Hydraulics

ENV3104 Hydraulics II Faculty of Engineering and Surveying

Study Guide 2011

Originally written by Prof Rod Smith Edited by Dr Malcolm Gillies Faculty of Engineering and Surveying University of Southern Queensland

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ENV3104 – Hydraulics II

Published by University of Southern Queensland Toowoomba Queensland 4350 Australia http://www.usq.edu.au © University of Southern Queensland, 2011

Copyrighted materials reproduced herein are used under the provisions of the Copyright Act 1968 as amended, or as a result of application to the copyright owner. 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 or otherwise without prior permission. Produced using Microsoft Word 2007 and the ICE Publishing Style

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Table of Contents

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Table of Contents Table of Contents ..................................................................................................... iii

GUIDE TO THIS COURSE ............................................................................. XI Presentation .............................................................................................................. xi Course Structure...................................................................................................... xii How to Pass this Course ........................................................................................ xiii Text Books ............................................................................................................. xiv

MODULE 1 – INTRODUCTORY CONCEPTS .................................................... 1 1.1 Objectives ............................................................................................................1 1.2 The Universal Truth .............................................................................................1 1.3 The Continuity Equation ......................................................................................2 1.4 Energy Equation ...................................................................................................2 1.5 The Momentum Equation ....................................................................................3 1.6 Application of the Energy and Momentum Equations ........................................5 1.7 A Systematic Approach to Solving Hydraulic Problems .....................................6 1.8 Notation................................................................................................................6 1.9 Self assessment Questions ...................................................................................7 1.10 Tutorial Problems...............................................................................................9 1.11 Solutions to Self assessment Questions .............................................................9

MODULE 2 – REVISION OF STEADY OPEN CHANNEL FLOW....................... 11 2.1 Objectives ..........................................................................................................11 2.2 Steady Open Channel Flow ...............................................................................11 2.3 Uniform Flow.....................................................................................................12 2.4 Solution of Steady Flow Problems ....................................................................13 2.5 Compound Channels ..........................................................................................14 2.6 Gradually Varied Flow.......................................................................................14 2.7 Self assessment Questions .................................................................................16 2.8 Tutorial Problems...............................................................................................18 2.9 Solutions to Self assessment Questions .............................................................18

MODULE 3 – INTRODUCTION TO UNSTEADY OPEN CHANNEL FLOW ......... 19 © University of Southern Queensland

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3.1 Objectives ..........................................................................................................19 3.2 Unsteady Free Surface Flows ............................................................................19 3.2.1 Gradually Varied Unsteady Flow .......................................................................... 20 3.2.2 Rapidly Varied Unsteady Flow ............................................................................. 20

3.3 Propagation of a Solitary Wave .........................................................................21 3.4 Self assessment Questions .................................................................................23 3.5 Solutions to Self assessment Questions .............................................................24

MODULE 4 – RAPIDLY VARIED UNSTEADY FLOW .................................... 25 4.1 Objectives ..........................................................................................................25 4.2 Positive Surge Waves ........................................................................................26 4.2.1 Alternative Form of Momentum Equations........................................................... 28

4.3 Negative Surge Waves .......................................................................................29 4.4 Solution of Problems Involving Surge Waves ...................................................30 4.5 Self assessment Questions .................................................................................30 4.6 Tutorial Problems...............................................................................................32 4.7 Solutions to Self assessment Questions .............................................................32

MODULE 5 – GRADUALLY VARIED UNSTEADY FLOW .............................. 33 5.1 Objectives ..........................................................................................................33 5.2 Equations of Motion ..........................................................................................33 5.2.1 Application to Unit Width of Channel .................................................................. 34 5.2.2 Alternative Derivation of the Backwater Equation ............................................... 35

5.3 Numerical Solutions...........................................................................................36 5.3.1 Characteristics ....................................................................................................... 36 5.3.2 Finite Difference Techniques ................................................................................ 36 5.3.3 Numerical Solution Schemes ................................................................................ 37 5.3.4 Boundary Conditions ............................................................................................. 37

5.4 Self assessment Questions .................................................................................39 5.5 Tutorial Problems...............................................................................................41 5.5.1 Answers to Tutorial Problems ............................................................................... 42

5.6 Solutions to Self assessment Questions .............................................................46

MODULE 6 – SEDIMENT TRANSPORT......................................................... 47 6.1 Objectives ..........................................................................................................47 6.2 Bed Form ...........................................................................................................47

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6.3 Threshold of Motion ..........................................................................................48 6.4 Armouring of Streams........................................................................................49 6.5 Resistance to Flow .............................................................................................49 6.6 Mechanics of Sediment Transport .....................................................................50 6.7 Sediment Transport Equations ...........................................................................50 6.7.1 Bed Load Formulae ............................................................................................... 50 6.7.2 Bed Shear Stress .................................................................................................... 51 6.7.3 Total Load Formulae ............................................................................................. 52

6.8 Worked Examples ..............................................................................................52 6.9 Self assessment Questions .................................................................................52 6.10 Tutorial Problems.............................................................................................53 6.11 Solutions to Self assessment Questions ...........................................................54

MODULE 7 – TRACTIVE FORCE DESIGN .................................................... 55 7.1 Objectives ..........................................................................................................55 7.2 Introductory Comments on Design of Erodible Channels .................................55 7.3 Tractive or Shear Force ......................................................................................56 7.4 Tractive Force Ratio ..........................................................................................57 7.5 Design Procedure ...............................................................................................58 7.6 Stable Hydraulic Section ....................................................................................58 7.7 Further Reading .................................................................................................60 7.8 Worked Examples ..............................................................................................60 7.9 Self assessment Questions .................................................................................62 7.10 Tutorial Problems.............................................................................................63 7.10.1 Answers to Tutorial Problems ............................................................................. 64

7.11 Solutions to Self assessment Questions ...........................................................64

MODULE 8 – VEGETATIVE LINED CHANNELS ........................................... 65 8.1 Objectives ..........................................................................................................65 8.2 Vegetated Channels ...........................................................................................65 8.3 Manning n ..........................................................................................................66 8.4 Design Procedure ...............................................................................................67 8.5 Permissible Velocity ..........................................................................................69 8.6 Worked Examples ..............................................................................................70 8.7 Self assessment Questions .................................................................................72

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8.8 Tutorial Problems...............................................................................................73 8.8.1 Answers to Tutorial Problems ............................................................................... 74

8.9 Solutions to Self assessment Questions .............................................................74

MODULE 9 – REVISION OF PIPELINE FLOW ............................................... 75 9.1 Objectives ..........................................................................................................75 9.2 Turbulent Flow...................................................................................................75 9.2.1 Other Friction Factor Diagrams ............................................................................. 76

9.3 Empirical Equations ...........................................................................................77 9.3.1 Hazen-Williams Equation ..................................................................................... 80 9.3.2 Manning Equation ................................................................................................. 80

9.4 Local Losses .......................................................................................................80 9.5 Series Parallel and Branched Pipelines ..............................................................81 9.6 Worked Examples ..............................................................................................81 9.7 Self assessment Questions .................................................................................82 9.8 Tutorial Problems...............................................................................................84 9.8.1 Answers to Tutorial Problems ............................................................................... 85

9.9 Solutions to Self assessment Questions .............................................................85

MODULE 10 – PIPE NETWORK ANALYSIS.................................................. 87 10.1 Objectives ........................................................................................................87 10.2 Introduction ......................................................................................................87 10.3 Generalised Pipe Flow Equation ......................................................................88 10.4 Hydraulic Analysis ...........................................................................................88 10.5 Hardy-Cross Loop and Node Balancing Methods ...........................................91 10.6 Linearisation ....................................................................................................92 10.7 Worked Examples ............................................................................................93 10.8 Self assessment Questions ...............................................................................93 10.9 Tutorial Problems.............................................................................................95 10.9.1 Answers to Tutorial Problems ............................................................................. 95

10.10 Solutions to Self assessment Questions .........................................................96

MODULE 11 – PUMP/PIPELINE SYSTEMS ................................................... 97 11.1 Objectives ........................................................................................................97 11.2 Introductory Concepts ......................................................................................97 11.3 Pump Affinity Laws and Specific Speed .........................................................98 © University of Southern Queensland

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11.4 Head-Discharge Characteristic Curve ..............................................................99 11.4.1 Theoretical H-Q Curves ...................................................................................... 99 11.4.2 Actual H-Q Curves .............................................................................................. 99

11.5 Pump-Pipeline System Design .......................................................................100 11.6 Design Based on Suction Considerations - Cavitation and NPSH ................102 11.7 Additional Reading and Worked Examples ...................................................104 11.8 Self assessment Questions .............................................................................104 11.9 Tutorial Problems...........................................................................................106 11.9.1 Answers to Tutorial Problems ........................................................................... 107

11.10 Solutions to Self assessment Questions .......................................................107

MODULE 12 – UNSTEADY PIPE FLOW (SURGE) ....................................... 109 12.1 Objectives ......................................................................................................109 12.2 Simple Pipeline/Valve System .......................................................................109 12.2.1 Worked Example ............................................................................................... 112

12.3 Surge Tank Operation ....................................................................................113 12.3.1 Worked Example ............................................................................................... 114

12.4 Self assessment Questions .............................................................................114 12.5 Tutorial Problems...........................................................................................116 12.5.1 Answers to Tutorial Problems ........................................................................... 117

12.6 Solutions to Self assessment Questions .........................................................117

MODULE 13 – UNSTEADY PIPE FLOW (WATER HAMMER) ...................... 119 13.1 Objectives ......................................................................................................119 13.2 Introductory Concepts ....................................................................................119 13.3 Compressible Fluid/Rigid Pipe ......................................................................120 13.4 Compressible Fluid/Elastic Pipe ....................................................................121 13.5 Governing Equations for Rapid and Instantaneous Closure ..........................122 13.6 Solution of Equations .....................................................................................124 13.7 Selection of Method of Analysis ....................................................................124 13.8 Self assessment Questions .............................................................................124 13.9 Tutorial Problems...........................................................................................125 13.9.1 Answers to Tutorial Problems ........................................................................... 126

13.10 Solutions to Self assessment Questions .......................................................126

MODULE 14 – GRAVITY PIPELINES ......................................................... 127 © University of Southern Queensland

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14.1 Objectives ......................................................................................................127 14.2 Types of Gravity Pipelines .............................................................................127 14.3 Types of Culvert Flow ...................................................................................128 14.4 Culvert Design ...............................................................................................128 14.5 Worked Examples – Application of the Energy Equation to a Culvert Flowing Full .........................................................................................................................129 14.6 Partially Full Pipe Flow .................................................................................129 14.7 Self assessment Questions .............................................................................130 14.8 Tutorial Problems...........................................................................................131 14.8.1 Answers to Tutorial Problems ........................................................................... 132

14.9 Solutions to Self assessment Questions .........................................................132

MODULE 15 – MINIMUM ENERGY LOSS STRUCTURES ............................ 133 15.1 Objectives ......................................................................................................133 15.2 Revision of Specific Energy Concepts ..........................................................133 15.2.1 Self assessment Questions on Specific Energy ................................................. 134 15.2.2 Tutorial Problems on Specific Energy............................................................... 135

15.3 Minimum Energy Loss Design ......................................................................135 15.3.1 Establishing Critical Flow ................................................................................. 135 15.3.2 Elevation ............................................................................................................ 136 15.3.3 Contraction ........................................................................................................ 137 15.3.4 Simultaneous Variation of Bed Elevation and Channel Width ......................... 138 15.3.5 Design ................................................................................................................ 139

15.4 Additional Readings.......................................................................................139 15.5 Tutorial Problems...........................................................................................140 15.5.1 Answers to Tutorial Problems ........................................................................... 141

15.6 Solutions to Self assessment Questions .........................................................142

MODULE 16 – MEASUREMENT AND CONTROL STRUCTURES................... 143 16.1 Objectives ......................................................................................................143 16.2 Gated Control Structures ................................................................................143 16.3 Sharp Crested Weirs ......................................................................................144 16.4 Long Based Weirs (Broad Crested Weirs) .....................................................144 16.5 Flumes ............................................................................................................144 16.6 Tutorial Problems...........................................................................................145

MODULE 17 – DIMENSIONAL ANALYSIS AND HYDRAULIC SIMILITUDE .. 147 © University of Southern Queensland

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17.1 Objectives ......................................................................................................147 17.2 Dimensional Analysis ....................................................................................147 17.3 The Index Method ..........................................................................................148 17.4 Buckingham Pi (П) Method ...........................................................................151 17.4.1 Worked Examples ............................................................................................. 153

17.5 Hydraulic Similitude ......................................................................................158 17.5.1 Worked Examples ............................................................................................. 161

17.6 Self assessment Questions .............................................................................165 17.7 Tutorial Problems...........................................................................................167 17.7.1 Answers to Tutorial Problems ........................................................................... 168

17.8 Solutions to Self assessment Questions .........................................................168

LIST OF REFERENCES .............................................................................. 169 APPENDIX A - SOME USEFUL EQUATIONS ............................................... 171 APPENDIX B – LIST OF SYMBOLS ............................................................ 173

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ENV3104 – Hydraulics II

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Guide to this Course

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Guide to this Course Presentation These notes have been written in the form of a guide through the set text books. The instructions serve to lay the foundation for each topic and to identify the essential material in the texts. Where a topic is not covered adequately in either text, appropriate material has been included in a Book of Readings. In that case this guide will chart your path through those readings. Supplementary notes have been included in this study guide to assist you in your work, correct deficiencies in the texts and readings and to convey the examiner's philosophy. You should read these notes in conjunction with the recommended readings. The objectives for each topic, the self assessment questions and the tutorial problems will also aid in the identification of the most important material. The core objective of this course is to equip you with the hydraulic tools necessary for the solution of problems in hydraulic engineering and to train you in the application of these tools. To meet this objective you should master all of the set problems.

Important Note Do not try to complete all modules of this subject strictly in the order given. Doing so will not prepare you for the material assessed in the assignments. Please consult Studydesk throughout the semester for an up-to-date version of the study schedule.

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ENV3104 – Hydraulics II

Course Structure Hydraulics II is divided into a total of 17 short modules, the figure below outlines the general structure of the course. In other courses you may be used to fewer modules, with the aim of covering a single module per week. This course is different, some of the modules are small and serve primarily as revision of the preceding course in this subject: ENV2301 Hydraulics I.

Hydraulics II

Part 1 Fundamentals Module 1

Part 2 Open Channels

Part 3 Pipelines

Module 2

Module 9

Module 1

Module 1

Part 4 Hydraulic Structures Module 15

Part 5 Dimensional Analysis Module 17

Module 1 Module 1

Unsteady Flow

Erodible Channels

Modules 3,4,5

Modules 6,7,8

Module 1

Module 1

Steady Flow Modules 9,10

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Culverts Module 14 Module 1

Pumped Piplines Module 11

Min Energy Loss Module 15 Module 1

Unsteady Flow Module 12,13

Measurement & Control Module 16 Module 1

Guide to this Course

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How to Pass this Course To gain the maximum benefit from this guide, you should approach the material as follows: 1. Follow the instructions given in each module. Read the text sections, readings and supplementary notes in the order in which they occur.

Reading from Text The details of the text sections and readings are indicated by the reading dog, along with comments indicating the relative importance of the material. Supplementary notes are presented as simple text. All of this supplementary material should be mastered. Interesting asides and anecdotes are presented in italics. 2. Prepare your own comprehensive notes for each of the modules as you progress through the course. Summarise the text, readings and study guide material, concentrating on those sections designated as essential knowledge. Your subsequent revision study is best done from these notes. 3. Attempt to solve each of the worked examples (provided in this guide or referred to in the texts) before reading through the given solution. These solutions should be fully understood before attempting the tutorial problems. 4. Complete all self assessment questions to test your understanding. 5. Attempt the tutorial problems.

Tutorial Problems Worked solutions from the texts and the tutorial problems from the texts are indicated by the pencil graphic as shown here. Often they require extra comments which can be found immediately below they are first mentioned in this guide.

Additional Tutorial Problems Some additional worked solutions and tutorial problems are provided in areas not covered by the problems from the texts. 6. If you are unable to complete any problem, re-read the appropriate text material and worked solutions then re-attempt the problem. If you are still unable to complete the problem seek the assistance of the course examiner immediately. The semester is short and there is much material to cover.

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Text Books As this book is merely a study guide, all students will be expected to have access to the specified texts. The text books contain the majority of the tutorial problems and will serve as a valuable reference for your future career. The following texts will be required: Chadwick, A., Morfett, J. and Borthwick, M. Hydraulics in Civil and Environmental Engineering. 4th Edition, E & F N Spon, 2004. Marriott, M. Nalluri and Featherstone’s Civil Engineering Hydraulics. 5th Edition, Wiley-Blackwell, 2009. This course also relies on a “book of readings” which will be referred to throughout each of the course modules. All page references provided in this study guide refer to the most recent version of these texts. However, previous versions are almost identical apart from changes in the page numbers. Where appropriate, page numbers have been given for both the Marriott (5th Ed.) and Nalluri and Featherstone (4th Ed.) versions of the second text above.

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Module 1 – Introductory Concepts

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Module 1 – Introductory Concepts

1.1 Objectives When you have mastered the material in this module you will be able to:  plan the solution path for any hydraulic problem;  apply the fundamental laws (continuity, energy and momentum) to simple steady flow problems; and  recognise the alternative systems of notation commonly used in hydraulic texts.

1.2 The Universal Truth The Universal Truth in applied hydraulics is that (almost) all problems can be solved using two of the three fundamental laws, viz:  the continuity equation (conservation of mass);  the energy equation (in certain circumstances known as the Bernoulli equation); and  the momentum equation (Newton's 2nd Law).

Most steady flow problems (except those involving forces) are solved using the continuity and energy equations. Unsteady flow problems usually require the continuity and momentum equations. Occasionally, for example, in the design of sluice gates, all three laws must be applied. The other common requirement is for an equation relating flow velocity and energy slope to the size and surface features of the conduit (pipe or channel). This may be an entirely empirical equation, such as the Manning equation, or a semi-empirical equation, such as Darcy-Weisbach. In each case the term describing the surface roughness of the conduit must be evaluated. Alternatively, solution of the problem may require evaluation of an empirical constant, such as the coefficient of discharge for an orifice or weir structure.

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Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 2.4 & 2.5 All of this material is revision of material covered in the preceeding course Hydraulics I. However absolute mastery of it is a requirement for success in this course. In particular note that these sections refer only to steady flows. Much of our interest in this course will be in the application of the fundamental laws to unsteady and sometimes spatially varied flows. Note also the general requirement for the inclusion of energy and momentum coefficients,  and  , respectively. Frequently these coefficients will be neglected, either because their value is close to unity or because the appropriate terms cancel in the particular problem. However you should remember to include them if you are uncertain that either of these cases apply.

1.3 The Continuity Equation In general terms conservation of mass implies: mass flow in - mass flow out = rate of change of mass stored in system For an incompressible fluid (that is, of constant density), volume can be substituted for mass giving:

Qin  Qout 

S t

1.1

where the Q terms are volumetric flow rates in m3s-1 and S is the change in volume in storage over time t. For a steady flow system S/t equals zero and Qin = Qout.

1.4 Energy Equation In general terms the statement of the energy equation is:

total total energy energy energy energy = energy + + lost extracted added in out

1.2

The convention in hydraulics is to express all of the energy terms as an energy per unit weight of fluid (or head of water). This involves dividing the energy term by the © University of Southern Queensland

Module 1 – Introductory Concepts

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product g. The symbol H will be used for these energy heads and the units will be m. For convenience the terms energy and head will be used interchangeably throughout these notes to mean energy per unit weight. The energy equation becomes: H in  H out  H lost  H extracted  H added

1.3

Both of the total energy terms include the elevation and pressure potential energies and the kinetic energy of the fluid. Thus at any point in a flow system the total energy will be:

P V 2 H z g 2g

1.4

where P is the hydrostatic pressure at the point,  is the fluid density, z is the elevation of the point and V is the mean flow velocity. For an open channel the total energy is:

H  yz

V 2 2g

1.5

where y is the depth of flow in the channel, z is the elevation of the channel bed and all other terms are as previously defined.

1.5 The Momentum Equation The momentum equation is derived from Newton's 2nd Law, which states that the sum of the external forces on a body (of fluid) is equal to mass of the body times its acceleration, or: F  ma

1.6

this is also equal to the rate of change of momentum and can be written as:

F 

dM dt

1.7

M is the momentum of the fluid and is equal to the product of the mass and velocity of the flow, mV.

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In the general unsteady case the momentum is a function of both distance x and time t. The rate of change of momentum is thus given by:

dM M M dx   dt t x dt

1.8

where the first term on the RHS is the temporal rate of change of momentum and the second term is the spatial rate of change. This can be approximated by: dM M   M      x dt t x  t 

1.9

The momentum M is equal to Vx, where A is the cross sectional area of the flow and x is the length of the element of fluid under consideration. The term M/t is the momentum flux and for an open channel or free surface flow is given by:

M mV   QV  AV 2 t t

1.10

where Q is the mass flow rate and is equal to m/t. Hence:

dM    AV x  QV x dt t x

1.11

For steady flows the momentum is time invariant, therefore the temporal rate of change is zero and we are concerned only with the spatial component of the momentum change. In this case:

F 

d  QV x dx

1.12

In some cases this may be further simplified to:

F  QV where the RHS simply equals the total change in momentum flux.

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Module 1 – Introductory Concepts

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1.6 Application of the Energy and Momentum Equations Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 2.6, 2.7 & 5.8 Essential knowledge. Again this is revision of material covered in Hydraulics I. Sections 2.6 and 2.7 will remind you of the range of applications for the Energy and Momentum equations. In Section 5.8 the development of the equations for the hydraulic jump (eqn 5.21) and the energy loss across a jump (eqn 5.25) are an important introduction to the unsteady applications in future modules.

Readings from Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Section 4.1 Figure 4.1 and the application of the energy equation to a simple pipeline flow problem (equation 4.1) are all that is required from this section now. You should ensure at this stage that you can differentiate between the energy line and the hydraulic grade line.

Worked Examples From Texts Chadwick et al. Hydraulics in Civil and Environmental Engineering Examples 2.1, 2.2, 2.3 & 2.4 Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Examples 3.1, 3.2 & 3.7 In addition, attempt to draw in the energy and hydraulic grade lines for Chadwick et al. 2.2 and Marriott 3.2.

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1.7 A Systematic Approach to Solving Hydraulic Problems Book of Readings – Reading 1.1 Section 1.6.2 from Gerhart, P.M. and Gross, R.J. 1985, Fundamentals of Fluid Mechanics. Addison-Wesley, Reading, MA. Engineering students generally have developed the art of problem solving to a high level, particularly in the structural and dynamics areas, through the extensive practice in the applied mechanics and related subjects. When faced with problems in another discipline such as hydraulics the art alone is inadequate. The science is required to allow the art to be applied across the disciplines. The approach to problem solving in fluid mechanics provided by Gerhart and Gross (1985) can be readily applied to the solution of problems in hydraulics. It is also easily translated to other subject areas. Some of the steps may appear trivial and for some problems this is certainly the case. Despite this it is recommended, that until such time that a systematic approach is second nature to you, that you formally apply each step to all problems. A further step may be added to those of Gerhart and Gross, as follows: Step 7. Evaluate your solution to ensure that it is consistent with the physical description of Step 1, the fundamental laws of Step 3 and the assumptions of Step 4. Revise as necessary.

1.8 Notation Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering List of Principal Symbols - page xviii Introductory notes – page xxvii Peruse these sections and remind yourself of the notation and units for the terms commonly used in hydraulics. A complication introduced by studying from multiple texts is that the notation used may not be consistent from one to another. As practicing engineers you must be able to contend with this variability. Following are some examples of common hydraulic variables for which different symbols are frequently used. In this guide I will tend to (but not always) use the notation used in the notes for Hydraulics I.

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Module 1 – Introductory Concepts

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Table 1.1 – Comparing notation between different texts

Quantity

Hydraulics I

Hydraulics II

Roughness height Friction factor Froude Number Reynolds Number Specific energy Specific weight

k/ f Fr Re E g

k f FR / NF NR

Chadwick et al. ks

Marriott





Fr Re Es g

F Re Es

E g

k



Note that Chadwick et al. define Sf as the slope of the hydraulic gradient. To avoid confusion between the hydraulic gradient terminology and the hydraulic grade line, it is better to define Sf as the slope of the energy line.

1.9 Self assessment Questions Before proceeding on to the tutorial problems, test yourself on the following few questions. Answers are provided at the end of the module.

1.1

1.2

Steady flow occurs when: (a)

conditions do not change with time at any point

(b)

conditions are the same at adjacent points at any instant

(c)

conditions change steadily with time

(d)

none of the above

Uniform flow occurs (a)

whenever the flow is steady

(b)

when conditions do not change with time at any point

(c)

when conditions are the same at adjacent points at any instant

(d)

conditions change steadily with time

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ENV3104 – Hydraulics II

The continuity equation: (a)

The continuity equation:

(b)

expresses the relation between energy and work

(c)

applies only to steady uniform flow

(d)

relates to the mass flow rate through a control volume

1.4 The continuity equation may take the form (a)

Q = PAV

(b)

A1 = A2

(c)

PAV = P2A2V2

(d)

AV = A2V2

1.5 The equation H  y  z 

V 2 2g

has the units:

(a)

kPa

(b)

m

(c)

kPa/kg

(d)

m/kg

z

1.6 The velocity head is: (a)

V2/2g

(b)

(c)

V

(d)

2 gH

1.7 The equation  F  QV requires for its application: (a)

steady flow

(b)

uniform flow

(c)

compressible fluid

(d)

frictionless fluid

1.8 For a hydraulic jump, the depth (m) conjugate to y = 3 m and V = 1.27 m/s is: (a)

0.1

(b)

0.3

(c)

1

(d)

3

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Module 1 – Introductory Concepts

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1.10 Tutorial Problems Answers to these problems are given at the rear of the text. 1.1

Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Problems on Chapter 3, Page 84 (4th Ed. p. 87/88) Question 5

1.2

Marriott

Ch 3, Question 8

Note that the answer given in the text for Question 8 is incorrect. The correct answer is a reaction force of 13.055 kN at an angle of 12.6 to the horizontal. For students who have recently completed ENV2103 Hydraulics I it is likely that you may have already attempted these questions.

1.11 Solutions to Self assessment Questions 1. (a)

2. (c)

3. (d)

4. (d)

5. (b)

6. (a)

7. (a)

8. (b)

. .

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ENV3104 – Hydraulics II

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Module 2 – Revision of Steady Open Channel Flow

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Module 2 – Revision of Steady Open Channel Flow

2.1 Objectives When you have mastered the material in this module you will be able to:  recognise the equations for steady uniform (normal) flow;  apply the equations to simple rigid bed channels to solve for either discharge, velocity, slope, roughness or channel geometry;  analyse steady flow in compound channels; and  calculate steady, gradually varied, water surface profiles.

Note Most of the material in this module is revision of material covered previously in Hydraulics I. Quickly read the text material indicated to refresh your knowledge. If you have trouble with the self assessment questions or the set problems you must undertake a more extensive revision before proceeding on to the next module. Pay extra attention to Section 2.6 -Gradually Varied Flow if you completed Hydraulics 1 in 2010 as this material was covered in greater detail in 2009 and previous.

2.2 Steady Open Channel Flow Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 5.1 to 5.5 Note particularly the flow classifications (Figure 5.2 p. 124) and the geometric properties of open channels (Table 5.1 p. 125).

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ENV3104 – Hydraulics II

2.3 Uniform Flow Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 5.6 & 15.2 Note that the development of equation 5.5 is simply the application of Newton's 2nd Law (or the momentum equation). In this case F = 0, that is, the external forces on the fluid element are in balance and that there is zero change in momentum. Equation 5.5 is a particularly important relationship and one that is used frequently in later modules. The useful result from Section 15.2 is the hydraulically optimum channel section (best hydraulic section from Hydraulics I ), where:

R

y 2

2.1

A number of equations (semi-empirical and empirical) have been proposed to describe the relationship between discharge and slope (friction loss) during steady flow in pipes and channels. All are of similar form and contain essentially the same terms. Expressed in the form usually applied to open channel flows the equations are: Chezy equation:

V  C RS 

Darcy-Weisbach equation:

 8g  V    f 

1

1

2

2

 RS 

2.2

1

2

1 12 2 3 S R n

Manning equation:

V

Hazen-Williams equation:

V  0.849CHW S 0.54 R0.63

2.3

2.4

2.5

where R is the hydraulic radius of the channel and is equal to the cross-section area divided by the wetted perimeter.

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Module 2 – Revision of Steady Open Channel Flow

13

The slope S is the rate of energy loss due to friction. In the case of steady uniform or normal flow, the energy slope Sf, the water surface slope and the bed slope So are all parallel. Hence bed slope So is used in the equations. For any non-uniform or unsteady flow these slopes are not parallel and only the energy slope Sf should be used in the equations. The coefficients C, f, n and CHW all describe the roughness of the channel wetted perimeter. The Chezy equation can be viewed as the general form of flow equation, encompassing all others. It is not widely used in practice. The Manning equation is the most commonly used open channel equation, largely because of the large body of empirical experience accumulated and the relative ease (not accuracy) of estimating the coefficient n. The relative merits of the flow equations will be discussed later in Module 9. In changing from the more usual form of the Darcy-Weisbach equation, viz:

hf 

fLV 2 2 gD

2.6

to that listed above it is necessary to replace the energy loss hf and diameter D with terms more appropriate to open channels. These transformations are: Sf 

hf L

and

R

D 4

2.7

You should attempt to prove the latter relationship for a circular conduit flowing full.

2.4 Solution of Steady Flow Problems Worked Examples From Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Examples 5.1 & 5.2 These examples illustrate the two main classes of problem: Example 5.1 when the geometry of the channel is known and either V (or Q), S or n is unknown; and Example 5.2 when V (or Q), S and n are known and a channel geometry (width b and depth y) is required. Note that in the latter case, if the bed width is not given as in 5.2, then there is an infinite number of pairs of b & y that will satisfy the flow equation. Not all of these combinations will be physically realistic. The solution procedure described in 5.2 will be developed further in later modules where consideration of the erodibility of the bed will impose constraints on the solution.

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ENV3104 – Hydraulics II

Worked Example From Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 8.1 An interesting example of the application of the Darcy-Weisbach equation to an open channel problem. The form of equation used is one where the Darcy-Weisbach equation is combined with the Colebrook-White equation for f. The Moody diagram could be used in lieu of this latter equation. Refer to Example 4.2 (p. 108-109) in Chadwick, Morfett and Borthwick.

2.5 Compound Channels Readings from Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Sections 8.3 & 8.4 Note the difference between a composite channel (different roughness on bed and banks) and a compound channel (made up of distinct sub-areas). Analysis of channels of compound section (section 8.4) is the important material here.

Worked Example From Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Example 5.3 Note that in Example 5.3 the subdivision of the channel into the component sub-areas using vertical divisions. The fluid to fluid boundaries at these divisions are not included as part of the wetted perimeters for the sub-areas. The velocity (or kinetic energy) coefficient  has a value significantly different from unity. However this does not affect the calculation of the discharge conveyed by the channel.

2.6 Gradually Varied Flow Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 5.10 Important material in this section, and that which will be developed in later modules, is the development of the general equation of gradually varied flow (equation 5.38) © University of Southern Queensland

Module 2 – Revision of Steady Open Channel Flow

15

and the classification of the flow profiles (Figure 5.20) particularly for channels of mild slope. If you follow the derivation through from equation 5.36 retaining the  term the resulting equation is: dy So  S f  dx 1   FR2

2.8

For  = 1 this reduces to equation 5.38: dy So  S f  dx 1  FR2

2.9

This is a point form of the equation, hence So, Sf and FR are the values of those variables at the point at which the rate of change of depth dy/dx is required. Note that this equation is derived by differentiation of the energy equation. In Module 5 an alternative derivation of this equation will be presented using the momentum equation. It is traditional in Hydraulics texts to present techniques for solving this equation over finite lengths of channel and using simple finite difference approximations. The forms of equation commonly used in these step methods are:

x  y

 

1   FR

So  S f

2

or

x 

E So  S f

2.10

where E is the change in specific energy over the distance x. Recall that the specific energy E is the energy measured relative to the channel bed and equals y + V2/2g. Note that in both of the above equations the variable terms S f and FR are average values over the incremental distance x.

Worked Example From Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Example 5.7 This solution describes the steps involved in calculating a gradually varied flow profile. The final part of the solution, that is, applying the step equation, has been written around a computer solution. Solution of this class of problem is improved by smaller increments of depth y.

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ENV3104 – Hydraulics II

2.7 Self assessment Questions 2.1

2.2

2.3

2.4

2.5

Manning n for a channel: (a)

decreases as the wetted perimeter decreases

(b)

varies with the product VR

(c)

increases as the roughness of the channel perimeter increases

(d)

has the dimensions [T/L]

The energy losses in open channel flow generally vary as the: (a)

first power of the roughness

(b)

inverse of the roughness

(c)

square of the velocity

(d)

inverse square of the hydraulic radius

In steady uniform flow the slope of the energy line: (a)

varies with distance downstream

(b)

is constant with time and distance

(c)

is constant with time only

(d)

diverges from the hydraulic grade line with distance downstream

: In gradually varied flow: (a)

the energy and hydraulic grade lines are parallel

(b)

the depth and velocity of flow vary gradually with distance

(c)

the depth and velocity of flow vary gradually with time

(d)

the energy increases gradually with time

In open channel flow: (a)

the hydraulic grade line is always parallel to the energy line

(b)

the energy line coincides with the free surface

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Module 2 – Revision of Steady Open Channel Flow

(c)

the energy and hydraulic grade lines coincide

(d)

the hydraulic grade line and free surface coincide

17

2.6 For a channel of semi-circular cross section and radius r (or diameter D) the wetted perimeter is: (a)

D

(b)

r

(c)

r2

(d)

2D2

2.7 The hydraulic radius R of the above channel is: (a)

D/4

(b)

2r

(c)

0.5

(d)

r

2.8 In an open channel of great width the hydraulic radius equals: (a)

y/2

(b)

2y/3

(c)

y

(d)

none of these

2.9 A long rectangular channel 2 m wide and 1 m deep has a bed slope of 0.001 and a Manning n of 0.02. The discharge (m3/s) conveyed by the channel is: (a)

2.0

(b)

1.0

(c)

0.2

(d)

10

2.10 The average shear stress o on the wetted perimeter of a channel is given by: (a)

qV

(b)

qV

(c)

qy

(d)

gRSo

2.11 At a point in a rectangular stream of width 30 m and depth 1.5 m the following slope information was obtained: dy/dx = 0.002, So = 0.003, and Sf = 0.002. Estimate the Froude Number at this point if the flow is steady.

2.12 Assuming steady flow, estimate the average friction slope over a 300 m reach of an approximately rectangular channel 30 m wide, given the following data: © University of Southern Queensland

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ENV3104 – Hydraulics II

y1 = 2 m

y2 = 1.75 m

V1 = 0.6 m/s

V2 = 0.69 m/s

So = 3°

(b)

2.8 Tutorial Problems Answers to these problems are given at the rear of the text. 2.1

Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Problems on Chapter 8 page 237-240 (4th Ed. p 253) Question 1, 2, 3, 9 and 20

2.2

Marriott

Ch 8, Question 2

2.3

Marriott

Ch 8, Question 3

2.4

Marriott

Ch 8, Question 9

Do not be concerned with the shape of the river channel Marriott

2.5

Ch 8, Question 20

Hint: the direct step method

2.9 Solutions to Self assessment Questions 1. (c)

2. (c)

3. (b)

4. (b)

5. (d)

6. (b)

7. (a)

8. (c)

9. (a)

10 (d)

11 FR = 0.71

12. Sf = 0.0532

.

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Module 2 – Introduction to Unsteady Open Channel Flow

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Module 3 – Introduction to Unsteady Open Channel Flow

3.1 Objectives When you have mastered the material in this module you will be able to:  describe the various forms of unsteady open channel (or free surface) flow;  calculate the speed of propagation of a small solitary wave; and  transform an unsteady flow system into an apparently steady flow system.

3.2 Unsteady Free Surface Flows This module starts from the proposition that all flows are unsteady. Unsteady flow is defined as flow whose depth, velocity or discharge vary with time. The material in this and the subsequent Modules 4 to 8 inclusive, refers to all flows with a free water surface. This includes those flows confined within natural or man made channels as well as unconstrained flows such as overland and estuarine flows. Steady flow is simply a special case of unsteady flow, in which the unsteady terms are zero. That is:

dQ dV dy   0 dt dt dt

3.1

The unsteady flow problem most commonly encountered in open channels deals with the motion of translatory waves. The translatory wave is a gravity wave that propagates in a channel due to some disturbance to or change in the flow. It results in an appreciable displacement of the water particles in a direction parallel to the flow. Another type of gravity wave is the oscillatory wave, in which the water particles oscillate in an orbit about a mean position but do not display appreciable displacement in the direction of the wave propagation. This is typical of the motion of a deep water ocean wave. At the beach break the wave becomes a translatory wave. For the purpose of analysis, unsteady flows are classified into two types, viz:  gradually varied unsteady flow; and  rapidly varied unsteady flow. © University of Southern Queensland

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ENV3104 – Hydraulics II

3.2.1 Gradually Varied Unsteady Flow In the gradually varied type the curvature of the wave profile is mild and the change of depth (or velocity) with time is gradual. The vertical component of the acceleration of the water particles is negligible in comparison with the total acceleration. The effect of channel bed friction is usually appreciable and should be taken into account. Examples of gradually varied unsteady flows are:  the passage of a flood hydrograph down a stream channel;  variable flows due to the slow operation of a control structure in a channel; and  tidal flows in estuaries. Spatially varied examples also occur, that is unsteady flows with a lateral inflow or outflow, viz:  the rise of an overland flow hydrograph due to the rainfall incident on the surface; and  surface irrigation hydraulics.

Gradually varied unsteady flows may be analysed using one, two or three dimensional models. Most problems can be satisfactorily handled by a relatively simple one dimensional approach, that is, by assuming the flows in the other two dimensions are negligible. Tidal flows are best handled by considering flows in the two horizontal dimensions. Examples requiring a full three dimensional model are rare and will not be considered further.

3.2.2 Rapidly Varied Unsteady Flow In the rapidly varied type of flow the curvature of the wave profile is large and the surface of the profile may become virtually discontinuous. The effect of bed friction is small in comparison with the dynamic effect of the flow. It is usually ignored particularly when analysing the motion of a wave at or near its source. Alternative expressions for rapidly varied unsteady flow are surge and moving hydraulic jump. Examples of rapidly varied unsteady flow are the surges of various kinds caused by the rapid operation of control structures. The extreme example is the dam break problem.

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Module 2 – Introduction to Unsteady Open Channel Flow

21

Case study – Tidal Bores An interesting example of rapidly varied flow An unusual example in nature of a surge or moving hydraulic jump is the hydraulic or tidal bore that occurs in some tidal rivers. This a steep wave that moves rapidly upstream causing a significant increase in depth after it has passed. It also causes a reversal in the flow direction. Ahead of the wave, flow is in the downstream direction while behind the wave, flow is in the upstream direction. Examples occur in the Chien Tang River in China, the Severn and Trent Rivers in England, the Bay of Fundy in Novia Scotia and the Amazon river where it is known locally as the “Pororoca”. [Refer to Lynch, DK (1982) Tidal Bores. Scientific American, 274(4), pages 134-144 and Chow, VT (1959) Open Channel Hydraulics. McGraw-Hill, New York, pages 557-559.] For more information on the Severn Bore you might search the internet for the many web sites on this subject or visit the web site of the UK Environment Agency at: http://www.environment-agency.gov.uk/homeandleisure/recreation/31439.aspx Rod Smith, the former examiner of this course visited the Severn River bore in 1998, his photographs are available at his web site at: http://www.usq.edu.au/users/smithrod/severn.htm

3.3 Propagation of a Solitary Wave Book of Readings – Reading 3.1 Section 18.6 from Chow V.T. 1959, Open Channel Hydraulics. McGraw-Hill, New York. Note the transformation of the real flow system of Figure 18.7(a) to an apparent steady flow system as seen through the eyes of an observer moving with the wave. This idea of transforming flow systems must be mastered as you will be using it extensively in the next module. It should not be entirely foreign to you. Each time you drive your car you are assessing the speed of other vehicles relative to your own (ie, the other vehicles are the flow and you are the moving observer). The simple rule is to subtract the velocity of the observer from all other velocities, remembering that velocities are vectors and have direction (sign) as well as magnitude. Other important material includes:  equation 18-46 for the celerity of the wave;  equation 18-54 for the wave velocity; and  the various figures 18.8 (a) to (d). © University of Southern Queensland

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ENV3104 – Hydraulics II

It is not clear in the article from Chow how the kinetic energy term at the wave is developed. This term comes from an application of the Continuity equation to the apparently steady flow system of Figure 18.7(b). If we call the velocity at this point V* then:

q  cy  V   y  H 

3.2

where q is an apparent steady discharge per unit width. Thus: V* 

cy yH

3.3

and: KE 

V 2 c2 y 2 2g 2g  y  H 

3.4

It is also important to be clear on the definition of terms in this section. Those that I prefer are: Flow Velocity V - the mean velocity of the water flow relative to the channel bed. Wave Celerity c - the velocity of the wave relative to the flow. [Celerity is from the Latin and means swiftness.] Wave Velocity Vw - the velocity of the wave relative to the channel bed. These definitions lead directly to the equation: Vw  V  c

3.5

Cautionary Note The notation used in the two set texts for this course varies from that defined above. Marriott (Nalluri and Featherstone) use c for the wave velocity Vw, i.e. equation 13.1 defined on page 316 (4th Ed. p. 341). They have no symbol for celerity as defined above and simply use gy . They also tend to use the words velocity and celerity interchangeably. This may be permissible providing the frame of reference is always stated. However it is better to avoid the confusion and restrict the words to the specific uses defined above. Chadwick et al. use V for the wave velocity Vw.

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Module 2 – Introduction to Unsteady Open Channel Flow

23

You will recall that the Froude Number of a flow is defined as:

FR 

V gy

3.6

It is interesting to compare this with the expression for wave celerity, viz:

c  gy

3.7

This leads to a new definition of Froude Number as the ratio of the flow velocity over the wave celerity. Flows will be sub-critical if V < c and super-critical if V > c.

Next time you are picnicking beside a stream, amuse yourself by using this knowledge and the associated Figures 18.8 from Article 1 to determine if the flow in the stream is sub- or super-critical. No computations are required. You only need to observe the direction of propagation of the upstream wave caused by throwing a stone into the stream.

3.4 Self assessment Questions . 3.1 A steady flow has a velocity of 0.6 m/s. The velocity of this flow (in ms -1) relative to an observer moving upstream at 1.3 m/s is: (a)

0.7

(b)

1.0

(c)

2.6

(d)

1.9

3.2 If the discharge in the previous question is 5 m3/s estimate the apparent discharge (in m3/s) as seen by the observer. (a)

15.53

(b)

5

(c)

0.23

(d)

8.33

3.3 The speed of an elementary wave in still water is given by: (a)

 gy 

(b)

2y d

(c)

2 gy

(d)

gy

2

13

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ENV3104 – Hydraulics II

3.4 In a channel with y = 1.2 m and V = 2.4 m/s an elementary wave will travel upstream with a velocity in m/s of: (a)

1.03

(b)

3.43

(c)

4.89

(d)

5.90

3.5 At a point in a rectangular stream 30 m wide the discharge is 15 m3/s and the depth of flow 1.2 m. The speed in m/s (relative to an observer standing on the bank) of the wave, that travels in the upstream direction, caused by a small disturbance is: (a)

0.42

(b)

3.43

(c)

3.85

(d)

3.01

3.6 At a point in a rectangular stream 30 m wide the discharge is 15 m3/s and the depth of flow 1.2 m. The apparent flow velocity in m/s relative to an observer moving downstream at a speed equal to the celerity of a solitary wave is: (a)

3.85

(b)

-3.01

(c)

-3.43

(d)

3.01

3.5 Solutions to Self assessment Questions 1. (d)

2. (a)

5. (d)

6. (b)

. .

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3. (d)

4. (a)

Module 4 – Rapidly Varied Unsteady Flow

25

Module 4 – Rapidly Varied Unsteady Flow

4.1 Objectives When you have mastered the material in this module you will be able to:  define the four types of surge wave and the situations in which they can occur;  apply the Continuity and Momentum equations to each type of surge; and  calculate the speed of transmission of a negative surge wave.

Note on Symbols The two texts and this study book all tend to use slightly different symbols for the various velocity terms, which if not understood can lead to confusion. The Table below gives a summary of the different symbols used.

Table 4.1 – Conventions for velocity and celerity between the texts

Study Book

Marriott (Nalluri & Featherstone)

Chadwick et al.

Solitary Waves Wave celerity Wave velocity

gy

c=

Vw = V c

gy

gy

c=

c=V

gy

-

Surge Waves Flow velocity

V & Vi

V & Vi

Vi

Wave celerity

ci

-

-

Wave velocity

Vw

c

V

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ENV3104 – Hydraulics II

4.2 Positive Surge Waves Readings from Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Sections 13.1, 13.3, 13.4 and 13.5 Mastery of all of this material is necessary. Solution of problems involving surge waves are not amenable to recipe approaches. Every problem is different. The essential steps in the solution (following the scheme outlined in Module 1) are: 1. sketch the unsteady flow system, indicating the flow directions and the movement of the wave; 2. transform the flow system to an apparent steady system as seen through the eyes of an observer moving at velocity Vw; 3. apply the continuity and momentum equations to the apparent steady flow system; 4. combine and rearrange the equations to a form to suit the unknowns required for a solution; and 5. solve.

The momentum equation states that: F  ma

4.1

which for unit width of steady flow in a rectangular channel leads directly to:

F   qV

4.2

where q is the discharge per unit width of channel and equals Vy. Application of the momentum equation to the apparent steady flow is foolproof if the following sign convention is followed: Forces in Forces Momentum Momentum the flow - opposing = flux leaving - flux entering direction flow section section

4.3

Applied to unit width of the system defined in Figure 13.5 of Marriott (Nalluri and Featherstone) this gives:

1 1 gy12  gy22  qV2  Vw   qV1  Vw  2 2

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4.4

Module 4 – Rapidly Varied Unsteady Flow

27

The only forces being considered are the hydrostatic forces at each end of the section (for the purpose of the analysis the bed is assumed horizontal and frictionless).

When considering unit width of channel the continuity equation (applied to Figure 13.5) also takes a slightly different form: y1 V1  Vw   y 2 V2  Vw 

4.5

where the LHS and RHS of the equation are equal to the apparent steady discharge q. Following the steps outlined in Section 13.4 these equations will result in equation 13.8: 1

 gy  y  y1   2   V1 Vw   2 2 2 y 1  

4.6

This equation is equivalent to: Vw  c1  V1

4.7

where c1 is the celerity of the surge wave relative to the flow V1 and is given by: 1

 gy  y  y1   2  c1   2 2 2 y1  

4.8

For small waves, that is when y1  y2 , this term reduces to

gy .

The equations that result from the application of the momentum equation to the various types of surge wave can be summarised as follows:

Downstream Positive Wave

Tidal Bore

Upstream Positive Wave

Vw  c1  V1

Vw  c1  V1

Vw  c1  V1

or

or

or

Vw  c2  V2

Vw  c2  V2

Vw  c2  V2

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ENV3104 – Hydraulics II

1

 gy  y  y1   2 c1   2 2  2 y1  

where:

 gy  y  y1   c2   1 2  2 y2  

and

4.9

1 2

4.10

To solve surge flow problems, these momentum equations must be applied in conjuction with the relevant continuity equation as given below:

4.2.1 Alternative Form of Momentum Equations The momentum equation can be presented in an alternative form as follows, starting with the form as derived on the previous page:

 gy  y  Vw   2 2 2 y1 

1 y1  2 



  V1 

4.11

Re-arrangement to make y2 the dependent variable gives:

Vw  V1 2  y1  y2  1 8 1  2 gy1   The term

Vw  V1 2 gy1

4.12

is the Froude number of the apparent steady flow at point 1,

hence:

y2 

y1 2

 1 8F 1 2 R1

4.13

which is identical to the equation for a hydraulic jump. However it must be remembered that this equation can only be applied to the transformed flow. This equation is in fact four equations, each on having its direct equivalent in the above table, depending on whether or not the wave is upstream or downstream and depending on which depth is the dependant variable. The hydraulic jump that occurs in steady flow is only a special case of a stationary surge (Vw = 0).

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Module 4 – Rapidly Varied Unsteady Flow

29

Readings from Text Chadwick et al Hydraulics in Civil and Environmental Engineering Section 5.11 (Sub-section headed Surge Waves only) This is the alternative analysis which starts from the equation for the hydraulic jump and applies it to the transformed flow system. The velocity substituted into the Froude Number term in equation 5.28a is the apparent steady flow velocity. The continuity equation must also be applied to the apparent steady system to obtain the solution to any problem. This method should only be used with caution.

4.3 Negative Surge Waves The analysis used for the positive surge waves in the preceding section can also be used to analyse negative surge waves. In both cases it will give a reasonable approximation of the wave characteristics immediately after formation of the wave. In the case of the +ve wave it will also give an indication of the transmission of the wave over some distance. Negative waves are unstable and decay rapidly as they are transmitted along a channel. Consider the surge to be composed of a number of small disturbances superimposed on each other. The celerity of each disturbance would be given by c  gy . Clearly the disturbance nearest the leading edge of the wave (and in the deepest water) would travel at the highest velocity. The leading edge of the surge would soon outdistance the trough and the wave flatten, eventually becoming a gradually varied wave.

Readings from Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Section 13.6 Mastery of all of this material is necessary. The alternative analysis for -ve waves presented in this section allows prediction of the transmission of these waves over relatively long distances. It should be noted that this analysis also assumes a horizontal and frictionless channel bed and its application will be limited by these assumptions. The outcome of this alternative analysis is the following series of equations which allow calculation of the flow velocity V and wave velocity Vw at any point in the wave. The choice of which equation to use depends on the type of negative wave you are dealing with and which depth and velocity are known.

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ENV3104 – Hydraulics II

Upstream Negative Waves

Downstream Negative Waves

V  V1  2 gy1  2 gy

V  2 gy  2 gy1  V1

Vw  3 gy  V1  2 gy1

Vw  3 gy  2 gy1  V1

V  V2  2 gy2  2 gy

V  2 gy  2 gy2  V2

Vw  3 gy  V2  2 gy2

Vw  3 gy  2 gy2  V2

4.4 Solution of Problems Involving Surge Waves Worked Examples From Texts Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Examples 13.1 to 13.4

4.5 Self assessment Questions

4.1

4.2

The fundamental laws used in the solution of surge problems are: (a)

the Continuity and Energy equations

(b)

the Continuity and Momentum equations

(c)

the Energy and Momentum equations

(d)

the Continuity and Bernoulli equations

An upstream positive wave:

(c)

can be caused by the rapid closure of a sluice gate in an open channel can be caused by the rapid opening of a sluice gate in an open channel decays rapidly with distance

(d)

causes a decrease in the depth of flow

(a) (b)

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Module 4 – Rapidly Varied Unsteady Flow

4.3

4.4

4.5

31

A positive surge wave in an open channel: (a)

travels slower than the initial flow velocity

(b)

can never travel upstream

(c)

causes an increase in the depth of flow

(d)

decays rapidly with distance travelled

A negative surge wave in an open channel: (a)

travels slower than the initial flow velocity

(b)

can never travel upstream

(c)

causes an increase in the depth of flow

(d)

decays rapidly with distance travelled

A tidal channel, which may be assumed to be rectangular, 40 m wide, conveys a steady freshwater discharge of 60 m3s-1 at a depth of 1.52 m. After a tidal bore propagates upstream the depth and velocity of flow are 2.66 m and -1.58 ms-1, respectively. Using only the continuity equation, determine the velocity of the bore relative to a stationary observer.

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ENV3104 – Hydraulics II

4.6 Tutorial Problems Answers to these problems are given at the rear of the text. 4.1-4.5 Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Problems on Chapter 13, page 326 (4th Ed. p. 360/61) Questions 1 to 5

4.7 Solutions to Self assessment Questions 1. (b)

2. (a)

5. Vw = 5 m/s

. .

© University of Southern Queensland

3. (c)

4. (d)

Module 5 – Gradually Varied Unsteady Flow

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Module 5 – Gradually Varied Unsteady Flow

5.1 Objectives When you have mastered the material in this module you should be able to:  understand the derivation of the equations of motion governing gradually varied unsteady free-surface flow;  apply the equations to the solution of practical problems involving, for example;  routing of a flood hydrograph through a natural stream, and  calculation of runoff due to rainfall incident on a plane surface.

5.2 Equations of Motion Book of Readings – Reading 5.1 Part Chapter 2 (pages 23 to 28) from Stephenson, D. and Meadows, M.E. 1986, Kinematic Hydrology and Modelling. Elsevier, Amsterdam. This a rigorous derivation of the free surface unsteady flow equations (or Governing equations). It is specifically for a broad overland flow and includes a lateral inflow qi m3 per unit length of channel. The proof for an irregular channel would not differ significantly. You should be able to follow the derivation but will not be expected to reproduce it. The final forms of the continuity and momentum equations as given by the text are: Continuity(Eq. 2.4):

Q A   qi x t

Momentum(Eq. 2.9):

Vq V V y V g  g S0  S f  i t x x A

5.1





5.2

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5.2.1 Application to Unit Width of Channel It is sometimes more convenient to talk in terms of unit width of channel. This is particularly so when analysing flow over a plane surface. In this case the equations are the Momentum equation: 1 V V V y Vr    So  S f  g t g x x gy

5.3

and the Continuity equation:

q y  r x t

5.4

where q is the discharge per unit width and equals Vy. Note that I prefer to use the symbol r for the lateral inflow per unit area of water surface rather than the symbol qi used in Article 5.1. Also note the difference in units between r and qi. Rearrangement of the Momentum Equation provides a useful illustration of the relationship between the various types of flow. S f  So



V V g x



1 V g t



Vr gy

5.5

Steady uniform (or normal) flow Steady non-uniform (or steady gradually-varied) flow Unsteady non-uniform (or unsteady gradually-varied) flow Unsteady non-uniform flow with lateral inflow Progressively eliminating terms from the right hand side reduces the equation from that for a fully unsteady flow system eventually to a normal flow equation. Similarly for steady flow the Continuity equation reduces to:

dq 0 dx

5.6

given that q = Vy then:

V

dy dV y 0 dx dx

5.7

To simplify solution of the equations governing unsteady flow, approximate (or simplified) forms of the Momentum equation are frequently used. The most common © University of Southern Queensland

Module 5 – Gradually Varied Unsteady Flow

35

of these is the Kinematic Wave Approximation. Here the Momentum equation is reduced to that applying to normal flow, viz: S0  S f

5.8

The argument used is that in certain circumstances the remaining terms are either small or cancel each other. The Continuity equation remains in full:

q y  r x t

5.9

Together these are known as the Kinematic equations.

With both the Governing and the Kinematic equations an expression is needed to evaluate the energy slope term Sf. This is usually the Manning equation, which for an overland flow can also be expressed in terms of unit width of channel as:

q

1 12 5 3 Sf y n

5.10

5.2.2 Alternative Derivation of the Backwater Equation From above the equations applicable to steady gradually varied flow are:

dy dV y 0 dx dx

Continuity:

V

and Momentum:

V dV dy   So  S f g dx dx

Combining and re-arranging results in:

5.11

dy So  S f  dx 1  FR2

5.12

5.13

Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Part Section 14.6 - Pages 502 to 504. An alternative derivation of the Governing equations (14.33b and 14.34b), included here for background knowledge only. The equations derived are frequently referred to as the St Venant equations, after their originator. In the form presented they apply to both natural and man made channels. Note the absence of any term to describe the lateral inflows (and outflows).

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5.3 Numerical Solutions 5.3.1 Characteristics Book of Readings – Reading 5.2 Part Chapter 5 (pages 81 to 83) from Stephenson, D. and Meadows, M.E. 1986, Kinematic Hydrology and Modelling. Elsevier, Amsterdam. Quickly read this material. Two aspects are important, viz: the characteristic equations (5.5) and the zones they define on the x - t solution space (Figure 5.1). The Characteristic equations are equivalent to: Vw  V  c

5.14

These define the motion of small disturbances (or changes) to the flow. For example, for a river discharging to the ocean the lower boundary condition will be the water surface elevation as determined by the tides. Any minor change in that elevation caused by the changing tide will propagate upstream with velocity Vw  V  c . That is, the backward characteristic as shown in Figure 5.1 of Stephenson and Meadows. Both the full governing equations and the kinematic equations are best solved using numerical techniques. In both cases the objective is to solve for the unknowns Q, V and y at particular points along the channel or stream and at particular times. For example, we may be interested in the variation in discharge with time at the downstream end or with the peak values at all points in the stream. To achieve this it is usually necessary to solve for the unknowns at all points in the x - t solution space.

5.3.2 Finite Difference Techniques Book of Readings – Reading 5.3 Part Chapter 5 (pages 86 to 88) from Stephenson, D. and Meadows, M.E. 1986, Kinematic Hydrology and Modelling. Elsevier, Amsterdam. Mastery of the concept of difference quotients and of the forward, backward and central difference approximations of any partial derivative is absolutely essential.

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5.3.3 Numerical Solution Schemes Book of Readings – Reading 5.4 Part Chapter 5 (pages 88 to 94) from Stephenson, D. and Meadows, M.E. 1986, Kinematic Hydrology and Modelling. Elsevier, Amsterdam. You must be able to set up the equations for any explicit solution scheme using the appropriate difference approximations. You should also be able to write the computer code for the explicit solution. Take particular note of the need to ensure a stable solution through appropriate grid sizes as expressed by the Courant condition (equation 5.23). Note here that Chadwick et al. use a slightly different form of the same expression for the time step, viz:

t 

x c

5.15

but we should remember they are using c for the wave speed where we would use Vw. You should be able to set up the equations for an implicit solution. However you will not be required to solve the equations. Note the requirement for the number of equations to equal the number of unknowns.

5.3.4 Boundary Conditions Both the explicit and implicit solution schemes require knowledge of certain initial (t = 0), upper boundary (x = 0) and lower boundary (x = L) conditions. Typically these might include the following: Initial Condition - values of the unknowns q, V and y at all points along the channel at time zero (that is, at all x when t = 0). Upstream or Upper Boundary - it is usually necessary to know the value of one unknown at the upstream boundary (that is, at all t when x = 0) for all t, for example, discharge into the channel may be available in the form of an inflow hydrograph (that is, Q vs t). Alternatively the upper boundary information may be presented in the form of an equation linking two of the unknowns, for example, as a rating curve (that is, an equation linking discharge with depth at that upper boundary). Downstream or Lower Boundary - again it is usually necessary to know the value of one unknown at all times or have an equation linking two unknowns at the lower boundary (that is, at all t when x = L). Typically these may be:  a tidal condition where y(t) is known at all times; or  a rating curve linking q with y. © University of Southern Queensland

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ENV3104 – Hydraulics II

Worked Example 5.1 The depths at a particular point in a stream measured at three times 5 minutes apart are 4.26, 4.38 and 4.44 m. Estimate the rate of change of depth at the central time using: a) a forward difference approximation; b) a backward difference approximation; and c) a central difference approximation. Solution let

yt  4.38 m

yt t  4.26 m

yt t  4.44 m

t  300 s

a)

Forward Diff

y yt t  yt 4.44  4.38   t t 300

= 0.0002 m/s

b)

Backward Diff

y yt  yt t 4.38  4.26   t t 300

= 0.0004 m/s

c)

Central Diff

y yt t  yt t 4.44  4.26   t 2t 600

= 0.0003 m/s

Worked Example 5.2 If the discharge at the same point in the above stream at time t is 80 m3/s and the width of the stream is 20 m, using only the continuity equation estimate the discharge 1 km downstream. Solution From the central difference above:

y  0.0003 m/s t

And the continuity equation (Eq. 5.1) adapted for steady flow:

Q y b  0 x t

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39

Q  20  0.0003  0 x Q  0.006 m3 /s/m length x

thus:

Using a forward difference in the x direction:

Q Qx x  Qx  x x Q  80 0.006  x x 1000 Qx x  80  0.006 1000 Qx x  74m3s-1

Worked Example From Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Example 14.4 This not a particularly easy example to follow, largely due to the different notation used. However it is a good example of the movement of a flood wave down a natural stream.

5.4 Self assessment Questions .

5.1

The discharges in a river at three points 100 m apart (in the downstream direction) are 65, 71 and 80 m3/s. Estimate the spatial rate of change of discharge Q/x at the central point using a forward difference approximation.

5.2

Repeat Question 1 using a backward difference approximation.

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ENV3104 – Hydraulics II

5.3

Repeat Question 1 using a central difference approximation.

5.4

The discharge in Question 1 is: (a)

decreasing with time

(b)

decreasing with distance downstream

(c)

increasing with time

(d)

increasing with distance upstream

5.5

If the depth of flow at the central point in Question 3 above is 5 m and the width of the river is 100 m, using only the Continuity equation estimate the depth of flow in 5 minutes time.

5.6

At a particular point in a channel the depth was seen to increase from 3 m to 3.2 m in a time of 20 min. Estimate y/t.

5.7

The discharge in Question 6 is: (a)

decreasing with distance upstream

(b)

decreasing with time

(c)

increasing with distance downstream

(d)

increasing with distance upstream

5.8

The tide at the mouth of a river 200 m wide is rising at the rate of 0.2 m/hour, estimate the difference between the discharge at the mouth and that 500 m upstream from the mouth. [Hint: use only the Continuity equation.]

5.9

You are observing the rising stage of a major flood in a large river. At your observation point it is seen that at a certain time the discharge is 2100 m3/s and that the water level is rising at the rate of 0.3 m/hour. The width of the river at this point, and for some distance upstream and downstream, is 800 m. You are asked to make a quick estimate of the present magnitude of the discharge at a point 8000 m upstream. Using only the Continuity equation, what is your estimate?

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Module 5 – Gradually Varied Unsteady Flow

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5.10 A stream conveying a flood peak of 50 m3/s (velocity 1.25 m/s and depth 1 m) is divided into 1 km long reaches for solution of the unsteady equations. The Courant requirement suggests that the largest value of t for which the solution will be stable is: (a)

60 s

(b)

240 s

(c)

3.5 min

(d)

5 min

5.5 Tutorial Problems 5.1

The task is to develop the algorithms for the explicit solution of the full Governing or St Venant equations when applied to the problem of runoff from a plane surface subject to a steady rate of rainfall excess. The initial and boundary conditions are as follows:  Initial condition (t = 0):

q = 0 and y = 0 for all x;

 Upstream boundary (x = 0):

q = 0 for all t;

 Downstream boundary (x = L):

So = Sf.

Develop the equations for the solution of q, v and y as follows: a. In the main body of the x-t solution space using central differences; b. At the upstream boundary using forward differences for the spatially variant terms; c. At the downstream boundary using backward differences for the spatially variant terms; and d. At time zero and any x using forward differences for the time variant terms. 5.2

Repeat Question 1(a) using an array terminology with the columns (x position) designated i and the row (t position) designated j and where the equations are applied at the point i, j.

5.3

Repeat Question 1, using the Kinematic equations rather than the full St Venant equations.

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5.4

ENV3104 – Hydraulics II

Repeat Question 3, using forward differences in the time direction and backward differences in the x direction throughout the solution space. (This is the first stage of Assignment 1 hence no solution is given for this problem. When you have formulated your solution you should send it to the examiner of the course for confirmation prior to proceeding with the rest of the assignment.)

5.5.1 Answers to Tutorial Problems

5.1

The terminology used in this solution is similar to that in Figure 5.7 of Reading 5.4, with an additional point O located immediately below point M (that is, at the same distance x as M but one time increment earlier) as indicated below:

a) The solution of Question 1(a) is illustrated using the continuity equation:

q y  r x t We need approximations for the two derivative terms q/x and y/t. The equations are to be applied at point M. At this point, q/x is approximated by taking the average rate of change of discharge q in the x direction, that is, by using the values of q at points R and L, giving: q q R  q L  x 2x

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43

Similarly, y/t is approximated by: y y P  yO  t 2 t

Substitution into the continuity equation and re-arrangement gives the first equation in the solution:

q  qL   y P   rM  R 2t  yO 2x   A similar process is followed to give approximations for the derivative terms in the momentum equation, with the result being:  V r V V  V L  y R  y L  V P   S o  S fM  M M  M R   2 gt  VO gy M g 2x 2x  

The solutions for 1 (b), 1 (c) and 1 (d) follow the same path except for one significant difference. In all cases, because we are near the boundary of the solution space, one of the points L, O or R does not exist (depending on which boundary we are on). In that case approximations in the direction perpendicular to the boundary are found across one grid square only. We should also be alert to the fact that, at time zero and at the upstream boundary, certain variables have zero values. b) At the upstream boundary we have:

As before:

but:

y y P  yO  t 2 t q q R  q M  x x

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ENV3104 – Hydraulics II

We also know that qM = 0, hence q q R  x x

Substitution into the continuity equation and re-arrangement gives:

q   y P   rM  R 2t  yO x   We do not have to apply the momentum equation to calculate VP at this boundary because the given boundary condition declared q to be zero. Hence: VP  0

A similar result occurs at the downstream boundary where the given boundary condition over-rides the momentum equation. c)

q  qL   y P   rM  M 2t  yO x   Because of the given boundary condition S f  S0 :

VP  d)

1 1/ 2 2 / 3 So yP n

y P  rM t VP is indeterminate because SfM is indeterminate at t = 0 because q and y are zero. This means that the Kinematic approximation must be used for the first time step, that is the momentum equation is ignored for the first time step. In all cases (except the initial condition) SfM is found from the Manning equation, viz:

VM 

1 1/2 2/3 S f yM n M

and:

q p  Vp y p

In solving these equations numerically it is assumed that we know the values of all terms subscripted L, M, R, or O and that only the values at point P are unknown.

© University of Southern Queensland

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5.2

45

The equations are to be applied at the point i, j.

Using this terminology, the approximation for q/x becomes: q qi 1, j  qi 1, j  x 2x

Similarly, y/t is approximated by: y y i , j 1  y i , j 1  t 2 t

Substitution into the continuity equation and re-arrangement gives the first equation in the solution: qi 1, j  qi 1, j  y i , j 1   ri , j  2x 

 2t  y i , j 1 

Again a similar process is followed to give approximations for the derivative terms in the momentum equation, with the result being:

 Vi , j ri , j Vi , j Vi 1, j  Vi 1, j  y i 1, j  yi 1, j  Vi , j 1   S o  S fi , j     2 gt  Vi , j 1 gyi , j g 2x 2x  

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ENV3104 – Hydraulics II

The solution for this problem follows a similar approach to Question 1 except that the Momentum equation does not apply and the velocity at point P is determined from the depth using the Manning equation.

5.3

a)

q  qL   y P   rM  R 2t  yO 2x  

b)

VP  0 q   y P   rM  R 2t  yO x  

c)

q  qL   y P   rM  M 2t  yO x  

d)

y P  rM t Except where specified VP is found from the Manning equation, viz:

VP 

1 1/ 2 2 / 3 So yP n

and:

q p  Vp y p

5.6 Solutions to Self assessment Questions 1. 0.09 m3s-1/m

2. 0.06 m3s-1/m

3. 0.075 m3s-1/m

5. 4.775 m

6. 0.01 m/min

7. (d)

8. Qu/s 20000 m3/h greater

. .

© University of Southern Queensland

9. 2633 m3s-1

4. (a)

10. (c)

Module 6 – Sediment Transport

47

Module 6 – Sediment Transport

6.1 Objectives When you have mastered the material in this module you should be able to:  define the threshold of motion of a particle at rest on the bed of a channel;  predict the particle size that will be present from the armouring of a stream;  discuss the mechanics of sediment transport; and  apply the sediment transport equations to the estimation of the sediment load of a channel or stream.

6.2 Bed Form Book of Readings – Reading 6.1 Part Chapter 10 (pages 406-408) Henderson, F.M. 1966, Open Channel Flow. Macmillan, New York. Important background material. The bed forms seen in the breaker zone at a surf beach and on the sand flats in estuaries are usually either ripples or dunes with superimposed ripples, as shown in Figure 10.1 (a) and (b) in the article from Henderson.

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6.3 Threshold of Motion Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 9.2 Important material in this reading is:  the application of Newton's 2nd Law to the exposed particle (in this case F = 0);  the development of the "entrainment function";  the Shields' diagram (Figure 9.3) ; and  equation 9.1c for the critical or threshold shear stress. Module 7 relies on mastery of this material. The entrainment function is essentially a Froude number (squared) thus the Shields' diagram can be viewed as a plot of the particle Froude number versus the particle Reynolds number. The entrainment function Fs is given by:

Fs 

o   s    gd

6.1

where o is the average shear stress on the bed of the channel;

s is the density of the solid particles; and d is the particle diameter. Marriott (Nalluri and Featherstone) uses an alternative expression: Fs 

o gd

6.2

where  is the submerged relative density of the particles and is given by; 

1



 s   

6.3

This is equivalent to the expression (Ss - 1) used in other publications. NOTE: This differs from Hydraulics 1 where Fs was instead used for the specific force © University of Southern Queensland

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Book of Readings – Reading 6.2 Part Chapter 10 (pages 411-413) Henderson, F.M. 1966, Open Channel Flow. Macmillan, New York Further background material giving an alternative experimental approach to the determination of the threshold of motion. The quality of the Shields' diagram presented here is significantly higher than those presented in the texts, as is the discussion of the diagram.

6.4 Armouring of Streams In many natural streams, over the course of time the smaller particles are flushed out of the surface of the bed leaving only the larger material. This is known as armouring of the bed. The entrainment function can be used to determine the minimum size of stone that will be left behind on the bed. At the threshold of motion the entrainment function will approximate 0.056 for a typical turbulent flow (that is, high NR), therefore:

Fs 

o  0.056  s    gd

6.4

substituting  0   gRS0 and assuming  s    /   1.65 gives: d  11RS0

6.5

where d is the diameter of the largest particles which will be removed from the bed (or the smallest particles which will remain behind).

6.5 Resistance to Flow Readings from Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Section 14.4 You should be aware of the range of flow resistance equations available and be able to apply them to a given situation. You will not be required to commit them to memory. The material in this section will not be examined.

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The equations presented under the heading of the 'total resistance approach' are all empirical expressions of similar form to the normal flow equations such as Manning. Here the value of the resistance coefficient and the exponents are varied to suit the experimental data. It is important that these equations only be used in situations similar to those for which they were derived.

6.6 Mechanics of Sediment Transport Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 9.3 This material is included as background for the next section on sediment transport equations. You require only a passing knowledge of this material.

6.7 Sediment Transport Equations Readings from Texts Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 9.4 and 9.5 Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Section 14.4 You should be aware of the range of transport equations available and be able to apply them to a given situation. You will not be required to commit them to memory. The material in this section will not be examined. The transport of sediment in channels or in natural streams is one of the grey areas of hydraulics. An inability to rigourously describe the very complex physical processes that control the entrainment, transport and deposition of both cohesive and noncohesive materials has led to the development of numerous empirical and semirational equations. The more important are summarised below.

6.7.1 Bed Load Formulae Understandably these formulae have a common form. The bed shear stress or excess shear stress (shear stress above the threshold value) is the central component. The other terms in the entrainment function also make an appearance. The three most common equations are: © University of Southern Queensland

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Du Boys: q s  K 3 o  o   cr 

6.6

Chadwick et al. Section 9.4 presents some example alternative expressions for K3. K3 is function of s, g and d which has led to a variety of expressions for qs, for example, the shields equation below.

Shields:

qs 

10qS o  2  o   cr 

 s   s    gd 2

6.7

Einstein-Brown:

qs 

  o  40     s    gd  gd 3   s   

3

6.8

Equivalent to Chadwick et al. equations 9.12 & 9.17.

In each of these equations qs is the volumetric rate of sediment transport per unit width of channel (m3s-1/m); and cr is the threshold shear stress. All other terms have their usual meanings.

6.7.2 Bed Shear Stress The bed shear stress o can be interpreted as the mean shear stress on the wetted perimeter in which case it will be evaluated using:

 0   gRS0

6.9

Alternatively it may be taken as the shear stress on the level portion of the wetted perimeter, that is, the actual bed shear stress and be given by:

 b   gyS0

6.10

For broad shallow channels (where R  y) these two expressions are the same. In all other cases the latter expression will give the higher shear stress and hence higher sediment load.

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The critical or threshold shear stress cr will be obtained from the shields diagram. For most real flows, that is, relatively high particle Reynolds Number, the value of the entrainment function will be 0.056, thus:

 cr  0.056  s    gd

6.11

6.7.3 Total Load Formulae The total (bed + suspended) load formulae tend to be more recent developments. The physical process is even more complex than for bed load transport alone, hence the results are not encapsulated into the single equations typical of the bed load case. The Ackers-White approach is a typical and probably the preferred total load equation.

6.8 Worked Examples Worked Examples From Texts Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 14.2 Chadwick et al. Hydraulics in Civil and Environmental Engineering Examples 9.2 and 9.3 In both examples 14.2 and 9.2 it is debatable whether to use R or y in the calculation of the bed shear stress (see Section 6.7 above). In example 9.3 concentrate on the solution using the Ackers-White equations.

6.9 Self assessment Questions .

6.1 The entrainment function is expressed as: (a) (c)

V gy

o 0.056  s   gd

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(b) (d)

o  s   gd o   s   gd

Module 6 – Sediment Transport

6.2

53

The threshold of motion of a particle occurs when: (a)

the particle Reynolds number is high

(b)

the Froude number of the flow is > 1

(c)

the entrainment function exceeds the Reynolds number

(d)

none of the above answers

6.3 A stream bed (slope 0.005 and n of 0.018) is observed to comprise particles larger than 10 mm. The normal mean flow velocity (m/s) of the stream is: (a)

3.26

(b)

2.52

(c)

1.63

(d)

1.26

6.10 Tutorial Problems Solutions to all the questions below are provided in the text 6.1

Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Problems on Chapter 14 (page 359) Question 1

6.2

Marriott

Ch 14, Question 2

6.3

Marriott

Ch 14, Question 5

Note that Question 5 requires use of the Chezy equation and equation 14.21 from Marriott to evaluate the Chezy C. 6.4

Chadwick et al. Hydraulics in Civil and Environmental Engineering Page 627 Problems on Chapter 9 Question 1. Part (b) of this question i included for interest only

6.5

Chadwick et al.

Ch 9, Question 2

This question is included for purposes of illustration only and will not be examined.

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6.11 Solutions to Self assessment Questions 1. (b)

2. (d) . .

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3. (d)

Module 7 – Tractive Force Design

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Module 7 – Tractive Force Design

7.1 Objectives When you have mastered the material in this module you should be able to:  design a channel (excavated through erodible material) such that it will not scour; and  predict the most efficient stable cross section for a channel.

7.2 Introductory Comments on Design of Erodible Channels In Module 2 it was noted that in the design of rigid bed (non-erodible) channels there is no limit imposed on the velocity of flow in the channel. Therefore, for a given discharge, slope and roughness, there is an infinite number of combinations of channel width and depth that will satisfy the normal flow equation used in the design. For channels excavated through earth materials, that is channels with the potential to erode, clearly there will be some upper limit on velocity, below which erosion or scouring of the channel wetted perimeter will not occur. The imposition of this upper limit on flow velocity (either directly or implied) means that there is now a single optimum solution to the design problem. This optimum is the channel cross section for which at the design discharge the bed material is at the threshold of motion. One consequence of imposing a limit on the flow velocity is that the allowable depth of flow will be shallow and the channel section will be broad. There are a number of methods available for the design of channels in erodible materials, viz:  the permissible velocity approach;  the rational approach;  the regime approach; and  the tractive force method.

The permissible velocity approach involves the specification of a maximum or erosion threshold velocity Vmax, based on knowledge of the erodibility of the material comprising the channel bed and banks. Older text books, such as Chow (1959) or © University of Southern Queensland

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Henderson (1966), usually contain a description of the method and tables for the selection of the permissible velocity. Marriott makes a fleeting reference to the method on page 190 (4th Ed. p199). The solution procedure for the permissible velocity approach (for known Q, side slope z and So) is as follows:  select values for n and Vmax;  solve the Manning equation for R (note that having nominated a value for Vmax there is now a unique solution for R);  using A = Q/V and P = A/R solve for the bed width b and depth y (again there is a unique solution for b and y).

The permissible velocity approach is not recommended for earthen channels but is widely used for vegetative lined channels, and this use will be described in Module 8. The regime and rational approaches are discussed on page 531 to 533 of Chadwick et al. and further discussion here is not required. These methods are also not recommended. The remainder of this module will be a description and application of the tractive force method.

7.3 Tractive or Shear Force The tractive force is the force exerted by the water on the wetted area of a channel. Following on from this definition is that of 'unit tractive force' which is the force per unit area or shear stress on the wetted perimeter. Although once commonly used, the term tractive force is now limited to this application. Shear stress is now the preferred terminology elsewhere. We have seen previously that the mean shear stress on the channel wetted perimeter is given by:

 0   gRS0

7.1

However this boundary shear stress is not uniformly distributed around the wetted perimeter. The maximum shear stress will occur on the bed and for a trapezoidal channel may be taken as gySo. On the channel sides the shear stresses will be lower and reach a maximum of approximately 0.75gySo (see Figure 15.4 (b) in Chadwick et al.).

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Chow (1959) suggests that the shear stress distribution varies with the shape of the channel. For the purpose of this course this variation will be ignored.

It is necessary to define four shear stress terms, viz:

b is the actual shear stress on the channel bed, given by:  b   gyS0

7.2

s is the actual shear stress on the channel sides, given by:  s  0.75 gyS0

7.3

cb is the critical or threshold shear stress on the channel bed and is calculated from the Shields' criterion, that is, equating the entrainment function Fs to a value of 0.056, giving:

 cb  0.056  s    gd

7.4

cs is the critical or threshold shear stress on the channel sides and is given by:  cs  K cb

7.5

where K is the tractive force ratio.

7.4 Tractive Force Ratio Book of Readings – Reading 7.1 Section 7.12 (pages 170 and 171) from Chow, V.T. 1959, Open Channel Hydraulics. McGraw-Hill, New York. The tractive force ratio is determined from a consideration of the threshold of motion of particles on the bed and sides of a channel. The derivation is included as important background to the topic and for the resulting expression for K. Note that Chow uses  for the side slope angle of the channel and  for the angle of internal friction of the bed material. The two texts and this study guide use the reverse notation, viz:  for the side slope angle and  for the angle of internal friction.

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7.5 Design Procedure The word design as used here simply means the selection of the channel cross section dimensions (bed width b and depth y) such that erosion of the channel bed or sides does not occur. This usually involves three steps, viz:  estimation of the critical bed shear cb and the tractive force ratio K;  calculation of the maximum depth y allowed from a consideration of the shear stresses; and  application of a normal flow equation such as Manning to give the width b required to convey the given discharge. The depth y is determined from either consideration of the bed:

 b   cb

7.6

which simply says that the actual shear stress on the bed must be equal or less than the threshold value. Or similarly for the sides of the channel:

 s   cs

7.7

These expressions lead directly to:

 gyS0  0.056  s    gd

7.8

0.75 gyS0  K  0.056  s    gd

7.9

and:

Thus:

y

0.056   s    d

 S0

or

y

K  0.056   s    d 0.75 S0

7.10

The depth adopted is the lesser of these two values. The resulting depth y will satisfy both expressions ensuring that erosion does not occur on the bed or sides.

7.6 Stable Hydraulic Section In a trapezoidal channel the condition of impending motion occurs at only one part of the wetted perimeter, either on the bed or the sides depending on the shear stress distribution and the value of K. Over the remainder of the wetted perimeter the shear stresses are less than the appropriate threshold value. © University of Southern Queensland

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Intuitively the maximum conveyance efficiency would result if a state of impending motion prevailed over the entire perimeter. The channel cross section where this applies is known as the "stable hydraulic section" and for a given discharge results in:  minimum area of section;  maximum value of the mean velocity; and  minimum top width.

If the side slope angle  of the channel is allowed to vary continuously from zero at the centre to  at the intersection with the water surface and the actual shear stresses at all points are equated to the corresponding threshold shear stresses then the following equation results:  x tan   y  cos   yo  yo 

7.11

which is the equation of a cosine curve giving depth y at distance x from the centre line of the channel in terms of the depth yo at the centre and where: y0   cb  gS0

7.12

The properties of the section are:

A

2 yo2 tan 

7.13

P

2 yo E sin 

7.14

R

yo cos  E

7.15

and:

where E is a complete elliptical integral of the second kind, values for which are given in Article 7.2 in the Book of Readings. For the values of  likely for most soils, E can be approximated by: E



1 2  1  sin   2 4 

7.16

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The velocity (and discharge) in the channel can be calculated in the usual manner, by application of the Manning equation. If the capacity of the channel is less than required then an appropriate width of level bed can be inserted. From the Manning equation the bed width b required is:

b

nQ S01/2 y05/3

7.17

where Q' is the difference between the required discharge and the capacity of the stable section.

The stable hydraulic section is the shape that a trapezoidal channel constructed in erodible material would be expected to develop over long time. This material on the stable section is include for interest only and will not be examined.

7.7 Further Reading Readings from Texts Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Section 8.5 Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 15.4 (pages 533-536) The material in both texts covers that presented in Sections 7.3 to 7.5 of these notes but in less detail.

7.8 Worked Examples Worked Example From Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 8.10 A simple example of the case where the critical tractive force is given rather than calculated from the Shields' criterion. This solution also assumes that the condition of impending motion will first occur on the sides of the channel. While this will be the case in most channels it is wise to also compare b with cb.

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Worked Example 7.1 Determine the stable hydraulic section for the channel in example 8.10 above, viz:

 cb  2.4Nm2

  30

Q  10 m3 /s

n  0.02

S0  0.0001

Solution y0   cb  gS0  2.4465 m

and:

2 yo2 A  20.734 m2 tan 

For   30 , Looking up the table in Reading 7.2, E  1.4675 thus:

R

y0 cos   1.4438 E

V

1 1/2 2/3 So R  0.63872 m/s n

Therefore:

Q  VA Q  13.243  13.2 m3 /s This is greater than the required discharge therefore the section could be reduced in width from the centre until the desired capacity is achieved. The top width W of the channel can be found from: W  b  2 x0

where xo is the distance from the point where y = yo to y = 0, that is:  x tan   cos  0  yo 

or:

 x tan       yo  2

Thus: x0 

 y0  6.6562  6.66 m 2 tan 

and:

W  13.3124  13.3 m

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It is interesting to compare the answers for the above problem to those for example 8.10.

"Stable" section

Marriott

Depth y (m)

2.45

1.44

Hydraulic rad. R (m)

1.44

1.13

0

9.95

Top width W (m)

13.31

15.71

Velocity V (m/s)

0.639

0.54

13.2

10.0

Bed width b (m)

Discharge Q (m3/s)

7.9 Self assessment Questions .

7.1 The threshold shear stress on the sides of an earthen channel can be expressed by:

7.2

(a)

0.75gySo

(b)

(s - )gd

(c)

0.056(s - )gd

(d)

K 0.056(s - )gd

The tractive force ratio is defined as:

(b)

the ratio of the side slope angle to the angle of internal friction of the soil, ie, / s/cs

(c)

s/b

(d)

cs/cb

(a)

7.3 Motion of a particle on the bed of a channel will occur if: (a)

b < cb

(b)

b > gySo

(c)

Fs > 0.056

(d)

K>1

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7.4 A channel 0.5 m deep is constructed in material with its threshold of motion defined by a shear stress of 12 Nm-1. The maximum allowable bed slope for this channel is: (a)

0.00035

(b)

0.0024

(c)

0.0035

(d)

0.024

7.10 Tutorial Problems 7.1

Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Problems on Chapter 8, page 239 (4th Ed. p 255) Question 13 The answer to this problem is given at the rear of the text.

7.2

The compound channel illustrated in below is used to transmit flood flows. The bed slope is 0.001. Determine the maximum non-eroding depth and the corresponding discharge. Justify any assumptions you make.

The bed material in each section is as listed below: Median size Friction angle (mm) (°)

Section AB

35

35

BC

35

35

CD

50

37

DE

20

30

EF

20

30

The Manning n for each section can be found from the Strickler equation: n  0.0385  d 

16

where d is the particle diameter in m.

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7.10.1 Answers to Tutorial Problems

7.1

Answer provided in Marriott

7.2

A maximum depth must be calculated for each section of the channel as follows: Depth (m)

Section AB

3.89

BC

3.22

CD

5.60

DE

1.85 + 1.5

EF

2.15 + 1.5

Adopt the lowest depth therefore: y = 3.22 m

and:

Q = 660 m3s-1

7.11 Solutions to Self assessment Questions 1. (d)

2. (d)

. .

© University of Southern Queensland

3. (c)

4. (b)

Module 8 – Vegetative Lined Channels

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Module 8 – Vegetative Lined Channels

8.1 Objectives When you have mastered the material in this module you should be able to:  estimate the Manning n for a vegetated surface; and  design and analyse the performance of channels with a vegetative lining.

8.2 Vegetated Channels Channels which carry flows only intermittently can be protected by a lining of vegetation (preferably a sod forming grass). The effect of this grass is twofold. First, the grass binds and covers the soil surface thus allowing a greater flow velocity. Second, the increased roughness caused by the vegetation will reduce the velocity for a given depth. Again the imposition of an upper limit on the flow velocity will result in a relatively broad shallow cross section. Examples of grassed lined channels are the waterways associated with soil conservation works. Urban floodways may have a grassed floodway in association with a concrete lined low-flow channel. Design is based on the permissible velocity approach, the value of the permissible velocity being a function of the vegetation (type, cover and physical condition) and of the soil. Two problems arise in the estimation of n for a grass lined channel. The Manning n will vary with:  the depth and velocity of flow; and  the condition of the vegetation (which will vary with growth stage, time of year and the grazing management or mowing of the grass).

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8.3 Manning n Book of Readings – Reading 8.1 From Ree, W.O. 1949, Hydraulic characteristics of vegetation for vegetated waterways. Agricultural Engineering, 30: 184-185. Important illustration of the variation of n with depth of flow for a particular grass. The variation with discharge would show a similar pattern. Note the large range of hydraulic resistance (n ranging from 0.04 – 0.4) depending on the depth of flow.

Book of Readings – Reading 8.2 Section 7-17 from Chow, V.T. 1959, Open Channel Flow. McGrawHill, New York. Useful background to the development of the relationships describing the Manning n for particular grasses. This data is provided for information only and should not be used in calculations unless care is taken to translate the units from the imperial (ft - s) system used in this article. The work by the US Soil Conservation Service, illustrated in the two readings, led to the development of standard curves for describing the vegetative retardance of fully submerged grasses. Five standard retardance classes were adopted, designated A to E in order of decreasing roughness or retardance. For each class the Manning n is plotted against the product VR. Subsequent Australian work has led to the development of guide-lines for the selection of appropriate retardance classes for the grasses commonly used in Australia. For water at constant temperature (constant viscosity) the product VR is essentially a Reynolds Number.

Book of Readings – Reading 8.3 Figure 1.9 from the Queensland Water Resources Commission. Farm Water Supplies Design Manual Vol 1 Farm Storages Standard n-VR curves and the Australian guide-lines for the selection of retardance class for particular grasses. Note the variation in recommended class with length and condition of the grass.

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8.4 Design Procedure The design must satisfy two main criteria, viz:  The velocity of flow does not exceed the permissible velocity nominated for the particular grass and soil in the waterway, that is, the channel must be stable.  The depth of flow does not exceed the height of the channel banks, that is, the channel must have sufficient capacity.

The highest flow velocity is most likely to occur when the vegetative retardance is at a minimum and overtopping occur when the retardance is at a maximum. In other words the stability of the waterway should be based on the lowest retardance expected and the capacity determined from the highest retardance expected.

The steps in the design (for known Q, z and So) are as follows:

1. Preliminary: For the given grass determine the range of conditions to be expected, the corresponding retardance curves and the permissible velocity Vmax.

2. Stability: Select the n-VR curve corresponding to the lowest retardance to be expected, then iteratively solve for R as follows: a) assume a value for n; b) with V = Vmax solve the Manning equation for R; c) calculate VR; c) from the n-VR curves read a new value for n; and d) repeat the above steps until the assumed and final values of n are the same.

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Using the value of R from the final iteration, calculate the channel width and depth from: A

Q Vmax

and

P

A R

8.1

where both A and P are functions of the unknowns b and y. The bed width so calculated is adopted as the final bed width.

3. Capacity: To check the capacity of the channel, the bed width calculated in step 2 is used in conjunction with the curve for the highest retardance to be expected. Unknowns under these conditions are the velocity of flow (which will be less than Vmax) and the depth y (which will be greater than that calculated in step 2). The iterative solution procedure is as follows: a) assume a flow depth y and calculate the area A and hydraulic radius R from the known geometry of the section; b) calculate velocity from V = Q/A and the value of VR; c) read n from the n-VR curve; d) solve the Manning equation for V; e) assume a new value for y and repeat the above steps until the velocity calculated in step (b) equals that in step (d).

The final depth is adopted as the maximum depth of flow likely to occur.

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8.5 Permissible Velocity Selection of the maximum permissible velocity must take into account the grass type and condition and the soil erodibility. The guide-lines used by the Australian soil conservation agencies are presented in Table 8.1.

Table 8.1 –Permissible Velocities for Channels Lined with Vegetation (after QDPI Soil Conservation Handbook)

Cover Kikuyu1 African Star Grass, Couch Grasses, Carpet Grass Rhodes Grass2 Native Grasses, Rhodes Grass on "black" soil, Other Tussock Grasses Lucerne, Sudan Grass

Slope (%) 0–5 5 – 10 over 10 0–5 5 – 10 over 10 0–5 5 – 10 over 10

Maximum Permissible Velocity (m/s) Erosion resistant Erodible soils soils 2.4 2.1 2.4 2.1 2.4 2.1 2.4 1.8 2.1 1.5 1.8 1.2 2.1 1.5 1.8 1.2 1.5 0.9

0–5

1.5

1.2

0–5

1.1

0.8

Notes: 1. The permissible velocity for Kikuyu in a well maintained sod chute can be considerably higher than the velocities shown in this table, which are intended for use in ordinary (less well maintained) waterways. 2. On the so called "black" soils Rhodes grass behaves as a tussock grass.

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8.6 Worked Examples Worked Example 8.1 Determine the waterway cross section required to safely convey a discharge of 25m3/s down a slope of 0.005 if the waterway is in an easily eroded soil, is to be planted with Kikuyu and will have side slopes of 3 (horiz) : 1 (vert).. Solution From Table 8.1 in Section 8.4 above the permissible velocity Vmax can be taken as 2.1 m/s. The guide-lines in Article 8.3 of the readings suggest the retardance for Kikuyu would range between B under conditions of maximum fertility to D under heavy grazing. Stability Determine width required for Vmax and retardance D. Assume n equal to 0.03 (equivalent to a VR of 2.0). From Manning: 3 2

3

 nV  0.03  2.1  2 R   1/2      0.84098  0.005   So 

which gives VR = 1.7661. The Manning n corresponding to a VR of 1.77 is 0.03, which is the same as the initial estimate. Therefore adopt R = 0.84098. For the particular channel:

A  b  3 y  y

and

A  Q / V  11.905 m2

P  b  2 y 10

and

P  A / R  14.156 m .

and:

Which results in the following two equations:

A   b  3 y  y  11.905

and

P  b  2 y 10  14.156

Simultaneous solution gives y = 1.1534 m and b = 6.8613 m. Adopt the bed width of 6.86 m.

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Capacity Determine the depth and velocity of flow given the width of 6.86 m and retardance B. Assume a depth of 1.5 m. This gives: A = 17.040 m2 P = 16.347 m R = 1.0424 m V = 1.4671 m/s VR = 1.5293 m2/s From retardance curve B and the above value of VR, n is taken as 0.043.

Solution of Manning equation gives V = 1.6906 m/s which is greater than that calculated above therefore assume a new value of y and repeat. As the velocity is too high the next estimate will be a smaller depth.

y (m)

1.5

1.4

2

A (m )

17.040

15.484

14.729

P (m)

16.347

15.714

15.398

R (m)

1.0424

0.98536

0.95655

V (m/s)

1.4671

1.6146

1.6973

VR (m2/s)

1.5293

1.5910

1.6236

0.043

0.042

0.042

1.6906

1.6671

1.6345

n V (m/s)

1.35

From the trend shown in the table the velocities should converge somewhere close to 1.38m, therefore adopt a depth of 1.38 m.

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8.7 Self assessment Questions

8.1

8.2

8.3

All other factors held constant the value of Manning n for vegetated, grass lined channels will: (a)

Increase linearly with the flow velocity

(b)

Remain constant for turbulent flow

(c)

Decrease as the velocity increases

(d)

Increase exponentially with the flow velocity

According to the Australian guidelines, classification of the retardance of a grass lined channel requires the following information: (a)

Grass species only

(b)

Grass length, species and design water velocity

(c)

Grass condition and species

(d)

Grass condition, species and length

One of the criteria for the design of vegetated lined channels is stability. This stability criteria states: (a)

The channel must have sufficient capacity for the design flow rate

(b)

The depth of flow does not overtop the channel banks

(c)

The velocity of flow does not exceed the maximum allowable velocity for the selected grass The dimensions of the channel must ensure that the flow velocity remains subcritical

(d)

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8.8 Tutorial Problems 8.1

A waterway in an erosion resistant soil is required to convey a discharge of 15 m3/s. Design this waterway if the slope is 3% and it is to be lined with couch grass and mown regularly. Side slopes are 3 (horiz) : 1 (vert).

8.2

In association with a particular urban development a stream is to be channelled in the form shown in the following figure.

The channel will consist of:  a narrow concrete lined section (n = 0.014) designed to carry the normal low streamflow; and  a broad shallow grassed waterway for the transmission of flood flows. The bed slope is 0.002 and the grass to be planted on the waterway will be Kikuyu. What is the peak discharge in the compound channel assuming the grass condition is retardance C and the maximum depth of flow in the waterway is 1.0 m?

8.3

A broad shallow grass-lined waterway 100 m wide and with side slopes 1:1 conveys a discharge of 90 m3/s down a slope of 0.021 at a depth of 0.5 m. At a point downstream an additional 30 m3/s is input to the waterway. Determine a suitable cross-section for this increased discharge.

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8.8.1 Answers to Tutorial Problems

8.1

If retardance C is assumed to apply at all times the results are a width of 15.5 m and depth of 0.37 m. Selecting retardance D as the worst condition will require a wider and shallower channel.

8.2

285 m3s-1

8.3

Two alternative answers i. Increase width to 133 m (y and V remain unchanged); ii. Increase velocity to say 2.1 m/s, b = 102 m and y = 0.56.

8.9 Solutions to Self assessment Questions 1. (c)

2. (d)

. .

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Module 9 – Revision of Pipeline Flow

9.1 Objectives When you have mastered the material in this module you will be able to:  evaluate the equations used for calculating the energy loss due to friction during turbulent flow in a pipe;  determine the equivalence between the various parameters used to describe the roughness of the wetted perimeter;  calculate the energy loss due to friction during turbulent flow in a simple pipeline; and  solve problems involving compound and branching pipelines.

Note Most of the material in this module is revision of material covered previously in Hydraulics I. Quickly read the text material indicated to refresh your knowledge. If you have trouble with the self assessment questions or the set problems your knowledge of pipeline hydraulics is inadequate and must be reinforced by a more extensive revision before proceeding on to the next module.

9.2 Turbulent Flow Readings from Texts Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 4.5 and 12.2 Marriott Nalluri & Featherstone’s Civil Engineering Hydraulics Section 4.1 Important here are:  the application of the energy equation;  the Darcy-Weisbach equation; and  the Colebrook-White equation and the Moody diagram. © University of Southern Queensland

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9.2.1 Other Friction Factor Diagrams The Moody diagram is not the only diagram available for the evaluation of the friction factor for a pipeline. In fact, the Moody diagram is applicable only for pipelines with a non-uniform, or in other words random (size and spacing) spot roughness geometry. It is only coincidental that the surfaces of commercially available pipes (concrete, steel, uPVC etc) approximate a non-uniform spot roughness and the Moody diagram applies.

Morris (1963) provided similar diagrams for five other roughness geometries. In each case the friction factor f is plotted as a function of two variables, that is:

f  fun  Re , Z 

9.1

where Z is a parameter describing the roughness geometry of the pipe surface. In the case of the Moody diagram the Z parameter is the relative roughness k/D. The other diagrams are for:

(1) Closely spaced uniform spot roughness. This is the sand grain roughness of Nikuradse, Z = k/D. (2) Corrugation. Applicable to corrugated pipe, Z = D/ where  is the wavelength of the corrugations. (3) Sharp edged strip. No commercial application, Z = D/. (4) Well spaced uniform spot roughness. An effective way of artificially roughening a pipe or channel surface, Z is determined from roughness height and spacing. (5) Circumferential grooves. A means of quantifying the energy loss due to the pipe joints in concrete drainage pipe, Z is a complex function of the groove width and spacing.

Remember that the Moody diagram is not a universal friction factor plot.

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9.3 Empirical Equations The Darcy-Weisbach equation was originally derived experimentally but its use was limited by the difficulty in evaluating the friction factor f (or ). It was some time later that the diagrams such as Moody became available. In the interim a multitude of empirical equations were developed, but only two remain in common use, viz, the Hazen-Williams equation:

V  0.849CHW S f 0.54 R0.63

9.2

and the Manning equation:

V

1 12 2 3 Sf R n

9.3

where the energy slope Sf is equal to hf/L and the hydraulic radius R equals D/4 for full circular pipes.

These equations survived because they have certain advantages:  they are simply expressed;  they can be solved directly; and  they have been widely used for many years.

However they both suffer from the following serious disadvantages:  they are accurate only for a limited range of fluids, temperatures and roughness geometries; and  they are dimensionally non-homogeneous and therefore care is required with the selection of units.

These limitations can be illustrated by comparing the equation predictions to the f-NR plots of the Moody diagram as shown in Figure 9.1 for the Hazen-Williams and Figure 9.2 for the Moody approximations.

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Figure 9.1 - Evaluation of the Hazen-Williams equation © University of Southern Queensland

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Figure 9.2 - Evaluation of the Manning equation - n = 0.01 and n = 0.02 © University of Southern Queensland

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9.3.1 Hazen-Williams Equation Figure 9.1 shows the comparison between the Hazen-Williams coefficient and the equivalent f values for a range of pipe diameters and flow velocities. [For given CHW, D and V, the slope Sf was calculated from the H-W equation and this was used in the Darcy-Weisbach equation to give the equivalent value for f.] In this case the points for a given CHW all fall on a single straight line. Two conclusions can be drawn from this plot:  In the transition or smooth turbulent zone the plot for CHW = 140 parallels approximately the relative roughness lines. This means that a given CHW is equivalent to a particular relative roughness not a roughness height and that CHW should increase with increasing diameter.  For rough pipes, a particular CHW does not represent any particular relative roughness but cuts diagonally across the relative roughness lines. The H-W equation does not describe rough pipes adequately and its use when CHW < 120 is hard to justify.

9.3.2 Manning Equation Figure 9.2 shows the f values equivalent to two values of the Manning n for various V and D. In these figures the plots for the various D plot as separate parallel lines. For a low n the lines cut across the relative roughness lines. Hence it can be concluded that the Manning n does not describe smooth turbulent flows adequately. For rough pipes the lines for constant n parallel the relative roughness lines, with the n of 0.02 being equivalent to a roughness height of 15.3 mm. The Manning equation is therefore applicable to rough pipes (n > 0.015).

9.4 Local Losses Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 4.6 Essential background knowledge.

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9.5 Series Parallel and Branched Pipelines Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 12.3 Mastery of the application of the continuity and energy equations to these cases is essential.

9.6 Worked Examples Worked Examples From Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Examples 4.2 (a), (b) & (d), 4.3, 4.6 and 4.7

Worked Example 9.1 Laboratory testing on a particular 100 mm diameter pipe showed it had a roughness height of 0.5 mm. What values for the Hazen-Williams coefficient and Manning n would you recommend for a typical flow velocity of 1.0 m/s. Solution Assume   1.13E-6 m2 /s then:

N R  VD   8.8496 104

Relative roughness k D  0.005 then from the Moody diagram f  0.0315

Calculate the friction slope using the Darcy-Weisbach equation:

Sf 

hf L



fLV 2 1 0.0315 12   0.016055 2 gD L 2  g  0.1

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Substitution into the re-arranged Hazen-Williams equation: CHW 

V 0.849 S f

0.54

 D 4 0.63

1 0.849  0.0160550.54  0.0250.63  112.04  112 

Similarly for the Manning equation:

n

S f 1 2  D 4

23

V

0.0160551 2  0.025   1  0.010833  0.0108

23

Hence, in this case the roughness height of 0.5 mm is equivalent to CHW of 112 and an n of 0.0108. Please note that these recommended values will vary with the pipe diameter and flow velocity.

9.7 Self assessment Questions .

9.1

For a pipeline the hydraulic grade line is: (a)

always above the energy line

(b)

always above the pipeline

(c)

always sloping downward in the direction of flow

(d)

below the energy line by an amount equal to the velocity head

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9.2

83

In turbulent flow in a rough pipe the friction factor is independent of the Reynolds number: (a)

in the transition zone

(b)

when the laminar sub-layer is larger than the roughness elements

(c)

when the Reynolds number is less than 4000

(d)

none of these answers

9.3 The friction factor in turbulent flow in smooth pipes depends upon:

9.4

(a)

V, D, L, , k

(b)

Q, L, , k

(c)

V, D, , k

(d)

V, D, 

The following quantities are computed for non-circular sections by replacing diameter with 4R: (a)

discharge and head loss

(b)

discharge and relative roughness

(c)

Reynolds number, relative roughness and head loss

(d)

velocity, Reynolds number and friction factor

9.5 One pipe system is said to be equivalent to another when the following two quantities are the same:

9.6

(a)

h, Q

(b)

L, Q

(c)

L, D

(d)

f, D

In parallel pipe problems:

(b)

the head losses through each pipe are added to obtain the total head loss the discharge is the same through all pipes

(c)

the head loss is the same through each pipe

(d)

a trial solution is not needed

(a)

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9.7 The friction factor for a pipeline 75 mm diameter, roughness height 0.2 mm conveying a discharge of 5 L/s is:

9.8

(a)

0.0275

(b)

0.025

(c)

0.0225

(d)

0.021

The Hazen-Williams coefficient CHW: (a)

increases as the pipe roughness decreases

(b)

decreases as the relative roughness decreases

(c)

increases with increasing Re

(d)

decreases as Q increases

9.8 Tutorial Problems 9.1

Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Problems on Chapter 4, pages 113 & 114 (4th Ed, p 118 & 119) Question 1

9.2

Marriott

Ch 4, Question 4

9.3

Marriott

Ch 4, Question 6

9.4

Field testing on a certain type of pipe showed it had a roughness height k of 0.4 mm. What value of the Hazen-Williams coefficient would you recommend for pipes of diameter 75, 150, 300, 600 & 1200 mm.

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9.8.1 Answers to Tutorial Problems Answers for questions 9.1 – 9.3 are given in Marriott 9.4

For a nominal velocity of 0.5 m/s: D (mm)

CHW

75

120

150

125

300

130

600

135

1200

140

9.9 Solutions to Self assessment Questions 1. (d)

2. (d)

3. (c)

4. (c)

5. (a)

6. (c)

7. (a)

8. (a)

. .

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© University of Southern Queensland

Module 10 – Pipe Network Analysis

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Module 10 – Pipe Network Analysis

10.1 Objectives When you have mastered the material in this module you will be able to:  apply the continuity and energy equations to and develop the solution matrix for a steady flow pipe network;  solve simple pipe networks using an appropriate method.

10.2 Introduction A network is a system of interconnected pipelines (and other hydraulic elements) so arranged that a change to any one element changes the discharge or pressure in all other elements. As well as pipelines, a network may include other elements such as: pumps, reservoirs and/or a variety of valves (gate, reflux, pressure reducing, etc). For any given steady input(s) to the network and equal steady supply (outputs) from the network, steady flows and pressures will establish in the network. The task of a network analysis is to predict those flows and pressures. A network may consist of only 5 or 6 elements, as in the simple examples presented in text books, or it may consist of hundreds of elements, as in real urban water supply networks. Real networks do not usually operate at steady flows and pressures. Water supply to a community is a dynamic process showing a significant variation in demand over all time periods, for example, daily and annually. The coverage in this course will not extend to the prediction of dynamic effects. It will be assumed that the system is to be designed to accommodate the peak water requirement. For many networks this steady flow assumption is entirely adequate. If it is not adequate, computer packages are available for modelling the dynamic behaviour of networks.

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10.3 Generalised Pipe Flow Equation For simplicity, a pipe flow equation of the following form will be assumed:

h f  rQn

10.1

where r and n are constants for each element in the network. This equation applies equally well to pumps and fittings as well as to the pipelines in the network. For example if the Darcy-Weisbach equation is used then the constants from 10.1 are: r

8 fL g 2 D 5

and

n2

10.2

In this case r is not constant but is a slowly changing variable due to the variation in f with changing NR.

For the Hazen-Williams equation the constants from 10.1 become: r

10.68 L 1.852 C HW D 4.87

and

n  1.852

10.3

Similarly the same generalised flow equation can be applied to fittings where the flow equation is: hf  k

V2  k  2 Q  2 g  2 gA 2 

10.4

It is also assumed for simplicity that the velocity heads are small compared to the pressures and hence can be ignored in any statement of the energy equation. This is equivalent to assuming the HGL (Hydraulic Grade Line) and energy line to be coincident.

10.4 Hydraulic Analysis To achieve a solution in a network analysis the following conditions must be satisfied: 1. External flows (inputs and outputs) must sum to zero, that is, external continuity must apply.

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2. Continuity must be satisfied at each junction or node in the network, that is, flows into the node equal flows out of the node. This includes internal (pipeline) flows and any external flow at the node. 3. The appropriate flow equation must be satisfied for each element in the network. 4. The algebraic sum of the energy losses (or gains) around each circuit or loop in the network must be zero, that is, the energy equation must be satisfied.

Consider the following simple example as illustrated in Figure 10.1.

Figure 10.1 – Simple Pipe Network

1.

External continuity gives: flow in = flow out 200  40  60  50  50

2.

Continuity at all nodes, five in total gives:

Q1  Q3  200 Q1  Q2  Q4  40 Q2  Q5  60 Q3  Q4  Q6  50 Q5  Q6  50 Because of the external continuity one of these equations is "dependent" upon the others. Hence there are 4 "independent" equations and 6 unknown discharges.

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Flow equation gives:

h fi  ri Qin 4.

where ri is known.

Energy around each loop gives: hf 1  hf 4  hf 3  0

and

hf 2  hf 5  hf 6  hf 4  0

Substitution of the flow equations into the two loop equations, assuming n = 2 results in:

r1Q12  r4Q42  r3Q32  0

and

r2Q22  r5Q52  r6Q62  r4Q42  0

These 2 equations in Q plus the 4 continuity equations give a total of 6 equations and 6 unknown discharges. The equations are: Q1  Q3  200

(i)

Q1  Q2  Q4  40

(ii)

Q2  Q5  60

(iii)

Q3  Q4  Q6  50

(iv)

r1Q12  r4Q42  r3Q32  0

(v)

r2Q22  r5Q52  r6Q62  r4Q42  0

(vi)

which is a set of soluble non-linear equations. The solution technique therefore will involve some technique employing successive approximations. The equations can be expressed in matrix form:

 1  1   0   0 r1Q1   0

0 1 1 0 0 r2 Q2

1 0 0 1  r3Q3 0

© University of Southern Queensland

0 1 0 1 r4 Q4  r4 Q4

0 0 1 0 0 r5Q5

0   Q1  200 0  Q2   40    0  Q3   60        1  Q4   50  0  Q5   0        r6 Q6  Q6   0 

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10.5 Hardy-Cross Loop and Node Balancing Methods Reading from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 12.4 (including examples 12.2 & 12.3) Mastery of this material is essential. Note the application of a consistent sign convention where flows (and head losses) that are in the same direction as the loop direction are deemed positive. An alternative expression of equation 12.4 (the correction equation for the loop balance) can be found by substitution of the flow equation, giving:

Q  

r Q 2 r Q 2 i

i

i

10.5

i

or more correctly

Q  

r Q Q 2 r Q i

i

i

i

10.6

i

This latter interpretation means that the sign of the terms in the summation in the numerator will be the same as the sign of the individual discharges. The denominator will always be positive. The correction equation for the node balance method (equation 12.5) provides a correction H for the energy H at each node. The corresponding correction to the energy loss in each element of the network is:

 h   H

10.7

Reading from Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Sections 5.2 and 5.3 This is an alternative description of the same material as in Chadwick et al.. You might find it helps your understanding of the topic.

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10.6 Linearisation In Section 10.4 it was shown that if we have NL loops and NJ nodes or junctions then the continuity and energy equations give NL+NJ non-linear equations to solve for NL+NJ-1 unknown discharges. For example, the matrix developed earlier (already less one of the node equations) is:

 1  1   0   0 r1Q1   0

0 1 1 0 0 r2 Q2

1 0 0 1  r3Q3 0

0 1 0 1 r4 Q4  r4 Q4

0 0 1 0 0 r5Q5

0   Q1  200 0  Q2   40    0  Q3   60        1  Q4   50  0  Q5   0        r6 Q6  Q6   0 

Here we have said that: h fi  ri Qi Qi

h fi  ri Qi Qi

or more correctly

10.8

This expression can be approximated by:

h fi  ri Qi* Qi

10.9

where Qi* is the absolute estimated or current value of Qi. This transforms the matrix into a set of linear equations which can be solved for Qi, for example:

 1  1   0   0 r1 Q1*   0

0 1 1 0 0

1 0 0 1  r3 Q3*

0 1 0 1 r4 Q4*

0 0 1 0 0

r2 Q2*

0

 r4 Q4*

r5 Q5*

  Q1  200       Q2   40   Q3   60        Q4   50   Q   0    5   *  r6 Q6  Q6   0  0 0 0 1 0

The equations are solved by Gaussian elimination for Qi. Averaging the calculated Qi with the Qi* from the previous iteration gives the values of Qi* for the next approximation:

newQi* 

Qi*  Qi 2

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Note that only the absolute magnitude of the average is used in the matrix, the sign (or direction) of the flow has already been taken into account in establishing the matrix. There is no requirement for the initial estimates to satisfy continuity - this matter is handled by inclusion of the continuity equations in the matrix. This method provides a more rapid convergence on a solution than does the HardyCross method and is more suited to machine calculation. Note that there is no requirement for the initial estimates of Q to satisfy continuity this has already been taken into account in establishing the matrix. However you should check to ensure that your solution satisfies continuity at each node.

10.7 Worked Examples Worked Examples From Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Examples 5.1, 5.2, 5.3, 5.4, 5.6 and 5.8 Examples 5.1 and 5.4 demonstrate the computational procedure for very basic loop and node balancing problems. Solutions 5.2, 5.3 and 5.6 are for networks with other elements such as pumps and valves. In all cases note the sign convention used. Note also that the magnitudes of the corrections Q and H reduce with successive iterations, failure to do so is a clear sign of an error in the computations. Example 5.8 shows the linearisation computational procedure for a very basic network.

10.8 Self assessment Questions

10.1 A pipeline 100 mm diameter and 1 km long is conveying a discharge of 25 L/s. If the pipeline is assessed as having a CHW of 135, the pipeline constant r is equal to (a)

89780

(b)

8978

(c)

897.8

(d)

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10.2 For the pipeline in Question 1 the exponent n on the discharge term in the flow equation is: (a)

2

(b)

1.852

(c)

0.54

(d)

0.63

10.3 The energy loss hf (m) for the pipeline in Question 1 is: (a)

0.969

(b)

969

(c)

96.9

(d)

9.69

10.4 In a pipe network a high value of r for a particular element means: (a)

a relatively low energy loss in that element

(b)

a relatively low flow rate in that element

(c)

the pipe is relatively short and of large diameter

(d)

there is little resistance to flow in that element

10.5 In a network of pipes: (a)

the sum of the energy losses around each circuit must be zero

(b)

the loss in all elements is the same

(c)

the sum of the energy losses for all elements joining at a node must be zero friction factors must be assumed for each pipe

(d)

10.6 The estimated flow in a particular element in a network is 0.013 m3/s (anticlockwise). If the flow correction calculated using a convention of clockwise +ve is -0.0021 m3/s then the corrected discharge (m3/s) in the element is: (a)

0.0109

(b)

0.0172

(c)

0.0088

(d)

0.0151

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10.7 The estimated energy loss in a pipe flowing out of a junction in a network is 25 m. If the correction h is - 2.3 m then the corrected energy loss (m) in the pipeline is: (a)

20.4

(b)

22.7

(c)

27.3

(d)

29.6

10.9 Tutorial Problems The answers to these problems are given on page 398 of the text. Note that the text reverses the naming of the pipe segments to ensure all answers are positive flows. Also note that some answers in the text are incorrect, see the “Answers section for further information” 10.1

Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Problems on Chapter 5, pages 142-145 (4th Ed, p 151 & 153) Question 7

10.2

Marriott, Ch 5, Question 9

10.3

Marriott, Ch 5, Question 10

10.4

Marriott, Ch 5, Question 11

10.5

Marriott, Ch 5, Question 7 – Repeat using the linearization method

10.9.1 Answers to Tutorial Problems All answers in Marriott are correct, previous versions of the text contain some errors. The correct answers are repeated below: 10.1

Answers in Marriott are correct, repeated here for those with previous versions: Pipe Discharge (l/s)

AB 104.81

BC

CD

ED

AE

BE

45.41

-4.61

44.61

95.21

-0.61

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10.2

Answers provided in text are correct

10.3

For previous versions of Nalluri and Featherstone the answer in the table for pipe FG should be GF and pipe GA should be AG otherwise the implied flow directions are incorrect.

10.4 Pipe

AB

Discharge (l/s)

10.5

95.3

BC

CD

90.1

30.1

DE

EF

-49.9

-44.7

Answer should be the same as for 10.1 (i.e. Q7 in the text)

10.10 Solutions to Self assessment Questions 1. (a)

2. (b)

3. (c)

5. (a)

6. (d)

7. (c)

. .

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4. (b)

BE 5.2

Module 11 – Pump/Pipeline Systems

97

Module 11 – Pump/Pipeline Systems

11.1 Objectives When you have mastered the material in this module you will be able to:  select a pump to meet a specified duty;  match a pump to a pipeline system and determine the operating point of the pump; and  determine the maximum discharge and suction lift for cavitation free operation of a pump.

11.2 Introductory Concepts Reading from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 7.1 and 7.2

Book of Readings – Reading 11.1 Chapter 12 from Stephenson, D. 1989, Pipeline Design for Water Engineers. Elsevier, Amsterdam. Parts of this reading are referred to later Both of these readings are essential background. Important here are the types of pumps and their operation; and the head, power and efficiency characteristic curves for roto-dynamic pumps. The remainder of this module will concentrate on rotodynamic pumps and in particular centrifugal pumps.

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11.3 Pump Affinity Laws and Specific Speed Reading from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 11.6 (pages 384 to 385, Hydraulic Machines) Important here are the non-dimensional groups (1, 2, etc). The non-dimensional groups (1, 2, etc), otherwise known as the Pump Affinity Laws, are developed from a consideration of hydraulic similitude (or similarity). Those students interested in the development of these groups are referred to Streeter and Wylie (1983). These non-dimensional groups are of primary interest to designers and manufacturers of pumps. For designers of pumping systems, the two important groups are 2 and 3. From these it can be concluded that for any pump: Q  constant ND 3

11.1

H  constant N 2 D2

11.2

and

where N is the rotational speed of the pump in RPM; and D is a characteristic dimension (usually the impeller diameter). These two groups can be used to predict the H-Q characteristic for a pump at any speed (or impeller diameter) given the head and discharge at another speed (or impeller diameter).

Note Non-dimensional groups and hydraulic similitude are covered in detail in Module 17

The Specific Speed of a pump is defined as the speed at which the pump delivers unit discharge against unit head. For a particular pump the specific speed is constant and is given by: Ns 

NQ 1/ 2 H 3/ 4

Specific speed is used to select the type of pump for a particular duty. © University of Southern Queensland

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Book of Readings – Reading 11.1 – Page 233-234 Chapter 12 from Stephenson, D. 1989, Pipeline Design for Water Engineers. Elsevier, Amsterdam. Pages 233 to 234 Important here is figure 12.4

11.4 Head-Discharge Characteristic Curve 11.4.1 Theoretical H-Q Curves Book of Readings – Reading 11.1 – Page 234-236 Chapter 12 from Stephenson, D. 1989, Pipeline Design for Water Engineers. Elsevier, Amsterdam. pages 234 to 236 Useful background on impeller dynamics

11.4.2 Actual H-Q Curves Book of Readings – Reading 11.1 – Page 236 Chapter 12 from Stephenson, D. 1989, Pipeline Design for Water Engineers. Elsevier, Amsterdam. page 236 – pump characteristic curves For a brief description of the form of the actual H-Q curves

Book of Readings – Reading 11.2 Kelly and Lewis Pump Catalogue.

Typical pump curves from catalogue of Kelly and Lewis Pumps. Complete understanding of the curves is required. You may also browse pump curves from other manufacturers on the internet (for example Tyco-Southern Cross, Batescew, Davey) © University of Southern Queensland

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The information usually provided on a set of pump characteristic curves is as follows:  H-Q curves - comprising a family of curves for various impeller diameters and a single speed or various speeds and a single impeller diameter. Note that pump speed and impeller diameter are infinitely variable within the limits for a particular machine and the pump can operate at speeds and with impeller diameters between those for which curves are provided.  Efficiency - pump efficiency will always fall below the ideal 100% due to the effects of friction, imperfect guidance of the fluid and leakage through the clearances between casing and impeller. Information is usually provided in the form of contours of equal efficiency.  Power - gives an indication of the power required to drive the pump. The power contours often coincide with the size of electric motors that are available.  NPSHR – can be presented in different ways, see Section 11.6 .

11.5 Pump-Pipeline System Design Reading from Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Sections 6.2 to 6.4 Mastery of this material is essential The important result from the application of the energy equation to a typical pumppipeline system is the equation: H m  H ST  losses

11.4

H m  H ST  hls  hld  local losses

11.5

or:

where Hs is the static head in the reservoir (or elevation z relative to the valve); Hm is the energy added by the pump (or pump head); HST is the static lift (the difference in elevation between the source and sink; and hls and hld are the energy losses in the suction and delivery pipelines, respectively.

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It is important to note that the above equality is satisfied at only one discharge. The pump head Hm is a variable, a function of Q as defined by the H-Q curve for the particular pump, impeller and speed and has the form shown in Figure 11.1. The RHS of the equation is also a variable. It comprises a (usually) constant term HST and the variable energy loss terms which are functions of Q2. Together they give the characteristic curve for the pipeline system (Figure 11.1). The natural operating point of the pump-pipeline system is found by solution of the equation, which for real pumps is best achieved graphically by superimposing the system curve over the pump curve as in Figure 11.1.

Pump (H-Q) curve

Head, H

(m)

Pipeline System Curve

Operating Point (duty point)

Static lift (Hst)

Discharge, Q

(L/s)

Figure 11.1 – Graphical Superposition of Pump and Pipeline Curves - Pump Operating Point

Variation of the operating point can be achieved by: 1. Varying the pump curve. At the design stage this is achieved by selecting a pump with a different characteristic. For a particular pump-pipeline system, it is achieved by varying either the pump speed or the impeller diameter. 2. Varying the system curve. This is achieved by varying the energy losses in the pipeline by: - enlargement or duplication of the pipeline; or - operation of a gate valve in the delivery line.

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11.6 Design Based on Suction Considerations - Cavitation and NPSH Reading from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 3.7 Background to the subject of cavitation.

Reading from Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Sections 6.5 Mastery of this material is essential The net positive suction head for a pump is concept for which no clear consistent definition is available. Consequently the expressions cited vary from one text to another. Some sources express NPSH as a pressure head while others as an energy head, hence a difference in the handling of the velocity head term. The definition and expressions used in the following notes are to be consistent with Marriott and the prerequisite hydraulics courses.

The net positive suction head (NPSH) is defined here as the minimum absolute pressure head required at the pump inlet (relative to the inlet) in excess of the vapour pressure of the fluid. By definition this is given by: NPSHR 

Ps  Pv g

11.6

where Ps is the absolute pressure at the pump inlet; Pv is the vapour pressure of the fluid; and Vs is the flow velocity at the pump inlet. The division by ρg is a conversion back to metres head of water This required head, designated NPHSR, is a function of the pump design and is supplied by the pump manufacturer (as contour lines of equal NPSHR on the pump characteristic curves). It varies with Q as illustrated in Figure 11.2.

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Application of the energy equation between the source and the inlet to the pump results in an alternative expression for NPSH:

P P NPSHA  a v  hs  hls g

Vs2  2g

11.7

where hs is the elevation difference between the source and the pump inlet; and hls is the energy loss in the suction pipeline. Pa is the absolute pressure at the source (upstream of the suction line) The velocity head term on the right side of the equation is required if the source represents a free water body such as a reservoir or stream. This head, designated NPSHA, is the head available at the pump inlet and is a function of the system. It comprises three constant terms (the pressures and the elevation term) and the loss term and optional velocity head term which vary with Q2 (Figure 11.2). Cavitation will occur whenever NPSHR  NPSHA. The discharge at which this occurs is best determined graphically by superposition of the two curves as in Figure 11.2.

NPSH

(m)

NPSHA (suction pipe)

NPSHR (pump manufacturer) Max. Cavitation free discharge

Discharge, Q

(L/s)

Figure 11.2 – Graphical Solution for Maximum Cavitation Free Discharge

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The discharge calculated here may be greater or less than that defined by the operating point of the pump as determined in Section 11.5 . If greater, the pump will operate safely at the natural operating point without risk of cavitation. If less, the operating point of the pump must be altered to reduce the discharge below that indicated by the NPSH calculations. This can be achieved by:  partially closing a gate valve on the delivery line;  reducing the pump speed; or  reducing the impeller diameter.

11.7 Additional Reading and Worked Examples Reading from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 3. An alternative description of the design of pump/pipeline systems which may help your understanding of the topic.

Worked Examples From Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Examples 6.3, 6.5, 6.6 and 6.8

11.8 Self assessment Questions

11.1 Specific speed is defined as the speed at which: (a)

a pump of unit size delivers unit discharge at unit head

(b)

a pump of unit size requires unit power for unit head

(c)

a pump delivers unit discharge at unit head

(d)

a pump delivers unit discharge at unit power

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11.2 A pump operating at a speed of 1470 RPM delivers a discharge of 60 L/s at a head of 21 m. The specific speed of this unit is: (a)

3.67

(b)

36.7

(c)

367

(d)

3760

11.3 At what speed (RPM) should the above pump be operated to give a discharge of 55 L/s (assuming that the head will also vary from that above)? (a)

1501

(b)

1470

(c)

1348

(d)

1276

11.4 What is the head in m corresponding to the discharge in Question 3? (a)

22.3

(b)

14.5

(c)

21

(d)

17.7

11.5 The flow from one reservoir to another at an elevation 25 m lower is boosted by a pump. Total head losses in the suction and delivery pipelines are 8 and 29 m, respectively. The head (m) delivered by the pump is: (a)

12

(b)

25

(c)

37

(d)

62

11.6 A KL 10 MHV-HHC pump operating at a speed of 1180 RPM delivers 250 L/s. From the manufacturers pump curves (given in Article 11.2 in the Book of Readings) specify the pump head, efficiency, motor power and NSPHR for that particular operating point.

11.7 Cavitation is caused by: (a)

high velocity

(b)

low barometric pressure

(c)

high pressure

(d)

low pressure

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11.8 Cavitation of a pump will occur if: (a)

the NSPHR is exceeded

(b)

the NSPHA > NSPHR

(c)

the NSPHR > NSPHA

(d)

the specific speed is exceeded

11.9 Tutorial Problems 11.1

Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Problems on Chapter 6, pages 164-166 (4th Ed, p 174-175) Question 4

11.2

Marriott

Ch 6, Question 7

11.3

Marriott

Ch 6, Question 8

11.4

Marriott

Ch 6, Question 9

11.5

A self-priming pump is to be mounted on a bore to be used for dewatering purposes. The static water level in the bore is 2 m below the eye of the pump and the water is to pumped to a drainage channel 1.5 m above the pump. Drawdown in the bore occurs at the rate of 1 m per 12 L/s of discharge. The pump is a KL 6" T6A3-B operating at 1470 RPM (characteristic curves given in Article 11.2). The pipeline friction constant r in the expression hf = rQ2 is 1/1800 for the suction line and 1/360 for the delivery line (where Q is in L/s). Assume atmospheric pressure is 10 m and the saturated vapour pressure of water at 15°C is 0.2 m.

Determine the maximum drawdown in the bore and the operating point of the pump for cavitation free operation.

Note that the values of r given above assume that Q is in L/s.

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11.9.1 Answers to Tutorial Problems Answers to problems 10.1 – 10.4 are given in the text 11.5

Assuming a small margin of safety is applied Drawdown

4m

Discharge

48 L/s

Head

14.5 m

11.10 Solutions to Self assessment Questions 1. (c)

2. (b)

5. (a)

6. 5.8 m,

69%

3. (c)

4. (d)

21 kW, 6.25 m

7. (d)

8. (c)

. .

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Module 12 – Unsteady Pipe Flow (Surge)

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Module 12 – Unsteady Pipe Flow (Surge)

12.1 Objectives When you have mastered the material in this module you will be able to:  apply rigid column theory to the analysis of mass oscillations in pipelines;  calculate the maximum allowable rates for valve open and closure; and  chart the fluctuations in water level in a surge tank.

12.2 Simple Pipeline/Valve System Reading from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 6.1 Mastery of this material is essential.

Application of the energy equation between the reservoir and the valve in Figure 6.1(a) from Chadwick et al. (ignoring minor losses) gives: H s  H v  h f  hi

12.1

where Hs is the static head in the reservoir (or elevation z relative to the valve); Hv is the total head at the valve and can be expressed as a function of velocity head, that is, kvV2/2g; hf is the friction loss in the pipeline and equals fLV2/2gD; and hi is the inertial head (called the surge pressure h by Chadwick et al.).

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The increase in head hi results from the deceleration (or acceleration) of the fluid in the pipeline and can be obtained from application of Newton's 2nd Law (F = ma), that is: pressure  area  mass  acceleration dV  ghi Ap   LAp dt

12.2

L dV g dt

12.3

Thus:

hi 

Substitution into the energy equation gives:

z

L dV V 2 fLV 2  kv  0 g dt 2 g 2 gD

12.4

Application of the energy equation across the valve gives an equation of the form of the standard orifice equation: Q  Cv Av  2 gH v 

1

2

12.5

where Q is the discharge in the pipeline, Av is the open area of the valve and Cv is a discharge coefficient for the valve. Continuity at the valve states that: Q  ApV  Cv AvVv

12.6

where Ap is the cross section area of the pipe and Vv is the velocity downstream of the valve.

Combining these equations (12.5 &12.6) and rearranging results in an expression for Hv:

V 2  Ap  Hv    2 g  Cv Av 

2

12.7

CvAv can be considered to be an effective open area for the valve, and which can be expressed as a fraction or percentage of the pipe area Ap.

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The above equations are required to be solved to determine appropriate rates of closure of the valve. Three closure strategies can be considered, viz: 1. Constant Deceleration. This gives a constant value for the inertial head, a variable head at the valve and a variable rate of closure of the valve. 2. Constant Valve Head. Requires a variable deceleration and variable inertial head. Closure of the valve occurs at a variable rate. 3. Constant Rate of Closure. Gives variable inertial and valve heads.

In all cases the head at the valve must be maintained below some specified safe level. If dV/dt is known the governing equation can be solved directly at any point in time using the flow velocity appropriate to that time. In other cases, such as where Hv is known, solution of the equations usually requires that the differential term dV/dt be replaced by a difference approximation V/t. The velocity V will then be taken as the value Vm at the middle of the time period t. Thus for any time period:

z

fLVm2 L V  Hv  0 g t 2 gD

12.8

The velocity Vm is given by:

Vm  Vi 

V 2

12.9

where Vi is the value at the start of the time period and will usually be known.

Analysis will show that the first 90% of closure can occur almost instantaneously without causing excessive inertial and valve heads. It is the last 10% of closure which must be controlled. Note that the above analysis and equations apply only during the act of opening or closing the valve. When the valve is in a fixed position and the flow is steady hi = 0 and the dynamic equation reduces to: H s  Hv  h f

12.10

And when the valve is wide open (again ignoring minor losses) Hv = 0 and: Hs  hf

12.11

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12.2.1 Worked Example Worked Example From Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Example 6.1

Worked Example 12.1 Using the details given in example 6.1 above, determine the pattern of adjustment of the effective open area of the valve as it is closed from the fully open to the point where velocity is reduced by 50%. Solution As the valve is closed the velocity in the pipe is reduced from 1.4 m/s to 0.7 m/s in 2 s. From the solution of example 6.1 in Chadwick et al.: dV/dt = 0.35 m2/s

hi = 53.5 m

Hs = 20 m

The solution is best presented in tabular form decreasing the velocity in 0.2 s steps. Because dV/dt is known (and constant) at all times this solution will determine the effective area at the end of each time increment.

t (s)

Vi (m/s)

0 0.2 0.4 0.6

1.40 1.33 1.26 1.19

18.032 16.184 14.436

55.468 57.316 59.064

100 4.03 3.76 3.50

0.8 1.0 1.2 1.4 1.6 1.8 2.0

1.12 1.05 0.98 0.91 0.84 0.77 0.70

12.787 11.239 9.7903 8.4417 7.1929 6.0440 4.9951

60.713 62.261 63.710 65.058 66.307 67.456 68.505

3.25 3.00 2.77 2.55 2.33 2.12 1.91

© University of Southern Queensland

hf (m)

Hv (m)

Effective Area (%)

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Notes: 0.02 1500  Vi 2 2  9.81 0.15  10.194Vi 2

hf 

and

H v  H s  hi  h f  73.5  h f

The effective open area of the valve expressed as a percentage of the pipe area is:

100Cv Av 100Vi  Ap 2 gH v The solution shows that the first 95% of the closure can occur almost instantaneously with the remaining 5% occurring at a controlled and decreasing rate over the remainder of the 4 s time.

12.3 Surge Tank Operation Readings from Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Sections 12.1 and 12.4 Mastery of this material is essential. Note that the dynamic equation for the surge tank case is essentially identical to that for the simpler pipeline/valve system, comprising an elevation term, the inertial head and two energy loss terms. The only difference in this case is that flow reversals occur hence the terms V, V, z and z may be positive or negative. Use of a consistent sign convention is essential. The three  terms in equation 12.22 of Marriott should be + ve when Vi is + ve and ve when Vi is -ve.

Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 12.6 and part Section 14.6 (pages 499-502) An alternative description which may help your understanding of the topic. Equations 12.10 and 12.11 of Chadwick et al. are an analytical solution of the surge tank equations for the case of zero friction. For a real system (with friction) the amplitude of the oscillations will always be less than that given by equation 12.10. However the wave length or period of the oscillations will be the same as given by equation 12.11. These equations provide a useful check of any numerical solution.

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The numerical solution technique presented in Section 14.6 is less rigorous and hence less suitable than that in Marriott. It requires a very much smaller value of t to give an accurate solution (see Chadwick et al figure 14.5). This approach is not recommended.

12.3.1 Worked Example Worked Example From Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Example 12.3

12.4 Self assessment Questions

12.1 Inertial head in a pipeline is a function of: (a)

f, L, V, g, D

(b)

, g, L, V

(c)

t, V, L, g

(d)

, g, t, V, D

12.2 The flow velocity in a 50 m long pipeline is reduced from 0.8 m/s to 0.1 m/s in 1.4 s. The inertial head (m) is: (a)

5.1

(b)

2.55

(c)

0.51

(d)

0.255

12.3 If the static head in Question 2 is 15 m and the pipe friction loss is 5V2 the head loss (m) at the valve is: (a)

17.6

(b)

12.5

(c)

15.0

(d)

17.5

12.4 Calculate the head at the valve at the point of complete closure, assuming the rate of deceleration remains constant.

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12.5 In a conventional surge tank system, if V, z and z are +ve and V is -ve then:

(c)

the water level in the surge tank is approaching the peak of an upsurge the water level in the tank is approaching the trough of a downsurge the water level is at the peak of an upsurge

(d)

the stated flow conditions are impossible

(a) (b)

12.6 In a conventional surge tank system, if V, z and z are -ve and V is +ve then:

(c)

the water level in the surge tank is approaching the peak of an upsurge the water level in the tank is approaching the trough of a downsurge the water level is at the peak of an upsurge

(d)

the stated flow conditions are impossible

(a) (b)

12.7 In a conventional surge tank system, if z is -ve and V, z and V are +ve then:

(c)

the water level in the surge tank is approaching the peak of an upsurge the water level in the tank is approaching the trough of a downsurge the water level is at the peak of an upsurge

(d)

the stated flow conditions are impossible

(a) (b)

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12.5 Tutorial Problems 12.1

A pipeline with a nominal diameter of 300 mm connects a reservoir to a terminal valve at the other end of the pipe, 5415 m distant. The flow velocity with valve wide open is 1.6 m/s and the frictional head loss is 35 m. Determine the maximum head at the valve and the time pattern of reduction of the effective area AvCv, if the valve is operated so that: i. the rate of deceleration is uniform and stops the flow in 90 s; and ii. the head at the valve is held at 40 m during the whole closure.

Hint: In part (i) use a t of 10 s and follow the worked solution given in Section 12.2.1 . In part (ii) use a V of 0.1 m/s.

12.2

An unrestricted (Fs = 0) surge tank 10 m diameter is at the downstream end of a 2 km long pipeline whose internal diameter is 2.5 m. With a steady flow of 30 m3/s the level of the water surface in the surge tank is 18.2 m below the level in the supply reservoir. Determine the height of and time to the first upsurge and the first downsurge if the valve downstream of the tank is closed completely and instantaneously.

Hint: Write a spreadsheet or calculator routine for the numerical solution of this problem and use a t of 10 s.

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12.5.1 Answers to Tutorial Problems

12.1

(i)

Max Hv is 44.8 m at point of complete closure Time (s) 30 60 90

(ii)

Effective Area (% of Ap) 4.4 1.9 0.0

Hv is constant at 40 m and time to closure is approx. 81 s

12.2 1st upsurge 1st downsurge

z (m) -18.2 0 11.8 0 -7.1

t (s) 0 53 135 236 314

12.6 Solutions to Self assessment Questions 1. (c)

2. (b)

3. (d)

5. (a)

6. (b)

7. (d)

4. 17.55

. .

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Module 13 – Unsteady Pipe Flow (Water Hammer)

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Module 13 – Unsteady Pipe Flow (Water Hammer)

13.1 Objectives When you have mastered the material in this module you will be able to:  apply elastic theory to the analysis of pressure transients in simple pipelines; and  use appropriate computer packages in the analysis of more complex problems.

13.2 Introductory Concepts Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 6.2 and Part 6.3 – General Description Essential background to the topic. Complete understanding of the mechanism for the propagation of pressure transients, as described in Figure 6.3, is required. From basic physics, the celerity c of sound waves in an elastic medium is given by: c

K



13.1

where K is the elastic modulus of the medium. The term celerity has its usual meaning, that is, the speed of the wave relative to the flow of the medium.

Extra Background Information For sound waves in air c  343 m/s (dry air at 20°) which is very much greater than the flow velocity of the air itself. Similarly for pressure transients in a pipeline. The wave may be propagated at a celerity of greater than 1200 m/s when the local velocity of the fluid would rarely exceed 5 m/s. Similarly to air the speed of sound in water is a function of temperature and is equal to 1482 m/s at 20°.

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13.3 Compressible Fluid/Rigid Pipe Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Part Section 6.3 – Simple equations for instantaneous alteration of valve setting in a rigid pipeline (p191) Essential knowledge. This is a simple analysis for the prediction of the pressure rise due to the instantaneous closure of a valve, that considers only the compressibility of the fluid and ignores the elasticity of the pipe. Consider the pressure wave shown in Figure 6.2(a) of Chadwick et al. (with the symbols altered to be consistent with the remainder of this study guide). Flow in the pipe upstream of the wave is V and downstream is V - V. The pressure wave is moving upstream with velocity Vw = c - V. This unsteady flow system is transformed to an apparent steady system as seen through the eyes of an observer moving with the wave (Figure 6.2b). Note that the flow velocity after the wave in this transformed system is given incorrectly and should be c – δV. Application of the continuity equation to this apparent steady system gives:

 A  c        A  c   V 

13.2

Ignoring products of  terms this reduces to:

V  c

13.3

Note here the error in equation 6.3 of Chadwick, et al.

Similarly application of the momentum equation results in:

 p   cV

13.4

If the velocity is reduced to zero, that is, V = V then the two equations reduce to;

V  c

13.5

 p   cV

13.6

and:

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which is equivalent to equation 6.4 of Chadwick et al.. Note that these latter equations are valid only for complete closure of a valve. You can see from the above equations that the change in density resulting from rapid closure will be quite small (being a function of V/c) but the increase in pressure will be substantial (being the product cV).

Introduction of an equation of state, that is, a stress-strain relationship linking p with  viz:

p  K

 

13.7

and combining with the two equations above gives the expression for c as: c

K



13.8

Worked Example From Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Example 6.2

13.4 Compressible Fluid/Elastic Pipe Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Part Section 6.3 – Equations for instantaneous valve closure in a rigid pipeline (p194-197) Essential knowledge. This is an extension of the analysis in the preceding Section 13.3 . Again it predicts the pressure rise due to instantaneous closure of a valve but we now include consideration of the elasticity of the pipe. The equation for pressure rise is the same as derived in Section 13.3 :

 p   cV

(or  p   cV for complete closure)

13.9

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However the elasticity of the pipe modifies (reduces) the celerity of the pressure wave and hence the magnitude of the pressure rise. Two equations of state (stress-strain relationships) are introduced that relate:

p with as given in 13.3 ) and p with the change in cross sectional area of the pipeline:

p 

D A Ee A

13.10

where D is the internal diameter of the pipeline, A is the cross-sectional area, e is the wall thickness of the pipeline, and E is the Young’s (or elastic) modulus of the pipe material.

The result is a new expression for the celerity of the pressure wave:

c

  K 1     1  KD  Ee  

13.11

13.5 Governing Equations for Rapid and Instantaneous Closure Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 6.4 You will be expected to follow the derivation of the governing equations but not reproduce the derivation.. The equations derived here are used to predict the movement and subsequent decay of the pressure transients caused by the valve closure. This completes the development of equations for the analysis of unsteady pipe flow problems. The momentum or dynamic equation developed here remains essentially the same as that used in the rigid pipe mass oscillations studied in the previous module. In both cases it comprises an elevation term, an inertial head term and energy loss terms. It is the continuity equation which increases in complexity as the problems become more complex.

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It is interesting to compare the governing equations for unsteady pipeline flow with those for unsteady open channel flow.

Momentum Equations: Open Channel: 1 V V V y    So  S f g t g x x

13.12

1 V V V H    S f g t g x x

13.13

Pipeline:

The two equations are essentially the same. The term in H in the pipeline equation contains both a pressure component and an elevation component, and is equivalent to y  x  S0 in the open channel equation. The second term in the pipeline equation is an acceleration which derives from the spatial change in momentum. This term is small and is frequently ignored, for example, in Chadwick et al..

Continuity Equations: Open Channel:

y V y y V 0 t x x

13.14

Pipeline:

H c 2 V H  V  V sin  0 t g x x

13.15

where c is as derived in Section 13.4 .

Again the pipeline and open channel equations are essentially the same. For an open channel y  c 2 g . The latter two terms in the pipeline equation are equivalent to V y  x  S0 and are directly analogous to the last term in the channel equation.

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13.6 Solution of Equations Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 14.6 (pages 513 to 518 only)

These sections are included for completeness. They will not be examined. However you should appreciate that the solution of the governing equations can be performed using the same Finite Difference techniques that were used in Module 5 for the solution of unsteady free surface flows. The main difference here is in the different boundary conditions.

13.7 Selection of Method of Analysis Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 6.5 The definitions of instantaneous, rapid and slow closure are essential knowledge. If the closure is slow then the rigid pipe analysis of Module 12 is adequate, otherwise the elastic pipe analysis presented in this module should be applied.

13.8 Self assessment Questions

13.1 A mass oscillation (surge) may be differentiated from a pressure transient (water hammer) by: (a)

the time taken for a pressure wave to traverse the pipe

(b)

the presence of a reservoir at one end of the pipe

(c)

the rate of deceleration of the flow

(d)

the relative compressibility of the fluid to the elasticity of the pipe

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13.2 The speed of a pressure wave through a pipe depends upon: (a)

the length of the pipe

(b)

the original head at the valve

(c)

the viscosity of the fluid

(d)

none of these answers

13.3 The valve on the end of a pipeline 300 m long is closed instantaneously. Taking K as 2.11E9 Nm-2 the decompression wave will arrive at the valve after a time (s) of: (a)

2

(b)

0.4

(c)

0.2

(d)

0.04

13.4 If the valve in Question 3 was closed at a steady rate over a time of 2 s this would be considered to be: (a)

an instantaneous closure

(b)

a rapid closure

(c)

a slow closure

(d)

none of the above answers

13.9 Tutorial Problems 13.1

Chadwick et al. Hydraulics in Civil and Environmental Engineering Pages 625 Problems on Chapter 6 Question 1

13.2

Chadwick et al.

Ch 6, Question 3

13.3

Chadwick et al.

Ch 6, Question 4 Answers for Q4 are incorrect

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13.9.1 Answers to Tutorial Problems 13.1

Given in text

13.2

Given in text

13.3

The answer to problem 4(a) is 822 m/s, and 4(c) 781 kN/m2.

13.10 Solutions to Self assessment Questions 1. (c)

2. (d)

. .

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4. (b)

Module 14 – Gravity Pipelines

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Module 14 – Gravity Pipelines

14.1 Objectives When you have mastered the material in this module you will be able to:  recognise the different types of gravity pipeline;  describe the types of flow commonly encountered in culverts and the circumstances under which they can occur; and  design pipe and box culverts.

14.2 Types of Gravity Pipelines Gravity pipelines are characterised by the presence of a free water surface at both ends (or at least the upstream end) of the pipeline. This means that the energy available for flow is simply the difference between the total energy at each of these free surfaces. Gravity pipelines can be classified according to function and form, viz:  sewers;  urban drainage pipelines;  culverts; and  inverted siphons.

Sewers and urban drainage pipelines usually consist of an inter-connected system of gravity pipelines, laid on grade and connected by inspection and inlet pits. Design commences at the downstream end with each section of the pipeline designed as an individual gravity pipeline. The tailwater or downstream water level for each pipeline is taken from the headwater or upstream water level from the previous (downstream) pipeline. The significant difference between these two types of pipeline is that urban drainage lines are usually designed to flow full whereas sewers usually flow partially full. The design of these pipelines will be covered for Civil and Environmental Engineering students in the course ENV4203 Public Health Engineering.

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A culvert is a relatively short gravity pipeline used to convey the flow in a natural or man-made channel under a road or similar structure. The pipeline is usually laid on a grade similar to that of the channel. The pipeline may flow full (as in an irrigation channel) or may flow partially full (as frequently occurs with the conveyance of natural flood flows). A variety of entrance and exit configurations may be used. Road culverts are frequently constructed with a rectangular cross section (box culverts). An inverted siphon is a longer pipeline used to convey the flow in an irrigation channel under a road, channel, drain or river whose level is lower than the original channel. In this case the pipeline is not laid on grade. Inverted siphons must be designed to flow full. Hydraulically there is little difference between the different types of gravity pipelines. For simplicity this module will concentrate on the hydraulic analysis and design of culverts. The extension of this material to the design of other gravity pipelines is a relatively simple matter.

14.3 Types of Culvert Flow Book of Readings – Reading 14.1 Section 17.8 from Chow, V.T. 1959, Open Channel Hydraulics. McGraw Hill, New York Important here is Figure 17.28 and the associated discussion.

14.4 Culvert Design Book of Readings – Reading 14.2 Boyd, M.J. 1985, Head-discharge relations for culverts. 21st IAHR Congress, Melbourne, 118-122 Mastery of this material is essential

Book of Readings – Reading 14.3 Monier Rocla. Pipe and culvert hydraulic manual. Mastery of this material is essential

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14.5 Worked Examples – Application of the Energy Equation to a Culvert Flowing Full Worked Examples From Book of Readings Reading 14.3 Monier Rocla. Design examples in Section 3.4

Pipe and culvert hydraulic manual.

Examples based on the use of the design flow chart presented in Reading 14.3.

Worked Example From Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 9.4 Example 4.9 illustrates the application of the energy equation to a culvert flowing full. It is a more rigorous application of the energy equation than that used by Boyd in his equation 10 (Article 14.2). Note here the values used for the entrance and exit loss coefficients. These are a function of the entrance and exit geometries and should be determined experimentally for the particular geometry (see also Section 4 in Article 14.3).

Worked Example From Text Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 8.27 Example 8.27 introduces the full range of possible flow conditions. Compare the approach here to that in Reading 14.3. In case (i) note the addition of an energy loss term which transforms yc = 1.5H to yc = 1.75H. In case (iii) yo is the normal depth of flow in the culvert.

14.6 Partially Full Pipe Flow Reading from Text Chadwick et al, Hydraulics in Civil and Environmental Engineering Section 4.7 An expansion of the material in Reading 14.3. Mastery of this material is essential. However you are not expected to remember the equations developed. Note that the depth d for the partially full flow is a "normal" depth.

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14.7 Self assessment Questions

14.1 Entrance control of flow through a culvert will most probably occur when: (a)

the inlet is submerged

(b)

normal depth in the culvert exceeds the culvert diameter

(c)

normal depth in the culvert is less than the critical depth

(d)

normal depth in the culvert is less than the culvert diameter

14.2 Outlet control will occur in a culvert when: (a)

the inlet is submerged

(b)

normal depth in the culvert exceeds the culvert diameter

(c)

normal depth in the culvert is less than the critical depth

(d)

normal depth in the culvert is less than the culvert diameter

14.3 A box culvert, internal size 1 m wide by 0.8 m deep, is required to convey a discharge of 3.75 m3s-1. If the slope is 0.01, the length 60 m and the Manning n 0.012, then the normal depth (m) of flow in the culvert is: (a)

1.05

(b)

0.95

(c)

0.85

(d)

0.75

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14.8 Tutorial Problems 14.1

A pipe culvert is required to convey a discharge of 4 m3s-1 a length of 45 m under a highway. The slope of the culvert is 0.025 and the Manning n of the pipes is 0.012. Determine a suitable diameter for the culvert if the tailwater depth is 1 m and the headwater depth is not to exceed 2.5 m. For the purpose of this problem nominal pipe diameters for RC culvert pipes can be assumed to increase in 75 mm increments for diameters between 150 and 600 mm and to increase in 150 mm increments for diameters larger than 600 mm. The actual internal diameters are typically 1.5% greater than the nominal diameter.

14.2

A 2.5 m diameter pipe culvert, square edged with vertical headwalls is laid on a slope of 0.02 over a length of 20 m. Headwater is 2.55 m above the pipe invert at entrance and the tailwater is 1.75 m above invert at outlet. Manning n is 0.015. Determine:

14.3

(i)

the type of flow;

(ii)

the discharge of the culvert; and

(iii)

the normal depth of flow in the barrel.

Repeat Question 14.2 above (i, ii & iii) if for this culvert n is changed to 0.025 and the headwater increased to 4 m.

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14.8.1 Answers to Tutorial Problems

14.1

1200 mm diameter with a headwater depth of 2.34 m (based on an actual internal diameter of 1219 mm)

14.2

14.3

The answers to this problem will vary considerably with the particular equations and coefficients used. The answers given are indicative only and are based on the equations given in Article 14.2. (i)

Inlet control, inlet not submerged

(ii)

13.5 to 14 m3s-1

(iii)

1.1 m

Same as for Q 14.2 (i)

Inlet control, inlet submerged

(ii)

20.6 to 21.5 m3s-1

(iii)

very close to pipe full flow

14.9 Solutions to Self assessment Questions 1. (c)

2. (b)

. .

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Module 15 – Minimum Energy Loss Structures

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Module 15 – Minimum Energy Loss Structures

15.1 Objectives When you have mastered the material in this module you will be able to:  calculate the maximum height of sill or the maximum width contraction to give critical flow conditions in a channel; and  design weir and culvert type structures using the minimum energy loss principle.

15.2 Revision of Specific Energy Concepts Reading from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 5.7 (p 138-150) and 5.9 (p 155-158) This is all revision of material introduced in Hydraulics I but not covered in the detail required as background to this module. Mastery of this material is essential as it provides the basis for the design procedure to be developed later in the module. In particular note Figure 5.11, the development of the general equation of critical flow (p144 to 147) and the equations 5.23, 5.24 and 5.25. You should be able to reproduce the development of these equations.

Worked Examples From Texts Chadwick et al. Hydraulics in Civil and Environmental Engineering Example 5.4 Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Examples 8.14 & 8.15

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15.2.1 Self assessment Questions on Specific Energy

15.1 Critical depth in a rectangular channel is expressed by: (a)

Vy

(b)

q g

(c)

gy

(d)

 q2  3    g

1

15.2 If the specific energy of a flow is 1.235 m, the maximum discharge per m width (m²/s) is: (a)

0.50

(b)

1.12

(c)

1.50

(d)

2.34

15.3 If the minimum possible specific energy for a flow is 0.755 m, the discharge per m width (m²/s) is: (a)

0.50

(b)

1.12

(c)

1.50

(d)

2.23

15.4 If the discharge per unit width of flow in a channel is 1.5 m²/s, the minimum specific energy (m) is: (a)

0.918

(b)

1.112

(c)

1.500

(d)

2.343

15.5 For a channel with constant q, critical flow conditions can be produced by: (a)

expanding the width of the channel

(b)

decreasing the width of the channel

(c)

raising the bed of the channel

(d)

simultaneously lowering the bed and decreasing the width

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15.2.2 Tutorial Problems on Specific Energy All answers are provided in the text 15.1

Chadwick et al. Hydraulics in Civil and Environmental Engineering Page 623 Problems on Chapter 5 Question 3

15.2

Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Problems on Chapter 8 page 239 (4th Ed. p 255) Question 15

15.3

Marriott

Ch 8 Question 16

15.3 Minimum Energy Loss Design 15.3.1 Establishing Critical Flow We have seen from the above material that for any channel, critical flow can be achieved in either one of two ways, viz:  by elevation of the channel bed; or  by contraction of the width of the channel.

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15.3.2 Elevation For a rectangular channel of constant width (constant q), elevation of the bed causes the specific energy E to vary (decrease) and the depth of flow over the elevated section also to decrease. On the specific energy diagram (Figure 15.1), this is equivalent to moving around the constant q line from the initial condition toward the point where critical depth occurs (assuming that the change in bed is gradual so as not to cause any energy loss additional to the existing energy grade). Critical flow will occur when the specific energy is reduced to the minimum for the particular discharge. This corresponds to a z which is a maximum allowable without affecting the upstream condition. The critical depth yc is calculated from:

q  yc     g 2

1 3

and

E min 

3 yc 2

15.1

y=E

Water Depth, y (m)

(1:1 line)

y0

Initial condition (normal depth y0 = yn)

yc

Critical depth (yc) y vs E for constant q

Emin

E0

Specific Energy, E (m) Figure 15.1 – Specific Energy Diagram - Effect of Elevation of the Bed

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15.3.3 Contraction Contracting the width of the channel means that the specific energy E is held constant while the specific discharge q is increased. This is equivalent to moving vertically downward on the specific energy diagram (Figure 15.2), from the initial condition toward the point where critical depth occurs. In this case critical flow will occur when q is a maximum for the particular value of E. At this point the contraction will be the maximum allowed without altering the upstream condition. Here:

yc 

2 Eo 3

and

q max  gyc3

15.2

Note that the critical depth for this case is greater than that occurring when the bed is elevated.

y=E

Water Depth, y (m)

(1:1 line)

Initial condition y0

(y0)

yc

Critical depth (yc) qmax q0

E0

Specific Energy, E (m) Figure 15.2 – Specific Energy Diagram - Effect of Channel Width Contraction

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15.3.4 Simultaneous Variation of Bed Elevation and Channel Width Once critical flow has been achieved it can be maintained by holding the channel geometry constant and the flow returned to its normal state by the reverse of the original change, that is, by lowering the bed or expanding the width. Again it is necessary for these changes to be gradual so as not to cause increased energy loss.

Alternatively, critical flow can be maintained and the energy level altered if the geometry changed by:  simultaneously expanding the width of the channel and further elevating the bed; or  simultaneously lowering the bed and further contracting the width.

Water Depth, y (m)

Both of these actions involve moving along the line defining critical flow (Figure 15.3) but in opposite directions. In both cases q, E and yc are varying in magnitude simultaneously.

Critical depth line

2    yc  Emin  3  

Contraction & depression

Expansion & elevation

Specific Energy, E (m) Figure 15.3 – Specific Energy Diagram - Simultaneous Contraction of the Width and Lowering of the Bed

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15.3.5 Design For a particular discharge and energy level, critical flow gives the minimum depth, maximum velocity and hence the minimum cross section for that flow. This fact and the strategies described above can now be used in the design of a class of very efficient "weir" and "culvert" type structures. In reality these are not structures but are shaped transitions in an open channel in which the normal flow is converted to a critical flow, held in the critical condition for an appropriate distance and then returned to the normal condition. For a minimum energy loss "culvert" the steps in the transition are:  contract the channel width to the maximum allowable to cause critical flow;  continue to contract the channel to the desired width and simultaneously lower the bed, thus maintaining critical flow but at a higher E and q;  hold the channel geometry constant for an appropriate distance; then  return the flow to its normal condition by the reverse of the above processes.

Worked Examples From Reading 15.1 Appendix 7 from MRDQ, Urban Road Design Manual

15.4 Additional Readings Book of Readings – Reading 15.2 Isaacs, L.T. 1990, “Design calculations for minimum energy loss culverts”. Trans. IEAust Civil Eng, CE32(2): 87-92. Useful extension of the above material written in the form of a design procedure.

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15.5 Tutorial Problems 15.1

An existing rectangular channel 4 m wide conveys a discharge of 15 m3s-1 at a depth of 2 m. A footbridge crosses this channel with the underside of the bridge 2.1 m above the channel bed. It is required to increase the discharge in the channel to 20 m3s-1. What modifications can be made to the channel to ensure that the footbridge is not inundated by this increased discharge? Calculate the height of the water surface under the footbridge following completion of these modifications. Hint: Before attempting to apply any modifications to the channel you must first estimate the flow depth, velocity and specific energy for the increased discharge of 20 m3s-1. This can be done using the Manning equation assuming S o0.5 n is constant but is more easily done using the Chezy equation where CS o0.5 is taken to be constant.

15.2

List the steps required in the design of a minimum energy loss "weir" type structure.

15.3

For the channel in the above worked solution, viz: b = 25 m

y = 1.2 m

V = 0.6 m/s

Q = 18 m3s-1

design a 'minimum energy loss weir' whose crest is 1 m above the original bed.

15.4

A weir designed using the 'minimum energy loss' concept has the following dimensions:

Distance D/S from crest (m)

Elevation

Width

(m)

(m)

00

630.293

301.5

50

629.805

75

100

628.193

75

Remarks Weir crest

D/S Channel bed

Estimate the design discharge for this structure if the normal depth of flow downstream of the structure is 2.4 m.

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15.5.1 Answers to Tutorial Problems 15.1

Elevation of the bed and width contraction are both feasible options: z = 0.645 m

Elevation

Water surface is 2.015 m above original bed Contraction

Width = 2.63 m Water surface is 1.8 m above original bed

15.2

i. Elevate the bed to give critical flow ii.

Simultaneously elevate the bed and expand the width to give the desired crest height or length while maintaining critical flow

iii. Hold the geometry constant over the crest of the weir iv. Return the flow to the normal condition by the reverse of the above processes

15.3

Station (m)

Width (m)

Bed Elevation (m)

0

2530

0.0

1

25.0

0.655

2

103.73

1.0

(Downstream is reverse of above)

15.4

Q = 91.7 m3s-1

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15.6 Solutions to Self assessment Questions 1. (d)

2. (d)

5. (c)

. .

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3. (b)

4. (a)

Module 16 – Measurement and Control Structures

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Module 16 – Measurement and Control Structures

16.1 Objectives When you have mastered the material in this module you will be able to design a range of open channel measurement and control structures, including:  sharp and broad crested weirs;  flumes; and  gated structures.

16.2 Gated Control Structures Reading from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 13.8 (including example 13.6) Essential knowledge. In particular note:  the application of the continuity and energy equations across the gate to give the equation (eqn 13.31) for free flowing (unsubmerged) conditions;  the use of the hydraulic jump equation (eqn 5.28) to determine if free flowing conditions apply or if the hydraulic jump will migrate upstream to drown the gate; and the change in the analysis required (equations 13.34 & 13.35) and the resulting larger value of YG when the hydraulic jump advances upstream and drowns the gate. Note also that in equation 13.35 the depth y3 will typically be the normal depth in the channel downstream.

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16.3 Sharp Crested Weirs Reading from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 13.2 (including example 13.1)  This is an expansion of material covered in Hydraulics I. Important here are:  application of the energy equation to give an expression for Qideal;  the discharge coefficient Cd;  application of the formulae for Q (you are not expected to remember these formulae); and  the correction for submergence of a weir (equation 13.7).

16.4 Long Based Weirs (Broad Crested Weirs) Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 5.9 and 13.3 (including example 13.2) Important concepts here are as for 16.2 above.

16.5 Flumes Readings from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 5.9 and 13.4 Background reading. Calibrated proprietry designs are usually employed in practice, an example of which (the cut throat flume) is described in the following reading.

Book of Readings – Reading 16.1 Section 7.6 from Kraatz, D.B. and Mahajan, I.K. 1975. Small Hydraulic Structures. FAO Irrigation and Drainage Paper 26/2. Particularly important here are the worked examples 1 & 2. © University of Southern Queensland

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16.6 Tutorial Problems 16.1

A rectangular channel 2 m wide, bed slope 0.001 and Manning n of 0.012, conveys discharges ranging up to 4 m3s-1. It is required to measure these discharges, but without increasing the upstream depth by more than 0.36 m. Prepare and evaluate alternative designs using: i. a rectangular sharp crested weir; ii. a broad crested weir; iii. a venturi flume; and iv. a short cut-throat flume.

It is permissible and may be necessary to alter the channel geometry in the vicinity of the measuring structure.

Note: There is an infinite number of feasible solutions to these problems, hence no answers are given.

16.2

An irrigation scheme is fed from a river via a diversion channel. The discharge into the channel is controlled by a vertical sluice gate. The irrigation channel is 4 m wide and is roughly faced with cemented rubble giving an estimated value of 0.028 for the Manning n. The bed slope is 0.002. Prepare a rating curve for the sluice gate (yG vs Q) and for the downstream depth (y3 vs Q) if the depth upstream is a constant 2.5 m. Assume that the maximum discharge is 15 m3s-1.

. .

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Module 17 – Dimensional Analysis and Hydraulic Similitude

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Module 17 – Dimensional Analysis and Hydraulic Similitude

17.1 Objectives When you have mastered the material in this module you will be able to:  explain the concepts of geometric, kinematic and dynamic similarity,  describe the important force ratios for common hydraulic structures,  formulate mathematical relationships for hydraulic systems using the index and Buckingham П methods of dimensional analysis, and  design undistorted scale models of hydraulic systems.

17.2 Dimensional Analysis Readings from Texts Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 11.1 to 11.4

Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Sections 9.1, and 9.2 Dimensional analysis can be described as the study of the relationships among the variables of a physical process which may be used to determine the outcome of the process. The applications of dimensional analysis include:  the conversion of different systems of measurement (e.g. metric to imperal units)  the development of mathematical models of processes, and  the design of physical scale models of processes.

While the units of measurement of quantities under different measurement systems will vary, the quantities themselves must have the same dimensions. For example the unit of measurement of velocity in the imperial system is the foot per second; in the © University of Southern Queensland

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metric system it is the metre per second. However the dimensions of velocity must be the same (LT−1) regardless of the system of measurement. This is the first application of dimensional analysis alluded to above. The concept of dimensional homogeneity described in Chadwick et al. (Section 11.3) is an extension of this fundamental principle. The Darcy-Weisbach and many other semi-empirical equations used in hydraulics have been formulated with the assistance of dimensional analysis, mainly because the process itself is too complicated to predict the outcome by analytical means alone. This is the second application of dimensional analysis described above. The third application of dimensional analysis relates to the design of scale models of processes. A scale model might be contemplated as a check against a poorly represented mathematical model of a process, or, more often than not, might be used instead of a mathematical model because the physical process is too complicated to reduce to a mathematical form. Many of the processes in hydraulics fit into this category. An example is the dispersion of a non-conservative point-source pollutant in a three-dimensional unsteady flow regime, such as that in large tidal estuaries. The dispersion process itself is quite complicated under this flow regime, and a physical scale model of the estuary might be constructed to examine the transport and dispersion of the pollutant under cyclic (tidal) discharge. Once the behaviour of the pollutant is verified by experimentation, the scale model might then be used, for example, to predict the effect of canal development on water quality in the estuary. In the context of determining a mathematical model of a process given the main variables of that process, there are two main methods of dimensional analyses described in the texts. These are:  the index method, and  the Buckingham П method As mentioned in the texts these are best illustrated by example.

17.3 The Index Method Worked Example 17.1 (From Text) Chadwick et al. Hydraulics in Civil and Environmental Engineering Example 11.1 page 374 In this example it is postulated that the thrust produced by a hydraulic machine is dependent on the flow velocity, the fluid density and viscosity and a characteristic length of the machine. This may be stated mathematically as: F  fn(  , u,  , L)

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Module 17 – Dimensional Analysis and Hydraulic Similitude

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where fn means is a function of and L is the characteristic length. If these are the only independent variables governing the dependent variable F, then this may be expressed as an explicit equation: F  C1 a u b  c Ld

17.2

where C1 is a constant and a, b, c, d are exponents which may be other than zero or one. Dimensional homogeneity requires that the dimensions of the left hand side of this equation must equal the dimensions of the right hand side. The dimensions of force are MLT−2, hence the dimensions of C1ρa, ub, μc,Ld collectively must also be MLT−2. The index method of dimensional analysis is used here to determine the exponents a, b, c and d. However in this case the number of independent variables (ρ, u, μ, L) is four (4) and the maximum number of dimensions is three (3) hence there are more unknowns than equations available. In the example this problem is overcome by assuming that the effect of the dynamic viscosity μ of the fluid is small compared to the effect of the other three independent variables, with the result that: F  C1 u 2 L2

c  c u c Lc

17.3

where C1 and c are unknowns which must be evaluated by experimentation. The fact that the independent variables on the right hand side of this equation can be arranged into two dimensionless groups is an important concept for the Buckingham П method which we shall investigate shortly.

The next example better illustrates the index method, in which of course the number of independent variables must be limited to three (3).

Worked Example 17.2 It is hypothesised that the discharge through a small sharp-edged orifice plate in a pipe is dependent on the density of the fluid, the orifice diameter and the pressure loss across the orifice. Determine a mathematical relationship for discharge using the index method of dimensional analysis. Solution As stated the discharge (Q) is a function of density (ρ), diameter of the orifice (d) and pressure loss (∆p): Q  fn   , D, p 

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Expressing this as an explicit equation: Q  C1 a Db  p 

c

where a, b and c are unknowns In dimensional form this becomes:



L3T 1  ML3

 L  ML a

b



1 2 c

T

According to dimensional homogeneity the units on the LHS and RHS must be equal therefore equate indices: For M

0ac

For L

3  3a  b  c

For T

1  2c

which is solved simultaneously to yield:

 a   1 / 2  b    2      c   1 / 2 

The functional form of the relationship is therefore: c Q  C1 a Db  p  Q  C1



Q  C1D 2

1 2 D2

1

 p  2

p



But it is often more convenient to express pressure losses in terms of metres head of water, H. Recall that p   gH . Next multiply the RHS by g g and attempt to replace the pressure p with H: Q  C1D 2

p



Q  C1 g D 2

Similarly replacing Diameter, D with Area A

Q  C1 g

4A





g g

p g H

Q  C2 A H

where H is the head loss across the orifice and D is the diameter of the orifice and C1 and C2 are constants which can be evaluated by experimentation. As expected, the © University of Southern Queensland

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form of this relationship is the same as that of the orifice equation Q  KA0 2 gH where the constant C2  K 2 g The index method is satisfactory when the number of independent variables is three (3) or less. In many physical systems however the number of independent variables of a process is much greater than three. The Buckingham П method of dimensional analysis overcomes this limitation.

17.4 Buckingham Pi (П) Method Readings from Texst Chadwick et al. Hydraulics in Civil and Environmental Engineering Section 11.5 Also revise section 11.4 for an introduction to П groups Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Section 9.4 The basis of the Buckingham П method is that the functional form of the process encompassing n variables can be expressed in terms of n − m dimensionless groups of these variable: fn  p1 , p2 , p3 , p4 , p5 ....... pn   0

17.4

Can be expressed as: fn  1 ,  2 ,  3 ,  4 ,  5 ....... n  m   0

17.5

where n is the total number of variables of the process and m is the number of dimensions of relevance to the process. For our purposes the relevant dimensions are M, L and T and hence m = 3. By grouping the variables into dimensionless blocks, the method effectively reduces the number of unknowns; a limiting factor in the index approach. These dimensionless groups are known as П groups. For example, if the outcome of a process (as represented by a dependent variable) involves five (5) independent variables then there are six (6) variables in total relevant to the process and there are (6 − 3) three (3) П groups required to represent that process.

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The Buckingham П procedure is perhaps better described by Munson Young and Okishi. (1994, p.399): a. Express each of the total n variables of the process in terms of basic dimensions M, L and T. b. Determine the required number of П groups. There will be n − 3 groups. c. Select 3 repeating variables from the group of independent variables. These will be combined with each of the remaining variables to form a П group. No two of these repeating variables can have the same dimensions and collectively all 3 dimensions must be represented in the 3 repeating variables. Those with the simplest dimensional form should be chosen as the repeating variables; i.e. select from the independent variables those with the smallest number of dimensions and smallest value of exponents for those dimensions as the repeating variables. If any two independent variables have the same dimensions their ratio will be one of the П groups. If an independent variable is dimensionless it is a П group. The dependent variable cannot be used as a repeating variable. d. Form a П group by multiplying one of the non-repeating variables by the product of the repeating variables each raised to an exponent that will make the combination dimensionless. Each П group will have the form pi p1a p2b p1c where pi is one of the non-repeating variables, p1, p2 and p3 are the repeating variables and a, b and c are determined so that the combination is dimensionless. e. Repeat the previous procedure for the remaining non-repeating variables. The resultant number of П groups must be n − 3. f. As a check against error, ensure that each of the П groups is dimensionless. g. Express the final form of the relationship between the dependent variable and the independent variables as a relationship among the П groups. The П groups can be manipulated by:  replacing any П group by any power of that group,  multiplying П groups together,  multiplying any П group by a constant, and  expressing a П group as a function of the other П groups.

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17.4.1 Worked Examples Worked Example 17.3 (From Text) Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 9.1 This example is reworked below to emphasise the basis of selection of the repeating variables.

Solution The hypothesis is: p  fn  L,  ,V , D,  , k 

There are six (6) independent variables and one (1) dependent variable making a total of seven (7) variables. The required number of П groups is 7 − 3 = 4; i.e. fn  1 ,  2 ,  3 ,  4   0

Expressing each of the variables in dimensional form: pressure loss

p :

F MLT 2   ML1T 2 2 A L

pipe length

L:

L

fluid density

:

MLT 3

velocity

V:

LT 1

D:

L

:

stress  time  ML1T 2  T  ML1T 1

k:

L

pipe diameter dynamic viscosity roughness height

The independent variables L, ρ, V, D and k have the simplest dimensional form, however no two can have the same dimensions so we must choose one of L, D and k. The diameter is chosen so that the repeating variables become ρ, D and V. Choosing ∆p as the first non-repeating variable: 1  p  a DbV c





1  ML1T 2 ML3



a

 L b  LT 1 

c

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Equate indices: For M

1 a  0

For L

1  3a  b  c  0

For T

2  c  0

 a   1 b    0      c   2

from which it follows that:

hence: 1  p  1D 0V 2 1 

p V 2

Now check to ensure that this П group is dimensionless: 1  p  1D 0V 2

   ML

   L   LT   M L  L  L T 

1  ML1T 2 ML3 1

1 2

T

1 3

1  M 11 L13 0 2 1  M 0 L0

1

1 2

0

0

2

T 2 2

T0

Hence the П group is dimensionless. Choose L as the next non-repeating variable:  2  L  a DbV c



 2   L  ML3



a

 L b  LT 1 

c

Equate indices:

from which it follows that:

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For M

a0

For L

1  3a  b  c  0

For T

c  0

 a   0 b    1     c   0 

2

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from which it follows that:  2  L  0 D 1V 0 2 

L D

This П group is obviously dimensionless as L and D possess the same units. The next non-repeating variable is k, but because it has the same dimensions as the pipe length L, the result will be the same as the previous calculation:  3  k  0 D 1V 0 3 

k D

The remaining non-repeating variable is μ:  4   a DbV c





 4  ML1T 1 ML3



a

 L b  LT 1 

c

Equate indices: For M

1 a  0

For L

1  3a  b  c  0

For T

1  c  0

 a   1 b    1     c   1

from which it follows that:

from which it follows that:  4   1D 1V 1

   ML

   L   LT   M L  L  L T 

 4  ML1T 1 ML3 4

1 1

T

1 3

 4  M 11 L1311  4  M 0 L0

1

1

1

1 1

1 1

T 11

T0

Hence the П group is dimensionless.

The functional form of the relationship is then: © University of Southern Queensland

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fn  1 ,  2 ,  3 ,  4   0  p L k  fn  , , , 2  V D D  DV

 0 

L k p    fn  , ,  2 V  D D  DV 

Rearranging and noting that the inverse of П4 is Reynolds (Re) for pipe flow: p

V

2



k L    fn  ,  D  D  DV 

p V 2 k    fn  , Re  L D D  p V 2 k    fn  , Re   gL  gD D  

Noting that the LHS is equivalent to the friction loss per unit length: hf L



p V 2 k    fn  , Re   gL gD D  

which is the form of the Darcy-Weisbach equation: hf L



hf 

fV 2 gD fLV 2 gD k  , Re  ) D 

where f is a function of relative roughness and Reynolds number ( f  fn 

Worked Example 17.4 (From Text) Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 9.2 In Example 9.2 there is a total of seven (7) variables and hence four (4) П groups. The repeating variables ρ, N and D are chosen. The dimensions of the repeating variables are ML−3, T−1 and L respectively. Hence all three dimensions are collectively represented. The functional form of the relationship among variables is: fn  1,  2,  3,  4,   0 gh p  ND 2  B Q fn  , , , 0 3 2 2    D ND N D

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If we compare these П groups with the pump affinity laws given previously without proof as equations 11.1 and 11.2 in module 11 (and repeated below as 17.8 and 17.9) we see that П2 and П3 are exactly the same as the discharge and head laws for dynamically similar (homologous) pumps. The expression for output power given in equation 17.10 is simply the product of the expressions for discharge and pump head: Q  constant ND3 gh p N 2 D2

 constant

P0   gh p Q

Q  ND 3

so

so

so

hp 

P0   g

17.8

N 2 D2 g

17.9

N 2 D2  ND 3 g

P0   gN 3 D 5

17.10

Realising that the proportionality sign can be replaced with a constant: P0  constant   gN 3 D5 P0

 gN 3 D5

 constant

17.11

It should not surprise you that the pump affinity laws are derived from dimensional analysis.

Worked Example 17.5 (From Text) Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 9.3 The relationship for discharge over a V notch weir using dimensional analysis is shown in Example 9.3: 1 3   2 2  Q  g H  gH 2  fn  1 5 , , ,   0    g2H 2   

Disregard the discussion at the end of this example relating to Reynolds and Weber number as we will return to this in the next section.

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Worked Example 17.6 (From Text) Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 9.4 The result for the sharp-crested rectangular weir in Example 9.4 is similar to that for the V notch in the previous problem with an obvious difference in the exponent for head on the weir.

Worked Example 17.6 (From Text) Chadwick et al. Hydraulics in Civil and Environmental Engineering Example 11.2(a) Example 11.2(a) in Chadwick et al. is similar to that of Example 9.1 in Marriott. Note that we will use the Buckingham П method of repeating variables and not the matrix method.

17.5 Hydraulic Similitude Reading from Text Chadwick et al. Hydraulics in Civil and Environmental Engineering Sections 11.2 (revise) 11.6 and 11.7 Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Section 9.5 An extension of dimensional analysis is its application in the design of scale models. A model of a real system (i.e. the prototype) is said to be geometrically similar to (or, to have geometric similitude with) the prototype when the ratio of lengths and the shape of the streamlines and boundary geometry between the model and prototype is the same; i.e. for geometric similitude:

Lm  constant LP

17.12

where the subscripts m and p relate to the model and prototype respectively.

Kinematic similitude is said to exist between model and prototype when the ratio of velocities and accelerations are the same:

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for kinematic similitude:

Vm  constant VP

159

17.13

Obviously kinematic similitude cannot be achieved without prior geometric similitude. The ultimate goal in model design is to achieve exact dynamic similitude between model and prototype so that the ratio of all forces is the same: for dynamic similitude:

Fm  constant FP

17.14

which is impossible without prior kinematic similitude.

In the preceding section the П groups of the relationship between the dependent variable and the various independent variables of a process represent, in one way or another, the terms defining the required geometric, kinematic and dynamic similitude which must exist between the input and output variables. Thus to design a scale model so as to achieve exact dynamic similitude, it follows that each and every one of the corresponding П groups must be the same between model and prototype. In practice exact dynamic similarity between model and prototype is very difficult to achieve. The П groups representing geometric and kinematic terms might be easily duplicated in the model, but it is more difficult to simultaneously reproduce all groups representing force ratios. However we might postulate that some forces will be more dominant than others depending on the situation being modelled. For example, you may recall from previous work in pipe flow that the effect of fluid viscosity (i.e. the viscous forces acting on fluid particles) is quite significant for laminar flow, less significant for smooth turbulent flow and hardly significant at all for rough turbulent flow. Thus in order to design a scale model of an estuary for example, then as long as the flow is rough turbulent (and most open channel flow involving water is), or nearly so, in model and prototype, then viscous forces will be small compared to other forces (i.e. gravity) and can be neglected. It follows that, while we may not be able to achieve exact dynamic similarity in all cases, it may be possible to design a scale model with approximate dynamic similitude by neglecting the effect of the less significant force ratios and design such that similitude is achieved among the more dominant force ratios. Some of the more important force ratios are listed in Table 17.1.

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Table 17.1 - Some common force ratios in fluid mechanics

П group

Name

Force Ratio

Applications

Reynolds number

inertia force ma  viscous force  A

generally of importance in all types of fluid dynamics problems

u gL

Froude number

inertia force ma  gravity force mg

free surface flows

u 2 p

Euler number

inertia force ma  pressure force pA

situations in which pressure differences are of interest

u 2 

Cauchy number

inertia force ma  compressibility force 

situations in which fluid compressibility is important

 uL 

u 2 

Mach number

u 2 L 

Weber number

inertia force ma  compressibility force 

inertia force ma  surface tension force  L

as for Cauchy number situations in which surface tension effects are important

Adapted from Munson et al. (1994, p.413) Some of the terms above have not been previously defined: κ is the bulk modulus of elasticity (Pa or N/m2) σ is the surface tension (N/m)  is volume (m3)

Note that L in Table 17.1 is a characteristic length which depends on the situation. For pipe flow the characteristic length is the pipe diameter; for open channel flow it is the hydraulic mean depth. Note that Marriott instead adopts the hydraulic radius for this characteristic length for open channel flow. However the difference between the two is minimal especially when the width of the channel is large compared to the depth of flow. There are numerous other dimensionless numbers used in hydraulics. For example the Strouhal number is the ratio of the time scale of the flow to the period of oscillation of cyclic flows, and is useful in modelling drag on bridge piers where pressure variations due to vortex shedding effects are significant, and in the modelling of estuary flows under tidal oscillations. Another is the Peclet number which is the ratio of advection to diffusion; important in studies involving the dispersion of pollutants in streams. The proof that the П groups in Table 17.1 are actually dimensionless force ratios is provided by dimensional analysis:

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For Reynolds number:

    

 L3 LT 2 ma ma  L4T 2  L2T 1  LT 1L  uL Re         A  du A    T 1L2 LT 1 2  L dL L

17.15

For Froude number:

    



1  L3 LT 2 ma  L4T 2 LT 2 LT 2 L LT Fr        mg g g L gL  L3 g  L3 g 2



2



u2 gL

17.16

And for Weber number:

 





1  L3 LT 2 ma  L4T 2  L3T 2  LT We      L L L  



2

L



u 2 L 

17.17

where σ is the surface tension which, for water at 16°C, is in the order of 0.0735 N/m. For fluids such as oil and ethanol surface tension is much lower at approx 0.030N/m and 0.022 N/m respectively while for mercury the surface tension is much higher at 0.470 N/m.

17.5.1 Worked Examples Worked Example 17.7 (From Text) Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 9.3 The П groups for discharge over a V notch sharp-crested weir were found to be: fn  1,  2,  3,  4,   0 1 3   2 2  Q  g H  gH 2  fn  1 5 , , ,   0    g2H 2   

It can be shown by dimensional analysis that the П1 group is a form of Froude number: Q 1 5 g2H 2



uL2 1 5 g2H 2



u 1 1 g2H 2

 Fr

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П2 is the Reynolds number: 1 3 g2H 2







 LT 2 



1 2

3

L2

1

3

 L2 T 1L2  L2T 1  LT 1L  uL      Re    

and П3 is the Weber number:







2 2 1  gH 2  LT L  L2T 2 L  LT       



2

L



u 2 L  We 

Thus to design a scale model of the V notch weir so as to achieve exact dynamic similarity, it would be necessary to ensure that firstly the notch half-angle (θ) is the same between model and prototype (for geometric similarity) and secondly that the Froude, Reynolds and Weber numbers are the same (for dynamic similitude). However if low viscosity fluids such as water are to be used at (fairly high) velocities typical of those in open channels, then the flow regime will most likely be rough turbulent, in which case Reynolds number will be high and hence viscous forces small. For this situation gravity forces are likely to dominate over viscous forces, and Froude number is far more influential than Reynolds number. On the other hand if a viscous fluid is to be used then Reynolds number will be small, particularly at low velocities, and hence viscous forces will be significant. At low velocities too, the inertia forces on the fluid particles are likely to be small compared to the surface tension forces, resulting in a small Weber number. Hence if the model is designed so that velocities are small and/or the fluid used is viscous and/or has a high surface tension, then all three force ratios will be important and the model must be designed accordingly. However, if water is used and velocities are relatively high (i.e. the model is not required to reproduce very small flow rates) then the Reynolds and Weber numbers will be high and therefore not important in model design and the dominant force ratio will be that of inertia to gravity; i.e. Froude number. This is not to say that viscous and surface tension forces will be zero for this case; they will still be evident but they will be very small provided Reynolds and Weber numbers are high, and an approximate dynamic similitude can be achieved by designing such that Froude number is the same between model and prototype.

Worked Example 17.8 (From Text) Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 9.4 The result for the rectangular weir in Example 9.4 is similar to the V notch, however for geometric similarity it is the ratio of weir height to head (or weir length) which must be preserved in the model.

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Worked Example 17.9 (From Text) Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 9.5 Example 9.5 involves the design of a scale model of a pipeline. It was shown by dimensional analysis in a previous example that the hydraulic gradient may be represented as: Sf 

V2 k   fn  , Re  gD D 

Since the characteristic length for pipe flow is the diameter, the П group

V2 is the gD

square of Froude number, but Froude number is not important for pressure flow in pipes as viscous rather than gravity forces are more dominant. Reynolds number is the dominant П group for this situation hence the model can be designed for approximate dynamic similarity by ensuring that Re is the same between model and prototype. Geometric similitude is achieved by preserving the ratio of relative roughness.

Worked Example 17.10 (From Text) Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 9.6 Oil is to be used for the prototype of the V notch weir in Example 9.6. Because oil is viscous, shear forces are likely to be significant, hence to achieve an approximate dynamic similarity both Froude and Reynolds numbers must be the same in model and prototype. Since it is not possible to reproduce the same Froude, Reynolds and Weber numbers simultaneously, an exact dynamic similitude cannot be achieved. Equating Reynolds number: 1 3   g2H 2    

1 3     g2H 2      oil 

     water

                   oil   water 3 H2

3 H2

noting that  =

 and g=constant 

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from which the required head for oil flow is calculated as 0.553 m. Equating Froude number:   Q  1 5  g2H 2    Q  5  2 H

    Q   1 5   oil  g 2 H 2

    Q   5   oil  H 2

     water

     water

from which the required oil discharge for Froude similarity is 317 L/s.

Worked Example 17.11 (From Text) Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Example 9.7 In Example 9.7 the model is to be used to reproduce relatively high velocity flows using a low viscosity fluid, hence the Weber and Reynolds numbers are not important and the model is designed using Froude scaling alone. The more general form of Froude number is used here with the hydraulic mean depth as the characteristic length. It is confirmed that the resultant flow regime in the model is turbulent, hence Reynolds number will be high and viscous forces small. Please note that the critical Reynolds numbers for open channel flow are approximately 500 (upper laminar) and 2000 (lower turbulent) which is different to that for pipe flow.

Further Worked Examples From Texts Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Examples 9.10 & 9.11 Chadwick et al. Hydraulics in Civil and Environmental Engineering Examples 11.2(b) & 11.4 An interesting result from Example 9.10 of Marriott is the specific speed equation from dimensional analysis. Froude scaling is used in Example 9.11 of Marriott for the design of a model spillway as the dominant force ratio is inertia to gravity in most open channel flow situations. Example 11.2(b) of Chadwick et al. is another example of Reynolds scaling while Example 11.4 is similar to that in Nalluri & Featherstone dealing with homologous pumps. In this subject our coverage will not extend to the design of distorted scale models.

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17.6 Self assessment Questions 17.1 Which of the following quantities is dimensionless? (a)

angular velocity

(b)

kinematic viscosity

(c)

relative density

(d)

specific weight

(e)

none of these answers

17.2 It is postulated that the dependent variable of a physical process is a function of six (6) independent variables. The number of П groups required to represent this process is: (a)

six

(b)

seven

(c)

four

(d)

three

(e)

none of these answers

17.3 With the index method of dimensional analysis, the number of independent variables is limited to: (a)

six

(b)

seven

(c)

four

(d)

three

(e)

none of these answers

17.4 In dimensional analysis, the Froude number may be defined as the ratio of inertia force to: (a)

viscous force

(b)

gravity force

(c)

surface tension

(d)

kinematic viscosity

(e)

none of these answers

17.5 In the dimensional analysis of rotodynamic pumps, the П group  ND 2 represents:  (a)

Froude number

(b)

Reynolds number

(c)

Weber number

(d)

Mach number

(e)

none of these answers

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17.6 In the dimensional analysis of flow over a sharp-crested V notch weir, the П  gH 2 group represents:  (a)

Froude number

(b)

Reynolds number

(c)

Weber number

(d)

Mach number

(e)

none of these answers

17.7 In the dimensional analysis of flow over a sharp-crested V notch weir, the П group

Q 1 5 2 g H2

represents:

(a)

Froude number

(b)

Reynolds number

(c)

Weber number

(d)

Mach number

(e)

none of these answers

17.8 A scale model of a sharp-crested weir is to be designed using oil as the fluid. The dominant П groups of this design will be: (a)

Froude and Reynolds numbers

(b)

Froude and Weber numbers

(c)

Froude, Reynolds and Weber numbers

(d)

Reynolds number

(e)

none of these answers

17.9 In which of the following model studies would Reynolds scaling alone be appropriate? (a)

dispersion of pollutant in estuaries

(b)

drag on bridge piers under vortex shedding

(c)

drag on a supersonic aircraft

(d)

drag on low-velocity marine vessel

(e)

none of these answers

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17.10 In which of the following model studies would Froude scaling alone be appropriate? (a)

dispersion of pollutant in estuaries

(b)

drag on bridge piers under vortex shedding

(c)

drag on a supersonic aircraft

(d)

drag on low-velocity marine vessel

(e)

none of these answers

17.7 Tutorial Problems 17.1

Chadwick, et al. Hydraulics in Civil and Environmental Engineering Problems on Chapter 11 (pages 629 & 630) Question 1 Hint: assume discharge a function of density ρ, gravitational acceleration g, head on weir H, dynamic viscosity μ, surface tension σ and length of weir L.

17.2

Chadwick, et al.

Ch 11 Question 3

Hint: “Specific speed” 17.3

Marriott, Nalluri & Featherstone’s Civil Engineering Hydraulics Problems on Chapter 9 (page 258) Question 4

17.4

Marriott, Ch 9

Question 6

17.5

In forced vortex motion, the vertical pressure variation (∆p) of a fluid is thought to depend on radial distance r, fluid density ρ and angular velocity ω. Determine the form of the relationship by dimensional analysis.

17.6

It is thought that the buoyant force Fb acting on a submerged body depends on the submerged volume of fluid υ, the fluid density ρ, the gravitational acceleration g and the surface tension σ of the fluid. Determine the form of the relationship by dimensional analysis.

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A spillway model is built to a scale of 1:25 across a flume which is 610 mm wide. The prototype is 11.4 m high and the maximum head expected over the spillway is 1.52 m. a. What spillway height and head should be used in the model? b. If the flow over the model at 61 mm head is 20 L/s what flow may be expected for the prototype under a head of 1.52 m? c. If the model shows a measured hydraulic jump of 26 mm, how high is the jump in the prototype? d. If the power dissipated in the hydraulic jump in the model is 112 W, what is the power dissipation in the prototype?

17.7.1 Answers to Tutorial Problems Answers to Q 17.1 – 17.4 can be found in the texts 17.5

p  C1  r 2 2

17.6

Fb  C1 g

17.7

(a)

456 mm

(b)

61 mm

(c)

62.5 m3/s

(d)

650 mm

(e)

8.75 MW

17.8 Solutions to Self assessment Questions 1. (c)

2. (c)

3. (d)

4. (b)

5. (b)

6. (c)

7. (a)

8. (c)

9. (d)

10. (e) . .

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List of References

169

List of References Boyd, M.J. 1985, “Head-discharge relations for culverts”, in Proceedings of 21st Congress International Association for Hydraulic Research, Melbourne, Australia, 1985, IAHR, Melbourne pp. 118–122. Chadwick, A., Morfett, J. & Borthwick, M. 2004, Hydraulics in Civil and Environmental Engineering, 4th edn, Spon Press, Great Britain Chow, V.T., 1959, Open Channel Hydraulics, McGraw-Hill, New York Gerhart, P.M. and Gross, R.J. 1985, Fundamentals of Fluid Mechanics. AddisonWesley, Reading, MA. Henderson, F.M. 1966, Open Channel Flow. Macmillan, New York Isaacs, L.T. 1990, "Design Calculations for Minimum Energy Loss Culverts," Trans. IEAust Civil Eng, Vol CE32, No.2, pp. 87-92 Kelly and Lewis Pumps 1987, Pump Product Catalogue, Thompsons, Kelly & Lewis Pty Ltd, Melbourne Kraatz, D.B. and Mahajan, I.K. 1975. Small Hydraulic Structures. FAO Irrigation and Drainage Paper 26/2. Marriott, M, 2009 Nalluri & Featherstone’s Civil Engineering Hydraulics, 5th Edition, Wiley-Blackwell, United Kingdom Monier Rochla, Pipe and Culvert Hydraulics Manual, pp.1-20 Morris, H.M. 1963, Applied Hydraulics in Engineering, The Ronald Press Co, New York MRDQ, Urban Road Design Manual, Volume 2. Munson, B.R., Young, D.F. and Okiishi, T.H. 1994, Fundamentals of Fluid Mechanics 2nd Ed., Wiley, New York. QDPI, Soil Conservation Handbook, Queensland Department of Primary Industries, Soil Conservation Service Branch, Brisbane. QWRC, Farm Water Supplies Design Manual Vol 1 Farm Storages, Queensland Water Resources Commisson. Ree, W.O. 1949, Hydraulic characteristics of vegetation for vegetated waterways. Agricultural Engineering, vol. 30: pp. 184-185. Streeter, V.L. & Wylie, E.B. 1983, Fluid Mechanics – First SI Edition, McGrawHill, Singapore

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Spiegal, M.R. 1968, Mathematical Handbook of Formulas and Tables, Schaum’s Outline Series, McGraw-Hill, New York. Stephenson, D. and Meadows, M.E. 1986, Kinematic Hydrology and Modelling. Elsevier, Amsterdam.

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Appendix A

171

Appendix A - Some Useful equations F 

M  QV t

A 1/ 2 2 / 3 Sf R n

Q

V

dM M M dx   dt t x dt

q

fLV 2 hf  2 gD

 0.849C HW S 0f .54 R 0.63 dy So  S f  dx 1  FR2

Fs 

1 1/ 2 5 / 3 Sf y n

y2 

y1 2

 1 8F 1 2 R1

o  s   gd

 o  0.056  s   gd

s  cs

b  cb

s = 0.75 gySo

b = gySo d = 11 RSo  sin 2   K  1   2  sin  

Q  

1/ 2

 rQ n  rnQ n1

Q/ND3 = const

h  

Ns = NQ½/H¾

H/N2D2 = const NPSHA 

n Q Q / h

Pa  Pv  hs  hls g

hi 

L dV g dt

Hs = Hv - hi + hf V2 Hv  2g

 Ap  C A  v v

   

2

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Figure A. 1 – Moody Diagram

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Appendix B - List of Symbols

173

Appendix B – List of Symbols Included here is a list of symbols used in this book. For a more comprehensive list see List of Principle Symbols in Chadwick et al.



(alpha) Coriolis coefficient for kinetic energy

A

area (m2)



(beta) Boussinesq coefficient for momentum

AP

pipe cross section area (m2)



(delta) difference

AV

open area of the valve (m2)



(epsilon) old term for pipe roughness height (m)

a

acceleration (m/s2)



(theta) angle OR repose angle of an erodible sediment

B

pump impeller width (m)



(kappa) bulk modulus of elasticity (N/m2)

b

channel bet width (m)



(lambda) wavelength of pipe corrugations (m)

C

Constant



(mu) dynamic viscosity (Pa.s)

Cd

Discharge coefficient



(nu) kinematic viscosity (m2/s)



(pi) Pi group for dimensional analysis



(pi) ≈3.14159265

c

wave celerity (m/s)



(rho) density (kg/m3)

D

diameter (m)

s

density of solid particles or sediment (kg/m)

d

particle diameter of sediment(m)

C HW

Hazen-Williams coefficient

CV

valve discharge coefficient



(sigma) surface tension (N/m)

E

specific energy (m) OR Young’s modulus of pipe material



(tau)shear stress (N/m2)

e

pipe wall thickness (m)



volume (m3)

F

force (N)



(phi) channel side slope angle

FR

Froude number



(omega) angular velocity (rad/s)

Fs

entrainment function

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f

Darcy-Weisbach friction coefficient

p

pressure (N/m2)

g

gravity (9.81 m/s2)

Q

discharge (m3/s)

H

Head (m)

q

unit discharge per metre width (m2/s)

hi

inertial head (m)

qi

lateral inflow (m2/s)

hp

pump head (m)

R

hydraulic radius (m)

Hs

static head (m)

Re

Reynolds number

Hv

total head at valve (m)

r

radius (m) OR lateral inflow (m2/s)

h

height (m)

S

volume (m3)

hf

friction loss head (m)

Sf

friction slope (m/m)

k

pipe roughness height (m)

S0

slope (m/m)

L

length (m)

t, T

time (s)

K

tractive force ratio OR elastic modulus of a pipe

u

velocity (m/s)

M

momentum (kg.m/s)

V

velocity (m/s)

m

mass (kg)

Vw

wave velocity (m/s)

N

pump speed (rad/s or rpm)

We

Webber number

n

Manning’s roughness

x

distance (m)

P

power (W) OR channel wetted perimeter (m)

y

water depth (m)

Pa

absolute pressure (Pa)

yc

critical depth (m)

Po

output power (W)

yn

normal depth (m)

Pv

vapour pressure (Pa)

z

elevation (m)

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