0131292935_COVER
9/8/05
3:07 PM
Page 1
FLUID MECHANICS FIFTH EDITION Authors
The fifth edition of this established text provides an excellent and comprehensive treatment of fluid mechanics that is concisely written and supported by good worked examples. This revision of a classic text presents relevant material for mechanical and civil engineers, as well as energy and environmental services engineers. It recognises the evolution of the subject and provides thorough coverage of both established theory and emerging topics.
Excellent coverage • All the latest developments and applications, including emerging specialisms • Strong coverage of the principles of fluid flow – fundamentals emphasized early in the text Emphasis on understanding • Good, clear explanations, together with extensive worked examples and tutorials – brought together to reinforce the reader’s understanding of all the key principles Helpful resources on accompanying website • With solutions to tutorials and simulations of fluid mechanics, presented through some 20 programs, all fully discussed in the text
Building on the success of previous editions, this fifth edition introduces the following new features: • New chapters on the impact of environmental change and the fluid mechanics required to both understand the consequences of climate change and develop renewable energy technologies • Updated chapter on dimensional analysis • Updated solutions manual • Extended website to include enhanced and additional simulations
Dr Janusz Gasiorek, formerly of South Bank University, London where he led the Fluid Mechanics group in Mechanical Engineering, with specialist research interest in rotodynamic machinery and fan engineering. Professor John Swaffield, Heriot–Watt University, has taught fluid mechanics for 30 years with specialist research in pressure transients, free surface unsteady flows and water conservation. Dr Lynne Jack, Heriot–Watt University, senior lecturer in energy systems and associated environmental impacts, with research interests in unsteady flow modelling and the implications for the built environment of climate change.
FIFTH EDITION
FLUID MECHANICS
Fluid Mechanics has become a textbook of choice with both students and lecturers, due to its:
Dr John Douglas, formerly of South Bank University, London.
COVER IMAGE: MARINE CURRENT TURBINES
TM
LTD
DOUGLAS GASIOREK SWAFFIELD JACK
Fluid Mechanics is ideal for use throughout a first degree course in all engineering disciplines where a good understanding of the subject is required. It is also suitable for conversion MSc courses requiring a fundamental treatment of fluid mechanics and will be a valuable resource for specialist Continuing Professional Development courses, including those offered by Distance Learning.
John F. Douglas Janusz M. Gasiorek John A. Swaffield www.pearson-books.com
Additional student support at www.pearsoned.co.uk/douglas
Lynne B. Jack
FM5_A01.fm Page i Friday, November 17, 2006 4:43 PM
Fluid Mechanics Visit the Fluid Mechanics, fifth edition Companion Website at www.pearsoned.co.uk/douglas to find valuable student learning material including: l
Simulations and computer programs for students, with clear instructions on how to use them to enhance study
FM5_A01.fm Page ii Friday, November 17, 2006 4:43 PM
We work with leading authors to develop the strongest educational materials in engineering, bringing cutting-edge thinking and best learning practice to a global market. Under a range of well-known imprints, including Prentice Hall, we craft high quality print and electronic publications which help readers to understand and apply their content, whether studying or at work. To find out more about the complete range of our publishing please visit us on the World Wide Web at: www.pearsoned.co.uk
FM5_A01.fm Page iii Friday, November 17, 2006 4:43 PM
Fluid Mechanics Fifth edition JOHN F. DOUGLAS M.Sc., Ph.D., A.C.G.I., D.I.C., C.Eng., M.I.C.E., M.I.Mech.E. Formerly of London South Bank University
JANUSZ M. GASIOREK B.Sc., Ph.D., C.Eng., M.I.Mech.E., M.C.I.B.S.E. Formerly of London South Bank University
JOHN A. SWAFFIELD F.R.S.E., B.Sc., M.Phil., Ph.D., C.Eng., M.R.Ae.S., F.C.I.W.E.M., F.C.I.B.S.E. William Watson Professor of Building Engineering and Head of the School of the Built Environment, Heriot-Watt University, Edinburgh
LYNNE B. JACK B.Eng., Ph.D., M.I.L.T. Senior Lecturer in Environmental Engineering, School of the Built Environment, Heriot-Watt University, Edinburgh
FM5_A01.fm Page iv Friday, November 17, 2006 4:43 PM
Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk First published by Pitman Publishing Limited 1979 Second Edition 1985 Third Edition published under the Longman imprint 1995 Fourth Edition 2001 Fifth Edition 2005 © J. F. Douglas, J. M. Gasiorek and J. A. Swaffield 1979, 2001 © J. F. Douglas, J. M. Gasiorek, J. A. Swaffield and Lynne B. Jack 2005 The rights of J. F. Douglas, J. M. Gasiorek, J. A. Swaffield and Lynne B. Jack to be identified as authors of this Work have been asserted in accordance with the Copyright, Designs and Patents Act 1988. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, 90 Tottenham Court Road, London W1T 4LP. ISBN-13: 978-0-13-129293-2 ISBN-10: 0-13-129293-5 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data Fluid mechanics / John F. Douglas ... [et al.].— 5th ed. p. cm. Includes bibliographical references and index. 1. Fluid mechanics. I. Douglas, John F. TA357.D68 2006 620.1’06—dc22 2005054617 10 9 8 7 6 5 4 3 2 10 09 08 07 06 Set by 35 in 10/12pt Times Roman Printed and bound by Ashford Colour Press Ltd, Gosport
FM5_A01.fm Page v Friday, November 17, 2006 4:43 PM
for Janusz M. Gasiorek 1927–2003 friend and guide
FM5_A01.fm Page vi Friday, November 17, 2006 4:43 PM
FM5_A01.fm Page vii Friday, November 17, 2006 4:43 PM
Contents Preface to the Fifth Edition xix Preface to the Fourth Edition xxi Preface to the Third Edition xxiv Preface to the Second Edition xxvi Preface to the First Edition xxvii Acknowledgements xxviii List of Computer Programs xxix List of Symbols xxxi
PART I ELEMENTS OF FLUID MECHANICS Chapter 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20
Fluids and their Properties 2
Fluids 4 Shear stress in a moving fluid 4 Differences between solids and fluids 5 Newtonian and non-Newtonian fluids 6 Liquids and gases 7 Molecular structure of materials 7 The continuum concept of a fluid 9 Density 10 Viscosity 11 Causes of viscosity in gases 12 Causes of viscosity in a liquid 13 Surface tension 14 Capillarity 15 Vapour pressure 16 Cavitation 17 Compressibility and the bulk modulus 17 Equation of state of a perfect gas 19 The universal gas constant 19 Specific heats of a gas 19 Expansion of a gas 20 Concluding remarks 22 Summary of important equations and concepts 22
FM5_A01.fm Page viii Friday, November 17, 2006 4:43 PM
viii
Contents
Chapter 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20
Statics of fluid systems 26 Pressure 27 Pascal’s law for pressure at a point 28 Variation of pressure vertically in a fluid under gravity 29 Equality of pressure at the same level in a static fluid 30 General equation for the variation of pressure due to gravity from point to point in a static fluid 32 Variation of pressure with altitude in a fluid of constant density 33 Variation of pressure with altitude in a gas at constant temperature 34 Variation of pressure with altitude in a gas under adiabatic conditions 35 Variation of pressure and density with altitude for a constant temperature gradient 38 Variation of temperature and pressure in the atmosphere 39 Stability of the atmosphere 41 Pressure and head 43 The hydrostatic paradox 44 Pressure measurement by manometer 45 Relative equilibrium 51 Pressure distribution in a liquid subject to horizontal acceleration 51 Effect of vertical acceleration 52 General expression for the pressure in a fluid in relative equilibrium 52 Forced vortex 56 Concluding remarks 57 Summary of important equations and concepts 57 Problems 57
Chapter 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13
Pressure and Head 24
Static Forces on Surfaces. Buoyancy 60
Action of fluid pressure on a surface 62 Resultant force and centre of pressure on a plane surface under uniform pressure 62 Resultant force and centre of pressure on a plane surface immersed in a liquid 63 Pressure diagrams 68 Force on a curved surface due to hydrostatic pressure 71 Buoyancy 73 Equilibrium of floating bodies 76 Stability of a submerged body 76 Stability of floating bodies 77 Determination of the metacentric height 78 Determination of the position of the metacentre relative to the centre of buoyancy 78 Periodic time of oscillation 81 Stability of a vessel carrying liquid in tanks with a free surface 82 Concluding remarks 85 Summary of important equations and concepts 85 Problems 85
FM5_A01.fm Page ix Friday, November 17, 2006 4:43 PM
Contents
PART II
CONCEPTS OF FLUID FLOW 88
Chapter 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14
Fluid flow 92 Uniform flow and steady flow 93 Frames of reference 93 Real and ideal fluids 94 Compressible and incompressible flow 94 One-, two- and three-dimensional flow 95 Analyzing fluid flow 96 Motion of a fluid particle 96 Acceleration of a fluid particle 98 Laminar and turbulent flow 100 Discharge and mean velocity 102 Continuity of flow 104 Continuity equations for three-dimensional flow using Cartesian coordinates 107 Continuity equation for cylindrical coordinates 109 Concluding remarks 109 Summary of important equations and concepts 110 Problems 110
Chapter 5 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17
Motion of Fluid Particles and Streams 90
The Momentum Equation and its Applications 112
Momentum and fluid flow 114 Momentum equation for two- and three-dimensional flow along a streamline 115 Momentum correction factor 116 Gradual acceleration of a fluid in a pipeline neglecting elasticity 119 Force exerted by a jet striking a flat plate 120 Force due to the deflection of a jet by a curved vane 123 Force exerted when a jet is deflected by a moving curved vane 124 Force exerted on pipe bends and closed conduits 126 Reaction of a jet 129 Drag exerted when a fluid flows over a flat plate 136 Angular motion 138 Euler’s equation of motion along a streamline 141 Pressure waves and the velocity of sound in a fluid 143 Velocity of propagation of a small surface wave 146 Differential form of the continuity and momentum equations 148 Computational treatment of the differential forms of the continuity and momentum equations 151 Comparison of CFD methodologies 155 Concluding remarks 162 Summary of important equations and concepts 162 Further reading 163 Problems 163
ix
FM5_A01.fm Page x Friday, November 17, 2006 4:43 PM
x
Contents
Chapter 6 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13 6.14 6.15 6.16 6.17 6.18 6.19
Mechanical energy of a flowing fluid 168 Steady flow energy equation 172 Kinetic energy correction factor 174 Applications of the steady flow energy equation 175 Representation of energy changes in a fluid system 178 The Pitot tube 180 Determination of volumetric flow rate via Pitot tube 181 Computer program VOLFLO 183 Changes of pressure in a tapering pipe 183 Principle of the venturi meter 185 Pipe orifices 187 Limitation on the velocity of flow in a pipeline 188 Theory of small orifices discharging to atmosphere 188 Theory of large orifices 192 Elementary theory of notches and weirs 193 The power of a stream of fluid 197 Radial flow 198 Flow in a curved path. Pressure gradient and change of total energy across the streamlines 199 Vortex motion 202 Concluding remarks 208 Summary of important equations and concepts 209 Problems 209
Chapter 7 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11
The Energy Equation and its Applications 166
Two-dimensional Ideal Flow 212
Rotational and irrotational flow 214 Circulation and vorticity 216 Streamlines and the stream function 218 Velocity potential and potential flow 220 Relationship between stream function and velocity potential. Flow nets 224 Straight line flows and their combinations 228 Combined source and sink flows. Doublet 236 Flow past a cylinder 241 Curved flows and their combinations 244 Flow past a cylinder with circulation. Kutta–Joukowsky’s law 249 Computer program ROTCYL 252 Concluding remarks 253 Summary of important equations and concepts 253 Problems 254
PART III DIMENSIONAL ANALYSIS AND SIMILARITY 256 Chapter 8 8.1 8.2
Dimensional Analysis 258
Dimensional analysis 260 Dimensions and units 260
FM5_A01.fm Page xi Friday, November 17, 2006 4:43 PM
Contents
8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11
Dimensional reasoning, hom*ogeneity and dimensionless groups 260 Fundamental and derived units and dimensions 261 Additional fundamental dimensions 263 Dimensions of derivatives and integrals 265 Units of derived quantities 266 Conversion between systems of units, including the treatment of dimensional constants 266 Dimensional analysis by the indicial method 269 Dimensional analysis by the group method 271 The significance of dimensionless groups 279 Concluding remarks 280 Summary of important equations and concepts 280 Further reading 280 Problems 281
Chapter 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12
Similarity 282
Geometric similarity 286 Dynamic similarity 286 Model studies for flows without a free surface. Introduction to approximate similitude at high Reynolds numbers 291 Zone of dependence of Mach number 293 Significance of the pressure coefficient 294 Model studies in cases involving free surface flow 295 Similarity applied to rotodynamic machines 297 River and harbour models 299 Groundwater and seepage models 305 Computer program GROUND, the simulation of groundwater seepage 310 Pollution dispersion modelling, outfall effluent and stack plumes 311 Pollutant dispersion in one-dimensional steady uniform flow 314 Concluding remarks 319 Summary of important equations and concepts 319 Further reading 320 References 320 Problems 321
PART IV BEHAVIOUR OF REAL FLUIDS 322 Chapter 10 Laminar and Turbulent Flows in Bounded Systems 324 10.1 10.2 10.3
xi
Incompressible, steady and uniform laminar flow between parallel plates 326 Incompressible, steady and uniform laminar flow in circular cross-section pipes 331 Incompressible, steady and uniform turbulent flow in bounded conduits 335
FM5_A01.fm Page xii Friday, November 17, 2006 4:43 PM
xii
Contents
10.4
Incompressible, steady and uniform turbulent flow in circular cross-section pipes 338 10.5 Steady and uniform turbulent flow in open channels 342 10.6 Velocity distribution in turbulent, fully developed pipe flow 343 10.7 Velocity distribution in fully developed, turbulent flow in open channels 352 10.8 Separation losses in pipe flow 352 10.9 Significance of the Colebrook–White equation in pipe and duct design 359 10.10 Computer program CBW 362 Concluding remarks 362 Summary of important equations and concepts 363 Further reading 363 Problems 364
Chapter 11
Boundary Layer 366
11.1 11.2 11.3
Qualitative description of the boundary layer 368 Dependence of pipe flow on boundary layer development at entry 370 Factors affecting transition from laminar to turbulent flow regimes 371 11.4 Discussion of flow patterns and regions within the turbulent boundary layer 372 11.5 Prandtl mixing length theory 374 11.6 Definitions of boundary layer thicknesses 377 11.7 Application of the momentum equation to a general section of boundary layer 378 11.8 Properties of the laminar boundary layer formed over a flat plate in the absence of a pressure gradient in the flow direction 379 11.9 Properties of the turbulent boundary layer over a flat plate in the absence of a pressure gradient in the flow direction 384 11.10 Effect of surface roughness on turbulent boundary layer development and skin friction coefficients 388 11.11 Effect of pressure gradient on boundary layer development 388 Concluding remarks 391 Summary of important equations and concepts 391 Further reading 392 Problems 392
Chapter 12 12.1 12.2 12.3 12.4 12.5 12.6 12.7
Incompressible Flow around a Body 394
Regimes of external flow 396 Drag 397 Drag coefficient and similarity considerations 401 Resistance of ships 403 Flow past a cylinder 407 Flow past a sphere 411 Flow past an infinitely long aerofoil 418
FM5_A01.fm Page xiii Friday, November 17, 2006 4:43 PM
Contents
xiii
12.8 Flow past an aerofoil of finite length 426 12.9 Wakes and drag 430 12.10 Computer program WAKE 435 Concluding remarks 436 Summary of important equations and concepts 436 Problems 436
Chapter 13 13.1 13.2 13.3 13.4 13.5
Compressible Flow around a Body 438
Effects of compressibility 440 Shock waves 445 Oblique shock waves 455 Supersonic expansion and compression 457 Computer program NORSH 459 Concluding remarks 459 Summary of important equations and concepts 460 Problems 460
PART V STEADY FLOW IN PIPES, DUCTS AND OPEN CHANNELS 462 Chapter 14 Steady Incompressible Flow in Pipe and Duct Systems 464 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10 14.11 14.12 14.13 14.14
General approach 466 Incompressible flow through ducts and pipes 467 Computer program SIPHON 470 Incompressible flow through pipes in series 471 Incompressible flow through pipes in parallel 473 Incompressible flow through branching pipes. The three-reservoir problem 475 Incompressible steady flow in duct networks 478 Resistance coefficients for pipelines in series and in parallel 486 Incompressible flow in a pipeline with uniform draw-off 490 Incompressible flow through a pipe network 490 Head balance method for pipe networks 491 Computer program HARDYC 492 The quantity balance method for pipe networks 494 Quasi-steady flow 497 Concluding remarks 503 Summary of important equations and concepts 503 Further reading 504 Problems 504
Chapter 15 15.1 15.2
Uniform Flow in Open Channels 508
Flow with a free surface in pipes and open channels 510 Resistance formulae for steady uniform flow in open channels 512
FM5_A01.fm Page xiv Friday, November 17, 2006 4:43 PM
xiv
Contents
15.3 15.4
Optimum shape of cross-section for uniform flow in open channels 517 Optimum depth for flow with a free surface in covered channels 521 Concluding remarks 524 Summary of important equations and concepts 525 Further reading 525 Problems 526
Chapter 16 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8 16.9 16.10 16.11 16.12 16.13 16.14
Specific energy and alternative depths of flow 530 Critical depth in non-rectangular channels 532 Computer program CRITNOR 534 Non-dimensional specific energy curves 535 Occurrence of critical flow conditions 536 Flow over a broad-crested weir 537 Effect of lateral contraction of a channel 538 Non-uniform steady flow in channels 541 Equations for gradually varied flow 542 Classification of water surface profiles 544 The hydraulic jump 547 Location of a hydraulic jump 549 Computer program CHANNEL 550 Annular water flow considerations 551 Concluding remarks 556 Summary of important equations and concepts 556 Further reading 557 Problems 558
Chapter 17 17.1 17.2 17.3 17.4 17.5 17.6 17.7 17.8 17.9
Non-uniform Flow in Open Channels 528
Compressible Flow in Pipes 560
Compressible flow. The basic equations 562 Steady isentropic flow in non-parallel-sided ducts neglecting friction 563 Mass flow through a venturi meter 564 Mass flow from a reservoir through an orifice or convergent–divergent nozzle 567 Conditions for maximum discharge from a reservoir through a convergent–divergent duct or orifice 568 The Laval nozzle 569 Normal shock wave in a diffuser 573 Compressible flow in a duct with friction under adiabatic conditions. Fanno flow 578 Isothermal flow of a compressible fluid in a pipeline 582 Concluding remarks 585 Summary of important equations and concepts 586 Problems 586
FM5_A01.fm Page xv Friday, November 17, 2006 4:43 PM
Contents
PART VI FLUID MECHANICS FOR ENVIRONMENTAL CHANGE 588 Chapter 18 Environmental Change and Renewable Energy Technologies 590 18.1 18.2 18.3 18.4
Environmental change 592 The application of wind turbines to electrical power generation 602 Wave energy conversion for electrical power generation 616 Tidal power 631 Concluding remarks 632 Summary of important concepts 633 Further reading 634 References 635
Chapter 19 Environmental Change and Rainfall Runoff Flow Modelling 636 19.1 19.2 19.3 19.4 19.5 19.6 19.7
Gradually varied unsteady free surface flow 638 Computer program UNSCHAN 646 Implicit four-point scheme 648 Flood routeing 650 The prediction of flood behaviour 652 Time-dependent urban stormwater routeing 657 Combined free surface and pressure surge analysis. Siphonic rainwater systems 660 Concluding remarks 669 Summary of important equations and concepts 669 Further reading 670 References 670
PART VII UNSTEADY FLOW IN BOUNDED SYSTEMS 672 Chapter 20 20.1 20.2 20.3 20.4 20.5 20.6 20.7 20.8
Pressure Transient Theory and Surge Control 674
Wave propagation velocity and its dependence on pipe and fluid parameters and free gas 682 Computer program WAVESPD 688 Simplification of the basic pressure transient equations 690 Application of the simplified equations to explain pressure transient oscillations 690 Surge control 695 Control of surge following valve closure, with pump running and surge tank applications 697 Computer program SHAFT 704 Control of surge following pump shutdown 706 Concluding remarks 711 Summary of important equations and concepts 711
xv
FM5_A01.fm Page xvi Friday, November 17, 2006 4:43 PM
xvi
Contents
Further reading 712 Problems 714
Chapter 21 Simulation of Unsteady Flow Phenomena in Pipe, Channel and Duct Systems 716 21.1 21.2 21.3 21.4 21.5
21.6 21.7 21.8
Development of the St Venant equations of continuity and motion 718 The method of characteristics 724 Network simulation 737 Computer program FM5SURG. The simulation of waterhammer 739 Computer programs FM5WAVE and FM5GUTT. The simulation of open-channel free surface and partially filled pipe flow, with and without lateral inflow 749 Simulation of low-amplitude air pressure transient propagation 755 Computer program FM5AIR. The simulation of unsteady air flow in pipe and duct networks 756 Entrained air flow analysis review 760 Concluding remarks 763 Summary of important equations and concepts 764 Further reading 764 References 765
PART VIII FLUID MACHINERY. THEORY, PERFORMANCE AND APPLICATION 766 Chapter 22 22.1 22.2 22.3 22.4 22.5
Introduction 770 One-dimensional theory 772 Isolated blade and cascade considerations 780 Departures from Euler’s theory and losses 788 Compressible flow through rotodynamic machines 794 Concluding remarks 798 Summary of important equations and concepts 798 Further reading 798 Problems 799
Chapter 23 23.1 23.2 23.3 23.4 23.5 23.6 23.7
Theory of Rotodynamic Machines 768
Performance of Rotodynamic Machines 800
The concept of performance characteristics 802 Losses and efficiencies 803 Dimensionless coefficients and similarity laws 809 Computer program SIMPUMP 815 Scale effects 816 Type number 817 Centrifugal pumps and fans 820
FM5_A01.fm Page xvii Friday, November 17, 2006 4:43 PM
Contents
23.8 23.9 23.10 23.11 23.12 23.13 23.14
Axial flow pumps and fans 822 Mixed flow pumps and fans 825 Water turbines 826 The Pelton wheel 827 Francis turbines 831 Axial flow turbines 836 Hydraulic transmissions 839 Concluding remarks 846 Summary of important equations and concepts 847 Problems 848
Chapter 24 24.1 24.2 24.3 24.4 24.5 24.6
Reciprocating pumps 852 Rotary pumps 863 Rotary gear pumps 864 Rotary vane pumps 865 Rotary piston pumps 866 Hydraulic motors 868 Concluding remarks 868 Summary of important equations and concepts 869 Problems 870
Chapter 25 25.1 25.2 25.3 25.4 25.5 25.6 25.7 25.8 25.9 25.10 25.11 25.12 25.13 25.14
Positive Displacement Machines 850
Machine–Network Interactions 872
Fans, pumps and fluid networks 874 Parallel and series pump operation 881 Fans in series and parallel 883 Fan and system matching. An application of the steady flow energy equation 888 Change in the pump speed and the system 892 Change in the pump size and the system 895 Changes in fan speed, diameter and air density 897 Jet fans 900 Computer program MATCH 908 Cavitation in pumps and turbines 909 Fan and pump selection 914 Fan suitability 918 Ventilation and airborne contamination as a criterion for fan selection 921 Computer program CONTAM 929 Concluding remarks 931 Summary of important equations and concepts 932 Further reading 933 Problems 933
xvii
FM5_A01.fm Page xviii Friday, November 17, 2006 4:43 PM
xviii
Contents
Appendix 1
Some Properties of Common Fluids 938
A1.1 Variation of some properties of water with temperature 938 A1.2 Variation of bulk modulus of elasticity of water with temperature and pressure 939 A1.3 Variation of some properties of air with temperature at atmospheric pressure 939 A1.4 Some properties of common liquids 939 A1.5 Some properties of common gases (at p = 1 atm, T = 273 K) 940 A1.6 International Standard Atmosphere 940 A1.7 Solubility of air in pure water at various temperatures 941 A1.8 Absolute viscosity of some common fluids 941
Appendix 2 Values of Drag Coefficient CD for Various Body Shapes 942 Index 943
Supporting resources Visit www.pearsoned.co.uk/douglas to find valuable online resources Companion Website for students l Simulations and computer programs for students, with clear instructions on how to use them to enhance study For instructors l Complete, downloadable Solutions Manual For more information please contact your local Pearson Education sales representative or visit www.pearsoned.co.uk/douglas
FM5_A01.fm Page xix Friday, November 17, 2006 4:43 PM
Preface to the Fifth Edition Fluid mechanics remains a core component of engineering education. While its applications may have changed from the preoccupations of earlier engineers, and while our ability to apply its principles has been transformed by modern computing capabilities, those principles remain unchanged. Indeed it is possible that early engineers, perhaps two millennia in the past, concerned with the efficient delivery of a water supply dependent upon open channel flows, would recognize some of our current concerns as to the representation of flood routing. This fifth edition of Fluid Mechanics therefore recognizes evolution in the application of fluid mechanics as well as the necessity to reinforce and underpin the student’s understanding of its fundamental precepts. The fifth edition retains its emphasis on fundamentals in the early Parts of the text. As in previous editions, fundamentals are reinforced with both ample worked examples and tutorial examples whose solutions are available on the supporting website. Similarly computing support is provided on the website with some 20 simulations that the student or lecturer may use to extend the scope of the material provided by allowing a wide range of applications to be modelled, ranging from contamination decay in a ventilated space to pressure surge in pipe and duct flow or unsteady free surface flows in long channels. Later Parts introduce more specialist topics, including as in previous editions rotodynamic machinery and unsteady flow. The authors believe that a continuation of this presentation is both appropriate and essential. However, at the start of the twenty-first century the text cannot avoid the issues of global climate change that now increasingly appear to be corroborated by environmental research. If that research is fully substantiated then engineers will have two main roles to play, namely providing alternative energy sources and power generation to allow a continuation of supply without exacerbating environmental change, and a role in mitigating the environmental consequences of climate change, through for example the management of flood risk. The fifth edition of Fluid Mechanics addresses these concerns in a major new Part consisting of two chapters that deal with environment change, the application of fluid mechanics to energy generation from renewable sources, including wind and wave power, the fundamentals of flow simulation necessary to support flood modelling, and the development of improved techniques for controlling and attenuating rainfall runoff. The emphasis is firmly on identifying the fluid mechanics principles that future engineers will require to deploy to contribute to our response to climate change. Thus the fifth edition continues the ethos that the fundamental principles of the subject must be fully understood to allow the later introduction of specialized content, and therefore will continue to be attractive across the range of courses that include fluid mechanics.
FM5_A01.fm Page xx Friday, November 17, 2006 4:43 PM
xx
Preface to the Fifth Edition
As in previous editions the authors wish to thank all their colleagues, both at Heriot Watt and internationally, who have contributed to the development of the text by their comments and suggestions following earlier editions. As in previous editions we have attempted to recognize and incorporate all the helpful suggestions we have received; however, any errors of understanding are ours. In particular at Heriot Watt we are grateful to Dr Ian McDougall, for again ensuring that our computing simulations are in a form suitable for dissemination via the website, and to Dr David Campbell for providing the worked solutions manual. Professor Julian Wolfram is to be thanked for his support in the development of the wave energy device sections, along with Professor Garry Pender, who contributed a review of flood modelling, and Dr Grant Wright for making available his work on siphonic roof drainage and the application of the McCormack technique, both in Chapter 19. Dr Steve Wallis and Dr Sylvain Neelz are thanked for their contribution to the pollution dispersion in channel flow content. Pauline Gillett, Assistant Editor for Science and Engineering and our main contact at Pearsons, is to be thanked for continuing the tradition of pressure tempered with infinite patience we have encountered in all our dealings with staff at Pearson, Longman and initially Pitman over the 30 year development of the text through five editions. However, in one essential respect this edition is different to any of its predecessors. Janusz Gasiorek, a founder of this series of texts, died in the summer of 2003. He is deeply missed, and his guidance during the four preceding editions will be impossible to replace. John, as he was known to all his friends and colleagues at London South Bank University, was committed to the educational approach represented by this text. He firmly believed in the importance of emphasising the fundamentals of the subject. Twenty-five years ago he recognized the central importance of computer-aided material and the need to include simulations to broaden the text. His initial work used the leading edge computing of its time and led directly to the emphasis in this edition on simulations that would have been impossible in 1980. We hope that this fifth edition, dedicated to his memory, will continue and reinforce that ethos. John A. Swaffield Lynne B. Jack Edinburgh, February 2005
FM5_A01.fm Page xxi Friday, November 17, 2006 4:43 PM
Preface to the Fourth Edition The study of fluid mechanics remains within the core of engineering education. Advances in the media available for the delivery of that process provide exciting challenges to the academic. The availability of fast, powerful and inexpensive computing and the multi-various opportunities presented by the web and Internet have the potential to transform fluid mechanics education. Always an experimental subject that traditionally relied heavily on laboratory demonstrations and experience, the opportunities offered by validated simulations are particularly appropriate, extending the student’s experience far beyond the constraints of laboratory space or equipment. While these changes are to be welcomed it remains essential that any fluid mechanics course or supporting text provides the fundamental underpinning that will allow the student, and later the practitioner, to recognize when a flow simulation, however ‘sophisticated’ the package, is less than accurate. Advances in media and particularly computing have therefore provided both the challenges and the solutions necessary to ensure that fluid mechanics remains at the centre of engineering education. However, the objectives of that educational process have also changed, particularly in the UK where fundamental reassessments of the academic and practice levels necessary for professional recognition have introduced differentiated courses. Current Engineering Council regulations will progressively reduce the percentage of graduates reaching Chartered Engineer status, while the introduction of BEng and MEng course requirements, incorporating Matching Sections to allow those unable to progress directly to an MEng qualification the opportunity to reach chartered status, will inevitably determine the partition of fluid mechanics into fundamental principles required by all and a range of more specialist topics that may be covered in greater depth. The Matching Section approach will also inevitably lead to post-university courses taken in many instances by part-time or distance learning routes, again offering both tremendous challenges to the course provider and the opportunity to fully utilize the advantages of the computing and delivery media advances already mentioned. The original aims of this text, dating back to the first edition and explicit by the third edition, clearly meet the needs of this changed educational landscape. The text has consistently emphasized the importance of a fundamental understanding of the principles of fluid mechanics, while at the same time providing specialist topics to be covered in greater depth, whether in the area of rotodynamic machinery or unsteady flow. The fundamental material may be seen as crossing the boundaries of the engineering disciplines committed to a coverage of fluid mechanics and it may be argued that the specialist areas chosen also meet this criterion, although in a more selective sense. The second and third editions experimented with the provision of computing simulations; however, the support infrastructure is only now fully available to allow the maximum benefit to be drawn from the provision, in this fourth edition, of a wider range of computing applications. In many ways the development
FM5_A01.fm Page xxii Friday, November 17, 2006 4:43 PM
xxii
Preface to the Fourth Edition
of this provision, from a separately purchased BBC Basic floppy disk in 1984 to the opportunity in the current edition to utilize simulations of a complexity unavailable in the early 1980s, is an allegory for the development of computing over this period. Clearly the lifetime of this edition will see continued exponential change in delivery systems so that it is probably not practical to predict the format of the fifth edition. While the third edition provided a hardcopy solutions manual, this will now be available for downloading, making the ‘Problem’ sections provided more attractive as a basis for student tutorial activity or distance learning education; applications already available to students at Heriot-Watt University through the extensive distance learning Masters course provision from the Department of Building Engineering. Thus the fourth edition retains the educational aims and objectives of earlier editions, while continuing to make full use of the available computing infrastructure. The content has been revised and extended – in the treatment of air and gas distribution networks to include time dependency, including the provision of simulations to extend any laboratory provision in this area, the inclusion of modelling and simulation considerations for both water and airborne pollution and groundwater seepage flow and the introduction of wind turbine coverage aimed at power generation. The impact of computing through computational fluid dynamics (CFD) is recognized with the emphasis placed firmly on the development of the fundamental principles, including the essential recognition that boundary equation definition, if not based on a full understanding of the flow condition, can lead to worthless predictions. Similarly the computational constraints defining stability have been reinforced. While all these are defined in terms of the exciting field of CFD, the text emphasizes that these considerations were always present in the simulations provided in this and earlier editions and may be found at the root of the cases described in the reworked coverage of unsteady flows across the whole spectrum of conditions from free surface wave attenuation to low-amplitude air pressure transient propagation and traditional waterhammer. As in previous editions the text emphasizes the linkage between theory and practice; engineering is still fundamentally about changing the rules and making things work. Examples throughout the text illustrate the application of theory. All the computing is presented in terms of a description of the calculation or simulation, followed by an example and an invitation to consider further several linked problems. The programs provided with the text fall into a number of natural categories, firstly relatively simple calculations, for example friction factor, lift coefficient or free surface flow depths; then calculations designed to provide solutions for steady state system operation, for example fan or pump operating points, free surface gradually varied flow surface profiles or groundwater seepage flow nets beneath dams; and finally unsteady flow simulations, whether for air distribution and recirculation networks or for waterhammer in response to changes in system operating conditions. This content both extends that provided in previous editions and enhances its presentation. The authors would again like to thank all their colleagues, both at Heriot-Watt University and at many other universities worldwide, who have contributed to the development of this series of editions directly or through informed comment, and all those students who have used the texts. In particular the authors wish to thank Dr Ruth Thomas and Dr Nils Tomes at Heriot-Watt University for the development of several of the third edition computer programs into a form suitable for both the Heriot-Watt distance learning MSc programme and for this text. From the Department of Building Engineering and Surveying Dr Ian McDougall contributed to the translation of the authors’ Fortran-based simulations into a form suitable for
FM5_A01.fm Page xxiii Friday, November 17, 2006 4:43 PM
Preface to the Fourth Edition
xxiii
the twenty-first century; Dr David Campbell is again to be thanked for the quality of the additional original artwork provided with the text and for collating the solutions manual into a form suitable for electronic transmission; and Dr Fan Wang provided the background to the CFD descriptions included in this fourth edition. Through their subtle but effective insistence Anna Faherty and Karen Sutherland, Commissioning Editors at Pearson Education, are ultimately responsible for the manuscript being produced almost to schedule and for this and the support of all our other colleagues, past and present, at Pitman, Longman and Pearson Education since 1974, we are grateful. Nevertheless any errors, factual or of understanding, remain the authors’ responsibility. Fluid mechanics is the most fascinating and exciting of the engineering disciplines and one that impinges on all our lives in a multitude of ways both recognized and taken for granted. The authors hope that this text will communicate some of that excitement to the reader. Finally it was with deep sadness that we learnt of the death in 1998 of John Douglas who initiated this series of editions in 1974. John was always committed to the educational concepts embodied in this text. His commitment to engineering education and his ability to introduce the fundamentals of fluids to students was exceptional, as evidenced by the parallel and highly successful ‘Solving Problems in Fluid Mechanics’ series. This text will we hope continue that commitment and is dedicated to his memory. J. M. Gasiorek J. A. Swaffield Edinburgh, January 2000
FM5_A01.fm Page xxiv Friday, November 17, 2006 4:43 PM
Preface to the Third Edition This third edition of Fluid Mechanics has retained the aims of the original text in providing a broad-based approach to the study of fluid flow together with a detailed and more advanced treatment of specialist topics that find wide application within the design and analysis of flow systems. The text repeats the previous mix of exposition and example shown to be successful by earlier editions. The study of fluid flow is one of the few areas within engineering that truly crosses the boundaries between the various engineering disciplines. It is of equal importance to mechanical, civil, chemical and process, aeronautical and environmental and building services engineers and is to be found as a fundamental building block in the education and formation of these engineers. While this has remained true, the techniques available to enable students to achieve an understanding of fluid mechanics have been revolutionized by the readily available access to computing facilities; facilities that have already advanced immeasurably since the second edition of this text was published. The use of the computer to aid understanding through the provision of interactive simulations, including the use of multi-media packages, will undoubtedly advance even more rapidly during the lifetime of this edition. Whereas the second edition was accompanied by an optional floppy disk containing the programs presented in the text, this is no longer appropriate. Instead the program listings, including a number of new or enhanced programs, have been presented in a format that will make them readily scannable and so usable on a wide range of machines. While the third edition text retains the philosophy and methodology introduced with the earlier editions, the content has been refined to both extend and, in the authors’ view, improve the presentation of existing material. In particular the text has been reordered to present earlier the fundamentals of dimensional analysis and the laws of similarity. The treatment given to the steady flow energy equation has been extended, together with a general enhancement of the analysis of air and gas flow networks. In this context the coverage of fans within rotodynamic machinery has been strengthened and new material covering the use of fans in ventilation, and the ventilation of tunnels by jet fans in particular, has been presented. A new chapter dealing with the mechanisms of mechanical and natural ventilation has been added to provide both a treatment of this important topic and a background to one of the most common applications for fan technology. As in previous editions current research has been utilized in the treatment of specialist topics, such as the jet fan tunnel ventilation and the unsteady flow analysis presentations. In the latter case the treatment presented in this edition seeks to emphasize the commonality of a range of unsteady flow analyses, from classical waterhammer to free surface waves and low-amplitude transient propagation in gas flows, by demonstrating the general development of the defining equations and the
FM5_A01.fm Page xxv Friday, November 17, 2006 4:43 PM
Preface to the Third Edition
xxv
identical solution by finite difference techniques, allied to computer simulation, once the appropriate terms have been identified for each application. Once again the authors would like to thank all their colleagues in the many universities in the UK and overseas who have contributed to this text by their support for, and comments on, the earlier editions. The authors are grateful to the staff at Longman, particularly Ian Francis and Chris Leeding, who have both supported us in completing this edition and shown considerable patience with the process. Nevertheless, any errors, factual or of understanding, remain ours. We have found fluid mechanics, in all its multi-disciplinary manifestations, to be the most stimulating of engineering areas; we hope that this text will communicate some of that experience and enthusiasm to students of this most demanding of engineering disciplines. J. F. Douglas J. M. Gasiorek J. A. Swaffield Edinburgh, December 1993
FM5_A01.fm Page xxvi Friday, November 17, 2006 4:43 PM
Preface to the Second Edition In the preparation of this second edition we have retained the aims of the original text, namely to provide a broad-based treatment of the essentials of fluid mechanics, while at the same time demonstrating the application of the subject, particularly to the study and solution of higher level problems in selected areas. In retaining this ‘applications’ approach we are both aware and pleased that this technique currently features in the UK Engineering Council statements on the training, education and ‘formation’ of engineers, strengthening our view that this is one of the most efficient and relevant methods of helping students in general to understand our subject. We believe that such an approach should also include the use of improved computer-based numerical solutions as these will become part of the engineer’s everyday activities. In the five years since the first edition was published there has been a significant change in the availability of and access to micro and other computers for both the student and the practising engineer. Computers and programs are of course not ends in themselves but rather they are powerful tools that we can utilize to dispense with many tedious and repetitive calculations, thereby allowing the study, within an educational framework, of problems of greater complexity and relevance, including time-dependent phenomena that were previously beyond our capability without recourse to simplifying assumptions. This second edition therefore includes a series of computer programs chosen to illustrate these aspects of computer application and to be of direct use to both student and practising engineer alike. While the programs have been written in BBC Basic they may be transferred with little difficulty to Apple, Commodore or Sinclair machines. A program cassette tape will also be available to support the text. None of this of course removes the necessity to provide a thorough basis for the subject and this remains one of the text’s main objectives. We have included new material in areas that have been found particularly interesting by our readers, as well as updating and refining the existing text. The treatment of incompressible flows around a body has been extended to include the study of wakes, while the coverage of fluid machinery has been strengthened by the inclusion of a major new chapter on positive displacement machines. The existing treatment of unsteady flow has been extended to allow the application of numerical modelling techniques to unsteady open channel or partially filled pipe flows. Taken together with the introduction of computing methods we view these additions as supporting and reaffirming the aims and objectives of the original text. Once again we would like to thank all our colleagues in many universities and polytechnics in the UK and overseas who have encouraged us by their positive response to and constructive comments on the first edition. All have helped us to formulate this new edition which we hope will fulfil a useful role for both the student and the practising engineer. J. F. Douglas J. M. Gasiorek J. A. Swaffield London, May 1984
FM5_A01.fm Page xxvii Friday, November 17, 2006 4:43 PM
Preface to the First Edition This is a textbook for all manner of engineers. Whether the reader is concerned with Civil, Mechanical or Chemical Engineering, Buiding Services or Environmental Engineering, the principles of fluid mechanics remain the same. Drawing on our joint experience of teaching students in all these disciplines, we have tried to set out these principles simply and clearly and to illustrate their application by examples drawn from the various branches of engineering. In the planning of this book we are indebted to our colleagues in other colleges, polytechnics and universities for the opportunity to study their syllabuses and examination papers which has enabled us to cover the general requirements of the Honours Degree and Professional examinations. We have also deliberately dealt with the elementary aspects of the subject very fully and so the book will meet the requirements of those studying for the Higher National Diploma or for the Higher Diploma or Higher Certificate of the Business and Technician Education Council (B.T.E.C.). For ease of reference the contents has been divided into Parts which are substantially self-contained and we hope that they will provide a convenient source of information for the practising engineer in his day to day activities. J. F. Douglas J. M. Gasiorek J. A. Swaffield
FM5_A01.fm Page xxviii Friday, November 17, 2006 4:43 PM
Acknowledgements We are grateful to the following for permission to reproduce copyright material: Figures 5.20 and 5.21 reproduced with permission of FLUENT Inc.; Figure 5.22 reproduced by permission of Building Research Establishment Ltd.; Figure 5.23 reproduced with permission of Computational Dynamics Ltd.; Figures 6.11 (a) and (b), 25.6, 25.12, 25.17, 25.28 and 25.35 reproduced with permission of Woods Air Movement Ltd.; Figures 9.3 (a) and (b) reproduced with permission of Dr Stephen Huntingdon, HR Wallingford Group Ltd; Figures 18.4 and 18.5 reproduced with permission of UK Climate Impacts Programme; Figure 18.7 reproduced with permission of British Wind Energy Association; Figure 18.24 reproduced with permission of Ocean Power Delivery Ltd; Figures 18.29 (a) and (b) reproduced with permission of Marine Current Turbines Ltd; Figures 19.7 (a) and (b) reproduced with permission of Prof. Garry Pender, Heriot-Watt University; Figures 19.9 and 19.10 reproduced with permission of the Belgian Building Research Institute, Brussels; Figure 19.11 reproduced with permission of Fullfow Group Ltd, © 2000 UV-system; Figure 20.12 (a) reproduced with permission of Thames Water; Figures 21.3, 21.4 and 21.5 reproduced from Pressure Surges in Pipe and Duct Systems by J.A. Swaffield and A.P. Boldy, with kind permission from Ashgate Publishing Group and Adrian P. Boldy. Chapters 8 and 9 photographs reproduced with permission of Dr Carl Trygve Stansberg, Marintek, Trondheim, Norway; Chapter 18 photograph reproduced courtesy of Windcluster 2000 Ltd; Chapter 19 photographs reproduced courtesy of Scottish Water and City of York Council. Parts I, V and VII photographs reproduced with permission of Thames Water; Part II photograph reproduced with permission of NEG Micon AS © NEG Micon; Part V photograph reproduced with permission of Scottish and Southern Energy plc; Part VI photograph courtesy of NASA Earth Observatory, http://earthobservatory.nasa.gov; Part VII photograph © Crown CopyrightMOD. Reproduced with permission of Her Majesty’s Stationery Office; Part VIII photograph reproduced with permission of Woods Air Movement Ltd. Whilst every effort has been made to trace the owners of copyright material, in a few cases this has proved impossible and we take this opportunity to offer our apologies to any copyright holders whose rights we may have unwittingly infringed.
FM5_A01.fm Page xxix Friday, November 17, 2006 4:43 PM
List of Computer Programs SECTION NO. 6.8
PROGRAM
DESCRIPTION
VOLFLO
7.11
ROTCYL
9.10 10.10
GROUND CBW
12.10
WAKE
13.5
NORSH
14.3
SIPHON
14.12
HARDYC
16.3
CRITNOR
16.13
CHANNEL
19.2
UNSCHAN
20.2
WAVESPD
20.7
SHAFT
21.4
FM5SURG
Flow summation for a circular or rectangular duct cross-section based on a velocity or pitot pressure traverse Stagnation points and lift coefficient calculation for a rotating cylinder Simulation of groundwater seepage Friction factor calculation based on the Colebrook–White equation for a circular section pipe or duct Drag on a body calculated from a traverse across its wake Parameter of state calculations across a normal shock Flow between reservoirs based on the steady flow energy equation, including a high point between reservoirs Hardy–Cross method applied to determine flow distribution in a network Normal and critical depth calculations for free surface flows in rectangular section open channels or partially filled circular section pipes Gradually varied flow profile calculations for free surface flows in rectangular section channels or partially filled circular section pipe flows Unsteady gradually varied flow prediction in long free surface channels using the McCormack method Pressure transient propagation velocity calculations, including fluid and pipe wall properties and free gas Surge tank surface oscillation predictions following turbine load rejection Pressure transient prediction in a three-pipe network, including boundary conditions representing valve closure, column separation and gas release
PAGE
183 252 310
362 435 459
470 492
534
550
646
688 703
739
FM5_A01.fm Page xxx Friday, November 17, 2006 4:43 PM
xxx
List of Computer Programs
SECTION NO. 21.5
PROGRAM
DESCRIPTION
FM5WAVE
21.5
FM5GUTT
21.7
FM5AIR
23.4
SIMPUMP
25.9
MATCH
25.14
CONTAM
Free surface wave attenuation in open channels or partially filled pipe flows, including circular, parabolic or rectangular cross sections Free surface profile prediction for unsteady flow in rainwater gutters of circular, parabolic, rectangular or trapezoidal cross-section Unsteady airflow prediction in circular or rectangular section distribution ductwork as a result of fan speed or control valve setting changes Application of the similarity laws for fans or pumps System operating point determination for fans and pumps, utilizing either pressure-flow data or non-dimensional pressure-flow coefficients Contamination decay in a ventilated space for one or more non-reacting contaminants and series alterations in ventilation rate, contamination generation or occupation parameters
PAGE
749
749
756 815
908
929
In each case the program background theory is presented, together with an application example and output and a series of suggested further investigations. In addition four further program listings are provided as solutions to end of chapter problems, namely CHAPTER Chapter 9
PROBLEM 18
Chapter 14
19 and 20
Chapter 20
19
DESCRIPTION Finite difference representation of seepage flow beneath a dam Quasi-steady discharge from a tank and fluid transfer between two reservoirs using a finite difference approach Surge shaft oscillation as an application of a finite difference approach
PAGE 321
506 714
FM5_A01.fm Page xxxi Friday, November 17, 2006 4:43 PM
List of Symbols a A b B c cp cv C Cc Cd CD Cf CL Cp Cr Cv d D e e E f f( ) F F( ) g h h H i I k K l L m M n N
acceleration, area, amplitude area, constant width, breadth, channel width width, breadth, constant chord length, velocity of sound, wave speed specific heat at constant pressure specific heat at constant volume constant, contaminant concentration coefficient of contraction coefficient of discharge coefficient of drag coefficient of friction coefficient of lift power coefficient Courant number coefficient of velocity diameter, depth drag, diameter, depth, diffusion coefficient base of natural logarithms error, internal energy per unit mass modulus of elasticity, energy friction factor, function or variable, frequency, force reflected pressure wave force, stress pressure wave gravitational acceleration vertical height, depth head loss head, enthalpy, building height, wave height hydraulic gradient, node identifier moment of inertia constant, radius of gyration, pipe wall roughness, concentration dependent rate coefficient, wave number bulk modulus, channel conveyance, buoyancy factor length lift, channel length mass, area ratio, doublet strength, hydraulic mean depth molecular weight, mass number of, polytropic index, Mannings channel roughness coefficient rotational speed
FM5_A01.fm Page xxxii Friday, November 17, 2006 4:43 PM
xxxii
List of Symbols
p P q Q r R R s S S0 Sox,y Sfx,y t T u U v, V vf vr vx vy vz vr vθ V w W x, y, z y Z
pressure force, power, wetted perimeter, contaminant flux flow rate per unit width or unit depth, lateral channel inflow volumetric flow rate radius, radial distance radius, reaction force, hydraulic radius, combined damping coefficient gas constant slope, distance, arbitrary coordinate within Cartesian system, slip surface, entropy, channel and friction slope longitudinal channel slope bed slope in x and y directions bed friction slope in x and y directions time, annular film thickness temperature, torque surface width, flow surface width, wave period velocity, peripheral blade velocity internal energy, velocity, wind velocity velocity velocity of flow relative velocity velocity component in x direction velocity component in y direction velocity component in z direction radial velocity tangential velocity volume, volume storage specific weight, Priessmann slot width weight, work orthogonal coordinates gas content (per cent), variable potential head, depth
α β γ Γ δ Δ Δx, Δy ε ζ η θ λ μ ν ρ σ τ φ Φ
angle, angular acceleration angle adiabatic index (cp cv), turbine damping coefficient, angle of yaw circulation difference, increment change in cell dimensions absolute roughness, eddy viscosity vorticity efficiency, free surface amplitude (from datum) angle multiplier in methods of characteristics, wavelength coefficient of dynamic viscosity, radiation damping coefficient coefficient of kinematic viscosity, Poisson’s ratio mass density relative density (specific gravity), surface tension, temporal multiplier shear stress shear strain, angle velocity potential
FM5_A01.fm Page xxxiii Friday, November 17, 2006 4:43 PM
List of Symbols
Ψ ω
stream function angular (rotational) velocity, stage variable
Fr Ma Re Str We
Froude number Mach number Reynolds number Strouhal number Weber number
L M T Θ
Dimensions of Dimensions of Dimensions of Dimensions of
length mass time temperature
xxxiii
FM5_C01a.fm Page xxxiv Tuesday, September 20, 2005 1:21 PM
Part I
Elements of Fluid Mechanics 1 Fluids and their Properties 2 2 Pressure and Head 24 3 Static Forces on Surfaces. Buoyancy 60
FM5_C01b.fm Page 1 Wednesday, September 21, 2005 9:51 AM
Fluid mechanics, as the name indicates, is that branch of applied mechanics that is concerned with the statics and dynamics of liquids and gases. The analysis of the behaviour of fluids is based upon the fundamental laws of applied mechanics that relate to the conservation of mass–energy and the force– momentum equation, together with other concepts and equations with which the student who has already studied solid-body mechanics will be familiar. There are, however, two major aspects of fluid mechanics which differ from solid-body mechanics. The first is the nature and properties of the fluid itself, which are very different from those of a solid. The second is that, instead of dealing with individual bodies or elements of known mass, we are frequently concerned with the behaviour of a continuous stream of fluid, without beginning or end.
Opposite: Water effects, image courtesy of Thames Water
A further problem is that it can be extremely difficult to specify either the precise movement of a stream of fluid or that of individual particles within it. It is, therefore, often necessary – for the purpose of theoretical analysis – to assume ideal, simplified conditions and patterns of flow. The results so obtained may then be modified by introducing appropriate coefficients and factors, determined experimentally, to provide a basis for the design of fluid systems. This approach has proved to be reasonably satisfactory – in so far as the theoretical analysis usually establishes the form of the relationship between the variables; the experimental investigation corrects for the factors omitted from the theoretical model and establishes a quantitative relationship.
FM5_C01b.fm Page 2 Wednesday, September 21, 2005 9:51 AM
Chapter 1
Fluids and their Properties Fluids Shear stress in a moving fluid Differences between solids and fluids 1.4 Newtonian and non-Newtonian fluids 1.5 Liquids and gases 1.6 Molecular structure of materials 1.7 The continuum concept of a fluid 1.8 Density 1.9 Viscosity 1.10 Causes of viscosity in gases 1.1 1.2 1.3
1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20
Causes of viscosity in a liquid Surface tension Capillarity Vapour pressure Cavitation Compressibility and the bulk modulus Equation of state of a perfect gas The universal gas constant Specific heats of a gas Expansion of a gas
FM5_C01b.fm Page 3 Wednesday, September 21, 2005 9:51 AM
This chapter will define the nature of fluids, stressing both the commonality with concepts of applied mechanics applied to solid-body systems and the fundamental differences that arise from the nature of fluids. The appropriate physical properties that define these differences and allow the
differentiation of fluids into gases and liquids, Newtonian and non-Newtonian, compressible and incompressible, will be identified. The application of the equation of state for perfect gases will be introduced. l l l
FM5_C01b.fm Page 4 Wednesday, September 21, 2005 9:51 AM
4
Chapter 1
Fluids and their Properties
1.1
FLUIDS
In everyday life, we recognize three states of matter: solid, liquid and gas. Although different in many respects, liquids and gases have a common characteristic in which they differ from solids: they are fluids, lacking the ability of solids to offer permanent resistance to a deforming force. Fluids flow under the action of such forces, deforming continuously for as long as the force is applied. A fluid is unable to retain any unsupported shape; it flows under its own weight and takes the shape of any solid body with which it comes into contact. Deformation is caused by shearing forces, i.e. forces such as F (Fig. 1.1), which act tangentially to the surfaces to which they are applied and cause the material originally occupying the space ABCD to deform to AB′C′D. This leads to the definition:
FIGURE 1.1 Deformation caused by shearing forces
A fluid is a substance which deforms continuously under the action of shearing forces, however small they may be. Conversely, it follows that: If a fluid is at rest, there can be no shearing forces acting and, therefore, all forces in the fluid must be perpendicular to the planes upon which they act.
1.2
SHEAR STRESS IN A MOVING FLUID
Although there can be no shear stress in a fluid at rest, shear stresses are developed when the fluid is in motion, if the particles of the fluid move relative to each other so that they have different velocities, causing the original shape of the fluid to become distorted. If, on the other hand, the velocity of the fluid is the same at every point, no shear stresses will be produced, since the fluid particles are at rest relative to each other. Usually, we are concerned with flow past a solid boundary. The fluid in contact with the boundary adheres to it and will, therefore, have the same velocity as the boundary. Considering successive layers parallel to the boundary (Fig. 1.2), the velocity of the fluid will vary from layer to layer as y increases. If ABCD (Fig. 1.1) represents an element in a fluid with thickness s perpendicular to the diagram, then the force F will act over an area A equal to BC × s. The force per
FM5_C01b.fm Page 5 Wednesday, September 21, 2005 9:51 AM
1.3
Differences between solids and fluids
5
FIGURE 1.2 Variation of velocity with distance from a solid boundary
unit area FA is the shear stress τ and the deformation, measured by the angle φ (the shear strain), will be proportional to the shear stress. In a solid, φ will be a fixed quantity for a given value of τ, since a solid can resist shear stress permanently. In a fluid, the shear strain φ will continue to increase with time and the fluid will flow. It is found experimentally that, in a true fluid, the rate of shear strain (or shear strain per unit time) is directly proportional to the shear stress. Suppose that in time t a particle at E (Fig. 1.1) moves through a distance x. If E is a distance y from AD then, for small angles, Shear strain, φ = xy, Rate of shear strain = xyt = (xt)y = uy, where u = xt is the velocity of the particle at E. Assuming the experimental result that shear stress is proportional to shear strain, then
τ = constant × uy.
(1.1)
The term uy is the change of velocity with y and may be written in the differential form dudy. The constant of proportionality is known as the dynamic viscosity µ of the fluid. Substituting into equation (1.1), du τ = µ −−− , dy
(1.2)
which is Newton’s law of viscosity. The value of µ depends upon the fluid under consideration.
1.3
DIFFERENCES BETWEEN SOLIDS AND FLUIDS
To summarize, the differences between the behaviours of solids and fluids under an applied force are as follows: 1.
2.
For a solid, the strain is a function of the applied stress, provided that the elastic limit is not exceeded. For a fluid, the rate of strain is proportional to the applied stress. The strain in a solid is independent of the time over which the force is applied and, if the elastic limit is not exceeded, the deformation disappears when the force is removed. A fluid continues to flow for as long as the force is applied and will not recover its original form when the force is removed.
FM5_C01b.fm Page 6 Wednesday, September 21, 2005 9:51 AM
6
Chapter 1
Fluids and their Properties
In most cases, substances can be classified easily as either solids or fluids. However, certain cases (e.g. pitch, glass) appear to be solids because their rate of deformation under their own weight is very small. Pitch is actually a fluid which will flow and spread out over a surface under its own weight – but it will take days to do so rather than milliseconds! Similarly, solids will flow and become plastic when subjected to forces sufficiently large to produce a stress in the material which exceeds the elastic limit. They will also ‘creep’ under sustained loading, so that the deformation increases with time. A plastic substance does not meet the definition of a true fluid, since the shear stress must exceed a certain minimum value before flow commences.
1.4
NEWTONIAN AND NON-NEWTONIAN FLUIDS
Even among substances commonly accepted as fluids, there is a wide variation in behaviour under stress. Fluids obeying Newton’s law of viscosity (equation (1.2)) and for which µ has a constant value are known as Newtonian fluids. Most common fluids fall into this category, for which shear stress is linearly related to velocity gradient (Fig. 1.3). Fluids which do not obey Newton’s law of viscosity are known as nonNewtonian and fall into one of the following groups:
FIGURE 1.3 Variation of shear stress with velocity gradient
1.
Plastic, for which the shear stress must reach a certain minimum value before flow commences. Thereafter, shear stress increases with the rate of shear according to the relationship n
du τ = A + B ⎛⎝ −−−⎞⎠ , dy where A, B and n are constants. If n = 1, the material is known as a Bingham plastic (e.g. sewage sludge).
FM5_C01b.fm Page 7 Wednesday, September 21, 2005 9:51 AM
1.6
2. 3. 4. 5. 6.
Molecular structure of materials
7
Pseudo-plastic, for which dynamic viscosity decreases as the rate of shear increases (e.g. colloidal solutions, clay, milk, cement). Dilatant substances, in which dynamic viscosity increases as the rate of shear increases (e.g. quicksand). Thixotropic substances, for which the dynamic viscosity decreases with the time for which shearing forces are applied (e.g. thixotropic jelly paints). Rheopectic materials, for which the dynamic viscosity increases with the time for which shearing forces are applied. Viscoelastic materials, which behave in a manner similar to Newtonian fluids under time-invariant conditions but, if the shear stress changes suddenly, behave as if plastic.
The above is a classification of actual fluids. In analyzing some of the problems arising in fluid mechanics we shall have cause to consider the behaviour of an ideal fluid, which is assumed to have no viscosity. Theoretical solutions obtained for such a fluid often give valuable insight into the problems involved, and can, where necessary, be related to real conditions by experimental investigation.
1.5
FIGURE 1.4 Behaviour of a fluid in a container
LIQUIDS AND GASES
Although liquids and gases both share the common characteristics of fluids, they have many distinctive characteristics of their own. A liquid is difficult to compress and, for many purposes, may be regarded as incompressible. A given mass of liquid occupies a fixed volume, irrespective of the size or shape of its container, and a free surface is formed (Fig. 1.4(a)) if the volume of the container is greater than that of the liquid. A gas is comparatively easy to compress. Changes of volume with pressure are large, cannot normally be neglected and are related to changes of temperature. A given mass of a gas has no fixed volume and will expand continuously unless restrained by a containing vessel. It will completely fill any vessel in which it is placed and, therefore, does not form a free surface (Fig. 1.4(b)).
1.6
MOLECULAR STRUCTURE OF MATERIALS
Solids, liquids and gases are all composed of molecules in continuous motion. However, the arrangement of these molecules, and the spaces between them, differ, giving rise to the characteristic properties of the three different states of matter. In solids, the molecules are densely and regularly packed and movement is slight, each molecule being restrained by its neighbours. In liquids, the structure is looser; individual molecules have greater freedom of movement and, although restrained to some degree by the surrounding molecules, can break away from this restraint, causing a change of structure. In gases, there is no formal structure, the spaces between molecules are large and the molecules can move freely. The molecules of a substance exert forces on each other which vary with their intermolecular distance. Consider, for simplicity, a monatomic substance in which each molecule consists of a single atom. An idea of the nature of the forces acting may be formed from observing the behaviour of such a substance on a macroscopic scale.
FM5_C01b.fm Page 8 Wednesday, September 21, 2005 9:51 AM
8
Chapter 1
Fluids and their Properties
FIGURE 1.5 (a) Variation of force with separation. (b) Variation of potential energy with separation
1.
2.
3.
If two pieces of the same material are far apart, there is no detectable force exerted between them. Thus, the forces between molecules are negligible when widely separated and tend to zero as the separation tends towards infinity. Two pieces of the same material can be made to weld together if they are forced into very close contact. Under these conditions, the forces between the molecules are attractive when the separation is very small. Very large forces are required to compress solids or liquids, indicating that a repulsive force between the molecules must be overcome to reduce the spacing between them.
It appears from these observations that interatomic forces vary with the distance of separation (Fig. 1.5(a)) and that there are two types of force, one attractive and the other repulsive. At small separations, the repulsive force is dominant; at larger separations, it becomes insignificant by comparison with the attractive force. These conclusions can also be expressed in terms of the potential energy, defined as the energy required to bring one atom from infinity to a distance r from the second atom. The potential energy is zero if the atoms are infinitely far apart and is positive if external energy is required to move the first atom towards the second. Since Fig. 1.5(a) is the graph of the force F between the atom vs. the distance of separation, the potential energy curve (Fig. 1.5(b)) will be the integral of this curve from ∞ to r, which is the shaded area in Fig. 1.5(a). At r0, there is a condition of minimum energy, corresponding to F = 0 and representing a position of stable equilibrium, accounting for the inherent stability of solids and liquids in which the molecules are sufficiently densely packed for this condition to exist. Figure 1.5(b) also indicates that a pair of atoms can be separated
FM5_C01b.fm Page 9 Wednesday, September 21, 2005 9:51 AM
1.7
The continuum concept of a fluid
9
completely, so that r = ∞, by the application of a finite amount of energy ∆E, which is called the dissociation or binding energy. Considering a large number of particles of a substance, each particle will have kinetic energy −12 mu2, where m is the mass of the particle and u its velocity. If a particle collides with a pair of particles, it will only cause them to separate if it can transfer to the pair energy in excess of ∆E. Thus, the possibility of stable pairs forming will depend on the average value of −12 mu2 in relation to ∆E. 1.
If the average value of −12 mu2 ∆E, no stable pairs can form. The system will behave as a gas, consisting of individual particles moving rapidly with no apparent tendency to aggregate or occupy a fixed space.
2.
If the average values of −12 mu2 ∆E, no dissociation of pairs is possible and the colliding particle may be captured by the pair. The system has the properties of a solid, forming a stable conglomeration of particles which can only be dissociated by supplying energy from outside (e.g. by heating to produce melting and, subsequently, boiling).
3.
If the average value of −12 mu2 ∆E, we have a system intermediate between (1) and (2), corresponding to the liquid state, since some particles will have values of −1 mu2 ∆ E, causing dissociation, while others will have values of −1 mu2 ∆E 2 2 and will aggregate.
Summing up, in a solid, the individual molecules are close packed and their movement is restricted to vibrations of small amplitude. The kinetic energy is small compared with the dissociation energy, so that the molecules do not become separated but retain the same relative conditions. In a liquid, the molecules are still close packed, but their movement is greater. Certain of the molecules will have sufficient kinetic energy to break through the surrounding molecules, so that the relative positions of the molecules can change from time to time. The material will cease to be rigid and can flow under the action of applied forces. However, the attraction between molecules is still sufficient to ensure that a given mass of liquid has a fixed volume and that a free surface will be formed. In a gas, the spacing between molecules is some ten times as great as in a liquid. The kinetic energy is far greater than the dissociation energy. The attractive forces between molecules are very weak and intermolecular effects are negligible, so that molecules are free to travel until stopped by a solid or a liquid boundary. A gas will, therefore, expand to fill a container completely, irrespective of volume.
1.7
THE CONTINUUM CONCEPT OF A FLUID
Although the properties of a fluid arise from its molecular structure, engineering problems are usually concerned with the bulk behaviour of fluids. The number of molecules involved is immense, and the separation between them is normally negligible by comparison with the distances involved in the practical situation being studied. Under these conditions, it is usual to consider a fluid as a continuum – a hypothetical continuous substance – and the conditions at a point as the average of a very large number of molecules surrounding that point within a distance which is large compared with the mean intermolecular distance (although very small in absolute
FM5_C01b.fm Page 10 Wednesday, September 21, 2005 9:51 AM
10
Chapter 1
Fluids and their Properties
terms). Quantities such as velocity and pressure can then be considered to be constant at any point, and changes due to molecular motion may be ignored. Variations in such quantities can also be assumed to take place smoothly, from point to point. This assumption breaks down in the case of rarefied gases, for which the ratio of the mean free path of the molecules to the physical dimensions of the problem is very much larger. In this book, fluids will be assumed to be continuous substances and, when the behaviour of a small element or particle of fluid is studied, it will be assumed that it contains so many molecules that it can be treated as part of this continuum. Properties of fluids The following properties of fluids are of general importance to the study of fluid mechanics. For convenience, a fuller list of the values of these properties for common fluids is given in Appendix 1, but typical values, SI units and dimensions in the MLT system (see Chapter 8) are given here.
1.8
DENSITY
The density of a substance is that quantity of matter contained in unit volume of the substance. It can be expressed in three different ways, which must be clearly distinguished.
1.8.1 Mass density Mass density ρ is defined as the mass of the substance per unit volume. As mentioned above, we are concerned, in considering this and other properties, with the substance as a continuum and not with the properties of individual molecules. The mass density at a point is determined by considering the mass δ m of a very small volume δV surrounding the point. In order to preserve the concept of the continuum, δV cannot be made smaller than x3, where x is a linear dimension which is large compared with the mean distance between molecules. The density at a point is the limiting value as δV tends to x3:
δm ρ = δlim −−−− . V→x δ V 3
Units: kilograms per cubic metre (kg m−3). Dimensions: ML−3. Typical values at p = 1.013 × 105 N m−2, T = 288.15 K: water, 1000 kg m−3; air, 1.23 kg m−3.
1.8.2 Specific weight Specific weight w is defined as the weight per unit volume. Since weight is dependent on gravitational attraction, the specific weight will vary from point to point, according to the local value of gravitational acceleration g. The relationship between w and ρ can be deduced from Newton’s second law, since
FM5_C01b.fm Page 11 Wednesday, September 21, 2005 9:51 AM
1.9
Viscosity
11
Weight per unit volume = Mass per unit volume × g w = ρg. Units: newtons per cubic metre (N m−3). Dimensions: ML−2 T −2. Typical values: water, 9.81 × 103 N m−3; air, 12.07 N m−3.
1.8.3 Relative density Relative density (or specific gravity) σ is defined as the ratio of the mass density of a substance to some standard mass density. For solids and liquids, the standard mass density chosen is the maximum density of water (which occurs at 4 °C at atmospheric pressure):
σ = ρsubstanceρH O at 4 °C . 2
For gases, the standard density may be that of air or of hydrogen at a specified temperature and pressure, but the term is not used frequently. Units: since relative density is a ratio of two quantities of the same kind, it is a pure number having no units. Dimensions: as a pure number, its dimensions are M0L0 T 0 = 1. Typical values: water, 1.0; oil, 0.9.
1.8.4 Specific volume In addition to these measures of density, the quantity specific volume is sometimes used, being defined as the reciprocal of mass density, i.e. it is used to mean volume per unit mass.
1.9
VISCOSITY
A fluid at rest cannot resist shearing forces, and, if such forces act on a fluid which is in contact with a solid boundary (Fig. 1.2), the fluid will flow over the boundary in such a way that the particles immediately in contact with the boundary have the same velocity as the boundary, while successive layers of fluid parallel to the boundary move with increasing velocities. Shear stresses opposing the relative motion of these layers are set up, their magnitude depending on the velocity gradient from layer to layer. For fluids obeying Newton’s law of viscosity, taking the direction of motion as the x direction and vx as the velocity of the fluid in the x direction at a distance y from the boundary, the shear stress in the x direction is given by dv τ x = µ −−−−x . dy
(1.3)
1.9.1 Coefficient of dynamic viscosity The coefficient of dynamic viscosity µ can be defined as the shear force per unit area (or shear stress τ) required to drag one layer of fluid with unit velocity past another layer a unit distance away from it in the fluid. Rearranging equation (1.3),
FM5_C01b.fm Page 12 Wednesday, September 21, 2005 9:51 AM
12
Chapter 1
Fluids and their Properties
dv Force Velocity Force × Time µ = τ −−− = −−−−−−−− −−−−−−−−−−−−− = −−−−−−−−−−−−−−−−−−−− dy Area Distance Area
or
Mass −−−−−−−−−−−−−−−−−−−−−−−. Length × Time
Units: newton seconds per square metre (N s m−2) or kilograms per metre per second (kg m−1 s−1). (But note that the coefficient of viscosity is often measured in poise (P); 10 P = 1 kg m−1 s−1.) Dimensions: ML−1T−1. Typical values: water, 1.14 × 10−3 kg m−1 s−1; air, 1.78 × 10−5 kg m−1 s−1.
1.9.2 Kinematic viscosity The kinematic viscosity ν is defined as the ratio of dynamic viscosity to mass density:
ν = µ ρ. Units: square metres per second (m2 s−1). (But note that kinematic viscosity is often measured in stokes (St); 104 St = 1 m2 s−1.) Dimensions: L2 T −1. Typical values: water, 1.14 × 10 −6 m2 s−1; air, 1.46 × 10 −5 m2 s−1.
1.10 CAUSES OF VISCOSITY IN GASES When a gas flows over a solid boundary, the velocity of flow in the x direction, parallel to the boundary, will change with the distance y, measured perpendicular to the boundary. In Fig. 1.6, the velocity in the x direction is vx at a distance y from the boundary and vx + δ vx at a distance y + δ y. As the molecules of gas are not rigidly constrained, and cohesive forces are small, there will be a continuous interchange of molecules between adjacent layers which are travelling at different velocities. Molecules moving from the slower layer will exert a drag on the faster, while those moving from the faster layer will exert an accelerating force on the slower. Assuming that the mass interchange per unit time is proportional to the area A under consideration, and inversely proportional to the distance δ y between them, Mass interchange per unit time = kAδy, where k is a constant of proportionality;
FIGURE 1.6
FM5_C01b.fm Page 13 Wednesday, September 21, 2005 9:51 AM
1.11
Causes of viscosity in a liquid
13
Change of velocity = δ vx ; Force exerted by one layer on the other = Rate of change of momentum = Mass interchange per unit time × Change of velocity
δv F = kA −−−−x ; δy δv Viscous shear stress, τ = FA = k −−−−x . δy Thus, from consideration of molecular mass interchange occurring in a gas, we arrive at Newton’s law of viscosity. If the temperature of a gas increases, the molecular interchange will increase. The viscosity of a gas will, therefore, increase as the temperature increases. According to the kinetic theory of gases, viscosity should be proportional to the square root of the absolute temperature; in practice, it increases more rapidly. Over the normal range of pressures, the viscosity of a gas is found to be independent of pressure, but it is affected by very high pressures.
1.11 CAUSES OF VISCOSITY IN A LIQUID While there will be shear stresses due to molecular interchange similar to those developed in a gas, there are substantial attractive, cohesive forces between the molecules of a liquid (which are very much closer together than those of a gas). Both molecular interchange and cohesion contribute to viscous shear stress in liquids. The effect of increasing the temperature of a fluid is to reduce the cohesive forces while simultaneously increasing the rate of molecular interchange. The former effect tends to cause a decrease of shear stress, while the latter causes it to increase. The net result is that liquids show a reduction in viscosity with increasing temperature which is of the form
µT = µ0 (1 + A1T + B1T 2 ),
(1.4)
where µT is the viscosity at T °C, µ0 is the viscosity at 0 °C and A1 and B1 are constants depending upon the liquid. For water, µ 0 = 0.0179 P, A1 = 0.033 68 and B1 = 0.000 221. When plotted, equation (1.4) gives a hyperbola, viscosity tending to zero as temperature tends to infinity. An alternative relationship is
µµ0 = A2 exp[B2(1T′ − 1T0)],
(1.5)
where A2 and B2 are constants and T′ is the absolute temperature. High pressures also affect the viscosity of a liquid. The energy required for the relative movement of the molecules is increased and, therefore, the viscosity increases with increasing pressure. The relationship depends on the nature of the liquid and is exponential, having the form
µp = µ 0 exp[C( p − p0)],
(1.6)
FM5_C01b.fm Page 14 Wednesday, September 21, 2005 9:51 AM
14
Chapter 1
Fluids and their Properties
where C is a constant for the liquid and µp is the viscosity at pressure p. For oils of the type used in oil hydraulic machinery, the increase in viscosity is of the order of 10 to 15 per cent for a pressure increase of 70 atm. Water, however, behaves rather differently from other fluids, since its viscosity only doubles for an increase in pressure from 1 to 1000 atm.
1.12 SURFACE TENSION Although all molecules are in constant motion, a molecule within the body of the liquid is, on average, attracted equally in all directions by the other molecules surrounding it, but, at the surface between liquid and air, or the interface between one substance and another, the upward and downward attractions are unbalanced, the surface molecules being pulled inward towards the bulk of the liquid. This effect causes the liquid surface to behave as if it were an elastic membrane under tension. The surface tension σ is measured as the force acting across the unit length of a line drawn in the surface. It acts in the plane of the surface, normal to any line in the surface, and is the same at all points. Surface tension is constant at any given temperature for the surface of separation of two particular substances, but it decreases with increasing temperature. The effect of surface tension is to reduce the surface of a free body of liquid to a minimum, since to expand the surface area molecules have to be brought to the surface from the bulk of the liquid against the unbalanced attraction pulling the surface molecules inwards. For this reason, drops of liquid tend to take a spherical shape in order to minimize surface area. For such a small droplet, surface tension will cause an increase of internal pressure p in order to balance the surface force. Considering the forces acting on a diametral plane through a spherical drop of radius r, the force due to internal pressure = p × π r 2, and the force due to surface tension around the perimeter = 2π r × σ. For equilibrium, pπ r 2 = 2π rσ or p = 2σr. Surface tension will also increase the internal pressure in a cylindrical jet of fluid, for which p = σr. In either case, if r is very small, the value of p becomes very large. For small bubbles in a liquid, if this pressure is greater than the pressure of vapour or gas in a bubble, the bubble will collapse. In many of the problems with which engineers are concerned, the magnitude of surface tension forces is very small compared with the other forces acting on the fluid and may, therefore, be neglected. However, these forces can cause serious errors in hydraulic scale models and through capillary effects. Surface tension forces can be reduced by the addition of detergents.
EXAMPLE 1.1
Air is introduced through a nozzle into a tank of water to form a stream of bubbles. If the bubbles are intended to have a diameter of 2 mm, calculate by how much the pressure of the air at the nozzle must exceed that of the surrounding water. Assume that σ = 72.7 × 10−3 N m−1.
Solution Excess pressure, p = 2σr.
FM5_C01b.fm Page 15 Wednesday, September 21, 2005 9:51 AM
1.13
Capillarity
15
Putting r = 1 mm = 10 −3 m, σ = 72.7 × 10 −3 N m−1. Excess pressure, p = (2 × 72.7 × 10−3)(1 × 10−3) = 145.4 N m−2.
1.13 CAPILLARITY If a fine tube, open at both ends, is lowered vertically into a liquid which wets the tube, the level of the liquid will rise in the tube (Fig. 1.7(a)). If the liquid does not wet the tube, the level of liquid in the tube will be depressed below the level of the free surface outside (Fig. 1.7(b)). If θ is the angle of contact between liquid and solid and d is the tube diameter (Fig. 1.7(a)),
FIGURE 1.7 Capillarity
Upward pull Component of Perimeter due to surface = surface tension × of = σ cos θ × π d. tension acting upwards tube
(1.7)
The atmospheric pressure is the same inside and outside the tube, and, therefore, the only force opposing this upward pull is the weight of the vertical-sided column of liquid of height H, since, by definition, there are no shear stresses in a liquid at rest. Therefore, in Fig. 1.7, there will be no shear stress on the vertical sides of the column of liquid under consideration. Weight of column raised = ρ g(π4)d 2H,
(1.8)
where ρ is the mass density of the liquid. Equating the upward pull to the weight of the column, from equations (1.7) and (1.8),
σ cos θ × π d = ρg(π4)d 2H, Capillary rise, H = 4σ cos θρgd.
FM5_C01b.fm Page 16 Wednesday, September 21, 2005 9:51 AM
16
Chapter 1
Fluids and their Properties
FIGURE 1.8 Capillary rise in glass tubes of circular crosssection
Capillary action is a serious source of error in reading liquid levels in fine-gauge tubes, particularly as the degree of wetting and, therefore, the contact angle θ are affected by the cleanness of the surfaces in contact. For water in a tube of 5 mm diameter, the capillary rise will be approximately 4.5 mm, while for mercury the corresponding figure would be −1.4 mm (Fig. 1.8). Gauge glasses for reading the level of liquids should have as large a diameter as is conveniently possible, to minimize errors due to capillarity.
1.14 VAPOUR PRESSURE Since the molecules of a liquid are in constant agitation, some of the molecules in the surface layer will have sufficient energy to escape from the attraction of the surrounding molecules into the space above the free surface. Some of these molecules will return and condense, but others will take their place. If the space above the liquid is confined, an equilibrium will be reached so that the number of molecules of liquid in the space above the free surface is constant. These molecules produce a partial pressure known as the vapour pressure in the space. The degree of molecular activity increases with increasing temperature, and, therefore, the vapour pressure will also increase. Boiling will occur when the vapour pressure is equal to the pressure above the liquid. By reducing the pressure, boiling can be made to occur at temperatures well below the boiling point at atmospheric
FM5_C01b.fm Page 17 Wednesday, September 21, 2005 9:51 AM
1.16
Compressibility and the bulk modulus
17
pressure: for example, if the pressure is reduced to 0.2 bar (0.2 atm), water will boil at a temperature of 60 °C.
1.15 CAVITATION Under certain conditions, areas of low pressure can occur locally in a flowing fluid. If the pressure in such areas falls below the vapour pressure, there will be local boiling and a cloud of vapour bubbles will form. This phenomenon is known as cavitation and can cause serious problems, since the flow of liquid can sweep this cloud of bubbles on into an area of higher pressure where the bubbles will collapse suddenly. If this should occur in contact with a solid surface, very serious damage can result due to the very large force with which the liquid hits the surface. Cavitation can affect the performance of hydraulic machinery such as pumps, turbines and propellers, and the impact of collapsing bubbles can cause local erosion of metal surfaces. Cavitation can also occur if a liquid contains dissolved air or other gases, since the solubility of gases in a liquid decreases as the pressure is reduced. Gas or air bubbles will be released in the same way as vapour bubbles, with the same damaging effects. Usually, this release occurs at higher pressures and, therefore, before vapour cavitation commences.
1.16 COMPRESSIBILITY AND THE BULK MODULUS All materials, whether solids, liquids or gases, are compressible, i.e. the volume V of a given mass will be reduced to V – δV when a force is exerted uniformly all over its surface. If the force per unit area of surface increases from p to p + δ p, the relationship between change of pressure and change of volume depends on the bulk modulus of the material: Bulk modulus = Change in pressureVolumetric strain. Volumetric strain is the change in volume divided by the original volume; therefore, Change in volume Change in pressure −−−−−−−−−−−−−−−−−−−−−−−−−−− = −−−−−−−−−−−−−−−−−−−−−−−−−−−−− , Original volume Bulk modulus −δVV = δpK, the minus sign indicating that the volume decreases as pressure increases. In the limit, as δp → 0, dp K = – V −−−− . dV
(1.9)
Considering unit mass of a substance, V = 1ρ.
(1.10)
FM5_C01b.fm Page 18 Wednesday, September 21, 2005 9:51 AM
18
Chapter 1
Fluids and their Properties
Differentiating, V dρ + ρ dV = 0 dV = −(Vρ) dρ. Substituting for V from equation (1.10), dV = −(1ρ 2) dρ.
(1.11)
Putting the values of V and dV obtained from equations (1.10) and (1.11) in equation (1.9),
dp K = ρ −−− . dρ
(1.12)
The value of K is shown by equation (1.12) to be dependent on the relationship between pressure and density and, since density is also affected by temperature, it will depend on how the temperature changes during compression. If the temperature is constant, conditions are said to be isothermal, while, if no heat is allowed to enter or leave during compression, conditions are adiabatic. The ratio of the adiabatic bulk modulus to the isothermal bulk modulus is equal to γ , the ratio of the specific heat of a fluid at constant pressure to that at constant volume. For liquids, γ is approximately unity and the two conditions need not be distinguished; for gases, the difference is substantial (for air, γ = 1.4). The concept of the bulk modulus is mainly applied to liquids, since for gases the compressibility is so great that the value of K is not a constant, but proportional to pressure and changes very rapidly. The relationship between pressure and mass density is more conveniently found from the characteristic equation of a gas (1.13). For liquids, the value of K is high and changes of density with pressure are small, but increasing pressure does bring the molecules of the liquid closer together, increasing the value of K. For water, the value of K will double if the pressure is increased from 1 to 3500 atm. An increase of temperature will cause the value of K to fall. For liquids, the changes in pressure occurring in many fluid mechanics problems are not sufficiently great to cause appreciable changes in density. It is, therefore, usual to ignore such changes and to treat liquids as incompressible. Where, however, sudden changes of velocity generate large inertial forces, high pressures can occur and compressibility effects cannot be disregarded in liquids (see Chapter 20). Gases may also be treated as incompressible if the pressure changes are very small, but, usually, compressibility cannot be ignored. In general, compressibility becomes important when the velocity of the fluid exceeds about one-fifth of the velocity of a pressure wave (e.g. the velocity of sound) in the fluid. Units: since volumetric strain is the ratio of two volumes, the units of bulk modulus will be the same as those of pressure, newtons per square metre (N m−2). Dimensions: ML−1T−2. Typical values: water, 2.05 × 109 N m−2; oil, 1.62 × 109 N m−2.
FM5_C01b.fm Page 19 Wednesday, September 21, 2005 9:51 AM
1.19
Specific heats of a gas
19
1.17 EQUATION OF STATE OF A PERFECT GAS The mass density of a gas varies with its absolute pressure p and absolute temperature T. For a perfect gas, p = ρRT,
(1.13)
where R is the gas constant for the gas concerned. Most gases at pressures and temperatures well removed from liquefaction follow this characteristic equation closely, but it does not apply to vapours. Units: the gas contant is measured in joules per kilogram per kelvin (J kg−1 K−1). Dimensions: L 2 T −2Θ −1. Typical values: air, 287 J kg−1 K−1; hydrogen, 4110 J kg−1 K−1.
1.18 THE UNIVERSAL GAS CONSTANT From equation (1.13) ρR is constant for a given value of pressure p and temperature T. By Avogadro’s hypothesis, all pure gases have the same number of molecules per unit volume at the same temperature and pressure, so that ρ is proportional to the molar mass M (kg kmol−1). Therefore, the quantity MR will be constant for all perfect gases, and is known as the universal gas constant R0. R0 = MR = 8.314 kJ kmol−1 K−1.
1.19 SPECIFIC HEATS OF A GAS Since pressure, temperature and density of a gas are interrelated, the amount of heat energy H required to raise the temperature of a gas from T1 to T2 will depend upon whether the gas is allowed to expand during the process, so that some of the energy supplied is used in doing work instead of raising the temperature of the gas. Two different specific heats are, therefore, given for a gas, corresponding to the two extreme conditions of constant volume and constant pressure. 1.
Specific heat at constant volume cv. For a temperature change from T1 to T2 at constant volume, Heat supplied per unit mass, H = cv(T2 − T1).
2.
Since there is no change in volume, no external work is done, so that the increase of internal energy per unit mass of gas is cv (T2 − T1) heat units. Specific heat at constant pressure cp. If the pressure is kept constant, the gas will expand as the temperature changes from T1 to T2: Heat supplied per unit mass = cp(T2 − T1) heat units.
FM5_C01b.fm Page 20 Wednesday, September 21, 2005 9:51 AM
20
Chapter 1
Fluids and their Properties
Only part of this energy is used to raise the temperature of the gas; the rest goes to external work. Thus, cp cv: cp(T2 − T1 ) = cv(T2 − T1) + External work (in heat units). It can be shown that R = (cp − cv), where R, cp and cv have the same units. Units: specific heat is measured in joules per kilogram per kelvin, as is R. Dimensions: L2 T −2 Θ−1. Typical values: air, cp = 1.005 kJ kg−1 K−1, cv = 0.718 kJ kg−1 K−1.
1.20 EXPANSION OF A GAS When a gas expands, the amount of work done will depend upon the relationship between pressure and volume, which, in turn, depends upon whether the gas receives or loses heat during the process. If a unit mass of a gas has a volume V1 at pressure p1 and volume V2 at pressure p2, as shown in Fig. 1.9, then, Work done Area under p–V = = during expansion curve between V1 and V2
V2
p dV.
V1
FIGURE 1.9 Expansion of a gas
1. If the expansion is isothermal, the absolute temperature T (in kelvin) of the gas remains unchanged and the characteristic equation p = ρRT becomes pρ = constant; or, putting V = volume of unit mass = 1ρ, pV = constant = p1V1 = RT, p = p1V1(1V ).
(1.14)
FM5_C01b.fm Page 21 Wednesday, September 21, 2005 9:51 AM
1.20
Expansion of a gas
21
From equation (1.14),
Work done per unit mass = p 1 V 1
V2
V1
dV −−− V
= p1V1 loge(V2 V1 ) = RT log e(V2 V1 ). 2. For any known relationship between pressure and mass density of the form pρ n = constant, putting V = 1ρ, pV n = p 1V n1 = constant.
(1.15)
Therefore, p = p 1V 1 V . n
–n
Work done by gas per unit mass =
V2
p dV
V1
= p1 V 1 n
V2
V
–n
dV
V1 (1 – n)
= [ p 1V 1 ( 1 – n ) ] [V 2 n
= ( 1 – n ) [ p 1V V –1
n 1
(1 – n) 2
(1 – n)
–V 1
]
– p 1 V 1 ],
or, since p 1V 1 = p 2V 2 , n
n
Work done by gas per unit mass = (p2V2 − p1V1)(1 − n) = ( p1V1 − p2V2)(n − 1) = R(T1 − T2)(n − 1).
(1.16)
3. If the compression is carried out adiabatically, no heat enters or leaves the system. Now, for any mode of compression, considering unit mass, Heat supplied = Change of internal energy + Work done (in heat units). Change of internal energy = cv(T2 − T1 ). Mechanical work done = ( p2V2 − p1V1)(1 − n). Thus, in general, if H is the heat supplied, H = cv(T2 − T1) + ( p2V2 − p1V1)(1 − n). Now,
R = (cp − cv) or cv = R(cpcv − 1).
FM5_C01b.fm Page 22 Wednesday, September 21, 2005 9:51 AM
22
Chapter 1
Fluids and their Properties
Also
R(T2 − T1) = ( p2V2 − p1V1).
Thus,
H = ( p2V2 − p1V1)(cpcv − 1) + ( p2V2 − p1V1)(1 − n).
For an adiabatic change, H = 0, so that ( p2V2 − p1V1)(cpcv − 1) = −( p2V2 − p1V1)(1 − n) = ( p2V2 − p1V1)(n − 1), and, therefore, n = cp cv = γ. Thus, for an adiabatic change, the relationship between pressure and density is given by pV γ = pρ γ = constant,
(1.17)
and, from (2), Work done by gas per unit mass = ( p1V1 − p2V2)(γ − 1) = R(T1 − T2)(γ − 1).
(1.18)
Concluding remarks The material presented in this chapter will be utilized in all sections of the text; in particular the influence of fluid viscosity will be of the utmost importance. The equation of state and the definition of compressible flows will also be used to differentiate flow conditions.
Summary of important equations and concepts 1.
2.
3.
The relationship between shear stress, viscosity and velocity gradient, equation (1.3), will recur throughout the text. While this text will be concerned with Newtonian fluids, as defined in Section 1.4, the reader should be familiar with the differentiation between these and other non-Newtonian fluid types. The fundamental fluid properties introduced must be understood and their dependence on temperature and pressure appreciated. In particular the defining differences between liquids and gases become essential in dealing with concepts of compressibility and time dependency, Section 1.5. A range of properties are introduced in this chapter whose importance will be returned to later under particular flow conditions: for example surface tension and its effects on capillary action; vapour pressure and its role in pressure surge analysis; and cavitation as a limit to pump operation and propellerturbine blade design.
FM5_C01b.fm Page 23 Wednesday, September 21, 2005 9:51 AM
Summary of important equations and concepts
4.
5.
The concept of compressibility, the differences between gas and liquid compressibility and the conditions under which flows may be considered incompressible must be understood, Section 1.16. The gas laws are essential to the later development of the concepts of gaseous fluid flow, Sections 1.17 to 1.20.
23
FM5_C02.fm Page 24 Wednesday, September 21, 2005 10:00 AM
Chapter 2
Pressure and Head 2.1 2.2 2.3 2.4 2.5 2.6
2.7 2.8 2.9
Statics of fluid systems Pressure Pascal’s law for pressure at a point Variation of pressure vertically in a fluid under gravity Equality of pressure at the same level in a static fluid General equation for the variation of pressure due to gravity from point to point in a static fluid Variation of pressure with altitude in a fluid of constant density Variation of pressure with altitude in a gas at constant temperature Variation of pressure with altitude in a gas under adiabatic conditions
2.10 Variation of pressure and density with altitude for a constant temperature gradient 2.11 Variation of temperature and pressure in the atmosphere 2.12 Stability of the atmosphere 2.13 Pressure and head 2.14 The hydrostatic paradox 2.15 Pressure measurement by manometer 2.16 Relative equilibrium 2.17 Pressure distribution in a liquid subject to horizontal acceleration 2.18 Effect of vertical acceleration 2.19 General expression for the pressure in a fluid in relative equilibrium 2.20 Forced vortex
FM5_C02.fm Page 25 Wednesday, September 21, 2005 10:00 AM
This chapter will consider and introduce the forces acting on, or generated by, fluids at rest. In particular the concept of pressure will be introduced, including its variation with depth of submergence, via the hydrostatic equation, its unique value at any particular depth in a continuous fluid and direction of application at that depth. The concept that the atmosphere dictates that all activities on the Earth’s
surface are effectively carried out submerged in a fluid will be stressed and the pressure variations within, and stability of, the atmosphere will be treated. This understanding of pressure will be used to introduce methods of pressure measurement that will be essential to the treatment, in later chapters, of fluids in motion. l l l
FM5_C02.fm Page 26 Wednesday, September 21, 2005 10:00 AM
26
Chapter 2
Pressure and Head
2.1
STATICS OF FLUID SYSTEMS
The general rules of statics apply to fluids at rest, but, from the definition of a fluid (Section 1.1), there will be no shearing forces acting and, therefore, all forces (such as F in Fig. 2.1(a)) exerted between the fluid and a solid boundary must act at right angles to the boundary. If the boundary is curved (Fig. 2.1(b)), it can be considered to be composed of a series of chords on each of which a force F1, F2, . . . , Fn acts perpendicular to the surface at the section concerned. Similarly, considering any plane drawn through a body of fluid at rest (Fig. 2.1(c)), the force exerted by one portion of the fluid on the other acts at right angles to this plane.
FIGURE 2.1 Forces in a fluid at rest
Shear stresses due to viscosity are only generated when there is relative motion between elements of the fluid. The principles of fluid statics can, therefore, be extended to cases in which the fluid is moving as a whole but all parts are stationary relative to each other. In the analysis of a problem it is usual to consider an element of the fluid defined by solid boundaries or imaginary planes. A free body diagram can be drawn for this element, showing the forces acting on it due to the solid boundaries or surrounding fluid. Since the fluid is at rest, the element will be in equilibrium, and the sum of the component forces acting in any direction must be zero. Similarly, the sum of the moments of the forces about any point must be zero. It is usual to test equilibrium by resolving along three mutually perpendicular axes and, also, by taking moments in three mutually perpendicular planes. Although a body or element may be in equilibrium, it can also be of interest to know what will happen if it is displaced from its equilibrium position. For example, in the case of a ship it is of the utmost importance to know whether it will overturn when it pitches or rolls or whether it will tend to right itself and return to its original position. There are three possible conditions of equilibrium:
1.
Stable equilibrium. A small displacement from the equilibrium position generates a force producing a righting moment tending to restore the body to its equilibrium position.
FM5_C02.fm Page 27 Wednesday, September 21, 2005 10:00 AM
2.2
2. 3.
Pressure
27
Unstable equilibrium. A small displacement produces an overturning moment tending to displace the body further from its equilibrium position. Neutral equilibrium. The body remains at rest in any position to which it is displaced.
These conditions are typified by the three positions of a cone on a horizontal surface shown in Fig. 2.2.
FIGURE 2.2 Types of equilibrium
2.2
PRESSURE
A fluid will exert a force normal to a solid boundary or any plane drawn through the fluid. Since problems may involve bodies of fluids of indefinite extent and, in many cases, the magnitude of the force exerted on a small area of the boundary or plane may vary from place to place, it is convenient to work in terms of the pressure p of the fluid, defined as the force exerted per unit area. If the force exerted on each unit area of a boundary is the same, the pressure is said to be uniform: Force exerted Pressure = −−−−−−−−−−−−−−−−−−−−−−−−−−− Area of boundary
F or p = −−. A
If, as is more commonly the case, the pressure changes from point to point, we consider the element of force δF normal to a small area δA surrounding the point under consideration:
δF Mean pressure, p = −−−−. δA In the limit, as δA → 0 (but remains large enough to preserve the concept of the fluid as a continuum),
δ F dF Pressure at a point, p = lim −−−− = −−−−. dA δ A→ 0 δ A Units: newtons per square metre (N m−2). (Note that an alternative metric unit is the bar; 1 bar = 105 N m−2.) Dimensions: ML−1T −2.
FM5_C02.fm Page 28 Wednesday, September 21, 2005 10:00 AM
28
Chapter 2
Pressure and Head
2.3
PASCAL’S LAW FOR PRESSURE AT A POINT
By considering the equilibrium of a small fluid element in the form of a triangular prism surrounding a point in the fluid (Fig. 2.3), a relationship can be established between the pressures px in the x direction, py in the y direction and ps normal to any plane inclined at any angle θ to the horizontal at this point. FIGURE 2.3 Equality of pressure in all directions at a point
If the fluid is at rest, px will act at right angles to the plane ABFE, py at right angles to CDEF and ps at right angles to ABCD. Since the fluid is at rest, there will be no shearing forces on the faces of the element and the element will not be accelerating. The sum of the forces in any direction must, therefore, be zero. Considering the x direction: Force due to px = px × Area ABFE = pxδ yδ z; Component of force due to ps = − ( ps × Area ABCD) sin θ
δy = – p s δ s δ z −−− = −psδ yδ z δs (since sin θ = δyδ s). As py has no compound in the x direction, the element will be in equilibrium if pxδ yδ z + (−psδ yδ z) = 0, px = ps.
(2.1)
Similarly, in the y direction, Force due to py = py × Area CDEF = pyδ xδ z; Component of force due to ps = −( ps × Area ABCD) cos θ
δx = – p s δ s δ z −−− = −psδ xδ z δs (since cos θ = δxδs).
FM5_C02.fm Page 29 Wednesday, September 21, 2005 10:00 AM
2.4
Variation of pressure vertically in a fluid under gravity
29
Weight of element = −Specific weight × Volume = −ρg × −12 δ xδ yδ z. As px has no component in the y direction, the element will be in equilibrium if pyδ xδ z + (−psδ xδ z) + (−ρg × −12 δ xδ yδ z) = 0. Since δx, δy and δz are all very small quantities, δxδyδz is negligible in comparison with the other two terms, and the equation reduces to py = ps .
(2.2)
Thus, from equations (2.1) and (2.2), ps = px = py.
(2.3)
Now ps is the pressure on a plane inclined at any angle θ ; the x, y and z axes have not been chosen with any particular orientation, and the element is so small that it can be considered to be a point. This proof may be extended to the z axis. Equation (2.3), therefore, indicates that the pressure at a point is the same in all directions. This is known as Pascal’s law and applies to a fluid at rest. If the fluid is flowing, shear stresses will be set up as a result of relative motion between the particles of the fluid. The pressure at a point is then considered to be the mean of the normal forces per unit area (stresses) on three mutually perpendicular planes. Since these normal stresses are usually large compared with shear stresses it is generally assumed that Pascal’s law still applies.
2.4
VARIATION OF PRESSURE VERTICALLY IN A FLUID UNDER GRAVITY
Figure 2.4 shows an element of fluid consisting of a vertical column of constant crosssectional area A and totally surrounded by the same fluid of mass density ρ. Suppose that the pressure is p1 on the underside at level z1 and p2 on the top at level z2. Since the fluid is at rest the element must be in equilibrium and the sum of all the vertical forces must be zero. The forces acting are: Force due to p1 on area A acting up = p1 A, Force due to p2 on area A acting down = p2 A, Force due to the weight of the element = mg = Mass density × g × Volume = ρgA(z2 − z1). Since the fluid is at rest, there can be no shear forces and, therefore, no vertical forces act on the side of the element due to the surrounding fluid. Taking upward forces as positive and equating the algebraic sum of the forces acting to zero, p1A − p2A − ρgA(z2 − z1) = 0,
FM5_C02.fm Page 30 Wednesday, September 21, 2005 10:00 AM
30
Chapter 2
Pressure and Head
FIGURE 2.4 Vertical variation of pressure
p2 − p1 = −ρg(z2 − z1).
(2.4)
Thus, in any fluid under gravitational attraction, pressure decreases with increase of height z.
EXAMPLE 2.1
A diver descends from the surface of the sea to a depth of 30 m. What would be the pressure under which the diver would be working above that at the surface assuming that the density of sea water is 1025 kg m−3 and remains constant?
Solution In equation (2.4), taking sea level as datum, z1 = 0. Since z2 is lower then z1 the value of z2 is −30 m. Substituting these values and putting ρ = 1025 kg m−3: Increase of pressure = p2 − p1 = −1025 × 9.81(−30 − 0) = 301.7 × 103 N m−2.
2.5
EQUALITY OF PRESSURE AT THE SAME LEVEL IN A STATIC FLUID
If P and Q are two points at the same level in a fluid at rest (Fig. 2.5), a horizontal prism of fluid of constant cross-sectional area A will be in equilibrium. The forces acting on this element horizontally are p1A at P and p2A at Q. Since the fluid is at rest, there will be no horizontal shear stresses on the sides of the element. For static equilibrium the sum of the horizontal forces must be zero: p1A = p2A,
FM5_C02.fm Page 31 Wednesday, September 21, 2005 10:00 AM
2.5
Equality of pressure at the same level in a static fluid
31
FIGURE 2.5 Equality of pressures at the same level
p1 = p2. Thus, the pressure at any two points at the same level in a body of fluid at rest will be the same. In mathematical terms, if (x, y) is the horizontal plane,
∂p −−− = 0 ∂x
and
∂p −−− = 0; ∂y
partial derivatives are used because pressure p could vary in three directions. Pressures at the same level will be equal even though there is no direct horizontal path between P and Q, provided that P and Q are in the same continuous body of fluid. Thus, in Fig. 2.6, P and Q are connected by a horizontal pipe, R and S being two points at the same level at the entrance and exit to the pipe. If the pressure is pP at P, pQ at Q, pR at R and pS at S, then, since R and S are at the same level, pR = pS; also
pR = pP + ρgz and pS = pQ + ρgz.
Substituting in equation (2.5), pP + ρgz = pQ + ρgz, pP = pQ.
FIGURE 2.6 Equality of pressures in a continuous body of fluid
(2.5)
FM5_C02.fm Page 32 Wednesday, September 21, 2005 10:00 AM
32
Chapter 2
Pressure and Head
2.6
GENERAL EQUATION FOR THE VARIATION OF PRESSURE DUE TO GRAVITY FROM POINT TO POINT IN A STATIC FLUID
Let p be the pressure acting on the end P of an element of fluid of constant crosssectional area A and p + δp be the pressure at the other end Q (Fig. 2.7). FIGURE 2.7 Variation of pressure in a stationary fluid
The axis of the element is inclined at an angle θ to the vertical, the height of P above a horizontal datum is z and that of Q is z + δz. The forces acting on the element are: pA acting at right angles to the end face at P along the axis of the element, ( p + δp)A acting at Q along the axis in the opposite direction; mg the weight of the element, due to gravity, acting vertically down = Mass density × Volume × Gravitational acceleration = ρ × Aδ s × g. There are also forces due to the surrounding fluid acting normal to the sides of the element, since the fluid is at rest, and, therefore, perpendicular to its axis PQ. For equilibrium of the element PQ, the resultant of these forces in any direction must be zero. Resolving along the axis PQ, pA − ( p + δp)A − ρgAδs cos θ = 0, δp = −ρgδs cos θ, or, in differential form, dp −−− = – ρ g cos θ . ds In the general three-dimensional case, s is a vector with components in the x, y and z directions. Taking the (x, y) plane as horizontal, if the axis of the element is also horizontal, θ = 90° and dp⎞ ∂p ∂p ⎛− = −−− = −−− = 0, −− ⎝ ds ⎠ θ = 90° ∂ x ∂ y
(2.6)
FM5_C02.fm Page 33 Wednesday, September 21, 2005 10:00 AM
2.7
Variation of pressure with altitude in a fluid of constant density
33
confirming the results of Section 2.5 that, in a static fluid, pressure is constant everywhere in a horizontal plane. It is for this reason that the free surface of a liquid is horizontal. If the axis of the element is in the vertical z direction, θ = 0° and dp⎞ ∂p ⎛− −− = −−− = – ρ g, ⎝ ds ⎠ θ = 0° ∂ z and, since ∂p∂x = ∂p∂y = 0, the partial derivative ∂p∂z can be replaced by the total differential dpdz, giving dp −−− = – ρ g, dz
(2.7)
which corresponds to the result obtained in Section 2.4. Also, considering any two horizontal planes a vertical distance z apart, Pressure at all points on lower plane = p,
∂p Pressure at all points on upper plane = p + z −−− , ∂z ∂p Difference of pressure = z −−−. ∂z Since the planes are horizontal, the pressure must be constant over each plane; therefore, ∂p∂z cannot vary horizontally. From equation (2.7), this implies that ρg shall be constant and, therefore, for equilibrium, the density ρ must be constant over any horizontal plane. Thus, the conditions for equilibrium under gravity are: 1. 2. 3.
The pressure at all points on a horizontal plane must be the same. The density at all points on a horizontal plane must be the same. The change of pressure with elevation is given by dpdz = −ρg.
The actual pressure variation with elevation is found by integrating equation (2.7):
dp = – ρ g dz
or
p2 – p1 = –
z2
ρ g dz,
(2.8)
z1
but this cannot be done unless the relationship between ρ and p is known.
2.7
VARIATION OF PRESSURE WITH ALTITUDE IN A FLUID OF CONSTANT DENSITY
For most problems involving liquids it is usual to assume that the density ρ is constant, and the same assumption can also be made for a gas if pressure differences are very small. Equation (2.8) can then be written
FM5_C02.fm Page 34 Wednesday, September 21, 2005 10:00 AM
34
Chapter 2
Pressure and Head
p = –ρ g
dz = –ρgz + constant,
or, for any two points at altitude z1 and z2 above datum, p2 − p1 = −ρg(z2 − z1 ).
VARIATION OF PRESSURE WITH ALTITUDE IN A GAS AT CONSTANT TEMPERATURE
2.8
The relation between pressure, density and temperature for a perfect gas is given by the equation pρ = RT. If conditions are assumed to be isothermal, so that temperature does not vary with altitude, ρ can be expressed in terms of p and the result substituted in equation (2.7): p ρ = −−−− , RT and, from equation (2.7), dp pg −−− = – ρ g = – −−−− , dz RT dp g −−− = – −−−− dz. p RT Integrating from p = p1 when z = z1, to p = p2 when z = z2, log e( p2p1) = −(gRT )(z2 − z1), p2 p1 = exp[−( gRT )(z2 − z1) ].
EXAMPLE 2.2
At an altitude z, of 11 000 m, the atmospheric temperature T is −56.6 °C and the pressure p is 22.4 kN m−2. Assuming that the temperature remains the same at higher altitudes, calculate the density of the air at an altitude of z2 of 15 000 m. Assume R = 287 J kg−1 K−1.
Solution Let p2 be the absolute pressure at z2. Since the temperature is constant, p2 p1 = exp [−( gRT )(z2 − z1) ].
FM5_C02.fm Page 35 Wednesday, September 21, 2005 10:00 AM
2.9
Variation of pressure with altitude in a gas under adiabatic conditions
35
Putting p 1 = 22.4 kN m −2 = 22.4 × 10 3 N m −2, z 1 = 11 000 m, z 2 = 15 000 m, R = 287 J kg−1 K−1, T = −56.6 °C = 216.4 K: 9.81 ( 15 000 – 11 000 ) 3 p 2 = 22.4 × 10 × exp – −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 287 × 216.4 = 22.4 × 103 × exp(−0.632) = 11.91 × 103 N m−2. Also, from the equation of state for a perfect gas (see equation (1.13)), p2 = ρ2RT and so Density of air at 15 000 m, ρ2 = p2 RT = (11.91 × 103)(287 × 216.4) = 0.192 kg m−3.
2.9
VARIATION OF PRESSURE WITH ALTITUDE IN A GAS UNDER ADIABATIC CONDITIONS
If conditions are adiabatic, the relationship between pressure and density is given by pργ = constant = p1 ρ γ1 , so that
ρ = ρ1( pp1)1γ. Substituting in equation (2.7),
ρ 1 g 1γ dp −−− = – −−− −γ p , p 1 dz 1 γ p 1 1 ⎞ – 1γ − p dp. dz = – ⎛ −−− ⎝ ρ1 g ⎠
Integrating from p = p1 when z = z1, to p = p2 when z = z2 , γ p ( γ –1 )γ p 1 1 − −−−−−−−−−−−−− z 2 – z 1 = – −−− ρ 1 g ( γ – 1 )γ
p2
p1
γ γ p 1 1 − ( p (2γ –1 )γ – p (1γ –1 )γ ) = – ⎛ −−−−−− ⎞ −−− ⎝ γ – 1⎠ ρ 1 g
p 1 ⎛ p 2⎞ ( γ –1 )γ γ −− –1 , = – ⎛ −−−−−− ⎞ −−−− ⎝ γ – 1⎠ ρ 1 g ⎝ p 1⎠ or, since p1ρ1 = RT, for any gas,
FM5_C02.fm Page 36 Wednesday, September 21, 2005 10:00 AM
36
Chapter 2
Pressure and Head
γ RT ⎛ p−2⎞ ( γ –1 )γ – 1 , z 2 – z 1 = – ⎛ −−−−−− ⎞ −−−−−−1 ⎝ − p 1⎠ ⎝ γ – 1⎠ g ⎛ p−−2⎞ ⎝ p 1⎠
( γ –1 )γ
g ( z 2 – z 1 ) ⎛ γ – 1⎞ = −−−−−− −−−−−−− −−−−−− + 1, RT 1 ⎝ γ ⎠
g ( z2 – z1 ) γ – 1 p2 −− = 1 – −−−−−−−−−−−−− ⎛⎝ −−−−−−⎞⎠ γ RT 1 p1
γ ( γ –1 )
.
(2.9)
This can be extended to any isentropic process for which pρ n = constant, to give g ( z2 – z1 ) n – 1 p2 −− = 1 – −−−−−−−−−−−−− ⎛⎝ −−−−−−⎞⎠ n RT 1 p1
n( n –1 )
.
(2.10)
The rate of change of temperature with altitude – the temperature lapse rate – can also be found for adiabatic conditions. From the characteristic equation, ρ = pRT and, since, from equation (2.7), dz = −dpρg, substituting for ρ, dz = −(RTgp) dp. For adiabatic conditions, γ
pργ = p1ρ 1 , or, since pρ = RT, p = p1(T1T )γ (1−γ ) and, differentiating, dp = −[γ (1 − γ )] p1 T γ1(1 – γ ) T −1(1−γ ) dT. Substituting these values of p and dp in equation (2.11): RT { – [ γ ( 1 – γ ) ]p T γ ( 1 – γ ) T –1 ( 1 – γ ) dT } dz = – −−−− −−−−−−−−−−−−−−−−−−−−−γ−1−(−1−–−−1γ−)−−−−–−− −−−−−−−−−−−−−−−−−−−−− p1 T 1 T γ (1 – γ ) g = [γ (1 − γ )](Rg) dT.
(2.11)
FM5_C02.fm Page 37 Wednesday, September 21, 2005 10:00 AM
2.9
Variation of pressure with altitude in a gas under adiabatic conditions
37
Temperature gradient, dT −−−− = – [ ( γ – 1 )γ ] ( gR ). dz
EXAMPLE 2.3
(2.12)
Calculate the pressure, temperature and density of the atmosphere at an altitude of 1200 m if at zero altitude the temperature is 15 °C and the pressure 101 kN m−2. Assume that conditions are adiabatic (γ = 1.4) and R = 287 J kg−1 K−1.
Solution From equation (2.9), g ( z2 – z1 ) γ – 1 p 2 = p 1 1 – −−−−−−−−−−−−− ⎛⎝ −−−−−−⎞⎠ RT 1 γ
γ ( γ –1 )
.
Putting p1 = 101 × 103 N m−2, z1 = 0, z2 = 1200 m, T1 = 15 °C = 288 K, γ = 1.4, R = 287 J kg−1 K−1: 9.81 × 1200 0.4 3 p 2 = 101 × 10 1 – −−−−−−−−−−−−−−−− ⎛⎝ −−−−⎞⎠ 287 × 288 1.4
1.40.4
Nm
–2
= 87.33 × 103 N m−2. From equation (2.12), Temperature gradient, dT −−−− = – [ ( γ – 1 )γ ] ( gR ) = −(0.41.4) × (9.81287) dz = −9.76 × 10−3 K m−1, dT T2 = T 1 + −−− (z 2 – z 1 ) dz = 288 − (9.76 × 10−3 × 1200) = 276.3 K = 3.3 °C. From the equation of state, Density at 1200 m,
ρ2 = p2 RT2 = (87.33 × 103)(287 × 276.3) = 1.101 kg m−3.
FM5_C02.fm Page 38 Wednesday, September 21, 2005 10:00 AM
38
Chapter 2
Pressure and Head
2.10 VARIATION OF PRESSURE AND DENSITY WITH ALTITUDE FOR A CONSTANT TEMPERATURE GRADIENT Assuming that there is a constant temperature lapse rate (i.e. dTdz = constant) with elevation in a gas, so that its temperature falls by an amount δT for a unit change of elevation, then, if T1 = temperature at level z1, T = temperature at level z, T = T1 − δ T(z − z1).
(2.13)
From equation (2.7), dpdz = −ρg and, since pρ = RT, putting ρ = pRT, dp g −−− = – p −−−− , dz RT dp g −−− = – −−−− dz. p RT Substituting for T from equation (2.13), dpp = −{gR[T1 − δ T(z − z1) ]}dz. Integrating between the limits p1 and p2 and z1 and z2, log e (p2 p1) = ( gRδT ) log e{[T1 − δ T(z2 − z1)]T1}, p2 p1 = [1 − (δ TT1)(z2 − z1)] gRδ T.
(2.14)
Comparing this with the result obtained in Section 2.9, and putting dT n–1 g δ T = – −−− = ⎛⎝ −−−−−−⎞⎠ ⎛⎝ −−⎞⎠ , dz n R we have gRδT = n(n − 1). Substituting in equation (2.14), p2 p1 = [1 − (gRT1)(z2 − z1)(n − 1)n]n(n−1) which agrees with equation (2.10). To find the corresponding change of density ρ, since pρ = RT,
ρ2 p2 T1 p2 T −− = −− × −−− = −− × −−−−−−−−−−−−−1−−−−−−−−−− ρ1 p1 T2 p1 T1 – δ T ( z2 – z1 ) and, substituting from equation (2.14) for p2p1,
ρ2 ρ1 = [1 − (δ TT1)(z2 − z1)] gRδT[1 − (δTT1)(z2 − z1)]−1 = [1 − (δ TT1)(z2 − z1)](gRδ T)−1.
(2.15)
FM5_C02.fm Page 39 Wednesday, September 21, 2005 10:00 AM
2.11
EXAMPLE 2.4
Variation of temperature and pressure in the atmosphere
39
Assuming that the temperature of the atmosphere diminishes with increasing altitude at the rate of 6.5 °C per 1000 m, find the pressure and density at a height of 7000 m if the corresponding values at sea level are 101 kN m−2 and 1.235 kg m−3 when the temperature is 15 °C. Take R = 287 J kg−1 K−1.
Solution From equation (2.14), p2 = p1[1 − (δTT1)(z2 − z1)] gRδ T. Putting p1 = 101 × 103 N m−2, δT = 6.5 °C per 1000 m = 0.0065 K m−1, T1 = 15 °C = 288 K, (z2 − z1) = 7000 m, R = 287 J kg−1 K−1: p2 = 101 × 103[1 − (0.0065288) × 7000]9.81(287×0.0065) = 40.89 × 103 N m−2. From the equation of state, Density, ρ2 = p2 RT2 = p2 R[T1 − δ T(z2 − z1)] = 40.89 × 103287(288 − 0.0065 × 7000) = 0.588 kg m−3.
2.11 VARIATION OF TEMPERATURE AND PRESSURE IN THE ATMOSPHERE A body of fluid which is of importance to the engineer is the atmosphere. In practice, it is never in perfect equilibrium and is subject to large incalculable disturbances. In order to provide a basis for the design of aircraft an International Standard Atmosphere has been adopted which represents the average conditions in Western Europe; the relations between altitude, temperature and density have been tabulated (Table 2.1). Essentially, the standard atmosphere comprises the troposphere – extending from sea level to 11 000 m – in which the temperature lapse rate is constant at approximately 0.0065 K m−1 and the pressure–density relationship is pρ n = constant, where n = 1.238. Above 11 000 m lies the stratosphere, in which conditions are assumed to be isothermal, with the temperature constant at −56 °C. Figure 2.8 shows the variation of pressure with altitude in the International Standard Atmosphere. The atmospheric pressure at sea level is assumed to be equivalent to 760 mm of mercury, the temperature 15 °C and the density 1.225 kg m−3. In the real atmosphere, the troposphere extends to an average of 11 000 m, but can vary from 7600 m at the poles to 18 000 m at the equator. While the temperature, in general, falls steadily with altitude, meteorological conditions can arise in the lower layers which produce temperature inversion – the temperature increasing with altitude. In the stratosphere, the temperature remains substantially constant up to approximately 32 000 m; it then rises to about 70 °C before falling again. Figure 2.9 shows typical values for pressure and temperature.
FM5_C02.fm Page 40 Wednesday, September 21, 2005 10:00 AM
40
Chapter 2
Pressure and Head
TABLE 2.1 International Standard Atmosphere
FIGURE 2.8 Variation of temperature with altitude in the International Standard Atmosphere
Altitude above sea level (m)
Absolute pressure (bar)
Absolute temperature (K)
Mass density (kg m−3)
Kinematic viscosity (m2 s−1 × 10−5)
0 1 000 2 000 4 000 6 000 8 000 10 000 11 500 12 000 14 000 16 000 18 000 20 000 22 000 24 000 26 000 28 000 30 000 32 000
1.013 25 0.898 8 0.795 0 0.616 6 0.472 2 0.356 5 0.265 0 0.209 8 0.194 0 0.141 7 0.103 5 0.075 65 0.055 29 0.040 47 0.029 72 0.021 88 0.016 16 0.011 97 0.008 89
288.15 281.7 275.2 262.2 249.2 236.2 223.3 216.7 216.7 216.7 216.7 216.7 216.7 218.6 220.6 222.5 224.5 226.5 228.5
1.225 0 1.111 7 1.006 6 0.819 4 0.660 2 0.525 8 0.413 6 0.337 5 0.311 9 0.227 9 0.166 5 0.121 6 0.088 91 0.064 51 0.046 94 0.034 26 0.025 08 0.018 41 0.013 56
1.461 1.581 1.715 2.028 2.416 2.904 3.525 4.213 4.557 6.239 8.540 11.686 15.989 22.201 30.743 42.439 58.405 80.134 109.620
FM5_C02.fm Page 41 Wednesday, September 21, 2005 10:00 AM
2.12
Stability of the atmosphere
41
FIGURE 2.9 Variation of temperature and pressure in the real atmosphere
Because of vertical currents, the composition of the air remains practically constant in both the troposphere and the stratosphere, except that there is a negligible amount of water vapour in the stratosphere and a slight reduction in the ratio of oxygen to nitrogen above an altitude of 20 km. Nine-tenths of the mass of the atmosphere is contained below 20 km and 99 per cent below 60 km.
2.12 STABILITY OF THE ATMOSPHERE We have seen that there are variations of density from point to point in the atmosphere when it is at rest. In practice, there are local disturbances due to air currents. There are also changes of density as a result of local thermal effects, which cause the movement of elements of air into regions where they are surrounded by air of slightly different density and temperature. If the density of the surrounding air is less than that of the newly arrived element, there is a tendency for the element to return to its original position – since the net upward force exerted by the surrounding fluid is less than the weight of the element. In Fig. 2.10, if ρ1 is the density of the surrounding air and ρ2 is the density of the air in the displaced element, Weight of element, mg = ρ2 gAδ z. Upward force due to surrounding fluid = δp × A = ρ1gδzA, and there is, therefore, a net downward force of ( ρ2 − ρ1)gδzA. As an element of fluid rises in the atmosphere, its pressure and temperature fall. Air is a poor conductor, and conditions are, therefore, approximately adiabatic. From
FM5_C02.fm Page 42 Wednesday, September 21, 2005 10:00 AM
42
Chapter 2
Pressure and Head
FIGURE 2.10 Stability of the atmosphere
equation (2.9), substituting γ = 1.414 and R = 287 J kg−1 K−1, the adiabatic temperature lapse rate in the element is δT = 0.01 K m−1. The natural temperature lapse rate δT′ occurring in the atmosphere is found to be of the order of 0.0065 K m−1. Since the lapse rates differ for the ascending element of air and the surrounding atmosphere the changes of density with altitude will also differ and can be calculated from equation (2.15). For example, if ρ1 = density of air at sea level, ρ2 = density of air at an elevation of 1000 m, R = 287 J kg−1 K−1, then: 1.
Assuming δT = 0.01 K m−1, for the air in the displaced element, g 9.81 −−−−−− – 1 = −−−−−−−−−−−−−−− – 1 = 2.418, RδT 287 × 0.01 and from equation (2.15), 0.01 × 1000 2.48 ρ 2 = ρ 1 ⎛⎝ 1 – −−−−−−−−−−−−−−−− ⎞⎠ , 288
2.
ρ2 −− = 0.9181 ρ1
for air in the element. Assuming δT = 0.0065 K m−1 for the surrounding atmosphere, g 9.81 −−−−−− – 1 = −−−−−−−−−−−−−−−−−− – 1 = 4.258, RδT 287 × 0.0065
ρ 2 ⎛ 0.0065 × 1000⎞ 4.258 −− = 1 – −−−−−−−−−−−−−−−−−−−− = 0.9018 ⎠ ρ1 ⎝ 288 for the surrounding air. Thus, the density of the ascending element expanding adiabatically decreases less rapidly than that of the surrounding air; the element eventually becomes denser than the surroundings and tends to fall back to its original level. The atmosphere, therefore, tends to be stable under normal conditions. If, however, the natural temperature lapse rate were to exceed the adiabatic lapse rate, equilibrium would be unstable and an element displaced upwards would continue to rise. Such conditions can arise in thundery weather.
FM5_C02.fm Page 43 Wednesday, September 21, 2005 10:00 AM
2.13
Pressure and head
43
2.13 PRESSURE AND HEAD In a fluid of constant density, dpdz = −ρg can be integrated immediately to give p = −ρgz + constant. In a liquid, the pressure p at any depth z, measured downwards from the free surface so that z = −h (Fig. 2.11), will be p = ρgh + constant and, since the pressure at the free surface will normally be atmospheric pressure patm , p = ρgh + patm.
(2.16)
It is often convenient to take atmospheric pressure as a datum. Pressures measured above atmospheric pressure are known as gauge pressures.
FIGURE 2.11 Pressure and head
Since atmospheric pressure varies with atmospheric conditions, a perfect vacuum is taken as the absolute standard of pressure. Pressures measured above perfect vacuum are called absolute pressures: Absolute pressure = Gauge pressure + Atmospheric pressure. Taking patm as zero, equation (2.16) becomes p = ρgh,
(2.17)
which indicates that, if g is assumed constant, the gauge pressure at a point X (Fig. 2.11) can be defined by stating the vertical height h, called the head, of a column of a given fluid of mass density ρ which would be necessary to produce this pressure. Note that when pressures are expressed as head, it is essential that the mass density ρ is given or the fluid named. For example, since from equation (2.17) h = pρg, a pressure of 100 kN m−2 can be expressed in terms of water ( ρH O = 103 kg m−3) as a head of (100 × 103)(103 × 9.81) = 10.19 m of water. Alternatively, in terms of mercury (relative density 13.6) a pressure of 100 kN m−2 will correspond to a head of (100 × 103)(13.6 × 103 × 9.81) = 0.75 m of mercury. 2
FM5_C02.fm Page 44 Wednesday, September 21, 2005 10:00 AM
44
Chapter 2
Pressure and Head
EXAMPLE 2.5
A cylinder contains a fluid at a gauge pressure of 350 kN m−2. Express this pressure in terms of a head of (a) water ( ρH O = 1000 kg m−3), (b) mercury (relative density 13.6). What would be the absolute pressure in the cylinder if the atmospheric pressure is 101.3 kN m−2? 2
Solution From equation (2.17), head, h = pρg. (a) Putting p = 350 × 103 N m−2, ρ = ρH O = 1000 kg m−3, 2
350 × 10 3 Equivalent head of water = −−−−3−−−−−−−−−− = 35.68 m. 10 × 9.81 (b) For mercury ρHg = σρH O = 13.6 × 1000 kg m−3, 2
350 × 10 3 Equivalent head of water = −−−−−−−−−−−−−−3−−−−−−−−−− = 2.62 m, 1.36 × 10 × 9.81 Absolute pressure = Gauge pressure + Atmospheric pressure = 350 + 101.3 = 451.3 kN m−2.
2.14 THE HYDROSTATIC PARADOX From equation (2.17) it can be seen that the pressure exerted by a fluid is dependent only on the vertical head of fluid and its mass density ρ; it is not affected by the weight of the fluid present. Thus, in Fig. 2.12 the four vessels all have the same base area A and are filled to the same height h with the same liquid of density ρ. Pressure on bottom in each case, p = ρgh, Force on bottom = Pressure × Area = pA = ρghA. Thus, although the weight of fluid is obviously different in the four cases, the force on the bases of the vessels is the same, depending on the depth h and the base area A. FIGURE 2.12 The hydrostatic paradox
FM5_C02.fm Page 45 Wednesday, September 21, 2005 10:00 AM
2.15
Pressure measurement by manometer
45
2.15 PRESSURE MEASUREMENT BY MANOMETER The relationship between pressure and head is utilized for pressure measurement in the manometer or liquid gauge. The simplest form is the pressure tube or piezometer shown in Fig. 2.13, consisting of a single vertical tube, open at the top, inserted into a pipe or vessel containing liquid under pressure which rises in the tube to a height depending on the pressure. If the top of the tube is open to the atmosphere, the pressure measured is ‘gauge’ pressure: Pressure at A = Pressure due to column of liquid of height h1, pA = ρgh1. Similarly, FIGURE 2.13 Pressure tube or piezometer
EXAMPLE 2.6
Pressure at B = pB = ρgh2. If the liquid is moving in the pipe or vessel, the bottom of the tube must be flush with the inside of the vessel, otherwise the reading will be affected by the velocity of the fluid. This instrument can only be used with liquids, and the height of the tube which can conveniently be employed limits the maximum pressure that can be measured.
What is the maximum gauge pressure of water that can be measured by means of a piezometer tube 2 m high? (Mass density of water ρH O = 103 kg m−3.) 2
Solution Since p = ρgh for maximum pressure, put ρ = ρH O = 103 and h = 2 m, giving 2
Maximum pressure, p = 103 × 9.81 × 2 = 19.62 × 103 N m−2.
The U-tube gauge, shown in Fig. 2.14, can be used to measure the pressure of either liquids or gases. The bottom of the U-tube is filled with a manometric liquid Q which is of greater density ρman and is immiscible with the fluid P, liquid or gas, of
FIGURE 2.14 U-tube manometer
FM5_C02.fm Page 46 Wednesday, September 21, 2005 10:00 AM
46
Chapter 2
Pressure and Head
density ρ, whose pressure is to be measured. If B is the level of the interface in the lefthand limb and C is a point at the same level in the right-hand limb, Pressure pB at B = Pressure pC at C. For the left-hand limb, pB = Pressure pA at A + Pressure due to depth h1 of fluid P = pA + ρgh1. For the right-hand limb, pC = Pressure pD at D + Pressure due to depth h 2 of liquid Q. But
pD = Atmospheric pressure = Zero gauge pressure,
and so pC = 0 + ρman gh 2. Since pB = pC , pA + ρgh1 = ρman gh 2, pA = ρman gh 2 − ρgh1.
EXAMPLE 2.7
(2.18)
A U-tube manometer similar to that shown in Fig. 2.14 is used to measure the gauge pressure of a fluid P of density ρ = 800 kg m−3. If the density of the liquid Q is 13.6 × 103 kg m−3, what will be the gauge pressure at A if (a) h1 = 0.5 m and D is 0.9 m above BC, (b) h1 = 0.1 m and D is 0.2 m below BC?
Solution (a) In equation (2.18), ρman = 13.6 × 103 kg m−3, ρ = 0.8 × 103 kg m−3, h1 = 0.5 m, h 2 = 0.9 m; therefore: pA = 13.6 × 103 × 9.81 × 0.9 − 0.8 × 103 × 9.81 × 0.5 = 116.15 × 103 N m−2. (b) Putting h1 = 0.1 m and h 2 = −0.2 m, since D is below BC: pA = 13.6 × 103 × 9.81 × (−0.2) − 0.8 × 103 × 9.81 × 0.1 = −27.45 × 103 N m−2, the negative sign indicating that pA is below atmospheric pressure.
FM5_C02.fm Page 47 Wednesday, September 21, 2005 10:00 AM
2.15
Pressure measurement by manometer
47
FIGURE 2.15 Measurement of pressure difference
In Fig. 2.15, a U-tube gauge is arranged to measure the pressure difference between two points in a pipeline. As in the previous case, the principle involved in calculating the pressure difference is that the pressure at the same level CD in the two limbs must be the same, since the fluid in the bottom of the U-tube is at rest. For the left-hand limb, pC = pA + ρga. For the right-hand limb pD = pB + ρg(b − h) + ρman gh. Since pC = pD, pA + ρga = pB + ρg(b − h) + ρman gh, Pressure difference = pA − pB = ρg(b − a) + hg(ρman − ρ).
EXAMPLE 2.8
(2.19)
A U-tube manometer is arranged, as shown in Fig. 2.15, to measure the pressure difference between two points A and B in a pipeline conveying water of density ρ = ρH O = 103 kg m−3. The density of the manometric liquid Q is 13.6 × 103 kg m−3, and point B is 0.3 m higher than point A. Calculate the pressure difference when h = 0.7 m. 2
Solution In equation (2.19), ρ = 103 kg m−3, ρman = 13.6 × 103 kg m−3, (b − a) = 0.3 m and h = 0.7 m. Pressure difference = pA − pB = 103 × 9.81 × 0.3 + 0.7 × 9.81(13.6 − 1) × 103 N m−2 = 89.467 × 103 N m−2.
FM5_C02.fm Page 48 Wednesday, September 21, 2005 10:00 AM
48
Chapter 2
Pressure and Head
FIGURE 2.16 U-tube with one leg enlarged
In both the above cases, if the fluid P is a gas its density ρ can usually be treated as negligible compared with ρman and the equations (2.18) and (2.19) can be simplified. In forming the connection from a manometer to a pipe or vessel in which a fluid is flowing, care must be taken to ensure that the connection is perpendicular to the wall and flush internally. Any burr or protrusion on the inside of the wall will disturb the flow and cause a local change in pressure so that the manometer reading will not be correct. Industrially, the simple U-tube manometer has the disadvantage that the movement of the liquid in both limbs must be read. By making the diameter of one leg very large as compared with the other (Fig. 2.16), it is possible to make the movement in the large leg very small, so that it is only necessary to read the movement of the liquid in the narrow leg. Assuming that the manometer in Fig. 2.16 is used to measure the pressure difference p1 − p2 in a gas of negligible density and that XX is the level of the liquid surface when the pressure difference is zero, then, when pressure is applied, the level in the right-hand limb will rise a distance z vertically. Volume of liquid transferred from left-hand leg to right-hand leg = z × (π 4)d 2; Fall in level of the left-hand leg Volume transferred z × ( π 4 )d 2 d 2 = −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− = −−−−−−−−−−−−−−−2−− = z ⎛ −−⎞ . ⎝ D⎠ ( π 4 )D Area of left-hand leg The pressure difference, p1 − p2, is represented by the height of the manometric liquid corresponding to the new difference of level: p1 − p2 = ρg[z + z(dD)2] = ρgz[1 + (dD)2], or, if D is large compared with d, p1 − p2 = ρgz. If the pressure difference to be measured is small, the leg of the U-tube may be inclined as shown in Fig. 2.17. The movement of the meniscus along the inclined leg, read off on the scale, is considerably greater than the change in level z: Pressure difference, p1 − p2 = ρgz = ρgx sin θ.
FM5_C02.fm Page 49 Wednesday, September 21, 2005 10:00 AM
2.15
Pressure measurement by manometer
49
FIGURE 2.17 U-tube with inclined leg
FIGURE 2.18 Inverted U-tube manometer
The manometer can be made as sensitive as may be required by adjusting the angle of inclination of the leg and choosing a liquid with a suitable value of density ρ to give a scale reading x of the desired size for a given pressure difference. The inverted U-tube shown in Fig. 2.18 is used for measuring pressure differences in liquids. The top of the U-tube is filled with a fluid, frequently air, which is less dense than that connected to the instrument. Since the fluid in the top is at rest, pressures at level XX will be the same in both limbs. For the left-hand limb, pXX = pA − ρga − ρman gh. For the right-hand limb, pXX = pB − ρg(b + h). Thus
pB − pA = ρg(b − a) + gh(ρ − ρman),
or, if A and B are at the same level, pB − pA = (ρ − ρman )gh. If the top of the tube is filled with air ρman is negligible compared with ρ and pB − pA = ρgh. On the other hand, if the liquid in the top of the tube is chosen so that ρman is very nearly equal to ρ, and provided that the liquids do not mix, the result will be a very sensitive manometer giving a large value of h for a small pressure difference.
FM5_C02.fm Page 50 Wednesday, September 21, 2005 10:00 AM
50
Chapter 2
Pressure and Head
EXAMPLE 2.9
An inverted U-tube of the form shown in Fig. 2.18 is used to measure the pressure difference between two points A and B in an inclined pipeline through which water is flowing ( ρH O = 103 kg m−3). The difference of level h = 0.3 m, a = 0.25 m and b = 0.15 m. Calculate the pressure difference pB − pA if the top of the manometer is filled with (a) air, (b) oil of relative density 0.8. 2
Solution In either case, the pressure at XX will be the same in both limbs, so that pXX = pA − ρga − ρman gh = pB − ρg(b + h), pB − pA = ρg(b − a) + gh(ρ − ρman). (a) If the top is filled with air ρman is negligible compared with ρ. Therefore, pB − pA = ρg(b − a) + ρgh = ρg(b − a + h). Putting ρ = ρH O = 103 kg m−3, b = 0.15 m, a = 0.25 m, h = 0.3 m: 2
pB − pA = 103 × 9.81(0.15 − 0.25 + 0.3) = 1.962 × 103 N m−2. (b) If the top is filled with oil of relative density 0.8, ρman = 0.8ρH O = 0.8 × 103 kg m−3. 2
pB − pA = ρg(b − a) + gh(ρ − ρman ) = 103 × 9.81(0.15 − 0.25) + 9.81 × 0.3 × 103(1 − 0.8) N m−2 = 103 × 9.81(−0.1 + 0.06) = −392.4 N m−2.
The manometer in its various forms is an extremely useful type of pressure gauge, but suffers from a number of limitations. While it can be adapted to measure very small pressure differences, it cannot be used conveniently for large pressure differences – although it is possible to connect a number of manometers in series and to use mercury as the manometric fluid to improve the range. A manometer does not have to be calibrated against any standard; the pressure difference can be calculated from first principles. However, for accurate work, the temperature should be known, since this will affect the density of the fluids. Some liquids are unsuitable for use because they do not form well-defined menisci. Surface tension can also cause errors due to capillary rise; this can be avoided if the diameters of the tubes are sufficiently large – preferably not less than 15 mm diameter. It is difficult to correct for surface tension, since its effect will depend upon whether the tubes are clean. A major disadvantage of the manometer is its slow response, which makes it unsuitable for measuring fluctuating pressures. Even under comparatively static conditions, slight fluctuations of pressure can make the liquid in the manometer oscillate, so that it is difficult to get a precise reading of the levels of the liquid in the gauge. These oscillations can be reduced by putting restrictions in the manometer connections. It is also essential that the pipes connecting the manometer to the pipe or vessel containing the liquid under pressure should be filled with this liquid and that there should be no air bubbles in the liquid.
FM5_C02.fm Page 51 Wednesday, September 21, 2005 10:00 AM
2.17
Pressure distribution in a liquid subject to horizontal acceleration
51
2.16 RELATIVE EQUILIBRIUM If a fluid is contained in a vessel which is at rest, or moving with constant linear velocity, it is not affected by the motion of the vessel; but if the container is given a continuous acceleration, this will be transmitted to the fluid and affect the pressure distribution in it. Since the fluid remains at rest relative to the container, there is no relative motion of the particles of the fluid and, therefore, no shear stresses, fluid pressure being everywhere normal to the surface on which it acts. Under these conditions the fluid is said to be in relative equilibrium.
2.17 PRESSURE DISTRIBUTION IN A LIQUID SUBJECT TO HORIZONTAL ACCELERATION Figure 2.19 shows a liquid contained in a tank which has an acceleration a. A particle of mass m on the free surface at O will have the same acceleration a as the tank and so will be subject to an accelerating force F. From Newton’s law, F = ma.
(2.20)
FIGURE 2.19 Effect of horizontal acceleration
Also, F is the resultant of the fluid pressure force R, acting normally to the free surface at O, and the weight of the particle mg, acting vertically. Therefore, F = mg tan θ.
(2.21)
Comparing equations (2.20) and (2.21), tan θ = ag
(2.22)
and is constant for all points on the free surface. Thus, the free surface is a plane inclined at a constant angle θ to the horizontal. Since the acceleration is horizontal, vertical forces are not changed and the pressure at any depth h below the surface will be ρgh. Planes of equal pressure lie parallel to the free surface.
FM5_C02.fm Page 52 Wednesday, September 21, 2005 10:00 AM
52
Chapter 2
Pressure and Head
2.18 EFFECT OF VERTICAL ACCELERATION If the acceleration is vertical, the free surface will remain horizontal. Considering a vertical prism of cross-sectional area A (Fig. 2.20), subject to an upward acceleration a, then at depth h below the surface, where the pressure is p, Upward accelerating force, F = Force due to p – Weight of prism = pA − ρghA. By Newton’s second law, F = Mass of prism × Acceleration = ρhA × a. Therefore, pA − ρghA = ρhAa, p = ρgh(1 + ag).
(2.23)
FIGURE 2.20 Effect of vertical acceleration
2.19 GENERAL EXPRESSION FOR THE PRESSURE IN A FLUID IN RELATIVE EQUILIBRIUM If ∂p∂x, ∂p∂y and ∂p∂z are the rates of change of pressure p in the x, y and z directions (Fig. 2.21) and ax , ay and az the accelerations,
∂p Force in x direction, Fx = p δ y δ z – ⎛ p + −−− δ x⎞ δ y δ z ⎝ ∂x ⎠ ∂p = – −−− δ x δ y δ z. ∂x By Newton’s second law, Fx = ρδxδyδz × ax ; therefore,
∂p – −−− = ρ a x . ∂x
(2.24)
FM5_C02.fm Page 53 Wednesday, September 21, 2005 10:00 AM
2.19
General expression for the pressure in a fluid in relative equilibrium
53
FIGURE 2.21 Relative equilibrium: the general case
Similarly, in the y direction,
∂p – −−− = ρ a y . ∂y
(2.25)
In the vertical z direction, the weight of the element ρgδxδyδz must be considered:
∂p F z = p δ x δ y – ⎛ p + −−− δ z⎞ δ x δ y – ρ g δ x δ y δ z ⎝ ∂z ⎠ ∂p = – −−− δ x δ y δ z – ρ g δ x δ y δ z. ∂z By Newton’s second law, Fz = ρδxδyδz × az ; therefore,
∂p – −−− = ρ ( g + a z ). ∂z
(2.26)
For an acceleration as in any direction in the x − z plane making an angle φ with the horizontal, the components of the acceleration are ax = as cos φ and az = as sin φ. Now
dp ∂ p dx ∂ p dz −−− = −−−−−−−− + −−− −−−. ds ∂ x ds ∂ z ds
(2.27)
For the free surface and all other planes of constant pressure, dpds = 0. If θ is the inclination of the planes of constant pressure to the horizontal, tan θ = dzdx. Putting dpds = 0 in equation (2.27),
∂ p dx ∂ p dz −−− −−− + −−− −−− = 0 ∂ x ds ∂ z ds dz ∂p ∂p −−− = tan θ = – −−− −−− . dx ∂x ∂z
FM5_C02.fm Page 54 Wednesday, September 21, 2005 10:00 AM
54
Chapter 2
Pressure and Head
Substituting from equations (2.24) and (2.26), tan θ = −ax(g + az),
(2.28)
or, in terms of as , a cos φ tan θ = – −−−−−−s−−−−−−−−−−−−. ( g + a s sin φ )
(2.29)
For the case of horizontal acceleration, φ = 0 and equation (2.29) gives tan θ = −asg, which agrees with equation (2.22). (Note effect of sign convention.) For vertical acceleration, φ = 90° giving tan θ = 0, indicating that the free surface remains horizontal. Since, for the two-dimensional case,
∂p ∂p dp = −−− dx + −−− dz, ∂x ∂z the pressure at a particular point in the fluid can be found by integration: p=
dp = −∂∂−x−p dx + −∂∂−−pz dz.
Substituting from equations (2.24) and (2.26) and assuming that ρ is constant: p=
(–ρa ) dx + [–ρ(g + a )] dz + constant x
z
= −ρ (xas cos φ − gz − zas sin φ) + constant, or, since xz = tan θ, p = −zρ (as tan θ cos φ − g − as sin φ) + constant,
(2.30)
where z is positive measured upwards from a horizontal datum fixed relative to the fluid.
EXAMPLE 2.10
A rectangular tank 1.2 m deep and 2 m long is used to convey water up a ramp inclined at an angle φ of 30° to the horizontal (Fig. 2.22). Calculate the inclination of the water surface to the horizontal when (a) the acceleration parallel to the slope on starting from the bottom is 4 m s−2, (b) the deceleration parallel to the slope on reaching the top is 4.5 m s−2. If no water is to be spilt during the journey what is the greatest depth of water permissible in the tank when it is at rest?
Solution The slope of the water surface is given by equation (2.29). (a) During acceleration, as = +4 m s−2.
FM5_C02.fm Page 55 Wednesday, September 21, 2005 10:00 AM
2.19
General expression for the pressure in a fluid in relative equilibrium
55
FIGURE 2.22 Acceleration up an inclined plane
– a cos φ 4 cos 30° tan θ A = −−−−−−s−−−−−−−−− = – −−−−−−−−−−−−−−−−−−−−−− g + a s sin φ 9.81 + 4 sin 30° = −0.2933
θA = 163° 39′′. (b) During retardation, as = −4.5 m s−2. ( – 4.5 ) cos 30° tan θ R = – −−−−−−−−−−−−−−−−−−−−−−−− = 0.5154 9.81 – 4.5 sin 30°
θR = 27° 16′′. Since 180° − θR θA, the worst case for spilling will be during retardation. When the water surface is inclined, the maximum depth at the tank wall will be Depth + −12 Length × tan θ, which must not exceed 1.2 m if the water is not to be spilt. Putting length = 2 m, tan θ = tan θR = 0.5154, Depth + (2.02) × 0.5154 = 1.2, Depth = 1.2 − 0.5154 = 0.6846 m.
The equations derived in this section indicate: 1. 2. 3.
if there is no horizontal acceleration, ax = 0 and then θ = 0 so that surfaces of constant pressure are horizontal; in free space, g will be zero so that tan θ = −axaz (surfaces of constant pressure will therefore be perpendicular to the resultant acceleration); since free surfaces of liquids are surfaces of constant pressure, their inclination will be determined by equation (2.29); thus, if ax and ay are zero, the free surface will be horizontal.
FM5_C02.fm Page 56 Wednesday, September 21, 2005 10:00 AM
56
Chapter 2
Pressure and Head
2.20 FORCED VORTEX A body of fluid, contained in a vessel which is rotating about a vertical axis with uniform angular velocity, will eventually reach relative equilibrium and rotate with the same angular velocity ω as the vessel, forming a forced vortex. The acceleration of any particle of fluid at radius r due to rotation will be −ω 2r perpendicular to the axis of rotation, taking the direction of r as positive outward from the axis. Thus, from equation (2.24), dp −−− = – ρω 2 r. dr Figure 2.23 shows a cylindrical vessel containing liquid rotating about its axis, which is vertical. At any point P on the free surface, the inclination θ of the free surface is given by equation (2.28): FIGURE 2.23 Forced vortex
a ω 2 r dz tan θ = – −−−−−x−−− = −−−−− = −−−. g dr g + az
(2.31)
The inclination of the free surface varies with r and, if z is the height of P above O, the surface profile is given by integrating equation (2.31):
−ω−g−−−r dr = −ω−2g−−−r− + constant. r
z=
2
2 2
(2.32)
Thus, the profile of the water surface is a paraboloid. Similarly other surfaces of equal pressure will also be paraboloids. If the container is closed and the fluid has no free surface, the paraboloid drawn to represent the imaginary free surface represents the variation of pressure head with radius. Thus, the pressure p at radius r is given by equation (2.32) as z = pρg = ω 2r 22g + constant, p = ρω 2r 22 + constant.
(2.33)
FM5_C02.fm Page 57 Wednesday, September 21, 2005 10:00 AM
Problems
57
Concluding remarks The definitions of pressure and head presented in this chapter are fundamental to the study of fluid mechanics. In particular the development of the hydrostatic equation and the proof that pressure at a depth in a continuous fluid is constant was central to the development of both flow pressure and velocity measuring techniques due to the equation’s application to the design and operation of pressure measuring manometers. The basic expressions derived for manometer use will be utilized later in the text in the treatment of flow measurement, Chapter 6. The treatment of the atmosphere stresses the obvious point that all our activities and structures must be seen as submerged within a fluid.
Summary of important equations and concepts 1.
2.
3.
4.
Pascal’s law, equation (2.3), stating that pressures at a depth are equal in all coordinate directions, coupled with the hydrostatic equation (2.4) that links depth, gravitational acceleration, density and pressure are the fundamental underpinning of this chapter’s treatment of hydrostatic forces. Equation (2.8) underlines the importance of the density–pressure relationship identified in Chapter 1. This becomes essential to the understanding of the atmospheric variations of pressure and density with altitude, Sections 2.7 to 2.12. Despite the predominance of electronic measurement of fluid flow conditions the application of hydrostatics to manometer measurement of pressure, and hence flow conditions, remains important. Sections 2.13 to 2.15 investigate these applications fully. Relative equilibrium is introduced in Section 2.16 and is used to discuss the effects of acceleration and the generation of a forced vortex, concepts returned to in more detail in Chapter 6.
Problems 2.1 Calculate the pressure in the ocean at a depth of 2000 m assuming that salt water is (a) incompressible with a constant density of 1002 kg m−3, (b) compressible with a bulk modulus of 2.05 GN m−2 and a density at the surface of 1002 kg m−3. [(a) 19.66 MN m−2, (b) 19.75 MN m−2] 2.2 What will be (a) the gauge pressure, (b) the absolute pressure of water at a depth of 12 m below the free surface? Assume the density of water to be 1000 kg m−3 and the atmospheric pressure 101 kN m−2. [(a) 117.72 kN m−2, (b) 218.72 kN m−2] 2.3 What depth of oil, specific gravity 0.8, will produce a pressure of 120 kN m−2? What would be the corresponding depth of water? [15.3 m, 12.2 m]
2.4 At what depth below the free surface of oil having a density of 600 kg m−3 will the pressure be equal to 1 bar? [17 m] 2.5 What would be the pressure in kilonewtons per square metre if the equivalent head is measured as 400 mm of (a) mercury of specific gravity 13.6, (b) water, (c) oil of specific weight 7.9 kN m −3, (d ) a liquid of density 520 kg m−3? [(a) 53.4 kN m−2, (b) 3.92 kN m−2, (c) 3.16 kN m−2, (d ) 2.04 kN m−2] 2.6 A mass of 50 kg acts on a piston of area 100 cm2. What is the intensity of pressure on the water in contact with the underside of the piston, if the piston is in equilibrium? [4.905 × 104 N m−2]
FM5_C02.fm Page 58 Wednesday, September 21, 2005 10:00 AM
58
Chapter 2
Pressure and Head
2.7 The pressure head in a gas main at a point 120 m above sea level is equivalent to 180 mm of water. Assuming that the densities of air and gas remain constant and equal to 1.202 kg m−3 and 0.561 kg m−3, respectively, what will be the pressure head in millimetres of water at sea level? [103 mm] 2.8 A manometer connected to a pipe in which a fluid is flowing indicates a negative gauge pressure head of 50 mm of mercury. What is the absolute pressure in the pipe in newtons per square metre if the atmospheric pressure is 1 bar. [93.3 kN m−2] 2.9 An open tank contains oil of specific gravity 0.75 on top of water. If the depth of oil is 2 m and the depth of water 3 m, calculate the gauge and absolute pressures at the bottom of the tank when the atmospheric pressure is 1 bar. [44.15 kN m−2, 144.15 kN m−2] 2.10 A closed tank contains 0.5 m of mercury, 2 m of water, 3 m of oil of density 600 kg m−3 and there is an air space above the oil. If the gauge pressure at the bottom of the tank is 200 kN m−2, what is the pressure of the air at the top of the tank? [96 kN m−2] 2.11 An inverted cone 1 m high and open at the top contains water to half its height, the remainder being filled with oil of specific gravity 0.9. If half the volume of water is drained away find the pressure at the bottom (apex) of the inverted cone. [9033 N m−2]
values are 15 °C and 101.5 kN m−2. Assuming that the temperature decreases uniformly with increasing altitude, estimate the temperature lapse rate and the pressure and density of the air at an altitude of 3000 m. [6.37 °C per 1000 m, 70.22 kN m−2, 0.91 kg m−3] 2.15 Show that the ratio of the atmospheric pressure at an altitude h1 to that at sea level may be expressed as ( pp0) = (TT0) n, a uniform temperature lapse rate being assumed. Find the ratio of the pressures and the densities at 10 700 m and at sea level taking the standard atmosphere as having a sea level temperature of 15 °C and a lapse rate of 6.5 °C per 1000 m to a minimum of −56.5 °C. [0.2337, 0.3082] 2.16 The barometric pressure of the atmosphere at sea level is equivalent to 760 mm of mercury and its temperature is 288 K. The temperature decreases with increasing altitude at the rate of 6.5 K per 1000 m until the stratosphere is reached in which the temperature remains constant at 216.5 K. Calculate the pressure in millimetres of mercury and the density in kilograms per cubic metre at an altitude of 14 500 m. Assume R = 287 J kg−1 K−1. [97.52 mm, 0.209 kg m−3] 2.17 In Fig. 2.24 fluid P is water and fluid Q is mercury. If the specific weight of mercury is 13.6 times that of water and the atmospheric pressure is 101.3 kN m−2, what is the absolute pressure at A when h1 = 15 cm and h2 = 30 cm? [59.8 kN m−2]
2.12 A hydraulic press has a diameter ratio between the two pistons of 8:1. The diameter of the larger piston is 600 mm and it is required to support a mass of 3500 kg. The press is filled with a hydraulic fluid of specific gravity 0.8. Calculate the force required on the smaller piston to provide the required force (a) when the two pistons are level, (b) when the smaller piston is 2.6 m below the larger piston. [(a) 536 N, (b) 626.2 N] 2.13 Show that the ratio of the pressures ( p2p1) and densities ( ρ2 ρ 1) for altitudes h2 and h1 in an isothermal atmosphere is given by
p2 ρ2 −− = −− = e –g ( h p1 ρ1
2
– h 1 )RT
. FIGURE 2.24
What increase in altitude is necessary in the stratosphere to halve the pressure? Assume a constant temperature of −56.5 °C and the gas constant R = 287 J kg−1 K−1. [4390 m] 2.14 From observation it is found that at a certain altitude in the atmosphere the temperature is −25 °C and the pressure is 45.5 kN m−2, while at sea level the corresponding
2.18 A U-tube manometer (Fig. 2.25) measures the pressure difference between two points A and B in a liquid of density ρ1. The U-tube contains mercury of density ρ2. Calculate the difference of pressure if a = 1.5 m, b = 0.75 m and h = 0.5 m if the liquid at A and B is water and ρ2 = 13.6ρ1. [54.4 kN m−2]
FM5_C02.fm Page 59 Wednesday, September 21, 2005 10:00 AM
Problems
59
end of 44 mm diameter. The left-hand limb and the bottom of the tube are filled with water and the top of the righthand limb is filled with oil of specific gravity 0.83. The free surfaces of the liquids are in the enlarged ends and the interface between the oil and water is in the tube below the enlarged end. What would be the difference in pressures applied to the free surfaces which would cause the oilwater interface to move 1 cm? [21 N m−2]
FIGURE 2.25 2.19 The top of an inverted U-tube manometer is filled with oil of specific gravity 0.98 and the remainder of the tube with water of specific gravity 1.01. Find the pressure difference in newtons per square metre between two points at the same level at the base of the legs when the difference of water level is 75 mm. [22 N m−2] 2.20 An inclined manometer is required to measure an air pressure difference of about 3 mm of water with an accuracy of ±3 per cent. The inclined arm is 8 mm diameter and the enlarged end is 24 mm diameter. The density of the manometer fluid is 740 kg m−3. Find the angle which the inclined arm must make with the horizontal to achieve the required accuracy assuming an acceptable readability of 0.5 mm. [12° 39′] 2.21 An inclined tube manometer consists of a vertical cylinder of 35 mm diameter to the bottom of which is connected a tube of 5 mm diameter inclined upwards at 15° to the horizontal. The manometer contains oil of relative density 0.785. The open end of the inclined tube is connected to an air duct while the top of the cylinder is open to the atmosphere. Determine the pressure in the air duct if the manometer fluid moves 50 mm along the inclined tube. What is the error if the movement of the fluid in the cylinder is ignored? [107.5 N m−2 7.85 N m−2] 2.22 A manometer consists of a U-tube, 7 mm internal diameter, with vertical limbs each with an enlarged upper
2.23 A vessel 1.4 m wide and 2.0 m long is filled to a depth of 0.8 m with a liquid of mass density 840 kg m−3. What will be the force in N on the bottom of the vessel (a) when being accelerated vertically upwards at 4 m s−1, (b) when the acceleration ceases and the vessel continues to move at a constant velocity of 7 m s−1 vertically upwards? [(a) 25 985 N, (b) 18 458 N] 2.24 A pipe 25 mm in diameter is connected to the centre of the top of a drum 0.5 m in diameter, the cylindrical axis of the pipe and the drum being vertical. Water is poured into the drum through the pipe until the water level stands in the pipe 0.6 m above the top of the drum. If the drum and pipe are now rotated about their vertical axis at 600 rev min−1 what will be the upward force exerted on the top of the drum? [13.26 kN] 2.25 A tube ABCD has the end A open to the atmosphere and the end D closed. The portion ABC is vertical while the portion CD is a quadrant of radius 250 mm with its centre at B, the whole being arranged to rotate about its vertical axis ABC. If the tube is completely filled with water to a height in the vertical limb of 300 mm above C find (a) the speed of rotation which will make the pressure head at D equal to the pressure head at C, (b) the value and position of the maximum pressure head in the curved portion CD when running at this speed. [(a) 84.6 rev min−1, (b) 0.362 m of water and 0.12 m below point D] 2.26 A closed airtight tank 4 m high and 1 m in diameter contains water to a depth of 3.3 m. The air in the tank is at a pressure of 40 kN m−2 gauge. What are the absolute pressures at the centre and circumference of the base of the tank when it is rotating about its vertical axis at a speed of 180 rev min−1? At this speed the water wets the top surface of the tank. [17.01 m absolute, 21.53 m absolute]
FM5_C03.fm Page 60 Wednesday, September 21, 2005 9:01 AM
Chapter 3
Static Forces on Surfaces. Buoyancy 3.1 3.2
3.3
3.4 3.5 3.6
Action of fluid pressure on a surface Resultant force and centre of pressure on a plane surface under uniform pressure Resultant force and centre of pressure on a plane surface immersed in a liquid Pressure diagrams Force on a curved surface due to hydrostatic pressure Buoyancy
Equilibrium of floating bodies Stability of a submerged body Stability of floating bodies Determination of the metacentric height 3.11 Determination of the position of the metacentre relative to the centre of buoyancy 3.12 Periodic time of oscillation 3.13 Stability of a vessel carrying liquid in tanks with a free surface 3.7 3.8 3.9 3.10
FM5_C03.fm Page 61 Wednesday, September 21, 2005 9:01 AM
The implication of the hydrostatic equation, together with the realization that pressures at any equal depth in a continuous fluid are both equal and act equally in all directions, leads to the treatment of static forces on submerged surfaces. This also defines buoyancy and the stability of floating bodies. This chapter will introduce the techniques available to
determine the forces acting on surfaces as a result of the applied fluid pressure, and will stress the difference between pressure, which is a scalar quantity acting equally in all directions at a particular depth, and the associated force, which is a vector quantity possessing both magnitude and direction. l l l
FM5_C03.fm Page 62 Wednesday, September 21, 2005 9:01 AM
62
Chapter 3
Static Forces on Surfaces. Buoyancy
3.1
FIGURE 3.1 Forces on a plane surface
ACTION OF FLUID PRESSURE ON A SURFACE
Since pressure is defined as force per unit area, when fluid pressure p acts on a solid boundary – or across any plane in the fluid – the force exerted on each small element of area δA will be pδA, and, since the fluid is at rest, this force will act at right angles to the boundary or plane at the point under consideration. In a body of fluid, the pressure p may vary from point to point, and the forces on each element of area will also vary. If the fluid pressure acts on or across a plane surface, all the forces on the small elements will be parallel (Fig. 3.1) and can be represented by a single force, called the resultant force, acting at right angles to the plane through a point called the centre of pressure. Resultant force, R = Sum of forces on all elements of area R = p1 δA1 + p2 δA2 + · · · + pn δAn = ∑ p δA,
FIGURE 3.2 Forces on a curved surface
where ∑ means ‘the sum of’. If the boundary is a curved surface, the elementary forces will act perpendicular to the surface at each point and will, therefore, not be parallel (Fig. 3.2). The resultant force can be found by resolution or by a polygon of forces, but will be less than ∑ pδA. For example, in the extreme case of the curved surface of a bucket filled with water (Fig. 3.3), the elementary forces acting radially on the vertical wall will balance and the resultant force will be zero. If this were not so, there would be an unbalanced horizontal force in some direction and the bucket would move of its own accord.
3.2 FIGURE 3.3 Forces on a cylindrical surface
RESULTANT FORCE AND CENTRE OF PRESSURE ON A PLANE SURFACE UNDER UNIFORM PRESSURE
The pressure p on a plane horizontal surface in a fluid at rest will be the same at all points, and will act vertically downwards at right angles to the surface. If the area of the plane surface is A, Resultant force = pA. It will act vertically downwards and the centre of pressure will be the centroid of the surface. For gases, the variation of pressure with elevation is small and so it is usually possible to assume that gas pressure on a surface is uniform, even though the surface may not be horizontal. The resultant force is then pA acting through the centroid of the plane surface.
FM5_C03.fm Page 63 Wednesday, September 21, 2005 9:01 AM
3.3
Resultant force and centre of pressure on a plane surface immersed in a liquid
63
FIGURE 3.4 Resultant force on a plane surface immersed in a fluid
3.3
RESULTANT FORCE AND CENTRE OF PRESSURE ON A PLANE SURFACE IMMERSED IN A LIQUID
Figure 3.4 shows a plane surface PQ of any area A totally immersed in a liquid of density ρ and inclined at an angle φ to the free surface. Considering one side only, there will be a force due to fluid pressure p acting on each element of area δA. The magnitude of p will depend on the vertical depth y of the element below the free surface. Taking the pressure at the free surface as zero, from equation (2.4), and measuring y downwards, p = ρgy; therefore, Force on element of area, δA = pδA = ρgyδA. Summing the forces on all such elements over the whole surface, since these forces are all perpendicular to the plane PQ, Resultant force, R = ∑ ρ g y δA. If we assume that ρ and g are constant, R = ρg ∑ yδA.
(3.1)
The quantity ∑ yδA is the first moment of area under the surface PQ about the free surface of the liquid and is equal to AD, where A = the area of the whole immersed surface PQ and D = the vertical depth to the centroid G of the immersed surface. Substituting in equation (3.1), Resultant force, R = ρgAD.
(3.2)
This resultant force R will act perpendicular to the immersed surface at the centre of pressure C at some vertical depth D below the free surface, such that the moment of R about any point will be equal to the sum of the moments of the forces on all the elements δA about the same point. Thus, if the plane of the immersed surface cuts the free surface at O, Moment of R about O = Sum of moments of forces on all elements of area δA about O, Force on any small element = ρgyδA = ρgs sin φ × δA, since y = s sin φ.
(3.3)
FM5_C03.fm Page 64 Wednesday, September 21, 2005 9:01 AM
64
Chapter 3
Static Forces on Surfaces. Buoyancy
Moment of force on element about O = pgs sin φ × δA × s = ρg sin φ × δA × s 2. Since ρ, g and φ are the same for all elements, Sum of the moments of all such forces about O = ρg sin φ ∑ s 2δA. Also R = ρgAD; therefore, Moment of R about O = ρgAD × OC = ρgAD(Dsin φ). Substituting in equation (3.3),
ρgAD(Dsin φ) = ρg sin φ ∑ s 2δA, D = sin2 φ (∑ s 2δA)AD, ∑ s 2δA = Second moment of area of the immersed surface about an axis in the free surface through O = IO = AkO2 , where kO = the radius of gyration of the immersed surface about O. Therefore, D = sin2 φ (IOAD) = sin2 φ ( kO2 /D ).
(3.4)
The values of IO and kO2 can be found if the second moment of area of the immersed surface IG about an axis through its centroid G parallel to the free surface is known by using the parallel axis rule, or,
AkO2 = AkG2 + A(Dsin φ)2.
Thus
D = sin2 φ [ kG2 + (Dsin φ)2D] = sin2 φ ( kG2 D) + D.
(3.5)
The geometrical properties of some common figures are given in Table 3.1. From equation (3.5) it can be seen that the centre of pressure will always be below the centroid G except when the surface is horizontal (φ = 0°). As the depth of immersion increases, the centre of pressure will move nearer to the centroid, since for the given surface the change of pressure between the upper and lower edge becomes proportionately smaller in comparison with the mean pressure, making the pressure distribution more uniform. The lateral position of the centre of pressure can be found by taking moments about the line OG, which is the line of intersection of the immersed surface with a vertical plane through G: R × d = Sum of moments of forces on small elements about OG = ∑ ρgδAyx. Putting
R = ρgAD, d = ( ∑δAπx)Ay.
FM5_C03.fm Page 65 Wednesday, September 21, 2005 9:01 AM
3.3
Resultant force and centre of pressure on a plane surface immersed in a liquid
TABLE 3.1 Geometrical properties of some common figures
Area A
65
Second moment of area IGG about axis GG through the centroid
If the area is symmetrical about a vertical plane through the centroid G, the moment of each small element on one side is balanced by that due to a similar element on the other side so that ∑ δAy = 0. Therefore, d = 0 and the centre of pressure will be on the axis of symmetry.
EXAMPLE 3.1
A trapezoidal opening in the vertical wall of a tank is closed by a flat plate which is hinged at its upper edge (Fig. 3.5). The plate is symmetrical about its centreline and is 1.5 m deep. Its upper edge is 2.7 m long and its lower edge is 1.2 m long. The free surface of the water in the tank stands 1.1 m above the upper edge of the plate. Calculate the moment about the hinge line required to keep the plate closed.
Solution The moment required to keep the plate closed will be equal and opposite to the moment of the resultant force R due to the water acting at the centre of pressure C, i.e. R × CB. From equation (3.2), R = ρgAD. Area of plate, A = −12 (2.7 + 1.2) × 1.5 = 2.925 m2.
FM5_C03.fm Page 66 Wednesday, September 21, 2005 9:01 AM
66
Chapter 3
Static Forces on Surfaces. Buoyancy
FIGURE 3.5 Trapezoidal sluice gate
To find the position of the centroid G, take moments of area about BB′, putting the vertical distance GB = y: A × y = Moment of areas BHE and FJB′ + Moment of EFJH = 2 × ( −12 × 1.5 × 0.75) × 0.5 + (1.2 × 1.5) × 0.75. 2.925y = 0.5625 + 1.35 = 1.9125, y = 0.654 m. Depth to the centre of pressure, D = y + OB = 0.654 + 1.1 = 1.754 m. Substituting in equation (3.2), Resultant force, R = 103 × 9.81 × 2.925 × 1.754 = 50.33 kN. From equation (3.4), Depth to centre of pressure C, D = sin2 φ (IOAD). Using the parallel axis rule for second moments of area, IO = Second moment of EFJH about O + Second moment of BEH and B′FJ about O 1.2 × 1.5 3 1.5 × 1.5 3 = ⎛ −−−−−−−−−−−−− + 1.2 × 1.5 × 1.85 2⎞ + ⎛ −−−−−−−−−−−−− + 1.5 × 0.75 × 1.6 2⎞ m 4 ⎝ ⎠ ⎝ ⎠ 12 36 = 9.5186 m4. As the wall is vertical, sin φ = 1; therefore, 9.5186 Depth to centre of pressure, D = −−−−−−−−−−−−−−−−−−− = 1.8553 m. 2.925 × 1.754 Moment about hinge = R × BC = 50.33(1.8553 − 1.1) = 38.01 kN m.
FM5_C03.fm Page 67 Wednesday, September 21, 2005 9:01 AM
3.3
EXAMPLE 3.2
Resultant force and centre of pressure on a plane surface immersed in a liquid
67
The angle between a pair of lock gates (Fig. 3.6) is 140° and each gate is 6 m high and 1.8 m wide, supported on hinges 0.6 m from the top and bottom of the gate. If the depths of water on the upstream and downstream sides are 5 m and 1.5 m, respectively, estimate the reactions at the top and bottom hinges.
FIGURE 3.6 Lock gate
Solution Figure 3.6(a) shows the plan view of the gates. F is the force exerted by one gate on the other and is assumed to act perpendicular to the axis of the lock if friction between the gates is neglected. P is the resultant of the water forces P1 and P2 (Fig. 3.6(b)) acting on the upstream and downstream faces of the gate, and R is the resultant of the forces R T and RB on the top and bottom hinges. Using equation (3.2) Upstream water force, P1 = ρgA1D1 = 103 × 9.81 × (5 × 1.8) × 2.5 = 220.725 × 103 N, Downstream water force, P2 = ρgA2z2 = 103 × 9.81 × (1.5 × 1.8) × 0.75 = 19.865 × 103 N, Resultant water force on one gate, P = P1 − P2 = (220.73 − 19.86) × 103 N = 200.87 × 103 N. The gates are rectangular, and so P1 and P2 will act at one-third of the depth of water (as shown in Fig. 3.6(b)), since, in equation (3.5), φ = 90°, D = d2, kG2 = d 212, where d is the depth of the gate immersed (see also Section 3.4). The height above the base at which the resultant force P acts can be found by taking moments. If P acts at a distance x from the bottom of the gate, then by taking moments about O,
FM5_C03.fm Page 68 Wednesday, September 21, 2005 9:01 AM
68
Chapter 3
Static Forces on Surfaces. Buoyancy
Px = P1 × (53) − P2 × (1.53) = (220.73 × 53 − 19.86 × 0.5) × 103 = 357.95 × 103 N m. x = (357.95 × 103 )(200.87 × 103 ) = 1.782 m. Assuming that F, R and P are coplanar, they will meet at a point, and, since F is assumed to be perpendicular to the axis of the lock on plan, both F and R are inclined to the gate as shown at an angle of 20° so that F = R and P = F sin 20° + R sin 20° = 2R sin 20°, P 200.87 × 10 3 R = −−−−−−−−−−−− = −−−−−−−−−−−−−−−−−− = 293.65 × 10 3 N. 2 sin 20° 2 × 0.342 If R is coplanar with P it acts at 1.78 m from the bottom of the gate. Taking moments about the bottom hinge, 4.8R T = 1.18R R T = 1.184.8 × 293.65 × 103 = 72.2 × 103 N = 72.2 kN, RB = R − R T = 293.65 − 72.2 = 221.45 kN.
3.4
PRESSURE DIAGRAMS
The resultant force and centre of pressure can be found graphically for walls and other surfaces of constant vertical height for which it is convenient to calculate the horizontal force exerted per unit width. In Fig. 3.7, ABC is the pressure diagram for the vertical wall of the tank containing a liquid, pressure being plotted horizontally against depth vertically. At the free surface A, the (gauge) pressure is zero. At depth y, p = ρgy. The relationship between p and y is linear and can be represented by the triangle ABC. The area of this triangle will be the product of depth (in metres) and pressure (in newtons per square metre), and will represent, to scale, the resultant force R on unit width of the immersed surface perpendicular to the plane of the diagram (in newtons per metre). Area of pressure diagram = −12 AB × BC = −12 H × ρgH.
FIGURE 3.7 Pressure diagram for a vertical wall
FM5_C03.fm Page 69 Wednesday, September 21, 2005 9:01 AM
3.4
Pressure diagrams
69
Therefore, Resultant force, R = ρgH 22 for unit width, and R will act through the centroid P of the pressure diagram, which is at a depth of −2 H from A. 3 This result could also have been obtained from equations (3.2) and (3.5), since, for unit width, R = ρgAD = ρ g(H × 1) × −12 H = ρgH 22, and, in equation (3.5), φ = 90°, sin φ = 1, D = H2, kG2 = H 212; therefore, H 212 H 2 D = −−−−−−−−− + −− = −H, H2 2 3 as before. If the plane surface is inclined and submerged below the surface, the pressure diagram is drawn perpendicular to the immersed surface (Fig. 3.8) and will be a straight line extending from p = 0 at the free surface to p = ρgH at depth H. As the immersed surface does not extend to the free surface, the resultant force R is represented by the shaded area, instead of the whole triangle, and acts through the centroid P of this area.
FIGURE 3.8 Pressure diagram for an inclined submerged surface
It is also possible to draw pressure diagrams in three dimensions for immersed areas of various shapes as, for example, the triangular sluice gate in Fig. 3.9. However, such diagrams do little more than provide assistance in visualizing the situation.
FIGURE 3.9 Pressure diagram for a triangular sluice gate
FM5_C03.fm Page 70 Wednesday, September 21, 2005 9:01 AM
70
Chapter 3
Static Forces on Surfaces. Buoyancy
EXAMPLE 3.3
A closed tank (Fig. 3.10), rectangular in plan with vertical sides, is 1.8 m deep and contains water to a depth of 1.2 m. Air is pumped into the space above the water until the air pressure is 35 kN m−2. If the length of one wall of the tank is 3 m, determine the resultant force on this wall and the height of the centre of pressure above the base.
FIGURE 3.10 Pressure diagram
Solution The air pressure will be transported uniformly over the whole of the vertical wall, and can be represented by the pressure diagram ABCD (Fig. 3.10(b)), the area of which represents the force exerted by the air per unit width of wall. Force due to air, RAir = (p × AB) × Width = 35 × 103 × 1.8 × 3 = 189 × 103 N, and, since the wall is rectangular and the pressure uniform, RAir will act at mid-height, which is 0.9 m above the base. The pressure due to the water will start from zero at the free surface, corresponding to the point E, and reach a value DF equal to ρgh at the bottom. The area of the triangular pressure diagram EFD represents the force exerted by the water per unit width: Force due to water, RH O =
−1 2
× ( ρgh × DE) × Width
=
−1 2
× 103 × 9.81 × 1.2 × 1.2 × 3
2
= 21.19 × 103 N, and, since the wall is rectangular, RH O will act at 1−3 h = 0.4 m from the base. 2
Total force due to both air and water, R = RAir + RH O 2
= (189 + 21.19) × 103 = 210.19 × 103 N.
FM5_C03.fm Page 71 Wednesday, September 21, 2005 9:01 AM
3.5
Force on a curved surface due to hydrostatic pressure
71
If x is the height above the base of the centre of pressure through which R acts, R × x = RAir × 0.9 + RH O × 0.4, 2
x = (189 × 0.9 + 21 × 0.4)210.19 = 0.85 m.
3.5
FORCE ON A CURVED SURFACE DUE TO HYDROSTATIC PRESSURE
If a surface is curved, the forces produced by fluid pressure on the small elements making up the area will not be parallel and, therefore, must be combined vectorially. It is convenient to calculate the horizontal and vertical components of the resultant force. This can be done in three dimensions, but the following analysis is for a surface curved in one plane only. In Fig. 3.11(a) and (b), AB is the immersed surface and R h and R v are the horizontal and vertical components of the resultant force R of the liquid on one side of the surface. In Fig. 3.11(a) the liquid lies above the immersed surface, while in Fig. 3.11(b) it acts below the surface. In Fig. 3.11(a), if ACE is a vertical plane through A, and BC is a horizontal plane, then, since element ACB is in equilibrium, the resultant force P on AC must equal the horizontal component Rh of the force exerted by the fluid on AB because there are no other horizontal forces acting. But AC is the projection of AB on a vertical plane; therefore,
FIGURE 3.11 Hydrostatic force on a curved surface
Horizontal component Rh = Resultant force on the projection of AB on a vertical plane. Also, for equilibrium, P and R h must act in the same straight line; therefore, the horizontal component R h acts through the centre of pressure of the projection of AB on a vertical plane.
FM5_C03.fm Page 72 Wednesday, September 21, 2005 9:01 AM
72
Chapter 3
Static Forces on Surfaces. Buoyancy
Similarly, in Fig. 3.11(b), element ABF is in equilibrium, and so the horizontal component R h is equal to the resultant force on the projection BF of the curved surface AB on a vertical plane, and acts through the centre of pressure of this projection. In Fig. 3.11(a), the vertical component R v will be entirely due to the weight of the fluid in the area ABDE lying vertically above AB. There are no other vertical forces, since there can be no shear forces on AE and BD because the fluid is at rest. Thus, Vertical component, R v = Weight of fluid vertically above AB, and will act vertically downwards through the centre of gravity G of ABDE. In Fig. 3.11(b), if the surface AB were removed and the space ABDE filled with the liquid, this liquid would be in equilibrium under its own weight and the vertical force on the boundary AB. Therefore, Vertical component, R v = Weight of the volume of the same fluid which would lie vertically above AB, and will act vertically upwards through the centre of gravity G of this imaginary volume of fluid. In the case of closed vessels under pressure, a free surface does not exist, but an imaginary free surface can be substituted at a level pρg above a point at which the pressure p is known, ρ being the mass density of the actual fluid. The resultant force R is found by combining the components vectorially. In the general case, the components in three directions may not meet at a point and, therefore, cannot be represented by a single force. However, in Fig. 3.11, if the surface is of uniform width perpendicular to the diagram, R h and R v will intersect at O. Thus, Resultant force, R = ( R 2h + R 2v ),
FIGURE 3.12 Resultant force on a cylindrical surface
EXAMPLE 3.4
and acts through O at an angle θ given by tan θ = R vR h. In the special case of a cylindrical surface, all the forces on each small element of area acting normal to the surface will be radial and will pass through the centre of curvature O (Fig. 3.12). The resultant force R must, therefore, also pass through the centre of curvature O.
A sluice gate is in the form of a circular arc of radius 6 m as shown in Fig. 3.13. Calculate the magnitude and direction of the resultant force on the gate, and the location with respect to O of a point on its line of action.
Solution Since the water reaches the top of the gate,
FM5_C03.fm Page 73 Wednesday, September 21, 2005 9:01 AM
3.6
Buoyancy
73
FIGURE 3.13 Sector gate
Depth of water, h = 2 × 6 sin 30° = 6 m, Horizontal component of force on gate = R h per unit length = Resultant force on PQ per unit length = ρg × h × h2 = ρgh 22 = (103 × 9.81 × 36)2 N m−1 = 176.58 kN m−1, Vertical component of force on gate = R v per unit length = Weight of water displaced by segment PSQ = (Sector OPSQ − ∆OPQ)ρ g = [ (60360) × π × 62 − 6 sin 30° × 6 cos 30°] × 103 × 9.81 N m−1 = 32.00 kN m−1, Resultant force on gate, R = ( R 2h + R 2v ) = (176.582 + 32.002) = 179.46 kN m−1. If R is inclined at an angle θ to the horizontal, tan θ = R vRh = 32.00176.58 = 0.181 22
θ = 10.27° to the horizontal. Since the surface of the gate is cylindrical, the resultant force R must pass through O.
3.6
BUOYANCY
The method of calculating the forces on a curved surface applies to all shapes of surface and, therefore, to the surface of a totally submerged object (Fig. 3.14). Considering any vertical plane VV through the body, the projected area of each of the
FM5_C03.fm Page 74 Wednesday, September 21, 2005 9:01 AM
74
Chapter 3
Static Forces on Surfaces. Buoyancy
FIGURE 3.14 Buoyancy
two sides on this plane will be equal and, as a result, the horizontal forces F will be equal and opposite. There is, therefore, no resultant horizontal force on the body due to the pressure of the surrounding fluid. The only force exerted by the fluid on an immersed body is vertical and is called the buoyancy or upthrust. It will be equal to the difference between the resultant forces on the upper and lower parts of the surface of the body. If ABCD is a horizontal plane, Upthrust on body = Upward force on lower surface ADEC − Downward force on upper surface ABCD = Weight of volume of fluid AECDGFH − Weight of volume of fluid ABCDGFH = Weight of volume of fluid ABCDE, Upthrust on body = Weight of fluid displaced by the body, and will act through the centroid of the volume of fluid displaced, which is known as the centre of buoyancy. This result is known as Archimedes’ principle. As an alternative to the proof given above, it can be seen that, if the body were completely replaced by the fluid in which it is immersed, the forces exerted on the boundaries corresponding to the original body would exactly maintain the substituted fluid in equilibrium. Thus, the upward force on the boundary must be equal to the downward force corresponding to the weight of the fluid displaced by the body. If a body is immersed so that part of its volume V1 is immersed in a fluid of density ρ1 and the rest of its volume V2 in another immiscible fluid of mass density ρ2 (Fig. 3.15),
FIGURE 3.15 Body immersed in two fluids
FM5_C03.fm Page 75 Wednesday, September 21, 2005 9:01 AM
3.6
Buoyancy
75
Upthrust on upper part, R1 = ρ1 gV1 acting through G1, the centroid of V1, Upthrust on lower part, R2 = ρ2 gV2 acting through G2, the centroid of V2, Total upthrust = ρ1 gV1 + ρ2 gV2. The positions of G1 and G2 are not necessarily on the same vertical line, and the centre of buoyancy of the whole body is, therefore, not bound to pass through the centroid of the whole body.
EXAMPLE 3.5
A rectangular pontoon has a width B of 6 m, a length l of 12 m, and a draught D of 1.5 m in fresh water (density 1000 kg m−3). Calculate (a) the weight of the pontoon, (b) its draught in sea water (density 1025 kg m−3) and (c) the load (in kilonewtons) that can be supported by the pontoon in fresh water if the maximum draught permissible is 2 m.
Solution When the pontoon is floating in an unloaded condition, Upthrust on immersed volume = Weight of pontoon. Since the upthrust is equal to the weight of the fluid displaced, Weight of pontoon = Weight of fluid displaced, W = ρgBlD. (a) In fresh water, ρ = 1000 kg m−3 and D = 1.5 m; therefore, Weight of pontoon, W = 1000 × 9.81 × 6 × 12 × 1.5 N = 1059.5 kN. (b) In sea water, ρ = 1025 kg m−3; therefore, Draught in sea water, D = WρgBl 1059.5 × 10 3 = −−−−−−−−−−−−−−−−−−−−−−−−−−−−− = 1.46 m. 1025 × 9.81 × 6 × 12 (c) For the maximum draught of 2 m in fresh water, Total upthrust = Weight of water displaced = ρgBlD = 1000 × 9.81 × 6 × 12 × 2 N = 1412.6 kN, Load that can be supported
= Upthrust − Weight of pontoon = 1412.6 − 1059.5 = 353.1 kN.
FM5_C03.fm Page 76 Wednesday, September 21, 2005 9:01 AM
76
Chapter 3
Static Forces on Surfaces. Buoyancy
3.7
EQUILIBRIUM OF FLOATING BODIES
When a body floats in vertical equilibrium in a liquid, the forces present are the upthrust R acting through the centre of buoyancy B (Fig. 3.16) and the weight of the body W = mg acting through its centre of gravity. For equilibrium, R and W must be equal and act in the same straight line. Now, R will be equal to the weight of fluid displaced, ρgV, where V is the volume of fluid displaced; therefore, V = mgρg = mρ. As explained in Section 2.1, the equilibrium of a body may be stable, unstable or neutral, depending upon whether, when given a small displacement, it tends to return to the equilibrium position, move further from it or remain in the displaced position. For a floating body, such as a ship, stability is of major importance. FIGURE 3.16 Body floating in equilibrium
3.8
STABILITY OF A SUBMERGED BODY
For a body totally immersed in a fluid, the weight W = mg acts through the centre of gravity of the body, while the upthrust R acts through the centroid of the body B, which is the centre of buoyancy. Whatever the orientation of the body, these two points will remain in the same positions relative to the body. It can be seen from Fig. 3.17 that a small angular displacement θ from the equilibrium position will generate a moment W × BG × θ. If the centre of gravity G is below the centre of buoyancy B (Fig. 3.17(a)), this will be a righting moment and the body will tend to return to its equilibrium position. However, if (as in Fig. 3.17(b)) the centre of gravity is above the centre of buoyancy, an overturning moment is produced and the body is unstable. Note that, as FIGURE 3.17 Stability of submerged bodies
FM5_C03.fm Page 77 Wednesday, September 21, 2005 9:01 AM
3.9
Stability of floating bodies
77
the body is totally immersed, the shape of the displaced fluid is not altered when the body is tilted and so the centre of buoyancy remains unchanged relative to the body.
3.9
STABILITY OF FLOATING BODIES
Figure 3.18(a) shows a body floating in equilibrium. The weight W = mg acts through the centre of gravity G and the upthrust R acts through the centre of buoyancy B of the displaced fluid in the same straight line as W. When the body is displaced through an angle θ (Fig. 3.18(b)), W continues to act through G; the volume of liquid remains unchanged since R = W, but the shape of this volume changes and its centre of gravity, which is the centre of buoyancy, moves relative to the body from B to B1. Since R and W are no longer in the same straight line, a turning moment proportional to W × θ is produced, which in Fig. 3.18(b) is a righting moment and in Fig. 3.18(d) is an overturning moment. If M is the point at which the line of action of the upthrust R cuts the original vertical through the centre of gravity of the body G, x = GM × θ, provided that the angle of tilt θ is small, so that sin θ = tan θ = θ in radians. The point M is called the metacentre and the distance GM is the metacentric height. Comparing Fig. 3.18(b) and (d) it can be seen that: 1. 2. 3.
FIGURE 3.18 Stable and unstable equilibrium
If M lies above G, a righting moment W × GM × θ is produced, equilibrium is stable and GM is regarded as positive. If M lies below G, an overturning moment W × GM × θ is produced, equilibrium is unstable and GM is regarded as negative. If M coincides with G, the body is in neutral equilibrium.
FM5_C03.fm Page 78 Wednesday, September 21, 2005 9:01 AM
78
Chapter 3
Static Forces on Surfaces. Buoyancy
Since a floating body can tilt in any direction, it is usual, for a ship, to consider displacement about the longitudinal (rolling) and transverse (pitching) axes. The position of the metacentre and the value of the metacentric height will normally be different for rolling and pitching.
3.10 DETERMINATION OF THE METACENTRIC HEIGHT The metacentric height of a vessel can be determined if the angle of tilt θ caused by moving a load P (Fig. 3.19) a known distance x across the deck is measured. Overturning moment due to movement of load P = Px.
(3.6)
If GM is the metacentric height and W = mg is the total weight of the vessel including P, FIGURE 3.19 Determination of metacentric height
Righting moment = W × GM × θ.
(3.7)
For equilibrium in the tilted position, the righting moment must equal the overturning moment so that, from equations (3.6) and (3.7), W × GM × θ = Px, Metacentric height, GM = PxWθ.
(3.8)
The true metacentric height is the value of GM as θ → 0.
3.11 DETERMINATION OF THE POSITION OF THE METACENTRE RELATIVE TO THE CENTRE OF BUOYANCY For a vessel of known shape and displacement, the position of the centre of buoyancy B is comparatively easily found and the position of the metacentre M relative to B can be calculated as follows. In Fig. 3.20, AC is the original waterline plane and B the
FIGURE 3.20 Height of metacentre above centre of buoyancy
FM5_C03.fm Page 79 Wednesday, September 21, 2005 9:01 AM
3.11
Determination of the position of the metacentre relative to the centre of buoyancy
79
centre of buoyancy in the equilibrium position. When the vessel is tilted through a small angle θ, the centre of buoyancy will move to B′ as a result of the alteration in the shape of the displaced fluid. A′C′ is the waterline plane in the displaced position. For small angles of tilt, BM = BB′θ. The movement of the centre of buoyancy, which is the centre of gravity of the displaced fluid, from B to B′ is the result of the removal of a volume of fluid corresponding to the wedge AOA′ and the addition of a wedge COC′. The total weight of fluid displaced remains unchanged, since it is equal to the weight of the vessel; therefore, Weight of wedge AOA′ = Weight of wedge COC′. If a is a small area in the waterline plane at a distance x from the axis of rotation OO, it will generate a small volume, shown shaded, when the vessel is tilted. Volume swept out by a = DD′ × a = axθ. Summing all such volumes and multiplying by the specific weight ρg of the liquid, Weight of wedge AOA′ =
x =AO
∑ ρ gax θ .
(3.9)
x =0
Similarly, Weight of wedge COC′ =
x = CO
∑ ρ gax θ .
(3.10)
x =0
Since there is no change in displacement, we have, from equations (3.9) and (3.10), x = AO
x = CO
x =0
x =0
ρ g θ ∑ ax = ρ g θ ∑ ax, ∑ax = 0. But ∑ax is the first moment of area of the waterline plane about OO; therefore the axis OO must pass through the centroid of the waterline plane. The distance BB′ can now be calculated, since the couple produced by the movement of the wedge AOA′ to COC′ must be equal to the couple due to the movement of R from B to B′. Moment about OO of the weight of fluid swept out by area a = ρgaxθ × x. Total moment due to altered displacement = ρgθ ∑ax2. Putting ∑ax2 = I = Second moment of area of waterline plane about OO, Total moment due to altered displacement = ρgθI,
(3.11)
Moment due to movement of R = R × BB′ = ρgV × BB′,
(3.12)
FM5_C03.fm Page 80 Wednesday, September 21, 2005 9:01 AM
80
Chapter 3
Static Forces on Surfaces. Buoyancy
where V = volume of liquid displaced. Equating equations (3.11) and (3.12),
ρgV × BB′ = ρgθI, giving
BB′ = θIV,
(3.13)
BM = BB′θ = IV.
(3.14)
The distance BM is known as the metacentric radius.
EXAMPLE 3.6
A cylindrical buoy (Fig. 3.21) 1.8 m in diameter, 1.2 m high and weighing 10 kN floats in salt water of density 1025 kg m−3. Its centre of gravity is 0.45 m from the bottom. If a load of 2 kN is placed on the top, find the maximum height of the centre of gravity of this load above the bottom if the buoy is to remain in stable equilibrium.
FIGURE 3.21 Stability of a cylindrical buoy
Solution In Fig. 3.21, let G be the centre of gravity of the buoy, G1 the centre of gravity of the load at a height Z1 above the bottom, and G′ the combined centre of gravity of the load and the buoy at a height Z′ above the bottom. When the load is in position, let V be the volume of salt water displaced and Z the depth of immersion of the buoy. Buoyancy force = Weight of salt water displaced = ρgV = ρg(π4)d 2Z. For equilibrium, the buoyancy force must equal the combined weight of the buoy and the load (W + W1); therefore, W + W1 = ρg(π 4)d 2Z, Depth of immersion, Z = 4(W + W1)ρgπ d 2 = 4(10 + 2) × 103(1025 × 9.81 × 1.82 × π ) = 0.47 m.
FM5_C03.fm Page 81 Wednesday, September 21, 2005 9:01 AM
3.12
Periodic time of oscillation
81
The centre of buoyancy B will be at the centre of gravity of the displaced water, so that OB = −12 Z = 0.235 m. If the buoy and the load are just in stable equilibrium, the metacentre M must coincide with the centre of gravity G′ of the buoy and load combined. The metacentric height G′M will then be zero and BG′ = BM. From equation (3.14), 1.8 2 I π d 464 BG′ = BM = −− = −−−−−2−−−−− = −−−−−−−−−−−−− = 0.431 m. π d z4 16 × 0.47 V Thus, the position of G′ is given by Z′ = −12 Z + BG′ = 0.235 + 0.431 = 0.666 m. The value of Z1 corresponding to this value of Z′ is found by taking moments about O: W1Z1 + 0.45W = (W + W1)Z′. Maximum height of load above bottom, ( W + W )Z′ – 0.45W Z 1 = −−−−−−−−−−−−−1−−−−−−−−−−−−−−−−−− W1 12 × 10 3 × 0.666 – 0.45 × 10 × 10 3 = −−−−−−−−−−−−−−−−−−−−−−−−−−−−−3−−−−−−−−−−−−−−−−−−− m = 1.746 m. 2 × 10
3.12 PERIODIC TIME OF OSCILLATION The displacement of a stable vessel through an angle θ from its equilibrium position produces a righting moment T which, from equation (3.7), is given by T = W × GM × θ, where W = mg is the weight of the vessel and GM is the metacentric height. This will produce an angular acceleration d2θdt2, and, if I is the mass moment of inertia of the vessel about its axis of rotation, d2 θ T W × GM × θ GM × θ g −−−−2 = −− = – −−−−−−−−−−−−−−−− −− = – −−−−−−−−2−−−−− , dt ( Wg )k 2 k I where k is the radius of gyration from its axis of rotation. The negative sign indicates that the acceleration is in the opposite direction to the displacement. Since this corresponds to simple harmonic motion, Displacement Periodic time, t = 2 π ⎛ −−−−−−−−−−−−−−−−−−−− ⎞ = 2 π ⎝ Acceleration ⎠
= 2π [k 2(GM × g)],
θ −−−−−−−−−−−−−−−−−−−−−−−2− GM × θ × ( gk ) (3.15)
FM5_C03.fm Page 82 Wednesday, September 21, 2005 9:01 AM
82
Chapter 3
Static Forces on Surfaces. Buoyancy
from which it can be seen that, although a large metacentric height will improve stability, it produces a short periodic time of oscillation, which results in discomfort and excessive stress on the structure of the vessel.
3.13 STABILITY OF A VESSEL CARRYING LIQUID IN TANKS WITH A FREE SURFACE The stability of a vessel carrying liquid in tanks with a free surface (Fig. 3.22) is affected adversely by the movement of the centre of gravity of the liquid in the tanks as the vessel heels. Thus, G1 will move to G′1 and G2 to G′2 . The distance moved is calculated in the same way as the movement BB′ of the centre of buoyancy, given by equation (3.13): G 1 G′1 = θ I1V1 and
G 2 G′2 = θI2 V2 ,
FIGURE 3.22 Vessel carrying liquid in tanks
where I1 and I2 are the second moments of area of the free surfaces, and V1 and V2 the volumes, of the liquid in the tanks. As a result of the movement of G1 and G2, the centre of gravity G of the whole vessel and contents will move to G′. If V is the volume of water displaced by the vessel and ρ is the mass density of water, Weight of vessel and contents = Weight of water displaced = ρgV. If the volume of liquid of density ρ1 in the tanks is V1 and V2, Weight of contents of the first tank
= ρ1 gV1,
Weight of contents of the second tank = ρ1 gV2.
FM5_C03.fm Page 83 Wednesday, September 21, 2005 9:01 AM
3.13
Stability of a vessel carrying liquid in tanks with a free surface
83
Taking moments to find the change in the centre of gravity of the vessel and contents,
ρgV × GG′ = ρ1 gV1 × G 1 G′1 + ρ1 gV2 × G 2 G′2 = ρ1 gV1 × θI1V1 + ρ1 gV2 × θI2 V2 , 1 GG′ = −− ( ρ1ρ)θ (I1 + I2 ). V In the tilted position, the new vertical through B′ intersects the original vertical through G at the metacentre M, but the weight W acts through G′ instead of G and its line of action cuts the original vertical at N, reducing the metacentric height from GM to NM. Effective metacentric height, NM = ZB + BM − (ZG + GN), 1 and, since BM = IV and GN = GG′θ = −− ( ρ1ρ)(I1 + I2 ), V 1 NM = ZB − ZG + −− [1 − (ρ1ρ)(I1 + I2)]. V
(3.16)
Thus, the effect of the liquid in the tank is to reduce the effective metacentric height and impair stability, provided that the liquid in the tanks has a free surface so that its centre of gravity moves as the vessel tilts. Lateral subdivision of the tanks improves stability by reducing the sum of the second moments of area I1, I2, etc.
EXAMPLE 3.7
FIGURE 3.23 Barge containing liquids
A barge (Fig. 3.23) has vertical sides and ends and a flat bottom. In plan view it is rectangular, 20 m long by 6 m wide, but with an additional semicircular portion of 3 m radius at one end. The empty barge weighs 200 kN and floats upright in fresh water. The part of the vessel which is rectangular in plan is divided by a wall into two
FM5_C03.fm Page 84 Wednesday, September 21, 2005 9:01 AM
84
Chapter 3
Static Forces on Surfaces. Buoyancy
compartments 3 m wide by 20 m long. These compartments form open-top tanks which are partly filled with liquid of relative density 0.8 to a depth of 0.8 m in one tank and 1.0 m in the other. The vessel rolls about a horizontal axis, but the flat end remains in a vertical plane. Ignoring the thickness of the material of the barge structure and assuming that the centre of gravity of the barge and contents is 0.45 m above the bottom, find the angle of roll.
Solution In order to be able to determine the angle of roll, we must first find the effective metacentric height from equation (3.16). For the whole vessel lb 3 π b 4 I OO = −−− + −−−−− = 20 × 6312 + π × 64128 m4 = 391.9 m4. 12 128 For each tank ICC = l × ( −12 b)312 = 20 × 3312 = 45 m4. Weight of barge = 200 kN. Weight of liquid load = 0.8 × 103 × 9.81(20 × 3 × 1 + 20 × 3 × 0.8) N = 846 kN. Total weight of barge and contents = 1046 kN. Area of waterline plane of vessel = 20 × 6 + −12 π × 32 = 134.1 m2. Volume of vessel submerged = Weight(Density × g) = 1046 × 103(103 × 9.81) = 106.8 m3. Depth submerged = 106.8134.1 = 0.80 m. Height of centre of buoyancy B above bottom = −12 Depth submerged = 0.4 m. Putting these values in equation (3.16) with ρ1ρ = 0.8, Effective metacentric height, NM = 0.4 − 0.45 + (391.9 − 0.8 × 2 × 45)106.8 = 2.95 m. The overturning moment is caused by the weight of the excess liquid in one tank, P = 0.8 × 103 × 9.81 × 20 × 3(1.0 − 0.8) = 94 kN. The centre of gravity of this excess liquid is 1.5 m from the centreline. Overturning moment due to excess liquid = P × 1.5, Righting moment = W × NM tan θ. Thus, for equilibrium, P × 1.5 = W × NM tan θ, tan θ = 94 × 1.5(1046 × 2.95) = 0.0457, Angle of roll, θ = 2°37′′.
FM5_C03.fm Page 85 Wednesday, September 21, 2005 9:01 AM
Problems
85
Concluding remarks Arising directly from the hydrostatic equation developed in Chapter 2, this chapter has demonstrated techniques available for determining the forces acting on submerged or partially submerged surfaces. The chapter has also stressed the relationship between pressure and force, and in particular has highlighted the fact that force is a vector quantity, calculated by reference to the applied pressure and the surface area normal to the force direction. This concept, although apparently obvious, will form the basis of later calculations of lift and drag on aerofoils, and the definition of appropriate lift and drag coefficients. The treatment of buoyancy and floating-body stability is a further demonstration of the application of the techniques appropriate to the analysis of solid-body mechanics.
Summary of important equations and concepts 1.
2. 3.
The integration necessary to determine the resultant force acting on a surface is emphasized, Section 3.2, and most importantly the concept of a centre of pressure through which this force acts is introduced, Section 3.3. It will be shown later that centre of pressure movement as flow becomes supersonic can affect wing stability and introduce the need for remedial action, either by control surface activation or by corresponding movement of the aircraft centre of gravity. The treatment of a range of hydrostatic force and moment examples illustrates the interface between hydrostatics and mechanics, Sections 3.4 and 3.5. The treatment of buoyancy and stability of floating bodies continues this linkage.
Problems 3.1 A circular lamina 125 cm in diameter is immersed in water so that the distance of its edge measured vertically below the free surface varies from 60 cm to 150 cm. Find the total force due to the water acting on one side of the lamina, and the vertical distance of the centre of pressure below the surface. [12 639 N, 1.1 m] 3.2 One end of a rectangular tank is 1.5 m wide by 2 m deep. The tank is completely filled with oil of specific weight 9 kN m−3. Find the resultant pressure on this vertical end and the depth of the centre of pressure from the top. [27 kN, 1.33 m]
3.5 A barge in the form of a closed rectangular tank 20 m long by 4 m wide floats in water. If the bottom is 1.5 m below the surface, what is the water force on one long side and at what level below the surface does it act? If a uniform pressure of 50 kN m−2 gauge is applied inside the barge what will be the new magnitude and point of action of the resultant force on the side? The deck is 0.2 m above water level. [220.73 kN, 1.0 m; 1479.27 kN, 0.6 m below surface] 3.6 A rectangular sluice door (Fig. 3.24) is hinged at the top at A and kept closed by a weight fixed to the door.
3.3 What is the position of the centre of pressure of a vertical semicircular plane submerged in a hom*ogeneous liquid with its diameter d at the free surface? [Depth 3π d32] 3.4 A culvert draws off water from the base of a reservoir. The entrance to the culvert is closed by a circular gate 1.25 m in diameter which can be rotated about its horizontal diameter. Show that the turning moment on the gate is independent of the depth of water if the gate is completely immersed and find the value of this moment. [1177 N m]
FIGURE 3.24
FM5_C03.fm Page 86 Wednesday, September 21, 2005 9:01 AM
86
Chapter 3
Static Forces on Surfaces. Buoyancy
The door is 120 cm wide and 90 cm long and the centre of gravity of the complete door and weight is at G, the combined weight being 9810 N. Find the height of the water h on the inside of the door which will just cause the door to open. [0.88 m] 3.7 A rectangular gate (Fig. 3.25) of negligible thickness, hinged at its top edge and of width b, separates two tanks in which there is the same liquid of density ρ. It is required that the gate shall open when the level in the left-hand tank falls below a distance H from the hinge. The level in the righthand tank remains constant at a height y above the hinge. Derive an expression for the weight of the gate in terms of H, Y, y, b and g. Assume that the weight of the gate acts at its centre of area.
purpose. If the two halves of the container are not secured together, what must be the mass of the upper hemisphere if it just fails to lift off the lower hemisphere? [12.5 kg] 3.11 A sluice gate (Fig. 3.26) consists of a quadrant of a circle of radius 1.5 m pivoted at its centre O. Its centre of gravity is at G as shown. When the water is level with the pivot O, calculate the magnitude and direction of the resultant force on the gate due to the water and the turning moment required to open the gate. The width of the gate is 3 m and it has a mass of 6000 kg.
FIGURE 3.26 [61.6 kN, 57°31′, 35.3 kN m]
FIGURE 3.25
3Y 2 ( y + H ) – H 3 W = 0.77 ρ gb −−−−−−−−−−−−−−−−−−−−−−− Y 3.8 A masonry dam 6 m high has the water level with the top. Assuming that the dam is rectangular in section and 3 m wide, determine whether the dam is stable against overturning and whether tension will develop in the masonry joints. Density of masonry 1760 kg m−3. [Stable, tension on the water face]
3.12 A sector-shaped sluice gate having a radius of curvature of 5.4 m is as shown in Fig. 3.27. The centre of curvature C is 0.9 m vertically below the lower edge A of the gate and 0.6 m vertically above the horizontal axis passing through O about which the gate is constructed to turn. The mass of the gate is 3000 kg per metre run and its centre of gravity is 3.6 m horizontally from the centre O. If the water level is 2.4 m above the lower edge of the gate, find per metre run (a) the resultant force acting on the axis at O, (b) the resultant moment about O.
3.9 A pair of lock gates, each 3 m wide, have their lower hinges at the bottom of the gates and their upper hinges 5 m from the bottom. The width of the lock is 5.5 m. Find the reaction between the gates when the water level is 4.5m above the bottom of one side and 1.5 m on the other. Assuming that this force acts at the same height as the resultant force due to the water pressure find the reaction forces on the hinges. [331 kN; 107.6 kN, 223.4 kN] 3.10 A spherical container is made up of two hemispheres, the joint between the two halves being horizontal. The sphere is completely filled with water through a small hole in the top. It is found that 50 kg of water are required for this
FIGURE 3.27 [(a) 39.2 kN, (b) 106 kN m]
FM5_C03.fm Page 87 Wednesday, September 21, 2005 9:01 AM
Problems 3.13 The face of a dam (Fig. 3.28) is curved according to the relation y = x 2 2.4, where y and x are in metres. The height of the free surface above the horizontal plane through A is 15.25 m. Calculate the resultant force F due to the fresh water acting on unit breadth of the dam, and determine the position of the point B at which the line of action of this force cuts the horizontal plane through A.
87
3.16 The shifting of a portion of cargo of mass 25 000 kg through a distance of 6 m at right angles to the vertical plane containing the longitudinal axis of a vessel causes it to heel through an angle of 5°. The displacement of the vessel is 5000 metric tonnes and the value of I is 5840 m4. The density of sea water is 1025 kg m−3. Find (a) the metacentric height and (b) the height of the centre of gravity of the vessel above the centre of buoyancy. [(a) 0.342 m, (b) 0.849 m] 3.17 A buoy floating in sea water of density 1025 kg m−3 is conical in shape with a diameter across the top of 1.2 m and a vertex angle of 60°. Its mass is 300 kg and its centre of gravity is 750 mm from the vertex. A flashing beacon is to be fitted to the top of the buoy. If this unit has a mass of 55 kg what is the maximum height of its centre of gravity above the top of the buoy if the whole assembly is not to be unstable? (The centre of volume of a cone of height h is at a distance −34 h from the vertex.) [1.25 m]
FIGURE 3.28 [1290 kN m−1, 14.15 m] 3.14 A steel pipeline conveying gas has an internal diamter of 120 cm and an external diameter of 125 cm. It is laid across the bed of a river, completely immersed in water and is anchored at intervals of 3 m along its length. Calculate the buoyancy force in newtons per metre and the upward force in newtons on each anchorage. Density of steel = 7900 kg m−3, density of water = 1000 kg m−3. [12 037 N m−1, 13 742 N] 3.15 The ball-operated valve shown in Fig. 3.29 controls the flow from a tank through a pipe to a lower tank, in which it is situated. The water level in the upper tank is 7 m above the 10 mm diameter valve opening. Calculate the volume of the ball which must be submerged to keep the valve closed.
3.18 A rectangular pontoon 10 m by 4 m in plan weighs 280 kN. A steel tube weighing 34 kN is placed longitudinally on the deck. When the tube is in a central position, the centre of gravity for the combined weight lies on the vertical axis of symmetry 250 mm above the water surface. Find (a) the metacentric height, (b) the maximum distance the tube may be rolled laterally across the deck if the angle of heel is not to exceed 5°. [(a) 1.02 m, (b) 0.82 m] 3.19 A rectangular tank 90 cm long and 60 cm wide is mounted on bearings so that it is free to turn on a longitudinal axis. The tank has a mass of 68 kg and its centre of gravity is 15 cm above the bottom. When the tank is slowly filled with water it hangs in stable equilibrium until the depth of water is 45 cm after which it becomes unstable. How far is the axis of the bearings above the bottom of the tank? [0.21 m] 3.20 A cylindrical buoy 1.35 m in diameter and 1.8 m high has a mass of 770 kg. Show that it will not float with its axis vertical in sea water of density 1025 kg m−3. If one end of a vertical chain is fastened to the base, find the pull required to keep the buoy vertical. The centre of gravity of the buoy is 0.9 m from its base. [GM = −0.42 m, 4632 N] 3.21 A solid cylinder 1 m in diameter and 0.8 m high is of uniform relative density 0.85. Calculate the periodic time of small oscillations when the cylinder floats with its axis vertical in still water. [2.90 s]
FIGURE 3.29 [110 cm3]
3.22 A ship has displacement of 5000 metric tonnes. The second moment of area of the waterline section about a fore and aft axis is 12 000 m4 and the centre of buoyancy is 2 m below the centre of gravity. The radius of gyration is 3.7 m. Calculate the period of oscillation. Sea water has a density of 1025 kg m−3. [10.94 s]
FM5_C04.fm Page 88 Tuesday, September 20, 2005 1:27 PM
Part II
Concepts of Fluid Flow 4 Motion of Fluid Particles and Streams 90
6 The Energy Equation and its Applications 166
5 The Momentum Equation and its Applications 112
7 Two-dimensional Ideal Flow 212
FM5_C04.fm Page 89 Tuesday, September 20, 2005 1:27 PM
The study of fluid motion is complicated by the introduction of viscosity-dependent shear forces that were absent in the preceding treatment of stationary fluids. In the majority of flow cases analysis relies upon a body of empirical work, supported by the concepts of dimensional analysis and similarity. In this part of the text we will establish the analytical techniques that will later be combined with the empirical representation of frictional forces to allow the study of ‘real’ fluid behaviour. In order to deal effectively with flowing fluids it is first necessary to identify flow categories, defined in predominantly mathematical terms, that will allow the appropriate analysis to be undertaken by identifying suitable and acceptable simplifications. Examples of the categories to be introduced include variation of the flow parameters with time (steady or unsteady) or variations along the flow path (uniform or nonuniform). Similarly, compressibility effects may be important in high-speed gas flows but may be ignored in many liquid flow situations. In parallel to setting up these flow categories it is also necessary to develop a series of mathematically expressed principles that will allow the variations in flow parameters as a result of the motion of the fluid to be predicted. The principles of continuity, energy and momentum are developed in this part of the text.
The steady flow energy equation is introduced and will be utilized later to describe the behaviour of real fluids by the inclusion of an empirically based friction term. The momentum equation will be introduced and its application illustrated both for fluid-to-solid boundary transfers, such as the calculation of forces acting on moving vanes or pipe nozzles, and for other flow situations, such as the formation of hydraulic jumps in open-channel flows. While the treatment of the behaviour of real fluid motion requires the introduction of viscous and, possibly, compressibility terms, the study of an ideal fluid freed from these constraints is useful and important, particularly in the consideration of flow patterns away from the influence of solid boundaries. Primarily a mathematical modelling tool, the study of ideal fluid flow has its roots in the work of eighteenthcentury hydrodynamicists and has applications now in aerodynamics as it allows the introduction of a further flow classification, namely rotational or irrotational flow. The study of ideal flow allows flow patterns around aerofoil sections to be considered and therefore naturally leads to considerations of lift and vorticity. Taken together with Part I, this portion of the text provides the foundation upon which the study and application of the behaviour of real fluids may be based.
Opposite: Offshore wind turbines, photo courtesy of the British Wind Energy Association, © NEG Micon
FM5_C04.fm Page 90 Tuesday, September 20, 2005 1:27 PM
Chapter 4
Motion of Fluid Particles and Streams 4.1 4.2 4.3 4.4 4.5 4.6 4.7
Fluid flow Uniform flow and steady flow Frames of reference Real and ideal fluids Compressible and incompressible flow One-, two- and three-dimensional flow Analyzing fluid flow
Motion of a fluid particle Acceleration of a fluid particle Laminar and turbulent flow Discharge and mean velocity Continuity of flow Continuity equations for threedimensional flow using Cartesian coordinates 4.14 Continuity equation for cylindrical coordinates 4.8 4.9 4.10 4.11 4.12 4.13
FM5_C04.fm Page 91 Tuesday, September 20, 2005 1:27 PM
The prediction of the conditions encountered by, and as a result of, fluids in motion presents a range of problems that must be resolved by reference to the fundamental laws of physics, coupled with the particular fluid properties identified in Chapter 1. The treatment presented in this chapter will lay the foundations for later analysis in that the various fluid flow regimes, whether time dependent or determined by the shear forces assumed to act on the boundaries of the fluid flow or the compressibility of the fluid,
will be identified, including the Reynolds numberdependent laminar and turbulent flow regimes. The presence of velocity profiles within any fluid flow will be emphasized, together with the importance of fluid viscosity in determining the detail conditions within the fluid flow. The application of the conservation of mass relationship across a control volume defined within a flowing fluid will be introduced, and the relationship linking mass flow to local or mean flow velocity values will be detailed. l l l
FM5_C04.fm Page 92 Tuesday, September 20, 2005 1:27 PM
92
Chapter 4
Motion of Fluid Particles and Streams
4.1
FLUID FLOW
The motion of a fluid is usually extremely complex. The study of a fluid at rest, or in relative equilibrium, was simplified by the absence of shear forces, but when a fluid flows over a solid surface or other boundary, whether stationary or moving, the velocity of the fluid in contact with the boundary must be the same as that of the boundary, and a velocity gradient is created at right angles to the boundary (see Section 1.2). The resulting change of velocity from layer to layer of fluid flowing parallel to the boundary gives rise to shear stresses in the fluid. Individual particles of fluid move as a result of the action of forces set up by differences of pressure or elevation. Their motion is controlled by their inertia and the effect of the shear stresses exerted by the surrounding fluid. The resulting motion is not easily analysed mathematically, and it is often necessary to supplement theory by experiment. If an individual particle of fluid is coloured, or otherwise rendered visible, it will describe a pathline, which is the trace showing the position at successive intervals of time of a particle which started from a given point. If, instead of colouring an individual particle, the flow pattern is made visible by injecting a stream of dye into a liquid, or smoke into a gas, the result will be a streakline or filament line, which gives an instantaneous picture of the positions of all the particles which have passed through the particular point at which the dye is being injected. Since the flow pattern may vary from moment to moment, a streakline will not necessarily be the same as a pathline. When using tracers or dyes it is essential to choose a material having a density and other physical properties as similar as possible to those of the fluid being studied. In analysing fluid flow, we also make use of the idea of a streamline, which is an imaginary curve in the fluid across which, at a given instant, there is no flow. Thus, the velocity of every particle of fluid along the streamline is tangential to it at that moment. Since there can be no flow through solid boundaries, these can also be regarded as streamlines. For a continuous stream of fluid, streamlines will be continuous lines extending to infinity upstream and downstream, or will form closed curves as, for example, round the surface of a solid object immersed in the flow. If conditions are steady and the flow pattern does not change from moment to moment, pathlines and streamlines will be identical; if the flow is fluctuating this will not be the case. If a series of streamlines are drawn through every point on the perimeter of a small area of the stream cross-section, they will form a streamtube (Fig. 4.1). Since there is no flow across a streamline, the fluid inside the streamtube cannot escape through its walls, and behaves as if it were contained in an imaginary pipe. This is a useful concept in dealing with the flow of a large body of fluid, since it allows elements of the fluid to be isolated for analysis.
FIGURE 4.1 A streamtube
FM5_C04.fm Page 93 Tuesday, September 20, 2005 1:27 PM
4.3
4.2
Frames of reference
93
UNIFORM FLOW AND STEADY FLOW
Conditions in a body of fluid can vary from point to point and, at any given point, can vary from one moment of time to the next. Flow is described as uniform if the velocity at a given instant is the same in magnitude and direction at every point in the fluid. If, at the given instant, the velocity changes from point to point, the flow is described as non-uniform. In practice, when a fluid flows past a solid boundary there will be variations of velocity in the region close to the boundary. However, if the size and shape of the cross-section of the stream of fluid are constant, the flow is considered to be uniform. A steady flow is one in which the velocity, pressure and cross-section of the stream may vary from point to point but do not change with time. If, at a given point, conditions do change with time, the flow is described as unsteady. In practice, there will always be slight variations of velocity and pressure, but, if the average values are constant, the flow is considered to be steady. There are, therefore, four possible types of flow:
1.
2.
3.
4.
Steady uniform flow. Conditions do not change with position or time. The velocity and cross-sectional area of the stream of fluid are the same at each cross-section: for example, flow of a liquid through a pipe of uniform bore running completely full at constant velocity. Steady non-uniform flow. Conditions change from point to point but not with time. The velocity and cross-sectional area of the stream may vary from cross-section to cross-section, but, for each cross-section, they will not vary with time: for example, flow of a liquid at a constant rate through a tapering pipe running completely full. Unsteady uniform flow. At a given instant of time the velocity at every point is the same, but this velocity will change with time: for example, accelerating flow of a liquid through a pipe of uniform bore running full, such as would occur when a pump is started up. Unsteady non-uniform flow. The cross-sectional area and velocity vary from point to point and also change with time: for example, a wave travelling along a channel.
4.3
FRAMES OF REFERENCE
Whether a given flow is described as steady or unsteady will depend upon the situation of the observer, since motion is relative and can only be described in terms of some frame of reference which is determined by the observer. If a wave travels along a channel, then to an observer on the bank the flow in the channel will appear to vary with time, and, therefore, be unsteady. If, however, the observer were travelling on the crest of the wave, conditions would not appear to the observer to change with time, and the flow would be steady according to the observer’s frame of reference.
FM5_C04.fm Page 94 Tuesday, September 20, 2005 1:27 PM
94
Chapter 4
Motion of Fluid Particles and Streams
The frame of reference adopted for describing the motion of a fluid is usually a set of fixed coordinate axes, but the analysis of steady flow is usually simpler than that of unsteady flow and it is sometimes useful to use moving coordinate axes to convert an unsteady flow problem to a steady flow problem. The normal laws of mechanics will still apply, provided that the movement of the coordinate axes takes place with uniform velocity in a straight line.
4.4
REAL AND IDEAL FLUIDS
When a real fluid flows past a boundary, the fluid immediately in contact with the boundary will have the same velocity as the boundary. As explained in Section 4.1, the velocity of successive layers of fluid will increase as we move away from the boundary. If the stream of fluid is imagined to be of infinite width perpendicular to the boundary, a point will be reached beyond which the velocity will approximate to the free stream velocity, and the drag exerted by the boundary will have no effect. The part of the flow adjoining the boundary in which this change of velocity occurs is known as the boundary layer. In this region, shear stresses are developed between layers of fluid moving with different velocities as a result of viscosity and the interchange of momentum due to turbulence causing particles of fluid to move from one layer to another. The thickness of the boundary layer is defined as the distance from the boundary at which the velocity becomes equal to 99 per cent of the free stream velocity. Outside this boundary layer, in a real fluid, the effect of the shear stresses due to the boundary can be ignored and the fluid can be treated as if it were an ideal fluid, which is assumed to have no viscosity and in which there are no shear stresses. If the fluid velocity is high and its velocity low, the boundary layer is comparatively thin, and the assumption that a real fluid can be treated as an ideal fluid greatly simplifies the analysis of the flow and still leads to useful results. Even in problems in which the effects of viscosity and turbulence cannot be neglected, it is often convenient to carry out the mathematical analysis assuming an ideal fluid. An experimental investigation can then be made to correct the theoretical analysis for the factors omitted and to bring the results obtained into agreement with the behaviour of a real fluid.
4.5
COMPRESSIBLE AND INCOMPRESSIBLE FLOW
All fluids are compressible, so that their density will change with pressure, but, under steady flow conditions and provided that the changes of density are small, it is often possible to simplify the analysis of a problem by assuming that the fluid is incompressible and of constant density. Since liquids are relatively difficult to compress, it is usual to treat them as if they were incompressible for all cases of steady flow. However, in unsteady flow conditions, high pressure differences can develop (see Chapter 20) and the compressibility of liquids must be taken into account. Gases are easily compressed and, except when changes of pressure and, therefore, density are very small, the effects of compressibility and changes of internal energy must be taken into account.
FM5_C04.fm Page 95 Tuesday, September 20, 2005 1:27 PM
4.6
4.6
One-, two- and three-dimensional flow
95
ONE-, TWO- AND THREE-DIMENSIONAL FLOW
Although, in general, all fluid flow occurs in three dimensions, so that velocity, pressure and other factors vary with reference to three orthogonal axes, in some problems the major changes occur in two directions or even in only one direction. Changes along the other axis or axes can, in such cases, be ignored without introducing major errors, thus simplifying the analysis. Flow is described as one-dimensional if the factors, or parameters, such as velocity, pressure and elevation, describing the flow at a given instant, vary only along the direction of flow and not across the cross-section at any point. If the flow is unsteady, these parameters may vary with time. The one dimension is taken as the distance along the central streamline of the flow, even though this may be a curve in space, and the values of velocity, pressure and elevation at each point along this streamline will be the average values across a section normal to the streamline. A one-dimensional treatment can be applied, for example, to the flow through a pipe, but, since in a real fluid the velocity at any cross-section will vary from zero at the pipe wall (Fig. 4.2) to a maximum at the centre, some correction will be necessary to compensate for this (see Chapter 10) if a high degree of accuracy is required. In two-dimensional flow it is assumed that the flow parameters may vary in the direction of flow and in one direction at right angles, so that the streamlines are curves lying in a plane and are identical in all planes parallel to this plane. Thus, the flow over a weir of constant cross-section (Fig. 4.3) and infinite width perpendicular to the plane of the diagram can be treated as two-dimensional. A real weir has a limited width, but it can be treated as two-dimensional over its whole width and then an end correction can be introduced to modify the result to allow for the effect of the disturbance produced by the end walls (see Chapter 16).
FIGURE 4.2 Velocity profiles for one-dimensional flow
FIGURE 4.3 Two-dimensional flow
FM5_C04.fm Page 96 Tuesday, September 20, 2005 1:27 PM
96
Chapter 4
Motion of Fluid Particles and Streams
A special case of two-dimensional flow occurs when the cross-section of the flow is circular and the flow parameters vary symmetrically about the axis. For example, ideally the velocity distribution in a circular pipe will be the same across any diameter, the velocity varying from zero at the wall to a maximum at the centre. Referred to orthogonal coordinate axes (x in the direction of motion, y and z in the plane of the cross-section) the flow is three-dimensional, but, since it is axisymmetric, it can be reduced to two-dimensional flow by using a system of cylindrical coordinates (x in the direction of flow and r the radius defining the position in the cross-section).
4.7
ANALYZING FLUID FLOW
One difficulty encountered in deciding how to investigate the flow of a fluid is that, in the majority of problems, we are dealing with an endless stream of fluid. We have to decide what part of this stream shall constitute the element or system to be studied and what shall be regarded as the surroundings which act upon this system. There are two main alternatives: 1.
We can study the behaviour of a specific element of the fluid of fixed mass. Such an element constitutes a closed system. Its boundaries are a closed surface which may vary with time, but always contain the same mass of fluid. At any instant, a free body diagram can be drawn showing the forces exerted by the surrounding fluid and any solid boundaries on this element. We can define the system to be studied as a fixed region in space, or in relation to some frame of reference, known as a control volume, through which the fluid flows, forming, in effect, an open system. The boundary of this system is its control surface and its shape does not change with time. The control volume for a particular problem is chosen arbitrarily for reasons of convenience of analysis. However, the control surface will usually follow solid boundaries where these are present, and where it cuts the flow direction it will do so at right angles. Where there are no solid boundaries the control volume may form a streamtube.
2.
4.8
MOTION OF A FLUID PARTICLE
Any particle or element of fluid will obey the normal laws of mechanics in the same way as a solid body. When a force is applied, its behaviour can be predicted from Newton’s laws, which state:
1. 2. 3.
A body will remain at rest or in a state of uniform motion in a straight line until acted upon by an external force. The rate of change of momentum of a body is proportional to the force applied and takes place in the direction of action of that force. Action and reaction are equal and opposite.
FM5_C04.fm Page 97 Tuesday, September 20, 2005 1:27 PM
4.8
Motion of a fluid particle
97
Since momentum is the product of mass and velocity, for an element of fixed mass Newton’s second law relates the change of velocity occurring in a given time (i.e. the acceleration) to the applied force. Working in a coherent system of units, such as SI, the proportionality becomes an equality and Newton’s second law can be written Change of velocity Force = Mass × −−−−−−−−−−−−−−−−−−−−−−−−−−−− Time = Mass × Acceleration. The relationships between the acceleration a, initial velocity v1, final velocity v2 and the distance moved s in time t are given by the equations of motion: v2 = v1 + at, s = v1t + −12 at2, v 22 = v 21 + 2as. In any body of flowing fluid, the velocity at a given instant will generally vary from point to point over any specified region, and if the flow is unsteady the velocity at each point may vary with time. In this field of flow, at any given time, a particle at point A will have a different velocity from that of a particle at point B. The velocities at A and B may also change with time. Thus the change of velocity δ v, which occurs when a particle moves from A to B through a distance δ s in time δ t, is given by Difference of Change of Total change velocity between velocity at = + of velocity A and B at the B occurring given instant in time δ t.
(4.1)
The velocity v depends on both distance s and time t. The rate of change of velocity with position at a given time is, therefore, expressed by the partial differential ∂ v∂ s, and the rate of change of velocity with time at a given point is expressed by the partial differential ∂ v∂ t. Since A and B are a distance δs apart,
∂v Difference of velocity between A and B at the given instant = −−− (δ s ). ∂s Also,
∂v Change of velocity at B in time t = −−− (δ t ). ∂t Thus, in symbols, equation (4.1) is
∂v ∂v dv = −−− (δ s ) + −−− (δ t ). ∂s ∂t
(4.2)
FM5_C04.fm Page 98 Tuesday, September 20, 2005 1:27 PM
98
Chapter 4
Motion of Fluid Particles and Streams
4.9
ACCELERATION OF A FLUID PARTICLE
The forces acting on a particle are related to the resultant acceleration δ vδ t of the particle by Newton’s second law. From equation (4.2) in the limit at δ t → 0, dv ∂ v ds ∂ v Acceleration in the direction of flow, a = −−− = −−− −−− + −−−. dt ∂ s dt ∂ t To denote that the derivative dvdt is obtained by following the motion of a single particle, it is written DvDt, and since dsdt = v, Dv ∂v ∂v a = −−−− = v −−− + −−−. Dt ∂s ∂t
(4.3)
The derivative DDt is known as the substantive derivative. The total acceleration, known as the substantive acceleration, is composed of two parts, as shown in equation (4.3): 1.
2.
the convective acceleration v(∂ v∂s) due to the movement of the particle from one point to another point at which the velocity at the given instant is different; the local or temporal acceleration ∂ v∂ t due to the change of velocity at every point with time.
For steady flow, ∂ v∂t = 0, while for uniform flow, ∂ v∂s = 0. We have so far assumed that the particle is accelerating in a straight line, but, if it is moving in a curved path, its velocity will be changing in direction and consequently there will be an acceleration perpendicular to its path, whether the velocity v is changing in magnitude or not. Figure 4.4 shows a particle moving from A to B along a curved path of length δ s subtending a small angle δθ at the centre of curvature. The change of velocity δ vn will be perpendicular to the direction of motion and, from the velocity diagram,
δ vn = vδθ = vδ sR.
FIGURE 4.4 Change of velocity for a circular path
FM5_C04.fm Page 99 Tuesday, September 20, 2005 1:27 PM
4.9
Acceleration of a fluid particle
99
Dividing by δ t, the time in which the change occurs, in the limit the acceleration perpendicular to the direction of motion is dv v ds a n = −−−−n = −− −−− dt R dt or, since ds −−− = v, dt an = v 2R. This is the convective term, and, if v has a component vn towards the instantaneous centre of curvature, there will be a temporal term ∂ vn ∂ t so that the substantive derivative is v2 ∂ vn a n = −− + −−−−. R ∂t In general, the motion of a fluid particle is three-dimensional and its velocity and acceleration can be expressed in terms of three mutually perpendicular components. Thus, if vx, vy and vz are the components of the velocity in the x, y and z directions, respectively, and ax, ay and az the corresponding components of acceleration, the velocity field is described by vx = vx(x, y, z, t), vy = vy(x, y, z, t), vz = vz(x, y, z, t), and the velocity v at any point is given by v = vxi + vyj + vzk, where i, j and k are the unit vectors in the x, y and z directions. The change of the component velocities in each direction as a particle moves in a fluid can now be calculated. Thus, in the x direction,
∂v ∂v ∂v ∂v δ v x = −−−−x (δ x ) + −−−−x (δ y ) + −−−−x (δ z ) + −−−−x (δ t ), ∂x ∂y ∂z ∂t and the acceleration in the x direction, in the limit as δ t → 0, will be
∂ v dx ∂ v dy ∂ v dz ∂ v Dv a x = −−−−−x = −−−−x −−− + −−−−x −−− + −−−−x −−− + −−−−x Dt ∂ x dt ∂ y dt ∂ z dt ∂ t or, since dxdt = vx, dydt = vy , dzdt = vz,
∂v ∂v ∂v ∂ vx Dv a x = −−−−−x = v x −−−−x + v y −−−−x + v z −−−−x + −−−−. Dt ∂x ∂y ∂z ∂t
(4.4)
Similarly,
∂v ∂v ∂v ∂v Dv a y = −−−−−y = v x −−−−y + vy −−−−y + vz −−−−y + −−−−y , Dt ∂x ∂y ∂z ∂t
(4.5)
∂v ∂v ∂ v ∂ vz Dv a z = −−−−−z = v x −−−−z + vy −−−−z + vz −−−−z + −−−−. Dt ∂x ∂y ∂z ∂t
(4.6)
FM5_C04.fm Page 100 Tuesday, September 20, 2005 1:27 PM
100
Chapter 4
Motion of Fluid Particles and Streams
The first three terms in each of equations (4.4) to (4.6) represent the convective acceleration and the final term the local or temporal acceleration.
4.10 LAMINAR AND TURBULENT FLOW Observation shows that two entirely different types of fluid flow exist. This was demonstrated by Osborne Reynolds in 1883 through an experiment in which water was discharged from a tank through a glass tube (Fig. 4.5). The rate of flow could be controlled by a valve at the outlet, and a fine filament of dye injected at the entrance to the tube. At low velocities, it was found that the dye filament remained intact throughout the length of the tube, showing that the particles of water moved in parallel lines. This type of flow is known as laminar, viscous or streamline, the particles of fluid moving in an orderly manner and retaining the same relative positions in successive cross-sections.
FIGURE 4.5 Reynolds’ apparatus
As the velocity in the tube was increased by opening the outlet valve, a point was eventually reached at which the dye filament at first began to oscillate and then broke up so that the colour was diffused over the whole cross-section, showing that the particles of fluid no longer moved in an orderly manner but occupied different relative positions in successive cross-sections. This type of flow is known as turbulent and is characterized by continuous small fluctuations in the magnitude and direction of the velocity of the fluid particles, which are accompanied by corresponding small fluctuations of pressure. When the motion of a fluid particle in a stream is disturbed, its inertia will tend to carry it on in the new direction, but the viscous forces due to the surrounding fluid will tend to make it conform to the motion of the rest of the stream. In viscous flow, the viscous shear stresses are sufficient to eliminate the effects of any deviation, but in turbulent flow they are inadequate. The criterion which determines whether flow will be viscous or turbulent is therefore the ratio of the inertial force to the viscous force acting on the particle. Suppose l is a characteristic length in the system under consideration, e.g. the diameter of a pipe or the chord of an aerofoil, and t is a typical time; then lengths, areas, velocities and accelerations can all be expressed in terms of l and t. For a small element of fluid of mass density ρ,
FM5_C04.fm Page 101 Tuesday, September 20, 2005 1:27 PM
4.10
Laminar and turbulent flow
101
Volume of element = k1l 3, Mass of element = k1ρl 3, Velocity of element, v = k2lt, Acceleration of element = k3lt2, where k1, k2 and k3 are constants. By Newton’s second law, Inertial force = Mass × Acceleration = k1ρl 3 × k3lt2 = k1k3ρl 2(lt)2 = (k1k3 k 22 )ρl 2v 2. Similarly, Viscous force = Viscous shear stress × Area on which stress acts. From Newton’s law of viscosity (equation (1.2)), Viscous shear stress = µ × Velocity gradient = µ(vk4l), where µ = coefficient of dynamic viscosity. Area on which shear stress acts = k5l 2. Therefore, Viscous force = µ(vk4l) × k5l 2 = (k5 k4)µ vl. The ratio k k k ρ l 2 v2 Inertial force ρ vl −−−−−−−−−−−−−−−−−−−− = −−1−−2−3−−−4 −−−−−−− = constant × −−−− . Viscous force µ k 2 k5 µ vl Thus, the criterion which determines whether flow is viscous or turbulent is the quantity ρ vlµ, known as the Reynolds number. It is a ratio of forces and, therefore, a pure number and may also be written as vlν, where ν is the kinematic viscosity (ν = µρ). Experiments carried out with a number of different fluids in straight pipes of different diameters have established that if the Reynolds number is calculated by making l equal to the pipe diameter and using the mean velocity C (Section 4.11), then, below a critical value of ρ Cdµ = 2000, flow will normally be laminar (viscous), any tendency to turbulence being damped out by viscous friction. This value of the Reynolds number applies only to flow in pipes, but critical values of the Reynolds number can be established for other types of flow, choosing a suitable characteristic length such as the chord of an aerofoil in place of the pipe diameter. For a given fluid flowing in a pipe of a given diameter, there will be a critical velocity of flow Cc corresponding to the critical value of the Reynolds number, below which flow will be viscous.
FM5_C04.fm Page 102 Tuesday, September 20, 2005 1:27 PM
102
Chapter 4
Motion of Fluid Particles and Streams
In pipes, at values of the Reynolds number above 2000, flow will not necessarily be turbulent. Laminar flow has been maintained up to Re = 50 000, but conditions are unstable and any disturbance will cause reversion to normal turbulent flow. In straight pipes of constant diameter, flow can be assumed to be turbulent if the Reynolds number exceeds 4000.
4.11 DISCHARGE AND MEAN VELOCITY The total quantity of fluid flowing in unit time past any particular cross-section of a stream is called the discharge or flow at that section. It can be measured either in terms of mass, in which case it is referred to as the mass rate of flow A and measured in units such as kilograms per second, or in terms of volume, when it is known as the volume rate of flow Q, measured in units such as cubic metres per second. In an ideal fluid, in which there is no friction, the velocity u of the fluid would be the same at every point of the cross-section (Fig. 4.2). In unit time, a prism of fluid would pass the given cross-section and, if the cross-sectional area normal to the direction of flow is A, the volume passing would be Au. Thus Q = Au. In a real fluid, the velocity adjacent to a solid boundary will be zero or, more accurately, equal to the wall velocity in the flow direction, a condition known as ‘no slip’, which will be true as long as the flow does not separate from the wall. For a pipe, the velocity profile would be as shown in Fig. 4.6(a) for laminar flow and Fig. 4.6(b) for turbulent flow.
FIGURE 4.6 Calculation of discharge for a circular section, note ‘no slip’ at wall
If u is the velocity at any radius r, the flow δ Q through an annular element of radius r and thickness δ r will be
δQ = Area of element × Velocity = 2π rδ r × u, and, hence,
ur dr. R
Q = 2π
(4.7)
FM5_C04.fm Page 103 Tuesday, September 20, 2005 1:27 PM
4.11
Discharge and mean velocity
103
If the relation between u and r can be established, this integral can be evaluated or the integration may be undertaken numerically, see Section 6.8, program VOLFLO. In many problems, the variation of velocity over the cross-section can be ignored, the velocity being assumed to be constant and equal to the mean velocity B, defined as volume rate of discharge Q divided by the area of cross-section A normal to the stream: Mean velocity, B = QA.
EXAMPLE 4.1
Air flows between two parallel plates 80 mm apart. The following velocities were determined by direct measurement. Distance from one plate (mm) Velocity (m s−1)
0 10 20 30 40 50 60 70 80 0 23 28 31 32 29 22 14 0
Plot the velocity distribution curve and calculate the mean velocity.
Solution Figure 4.7 shows the velocity distribution curve. The area enclosed by the curve represents the product of velocity and distance, and since the two plates are parallel Discharge per unit width Mean velocity, B = −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−. Distance between plates FIGURE 4.7 Velocity distribution curve
The area under the graph may be determined by the mid-ordinate method, taking values from Fig. 4.7, B = (∑Mid-ordinates8) = (17.5 + 26.0 + 29.6 + 31.9 + 30.7 + 25.4 + 18.1 + 7.7)8 = 23.36 m s−1.
FM5_C04.fm Page 104 Tuesday, September 20, 2005 1:27 PM
104
Chapter 4
Motion of Fluid Particles and Streams
4.12 CONTINUITY OF FLOW Except in nuclear processes, matter is neither created nor destroyed. This principle of conservation of mass can be applied to a flowing fluid. Considering any fixed region in the flow (Fig. 4.8) constituting a control volume, Increase of mass Mass of fluid Mass of fluid entering of fluid in the = leaving per unit + per unit time control volume time per unit time. FIGURE 4.8 Continuity of flow
For steady flow, the mass of fluid in the control volume remains constant and the relation reduces to Mass of fluid entering Mass of fluid leaving = per unit time per unit time. Applying this principle to steady flow in a streamtube (Fig. 4.9) having a crosssectional area small enough for the velocity to be considered as constant over any given cross-section, for the region between sections 1 and 2, since there can be no flow through the walls of a streamtube, Mass entering per unit time = Mass leaving per unit time at section 1 at section 2. Suppose that at section 1 the area of the streamtube is δA1, the velocity of the fluid u1 and its density ρ1, while at section 2 the corresponding values are δA2, u2 and ρ2; then Mass entering per unit time at 1 = ρ1δA1u1, Mass leaving per unit time at 2 = ρ2δA2u2. FIGURE 4.9 Continuous flow through a streamtube
FM5_C04.fm Page 105 Tuesday, September 20, 2005 1:27 PM
4.12
Continuity of flow
105
Then, for steady flow,
ρ1δA1u1 = ρ2δA2u2 = Constant.
(4.8)
This is the equation of continuity for the flow of a compressible fluid through a streamtube, u1 and u2 being the velocities measured at right angles to the crosssectional areas δA1 and δA2. For the flow of a real fluid through a pipe or other conduit, the velocity will vary from wall to wall. However, using the mean velocity B, the equation of continuity for steady flow can be written as
ρ1A1B1 = ρ2A2B2 = A,
(4.9)
where A1 and A2 are the total cross-sectional areas and A is the mass rate of flow. If the fluid can be considered as incompressible, so that ρ1 = ρ2, equation (4.9) reduces to A1B1 = A2B2 = Q.
(4.10)
The continuity of flow equation is one of the major tools of fluid mechanics, providing a means of calculating velocities at different points in a system. The continuity equation can also be applied to determine the relation between the flows into and out of a junction. In Fig. 4.10, for steady conditions, Total inflow to junction = Total outflow from junction,
ρ1Q1 = ρ2Q2 + ρ3Q3. FIGURE 4.10 Applications of the continuity equation
For an incompressible fluid, ρ 1 = ρ 2 = ρ 3 so that Q1 = Q2 + Q3 or
A1C1 = A2C2 + A3C3.
In general, if we consider flow towards the junction as positive and flow away from the junction as negative, then for steady flow at any junction the algebraic sum of all the mass flows must be zero: ∑ ρQ = 0.
FM5_C04.fm Page 106 Tuesday, September 20, 2005 1:27 PM
106
Chapter 4
Motion of Fluid Particles and Streams
EXAMPLE 4.2
Water flows from A to D and E through the series pipeline shown in Fig. 4.11. Given the pipe diameters, velocities and flow rates below, complete the tabular data for this system.
FIGURE 4.11 Relations between discharge, diameter and velocity
Pipe AB BC CD DE
Diameter (mm) d1 = 50 d2 = 75 d3 = ? d4 = 30
Flow rate (m3 s−1)
Velocity (m s−1)
Q1 = ? Q2 = ? Q3 = 2Q4 Q4 = 0.5Q3
C1 = ? C2 = 2.0 C3 = 1.5 C4 = ?
Solution Adding area A = (227)d 24 to the data table and noting that Q = AC and that Q1 = Q2 = (Q 3 + Q 4) = 1.5Q3 allows the table to be completed as (additions in bold),
Diameter (mm)
Area (m2)
Flow rate (m3 s−1)
Velocity (m s−1)
d 1 = 50
1.9643 × 10−3
Q1 = Q2 = 8.839 × 10−3
C1 = C2 A2A1 = 2.0 × 4.41961.9643 = 4.27
d 2 = 75
4.4196 × 10−3
Q2 = 2.0 × 4.4196 × 10−3 = 8.839 × 10−3
C2 = 2.0
Q3 = Q21.5 = 5.893 × 10−3
C3 = 1.5
Q4 = 0.5Q3 = 0.5 × 5.893 × 10−3 = 2.947 × 10−3
C4 = Q 4A4 = 2.9470.7071 = 4.17
d 3 = [Q3(v3 π4)] 0.5 = (5.893 × 10 −31.5 × 0.786)0.5 = 0.707 d 4 = 30
0.707 × 10 −3
(The calculation route is as follows: calculate areas where possible and then Q2 and hence Q1 and C1. From Q2 calculate Q 3 and Q 4 and hence C4. Calculate d 3 from Q 3 and C3.)
FM5_C04.fm Page 107 Tuesday, September 20, 2005 1:27 PM
4.13
Continuity equations for three-dimensional flow using Cartesian coordinates
107
4.13 CONTINUITY EQUATIONS FOR THREE-DIMENSIONAL FLOW USING CARTESIAN COORDINATES The control volume ABCDEFGH in Fig. 4.12 is taken in the form of a small rectangular prism with sides δx, δy and δ z in the x, y and z directions, respectively. The mean values of the component velocities in these directions are vx, vy and vz. Considering flow in the x direction, Mass inflow through ABCD in unit time = ρ vxδyδz.
FIGURE 4.12 Continuity in three dimensions
In the general case, both mass density ρ and velocity vx will change in the x direction and so
∂ Mass outflow through EFGH in unit time = ρ vx + −−− ( ρ v x ) δ x δ y δ z. ∂x Thus,
∂ Net outflow in unit time in x direction = −−− ( ρ vx ) δ x δ y δ z. ∂x Similarly,
∂ Net outflow in unit time in y direction = −−− ( ρ vy ) δ x δ y δ z, ∂y ∂ Net outflow in unit time in z direction = −−− ( ρ vz ) δ x δ y δ z. ∂z Therefore,
∂ ∂ ∂ Total net outflow in unit time = −−− ( ρ vx ) + −−− ( ρ vy ) + −−− ( ρ vz ) δ x δ y δ z. ∂x ∂y ∂z
FM5_C04.fm Page 108 Tuesday, September 20, 2005 1:27 PM
108
Chapter 4
Motion of Fluid Particles and Streams
Also, since ∂ρ∂t is the change in mass density per unit time,
∂ρ Change of mass in control volume in unit time = − −−− δ x δ y δ z ∂t (the negative sign indicating that a net outflow has been assumed). Then, Total net outflow in unit time = change of mass in control volume in unit time
∂ ∂ ∂ ∂ρ −−− ( ρ vx ) + −−− ( ρ vy ) + −−− ( ρ vz ) δ x δ y δ z = – −−− δ x δ y δ z ∂x ∂y ∂z ∂t or
∂ ∂ ∂ ∂ρ −−− ( ρ v x ) + −−− ( ρ v y ) + −−− ( ρ v z ) = – −−−. ∂x ∂y ∂z ∂t
(4.11)
Equation (4.11) holds for every point in a fluid flow whether steady or unsteady, compressible or incompressible. However, for incompressible flow, the density ρ is constant and the equation simplifies to
∂ vx ∂ vy ∂ vz −−−− + −−−− + −−−− = 0. ∂x ∂y ∂ z
(4.12)
For two-dimensional incompressible flow this will simplify still further to
∂ vx ∂ vy −−−− + −−−− = 0. ∂x ∂y
EXAMPLE 4.3
(4.13)
The velocity distribution for the flow of an incompressible fluid is given by vx = 3 − x, vy = 4 + 2y, vz = 2 − z. Show that this satisfies the requirements of the continuity equation.
Solution For three-dimensional flow of an incompressible fluid, the continuity equation simplifies to equation (4.12):
∂ vx −−−− = –1, ∂x
∂v −−−−y = +2, ∂y
∂ vz −−−− = – 1, ∂z
and, hence,
∂ vx ∂ vy ∂ vz −−−− + −−−− + −−−− = – 1 + 2 – 1 = 0, ∂x ∂ y ∂ z which satisfies the requirement for continuity.
FM5_C04.fm Page 109 Tuesday, September 20, 2005 1:27 PM
Concluding remarks
109
4.14 CONTINUITY EQUATION FOR CYLINDRICAL COORDINATES The form of the continuity equation for a system of cylindrical coordinates r, θ and z, in which r and θ are measured in a plane corresponding to the x−y plane for Cartesian coordinates, can be found by using the relations between polar and Cartesian coordinates: x 2 + y 2 = r 2, ( yx) = tan θ, vx = vr cos θ − vθ sin θ, vy = vr sin θ + vθ cos θ,
∂ ∂ ∂ r ∂ ∂θ −−− = −−− −−− + −−− −−−, ∂x ∂ r ∂x ∂θ ∂x
∂ ∂ ∂ r ∂ ∂θ −−− = −−− −−− + −−− −−− . ∂ y ∂ r ∂y ∂θ ∂y
This results in equation (4.12) becoming 1 ∂ v ∂v 1 ∂ − −−− ( rv r ) + − −−−−θ + −−−−z = 0. r ∂θ ∂ z r ∂r
(4.14)
In the case of two-dimensional flow, this can be simplified further. Putting ∂ vz ∂ z = 0 and writing
∂v ∂ −−− ( rv ) = ⎛ r −−−−r + v r⎞ , ⎝ ∂r ⎠ ∂r r equation (4.14) becomes v r ∂ v r 1 ∂ vθ −− + −−−− + − −−−− = 0. r ∂ r r ∂θ
Concluding remarks This chapter has provided the frame of reference for much of the later material in this text. The classification of flows based on time or distance dependence, together with the influence of viscosity via Reynolds number, is fundamental. In defining the flow into laminar and turbulent regimes much of the observed fluid flow behaviour becomes understandable. The presence of velocity profiles across a fluid stream between boundaries is similarly fundamental as it heralds the later work on the development of the boundary layer and the recognition of the condition of ‘no slip’ at fluidsurface interfaces. The introduction of the continuity equation, whether in its volumetric or its more widely applicable mass flow form, provides one of the recurring tools for fluid flow analysis, which, when the storage terms are introduced, will find application
FM5_C04.fm Page 110 Tuesday, September 20, 2005 1:27 PM
110
Chapter 4
Motion of Fluid Particles and Streams
throughout this text for both compressible and incompressible flows under steady or unsteady conditions.
Summary of important equations and concepts 1.
2.
3.
4.
Chapter 4 introduces definitions of flow conditions, such as steady and unsteady, relating these to changes in flow condition, Section 4.2, together with the concept of the boundary layer, Section 4.4, which will become central to later chapters. Movement in more than one dimension is introduced, with examples, emphasizing the simplifications possible if flow can be considered as one-dimensional, Sections 4.6 to 4.9. The classification of flows into laminar and turbulent regimes, following Reynolds, is an essential concept and one that will be returned to continuously in later chapters. It is important to recognize that the ratio of forces represented by the Reynolds number applies to a whole range of flow geometries, not restricted to pipeflow, Section 4.10. The concept of continuity of mass flow is established and shown in the special case of incompressible flow to reduce to volumetric flow continuity that allows mean velocities to be calculated in series and parallel pipe networks, equations (4.9) and (4.10).
Problems 4.1 The velocity of a fluid varies with time t. Over the period from t = 0 to t = 8 s the velocity components are u = 0 m s−1 and v = 2 m s−1, while from t = 8 s to t = 16 s the components are u = 2 m s−1 and v = −2 m s−1. A dye streak is injected into the flow at a certain point commencing at time t = 0 and the path of a particle of fluid is also traced from that point starting at t = 0. Draw to scale the streakline, the pathline of the particle and the streamlines at time t = 12 s. 4.2 The velocity distribution for a two-dimensional field of flow is given by 2 u = −−−−−− m s –1 3+t
and
t2 v = 2 – −−− m s –1 . 32
For the period of time from t = 0 to t = 12 s draw a streakline for an injection of dye through a certain point A and a pathline for a particle of fluid which was at A when t = 0. Draw also the streamlines for t = 6 s and t = 12 s. 4.3 A nozzle is formed so that its cross-sectional area converges linearly along its length. The inside diameters are 75 mm and 25 mm at inlet and exit and the length of the nozzle is 300 mm. What is the convective acceleration at a section halfway along the length of the nozzle if the discharge is constant at 0.014 m3 s−1? [337.94 m s−2]
4.4 During a wind tunnel test on a sphere of radius r = 150 mm it is found that the velocity of flow u along the longitudinal axis of the tunnel passing through the centre of the sphere at a point upstream which is a distance x from the centre of the sphere is given by r3 u = U 0 ⎛ 1 – −−3⎞ ⎝ x⎠ where U0 is the mean velocity of the undisturbed airstream. If U0 = 60 m s−1 what is the convective acceleration when the distance x is (a) 300 mm, (b) 150 mm? [(a) 3937.5 m s−2, (b) 0] 4.5 The velocity along the centreline of a nozzle of length L is given by x u = 2t ⎛ 1 – 0.5 −−⎞ ⎝ L⎠
2
where u is the velocity in metres per second, t is the time in seconds from the commencement of flow, x is the distance from the inlet to the nozzle. Find the convective acceleration and the local acceleration when t = 3 s, x = --12- L and L = 0.8 m. [18.99 m s−2, 1.125 m s−2] 4.6 Water flows through a pipe 25 mm in diameter at a velocity of 6 m s−1. Determine whether the flow will be laminar or turbulent assuming that the dynamic viscosity
FM5_C04.fm Page 111 Tuesday, September 20, 2005 1:27 PM
Problems
111
of water is 1.30 × 10−3 kg m−1 s−1 and its density 1000 kg m−3. If oil of specific gravity 0.9 and dynamic viscosity 9.6 × 10−2 kg m−1 s−1 is pumped through the same pipe, what type of flow will occur? [ Turbulent, Re = 115 385; laminar, Re = 1406]
4.13 During a test on a circular duct 2 m in diameter it was found that the fluid velocity was zero at the duct surface and 6 m s−1 on the axis of the duct when the flow rate was 9 m3 s−1. Assuming the velocity distribution to be given by
4.7 An air duct is of rectangular cross-section 300 mm wide by 450 mm deep. Determine the mean velocity in the duct when the rate of flow is 0.42 m3 s−1. If the duct tapers to a cross-section 150 mm wide by 400 mm deep, what will be the mean velocity in the reduced section assuming that the density remains unchanged? [3.11 m s−1, 7.0 m s−1]
u = c1 − c2 r n,
4.8 A sphere of diameter 300 mm falls axially down a 305 mm diameter vertical cylinder which is closed at its lower end and contains water. If the sphere falls at a speed of 150 mm s−1, what is the mean velocity relative to the cylinder wall of the water in the gap surrounding the midsection of the sphere? [4.46 m s−1] 4.9 The air entering a compressor has a density of 1.2 kg m−3 and a velocity of 5 m s−1, the area of the intake being 20 cm2. Calculate the mass flow rate. If air leaves the compressor through a 25 mm diameter pipe with a velocity of 4 m s−1, what will be its density? [12 × 10−3 kg s−1, 6.11 kg m−3] 4.10 Water flows through a pipe AB 1.2 m in diameter at 3 m s−1 and then passes through a pipe BC which is 1.5 m in diameter. At C the pipe forks. Branch CD is 0.8 m in diameter and carries one-third of the flow in AB. The velocity in branch CE is 2.5 m s−1. Find (a) the volume rate of flow in AB, (b) the velocity in BC, (c) the velocity in CD, (d ) the diameter of CE. [ (a) 3.393 m3 s−1, (b) 1.92 m s−1, (c) 2.25 m s−1, (d ) 1.073 m] 4.11 A closed tank of fixed volume is used for the continuous mixing of two liquids which enter at A and B and are discharged completely mixed at C. The diameter of the inlet pipe at A is 150 mm and the liquid flows in at the rate of 56 dm3 s−1 and has a specific gravity of 0.93. At B the inlet pipe is of 100 mm diameter, the flow rate is 30 dm3 s−1 and the liquid has a specific gravity of 0.87. If the diameter of the outlet pipe at C is 175 mm, what will be the mass flow rate, velocity and specific gravity of the mixture discharged? [78.18 kg s−1, 3.58 m s−1, 0.909] 4.12 In a 0.6 m diameter duct carrying air the velocity profile was found to obey the law u = −5r 2 + 0.45 m s−1 where u is the velocity at radius r. Calculate the volume rate of flow of the air and the mean velocity. [0.0636 m3 s−1, 0.225 m s−1]
where u is the fluid velocity at any radius r, determine the values of the constants c1, c2 and n, specifying the units of c1 and c2. Evaluate the mean velocity and determine the radial position at which a Pitot tube must be placed to measure this mean velocity. [6 m s−1, 6 m −0.825 s−1, 1.825; 2.86 m s−1, 0.701 m] 4.14 Air flows through a rectangular duct which is 30 cm wide by 20 cm deep in cross-section. To determine the volume rate of flow experimentally the cross-section is divided into a number of imaginary rectangular elements of equal area and the velocity measured at the centre of each element with the following results:
Distance from bottom of duct (cm)
Distance from side of duct (cm) 3
9
15
21
27
18 14 10 6 2
1.6 1.9 2.1 2.0 1.8
2.0 3.4 6.8 3.5 2.0
2.2 6.9 10.0 7.0 2.3
2.0 3.7 7.0 3.8 2.1
1.7 2.0 2.3 2.1 1.9
Calculate the volume rate of flow and the mean velocity in the duct. [0.202 m3 s−1, 3.364 m s−1] 4.15 If a two-dimensional flow field were to have velocity components u = U(x 3 + xy 2) and v = U(y 3 + yx 2) would the continuity equation be satisfied?
[Yes]
4.16 Determine whether the following expressions satisfy the continuity equation: (a) u = 10xt, v = −10yt, ρ = constant
[Yes]
(b) u = U ( y δ )17 , v = 0, ρ = constant.
[Yes]
FM5_C05.fm Page 112 Wednesday, September 21, 2005 9:20 AM
Chapter 5
The Momentum Equation and its Applications 5.1 5.2
5.3 5.4 5.5 5.6 5.7 5.8 5.9
Momentum and fluid flow Momentum equation for two- and three-dimensional flow along a streamline Momentum correction factor Gradual acceleration of a fluid in a pipeline neglecting elasticity Force exerted by a jet striking a flat plate Force due to the deflection of a jet by a curved vane Force exerted when a jet is deflected by a moving curved vane Force exerted on pipe bends and closed conduits Reaction of a jet
5.10 Drag exerted when a fluid flows over a flat plate 5.11 Angular motion 5.12 Euler’s equation of motion along a streamline 5.13 Pressure waves and the velocity of sound in a fluid 5.14 Velocity of propagation of a small surface wave 5.15 Differential form of the continuity and momentum equations 5.16 Computational treatment of the differential forms of the continuity and momentum equations 5.17 Comparison of CFD methodologies
FM5_C05.fm Page 113 Wednesday, September 21, 2005 9:20 AM
The analysis of fluid flow phenomena fundamentally depends upon the application of Newton’s laws of motion, together with a recognition of the special properties of fluids in motion. The momentum equation relates the sum of the forces acting on a fluid element to its acceleration or rate of change of momentum in the direction of the resultant force. This relationship is, perhaps, when taken with the conservation of mass and the energy equation, the foundation upon which all fluid flow analysis is based. This chapter will introduce the application of the momentum equation to a range of fluid flow conditions, including forces exerted upon and by a fluid as a result of changes in direction and
impact upon both stationary and moving surfaces, as well as introducing the application of the momentum equation to determine engine thrust as a result of changes to fluid momentum. The application of the momentum equation to the prediction of the rate of propagation of pressure or surface wave discontinuities will be presented. By utilizing the momentum equation, together with the conservation of mass, this chapter will also introduce Euler’s equation for motion along a streamline under general conditions. Bernoulli’s equation, the special form of Euler’s equation applicable to incompressible inviscid flows, will be introduced and its application demonstrated. l l l
FM5_C05.fm Page 114 Wednesday, September 21, 2005 9:20 AM
114
Chapter 5
The Momentum Equation and its Applications
5.1
MOMENTUM AND FLUID FLOW
In mechanics, the momentum of a particle or object is defined as the product of its mass m and its velocity v: Momentum = mv. The particles of a fluid stream will possess momentum, and, whenever the velocity of the stream is changed in magnitude or direction, there will be a corresponding change in the momentum of the fluid particles. In accordance with Newton’s second law, a force is required to produce this change, which will be proportional to the rate at which the change of momentum occurs. The force may be provided by contact between the fluid and a solid boundary (e.g. the blade of a propeller or the wall of a bend in a pipe) or by one part of the fluid stream acting on another. By Newton’s third law, the fluid will exert an equal and opposite force on the solid boundary or body of fluid producing the change of velocity. Such forces are known as dynamic forces, since they arise from the motion of the fluid and are additional to the static forces (see Chapter 3) due to pressure in a fluid; they occur even when the fluid is at rest. To determine the rate of change of momentum in a fluid stream consider a control volume ABCD (Fig. 5.1). As the fluid flow is assumed to be steady and nonuniform in nature the continuity of mass flow across the control volume may be expressed as FIGURE 5.1 Momentum in a flowing fluid
ρ2A2v2 = ρ1A1v1 = A,
(5.1)
i.e. there is no storage within the control volume and A is the fluid mass flow. The rate at which momentum exits the control volume across boundary CD may be defined as
ρ2A2v2v2. Similarly the rate at which momentum enters the control volume across AB may be expressed as
ρ1A1v1v1. Thus the rate of change of momentum across the control volume may be seen to be
ρ2 A2 v2 v2 − ρ1A1v1 v1
(5.2)
FM5_C05.fm Page 115 Wednesday, September 21, 2005 9:20 AM
5.2
Momentum equation for two- and three-dimensional flow along a streamline
115
or, from the continuity of mass flow equation,
ρ1A1v1(v2 − v1) = A(v2 − v1) = Mass flow per unit time × Change of velocity.
(5.3)
Note that this is the increase of momentum per unit time in the direction of motion, and according to Newton’s second law will be caused by a force F, such that F = A(v2 − v1).
(5.4)
This is the resultant force acting on the fluid element ABCD in the direction of motion. By Newton’s third law, the fluid will exert an equal and opposite reaction on its surroundings.
MOMENTUM EQUATION FOR TWO- AND THREEDIMENSIONAL FLOW ALONG A STREAMLINE
5.2
In Section 5.1, the momentum equation (5.4) was derived for one-dimensional flow in a straight line, assuming that the incoming and outgoing velocities v1 and v2 were in the same direction. Figure 5.2 shows a two-dimensional problem in which v1 makes an angle θ with the x axis, while v2 makes a corresponding angle φ. Since both momentum and force are vector quantities, they can be resolved into components in the x and y directions and equation (5.4) applied. Thus, if Fx and Fy are the components of the resultant force on the element of fluid ABCD, FIGURE 5.2 Momentum equation for two-dimensional flow
Fx = Rate of change of momentum of fluid in x direction = Mass per unit time × Change of velocity in x direction = A(v2 cos φ − v1 cos θ ) = A(vx2 − vx1). Similarly, Fy = A(v2 sin φ − v1 sin θ ) = A(vy2 − vy1).
FM5_C05.fm Page 116 Wednesday, September 21, 2005 9:20 AM
116
Chapter 5
The Momentum Equation and its Applications
These components can be combined to give the resultant force, F = ( F 2x + F 2y ). Again, the force exerted by the fluid on the surroundings will be equal and opposite. For three-dimensional flow, the same method can be used, but the fluid will also have component velocities vz1 and vz2 in the z direction and the corresponding rate of change of momentum in this direction will require a force Fz = A(vz2 − vz1). To summarize the position, we can say, in general, that Total force exerted on Rate of change of momentum the fluid in a control in the given direction of = volume in a given the fluid passing through direction the control volume, F = A(vout − vin). The value of F is positive in the direction in which v is assumed to be positive. For any control volume, the total force F which acts upon it in a given direction will be made up of three component forces: F1 = Force exerted in the given direction on the fluid in the control volume by any solid body within the control volume or coinciding with the boundaries of the control volume. F2 = Force exerted in the given direction on the fluid in the control volume by body forces such as gravity. F3 = Force exerted in the given direction on the fluid in the control volume by the fluid outside the control volume. Thus, F = F1 + F2 + F3 = A(vout − vin).
(5.5)
The force R exerted by the fluid on the solid body inside or coinciding with the control volume in the given direction will be equal and opposite to F1 so that R = −F1.
5.3
MOMENTUM CORRECTION FACTOR
The momentum equation (5.5) is based on the assumption that the velocity is constant across any given cross-section. When a real fluid flows past a solid boundary, shear stresses are developed and the velocity is no longer uniform over the cross-section. In a pipe, for example, the velocity will vary from zero at the wall to a maximum at the centre. The momentum per unit time for the whole flow can be found by summing the momentum per unit time through each element of the cross-section, provided that
FM5_C05.fm Page 117 Wednesday, September 21, 2005 9:20 AM
5.3
Momentum correction factor
117
these are sufficiently small for the velocity perpendicular to each element to be taken as uniform. Thus, if the velocity perpendicular to the element is u and the area of the element is δA, Mass passing through element in unit time = ρδA × u, Momentum per unit time passing through element = Mass per unit time × Velocity = ρδAu × u = ρu2δA, Total momentum per unit time passing whole = cross-section
ρu dA. 2
(5.6)
To evaluate this integral, the velocity distribution must be known. If we consider turbulent flow through a pipe of radius R (Fig. 5.3), the velocity u at any distance y from the pipe wall is given approximately by Prandtl’s one-seventh power law: u = umax(yR)17,
FIGURE 5.3 Calculation of momentum correction factor
the maximum velocity, umax, occurring at the centre of the pipe. Since the velocity is constant at any radius r = R − y, it is convenient to take the element of area δA in equation (5.6) as an annulus of radius r and width δr,
δA = 2π rδ r, and, from equation (5.6), for the whole cross-section,
ρu dA R
Total momentum per unit time =
2
=
R
ρ u 2max ( yR )
27
2 πρ 2 π r dr = ⎛ −−−2−−⁄ 7 ⎞ u 2max ⎝R ⎠
y R
27
r dr.
(5.7)
Since r = R − y, dr = −dy, and so, substituting for r and dr in equation (5.7) and changing the limits (because y = 0 when r = R),
FM5_C05.fm Page 118 Wednesday, September 21, 2005 9:20 AM
118
Chapter 5
The Momentum Equation and its Applications
2 πρ u 2max Total momentum per unit time = −−−−−−27 −−−−− R 2 πρ u 2max = −−−−−−27 −−−−− R
y 0
27
( R – y ) ( – dy )
R
2 πρ u 2max ⎛ 7 167 7 97 ⎞ 0 ( y 97 – Ry 27 ) dy = −−−−−−27 −−−−− ⎝ −−−y – −Ry ⎠ 16 9 R R R 0
2 πρ u 2max 167 ⎛ 7 7 ⎞ 49 = −−−−−−27 −−−−− R − – −− = −−− πρ R 2 u 2max . ⎝9 − 16 ⎠ 72 R
(5.8)
In practice, it is usually more convenient to use the mean velocity B instead of the maximum velocity umax: Total volume per unit time passing section Mean velocity, B = -----------------------------------------------------------------------------------------------------------Total area of cross-section 1 = −−−−−2 πR
uδA. R
Putting u = umax(yR)17 and δA = 2π rδ r, 1 B = −−−−−2 πR
R
y u max ⎛ −−⎞ ⎝ R⎠ 0
17
2u max 2 π r dr = −−−− −−− R 157
y R
17
r dr.
Putting r = R − y, dr = −dy, and changing the limits, 2u max B = −−−− −−− R 157
2u max y 17 ( R – y ) ( – dy ) = −−−− −−− R 157 R
(y 0
87
– Ry 17 ) dy
R
2u max ⎛ 7 157 7 87 ⎞ 0 49 −−− −−−y – −Ry ⎠ = −−−u max , = −−−− 60 R 157 ⎝ 15 8 R 60 u max = −−−B. 49
(5.9)
Substituting from equation (5.9) in equation (5.8), 49 60 2 Total momentum per unit time = −−− πρ R 2 ⎛ −−−⎞ B 2 = 1.02ρπR2B2, ⎝ 72 49⎠
(5.10)
or, since ρπR2B = mass per unit time, Momentum per unit time = 1.02 × Mass per unit time × Mean velocity. If the momentum per unit time of the stream had been calculated from the mean velocity without considering the velocity distribution, the value obtained would have been ρπR2B2. To take the velocity distribution into account, a momentum correction factor β must be introduced, so that, for the whole stream, True momentum per unit time = β × Mass per unit time × Mean velocity. The value of β depends upon the shape of the cross-section and the velocity distribution.
FM5_C05.fm Page 119 Wednesday, September 21, 2005 9:20 AM
5.4
Gradual acceleration of a fluid in a pipeline neglecting elasticity
119
GRADUAL ACCELERATION OF A FLUID IN A PIPELINE NEGLECTING ELASTICITY
5.4
It is frequently the case that the velocity of the fluid flowing in a pipeline has to be changed, thus causing the momentum of the whole mass of fluid in the pipeline to change. This will require the action of a force, which can be calculated from the rate of change of momentum of the mass of fluid and is produced as a result of a change in the pressure difference between the ends of the pipeline. In the case of liquids flowing in rigid pipes, an approximate value of the change of pressure can be obtained by neglecting the effects of elasticity – provided that the acceleration or deceleration is small.
EXAMPLE 5.1
Water flows through a pipeline 60 m long at a velocity of 1.8 m s−1 when the pressure difference between the inlet and outlet ends is 25 kN m−2. What increase of pressure difference is required to accelerate the water in the pipe at the rate of 0.02 m s−2? Neglect elasticity effects.
Solution Let A = cross-sectional area of the pipe, l = length of pipe, ρ = mass density of water, a = acceleration of water, δ p = increase in pressure at inlet required to produce acceleration a. As this is not a steady flow problem, consider a control mass comprising the whole of the water in the pipe. By Newton’s second law, Force due to δ p in Rate of change of momentum of water in = direction of motion the whole pipe = Mass of water in pipe × Acceleration,
(I)
Force due to δ p = Cross-sectional area × δp = Aδp, Mass of water in pipe = Mass density × Volume = ρAl. Substituting in (I), Aδ p = ρAla,
δ p = ρla = 103 × 60 × 0.02 N m−2 = 1.2 kN m−2. In this example, the change of pressure difference is small because the acceleration is small, but very large pressures can be developed by sudden accelerations or decelerations, such as may occur when valves are shut suddenly. The elasticity of the fluid and of the pipe must then be taken into account, as explained in Chapter 20.
FM5_C05.fm Page 120 Wednesday, September 21, 2005 9:20 AM
120
Chapter 5
The Momentum Equation and its Applications
5.5
FORCE EXERTED BY A JET STRIKING A FLAT PLATE
Consider a jet of fluid striking a flat plate that may be perpendicular or inclined to the direction of the jet, or indeed may be moving in the initial direction of the jet (Fig. 5.4). A control volume encapsulating the approaching jet and the plate may be established, this control volume being fixed relative to the plate and therefore moving with it. It is helpful to consider components of the velocity and force vectors perpendicular and parallel to the surface of the plate. In cases where the plate is itself in motion the most helpful technique is to reduce the plate, and therefore the associated control volume, to rest by the superimposition on the system of an equal but opposite plate velocity, as illustrated in Fig. 5.4. This reduces all the cases illustrated to a simple consideration of a jet striking a stationary plain surface, the initial motion of the surface being reflected in the amendment of the jet velocity relative to the plate. In each of the cases illustrated the impingement of the jet on the plate surface reduces the jet velocity component normal to the plate surface to zero. In general terms the jet velocity thus destroyed may be expressed as vnormal = (v − u) cos θ.
FIGURE 5.4 Force exerted on a flat plate
FM5_C05.fm Page 121 Wednesday, September 21, 2005 9:20 AM
5.5
Force exerted by a jet striking a flat plate
121
The mass flow entering the control volume is also affected by the superposition of a velocity equal and opposite to the plate velocity and may be expressed as A = ρA(v − u) cos θ, which reduces to A = ρAv if the plate is stationary. (Note that the sign convention adopted is positive in the direction of the initial jet velocity.) Thus the rate of change of momentum normal to the plate surface is given by dMomentumdt = ρA(v − u)(v − u) cos θ.
Clearly this expression reduces to dMomentumdt = ρAv 2 cos θ if the plate is stationary, and dMomentumdt = ρAv 2 if the plate is both stationary and perpendicular to the initial jet direction. (Note that the plate velocity, represented by the superposition of an equal and opposite velocity on the system as a whole to bring the control volume to rest, appears in both the mass flow and relative jet velocity terms.) There will therefore be a force exerted upon the plate equal to the rate of momentum destroyed normal to the plate, given in the general case by an expression of the form Force normal to plate = ρA(v − u)(v − u) cos θ. There will be an equal and opposite reaction force exerted on the jet by the plate. In a direction parallel to the plate, the force exerted will depend upon the shear stress between the fluid and the surface of the plate. For an ideal fluid there would be no shear stress and hence no force parallel to the plate. The fluid would flow out over the plate so that the total momentum in unit time parallel to the plate remained unchanged.
EXAMPLE 5.2
A jet of water from a fixed nozzle has a diameter of 25 mm and strikes a flat plate inclined to the jet direction. The velocity of the jet is 5 m s−1, and the surface of the plate may be assumed frictionless. (a) Indicate in tabular form the reduction in the force normal to the plate surface as the inclination of the plate to the jet varies from 90° to 0°. (b) Indicate in tabular form the force normal to the plate surface as the plate
FM5_C05.fm Page 122 Wednesday, September 21, 2005 9:20 AM
122
Chapter 5
The Momentum Equation and its Applications
velocity changes from 2 m s−1 to −2 m s−1 in the direction of the jet, given that the plate is itself perpendicular to the approaching jet.
Solution For each case the control volume taken is fixed relative to the plate. If the plate is in motion the control volume is brought to rest by the superposition of an equal and opposite velocity on the system as a whole. As a force applied normal to the plate is required in each case, the components of velocity and force normal and parallel to the plate surface are considered. From equation (5.5) the gravity force is negligible and, if the fluid jet is assumed to be parallel sided and passing through a region at atmospheric pressure, there is no force exerted on the jet by fluid outside the control volume. Thus the force exerted normal to the plate in the general case is given by Force normal to plate = ρA(v − u)(v − u) cos θ, which may be utilized in the following tables drawn up to answer (a) and (b) above. The cross-sectional area of the jet is A = π 0.02524 = 4.9 × 10− 4 and the density of the water jet ρ = 1000.0 kg m−3. The jet velocity v = 5.0 m s−1.
TABLE (A) Variation of force exerted normal to the plate with plate angle
θ (deg) 0 15 30 45 60 75 90
TABLE (B) Variation of force exerted normal to the plate with plate velocity
θ (deg) 0 0 0 0 0
v cos θ (m s−1)
ρAv (kg s−1)
Force = ρAv 2 cos θ (N)
5.00 4.83 4.33 3.54 2.50 1.29 0.00
2.46 2.46 2.46 2.46 2.46 2.46 2.46
12.28 11.86 10.63 8.68 6.14 3.18 0.00
v (m s−1)
u (m s−1)
v−u (m s−1)
ρA(v − u) (kg s−1)
Force = ρA(v − u)2 (N)
5.0 5.0 5.0 5.0 5.0
2.0 1.0 0.0 −1.0 −2.0
3.0 4.0 5.0 6.0 7.0
1.47 1.96 2.46 2.94 3.43
4.41 7.84 12.28 17.64 24.01
FM5_C05.fm Page 123 Wednesday, September 21, 2005 9:20 AM
5.6
Force due to the deflection of a jet by a curved vane
123
FORCE DUE TO THE DEFLECTION OF A JET BY A CURVED VANE
5.6
Both velocity and momentum are vector quantities and, therefore, even if the magnitude of the velocity remains unchanged, a change in direction of a stream of fluid will give rise to a change of momentum. If the stream is deflected by a curved vane (Fig. 5.5), entering and leaving tangentially without impact, a force will be exerted between the fluid and the surface of the vane to cause this change of momentum. It is usually convenient to calculate the components of this force parallel and perpendicular to the direction of the incoming stream by calculating the rate of change of momentum in these two directions. The components can then be combined to give the magnitude and direction of the resultant force which the vane exerts on the fluid, and the equal and opposite reaction of the fluid on the vane.
FIGURE 5.5 Force exerted on a curved vane
EXAMPLE 5.3
A jet of water from a nozzle is deflected through an angle θ = 60° from its original direction by a curved vane which it enters tangentially (see Fig. 5.5) without shock with a mean velocity C1 of 30 m s−1 and leaves with a mean velocity C2 of 25 m s−1. If the discharge A from the nozzle is 0.8 kg s−1, calculate the magnitude and direction of the resultant force on the vane if the vane is stationary.
Solution The control volume will be as shown in Fig. 5.5. The resultant force R exerted by the fluid on the vane is found by determining the component forces Rx and Ry in the x and y directions, as shown. Using equation (5.5), Rx = −F1 = F2 + F3 − A(vout − vin)x . Neglecting force F2 due to gravity and assuming that for a free jet the pressure is constant everywhere, so that F3 = 0, Rx = A(vin − vout)x ,
(I)
and, similarly, Ry = A(vin − vout)y .
(II)
FM5_C05.fm Page 124 Wednesday, September 21, 2005 9:20 AM
124
Chapter 5
The Momentum Equation and its Applications
Since the nozzle and vane are fixed relative to each other, Mass per unit Mass per unit time entering = A = time leaving control volume nozzle. In the x direction, vin = Component of C1 in x direction = C1, vout = Component of C2 in x direction = C2 cos θ. Substituting in (I), Rx = A(C1 − C2 cos θ ).
(III)
Putting A = 0.8 kg s−1, C1 = 30 m s−1, C2 = 25 m s−1, θ = 60°, Rx = 0.8(30 − 25 cos 60°) = 14 N. In the y direction, vin = Component of C1 in y direction = 0, vout = Component of C2 in y direction = C2 sin θ. Thus, from (II), Ry = AC2 sin θ.
(IV)
Putting in the numerical values, Ry = 0.8 × 25 sin 60° = 17.32 N. Combining the rectangular components Rx and Ry, Resultant force exerted = ( R 2x – R 2y ) by fluid on vane, R = ( 14 2 + 17.32 2 ) = 22.27 N. This resultant force R will be inclined to the x direction at an angle φ = tan−1(Ry Rx) = tan−1(17.3214) = 51°3′.
5.7
FORCE EXERTED WHEN A JET IS DEFLECTED BY A MOVING CURVED VANE
If a jet of fluid is to be deflected by a moving curved vane without impact at the inlet to the vane, the relation between the direction of the jet and the tangent to the curve of the vane at inlet must be such that the relative velocity of the fluid at inlet is tangential to the vane. The force in the direction of motion of the vane will be equal to the rate of change of momentum of the fluid in the direction of motion, i.e. the mass
FM5_C05.fm Page 125 Wednesday, September 21, 2005 9:20 AM
5.7
Force exerted when a jet is deflected by a moving curved vane
125
deflected per second multiplied by the change of velocity in that direction. The force at right angles to the direction of motion will be equal to the mass deflected per second times the change of velocity at right angles to the direction of motion.
EXAMPLE 5.4
A jet of water 100 mm in diameter leaves a nozzle with a mean velocity C1 of 36 m s−1 (Fig. 5.6) and is deflected by a series of vanes moving with a velocity u of 15 m s−1 in a direction at 30° to the direction of the jet, so that it leaves the vane with an absolute mean velocity C2 which is at right angles to the direction of motion of the vane. Owing to friction, the velocity of the fluid relative to the vane at outlet Cr2 is equal to 0.85 of the relative velocity Cr1 at inlet. Calculate (a) the inlet angle α and outlet angle β of the vane which will permit the fluid to enter and leave the moving vane tangentially without shock, and (b) the force exerted on the series of vanes in the direction of motion u.
FIGURE 5.6 Force exerted on a series of moving vanes
Solution If the absolute velocity C2 is to be at right angles to the direction of motion, the vane must turn the fluid so that it leaves with a relative velocity Cr2 , which has a component velocity equal and opposite to u as shown in the outlet velocity triangle (Fig. 5.6). (a) To determine the inlet angle α, consider the inlet velocity triangle. The velocity of the fluid relative to the vane at inlet, Cr1, must be tangential to the vane and make an angle α with the direction of motion, tan α = CDBC = C1 sin 30°(C1 cos 30° − u). Putting C1 = 36 m s−1 and u = 15 m s−1, tan α = 36 × 0.5(36 × 0.866 − 15) = 1.113,
α = 48°3′. To determine the outlet angle β, if C2 has no component in the direction of motion, the outlet velocity triangle is right angled, cos β = uCr2, but Cr2 = 0.85Cr1 and, from the inlet triangle, Cr1 = CDsin α = C1 sin 30°sin α.
FM5_C05.fm Page 126 Wednesday, September 21, 2005 9:20 AM
126
Chapter 5
The Momentum Equation and its Applications
Therefore 15 × 0.744 u sin α cos β = −−−−−−−−−−−−−−−−−−− = −−−−−−−−−−−−−−−−−−−−− = 0.729, 0.85v 1 sin30° 0.85 × 36 × 0.5
β = 43°11′. (b) Since the jet strikes a series of vanes, perhaps mounted on the periphery of a wheel, so that as each vane moves on its place is taken by the next in the series, the average length of the jet does not alter and the whole flow from the nozzle of diameter d is deflected by the vanes. Neglecting the force due to gravity and assuming a free jet that does not fill the space between the vanes completely, so that the pressure is constant everywhere, the component forces in the x and y directions (Fig. 5.6) can be found from equation (5.5) putting R = −F1 and F2 = F3 = 0. In the direction of motion, which is the x direction, Rx = A(vin − vout)x
(I)
Mass per unit time Mass per unit time entering control = A = leaving nozzle volume = ρ(π4)d 2C1, vin = Component of C1 in x direction = C1 cos 30°, vout = Component of C2 in x direction = C2 cos 90° = 0. Substituting in (I), Force on vanes in = Rx = ρ(π4)d 2C1 × C1 cos 30°. direction of motion Putting in the numerical values, Force on vanes in 2 direction of motion = 1000 × (π4)(0.1) × 36 × 36 × 0.866 N = 8816 N.
5.8
FORCE EXERTED ON PIPE BENDS AND CLOSED CONDUITS
Figure 5.7 shows a bend in a pipeline containing fluid. When the fluid is at rest, it will exert a static force on the bend because the lines of action of the forces due to pressures p1 and p2 do not coincide. If the bend tapers, the magnitude of the static forces will also be affected. When the fluid is in motion, its momentum will change as it passes round the bend due to the change in its direction and, if the pipe tapers, any consequent change in magnitude of its velocity. There must, therefore, be an additional force acting between the fluid and the pipe.
FM5_C05.fm Page 127 Wednesday, September 21, 2005 9:20 AM
5.8
EXAMPLE 5.5
Force exerted on pipe bends and closed conduits
127
A pipe bend tapers from a diameter of d1 of 500 mm at inlet (see Fig. 5.7) to a diameter of d2 of 250 mm at outlet and turns the flow through an angle θ of 45°. Measurements of pressure at inlet and outlet show that the pressure p1 at inlet is 40 kN m−2 and the pressure p2 at outlet is 23 kN m−2. If the pipe is conveying oil which has a density ρ of 850 kg m−3, calculate the magnitude and direction of the resultant force on the bend when the oil is flowing at the rate of 0.45 m3 s−1. The bend is in a horizontal plane.
FIGURE 5.7 Force on a tapering bend
Solution Referring to Fig. 5.7, take the x direction parallel to the incoming velocity C1 and the y direction as shown. The control volume is bounded by the inside wall of the bend and the inlet and outlet sections 1 and 2. Mass per unit time entering control volume = ρQ. The forces acting on the fluid will be F1 exerted by the walls of the pipe, F2 due to gravity (which will be zero), and F3 due to the pressures p1 and p2 of the fluid outside the control volume acting on areas A1 and A2 at sections 1 and 2. The force exerted by the fluid on the bend will be R = −F1. Using equation (5.5), putting F2 = 0 and resolving in the x direction: (F1 + F3)x = A(vout − vin)x and, since Rx = −(F1)x, Rx = (F3)x − A(vout − vin)x.
(I)
Now (F3)x = p1A1 − p2A2 cos θ, vout = Component of C2 in x direction = C2 cos θ, vin = Component of C1 in x direction = C1. Substituting in (I), Rx = p1A1 − p2A2 cos θ − ρQ(C2 cos θ − C1). Resolving in the y direction, (F1 + F3)y = A(vout − vin)y
(II)
FM5_C05.fm Page 128 Wednesday, September 21, 2005 9:20 AM
128
Chapter 5
The Momentum Equation and its Applications
and, since Ry = −(F1)y , Ry = (F3)y − A(vout − vin)y .
(III)
Now, (F3)y = 0 + p2A2 sin θ, vout = Component of C2 in y direction = −C2 sin θ, vin = Component of C1 in y direction = 0. Substituting in (III), Ry = p2A2 sin θ + ρQC2 sin θ.
(IV)
For the given problem, A1 = (π 4) d 21 = (π 4)(0.5)2 = 0.196 35 m2, A2 = (π 4) d 22 = (π 4)(0.25)2 = 0.049 09 m2, Q = 0.45 m3 s−1, C1 = QA1 = 0.450.196 35 = 2.292 m s−1, C2 = QA2 = 0.450.049 09 = 9.167 m s−1. Putting ρ = 850 kg m−3, θ = 45°, p1 = 40 kN m−2, p2 = 23 kN m−2, and substituting in equation (II), Rx = 40 × 103 × 0.196 35 − 23 × 103 × 0.049 09 cos 45° − 850 × 0.45(9.167 cos 45° − 2.292) N = 103(7.855 − 0.798 − 1.603) N = 5.454 × 103 N. Substituting in equation (IV), Ry = 23 × 103 × 0.049 09 sin 45° + 850 × 0.45 × 9.167 sin 45° N = 10 3(0.798 + 2.479) N = 3.277 × 103 N. Combining the x and y components, Resultant force on bend, R = (Rx2 + Ry2 ) = (5.4542 + 3.2772) kN = 6.362 kN. The inclination of R to the x direction is given by
φ = tan−1(Ry Rx) = tan−1(3.2775.454) = 31°.
FM5_C05.fm Page 129 Wednesday, September 21, 2005 9:20 AM
5.9
Reaction of a jet
129
REACTION OF A JET
5.9
Whenever the momentum of a stream of fluid is increased in a given direction in passing from one section to another, there must be a net force acting on the fluid in that direction, and, by Newton’s third law, there will be an equal and opposite force exerted by the fluid on the system which is producing the change of momentum. A typical example is the reaction force exerted when a fluid is discharged in the form of a high-velocity jet, and which is applied to the propulsion of ships and aircraft through the use of propellers, pure jet engines and rocket motors. The propulsive force can be determined from the application of the linear momentum equation (5.5) to flow through a suitable control volume.
EXAMPLE 5.6
A jet of water of diameter d = 50 mm issues with velocity C = 4.9 m s−1 from a hole in the vertical side of an open tank which is kept filled with water to a height of 1.5 m above the centre of the hole (Fig. 5.8). Calculate the reaction of the jet on the tank and its contents (a) when it is stationary, (b) when it is moving with a velocity u = 1.2 m s−1 in the opposite direction to the jet while the velocity of the jet relative to the tank remains unchanged. In the latter case, what would be the work done per second?
FIGURE 5.8 Reaction of a jet
Solution Take the control volume shown in Fig. 5.8. In equation (5.5), the direction under consideration will be that of the issuing jet, which will be considered as positive in the direction of motion of the jet: therefore, F2 = 0, and, if the jet is assumed to be at the same pressure as the outside of the tank, F3 = 0. Force exerted by fluid system in = −F1 = −A(vout − vin), direction of motion, R or in words, Reaction force in direction Mass discharged Increase of velocity × = opposite to that per unit time in direction of jet. of the jet
(I)
FM5_C05.fm Page 130 Wednesday, September 21, 2005 9:20 AM
130
Chapter 5
The Momentum Equation and its Applications
In the present problem, Mass discharged = ρ(π 4)d 2C = 1000 × (π4)(0.05)2 × 4.9 kg s−1 per unit time, A = 9.62 kg s−1. (a) If the tank is stationary, vout = C = 4.9 m s−1, vin = Component of velocity of the free surface in the direction of the jet = 0. Substituting in equation (I), Reaction of jet on tank = 9.62 × (4.9 − 0) N = 47.14 N in the direction opposite to that of the jet. (b) If the tank is moving with a velocity u in the opposite direction to that of the jet, the effect is to superimpose a velocity of −u on the whole system: vout = C − u, vin = −u, vout − vin = C. Thus, the reaction of the jet R remains unaltered at 47.14 N. Work done per second = Reaction × Velocity of tank = R × u = 47.14 × 1.2 = 56.57 W.
A rocket motor is, in principle, a simple form of engine in which the thrust is developed as the result of the discharge of a high-velocity jet of gas produced by the combustion of the fuel and oxidizing agent. Both the fuel and the oxidant are carried in the rocket and so it can operate even in outer space. It does not require atmospheric air, either for combustion or for the jet to push against; the thrust is entirely due to the reaction developed from the momentum per second discharged in the jet.
EXAMPLE 5.7
The mass of a rocket mr is 150 000 kg and, when ready to launch, it carries a mass of fuel mf0 of 300 000 kg. The initial thrust of the rocket motor is 5 MN and fuel is consumed at a constant rate A. The velocity Cr of the jet relative to th
