6 1 5slugs/ft3 65ft/s2 65ft6 5 5lbf/ft2 6 1 5slugs/(ft # s2 ) 6 But, from Eq. (1.3), 1 slug 5 1 lbf ∙ s2/ft, so that 5slugs/(ft # s2 ) 6 5

5lbf # s2/ft6 5 5lbf/ft2 6 5ft # s2 6

Ans. (c)

All terms have the unit of pounds-force per square foot. No conversion factors are needed in the BG system either.

There is still a tendency in English-speaking countries to use pound-force per square inch as a pressure unit because the numbers are more manageable. For example, standard atmospheric pressure is 14.7 lbf/in2 5 2116 lbf/ft2 5 101,300 Pa. The pascal is a small unit because the newton is less than 14 lbf and a square meter is a very large area.

Consistent Units

Note that not only must all (fluid) mechanics equations be dimensionally homogeneous, one must also use consistent units; that is, each additive term must have the same units. There is no trouble doing this with the SI and BG systems, as in  Example 1.3, but woe unto those who try to mix colloquial English units. For example, in Chap. 9, we often use the assumption of steady adiabatic compressible gas flow: h 1 12V 2 5 constant

1.4 Dimensions and Units

13

where h is the fluid enthalpy and V2/2 is its kinetic energy per unit mass. Colloquial thermodynamic tables might list h in units of British thermal units per pound mass (Btu/lb), whereas V is likely used in ft/s. It is completely erroneous to add Btu/lb to ft2/s2. The proper unit for h in this case is ft ∙ lbf/slug, which is identical to ft2/s2. The conversion factor is 1 Btu/lb < 25,040 ft2/s2 5 25,040 ft ∙ lbf/slug.

Homogeneous versus Dimensionally Inconsistent Equations

All theoretical equations in mechanics (and in other physical sciences) are dimensionally homogeneous; that is, each additive term in the equation has the same dimensions. However, the reader should be warned that many empirical formulas in the engineering literature, arising primarily from correlations of data, are dimensionally inconsistent. Their units cannot be reconciled simply, and some terms may contain hidden variables. An example is the formula that pipe valve manufacturers cite for liquid volume flow rate Q (m3/s) through a partially open valve: Q 5 CV a

Table  1.3 Convenient Prefixes for  Engineering Units Multiplicative factor

Prefix

Symbol

1012 109 106 103 102 10 1021 1022 1023 1026 1029 10212 10215 10218

tera giga mega kilo hecto deka deci centi milli micro nano pico femto atto

T G M k h da d c m μ n p f a

Convenient Prefixes in Powers of 10

¢p 1/2 b SG

where Dp is the pressure drop across the valve and SG is the specific gravity of the liquid (the ratio of its density to that of water). The quantity CV is the valve flow coefficient, which manufacturers tabulate in their valve brochures. Since SG is dimensionless {1}, we see that this formula is totally inconsistent, with one side being a flow rate {L3/T} and the other being the square root of a pressure drop {M1/2/L1/2T}. It follows that CV must have dimensions, and rather odd ones at that: {L7/2/M1/2}. Nor is the resolution of this discrepancy clear, although one hint is that the values of CV in the literature increase nearly as the square of the size of the valve. The presentation of experimental data in homogeneous form is the subject of dimensional analysis (Chap.  5). There we shall learn that a homogeneous form for the valve flow relation  is Q 5 Cd Aopeninga

¢p 1/2 b ρ

where ρ is the liquid density and A the area of the valve opening. The discharge coefficient Cd is dimensionless and changes only moderately with valve size. Please believe—until we establish the fact in Chap.  5—that this latter is a much better formulation of the data. Meanwhile, we conclude that dimensionally inconsistent equations, though they occur in engineering practice, are misleading and vague and even dangerous, in the sense that they are often misused outside their range of applicability. Engineering results often are too small or too large for the common units, with too many zeros one way or the other. For example, to write p 5 114,000,000 Pa is long and awkward. Using the prefix “M” to mean 106, we convert this to a concise p 5 114 MPa (megapascals). Similarly, t 5 0.000000003 s is a proofreader’s nightmare compared to the equivalent t 5 3 ns (nanoseconds). Such prefixes are common and convenient, in both the SI and BG systems. A complete list is given in Table 1.3.

14

Chapter 1 Introduction

EXAMPLE 1.4 In 1890 Robert Manning, an Irish engineer, proposed the following empirical formula for the average velocity V in uniform flow due to gravity down an open channel (BG units): V5

1.49 2/3 1/2 R S n

(1)

where R 5 hydraulic radius of channel (Chaps. 6 and 10) S 5 channel slope (tangent of angle that bottom makes with horizontal) n 5 Manning’s roughness factor (Chap. 10) and n is a constant for a given surface condition for the walls and bottom of the channel. (a) Is Manning’s formula dimensionally consistent? (b) Equation (1) is commonly taken to be valid in BG units with n taken as dimensionless. Rewrite it in SI form.

Solution • Assumption: The channel slope S is the tangent of an angle and is thus a dimensionless ratio with the dimensional notation {1}—that is, not containing M, L, or T. • Approach (a): Rewrite the dimensions of each term in Manning’s equation, using brackets {}: 5V6 5 e

1.49 L 1.49 f 5R2/3 6 5S1/2 6 or  e f 5 e f 5L2/3 6 516 n n T

This formula is incompatible unless {1.49/n} 5 {L1/3/T}. If n is dimensionless (and it is never listed with units in textbooks), the number 1.49 must carry the dimensions of {L1/3/T}. Ans. (a) • Comment (a): Formulas whose numerical coefficients have units can be disastrous for engineers working in a different system or another fluid. Manning’s formula, though popular, is inconsistent both dimensionally and physically and is valid only for water flow with certain wall roughnesses. The effects of water viscosity and density are hidden in the numerical value 1.49. • Approach (b): Part (a) showed that 1.49 has dimensions. If the formula is valid in BG units, then it must equal 1.49 ft1/3/s. By using the SI conversion for length, we obtain (1.49 ft1/3/s) (0.3048 m/ft) 1/3 5 1.00 m1/3/s Therefore, Manning’s inconsistent formula changes form when converted to the SI system: SI units: V 5

1.0 2/3 1/2 R S n

Ans. (b)

with R in meters and V in meters per second. • Comment (b): Actually, we misled you: This is the way Manning, a metric user, first proposed the formula. It was later converted to BG units. Such dimensionally inconsistent formulas are dangerous and should either be reanalyzed or treated as having very limited application.

1.6 Thermodynamic Properties of a Fluid

15

1.5 Properties of the Velocity Field

In a given flow situation, the determination, by experiment or theory, of the properties of the fluid as a function of position and time is considered to be the solution to the problem. In almost all cases, the emphasis is on the space–time distribution of the fluid properties. One rarely keeps track of the actual fate of the specific fluid particles. This treatment of properties as continuum-field functions distinguishes fluid mechanics from solid mechanics, where we are more likely to be interested in the trajectories of individual particles or systems.

The Velocity Field

Foremost among the properties of a flow is the velocity field V(x, y, z, t). In fact, determining the velocity is often tantamount to solving a flow problem, since other properties follow directly from the velocity field. Chapter 2 is devoted to the calculation of the pressure field once the velocity field is known. Books on heat transfer (for example, Ref. 20) are largely devoted to finding the temperature field from known velocity fields. In general, velocity is a vector function of position and time and thus has three components u, v, and w, each a scalar field in itself: V( x, y, z, t) 5 iu(x, y, z, t) 1 jv(x, y, z, t) 1 kw(x, y, z, t)

(1.4)

The use of u, v, and w instead of the more logical component notation Vx, Vy, and Vz is the result of an almost unbreakable custom in fluid mechanics. Much of this textbook, especially Chaps. 4, 7, 8, and 9, is concerned with finding the distribution of the velocity vector V for a variety of practical flows.

The Acceleration Field

The acceleration vector, a 5 dV/dt, occurs in Newton’s law for a fluid and thus is very important. In order to follow a particle in the Eulerian frame of reference, the final result for acceleration is nonlinear and quite complicated. Here we only give the formula: a5

dV 0V 0V 0V 0V 5 1u 1v 1w dt 0t 0x 0y 0z

(1.5)

where (u, v, w) are the velocity components from Eq. (1.4). We shall study this formula in detail in Chap. 4. The last three terms in Eq. (1.5) are nonlinear products and greatly complicate the analysis of general fluid motions, especially viscous flows.

1.6 Thermodynamic Properties of a Fluid

While the velocity field V is the most important fluid property, it interacts closely with the thermodynamic properties of the fluid. We have already introduced into the discussion the three most common such properties: 1. Pressure p 2. Density ρ 3. Temperature T

16

Chapter 1 Introduction

These three are constant companions of the velocity vector in flow analyses. Four other intensive thermodynamic properties become important when work, heat, and energy balances are treated (Chaps. 3 and 4): 4. 5. 6. 7.

Internal energy û Enthalpy h 5 û 1 p/ρ Entropy s Specific heats cp and cv

In addition, friction and heat conduction effects are governed by the two so-called transport properties: 8. Coefficient of viscosity μ 9. Thermal conductivity k All nine of these quantities are true thermodynamic properties that are determined by the thermodynamic condition or state of the fluid. For example, for a single-phase substance such as water or oxygen, two basic properties such as pressure and temperature are sufficient to fix the value of all the others: ρ 5 ρ(p, T)   h 5 h(p, T)   μ 5 μ(p, T) and so on for every quantity in the list. Note that the specific volume, so important in thermodynamic analyses, is omitted here in favor of its inverse, the density ρ. Recall that thermodynamic properties describe the state of a system—that is, a collection of matter of fixed identity that interacts with its surroundings. In most cases here the system will be a small fluid element, and all properties will be assumed to be continuum properties of the flow field: ρ 5 ρ(x, y, z, t), and so on. Recall also that thermodynamics is normally concerned with static systems, whereas fluids are usually in variable motion with constantly changing properties. Do the properties retain their meaning in a fluid flow that is technically not in equilibrium? The answer is yes, from a statistical argument. In gases at normal pressure (and even more so for liquids), an enormous number of molecular collisions occur over a very short distance of the order of 1 μm, so that a fluid subjected to sudden changes rapidly adjusts itself toward equilibrium. We therefore assume that all the thermodynamic properties just listed exist as point functions in a flowing fluid and follow all the laws and state relations of ordinary equilibrium thermodynamics. There are, of course, important nonequilibrium effects such as chemical and nuclear reactions in flowing fluids, which are not treated in this text.

Pressure

Pressure is the (compression) stress at a point in a static fluid (Fig.  1.3). Next to velocity, the pressure p is the most dynamic variable in fluid mechanics. Differences or gradients in pressure often drive a fluid flow, especially in ducts. In low-speed flows, the actual magnitude of the pressure is often not important, unless it drops so low as to cause vapor bubbles to form in a liquid. For convenience, we set many such problem assignments at the level of 1 atm 5 2116 lbf/ft2 5 101,300 Pa. High-speed (compressible) gas flows (Chap.  9), however, are indeed sensitive to the magnitude of pressure.

1.6 Thermodynamic Properties of a Fluid

Temperature

17

Temperature T is related to the internal energy level of a fluid. It may vary considerably during high-speed flow of a gas (Chap. 9). Although engineers often use Celsius or Fahrenheit scales for convenience, many applications in this text require absolute (Kelvin or Rankine) temperature scales: °R 5 °F 1 459.69 K 5 °C 1 273.16 If temperature differences are strong, heat transfer may be important [20], but our concern here is mainly with dynamic effects.

Density

The density of a fluid, denoted by ρ (lowercase Greek rho), is its mass per unit volume. Density is highly variable in gases and increases nearly proportionally to the pressure level. Density in liquids is nearly constant; the density of water (about 1000 kg/m3) increases only 1 percent if the pressure is increased by a factor of 220. Thus most liquid flows are treated analytically as nearly “incompressible.” In general, liquids are about three orders of magnitude more dense than gases at atmospheric pressure. The heaviest common liquid is mercury, and the lightest gas is hydrogen. Compare their densities at 208C and 1 atm: Mercury: ρ 5 13,580 kg/m3   Hydrogen: ρ 5 0.0838 kg/m3 They differ by a factor of 162,000! Thus, the physical parameters in various liquid and gas flows might vary considerably. The differences are often resolved by the use of dimensional analysis (Chap.  5). Other fluid densities are listed in Tables A.3 and A.4 (in App. A), and in Ref. 21.

Specific Weight

The specific weight of a fluid, denoted by γ (lowercase Greek gamma), is its weight per unit volume. Just as a mass has a weight W 5 mg, density and specific weight are simply related by gravity: γ 5 ρg (1.6) The units of γ are weight per unit volume, in lbf/ft3 or N/m3. In standard earth gravity, g 5 32.174 ft/s2 5 9.807 m/s2. Thus, for example, the specific weights of air and water at 208C and 1 atm are approximately γair 5 (1.205 kg/m3 ) (9.807 m/s2 ) 5 11.8 N/m3 5 0.0752 lbf/ft3 γwater 5 (998 kg/m3 ) (9.807 m/s2 ) 5 9790 N/m3 5 62.4 lbf/ft3 Specific weight is very useful in the hydrostatic pressure applications of Chap.  2. Specific weights of other fluids are given in Tables A.3 and A.4.

Specific Gravity

Specific gravity, denoted by SG, is the ratio of a fluid density to a standard reference fluid, usually water at 48C (for liquids) and air (for gases): SGgas 5 SGliquid 5

ρgas ρair

5

ρliquid ρwater

ρgas 1.205 kg/m3

5

ρliquid 1000 kg/m3

(1.7)

18

Chapter 1 Introduction

For example, the specific gravity of mercury (Hg) is SGHg 5 13,580/1000 < 13.6. Engineers find these dimensionless ratios easier to remember than the actual numerical values of density of a variety of fluids.

Potential and Kinetic Energies

In thermostatics the only energy in a substance is that stored in a system by molecular activity and molecular bonding forces. This is commonly denoted as internal energy û. A commonly accepted adjustment to this static situation for fluid flow is to add two more energy terms that arise from newtonian mechanics: potential energy and kinetic energy. The potential energy equals the work required to move the system of mass m from the origin to a position vector r 5 ix 1 jy 1 kz against a gravity field g. Its value is 2mg ? r, or 2g ? r per unit mass. The kinetic energy equals the work required to change the speed of the mass from zero to velocity V. Its value is 12 mV 2 or 12 V 2 per unit mass. Then by common convention the total stored energy e per unit mass in fluid mechanics is the sum of three terms: e 5 uˆ 1 12V 2 1 (–g ? r)

(1.8)

Also, throughout this book we shall define z as upward, so that g 5 2gk and g ? r  5 2gz. Then Eq. (1.8) becomes e 5 uˆ 1 12V 2 1 gz

(1.9)

The molecular internal energy û is a function of T and p for the single-phase pure substance, whereas the potential and kinetic energies are kinematic quantities.

State Relations for Gases

Thermodynamic properties are found both theoretically and experimentally to be related to each other by state relations that differ for each substance. As mentioned, we shall confine ourselves here to single-phase pure substances, such as water in its liquid phase. The second most common fluid, air, is a mixture of gases, but since the mixture ratios remain nearly constant between 160 and 2200 K, in this temperature range air can be considered to be a pure substance. All gases at high temperatures and low pressures (relative to their critical point) are in good agreement with the perfect-gas law p 5 ρRT  R 5 cp 2 cv 5 gas constant

(1.10)

where the specific heats cp and cv are defined in Eqs. (1.14) and (1.15). Since Eq. (1.10) is dimensionally consistent, R has the same dimensions as specific heat, {L2T22Θ21}, or velocity squared per temperature unit (kelvin or degree Rankine). Each gas has its own constant R, equal to a universal constant Λ divided by the molecular weight Rgas 5

¶ Mgas

(1.11)

1.6 Thermodynamic Properties of a Fluid

19

where Λ 5 49,700 ft-lbf/(slugmol ∙ 8R) 5 8314 J/(kmol ∙ K). Most applications in this book are for air, whose molecular weight is M 5 28.97/mol: 49,700 ft # lbf/(slugmol # °R) ft # lbf ft2 m2 Rair 5 5 1716 5 1716 2 (1.12) 5 287 2 # # 28.97/mol slug °R s °R s K Standard atmospheric pressure is 2116 lbf/ft2 5 2116 slug/(ft ∙ s2), and standard temperature is 608F 5 5208R. Thus standard air density is 2116 slug/(ft # s2 ) ρair 5 5 0.00237 slug/ft3 5 1.22 kg/m3 (1.13) 31716 ft2/(s2 # °R) 4 (520°R) This is a nominal value suitable for problems. For other gases, see Table A.4. Most of the common gases—oxygen, nitrogen, hydrogen, helium, argon—are nearly ideal. This is not so true for steam, whose simplified temperature-entropy chart is shown in Fig.  1.3. Unless you are sure that the steam temperature is “high” and the pressure “low,” it is best to use the Steam Tables to make accurate calculations. One proves in thermodynamics that Eq. (1.10) requires that the internal molecular energy û of a perfect gas vary only with temperature: û 5 û(T ). Therefore, the specific heat cv also varies only with temperature: cv 5 a

0û dû b 5 5 cv (T ) 0T ρ dT

dû 5 cv (T )dT

or

(1.14)

In like manner h and cp of a perfect gas also vary only with temperature: p h 5 û 1 5 û 1 RT 5 h(T ) ρ 0h dh b 5 5 cp (T ) 0T p dT dh 5 cp ( T )dT

cp 5 a

(1.15)

800 700 p = 22,060 kPa = 10,000 = 1,000 = 100 = 10

Fig. 1.3 Temperature-entropy chart for steam. The critical point is pc 5 22,060 kPa, Tc 5 3748C, Sc 5 4.41 kJ/(kg ∙ K). Except near the critical point, the smooth isobars tempt one to assume, often incorrectly, that the ideal-gas law is valid for steam. It is not, except at low pressure and high temperature: the upper right of the graph.

Temperature, °C

600 500 400

Critical point

300 200 Two-phase region 100 0

1

2

3

4

5

6

7

Entropy, kJ/(kg . K)

8

9

10

11

20

Chapter 1 Introduction

The ratio of specific heats of a perfect gas is an important dimensionless parameter in compressible flow analysis (Chap. 9) k5

cp 5 k(T ) \$ 1 cv

(1.16)

As a first approximation in airflow analysis we commonly take cp, cv, and k to be constant: kair < 1.4 R cv 5 < 4293 ft2/(s2 # °R) 5 718 m2/(s2 # K) k21 kR < 6010 ft2/(s2 # °R) 5 1005 m2/(s2 # K) cp 5 k21

(1.17)

Actually, for all gases, cp and cv increase gradually with temperature, and k decreases gradually. Experimental values of the specific-heat ratio for eight common gases are shown in Fig. 1.4. Nominal values are in Table A.4.

1.7 Ar 1.6

Atmospheric pressure 1.5

H2

1.4 cp k= c υ

CO

1.3

O2

Air and N2

Steam 1.2 CO2 1.1

Fig. 1.4 Specific-heat ratio of eight common gases as a function of temperature. (Data from Ref. 22.)

1.0

0

1000

2000

3000

Temperature, °R

4000

5000

1.6 Thermodynamic Properties of a Fluid

21

Many flow problems involve steam. Typical steam operating conditions are often close to the critical point, so that the perfect-gas approximation is inaccurate. Then we must turn to the steam tables, either in tabular or CD-ROM form [23] or as online software [24]. Most online steam tables require a license fee, but the  writer, in Example 1.5 that follows, suggests a free online source. Sometimes the error of using the perfect-gas law for steam can moderate, as the following example shows.

EXAMPLE 1.5 Estimate ρ and cp of steam at 100 lbf/in2 and 4008F, in English units, (a) by the perfect-gas approximation and (b) by the ASME Steam Tables [23].

Solution • Approach (a)—the perfect-gas law: Although steam is not an ideal gas, we can estimate these properties with moderate accuracy from Eqs. (1.10) and (1.17). First convert pressure from 100 lbf/in2 to 14,400 lbf/ft2, and use absolute temperature, (4008F 1 460) 5 8608R. Then we need the gas constant for steam, in English units. From Table A.4, the molecular weight of H2O is 18.02, whence Rsteam 5

¶English MH2O

5

49,700 ft # lbf/(slugmol °R) ft # lbf 5 2758 18.02/mol slug °R

Then the density estimate follows from the perfect-gas law, Eq. (1.10): ρ
> L U L/8

V

Iceberg

P7.101

CD T, VT

V

45°

7L / 8

P7.104 The Russian Typhoon-class submarine is 170 m long, with a maximum diameter of 23 m. Its propulsor can deliver up to 80,000 hp to the seawater. Model the submarine as an 8:1 ellipsoid and estimate the maximum speed, in knots, of this ship. P7.105 A ship 50 m long, with a wetted area of 800 m2, has the hull shape tested in Fig. 7.19. There are no bow or stern bulbs. The total propulsive power available is 1 MW. For seawater at 208C, plot the ship’s velocity V kn versus power P for 0 , P , 1 MW. What is the most efficient setting? P7.106 For the kite-assisted ship of Prob. P7.85, again neglect wave drag and let the wind velocity be 30 mi/h. Estimate the kite area that would tow the ship, unaided by the propeller, at a ship speed of 8 knots. P7.107 The largest flag in Rhode Island stands outside Herb Chambers’ auto dealership, on the edge of Route I-95 in Providence. The flag is 50 ft long, 30 ft wide, weighs 250 lbf, and takes four strong people to raise it or lower it. Using Prob. P7.40 for input, estimate (a) the wind speed, in mi/h, for which the flag drag is 1000 lbf and (b) the flag drag when the wind is a low-end category 1 hurricane, 74 mi/h. [Hint: Providence is at sea level.] P7.108 The data in Fig. P7.108 are for the lift and drag of a spinning sphere from Ref. 45. Suppose that a tennis ball (W < 0.56 N, D < 6.35 cm) is struck at sea level with initial velocity V0 5 30 m/s and “topspin” (front of the ball rotating downward) of 120 r/s. If the initial height of the ball is 1.5 m, estimate the horizontal distance traveled before it strikes the ground.

Problems

513

0.8 0.7

CD

0.6 0.5 CL

0.4 CL

0.3 0.2

ω

V

CD

0.1

P7.108 Drag and lift coefficients for a rotating sphere at ReD < 105, from Ref. 45. (Reproduced by permission of the American Society of Mechanical Engineers.)

0.0 – 0.1

0.0

1.0

P7.109 The world record for automobile mileage, 12,665 miles per gallon, was set in 2005 by the PAC-CAR II in Fig. P7.109, built by students at the Swiss Federal Institute of Technology in Zurich [52]. This little car, with an empty weight of 64 lbf and a height of only 2.5 ft, traveled a 21-km course at 30 km/h to set the record. It has a reported drag coefficient

2.0

3.0 ωR V

4.0

5.0

6.0

of 0.075 (comparable to an airfoil), based upon a frontal area of 3 ft2. (a) What is the drag of this little car when on the course? (b) What horsepower is required to propel it? (c) Do a bit of research and explain why a value of miles per gallon is completely misleading in this particular case. P7.110 A baseball pitcher throws a curveball with an initial velocity of 65 mi/h and a spin of 6500 r/min about a vertical axis. A baseball weighs 0.32 lbf and has a diameter of 2.9 in. Using the data of Fig. P7.108 for turbulent flow, estimate how far such a curveball will have deviated from its straight-line path when it reaches home plate 60.5 ft away. *P7.111 A table tennis ball has a mass of 2.6 g and a diameter of 3.81 cm. It is struck horizontally at an initial velocity of 20 m/s while it is 50 cm above the table, as in Fig. P7.111. For sea-level air, what spin, in r/min, will cause the ball to strike the opposite edge of the table, 4 m away? Make an analytical estimate, using Fig. P7.108, and account for the fact that the ball decelerates during flight. P7.112 A smooth wooden sphere (SG 5 0.65) is connected by a thin rigid rod to a hinge in a wind tunnel, as in Fig. P7.112. Air at 208C and 1 atm flows and levitates the sphere. (a) Plot the angle θ versus sphere diameter d in the range 1 cm # d # 15 cm. (b) Comment on the feasibility of this configuration. Neglect rod drag. P7.109 The world’s best mileage set by PAC-Car II of ETH Zurich.

514

Chapter 7 Flow Past Immersed Bodies 4m 20 m/s

ω?

50 cm *P7.118 ?

P7.119 P7.111

Rod, L = 50 cm

P7.120

U = 12 m/s

P7.112

θ

Hinge

P7.113 An automobile has a mass of 1000 kg and a drag area CDA 5 0.7 m2. The rolling resistance of 70 N is approximately constant. The car is coasting without brakes at 90 km/h as it begins to climb a hill of 10 percent grade (slope 5 tan21 0.1 5 5.718). How far up the hill will the car come to a stop? P7.114 The deep submergence vehicle ALVIN is 23 ft long and 8.5 ft wide. It weighs about 36,000 lbf in air and ascends (descends) in the seawater due to about 360 lbf of positive (negative) buoyancy. Noting that the front face of the ship is quite different for ascent and descent, (a) estimate the velocity for each direction, in meters per minute. (b) How long does it take to ascend from its maximum depth of 4500 m?

Lifting bodies—airfoils P7.115 The Cessna Citation executive jet weighs 67 kN and has a wing area of 32 m2. It cruises at 10 km standard altitude with a lift coefficient of 0.21 and a drag coefficient of 0.015. Estimate (a) the cruise speed in mi/h and (b) the horsepower required to maintain cruise velocity. P7.116 An airplane weighs 180 kN and has a wing area of 160 m2 and a mean chord of 4 m. The airfoil properties are given by Fig. 7.25. If the plane is designed to land at V0 5 1.2Vstall, using a split flap set at 608, (a) what is the proper landing speed in mi/h? (b) What power is required for takeoff at the same speed? P7.117 The Transition® auto-car in Fig. 7.30 has a weight of 1200 lbf, a wingspan of 27.5 ft, and a wing area of 150 ft2, with

P7.121

a symmetrical airfoil, CDq < 0.02. Assume that the fuselage and tail section have a drag-area comparable to the Toyota Prius [21], CDA < 6.24 t2. If the pusher propeller provides a thrust of 250 lbf, how fast, in mi/h, can this carplane fly at an altitude of 8200 ft? Suppose that the airplane of Prob. P7.116 is fitted with all the best high-lift devices of Fig. 7.28. What is its minimum stall speed in mi/h? Estimate the stopping distance if the plane lands at V0 5 1.25Vstall with constant CL 5 3.0 and CD 5 0.2 and the braking force is 20 percent of the weight on the wheels. A transport plane has a mass of 45,000 kg, a wing area of 160 m2, and an aspect ratio of 7. Assume all lift and drag due to the wing alone, with CD∞ 5 0.020 and CL,max 5 1.5. If the aircraft flies at 9000 m standard altitude, make a plot of drag (in N) versus speed (from stall to 240 m/s) and determine the optimum cruise velocity (minimum drag per unit speed). Show that if Eqs. (7.70) and (7.71) are valid, the maximum lift-to-drag ratio occurs when CD 5 2CD∞. What are (L/D)max and α for a symmetric wing when AR 5 5 and CD∞ 5 0.009? In gliding (unpowered) flight, the lift and drag are in equilibrium with the weight. Show that if there is no wind, the aircraft sinks at an angle tan θ
0 dp < 0 Subsonic nozzle

607

Supersonic Ma > 1

dV > 0 dp < 0 Supersonic nozzle

dV < 0 dp > 0 Supersonic diffuser

Fig. 9.5 Effect of Mach number on property changes with area change in duct flow.

Momentum Sound speed:

dp 1 V dV 5 0 ρ

(9.39)

2

dp 5 a dρ

Now eliminate dp and dρ between Eqs. (9.38) and (9.39) to obtain the following relation between velocity change and area change in isentropic duct flow: dp dA 1 dV 5 52 2 V A Ma2 2 1 ρV

(9.40)

Inspection of this equation, without actually solving it, reveals a fascinating aspect of compressible flow: Property changes are of opposite sign for subsonic and supersonic flow because of the term Ma2 2 1. There are four combinations of area change and Mach number, summarized in Fig. 9.5. From earlier chapters we are used to subsonic behavior (Ma , 1): When area increases, velocity decreases and pressure increases, which is denoted a subsonic diffuser. But in supersonic flow (Ma . 1), the velocity actually increases when the area increases, a supersonic nozzle. The same opposing behavior occurs for an area decrease, which speeds up a subsonic flow and slows down a supersonic flow. What about the sonic point Ma 5 1? Since infinite acceleration is physically impossible, Eq. (9.40) indicates that dV can be finite only when dA 5 0—that is, a minimum area (throat) or a maximum area (bulge). In Fig. 9.6 we patch together a throat section and a bulge section, using the rules from Fig. 9.5. The throat or converging–diverging section can smoothly accelerate a subsonic flow through sonic to supersonic flow, as in Fig. 9.6a. This is the only way a supersonic flow can be created by expanding the gas from a stagnant reservoir. The bulge section fails; the bulge Mach number moves away from a sonic condition rather than toward it. Although supersonic flow downstream of a nozzle requires a sonic throat, the opposite is not true: A compressible gas can pass through a throat section without becoming sonic.

608

Chapter 9 Compressible Flow Amax Amin

Fig. 9.6 From Eq. (9.40), in flow through a throat (a) the fluid can accelerate smoothly through sonic and supersonic flow. In flow through the bulge (b) the flow at the bulge cannot be sonic on physical grounds.

Perfect-Gas Area Change

Subsonic

Ma = 1 Supersonic

Subsonic:

Ma < 1 Subsonic:

(Supersonic:

Supersonic) Ma > 1

(a)

(b)

We can use the perfect-gas and isentropic flow relations to convert the continuity relation (9.37) into an algebraic expression involving only area and Mach number, as follows. Equate the mass flow at any section to the mass flow under sonic conditions (which may not actually occur in the duct): ρVA 5 ρ*V*A* ρ* V* A 5 ρ V A*

or

(9.41)

Both terms on the right are functions only of Mach number for isentropic flow. From Eqs. (9.28) and (9.32) 1/(k21) ρ* ρ0 ρ* 2 1 5 5 e c 1 1 (k 2 1) Ma2 d f ρ ρ0 ρ k11 2

(9.42)

From Eqs. (9.26) and (9.32) we obtain (kRT*) 1/2 (kRT) 1/2 T* 1/2 T0 1/2 V* 5 5 a b a b V V V T0 T 5

1/2 1 2 1 e c 1 1 (k 2 1) Ma2 d f Ma k 1 1 2

(9.43)

Combining Eqs. (9.41) to (9.43), we get the desired result: 1

2

A 1 1 1 2 (k 2 1) Ma (1/2)(k11)/(k21) d 5 c 1 A* Ma 2 (k 1 1)

(9.44)

For k 5 1.4, Eq. (9.44) takes the numerical form A 1 (1 1 0.2 Ma2 ) 3 5 A* Ma 1.728

(9.45)

which is plotted in Fig. 9.7. Equations (9.45) and (9.34) enable us to solve any one-dimensional isentropic airflow problem given, say, the shape of the duct A(x) and the stagnation conditions and assuming that there are no shock waves in the duct.

9.4 Isentropic Flow with Area Changes

609

4.0 3.5

p0 p

A A*

3.0

ρ 0 ρ

2.5

T0 T

2.0 1.5 1.0

Fig. 9.7 Area ratio and fluid properties versus Mach number for isentropic flow of a perfect gas with k 5 1.4.

0.5 0.0

0

0.5

1

1.5 Mach number

2

2.5

3

Figure 9.7 shows that the minimum area that can occur in a given isentropic duct flow is the sonic, or critical, throat area. All other duct sections must have A greater than A*. In many flows a critical sonic throat is not actually present, and the flow in the duct is either entirely subsonic or, more rarely, entirely supersonic.

Choking

From Eq. (9.41) the inverse ratio A*/A equals ρV/(ρ*V*), the mass flow per unit area at any section compared with the critical mass flow per unit area. From Fig. 9.7 this inverse ratio rises from zero at Ma 5 0 to unity at Ma 5 1 and back down to zero at large Ma. Thus, for given stagnation conditions, the maximum possible mass flow passes through a duct when its throat is at the critical or sonic condition. The duct is then said to be choked and can carry no additional mass flow unless the throat is widened. If the throat is constricted further, the mass flow through the duct must decrease. From Eqs. (9.32) and (9.33) the maximum mass flow is # mmax 5 ρ*A*V* 5 ρ0 a 5 k1/2 a

1/(k21) 1/2 2 2k b RT0 b A* a k11 k11

(1/2)(k11)/(k21) 2 A*ρ0 (RT0 ) 1/2 b k11

(9.46a)

For k 5 1.4 this reduces to 0.6847p0 A* # mmax 5 0.6847A*ρ0 (RT0 ) 1/2 5 (RT0 ) 1/2

(9.46b)

For isentropic flow through a duct, the maximum mass flow possible is proportional to the throat area and stagnation pressure and inversely proportional to the square root of the stagnation temperature. These are somewhat abstract facts, so let us illustrate with some examples.

610

Chapter 9 Compressible Flow

The Local Mass Flow Function

Equations (9.46) give the maximum mass flow, which occurs at the choking condition (sonic exit). They can be modified to predict the actual (nonmaximum) mass flow at any section where local area A and pressure p are known.1 The algebra is convoluted, so here we give only the final result, expressed in dimensionless form: # p 2/k p (k21)/k m 1RT0 2k Mass flow function 5 5 a b c1 2 a b d p0 A p0 B k 2 1 p0

(9.47)

We stress that p and A in this relation are the local values at position x. As p/p0 falls, this function rises rapidly and then levels out at the maximum of Eqs. (9.46). A few values may be tabulated here for k 5 1.4: p/p0

1.0

0.98

0.95

0.9

0.8

0.7

0.6

Function

0.0

0.1978

0.3076

0.4226

0.5607

0.6383

0.6769

#0.5283 0.6847

Equation (9.47) is handy if stagnation conditions are known and the flow is not choked. When A/A* is known and the Mach number is unknown, no algebraic solution of Eq. (9.44) is known to the writer. One could interpolate in Table B.1 or simply iterate Eq. (9.44) with a calculator. But Excel can iterate Eq. (9.44) for subsonic flow in its direct form: Subsonic flow: Ma 5

A* 1 1 0.5(k 2 1)Ma2 0.5(k11)/(k21) c d A 0.5(k 1 1)

(9.48)

Make a subsonic guess for Ma on the right side and then replace it with the value calculated on the left side. For example, suppose A/A* 5 2.035, corresponding to Ma 5 0.300. A poor guess of Ma 5 0.5 in Eq. (9.44) leads to a better Ma 5 0.329, then 0.303, then 0.300. For supersonic flow, iteration of Eq. (9.44) diverges. Instead, simply try different Mach numbers in Eq. (9.44) until the proper area is achieved. For example, suppose A/A* 5 3.183, corresponding to Ma 5 2.70. A poor guess of Ma 5 2.4 yields A/A* 5 2.403, 24 percent low. Improve the guess to Ma 5 2.8 to give A/A* 5 3.500, or 10 percent high. Interpolate to Ma 5 2.72, A/A* 5 3.244, 2 percent high. Finally settle on Ma 5 2.70, correct. These calculations simply require that you retype your guess for Ma, check the error, and convergence only depends upon your cleverness. Note that two solutions are possible for a given A/A*, one subsonic and one supersonic. The proper solution cannot be selected without further information, such as known pressure or temperature at the given duct section. EXAMPLE 9.4 Air flows isentropically through a duct. At section 1 the area is 0.05 m2 and V1 5 180 m/s, # p1 5 500 kPa, and T1 5 470 K. Compute (a) T0, (b) Ma1, (c) p0, and (d ) both A* and m. 2 If at section 2 the area is 0.036 m , compute Ma2 and p2 if the flow is (e) subsonic or (f )  supersonic. Assume k 5 1.4.

1

The author is indebted to Georges Aigret, of Chimay, Belgium, for suggesting this useful function.

9.4 Isentropic Flow with Area Changes

611

Solution Part (a)

A general sketch of the problem is shown in Fig. E9.4. With V1 and T1 known, the energy equation (9.23) gives

Possibly supersonic

Subsonic Throat

V1 = 180 m/s

Assume isentropic flow

p1 = 500 kPa T1 = 470 K 2E A2 = 0.036 m2

2F A2 = 0.036 m2

1

E9.4

A1 = 0.05 m2

T0 5 T1 1

Part (b)

V21 (180) 2 5 470 1 5 486 K 2cp 2(1005)

The local sound speed a1 5 2kRT1 5 3 (1.4) (287) (470) 4 1/2 5 435 m/s. Hence Ma1 5

Part (c)

V1 180 5 5 0.414 a1 435

Ans. (b)

With Ma1 known, the stagnation pressure follows from Eq. (9.34): p0 5 p1 (1 1 0.2 Ma21 ) 3.5 5 (500 kPa) 31 1 0.2(0.414) 2 4 3.5 5 563 kPa

Part (d)

Ans. (a)

Ans. (c)

Similarly, from Eq. (9.45), the critical sonic throat area is 31 1 0.2(0.414) 2 4 3 A1 (1 1 0.2 Ma21 ) 3 5 5 5 1.547 A* 1.728 Ma1 1.728(0.414) or

A* 5

A1 0.05 m2 5 5 0.0323 m2 1.547 1.547

Ans. (d )

This throat must actually be present in the duct if the flow is to become supersonic. We now know A*. So to compute the mass flow we can use Eqs. (9.46), which remain valid, based on the numerical value of A*, whether or not a throat actually exists: p0 A* (563,000) (0.0323) # m 5 0.6847 5 0.6847 5 33.4 kg/s 1RT0 1(287) (486)

Ans. (d )

Or we could fare equally well with our new “local mass flow” formula, Eq. (9.47), using, say, the pressure and area at section 1. Given p1/p0 5 500/563 5 0.889, Eq. (9.47) yields kg 2(1.4) # 1287(486) # m Ans. (d ) 5 (0.889) 2/1.4 31 2(0.889) 0.4/1.4 4 5 0.444  m 533.4 s 563,000(0.05) A 0.4

Part (e)

For subsonic flow upstream of the throat at section 2E, the area ratio is A2/A* 5 0.036/0.0323 5 1.115, corresponding to the left side of Fig. 9.7 or the subsonic numbers

612

Chapter 9 Compressible Flow in Table B.1, neither of which is very accurate. Guess Ma2 at section 2E, from Fig. 9.7, at about 0.70. Enter this guess into Eq. (9.48) and repeat. The Excel table is: Ma – guess

Ma 2 Eq. (9.48)

A/A*

0.700 0.687 0.680 0.677 0.675 0.674

0.687 0.680 0.677 0.675 0.674 0.674

1.115 1.115 1.115 1.115 1.115 1.115

The (slowly) converged subsonic Mach number is Ma2 5 0.674

Ans. (e)

The pressure is given by the isentropic relation p2 5

po 31 1 0.2(0.674) 2 4 3.5

5

563 kPa 5 415 kPa 1.356

Ans. (e)

Part (e) does not require a throat, sonic or otherwise; the flow could simply be contracting subsonically from A1 to A2.

Part (f)

For supersonic flow at section 2F, again the area ratio is 0.036/0.0323 5 1.115. On the right side of Fig. 9.7 we estimate Ma2 < 1.5. The table from Eq. (9.44) is Ma – guess A/A* 2 Eq. (9.44) A/A* 1.5000 1.4000 1.4001

1.1762 1.1149 1.1150

1.1150 1.1150 1.1150

We were lucky that this Mach number is easy to guess: Ma2 5 1.4001

Ans. ( f )

Again the pressure is given by the isentropic relation at this new Mach number: p2 5

po 31 1 0.2(1.4001) 2 4 3.5

5

563 kPa 5 177 kPa 3.183

Ans. ( f )

Note that the supersonic-flow pressure level is much less than p2 in part (e), and a sonic throat must have occurred between sections 1 and 2F.

EXAMPLE 9.5 It is desired to expand air from p0 5 200 kPa and T0 5 500 K through a throat to an exit Mach number of 2.5. If the desired mass flow is 3 kg/s, compute (a) the throat area and the exit (b) pressure, (c) temperature, (d ) velocity, and (e) area, assuming isentropic flow, with k 5 1.4.

9.5 The Normal Shock Wave

613

Solution The throat area follows from Eq. (9.47), because the throat flow must be sonic to produce a supersonic exit: # 3.03287(500) 4 1/2 m (RT0 ) 1/2 1 A* 5 5 5 0.00830 m2 5 πD*2 0.6847p0 0.6847(200,000) 4 Dthroat 5 10.3 cm

or

Ans. (a)

With the exit Mach number known, the isentropic flow relations give the pressure and temperature: pe 5

p0 31 1 0.2(2.5) 2 4 3.5 Te 5

5

T0 1 1 0.2(2.5) 2

200,000 5 11,700 Pa 17.08

5

500 5 222 K 2.25

Ans. (b) Ans. (c)

The exit velocity follows from the known Mach number and temperature: Ve 5 Mae (kRTe ) 1/2 5 2.531.4(287) (222) 4 1/2 5 2.5(299 m/s) 5 747 m/s

Ans. (d )

The exit area follows from the known throat area and exit Mach number and Eq. (9.45): 31 1 0.2(2.5) 2 4 3 Ae 5 5 2.64 A* 1.728(2.5) or

Ae 5 2.64A* 5 2.64(0.0083 m2 ) 5 0.0219 m2 5 14πD2e

or

De 5 16.7 cm

Ans. (e)

One point might be noted: The computation of the throat area A* did not depend in any way on the numerical value of the exit Mach number. The exit was supersonic; therefore the throat is sonic and choked, and no further information is needed.

9.5 The Normal Shock Wave

Shock waves are nearly discontinuous changes in a supersonic flow. They can occur due to a higher downstream pressure, a sudden change in flow direction, blockage by a downstream body, or the result of an explosion. The simplest algebraically is a one-dimensional change, or normal shock wave, shown in Fig. 9.8. We select a control volume just before and after the wave. The analysis is identical to that of Fig. 9.1; that is, a shock wave is a fixed strong pressure wave. To compute all property changes rather than just the wave speed, we use all our basic one-dimensional steady flow relations, letting section 1 be upstream and section 2 be downstream: Continuity:

ρ1V1 5 ρ2V2 5 G 5 const

Momentum:

p1 2 p2 5

Energy:

1 2 2 V1

Perfect gas: Constant cp:

h1 1

ρ2V22

2

ρ1V21

1 2 2 V2

5 h2 1 5 h0 5 const p1 p2 5 ρ 1 T1 ρ 2 T2

h 5 cpT    k 5 const

(9.49a) (9.49b) (9.49c) (9.49d) (9.49e)

614

Chapter 9 Compressible Flow Fixed normal shock Isoenergetic T01 = T02 1 Isentropic upstream s = s1

Fig. 9.8 Flow through a fixed normal shock wave.

Ma1 > 1

2

Ma2 < 1

Isentropic downstream s = s2 > s1 A*2 > A*1 p02 < p01

Thin control volume A1 ≈ A2

Note that we have canceled out the areas A1 < A2, which is justified even in a variable duct section because of the thinness of the wave. The first successful analyses of these normal shock relations are credited to W. J. M. Rankine (1870) and A. Hugoniot (1887), hence the modern term Rankine-Hugoniot relations. If we assume that the upstream conditions (p1, V1, ρ1, h1, T1) are known, Eqs. (9.49) are five algebraic relations in the five unknowns (p2, V2, ρ2, h2, T2). Because of the velocitysquared term, two solutions are found, and the correct one is determined from the second law of thermodynamics, which requires that s2 . s1. The velocities V1 and V2 can be eliminated from Eqs. (9.49a) to (9.49c) to obtain the Rankine-Hugoniot relation: 1 1 1 ( p2 2 p1) a 1 b (9.50) ρ2 ρ1 2 This contains only thermodynamic properties and is independent of the equation of state. Introducing the perfect-gas law h 5 cpT 5 kp/[(k 2 1)ρ], we can rewrite this as h2 2 h1 5

ρ2 1 1 βp2/p1 k11 5     β 5 ρ1 β 1 p2/p1 k21

(9.51)

We can compare this with the isentropic flow relation for a very weak pressure wave in a perfect gas: ρ2 p2 1/k 5a b (9.52) p1 ρ1 Also, the actual change in entropy across the shock can be computed from the perfectgas relation: p2 ρ1 k s2 2 s1 5 ln c a b d cυ p1 ρ2

(9.53)

Assuming a given wave strength p2/p1, we can compute the density ratio and the entropy change and list them as follows for k 5 1.4:

9.5 The Normal Shock Wave r2/r1

p2 p1

Eq. (9.51)

Isentropic

s2 2 s1 cυ

0.5 0.9 1.0 1.1 1.5 2.0

0.6154 0.9275 1.0 1.00704 1.3333 1.6250

0.6095 0.9275 1.0 1.00705 1.3359 1.6407

20.0134 20.00005 0.0 0.00004 0.0027 0.0134

615

We see that the entropy change is negative if the pressure decreases across the shock, which violates the second law. Thus a rarefaction shock is impossible in a perfect gas.2 We see also that weak shock waves (p2/p1 # 2.0) are very nearly isentropic.

Mach Number Relations

For a perfect gas all the property ratios across the normal shock are unique functions of k and the upstream Mach number Ma1. For example, if we eliminate ρ2 and V2 from Eqs. (9.49a) to (9.49c) and introduce h 5 kp/[(k 2 1)ρ], we obtain 2ρ1V21 p2 1 5 2 (k 2 1) d c p1 p1 k11

(9.54)

But for a perfect gas ρ1V21/p1 5 kV21/(kRT1 ) 5 k Ma21, so that Eq. (9.54) is equivalent to p2 1 32k Ma21 2 (k 2 1) 4 5 p1 k11

(9.55)

From this equation we see that, for any k, p2 . p1 only if Ma1 . 1.0. Thus for flow through a normal shock wave, the upstream Mach number must be supersonic to satisfy the second law of thermodynamics. What about the downstream Mach number? From the perfect-gas identity ρV2 5 kp Ma2, we can rewrite Eq. (9.49b) as p2 1 1 k Ma21 5 p1 1 1 k Ma22

(9.56)

which relates the pressure ratio to both Mach numbers. By equating Eqs. (9.55) and (9.56) we can solve for Ma22 5

(k 2 1) Ma21 1 2 2k Ma21 2 (k 2 1)

(9.57)

Since Ma1 must be supersonic, this equation predicts for all k . 1 that Ma2 must be subsonic. Thus a normal shock wave decelerates a flow almost discontinuously from supersonic to subsonic conditions. 2

This is true also for most real gases; see Ref. 9, Sec. 7.3.

616

Chapter 9 Compressible Flow

Further manipulation of the basic relations (9.49) for a perfect gas gives additional equations relating the change in properties across a normal shock wave in a perfect gas: ρ2 V1 (k 1 1) Ma21 5 5 2 ρ1 V2 (k 2 1) Ma1 1 2 T2 2k Ma21 2 (k 2 1) 5 32 1 (k 2 1) Ma21 4 T1 (k 1 1) 2 Ma21

(9.58)

T02 5 T01 k/(k21) 1/(k21) ρ02 p02 (k 1 1) Ma21 k11 d 5 5 c d c p01 ρ01 2 1 (k 2 1) Ma21 2k Ma21 2 (k 2 1)

Of additional interest is the fact that the critical, or sonic, throat area A* in a duct increases across a normal shock: A*2 A*1

5

Ma2 2 1 (k 2 1) Ma21 (1/2)(k11)/(k21) c d Ma1 2 1 (k 2 1) Ma22

(9.59)

All these relations are given in Table B.2 and plotted versus upstream Mach number Ma1 in Fig. 9.9 for k 5 1.4. We see that pressure increases greatly while temperature and density increase moderately. The effective throat area A* increases slowly at first and then rapidly. The failure of students to account for this change in A* is a common source of error in shock calculations. The stagnation temperature remains the same, but the stagnation pressure and density decrease in the same ratio; in other words, the flow across the shock is adiabatic but nonisentropic. Other basic principles governing the behavior of shock waves can

6

A*2

5

A*1

p2 p1

V1 ρ2 = V2 ρ1

4

T2 T1

3

2

p02 ρ02 p01 = ρ 01

1 Ma2

Fig. 9.9 Change in flow properties across a normal shock wave for k 5 1.4.

0

1

1.5

2

2.5 Ma1

3

3.5

4

9.5 The Normal Shock Wave

617

be summarized as follows: 1. The upstream flow is supersonic, and the downstream flow is subsonic. 2. For perfect gases (and also for real fluids except under bizarre thermodynamic conditions) rarefaction shocks are impossible, and only a compression shock can exist. 3. The entropy increases across a shock with consequent decreases in stagnation pressure and stagnation density and an increase in the effective sonic throat area. 4. Weak shock waves are very nearly isentropic. Normal shock waves form in ducts under transient conditions, such as in shock tubes, and in steady flow for certain ranges of the downstream pressure. Figure 9.10a shows a normal shock in a supersonic nozzle. Flow is from left to right. The oblique wave pattern to the left is formed by roughness elements on the nozzle walls and indicates

(a )

Fig. 9.10 Normal shocks form in both internal and external flows. (a) Normal shock in a duct; note the Mach wave pattern to the left (upstream), indicating supersonic flow. (Courtesy of U.S. Air Force Arnold Engineering Development Center.) (b) Supersonic flow past a blunt body creates a normal shock at the nose; the apparent shock thickness and body-corner curvature are optical distortions. (Courtesy of U.S. Army Ballistic Research Laboratory, Aberdeen Proving Ground.)

(b )

618

Chapter 9 Compressible Flow

that the upstream flow is supersonic. Note the absence of these Mach waves (see Sec. 9.10) in the subsonic flow downstream. Normal shock waves occur not only in supersonic duct flows but also in a variety of supersonic external flows. An example is the supersonic flow past a blunt body shown in Fig. 9.10b. The bow shock is curved, with a portion in front of the body that is essentially normal to the oncoming flow. This normal portion of the bow shock satisfies the property change conditions just as outlined in this section. The flow inside the shock near the body nose is thus subsonic and at relatively high temperature T2 . T1, and convective heat transfer is especially high in this region. Each nonnormal portion of the bow shock in Fig. 9.10b satisfies the oblique shock relations to be outlined in Sec. 9.9. Note also the oblique recompression shock on the sides of the body. What has happened is that the subsonic nose flow has accelerated around the corners back to supersonic flow at low pressure, which must then pass through the second shock to match the higher downstream pressure conditions. Note the fine-grained turbulent wake structure in the rear of the body in Fig. 9.10b. The turbulent boundary layer along the sides of the body is also clearly visible. The analysis of a complex multidimensional supersonic flow such as in Fig. 9.10 is beyond the scope of this book. For further information see, for example, Ref. 9, Chap. 9, or Ref. 5, Chap. 16.

Moving Normal Shocks

The preceding analysis of the fixed shock applies equally well to the moving shock if we reverse the transformation used in Fig. 9.1. To make the upstream conditions simulate a still fluid, we move the shock of Fig. 9.8 to the left at speed V1; that is, we fix our coordinates to a control volume moving with the shock. The downstream flow then appears to move to the left at a slower speed V1 2 V2 following the shock. The thermodynamic properties are not changed by this transformation, so that all our Eqs. (9.50) to (9.59) are still valid. EXAMPLE 9.6

1 2

3

Air flows from a reservoir where p 5 300 kPa and T 5 500 K through a throat to section 1 in Fig. E9.6, where there is a normal shock wave. Compute (a) p1, (b) p2, (c) p02, (d ) A*2 , (e) p03, (f ) A*3 , (g) p3, and (h) T03.

1 m2 2 m2

E9.6

3 m2

Solution System sketch: This is shown in Fig. E9.6. Between sections 1 and 2 is a normal shock. Assumptions: Isentropic flow before and after the shock. Lower p0 and ρ0 after the shock. Approach: After first noting that the throat is sonic, work your way from 1 to 2 to 3. Property values: For air, R 5 287 m2/(s2 ? K), k 5 1.40, and cp 5 1005 m2/(s2 ? K). The inlet stagnation pressure of 300 kPa is constant up to point 1. • Solution step (a): A shock wave cannot exist unless Ma1 is supersonic. Therefore the throat is sonic and choked: Athroat 5 A*1 5 1 m2. The area ratio gives Ma1 from Eq. (9.45) for k 5 1.4:

• • • •

A1 2 m2 1 (1 1 0.2 Ma21 ) 3 5 5 2.0 5      solve for     Ma1 5 2.1972 2 A*1 Ma1 1.728 1m

9.5 The Normal Shock Wave

619

Such four-decimal-place accuracy might require iteration or the use of Excel. Linear interpolation in Table B.1 would give Ma1 < 2.194, quite good also. The pressure at section 1 then follows from the isentropic relation, Eq. (9.28): p1 5 • Steps (b, c, d): p2 5

p01 (1 1

0.2Ma21 ) 3.5

5

300 kPa 31 1 0.2(2.194) 2 4 3.5

5 28.2 kPa

Ans. (a)

The pressure p2 is found from the normal shock Eq. (9.55) or Table B.2:

p1 28.2 kPa 32k Ma21 2(k 2 1)4 5 32(1.4) (2.194) 2 2 (1.4 21)4 5154 kPa Ans. (b) k11 (1.4 1 1)

Similarly, for Ma1 < 2.20, Table B.2 gives p02/p01 < 0.628 (Excel gives 0.6294) and A*2/A*1 < 1.592 (Excel gives 1.5888). Thus, to good accuracy, p02 < 0.628p01 5 0.628(300 kPa) < 188 kPa

Ans. (c)

A*2 5 1.59 A*1 5 1.59(1.0 m2 ) < 1.59 m2

Ans. (d )

• Comment: To calculate A*2 directly, without Table B.2, you would need to pause and calculate Ma2 < 0.547 from Eq. (9.57), since Eq. (9.59) involves both Ma1 and Ma2. • Step (e, f): The flow from 2 to 3 is isentropic (but at higher entropy than upstream of the shock); therefore

• Steps (g, h):

p03 5 p02 < 188 kPa

Ans. (e)

A*3 5 A*2 < 1.59 m2

Ans. (f )

The flow is adiabatic throughout, so the stagnation temperature is constant: T03 5 T02 5 T01 5 500 K

Ans. (h)

Next, the area ratio, using the new sonic area, gives the Mach number at section 3: A3 3 m2 1 (1 1 0.2 Ma23 ) 3 5 5 1.89 5    solve for   Ma3 < 0.33 A*3 Ma3 1.728 1.59 m2 Excel would yield Ma3 5 0.327. Finally, with p02 known, Eq. (9.28) yields p3: p3 5

p02 (1 1 0.2

Ma23 ) 3.5

a1

Vn1 > a1

Fig. 9.20 Geometry of flow through an oblique shock wave.

Ans.

θ

Vn2 < a2 β

Deflection angle

646

Chapter 9 Compressible Flow

It is convenient to analyze the flow by breaking it up into normal and tangential components with respect to the wave, as shown in Fig. 9.20. For a thin control volume just encompassing the wave, we can then derive the following integral relations, canceling out A1 5 A2 on each side of the wave: ρ1Vn1 5 ρ2Vn2

Continuity:

p1 2 p2 5

Normal momentum:

2

(9.80a) 2 ρ1V n1

(9.80b)

0 5 ρ1Vn1 (Vt 2 2 Vt1 )

Tangential momentum: Energy:

2 ρ2V n2

h1 1

1 2 2 V n1

1

1 2 2 V t1

5 h2 1

1 2 2 V n2

1

(9.80c) 1 2 2 V t2

5 h0

(9.80d)

We see from Eq. (9.80c) that there is no change in tangential velocity across an oblique shock: Vt2 5 Vt1 5 Vt 5 const (9.81) Thus tangential velocity has as its only effect the addition of a constant kinetic energy 1 2 2 V t to each side of the energy equation (9.80d). We conclude that Eqs. (9.80) are identical to the normal shock relations (9.49), with V1 and V2 replaced by the normal components Vn1 and Vn2. All the various relations from Sec. 9.5 can be used to compute properties of an oblique shock wave. The trick is to use the “normal” Mach numbers in place of Ma1 and Ma2: Man1 5

Vn1 5 Ma1 sin β a1 (9.82)

Vn2 Man2 5 5 Ma2 sin ( β 2 θ) a2

Then, for a perfect gas with constant specific heats, the property ratios across the oblique shock are the analogs of Eqs. (9.55) to (9.58) with Ma1 replaced by Man1: p2 1 32k Ma21 sin2 β 2 (k 2 1) 4 5 p1 k11

(9.83a)

ρ2 (k 1 1) Ma21 sin2 β tan β Vn1 5 5 5 ρ1 tan (β 2 θ) Vn2 (k 2 1) Ma21 sin2 β 1 2

(9.83b)

2k Ma21 sin2 β 2 (k 2 1) T2 5 32 1 (k 2 1) Ma21 sin2 β 4 T1 (k 1 1) 2 Ma21 sin2 β T02 5 T01 k/(k21) 1/(k21) p02 (k 1 1) Ma21 sin2 β k11 5 c d c d 2 2 2 2 p01 2 1 (k 2 1) Ma1 sin β 2k Ma1 sin β 2 (k 2 1)

Ma2n2 5

(k 2 1) Ma2n1 1 2 2k Ma2n1 2 (k 2 1)

(9.83c) (9.83d) (9.83e) (9.83f )

All these are tabulated in the normal shock Table B.2. If you wondered why that table listed the Mach numbers as Man1 and Man2, it should be clear now that the table is also valid for the oblique shock wave.

9.9 Mach Waves and Oblique Shock Waves

647

Thinking all this over, we realize with hindsight that an oblique shock wave is the flow pattern one would observe by running along a normal shock wave (Fig. 9.8) at a constant tangential speed Vt. Thus the normal and oblique shocks are related by a galilean, or inertial, velocity transformation and therefore satisfy the same basic equations. If we continue with this run-along-the-shock analogy, we find that the deflection angle θ increases with speed Vt up to a maximum and then decreases. From the geometry of Fig. 9.20, the deflection angle is given by θ 5 tan21

Vt Vt 2 tan21 Vn2 Vn1

(9.84)

If we differentiate θ with respect to Vt and set the result equal to zero, we find that the maximum deflection occurs when Vt /Vn1 5 (Vn2/Vn1)1/2. We can substitute this back into Eq. (9.84) to compute θmax 5 tan21 r1/2 2 tan21 r21/2      r 5

Vn1 Vn2

(9.85)

For example, if Man1 5 3.0, from Table B.2 we find that Vn1/Vn2 5 3.8571, the square root of which is 1.9640. Then Eq. (9.85) predicts a maximum deflection of tan21 1.9640 2 tan21 (1/1.9640) 5 36.038. The deflection is quite limited even for infinite Man1: From Table B.2 for this case Vn1/Vn2 5 6.0, and we compute from Eq. (9.85) that θmax 5 45.588. This limited-deflection idea and other facts become more evident if we plot some of the solutions of Eqs. (9.83). For given values of V1 and a1, assuming as usual that k 5 1.4, we can plot all possible solutions for V2 downstream of the shock. Figure 9.21 does this in velocity-component coordinates Vx and Vy, with x parallel to V1. Such a plot is called a hodograph. The heavy dark line that looks like a fat airfoil is the locus, or shock polar, of all physically possible solutions for the given Ma1. The two dashed-line fishtails are solutions that increase V2; they are physically impossible because they violate the second law. Examining the shock polar in Fig. 9.21, we see that a given deflection line of small angle θ crosses the polar at two possible solutions: the strong shock, which greatly decelerates the flow, and the weak shock, which causes a much milder deceleration. Vy Weak wave angle

Rarefaction shock impossible by second law

θmax β Weak shock

Fig. 9.21 The oblique shock polar hodograph, showing double solutions (strong and weak) for small deflection angle and no solutions at all for large deflection.

θ

Normal shock

Strong shock Vx V1

Mach wave (V2 = V1)

648

Chapter 9 Compressible Flow

The flow downstream of the strong shock is always subsonic, while that of the weak shock is usually supersonic but occasionally subsonic if the deflection is large. Both types of shock occur in practice. The weak shock is more prevalent, but the strong shock will occur if there is a blockage or high-pressure condition downstream. Since the shock polar is only of finite size, there is a maximum deflection θmax, shown in Fig. 9.21, that just grazes the upper edge of the polar curve. This verifies the kinematic discussion that led to Eq. (9.85). What happens if a supersonic flow is  forced to deflect through an angle greater than θmax? The answer is illustrated in Fig. 9.22 for flow past a wedge-shaped body. In Fig. 9.22a the wedge half-angle θ is less than θmax, and thus an oblique shock forms at the nose of wave angle β just sufficient to cause the oncoming supersonic stream to deflect through the wedge angle θ. Except for the usually small effect of boundary layer growth (see, for example, Ref. 19, Sec. 7–5.2), the Mach number Ma2 is constant along the wedge surface and is given by the solution of Eqs. (9.83). The pressure, density, and temperature along the surface are also nearly constant, as predicted by Eqs. (9.83). When the flow reaches the corner of the wedge, it expands to higher Mach number and forms a wake (not shown) similar to that in Fig. 9.10. In Fig. 9.22b the wedge half-angle is greater than θmax, and an attached oblique shock is impossible. The flow cannot deflect at once through the entire angle θmax, yet somehow the flow must get around the wedge. A detached curve shock wave forms in front of the body, discontinuously deflecting the flow through angles smaller than θmax. The flow then curves, expands, and deflects subsonically around the wedge, becoming sonic and then supersonic as it passes the corner region. The flow just inside each point on the curved shock exactly satisfies the oblique shock relations (9.83) for

Weak shock family above sonic line Ma > 1 Sonic line Strong shock family below sonic line Ma 2

Ma < 1

θ >θ max

θ 1

Ma1 > 1 Ma < 1 Ma 2 Sonic line Ma > 1

(a) (b)

Fig. 9.22 Supersonic flow past a wedge: (a) small wedge angle, attached oblique shock forms; (b) large wedge angle, attached shock not possible, broad curved detached shock forms.

9.9 Mach Waves and Oblique Shock Waves

649

that particular value of β and the given Ma1. Every condition along the curved shock is a point on the shock polar of Fig. 9.21. Points near the front of the wedge are in the strong shock family, and points aft of the sonic line are in the weak shock family. The analysis of detached shock waves is extremely complex [13], and experimentation is usually needed, such as the shadowgraph optical technique of Fig. 9.10. The complete family of oblique shock solutions can be plotted or computed from Eqs. (9.83). For a given k, the wave angle β varies with Ma1 and θ, from Eq. (9.83b). By using a trigonometric identity for tan (β 2 θ) this can be rewritten in the more convenient form tan θ 5

2 cot β (Ma21 sin2 β 2 1) Ma21 (k 1 cos 2β) 1 2

(9.86)

All possible solutions of Eq. (9.86) for k 5 1.4 are shown in Fig. 9.23. For deflections θ , θmax there are two solutions: a weak shock (small β) and a strong shock (large β), as expected. All points along the dash–dot line for θmax satisfy Eq. (9.85). A dashed line has been added to show where Ma2 is exactly sonic. We see that there is a narrow region near maximum deflection where the weak shock downstream flow is subsonic. For zero deflections (θ 5 0) the weak shock family satisfies the wave angle relation 1 β 5 μ 5 sin21 (9.87) Ma1 Thus weak shocks of vanishing deflection are equivalent to Mach waves. Meanwhile the strong shocks all converge at zero deflection to the normal shock condition β 5 908.

50° k = 1.4 Ma1 = ∞ 10 40° 6

Deflection angle θ

4 3

30°

2.5

2

20°

1.8

Fig. 9.23 Oblique shock deflection versus wave angle for various upstream Mach numbers, k 5 1.4: dash–dot curve, locus of θmax, divides strong (right) from weak (left) shocks; dashed curve, locus of sonic points, divides subsonic Ma2 (right) from supersonic Ma2 (left).

1.6 10° 1.4 1.2 0°

30°

60° Wave angle β

90°

650

Chapter 9 Compressible Flow

Two additional oblique shock charts are given in App. B for k 5 1.4, where Fig. B.1 gives the downstream Mach number Ma2 and Fig. B.2 the pressure ratio p2/p1, each plotted as a function of Ma1 and θ. Additional graphs, tables, and computer programs are given in Refs. 20 and 21.

Very Weak Shock Waves

For any finite θ the wave angle β for a weak shock is greater than the Mach angle μ. For small θ Eq. (9.86) can be expanded in a power series in tan θ with the following linearized result for the wave angle: sin β 5 sin μ 1

k11 tan θ 1 p 1 2(tan2 θ) 1 p 4 cos μ

(9.88)

For Ma1 between 1.4 and 20.0 and deflections less than 68 this relation predicts β to within 18 for a weak shock. For larger deflections it can be used as a useful initial guess for iterative solution of Eq. (9.86). Other property changes across the oblique shock can also be expanded in a power series for small deflection angles. Of particular interest is the pressure change from Eq. (9.83a), for which the linearized result for a weak shock is p2 2 p1 k Ma21 5 tan θ 1 p 1 2(tan2 θ) 1 p p1 (Ma21 2 1) 1/2

(9.89)

The differential form of this relation is used in the next section to develop a theory for supersonic expansion turns. Figure 9.24 shows the exact weak shock pressure jump 3.0 k = 1.4

Ma1 = 10 8 2.0 6 p2 – p1 p1

4 3 1.0 2

Fig. 9.24 Pressure jump across a weak oblique shock wave from Eq. (9.83a) for k 5 1.4. For very small deflections Eq. (9.89) applies.

Eq. (9.89), Ma1 = 2 0

0

10°

5° Flow deflection θ

15°

9.9 Mach Waves and Oblique Shock Waves

651

computed from Eq. (9.83a). At very small deflections the curves are linear with slopes given by Eq. (9.89). Finally, it is educational to examine the entropy change across a very weak shock. Using the same power series expansion technique, we can obtain the following result for small flow deflections: s2 2 s1 (k2 2 1)Ma61 5 tan3 θ 1 p 1 2(tan4 θ) 1 p cp 12(Ma21 2 1) 3/2

(9.90)

The entropy change is cubic in the deflection angle θ. Thus weak shock waves are very nearly isentropic, a fact that is also used in the next section.

EXAMPLE 9.17 Ma2 Ma1 = 2.0

β

p1 = 10 lbf/in2 10°

Air at Ma 5 2.0 and p 5 10 lbf/in2 absolute is forced to turn through 108 by a ramp at the body surface. A weak oblique shock forms as in Fig. E9.17. For k 5 1.4 compute from exact oblique shock theory (a) the wave angle β, (b) Ma2, and (c) p2. Also use the linearized theory to estimate (d ) β and (e) p2.

Solution using Excel E9.17

With Ma1 5 2.0 and θ 5 108 known, we can estimate β < 408 6 28 from Fig. 9.23. For more accuracy, we can set an Excel iteration, using improved guesses for β. Begin with a guess of 408 and zero in on the correct wave angle. The writer’s guesses are as follows:

1 2 3 4 5

A

B

β – guess

θ – Eq. (9.86)

40.00 38.00 39.00 39.30 39.32

10.623 8.767 9.710 9.987 10.006

The iteration converges to

β 5 39.328

Ans. (a)

The normal Mach number upstream is thus Man1 5 Ma1 sin β 5 2.0 sin 39.32° 5 1.267 With Man1 we can use the normal shock relations (Table B.2) or Fig. 9.9 or Eqs. (9.56) to (9.58) to compute Man2 5 0.8031

p2 5 1.707 p1

Thus the downstream Mach number and pressure are Ma2 5

Man2 0.8031 5 5 1.64 sin ( β 2 θ) sin (39.32° 2 10°)

p2 5 (10 lbf/in2 absolute) (1.707) 5 17.07 lbf/in2 absolute

Ans. (b) Ans. (c)

652

Chapter 9 Compressible Flow Notice that the computed pressure ratio agrees with Figs. 9.24 and B.2. For the linearized theory the Mach angle is μ 5 sin21 (1/2.0) 5 308. Equation (9.88) then estimates that sin β < sin 30° 1

2.4 tan 10° 5 0.622 4 cos 30°

β < 38.5°

or

Ans. (d )

Equation (9.89) estimates that p2 1.4 (2) 2 tan 10° 1

653

Ma > 1

Ma increases

Fig. 9.25 Some examples of supersonic expansion and compression: (a) gradual isentropic compression on a concave surface, Mach waves coalesce farther out to form oblique shock; (b) gradual isentropic expansion on convex surface, Mach waves diverge; (c) sudden compression, nonisentropic shock forms; (d ) sudden expansion, centered isentropic fan of Mach waves forms.

(a)

(b)

Oblique shock

Mach waves Ma2 < Ma1

Ma1 > 1

Ma > 1

Ma increases

(c)

(d)

the limit, β < μ 5 sin21

1 Ma

(9.91a)

dp k Ma2 < dθ p (Ma2 2 1) 1/2

(9.91b)

Since the flow is nearly isentropic, we have the frictionless differential momentum equation for a perfect gas: dp 5 2ρV dV 5 2kp Ma2

dV V

(9.92)

Combining Eqs. (9.91a) and (9.92) to eliminate dp, we obtain a relation between turning angle and velocity change: dθ 5 2(Ma2 2 1) 1/2

dV V

(9.93)

This can be integrated into a functional relation for finite turning angles if we can relate V to Ma. We do this from the definition of Mach number: V 5 Ma a or

d Ma da dV 5 1 a V Ma

(9.94)

654

Chapter 9 Compressible Flow

Finally, we can eliminate da/a because the flow is isentropic and hence a0 is constant for a perfect gas: a 5 a0 31 1 12 (k 2 1) Ma2 4 21/2

212 (k 2 1) Ma d Ma da 5 a 1 1 12 (k 2 1) Ma2

or

(9.95)

Eliminating dV/V and da/a from Eqs. (9.93) to (9.95), we obtain a relation solely between turning angle and Mach number: dθ 5 2

(Ma2 2 1) 1/2 11

1 2 (k

d Ma 2 1) Ma Ma 2

(9.96)

Before integrating this expression, we note that the primary application is to expansions: increasing Ma and decreasing θ. Therefore, for convenience, we define the Prandtl-Meyer angle ω(Ma), which increases when θ decreases and is zero at the sonic point: dω 5 2dθ    ω 5 0    at    Ma 5 1

(9.97)

Thus we integrate Eq. (9.96) from the sonic point to any value of Ma:

#

0

ω

dω 5

#

Ma

1

2

(Ma 2 1) 1/2 11

1 2 (k

d Ma 2 1) Ma Ma 2

(9.98)

The integrals are evaluated in closed form, with the result, in radians, ω(Ma) 5 K1/2 tan21 a

Ma2 2 1 1/2 b 2 tan21 (Ma2 2 1) 1/2 K

K5

where

(9.99)

k11 k21

This is the Prandtl-Meyer supersonic expansion function, which is plotted in Fig. 9.26 and tabulated in Table B.5 for k 5 1.4, K 5 6. The angle ω changes rapidly at first and then levels off at high Mach number to a limiting value as Ma → `: ωmax 5

π 1/2 (K 2 1) 5 130.45°    if    k 5 1.4 2

(9.100)

Thus a supersonic flow can expand only through a finite turning angle before it reaches infinite Mach number, maximum velocity, and zero temperature. Gradual expansion or compression between finite Mach numbers Ma1 and Ma2, neither of which is unity, is computed by relating the turning angle Dω to the difference in Prandtl-Meyer angles for the two conditions ¢ω1S2 5 ω(Ma2 ) 2 ω(Ma1 )

(9.101)

The change Dω may be either positive (expansion) or negative (compression) as long as the end conditions lie in the supersonic range. Let us illustrate with an example.

9.10 Prandtl-Meyer Expansion Waves

655

140° Ma → ∞: ω = 130.45°

120°

100°

80°

ω 60°

40°

20° k = 1.4

Fig. 9.26 The Prandtl-Meyer supersonic expansion from Eq. (9.99) for k 5 1.4.

0 1

4

8 12 Mach number

16

20

EXAMPLE 9.18 Air (k 5 1.4) flows at Ma1 5 3.0 and p1 5 200 kPa. Compute the final downstream Mach number and pressure for (a) an expansion turn of 208 and (b) a gradual compression turn of 208.

Solution using Excel Part (a)

The isentropic stagnation pressure is p0 5 p1 31 1 0.2(3.0) 2 4 3.5 5 7347 kPa and this will be the same at the downstream point. For Ma1 5 3.0 we find from Table B.5 or Eq. (9.99) that ω1 5 49.7578. The flow expands to a new condition such that ω2 5 ω1 1 ¢ω 5 49.757° 1 20° 5 69.757° Inversion of Eq. (9.99), to find Ma when ω is given, requires iteration, and Excel is well suited for this job. Hard to read, but Fig. 9.26 indicates ω < 4. Make a guess of ω 5 4 and program Eq. (9.99) into an Excel cell. The writer’s improved guesses are shown.

1 2 3 4

A

B

Ma – guess 4.00 4.20 4.30 4.32

ω 2 Eq. (9.99) 65.78 68.33 69.54 69.78

656

Chapter 9 Compressible Flow The iteration converges to

Ma2 5 4.32

Ans. (a)

The isentropic pressure at this new condition is p2 5

Part (b)

p0 31 1 0.2(4.32) 2 4 3.5

5

7347 5 31.9 kPa 230.1

Ans. (a)

The flow compresses to a lower Prandtl-Meyer angle: ω2 5 49.757° 2 20° 5 29.757° Again from Eq. (9.99), Table B.5, or Excel we compute that Ans. (b)

Ma2 5 2.125 p2 5

p0 31 1 0.2(2.125) 2 4 3.5

5

7347 5 773 kPa 9.51

Ans. (b)

Similarly, we compute density and temperature changes by noticing that T0 and ρ0 are constant for isentropic flow.

Application to Supersonic Airfoils

The oblique shock and Prandtl-Meyer expansion theories can be used to patch together a number of interesting and practical supersonic flow fields. This marriage, called shock expansion theory, is limited by two conditions: (1) Except in rare instances the flow must be supersonic throughout, and (2) the wave pattern must not suffer interference from waves formed in other parts of the flow field. A very successful application of shock expansion theory is to supersonic airfoils. Figure 9.27 shows two examples, a flat plate and a diamond-shaped foil. In contrast to subsonic flow designs (Fig. 8.21), these airfoils must have sharp leading edges, which form attached oblique shocks or expansion fans. Rounded supersonic leading edges would cause detached bow shocks, as in Fig. 9.19 or 9.22b, greatly increasing the drag and lowering the lift. In applying shock expansion theory, one examines each surface turning angle to see whether it is an expansion (“opening up”) or compression (obstruction) to the surface flow. Figure 9.27a shows a flat-plate foil at an angle of attack. There is a leading-edge shock on the lower edge with flow deflection θ 5 α, while the upper edge has an expansion fan with increasing Prandtl-Meyer angle Dω 5 α. We compute p3 with expansion theory and p2 with oblique shock theory. The force on the plate is thus F 5 (p2 2 p3)Cb, where C is the chord length and b the span width (assuming no wingtip effects). This force is normal to the plate, and thus the lift force normal to the stream is L 5 F cos α, and the drag parallel to the stream is D 5 F sin α. The dimensionless coefficients CL and CD have the same definitions as in low-speed flow, Eqs. (7.66), except that the perfect-gas law identity 12ρV 2 ; 12 kp Ma2 is very useful here: CL 5

1 2 kp∞

L D     CD 5 1 2 2 Ma∞ bC 2 kp∞ Ma∞ bC

(9.102)

9.10 Prandtl-Meyer Expansion Waves Expansion fan

657

Oblique shock

Ma3 > Ma∞

α

p3 < p∞ p03 = p0∞

Ma∞ p∞ p0∞

Vortex sheet

Ma2 < Ma∞ p2 > p∞ Oblique shock

p02 < p0∞ Expansion fan

(a)

α

Fig. 9.27 Supersonic airfoils: (a) flat plate, higher pressure on lower surface, drag due to small downstream component of net pressure force; (b) diamond foil, higher pressures on both lower surfaces, additional drag due to body thickness.

p3 > p∞ p5 < p3

Ma∞ p∞

p2 > p3

p4 > p5 p4 < p2

(b)

The typical supersonic lift coefficient is much smaller than the subsonic value CL < 2πα, but the lift can be very large because of the large value of 12 ρV 2 at supersonic speeds. At the trailing edge in Fig. 9.27a, a shock and fan appear in reversed positions and bend the two flows back so that they are parallel in the wake and have the same pressure. They do not have quite the same velocity because of the unequal shock strengths on the upper and lower surfaces; hence a vortex sheet trails behind the wing. This is very interesting, but in the theory you ignore the trailing-edge pattern entirely, since it does not affect the surface pressures: The supersonic surface flow cannot “hear” the wake disturbances. The diamond foil in Fig. 9.27b adds two more wave patterns to the flow. At this particular α less than the diamond half-angle, there are leading-edge shocks on both surfaces, the upper shock being much weaker. Then there are expansion fans on each shoulder of the diamond: The Prandtl-Meyer angle change Dω equals the sum of the leading-edge and trailing-edge diamond half-angles. Finally, the trailing-edge pattern is similar to that of the flat plate (9.27a) and can be ignored in the calculation. Both lower-surface pressures p2 and p4 are greater than their upper counterparts, and the lift is nearly that of the flat plate. There is an additional drag due to thickness because p4 and p5 on the trailing surfaces are lower than their counterparts p2 and p3. The diamond drag is greater than the flat-plate drag, but this must be endured in practice to achieve a wing structure strong enough to support these forces.

658

Chapter 9 Compressible Flow

The theory sketched in Fig. 9.27 is in good agreement with measured supersonic lift and drag as long as the Reynolds number is not too small (thick boundary layers) and the Mach number not too large (hypersonic flow). It turns out that for large ReC and moderate supersonic Ma` the boundary layers are thin and separation seldom occurs, so that the shock expansion theory, although frictionless, is quite successful. Let us look now at an example.

EXAMPLE 9.19 A flat-plate airfoil with C 5 2 m is immersed at α 5 88 in a stream with Ma` 5 2.5 and p` 5 100 kPa. Compute (a) CL and (b) CD, and compare with low-speed airfoils. Compute (c) lift and (d ) drag in newtons per unit span width.

Solution Instead of using a lot of space outlining the detailed oblique shock and Prandtl-Meyer expansion computations, we list all pertinent results in Fig. E9.19 on the upper and lower surfaces. Using the theories of Secs. 9.9 and 9.10, you should verify every single one of the calculations in Fig. E9.19 to make sure that all details of shock expansion theory are well understood. ω = 8° = α ω 3 = 47.124° Ma3 = 2.867 p03 = p0∞ = 1709 kPa p03 p3 = 30.05 p3 = 56.85 kPa

8° Ma∞ = 2.5 p∞ = 100 kPa p0∞ = 1709 kPa ω∞ = 39.124°

θ = α = 8° β = 30.01° Ma2 = 2.169 p2 p∞ = 1.657 p2 = 165.7 kPa

Do not compute

E9.19 The important final results are p2 and p3, from which the total force per unit width on the plate is F 5 ( p2 2 p3 )bC 5 (165.7 2 56.85) (kPa) (1 m) (2 m) 5 218 kN The lift and drag per meter width are thus L 5 F cos 8° 5 216 kN D 5 F sin 8° 5 30 kN

Ans. (c) Ans. (d )

These are very large forces for only 2 m2 of wing area. From Eq. (9.102) the lift coefficient is CL 5

216 kN 5 0.246 kPa) (2.5) 2 (2 m2 )

1 2 (1.4) (100

Ans. (a)

9.10 Prandtl-Meyer Expansion Waves

659

The comparable low-speed coefficient from Eq. (8.67) is CL 5 2π sin 88 5 0.874, which is 3.5 times larger. From Eq. (9.102) the drag coefficient is CD 5

1 2 (1.4) (100

30 kN 5 0.035 kPa) (2.5) 2 (2 m2 )

Ans. (b)

From Fig. 7.25 for the NACA 0009 airfoil, CD at α 5 88 is about 0.009, or about 4 times smaller. Notice that this supersonic theory predicts a finite drag in spite of assuming frictionless flow with infinite wing aspect ratio. This is called wave drag, and we see that the d’Alembert paradox of zero body drag does not occur in supersonic flow.

Thin-Airfoil Theory

In spite of the simplicity of the flat-plate geometry, the calculations in Example 9.19 were laborious. In 1925 Ackeret [28] developed simple yet effective expressions for the lift, drag, and center of pressure of supersonic airfoils, assuming small thickness and angle of attack. The theory is based on the linearized expression (9.89), where tan θ < surface deflection relative to the free stream and condition 1 is the free stream, Ma1 5 Ma`. For the flat-plate airfoil, the total force F is based on p2 2 p3 p2 2 p∞ p3 2 p∞ 5 2 p∞ p∞ p∞ 5

k Ma2q (Ma2q 2 1) 1/2

3α 2 (2α) 4

(9.103)

Substitution into Eq. (9.102) gives the linearized lift coefficient for a supersonic flatplate airfoil: CL
1 C–3

yn yc (c)

Mild S0 < Sc

Fr < 1 M–1 Fr < 1 M–2 Fr > 1 M–3

(d )

Horizontal S0 = 0 yn = ∞

yc

Fr < 1 Fr > 1

Fr < 1

Fig. 10.14 Gradually varied flow for five classes of channel slope, showing the 12 basic solution curves.

Adverse (e) S0 < 0 yn = imaginary

yc

H–2 H–3

A–2 A–3

Fr > 1

10.6 Gradually Varied Flow Slope class S0 S0 S0 S0 S0

. 5 , 5 ,

Sc Sc Sc 0 0

Slope notation Steep Critical Mild Horizontal Adverse

Depth class yc yc yc yn yn

. 5 , 5 5

yn yn yn ` imaginary

711

Solution curves S-1, S-2, S-3 C-1, C-3 M-1, M-2, M-3 H-2, H-3 A-2, A-3

The solution letters S, C, M, H, and A obviously denote the names of the five types of slopes. The numbers 1, 2, 3 relate to the position of the initial point on the solution curve with respect to the normal depth yn and the critical depth yc. In type 1 solutions, the initial point is above both yn and yc, and in all cases the water depth solution y(x) becomes even deeper and farther away from yn and yc. In type 2 solutions, the initial point lies between yn and yc, and if there is no change in S0 or roughness, the solution tends asymptotically toward the lower of yn or yc. In type 3 cases, the initial point lies below both yn and yc, and the solution curve tends asymptotically toward the lower of these. Figure 10.14 shows the basic character of the local solutions, but in practice, of course, S0 varies with x, and the overall solution patches together the various cases to form a continuous depth profile y(x) compatible with a given initial condition and a given discharge Q. There is a fine discussion of various composite solutions in Ref. 2, Chap. 9; see also Ref. 22, Sec. 12.7.

Numerical Solution

The basic relation for gradually varied flow, Eq. (10.49), is a first-order ordinary differential equation that can be easily solved numerically. For a given constant-volume flow rate Q, it may be written in the form dy S0 2 n2Q2/(α2A2R4/3 h ) 5 2 3 dx 1 2 Q b0/(gA )

(10.51)

subject to an initial condition y 5 y0 at x 5 x0. It is assumed that the bottom slope S0(x) and the cross-sectional shape parameters (b0, P, A) are known everywhere along the channel. Then one may solve Eq. (10.51) for local water depth y(x) by any standard numerical method. The author uses an Excel spreadsheet for a personal computer. Step sizes Dx may be selected so that each change Dy is limited to no greater than, say, 1 percent. The solution curves are generally well behaved unless there are discontinuous changes in channel parameters. Note that if one approaches the critical depth yc, the denominator of Eq. (10.51) approaches zero, so small step sizes are required. It helps physically to know what type of solution curve (M-1, S-2, or the like) you are proceeding along, but this is not mathematically necessary.

EXAMPLE 10.9 Let us extend the data of Example 10.5 to compute a portion of the profile shape. Given is a wide channel with n 5 0.022, S0 5 0.0048, and q 5 50 ft3/(s ? ft). If y0 5 3 ft at x 5 0, how far along the channel x 5 L does it take the depth to rise to yL 5 4 ft? Is the 4-ft depth position upstream or downstream in Fig. E10.9a?

712

Chapter 10 Open-Channel Flow

Solution In Example 10.5 we computed yc 5 4.27 ft. Since our initial depth y 5 3 ft is less than yc, we know the flow is supercritical. Let us also compute the normal depth for the given slope S0 by setting q 5 50 ft3/(s ? ft) in the Chézy formula (10.19) with Rh 5 yn: q5

1.486 α 1/2 1/2 3yn (1 ft) 4y2/3 5 50 ft3/(s # ft) AR2/3 h S0 5 n (0.0048) n 0.022 y = c 4.27 ft y = n 4.14 ft

y0 = 3 ft

y L = 4 ft S0 = 0.0048

L=? x=0

E10.9a

x=L

yn < 4.14 ft

Solve for:

Thus both y(0) 5 3 ft and y(L) 5 4 ft are less than yn, which is less than yc, so we must be on an S-3 curve, as in Fig. 10.14a. For a wide channel, Eq. (10.51) reduces to dy S0 2 n2q2/(α2y10/3 ) 5 dx 1 2 q2/(gy3 )