Water-resources engineering [3rd ed] 9780132833219, 0273785915, 9780273785910, 0132833212

This in-depth review of water-resources engineering essentials focuses on both fundamentals and design applications.KEY

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Water-resources engineering [3rd ed]
 9780132833219, 0273785915, 9780273785910, 0132833212

Table of contents :
Cover......Page 1
Contents......Page 6
Preface......Page 16
1.2 The Hydrologic Cycle......Page 20
1.3.1 Water-Control Systems......Page 24
1.3.2 Water-Use Systems......Page 25
1.3.3 Supporting Federal Agencies in the United States......Page 26
Problem......Page 27
2.2.1 Steady-State Continuity Equation......Page 28
2.2.2 Steady-State Momentum Equation......Page 29
2.2.3 Steady-State Energy Equation......Page 41
2.2.3.1 Energy and hydraulic grade lines......Page 44
2.2.3.3 Head losses in transitions and fittings......Page 46
2.2.3.4 Head losses in noncircular conduits......Page 50
2.2.3.5 Empirical friction-loss formulae......Page 51
2.2.4 Water Hammer......Page 54
2.3 Pipe Networks......Page 58
2.3.1 Nodal Method......Page 59
2.3.2 Loop Method......Page 61
2.4 Pumps......Page 65
2.4.1 AffinityLaws......Page 70
2.4.2.1 Commercially available pumps......Page 72
2.4.2.2 System characteristics......Page 73
2.4.2.3 Limits on pump location......Page 74
2.4.3 Multiple-Pump Systems......Page 77
2.4.4 Variable-Speed Pumps......Page 79
Problems......Page 81
3.2 Water Demand......Page 89
3.2.1.1 Estimation of per-capita demand......Page 90
3.2.1.2 Estimation of population......Page 91
3.2.2 Temporal Variations in Water Demand......Page 95
3.2.3 Fire Demand......Page 96
3.2.4 Design Flows......Page 98
3.3.1 Pipelines......Page 100
3.3.1.1 Minimumsize......Page 101
3.3.1.3 Pipe materials......Page 102
3.3.4 Meters......Page 104
3.3.5 Fire Hydrants......Page 105
3.3.6 Water-Storage Reservoirs......Page 106
3.4 Performance Criteria for Water-Distribution Systems......Page 109
3.4.3 Water Quality......Page 110
3.4.4 Network Analysis......Page 111
3.5 Building Water-Supply Systems......Page 112
3.5.2 Specification of Minimum Pressures......Page 113
3.5.3 Determination of Pipe Diameters......Page 115
Problems......Page 120
4.2.1 Steady-State Continuity Equation......Page 122
4.2.2 Steady-State Momentum Equation......Page 123
4.2.2.1 Darcy–Weisbach equation......Page 125
4.2.2.2 Manning equation......Page 129
4.2.2.3 Other equations......Page 138
4.2.2.4 Velocity distribution......Page 139
4.2.3 Steady-State Energy Equation......Page 140
4.2.3.2 Specific energy......Page 144
4.3.1 Profile Equation......Page 151
4.3.2 Classification of Water-Surface Profiles......Page 153
4.3.3 Hydraulic Jump......Page 158
4.3.4 Computation of Water-Surface Profiles......Page 162
4.3.4.1 Direct-integration method......Page 164
4.3.4.2 Direct-step method......Page 166
4.3.4.3 Standard-step method......Page 167
4.3.4.4 Practical considerations......Page 169
4.3.4.5 Profiles across bridges......Page 173
Problems......Page 178
5.1 Introduction......Page 185
5.2.1 Best Hydraulic Section......Page 186
5.2.2 Boundary Shear Stress......Page 189
5.2.3 Cohesive versus Noncohesive Materials......Page 191
5.2.4 Bends......Page 196
5.2.6 Freeboard......Page 197
5.3 Design of Channels with Rigid Linings......Page 199
5.4 Design of Channels with Flexible Linings......Page 201
5.4.1 General Design Procedure......Page 202
5.4.2 Vegetative Linings and Bare Soil......Page 206
5.4.3 RECP Linings......Page 216
5.4.4 Riprap, Cobble, and Gravel Linings......Page 218
5.4.5 Gabions......Page 222
5.5 CompositeLinings......Page 224
Problems......Page 227
6.2.1 Residential Sources......Page 230
6.2.2 Nonresidential Sources......Page 231
6.2.3 Inflow and Infiltration (I/I)......Page 232
6.2.4 Peaking Factors......Page 233
6.3 Hydraulics of Sewers......Page 235
6.3.1 Manning Equation with Constant n......Page 237
6.3.2 Manning Equation with Variable n......Page 239
6.3.3 Self-Cleansing......Page 242
6.3.5 Design Computations for Diameter and Slope......Page 243
6.3.6 Hydraulics of Manholes......Page 246
6.4.2 Pipe Material......Page 248
6.4.6 Manholes......Page 250
6.4.8 Force Mains......Page 252
6.4.9 Hydrogen-Sulfide Control......Page 253
6.5 Design Computations......Page 255
6.5.1.2 Minimum slope for self-cleansing......Page 256
6.5.2 Procedure for System Design......Page 259
Problems......Page 266
7.2.1 Hydraulics......Page 269
7.2.1.1 Submerged entrances......Page 271
7.2.1.2 Unsubmerged entrances......Page 278
7.2.2 Design Constraints......Page 281
7.2.3 Sizing Calculations......Page 283
7.2.3.1 Fixed-headwater method......Page 284
7.2.3.2 Fixed-flow method......Page 288
7.2.4 Roadway Overtopping......Page 290
7.2.5 Riprap/Outlet Protection......Page 293
7.3 Gates......Page 294
7.3.1 Free Discharge......Page 295
7.3.2 Submerged Discharge......Page 298
7.3.3 Empirical Equations......Page 300
7.4.1.1 Rectangular weirs......Page 301
7.4.1.2 V-notchweirs......Page 307
7.4.1.3 Compound weirs......Page 310
7.4.1.4 Other types of sharp-crested weirs......Page 312
7.4.2.1 Rectangular weirs......Page 313
7.4.2.2 Compound weirs......Page 316
7.4.2.3 Gabionweirs......Page 317
7.5.1 Uncontrolled Spillways......Page 318
7.5.2 Controlled (Gated) Spillways......Page 326
7.5.2.1 Gates seated on the spillway crest......Page 327
7.5.2.2 Gates seated downstream of the spillway crest......Page 328
7.6.1 Type Selection......Page 331
7.6.2 Design Procedure......Page 333
7.7 Dams and Reservoirs......Page 337
7.7.1 Types of Dams......Page 338
7.7.2 Reservoir Storage......Page 341
7.7.2.1 Sediment accumulation......Page 342
7.7.2.2 Determination of storage requirements......Page 345
7.7.3.1 Turbines......Page 347
7.7.3.2 Turbine performance......Page 352
7.7.3.3 Feasibility of hydropower......Page 353
Problems......Page 354
8.1 Introduction......Page 363
8.2.1 Discrete Probability Distributions......Page 364
8.2.2 Continuous Probability Distributions......Page 365
8.2.3 Mathematical Expectation and Moments......Page 366
8.2.4 Return Period......Page 369
8.2.5.1 Binomial distribution......Page 370
8.2.5.2 Geometric distribution......Page 372
8.2.5.3 Poisson distribution......Page 373
8.2.5.4 Exponential distribution......Page 375
8.2.5.5 Gamma/Pearson Type III distribution......Page 376
8.2.5.6 Normal distribution......Page 379
8.2.5.7 Log-normal distribution......Page 381
8.2.5.8 Uniform distribution......Page 382
8.2.5.9 Extreme-value distributions......Page 383
8.2.5.10 Chi-square distribution......Page 390
8.3.1.1 Probability distribution of observed data......Page 391
8.3.1.2 Hypothesis tests......Page 395
8.3.2.1 Method of moments......Page 398
8.3.2.2 Maximum-likelihood method......Page 401
8.3.2.3 Method of L-moments......Page 402
8.3.3 Frequency Analysis......Page 406
8.3.3.1 Normal distribution......Page 407
8.3.3.2 Log-normal distribution......Page 408
8.3.3.3 Gamma/Pearson Type III distribution......Page 409
8.3.3.4 Log-Pearson Type III distribution......Page 410
8.3.3.5 Extreme-value Type I distribution......Page 412
8.3.3.6 General extreme-value (GEV) distribution......Page 413
8.4 Uncertainty Analysis......Page 414
Problems......Page 416
9.2 Rainfall......Page 420
9.2.1 Measurement of Rainfall......Page 422
9.2.2 Statistics of Rainfall Data......Page 424
9.2.2.2 Secondary estimation of IDF curves......Page 429
9.2.3 Spatial Averaging and Interpolation of Rainfall......Page 435
9.2.4.1 Return period......Page 440
9.2.4.4 Temporal distribution......Page 441
9.2.4.5 Spatial distribution......Page 447
9.2.5 Extreme Rainfall......Page 448
9.2.5.2 Statistical estimation method......Page 449
9.2.5.4 Probable maximum storm......Page 451
9.3.1 Interception......Page 452
9.3.3 Infiltration......Page 456
9.3.3.1 The infiltration process......Page 458
9.3.3.2 Horton model......Page 461
9.3.3.3 Green–Ampt model......Page 466
9.3.3.4 NRCS curve-number model......Page 472
9.3.3.5 Comparison of infiltration models......Page 479
9.3.4 Rainfall Excess on Composite Areas......Page 480
9.4 Baseflow......Page 483
Problems......Page 487
10.2 Mechanisms of Surface Runoff......Page 492
10.3.1.1 Kinematic-wave equation......Page 493
10.3.1.2 NRCS method......Page 497
10.3.1.4 Izzard equation......Page 500
10.3.1.5 Kerby equation......Page 501
10.3.2 Channel Flow......Page 503
10.3.3 Accuracy of Estimates......Page 505
10.4.1 The Rational Method......Page 506
10.4.2 NRCS-TR55 Method......Page 511
10.5.1 Unit-Hydrograph Theory......Page 514
10.5.2 Instantaneous Unit Hydrograph......Page 520
10.5.3 Unit-Hydrograph Models......Page 521
10.5.3.1 Snyder unit-hydrograph model......Page 522
10.5.3.2 NRCS dimensionless unit hydrograph......Page 525
10.5.4 Time-Area Models......Page 528
10.5.5 Kinematic-Wave Model......Page 533
10.5.6 Nonlinear-Reservoir Model......Page 534
10.5.7 Santa Barbara Urban Hydrograph Model......Page 536
10.5.8 Extreme Runoff Events......Page 538
10.6.1.1 Modified Puls method......Page 539
10.6.1.2 Muskingum method......Page 543
10.6.2 Hydraulic Routing......Page 550
10.7.1 Event-Mean Concentrations......Page 552
10.7.2.1 USGS model......Page 554
10.7.2.2 EPA model......Page 556
Problems......Page 558
11.2 Street Gutters......Page 564
11.3 Inlets......Page 568
11.3.1 CurbInlets......Page 569
11.3.2 Grate Inlets......Page 573
11.3.3 Combination Inlets......Page 579
11.3.4 Slotted Inlets......Page 584
11.4 Roadside and Median Channels......Page 585
11.5 Storm Sewers......Page 586
11.5.1 Calculation of Design Flow Rates......Page 587
11.5.2 Pipe Sizing and Selection......Page 590
11.5.3 Manholes......Page 595
11.5.4 Determination of Impervious Area......Page 596
11.5.5 System-Design Computations......Page 597
11.5.6 Other Design Considerations......Page 602
Problems......Page 603
12.2.2 Quality Control......Page 605
12.3.1 Storage Impoundments......Page 606
12.3.1.1 Detentionbasins—Designparameters......Page 607
12.3.1.2 Wet detention basins......Page 609
12.3.1.3 Dry detention basins......Page 611
12.3.1.4 Design of outlet structures......Page 612
12.3.1.5 Design for flood control......Page 618
12.3.2 Infiltration Basins......Page 622
12.3.3 Swales......Page 624
12.3.3.1 Retention swales......Page 625
12.3.3.2 Biofiltration swales......Page 626
12.3.5 Bioretention Systems......Page 629
12.3.6 Exfiltration Trenches......Page 631
12.3.6.1 General design guidelines......Page 632
12.3.6.2 Design for flood control......Page 633
12.3.6.3 Design for water-quality control......Page 635
12.3.7 Subsurface Exfiltration Galleries......Page 636
12.4.2 Structural SCMs......Page 637
Problems......Page 638
13.2 Penman–Monteith Equation......Page 643
13.2.1 Aerodynamic Resistance......Page 644
13.2.2 Surface Resistance......Page 645
13.2.3.1 Shortwave radiation......Page 646
13.2.3.2 Longwave radiation......Page 648
13.2.4 Soil Heat Flux......Page 649
13.2.6 Psychrometric Constant......Page 650
13.2.9 Actual Vapor Pressure......Page 651
13.2.10 Air Density......Page 652
13.3 Application of the PM Equation......Page 653
13.4 Potential Evapotranspiration......Page 656
13.5 Reference Evapotranspiration......Page 657
13.5.1 FAO56-Penman–Monteith Method......Page 658
13.5.2 ASCE Penman–Monteith Method......Page 662
13.5.3 Evaporation Pans......Page 663
13.5.4 Empirical Methods......Page 667
13.6.1 Index-of-Dryness Method......Page 670
13.6.3 Remote Sensing......Page 672
Problems......Page 673
14.1 Introduction......Page 675
14.2 Darcy’s Law......Page 681
14.2.1.1 Empirical formulae......Page 685
14.2.1.3 Anisotropic properties......Page 689
14.2.1.4 Stochastic properties......Page 693
14.3 General Flow Equation......Page 695
14.4.1 Unconfined Aquifers......Page 700
14.4.2 Confined Aquifers......Page 706
14.5 Flow in the Unsaturated Zone......Page 710
Problems......Page 715
15.2.1 Unconfined Flow Between Two Reservoirs......Page 719
15.2.2 Well in a Confined Aquifer......Page 721
15.2.3 Well in an Unconfined Aquifer......Page 725
15.2.4 Well in a Leaky Confined Aquifer......Page 728
15.2.5 Well in an Unconfined Aquifer with Recharge......Page 732
15.2.6 Partially Penetrating Wells......Page 733
15.3.1 Well in a Confined Aquifer......Page 737
15.3.2 Well in an Unconfined Aquifer......Page 747
15.3.3 Well in a Leaky Confined Aquifer......Page 755
15.4 Principle of Superposition......Page 760
15.4.1 Multiple Wells......Page 761
15.4.2 Well in Uniform Flow......Page 763
15.5.1 Constant-Head Boundary......Page 765
15.5.2 Impermeable Boundary......Page 769
15.6 Saltwater Intrusion......Page 771
Problems......Page 780
16.2 Design of Wellfields......Page 790
16.3.1 Delineation of Wellhead Protection Areas......Page 793
16.3.2 Time-of-Travel Approach......Page 794
16.4.1 Types of Wells......Page 796
16.4.2 Design of Well Components......Page 797
16.4.2.2 Screenintake......Page 798
16.4.2.3 Gravel pack......Page 802
16.4.2.4 Pump......Page 803
16.4.2.5 Other considerations......Page 804
16.4.3 Performance Assessment......Page 807
16.4.4 Well Drilling......Page 812
16.5.1 Pumping Well......Page 813
16.5.2 Observation Wells......Page 814
16.5.3 Field Procedures......Page 815
16.6 Design of Slug Tests......Page 817
16.7 Design of Exfiltration Trenches......Page 822
16.8 Seepage Meters......Page 827
Problems......Page 828
17.2 Planning Process......Page 834
17.3 Economic Feasibility......Page 837
17.3.1.1 Single-payment factors......Page 838
17.3.1.3 Arithmetic-gradient factors......Page 839
17.3.1.4 Geometric-gradient factors......Page 840
17.3.2.1 Present-worth analysis......Page 842
17.3.2.3 Rate-of-return analysis......Page 844
17.3.2.4 Benefit–cost analysis......Page 847
Problems......Page 848
A.1 Units......Page 850
A.2 Conversion Factors......Page 851
B.2 Organic Compounds Found in Water......Page 853
B.3 Air at Standard Atmospheric Pressure......Page 855
C.1 Areas Under Standard Normal Curve......Page 856
C.2 Frequency Factors for Pearson Type III Distribution......Page 858
C.3 Critical Values of the Chi-Square Distribution......Page 860
C.4 Critical Values for the Kolmogorov–Smirnov Test Statistic......Page 861
D.1 Error Function......Page 862
D.2.2.1 Bessel function of the first kind of order n......Page 863
D.2.2.5 Tabulated values of useful Bessel functions......Page 864
D.3 Gamma Function......Page 867
D.4 Exponential Integral......Page 868
E.2 Ductile-Iron Pipe......Page 869
E.4 Physical Properties of Common Pipe Materials......Page 870
F.1 Definition of Soil Groups......Page 871
F.2 Terminology......Page 872
Bibliography......Page 873
A......Page 931
C......Page 932
D......Page 935
E......Page 937
F......Page 938
G......Page 939
H......Page 941
I......Page 942
L......Page 943
N......Page 944
O......Page 945
P......Page 946
R......Page 948
S......Page 950
T......Page 954
U......Page 955
W......Page 956
Z......Page 958

Citation preview

WATER-RESOURCES ENGINEERING

Third Edition

David A. Chin International Edition contributions by Asis Mazumdar Pankaj Kumar Roy

Vice President and Editorial Director, ECS: Marcia J. Horton Executive Editor: Holly Stark Editorial Assistant: Carlin Heinle Executive Marketing Manager: Tim Galligan Marketing Assistant: Jon Bryant Permissions Project Manager: Karen Sanatar Senior Managing Editor: Scott Disanno Publisher, International Edition: Angshuman Chakraborty Acquisitions Editor, International Edition: Sandhya Ghoshal Publishing Administrator and Business Analyst, International Edition: Shokhi Shah Khandelwal

Senior Print and Media Editor, International Edition: Ashwitha Jayakumar Project Editor, International Edition: Daniel Luiz Publishing Administrator, International Editions: Hema Mehta Senior Manufacturing Controller, Production, International Editions: Trudy Kimber Production Project Manager / Editorial Production Manager: Greg Dulles Cover Designer: Jodi Notowitz Cover Photo: United States Bureau of Reclamation Full-Service Project Management: Abinaya Rajendran, Integra Software Services, Pvt. Ltd.

Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoninternationaleditions.com © Pearson Education Limited 2013

The right of Davis A. Chin to be identified as author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act 1988. Authorized adaptation from the United States edition, entitled Water-Resources Engineering, 3rd edition, ISBN 978-0-13-283321-9 by David A. Chin published by Pearson Education © 2013. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. Microsoft and/or its respective suppliers make no representations about the suitability of the information contained in the documents and related graphics published as part of the services for any purpose. All such documents and related graphics are provided “as is” without warranty of any kind. Microsoft and/or its respective suppliers hereby disclaim all warranties and conditions with regard to this information, including all warranties and conditions of merchantability, whether express, implied or statutory, fitness for a particular purpose, title and non-infringement. In no event shall Microsoft and/or its respective suppliers be liable for any special, indirect or consequential damages or any damages whatsoever resulting from loss of use, data or profits, whether in an action of contract, negligence or other tortious action, arising out of or in connection with the use or performance of information available from the services. The documents and related graphics contained herein could include technical inaccuracies or typographical errors. Changes are periodically added to the information herein. Microsoft and/or its respective suppliers may make improvements and/or changes in the product(s) and/or the program(s) described herein at any time. Partial screen shots may be viewed in full within the software version specified. R and Windows R are registered trademarks of the Microsoft Corporation in the U.S.A. and other countries. This book is Microsoft not sponsored or endorsed by or affiliated with the Microsoft Corporation.

ISBN 10: 0-273-78591-5 ISBN 13: 978-0-273-78591-0 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library 10 9 8 7 6 5 4 3 2 1 14 13 12 11 10 Typeset in 10/11 Times by Integra Software Services Private Limited, Pondicherry. Printed and bound by Courier Kendalville in The United States of America The publisher’s policy is to use paper manufactured from sustainable forests.

To Andrew and Stephanie. “But those who hope in the Lord will renew their strength. They will soar on wings like eagles; they will run and not grow weary, they will walk and not be faint.” Isaiah 40:31

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Contents Preface

15

1 Introduction 1.1 Water-Resources Engineering . . . . . . . . . . . . . . . . 1.2 The Hydrologic Cycle . . . . . . . . . . . . . . . . . . . . . 1.3 Design of Water-Resource Systems . . . . . . . . . . . . . 1.3.1 Water-Control Systems . . . . . . . . . . . . . . . . 1.3.2 Water-Use Systems . . . . . . . . . . . . . . . . . . 1.3.3 Supporting Federal Agencies in the United States Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2 Fundamentals of Flow in Closed Conduits 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Single Pipelines . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Steady-State Continuity Equation . . . . . . . . 2.2.2 Steady-State Momentum Equation . . . . . . . 2.2.3 Steady-State Energy Equation . . . . . . . . . . 2.2.3.1 Energy and hydraulic grade lines . . . 2.2.3.2 Velocity profile . . . . . . . . . . . . . 2.2.3.3 Head losses in transitions and fittings 2.2.3.4 Head losses in noncircular conduits . 2.2.3.5 Empirical friction-loss formulae . . . 2.2.4 Water Hammer . . . . . . . . . . . . . . . . . . 2.3 Pipe Networks . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Nodal Method . . . . . . . . . . . . . . . . . . . 2.3.2 Loop Method . . . . . . . . . . . . . . . . . . . 2.3.3 Application of Computer Programs . . . . . . . 2.4 Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Affinity Laws . . . . . . . . . . . . . . . . . . . 2.4.2 Pump Selection . . . . . . . . . . . . . . . . . . 2.4.2.1 Commercially available pumps . . . . 2.4.2.2 System characteristics . . . . . . . . . 2.4.2.3 Limits on pump location . . . . . . . . 2.4.3 Multiple-Pump Systems . . . . . . . . . . . . . 2.4.4 Variable-Speed Pumps . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3 Design of Water-Distribution Systems 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . 3.2 Water Demand . . . . . . . . . . . . . . . . . . . . 3.2.1 Per-Capita Forecast Model . . . . . . . . . 3.2.1.1 Estimation of per-capita demand 3.2.1.2 Estimation of population . . . . 3.2.2 Temporal Variations in Water Demand . . 3.2.3 Fire Demand . . . . . . . . . . . . . . . . 3.2.4 Design Flows . . . . . . . . . . . . . . . . 3.3 Components of Water-Distribution Systems . . . 3.3.1 Pipelines . . . . . . . . . . . . . . . . . . . 3.3.1.1 Minimum size . . . . . . . . . . . 3.3.1.2 Service lines . . . . . . . . . . . 3.3.1.3 Pipe materials . . . . . . . . . .

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Contents

3.3.2 Pumps . . . . . . . . . . . . . . . . . . . . . . 3.3.3 Valves . . . . . . . . . . . . . . . . . . . . . . 3.3.4 Meters . . . . . . . . . . . . . . . . . . . . . . 3.3.5 Fire Hydrants . . . . . . . . . . . . . . . . . . 3.3.6 Water-Storage Reservoirs . . . . . . . . . . . 3.4 Performance Criteria for Water-Distribution Systems 3.4.1 Service Pressures . . . . . . . . . . . . . . . . 3.4.2 Allowable Velocities . . . . . . . . . . . . . . 3.4.3 Water Quality . . . . . . . . . . . . . . . . . . 3.4.4 Network Analysis . . . . . . . . . . . . . . . . 3.5 Building Water-Supply Systems . . . . . . . . . . . . 3.5.1 Specification of Design Flows . . . . . . . . . 3.5.2 Specification of Minimum Pressures . . . . . 3.5.3 Determination of Pipe Diameters . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Fundamentals of Flow in Open Channels 4.1 Introduction . . . . . . . . . . . . . . . . . . . 4.2 Basic Principles . . . . . . . . . . . . . . . . . 4.2.1 Steady-State Continuity Equation . . . 4.2.2 Steady-State Momentum Equation . . 4.2.2.1 Darcy–Weisbach equation . . 4.2.2.2 Manning equation . . . . . . 4.2.2.3 Other equations . . . . . . . 4.2.2.4 Velocity distribution . . . . . 4.2.3 Steady-State Energy Equation . . . . . 4.2.3.1 Energy grade line . . . . . . 4.2.3.2 Specific energy . . . . . . . . 4.3 Water-Surface Profiles . . . . . . . . . . . . . 4.3.1 Profile Equation . . . . . . . . . . . . . 4.3.2 Classification of Water-Surface Profiles 4.3.3 Hydraulic Jump . . . . . . . . . . . . . 4.3.4 Computation of Water-Surface Profiles 4.3.4.1 Direct-integration method . 4.3.4.2 Direct-step method . . . . . 4.3.4.3 Standard-step method . . . . 4.3.4.4 Practical considerations . . . 4.3.4.5 Profiles across bridges . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . .

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103 103 103 104 105 108 109 109 109 110 111 112 112 114 119

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121 121 121 121 122 124 128 137 138 139 143 143 150 150 152 157 161 163 165 166 168 172 177

5 Design of Drainage Channels 5.1 Introduction . . . . . . . . . . . . . . . . . . . . 5.2 Basic Principles . . . . . . . . . . . . . . . . . . 5.2.1 Best Hydraulic Section . . . . . . . . . . 5.2.2 Boundary Shear Stress . . . . . . . . . . 5.2.3 Cohesive versus Noncohesive Materials 5.2.4 Bends . . . . . . . . . . . . . . . . . . . . 5.2.5 Channel Slopes . . . . . . . . . . . . . . 5.2.6 Freeboard . . . . . . . . . . . . . . . . . 5.3 Design of Channels with Rigid Linings . . . . . 5.4 Design of Channels with Flexible Linings . . . . 5.4.1 General Design Procedure . . . . . . . . 5.4.2 Vegetative Linings and Bare Soil . . . . 5.4.3 RECP Linings . . . . . . . . . . . . . . . 5.4.4 Riprap, Cobble, and Gravel Linings . . . 5.4.5 Gabions . . . . . . . . . . . . . . . . . .

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184 184 185 185 188 190 195 196 196 198 200 201 205 215 217 221

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5.5 Composite Linings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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6 Design of Sanitary Sewers 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Quantity of Wastewater . . . . . . . . . . . . . . . . . . 6.2.1 Residential Sources . . . . . . . . . . . . . . . . 6.2.2 Nonresidential Sources . . . . . . . . . . . . . . 6.2.3 Inflow and Infiltration (I/I) . . . . . . . . . . . . 6.2.4 Peaking Factors . . . . . . . . . . . . . . . . . . 6.3 Hydraulics of Sewers . . . . . . . . . . . . . . . . . . . 6.3.1 Manning Equation with Constant n . . . . . . . 6.3.2 Manning Equation with Variable n . . . . . . . 6.3.3 Self-Cleansing . . . . . . . . . . . . . . . . . . . 6.3.4 Scour Prevention . . . . . . . . . . . . . . . . . 6.3.5 Design Computations for Diameter and Slope . 6.3.6 Hydraulics of Manholes . . . . . . . . . . . . . 6.4 System Design Criteria . . . . . . . . . . . . . . . . . . 6.4.1 System Layout . . . . . . . . . . . . . . . . . . . 6.4.2 Pipe Material . . . . . . . . . . . . . . . . . . . 6.4.3 Depth of Sanitary Sewer . . . . . . . . . . . . . 6.4.4 Diameter and Slope of Pipes . . . . . . . . . . . 6.4.5 Hydraulic Criteria . . . . . . . . . . . . . . . . . 6.4.6 Manholes . . . . . . . . . . . . . . . . . . . . . 6.4.7 Pump Stations . . . . . . . . . . . . . . . . . . . 6.4.8 Force Mains . . . . . . . . . . . . . . . . . . . . 6.4.9 Hydrogen-Sulfide Control . . . . . . . . . . . . 6.4.10 Combined Sewers . . . . . . . . . . . . . . . . . 6.5 Design Computations . . . . . . . . . . . . . . . . . . . 6.5.1 Design Aids . . . . . . . . . . . . . . . . . . . . 6.5.1.1 Manning’s n . . . . . . . . . . . . . . . 6.5.1.2 Minimum slope for self-cleansing . . 6.5.2 Procedure for System Design . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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229 229 229 229 230 231 232 234 236 238 241 242 242 245 247 247 247 249 249 249 249 251 251 252 254 254 255 255 255 258 265

7 Design of Hydraulic Structures 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . 7.2 Culverts . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Hydraulics . . . . . . . . . . . . . . . . . . 7.2.1.1 Submerged entrances . . . . . . 7.2.1.2 Unsubmerged entrances . . . . . 7.2.2 Design Constraints . . . . . . . . . . . . . 7.2.3 Sizing Calculations . . . . . . . . . . . . . 7.2.3.1 Fixed-headwater method . . . . 7.2.3.2 Fixed-flow method . . . . . . . . 7.2.3.3 Minimum-performance method 7.2.4 Roadway Overtopping . . . . . . . . . . . 7.2.5 Riprap/Outlet Protection . . . . . . . . . . 7.3 Gates . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 Free Discharge . . . . . . . . . . . . . . . 7.3.2 Submerged Discharge . . . . . . . . . . . 7.3.3 Empirical Equations . . . . . . . . . . . . 7.4 Weirs . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Sharp-Crested Weirs . . . . . . . . . . . . 7.4.1.1 Rectangular weirs . . . . . . . . 7.4.1.2 V-notch weirs . . . . . . . . . . .

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268 268 268 268 270 277 280 282 283 287 289 289 292 293 294 297 299 300 300 300 306

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7.4.1.3 Compound weirs . . . . . . . . . . . . . . . . . 7.4.1.4 Other types of sharp-crested weirs . . . . . . . 7.4.2 Broad-Crested Weirs . . . . . . . . . . . . . . . . . . . . 7.4.2.1 Rectangular weirs . . . . . . . . . . . . . . . . 7.4.2.2 Compound weirs . . . . . . . . . . . . . . . . . 7.4.2.3 Gabion weirs . . . . . . . . . . . . . . . . . . . 7.5 Spillways . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 Uncontrolled Spillways . . . . . . . . . . . . . . . . . . . 7.5.2 Controlled (Gated) Spillways . . . . . . . . . . . . . . . 7.5.2.1 Gates seated on the spillway crest . . . . . . . 7.5.2.2 Gates seated downstream of the spillway crest 7.6 Stilling Basins . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Type Selection . . . . . . . . . . . . . . . . . . . . . . . . 7.6.2 Design Procedure . . . . . . . . . . . . . . . . . . . . . . 7.7 Dams and Reservoirs . . . . . . . . . . . . . . . . . . . . . . . . 7.7.1 Types of Dams . . . . . . . . . . . . . . . . . . . . . . . . 7.7.2 Reservoir Storage . . . . . . . . . . . . . . . . . . . . . . 7.7.2.1 Sediment accumulation . . . . . . . . . . . . . 7.7.2.2 Determination of storage requirements . . . . 7.7.3 Hydropower . . . . . . . . . . . . . . . . . . . . . . . . . 7.7.3.1 Turbines . . . . . . . . . . . . . . . . . . . . . . 7.7.3.2 Turbine performance . . . . . . . . . . . . . . 7.7.3.3 Feasibility of hydropower . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Probability and Statistics in Water-Resources Engineering 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Probability Distributions . . . . . . . . . . . . . . . . . . . 8.2.1 Discrete Probability Distributions . . . . . . . . . . 8.2.2 Continuous Probability Distributions . . . . . . . . 8.2.3 Mathematical Expectation and Moments . . . . . 8.2.4 Return Period . . . . . . . . . . . . . . . . . . . . . 8.2.5 Common Probability Functions . . . . . . . . . . . 8.2.5.1 Binomial distribution . . . . . . . . . . . 8.2.5.2 Geometric distribution . . . . . . . . . . 8.2.5.3 Poisson distribution . . . . . . . . . . . . 8.2.5.4 Exponential distribution . . . . . . . . . . 8.2.5.5 Gamma/Pearson Type III distribution . . 8.2.5.6 Normal distribution . . . . . . . . . . . . 8.2.5.7 Log-normal distribution . . . . . . . . . . 8.2.5.8 Uniform distribution . . . . . . . . . . . . 8.2.5.9 Extreme-value distributions . . . . . . . . 8.2.5.10 Chi-square distribution . . . . . . . . . . 8.3 Analysis of Hydrologic Data . . . . . . . . . . . . . . . . . 8.3.1 Estimation of Population Distribution . . . . . . . 8.3.1.1 Probability distribution of observed data 8.3.1.2 Hypothesis tests . . . . . . . . . . . . . . 8.3.1.3 Model selection criteria . . . . . . . . . . 8.3.2 Estimation of Population Parameters . . . . . . . . 8.3.2.1 Method of moments . . . . . . . . . . . . 8.3.2.2 Maximum-likelihood method . . . . . . . 8.3.2.3 Method of L-moments . . . . . . . . . . . 8.3.3 Frequency Analysis . . . . . . . . . . . . . . . . . . 8.3.3.1 Normal distribution . . . . . . . . . . . . 8.3.3.2 Log-normal distribution . . . . . . . . . . 8.3.3.3 Gamma/Pearson Type III distribution . .

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309 311 312 312 315 316 317 317 325 326 327 330 330 332 336 337 340 341 344 346 346 351 352 353

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362 362 363 363 364 365 368 369 369 371 372 374 375 378 380 381 382 389 390 390 390 394 397 397 397 400 401 405 406 407 408

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8.3.3.4 Log-Pearson Type III distribution . . . . . 8.3.3.5 Extreme-value Type I distribution . . . . . 8.3.3.6 General extreme-value (GEV) distribution 8.4 Uncertainty Analysis . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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9 Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Rainfall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Measurement of Rainfall . . . . . . . . . . . . . . . . . . . . 9.2.2 Statistics of Rainfall Data . . . . . . . . . . . . . . . . . . . 9.2.2.1 Rainfall statistics in the United States . . . . . . . 9.2.2.2 Secondary estimation of IDF curves . . . . . . . . 9.2.3 Spatial Averaging and Interpolation of Rainfall . . . . . . . 9.2.4 Design Rainfall . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.4.1 Return period . . . . . . . . . . . . . . . . . . . . 9.2.4.2 Rainfall duration . . . . . . . . . . . . . . . . . . . 9.2.4.3 Rainfall depth . . . . . . . . . . . . . . . . . . . . 9.2.4.4 Temporal distribution . . . . . . . . . . . . . . . . 9.2.4.5 Spatial distribution . . . . . . . . . . . . . . . . . . 9.2.5 Extreme Rainfall . . . . . . . . . . . . . . . . . . . . . . . . 9.2.5.1 Rational estimation method . . . . . . . . . . . . 9.2.5.2 Statistical estimation method . . . . . . . . . . . . 9.2.5.3 World-record precipitation amounts . . . . . . . . 9.2.5.4 Probable maximum storm . . . . . . . . . . . . . . 9.3 Rainfall Abstractions . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 Interception . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 Depression Storage . . . . . . . . . . . . . . . . . . . . . . . 9.3.3 Infiltration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.3.1 The infiltration process . . . . . . . . . . . . . . . 9.3.3.2 Horton model . . . . . . . . . . . . . . . . . . . . 9.3.3.3 Green–Ampt model . . . . . . . . . . . . . . . . . 9.3.3.4 NRCS curve-number model . . . . . . . . . . . . 9.3.3.5 Comparison of infiltration models . . . . . . . . . 9.3.4 Rainfall Excess on Composite Areas . . . . . . . . . . . . . 9.4 Baseflow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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419 419 419 421 423 428 428 434 439 439 440 440 440 446 447 448 448 450 450 451 451 455 455 457 460 465 471 478 479 482 486

10 Fundamentals of Surface-Water Hydrology II: Runoff 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . 10.2 Mechanisms of Surface Runoff . . . . . . . . . . . 10.3 Time of Concentration . . . . . . . . . . . . . . . 10.3.1 Overland Flow . . . . . . . . . . . . . . . 10.3.1.1 Kinematic-wave equation . . . . 10.3.1.2 NRCS method . . . . . . . . . . 10.3.1.3 Kirpich equation . . . . . . . . . 10.3.1.4 Izzard equation . . . . . . . . . . 10.3.1.5 Kerby equation . . . . . . . . . . 10.3.2 Channel Flow . . . . . . . . . . . . . . . . 10.3.3 Accuracy of Estimates . . . . . . . . . . . 10.4 Peak-Runoff Models . . . . . . . . . . . . . . . . 10.4.1 The Rational Method . . . . . . . . . . . . 10.4.2 NRCS-TR55 Method . . . . . . . . . . . . 10.5 Continuous-Runoff Models . . . . . . . . . . . . 10.5.1 Unit-Hydrograph Theory . . . . . . . . . 10.5.2 Instantaneous Unit Hydrograph . . . . . .

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491 491 491 492 492 492 496 499 499 500 502 504 505 505 510 513 513 519

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10.5.3 Unit-Hydrograph Models . . . . . . . . . . . . 10.5.3.1 Snyder unit-hydrograph model . . . . 10.5.3.2 NRCS dimensionless unit hydrograph 10.5.3.3 Accuracy of unit-hydrograph models 10.5.4 Time-Area Models . . . . . . . . . . . . . . . . 10.5.5 Kinematic-Wave Model . . . . . . . . . . . . . 10.5.6 Nonlinear-Reservoir Model . . . . . . . . . . . 10.5.7 Santa Barbara Urban Hydrograph Model . . . 10.5.8 Extreme Runoff Events . . . . . . . . . . . . . 10.6 Routing Models . . . . . . . . . . . . . . . . . . . . . . 10.6.1 Hydrologic Routing . . . . . . . . . . . . . . . . 10.6.1.1 Modified Puls method . . . . . . . . . 10.6.1.2 Muskingum method . . . . . . . . . . 10.6.2 Hydraulic Routing . . . . . . . . . . . . . . . . 10.7 Water-Quality Models . . . . . . . . . . . . . . . . . . . 10.7.1 Event-Mean Concentrations . . . . . . . . . . . 10.7.2 Regression Equations . . . . . . . . . . . . . . 10.7.2.1 USGS model . . . . . . . . . . . . . . 10.7.2.2 EPA model . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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520 521 524 527 527 532 533 535 537 538 538 538 542 549 551 551 553 553 555 557

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563 563 563 567 568 572 578 583 584 585 586 589 594 595 596 601 602

12 Design of Stormwater-Management Systems 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Performance Goals . . . . . . . . . . . . . . . . . . . . 12.2.1 Quantity Control . . . . . . . . . . . . . . . . . 12.2.2 Quality Control . . . . . . . . . . . . . . . . . . 12.3 Design of Stormwater Control Measures . . . . . . . . 12.3.1 Storage Impoundments . . . . . . . . . . . . . . 12.3.1.1 Detention basins—Design parameters 12.3.1.2 Wet detention basins . . . . . . . . . . 12.3.1.3 Dry detention basins . . . . . . . . . . 12.3.1.4 Design of outlet structures . . . . . . 12.3.1.5 Design for flood control . . . . . . . . 12.3.2 Infiltration Basins . . . . . . . . . . . . . . . . . 12.3.3 Swales . . . . . . . . . . . . . . . . . . . . . . . 12.3.3.1 Retention swales . . . . . . . . . . . . 12.3.3.2 Biofiltration swales . . . . . . . . . . . 12.3.4 Vegetated Filter Strips . . . . . . . . . . . . . .

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604 604 604 604 604 605 605 606 608 610 611 617 621 623 624 625 628

11 Design of Stormwater-Collection Systems 11.1 Introduction . . . . . . . . . . . . . . . . . 11.2 Street Gutters . . . . . . . . . . . . . . . . 11.3 Inlets . . . . . . . . . . . . . . . . . . . . . 11.3.1 Curb Inlets . . . . . . . . . . . . . . 11.3.2 Grate Inlets . . . . . . . . . . . . . 11.3.3 Combination Inlets . . . . . . . . . 11.3.4 Slotted Inlets . . . . . . . . . . . . 11.4 Roadside and Median Channels . . . . . . 11.5 Storm Sewers . . . . . . . . . . . . . . . . 11.5.1 Calculation of Design Flow Rates . 11.5.2 Pipe Sizing and Selection . . . . . . 11.5.3 Manholes . . . . . . . . . . . . . . 11.5.4 Determination of Impervious Area 11.5.5 System-Design Computations . . . 11.5.6 Other Design Considerations . . . Problems . . . . . . . . . . . . . . . . . . . . . .

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12.3.5 Bioretention Systems . . . . . . . . . . . . 12.3.6 Exfiltration Trenches . . . . . . . . . . . . 12.3.6.1 General design guidelines . . . . 12.3.6.2 Design for flood control . . . . . 12.3.6.3 Design for water-quality control 12.3.7 Subsurface Exfiltration Galleries . . . . . 12.4 Selection of SCMs for Water-Quality Control . . 12.4.1 Nonstructural SCMs . . . . . . . . . . . . 12.4.2 Structural SCMs . . . . . . . . . . . . . . . 12.4.3 Other Considerations . . . . . . . . . . . . 12.5 Major Drainage System . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

11

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628 630 631 632 634 635 636 636 636 637 637 637

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642 642 642 643 644 645 645 647 648 649 649 650 650 650 651 652 655 656 657 661 662 666 669 669 671 671 672 672

14 Fundamentals of Groundwater Hydrology I: Governing Equations 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Darcy’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.1 Hydraulic Conductivity . . . . . . . . . . . . . . . . . . 14.2.1.1 Empirical formulae . . . . . . . . . . . . . . 14.2.1.2 Classification . . . . . . . . . . . . . . . . . . 14.2.1.3 Anisotropic properties . . . . . . . . . . . . . 14.2.1.4 Stochastic properties . . . . . . . . . . . . . . 14.3 General Flow Equation . . . . . . . . . . . . . . . . . . . . . . 14.4 Two-Dimensional Approximations . . . . . . . . . . . . . . . 14.4.1 Unconfined Aquifers . . . . . . . . . . . . . . . . . . . 14.4.2 Confined Aquifers . . . . . . . . . . . . . . . . . . . . 14.5 Flow in the Unsaturated Zone . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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674 674 680 684 684 688 688 692 694 699 699 705 709 714

13 Estimation of Evapotranspiration 13.1 Introduction . . . . . . . . . . . . . . . . . 13.2 Penman–Monteith Equation . . . . . . . . 13.2.1 Aerodynamic Resistance . . . . . . 13.2.2 Surface Resistance . . . . . . . . . 13.2.3 Net Radiation . . . . . . . . . . . . 13.2.3.1 Shortwave radiation . . . 13.2.3.2 Longwave radiation . . . 13.2.4 Soil Heat Flux . . . . . . . . . . . . 13.2.5 Latent Heat of Vaporization . . . . 13.2.6 Psychrometric Constant . . . . . . 13.2.7 Saturation Vapor Pressure . . . . . 13.2.8 Vapor-Pressure Gradient . . . . . . 13.2.9 Actual Vapor Pressure . . . . . . . 13.2.10 Air Density . . . . . . . . . . . . . 13.3 Application of the PM Equation . . . . . . 13.4 Potential Evapotranspiration . . . . . . . . 13.5 Reference Evapotranspiration . . . . . . . 13.5.1 FAO56-Penman–Monteith Method 13.5.2 ASCE Penman–Monteith Method 13.5.3 Evaporation Pans . . . . . . . . . . 13.5.4 Empirical Methods . . . . . . . . . 13.6 Actual Evapotranspiration . . . . . . . . . 13.6.1 Index-of-Dryness Method . . . . . 13.6.2 Crop-Coefficient Method . . . . . 13.6.3 Remote Sensing . . . . . . . . . . . 13.7 Selection of ET Estimation Method . . . . Problems . . . . . . . . . . . . . . . . . . . . . .

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15 Fundamentals of Groundwater Hydrology II: Applications 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Steady-State Solutions . . . . . . . . . . . . . . . . . . 15.2.1 Unconfined Flow Between Two Reservoirs . . 15.2.2 Well in a Confined Aquifer . . . . . . . . . . . 15.2.3 Well in an Unconfined Aquifer . . . . . . . . . 15.2.4 Well in a Leaky Confined Aquifer . . . . . . . . 15.2.5 Well in an Unconfined Aquifer with Recharge . 15.2.6 Partially Penetrating Wells . . . . . . . . . . . . 15.3 Unsteady-State Solutions . . . . . . . . . . . . . . . . . 15.3.1 Well in a Confined Aquifer . . . . . . . . . . . 15.3.2 Well in an Unconfined Aquifer . . . . . . . . . 15.3.3 Well in a Leaky Confined Aquifer . . . . . . . . 15.3.4 Other Solutions . . . . . . . . . . . . . . . . . . 15.4 Principle of Superposition . . . . . . . . . . . . . . . . 15.4.1 Multiple Wells . . . . . . . . . . . . . . . . . . . 15.4.2 Well in Uniform Flow . . . . . . . . . . . . . . 15.5 Method of Images . . . . . . . . . . . . . . . . . . . . . 15.5.1 Constant-Head Boundary . . . . . . . . . . . . 15.5.2 Impermeable Boundary . . . . . . . . . . . . . 15.5.3 Other Applications . . . . . . . . . . . . . . . . 15.6 Saltwater Intrusion . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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718 718 718 718 720 724 727 731 732 736 736 746 754 759 759 760 762 764 764 768 770 770 779

16 Design of Groundwater Systems 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . 16.2 Design of Wellfields . . . . . . . . . . . . . . . . . 16.3 Wellhead Protection . . . . . . . . . . . . . . . . . 16.3.1 Delineation of Wellhead Protection Areas 16.3.2 Time-of-Travel Approach . . . . . . . . . 16.4 Design and Construction of Water-Supply Wells . 16.4.1 Types of Wells . . . . . . . . . . . . . . . . 16.4.2 Design of Well Components . . . . . . . . 16.4.2.1 Casing . . . . . . . . . . . . . . . 16.4.2.2 Screen intake . . . . . . . . . . . 16.4.2.3 Gravel pack . . . . . . . . . . . . 16.4.2.4 Pump . . . . . . . . . . . . . . . 16.4.2.5 Other considerations . . . . . . . 16.4.3 Performance Assessment . . . . . . . . . . 16.4.4 Well Drilling . . . . . . . . . . . . . . . . . 16.5 Design of Aquifer Pumping Tests . . . . . . . . . 16.5.1 Pumping Well . . . . . . . . . . . . . . . . 16.5.2 Observation Wells . . . . . . . . . . . . . 16.5.3 Field Procedures . . . . . . . . . . . . . . 16.6 Design of Slug Tests . . . . . . . . . . . . . . . . . 16.7 Design of Exfiltration Trenches . . . . . . . . . . 16.8 Seepage Meters . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

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789 789 789 792 792 793 795 795 796 797 797 801 802 803 806 811 812 812 813 814 816 821 826 827

17 Water-Resources Planning 17.1 Introduction . . . . . . . . . . . . . . . . 17.2 Planning Process . . . . . . . . . . . . . . 17.3 Economic Feasibility . . . . . . . . . . . 17.3.1 Compound-Interest Factors . . . 17.3.1.1 Single-payment factors 17.3.1.2 Uniform-series factors .

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833 833 833 836 837 837 838

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17.3.1.3 Arithmetic-gradient factors 17.3.1.4 Geometric-gradient factors 17.3.2 Evaluating Alternatives . . . . . . . 17.3.2.1 Present-worth analysis . . . 17.3.2.2 Annual-worth analysis . . . 17.3.2.3 Rate-of-return analysis . . 17.3.2.4 Benefit–cost analysis . . . . Problems . . . . . . . . . . . . . . . . . . . . . . .

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838 839 841 841 843 843 846 847

A Units and Conversion Factors A.1 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.2 Conversion Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

849 849 850

B Fluid Properties B.1 Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.2 Organic Compounds Found in Water . . . . . . . . . . . . . . . . . . . . . . B.3 Air at Standard Atmospheric Pressure . . . . . . . . . . . . . . . . . . . . .

852 852 852 854

C Statistical Tables C.1 Areas Under Standard Normal Curve . . . . . . . . . . . . C.2 Frequency Factors for Pearson Type III Distribution . . . C.3 Critical Values of the Chi-Square Distribution . . . . . . . C.4 Critical Values for the Kolmogorov–Smirnov Test Statistic

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855 855 857 859 860

D Special Functions D.1 Error Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D.2 Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D.2.2 Evaluation of Bessel Functions . . . . . . . . . . . . . . . . . . . D.2.2.1 Bessel function of the first kind of order n . . . . . . . . D.2.2.2 Bessel function of the second kind of order n . . . . . . D.2.2.3 Modified Bessel function of the first kind of order n . . D.2.2.4 Modified Bessel function of the second kind of order n D.2.2.5 Tabulated values of useful Bessel functions . . . . . . . D.3 Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D.4 Exponential Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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861 861 862 862 862 862 863 863 863 863 866 867

E Pipe Specifications E.1 PVC Pipe . . . . . . . . . . . . . . . . . . . . . . E.2 Ductile-Iron Pipe . . . . . . . . . . . . . . . . . E.3 Concrete Pipe . . . . . . . . . . . . . . . . . . . E.4 Physical Properties of Common Pipe Materials

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868 868 868 869 869

F Unified Soil Classification System F.1 Definition of Soil Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

870 870 871

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Index

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Preface Water-resources engineers design systems to control the quantity, quality, timing, and distribution of water to support human habitation and the needs of the environment. Water-supply and flood-control systems are commonly regarded as essential infrastructure for developed areas, and as such water-resources engineering is a core specialty area in civil engineering. Water-resources engineering is also a specialty area in environmental engineering, particularly with regard to the design of water-supply systems, wastewater-collection systems, and water-quality control in natural systems. The technical and scientific bases for most water-resources applications are in the areas of hydraulics and hydrology, and this text covers these areas with depth and rigor. The fundamentals of closed-conduit flow, open-channel flow, surface-water hydrology, groundwater hydrology, and water-resources planning and management are all covered in detail. Applications of these fundamentals include the design of water-distribution systems, hydraulic structures, sanitary-sewer systems, stormwater-management systems, and water-supply wellfields. The design protocols for these systems are guided by the relevant ASCE, WEF, and AWWA manuals of practice, as well as USFHWA design guidelines for urban and transportationrelated drainage structures, and USACE design guidelines for hydraulic structures. The topics covered in this book constitute the technical background expected of water-resources engineers. This text is appropriate for undergraduate and first-year graduate courses in hydraulics, hydrology, and water-resources engineering. Practitioners will also find the material in this book to be a useful reference on appropriate design protocols. The book has been organized in such a way as to sequentially cover the theory and design applications in each of the key areas of water-resources engineering. The theory of flow in closed conduits is covered in Chapter 2, including applications of the continuity, momentum, and energy equations to flow in closed conduits, calculation of water-hammer pressures, flows in pipe networks, affinity laws for pumps, pump performance curves, and procedures for pump selection and assessing the performance of multi-pump systems. The design of public water-supply systems and building water-supply systems are covered in Chapter 3, which includes the estimation of water demand, design of pipelines, pipeline appurtenances, service reservoirs, performance criteria for water-distribution systems, and several practical design examples. The theory of flow in open channels is covered in Chapter 4, which includes applications of the continuity, momentum, and energy equations to flow in open channels, and computation of water-surface profiles. The design of drainage channels is covered in Chapter 5, which includes the application of design standards for determining the appropriate channel dimensions for various channel linings, including vegetative and non vegetative linings. The design of sanitary-sewer systems is covered in Chapter 6, which includes design approaches for estimating the quantity of wastewater to be handled by sewers; sizing sewer pipes based on self-cleansing and capacity using the ASCE-recommended tractive-force method; and the performance of manholes, force mains, pump stations, and hydrogen-sulfide control systems are also covered. Design of the most widely used hydraulic structures is covered in Chapter 7, which includes the design of culverts, gates, weirs, spillways, stilling basins, and dams. This chapter is particularly important since most water-resources projects rely on the performance of hydraulic structures to achieve their objectives. The bases for the design of water-resources systems are typically rainfall and/or surface runoff, which are random variables that must generally be specified probabilistically. Applications of probability and statistics in water-resources engineering are covered in detail in Chapter 8, with particular emphasis on the analysis of hydrologic data and uncertainty analysis in predicting hydrologic variables. The fundamentals of surface-water hydrology are covered in Chapters 9 and 10. These chapters cover the statistical characterization of rainfall for design applications, methodologies for estimating peak runoff and runoff hydrographs, methodologies for routing runoff hydrographs through detention basins, and methods for estimating the quality of surface runoff. The design of stormwater-collection systems is covered in Chapter 11, 15

16

Preface

including the design of stormwater inlets and storm sewers. Stormwater-management systems are designed to treat stormwater prior to discharge into receiving waters, and the design of these systems is covered in Chapter 12. Several state-of-the-art design examples for the most commonly used stormwater-control measures are provided, including the design of infiltration basins, swales, filter strips, bioretention systems, and exfiltration trenches. The estimation of evapotranspiration, which is usually the dominant component of seasonal and annual water budgets in arid areas and a core component in the design of irrigation systems, is covered in Chapter 13. The fundamentals of groundwater hydrology are covered in Chapters 14 and 15, including an exposition on Darcy’s law, derivation of the general groundwater flow equation, practical solutions to the groundwater flow equation, and methods to assess and control saltwater intrusion in coastal aquifers. Applications of groundwater hydrology to the design of wellfields, the delineation of wellhead protection areas, and the design of wells, aquifer pumping tests, slug tests, and exfiltration trenches are all covered. Water-resources planning typically includes identifying alternatives and ranking the alternatives based on specified criteria. Chapter 17 covers the conventional approaches for identifying and ranking alternatives and the bases for the economic evaluation of these alternatives. In summary, this book provides an in-depth coverage of the subject areas that are fundamental to the practice of water-resources engineering. A firm grasp of the material covered in this book along with complementary practical experience are the foundations on which water-resources engineering is practiced at the highest level. Throughout the entire textbook, equations contained within boxes represent derived equations that are particularly useful in engineering applications. In contrast, equations without boxes are typically intermediate equations within an analysis leading to a derived useful equation. This book is a reflection of the author’s belief that water-resources engineers must gain a firm understanding of the depth and breadth of the technical areas that are fundamental to their discipline, and by so doing will be more innovative, view water-resource systems holistically, and be technically prepared for a lifetime of learning. On the basis of this vision, the material contained in this book is presented mostly from first principles, is rigorous, is relevant to the practice of water-resources engineering, and is reinforced by detailed presentations of design applications. Many persons have contributed in various ways to this book and to my understanding of water-resources engineering, and to recognize all of those who have helped me along the way would be a book onto itself. However, special recognition is deserved by Professor LaVere Merritt of Brigham Young University for his expert advice and detailed review of the chapter on design of sanitary sewers and Professor Dixie Griffin of Louisiana Tech for his extensive feedback and constructive comments throughout the years on the present and previous editions of this book. I am also grateful to Professors John Miknis of Pennsylvania State University, Jacob Ogaard of the University of Iowa, Francisco Olivera of Texas A&M University, and Ken Lee of the University of Massachusetts Lowell, for reviewing this book. What’s New in the Third Edition The third edition of this book contains much new and updated material and is significantly reorganized relative to the previous edition. The most notable changes are as follows: • The book contains 17 chapters compared to 7 chapters in the previous edition. In the previous edition, most of the chapters were quite long and contained both theory and practical examples. In the present edition, theory-oriented chapters have been separated from practice-oriented chapters. The material in all chapters has been revised and updated, with some chapters being almost entirely rewritten as described below. • Coverage of the design of drainage channels (Chapter 5) has been completely rewritten. Subsequent to the previous edition of the book, the Federal Highway Administration thoroughly revised their urban drainage design manual, Hydraulic Engineering Circular No.22 (HEC-22), which provides the primary design guidelines for the design of drainage channels in the United States. The updated chapter in this book is consistent with the latest edition of HEC-22. Appendix F describing the unified soil classification system has been added to support the design applications contained in this chapter.

Preface

17

• Coverage of the design of sanitary-sewers (Chapter 6) has been completely rewritten to be consistent with the latest version of the ASCE Manual of Practice No.60 (MOP 60) on the design of sanitary sewers. The latest version of MOP 60 is a significant departure from previous versions of MOP 60 in that the tractive-force design approach is now recommended as the preferred approach for designing sanitary sewers. The updated chapter emphasizes the tractive-force approach and contains the key design aids provided in ASCE Manual of Practice No.60. • Coverage of the design of stormwater-management systems (Chapter 12) has been significantly revised and updated. Over the past several years, much has been learned about the performance and design of various stormwater control measures (SCMs) and the latest design approaches to these systems are incorporated in the revised chapter. These design approaches are consistent with the latest version of ASCE Manual of Practice No.87. • In addition to updating the coverage on most topics covered in the book, several new topics have been added. For example, coverage of water hammer, variable-speed pumps, water-surface profiles across bridges, design of dams and reservoirs, and uncertainty analysis have all been added. • Many new end-of-chapter problems have been added to support the revised coverage in the book, and several problems from the previous edition have been removed or modified. Instructor resources for the International Edition are available at http://www.pearsoninternationaleditions.com/chin. In summary, this new edition reflects the state-of-the-art of water-resources engineering and is intended provide the necessary competencies expected by the profession. The redesigned chapters, which are shorter in length than previous chapters, are intended to provide a more focused treatment of individual cognate topics and hence contribute to more effective learning by those using this textbook.

David A. Chin

The publishers would like to thank D. Nagesh Kumar of the Indian Institute of Science, Bangalore for reviewing the content of the International Edition.

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C H A P T E R

1

Introduction 1.1 Water-Resources Engineering Water-resources engineering is an area of professional practice that includes the design of systems to control the quantity, quality, timing, and distribution of water to meet the needs of human habitation and the environment. Aside from the engineering and environmental aspects of water-resource systems, their feasibility from legal, economic, financial, political, and social viewpoints must generally be considered in the development process. In fact, the successful operation of an engineered system usually depends as much on nonengineering analyses (e.g., economic and social analyses) as on sound engineering design. Examples of water-resource systems include domestic, commercial, and industrial water supply, wastewater treatment, irrigation, drainage, flood control, salinity control, sediment control, pollution abatement, and hydropower-generation systems. The waters of the earth are found on land, in the oceans, and in the atmosphere, and the core science of water-resources engineering is hydrology, which deals with the occurrence, distribution, movement, and properties of water on earth. Engineering hydrologists are primarily concerned with water on land and in the atmosphere, from its deposition as atmospheric precipitation to its inflow into the oceans and its vaporization into the atmosphere. Water-resources engineering is commonly regarded as a subdiscipline of civil engineering, and several other specialty areas are encompassed within the field of water-resources engineering. For example, the specialty area of groundwater hydrology is concerned with the occurrence and movement of water below the surface of the earth; surface-water hydrology and climatology are concerned with the occurrence and movement of water above the surface of the earth; hydrogeochemistry is concerned with the chemical changes in water that is in contact with earth materials; erosion, sedimentation, and geomorphology are concerned with the effects of sediment transport on landforms; and water policy, economics, and systems analyses are concerned with the political, economic, and environmental constraints in the design and operation of water-resource systems. The quantity and quality of water are inseparable issues in design, and the modern practice of water-resources engineering demands that practitioners be technically competent in understanding the physical processes that govern the movement of water, the chemical and biological processes that affect the quality of water, the economic and social considerations that must be taken into account, and the environmental impacts associated with the construction and operation of water-resource projects. 1.2 The Hydrologic Cycle The hydrologic cycle is defined as the pathway of water as it moves in its various phases through the atmosphere, to the earth, over and through the land, to the ocean, and back to the atmosphere. The movement of water in the hydrologic cycle is illustrated in Figure 1.1. A description of the hydrologic cycle can start with the evaporation of water from the oceans, which is driven by energy from the sun. The evaporated water, in the form of water vapor, rises by convection, condenses in the atmosphere to form clouds, and precipitates onto land and ocean surfaces, predominantly as rain or snow. Rainfall on land surfaces is partially intercepted by surface vegetation, partially stored in surface depressions, partially infiltrated into the ground, and partially flows over land into drainage channels and rivers that ultimately lead back to the ocean. Rainfall that is intercepted by surface vegetation is eventually evaporated into the atmosphere; water held in depression storage either evaporates or infiltrates into the ground; and water that infiltrates into the ground contributes to the recharge of groundwater, which either is utilized by plants, evaporates, is stored, or becomes subsurface flow that ultimately emerges as recharge to streams or directly to the ocean. Snowfall in 19

20

Chapter 1

Introduction

FIGURE 1.1: Hydrologic cycle

Sun Hydrologic Cycle

Snowfall

Condensation and cloud formation

Snowcap

Rainfall Evapotranspiration

Snowmelt Mountain

Evaporation

Surface water Surface runoff Infiltration and percolation Groundwater flow

Ocean

mountainous areas typically accumulates in the winter and melts in the spring, thereby contributing to larger-than-average river flows during the spring. A typical example of the headwater of a river being fed by snowmelt is shown in Figure 1.2. Groundwater is defined as the water below the land surface, and water above the land surface (in liquid form) is called surface water. In urban areas, the ground surface is typically much more impervious than in rural areas, and surface runoff in urban areas is mostly controlled by constructed drainage systems. Surface waters and groundwaters in urban areas also tend to be significantly influenced by the water-supply and wastewater removal systems that are an integral part of the urban landscape. Since human-made systems are part of the hydrologic cycle, it is the responsibility of the water-resources engineer to ensure that systems constructed for water use and control are in harmony with the needs of the natural environment and the natural hydrologic cycle. The quality of water varies considerably as it moves through the hydrologic cycle, with contamination potentially resulting from several sources. Classes of contaminants commonly FIGURE 1.2: Snowmelt contribution to river flow

Section 1.2

The Hydrologic Cycle

21

TABLE 1.1: Classes of Water Contaminants

Contaminant class

Example

Oxygen-demanding wastes Infectious agents Plant nutrients Organic chemicals Inorganic chemicals

Plant and animal material Bacteria and viruses Fertilizers, such as nitrates and phosphates Pesticides, detergents, oil, grease Acids from coal mine drainage, inorganic chemicals such as iron from steel plants Clay silt on streambed, which may reduce or even destroy life forms living at the solid-liquid interface Waste products from mining and processing of radioactive material, radioactive isotopes after use Cooling water used in steam generation of electricity

Sediment from land erosion Radioactive substances Heat from industry

found in water, along with some examples, are listed in Table 1.1. The effects of the quantity and quality of water on the health of terrestrial ecosystems and the value of these ecosystems in the hydrologic cycle are often overlooked. For example, the modification of free-flowing rivers for energy or water supply, and the drainage of wetlands, can have a variety of adverse effects on aquatic ecosystems, including losses in species diversity, floodplain fertility, and biofiltration capability. Also, excessive withdrawals of groundwater in coastal areas can cause saltwater intrusion and hence deterioration of subsurface water quality. On a global scale, the distribution of the water resources of the earth is given in Table 1.2, where it is clear that the vast majority of the earth’s water resources (97%) is contained in the oceans, with most of the freshwater being contained in groundwater and polar ice. Groundwater is a particularly important water resource, comprising more than 98% of the liquid freshwater available on earth; less than 2% of liquid freshwater is contained in streams and lakes. The amount of water stored in the atmosphere is relatively small, although

TABLE 1.2: Estimated World Water Quantities

Item

Volume (*103 km3 )

Oceans Groundwater Fresh Saline Soil moisture Polar ice Other ice and snow Lakes Fresh Saline Marshes Rivers Biological water Atmospheric water

1,338,000

Total water Freshwater

1,385,984.61 35,029.21

10,530 12,870 16.5 24,023.5 340.6 91 85.4 11.47 2.12 1.12 12.9

Percent total water (%)

Percent freshwater (%)

96.5 0.76 0.93 0.0012 1.7 0.025 0.007 0.006 0.0008 0.0002 0.0001 0.001 100 2.5

30.1 0.05 68.6 1.0 0.26 0.03 0.006 0.003 0.04 100

Source: USSR National Committee for the International Hydrological Decade (1978).

22

Chapter 1

Introduction TABLE 1.3: Fluxes in Global Hydrologic Cycle

Component

Oceanic flux (mm/yr)

Terrestrial flux (mm/yr)

1270 1400 —

800 484 316

Precipitation Evaporation Runoff to ocean (rivers plus groundwater)

Source: USSR National Committee for the International Hydrological Decade (1978).

the flux of water into and out of the atmosphere dominates the hydrologic cycle. The typical residence time for atmospheric water is on the order of a week, the typical residence time for soil moisture is on the order of weeks to months, and the typical residence time in the oceans is on the order of tens of thousands of years (Wood, 1991). The estimated fluxes of precipitation, evaporation, and runoff within the global hydrologic cycle are given in Table 1.3. These data indicate that the global average precipitation over land is on the order of 800 mm/yr (31 in./yr), of which 484 mm/yr (19 in./yr) is returned to the atmosphere as evaporation and 316 mm/yr (12 in./yr) is returned to the ocean via surface runoff. On a global scale, large variations from these average values are observed. In the United States, for example, the highest annual rainfall is found at Mount Wai’ale’ale on the Hawaiian island of Kauai with an annual rainfall of 1168 cm (460 in.), while Greenland Ranch in Death Valley, California, has the lowest annual average rainfall of 4.5 cm (1.8 in.). Two of the most widely used climatic measures are the mean annual rainfall and the mean annual potential evapotranspiration. A climatic spectrum appropriate for subtropical and midlatitudinal regions is given in Table 1.4, and water-resource priorities tend to differ substantially between climates. For example, forecasting and planning for drought conditions is particularly important in semiarid climates, whereas droughts are barely noticeable in very humid areas. Climatic conditions also have a controlling influence on the water budget as shown in Figure 1.3. In humid areas, annual rainfall is high and surface runoff, groundwater recharge, and evapotranspiration (ET) are all significant processes. In comparison, in arid areas annual rainfall is low, evapotranspiration is the dominant process, surface runoff is also important, and groundwater recharge is almost negligible. In semiarid climates, evapotranspiration remains the dominant process, although surface runoff and groundwater recharge are both important processes. On regional scales, water resources are managed within topographically defined areas called watersheds or basins. These areas are typically enclosed by topographic high points in the land surface, and within these bounded areas the path of the surface runoff can usually be controlled with a reasonable degree of coordination. Human activities such as land-use changes, dam construction and reservoir operation, surface-water and groundwater TABLE 1.4: Climate Spectrum

Climate Superarid Hyperarid Arid Semiarid Subhumid Humid Hyperhumid Superhumid

Mean annual precipitation (mm)

Mean annual evapotranspiration (mm)

0.2

r

0.50 0.12 0.03 K

Reentrant flow

V

0.8 –1

K Pipe exit

V 1

V

1

D2/D1 Contraction D1

θ

D2

0.0 0.20 0.40 0.60 0.80 0.90

V2

D1/D2 Expansion V1

θ

D

0.0 0.20 0.40 0.60 0.80

D2

Vanes 90° miter bend

K θ = 20°

K θ = 180° 1.00 0.87 0.70 0.41 0.15

0.30 0.25 0.15 0.10

K = 1.1

With vanes

K = 0.2

90° smooth bend 1 2 4 6

r

K θ = 180° 0.50 0.49 0.42 0.27 0.20 0.10

Without vanes

r/D

D

K θ = 60° 0.08 0.08 0.07 0.06 0.06 0.06

K θ = 90° 0.35 0.19 0.16 0.21 K

Threaded pipe fittings

4 –10 5.0 0 – 0.2 5.6 2.2

Globe valve — wide open Angle valve — wide open Gate valve — wide open Gate valve — half open Return bend Tee straight-through flow side-outlet flow 90° elbow 45° elbow

h0 =

0.4 – 0.9 1.7 – 2.0 0.9–1.5 0.3 – 0.4



K

V2 2g

(2.74)

where K is a loss coefficient that is specific to each fitting and transition, and V is the average velocity at a defined location within the transition or fitting. The loss coefficients for several fittings and transitions are shown in Figure 2.7. Most pipelines will have fittings and

Section 2.2

Single Pipelines

47

transitions. Common transitions include intakes, bends, and exits. Common fittings include various types of valves, such as gate valves that are used to open and close pipelines, and globe valves that are used to regulate the flow in pipelines. Faucets at the end of pipes in household plumbing are globe valves. Head losses in transitions and fittings are also called local head losses or minor head losses. The latter term should be avoided, however, since in some cases these head losses are a significant portion of the total head loss in a pipe. Detailed descriptions of local head losses in various valve geometries can be found in Mott (1994), and additional data on local head losses in pipeline systems can be found in Brater and colleagues (1996). There is considerable uncertainty in the loss coefficients given in Figure 2.7, since loss coefficients generally vary with surface roughness, Reynolds number, and the details of the design. The loss coefficients of two seemingly identical valves from two different manufacturers can differ by a factor of 2 or more. Therefore, the particular manufacturer’s data should be consulted in the final design of piping systems rather than relying on the representative values in textbooks and handbooks. Although local head losses are commonly assumed to occur within a fitting or transition, these appurtenances typically influence the flow for several pipe diameters downstream. This is why most flow meter manufacturers recommend installing flow meters at least 10–20 pipe diameters downstream of any elbows or valves—this allows the swirling turbulent eddies generated by the elbow or valve to largely disappear and the velocity profile to become fully developed before entering the flow meter. EXAMPLE 2.6 A pump is to be selected that will pump water from a well into a storage reservoir. In order to fill the reservoir in a timely manner, the pump is required to deliver 5 L/s when the water level in the reservoir is 5 m above the water level in the well. Find the head that must be added by the pump. The pipeline system is shown in Figure 2.8. Assume that the local loss coefficient for each of the bends is equal to 0.25 and that the temperature of the water is 20◦ C. FIGURE 2.8: Pipeline system

15 m Reservoir 5m

5m

2m 5m

P

3m

100-mm PVC (pump to reservoir) 50-mm PVC (well to pump) Well

Solution Taking the elevation of the water surface in the well to be equal to 0 m, and proceeding from the well to the storage reservoir (where the head is equal to 5 m), the energy equation (Equation 2.65) can be written as 0 −

V12

V2 V2 V2 f L V2 f L V2 − 1 1 1 − K1 1 + hp − 2 2 2 − (K2 + K3 ) 2 − 2 = 5 2g D1 2g 2g D2 2g 2g 2g

where V1 and V2 are the velocities in the 50-mm (= D1 ) and 100-mm (= D2 ) pipes, respectively; L1 and L2 are the corresponding pipe lengths; f1 and f2 are the corresponding friction factors; K1 , K2 , and K3 are the loss coefficients for each of the three bends; and hp is the head added by the pump. The cross-sectional areas of each of the pipes, A1 and A2 , are given by

48

Chapter 2

Fundamentals of Flow in Closed Conduits π 2 π D1 = (0.05)2 = 0.001963 m2 4 4 π 2 π A2 = D2 = (0.10)2 = 0.007854 m2 4 4 When the flow rate, Q, is 5 L/s, the velocities V1 and V2 are given by Q 0.005 = V1 = = 2.54 m/s A1 0.001963 A1 =

V2 =

Q 0.005 = = 0.637 m/s A2 0.007854

PVC pipe can be considered smooth (ks L 0) and the friction factor, f , can be estimated using the Swamee–Jain equation as 0.25 f =   5.74 2 log10 Re0.9 where Re is the Reynolds number. At 20◦ C, the kinematic viscosity, ν, is equal to 1.00 * 10−6 m2 /s and for the 50-mm pipe (2.54)(0.05) V D = 1.27 * 105 Re1 = 1 1 = ν 1.00 * 10−6 which leads to 0.25 f1 =  2 = 0.0170 5.74 log10 (1.27 * 105 )0.9 and for the 100-mm pipe Re2 =

(0.637)(0.10) V2 D2 = = 6.37 * 104 ν 1.00 * 10−6

which leads to f2 = 

0.25 5.74 log10 (6.37 * 104 )0.9

2 = 0.0197

Substituting the values of these parameters into the energy equation yields     (0.0197)(22) 2.542 0.6372 (0.0170)(8) 0− 1 + + 0.25 + hp − + 0.25 + 0.25 + 1 =5 0.05 (2)(9.81) 0.10 (2)(9.81) which leads to hp = 6.43 m Therefore the head to be added by the pump is 6.43 m.

Local head losses in a pipeline can be neglected when the total friction loss is much greater than the sum of the local head losses. Rules of thumb that have been suggested are that local head losses can be neglected when the sum of the local head losses is less than 10% of the total friction loss (e.g., Jones, 2011), or local head losses can be neglected when there is a length of at least 1000 diameters between each local head loss (e.g., Streeter et al., 1998). Both of these rules have merit. The hydrodynamic entry length of a pipe is defined as the distance from a pipe entrance to where the wall shear stress (and thus the friction factor) reaches within 2% of the fully developed value; in other words, it is the distance required for fully turbulent flow to develop within the pipe. In many pipe flows of practical interest, the hydrodynamic entry length is approximately equal to 10 pipe diameters. Over distances on the order of the hydrodynamic entry length, frictional losses are greater than predicted by assuming fully turbulent flow from the pipe entrance; however, over distances much longer than the hydrodynamic pipe length entrance effects are negligible.

Section 2.2

2.2.3.4

Single Pipelines

49

Head losses in noncircular conduits

Most pipelines are of circular cross section, but flow of water in noncircular conduits is commonly encountered. The hydraulic radius, R, of a conduit of any shape is defined by the relation A (2.75) R= P where A is the cross-sectional area of the conduit and P is the wetted perimeter. For circular conduits of diameter, D, the hydraulic radius is given by R=

π D2 /4 D = πD 4

(2.76)

D = 4R

(2.77)

or Using the hydraulic radius, R, as the length scale of a closed conduit instead of D, the frictional head losses, hf , in noncircular conduits can be estimated using the Darcy–Weisbach equation for circular conduits by simply replacing D by 4R, which yields hf =

fL V 2 4R 2g

(2.78)

where the friction factor, f , is calculated using a Reynolds number, Re, defined by Re =

ρV(4R) μ

(2.79)

and a relative roughness defined by ks /4R. Characterizing a noncircular conduit by the hydraulic radius, R, is necessarily approximate, since the geometric characteristics of conduits of arbitrary cross section cannot be described with a single parameter. Secondary currents that are generated across a noncircular conduit cross section to redistribute the shears are another reason why noncircular conduits cannot be completely characterized by the hydraulic radius (Liggett, 1994). However, using the hydraulic radius as a basis for calculating frictional head losses in noncircular conduits is usually accurate to within 15% for turbulent flow (Munson et al., 1994; White, 1994). This approximation is much less accurate for laminar flows, where the accuracy is on the order of ;40% (White, 1994). Characterization of noncircular conduits by the hydraulic radius can be used for rectangular conduits where the ratio of sides, called the aspect ratio, does not exceed about 8:1 (Olson and Wright, 1990), although some engineers state that aspect ratios must be less than 4:1 (e.g., Potter and Wiggert, 2001).

EXAMPLE 2.7 Water flows through a rectangular concrete culvert of width 2 m and depth 1 m. If the length of the culvert is 10 m and the flow rate is 6 m3 /s, estimate the head loss through the culvert. Assume that the culvert flows full. Solution The head loss can be calculated using Equation 2.78. The hydraulic radius, R, is given by R=

A (2)(1) = = 0.333 m P 2(2 + 1)

and the mean velocity, V, is given by V=

Q 6 = = 3 m/s A (2)(1)

50

Chapter 2

Fundamentals of Flow in Closed Conduits At 20◦ C, ν = 1.00 * 10−6 m2 /s, and therefore the Reynolds number, Re, is given by Re =

(3)(4 * 0.333) V(4R) = 4.00 * 106 = ν 1.00 * 10−6

A median equivalent sand roughness for concrete can be taken as ks = 1.6 mm (Table 2.1), and therefore the relative roughness, ks /4R, is given by 1.6 * 10−3 ks = = 0.00120 4R 4(0.333) Substituting Re and ks /4R into the Swamee–Jain equation (Equation 2.39) for the friction factor gives f = 

0.25 

5.74 ks /4R + log 3.7 Re0.9

= ⎡ 2

0.25

⎤2 0.00120 5.74 ⎣log ⎦ + 3.7 (4.00 * 106 )0.9 

which yields f = 0.0206 The frictional head loss in the culvert, hf , is therefore given by the Darcy–Weisbach equation as hf =

fL V 2 (0.0206)(10) 32 = = 0.0709 m 4R 2g (4 * 0.333) 2(9.81)

The head loss in the culvert can therefore be estimated as 7.1 cm.

2.2.3.5

Empirical friction-loss formulae

Friction losses in pipelines should generally be calculated using the Darcy–Weisbach equation. However, a minor inconvenience in using this equation to relate the friction loss to the flow velocity results from the dependence of the friction factor on the flow velocity; therefore, the Darcy–Weisbach equation must be solved simultaneously with the Colebrook equation. In modern engineering practice, computer hardware and software make this a very minor inconvenience. In earlier years, however, this was considered a real problem, and various empirical head-loss formulae were developed to relate the head loss directly to the flow velocity. Those most commonly used are the Hazen–Williams formula and the Manning formula. The Hazen–Williams formula (Williams and Hazen, 1920) is applicable only to the flow of water in pipes and is given by V = 0.849CH R0.63 S0.54 f

(2.80)

where V is the flow velocity [m/s], CH is the Hazen–Williams roughness coefficient [dimensionless], R is the hydraulic radius [m], and Sf is the slope of the energy grade line [dimensionless], defined by hf (2.81) Sf = L where hf [L] is the head loss due to friction over a length L [L] of pipe. Values of CH for a variety of commonly used pipe materials are given in Table 2.2, where the value of CH typically varies in the range of 70–150 depending on the pipe material, diameter, and age. Combining Equations 2.80 and 2.81 yields the following expression for the frictional head loss:   V 1.85 L hf = 6.82 1.17 (2.82) CH D

Section 2.2

Single Pipelines

51

TABLE 2.2: Pipe Roughness Coefficients

CH Pipe material Ductile and cast iron: New, unlined Old, unlined Cement lined and seal coated Steel: Welded and seamless Riveted Mortar lining Asbestos cement Concrete Vitrified clay pipe (VCP) Polyvinyl chloride (PVC) Corrugated metal pipe (CMP)

n

Range

Typical

Range

Typical

120–140 40–100 100–140

130 80 120

— — 0.011–0.015

0.013 0.025 0.013

80–150 100–140 120–145

120 110 130 140 120 110 140 —

— 0.012–0.018 — — 0.011–0.015 0.012–0.014 0.007–0.011 —

0.012 0.015 — 0.011 0.012 — 0.009 0.025

100–140 110–140 135–150 —

Sources: Cruise et al., (2007); Velon and Johnson (1993); Wurbs and James (2002).

where D is the diameter of the pipe [m]. The Hazen–Williams equation is commonly assumed to be applicable to the flow of water at 16◦ C in pipes with diameters in the range of 50–1850 mm (2–72 in.), and flow velocities less than 3 m/s (10 ft/s) (e.g., Mott, 1994). Street and colleagues (1996) and Liou (1998) have shown that the Hazen–Williams coefficient has a strong Reynolds number dependence, and is mostly applicable where the pipe is relatively smooth and in the early part of its transition to rough flow. Furthermore, Jain and colleagues (1978) have shown that an error of up to 39% can be expected in the evaluation of the velocity by the Hazen–Williams formula over a wide range of diameters and slopes. In spite of these cautionary notes, the Hazen–Williams formula is frequently used in the United States for the design of large water-supply pipes without regard to its limited range of applicability, a practice that can have very detrimental effects on pipe design and could potentially lead to litigation (e.g., Bombardelli and Garc´ıa, 2003). In some cases, engineers have calculated correction factors for the Hazen–Williams roughness coefficient to account for these errors (e.g., Valiantzas, 2005). The attractiveness of the Hazen–Williams formula to practicing engineers is likely related to the relatively large database for CH values compared with the relatively small database for equivalent sand roughnesses, ks , required for application of the preferred Darcy–Weisbach equation. This reality can be partially addressed by determining the ks values corresponding to measured CH values under the given experimental conditions; the ks values remain constant under all flows and pipe sizes while the CH values do not. The relationship between ks and CH can be approximated by ks = D(3.320 − 0.021CH D0.01 )2.173 exp(−0.04125CH D0.01 )

(2.83)

where ks is the equivalent sand roughness [m], and D is the diameter of the pipe [m] used in determining the Hazen–Williams coefficient CH by experimentation; Equation 2.83 is approximately valid for 100 … CH D0.01 … 155 (Travis and Mays, 2007). In cases where D is unknown, a reasonable relationship between ks and CH can be determined by assuming that D is equal to 300 mm (12 in.). A second empirical formula that is sometimes used to describe flow in pipes is the Manning formula, which is given by V=

1 2 12 R 3 Sf n

(2.84)

52

Chapter 2

Fundamentals of Flow in Closed Conduits

where V, R, and Sf have the same meaning and units as in the Hazen–Williams formula, and n is the Manning roughness coefficient. Values of n for a variety of commonly used pipe materials are given in Table 2.2. Combining Equations 2.84 and 2.81 yields the following expression for the frictional head loss: hf = 6.35

n2 LV 2

(2.85)

4

D3

The Manning formula applies only to rough turbulent flows, where the frictional head losses are controlled by the relative roughness. Such conditions are delineated by Equation 2.37. EXAMPLE 2.8 Water flows at a velocity of 1 m/s in a 150-mm diameter new ductile-iron pipe. Estimate the head loss over 500 m using: (a) the Hazen–Williams formula, (b) the Manning formula, and (c) the Darcy– Weisbach equation. Compare your results and assess the validity of each head-loss equation. Solution (a) The Hazen–Williams roughness coefficient, CH , can be taken as 130 (Table 2.2), L = 500 m, D = 0.150 m, V = 1 m/s, and therefore the head loss, hf , is given by Equation 2.82 as hf = 6.82

L



D1.17

   V 1.85 1 1.85 500 = 6.82 = 3.85 m CH (0.15)1.17 130

(b) The Manning roughness coefficient, n, can be taken as 0.013 (approximation from Table 2.2), and therefore the head loss, hf , is given by Equation 2.85 as hf = 6.35

n2 LV 2 4

= 6.35

(0.013)2 (500)(1)2

D3

4

= 6.73 m

(0.15) 3

(c) The equivalent sand roughness, ks , can be taken as 0.26 mm (Table 2.1), and the Reynolds number, Re, is given by VD (1)(0.15) = 1.5 * 105 Re = = ν 1.00 * 10−6

where ν = 1.00 * 10−6 m2 /s at 20◦ C. Substituting ks , D, and Re into the Colebrook equation yields the friction factor, f , where     ks 0.26 2.51 2.51 1 = −2 log = −2 log + + 3.7D 3.7(150) f Re f 1.5 * 105 f

which yields f = 0.0238 The head loss, hf , is therefore given by the Darcy–Weisbach equation as hf = f

500 12 L V2 = 0.0238 = 4.04 m D 2g 0.15 2(9.81)

It is reasonable to assume that the Darcy–Weisbach equation yields the most accurate estimate of the head loss. In this case, the Hazen–Williams formula gives a head loss 5% less than the Darcy–Weisbach equation, and the Manning formula yields a head loss 67% higher than the Darcy–Weisbach equation. From the given data, Re = 1.5 * 105 , D/ks = 150/0.26 = 577, and Equation 2.37 gives the limit of rough turbulent flow as 1.5 * 105 1 Re = = 200(D/ks ) 200(577) f

'

f = 0.591

Section 2.2

Single Pipelines

53

Since the actual friction factor (= 0.0238) is much less than the minimum friction factor for rough turbulent flow (= 0.591), the flow is not in the rough turbulent regime and the Manning equation is not valid. Since the pipe diameter (= 150 mm) is between 50 mm and 1850 mm, and the velocity (= 1 m/s) is less than 3 m/s, the Hazen–Williams formula can be taken as valid. The Darcy–Weisbach equation is unconditionally valid. Given these results, it is not surprising that the Darcy–Weisbach and Hazen–Williams formulae are in close agreement, with the Manning equation giving a significantly different result. These results indicate why application of the Manning equation to closed-conduit flows is strongly discouraged.

In spite of general acceptance that the combined Darcy–Weisbach and Colebrook equations provide the most accurate description of flow in pipes, approximations continue to be developed to circumvent the relatively minor computational inconveniences of using these equations. 2.2.4

Water Hammer

The sudden closure of a valve generates a pressure wave that can sometimes cause severe structural damage to the pipeline. This process is called water hammer. Consider the situation in Figure 2.9, where the flow of a fluid in a pipe is suddenly halted by the rapid closure of a valve, and the pipe walls are rigid, such that they do not flex in response to pressure changes. Before the valve is closed, the velocity in the pipe is V, the fluid pressure is p0 , and the density of the fluid is ρ0 . As the valve is suddenly closed, the fluid adjacent to the valve is immediately halted and the effect of valve closure is propagated upstream by a pressure wave that moves with a velocity c. Behind the pressure wave, the velocity is equal to zero, the fluid pressure is p0 + p, and the fluid density is ρ0 + ρ, whereas in front of the pressure wave the velocity is V, the pressure is p0 , and the density is ρ0 . For the control volume illustrated in Figure 2.9, the momentum equation in the flow direction can be expressed in the conventional form as 

Fx =

  ⭸ ρvx dV + ρvx v · n dA ⭸t V A

(2.86)

 where x is the coordinate in the flow direction, Fx is the sum of the forces in the x direction that are acting of the fluid in the control volume, vx is the x component of the fluid velocity, v is the fluid velocity vector, n is the unit normal pointing out of the control volume, and V and A represent the volume and surface area of the control volume, respectively. Neglecting the shear resistance on the pipe boundary, Equation 2.86 can be written as p0 A − (p0 + p)A =

⭸ [ρ0 V(L − ct)A] + ρ0 V(−VA) ⭸t

(2.87)

where A is the cross-sectional area of the pipe and L is the length of the control volume. Simplifying Equation 2.87 gives

FIGURE 2.9: Pressure wave

L ct

po Flow velocity V

ρ

c

po + Δ p

ρ + Δρ

V=0

V Control volume

Rapidly closed value

Pressure wave

x

54

Chapter 2

Fundamentals of Flow in Closed Conduits

(−p)A =

⭸ (ρ0 LAV − ρ0 ctAV) − ρ0 V 2 A ⭸t

pA = ρ0 cAV + ρ0 V 2 A p = ρ0 cV + ρ0 V 2

(2.88)

which relates the pressure increase, p, associated with sudden valve closure to the fluid and flow properties. The pressure change p is commonly called the water-hammer pressure, or the Joukowsky∗ pressure rise after Nicolai E. Joukowsky (1847–1921), who first demonstrated its validity in 1897 (Douglas et al., 2001). In most cases, the velocity of the pressure wave, c, is much greater than the fluid velocity, V, and Equation 2.88 can be approximated as p = ρ0 cV

(2.89)

This equation is commonly known as the Joukowsky equation, but it is sometimes called either the Joukowsky–Frizell or the Allievi equation (e.g., Tijsseling and Anderson, 2007). Applying the continuity equation to the control volume shown in Figure 2.9 requires that   ⭸ ρ dV + ρv · n dA = 0 (2.90) ⭸t V A which in this case yields ⭸ [ρ0 (L − ct)A + (ρ0 + ρ)(ct)A] + ρ0 (−V)A = 0 ⭸t ⭸ [ρ0 LA + ρctA] − ρ0 VA = 0 ⭸t ρcA − ρ0 VA = 0 which simplifies to

ρ V = ρ0 c

(2.91)

(2.92)

Recalling the definition of the bulk modulus of elasticity, Ev , as Ev =

p ρ/ρ0

(2.93)

VEv p

(2.94)

then Equations 2.92 and 2.93 combine to give c=

and substituting Equation 2.89 into Equation 2.94 yields wave speed, c, in terms of the fluid properties as  Ev (2.95) c= ρ0 The wave speed, c, is also equal to the speed of sound in the fluid, since sound waves are pressure waves. In the case of water at 20◦ C, Ev = 2.15 * 106 kPa, ρ0 = 998 kg/m3 , and therefore c = 1470 m/s (3290 mph). Clearly, the approximation that V V c is reasonable for water in all practical cases.

∗ Also spelled Zhukovsky in some publications (e.g., Simon and Korom, 1997).

Section 2.2

Single Pipelines

55

The foregoing analysis has assumed that the pipe walls are rigid. If the pipe walls are slightly deformable, the wave speed, c, can be shown to be given by (Roberson et al., 1998)  c=

Ev /ρ0 1 + (Ev D/eE)

(2.96)

where D is the diameter of the pipe, e is the wall thickness, and E is the modulus of elasticity of the pipe-wall material. Properties of pipe materials commonly used in engineering applications are given in Appendix E. According to Equation 2.96, in pipes that are not rigid the wave speed, c, decreases as the pipe diameter increases and the wall thickness decreases. In general, the wave speed in a non-rigid pipe (e.g., PVC) is less than the wave speed in a rigid pipe (e.g., steel) with the same diameter and wall thickness. For normal pipe dimensions, the speed, c, of a pressure wave in a water pipe is usually in the range of 600–1200 m/s (1340– 2680 mph), but is always less than 1470 m/s (3290 mph), which is the speed of a pressure wave in free water. The propagation of a pressure wave generated by sudden valve closure is illustrated in Figure 2.10, where a fluid (water) source which maintains an approximately constant pressure p0 is located at a distance L upstream of the valve. The initial condition, at the instant of valve closure, is shown in Figure 2.10(a), where the pressure head in the pipeline decreases with distance from the source due to frictional effects. During the time interval 0 < t < L/c, the pressure wave propagates upstream, as illustrated in Figure 2.10(b), and at t = L/c FIGURE 2.10: Propagation of pressure wave

Hydraulic grade line po

Source Pipeline V

(a) L c

Δp

po

Closed valve

V

V=o

(b)

Pressure wave c

Δp

po V

V=o

(c)

po V

c

Δp

(d) V=o

po V

(e)

Δp

c

V=o

po V (f)

56

Chapter 2

Fundamentals of Flow in Closed Conduits

the pressure wave reaches the source. At this instant, the velocity in the pipe is zero, the pressure in the pipe is p0 + p, and the pressure at the source is p0 . This abrupt pressure change is the same as that which occurred at the valve when the valve was suddenly closed. To equalize the pressure difference between the source and the pipe, the fluid flows with a velocity, V, from the pipe into the source. This causes a pressure wave to reflect back in the direction of the valve (L/c < t < 2L/c), as illustrated in Figure 2.10(c). At t = 2L/c, the wave arrives at the valve, where the velocity must equal zero. Since the velocity, V, in the pipe is directed toward the source, this causes a sudden pressure drop of p, creating another pressure wave where the pressure on the valve side of the wave is p less than the pressure on the source side of the wave. The pressure drop, p, has the same magnitude as the pressure rise when the valve was closed, since the abrupt change in flow velocity, V, is the same in both cases. During time 2L/c < t < 3L/c, this wave travels toward the source, as illustrated in Figure 2.10(d). At t = 3L/c, the pressure wave reaches the source, at which time the pressure in the pipe is p less than the pressure in the source. This pressure difference is the same that occurred at the valve when the valve was closed. To equalize the pressures, fluid enters the pipe with velocity V. During the time interval 3L/c < t < 4L/c, a pressure wave propagates back in the direction of the closed valve, as illustrated in Figure 2.10(e). At t = 4L/c, the pressure wave arrives back at the valve, Figure 2.10(f), and conditions are now exactly the same as immediately after the valve was closed. The cycle then repeats itself until the pressure wave is dissipated by frictional effects. If the valve is closed gradually rather than instantaneously, the pressure increase also occurs gradually. As long as the valve is completely closed in a time less than 2L/c, however, the maximum pressure occurring as a result of valve closure is still equal to p. The critical time of closure or pipe period, tc , is therefore equal to tc =

2L c

(2.97)

If the valve is closed in a time longer than tc , then any pressure increases generated in the pipe are damped by the open valve. Valve closures in less than the pipe period are frequently referred to as rapid valve closures, while valve closures taking longer than the pipe period are called slow valve closures. Rapid valve closure is sometimes defined as closure times less than 10tc , and slow closure corresponding to closure times greater than 10tc (e.g., Chadwick and Morfett, 1998). The pipe period, tc , depends on the pipe elasticity via Equation 2.96, hence non-rigid pipes will have longer pipe periods than rigid pipes. EXAMPLE 2.9 Estimate the maximum water-hammer pressure generated in a rigid pipe where the initial water velocity is 2.5 m/s, the pipe is 5 km long, and a valve at the downstream end of the pipe is closed in 4 seconds. Assume a water temperature of 25◦ C. Solution At 25◦ C, Ev = 2.22 * 106 kPa, and ρ0 = 997.0 kg/m3 (Table B.1, Appendix B). The speed of the pressure wave, c, is given by Equation 2.95 as   Ev 2.22 * 109 = 1490 m/s = c= ρ0 997 The critical time of closure, tc , is given by Equation 2.97 as 2(5000) 2L = = 6.71 s c 1490 Since the closure time of 4 seconds is less than tc , then the maximum pressure increase in the pipe, p, is given by Equation 2.89 as tc =

p = ρ0 cV = (997)(1490)(2.5) = 3.71 * 106 Pa = 3710 kPa This pressure increase is significantly higher than the pressures normally encountered in waterdistribution systems, which are typically less than 620 kPa (90 psi).

Section 2.3

Pipe Networks

57

Pipelines in water-distribution systems experience transient pressures when valves are suddenly closed and/or pumps are suddenly stopped. The possible occurrence of damaging water-hammer pressures are controlled to some extent by keeping the pipeline velocities low. A variety of approaches can be taken to mitigate the effects of water hammer in pipeline systems. In some cases, valves that prevent rapid closure can be used, while in other cases pressure-relief valves, surge tanks, or air chambers are more practical. Pressure relief valves, which are also called surge-relief valves, are designed to open when the pressure in the pipeline at the valve exceeds a specified value. Surge tanks are large open tanks directly connected to the pipeline and are commonly used in hydropower systems. An air chamber is a hydropneumatic tank (containing both water and air) that is connected to the pipeline, often at the discharge side of pumps. 2.3 Pipe Networks Pipe networks are commonly encountered in the context of water-distribution systems. The performance criteria of these systems are typically specified in terms of minimum flow rates and minimum pressure heads that must be maintained at the specified points in the network. Analyses of pipe networks are usually within the context of: (1) designing a new network, (2) designing a modification to an existing network, and/or (3) evaluating the reliability of an existing or proposed network. The procedure for analyzing a pipe network usually aims at finding the flow distribution within the network, with the pressure distribution being derived from the flow distribution using the energy equation. A typical pipe network is illustrated in Figure 2.11, where the boundary conditions consist of inflows, outflows, and constant-head boundaries such as storage reservoirs. Inflows are typically from water-treatment facilities, and outflows from consumer withdrawals or for fire-fighting. All outflows are assumed to occur at network junctions. The basic equations to be satisfied in pipe networks are the continuity and energy equations. The continuity equation requires that, at each junction in the network, the sum of the outflows is equal to the sum of the inflows. This requirement is expressed by the relation 

NP(j)

Qij − Fj = 0,

j = 1, . . . , NJ

(2.98)

i=1

where NP(j) is the number of pipes meeting at junction j; Qij is the flow rate in pipe i at junction j (inflows positive); Fj is the external flow rate (outflows positive) at junction j; and NJ is the total number of junctions in the network. The energy equation requires that the heads at each of the junctions in the pipe network be consistent with the head losses in the pipelines connecting the junctions. There are two principal methods of calculating the flows in pipe networks: the nodal method and the loop method. In the nodal method, the energy equation is expressed in terms of the heads at the network nodes (= junctions), while in the loop method the energy equation is expressed in terms of the flows in closed loops within the pipe network. FIGURE 2.11: Typical pipe network

Qa2

Q3

Q1 Q4

A D

Qa1

A

Qb2

Q5

B

Qb1

B C

Loop Node

(a)

Qd1 Q2

D

Qd2

C

(b)

Qc2

Qc1

58

Chapter 2

Fundamentals of Flow in Closed Conduits

2.3.1

Nodal Method

In the nodal method, the energy equation is written for each pipeline in the network as  h2 = h1 −

  Q|Q| fL Q + + Km hp D |Q| 2gA2

(2.99)

where h1 and h2 are the heads at the upstream and downstream ends of a pipe, respectively; the terms in parentheses measure the friction loss and local losses, respectively; and hp is the head added by pumps in the pipeline. The energy equation given by Equation 2.99 has been modified to account for the fact that the flow direction is commonly unknown, in which case a positive flow direction in each pipeline must be assumed, and a consistent set of energy equations stated for the entire network. Equation 2.99 assumes that the positive flow direction is from node 1 to node 2. Application of the nodal method in practice is usually limited to relatively simple networks.

EXAMPLE 2.10 The high-pressure ductile-iron pipeline shown in Figure 2.12 becomes divided at point B and rejoins at point C. The pipeline characteristics are given in the following tables:

Pipe

Diameter (mm)

Length (m)

Location

Elevation (m)

1 2 3 4

750 400 500 700

500 600 650 400

A B C D

5.0 4.5 4.0 3.5

If the flow rate in Pipe 1 is 2 m3 /s and the pressure at point A is 900 kPa, calculate the pressure at point D. Assume that the flows are fully turbulent in all pipes.

FIGURE 2.12: Pipe network

pA = 900 kPa

pD = ? 2

Flow A

1

C

B

4

D

3

Solution The equivalent sand roughness, ks , of ductile-iron pipe is 0.26 mm, and the pipe and flow characteristics are as follows:

Pipe

Area (m2 )

Velocity (m/s)

ks /D

f

1 2 3 4

0.442 0.126 0.196 0.385

4.53 — — 5.20

0.000347 0.000650 0.000520 0.000371

0.0154 0.0177 0.0168 0.0156

Section 2.3

Pipe Networks

59

where it has been assumed that the flows are fully turbulent. Taking γ = 9.79 kN/m3 , the head at location A, hA , is given by V2 p 900 4.532 hA = A + 1 + zA = + + 5 = 98.0 m γ 2g 9.79 (2)(9.81) and applying the energy equation to each pipe gives the following relationships: Pipe 1:

(0.0154)(500) 4.532 f L V2 hB = hA − 1 1 1 = 98.0 − D1 2g 0.75 (2)(9.81) = 87.3 m

(2.100)

Q22 f L Q22 (0.0177)(600) = 87.3 − Pipe 2: hC = hB − 2 2 2 D2 2gA 0.40 (2)(9.81)(0.126)2 2 = 87.3 − 85.2Q22

(2.101)

Q23 f L Q23 (0.0168)(650) = 87.3 − Pipe 3: hC = hB − 3 3 2 D3 2gA 0.50 (2)(9.81)(0.196)2 3 = 87.3 − 29.0Q23

(2.102)

Q24 f L Q24 (0.0156)(400) Pipe 4: hD = hC − 4 4 = hC − 2 D4 2gA 0.70 (2)(9.81)(0.385)2 4 = hC − 3.07Q24

(2.103)

and the continuity equations at the two pipe junctions are Junction B: Q2 + Q3 = 2 m3 /s

(2.104)

Junction C: Q2 + Q3 = Q4

(2.105)

Equations 2.101 to 2.105 are five equations in five unknowns: hC , hD , Q2 , Q3 , and Q4 . Equations 2.104 and 2.105 indicate that Q4 = 2 m3 /s Combining Equations 2.101 and 2.102 leads to 87.3 − 85.2Q22 = 87.3 − 29.0Q23 and therefore Q2 = 0.583Q3 Substituting Equation 2.106 into Equation 2.104 yields 2 = (0.583 + 1)Q3 or Q3 = 1.26 m3 /s and from Equation 2.106 Q2 = 0.74 m3 /s According to Equation 2.102 hC = 87.3 − 29.0Q23 = 87.3 − 29.0(1.26)2 = 41.3 m and Equation 2.103 gives hD = hC − 3.07Q24 = 41.3 − 3.07(2)2 = 29.0 m Therefore, since the total head at D, hD , is equal to 29.0 m, then V2 p p 5.202 29.0 = D + 4 + zD = D + + 3.5 γ 2g 9.79 (2)(9.81)

(2.106)

60

Chapter 2

Fundamentals of Flow in Closed Conduits which yields pD = 236 kPa Therefore, the pressure at location D is 236 kPa. This problem has been solved by assuming that the flows in all pipes are fully turbulent. This is generally not known a priori, and therefore a complete solution would require that the Colebrook equation also be satisfied in all pipes.

2.3.2

Loop Method

In the loop method, the energy equation is written for each loop of the network, in which case the algebraic sum of the head losses within each loop is equal to zero. This requirement is expressed by the relation 

NP(i)

(hL,ij − hp,ij ) = 0,

i = 1, . . . , NL

(2.107)

j=1

where NP(i) is the number of pipes in loop i, hL,ij is the head loss in pipe j of loop i, hp,ij is the head added by any pumps that may exist in line ij, and NL is the number of loops in the network. Combining Equations 2.98 and 2.107 with an expression for calculating the head losses in pipes, such as the Darcy–Weisbach equation, and the pump characteristic curves, which relate the head added by the pump to the flow rate through the pump, yields a complete mathematical description of the flow problem. Solution of this system of flow equations is complicated by the fact that the equations are nonlinear, and numerical methods must be used to solve for the flow distribution in the pipe network. Hardy Cross method. The Hardy Cross method (Cross, 1936) is a simple technique for manual solution of the loop system of equations governing flow in pipe networks. This iterative method was developed before the advent of computers, and much more efficient algorithms are now used for numerical computations. However, the Hardy Cross method is presented here to illustrate the manual iterative solution of the loop equations in pipe networks, which is sometimes feasible. The Hardy Cross method assumes that the head loss, hL , in each pipe is proportional to the discharge, Q, raised to some power n, in which case hL = rQn

(2.108)

where typical values of n range from 1 to 2, where n = 1 corresponds to viscous flow and n = 2 to fully turbulent flow. The proportionality constant, r, depends on which head-loss equation is used and the types of losses in the pipe. If all head losses are due to friction and the Darcy–Weisbach equation is used to calculate the head losses, then r is given by r=

fL 2gA2 D

(2.109)

ˆ and Q is the error in this estimate, and n = 2. If the flow in each pipe is approximated as Q, ˆ then the actual flow rate, Q, is related to Q and Q by ˆ + Q Q=Q

(2.110)

and the head loss in each pipe is given by hL = rQn ˆ + Q)n = r(Q   n(n − 1) ˆ n + nQ ˆ n−1 Q + ˆ n−2 (Q)2 + · · · + (Q)n =r Q Q 2

(2.111)

Section 2.3

Pipe Networks

61

If the error in the flow estimate, Q, is small, then the higher-order terms in Q can be neglected and the head loss in each pipe can be approximated by ˆ n + rnQ ˆ n−1 Q hL L rQ

(2.112)

This relation approximates the head loss in the flow direction. However, in working with pipe networks, it is required that the algebraic sum of the head losses in any loop of the network (see Figure 2.11) must be equal to zero. We must therefore define a positive flow direction (such as clockwise), and count head losses as positive in pipes when the flow is in the positive direction and negative when the flow is opposite to the selected positive direction. Under these circumstances, the sign of the head loss must be the same as the sign of the flow direction. Further, when the flow is in the positive direction, positive values of Q require a positive correction to the head loss; when the flow is in the negative direction, positive values in Q also require a positive correction to the calculated head loss. To preserve the algebraic relation among head loss, flow direction, and flow error (Q), Equation 2.112 for each pipe can be written as ˆ Q| ˆ n−1 + rn|Q| ˆ n−1 Q hL = rQ|

(2.113)

where the approximation has been replaced by an equal sign. On the basis of Equation 2.113, the requirement that the algebraic sum of the head losses around each loop be equal to zero can be written as  j=1



NP(i)

NP(i)

rij Qj |Qj |n−1 + Qi

rij n|Qj |n−1 = 0,

i = 1, . . . , NL

(2.114)

j=1

where NP(i) is the number of pipes in loop i, rij is the head-loss coefficient in pipe j (in loop i), Qj is the estimated flow in pipe j, Qi is the flow correction for the pipes in loop i, and NL is the number of loops in the entire network. The approximation given by Equation 2.114 assumes that there are no pumps in the loop, and that the flow correction, Qi , is the same for each pipe in each loop. Solving Equation 2.114 for Qi leads to NP(i) Qi = −

rij Qj |Qj |n−1

j=1 NP(i) j=1

(2.115) nrij |Qj |n−1

This equation forms the basis of the Hardy Cross method. The steps to be followed in using the Hardy Cross method to calculate the flow distribution in pipe networks are: Step 1: Assume a reasonable distribution of flows in the pipe network. This assumed flow distribution must satisfy continuity. Step 2: For each loop, i, in the network, calculate the quantities rij Qj |Qj |n−1 and nrij |Qj |n−1 for each pipe in the loop. Calculate the flow correction, Qi , using Equation 2.115. Add the correction algebraically to the estimated flow in each pipe. [Note: Values of rij occur in both the numerator and denominator of Equation 2.115; therefore, values proportional to the actual rij may be used to calculate Qi .] Step 3: Repeat Step 2 until the corrections (Qi ) are acceptably small. The application of the Hardy Cross method is best demonstrated by an example.

62

Chapter 2

Fundamentals of Flow in Closed Conduits

EXAMPLE 2.11 Compute the distribution of flows in the pipe network shown in Figure 2.13(a), where the head loss in each pipe is given by hL = rQ2 and the relative values of r are shown in Figure 2.13(a). The flows are taken as dimensionless for the sake of illustration. FIGURE 2.13: Flows in pipe network

r=1

20

r=6

r=3

50

20

15

1

35

r=2

50

2 II

70

35 I

100

r=5

4

100

30

30

(a)

3

30

(b)

Solution The first step is to assume a distribution of flows in the pipe network that satisfies continuity. The assumed distribution of flows is shown in Figure 2.13(b), along with the positive-flow directions in each of the two loops. The flow correction for each loop is calculated using Equation 2.115. Since n = 2 in this case, the flow correction formula becomes NP(i)

rij Qj |Qj | j=1 Qi = −  NP(i) 2rij |Qj | j=1

The calculation of the numerator and denominator of this flow correction formula for loop I is tabulated as follows: Loop

Pipe

Q

rQ|Q|

2r|Q|

I

4–1 1–3 3–4

70 35 −30

29,400 3675 −4500 28,575

840 210 300 1350

The flow correction for loop I, QI , is therefore given by QI = −

28,575 = −21.2 1350

and the corrected flows are Loop

Pipe

Q

I

4–1 1–3 3–4

48.8 13.8 −51.2

Moving to loop II, the calculation of the numerator and denominator of the flow correction formula for loop II is given by

Section 2.3 Loop

Pipe

Q

rQ|Q|

2r|Q|

II

1–2 2–3 3–1

15 −35 −13.8

225 −2450 −574 −2799

30 140 83 253

Pipe Networks

63

The flow correction for loop II, QII , is therefore given by QII = −

−2799 = 11.1 253

and the corrected flows are Loop

Pipe

Q

II

1–2 2–3 3–1

26.1 −23.9 −2.7

This procedure is repeated in the following table until the calculated flow corrections do not affect the calculated flows, to the level of significant digits retained in the calculations. Iteration

Loop

Pipe

Q

2

I

4–1 1–3 3–4

II

3

I

II

4

I

II

1–2 2–3 3–1

4–1 1–3 3–4

1–2 2–3 3–1

4–1 1–3 3–4

1–2 2–3 3–1

rQ|Q|

2r|Q|

48.8 2.7 −51.2

14,289 22 −13,107 1204

586 16 512 1114

26.1 −23.9 −1.6

681 −1142 −8 −469

52 96 10 157

13,663 6 −13,666 3

573 8 523 1104

847 −874 6 −21

58 84 8 150

13,662 7 −13,668 1

573 9 523 1104

853 −865 7 −5

58 83 9 150

47.7 1.4 −52.3

29.1 −20.9 1.4

47.7 1.5 −52.3

29.2 −20.8 1.5

Q

−1.1

Corrected Q 47.7 1.6 −52.3

29.1 −20.9 1.4 3.0 47.7 1.4 −52.3 0.0 29.2 −20.8 1.5 0.1 47.7 1.5 −52.3 0.0 29.2 −20.8 1.5 0.0

64

Chapter 2

Fundamentals of Flow in Closed Conduits The final flow distribution, after four iterations, is given by Pipe

Q

1–2 2–3 3–4 4–1 1–3

29.2 −20.8 −52.3 47.7 −1.5

It is clear that the final results are fairly close to the flow estimates after only one iteration.

As the above example illustrates, complex pipe networks can generally be treated as a combination of simple loops, with each loop balanced in turn until compatible flow conditions exist in all loops. Typically, after the flows have been computed for all pipes in a network, the elevation of the hydraulic grade line and the pressure are computed for each junction node. These pressures are then assessed relative to acceptable operating pressures. 2.3.3

Application of Computer Programs

In practice, analyses of complex water-distribution networks are usually done using computer programs that solve the system of continuity and energy equations that govern the flows in the network pipelines. These computer programs, such as EPANET 2 (Rossman, 2000), generally use algorithms that are computationally more efficient than the Hardy Cross method, such as the linear theory method, the Newton–Raphson method, and the gradient algorithm (Lansey and Mays, 1999). All algorithms should lead to the same result, although the speed of convergence to the result is different. The methods described in this text for computing steady-state flows and pressures in water-distribution systems are useful for assessing the performance of systems under normal operating conditions. Sudden changes in flow conditions, such as pump shutdown/startup and valve opening/closing, cause hydraulic transients that can produce significant increases in water pressure associated with water hammer. The analysis of transient conditions requires a computer program to perform a numerical solution of the one-dimensional continuity and momentum equation for flow in pipelines, and is an essential component of water-distribution system design (Wood, 2005). Transient conditions will be most severe at pump stations and control valves, high-elevation areas, locations with low static pressures, and locations that are far from elevated storage reservoirs. 2.4 Pumps Pumps are hydraulic machines that convert mechanical energy to fluid energy. They can be classified into two main categories: (1) positive-displacement pumps, and (2) rotodynamic or kinetic pumps. Positive-displacement pumps deliver a fixed quantity of fluid with each revolution of the pump rotor, such as with a piston or cylinder, while rotodynamic pumps add energy to the fluid by accelerating it through the action of a rotating impeller. Rotodynamic pumps are far more common in engineering practice and will be the focus of this section. Three types of rotodynamic pumps commonly encountered are: centrifugal pumps, axial-flow pumps, and mixed-flow pumps. In centrifugal pumps, the flow enters the pump chamber along the axis of the impeller and is discharged radially by centrifugal action, as illustrated in Figure 2.14(a). In axial-flow pumps, the flow enters and leaves the pump chamber along the axis of the impeller, as shown in Figure 2.14(b). In mixed-flow pumps, outflows have both radial and axial components. Schematic diagrams of typical centrifugal and axial-flow pump installations are illustrated in Figure 2.15. Key components of the centrifugal pump are a foot valve installed in the suction pipe to prevent water from leaving the pump when it is stopped and a check valve in the discharge pipe to prevent backflow if there is a power failure. If the suction line is empty prior to starting the pump, then the suction line must be primed (filled) prior to startup. Unless the water is known to be very clean, a strainer should be installed at the inlet to the suction piping. The pipe size of the suction line should never be smaller than the inlet connection on the pump; if a reducer

Section 2.4 FIGURE 2.14: Types of pumps

Pumps

65

A Impeller

Flow

Flow

Eye of impeller

Eye of impeller

Casing

View A–A

A (a) Centrifugal pump

ω Flow

Input shaft

Outlet vane Inlet vane

Impeller

(b) Axial-flow pump

is required, it should be of the eccentric type, since concentric reducers place part of the supply pipe above the pump inlet where an air pocket could form. The discharge line from the pump should contain a valve close to the pump to allow service or pump replacement. A typical centrifugal pump is shown in Figure 2.16(a) and a typical axial-flow pump is shown in Figure 2.16(b). For the centrifugal pump shown in Figure 2.16(a), the inflow enters into the lower pipe opening and the outflow exits from the upper pipe opening, and the motor driving the pump is shown in the background. For the axial-flow pump shown in Figure 2.16(b), the inflow enters below the water level and the outflow exits horizontally, just downstream of the motor which juts out at the pipe bend. The pumps illustrated in Figure 2.15 are both single-stage pumps, which means that they have only one impeller. In multistage pumps, two or more impellers are arranged in series in such a way that the discharge from one impeller enters the eye of the next impeller. If a pump has three impellers in series, it is called a three-stage pump. Multistage pumps are typically used when large pumping heads are required, and are commonly used when extracting water from deep underground sources. The performance of a pump is measured by the head added by the pump and the pump efficiency. The head added by the pump, hp , is equal to the difference between the total head on the discharge side of the pump and the total head on the suction side of the pump, and is sometimes referred to as the total dynamic head (TDH). The efficiency of the pump, η, is defined by η=

power delivered to the fluid power supplied to the shaft

(2.116)

66

Chapter 2

Fundamentals of Flow in Closed Conduits

FIGURE 2.15: Schematic illustrations of centrifugal and axial-flow pump installations

Reducer (if necessary)

P U M P

Flow Check valve

Gate valve (service)

Foot valve Flow

Strainer (a) Centrifugal pump

Motor

Flow Guide vanes Impeller

Flow

Intake

(b) Axial-flow pump

FIGURE 2.16: (a) Centrifugal pump; and (b) axial-flow pump

(a)

(b)

Section 2.4

Pumps

67

Pumps are inefficient for a variety of reasons, such as frictional losses as the fluid moves over the solid surfaces, separation losses, leakage of fluid between the impeller and the casing, and mechanical losses in the bearings and sealing glands of the pump. The pump-performance parameters, hp and η, can be expressed in terms of the fluid properties and the physical characteristics of the pump by the functional relation or

ghp

η = f1 (ρ, μ, D, ω, Q)

(2.117)

where the energy added per unit mass of fluid, ghp , is used instead of hp (to remove the effect of gravity), f1 is an unknown function, ρ and μ are the density and dynamic viscosity of the fluid, respectively, Q is the flow rate through the pump, D is a characteristic dimension of the pump (usually the inlet or outlet diameter), and ω is the angular speed of the pump impeller. Equation 2.117 is a functional relationship between six variables in three dimensions. According to the Buckingham pi theorem, this relationship can be expressed as a relation between three dimensionless groups as follows:   ghp Q ρωD2 or η = f2 , (2.118) μ ω2 D2 ωD3 where ghp /ω2 D2 is called the head coefficient, and Q/ωD3 is called the flow coefficient (Douglas et al., 1995) or discharge coefficient (Cruise et al., 2007). In most cases, the flow through the pump is fully turbulent and viscous forces are negligible relative to the inertial forces. Under these circumstances, the viscosity of the fluid is neglected and Equation 2.118 becomes   ghp Q (2.119) or η = f 3 ω2 D2 ωD3 This relationship describes the performance of all (geometrically similar) pumps in which viscous effects are negligible, but the exact form of the function in Equation 2.119 depends on the geometry of the pump. A series of pumps having the same shape (but different sizes) are expected to have the same functional relationships between ghp /(ω2 D2 ) and Q/(ωD3 ) as well as η and Q/(ωD3 ). A class of pumps that have the same shape (i.e., are geometrically similar) is called a homologous series, and the performance characteristics of a homologous series of pumps are described by curves such as those in Figure 2.17. Pumps are selected to meet specific design conditions and, since the efficiency of a pump varies with the operating condition, it is usually desirable to select a pump that operates at or near the point of maximum efficiency, indicated by the point P in Figure 2.17. The point of maximum efficiency of a pump is commonly called the best-efficiency point (BEP), and sometimes the nameplate or design point. Maintaining operation near the BEP will allow a pump to function for years with little maintenance, and as the operating point moves away from the BEP, pump thrust and radial loads increase, which increases the wear on the pump bearings and shaft. For these reasons, it is generally recommended that pump operation should be maintained between 70% and 130% of the BEP flow rate (Lansey and El-Shorbagy, 2001). At the BEP in Figure 2.17, FIGURE 2.17: Performance curves of a homologous series of pumps

hmax K1

P

ghp

Efficiency curve h

ω2D 2

K2 Q ωD 3

68

Chapter 2

Fundamentals of Flow in Closed Conduits

ghp = K1 ω2 D2

Q = K2 ωD3

and

(2.120)

Eliminating D from these equations yields & 'K ' 2 =( 3 3 (ghp ) 4 K12 1

ωQ 2

(2.121)

The term on the right-hand side of this equation is a constant for a homologous series of pumps and is denoted by the specific speed, ns , defined by 1

ns =

ωQ 2 3

(2.122)

(ghp ) 4

Typical SI units used in calculating the specific speed, ns , are ω in rad/s, Q in m3 /s, g in m/s2 , and hp in m; however, since ns is dimensionless, any consistent set of units can be used. The specific speed, ns , is based on the parameters at the most efficient operating point of a homologous series, and so ns is the basis for selecting the appropriate homologous series for any desired set of operating conditions. The nomenclature of calling ns the specific speed is somewhat unfortunate, since ns is dimensionless and hence does not have units of speed. The specific speed, ns , is also called the shape number (Hwang and Houghtalen, 1996; Wurbs and James, 2002) or the type number (Douglas et al., 2001), where the latter terms are perhaps preferable in that ns is more used to select the shape or type of the required pump rather than the speed of the pump. In lieu of defining the specific speed, ns by Equation 2.122, it is also common practice to define the specific speed, Ns , as 1

Ns =

ωQ 2 3

(2.123)

hp4 where Ns is not dimensionless, and so when Equation 2.123 is used to define the specific speed, the units of ω, Q, and hp must be explicitly specified. In the United States, ω is normally expressed in revolutions per minute (rpm), Q is in gallons per minute (gpm), and hp is in feet (ft). In the United Kingdom, ω is normally expressed in revolutions per minute (rpm), Q is in liters per second (L/s), and hp is in meters (m). Be cautious, although Ns has dimensions, the units are seldom stated in practice. The required pump operating point gives the flow rate, Q, and head, hp , required from the pump; the rotational speed, ω, is determined by the synchronous speeds of available motors; and the specific speed calculated from the required operating point is the basis for selecting the appropriate pump type. Since the specific speed is independent of the size of a pump, and all homologous pumps (of varying sizes) have the same specific speed, then the calculated specific speed at the desired operating point indicates the type of pump that must be selected to ensure optimal efficiency. The types of pump that give the maximum efficiency for given specific speeds, ns , are listed in Table 2.3 along with typical flow rates delivered by the pumps. Table 2.3 indicates that centrifugal pumps have low specific speeds, ns < 1.5; mixed-flow pumps have medium specific speeds, 1.5 < ns < 3.7; and axial-flow pumps have high specific speeds, ns > 3.7. This indicates that centrifugal pumps are most efficient at delivering low flows at high heads, while axial-flow pumps are most efficient at delivering high flows at low heads. The efficiencies of radial-flow (centrifugal) pumps increase with increasing specific speed, while the efficiencies of mixed-flow and axial-flow pumps decrease with increasing specific speed. Pumps with specific speeds less than 0.3 tend to be inefficient (Finnemore and Franzini, 2002). Since axial-flow pumps are most efficient at delivering high flows at low heads, this type of pump is commonly used to move large volumes of water through major canals, and an example of

Section 2.4

Pumps

69

TABLE 2.3: Pump Selection Guidelines

Range of specific speeds, ns (dimensionless)

Type of pump Centrifugal Mixed flow Axial flow

Typical flow rates (L/s) (gpm) 300

0.15–1.5 1.5–3.7 3.7–5.5

Typical efficiencies (%)

4750

70–94 90–94 84–90

FIGURE 2.18: Axial-flow pump operating in a canal Source: South Florida Water Management District.

this application is shown in Figure 2.18, where there are three axial-flow pumps operating in parallel, and these pumps are driven by motors housed in the pump station. Most pumps are driven by standard electric motors. The standard speed of AC synchronous induction motors at 60 Hz and 220 to 440 volts is given by Synchronous speed (rpm) =

3600 no. of pairs of poles

(2.124)

A common problem is that, for the motor speed chosen, the calculated specific speed does not exactly equal the specific speed of available pumps. In these cases, it is recommended to either choose a pump with a specific speed that is close to and greater than the required specific speed or use a variable-speed motor in the pump. The efficiencies of variable-speed motors (which require a converter) can be significantly lower for lower speeds, which may in turn significantly affect the overall efficiency of the pump (Ulanicki et al., 2008). In rare cases, a new pump may be designed to meet the design conditions exactly; however, this is usually very costly and only justified for very large pumps. 2.4.1

Affinity Laws

The performance curves for a homologous series of pumps are illustrated in Figure 2.17. Any two pumps in the homologous series are expected to operate at the same efficiency when 

Q ωD3



 =

1

Q ωD3



 and 2

hp ω2 D2



 =

1

hp ω2 D2

 (2.125) 2

where the subscripts 1 and 2 designate different pumps (i.e., Pump 1 and Pump 2) within the same homologous series. The relationships given in Equation 2.125 are sometimes called the affinity laws for homologous pumps. An affinity law for the power delivered to the fluid, P, can be derived from the affinity relations given in Equation 2.125, since P is defined by P = γ Qhp

(2.126)

70

Chapter 2

Fundamentals of Flow in Closed Conduits

which leads to the following derived affinity relation: 

P 3 ω D5



 =

1

P 3 ω D5

 (2.127) 2

In accordance with the dimensional analysis of pump performance, Equation 2.118, the affinity laws for scaling pump performance within any homologous series are valid as long as viscous effects are negligible. The effect of viscosity is measured by the Reynolds number, Re, defined by ρωD2 Re = (2.128) μ and scale effects are negligible when Re > 3 * 105 (Gerhart et al., 1992). In lieu of stating a Reynolds number criterion for scale effects to be negligible, it is sometimes stated that larger pumps are more efficient than smaller pumps and that the scale effect on efficiency is given by (Stepanoff, 1957; Moody and Zowski, 1989) 1 − η2 = 1 − η1



D1 D2

1 4

(2.129)

where η1 and η2 are the efficiencies of homologous pumps of diameters D1 and D2 , respectively. The effect of changes in flow rate on efficiency is sometimes estimated using the relation   0.94 − η2 Q1 0.32 = (2.130) 0.94 − η1 Q2 where Q1 and Q2 are corresponding homologous flow rates. EXAMPLE 2.12 A homologous series of centrifugal pumps has a specific speed of 1.1 and are driven by 2400-rpm motors. For a 400-mm size within this series, the manufacturer claims that the best efficiency of 85% occurs when the flow rate is 500 L/s and the head added by the pump is 89.5 m. What would be the best-efficiency operating point for a 300-mm size within this homologous series, and estimate the corresponding efficiency. Solution From the given data: ns = 1.1, ω1 = ω2 = 2400 rpm, D1 = 400 mm, D2 = 300 mm, Q1 = 500 L/s, hp1 = 89.5 m, and η1 = 85%. Applying the affinity relationships given by Equation 2.125 requires that     Q2 Q1 = (ω1 )(D1 )3 (ω2 )(D2 )3     Q2 500 = (2400)(400)3 (2400)(300)3 which yields Q2 = 210 L/s. Also,





hp1 (ω1 )2 (D1 )2

89.5 (2400)2 (400)2



 =



 =

hp2



(ω2 )2 (D2 )2 hp2



(2400)2 (300)2

which yields hp2 = 50.3 m. Therefore, the best-efficiency operating point for a 300-mm pump in the given homologous series is at Q = 210 L/s and hp = 50.3 m. The efficiency at this operating point can be estimated using Equation 2.129 which gives

Section 2.4 1 − η2 = 1 − η1 1 − η2 = 1 − 0.85

 

D1 D2

Pumps

71

1

400 300

4

1 4

which yields η2 = 0.84. The efficiency can also be estimated by Equation 2.130 which gives   0.94 − η2 Q1 0.32 = 0.94 − η1 Q2   500 0.32 0.94 − η2 = 0.94 − 0.85 210 which yields η2 = 0.82. Based on these results, it is estimated that the 300-mm pump will have an efficiency somewhere in the range of 82%–84%.

2.4.2

Pump Selection

In selecting a pump for any application, consideration must be given to the required pumping rate, commercially available pumps, characteristics of the system in which the pump operates, and the physical limitations associated with pumping water. 2.4.2.1

Commercially available pumps

Specification of a pump generally requires selection of a manufacturer, model (homologous) series, size (D), and rotational speed (ω). For each model series, size, and rotational speed, pump manufacturers usually provide a performance curve or characteristic curve that shows the relationship between the head, hp , added by the pump and the flow rate, Q, through the pump. A typical example of a set of pump characteristic curves (hp versus Q) provided by a manufacturer for a homologous series of pumps is shown in Figure 2.19. In this case, FIGURE 2.19: Pump performance curve Source: Goulds Pumps (www.gouldspumps.com).

Goulds Pumps ITT Industries

RPM 885 Model: 3409 Size: 14 16 -17 Imp. Dwg: 08217815 Pattern: P-2599 Eye Area: 171.0 in2

ft 17.5 in

80

55% 65%

16.6 in

70

15.7 in

60

14.8 in

73% 79%

83% 85% 86%

85% 83% 79%

13.9 in 13 in 12 in

50 40

A-7874-8

CDS

CENTRIFUGAL PUMP CHARACTERISTICS

m 25

20

15

73% (Efficiency) 10

30 0ft (Pump curve)

20

5

10

7ft

10 ft

16 ft (Required NPSH)

hp 17.5 in

(Power curve)

75

15.7 in

50 25 0

0

1000

0

250

2000 500

3000 750

4000 1000

40

13.9 in

12.1 in 5000

6000

1250

kW 60

20 7000 1500

8000 gpm(US) 1750 m3/hr

0

72

Chapter 2

Fundamentals of Flow in Closed Conduits

the homologous series of pumps (Model 3409) has impeller diameters ranging from 12.1 in. (307 mm) to 17.5 in. (445 mm) with a rotational speed of 885 revolutions per minute. Superimposed on the characteristic curves are constant-efficiency lines for efficiencies ranging from 55% to 86%, and (dashed) isolines of required net positive suction head, which is the minimum allowable difference between the head on the suction side of the pump and the pressure head at which water vaporizes (i.e., saturation vapor pressure). In Figure 2.19, the required net positive suction head ranges from 16 ft (4.9 m) for higher flow rates down to approximately zero, which is indicated by a bold line that meets the 55% efficiency contour. Also shown in Figure 2.19, below the characteristic curves, is the power delivered to the pump (in kW) for various flow rates and impeller diameters. This power input to the pump shaft is called the brake horsepower (BHP) and is defined as BHP =

γ Qhp η

(2.131)

where BHP is in kW, γ is the specific weight of water in kN/m3 , Q is the discharge rate in m3 /s, hp is the head added by the pump in m, and η is the pump efficiency, which is dimensionless. Key points on any pump characteristic curve are the shutoff head, which is the head added by the pump when the discharge is equal to zero, and the rated capacity, which is the discharge when the pump is operating at maximum efficiency. 2.4.2.2

System characteristics

The goal in pump selection is to select a pump that operates at a point of maximum efficiency and with a net positive suction head that exceeds the minimum allowable value. Pumps are placed in pipeline systems such as that illustrated in Figure 2.20, in which case the energy equation for the pipeline system requires that the head, hp , added by the pump is given by 2

hp = z + Q

 

 Km fL + 2 2gA D 2gA2

 (2.132)

where z is the difference in elevation between the water surfaces of the source and destination reservoirs, the first term in the square brackets is the sum of the head losses due to friction, and the second term is the sum of the local head losses. Equation 2.132 gives the required relationship between hp and Q for the pipeline system, and this relationship is commonly called the system curve. Because the flow rate and head added by the pump must satisfy both the system curve and the pump characteristic curve, Q and hp are determined by simultaneous solution of Equation 2.132 and the pump characteristic curve. The resulting (solution) values of Q and hp identify the operating point of the pump. The location of the operating point on the performance curve is illustrated in Figure 2.21. FIGURE 2.20: Pipeline system

Q

Δz P Pump

Q

Section 2.4

73

Operating point

Pump head, hp

FIGURE 2.21: Operating point in pipeline system

Pumps

System curve Pump characteristic curve

Δz 0

2.4.2.3

Flow rate, Q

Limits on pump location

If the absolute pressure on the suction side of a pump falls below the saturation vapor pressure of water, the water will begin to vaporize, and this process of vaporization is called cavitation. Cavitation is usually a transient phenomenon that occurs as water enters the low-pressure suction side of a pump and experiences the even lower pressures adjacent to the rotating pump impeller. As the water containing vapor cavities moves toward the highpressure environment of the discharge side of the pump, the vapor cavities are compressed and ultimately implode, creating small, localized high-velocity jets that can cause considerable damage to the pump machinery. Collapsing vapor cavities have been associated with jet velocities on the order of 110 m/s (360 ft/s), and pressures of up to 800 MPa (116,000 psi) when the jets strike a solid wall (Knapp et al., 1970; Finnemore and Franzini, 2002). The damage caused by collapsing vapor cavities usually manifests itself as pitting of the metal casing and impeller, reduced pump efficiency, and excessive vibration of the pump. The noise generated by imploding vapor cavities resembles the sound of gravel going through a centrifugal pump. Since the saturation vapor pressure increases with temperature, a system that operates satisfactorily without cavitation during the winter may have problems with cavitation during the summer. The potential for cavitation is measured by the available net positive suction head, NPSHA , defined as the difference between the head on the suction side of the pump (at the inlet to the pump) and the head when cavitation begins, hence  NPSHA =

ps V2 + s + zs γ 2g



 −

pv + zs γ

 =

ps V2 pv + s − γ 2g γ

(2.133)

where ps , Vs , and zs are the pressure, velocity, and elevation, respectively, of the fluid at the suction side of the pump, and pv is the saturation vapor pressure of water at the temperature of the water. In cases where water is being pumped from a reservoir, the NPSHA can be calculated by applying the energy equation between the reservoir and the suction side of the pump; in this case the available net positive suction head is given by NPSHA =

p0 pv − zs − hL − γ γ

(2.134)

where p0 is the pressure at the surface of the reservoir (usually atmospheric), zs is the difference in elevation between the suction side of the pump and the water surface in the source reservoir (called the suction lift or static suction head or static head), hL is the head loss in the pipeline between the source reservoir and suction side of the pump (including local losses), and pv is the saturation vapor pressure. In applying either Equation 2.133 or 2.134 to calculate NPSHA , care must be taken to use a consistent measure of the pressures, using either gage pressures or absolute pressures. Absolute pressures are usually more convenient, since the vapor pressure is typically expressed as an absolute pressure. A pump requires a minimum NPSHA to prevent the onset of cavitation within the pump, and this minimum NPSHA is called the required net positive suction head, NPSHR ,

74

Chapter 2

Fundamentals of Flow in Closed Conduits

which is generally a function of the head, hp , added by the pump. This relationship is commonly expressed in terms of the constant cavitation number, σ , where NPSHR = σ hp

(2.135)

Pump manufacturers either present curves showing the relationship between NPSHR and hp or they provide values of σ for each pump. EXAMPLE 2.13

Water at 20◦ C is being pumped from a lower to an upper reservoir through a 200-mm pipe in the system shown in Figure 2.20. The water-surface elevations in the source and destination reservoirs differ by 5.2 m, and the length of the steel pipe (ks = 0.046 mm) connecting the reservoirs is 21.3 m. The pump is to be located 1.5 m above the water surface in the source reservoir, and the length of the pipeline between the source reservoir and the suction side of the pump is 3.5 m. The performance curves of the 885-rpm homologous series of pumps being considered for this system are given in Figure 2.19. If the desired flow rate in the system is 315 L/s (= 5000 gpm), what size and specific-speed pump should be selected? Assess the adequacy of the pump location based on a consideration of the available net positive suction head. The pipe intake loss coefficient can be taken as 0.1. Solution For the system pipeline: L = 21.3 m, D = 200 mm = 0.2 m, ks = 0.046 mm, and (neglecting local head losses) the energy equation for the system is given by fL Q2 2gA2 D

hp = 5.2 +

(2.136)

where hp is the head added by the pump (in meters), f is the friction factor, Q is the flow rate through the system (in m3 /s), and A is the cross-sectional area of the pipe (in m2 ) given by π π (2.137) A = D2 = (0.2)2 = 0.03142 m2 4 4 The friction factor, f , can be calculated using the Swamee–Jain formula (Equation 2.39), f = 

0.25 

log

ks 5.74 3.7D + Re0.9

2

(2.138)

where Re is the Reynolds number given by Re =

QD VD = ν Aν

(2.139)

and ν = 1.00 * 10−6 m2 /s at 20◦ C. Combining Equations 2.138 and 2.139 with the given data yields f = ⎡

0.25

⎞⎤2 ⎟⎥ ⎥ 0.9 ⎟ ⎠⎥ ⎦

(2.140)

4[log(6.216 * 10−5 + 4.32 * 10−6 Q−0.9 )]2

(2.141)



⎢ ⎜ 4.6 * 10−5 ⎢ ⎢log ⎜ ⎝ 3.7(0.2) +  ⎣

5.74 Q(0.2) (0.03142)(1.00 * 10−6 )

which simplifies to f =

1

Combining Equations 2.136 and 2.141 gives the following relation: hp = 5.2 + = 5.2 +

(21.3) 2(9.81)(0.03142)2 (0.2)(4)[log(6.216 * 10−5 + 4.32 * 10−6 Q−0.9 )]2 1375Q2 [log(6.216 * 10−5 + 4.32 * 10−6 Q−0.9 )]2

Q2 (2.142)

Section 2.4

Pumps

75

This relation is applicable for hp in meters and Q in m3 /s. Since 1 m = 3.281 ft and 1 m3 /s = 15,850 gpm, Equation 2.142 can be put in the form ) *2 Q 1375 15,850 hp = 5.2 +    3.281 ) *−0.9  2 Q −5 −6 + 4.32 * 10 log 6.216 * 10 15,850 which simplifies to hp = 17.1 +

1.79 * 10−5 Q2 [log(6.216 * 10−5 + 7.96 * 10−3 Q−0.9 )]2

(2.143)

where hp is in ft and Q in gpm. Equation 2.143 is the “system curve” which relates the head added by the pump to the flow rate through the system, as required by the energy equation. Since the pump characteristic curve must also be satisfied, the operating point of the pump is at the intersection of the system curve (Equation 2.143) and the pump characteristic curve given in Figure 2.19. The system curve and the pump curves are both plotted in Figure 2.22, and the operating points for the various pump sizes are listed in the following table: Pump Size (in.)

Operating Point (gpm)

12.1 13 13.9 14.8 15.7 16.6 17.5

2900 3400 3900 4400 4850 5450 5850

Since the desired flow rate in the system is 5000 gpm (315 L/s), the 16.6-in. (42.2-cm) pump should be selected. This 16.6-in. pump will deliver 5450 gpm (344 L/s) when all system valves are open, and can be throttled down to 5000 gpm as required. If a closer match between the desired flow rate and the operating point is desired for the given system, then an alternative series of homologous pumps should be considered. For the selected 16.6-in. pump, the maximum efficiency point is at Q = 5000 gpm, hp = 52 ft (15.8 m), and ω = 885 rpm; hence the specific speed, Ns , of the selected pump (in U.S. Customary units) is given by 1

Ns =

ωQ 2

1

=

(885)(5000) 2

= 3232 3 3 (52) 4 hp4 Comparing this result with the pump-selection guidelines in Table 2.3 confirms that the pump being considered must be a centrifugal pump. The available net positive suction head, NPSHA , is defined by Equation 2.134 as p pv NPSHA = 0 − zs − hL − (2.144) γ γ Atmospheric pressure, p0 , can be taken as 101 kPa; the specific weight of water, γ , is 9.79 kN/m3 ; the suction lift, zs , is 1.5 m; and at 20◦ C, the saturated vapor pressure of water, pv , is 2.34 kPa. The head loss, hL , is estimated as   fL V 2 (2.145) hL = 0.1 + D 2g where the entrance loss at the pump intake is 0.1 V 2 /2g. For a flow rate, Q, equal to 5450 gpm (344 L/s), Equation 2.141 gives the friction factor, f , as 1 f = 4[log(6.216 * 10−5 + 4.32 * 10−6 Q−0.9 )]2 =

1 4[log(6.216 * 10−5 + 4.32 * 10−6 (0.344)−0.9 )]2

= 0.00366

76

Chapter 2

Fundamentals of Flow in Closed Conduits

FIGURE 2.22: Pump operating points

Goulds Pumps ITT Industries

RPM 885 Model: 3409 Size: 14 16 -17 Imp. Dwg: 08217815 Pattern: P-2599 Eye Area: 171.0 in2

ft 17.5 in

80

55% 65% 73%

16.6 in

70

15.7 in

60

14.8 in

79%

83%

85% 86%

85% 83% 79% 73%

13.9 in 13 in 12 in

50 40 30

A-7874-8

CDS

Centrifugal pump characteristics

10

20 System curve

5

10 ft 16 ft

7ft

hp 17.5 in

75 15.7 in

50 25 0

1000

0

250

2000 500

3000

4000

750

5000

1000

6000

1250

kW 60 40

13.9 in

12.1 in 0

20

15

0ft

10

m 25

20 7000 1500

8000 gpm(US)

0

1750 m3/hr

and the average velocity of flow in the pipe, V, is given by V=

0.344 Q = = 10.9 m/s A 0.03142

Substituting f = 0.00366, L = 3.5 m, D = 0.2 m, and V = 10.9 m/s into Equation 2.145 yields   (0.00366)(3.5) 10.92 hL = 0.1 + = 3.44 m 0.2 2(9.81) and hence the available net positive suction head, NPSHA (Equation 2.144) is NPSHA =

2.34 101 − 1.5 − 3.44 − = 5.13 m 9.79 9.79

According to the pump properties given in Figure 2.22, the required net positive suction head, NPSHR , for the 16.6-in. pump at the operating point is 12 ft (= 3.66 m). Since the available net positive suction head (5.13 m) is greater than the required net positive suction head (3.66 m), the pump location relative to the intake reservoir is adequate and cavitation problems are not expected.

2.4.3

Multiple-Pump Systems

In cases where a single pump is inadequate to achieve a desired operating condition, multiple pumps can be used. Combinations of pumps are referred to as pump systems, and the pumps within these systems are typically arranged either in series or in parallel. The characteristic curve of a pump system is determined by the arrangement of pumps. Consider the case of two identical pumps in series, illustrated in Figure 2.23(a). The flow through each pump is equal to Q, and the head added by each pump is hp . For the two-pump system, the flow through the system is equal to Q and the head added by the system is 2hp . Consequently, the characteristic curve of the two-pump (in series) system is related to the characteristic curve of each pump in that for any flow, Q, the head added by the system is twice the head added by a single pump, and the relationship between the single-pump

Section 2.4 FIGURE 2.23: Pumps in series

hp

Pumps

77

EGL Two pumps in series

hp hp Q

Q

Q

P

P

One pump

Q

Pump system

Q

(a)

(b)

characteristic curve and the two-pump characteristic curve is illustrated in Figure 2.23(b). This analysis can be extended to cases where the pump system contains n identical pumps in series, in which case the n-pump characteristic curve is derived from the single-pump characteristic curve by multiplying the ordinate of the single-pump characteristic curve (hp ) by n. Pump systems that include multiple pumps in series are called multistage pump systems, and pumps that include multiple smaller pumps within a single housing are called multistage pumps. Multistage pump systems are commonly used in applications involving unusually high heads, and multistage pumps are commonly used to pump water from deep boreholes. The case of two identical pumps arranged in parallel is illustrated in Figure 2.24. In this case, the flow through each pump is Q and the head added is hp ; therefore, the flow through the two-pump system is equal to 2Q, while the head added is hp . Consequently, the characteristic curve of the two-pump system is derived from the characteristic curve of the individual pumps by multiplying the abscissa (Q) by two. This is illustrated in Figure 2.24(b). In a similar manner, the characteristic curves of systems containing n identical pumps in parallel can be derived from the single-pump characteristic curve by multiplying the abscissa (Q) by n. Pumps in parallel are used in cases where the desired flow rate is beyond the range of a single pump and also to provide flexibility in pump operations, since some pumps in the system can be shut down during low-demand conditions or for service. This arrangement is common in sewage pump stations and water-distribution systems, where flow rates vary significantly during the course of a day. When pumps are placed either in series or in parallel, it is usually desirable that these pumps be identical; otherwise, the pumps will be loaded unequally and the efficiency of the pump system will be less than optimal. In cases where nonidentical pumps are placed in series, the characteristic curve of the pump system is obtained by summing the heads added by the individual pumps for a given flow rate. In cases where nonidentical pumps are placed in parallel, the characteristic curve of the pump system is obtained by summing the flow rates through the individual pumps for a given head.

FIGURE 2.24: Pumps in parallel

hp

EGL

Q P 2Q

Q P

2Q

Two pumps in parallel hp One pump

Pump system

Q

(a)

(b)

78

Chapter 2

Fundamentals of Flow in Closed Conduits

EXAMPLE 2.14 If a pump has a performance curve described by the relation hp = 12 − 0.1Q2 then what is the performance curve for: (a) a system having three of these pumps in series; and (b) a system having three of these pumps in parallel? Solution (a) For a system with three pumps in series, the same flow, Q, goes through each pump, and each pump adds one-third of the head, Hp , added by the pump system. Therefore, Hp = 12 − 0.1Q2 3 and the characteristic curve of the pump system is Hp = 36 − 0.3Q2 (b) For a system consisting of three pumps in parallel, one-third of the total flow, Q, goes through each pump, and the head added by each pump is the same as the total head, Hp added by the pump system. Therefore  2 Q Hp = 12 − 0.1 3 and the characteristic curve of the pump system is Hp = 12 − 0.011Q2

2.4.4

Variable-Speed Pumps

In variable-speed pumps, the rotational speed can be adjusted, in contrast to having only on– off modes. For any homologous (i.e., geometrically similar) series of pumps, the performance characteristics of any pump within the series is given by   ghp Q =f ω2 D2 ωD3 where f (Q/ωD3 ) is a function that is unique to a homologous series and is the same for all pumps within the series. For a fixed pump size, D, for two different motor speeds, ω1 and ω2 , there will be different performance curves corresponding to the different speeds, and these performance curves can be expressed as   h1 Q1 =f ω1 D3 ω12 D2 and



h2 ω22 D2

=f

Q2 ω2 D3



where h1 and Q1 are the corresponding head and flow rate when the rotational speed is ω1 , and h2 and Q2 are corresponding head and flow rate when the rotational speed is ω2 . Since the pumps are part of a homologous series, then, neglecting scale effects, the function f is fixed, and therefore when (for a fixed D) Q1 Q2 = ω1 ω2

(2.146)

Section 2.4

79

ω2 > ω1 Pump head, hp

FIGURE 2.25: Pump curves for variable-speed motor

Pumps

h2 h1

P2 @ ω2

P1 @ ω1 Q1 Q2 Flow rate, Q

then

h1 ω12

=

h2

(2.147)

ω22

These relationships are used to relate the performance curve for any pump operating at ω1 to the performance curve of the same pump operating at ω2 as shown in Figure 2.25. P1 is a point on the ω1 operating curve with coordinates (Q1 ,h1 ), and P2 is a point on the ω2 operating curve with coordinates (Q2 ,h2 ), then according to Equations 2.146 and 2.147 these coordinates are related by  Q2 = Q1

ω2 ω1



 and

h2 = h1

ω2 ω1

2 (2.148)

It should be noted from these results that increased flows will generally be attained with increased rotational speed, and that variable-speed pumps provide an option for adjusting the operating point in pump-pipeline systems. EXAMPLE 2.15 A pump with a 1200-rpm motor has a performance curve of hp = 12 − 0.1Q2 where hp is in meters and Q is in cubic meters per minute. If the speed of the motor is changed to 2400 rpm, estimate the new performance curve. Solution From the given data: ω1 = 1200 rpm, ω2 = 2400 rpm, and the affinity laws (Equation 2.148) give that Q1 =

h1 =

ω1 1200 Q = 0.5Q2 Q = ω2 2 2400 2 ω12 ω22

h2 =

12002 h2 = 0.25h2 24002

Since the performance curve of the pump at speed ω1 is given by h1 = 12 − 0.1Q21 the performance curve at speed ω2 is given by 0.25h2 = 12 − 0.1(0.5Q2 )2 which leads to

h2 = 48 − 0.1Q22

80

Chapter 2

Fundamentals of Flow in Closed Conduits The performance curve of the pump with a 2400-rpm motor is therefore given by hp = 48 − 0.1Q2 Comparing the performance curve of the pump with a 2400-rpm motor with that of the pump with a 1200-rpm motor, it is noted that the shutoff head is considerably higher in the 2400-rpm pump (48 m) compared with the shutoff head in the 1200 rpm pump (12 m).

Problems 2.1. Water is pumped from a reservoir to another reservoir whose surface elevations are 400 m and 450 m, respectively. The connecting pipeline is 1.5 km long and 300 mm in diameter. If the pressure during pumping is 392 kPa at midlength of the pipe at elevation 425 m, what is the power? Assume the coefficient of friction is 0.025. 2.2. A pipeline of 0.6 m diameter is 1.5 km long. To augment the discharge, another pipeline of the same diameter is introduced parallel to the first in the second half of the length. Find the increase in discharge if the friction factor is 0.04 and the head at the inlet is 30 m. 2.3. The velocity distribution in a pipe is given by the equation   2  r (2.149) v(r) = V0 1 − R where v(r) is the velocity at a distance r from the centerline of the pipe, V0 is the centerline velocity, and R is the radius of the pipe. Calculate the average velocity and flow rate in the pipe in terms of V0 . 2.4. Calculate the energy and the momentum correction factor, α and β, for the velocity distribution given in Equation 2.149. 2.5. A pipeline of 500 mm is 1,200 m long. To increase the discharge, another line of the same diameter is introduced parallel to the first in the second half of the length. Neglect minor losses, find the increase in discharge if the head at inlet is 350 mm using f equals 0.03. 2.6. Water at 20◦ C flows at a velocity of 2 m/s in a 250-mm diameter horizontal ductile-iron pipe. Estimate the friction factor in the pipe, and state whether the pipe is hydraulically smooth or rough. Compare the friction factors derived from the Moody diagram, the Colebrook equation, and the Swamee–Jain equation. Estimate the change in pressure over 100 m of pipeline. How would the friction factor and pressure change be affected if the pipe were not horizontal but 1 m lower at the downstream section? 2.7. A straight pipe has a diameter of 25 mm, a roughness height of 0.1 mm, and is inclined upward at an angle of 10◦ . The water temperature is 20◦ C. (a) If the pressure at a given section of the pipe is to be maintained at 550 kPa, determine the pressure at a section 100 m downstream for flows of 2 L/min, and 20 L/min (two answers are required here). (b) For a flow of 20 L/min, compare the

value of the friction factor calculated using the Colebrook equation with the friction factor calculated using the Swamee–Jain equation; does your result support Swamee–Jain’s assertion that the results are within 1% of each other? 2.8. Show that the Colebrook equation can be written in the (slightly) more convenient form: f =

2.9.

2.10.

2.11.

2.12.

2.13. 2.14.

2.15.

0.25 {log[(ks /D)/3.7 + 2.51/(Re f )]}2

Why is this equation termed “(slightly) more convenient”? If you had your choice of estimating the friction factor either from the Moody diagram or from the Colebrook equation, which one would you pick? Explain your reasons. A pipeline 200 mm in diameter and 2.5 km long is used to pump fluid at a rate of 50 L/s and whose kinematic viscosity is 1 stokes. The center of the pipe line at the upper end is 30 m above the lower end. The discharge at the upper end is atmospheric. Find the pressure at the lower end and draw the hydraulic grade line (HGL) and total energy line (TEL). Two reservoirs with a surface level difference of 15 m are to be connected by 5 km long, 1.5 m diameter pipe. What will the discharge be when a cast iron pipe of roughness (ks ) = 0.25 mm is used? Neglect all losses and use the Karman-Prandtl equation. A pipe 250 mm in diameter and 220 m long discharges oil from a tank into atmosphere. At the mid-point of the pipe, the pressure is 1.5 times the atmospheric pressure. The specific gravity of the oil is 0.9. The friction factor is 0.01. Estimate the discharge rate of oil and the pressure at the inlet of the pipe. Repeat Problem 2.12 using the Swamee–Jain approximation (Equation 2.44). Use the velocity distribution given in Problem 2.3 to estimate the kinetic energy correction factor, α, for turbulent pipe flow. The velocity profile, v(r), for turbulent flow in smooth pipes is sometimes estimated by the seventh-root law, originally proposed by Blasius (1913): 1  r 7 v(r) = V0 1 − R

Problems

2.16.

2.17.

2.18.

2.19.

2.20.

where V0 is the maximum (centerline) velocity and R is the radius of the pipe. Estimate the energy and momentum correction factors corresponding to the seventhroot law. What will be the percentage increase in the discharge if the cast iron pipe is replaced by a steel pipe of roughness ks = 0.035 mm? Use data from Problem. 2.11. A valve of the end of a frictionless pipe 600 m long is closed in five equal steps each of 2L c . The initial head at the valve which discharges to the atmosphere is 100 m and the initial velocity in the pipe is 1 m/s. Determine the head at the valve after 1 and 2 s when the wave speed (c) is 1200 m/s. Water flows from tank A to tank B. The difference in water level between the two tanks is 5 m. The tanks are connected with a pipe of 200 mm diameter for 32 m length (friction factor equals 0.02), followed by pipe 100 mm in diameter (friction factor equals 0.015) for 32 m. There are two 90◦ bends in each pipe (k equals 0.50 each) and the coefficient of contraction (Cc ) equals 0.75. If the junction of the two pipes is 3 m below the top water level, find the pressure heads at the junction. Two reservoirs open to atmosphere are connected by a pipe 1 km long. The pipe goes over a hill whose height is 6.5 m above the level of water in the upper reservoir. The pipe diameter is 350 mm. The difference in water levels in the two reservoirs is 13.5 m. If the absolute pressure of water anywhere in the pipe is not allowed to fall below 1.5 m of water in order to prevent vapour formation, estimate the length of pipe in the portion between the upper reservoir and the hill summit, and also the discharge through the pipe. Neglect bend losses and use a friction factor of 0.035. The top floor of an office building is 40 m above street level and is to be supplied with water from a municipal pipeline buried 1.5 m below street level. The water pressure in the municipal pipeline is 450 kPa, the sum of the local loss coefficients in the building pipes is 10.0, and the flow is to be delivered to the top floor at 20 L/s through a 150-mm diameter PVC pipe. The length of the pipeline in the building is 60 m, the water temperature is 20◦ C, and the water pressure on the top floor must be at least 150 kPa. Will a booster pump be required for the building? If so, what power must be supplied by the pump?

2.21. Water is pumped from a supply reservoir to a ductileiron water-transmission line, as shown in Figure 2.26. The high point of the transmission line is at point A, 1 km downstream of the supply reservoir, and the low point of the transmission line is at point B, 1 km downstream of A. If the flow rate through the pipeline is 1 m3 /s, the diameter of the pipe is 750 mm, and the pressure at A is to be 350 kPa, then: (a) estimate the head that must be added by the pump; (b) estimate

A

Elev. 10 m

Elev. 7 m Supply reservoir

81

Transmission line

P Pump Elev. 4 m

B

FIGURE 2.26: Water pumped into transmission line

the power supplied by the pump; and (c) calculate the water pressure at B. 2.22. A pipeline is to be run from a water-treatment plant to a major suburban development 3 km away. The average daily demand for water at the development is 0.0175 m3 /s, and the peak demand is 0.578 m3 /s. Determine the required diameter of ductile-iron pipe such that the flow velocity during peak demand is 2.5 m/s. Round the pipe diameter upward to the nearest 25 mm (i.e., 25 mm, 50 mm, 75 mm, . . .). The water pressure at the development is to be at least 340 kPa during average demand conditions, and 140 kPa during peak demand. If the water at the treatment plant is stored in a groundlevel reservoir where the level of the water is 10.00 m NGVD and the ground elevation at the suburban development is 8.80 m NGVD, determine the pump power (in kilowatts) that must be available to meet both the average daily and peak demands. 2.23. Water exits a reservoir through a 125-m long 5-cm diameter horizontal cast-iron pipe as shown in Figure 2.27. The pipe entrance is sharp-edged, the water flows through a turbine, and the discharge is to the atmosphere. If the flow rate is 4 L/s, what power is extracted by the turbine? Describe a practical situation in which you might encounter this type of problem.

Turbine

40 m

Open globe valve T 125 m

FIGURE 2.27: Flow out of a reservoir

2.24. Draw the hydraulic grade line (HGL) and total energy line (TEL) using data from question no. 2.1, with the coefficient of friction as 0.0025. 2.25. Water flows at 10 m3 /s in a 2 m * 2 m square reinforcedconcrete pipe. If the pipe is laid on a (downward) slope of 0.002, what is the change in pressure in the pipe over a distance of 500 m?

82

Chapter 2

Fundamentals of Flow in Closed Conduits

2.26. Derive the Hazen–Williams head-loss relation, Equation 2.82, starting from Equation 2.80. 2.27. Compare the Hazen–Williams formula for head loss (Equation 2.82) with the Darcy–Weisbach equation for head loss (Equation 2.33) to determine the expression for the friction factor that is assumed in the Hazen– Williams formula. Based on your result, identify the type of flow condition (rough, smooth, or transition) incorporated in the Hazen–Williams formula. 2.28. Derive the Manning head-loss relation, Equation 2.85. 2.29. Compare the Manning formula for head loss (Equation 2.85) with the Darcy–Weisbach equation for head loss (Equation 2.33) to determine the expression for the friction factor that is assumed in the Manning formula. Based on your result, identify the type of flow condition (rough, smooth, or transition) incorporated in the Manning formula. 2.30. Determine the relationship between the Hazen– Williams roughness coefficient and the Manning roughness coefficient. 2.31. Given a choice between using the Darcy–Weisbach, Hazen–Williams, and Manning equations to estimate the friction losses in a pipeline, which equation would you choose? Why? 2.32. Water is flowing in a steel pipe of roughness ks equal to 0.045 mm having a kinematic viscosity of 10−6 m2 /s with a mean velocity of 1.5 m/s. The head loss is to be limited to 5.5 m per 200 m length of pipe. Estimate the diameter of the pipe. Assume that the friction factor in the Darcy–Weisbach equation is to be calculated using the Colebrook equation described in Equation 2.35. 2.33. Plot the water-hammer pressure versus time at the midpoint of the pipeline shown in Figure 2.10. Assume that the valve is closed instantaneously. 2.34. Water flows in a 2,600 m long pipe of diameter 600 mm with a velocity of 1.8 m/s. At the end of the pipe, a valve is provided. Find the rise in pressure if the valve is closed in 30 seconds. Use a velocity of pressure wave of 1,500 m/s. 2.35. If in the Problem 2.34, the valve is closed in 5 seconds, find the rise in pressure behind the valve. Assume that the pipe is rigid. 2.36. Water at 20◦ C flows in a 150-m long 50-mm diameter ductile-iron pipe at 4 m/s. The thickness of the pipe wall is 1.5 mm and the modulus of elasticity of ductile iron is 1.655 * 105 MN/m2 . What is the maximum waterhammer pressure that can occur? 2.37. The pipeline described in Problem 2.36 is replaced by a 50-mm diameter PVC pipe with a wall thickness of 2 mm and a modulus of elasticity of 1.7 * 104 MN/m2 . How will this affect the maximum water-hammer pressure?

2.38. Reservoirs A, B, and C are connected as shown in Figure 2.28. The water elevations in reservoirs A, B, and C are 100 m, 80 m, and 60 m, respectively. The three pipes connecting the reservoirs meet at the junction J, with pipe AJ being 900 m long, BJ 800 m long, CJ 700 m long, and the diameter of all pipes equal to 850 mm. If all pipes are made of ductile iron and the water temperature is 20◦ C, find the flow into or out of each reservoir. Elev. 100 m A

LAJ

=

Elev. 80 m

900 m B

LBJ = 800 m J

C

LJC = 700 m

Elev. 60 m

FIGURE 2.28: Connected reservoirs

2.39. The water-supply network shown in Figure 2.29 has constant-head elevated storage tanks at A and B, with inflows and withdrawals at C and D. The network is on flat terrain, and the pipeline characteristics are as follows:

Pipe

L (km)

D (mm)

AD BC BD AC

1.0 0.8 1.2 0.7

400 300 350 250

If all pipes are made of ductile iron, calculate the inflows/outflows from the storage tanks. Assume that the flows in all pipes are fully turbulent. Elev. 25 m

Reservoir

A

C

0.2 m3/s Elev. 20 m

0.2 m3/s

Reservoir D

B

FIGURE 2.29: Water-supply network

Problems Flow

A C

P

Flow

F

Flow

G

Flow

D

B

Flow

E

83

FIGURE 2.30: Pipe system

2.40. Water is handled by a system of pipes as shown in Figure 2.30 and the pipe characteristics are as follows:

Pipe AC, BC CD DE DF DG

Length (m)

Diameter (m)

f

100 300 500 400 500

0.50 0.75 0.30 0.25 0.30

0.0055 0.0050 0.0060 0.0060 0.0060

The elevation of outlets E, F, and G is 100 m above the elevation of inlets A and B. All outlets and inlets are at atmospheric pressure. If the mean velocity in the pipes AC and BC is 2.5 m/s, calculate the flow rate through the pump P, the pressure difference across the pump, and the power consumed by the pump. Take the pump efficiency as 76%. 2.41. Consider the pipe network shown in Figure 2.31. The Hardy Cross method can be used to calculate the pressure distribution in the system, where the friction loss, hf , is estimated using the equation hf = rQn

0.5 m3/s

A

0.5 m3/s F

E

You can assume that the flow in each pipe is hydraulically rough. 2.42. A portion of a municipal water-distribution network is shown in Figure 2.32, where all pipes are made of ductile iron and have diameters of 300 mm. Use the Hardy Cross method to find the flow rate in each pipe. If the pressure at point P is 500 kPa and the distribution network is on flat terrain, determine the water pressures at each pipe intersection. 0.05 m3/s

150 m

Pipe

D (mm)

AB BC CD DE EF FA BE

1000 750 800 700 900 900 950

300 325 200 250 300 250 350

0.1 m3/s

150 m

0.07 m3/s

100 m

P 150 m

150 m

100 m 0.06 m3/s

L (m)

D

FIGURE 2.31: Two-loop pipe network

100 m

and all pipes are made of ductile iron. What value of r and n would you use for each pipe in the system? The pipeline characteristics are as follows:

C

B

100 m 150 m

150 m

0.06 m3/s

FIGURE 2.32: Four-loop pipe network

2.43. Water is delivered at a rate of 48,000 m3 /d at node C of the distribution system shown in Figure 2.33, and the water is withdrawn at a rate of 8000 m3 /d at each of the nodes. All pipes are made of steel with roughness heights of 0.05 mm and are of diameter 1120 mm. The pipe lengths are as follows: CD = 5 km, DE = 12 km, EF = 9 km, FC = 6 km, FG = 7 km, GH = 8 km, and HC = 10 km. Determine the flow distribution in the pipes.

84

Chapter 2

Fundamentals of Flow in Closed Conduits 48,000 m3/d

8000 m3/d

8000 m3/d

C

H

D

8000 m3/d

G

8000 m3/d

8000 m3/d

8000 m3/d

F

E

FIGURE 2.33: Water-supply system

2.44. What is the constant that can be used to convert the specific speed in SI units (Equation 2.122) to the specific speed in U.S. Customary units (Equation 2.123)? 2.45. What is the highest synchronous speed for a motor driving a pump? 2.46. Derive the affinity relationship for the power delivered to a fluid by two homologous pumps. (Note: This affinity relation is given by Equation 2.127.) 2.47. A centrifugal pump with impeller diameter of 250 mm has a shut-off head of 8.0 m of water operated at 1200 rpm. At the same speed, best efficiency occurs at discharge of 22.22 L/s, where the head is 7.07 m of water. Develop a curve-fit for the pump at 1,200 rpm and 1,800 rpm. Assume that cavitation will not occur. 2.48. A pump is required to deliver 150 L/s (;10%) through a 300-mm diameter PVC pipe from a well to a reservoir. The water level in the well is 1.5 m below the ground surface and the water surface in the reservoir is 2 m above the ground surface. The delivery pipe is 300 m long, and local losses can be neglected. A pump manufacturer suggests using a pump with a performance curve given by hp = 6 − 6.67 * 10−5 Q2

where hp is in meters and Q in L/s. Is this pump adequate? 2.49. The pumped-storage system illustrated in Figure 2.34 is designed to exchange 2 m3 /s through a 1220-mm diameter 3.2-km long ductile-iron pipeline lined with bitumen. The elevation difference between the water surfaces in the upper and lower reservoirs is 61 m, and the pump/turbine is to operate for 8 hours during the day as a turbine (to generate electricity) and 8 hours during the night as a pump (to return the water to the upper reservoir). The pump efficiency is 85%, the turbine efficiency is 90%, the cost of pumping is $0.06/kWh, and the revenue from turbine operations (selling electricity) is $0.12/kWh. Determine the annual profit from this operation. Neglect the effect of storage on reservoir elevations. If the pump performance curve is given by hp = 80 − 3.5Q2 where hp is in meters and Q is in m3 /s, estimate the change in profit when the elevation difference is 65 m. For an elevation difference of 65 m, assume that the flow is fully turbulent. 2.50. A pump is to be selected to deliver water from a well to a treatment plant through a 300-m long pipeline. The temperature of the water is 20◦ C, the average elevation of the water surface in the well is 5 m below the ground surface, the pump is 50 cm above the ground surface, and the water surface in the receiving reservoir at the watertreatment plant is 4 m above the ground surface. The delivery pipe is made of ductile iron (ks = 0.26 mm) with a diameter of 800 mm. If the selected pump has a performance curve of hp = 12 − 0.1Q2 , where Q is in m3 /s and hp is in m, then what is the flow rate through the system? Calculate the specific speed of the required pump (in U.S. Customary units), and state what type of pump will be required when the speed of the pump motor is 1200 rpm. Neglect local losses. 2.51. Water is to be pumped out of a well and stored in an above-ground reservoir. The water surface in the well is 3 m below the ground surface, water is to be pumped through a 100-m long 50-mm diameter galvanized iron line and exit 19.3 m above the ground, and water is to be delivered to the upper reservoir at a rate of at

Upper reservoir

Pipeline 61 m

Pump/turbine

Lower reservoir

FIGURE 2.34: Pumped storage system

Problems

Bell & Gossett %

59

%

59.5%

200

60

59

55

%

%

1½ × 1½ × 7B 3500 rpm 57

%

%

220

53

%

50

45

%

35

7.0 in.

40

240 70

%

Series 80-SC 25 %

260

6.5 in.

%

55 % 5 3%

140 5.5 in.

120 5.0 in.

100

45 %

80 20 10

10

60

30

40 NPSH

20 0

NPSH in meters

57 6.0 in.

NPSH in feet

30

Total head in feet

40

160

% 50

Total head in meters

180 50

85

20

REQ

0 0

20

40

60

80

100

5

10

120

140

0 180

160

0

Capacity in U.S. gallons per minute 0

5

10

15 20 25 Capacity in cubic meters/hr

35

30

40

FIGURE 2.35: Pump performance curves

least 370 L/min when the reservoir is empty. The sum of the local loss coefficients in the system is 1.8. A local pump salesman suggests a pump model with performance curves shown in Figure 2.35. Determine if this pump will meet the demands of the project and, if so, the pump size that is required. What is the maximum height that the pump can be placed above ground? 2.52. A pump lifts water through a 100-mm diameter ductile-iron pipe from a lower to an upper reservoir (Figure 2.36). If the difference in elevation between the reservoir surfaces is 10 m, and the performance curve of the 2400-rpm pump is given by

hp = 15 − 0.1Q2 where hp is in meters and Q in L/s, then estimate the flow rate through the system. If the pump manufacturer gives the required net positive suction head under these operating conditions as 1.5 m, what is the maximum height above the lower reservoir that the pump can be placed and maintain the same operating conditions? 2.53. Water is being pumped from reservoir A to reservoir F through a 30-m long PVC pipe of diameter 150 mm (see Figure 2.37). There is an open gate valve located at C; 90◦ bends (threaded) located at B, D, and E; and the pump performance curve is given by hp = 20 − 4713Q2

100 m

where hp is the head added by the pump in meters and Q is the flow rate in m3 /s. The specific speed of the

Upper reservoir

P

10 m E

3m C

B P

1m Lower reservoir

FIGURE 2.36: Water pumped from lower to upper reservoir

F 10 m D

3m

A

FIGURE 2.37: Water-delivery system

86

Chapter 2

Fundamentals of Flow in Closed Conduits 2.55. If the performance curve of a certain pump model is given by hp = 30 − 0.05Q2

pump (in U.S. Customary units) is 3000. Assuming that the flow is turbulent (in the smooth, rough, or transition range) and the temperature of the water is 20◦ C, (a) write the energy equation between the upper and lower reservoirs, accounting for entrance, exit, and local losses between A and F; (b) calculate the flow rate and velocity in the pipe; (c) if the required net positive suction head at the pump operating point is 3.0 m, assess the potential for cavitation in the pump (for this analysis you may assume that the head loss in the pipe is negligible between the intake and the pump); and (d) use the affinity laws to estimate the pump performance curve when the motor on the pump is changed from 800 rpm to 1600 rpm. 2.54. The performance curve of a Goulds Model 3656 irrigation pump is shown in Figure 2.38. The pump is to be used to deliver water at 20◦ C from a pond to the center of a large field located 100 m from the pond. The desired pump is expected to deliver a maximum flow rate of 380 L/min through 107 m of 6-cm steel pipe having an equivalent sand roughness of 0.01 mm. The water level in the pond during the irrigation season is 10.00 m and the ground elevation of the field is 15.00 m.

where hp is in meters and Q is in L/s, what is the performance curve of a pump system containing n of these pumps in series? What is the performance curve of a pump system containing n of these pumps in parallel? 2.56. A pump is placed in a pipe system in which the energy equation (system curve) is given by hp = 15 + 0.03Q2 where hp is the head added by the pump in meters and Q is the flow rate through the system in L/s. The performance curve of the pump is hp = 20 − 0.08Q2 What is the flow rate through the system? If the pump is replaced by two identical pumps in parallel, what would be the flow rate in the system? If the pump is replaced by two identical pumps in series, what would be the flow rate in the system? 2.57. A 20-km long 1120-mm diameter steel pipe with an estimated roughness height of 0.05 mm is to deliver water from a water-supply reservoir through a system of parallel pumps as shown in Figure 2.39. The intake at A and the exit from the pump system at B both are at an elevation of 5 m, the water level in the reservoir is 2 m above the intake at A, and the elevation at the delivery point, C, is 15 m. There is negligible head loss due to friction between

(a) Which of the pumps shown in Figure 2.38 would you select for the job? (b) Using the efficiency of the pump under operating conditions, calculate the size of the motor in kilowatts that must be used to drive the pump. (c) What is the maximum elevation above the pond that the pump could be located?

m

ft

30%

Total dynamic head

15

NOTE: Not recommended for operation beyond printed N-Q curve.

Model 3656 / Size 2 × 2-5 RPM 3450 Curve CN346R00 IMP. DWG. NO. 119-86 43% 50% 55%

60%

4 in.

50

64%

3.75 in.

40

64% 3.5 in.

10

60% 55%

30

50%

3.25 in.

43%

20

3.0 in.

5 NPSHR - ft

10

5 ft

16 ft 6 ft 7 ft

0

0

0 0

20

40 10

60

80

100

20 Capacity

9 ft 120

140 30

FIGURE 2.38: Goulds Model 3656 pump curves

160

180 gpm 3 40 m /h

Problems

C Elev. 2.00 m p ≥ 350 kPa

C

P B

A

P

P

87

Elev. 3.00 m

B

A

P

D Elev. 5.00 m p ≥ 350 kPa

P

FIGURE 2.39: Parallel-pump system FIGURE 2.40: Water-supply system

A and B, and the performance curve for each pump in the system is given by hp = 65 − 7.6 * 10−8 Q2 where hp is the head added by the pump in m and Q is the flow through the pump in m3 /d. When the system is delivering 48,000 m3 /d at C, the required pressure at C is 448 kPa. (a) Determine how many pumps are required. (b) Assuming that the flow is fully turbulent, what is the actual flow rate at C when using the number of pumps determined in Part (a)? The temperature of the water is 20◦ C. 2.58. The water-supply system shown in Figure 2.40 is to be constructed such that water is delivered from a reservoir at A to two communities located at C and D. The pipe lengths, diameters, and demand flow rates are as follows:

of at least 350 kPa at C and D. All pipes are to be made of ductile iron. (a) Determine the minimum power that must be delivered by the pump. (b) A pump manufacturer will be able to match the minimum-power operating condition by providing several “micro-pumps” in series where each micro-pump has a performance characteristic given by hp = 0.455D − 4000Q2 where hp is the head added in m, D is the size of the micro-pump in cm, and Q is the flow in m3 /s. The manufacturer can deliver any size, D, in the range of 40–50 cm. How many micro-pumps will be needed and of what size? 2.59. The performance curve for a variable-speed pump operating at 600 rpm is given by

Line

Length (km)

Diameter (mm)

Flow (L/s)

hp = 6 − 0.05Q2

AB BC BD

1.05 2.80 2.50

200 150 150

27 12 15

where hp is the head added by the pump in m and Q is the flow rate in m3 /min. This pump is installed in a system where energy considerations require a system curve given by

The water-surface elevation of the supply reservoir is 3.00 m, and the elevations of the delivery pipes at C and D are 2.00 m and 5.00 m, respectively. Under the given demand conditions, it is desired to have water pressures

hp = 3 + 0.042Q2 Find the flow rate in the system when the pump is operating at 600 rpm and compare this with the flow rate when the pump is operating at 1200 rpm.

C H A P T E R

3

Design of Water-Distribution Systems 3.1 Introduction Water-distribution systems move water from water sources to treatment plants, and from treatment plants to homes, offices, industries, and other water consumers. The major components of a water-distribution system include pipelines, pumps, storage facilities, valves, and meters. The primary objectives in designing a water-distribution system are to: (1) supply each customer with water at an adequate rate and at an adequate pressure; (2) deliver water that meets water-quality criteria for drinking; and (3) have sufficient capacity and reserve storage for fire protection and emergency conditions. Water-distribution systems should be designed to meet these objectives in a cost-effective manner while taking into account safety, applicable regulations, environmental impact, societal concerns, and sustainability. This chapter presents the protocol for designing municipal water-distribution systems. Methods for estimating water demand, design of the functional components of distribution systems, network analysis, and the operational criteria for municipal water-distribution systems are all covered. 3.2 Water Demand A water-supply system must satisfy the water demand of the population being served and the fire flows needed to protect life and property, while providing due consideration to the proximity of the service area to the source(s) of water. The water demand at the end of the design life is usually the basis for system design, and so water-demand forecasting is generally required. There are a variety of methods that are used to forecast water demand, and the appropriate method depends on the particular situation. Most forecasting methods can be categorized as per-capita models, extrapolation models, disaggregation models, multiple-regression models, and land-use models (AWWA, 2007). Brief descriptions of these models are given below. Per-Capita Models. These models simply estimate the average consumption per capita (i.e., per person) and multiply this per-capita consumption by a projected population at the end of the design life to estimate the average total demand. In some cases, a trend is also applied to the per-capita consumption. Per-capita models produce satisfactory results as long as the population forecast is accurate, and the consumer mix does not change substantially. Extrapolation Models. Extrapolation models plot the annual or monthly water consumption as a function of time or population and then extrapolate this relationship into the future. This approach can be applied separately to various water-use components, such as residential and nonresidential use. Disaggregation Models. Water use is disaggregated into basic segments such as single-family residential, multifamily residential, institutional, commercial, industrial, and public facilities. The consumption per unit within each segment is estimated (e.g., in m3 day−1 unit−1 ) and multiplied by the projected number of units at the end of the design period. 88

Section 3.2

Water Demand

89

Multiple-Regression Models. These empirical models relate the water demand to a variety of independent variables such as population, number of households or dwelling units, household income, lot sizes, land use, employment, and various weather variables. Land-Use Models. These models base their water-use forecasts on the projected uses of residential, commercial, industrial, and public lands within the service area. Water-use projections are developed for each land-use segment. The development of water-demand forecast models involves highly specialized activities and usually require close coordination with the local planning department. The per-capita model can be used to illustrate the methodology for developing and applying a water-demand forecast model. 3.2.1

Per-Capita Forecast Model

Estimation of the water demand using a per-capita model requires prediction of population, development, and per-capita water usage in the service area at the end of the design period. The average water demand from various sectors of the population is usually taken as equal to the predicted population in that sector multiplied by the per-capita demand in that sector. This can be expressed as follows: Q=

N    qi * Pi

(3.1)

i=1

where Q is the average water-demand rate [L3 T−1 ], q is the average per-capita demand rate [L3 T−1 person−1 ], P is the population [persons], the subscript i indicates the demand sector, and N is the number of sectors. In some cases a single lumped per-capita demand is used to incorporate the demand from all sectors of the service area, in which case the average water-demand rate, Q, is estimated using the relation Q=q * P

(3.2)

where q is the lumped per-capita demand rate [L3 T−1 person−1 ], and P is the total population of the service area. Methods for estimating the per-capita demand rate and the population are given in the following sections. 3.2.1.1

Estimation of per-capita demand

There are usually several categories (sectors) of water demand within any populated area, and these sources of demand can be broadly grouped into residential, commercial, industrial, and public. Residential water use is associated with houses and apartments where people live; commercial water use is associated with retail businesses, offices, hotels, and restaurants; industrial water use is associated with manufacturing and processing operations; and public water use includes governmental facilities that use water. Large industrial requirements are typically satisfied by sources other than the public water supply. A typical distribution of per-capita water use for an average city in the United States is given in Table 3.1. These rates vary from city to city as a result of differences in local conditions that are mostly unrelated to the efficiency of water use. Generally, high per-capita rates are found in water-supply systems servicing large industrial or commercial sectors, affluent communities, arid and semiarid areas, and/or communities without water meters. Per-capita demand is commonly assumed to be constant in time; however, there are a variety of factors that can cause the per-capita demand in a service area to change over time. Such factors include pricing policy, distribution of housing types, regulatory changes, and changes in per-capita income (Polebitski et al., 2011). Careful consideration should be given to these factors before assuming that the per-capita water demand remains constant.

90

Chapter 3

Design of Water-Distribution Systems TABLE 3.1: Typical Distribution of Per-Capita Water Demand

Category

Average use (L/d) (gal/d)

Percent of total

Residential Commercial Industrial Public Loss

380 115 85 65 40

100 30 22 17 11

56 17 12 9 6

Total

685

180

100

Source: Solley (1998).

3.2.1.2

Estimation of population

In planning water-supply projects, future populations within the service area must be estimated. Prediction of population growth and development requires a variety of considerations, including urban planning constraints, location of urban growth boundaries, changes in transportation networks, and new land-use policies. The simplest models of population forecasting treat the population as a whole, fit empirical growth functions to historical population data, and forecast future populations based on past trends. The most complex models disaggregate the population into various groups and forecast the growth of each group separately (e.g., Polebitski et al., 2011). High levels of disaggregation have the advantage of making forecast assumptions very explicit, but these models tend to be complex and require more data than empirical models that treat the population as a whole. Over relatively short time horizons, on the order of 10 years or less, detailed disaggregation models may not be any more accurate than using empirical extrapolation models of the population as a whole. Short-term population projections. Populated areas tend to grow at varying rates, as illustrated in Figure 3.1. In the early stages of growth, there are usually wide-open spaces and the population, P, tends to grow geometrically according to the relation dP = k1 P dt

(3.3)

where k1 is a growth constant. Integrating Equation 3.3 gives the following expression for the population as a function of time: P(t) = P0 ek1 t

Psat Declining growth phase, Population, P

FIGURE 3.1: Growth phases in populated areas

Arithmetic growth phase,

Geometric growth phase, Time, t

dP dt

dP dt

dP dt

 k3 (Psat − P )

 k2

 k1P

(3.4)

Section 3.2

Water Demand

91

where P0 is the population at some initial time designated as t = 0. Beyond the initial geometric growth phase, the rate of growth begins to level off and the following arithmetic growth relation may be more appropriate: dP = k2 (3.5) dt where k2 is an arithmetic growth constant. Integrating Equation 3.5 gives the following expression for the population as a function of time: P(t) = P0 + k2 t

(3.6)

where P0 is the population at t = 0. Ultimately, the growth of population centers becomes limited by the resources available to support the population, and further growth is influenced by the saturation population of the area, Psat , and the population growth is described by a relation such as dP = k3 (Psat − P) (3.7) dt where k3 is a constant. This phase of growth is called the declining-growth phase. Almost all communities have zoning regulations that control the use of both developed and undeveloped areas within their jurisdiction (sometimes called a land-use master plan), and a review of these regulations will yield an estimate of the saturation population of an area. Integrating Equation 3.7 gives the following expression for the population as a function of time: P(t) = Psat − (Psat − P0 )e−k3 t

(3.8)

where P0 is the population at t = 0. The time scale associated with each growth phase is typically on the order of 10 years, although the actual duration of each phase can deviate significantly from this number. The duration of each phase is important in that population extrapolation using a single-phase equation can only be justified for the duration of that growth phase. Consequently, singlephase extrapolations are typically limited to 10 years or less, and these population predictions are termed short-term projections. Long-term population projections. Extrapolation beyond 10 years, called long-term projections, can be done by a variety of methods. A popular method is fitting an S-shaped curve to the historical population trends and then extrapolating using the fitted equation. The most commonly fitted S-curve is the so-called logistic curve, which is mostly suited to large cities and described by the equation P(t) =

Psat 1 + aebt

(3.9)

where a and b are constants. The parameters in Equation 3.9 can be estimated using the populations P0 , P1 , and P2 at times t0 , t1 , and t2 (in years), and the parameter-estimation equations are as follows Psat =

2P0 P1 P2 − P 21 (P0 + P2 ) P0 P2 − P 21

Psat − P0 P0   P0 (Psat − P1 ) 1 ln b= t P1 (Psat − P0 ) a=

(3.10) (3.11) (3.12)

where t is the time interval between the measurement times such that t = t1 − t0 = t2 − t1 . Using these parameters, the time t in Equation 3.9 is the time in years after t0 (i.e., t = 0 at t0 ).

92

Chapter 3

Design of Water-Distribution Systems

Utilization of Equation 3.10 is inappropriate when the calculated value of Psat is either negative or less than the latest value of population (i.e., P2 ). In an alternative approach, Psat can be assigned a value based on physical and/or political constraints of the area under consideration, and a and b can be calculated using Equations 3.11 and 3.12. Under this circumstance the logistic curve is forced to go through the points (P0 ,t0 ) and (P1 ,t1 ) and meet the condition that P → Psat as t → q. The conventional approach to fitting population equations to historical data is to plot the historical data, observe the trend in the data, and fit the curve that best matches the population trend. Using extrapolation methods, errors less than 10% can be expected for planning periods shorter than 10 years, and errors greater than 50% can be expected for planning periods longer than 20 years (Sykes, 1995).

EXAMPLE 3.1 You are in the process of designing a water-supply system for a town, and the design life of your system is to end in the year 2040. The population in the town has been measured every 10 years since 1940 by the U.S. Census Bureau, and the reported populations are tabulated below. Estimate the population in the town using (a) graphical extension, (b) arithmetic growth projection, (c) geometric growth projection, (d) declining growth projection (assuming a saturation concentration of 600,000 people), and (e) logistic curve projection. Year

Population

1940 1950 1960 1970 1980 1990 2000 2010

125,000 150,000 150,000 185,000 185,000 210,000 280,000 320,000

Solution The population trend is plotted in Figure 3.2, where a geometric growth rate approaching an arithmetic growth rate is indicated. (a) A growth curve matching the trend in the measured populations is indicated in Figure 3.2. Graphical extension to the year 2040 leads to a population estimate of 530,000 people. FIGURE 3.2: Population trend

600

(c)

530,000

Population (thousands)

500 400

(a) (b) (e) (d)

Population trend

350

C

300

B A

250 200

Census data

150 100 1940

1960

1980

2000 Year

2020

2040

Section 3.2

Water Demand

93

(b) Arithmetic growth is described by Equation 3.6 as P(t) = P0 + k2 t

(3.13)

where P0 and k2 are constants. Consider the arithmetic projection of a line passing through points B and C on the approximate growth curve shown in Figure 3.2. At point B, t = 0 (year 2000) and P = 270,000; at point C, t = 10 (year 2010) and P = 330,000. Applying these conditions to Equation 3.13 yields P = 270, 000 + 6000 t (3.14) In the year 2040, t = 40 years and the population estimate given by Equation 3.14 is P = 510,000 people (c) Geometric growth is described by Equation 3.4 as P = P0 ek1 t

(3.15)

where k1 and P0 are constants. Using points A and C in Figure 3.2 as a basis for projection, then at t = 0 (year 1990), P = 225,000, and at t = 20 (year 2010), P = 330,000. Applying these conditions to Equation 3.15 yields P = 225, 000e0.0195t

(3.16)

In the year 2040, t = 50 years and the population estimate given by Equation 3.16 is P = 597,000 people (d) Declining growth is described by Equation 3.8 as P(t) = Psat − (Psat − P0 )e−k3 t

(3.17)

where P0 and k3 are constants. Using points A and C in Figure 3.2, then at t = 0 (year 1990), P = 225,000, at t = 20 (year 2010), P = 330,000, and Psat = 600,000. Applying these conditions to Equation 3.17 yields (3.18) P = 600,000 − 375,000e−0.0164t In the year 2040, t = 50 years and the population estimate given by Equation 3.18 is P = 434,800 people (e) The logistic curve is described by Equation 3.9, and the parameters of this curve can be estimated using Equations 3.10 to 3.12 where P0 = 225,000, P1 = 270,000, P2 = 330,000, and t = 10 years. Substituting these data into Equation 3.10 (and expressing the population in thousands) gives Psat =

2P0 P1 P2 − P 21 (P0 + P2 ) 2(225)(270)(330) − (270)2 (225 + 330) = 2 (225)(330) − (270)2 P0 P2 − P 1 = −270 thousand people

Since the calculated value of Psat is negative, Equation 3.10 is not appropriate for estimating Psat . Therefore, take the estimated value of Psat = 600,000 as utilized in the declining growth model in Part (d). Hence, Equations 3.11 and 3.12 give the logistic-curve parameters a and b as 600 − 225 Psat − P0 = 1.67 = P0 225     P0 (Psat − P1 ) 225(600 − 270) 1 1 ln ln b= = = −0.0310 t P1 (Psat − P0 ) 10 270(600 − 225) a=

The logistic curve for predicting the population is given by Equation 3.9 as P=

600, 000 Psat = bt 1 + ae 1 + 1.67e−0.0310t

(3.19)

94

Chapter 3

Design of Water-Distribution Systems In 2040, t = 50 years and the population estimate given by Equation 3.19 is P = 443,000 people These results indicate that the population projection in 2040 is quite uncertain, with estimates ranging from 597,000 for geometric growth to 434,800 for declining growth. The projected results are compared graphically in Figure 3.2. Closer inspection of the predictions indicates that the declining and logistic growth models are limited by the specified saturation population of 600,000, while the geometric growth model is not limited by saturation conditions and produces the highest projected population.

The projected average water demand in any future year is calculated as the product of the projected population and the per-capita water demand in accordance with Equation 3.2. Temporal variations from the average demand within the projection year are calculated by multiplying the average demand by demand factors as described in the following section. 3.2.2

Temporal Variations in Water Demand

Water demand generally fluctuates on an hourly basis with the temporal pattern of fluctuation depending on several factors, such as land use, temperature, humidity, time since last rainfall, season, and day of the week. Water demand is generally below the average daily demand in the early-morning hours and above the average daily demand during the midday hours. Examples of daily cycles in water demand in residential and commercial areas of Boston, Massachusetts, are shown in Figure 3.3. On a typical day in most communities, water use is lowest at night (11 p.m. to 5 a.m.) when most people are asleep. Water use rises rapidly in the morning (5 a.m. to 11 a.m.) followed by moderate usage through midday (11 a.m. to 6 p.m.). Use then increases in the evening (6 p.m. to 10 p.m.) and drops rather quickly around 10 p.m. Overall, water-use patterns within a typical 24-hour period are characterized by demands that are 25%–40% of the average daily demand during the hours between midnight and 6 a.m. and 150%–175% of the average daily demand during the morning or evening peak periods (Velon and Johnson, 1993). The range of demand conditions that is to be expected in water-distribution systems is specified by demand factors or peaking factors that express the ratio of the demand under certain conditions to the average daily demand. Typical demand factors for various conditions are given in Table 3.2, where the maximum daily demand is defined as the demand on the day of the year that uses the most volume of water, and the maximum hourly demand is defined as the demand during the hour that uses the most volume of water. The demand factors in Table 3.2 should serve only as guidelines, with the actual demand factors in any distribution system being best estimated from local measurements. In small water systems, demand factors may be significantly higher than those shown in Table 3.2. FIGURE 3.3: Typical daily cycles in water demand

2.0 1.8

Source: Shvartser et al. (1993).

Residential area

Demand factor

1.6

Commercial area

1.4 1.2 1.0 0.8 0.6 0.4 0.2

0

2

4

6

8

10

12 Hour

14

16

18

20

22

24

Section 3.2

Water Demand

95

TABLE 3.2: Typical Demand Factors

Demand Daily average in maximum month Daily average in maximum week Maximum daily Maximum hourly Minimum hourly

3.2.3

Minimum

Maximum

Typical

1.10 1.20 1.50 2.00 0.20

1.50 1.60 3.00 4.00 0.60

1.20 1.40 1.80 3.25 0.30

Fire Demand

Besides the fluctuations in demand that occur under normal operating conditions, waterdistribution systems are usually designed to accommodate the large (short-term) water demands associated with fighting fires. Although there is no legal requirement that a governing body size its water-distribution system to provide fire protection, the governing bodies of most communities provide water for fire protection for reasons that include protection of the tax base from destruction by fire, preservation of jobs, preservation of human life, and reduction of human suffering. Flow rates required to fight fires can significantly exceed the maximum flow rates in the absence of fires, particularly in small water systems. In fact, for communities with populations less than 50,000, the need for fire protection is typically the determining factor in sizing water mains, storage facilities, and pumping facilities (AWWA, 2003c). In contrast to urban water systems, many rural water systems are designed to serve only domestic water needs, and fire-flow requirements are not considered in the design of these systems (AWWA, 2003c). Numerous methods have been proposed for estimating fire flows, the most popular of which was proposed by the Insurance Services Office, Inc. (ISO, 1980), which is an organization representing the fire-insurance underwriters. The required fire flow for individual buildings can be estimated using the formula (ISO, 1980) NFFi = Ci Oi (X + P)i

(3.20)

where NFFi is the needed fire flow at location i, Ci is the construction factor based on the size of the building and its construction, Oi is the occupancy factor reflecting the kinds of materials stored in the building (values range from 0.75 to 1.25), and (X + P)i is the sum of the exposure factor (Xi ) and communication factor (Pi ) that reflect the proximity and exposure of other buildings (values range from 1.0 to 1.75). The construction factor, Ci [L/min], is the portion of the NFF attributed to the size of the building and its construction and is given by  (3.21) Ci = 220F Ai where Ai is the effective floor area [m2 ], typically equal to the area of the largest floor in the building plus 50% of the area of all other floors; and F is a coefficient [dimensionless] based on the class of construction, given in Table 3.3. The maximum value of Ci calculated using Equation 3.21 is limited by the following: 30,000 L/min (8000 gpm∗ ) for construction classes 1 and 2; 23,000 L/min (6000 gpm) for construction classes 3, 4, 5, and 6; and 23,000 L/min (6000 gpm) for a one-story building of any class of construction. The minimum value of Ci is 2000 L/min (500 gpm), and the calculated value of Ci should be rounded to the nearest 1000 L/min (250 gpm). The occupancy factors, Oi , for various classes of buildings are given in Table 3.4. Detailed tables for estimating the exposure and communication factors, (X + P)i ,

∗ gpm = gallons per minute

96

Chapter 3

Design of Water-Distribution Systems TABLE 3.3: Construction Coefficient, F

Class of construction

Description

F

1 2 3 4 5 6

Frame Joisted masonry Noncombustible Masonry, noncombustible Modified fire resistive Fire resistive

1.5 1.0 0.8 0.8 0.6 0.6

Source: AWWA (1992). TABLE 3.4: Occupancy Factors, Oi

Combustibility class C-1 Noncombustible C-2 Limited combustible C-3 Combustible C-4 Free burning C-5 Rapid burning

Examples

Oi

Steel or concrete products storage Apartments, churches, offices Department stores, supermarkets Auditoriums, warehouses Paint shops, upholstering shops

0.75 0.85 1.00 1.15 1.25

Source: AWWA (1992).

can be found in AWWA (1992), and values of (X + P)i are typically on the order of 1.4. The NFF calculated using Equation 3.20 should not exceed 45,000 L/min (12,000 gpm), nor be less than 2000 L/min (500 gpm). According to AWWA (1992), 2000 L/min (500 gpm) is the minimum amount of water with which any fire can be controlled and suppressed safely and effectively. The NFF should be rounded to the nearest 1000 L/min (250 gpm) if less than 9000 L/min (2500 gpm), and to the nearest 2000 L/min (500 gpm) if greater than 9000 L/min (2500 gpm). For one- and two-family dwellings not exceeding two stories in height, the NFF listed in Table 3.5 should be used. For other habitable buildings not listed in Table 3.5, the NFF should not exceed 13,000 L/min (3500 gpm). Usually the local water utility will have a policy on the upper limit of fire protection that it will provide to individual buildings. Those wanting higher fire flows need to either provide their own system or reduce fire-flow requirements by installing sprinkler systems, fire walls, or fire-retardant materials. Estimates of the needed fire flow calculated using Equation 3.20 are used to determine the fire-flow requirements of the water-supply system, where the needed fire flow is calculated at several representative locations in the service area, and it is assumed that only one building is on fire at any time. The design duration of the fire should follow the guidelines in Table 3.6. If these durations cannot be maintained, insurance rates are typically increased accordingly.

TABLE 3.5: Needed Fire Flow for One- and Two-Family Dwellings

Distance between buildings (m) (ft) >30 9.5–30 3.5–9.5 100 30–100 10–30 20 >20

110

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Design of Water-Distribution Systems

distribution systems are the quality of the treated water fed into the system; the material and condition of the pipes, valves, and storage facilities that make up the system; and the amount of time that the water is kept in the system (AWWA, 2003c; Grayman et al., 2000). Key processes that affect water quality within the distribution system usually include the loss of disinfection residual, with resulting microbial regrowth, and the formation of disinfection by-products such as trihalomethanes. Water-quality deterioration is often proportional to the time the water is resident in the distribution system. The longer the water is in contact with the pipe walls and is held in storage facilities, the greater the opportunity for water-quality changes. Generally, a hydraulic detention time of less than 7 days in the distribution system is recommended (AWWA, 2003c). The velocity of flow in many water mains is very low, particularly in dead-end mains or in areas of low water consumption. As a result, corrosion products and other solids tend to settle on the pipe bottom. These deposits can be a source of color, odor, and taste in the water when they are stirred up by an increase in flow velocity or a reversal of flow in the distribution system. To prevent these sediments from accumulating and causing water-quality problems, pipe flushing is a typical maintenance routine. Flushing involves opening a hydrant located near the problem area, and keeping it open as long as needed to flush out the sediment, which typically requires the removal of up to three pipe volumes (AWWA, 2003c). Experience will teach an operator how often or how long certain areas should be flushed. Some systems find that dead-end mains must be flushed as often as weekly to avoid customer complaints of rusty water. The flow velocity required for effective flushing is in the range of 0.75–1.1 m/s (2.5–3.5 ft/s), with velocities limited to less than 3.1–3.7 m/s (10–12 ft/s) to avoid excessive scouring (AWWA, 2003c). If flushing proves to be inadequate for cleaning mains, air purging or cleaning devices, such as swabs or pigs, may need to be used. 3.4.4

Network Analysis

Methodologies for analyzing pipe networks were discussed in Section 2.3, and these methods can be applied to any given pipe network to calculate the pressure and flow distribution under a variety of demand conditions. In complex pipe networks, the application of computer programs to analyze pipe networks is standard practice. Computer programs allow engineers to easily calculate the hydraulic performance of complex networks, the age of water delivered to consumers, and the origin of the delivered water. Water age, measured from the time the water enters the system to the time the water exits the system, gives an indication of the overall quality of the delivered water. Steady-state analyses are usually adequate for assessing the performance of various components of the distribution system, including the pipelines, storage tanks, and pumping systems, while time-dependent (transient) simulations are useful in assessing the response of the system over short time periods (days or less), evaluating the operation of pumping stations and variable-level storage tanks, performing energy consumption and cost studies, and water-quality modeling. Modelers frequently refer to time-dependent simulations as extended-period simulations. An important part of analyzing large water-distribution systems is the skeletonizing of the system, which consists of representing the full water-distribution system by a subset of the system that includes only the most important elements. For example, consider the case of a water supply to the subdivision shown in Figure 3.11(a), where the system shown includes the service connections to the houses. A slight degree of skeletonization could be achieved by omitting the household service pipes (and their associated head losses) from consideration and accounting for the water demands at the tie-ins, as shown in Figure 3.11(b). This reduces the number of junctions from 48 to 19. Further skeletonization can be achieved by modeling just 4 junctions, consisting of the ends of the main piping and the major intersections, as shown in Figure 3.11(c). In this case, the water demands are associated with the nearest junctions to each of the service connections, and the dashed lines in Figure 3.11(c) indicate the service areas for each junction. A further level of skeletonization is shown in Figure 3.11(d), where the water supply to the entire subdivision is represented by a single node, at which the water demand of the subdivision is attributed. Clearly, further levels of skeletonization could be possible in large water-distribution systems. As a general guideline, larger systems permit more degrees of skeletonization without introducing significant error in the flow conditions of main distribution pipes.

Section 3.5

Building Water-Supply Systems

111

FIGURE 3.11: Skeletonizing a water-distribution system Source: Haestad Methods, 1997, Practical Guide: Hydraulics and Hydrology, pp. 61–62. Copyright © 1997 by Haestad Methods, Inc. Reprinted with permission.

(a)

(b)

(c)

(d)

The results of a pipe-network analysis should generally include pressures and/or hydraulic grade line elevations at all nodes, flow, velocity, and head loss through all pipes, as well as flow rates into and out of all storage facilities. These results are used to assess the hydraulic performance and reliability of the network, and they are to be compared with the guidelines and specifications required for acceptable performance. 3.5 Building Water-Supply Systems Water supply to individual buildings is typically provided by a service line that is connected to the water main that abuts the building. An isometric view of a typical residential water-supply system is shown in Figure 3.12. In this system, the service line first connects to a water meter that measures the flow entering the building, and a backflow preventer is typically provided to prevent backflows from the building into the water main. Backflows of possibly contaminated water can be caused by in-building sources such as pumps, boilers, and heat-exchange equipment. Water heaters are used to heat some of the water supply, and branches from the hot and cold water lines provide water to different parts of the building. The different parts of the building illustrated in Figure 3.12 correspond to delivery points A, B, and C. Fixtures such as toilets, sinks, showers, and washing machines are located near these delivery points. The basic hydraulic design problem is to size the pipes to ensure that water is delivered to all water-supply fixtures in the building with an adequate flow rate and at an adequate pressure. Guidelines for minimum flow rates and pressures at various fixtures are typically stated in the applicable local plumbing code. A typical design procedure for building watersupply systems is as follows: Step 1: Estimate the minimum daily pressure at the water-supply main. This pressure is best determined from direct measurements of pressure in the existing water main, and such measurements are usually available from the utility that serves the area. Alternately, a hydraulic analysis of the existing water-distribution system might be necessary to estimate the water-main pressure at the building location. Typical minimum daily pressures in water mains are on the order of 350 kPa (50 psi). Step 2: Estimate the design flow in the service line and the principal pipeline branches in the building.

112

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FIGURE 3.12: Isometric view of residential water-supply system

A

B

Hot water C Cold water

z Service line Water heater

y x

Backflow preventer Meter Main

Step 3: Specify the minimum required pressures at the fixtures connected to the principal branches. Step 4: Determine the minimum pipe diameter in each branch that will provide adequate pressures at all the fixtures connected to that branch. The procedures to be followed in Steps 2–4 are given in more detail in the following sections. 3.5.1

Specification of Design Flows

Each principal branch in a building water-supply system will have its own design flow based on the number and types of fixtures to be connected to that branch. The principal branches are laid out to support the architectural design and function of the building, and the locations and types of fixtures connected to each of the branches are identified. The water demand at each fixture is expressed in terms of water-supply fixture units (WSFU), which is an abstract number that takes into account both the flow rate to be delivered to the fixture and the frequency of use of the fixture. A commonly used relationship between the type of fixture and the number of fixture units assigned to that fixture is shown in Table 3.9 (Uniform Plumbing Code, 2009). For each branch in the building, the total number of fixture units to be supplied by that branch is determined by summing up the fixture units connected to that branch. The fixture-unit approach is generally preferred over summing the flow rates to be delivered at each fixture since the more fixtures there are the less likelihood that all the fixtures would be in operation at the same time. Once the number of fixture units to be supported by each branch is determined, the fixture units are converted into design flow rates using a curve similar to that shown in Figure 3.13. Curves relating design flow rates to fixture units, such as Figure 3.13, are called Hunter curves after Roy B. Hunter, who first suggested this relationship (Hunter, 1940). Hunter curves are included in most local plumbing codes, although some research has indicated that the peak flows estimated from fixture units provide conservative estimates of peak flows (AWWA, 2004). 3.5.2

Specification of Minimum Pressures

For each principal branch in a building water-supply system, the minimum pressure required at the locations where the fixtures are connected must be specified. Attainment of these pressures at the fixture connections is necessary to yield the required minimum flow rates through the fixtures, and these minimum flow rates are essential for keeping the fixtures

Section 3.5

Building Water-Supply Systems

113

TABLE 3.9: Water-Supply Fixture Units (WSFU)

Load values in WSFU Fixture

Occupancy

Supply control

Cold

Hot

Total

Bathroom group Bathroom group Bathtub Bathtub Bidet Combination fixture

Private Private Private Public Private Private

Flush tank Flush valve Faucet Faucet Faucet Faucet

2.7 6 1 3 1.5 2.25

1.5 3 1 3 1.5 2.25

3.6 8 1.4 4 2 3

Dishwashing machine Drinking fountain Kitchen sink Kitchen sink Laundry trays (1 to 3)

Private Offices, etc. Private Hotel, restaurant Private

Automatic 9.5-mm (3/8-in.) valve Faucet Faucet Faucet

— 0.25 1 3 1

1.4 — 1 3 1

1.4 0.25 1.4 4 1.4

Lavatory Lavatory Service sink Shower head Shower head

Private Public Offices, etc. Public Private

Faucet Faucet Faucet Mixing valve Mixing valve

0.5 1.5 2.25 3 1

0.5 1.5 2.25 3 1

0.7 2 3 4 1.4

Urinal Urinal Urinal

Public Public Public

25-mm (1-in.) flush valve 10 19-mm (3/4-in.) flush valve 5 Flush tank 3

— — —

10 5 3

Washing machine, 3.6 kg (8 lb) Private Washing machine, 3.6 kg (8 lb) Public Washing machine, 6.8 kg (15 lb) Public

Automatic Automatic Automatic

1 2.25 3

1 2.25 3

1.4 3 4

Water closet Water closet Water closet Water closet Water closet

Flush valve Flush tank Flush valve Flush tank Flushometer tank

6 2.2 10 5 2

— — — — —

6 2.2 10 5 2

Private Private Public Public Public or private

Source: International Plumbing Code (2012). FIGURE 3.13: Relationship between demand and fixture units

500

Demand (L /min)

400 Predominantly flush valves 300

200 Predominantly flush tanks 100

0

0

20

40

60

80

100 120 140 106 180 200 220 240 Fixture Units

114

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Design of Water-Distribution Systems TABLE 3.10: Typical Minimum Flow Rates and Pressures in Building Fixtures

Flow rate (L/min)

Pressure (kPa)

Flow rate (gpm)

Pressure (psi)

15 8 15 10.4 2.8 15

138 138 55 55 55 55

4 2 4 2.75 0.75 4

20 20 8 8 8 8

8 8

55 55

2 2

8 8

Shower head Shower head, temperature controlled Sink, residential Sink, service

11 11 9.5 11

55 138 55 55

3 3 2.5 3

8 20 8 8

Urinal, valve Water closet, blow out, flushometer valve Water closet, siphonic, flushometer valve

45 95 95

172 310 241

12 25 25

25 45 35

Water closet, tank, close coupled Water closet, tank, one piece

11 23

138 138

3 6

20 20

Fixture Bathtub Bidet Combination fixture Dishwasher, residential Drinking fountain Laundry tray Lavatory, private Lavatory, public

Source: International Plumbing Code (2012).

clean and sanitary. Recommended minimum pressures at various types of fixtures and their associated flow rates are given in Table 3.10. 3.5.3

Determination of Pipe Diameters

Pipe diameters in each branch of a building water supply system are selected so that the required pressure heads at the fixture connections are attained. To determine the required pipe diameters, the energy equation is applied between the water main and the delivery point in each branch pipe such that p0 − γ



   V12 p1 + + z = hf + hl γ 2g

(3.22)

where p0 is the pressure in the water main [FL−2 ], γ is the specific weight of water [FL−3 ], p1 is the pressure at the delivery point of the branch [FL−2 ], V1 is the flow velocity in the −1 ], g is gravity [LT−2 ], z is the height of the delivery point above the water branch [LT main [L], hf is the sum  of the friction losses in the pipes between the water main and the delivery point [L], and hl is the sum of the local head losses between the water main and the delivery point [L]. For each branch, the design fixture is usually the fixture with the greatest head loss from the main and/or the fixture with the greatest elevation difference (z) from the main and/or the fixture that requires the largest pressure. It is usually assumed that the total flow to be accommodated by a branch occurs up to the location of the design fixture for that branch. In some cases, several design fixtures might need to be tried to determine the one(s) that control specification of the branch-pipe diameter. The fixture requiring the largest branch-pipe diameter is called the critical fixture. The head loss in each pipeline is best described by the Darcy–Weisbach equation (Equation 2.33); however, it is also common practice in the United States to use the Hazen– Williams equation (Equation 2.82) to calculate head losses in building water-supply pipes. These head-loss equations can be put in the following convenient forms

Section 3.5

Building Water-Supply Systems



 ⎪ fL ⎪ ⎪ 0.230 Q2 , Darcy–Weisbach equation ⎪ ⎪ D5 ⎨   hf = ⎪ L ⎪ ⎪ ⎪ 85.4 Q1.85 , Hazen–Williams equation ⎪ 1.85 D4.87 ⎩ CH

115

(3.23)

where hf is the lead loss [m], f is the friction factor [dimensionless], L is the length of the pipeline [m], D is the diameter of the pipeline [cm], Q is the flow rate in the pipeline [L/min], and CH is the Hazen–Williams coefficient [dimensionless]. Design codes commonly incorporate graphical plots of these equations; however, if the friction-loss parameters (f or CH ) can be estimated from available data, it is generally preferable to use the analytic head-loss relationships directly rather than read head losses from plots of these equations. A typical roughness height used in the Darcy–Weisbach equation for copper tubing and PVC pipe is 0.0015 mm (1.5 μm). Typical values of CH are 130 for copper tubing and 140 for PVC pipe. Local head losses caused by valves and fittings in building water-supply systems can account for a significant portion of the total head losses. Local head losses include losses at the water meter, backflow preventer, transitions (e.g., “tees” and “ells”), and valves. Even when valves are open, they can cause significant head losses. It is common practice to express local head losses in terms of an equivalent pipe length that would cause the same head loss due to friction as the local head loss, and such relationships are shown in Table 3.11 for standard fittings of various sizes. It is apparent from Table 3.11 that the magnitude of the local head loss depends on the size of the fitting, with globe valves generally causing the highest local head losses. The minimum allowable diameter of service lines is typically 19 mm (3/4 in.), and diameters of service lines in residential and small commercial buildings rarely exceed 50 mm (2 in.). The diameters of pipelines within buildings are generally less than or equal to the diameter of the service line. The standard commercial diameters of building water-supply TABLE 3.11: Head Loss in Standard Fittings in Terms of Equivalent Pipe Lengths

90◦ Ell

45◦ Ell

90◦ Tee

Gate valve

Globe valve

Angle valve

(in.)

(m)

(m)

(m)

(m)

(m)

(m)

0.305

0.183

0.457

0.061

2.438

1.219

0.610

0.366

0.914

0.122

4.572

2.438

19.1

3 8 1 2 3 4

0.762

0.457

1.219

0.152

6.096

3.658

25.4

1

0.914

0.549

1.524

0.183

7.620

4.572

32

1.219

0.732

1.829

0.244

10.668

5.486

38

1 14 1 12

1.524

0.914

2.134

0.305

13.716

6.706

51

2

2.134

1.219

3.048

0.396

16.764

8.534

64

2 12

2.438

1.524

3.658

0.488

19.812

10.363

Fitting Diameter (mm) 9.5 12.7

76

3

3.048

1.829

4.572

0.610

24.384

12.192

102

4

4.267

2.438

6.401

0.823

38.100

16.764

127

5

5.182

3.048

7.620

1.006

42.672

21.336

152

6

6.096

3.658

9.144

1.219

50.292

24.384

Source: Uniform Plumbing Code (2012).

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Design of Water-Distribution Systems

lines are those shown in Table 3.11. Minimum allowable pipe sizes within buildings are typically either 13 mm (1/2 in.) or 19 mm (3/4 in.), depending on the types of fixtures served by the connected pipe. After pipe sizes are selected, the flow velocities under design conditions should be checked to ensure that they are in an acceptable range. Pipe flows with velocities above 3 m/s (10 ft/s) are usually too noisy, and flows with velocities above 1.8 m/s (6 ft/s) might be too noisy for acoustical-critical locations (Stein et al., 2006). The recommended design procedure is illustrated in the following example.

EXAMPLE 3.6 The water-supply system for a two-story factory building is shown in Figure 3.14, where Type L copper pipe is to be used for the service pipe and all pipelines in the building. The minimum daily pressure in the water main is 380 kPa (55 psi), and the diameter of the service line is 64 mm (2 1/2 in.). Based on manufacturers data, the tap into the water main is expected to cause a pressure drop of 10 kPa (1.5 psi), the meter is expected to cause a pressure drop of 76 kPa (11 psi), and the backflow preventer is expected to cause a pressure drop of 62 kPa (9 psi). The distribution of demand in the building is as follows:

Type

Pipe

Fixture units (WSFU)

Flow (L/min)

Service

AB

288

409

Cold

BC CF CE

264 132 132

396 291 291

Hot

BC’ C’F’ C’E’

24 12 12

144 45 45

The critical fixtures on both floors of the building require a minimum allowable pressure of 103 kPa (15 psi). All valves are expected to perform as gate valves, and all ells and tees are at 90◦ . Determine the required pipe diameters for the cold-water lines. Solution The calculation to determine the pressures at the critical fixtures is an iterative process in which different pipe sizes are tried until the calculated pressures at all the critical fixtures are greater than their minimum required values. The results of the design calculations are summarized in Table 3.12. The calculation procedures for lines AB and BC are typical and are given below. Line AB: Line AB is the service line. At the beginning of this line the (main) pressure, p0 , is 380 kPa, and so 380 p = 38.82 m Starting Head = 0 = γ 9.79 The flow in pipe AB is the total building demand, equal to 409 L/min, the length is 17.0 m, and the diameter is 64 mm. Based on these data, the velocity in the pipe (= Q/A) is 2.12 m/s, which is less than 3 m/s and is therefore acceptable. The fitting length associated with the 3 valves and the tee fitting at the end of the 64-mm line can be derived using the data in Table 3.11 which yields the following result: Fitting Length = 3 * valve loss + 1 * tee loss = 3(0.488) + 1(3.658) = 5.12 m which gives a total equivalent pipe length of 17.0 m + 5.12 m = 22.12 m. Other losses in the service line are given in terms of pressure losses at the water-main tap (= 10 kPa), meter (= 76 kPa), and backflow preventer (= 62 kPa). Hence, Other Losses =

10 + 76 + 62 total pressure loss = = 15.12 m γ 9.79

Section 3.5

Building Water-Supply Systems

117

FIGURE 3.14: Factory water-supply system

Building 46 m D

E

Legend: Floor 2

Valve Tee Ell

D1

E’ 46 m

Cold water pipe Hot water pipe

4.0 m 4.0 m 46 m C

F Floor 1

46 m C’

F’

2.5 m Backflow preventer Main A

Meter

2.5 m

B 1.7 m

1m B’

17 m

Water heater

TABLE 3.12: Results of Design Calculations Starting Fitting Total Friction Other Elev. Terminal Terminal head Flow Length Diameter Velocity length length loss losses diff. head pressure Pipe (m) (L/min) (m) (mm) (m/s) (m) (m) (m) (m) (m) (m) (kPa) AB BC CF CD DE

38.82 22.33 21.97 21.97 21.26

409 396 291 291 291

17.0 2.5 46.0 4.0 46.0

64 64 51 51 51

2.12 2.05 2.37 2.37 2.37

5.12 3.66 3.05 3.05 3.05

22.12 6.16 49.05 7.05 49.05

1.37 0.36 4.92 0.71 4.92

15.12 0 0 0 0

0 2.5 0 4.0 0

22.33 21.97 17.05 21.26 16.34

216 188 164 166 157

The friction loss in the pipe is calculated using the Darcy–Weisbach equation. For copper tubing, it can be assumed that ks = 0.0015 mm and with D = 64 mm, V = 2.12 m/s, and ν = 10−6 m2 /s (at 20◦ C), the Reynolds number, Re, in pipe AB is given by VD (2.12)(0.064) = 135,600 = ν 10−6 The friction factor, f , is calculated using the Swamee–Jain equation (Equation 2.39) as Re =

0.25

f = 

log

ks 5.74 3.7D + Re0.9

 2

0.25

= 

log

0.0015 + 5.74 3.7(64) 135,6000.9

= 0.0173 2

118

Chapter 3

Design of Water-Distribution Systems and the head loss due to friction, hf , is given by the Darcy–Weisbach equation as hf =

fL V 2 (0.0173)(22.12) (2.12)2 = = 1.37 m D 2g (0.064) 2(9.81)

The terminal head in pipe AB is then given by Terminal Head = starting head − friction loss − other losses = 38.82 m − 1.37 m − 15.12 m = 22.33 m The terminal pressure, p, is related to the terminal head by the relationship     2.122 V2 = (9.79) 22.33 − 0 − = 216 kPa p = γ terminal head − z − 2g 2(9.81) Line BC: Line BC originates at the end of line AB. The flow in pipe BC is 396 L/min and the length is 2.5 m. Taking the diameter as 51 mm yields a velocity (= Q/A) of 3.23 m/s, which is greater than 3 m/s and is therefore unacceptable. Taking the diameter as 64 mm yields a velocity of 2.05 m/s, which is less than 3 m/s and therefore acceptable. The starting head in line BC is equal to the terminal head in pipe AB which was calculated previously as 22.33 m. The fitting length associated with the tee at the end of the 64-mm pipe is derived from Table 3.11 which gives, Fitting Length = tee loss = 3.658 m and so the total equivalent pipe length of BC is 2.5 m + 3.658 m = 6.158 m L 6.16 m. The Reynolds number, Re, in pipe BC is Re =

(2.05)(0.064) VD = 13,100 = ν 10−6

and the friction factor, f , is f = 

0.25

ks + 5.74 log 3.7D Re0.9

=  2

0.25

= 0.0174 2

0.0015 + 5.74 log 3.7(64) 13,1000.9

and the head loss due to friction, hf , is hf =

(0.0174)(6.16) (2.05)2 fL V 2 = = 0.36 m D 2g (0.064) 2(9.81)

The terminal head in pipe BC is given by Terminal Head = starting head − friction loss = 22.33 m − 0.36 m = 21.97 m Since the change in elevation, z, between B and C is 2.5 m, the terminal pressure, p, is line BC is given by     2.052 V2 = (9.79) 21.97 − 2.5 − = 188 kPa p = γ terminal head − z − 2g 2(9.81) Calculations to determine the pressures at the ends of the other cold-water lines are done in the same manner as for lines AB and BC. Since the objective of the design is to achieve a minimum pressure of 103 kPa at terminal locations E and F, diameter adjustments show that it is acceptable to use 51-mm (2 in.) pipe for all lines except AB and BC, where 64-mm (2 1/2 in.) lines should be used. With this water-supply system, the expected pressures at the critical fixtures are 164 kPa on the first floor and 157 kPa on the second floor, both of which exceed the minimum required pressure of 103 kPa.

In cases where the water-main pressure is inadequate to provide adequate pressures throughout a building, supplementary components to the building water-supply system are used. Common supplementary building systems for augmenting water pressure are gravity tanks, direct-feed booster pumps, and hydro-pneumatic systems. These are described briefly as follows:

Problems

119

Gravity Tanks. Gravity-tank systems are widely used throughout the world. This system uses a roof tank to provide stable pressures and storage to supplement the available water supply. One or two pumps are used to pump water to the roof tank, which is typically sized to cover the peak building demands during the day. Water-level controls mounted in the tank control on/off pump operation to maintain the water level in the tank. A gravity tank system is recommended for buildings with long periods of low water demands, and is ideal for many commercial and residential applications. Direct-Feed Booster Pumps. Booster-pump systems are the most widely used type of supplementary system in the United States. In these systems, pumps are directly connected to the building water supply for the sole purpose of increasing the water pressure within the building. A booster-pump system is usually recommended for buildings where there is a continuous water demand, such as in hotels and hospitals. Hydro-Pneumatic Systems. Hydro-pneumatic systems use pressurized air and stored water contained in the same tank. A pump supplies water to the tank, with the water being contained in an expandable bladder surrounded by compressed air. As the volume of water stored in the bladder changes, the volume of air in the pressurized tank adjusts, thereby maintaining the water pressure in the pipes that are connected to the bladder. Problems 3.1. For a town, the water supply is at 150 L/d/person. Suggest a suitable breakup of the consumption on different heads. 3.2. The design life of a planned municipal water-distribution system is to end in 2030, and the population in the town has been measured every 10 years since 1920 by the U.S. Census Bureau. The reported populations are tabulated below, and it is estimated that the saturation population of the town is 100,000 people. Estimate the population in the town in 2030 using: (a) graphical extension, (b) arithmetic growth projection, (c) geometric growth projection, (d) declining growth projection, and (e) logistic curve projection.

Year

Population

1920 1930 1940 1950 1960 1970 1980 1990

25,521 30,208 30,721 37,253 38,302 41,983 56,451 64,109

3.3. In two periods of 10 years each, a city population has grown from 25,000 to 40,000 and then to 70,000. Determine the (a) saturation population, (b) equation of the logistic curve, and (c) expected population after 10 years. 3.4. A 300 mm diameter pipe is required for the distribution main of a city water supply. As pipes above 250 mm diameter are not available, two parallel mains of the

3.5.

3.6. 3.7.

3.8.

3.9.

same diameter will be laid. Determine the diameter of the parallel mains. A water supply scheme has to be designed for a city with a population of 150,000. Estimate the important kinds of draft that may be required for an average water consumption of 150 L/d/person. What is the maximum fire flow and corresponding duration that can be used for any building? Water is to be supplied in a four-story housing building having four flats on each floor. The average number of persons living per flat is 5 and each flat is provided with a toilet and a kitchen. The municipal water supply in the area is restricted to 3 hours in the morning and 3 hours in the evening. Design the sizes of the various units that are to be installed to ensure continuous tank supply. The living standards require an average daily demand of 150 L/d/person. For a city of a population of 100,000, estimate the coincident draft with an average water consumption of 200 L/d/person. Water meant to supply a town is to be stored in a large tank whose water surface elevation is 85.37 m. This tank is being supplied by a pipeline from a junction J. The friction loss in meters of water in this pipe is 0.6 Q1.95 , where Q is the discharge in m3 /s. Junction J receives water under pressure and without leakage from (a) a large reservoir I whose water surface elevation is constant at 100 m through a pipe (hf = 0.35 Q2 , per the Darcy-Weisbach equation); and (b) a second large reservoir II with its constant water level at 96 m through a pipe (hf = 0.46 Q1.85 , per the Hazen-Williams equation). Neglecting losses other than the pipe friction, determine the incoming flow into the tank in the town.

120

Chapter 3

Design of Water-Distribution Systems C 20 m

B P 20 m 3m

A

P

FIGURE 3.15: Water-distribution system

3.10. Calculate the volume of storage required for the elevated storage reservoir in the water-supply system described in Problem 3.7. 3.11. Water is to be distributed to 70,000 people from a service reservoir using the transmission pipes shown schematically in Figure 3.15. The average per-capita demand is estimated to be 600 L/d/person, with 50,000 people supplied from the connection at point B and 20,000 people supplied from the connection at point C. The transmission lines consist of 1200-mm diameter steel pipe with a typical roughness height of 1 mm, the length of pipe from A to B is 5 km, and the length of pipe from B to C is 7 km. It is estimated that the water demand will fluctuate such that the maximum-day factor is 1.8 and the maximum-hour factor is 3.25. The design fire flow in the service area that is connected to B is 15,000 L/min, and the fire flow for the service area connected to C is 10,000 L/min. Under design conditions, the pressure at B is to be no less than 550 kPa, and the pressure at C no less than 480 kPa. The two pumps within the system are to be selected such that they operate at their maximum efficiency, and these pumps will both be driven by 1200-rpm motors. (a) What are the design flows for lines AB and BC? (b) What specific speeds and pump types are required? Explain clearly what you would do with this information. (c) What volume of storage reservoir is required? 3.12. Design flow rates in household plumbing are based on the number and types of plumbing fixtures as measured by “fixture units.” The Hunter curve is used to relate fixture units to flow rate. Water supply is being provided in a seven-story building through an overhead tank installed at a height of 1.2 m above the terrace. The

height of each story is 3 m and water taps are installed at a height of 1.2 m at each floor. One supply pipe from the tank is serving two flats on each floor, and each flat is provided with the following water supply fixtures (a) a water closet-flushing tank type at 8.0 m, (b) a wash basin at 6 m, (c) a bath taps with shower at 4.5 m, and (d) a kitchen sink at 1.5 m (at mentioned distance from the down feed). Design the pipe diameters for the main as well as the branch lines. Suitable values of fixture units and their discharges rates for the fixtures may be assumed. 3.13. Design the hot-water pipes for the building given in Example 3.6. 3.14. Water is to be delivered from a public water supply to a two-story building. Under design conditions, each floor of the building is to be simultaneously supplied with 200 L/min. The pipes in the building plumbing system are to be made of galvanized iron. The length of pipe from the public water supply to the delivery point on the first floor is 20 m, the length of pipe from the delivery point on the first floor to the delivery point on the second floor is 5 m, the water delivery point on the first floor is 2 m above the water main connection, and the delivery point on the second floor is 3 m above the delivery point on the first floor. If the water pressure at the water main (meter) is 380 kPa, what is the minimum diameter pipe in the building plumbing system to ensure that the pressure is at least 240 kPa on the second floor? Neglect minor losses and consider pipe diameters in increments of 1/4 in., with the smallest allowable diameter being 1/2 in. For the selected diameter under design conditions, what is the water pressure on the first floor?

C H A P T E R

4

Fundamentals of Flow in Open Channels 4.1 Introduction In open-channel flows the water surface is exposed to the atmosphere. This type of flow is typically found in sanitary sewers, drainage conduits, canals, and rivers. Open-channel flow, sometimes referred to as free-surface flow, is more complicated than closed-conduit flow, since the location of the free surface is not constrained, and the depth of flow depends on such factors as the discharge and the shape and slope of the channel. Flows in conduits with closed sections, such as pipes, may be classified as either open-channel flow or closed-conduit flow, depending on whether the conduit is flowing full. A closed pipe flowing partially full is an open-channel flow, since the water surface is exposed to the atmosphere. Open-channel flow is said to be steady if the depth of flow at any specified location does not change with time; if the depth of flow changes with time, the flow is called unsteady. Most open-channel flows are analyzed under steady-flow conditions. The flow is said to be uniform if the depth of flow is the same at every cross section of the channel; if the depth of flow varies along the channel, the flow is called nonuniform or varied. Uniform flow can be either steady or unsteady, depending on whether the flow depth changes with time; however, uniform flows are practically nonexistent in nature. More commonly, open-channel flows are either steady nonuniform flows or unsteady nonuniform flows. Open channels are classified as either prismatic or nonprismatic. Prismatic channels are characterized by an unvarying shape of the cross section, constant bottom slope, and relatively straight alignment. In nonprismatic channels, the cross section, alignment, and/or bottom slope change along the channel. Constructed drainage channels such as pipes and canals tend to be prismatic, while natural channels such as rivers and creeks tend to be nonprismatic. This chapter covers the basic principles of open-channel flow and derives the most useful forms of the continuity, momentum, and energy equations. These equations are then applied to the computation of water-surface profiles. 4.2 Basic Principles The governing equations of flow in open channels are the continuity, momentum, and energy equations. Any flow in an open channel must satisfy all three of these equations. Analysis of open-channel flow can usually be accomplished with the control-volume form of the governing equations, and the most useful forms of these equations for steady open-channel flows are derived in the following sections. 4.2.1

Steady-State Continuity Equation

Consider the case of steady nonuniform flow in the open channel illustrated in Figure 4.1. The flow enters and leaves the control volume normal to the control surfaces, with the inflow velocity distribution denoted by v1 and the outflow velocity distribution by v2 ; both the inflow and outflow velocities vary across the control surfaces. The steady-state continuity equation can be written as   ρv1 dA = ρv2 dA (4.1) A1

A2

where ρ is the density of the fluid, which can be taken as constant for most applications involving water. Defining V1 and V2 as the average velocities across A1 and A2 , respectively, where 121

122

Chapter 4

Fundamentals of Flow in Open Channels

FIGURE 4.1: Flow in an open channel

1 2 Water surface

Flow v1

v2 Control volume

Velocity distribution

Boundary of control volume

V1 = and

1 A1

1 V2 = A2

 v1 dA

(4.2)

v2 dA

(4.3)

A1

 A2

then, for an incompressible fluid (ρ = constant) such as water, the steady-state continuity equation (Equation 4.1) can be written as V1 A1 = V2 A2

(4.4)

which is the same expression that was derived for steady flow of an incompressible fluid in closed conduits. 4.2.2

Steady-State Momentum Equation

Consider the case of steady nonuniform flow in the open channel illustrated in Figure 4.2. The steady-state momentum equation for the control volume shown in Figure 4.2 is given by   Fx = ρvx v · n dA (4.5) A

where Fx represents the forces in the flow direction, x; A is the surface area of the control volume; vx is the flow velocity in the x-direction, v is the velocity vector, and n is a unit normal directed outward from the control volume. Since the velocities normal to the control surface are nonzero only for the inflow and outflow surfaces, Equation 4.5 can be written as

FIGURE 4.2: Steady nonuniform flow in an open channel

1 Hydrostatic pressure distribution

Control volume 2

y1

γAΔx Water surface

y2

τ PΔ 0 x z1

Q

Δx z2

Datum

θ x

Section 4.2





Basic Principles

123



Fx = A2

ρv2x dA −

A1

ρv2x dA

(4.6)

where A1 and A2 are the upstream and downstream areas of the control volume, respectively. If the velocity is uniformly distributed (i.e., constant) across the control surface, then Equation 4.6 becomes  Fx = ρv22 A2 − ρv21 A1 (4.7) where v1 and v2 are the velocities on the upstream and downstream faces of the control volume, respectively. In reality, velocity distributions in open channels are never uniform, and so it is convenient to define a momentum correction coefficient, β, by the relation  v2 dA A β= (4.8) V2A where V is the mean velocity across the channel section of area A. The momentum correction coefficient, β, is sometimes called the Boussinesq coefficient, or simply the momentum coefficient. Applying the definition of the momentum correction coefficient to Equation 4.6 leads to the following form of the momentum equation:  Fx = ρβ2 V22 A2 − ρβ1 V12 A1 (4.9) where β1 and β2 are the momentum correction coefficients at the upstream and downstream faces of the control volume, respectively. Values of β are typically in the range of 1.03–1.07 for turbulent flow in prismatic channels, and typically in the range of 1.05–1.17 for turbulent flow in natural streams. Since the continuity equation requires that the discharge, Q, is the same at each cross section, then Q = A1 V1 = A2 V2 (4.10) and the momentum equation (Equation 4.9) can be written as  Fx = ρβ2 QV2 − ρβ1 QV1

(4.11)

By definition, values of β must be greater than or equal to unity. In practice, however, deviations of β from unity are second-order corrections that are small relative to the uncertainties in the other terms in the momentum equation. By assuming β1 L β2 = 1 the momentum equation can be written as  Fx = ρQV2 − ρQV1 = ρQ(V2 − V1 )

(4.12)

Considering the forces acting on the control volume shown in Figure 4.2, then Equation 4.12 can be written as γ Ax sin θ − τ0 Px + γ A(y1 − y2 ) = ρQ(V2 − V1 )

(4.13)

where γ is the specific weight of the fluid; A is the average cross-sectional area of the control volume; x is the length of the control volume; θ is the inclination of the channel; τ0 is the average shear stress on the channel boundary within the control volume; P is the average (wetted) perimeter of the channel cross section; and y1 and y2 are the upstream and downstream depths, respectively, at the control volume. The three force terms on the left-hand side of Equation 4.13 are the component of the weight of the fluid in the direction of flow, the shear force exerted by the channel boundary on the moving fluid, and the net

124

Chapter 4

Fundamentals of Flow in Open Channels

hydrostatic force. If z1 and z2 are the elevations of the bottom of the channel at the upstream and downstream faces of the control volume, then sin θ =

z1 − z2 x

(4.14)

Combining Equations 4.13 and 4.14 and rearranging leads to τ0 = −γ

A y A V V A z − γ − γ P x P x P g x

(4.15)

where z, y, and V are defined by z = z2 − z1 ,

y = y2 − y1 ,

V = V2 − V1

(4.16)

and V(= Q/A) is the average velocity in the control volume. The ratio A/P is commonly called the hydraulic radius, R, where A R= (4.17) P Combining Equations 4.15 and 4.17 and taking the limit as x → 0 yields   z y V V + lim + lim τ0 = −γ R lim g x→0 x x→0 x x→0 x   dy V dV dz + + = −γ R dx dx g dx   d V2 = −γ R z + y + dx 2g

(4.18)

The term in brackets is the mechanical energy per unit weight of the fluid, E, defined as E=y + z +

V2 2g

(4.19)

It should be noted that the mechanical energy per unit weight of a fluid element is usually defined as p /γ + z + V 2 /2g, where z is elevation of the fluid element relative to a defined datum and p is the pressure at the location of this fluid element. If the pressure distribution is hydrostatic across the channel cross section, then p /γ + z = constant = y + z, where y is the water depth and z is the elevation of the bottom of the channel. The mechanical energy per unit weight, E, can therefore be written as y + z + V 2 /2g. A plot of E versus the distance along the channel is called the energy grade line. The momentum equation, Equation 4.18, can now be written as dE (4.20) τ0 = −γ R dx or τ0 = γ RSf (4.21) where Sf is equal to the slope of the energy grade line, which is taken as positive when it slopes downward in the direction of flow. 4.2.2.1

Darcy–Weisbach equation

In practical applications, it is useful to express the average shear stress, τ0 , on the boundary of the channel in terms of the flow and surface-roughness characteristics. A functional expression for the average shear stress, τ0 , can be expressed in the following form τ0 = f0 (V, R, ρ, μ, ,   , m, s)

(4.22)

Section 4.2

Basic Principles

125

where V is the mean velocity in the channel [LT−1 ], R is the hydraulic radius [L], ρ is the fluid density [ML−3 ], μ is the dynamic viscosity of the fluid [FL−2 T],  is the characteristic size of the roughness projections on the channel boundary [L],   is the characteristic spacing of the roughness projections [L], m is a form factor that describes the shape of the roughness elements [dimensionless], and s is a channel shape factor that describes the shape of the channel cross section [dimensionless]. In accordance with the Buckingham pi theorem, the functional relationship given by Equation 4.22 between nine variables in three dimensions can also be expressed as a relation between six nondimensional groups as follows:  τ0   = f1 Re, , , m, s (4.23) 4R 4R ρV 2 where Re is the Reynolds number defined by the relation Re =

ρV(4R) μ

(4.24)

and the variable 4R is used instead of R for convenience in subsequent analyses. The relationship given by Equation 4.23 is as far as dimensional analysis goes, and experimental data are necessary to determine an empirical relationship between the nondimensional groups. The problem of determining an empirical expression for the boundary shear stress in open-channel flow is similar to the problem faced in determining an empirical expression for the boundary shear stress in pipe flow, where 4R for circular conduits is equal to the pipe diameter. If the influences of the shape of the cross section and the arrangement of roughness elements on the boundary shear stress, τ0 , are small relative to the influences of the size of the roughness elements and the viscosity of the fluid, and the flow is steady and uniform along the channel, then the shear stress can be expressed in the following functional form:

 τ0 1 f Re, (4.25) = 8 4R ρV 2 where the function f can be expected to closely approximate to the Darcy friction factor in pipes. In reality, the friction factor, f , in Equation 4.25 has been observed to be a function of channel shape, with the influence of shape decreasing roughly in the order of rectangular, triangular, trapezoidal, and circular channels (Chow, 1959). As channels become very wide or otherwise depart radically from the shape of a circle or semicircle, the friction factors derived from pipe experiments become less applicable to open channels (Daily and Harleman, 1966). Myers (1991) has shown that friction factors in wide rectangular open channels are as much as 45% greater than in narrow sections with the same Re and /4R. The question of how to account for the shape of an open channel in estimating the friction factor remains open. Also, the assumption that the friction factor is independent of the arrangement and shape of the roughness projections has been shown to be invalid in gravel-bed streams with high boulder concentrations (Ferro, 1999). The transition from laminar to turbulent flow in open channels occurs at a Reynolds number of about 600, and it is convenient to define three types of turbulent flow: smooth, transition, and rough. The flow is classified as “smooth” when the roughness projections on the channel boundary are submerged within a laminar sublayer, in which case the friction factor in open channels depends only on the Reynolds number, Re, and can be estimated by (Henderson, 1966) ⎧ 1 ⎪ ⎪ 1.78Re 8 , Re < 105 (4.26) ⎪ ⎨  1 = 2.51 ⎪ −2.0 log f ⎪ , (4.27) Re > 105 ⎪ 10 ⎩ Re f

126

Chapter 4

Fundamentals of Flow in Open Channels

´ an ´ equations for flow in These relations are the same as the Blasius and Prandtl–von Karm pipes. However, owing to the free surface and the interdependence of the hydraulic radius, discharge, and slope, the relationship between f and Re in open-channel flow is not identical to that for pipe flow. The flow is classified as “rough” when the roughness projections on the channel boundary extend out of the laminar sublayer, creating sufficient turbulence that the friction factor depends only on the relative roughness. “Rough” flow is also commonly referred to as fully turbulent flow, although “smooth” and “transitional” flow are also turbulent. Under rough-flow conditions, the friction factor can be estimated by (ASCE, 1963)

1 ks = −2 log10 (4.28) 12R f where ks is the equivalent sand roughness of the open channel. Equation 4.28 is derived from the integration of the Nikuradse velocity distribution for rough flow over a trapezoidal open channel cross section (Keulegan, 1938), and gives a higher friction factor than the ´ an ´ equation that is used in pipe flow. In the transition region between Prandtl–von Karm (hydraulically) smooth and rough flow, the friction factor depends on both the Reynolds number and the relative roughness and can be approximated by (ASCE, 1963)  2.5 1 ks = −2 log10 + (4.29) 12R f Re f This relation differs slightly from the Colebrook equation for transition flow in closed conduits, but is still called the Colebrook equation and is commonly applied in both smooth and rough flow in open channels. Caution should be exercised in applying Equation 4.29 to smooth-flow conditions in open channels since under these conditions Equation 4.29 asymp´ an ´ equation (Equation 4.27), which is known to deviate by totes to the Prandtl–von Karm up to 20% from measurements (Cheng et al., 2011). Equation 4.29 was originally suggested by Henderson (1966) for wide open channels (width/depth Ú 10), and others have suggested similar formulations with different constants (e.g., Yen, 1991). The three types of flow (smooth, transition, rough) can be delineated using the shear velocity Reynolds number, also known as the roughness Reynolds number, ks u∗ /ν, where u∗ is the shear velocity defined by   τ0 = gRSf (4.30) u∗ = ρ and ν is the kinematic viscosity of the fluid. The turbulent flow are as follows: ⎧ ⎪ ⎨70,

approximate ranges for the three types of smooth transition rough

(4.31)

There is still some debate on defining the transition-flow region by the limits in Equation 4.31; for example, Henderson (1966) defines the transition region as 4 … u∗ ks /ν … 100, Yang (1996) defines it as 5 … u∗ ks /ν … 70, and Rubin and Atkinson (2001) define it as 5 … u∗ ks /ν … 80. The Colebrook friction-factor equation given by Equation 4.29 is commonly applied across all three flow regimes. For shallow flows in rock-bedded channels, conditions commonly found in mountain streams, the larger rocks produce most of the resistance to flow. In these cases, where the size of the roughness elements are comparable to the flow depth, the roughness elements are called macroscale roughness, the drag force is caused more by pressure (form) drag than friction drag, and the shape and arrangement of the macroscale roughness elements can have a significant effect on the friction coefficient (Canovaro et al., 2007). Limerinos (1970) has shown that the friction factor, f , can be estimated from the size of the rock in the stream bed using the relation

Section 4.2

1 = 1.2 + 2.03 log10 f



R d84

Basic Principles

127

(4.32)

where d84 is the 84-percentile size of the rocks on the stream bed. Several alternatives to Equation 4.32 have been proposed for rock-bedded mountain streams, and a thorough review of alternative formulations can be found in Ferguson (2007) and Yen (2002). An extensive review of data in coarse-bed channels where d50 > 2 mm (the minimum size to be classified as gravel) showed that Equation 4.28 could be used to estimate the friction factor ´ ´ 2008). by taking ks equal to 2.4d90 , 2.8d84 , or 6.1d50 (Lopez and Barragan, For channels lined with submerged vegetation, the sources of drag within the channel are fundamentally different from the boundary drag in channels lined with sediment, gravel, or rocks. In the case of submerged vegetation, the friction factor, f , can be estimated by (Cheng, 2011)

1 kv 8 f = 0.40 (4.33) h where kv is a roughness length scale [L], defined by kv =

π λ D 41 − λ

(4.34)

where λ is the fraction of the channel bottom occupied by vegetation stems [dimensionless], D is the average stem diameter [L], and h is the depth of water above the vegetation. Equation 4.33 provides a good fit to the observed data for 1 * 10−5 … kv /h … 3 * 10−2 . Regardless of the empirical equation used to estimate the friction factor, f , combining Equations 4.21 and 4.25 leads to the following form of the momentum equation, which is most commonly used in practice:  8g  V= RSf (4.35) f where f is a function of the relative roughness and/or Reynolds number of the flow. Equation 4.35 is the same as the Darcy–Weisbach equation derived previously for pipe flow. In most practical applications of Equation 4.35, the friction factor, f , is estimated using the Colebrook equation (Equation 4.29). In cases where the flow is uniform, the slope of the energy grade line, Sf , is equal to the slope of the channel, S0 , since under these conditions  d V2 dz = S0 Sf = − y + z + =− dx 2g dx where the depth, y, and average velocity, V, are constant and independent of x under uniform-flow conditions. In many practical cases, determination of the flow conditions in open-channel flow requires simultaneous solution of Equations 4.29 and 4.35. In these cases, it is convenient to express the momentum equation, Equation 4.35, in the form V f = 8gRS0 (4.36) and the friction factor, f , given by Equation 4.29, can be expressed in the form ⎛ ⎞  ⎜ k ⎟ ks 2.5 0.625ν 1 ⎜ s ⎟ = −2 log10 ⎜ + + ⎟ = −2 log10 ρV(4R) ⎠ ⎝ 12R 12R f RV f f μ

(4.37)

128

Chapter 4

Fundamentals of Flow in Open Channels

Combining Equations 4.36 and 4.37 yields ⎛ ⎞ k 0.625ν s ⎠ + 3 Q = VA = −2A 8gRS0 log10 ⎝ 12R R 2 8gS0

(4.38)

This derived relationship is particularly useful in relating the flow rate, Q, to the flow area, A, and hydraulic radius, R, for a given channel slope, S0 , and roughness height, ks . EXAMPLE 4.1 Water flows at a depth of 1.83 m in a trapezoidal, concrete-lined section (ks = 1.5 mm) with a bottom width of 3 m and side slopes of 2 : 1 (H : V). The slope of the channel is 0.0005 and the water temperature is 20◦ C. Assuming uniform-flow conditions, estimate the average velocity and flow rate in the channel. Solution The flow in the channel is illustrated in Figure 4.3. FIGURE 4.3: Flow in a trapezoidal channel

T = 3 + 4y

1 2

y = 1.83 m

3m

From the given data and from the dimensions of the channel shown in Figure 4.3: S0 = 0.0005, A = 12.2 m2 , P = 11.2 m, and 12.2 A = = 1.09 m R= P 11.2 For concrete, ks = 1.5 mm = 0.0015 m, and ν = 1.00 * 10−6 m2 /s at 20◦ C. Substituting these data into Equation 4.38 gives the flow rate, Q, as ⎛ ⎞ k 0.625ν s ⎠ + 3 Q = −2A 8gRS0 log10 ⎝ 12R R 2 8gS0 ⎤ ⎡ −6 ) 0.0015 0.625(1.00 * 10 ⎦ = −2(12.2) 8(9.81)(1.09)(0.0005) log10 ⎣ + 3 12(1.09) (1.09) 2 8(9.81)(0.0005) = 19.8 m3 /s and the average velocity, V, is given by V=

19.8 Q = = 1.62 m/s A 12.2

Therefore, for the given flow depth in the channel, the flow rate is 19.8 m3 /s and the average velocity is 1.62 m/s.

4.2.2.2

Manning equation

To fully appreciate the advantage of using the Darcy–Weisbach equation (Equation 4.35) compared with other flow equations used in practice, some historical perspective is needed. The Darcy–Weisbach equation is based primarily on the pipe experiments of Nikuradse and Colebrook, which were conducted between 1930 and 1940; however, observations on rivers ´ ∗ proposed the following and other large open channels began much earlier. In 1775, Chezy ∗ Antoine de Chezy ´ (1718–1798) was a French engineer.

Section 4.2

expression for the mean velocity in an open channel  V = C RSf

Basic Principles

129

(4.39)

where C was referred to as the Ch´ezy coefficient. Equation 4.39 has exactly the same form as Equation 4.35 and was derived in the same way, except that the functional dependence of the ´ coefficient on the Reynolds number and the relative roughness was not considered. Chezy ´ coefficient is related to the friction factor by Comparing Equations 4.35 and 4.39, the Chezy  8g C= (4.40) f In 1869, Ganguillet and Kutter (1869) published an elaborate formula for C that became widely used. In 1890, Manning (1890) demonstrated that the data used by Ganguillet and Kutter were fitted just as well by a simpler formula in which C varies as the sixth root of R, where 1 R6 (4.41) C= n and n is a coefficient that is characteristic of the surface roughness alone. Since C is not a dimensionless quantity, values of n were specified to be consistent with length units measured in meters and time in seconds. If the length units are measured in feet, then Equation 4.41 becomes 1 R6 C = 1.486 (4.42) n where 1.486 is the cube root of 3.281, the number of feet in a meter. When either ´ equation, the resulting expression is called Equation 4.41 or 4.42 is combined with the Chezy the Manning equation or Strickler equation (in Europe) and is given by ⎧ 1 2 1 ⎪ ⎪ ⎪ R3 S2 ⎨ n V= ⎪ 1.486 2 1 ⎪ ⎪ ⎩ R3 S2 n

(SI units) (4.43) (U.S. Customary units)

where S = Sf = S0 under uniform flow conditions, and 1/n is called the Strickler coefficient when Equation 4.43 is called the Strickler equation (Douglas et al., 2001). The coefficient 1.486 in Equation 4.43 is much too precise considering the accuracy with which n is known and should not be written any more precisely than 1.49 or even 1.5 (Henderson, 1966). Normal depth is a commonly used term in open-channel hydraulics. It can be defined as the depth of flow that will occur in a channel of constant bed slope and roughness, provided the channel is sufficiently long and the flow is undisturbed (Knight et al., 2010). When the Darcy–Weisbach or Manning equation is used to calculate the depth of flow for a given discharge, channel roughness, and channel slope (for Sf = S0 ), the calculated depth of flow is taken to be the normal depth. Theoretical basis of Manning’s n. Since the scientific foundation of the Darcy–Weisbach equation is sounder than the empirical foundation of the Manning equation, the functional dependencies of Manning’s n can be identified by comparing the Manning and Darcy– Weisbach equations. By equating the mean velocities, V, calculated by these two equations (i.e., by Equations 4.35 and 4.43) it can be directly shown that the roughness coefficient, n, and friction factor, f , are related by  1 R6 8g = (4.44) n f

130

Chapter 4

Fundamentals of Flow in Open Channels

and substituting the expression for the friction factor under fully turbulent conditions (Equation 4.28) into Equation 4.44 and rearranging yields

1 R 6 ks n

= 1 R ks6 2.0 log 12 ks 1 8g

(4.45)

Taking g = 9.81 m/s2 , Equation 4.45 is plotted in Figure 4.4, which illustrates that value of 1

n/ks6 is effectively constant over a wide range of values of R/ks , where n being a constant is an essential assumption of the Manning equation. Equation 4.45 (with g = 9.81 m/s2 ) gives a minimum value of n 1

= 0.039

(4.46)

ks6 1

at R/ks = 34 and n/ks6 is within ;5% of a constant value over a range of R/ks given by 4 < R/ks < 500 as shown by Yen (1991). This range of R/ks for the validity of the Manning equation differs somewhat from the range given by Hager (1999) as 3.6 < R/ks < 360. It is important to note that the R/ks criterion for the validity of the Manning equation relies on the assumption that the flow is fully turbulent, that is, it is in the rough-flow regime. According to the flow-regime delineation given by Equation 4.31, the fully turbulent criterion can be taken as u∗ ks > 70 (4.47) ν where u∗ is the shear velocity [LT−1 ] defined as gRSf , g is gravity [LT−2 ], R is the hydraulic radius [L], Sf is the head loss per unit length [dimensionless], and ν is the kinematic viscosity of water [L2 T−1 ]. Since ν = 1.00 * 10−6 m2 /s at 20◦ C and g = 9.81 m/s2 , Equation 4.47 can be put in the more convenient form  ks RSf > 2.2 * 10−5 (4.48) where ks and R are expressed in meters. 1 In summary, the validity of the Manning equation relies on two assumptions: (1) n/ks6 is constant; and (2) the flow is fully turbulent. These assumptions require that 4 < R/ks < 500 and that u∗ ks /ν > 70, respectively. If these two conditions for the validity of the Manning FIGURE 4.4: Variation of

0.10

1

n/ks6 in fully turbulent flow

1

n/ks6

R/ks = 4

R/ks = 500

1

n/ks 6 = 0.039

0.01

1

10

100 R/ks

1000

Section 4.2

Basic Principles

131

equation are met, then the estimated velocities using the Manning equation and the Darcy– Weisbach equation are approximately the same, provided that Manning’s n is taken as 0.039ks .

EXAMPLE 4.2 Water flows at a depth of 3.00 m in a trapezoidal, concrete-lined section with a bottom width of 3 m and side slopes of 2 : 1 (H : V). The slope of the channel is 0.0005 and the water temperature is 20◦ C. Assess the validity of using the Manning equation with n = 0.014. Solution From the given data: n = 0.014, S0 = 0.0005, A = 27 m2 , P = 16.42 m, and R = A/P = 1.64 m. Assuming that the Manning equation is valid, then Equation 4.46 gives

ks =

6

n 0.014 6 = = 0.0021 m 0.039 0.039

R 1.64 = 781 = ks 0.0021 ks RS0 = 0.0021 (1.64)(0.0005) = 6.01 * 10−5 √ 1 Since ks RS > 2.2 * 10−5 the flow is fully turbulent; however, since R/ks > 500 the value of (n/ks ) 6 1 cannot be taken as a constant equal to 0.039 since (n/ks ) 6 will depend on the value of R/ks . Based on these results, the Manning-equation formulation is not strictly valid since the Manning roughness coefficient, n, cannot be taken as a constant and independent of the depth of flow. The Darcy–Weisbach equation would be more appropriate in this case. Unfortunately, it is still common practice to assume that the Manning equation is valid, without verifying that the fully turbulent and constant-n conditions are met.

The theoretical analysis presented here indicates that when Manning’s equation is applicable the n value is related to the equivalent sand roughness by Equation 4.46, which can be expressed in the form 1

n = 0.039ks6

(4.49)

where ks is in meters. Since natural and constructed channels have varying roughness characteristics, the relationship between the roughness characteristics and ks remains to be established. It should be noted that the sixth-root relationship between the roughness height, ks , and the roughness coefficient, n, means that large relative errors in estimating ks result in much smaller relative errors in estimating n. For example, a thousandfold change in the roughness height results in about a threefold change in n. Several investigators have suggested relationships between Manning’s n and a characteristic roughness height in the form 1

n = αdp6

(4.50)

where α is a constant and dp is a percentile grain size of the channel lining material. These approaches are theoretically consistent with Equation 4.49, since alternate definitions of the roughness, ks , would naturally be associated with different definitions of the roughness height, dp . These and similar semiempirical relationships that relate Manning’s n to roughness and relative roughness are shown in Table 4.1. Some of the formulas for estimating Manning’s n given in Table 4.1 express n as a function of both the characteristic grain size, dp , and the hydraulic radius, R, and such relationships are particularly applicable at low flow depths in channels lined with coarse-grained materials. Analyses of experimental data on riprap-lined channels have shown that it is appropriate to use the formulae relating n to dp when R/d50 > 3, otherwise n will depend on both R and dp (Froehlich, 2012) and an appropriate functional relationship that includes both of these variables should be used.

132

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Fundamentals of Flow in Open Channels TABLE 4.1: Semiempirical Expressions for Manning’s n

Expression for n

Conditions

Reference

1 6 0.047d50

Uniform sand

Strickler (1923)



Keulegan (1938)

Sand mixtures

Meyer–Peter and Muller (1948)

Gravel-lined canals

Lane and Carlson (1953)

Riprap-lined channels

Maynord (1991)



Limerinos (1970)

Gravel streams with S0 > 0.004

Bathurst (1985)

Wide channels

Dingman (1984)

Riprap-lined channels

Froehlich (2012)

1 6 0.0025d90 1 6 0.0039d90 1 6 0.047d75 1 6 0.046d50 1

R6 [7.69 ln(R/d84 ) + 63.4] 1

R6 [7.64 ln(R/d84 ) + 65.3] 1 R6

[7.83 ln(R/d84 ) + 72.9] 1 R6

[7.64 ln(R/d50 ) + 15.5]

Typical values of Manning’s n. Manning’s n can be estimated using typical values or by using empirical equations that are not of the form given by Equation 4.49. The use of typical n values is usually appropriate in lined artificial channels such as concrete-lined channels; however, in natural channels using typical values is less certain and other empirical models might be more appropriate. Typical values of the roughness coefficient, n, used in engineering practice are given in Table 4.2, where lower values of n are for surfaces in good condition, and higher values are for surfaces in poor condition. Other empirical equations for estimating Manning’s n. Several specialized equations have been proposed for estimating Manning’s n. For example, Dingman and Sharma (1997) have proposed the following empirical equation for estimating Manning’s n in natural channels, n = 0.217A−0.173 R0.267 S0.156 f

(4.51)

where A is the cross-sectional flow area [m2 ], R is the hydraulic radius [m], and Sf is the energy slope [dimensionless]. Equation 4.51 is applicable where flows exceed 1 m3 /s (35 ft3 /s), and has been found to perform acceptably in Australian rivers (Lang et al., 2003; Harman et al., 2008). For flow over angular riprap with slopes in the range of 3%–33%, such as down rock chute channels (Ferro, 2000), Manning’s n can be estimated using the relation (Rice et al., 1998) 3 n = 0.0292(S0 d50 ) 20 (4.52) where S0 is the longitudinal slope of the channel, and d50 is the 50-percentile size of the riprap. Nikora and others (2008) analyzed measurements from several New Zealand streams with submerged patches of vegetation and found that Manning’s n can be estimated using the relation 

 Wc h n = 0.025 exp 3.0 (4.53) y W where h is the patch-averaged height of the vegetation [L], Wc is the mean width of the vegetation patches in a cross section [L], y is the mean flow depth [L], and W is the mean flow

Section 4.2

Basic Principles

133

TABLE 4.2: Manning Coefficient for Open Channels

Channel type Lined channels: Brick, glazed Brick Concrete, float finish Asphalt Rubble or riprap Concrete, concrete bottom Gravel bottom with riprap Vegetal Excavated or dredged channels: Earth, straight, and uniform Earth, winding, fairly uniform Rock Dense vegetation Unmaintained Natural channels: Clean, straight Clean, irregular Weedy, irregular Brush, irregular Floodplains: Pasture, no brush Brush, scattered Brush, dense Timber and brush

Manning n

Range

0.013 — 0.015 — — 0.030 0.033 —

0.011–0.015 0.012–0.018 0.011–0.020 0.013–0.02 0.020–0.035 0.020–0.035 0.023–0.036 0.030–0.40

0.027 0.035 0.040 — 0.080

0.022–0.033 0.030–0.040 0.035–0.050 0.05–0.12 0.050–0.12

0.030 0.040 0.070 —

0.025–0.033 0.033–0.045 0.050–0.080 0.07–0.16

0.035 0.050 0.100 —

0.030–0.050 0.035–0.070 0.070–0.160 0.10–0.20

Sources: ASCE (1982); Wurbs and James (2002); Bedient and Huber (2002).

width [L]. Fathi-Moghadam and colleagues (2011) performed full-scale laboratory experiments of open-channel flow with submerged vegetation and found results that were closely described (R2 = 0.92) by n = 0.092V

−0.33

−0.51 0.32 y A h a

(4.54)

where V is the average velocity of flow [m/s], y is the flow depth [L], h is the height of vegetation [L], A is the plan area of the channel covered by vegetation [L2 ], and a is the canopy area associated by each individual element of vegetation [L2 ]. Experimental conditions limit the application of Equation 4.54 to channels with aspect ratios (width/depth) in the range of 4–40, and Equation 4.54 has been applied successfully to estimating Manning’s n in floodplains. Based on a review of the literature, Johnson (1996) indicated that Manning’s n estimated from field measurements typically have errors in the range of 5%–35%. Manning’s n in compound channels. Channels in which the perimeter roughness varies significantly within the channel are called composite channels. Channels having subsections with different shapes are called compound channels, and compound channels in which the perimeter roughness varies between sections are called compound-composite channels. In these types of channels, it is sometimes necessary to identify an effective Manning’s n to describe the overall channel roughness such that the Manning equation accurately describes the relationship between flow rate and normal depth for a given slope of the channel. This effective channel roughness, commonly called the composite roughness or equivalent roughness, ne , of the channel is usually computed by first subdividing the channel section into N smaller sections, where the ith section has a roughness, ni , wetted perimeter, Pi , and

134

Chapter 4

Fundamentals of Flow in Open Channels TABLE 4.3: Composite-Roughness Formulae

Formula∗ ⎛

Assumption ⎞2

N 

3

3 2

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

⎜ P i ni ⎜ ⎜ i=1 ne = ⎜ ⎜ ⎜ P ⎝ ⎛

N 

Total cross-sectional mean velocity equal to subarea mean velocity

Horton (1933a), Einstein (1934)

Total resistance force equal to the sum of subarea resistance forces

Pavlovskii (1931)

Friction slope and velocity same in all subsections

Einstein and Banks (1951)

Friction slope same in all subsections and total discharge is sum of subarea discharges

Lotter (1933)

Logarithmic velocity distribution over depth y for wide channel

Krishnamurthy and Christensen (1972)†

Total shear force equal to sum of subsection shear forces and friction slope same for all subsections

Cox (1973)

Total shear velocity equal to weighted sum of subsection shear velocities

Yen (1991)

⎞1 2

Pi n2i ⎟ ⎟

⎜ ⎜ ⎜ i=1 ne = ⎜ ⎜ ⎜ P ⎝ ⎛

Reference

⎟ ⎟ ⎟ ⎟ ⎠ ⎞2

N 

3 2

⎜ P i ni ⎜ ⎜ i=1 ne = ⎜ ⎜ ⎜ P ⎝

3

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

5

ne =

PR 3

5

n  Pi Ri3 ni i=1

N 

3

Pi yi2 ln ni

ln ne = i=1 N 

3 2

Pi yi

i=1

 ne = 1 N An A i=1 i i

ne =

1 1 N P R 3 n 1 i=1 i i i PR 3

Note: ∗ P, R, and A are the perimeter, hydraulic radius, and area of the entire cross-section, respectively. † y is the average flow depth in Section i. i

hydraulic radius, Ri , and several commonly used formulae for calculating ne based on subsection properties are listed in Table 4.3. The equations in Table 4.3 are based on differing assumptions on the distribution of flow in the compound channel. For example, the equation proposed by Horton (1933a) and Einstein (1934) assumes that each subdivided area has the same mean velocity, the equation proposed by Lotter (1933) assumes that the total flow is equal to the sum of the flows in the subdivided areas, and the equation proposed by Pavlovskii (1931) assumes that the total resisting force is equal to the sum of the resisting forces in the subdivided areas. Several additional formulae for estimating the composite

Section 4.2

Basic Principles

135

roughness in compound channels can be found in Yen (2002). It is interesting to note that the equations for estimating composite roughness require both the hydraulic radius and wetted perimeter, and hence depend on the way the subsections are divided and the relative amount of wetted perimeter of the subsections. However, according to Yen (2002), the differences between the composite roughness equations far exceed the differences due to subarea division methods. Motayed and Krishnamurthy (1980) used data from 36 natural-channel cross sections in Maryland, Georgia, Pennsylvania, and Oregon to assess the relative performance of Einstein and Banks (1951), Lotter (1933), and Krishnamurthy and Christensen (1972) formulae (shown in Table 4.3) and concluded that the formula proposed by Lotter (1933) performed best. In further support of the Lotter (1933) equation, it has been shown that, in some cases, this is the only composite-roughness equation that provides accurate convergence to the normal depth under some gradually-varied flow conditions in compound channels (Younis et al., 2009). In a broader context, the very limited data available in the literature show considerable scattering, and they are insufficient to identify which composite-roughness equations are most promising (Yen, 2002). Under these circumstances, it is prudent to determine the composite roughness using all of the proposed models and select a representative composite roughness based on these estimates. A composite roughness is usually necessary in analyzing flood flows in which a river bank is overtopped and the flow extends into the adjacent floodplain. The roughness elements in the floodplain, typically consisting of shrubs, trees, and possibly houses and cars, are usually much larger than the roughness elements in the river channel. Typical values of the Manning roughness coefficient in floodplains originally recommended by Chow (1959) and still widely used are shown in Table 4.4. The roughness coefficients given in Table 4.4 should be used with caution, since it is well known that, in cases where the heights of the roughness elements are comparable in magnitude to the flow depth, and where the elements themselves may be flexible, such as tall grasses, the Manning roughness coefficient depends on the flow velocity, flow depth, and the flexibility and density of the roughness elements (e.g., Carollo et al., 2005). In cases of rigid vegetation (e.g., trees), flow depth and vegetation density have a significant effect on the Manning roughness coefficient (Musleh and Cruise, 2006). In general, where the roughness elements cover most or all of the flow depth and are of sufficient size to offer significant resistance to flow, the losses are considered to be due to drag forces rather TABLE 4.4: Manning Roughness Coefficient for Floodplains

Surface Pasture, no brush Short grass High grass Cultivated areas No crop Mature row crops Mature field crops Brush Scattered brush, heavy weeds Light brush and trees, in winter Light brush and trees, in summer Medium to dense brush, in winter Medium to dense brush, in summer Trees Dense willows, summer, straight Cleared land with tree stumps, no sprouts Cleared land with tree stumps, heavy growth of sprouts Heavy stand of timber, a few down trees, little undergrowth, flood stage below branches Heavy stand of timber, a few down trees, little undergrowth, flood stage reaching branches

Minimum

Maximum

Typical

0.025 0.030

0.035 0.050

0.030 0.035

0.020 0.025 0.030

0.040 0.045 0.050

0.030 0.035 0.040

0.035 0.035 0.040 0.045 0.070

0.070 0.060 0.080 0.110 0.160

0.050 0.050 0.060 0.070 0.100

0.110 0.030 0.030

0.200 0.050 0.050

0.150 0.040 0.040

0.080

0.120

0.100

0.100

0.160

0.120

136

Chapter 4

Fundamentals of Flow in Open Channels

than to boundary shear as in the traditional approach. Under these circumstances, it is not unusual for the water-surface elevation in the portion of the channel with tall vegetation to be greater than the water-surface elevation in the section of the channel with submerged roughness elements (Wang and Wang, 2007). Manning’s n in floodplains are best regarded as uncertain variables whose standard deviations increase with increasing n value. For n values on the order of 0.02, 0.05, and 0.1, standard deviations on the order of 0.005, 0.015, and 0.05 are recommended, indicating standard deviations that are 25%–50% of the estimated value of n (U.S. Army Corps of Engineers, 1996). It has been shown that failure to take into account the uncertainty in Manning’s n can lead to an underestimation of flood risk (Guganesharajah et al., 2006). In cases where depth and flow measurements are available from historical floods, it is conventional practice to estimate n values in channel and floodplain sections from the calibration of flow models to match the historical observations (e.g., Ballasteros et al., 2011). EXAMPLE 4.3 The floodplain shown in Figure 4.5 can be divided into sections with approximately uniform roughness characteristics. The Manning’s n values for each section are as follows: Section

n

1 2 3 4 5 6 7

0.040 0.030 0.015 0.013 0.017 0.035 0.060

Use the formulae in Table 4.3 to estimate the composite roughness. Solution From the given shape of the floodplain (Figure 4.5), the following geometric characteristics are derived: Section, i

Pi (m)

Ai (m2 )

Ri (m)

ni

yi (m)

1 2 3 4 5 6 7

8.25 100 6.71 15.0 6.71 150 8.25

8.00 200 21 75 21 300 8.00

0.97 2.00 3.13 5.00 3.13 2.00 0.97

0.040 0.030 0.015 0.013 0.017 0.035 0.060

1.00 2.00 3.50 5.00 3.50 2.00 1.00

295

633

It should be noted that the internal waterlines dividing the subsections are not considered a part of the wetted perimeter in computing the subsection hydraulic radius, Ri . This is equivalent to assuming that the internal shear stresses at the dividing waterlines are negligible compared with the bottom shear stresses. For the given shape of the floodplain, the total perimeter, P, of the (compound) channel FIGURE 4.5: Flow in a floodplain

100 m 1

1

150 m 2m

2 3

4

4

5

1

1 2

2

15 m

3m

6

7

1 4

Section 4.2

Basic Principles

137

is 295 m, the total area, A, is 633 m2 , and hence the hydraulic radius, R, of the compound section is given by A 633 R= = = 2.15 m P 295 Substituting these data into the formulae listed in Table 4.3 yields the following results: Formula

ne

Horton/Einstein Pavlovskii Einstein and Banks Lotter Krishnamurthy and Christensen Cox Yen

0.033 0.033 0.033 0.022 0.026 0.030 0.031

Average

0.030

It is apparent from this example that estimates of ne can vary significantly. A conservative (high) estimate of the composite roughness is 0.033, and the average composite roughness predicted by the models is 0.030.

Using a composite roughness to calculate the flow rate in a compound channel is not entirely satisfactory, since the large-scale turbulence generated by the velocity shear between the main channel and overbank portion of the channel is not accounted for in the Manning equation (Bousmar and Zech, 1999; Rezaei and Knight, 2011). This consideration can be addressed by using a two-dimensional model to delineate the floodplain; however, there is no general rule that two-dimensional models produce more accurate floodplain delineations than one-dimensional models, especially since predicted floodplain areas depend also on the quality of topographic data and description of the river geometry (Cook and Merwade, 2009). The topography of floodplains can be estimated fairly accurately using LIDAR (Light Detection and Ranging) data and less accurately using most public-domain digital terrain models. Use of remote sensing data. The Manning equation is commonly used to estimate flow in rivers based on the cross-sectional geometry, slope, and a calibrated or estimated roughness coefficient. This approach may not be feasible in rivers at inaccessible locations, where estimating flows using remotely sensed hydraulic information might be the only practical alternative. Such estimates based on aerial or satellite observations of water-surface width and maximum channel width, and channel slope data obtained from topographic maps, have been shown to provide discharge estimates within a factor of 1.5–2 (Bjerklie et al., 2005a; Ashmore and Sauks, 2006). Related methods have also been developed for estimating bankfull velocity and discharge in rivers using remotely sensed river morphology information (Bjerklie, 2007). 4.2.2.3

Other equations

Discharge measurements from a wide range of river sizes and morphologies suggest that, in natural rivers, the exponent of the slope term in the Manning equation is closer to 0.33 than 0.5, as used in the conventional Manning equation (Bjerklie et al., 2005b). This discrepancy ´ assumption that the surface resistance is proportional to can possibly be linked to the Chezy the velocity squared, which is true only if the flow boundary does not change as the velocity is varied (Leopold et al., 1960). This condition is generally true for pipe flow, but is not true for open channels where the extent of the boundary changes substantially with discharge. Interaction between flow and channel geometry is particularly prevalent in alluvial channels, where special methods are sometimes used to estimate discharge from measured water level,

138

Chapter 4

Fundamentals of Flow in Open Channels

channel cross section, and energy slope (e.g., Yang and Tan, 2008). For gravel-bed rivers and mountain streams, an alternative equation that performs better than the Manning equation has been reported as Q = 6.04AR0.82 S0.26 (4.55) f where A is the cross-sectional area of the channel [m2 ], R is the hydraulic radius [m], and Sf ´ is the friction slope [dimensionless] (Lopez et al., 2007). A particularly appealing feature of Equation 4.55 is that it does not require explicit specification of a resistance coefficient. 4.2.2.4

Velocity distribution

In wide channels, lateral boundaries have negligible effects on the velocity distribution in the central portion of the channel. A wide channel is typically defined as one having a width that exceeds 10 times the flow depth. The velocity distribution, v(y), in wide open channels is typically assumed to be a function of the logarithm of the distance from the bottom of the channel, and is frequently approximated by the relation (Vanoni, 1941), v(y) = V +

y 1 gdS0 1 + 2.3 log κ d

(4.56)

´ an ´ constant (L 0.4) [dimenwhere V is the depth-averaged velocity [LT−1 ], κ is the von Karm sionless], d is the depth of flow [L], S0 is the slope of the channel [dimensionless], and y is the distance from the bottom of the channel [L]. As an alternative to Equation 4.56, the velocity distribution, v(y), is sometimes approximated by the power-law relationship

1 y 7 v(y) = Vmax d

(4.57)

where Vmax is the maximum velocity [LT−1 ], which is assumed to occur at the water surface where y = d. Reasons for using Equation 4.57 in lieu of Equation 4.56 are its simplicity and its closeness of fit to the logarithmic distribution (Knight et al., 2010). In channels that are not wide, the geometry of the lateral boundaries must be considered in estimating the velocity distribution, leading to more complex expressions (e.g., Wilkerson and McGahan, 2005; Maghrebi and Ball, 2006). Significant deviations from the logarithmic velocity distribution have been observed in vegetated floodplains (e.g., Yang et al., 2007). Equation 4.56 indicates that the average velocity, V, in wide open channels occurs at y/d = 0.368, or at a distance of 0.368d above the bottom of the channel. This result is commonly approximated by the relation V = v(0.4d)

(4.58)

The average velocity, V, can also be related to the velocities at two depths using Equation 4.56, which yields v(0.2d) + v(0.8d) V= (4.59) 2 It is standard practice of the U.S. Geological Survey (USGS) to use measurements at 0.2d and 0.8d with Equation 4.59 to estimate the average velocity in channel sections with depths greater than 0.75 m (2.5 ft) and to use measurements at 0.4d with Equation 4.58 to estimate the average velocity in sections with depths less than 0.75 m (2.5 ft). USGS velocity measurements at designated depths are usually collected using acoustic Doppler velocimeters (ADVs) and acoustic Doppler current profilers (ADCPs) (Muste et al., 2007; Rehmel, 2007; Le Coz et al., 2008). If the velocity profile in a stream is known, the mean velocity can be related to the surface velocity using the velocity profile equation. This permits flow rates in streams to be estimated using noncontact methods which combine remote measurements of the surface velocity with cross-sectional geometry to estimate the stream discharge. Field experiments

Section 4.2

Basic Principles

139

using microwave and ultra high frequency (UHF) Doppler radars (including radar guns) to measure surface-water velocities have yielded discharge estimates that are of comparable accuracy to conventional contact methods measuring stream discharge (Costa et al., 2006; Fulton and Ostrowski, 2008). This noncontact approach is particularly appealing when standard contact methods of measuring discharge are dangerous or cannot be obtained. The velocity distribution given by Equation 4.56 indicates that the maximum velocity occurs at the water surface. However, the maximum velocity in open channels sometimes occurs below the water surface, presumably as a result of air drag. In the Mississippi River, the maximum velocity has been observed to occur as much as one-third the water depth below the water surface (Gordon, 1992). Some field measurements have indicated that the ratio of maximum velocity to average velocity remains approximately constant over a wide range of flows, but different for each channel (Fulton and Ostrowski, 2008). Velocity measurements in open channels are usually combined to estimate the volumetric flow rate, Q, at a cross section of an open channel according to the relation Q=

N 

V i Ai

(4.60)

i=1

where N is the number of subareas across the channel and V i is the average velocity over subarea Ai . Typically, the subareas, Ai , are vertical sections across a channel where the average velocity in each section is estimated from measurements of the vertical velocity profile in the center of the section. Subareas are selected to contain no more than 5%–10% of total flow. Measured discharges are commonly expressed as a function of the corresponding watersurface elevation (= stage) to yield a stage-discharge curve or rating curve. Rating curves are commonly used to relate channel flows to stage measurements. Whereas rating curves can be fairly accurate under steady uniform conditions, unsteady nonuniform conditions can cause significant deviations from the rating curve. Under such conditions, expressing discharge as a function of stage measurements at two locations along the channel can improve discharge estimates (e.g., Schmidt and Yen, 2008). In natural channels, the major source of uncertainty in estimating streamflow from rating curves is the change in channel dimensions over time, which can be caused by bed scour/deposition, bank erosion, vegetation changes, and debris deposition. Such effects require recalibration of the rating curve to control error. Some innovative techniques, such as using air bubbles released at the bottom of the channel, can provide direct measures of volumetric flow rates within vertical sections (e.g., Yannopoulos et al., 2008). 4.2.3

Steady-State Energy Equation

The steady-state energy equation for the control volume shown in Figure 4.2 is  dQh dW − = ρe v · n dA dt dt A

(4.61)

where Qh is the heat added to the fluid in the control volume, W is the work done by the fluid in the control volume, A is the surface area of the control volume, ρ is the density of the fluid in the control volume, and e is the internal energy per unit mass of fluid in the control volume given by v2 e = gz + + u (4.62) 2 where z is the elevation of a fluid mass having a velocity, v, and internal energy, u. The normal stresses on the inflow and outflow boundaries of the control volume are equal to the pressure, p, with shear stresses tangential to the control-volume boundaries. As the fluid moves with velocity, v, the power expended by the fluid is given by  dW = pv · n dA (4.63) dt A

140

Chapter 4

Fundamentals of Flow in Open Channels

No work is done by the shear forces, since the velocity is equal to zero on the channel boundary and the flow direction is normal to the direction of the shear forces on the inflow and outflow boundaries. Combining Equation 4.63 with the steady-state energy equation (Equation 4.61) leads to

 dQh p = + e v · n dA (4.64) ρ dt ρ A Substituting the definition of the internal energy, e, (Equation 4.62) into Equation 4.64 gives the following form of the energy equation:   dQh v2 = ρ h + gz + v · n dA (4.65) dt 2 A where h is the enthalpy of the fluid defined by h=

p + u ρ

(4.66)

# then the energy Denoting the rate at which heat is being added to the fluid system by Q, equation becomes   2 v # = Q ρ h + gz + v · n dA (4.67) 2 A Considering the term h + gz, then

p p + u + gz = g + z + u ρ γ

h + gz =

(4.68)

where γ is the specific weight of the fluid. Equation 4.68 indicates that h + gz can be assumed to be constant across the inflow and outflow control surfaces, since the hydrostatic pressure distribution across the inflow/outflow boundaries guarantees that p/γ + z is constant across the boundaries, and the internal energy, u, depends only on the temperature, which can be assumed constant across each boundary. Since v · n is equal to zero over the impervious boundaries in contact with the fluid system, Equation 4.67 simplifies to   v2 # = (h1 + gz1 ) Q v · n dA ρv · n dA + ρ 2 A1 A1   v2 + (h2 + gz2 ) v · n dA (4.69) ρv · n dA + ρ 2 A2 A2 where the subscripts 1 and 2 refer to the inflow and outflow boundaries, respectively. Equation 4.69 can be further simplified by noting that the assumption of steady state requires that the rate of mass inflow, m, # to the control volume is equal to the rate of mass outflow, where   m # = ρv · n dA = − ρv · n dA (4.70) A2

A1

where the negative sign comes from the fact that the unit normal points out of the control volume. Also, the kinetic energy correction factors, α1 and α2 , can be defined by the equations  V3 v3 ρ (4.71) dA = α1 ρ 1 A1 2 2 A1  ρ A2

V3 v3 dA = α2 ρ 2 A2 2 2

(4.72)

Section 4.2

Basic Principles

141

where A1 and A2 are the areas of the inflow and outflow boundaries, respectively, and V1 and V2 are the corresponding mean velocities across these boundaries. The kinetic energy correction factor, α, is sometimes called the Coriolis coefficient or the energy coefficient. The kinetic energy correction factors, α1 and α2 , are determined by the velocity profile across the flow boundaries. Values of α are typically in the range of 1.1–1.2 for turbulent flow in prismatic channels, and typically in the range of 1.1–2.0 for turbulent flow in natural channels. Combining Equations 4.69 to 4.72 leads to V13 V3 A1 + (h2 + gz2 )m # + α2 ρ 2 A2 (4.73) 2 2 where the negative signs come from the fact that the unit normal points out of the inflow boundary, making v · n negative for the inflow boundary in Equation 4.69. Invoking the steady-state continuity equation # = −(h1 + gz1 )m Q # − α1 ρ

ρV1 A1 = ρV2 A2 = m # and combining Equations 4.73 and 4.74 leads to ⎡  ⎤ V12 V22 # =m ⎦ Q # ⎣ h2 + gz2 + α2 − h1 + gz1 + α1 2 2 which can be put in the form   # V12 V22 p2 u2 p1 u1 Q + + z2 + α2 + + z1 + α1 = − γ g 2g γ g 2g mg #

(4.74)

(4.75)

(4.76)

where p1 is the pressure at elevation z1 on the inflow boundary and p2 is the pressure at elevation z2 on the outflow boundary. Equation 4.76 can be further rearranged into the form ⎤ ⎡   # V12 V22 p2 1 p1 Q ⎦ + α1 + z1 = + α2 + z2 + ⎣ (u2 − u1 ) − (4.77) γ 2g γ 2g g mg # The energy loss per unit weight or head loss, hL , is defined by the relation hL =

# 1 Q (u2 − u1 ) − g mg #

Combining Equations 4.77 and 4.78 leads to a useful form of the energy equation:   V12 V22 p1 p2 + α1 + z1 = + α2 + z2 + hL γ 2g γ 2g

(4.78)

(4.79)

The head, h, of the fluid at any cross section is defined by the relation h=

p V2 + α + z γ 2g

(4.80)

where p is the pressure at elevation z, γ is the specific weight of the fluid, α is the kinetic energy correction factor, and V is the average velocity across the channel. The head, h, measures the average mechanical energy per unit weight of the fluid flowing across a channel cross section, where the piezometric head, p/γ + z, is taken to be constant across the section, assuming a hydrostatic pressure distribution normal to the direction of the flow. In this case, the piezometric head can be written in terms of the invert (i.e., bottom) elevation of the cross section, z0 , and the depth of flow, y, as p + z = y cos θ + z0 γ

(4.81)

142

Chapter 4

Fundamentals of Flow in Open Channels

where θ is the angle that the channel makes with the horizontal. Combining Equations 4.80 and 4.81 leads to the following expression for the head, h, at a flow boundary of a control volume: V2 + z0 h = y cos θ + α (4.82) 2g and therefore the energy equation (Equation 4.79) can be written as 

V2 y1 cos θ + α1 1 + z1 2g





V2 = y2 cos θ + α2 2 + z2 2g

+ hL

(4.83)

where y1 and y2 are the flow depths at the upstream and downstream sections of the control volume respectively, and z1 and z2 are the corresponding invert elevations. Equation 4.83 is the most widely used form of the energy equation in practice and can be written in the summary form h1 = h2 + hL

(4.84)

where h1 and h2 are the heads at the inflow and outflow boundaries of the control volume, respectively. A rearrangement of the energy equation (Equation 4.83) gives y2 cos θ + α2

V2 V22 = y1 cos θ + α1 1 + (z1 − z2 ) − hL 2g 2g

(4.85)

which can also be written in the more compact form 

V2 y cos θ + α 2g

1 = (S − S0 cos θ )L

(4.86)

2

where L is the distance between the inflow and outflow sections of the control volume, S is the head loss per unit length (= slope of the energy grade line) given by hL L

(4.87)

z1 − z2 L cos θ

(4.88)

S= and S0 is the slope of the channel defined as S0 =

In contrast to our usual definition of slopes, downward slopes are generally taken as positive in open-channel hydraulics. The relationship between S0 and cos θ is shown in Table 4.5, where it is clear that for open-channel slopes less than 0.1 (10%), the error in assuming that cos θ = 1 is less than 0.5%. Since this error is usually less than the uncertainty in other terms in the energy equation, the energy equation (Equation 4.86) is frequently written as TABLE 4.5: S0 versus cos θ

S0 0.001 0.01 0.1 1

cos θ 0.9999995 0.99995 0.995 0.707

Section 4.2



V2 y + α 2g

Basic Principles

143

1 = (S − S0 )L,

S0 < 0.1

(4.89)

2

and the range of slopes corresponding to this approximation is frequently omitted. The slopes of rivers and canals in plain areas are usually on the order of 0.01%–1%, while the slopes of mountain streams are typically on the order of 5%–10% (Montes, 1998; Wohl, 2000; Benke and Cushing, 2005). Channels with slopes in excess of 1% are commonly regarded as steep (Hunt, 1999). In cases where the head loss is entirely due to frictional resistance, the energy equation is written as 

V2 y + α 2g

1 = (Sf − S0 )L,

S0 < 0.1

(4.90)

2

where Sf is the frictional head loss per unit length. Equation 4.90 is sometimes written in the expanded form 

V2 y1 + α1 1 + z1 2g





V2 = y2 + α2 2 + z2 2g

+ hf ,

S0 < 0.1

(4.91)

where hf is the head loss due to friction. Equation 4.91 is superficially similar to the Bernoulli equation, but the two equations are fundamentally different, because the Bernoulli equation is derived from the momentum equation (not the energy equation) and does not contain a head-loss term. 4.2.3.1

Energy grade line

The head at each cross section of an open channel, h, is given by h=y + α

V2 + z0 , 2g

S0 < 0.1

(4.92)

where y is the depth of flow, V is the average velocity over the cross section, z0 is the elevation of the bottom of the channel, and S0 is the slope of the channel. As stated previously, in most cases the slope restriction (S0 < 0.1) is met, and this restriction is not explicitly stated in the definition of the head. When the head, h, at each section is plotted versus the distance along the channel, this curve is called the energy grade line. The point on the energy grade line corresponding to each cross section is located a distance αV 2 /2g vertically above the water surface; between any two cross sections the elevation of the energy grade drops by a distance equal to the head loss, hL , between the two sections. The energy grade line is useful in visualizing the state of a fluid as it flows along an open channel and is especially useful in visualizing the performance of hydraulic structures in open-channel systems. 4.2.3.2

Specific energy

The specific energy, E, of a fluid is defined as the mechanical energy per unit weight of the fluid measured relative to the bottom of the channel and is given by V2 2g

(4.93)

Q2 2gA2

(4.94)

E=y + α or E=y + α

144

Chapter 4

Fundamentals of Flow in Open Channels

where Q is the volumetric flow rate and A is the cross-sectional flow area. The specific energy, E, was originally introduced by Bakhmeteff (1912; 1932) and appears explicitly in the energy equation, Equation 4.90, which can be written in the form E1 − E2 = hL + z

(4.95)

where E1 and E2 are the specific energies at the upstream and downstream sections, respectively; hL is the head loss between sections; and z is the change in elevation between the upstream and downstream sections. In many cases of practical interest, E1 , hL , and z can be calculated from the given upstream flow conditions and channel geometry, E2 can be calculated using Equation 4.95, and the downstream depth of flow, y, can be calculated from E2 using Equation 4.94. For a given shape of the channel cross section and flow rate, Q, the specific energy, E, depends only on the depth of flow, y. The typical relationship between E and y given by Equation 4.94, for a constant value of Q, is shown in Figure 4.6. The salient features of Figure 4.6 are: (1) there is more than one possible flow depth for a given specific energy; and (2) the specific energy curve is asymptotic to the line y=E

(4.96)

In accordance with the energy equation (Equation 4.95), the specific energy at any cross section can be expressed in terms of the specific energy at an upstream section, the change in the elevation of the bottom of the channel, and the head loss between the upstream and downstream sections. The fact that there can be more than one possible depth for a given specific energy leads to the question of which depth will exist. The specific energy diagram, Figure 4.6, indicates that there is a depth, yc , at which the specific energy is a minimum. At this point, Equation 4.94 indicates that dE Q2 dA =0=1 − dy gA3c dy

(4.97)

where Ac is the flow area corresponding to y = yc , and the kinetic energy correction factor, α, has been taken equal to unity. Referring to the general open-channel cross section shown in Figure 4.7, it is clear that for small changes in the flow depth, y, dA = T dy

(4.98)

where dA is the increase in flow area resulting from a change in depth, dy, and T is the top width of the channel when the flow depth is y. Equation 4.98 can be written as dA =T dy FIGURE 4.6: Typical specific energy diagram

Depth of flow, y

y

=

E

y2

yc y1 Q = constant Emin Specific energy, E

(4.99)

Section 4.2 FIGURE 4.7: Typical channel cross section

Basic Principles

145

Top width,T dy

dA

y

which can be combined with Equation 4.97 to yield the following relationship under criticalflow conditions: Q2 A3 = c g Tc

(4.100)

where Tc is the top width of the channel corresponding to y = yc . Equation 4.100 forms the basis for calculating the critical flow depth, yc , in a channel for a given flow rate, Q, since the left-hand side of the equation is known and the right-hand side is a function of yc for a channel of given shape. In most cases, an iterative solution for yc is required and a variety of computational approaches have been suggested (e.g., Patil et al., 2005; Kanani et al., 2008). The specific energy under critical-flow conditions, Ec , is then given by Ec = yc +

Ac 2Tc

(4.101)

Defining the hydraulic depth, D, by the relation A T

(4.102)

V2 Q2 T = gD gA3

(4.103)

D= then a Froude number, Fr, can be defined by Fr2 =

Combining this definition of the Froude number with the critical-flow condition given by Equation 4.100 leads to the relation Frc = 1

(4.104)

where Frc is the Froude number under critical-flow conditions. When y > yc , Equation 4.103 indicates that Fr < 1, and when y < yc , Equation 4.103 indicates that Fr > 1. Flows where y < yc are called supercritical; where y > yc , flows are called subcritical. It is apparent from the specific energy diagram, Figure 4.6, that when the flow conditions are close to critical, a relatively large change of depth occurs with small variations in specific energy. Flow under these conditions is unstable, and excessive wave action or undulations of the water surface usually occur. Experiments in rectangular channels have shown that these instabilities can be avoided if Fr < 0.86 or Fr > 1.13 (U.S. Army Corps of Engineers, 1995). The above derivation of the critical-flow condition assumes that the channel slope is small (cos θ L1 or slope < 10%). In the atypical cases where the channel slope is large (slope Ú 10%), the critical condition is given by (Kanani et al., 2008) αQ2 T =1 gA3 cos θ

(4.105)

Curiously, there exists a particular channel section in which the specific energy does not change with flow depth and hence a critical flow depth does not exist. This channel section is called a singular open-channel section and the geometry of this section can be found in Swamee and Rathie (2007).

146

Chapter 4

Fundamentals of Flow in Open Channels

EXAMPLE 4.4 Determine the critical depth for water flowing at 10 m3 /s in a trapezoidal channel with bottom width 3 m and side slopes of 2 : 1 (H : V). Solution The channel cross section is illustrated in Figure 4.8, where the depth of flow is y, and the top width, T, and flow area, A, are given by T = 3 + 4y A = 3y + 2y2 Under critical-flow conditions

Q2 A3 = c g Tc

and since Q = 10 m3 /s and g = 9.81 m/s2 , under critical-flow conditions (3yc + 2y2c )3 102 = 9.81 (3 + 4yc ) Solving for yc yields

yc = 0.855 m

Therefore, the critical depth of flow in the channel is 0.855 m. Flow under this condition is unstable and is generally avoided in design applications. FIGURE 4.8: Trapezoidal cross section

T = 3 + 4y

1

y

2 3m

The critical-flow condition described by Equation 4.100 can be simplified considerably in the case of flow in rectangular channels, where it is convenient to deal with the flow per unit width, q, given by Q q= (4.106) b where b is the width of the channel. The flow area, A, and top width, T, are given by A = by

(4.107)

T=b

(4.108)

The critical-flow condition given by Equation 4.100 then becomes (byc )3 (qb)2 = g b

(4.109)

which can be solved to yield the critical flow depth  yc =

q2 g

1 3

(4.110)

and corresponding critical energy Ec = yc +

q2 2gy2c

(4.111)

Section 4.2

Basic Principles

147

Combining Equations 4.110 and 4.111 leads to the following simplified form of the minimum specific energy in rectangular channels: Ec =

3 yc 2

(4.112)

The specific energy diagram illustrated in Figure 4.9 for a rectangular channel is similar to the nonrectangular case shown in Figure 4.6, with the exception that the specific energy curve for a rectangular channel corresponds to a fixed value of q rather than a fixed value of Q in a nonrectangular channel. It is apparent from Figure 4.9 that the specific energy curve shifts upward and to the right for increasing values of q, which demonstrates that under critical flow conditions the channel yields the maximum flow rate for a given specific energy. This is true for both rectangular and nonrectangular sections, since for nonrectangular channels the specific energy curve also shifts to the right with increasing flow rate. The specific energy diagram in Figure 4.9 is particularly useful in understanding what happens to the flow in a rectangular channel when there is a constriction, such as when the channel width is narrowed to accommodate a bridge. Suppose that the flow per unit width upstream of the constriction is q1 and the depth of flow at this location is y1 . If the channel is constricted so that the flow per unit width becomes q2 , then, provided that the head loss in the constriction is minimal, Figure 4.9 indicates that there are two possible flow depths in the constricted section, y2 and y2 . Neglecting the head loss in the constriction is reasonable if the constriction is smooth and takes place over a relatively short distance. Figure 4.9 indicates that it is physically impossible for the flow depth in the constriction to be y2 , since this would require the flow per unit width, q, to increase and then decrease to reach y2 . Since the flow per unit width increases monotonically in a constriction, the flow depth can only go from y1 to y2 . A similar case arises when the flow depth upstream of the constriction is y1 and the possible flow depths at the constriction are y2 and y2 . In this case, only y2 is possible, since an increase and then a decrease in q would be required to achieve a flow depth of y2 . Based on this analysis, it is clear that if the flow upstream of the constriction is subcritical (y > yc or Fr < 1), then the flow in the constriction must be either subcritical or critical, and if the flow upstream of the constriction is supercritical (y < yc or Fr > 1), then the flow in the constriction must be either supercritical or critical. These results also mean that if two flow depths are possible in a constriction, then the flow depth that is closest to the upstream flow depth will occur in the constriction. This can be called the closest-depth rule. If the flow upstream of the constriction is critical, then flow through the constriction is not possible under the existing flow conditions and the upstream flow conditions must necessarily change upon installation of the constriction in the channel. Regardless of whether the flow upstream of the constriction is subcritical or supercritical, Figure 4.9 indicates that a maximum constriction will cause critical flow to occur at the constriction. A larger constriction (smaller opening) than that which causes critical flow to occur at the constriction is not possible based on the available specific energy upstream of the FIGURE 4.9: Specific energy diagram for rectangular channel

Depth of flow, y

y



E

q increases

1'

y1' y2'

2' yc y2 y1

2 1 Ec

E1 (⫽ E2)

Specific energy, E

q ⫽ q3 q ⫽ q2 q ⫽ q1

148

Chapter 4

Fundamentals of Flow in Open Channels

constriction. Any further constriction will result in changes in the upstream flow conditions to maintain critical flow at the constriction. Under these circumstances, the flow is said to be choked. Choked flow can result either from narrowing the channel or from raising the invert elevation of the channel. Narrowing the channel increases the flow rate per unit width, q, while maintaining the available specific energy, E, whereas raising the channel invert maintains q and reduces E. EXAMPLE 4.5 A rectangular channel 1.30 m wide carries 1.10 m3 /s of water at a depth of 0.85 m. (a) If a 30-cm wide pier is placed in the middle of the channel, find the elevation of the water surface at the constriction. (b) What is the minimum width of the constriction that will not cause a rise in the upstream water surface? Solution (a) The cross sections of the channel upstream of the constriction and at the constriction are shown in Figure 4.10. Neglecting the head loss between the constriction and the upstream section, the energy equation requires that the specific energy at the constriction be equal to the specific energy at the upstream section. Therefore, V12

y1 +

2g

= y2 +

V22 2g

where Section 1 refers to the upstream section and Section 2 refers to the constricted section. In this case, y1 = 0.85 m, and V1 =

Q 1.10 = 1.00 m/s = A1 (0.85)(1.30)

The specific energy at Section 1, E1 , is V2 1.002 = 0.901 m E1 = y1 + 1 = 0.85 + 2g 2(9.81) Equating the specific energies at Sections 1 and 2 yields 0.901 = y2 + = y2 +

Q2 2gA2 1.102 2(9.81)[(1.30 − 0.30)y2 ]2

which simplifies to y2 +

FIGURE 4.10: Constriction in rectangular channel

0.0617 y22

= 0.901

0.3 m Pier

y2

0.85 m

1.30 m

1.30 m

Upstream of constriction

At constriction

Section 4.2

Basic Principles

149

There are three solutions to this cubic equation: y2 = 0.33 m, 0.80 m, and −0.23 m. Of the two positive depths, we must select the depth corresponding to the same flow condition as upstream. √ At the upstream section, Fr1 = V1 / gy1 = 0.35; therefore, the upstream flow is subcritical and the flow at the constriction must also be subcritical. The flow depth must therefore be y2 = 0.80 m √ where Fr2 = V2 / gy2 = 0.49. The other depth (y2 = 0.33 m, Fr2 = 1.9) is supercritical and cannot be achieved. The closest-depth rule could also be invoked to select the flow depth in the constriction. The closest-depth rule requires that the flow depth in the constriction be 0.80 m, since there are two possible flow depths (0.80 m, 0.33 m) and 0.80 m is closest to the upstream flow depth (0.85 m). (b) The minimum width of constriction that does not cause the upstream depth to change is associated with critical-flow conditions at the constriction. Under these conditions (for a rectangular channel)  1 3 3 q2 3 E1 = E2 = Ec = yc = 2 2 g If b is the width of the constriction that causes critical flow, then  1 3 (Q/b)2 3 E1 = 2 g From the given data: E1 = 0.901 m, and Q = 1.10 m3 /s. Therefore  1 3 1.102 3 0.901 = 2 2 b (9.81) which gives b = 0.75 m If the constricted channel width is any less than 0.75 m, the flow will be choked and the upstream flow depth will increase—in other words, the flow will “back up.” Note: Part (a) of this problem required finding alternate flow depths as the cubic roots of the specific energy equation, which can be easily done using a programmable calculator. Alternatively, the following equations could be used to provide a direct solution of the specific energy equation (Abdulrahman, 2008) ⎧ ⎫ ⎡ ⎡ ⎤⎤⎪ ⎪ ⎡ ⎤ ⎪ ⎪  ⎪ ⎪

3 ⎨ ⎥⎬ ⎢ 2 2 E 2 ysub c ⎢1 ⎥ −1 −1 ⎥ ⎢ ⎦ − π⎦ (4.113) = cos 2 sin ⎣ cos ⎣ cos ⎣− ⎦⎪ ⎪ E 3 3 E 3 ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎧ ⎪ ⎪ ⎪ ⎨

⎡ ⎤⎫ ⎡ ⎡ ⎤⎤ ⎪ ⎪  ⎪

3 ⎬ ⎢ 2 2 ⎥⎥ ysup 1 E c ⎢ −1 −1 ⎣− ⎦⎦⎥ = cos 2 sin ⎢ cos cos ⎣ ⎣ ⎦ ⎪ ⎪ E 3 3 E ⎪ ⎪ ⎪ ⎪ ⎩ ⎭

(4.114)

where ysub and ysup are the subcritical and supercritical flow depths, respectively, E is the actual specific energy, and Ec is the specific energy under critical-flow conditions.

The specific energy analyses covered in this section assume negligible head losses; the kinetic energy correction factor, α, is approximately equal to unity; and the vertical pressure distribution is hydrostatic. Although these approximations are valid in many cases, in diverging transitions with angles exceeding 8◦ , flows tend to separate from the side of the channel,

150

Chapter 4

Fundamentals of Flow in Open Channels TABLE 4.6: Eddy Loss Coefficients, C

Eddy loss coefficient, C Transition

Expansion

Contraction

0.0 0.3 0.5 0.8

0.0 0.1 0.3 0.6

None or very gradual Gradual Typical bridge sections Abrupt

causing large head losses and substantial increases in α (Montes, 1998). Under these conditions, the assumption of a constant specific energy and a value of α approximately equal to unity are not justified. To minimize head losses in transitions, the U.S. Department of Agriculture (USDA, 1977) recommends that channel sides not converge at an angle greater than 14◦ or diverge at an angle greater than 12.5◦ . In cases where the vertical pressure distribution is significantly nonhydrostatic, such as for flow over a curved surface, an alternative specific-energy formulation must be used (e.g., Chanson, 2006). For the general case of contractions and expansions in open channels, the head loss, he , is usually expressed in the form (U.S. Army Corps of Engineers, 2010) $ $ $ V2 V12 $$ $ 2 − α1 $ he = C $α2 $ 2g 2g $

(4.115)

where C is either an expansion or a contraction coefficient, α1 and α2 are the energy coefficients at the upstream and downstream sections, respectively, and V1 and V2 are the average velocities at the upstream and downstream sections, respectively. The head loss, he , given by Equation 4.115 is commonly referred to as the eddy loss. Typical values of eddy loss coefficient, C, for expansions and contractions are given in Table 4.6. 4.3 Water-Surface Profiles 4.3.1

Profile Equation

The equation describing the water-surface profile in an open channel can be derived from the energy equation, Equation 4.90, which is of the form 

S0 − Sf =

V2  y + α 2g x

(4.116)

where S0 is the slope of the channel, Sf is the slope of the energy grade line, y is the depth of flow, α is the kinetic energy correction factor, V is the average velocity, x is a coordinate measured along the channel (the flow direction defined as positive), and x is the distance between the upstream and downstream sections. Equation 4.116 can be further rearranged into % 2&  α V2g y S0 − Sf = + (4.117) x x Taking the limit of Equation 4.117 as x → 0, and invoking the definition of the derivative, yields

Section 4.3

Water-Surface Profiles

% 2&  α V2g y S0 − Sf = lim + lim x x→0 x x→0  2 dy d V = + α dx dx 2g  dy d V 2 dy = + α dx dy 2g dx ⎡ ⎤  dy ⎣ d Q2 ⎦ = 1 + α dx dy 2gA2   dy Q2 dA = 1 − α dx gA3 dy

151

(4.118)

where Q is the (constant) flow rate and A is the (variable) cross-sectional flow area in the channel. Recalling that dA =T (4.119) dy where T is the top width of the channel and the hydraulic depth, D, of the channel is defined as A (4.120) D= T then the Froude number, Fr, of the flow can be written as  √ V dA Q T Q Fr = = = gD A gA gA3 dy or Fr2 =

(4.121)

Q2 dA gA3 dy

(4.122)

Combining Equations 4.118 and 4.122 and rearranging yields S0 − Sf dy = dx 1 − αFr2

(4.123)

This differential equation describes the water-surface profile in open channels, and its origi´ nal derivation has been attributed to Belanger (1828; see Chanson, 2009). To appreciate the utility of Equation 4.123, consider the relative magnitudes of the channel slope, S0 , and the friction slope, Sf . According to the Manning equation, for any given flow rate, ⎛

2 3

⎞2

Sf ⎜ An Rn ⎟ =⎝ 2 ⎠ S0 AR 3

(4.124)

where An and Rn are the cross-sectional area and hydraulic radius under normal flow conditions (Sf = S0 ), and A and R are the actual cross-sectional area and hydraulic radius of the 2

2

flow. Since AR 3 > An Rn3 when y > yn , Equation 4.124 indicates that Sf > S0

when

y < yn ,

and

Sf < S0

when

y > yn

(4.125)

152

Chapter 4

Fundamentals of Flow in Open Channels

or

S0 − Sf < 0 when

y < yn ,

and

S0 − Sf > 0

when

y > yn

(4.126)

It has already been shown that Fr > 1 or

1 − Fr2 < 0

when

when

y < yc ,

y < yc ,

and

and

Fr < 1

when

1 − Fr2 > 0

y > yc

when

y > yc

(4.127) (4.128)

Based on Equations 4.126 and 4.128, the sign of the numerator in Equation 4.123 is determined by the magnitude of the flow depth, y, relative to the normal depth, yn , and the sign of the denominator is determined by the magnitude of the flow depth, y, relative to the critical depth, yc (assuming α L 1). Therefore, the slope of the water surface, dy/dx, is determined by the relative magnitudes of y, yn , and yc . 4.3.2

Classification of Water-Surface Profiles

In hydraulic engineering, channel slopes are classified based on the relative magnitudes of the normal depth, yn , and the critical depth, yc . The hydraulic classification of slopes is shown in Table 4.7 and is illustrated in Figure 4.11. The range of flow depths for each slope can be divided into three zones, delimited by the normal and critical flow depths, where the highest zone above the channel bed is called Zone 1, the intermediate zone is Zone 2, and the lowest zone is Zone 3. Water-surface profiles are classified based on both the type of slope (e.g., type M) and the zone in which the water surface is located (e.g., Zone 2). Therefore, each water-surface profile is classified by a letter and number: for example, an M2 profile indicates a mild slope (yn > yc ) and the actual depth, y, is in Zone 2 (yn < y < yc ). In the case of nonsustaining slopes (Horizontal, Adverse), there is no Zone 1, as the normal depth is infinite in horizontal channels and is nonexistent in channels with adverse slope. There is no Zone 2 in channels with critical slope, as yn = yc . Profiles in Zone 1 normally occur upstream of control structures such as dams and weirs, and these profiles are sometimes classified as backwater curves. Profiles in Zone 2, with the exception of S2 profiles, occur upstream of free overfalls, while S2 curves generally occur at the entrance to steep channels leading from a reservoir. Profiles in Zone 2 are sometimes classified as drawdown curves. Profiles in Zone 3 normally occur on mild and steep slopes downstream of control structures such as gates. TABLE 4.7: Hydraulic Classification of Slopes

Name Mild Steep Critical Horizontal Adverse

Type

Condition

M S C H A

yn > yc yn < yc yn = yc yn = q S0 < 0

EXAMPLE 4.6 Water flows in a trapezoidal channel in which the bottom width is 5 m and the side slopes are 1.5:1 (H:V). The channel lining has an estimated Manning’s n of 0.04, and the longitudinal slope of the channel is 1%. If the flow rate is 60 m3 /s and the depth of flow at a gaging station is 4 m, classify the water-surface profile, state whether the depth increases or decreases in the downstream direction, and calculate the slope of the water surface at the gaging station. On the basis of this water-surface slope, estimate the depth of flow 100 m downstream of the gaging station. Solution To classify the water-surface profile, the normal and critical flow depths must be calculated and contrasted with the actual flow depth of 4 m. To calculate the normal flow depth, apply the Manning equation

Section 4.3 FIGURE 4.11: Slope classifications

1

dy

Horizontal

dx

2

dy

yn 3

dx

yc

dy

Water-Surface Profiles

153

⫽ ⫹, M1 profile ⫽ ⫺, M2 profile ⫽ ⫹, M3 profile

dx

Mild slope

1 Horizontal

2 3

yn

yc

dy dx ⫽ ⫹, S1 dy pro dx ⫽ file ⫺, S2 dy pro dx ⫽ file ⫹

, S3

Steep slope

1

pro

file

Horizontal

3 yn ⫽ y c

dy dx dy dx

⫽ ⫹, C1

⫽ ⫹, C3

profile

profile

Critical slope

2 3

yn ⫽ ∞

dy dx dy

yc

dx

⫽ ⫺, H2 profile ⫽ ⫹, H3 profile

Horizontal slope dy dx

yc

2 3

⫽ ⫺,

dy dx

rofile

A2 p

⫽ ⫹,

rofile

A3 p

Adverse slope 5 2 1 1 1 An3 21 Q = An Rn3 S02 = S n n 23 0 Pn

where Q is the flow rate (= 60 m3 /s), An and Pn are the areas and wetted perimeters under normal flow conditions, and S0 is the longitudinal slope of the channel (= 0.01). The Manning equation can be written in the more useful form  3 Qn A5n = S0 Pn2 where the left-hand side is a function of the normal flow depth, yn , and the right-hand side is in terms of given data. Substituting the given data leads to A5n Pn2

 =

(60)(0.04) √ 0.01

3 = 13, 824

154

Chapter 4

Fundamentals of Flow in Open Channels Since An = (b + myn )yn = (5 + 1.5yn )yn and

Pn = b + 2 1 + m2 yn = 5 + 2 1 + 1.52 yn = 5 + 3.61yn

then the Manning equation can be written as (5 + 1.5yn )5 y5n = 13, 824 (5 + 3.61yn )2 which yields yn = 2.25 m When the flow conditions are critical, then A3c Q2 = Tc g where Ac and Tc are the area and top width, respectively. The left-hand side of this equation is a function of the critical flow depth, yc , and the right-hand side is in terms of the given data. Thus A3c 602 = 367 = Tc 9.81 Since Ac = (b + myc )yc = (5 + 1.5yc )yc and Tc = b + 2myc = 5 + 2(1.5)yc = 5 + 3yc then the critical-flow equation can be written as (5 + 1.5yc )3 y3c = 367 5 + 3yc which yields yc = 1.99 m Since yn (= 2.25 m) >yc (= 1.99 m), then the slope is mild. Also, since y (= 4 m) > yn > yc , the water surface is in Zone 1 and therefore the water-surface profile is an M1 profile. This classification requires that the flow depth increases in the downstream direction. The slope of the water surface is given by (assuming α = 1) S 0 − Sf dy = dx 1 − Fr2 where Sf is the slope of the energy grade line and Fr is the Froude number. According to the Manning equation, Sf can be estimated by  2 nQ Sf = 2 AR 3 and when the depth of flow, y, is 4 m, then A = (b + my)y = (5 + 1.5 * 4)(4) = 44 m2 P = b + 2 1 + m2 y = 5 + 2 1 + 1.52 (4) = 19.4 m R=

44 A = = 2.27 m P 19.4

and therefore Sf is estimated to be ⎡

⎤2 (0.04)(60) ⎦ = 0.000997 Sf = ⎣ 2 (44)(2.27) 3

Section 4.3

Water-Surface Profiles

155

The Froude number, Fr, is given by Fr2 =

(Q/A)2 Q2 T V2 = = gD g(A/T) gA3

where the top width, T, is given by T = b + 2my = 5 + 2(1.5)(4) = 17 m and therefore the Froude number is given by Fr2 =

(60)2 (17) = 0.0732 (9.81)(44)3

Substituting the values for S0 (= 0.01), Sf (= 0.000997), and Fr2 (= 0.0732) into the profile equation yields 0.01 − 0.000997 dy = = 0.00971 dx 1 − 0.0732 The depth of flow, y, at a location 100 m downstream from where the flow depth is 4 m can be estimated by dy (100) = 4 + (0.00971)(100) = 4.97 m y=4 + dx The estimated flow depth 100 m downstream could be refined by recalculating dy/dx for a flow depth of 4.97 m and then using an averaged value of dy/dx to estimate the flow depth 100 m downstream.

Lake discharge problem. In some cases, the slope classification of a channel can play an important role in determining the flow rate in the channel. Such a circumstance is illustrated in Figure 4.12, where water is discharged from a large reservoir, such as a lake, into a channel at A, the flow is normal at C, and B is in the transition region. If H is the head at the channel entrance and the channel slope is steep (such that yn < yc ), then the depth of flow in the transition region must pass through yc such that H = yc + and

Q2 2gA2c

(4.129)

Q2 Tc =1 gA3c

(4.130)

where Equation 4.129 is required for conservation of energy and Equation 4.130 is the condition for critical flow. In contrast, if the channel slope is mild (such that yn > yc ), then the depth of flow in the transition region is approximately equal to yn such that FIGURE 4.12: Lake discharge problem

A

Ve2 2g B

Lake or Reservoir H d

C S0 = Mild slope yc S0 = Steep slope

Slop

e, S

0

yn

156

Chapter 4

Fundamentals of Flow in Open Channels

H = yn +

Q2 2gA2n

(4.131)

and Q=

2 1 1 An Rn3 S02 n

(4.132)

where Equation 4.131 is required for conservation of energy and Equation 4.132 is the condition for normal flow. To determine the actual flow rate, Equations 4.129 and 4.130 are first solved simultaneously for Q and yc , and then the calculated value of Q is used in the Manning equation to calculate yn . If yn < yc , a steep slope is confirmed and the calculated Q is the actual Q. If yn > yc , then the slope is mild and the actual Q is determined by simultaneous solution of Equations 4.131 and 4.132. It is relevant to note that when the slope is mild, a M1 profile is required between the upstream reservoir and the downstream channel. However, this would require that the flow depth increase in the downstream direction (as per Figure 4.11), which would not be possible in the present case. As a consequence, the depth of flow goes immediately to the normal flow depth. The problem illustrated in Figure 4.12 is commonly referred to as the lake discharge problem, and the aforementioned solution works in most cases. In cases where the downstream channel is mild and the channel is not sufficiently long, uniform flow might not be established in the channel, and the uniform-flow equation, Equation 4.132, must be replaced by the (energy) equation for gradually varied flow. In this case, the channel is called hydraulically short. Conversely, if the channel is mild and sufficiently long for uniform flow to be established, then the channel is hydraulically long. For steep slopes, the attainment of normal flow in the downstream channel is not required for Equations 4.129 and 4.130 to be valid, only that the downstream flow be supercritical. EXAMPLE 4.7 The target water-surface elevation of a reservoir is 50.05 m, and the reservoir discharges into a trapezoidal canal that has a bottom width of 2 m, side slopes of 3:1 (H:V), a longitudinal slope of 1%, and an estimated Manning’s n of 0.020. The elevation of the bottom of the canal at the reservoir discharge location is 47.01 m. Determine the discharge from the reservoir. Solution The depth at the discharge location is 50.05 m − 47.01 m = 3.04 m. Since the velocity head in a reservoir can be assumed to be negligible, then H = 3.04 m. From the given channel dimensions: b = 2 m, m = 3, S0 = 0.01, and n = 0.020. Assuming that the slope is hydraulically steep (i.e., yn < yc ), Equations 4.129 and 4.130 require that H = yc +

Q2 2gA2c

and

Q2 T c gA3c

=1

Eliminating Q from these equations yields the more convenient combined form as  gA3c 1 Ac H = yc + ' H = yc + 2 Tc 2T c 2gAc Substituting the geometric properties for a trapezoidal channel gives H = yc + 3.04 = yc +

byc + my2c 2[b + 2myc ] 2yc + 3y2c 2[2 + 2(3)yc ]

which yields yc = 2.37 m. The corresponding value of Q is then given by  Q=

' ( &3 % ( 2 ( 3 (9.81) 2y + 3y c c gAc ) = = 78.3 m3 /s Tc 2 + 2(3)yc

Section 4.3

Water-Surface Profiles

157

Determine the normal depth of flow, yn , corresponding to Q = 78.3 m3 /s by applying the Manning equation which requires that &5 % 5 3 byn + my2n 1 1 1 An3 12 S = S02 Q= 2 2 0 % & n 3 n 3 Pn b + 2yn 1 + m2 % 1 78.3 = 0.020 %

2yn + 3y2n

&5



3 1

2 + 2yn 1 + 32

2 & 2 (0.01) 3

which yields yn = 1.93 m. Since yn < yc (i.e., 1.93 m < 2.37 m), the slope is steep and so the initial assumption of a steep slope is validated. The discharge from the reservoir is 78.3 m3 /s.

4.3.3

Hydraulic Jump

In some cases, supercritical flow must necessarily transition into subcritical flow, even though such a transition does not appear to be possible within the context of the water-surface profiles discussed in the previous section. An example is a case where water is discharged as supercritical flow from under a vertical gate into a body of water flowing under subcritical conditions. If the slope of the channel is mild, then the water emerging from under the gate must necessarily follow an M3 profile, where the depth increases with distance downstream. However, if the depth downstream of the gate is subcritical, then it is apparently impossible for this flow condition to be reached, since this would require a continued increase in the water depth through the M2 zone, which according to Figure 4.11 is not possible. In reality, this transition is accomplished by an abrupt localized change in water depth called a hydraulic jump, which is illustrated in Figure 4.13(a), with the corresponding transition in the specific energy diagram shown in Figure 4.13(b). The supercritical (upstream) flow depth is y1 , the subcritical (downstream) flow depth is y2 , and the energy loss between the upstream and downstream sections is E. If the energy loss were equal to zero, then the downstream depth, y2 , could be calculated by equating the upstream and downstream specific energies. However, the transition between supercritical and subcritical flow is generally a turbulent process with a significant energy loss that cannot be neglected. Applying the momentum equation to the control volume between Sections 1 and 2 leads to P1 − P2 = ρQ(V2 − V1 )

(4.133)

FIGURE 4.13: Hydraulic jump and corresponding specific energy diagram

2 1 Supercritical flow

y2 y1

(a)

Subcritical flow

Depth of flow, y

where P1 and P2 are the hydrostatic pressure forces at Sections 1 and 2, respectively; Q is the flow rate; and V1 and V2 are the average velocities at Sections 1 and 2, respectively. Equation 4.133 neglects the friction forces exerted by the channel boundary within the control volume, where neglecting channel friction is justified by the assumption that over a short distance the friction force will be small compared to the difference in upstream and downstream hydrostatic forces. The momentum equation, Equation 4.133, can be written as

y2

2 ⌬E 1

y1

q ⫽ constant E2 E 1 Specific energy, E (b)

158

Chapter 4

Fundamentals of Flow in Open Channels

γ y1 A1 − γ y2 A2 = ρQ

Q Q − A2 A1

(4.134)

where y1 and y2 are the distances from the water surface to the centroids of Sections 1 and 2, and A1 and A2 are the cross-sectional areas at Sections 1 and 2, respectively. Equation 4.134 can be rearranged as Q2 Q2 + A1 y1 = + A2 y2 (4.135) gA1 gA2 or Q2 + Ay = constant gA

(4.136)

The term on the left-hand side is called the specific momentum, and Equation 4.136 states that the specific momentum remains constant across the hydraulic jump. In the case of a trapezoidal channel, the specific momentum equation, Equation 4.136, can be put in the form by21 my31 by2 my32 Q2 Q2 + + = 2 + + 2 3 gy1 (b + my1 ) 2 3 gy2 (b + my2 )

(4.137)

where b is the bottom width of the channel, and m is the side slope. Equation 4.137 is a fifthorder polynomial in either y1 or y2 ; solutions are most easily found using a programmable calculator with a built-in equation solver or a short computer program such as proposed by Das (2007). In the case of a rectangular channel, Equation 4.136 can be put in the form y2 y2 q2 q2 + 1 = + 2 gy1 2 gy2 2

(4.138)

where q is the flow per unit width. Equation 4.138 can be solved for y2 to yield ⎞ ⎛  y1 ⎝ 8q2 ⎠ y2 = −1 + 1 + 2 gy31

(4.139)

which can also be written in the following nondimensional form:

 y2 1 −1 + 1 + 8Fr21 = y1 2

(4.140)

where y2 /y1 is commonly called the sequent depth ratio, and Fr1 is the upstream Froude number defined by V1 q Fr1 = = (4.141) gy1 y1 gy1 Equation 4.140 was originally derived by Bresse (1860) and is sometimes called Bresse’s equa´ tion, although some have argued that this equation was originally derived by Belanger (1841) and should be called B´elanger’s equation (e.g., Chanson, 2009). The depths upstream and downstream of a hydraulic jump, y1 and y2 , are called the conjugate depths of the hydraulic jump, with y1 sometimes called the initial depth and y2 called the sequent depth. Experimental measurements have shown that Equation 4.140 yields values of y2 to within 1% of observed values (Streeter et al., 1998), where the neglect of shear forces in the momentum equation causes theoretical values of y2 to be slightly larger than

Section 4.3

Water-Surface Profiles

159

observed values of y2 (Beirami and Chamani, 2006). Shear forces can be significant for very rough channels, in which cases roughness effects can be accounted for by using the empirical relation (Carollo et al., 2009):

√ y2 ks (Fr1 − 1)0.963 − 1 = 2 exp − (4.142) y1 yc where ks is the roughness height of the channel, and yc is the critical flow depth. Equation 4.142 is applicable to hydraulic jumps in both smooth (ks = 0) and rough (ks Z 0) channels, and produces predictions of y2 /y1 that are as good or better than the Bresse equation (Equation 4.140) under smooth-channel conditions, and better than the Bresse equation under rough-channel conditions. The theoretical relationship between conjugate depths of a hydraulic jump in a horizontal rectangular channel, Equation 4.140, can also be used for hydraulic jumps on sloping channels, provided that the channel slope is less than about 5%. For larger channel slopes, the component of the weight of the fluid in the direction of flow becomes significant and must be incorporated into the momentum equation from which the hydraulic-jump equation is derived. Other hydraulic-jump equations have been developed for unusual cases, such as hydraulic jumps occurring on inclined contracting channels (Jan and Chang, 2009) and expanding channels (Kordi and Abustan, 2012). In cases where a hydraulic jump occurs in a closed conduit, the hydraulic-jump equations presented in this section are applicable provided that the sequent depth does not exceed the height of the closed conduit. In this context, the hydraulic jump is called a complete hydraulic jump or a free-surface hydraulic jump, and in cases where the sequent depth is greater than the conduit height the hydraulic jump is called an incomplete hydraulic jump or a pressure hydraulic jump. Although complete hydraulic jumps are far more common in engineering design, several specialized hydraulicjump equations have been developed for cases of incomplete hydraulic jumps (e.g., Lowe et al., 2011). The lengths of hydraulic jumps are around 6y2 for 4.5 < Fr1 < 13, and somewhat smaller outside this range. The length, L, of a hydraulic jump can also be estimated by (Hager, 1991) L Fr1 − 1 = 220 tanh y1 22

(4.143)

and the length, Lt , of the transition region between the end of the hydraulic jump and fully developed open-channel flow can be estimated by (Wu and Rajaratnam, 1996) Lt = 10y2

(4.144)

The length of a hydraulic jump is an important variable that is often used to define the downstream limit beyond which no bed protection or special channel provisions are necessary. Physical characteristics of hydraulic jumps in relation to the upstream Froude number, Fr1 , are listed in Table 4.8, and it is noteworthy that air entrainment commences when Fr1 > 1.7 (Novak, 1994). A steady well-established jump with 4.5 < Fr1 < 9.0 is often used as an energy dissipator downstream of a dam or spillway and can also be used to mix chemicals. TABLE 4.8: Characteristics of Hydraulic Jumps

Fr1

Energy dissipation

Characteristics

1.0–1.7 1.7–2.5 2.5–4.5 4.5–9.0 >9.0

1 ⎪ ⎩ n

(4.170)

(4.171)

where S0 and n are the slope and Manning roughness of the channel upstream of the free overfall respectively, and yb is the depth of flow at the brink of the free overfall. Free overfalls are control sections that are frequently found in both artificial and natural channels, for example, waterfalls. Some control sections do not require that critical-flow conditions

Section 4.3

Water-Surface Profiles

169

exist at the control section. For example, at some hydraulic structures, the fixed relationship between flow rate and depth is determined by the geometry of the structure and downstream flow conditions (e.g., at gates and culverts). Length of backwater profile. The length of a backwater profile is commonly defined as the distance from the control section to the section where the depth of flow is within 10% of the normal depth of flow. For prismatic channels this length, L, can be estimated by the relation (Samuels, 1989) yn L = 0.7 (4.172) S0 where yn is the normal depth of flow and S0 is the slope of the channel. However, this approximation is not a substitute for calculating the backwater profile. Steady-state assumption. The computation of a water-surface profile is typically done to determine the water-surface elevations expected along a channel during a specified flood event, such as the 100-year flood event that is used to delineate a floodplain. Although the flow in a channel during any flood event is unsteady, it is typically assumed that the peak discharge rate occurs at the same time for the entire length of the channel and that the discharge rate changes along the channel only at major tributaries. Natural channels. In natural channels the standard-step method is used with the energy equation given by Equation 4.159, repeated here as 

1 V2 y + α 2g 2 $ $ L = $ 2 V12 $$ C $ V2 − α1 $ − S0 Sf + $α2 L $ 2g 2g $

(4.173)

For channels without obstructions or transitions, the middle term in the denominator (the “eddy loss” term) is commonly neglected, but this term should otherwise be taken into account, such as at bridges and constructed channel transitions. The channel slope, S0 , will usually change between intervals along the channel; in such cases it is convenient to express the slope in terms of the bottom elevations of the channel, where S0 =

z1 − z2 L

(4.174)

where z1 and z2 are the bottom elevations of the upstream and downstream sections, respectively. In the usual case where the elevation of the bottom of the channel varies across the section, the bottom elevation at the deepest point is used as the elevation of the bottom of the channel. In most cases of practical interest, the backwater profile is described by the distribution of the water-surface elevation (also known as the stage) along the channel rather than the distribution of depth along the channel. In these cases, the stage, Zi , at the ith section can be calculated as the bottom elevation of the channel plus the depth of flow, Zi = zi + yi

(4.175)

where zi and yi are the bottom elevation and (maximum) flow depth at the ith section, respectively. When major flood flows are considered, the channel sections are typically compound sections where the kinetic energy correction deviates significantly from unity and must be calculated at each section. This is described in more detail below. Flow calculations. A variation of the backwater computation procedure is required when the upstream and downstream stages are given and the objective is to calculate the flow rate in the channel. This situation usually occurs in the context of high-water marks being left by

170

Chapter 4

Fundamentals of Flow in Open Channels

a flood, and the flood flow is to be determined using the high-water marks. In principle, this problem can be solved by varying the flow rate in the energy equation until the best match is achieved between the predicted and measured stages. A technique that is sometimes used to perform these calculations is the slope-area method (Dalrymple and Benson, 1967). However, field evidence indicates that the slope-area method and the one-dimensional energy equation are likely to overestimate the flow rate in floodplains. Alternative estimation methods that account for secondary flows caused by the shear between the floodplain and the main channel might provide improved estimates of the flow rate (e.g., Kordi and Abustan, 2011). Compound channels. For flows in compound channels, such as when the flow section includes a floodplain area, the kinetic energy correction factor, α, can deviate significantly from unity and so its value must be calculated and used in the backwater computations. From the definition of α (see Equation 4.71), for a compound channel, α at any section can be estimated by N  3 1 i=1 Vi Ai 3 α= v dA L (4.176)  AV 3 A V3 N i=1 Ai where A is the total flow area, V is the average velocity over the total flow area, v is the local velocity at any point in the flow area, Ai is a subarea in the total flow area, Vi is the average velocity over Ai , and N is the number of subareas. If it is assumed that the slope of the energy grade line is the same in all subareas, then Equation 4.176 can be expressed in the form % N

&2

α=% N

&3

i=1 Ai i=1 Ki

N  K3 i

i=1

A2i

(4.177)

where Ki is the conveyance of the channel in subarea i, where the conveyance has been defined previously by Equation 4.152. In compound channels that have very wide and flat overbanks and where the kinetic energy correction factor depends on the depth of flow, multiple critical depths are possible (Jain, 2001; U.S. Army Corps of Engineers, 2010). This possibility can be seen by differentiating the specific energy with respect to depth, dE/dy, while maintaining α as a function of y which gives dE αQ2 T Q2 dα =1 − (4.178) + dy gA3 2gA2 dy This relationship can be conveniently represented by defining the compound-channel Froude number, Fr∗ , as αQ2 T Q2 dα (4.179) Fr∗ = + 3 gA 2gA2 dy in which case, critical conditions occur when Fr∗ = 1. The occurrence of multiple critical depths (where Fr∗ = 1) in compound channels has been validated experimentally by Blalock and Sturm (1981). EXAMPLE 4.12 Consider the compound channel shown in Figure 4.15. For a flow rate of 165 m3 /s, determine the kinetic energy factor, α, and compound-channel Froude number, Fr∗ , as a function of the flow depth. Is there more than one critical flow depth? Solution The compound channel can be subdivided into three parts: the left overbank, the main channel, and the right overbank, and the dimensions in these parts of the channel will be denoted by subscripts 1, 2, and 3, respectively. From the given data: Q = 165 m3 /s, w1 = 180 m, n1 = 0.090, w2 = 23 m, n2 = 0.025, w3 = 130 m, n3 = 0.070, and z = 2.52 m. Denoting the depth of flow in the

Section 4.3 FIGURE 4.15: Compound channel

Water-Surface Profiles

171

23 m 180 m

130 m

n = 0.070

n = 0.025

n = 0.090 2.52 m

main channel as y2 , the computations can be conveniently separated into cases where y2 … 2.52 m and y2 > 2.52 m. Case 1, y2 … 2.52 m: In this case the flow is contained entirely in the main channel and the flow variables as a function of y2 are as follows: A = w2 y2 = 23y2 P = w2 + 2y2 = 23 + 2y2 5

K=

5

(23y2 ) 3 1 A3 1 = 2 n2 P 3 0.025 (23 + 2y ) 23 2

α=1 Fr∗ =

αQ2 T (1)(163)2 (23) 5.120 = = gA3 (9.81)(23y2 )3 y32

(4.180)

Since the velocity in each part of the compound channel is taken as being equal to the mean velocity in that part of the channel, this requires that α = 1 when the flow is confined to the main channel. To determine the critical flow depth when the flow is confined to the main channel, set Fr∗ = 1 in Equation 4.180 and solve for y2 . This yields a critical flow depth, yc , of 1.72 m. Case 2, y2 > 2.52 m: In this case the flow is contained in all three parts of the compound channel and the flow variables as a function of y2 are as follows: A1 = w1 y1 = 180(y2 − 2.52) P1 = w1 + y1 = 180 + (y2 − 2.52) 5

5

3 3 1 A1 1 A1 = K1 = n1 32 0.090 23 P1 P1

A2 = w2 y2 = 23y2 P2 = w2 + 2z = 23 + 2(2.52) = 28.04 m 5

5

1 A23 1 A23 = K2 = 2 n2 3 0.025 23 P2 P2 A3 = w3 y3 = 130(y2 − 2.52) P3 = w3 + y3 = 130 + (y2 − 2.52) 5

5

3 3 1 A3 1 A3 = K3 = 2 2 n3 3 0.070 3 P3 P3

Fundamentals of Flow in Open Channels 4.00

4.00

3.50

3.50

3.00 2.50 2.00 1.50 1.00 0.50 0.00 0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

3.00 2.50 2.00 1.50 1.00

Fr* = 1

FIGURE 4.16: Flow properties in compound channel

Depth of flow in main channel, y2 (m)

Chapter 4

Depth of flow in main channel, y2 (m)

172

0.50 0.00 0.0

Kinetic energy correction factor,

2.0

4.0

6.0

8.0

10.0

Compound-channel Froude number, Fr*

A = A1 + A2 + A3 K = K1 + K2 + K3 ⎛ ⎞ 3 K33 K23 A2 ⎝ K1 ⎠ α= + + K3 A21 A22 A23 α dα L dy y2

(4.181)

(4.182)

T = w1 + w2 + w3 = 180 + 23 + 130 = 333 m Fr∗ =

αQ2 T Q2 dα + 3 gA 2gA2 dy

(4.183)

Using these relationships, α can be calculated by Equation 4.181 for any specified value of y2 greater than 2.52 m, the values of dα/dy can be estimated by Equation 4.182 which is the incremental value of α divided by the corresponding incremental value of y2 , and Fr∗ can be calculated by Equation 4.183. The calculated values of α and Fr∗ as a function of y2 are shown in Figure 4.16, where increments of y2 = 0.01 m were used in estimating dα/dy. It is clear from these results that critical-flow conditions occur (Fr∗ = 1) when y2 L 2.52+ m and y2 = 2.91 m. These results show that α is significantly larger when the flow is contained in the entire compound channel compared to when the flow is confined only to the main channel. Also, there are three critical depths of flow, and the initial breaching of the main channel is accompanied by a sudden transition from subcritical to supercritical flow.

4.3.4.5

Profiles across bridges

Although the energy equation given by Equation 4.154 is usually adequate for calculating water-surface profiles along natural channels, inclusion of expansion or contraction losses as given in Equation 4.159 is usually necessary when there are sudden changes in the channel cross section, such as at bridge constrictions and other constructed channel transitions. The contraction and expansion reaches immediately upstream and downstream of a bridge are illustrated in Figure 4.17, and an appropriate methodology for calculating the backwater profile through bridges has been developed by the U.S. Army Corps of Engineers (2010). Section 1 is at the end of the expansion reach, Sections 2 and 3 are immediately downstream and upstream of the bridge (at the toe of the embankment), sections BD and BU are the cross sections inside the bridge structure at the downstream and upstream end of the bridge (accounting for constrictions caused by bridge deck, abutments, and bridge piers), Section 4 is at the beginning of the upstream contraction reach, and AB and CD indicate the intrusion of the bridge into the channel section. The length of the expansion reach, Le , can be estimated using the expansion ratio, ER, such that

Section 4.3 FIGURE 4.17: Cross sections at bridge Source: U.S. Army Corps of Engineers (2010).

Water-Surface Profiles

173

4

1 CR

Contraction reach Lc

3

BU BD A

Le

B

C

D

2

Expansion reach Idealized flow transition

ER 1

1

Le = ER * LO

(4.184)

where LO is the average obstruction length (which can be taken as the average of AB and CD in Figure 4.17), and values of ER are typically in the range of 1–3. The length of the contraction, Lc , can be estimated as 1.0–1.5 times the average obstruction length. Using the channel characteristics at the aforementioned section locations, the standard-step method is applied sequentially in the following section intervals: 1 to 2, 2 to BD, BD to BU, BU to 3, and 3 to 4. EXAMPLE 4.13 The water-surface elevations during 100-year flood conditions are to be determined within the floodplain of a river. The 100-year flow rate is 300 m3 /s, and a segment of the river that includes a bridge is shown in Figure 4.18. Section 1 is a downstream location where the water-surface elevation is controlled at 88.00 m relative to the North American Vertical Datum of 1988 (NAVD88), Section 2 at the downstream toe of the bridge embankment, Sections BD and BU are within the bridge structure at the downstream and upstream ends, respectively, Section 3 is at the upstream toe of the bridge embankment, and Section 4 is 200 m upstream of Section 3. The bridge contains 5 piers, each having a width of 2 m and covering a total width of 10 m under the bridge. The bottom elevations at Sections 1, 2, 3, and 4 are measured as 83.01 m, 83.21 m, 83.21 m, and 83.31 m, respectively, and Manning’s n in the main channel and floodplain are estimated as 0.030 and 0.050, respectively. Determine the water-surface elevations at all the given sections. Solution The energy equation to be satisfied between adjacent channel sections is given by Equation 4.173 and can be expressed as i  2 y + αV 2g i+1$ $ (4.185) L = $ $ % & 2 Vi+1 Vi2 $ $ zi −zi+1 C Sf + L $αi+1 2g − αi 2g $ − L $ $ where i and i + 1 denote the upstream and downstream sections, respectively, L is the distance between the sections, y is the depth of flow (in the main channel), α is the kinetic energy correction factor calculated using Equation 4.177, V is the average velocity, Sf is the average friction slope between the upstream and downstream sections, C is the energy loss factor taken as 0.5 for expanding flow and

174

Chapter 4

FIGURE 4.18: Flow-through bridge

Fundamentals of Flow in Open Channels 2.5 m

60 m

150 m

170 m

2.5 m Sections 1 and 4

10 m

10 m

Sections 2 and 3 L

5m

10 m

R

5m

Sections BD and BU

R

L Flow

4 200 m 3

4m

BU 20 m BD 2

4m

400 m

1 Channel width

0.3 for contracting flow, and z is the bottom elevation of the main channel. The solution procedure is to substitute the known downstream conditions into Equation 4.185 along with the upstream conditions expressed in terms of the unknown upstream depth of flow. This equation is then solved for the upstream depth of flow. The process is then repeated with the next upstream/downstream pair.

Section 4.3

Water-Surface Profiles

175

Step 1: Check expansion and contraction intervals. Before proceeding with the calculations, the adequacy of the distance between Sections 1 and 2 to account for expansion must be assessed. The average obstruction length, LO , at Section 2 is (140 m + 160 m)/2 = 150 m, and the assumed expansion distance, Le , is 400 m, and so the assumed expansion ratio, ER, is given by ER =

1440 Le = = 2.67 LO 150

Since ER is typically in the range 1–3, Le = 400 m is appropriate. The ratio of the contraction length between Sections 3 and 4, Lc (= 200 m), and the average obstruction length, LO (= 150 m), is given by Lc 200 = 1.3 = LO 150 Since this ratio is typically in the range 1–1.5, the section interval of 200 m (between Sections 3 and 4) allowed for contraction is adequate. Step 2: Initialize variables at Section 1. The key flow variables at any section are the depth of flow, y, flow area, A, wetted perimeter, P, and the conveyance, K. At each section, these variables will be denoted by yij , Aij , Pij , and Kij , respectively, where i denotes the section and j denotes the portion of the section, where j = 1, 2, 3, denote the left-overbank, main channel, and right-overbank portions of the channel, respectively. At Section 1, the widths of these portions are w11 = 150 m, w12 = 60 m, and w13 = 170 m, respectively. From the given data, n11 = n13 = 0.050, n12 = 0.030, Q = 300 m3 /s, y11 = 88.0 − 83.01 − 2.5 = 2.49 m A11 = w11 y11 = (150)(2.49) = 373.5 m2 P11 = w11 + y11 = 150 + 2.49 = 152.49 m 5

5

3 1 A11 1 373.5 3 = = 13573 m3 /s K11 = n11 32 0.050 152.49 23 P11

y12 = 88.0 − 83.01 = 4.99 m A12 = w12 y12 = (60)(4.99) = 299.4 m2 P12 = w12 + 2z = 60 + 2(2.5) = 65 m 5

5

3 1 A12 1 299.4 3 = = 27628 m3 /s K12 = 2 n12 3 0.030 65 23 P12

y13 = y11 = 2.49 m A13 = w13 y13 = (170)(2.49) = 423.3 m2 P13 = w13 + y13 = 170 + 2.49 = 172.49 m 5

5

3 1 A13 1 423.3 3 = = 15403 m3 /s K13 = 2 n13 3 0.050 172.49 23 P13

A1 = A11 + A12 + A13 = 373.5 + 299.4 + 423.3 = 1096.2 m2

176

Chapter 4

Fundamentals of Flow in Open Channels K1 = K11 + K12 + K13 = 13573 + 27628 + 15403 = 56604 m3 /s ⎛ ⎞ 3 3 K1i A21  ⎝ ⎠ = 1.81 α1 = K13 i=1 A21i  Sf 1 = V1 =

Q K1

2

=

300 2 = 0.0000281 56604

Q 300 = 0.27 m/s = A1 1096.2

Step 3: Specify variables at Section 2. From the given data, w21 = 10 m, w22 = 60 m, w23 = 10 m, n21 = n23 = 0.050, and n22 = 0.030. The flow variables at Section 2 can all be expressed in terms of the flow depth in the main channel at Section 2. Denoting this flow depth as y22 , the variables are as follows: y21 = y22 − 2.5 A21 = w21 y21 = (10)y21 P21 = w21 = 10 m 5 3 1 A21 K21 = 2 n21 3 P21

y22 = y22 A22 = w22 y22 = (60)y22 P22 = w22 + 2z = 60 + 2(2.5) = 65 m 5 3 1 A22 K22 = n22 23 P22

y23 = y21 A23 = w23 y23 = 10y23 P23 = w23 = 10 m 5 3 1 A23 K23 = n23 23 P23

A2 = A21 + A22 + A23 K2 = K21 + K22 + K23 ⎛ ⎞ 3 3 K2i A22  ⎝ ⎠ α2 = K23 i=1 A22i

Sf 2 = V2 =

Q 2 K2

Q A2

Problems

177

Step 4: Determine the flow depth at Section 2. From the given data, the invert elevations at Section 1 and Section 2 are given by z1 = 83.01 m and z2 = 83.21 m, respectively, and L = 400 m, hence the bottom slope, S0 , between Sections 1 and 2 is z − z1 83.21 − 83.01 = 0.0005 = S0 = 2 L12 400 Taking C = 0.5, since the flow is expanding between Sections 1 and 2, and substituting the above-calculated flow variables at Sections 1 and 2 into Equation 4.185 yields y22 = 4.80 m. The corresponding water-surface elevation, Z2 , is given by Z2 = y22 + z2 = 4.80 + 83.21 = 88.01 m Step 5: Determine water-surface elevations at all other sections. Steps 2–4 are repeated for all other section intervals to determine the water-surface elevations at Sections BD, BU, 3, and 4. The results of these calculations are as follows:

Section

Depth in Main Channel (m)

Stage (m)

1 2 BD BU 3 4

4.99 4.80 4.79 4.80 4.83 4.79

88.01 88.01 88.00 88.01 88.04 88.10

The approach in this example is appropriate for the usual case of subcritical flow through bridges. In unusual cases where the flow is supercritical or the flow is choked at the bridge, alternative approaches should be used (e.g., USACE, 2010).

Problems 4.1. A trapezoidal channel having an area of 40 m2 has a side slope of 1.5: 1 (H: V) and longitudinal slope of 1 in 1,000. Find the dimensions of the section when it is most economical. Determine the discharge of the most economical section, if Chezy’s constant, C = 60. 4.2. Find the force per unit length of a wall, the pressure coefficient at the base of the wall, and the force in the excess of the hydrostatic value if the pressure of water varies from 0 to 275 mm below free surface. The pressure values are 0, 51, 104, 159, 216, and 275 mm with a distance of 50 mm interval to reach 250 mm at the base of wall. 4.3. The approximate velocity distribution in a very wide river 3.5 m deep may be found with the equation u = 1 + 2(y/h)1/2 . Find V, α, and β. 4.4. A trapezoidal channel is to be excavated at a site where permit restrictions require that the channel have a bottom width of 5 m, side slopes of 1.5 : 1 (H:V), and a depth of flow of 1.8 m. If the soil material erodes when the shear stress on the perimeter of the channel exceeds 3.5 Pa, determine the appropriate slope and corresponding flow capacity of the channel. Use the Darcy–Weisbach equation and assume that the excavated channel has an equivalent sand roughness of 3 mm. 4.5. A trapezoidal channel having a side slope of 2H: 1V with a uniform bed slope of 1 * 10−3 carries a discharge of

60 m3 /s at a depth of 2.5 m. Calculate the bottom width. Determine the average and maximum shear stress on the bed and sides of this channel with n as 0.015. 4.6. Given that hydraulically rough flow conditions occur in open channels when u∗ ks /ν Ú 70, show that this condition can be expressed in terms of Manning parameters as n6 RS0 Ú 7.9 * 10−14 If a concrete-lined rectangular channel with a bottom width of 5 m is constructed on a slope of 0.05%, and Manning’s n is estimated to be 0.013, determine the minimum flow depth for hydraulically rough flow conditions to exist. 4.7. Water flows at a depth of 2.20 m in a trapezoidal, concrete-lined section (ks = 2 mm) with a bottom width of 3.6 m and side slopes of 2:1 (H:V). The longitudinal slope of the channel is 0.0006 and the water temperature is 20◦ C. Assuming uniform-flow conditions, estimate the average velocity and flow rate in the channel. Use both the Darcy–Weisbach and Manning equations and compare your answers. 4.8. Water flows in a trapezoidal channel that has a bottom width of 5 m, side slopes of 2:1 (H:V), and a longitudinal

178

Chapter 4

Fundamentals of Flow in Open Channels

slope of 0.0001. The channel has an equivalent sand roughness of 1 mm. Calculate the uniform flow depth in the channel when the flow rate is 18 m3 /s. Is the flow hydraulically rough, smooth, or in transition? Would the Manning equation be valid in this case? 4.9. Show that Manning’s n can be expressed in terms of the Darcy friction factor, f , by the following relation: 1

1

f 2 R6 8.86 where R is the hydraulic radius of the flow. Does this relationship conclusively show that n is a function of the flow depth? 4.10. A trapezoidal lined brick channel (ks equals 3 mm) has a bottom width of 2.5 m and depth of flow of 0.9 m. The side slope of the lined channel is 2H: 1V. The longitudinal slope of the channel is 3 * 10−4 . Estimate (a) the average shear stress, (b) the hydrodynamic nature of the surface, (c) Chezy and Manning’s constant, and (d) the discharge. 4.11. It has been shown that in fully turbulent flow Manning’s n can be related to the height, d, of the roughness projections by the relation n=

n=

equivalent sand roughness of 3 mm, and is laid on a slope of 0.1%. Determine the minimum flow depth for fully turbulent conditions to exist. Can the Manning equation be used at this flow depth? 4.15. A rectangular channel of 6 m wide carries a discharge of 20 m3 /s. The depth of flow of water at a certain section is 0.8 m. If a hydraulic jump takes place on the downstream side, find the depth of flow after the jump and loss of energy per kg of water. 4.16. The roadside gutter shown in Figure 4.19 has a curb depth of 15 cm, a cross-slope of 2%, a longitudinal slope of 1%, and an estimated surface roughness height of 1 mm. (a) Determine the flow capacity of the gutter using the Darcy–Weisbach equation; (b) determine the flow capacity using the Manning equation; (c) assess the validity of the Manning equation in this case; and (d) account for the discrepancy in the gutter capacity estimated using the Darcy–Weisbach and Manning equations. [Hint: The gutter capacity is equal to the flow rate when the water level is at the curb.] ks = 1 mm

1 0.039d 6

where d is in meters. If the estimated roughness height in a channel is 30 mm, then determine the percentage error in n resulting from a 70% error in estimating d.

15 cm Slope = 2%

1

4.12. Show that the minimum value of n/ks6 given by Equation 4.45 is n = 0.039 1 ks6 1 Determine the range of R/ks in which n/ks6 does not deviate by more than 5% from the minimum value. [Hint: You may need to use the relation log x = 0.4343 ln x.] 4.13. Stages are measured by two recording gages 100 m apart along a constructed water-supply channel. The channel has a bottom width of 5 m and side slopes of 3:1 (H:V). The bottom elevations of the channel at the upstream and downstream gage locations are 24.01 m and 23.99 m, respectively. At a particular instance, the upstream and downstream stages are 25.01 m and 24.95 m, respectively, and the flow is estimated as 15;2 m3 /s. (a) Derive an expression for Manning’s n as a function of the estimated flow rate; (b) estimate Manning’s n and the roughness height in the channel between the two measurement stations; and (c) quantitatively assess the sensitivity of the flow rate to the channel roughness. 4.14. Show that the turbulence condition u∗ ks /ν > 70 can be put in the form ks RS0 > 2.2 * 10−5 A trapezoidal concrete channel with a bottom width of 3 m and side slopes of 2:1 (H:V) is estimated to have an

FIGURE 4.19: Flow in gutter

4.17. A trapezoidal irrigation channel is to be excavated to supply water to a farm. The design flow rate is 1.8 m3 /s, the side slopes are 2:1 (H:V), the longitudinal slope of the channel is 0.1%, Manning’s n is 0.025, and the geometry of the channel is to be such that the length of each channel side is equal to the bottom width. (a) Specify the dimensions of the channel required to accommodate the design flow under normal conditions; and (b) if the channel lining can resist an average shear stress of up to 4 Pa, under what flow conditions is the channel lining stable? 4.18. Water flows in a concrete trapezoidal channel with a bottom width of 3 m, side slopes of 2:1 (H:V), and a longitudinal slope of 0.1%. The Manning roughness coefficient is estimated to be 0.015. (a) What roughness height is characteristic of this channel? (b) For what range of flow depths and corresponding flow rates can n be assumed to be approximately constant and the Manning equation applicable? (c) If the flow rate in the channel is 100 m3 /s, what Manning’s n should be used and what is the corresponding flow depth? How does this flow depth compare with that obtained by using n = 0.015? 4.19. A natural stream has a cross section that is approximately trapezoidal with a bottom width of 10 m and

Problems side slopes of 2.5:1 (H:V). The longitudinal slope of the channel is estimated to be 0.1%. When the flow in the stream is 150 m3 /s the flow is observed to be approximately uniform with a flow depth of 5 m. (a) Estimate Manning’s n and the roughness height in the channel. (b) Assess the validity of the Manning equation in this case. (c) If flow conditions change such that the depth of flow increases by 50%, what Manning’s n would you use? 4.20. Derive the equation for equivalent Manning roughness in a compound channel proposed by Horton (1933) and Einstein (1934). 4.21. Derive the equation for equivalent Manning roughness in a compound channel proposed by Lotter (1933). 4.22. The sections in the floodplain shown in Figure 4.5 have the following values of Manning’s n: Section

n

1 2 3 4 5 6 7

0.040 0.030 0.015 0.013 0.017 0.035 0.060

Left floodplain Main channel Right floodplain

4.25. Use Equation 4.56 to show that the depth-averaged velocity in an open channel with depth, d, can be estimated by averaging the velocities at 0.2d and 0.8d from the bottom of the channel. 4.26. In a trapezoidal channel (bed slope of 1 * 10−4 ) with a bottom width of 15 m and side slope of 1:1, 50 m3 /s of water is discharged. Find the normal depth of flow. If n is 0.02, is the flow subcritical or supercritical? 4.27. A horizontal rectangular channel discharges water with a velocity of 5 m/s through a sluice gate with a depth of flow of 0.56 m. The width of the channel is 6 m. Determine whether a hydraulic jump will occur, and if so, find its height. Calculate the loss of energy per kg of water. What is the associated energy loss in the hydraulic jump? Take the specific weight for water as 1,000 kg/m3 . 4.28. Use the Darcy–Weisbach equation to show that the head loss per unit length, S, between any two sections in an open channel can be estimated by the relation 2

S=

If the depth of flow in the floodplain is 5 m, use the formulae in Table 4.3 to estimate the composite roughness. 4.23. Consider the drainage channel and adjacent floodplains shown in Figure 4.20. The Manning roughness coefficients are given by: Section

179

4.29.

4.30.

n 0.040 0.016 0.050 4.31.

and the longitudinal slope is 0.5%. Field tests have shown that the Horton (1933) equation best describes the composite roughness of the channel. Find the capacity of the main channel and the lateral extent of the floodplains for a 100-year flow of 1590 m3 /s. 4.24. Use Equation 4.56 to show that the velocity in an open channel is equal to the depth-averaged velocity at a distance of 0.368d from the bottom of the channel, where d is the depth of flow.

4.32.

f V 4R 2g

where f , R, and V are the average friction factor, hydraulic radius, and flow velocity, respectively, between the upstream and downstream sections. Estimate the length of the back water curve caused by an affux of 1.8 m in a rectangular channel of width 40 m and depth 2.5 m. Use a bed slope of 1 in 2,000 and Manning’s constant n as 0.03. Water flowing in a rectangular channel 2.6 m wide is lined with rough concrete with a bed slope of 4 * 10−4 . If the normal depth of flow is 1.25 m and n is 0.015, calculate the (a) conveyance of the channel, (b) discharge, (c) Froude number, and (d) average bed shear stress. Water flows through a rectangular channel 10 m wide and 2 m deep with a velocity of 1.5 m/s. The flow of water through the channel of bed slope of 1 in 200 is regulated in such a way that energy line has a slope of 1 in 5,000. Find the rate of change of the depth of water. The specific energy for a 4 m wide rectangular channel is estimated as 3.5 m. If the flow rate of water through the channel is 15 m3 /s, determine the alternate depths of flow.

Left floodplain

Right floodplain

S ⫽ 1%

S ⫽ 4%

Main channel 3m

1 3

1 2

30 m FIGURE 4.20: Flow in an open channel

180

Chapter 4

Fundamentals of Flow in Open Channels

4.33. Consider the flow conditions in the concrete channel shown in Figure 4.21, where the flow rate in the channel is 16 m3 /s. The bottom width of the channel is to be contracted at a short distance downstream of the section shown in Figure 4.21. (a) State the equations that must be satisfied in order for choking to occur at the downstream section; simplify your equations as much as possible; and (b) Will the flow be choked at the downstream section when the bottom width of the channel is reduced to zero? What is the depth of flow in the downstream section under this condition?

1

2m

3m

3 10 m

FIGURE 4.21: Flow in pre contracted section

4.34. A lined rectangular concrete drainage channel is 10.0 m wide and carries a flow of 8 m3 /s. In order to pass the flow under a roadway, the channel is contracted to a width of 6 m. Under design conditions, the depth of flow just upstream of the contraction is 1.00 m, and the contraction takes place over a distance of 7 m. (a) If the energy loss in the contraction is equal to V12 /2g, where V1 is the average velocity upstream of the contraction, what is the depth of flow in the constriction? (b) Does consideration of energy losses have a significant effect on the depth of flow in the constriction? (c) If the width of the constriction is reduced to 4.50 m and a flow of 8 m3 /s is maintained, determine the depth of flow within the constriction (include energy losses). (d) If reducing the width of the constriction to 4.50 m influences the upstream depth, determine the new upstream depth. 4.35. A 3 m wide rectangular channel (bed slope 1 in 1,000) carries 10 m3 /s of water at a depth of 2.0 m. The discharge has to be increased to a maximum by changing the dimensions of the section for the same crosssectional area, bed slope, and roughness of the channel. Find the increase in discharge using Chezy’s constant C as 60. 4.36. Show that the critical step height required to choke the flow in a rectangular open channel is given by Fr21 3 2 zc − Fr13 =1 + y1 2 2 where zc is the critical step height, y1 is the flow depth upstream of the step, and Fr1 is the Froude number upstream of the step. Use this equation to verify your answer to Problem 4.35. 4.37. Water flows in a rectangular channel of width 5.5 m and the depth of uniform flow 0.75 m. If the channel depth is increased to 1 m by constructing a weir, find whether

a hydraulic jump takes place using Manning’s constant n as 0.02 and bed slope as 0.003. 4.38. Water flows at 18 m3 /s in a trapezoidal channel with a bottom width of 5 m and side slopes of 2 : 1 (H : V). The depth of flow in the channel is 2 m. If a bridge pier of width 50 cm is placed in the middle of the channel, what is the depth of flow adjacent to the pier? What is the maximum width of a pier that will not cause a rise in the water surface upstream of the pier? 4.39. Water flows at 15 m3 /s in a trapezoidal channel with a bottom width of 4.5 m and side slopes of 1.5 : 1 (H : V). The depth of flow in the channel is 1.9 m. If a step of height 15 cm is placed in the channel, what is the depth of flow over the step? What is the maximum height of the step that will not cause a rise in the water surface upstream of the step? 4.40. A wide rectangular channel (slope 5 * 10−4 ) connects two reservoirs 1.2 km apart. The upstream reservoir level is constant at an elevation of 105 m, and the elevation of the canal invert at the intake is 102 m. The intake is free and the loss of energy at the intake can be neglected. If n is 0.02, what should the downstream reservoir level be for uniform flow in the channel? 4.41. A float-finished trapezoidal concrete channel has a longitudinal slope of 0.05%, side slopes of 2:1 (H:V), and a bottom width of 5 m. The design flow rate in the channel is 7 m3 /s. (a) Verify the validity of the Manning equation and calculate the normal depth of flow in the channel. (b) If the bottom width of the channel abruptly changes from 5 m to 4 m, what is the head loss in the contraction and flow depth at the contracted section? What is the flow depth at the contracted section if the head loss in the contraction is neglected? Assess the importance of including head loss in your analysis. 4.42. A rectangular channel has a width of 30 m, a longitudinal slope of 0.5%, and an estimated Manning’s n of 0.025. The flow rate is 100 m3 /s at a particular section where the depth of flow is observed to be 3.000 m. Temporary construction requires that the channel be contracted to a width of 20 m over a distance of 40 m, and then returned to its original width of 30 m over a distance of 40 m. All sections are rectangular. Determine the depths of flow in the contracted and downstream sections when: (a) all energy losses between sections are neglected; and (b) friction, contraction, and expansion losses are taken into account. [Note: To simplify the computations you can assume that the friction slope is the same at all three sections.] Based on your results, evaluate the impact of accounting for energy losses on the estimated difference between the water stages at the upstream and downstream sections. 4.43. Water approaches a bridge constriction in a horizontal rectangular concrete-lined channel of width 10 m and, to accommodate the bridge, the channel abruptly contracts to a width of 7 m and then expands (abruptly) back to a width of 10 m. Under design conditions, the flow rate in

Problems

4.44.

4.45.

4.46.

4.47.

4.48.

4.49.

4.50.

the channel is 20 m3 /s and the flow depth in the approach channel is 2 m. (a) Taking into account contraction and expansion head losses, what is the flow depth at the bridge constriction and the flow depth downstream of the bridge? (b) What is the percentage error in the calculated flow depths if energy losses are neglected? Water flows at 36 m3 /s in a rectangular channel of width 10 m and Manning’s n of 0.030. If the depth of flow at a channel section is 3 m and the slope of the channel is 0.001, classify the water-surface profile. What is the slope of the water surface at the observed section? Would the water-surface profile be much different if the depth of flow were equal to 2 m? Water flows at 4.5 m3 /s/m in a horizontal rectangular channel. The head loss is found to be 5.5 m. If a hydraulic jump takes place in the channel, determine the sequent depths of the jump. Water flows in a trapezoidal channel where the bottom width is 6 m and side slopes are 2:1 (H: V). The channel lining has an estimated Manning’s n of 0.045, and the slope of the channel is 1.5%. When the flow rate is 80 m3 /s, the depth of flow at a gaging station is 5 m. Classify the water-surface profile, state whether the depth increases or decreases in the downstream direction, and calculate the slope of the water surface at the gaging station. On the basis of this water-surface slope, estimate the depths of flow 100 m downstream and 100 m upstream of the gaging station. Water flows in a trapezoidal channel (1.5H: 1V) of bottom width of 4 m, depth of 1.25 m, and a bed slope of 10−4 . Find the mean velocity of flow and indicate whether the boundary is smooth, rough, or a transition. Assume the equivalent sand roughness of the channel and kinematic viscosity as 2 mm and 1.1 * 10−6 m2 /s respectively. Derive an expression relating the conjugate depths in a hydraulic jump when the slope of the channel is equal to S0 . [Hint: Assume that the length of the jump is equal to 5y2 and that the shape of the jump between the upstream and downstream depths can be approximated by a trapezoid.] Water flows at 20 m3 /s of through a sluice gate in a 3.2 m wide rectangular channel at a depth of 0.7 m. Examine the type of hydraulic jump formed when the tailwater is 3.5 m. Assume the coefficient of contraction as 0.6. The head loss, hL , across a hydraulic jump is described by the equation: V2 V2 y1 + 1 = y2 + 2 + hL 2g 2g where the subscripts 1 and 2 refer to the conditions upstream and downstream of the jump respectively. Show that the dimensionless head loss, hL /y1 , is given by 

2  Fr21 hL y2 y1 =1 − + 1 − y1 y1 2 y2 where Fr1 is the upstream Froude number.

181

4.51. Water flows at 20 m3 /s at a depth of 1 m in a trapezoidal channel having a bottom width of 1 m and side slopes of 2:1 (H:V). (a) Is it possible for a hydraulic jump to occur in the channel? (b) If a hydraulic jump occurs between the channel section and a downstream rectangular section of bottom width 5 m, determine the downstream flow depth and the power loss in the jump. 4.52. Show that the hydraulic jump equation for a trapezoidal channel is given by by21 2

+

my31 3

+

my32 by22 + + 2

4.53.

4.54.

4.55.

4.56.

4.57.

4.58.

3

Q2 = gy1 (b + my1 ) Q2 gy2 (b + my2 )

where b is the bottom width of the channel, m is the side slope of the channel, y1 and y2 are the conjugate depths, and Q is the volumetric flow rate. Water flows in a horizontal trapezoidal channel at 21 m3 /s, where the bottom width of the channel is 2 m, side slopes are 1 : 1, and the depth of flow is 1 m. Calculate the downstream depth required for a hydraulic jump to form at this location. What would be the energy loss in the jump? A horizontal rectangular flume allows water to flow with a velocity of 6.5 m/s at a depth of 1.2 m. Verify whether the flow is supercritical. Calculate the conjugate depth and energy loss in the jump. Water flows at 10 m3 /s in a rectangular channel of width 5.5 m. The slope of the channel is 0.15% and the Manning roughness coefficient is 0.038. Estimate the depth 100 m upstream of a section where the flow depth is 2.2 m using: (a) the direct-integration method, and (b) the standard-step method. Approximately how far upstream of this section would you expect to find uniform flow? Water flows at 5 m3 /s in a 4-m-wide rectangular channel that is laid on a slope of 4%. If the channel has a Manning’s n of 0.05 and the depth at a given section is 1.5 m, how far upstream or downstream is the depth equal to 1 m? If the depth of flow in the channel described in Problem 4.37 is measured as 0.75 m, find the distance upstream of the weir where the depth is 2.2 m. A trapezoidal canal has a longitudinal slope of 1%, side slopes of 3 : 1 (H : V), a bottom width of 3.00 m, a Manning’s n of 0.015, and carries a flow of 20 m3 /s. The depth of flow at a gaging station is observed to be 1.00 m. Answer the following questions: (a) What is the normal depth of flow in the channel? (b) What is the critical depth of flow in the channel? (c) Classify the slope of the channel and the water surface profile at the gaging station. (d) How far from the gaging station is the depth of flow equal to 1.1 m? Does this depth occur upstream or downstream of the gaging station?

182

Chapter 4

Fundamentals of Flow in Open Channels 4.63. Consider the main channel and adjacent 100-year floodplain shown in Figure 4.23 where Manning’s n in the main channel and floodplain are 0.050 and 0.100, respectively, and the slope of the channel is 0.75%. Consider a design flow of 110 m3 /s. (a) Calculate the change in the normal depth of flow if developers are allowed to fill in 15 m of the width of the 100-m wide floodplain. (b) As an alternative to development within the floodplain, a hydraulic structure will be installed that will cause the stage at a particular section to be 54.50 m NAVD88 at a location 150 m downstream when the flow is 110 m3 /s. Estimate how far upstream of the structure you would have to go to have the same elevation as you would have in the filled-in floodway.

(e) If the bottom of the channel just downstream of the gaging station is raised by 0.20 m, determine the resulting depth of flow at the downstream section. The bottom width of the channel remains constant at 3 m. 4.59. The depth of the channel on both sides of the section is 1 m from the free water surface. With a bottom width of 3.0 m with side slope of 1:1. The middle section is being 4 m from the free water surface with a width of 6 m and side slope of 1:1. Assume that n is 0.022 and S is 1 * 10−4 for the entire section. Find the total discharge carried by the channels. 4.60. The flow conditions at Stations 1 and 2 in a rectangular channel are shown in Figure 4.22, where Station 1 is 100 m upstream of Station 2. If the flow rate in the channel is 2.5 m3 /s and the longitudinal slope is 0.5%, estimate Manning’s n in the channel. 4.61. Water flows at 11 m3 /s in a rectangular channel of width 5 m. The slope of the channel is 0.1 and the Manning roughness coefficient is equal to 0.035. If the depth of flow at a selected section is 2 m, calculate the upstream depths at 20 m intervals along the channel until the depth of flow is within 5% of the uniform flow depth. 4.62. Water flows in an open channel whose slope is 0.05%. The Manning roughness coefficient of the channel lining is estimated to be 0.040 when the flow rate is 250 m3 /s. At a given section of the channel, the cross section is trapezoidal with a bottom width of 12 m, side slopes of 2:1 (H:V), and a depth of flow of 8 m. Use the standard-step method to calculate the depth of flow 100 m upstream from this section where the cross section is trapezoidal with a bottom width of 16 m and side slopes of 3:1 (H:V).

4.64. At a particular river cross section, the elevation of the bottom of the channel is 102.05 m and the elevation of the river bank is 105.27 m. The river has an approximately trapezoidal cross section with side slopes of 3:1 (H:V) and a bottom width of 20 m. The longitudinal slope of the channel and the adjacent floodplain is 2%, and the Manning roughness of the channel is estimated as 0.07. It is planned to place a bridge 100 m downstream of the river section, and over this distance the river will be made to transition to a bottom width of 10 m while maintaining the same side slope. Under design conditions, the flow in the river is 12 m3 /s and the depth of flow at the upstream section is 1.60 m. (a) What will be the depth of flow at the bridge section? (b) Will the floodplain be flooded at the bridge location? 4.65. Show that the energy equation for open-channel flow between stations B (upstream) and A (downstream) can be written in the form

1m

0.9 m 5m

5m

Station 1

Station 2

FIGURE 4.22: Flow in upstream and downstream sections

100 m

3m

10 m FIGURE 4.23: Floodplain

Elevation = 50.00 m NAVD88

Problems 

V2 Z + α 2g

B

183

n is 0.01, and is to be designed such that any hydraulic jump would occur at the midpoint of the stilling basin. As a designer, what length of stilling basin would you specify? 4.67. A trapezoidal drainage channel of bottom width 5 m, side slopes 2:1 (H:V), Manning’s n of 0.018, and longitudinal slope of 0.1% terminates at a gate where the relationship between the flow through the gate, Q, headwater elevation (HW), tailwater elevation (TW), and gate opening (h) is given by √ Q = 13.3h HW − TW

= Sf L A

where Z is the water surface elevation, α is the energy coefficient, V is the average velocity, Sf is the average friction slope between stations A and B, and L is the distance between stations A and B. A backwater curve is being computed in the stream between two sections A and B, 140 m apart. The hydraulic properties of the cross sections are shown in Table 4.9. For a flow in the channel of 280 m3 /s and a Manning’s n of 0.040, the water elevation at Station A is 517.4 m. Compute the surface elevation at Station B. Just upstream of Station B, the flow is partially obstructed by a large bridge pier in the channel, presenting an obstruction 2.50 m wide normal to the direction of flow. The channel at this location can be considered roughly rectangular in cross section, with the bottom at Station B at elevation 515.10 m. Compute the watersurface elevation adjacent to the pier. 4.66. Under design conditions, flow exits at the bottom of a 10-m-wide spillway at a rate of 220 m3 /s, at a stage of 13.5 m, and at a depth of 1 m. Downstream of the spillway is a river at a stage of 21.5 m. The channel connecting the spillway exit to the river is to be horizontal and rectangular with a width of 10 m. This connecting channel (called a stilling basin) is to be constructed such that Manning’s

where Q is in m3 /s, and HW, TW, and h are in meters. If the elevation of the bottom of the channel at the gate location is 0.00 m, the tailwater elevation is 1.00 m and the flow in the channel is 20 m3 /s, estimate the minimum gate opening such that the water surface elevation 100 m upstream of the gate does not exceed 2.20 m. [Note: The depth of flow 100 m upstream of the gate is not 2.20 m.] 4.68. For the compound channel given in Example 4.12 and for a flow rate of 165 m3 /s, calculate the kinetic energy correction factor and the compound-channel Froude number corresponding to a depth of flow in the main channel of: (a) 1.25 m, and (b) 3.25 m. Classify the corresponding flows as either subcritical or supercritical. 4.69. For the bridge and flow conditions given in Example 4.13, calculate the water-surface elevation (i.e., stage) at Section BD.

TABLE 4.9

Area (m2 )

Wetted perimeter (m)

Water surface elevation (m)

Section A

Section B

Section A

Section B

518.5 518.2 517.9 517.6 517.2 516.9

— 181.86 166.81 152.13 137.82 123.88

118.45 108.42 98.66 86.96 78.18 —

— 52.21 50.84 48.77 47.40 46.02

36.27 35.05 33.83 32.77 31.70 —

C H A P T E R

5

Design of Drainage Channels 5.1 Introduction Constructed open channels frequently serve as drainageways in stormwater-management systems. A common feature of drainage channels is that they are usually dry when there is no surface runoff and are designed to accommodate the peak runoff rate from a design storm. Drainage channels are commonly found alongside roadways and are therefore closely linked to transportation infrastructure. Constructed drainage channels are designed to be either unlined or lined. Unlined channels, which are sometimes called earthen channels, are simply excavated channels in the ground through which water flows, and lined channels are excavated channels in which various lining materials are placed on the bottom and sides of the channel to provide stability and prevent erosion. Lining materials are classified as either rigid or flexible. Rigid linings include concrete, stone masonry, soil cement, grouted riprap, and precast interlocking blocks. Flexible linings include riprap, gravel, vegetation, manufactured mats, or combinations of these materials. Rigid linings are called “rigid” because they tend to crack when deflected, and flexible linings are called “flexible” because they are able to conform to changes in channel shape while maintaining the overall integrity of the channel lining. Channels with rigid linings are sometimes called rigid-boundary channels, and channels with flexible linings are sometimes called flexible-boundary channels. Channels with rigid lining are preferred for in a variety of cases, such as to (1) transport water at high velocities to reduce construction and excavation costs, (2) decrease seepage losses, (3) decrease operation and maintenance costs, and (4) ensure the stability of the channel section. All channels carrying supercritical flow should be lined with concrete and continuously reinforced both longitudinally and laterally. Since channels with rigid linings are capable of high conveyance and high-velocity flow, flood-control channels with rigid linings are often used to reduce the amount of land required for a surface-drainage system. When land is costly or unavailable because of restrictions, use of rigid-channel linings is preferred. A concrete-lined channel under construction using prefabricated concrete panels is shown in Figure 5.1. Channels with rigid linings are highly susceptible to failure from structural instability caused by freeze-thaw, swelling, and excessive soil pore-water pressures, and rigid linings tend to fail when a portion of the lining is damaged. Construction of rigid linings FIGURE 5.1: Concrete-lined canal

184

Section 5.2

Basic Principles

185

requires specialized equipment using relatively costly material. As a result, the cost of rigid linings is high. Prefabricated linings can be a less expensive alternative if shipping distances are not excessive. In contrast to channels with rigid linings, channels with flexible linings are generally less expensive, permit infiltration and exfiltration, filter out contaminants, provide better habitat opportunities for local flora and fauna, and have a natural appearance. However, channels with flexible linings have the disadvantage of being limited in the magnitude of the erosive force that they can sustain without damage to either the channel or the lining. Flexible linings are widely used as temporary channel linings for erosion control during construction or reclamation of disturbed areas. 5.2 Basic Principles When designed for stormwater management, the design flow rate to be accommodated by the channel is typically the peak flow rate from a runoff event with a specified return period. The objective in designing a drainage channel is to identify a shape, dimensions, and lining material that will safely accommodate the design flow rate at a reasonable cost while preventing the erosion of the material on the channel boundary. 5.2.1

Best Hydraulic Section

The best hydraulic section, also known as the most economic section (Marriott, 2009), can be determined by requiring that the flow area of the channel be minimized while maintaining the hydraulic capacity. Consider the Manning equation given by Q=

2 1 1 AR 3 S02 n

(5.1)

where Q is the flow rate in the channel [m3 /s], A is the cross-sectional flow area [m2 ], R is the hydraulic radius (= A/P) [m], P is the wetted perimeter [m], and S0 is the longitudinal slope of the channel [dimensionless]. Equation 5.1 can be rearranged into the form ⎛

⎞3 5

2 ⎜ Qn ⎟ A = ⎝ 1 ⎠ P5 S02

(5.2)

which demonstrates that for given values of Q, n, and S0 , the flow area, A, is proportional to the wetted perimeter, P, and minimizing A also minimizes P. The best hydraulic section is defined as the section that minimizes the flow area for given values of Q, n, and S0 . To illustrate the process of determining the best hydraulic section, consider the trapezoidal section shown in Figure 5.2, where the shape parameters are the bottom width, b, and the side slope m:1 (H:V). The flow area, A, and the wetted perimeter, P, are given by A = by + my2  P = b + 2y m2 + 1

FIGURE 5.2: Trapezoidal section

T = b + 2 my Freeboard y

A

b

1 m

(5.3) (5.4)

186

Chapter 5

Design of Drainage Channels

where y is the depth of flow in the channel. Eliminating b from Equations 5.3 and 5.4 leads to

 A = P − 2y m2 + 1 y + my2 (5.5) and eliminating A from Equations 5.2 and 5.5 yields

 2 P − 2y m2 + 1 y + my2 = cP 5 where c is a constant given by



(5.6)

⎞3 5

⎜ Qn ⎟ c=⎝ 1 ⎠ S02

(5.7)

Holding m constant in Equation 5.6, taking the partial derivative of each term with respect to y, and setting ⭸P/⭸y equal to zero leads to  P = 4y 1 + m2 − 2my (5.8) Similarly, holding y constant in Equation 5.6, taking the partial derivative of each term with respect to m, and setting ⭸P/⭸m equal to zero leads to √ 3 m= L 0.577 (5.9) 3 On the basis of Equations 5.8 and 5.9, the best hydraulic section for a trapezoid is one having the following geometric characteristics: √ √ √ 3 y, A = 3y2 b=2 (5.10) P = 2 3y, 3 where it is apparent that P = 3b, indicating that the (wetted) sides of the channel have the same length as the bottom. The best side slope, indicated by Equation 5.9, makes an angle of 60◦ with the horizontal. In cases where the side slopes are controlled by the angle of repose of the soil surrounding the channel, Equation 5.8 is used with a specified side slope, m, to find the best bottom-width-to-depth ratio. Under these circumstances, combining Equations 5.8 and 5.4 yields the bottom-width-to-depth ratio of the best hydraulic section as 

b =2 1 + m2 − m y

(5.11)

Utilization of Equation 5.11 is usually necessary in lieu of the channel dimensions given in Equation 5.10 (which correspond to side slopes of 0.577:1 or 60◦ ), since the recommended side slopes for excavated channels are usually less than 1.5:1 (H:V) or 33.7◦ . The above procedure can also be applied to other channel shapes, and the flow area, A, wetted perimeter, P, and top width, T, of the best hydraulic sections for a variety of channel shapes are given in Table 5.1. It is important to note that all the best hydraulic sections are not equally efficient. For example, the best hydraulic section for trapezoidal channels is more efficient than the best hydraulic section for rectangular channels, since for a given discharge the trapezoidal channel has a smaller flow area and wetted perimeter than a rectangular channel. Optimized channel geometries other than those listed in Table 5.1 have also been determined; for example, the optimal section with a horizontal bottom and parabolic sides was determined by Das (2007a), and the optimal section for a power-law channel was determined by Hussein (2008). Trapezoidal channels in which the bottom vertices are rounded rather than sharp are extensively used in India, and specifications for the best hydraulic sections of these types of channels can be found in Froehlich (2008).

Section 5.2

Basic Principles

187

TABLE 5.1: Geometric Characteristics of Best Hydraulic Sections

Shape

Best geometry

A √ 2 3y

P √ 2 3y

T √ 4 3y 3

4y √ 2 2y

2y

Trapezoid

Half of a hexagon

Rectangle

Half of a square

2y2

Half of a square

y2

Triangle Semicircle

π y2 2



Parabola

4 3



√ 2 2y

8 3

πy √ 2y

2y 2y √ 2 2y

The most efficient shape of all best-hydraulic sections is the semicircle; however, this shape is not practical to construct. The semicircular shape can be approximated by a compound section that is composed of a lower trapezoidal section and an upper rectangular section as shown in Figure 5.3. Such a compound section can be easily constructed in rocks and is more efficient than a trapezoidal section. The best compound section requires that m = 1, and the channel dimensions are given by (Abdulrahman, 2007) α=

 1 m + 1 − 1 + m2 m

b = 2(1 − αm)y

(5.13)

1

y=

24 3

(2 − α 2 m) 8

(5.12)

nQ √ S0

3 8

(5.14)

where y is the normal depth of flow in the channel derived using the Manning equation. In cases where it is not practical to use m = 1 in the compound section, Equations 5.12 to 5.14 give the most efficient compound section for any given value of m. Although the best hydraulic section appears to be the most economical in terms of excavation and channel lining, it is important to note that this section might not always be the most economical section for one or more of the following reasons:  The flow area does not include freeboard, and therefore is not the total area to be excavated.  It may not be possible to excavate a stable best hydraulic section in the native soil.  The depth of the channel might be limited due to high water table or underlying bedrock.  The best hydraulic section might cause unstable flow conditions (e.g., near-critical flow).  For lined channels, the cost of lining may be comparable to the excavation costs. FIGURE 5.3: Most efficient compound section

b + 2mαy

(1−α)y y

αy m:1 (H:V)

b

188

Chapter 5

Design of Drainage Channels

 Large channel widths and mild side slopes will result in high costs of right-of-way and structures such as bridges.  The ease of access to the site and the cost of disposing of removed material may affect the economics of the channel design. To address these constraints and to identify the most economical channel section, numerical optimization methods are sometimes used (e.g., Bhattacharjya, 2006). When such optimization methods are used, the most economical channel will usually depend on the particular site constraints. For example, if it is assumed that the channel construction cost is composed of land-acquisition costs, lining-material costs, and excavation costs, then it can be shown that the least-cost trapezoidal channel will generally be narrower and deeper than the most efficient section (Blackler and Guo, 2009). In contrast, if only excavation costs are considered and it is assumed that excavation costs increase linearly with the depth of excavation, then the least cost trapezoidal channel will generally be wider and shallower than the most efficient section (Swamee et al., 2001). EXAMPLE 5.1 A triangular concrete-lined drainage channel is to be designed on a slope of 0.5% and accommodate a peak runoff rate of 0.15 m3 /s. Determine the optimal dimensions of the channel. Assume n = 0.013 for the concrete lining. Solution From the given data: Q = 0.15 m3 /s, S0 = 0.5% = 0.005, and n = 0.013. According to Table 5.1, the optimal triangular channel has the shape of half of a square, which means that the side slopes are at 45◦ to the horizontal. The flow area, A, and wetted perimeter, P, can be expressed in terms of the flow depth, y, as A = y2 √ P = 2 2y and hence the Manning equation gives 5

Q=

2 1 1 1 A 3 12 AR 3 S02 = S n n P 23 0 5

0.15 =

1 (y2 ) 3 1 (0.005) 2 0.013 (2√2y) 23

which yields y = 0.337 m. Therefore the most efficient triangular channel will have side slopes of 45◦ and a maximum flow depth (at the vertex) of approximately 0.34 m.

The analyses presented here assume that Manning’s n is constant and does not vary with flow depth. However, in cases where different parts of the channel boundary have different roughness characteristics, the effective Manning’s n varies with flow depth and the most efficient section will generally differ from that obtained by assuming a constant n. In these cases, more specialized optimization methods should be used (e.g., Das, 2008; Froehlich, 2011b). 5.2.2

Boundary Shear Stress

Under uniform-flow conditions it has been shown (see Section 4.2.2) that the average shear stress, τ0 [FL−2 ], on the perimeter of a channel is given by τ0 = γ RS0

(5.15)

where γ is the specific weight of the fluid flowing in the channel [FL−3 ], R is the hydraulic radius of the flow area [L], and S0 is the slope of the channel [dimensionless]. The average shear stress, τ0 , is also called the unit tractive force. The boundary shear stress is not uniformly distributed around the perimeter of a channel and depends on the shape of the cross section, the structure of secondary flow cells (transverse to the main-flow direction), and the nonuniformity in the boundary roughness (Zarrati et al., 2008). The typical distribution of shear stress around the perimeter of a trapezoidal section under steady-flow conditions is shown in

Section 5.2 FIGURE 5.4: Shear-stress distribution in a trapezoidal channel

Basic Principles

1

189

1 m

m y b

τs

τs τb

Figure 5.4. In trapezoidal sections, the most commonly used shape for drainage channels, the maximum shear stress on the bottom of the channel, τb [FL−2 ], occurs at the center of the channel bottom and can be approximated by (Lane, 1955) τb = γ yS0

(5.16)

where y is the flow depth [L]. For a trapezoidal channel of bottom width b, Equation 5.16 provides a fairly accurate estimate of the maximum shear stress on the bottom of the channel when b/y Ú 4, and in cases where b/y < 4, Equation 5.16 provides a conservative estimate of the maximum bottom shear stress (USFHWA, 2005). The maximum shear stress on the sides of a trapezoidal channel, τs [FL−2 ], occurs approximately two-thirds of the way down the sides of the channel and can be approximated by τs = Ks τb

(5.17)

where Ks is the side-shear-stress factor [dimensionless]. If the side slope is m:1 (H:V), then Ks for trapezoidal channels can be estimated using the relation (Anderson et al., 1970) ⎧ ⎪ m … 1.5 ⎨0.77, (5.18) Ks = 0.066m + 0.67, 1.5 < m < 5 ⎪ ⎩1.0, m Ú 5 Estimation of the maximum shear stresses exerted on the bottom and sides of a channel is fundamental to designing open channels, since channels are stable when the perimeter shear stress is everywhere less than or equal to the shear stress required to move or dislodge the material on the perimeter of the channel. The shear stress required to dislodge the perimeter material is commonly called the maximum permissible shear stress, the critical shear stress, or simply the permissible shear stress. The maximum permissible shear stress is specific to the lining material in the channel. EXAMPLE 5.2 A trapezoidal drainage channel has a bottom width of 2 m, side slopes of 3:1 (H:V), and a longitudinal slope of 0.7%. Under design conditions the flow depth is 0.7 m. Estimate the maximum shear stress on the bottom and sides of the channel. Would a lining material with a permissible shear stress of 45 Pa be adequate to protect the channel from scour? Solution From the given data: b = 2 m, m = 3, S0 = 0.7% = 0.007, and y = 0.7 m. Taking γ = 9790 N/m3 (at 20◦ C), the maximum shear stress on the bottom of the channel, τb , is given by Equation 5.16 as τb = γ yS0 = (9790)(0.7)(0.007) = 48.0 Pa The side-shear-stress factor, Ks , is given by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 and so the maximum shear stress on the sides of the channel is given by Equation 5.17 as

190

Chapter 5

Design of Drainage Channels τs = Ks τb = (0.868)(48.0) = 41.7 Pa Therefore, a lining material with a permissible shear stress of 45 Pa would only protect the sides of the channel from scour. A lining material with a permissible shear stress greater than 48 Pa would be required to protect the bottom of the channel.

5.2.3

Cohesive versus Noncohesive Materials

Sediments and soils are generally classified as either cohesive or noncohesive, with cohesive sediments defined as having a plasticity index (PI) greater than or equal to 10, and noncohesive sediments having a PI less than 10. For noncohesive sediments on the boundary of an open channel with flowing water, the forces resisting motion are primarily associated with the weight of the sediment particles. In contrast, cohesive sediments contain appreciable fractions of silt or clay and resist motion mainly by cohesion rather than weight. Channels excavated in cohesive soils or lined with cohesive sediments are characterized by a single maximum permissible shear stress on the boundary of the channel. However, in the case of noncohesive sediments, the permissible shear stress on the side sediments is generally less than the permissible stress on the bottom sediments, since the downslope weight component of the side-sediment grains creates an additional force tending to move the side sediments. The permissible shear stress on noncohesive bottom sediments (e.g., sands and gravels) can be estimated from experimental results based on dimensional analysis. If the critical shear stress required to initiate motion is τc [FL−2 ], then the following functional relationship can be proposed (5.19) τc = f1 (γs − γ , ds , ρ, μ) where γs is the specific weight of the sediment particles [FL−3 ], γ is the specific weight of water [FL−3 ], ds is the characteristic sediment size [L], ρ is the density of water [ML−3 ], and μ is the dynamic viscosity of water [FL−2 T]. The functional relationship given by Equation 5.19 is between 5 variables in 3 dimensions (M,L,T), and so according to the Buckingham pi theorem Equation 5.19 can be expressed as a functional relationship between 5 − 3 = 2 dimensionless groups such as τc = f2 (γs − γ )ds



u∗c ds ν

 (5.20)

where f2 denotes a functional relationship, u∗c is the friction velocity corresponding to τc and  defined as τc (5.21) u∗c = ρ and ν is the kinematic viscosity of water, defined as μ/ρ. The relationship between the dimensionless groups in Equation 5.20 was originally proposed by Sheilds (1936), and the estimated relationship along with the results of several laboratory experiments are shown in Figure 5.5, where the estimated relationship is given by the solid line. The plot shown in Figure 5.5 is commonly known as the Sheilds diagram, where ds is the median size of the sediment particles. The nondimensional shear stress, τ∗ , and the (nondimensional) boundary Reynolds number, Re∗ , are defined as τ∗ =

τc (γs − γ )ds

and

Re∗ =

u∗c ds ν

(5.22)

so that the Sheilds diagram (Figure 5.5) gives the relationship between τ∗ and Re∗ . The auxiliary scale on the Sheilds diagram is included to facilitate calculating τc when the sediment  and fluid properties are known. To find τc , the value of (ds /ν) 0.1[(γs /γ ) − 1]gds is first calculated (using any consistent set of units), and then the value of τ∗ is determined from the intersection of the corresponding isoline with the Shields curve. The permissible stress on the bottom sediments is then equated to the derived critical stress.

Section 5.2

Basic Principles

FIGURE 5.5: Sheilds diagram

191

γs

(g/cm3)

Material

Source: ASCE (2006b).

*

Dimensionless shear stress, τ =

τc (γs – γ)d s

Amber Lignite Barite Fully developed turbulent velocity profile 1.0 0.8 0.6 0.5 0.4

Turbulent boundary layer

0.3 0.2

Value of 2

4

6 8 1

ds ν

(Shields)

Granite Sand (Casey) + Sand (Kramer) × Sand (U.S.WES.) Sand (Gilbert) Sand (White) Sand in air (White) Steel shot (White)

1.06 1.27 2.7 4.25 2.65 2.65 2.65 2.65 2.61 2.10 7.9

γs 0.1 ( γ _ 1) gds

2

4

6 100

2

4

6

1000

0.1 0.08 0.06 0.05 0.04 0.03 0.02 0.2

×

×

+

++ +×

Shields curve 0.4 0.6

1.0

2

4

6

8 10

20

40

60 100

200

500

1000

u d Boundary Reynolds Number, Re* = * s ν

EXAMPLE 5.3 The bottom sediments of an open channel consist mainly of coarse sand with a median grain size of 1 mm and a specific gravity of 2.65. If the temperature of the water is 20◦ C, estimate the permissible shear stress on the bottom of the channel. Solution From the given data: ds = 1 mm = 0.001 m and SG = 2.65. At 20◦ C, γ = 9790 kN/m3 and ν = 1.00 * 10−6 m2 /s. Hence,              γs ds  0.001 0.1 − 1 gds = 0.1 2.65 − 1 (9.81)(0.001) = 40 −6 ν γ 1.00 * 10 Using this value (40) in the Sheilds diagram (Figure 5.5) yields an intersection point of around τ∗ = 0.035 and Re∗ = 22. Using the definition of τ∗ given by Equation 5.22, the critical shear stress, τc , is given by τc = τ∗ (γs − γ )ds = τ∗ (SG − 1)γ ds = (0.035)(2.65 − 1)(9790)(0.001) = 0.57 Pa Therefore, the (maximum) permissible stress on the bottom of the channel is 0.57 Pa.

The Sheilds diagram is appropriate for assessing the stability of bottom sediments under uniform-flow conditions. Under non-uniform-flow conditions, alternative formulations have been proposed but are not in common usage (e.g., Hoan et al., 2011). Consider a particle of a noncohesive material with a (submerged) weight, wp [F], on the bottom of a channel. This particle resists the shear force of the flowing fluid by the friction force between the particle and surrounding particles on the bottom of the channel. The friction force is given by μp wp , where μp is the coefficient of friction between particles on the bottom of the channel [dimensionless]. When particle motion on the bottom of the channel is incipient, the shear stress on the bottom of the channel is equal to the permissible shear stress, τp [FL−2 ], and (5.23) Ap τp = μp wp

192

Chapter 5

Design of Drainage Channels

where Ap is the effective surface area of a particle on the bottom of the channel [L2 ]. The coefficient of friction between particles on the bottom of the channel is related to the angle of repose, α [degrees], of the particle material by μp = tan α

(5.24)

Combining Equations 5.23 and 5.24 leads to the following expression for the permissible shear stress on the bottom of the channel: wp τp = tan α (5.25) Ap For particles on the side of the channel, the force on each particle consists of the shear force exerted by the flowing fluid, which acts in the direction of flow, plus the component of the particle weight that acts down the side of the channel. Therefore the total force, Fp [F], tending to move a particle on the side of the channel is given by  (5.26) Fp = (τs Ap )2 + (wp sin θ )2 where τs is the shear stress exerted by the flowing fluid on the side of the channel, and θ is the angle that the side of the channel makes with the horizontal [degrees]. When motion is incipient on the side of the channel, the force tending to move the particle is equal to the frictional force, Ff [F], which is equal to the component of the particle weight normal to the side of the channel, multiplied by the coefficient of friction (tan α), in which case Ff = wp cos θ tan α

(5.27)

When motion is incipient, Fp = Ff , and the shear stress on the side of the channel is equal to the permissible shear stress on the side of the channel denoted by τps . Equations 5.26 and 5.27 combine to yield  wp cos θ tan α = (τps Ap )2 + (wp sin θ )2 which can be put in the form τps

 wp tan2 θ = cos θ tan α 1 − Ap tan2 α

(5.28)

Combining Equations 5.28 and 5.25 gives the ratio of the permissible shear stress on the side of the channel to the permissible shear stress on the bottom of the channel, called the tractive force ratio, K [dimensionless], as τps K= = τp

 1 −

sin2 θ sin2 α

(5.29)

Equation 5.29 is generally applicable to noncohesive lining materials and is generally used to estimate τps based on given values of τp , θ , and α. In practical applications, τp is a property of the lining material, θ is the side-slope angle, and α is a property of the size distribution and angularity of the noncohesive material. Estimates of α [degrees] for stone linings can be obtained using relationships such as those given in Table 5.2 (Froehlich, 2011), where the appropriate estimation equation depends on which percentile stone sizes are known. The characteristic stone sizes are d50 and d85 , which represent the 50- and 85-percentile stone sizes, respectively. If both d50 and d85 are known then Equation 1 in Table 5.2 should be used, if only d50 is known then Equation 2 should be used, and if neither d50 nor d85 is known then Equation 3 should be used. It is apparent from Table 5.2 that the angularity of noncohesive lining materials can have a significant effect on the angle of repose of these materials. Other relationships between stone size and angle of repose are also used, such as

Section 5.2

Basic Principles

193

TABLE 5.2: Angles of Repose of Stone Linings

Equation number

Parameter, α0 (degrees)

Angle of repose, α (degrees) 0.125 d85 α0 d50

1

Round stone

Subround or Subangular stone

Angular stone

30.9◦

33.4◦

37.1◦

2

α0 (d50 )0.00778

31.5◦

34.3◦

37.9◦

3

α0

31.8◦

34.6◦

38.4◦

the empirical relationship between median stone size, d50 , and angle of repose, α, advocated by the U.S. Federal Highway Administration (2005), as shown in Figure 5.6. The angles of repose given by Table 5.2 and Figure 5.6 are not generally in agreement, with the difference primarily due to the gradation of the stones. The stones represented by Table 5.2 are typical of “open-graded” stones with a wide range of sizes, while the stones represented by Figure 5.6 are typical of “narrowly graded” stones with a relatively small range of sizes that are derived from quarried rock. The angles of repose estimated by either Table 5.2 or Figure 5.6 are appropriate for larger stone linings that are typically used to armor the boundaries of open channels. In cases where native noncohesive soils form the channel perimeter (i.e., no additional stone lining is provided), the angles of repose shown in Figure 5.7 are commonly used in practice. The native-soil particle sizes used in Figure 5.7 are the 75-percentile sizes, represented as d75 . FIGURE 5.6: Angle of repose of gravel and riprap linings Source: USFHWA (2005).

5 1.

7 1. 0

6

0.

0.

3 0.

2

15

0.

0.

01

07

0.

0.

0.

06

03 0.

43

0.

02

Median stone size, d 50 (ft)

ed

39

k

Cr us h ang ula r

37

Very

nded

35

33

Very ro u

Angle of repose, α (degrees)

ro c

41

31

00

0 10

20

30 40 50 60 100

200 300 400 600

Median stone size, d 50 (mm)

194

Chapter 5

Design of Drainage Channels

FIGURE 5.7: Angle of repose of noncohesive material

0.1 42

Source: USBR (1953).

40

Particle size, d75 (in.) 0.2

0.3

0.6

0.4

0.8 1.0

2.0

3.0

4.0

40 50 60 70

100

Angle of repose, α (degrees)

38 36

lar

gu

34

n ya

n ly a

te

era

32

d Mo

30 28

tl

g

o

M

24

y

el

at

r de

ed

nd

u ro

y htl

Sli

26

lar

gu

n ya

gh

Sli

lar

gu

r

Ve

y

r Ve

ed

nd

u ro

d

de

n ou

r

22 20

3

4

5

6

7 8 9 10 20 30 Particle size, d75 (mm)

EXAMPLE 5.4 A trapezoidal channel with bottom width 3 m and side slopes 2:1 (H:V) is lined with stones of median size 200 mm and 85-percentile size 280 mm. The stones are angular and the permissible shear stress of the stones are estimated to be 170 Pa. Determine the permissible shear stress on the bottom and sides of the channel. Solution From the given data: b = 3 m, m = 2, d50 = 200 mm, and d85 = 280 mm. The permissible shear stress on the bottom of the channel, τp , is 170 Pa. The side-slope angle, θ, is given by     1 1 = tan−1 = 26.6◦ θ = tan−1 m 2 Since no information is given about the grading of the stone lining (open versus narrow), the angle of repose, α, will be estimated using both Table 5.2 and Figure 5.6. The smaller value of α will be used in design, since this gives the more conservative (i.e., lower) estimate of the tractive force ratio. The value of α estimated using Table 5.2 is 0.125   d85 280 0.125 = (37.1) = 38.7◦ α = α0 d50 200 For d50 = 200 mm, Figure 5.6 gives α = 41.5◦ for very angular stone and α = 39◦ for very rounded stone. Since the stone in this case is characterized as “angular,” we can estimate the angle of repose by the average value of α = 40◦ . Comparing the estimates of α derived from Table 5.2 (38.7◦ ) and Figure 5.6 (40◦ ), the more conservative estimate of α = 38.7◦ will be used. The tractive force ratio, K, is therefore given by Equation 5.29 as    sin2 (26.6) = 0.698 K = 1 − sin2 (38.7) Since the permissible shear stress on bottom of the channel, τp , is 170 Pa, the permissible shear stress on the side of the channel, τps , is given by τps = Kτp = (0.698)(170) = 119 Pa

Section 5.2

5.2.4

Basic Principles

195

Bends

Flow around a bend generates secondary currents, which impose higher shear stresses on the perimeter of a channel compared with straight sections. The maximum shear stress, τr [FL−2 ], on the perimeter of a channel section in a bend is given by τr = Kr τb

(5.30)

where Kr [dimensionless] is a factor that depends on the ratio of the channel curvature, rc [L], to the top (water-surface) width, T [L], and τb [FL−2 ] is the shear stress on the bottom of the channel, as given by Equation 5.16 for a trapezoidal channel. The functional relationship between Kr and rc /T can be estimated by (USFHWA, 2005) ⎧ rc ⎪ ⎪ 2.0, … 2 ⎪ ⎪ ⎪  2 T   ⎨ r r r Kr = 2.38 − 0.206 c + 0.0073 c , 2 < c < 10 (5.31) ⎪ T T T ⎪ ⎪ rc ⎪ ⎪ ⎩1.05, Ú 10 T The increased shear stress caused by a bend persists downstream of the bend for a distance Lp [m] given by (Nouh and Townsend, 1979) ⎛ ⎞ 7 6 Lp R ⎠ = 0.604 ⎝ R nb

(5.32)

where R is the hydraulic radius of the flow area [m], and nb is the Manning roughness coefficient in the bend [dimensionless]. EXAMPLE 5.5 A trapezoidal drainage channel has a bottom width of 2.5 m, side slopes of 3:1 (H:V), longitudinal slope of 0.8%, and a design flow depth of 1.00 m. The channel is to have a bend segment with a radius of curvature of 20 m and the estimated Manning’s n of the lining material is 0.020. Determine the maximum shear stress on the lining that is expected within the bend, and how far downstream of the bend this shear stress is expected to persist. Solution From the given data: b = 2.5 m, m = 3, S0 = 0.008, y = 1.00 m, rc = 20 m, and nb = 0.020. Taking γ = 9790 N/m2 (at 20◦ C), the maximum shear stress on the lining just upstream of the bend is given by τb = γ yS0 = (9790)(1.00)(0.008) = 78.3 Pa The top width, T, and hydraulic radius, R, of the flow area in the channel is given by T = b + 2my = 2.5 + 2(3)(1.00) = 8.5 m R=

(2.5)(1.00) + (3)(1.00)2 by + my2   = = 0.623 m b + 2y 1 + m2 2.5 + 2(1.00) 1 + 32

and hence 20 rc = = 2.35 T 8.5 Using Equation 5.31, the bend factor, Kr , can be estimated by  2   rc rc + 0.0073 = 2.38 − 0.206(2.35) + 0.0073(2.35)2 = 1.93 Kr = 2.38 − 0.206 T T Therefore, the maximum shear stress expected in the bend, τr , is given by Equation 5.30 as τr = Kr τb = (1.93)(78.3) = 151 Pa

196

Chapter 5

Design of Drainage Channels This increased shear stress persists for a distance Lp downstream of the bend, where Lp is given by Equation 5.32 as ⎛ ⎞ ⎛ ⎞ 7 7 6 6 R 0.623 ⎠ = 0.604(0.623) ⎝ ⎠ = 10.8 m Lp = 0.604R ⎝ nb 0.020 It is apparent from the results of this example that the maximum shear stress in a bend can be significantly higher than in the straight segment of the channel; the factor is 1.93 in this example. Also, the increased shear stress can persist for a significant distance downstream, approximately 4.3 channel bottom-widths in the present example.

5.2.5

Channel Slopes

The relevant slopes in channel design are the longitudinal slope and the side slope. Longitudinal slopes are constrained by both the ground slope and the maximum allowable shear stress on the channel lining. Excavation is usually minimized by laying the channel on a slope equal to the slope of the ground surface, and the channel is sized so that the permissible stress on the lining is not exceeded. The allowable side slopes are influenced by the material in which the channel is excavated. Typical maximum side slopes for channels excavated in various types of materials are shown in Table 5.3. In deep cuts, side slopes are often steeper above the water surface than below the water surface. If a channel is lined with concrete, then construction of side slopes greater than 1:1 usually require the use of forms, and for side slopes greater than 0.75:1 (H:V) the linings must be designed to withstand earth pressures. The U.S. Bureau of Reclamation recommends a 1.5:1 (H:V) slope for the usual sizes of concrete-lined canals (USBR, 1978), and the U.S. Federal Highway Administration recommends that side slopes in roadside and median channels not exceed 3:1 (USFHWA, 2005). If channel sides are to be mowed, slopes of 3:1 or less should generally be used. EXAMPLE 5.6 A median channel adjacent to a major highway is to be excavated in a soft clay and lined with grass. If the longitudinal slope of the highway is 1%, what would you select as the longitudinal slope and side slopes of the median channel? Solution The longitudinal slope of the median channel should be 1% to minimize excavation, and the side slopes should be 3:1 to facilitate mowing and comply with the recommendation of the U.S. Federal Highway Administration.

5.2.6

Freeboard

The freeboard is defined as the vertical distance between the water surface and the top of the channel when the channel is carrying the design flow rate at normal depth. In lined channels, freeboard is sometimes defined as the vertical distance between the water surface and the top of the lining. Freeboard is provided to account for the uncertainty in the design, construction, and operation of the channel. At a minimum, the freeboard should be sufficient to prevent TABLE 5.3: Steepest Recommended Side Slopes in Various Types of Material

Material Firm rock Fissured rock Earth with concrete lining Stiff clay Earth with stone lining Firm clay, soft clay, gravelly loam Loose sandy soils Very sandy soil, sandy loam, porous clay

Side slope (H:V) 0:1–0.25:1 0.5:1 0.5:1–1:1 0.75:1 1:1 1.5:1 2:1–2.5:1 3:1

Section 5.2

Basic Principles

197

waves or fluctuations in the water surface from overflowing the sides. A commonly used criterion is that the height of freeboard should at least be equal to the velocity head plus 15 cm (6 in.), hence V2 F = 0.15 + (5.33) 2g where F is the freeboard [m], V is the velocity in the channel under design conditions [m/s], and g is gravity [m/s2 ]. The minimum recommended freeboard is usually 30 cm (1 ft) (ASCE, 1992), which corresponds to a velocity, V, of 1.72 m/s (5.64 ft/s) in Equation 5.33. For very shallow channels, such as roadside channels and minor stormwater diversions, flow depths are often less than 30 cm (1 ft); in these cases a minimum reasonable freeboard equal to 15 cm (6 in.) or one-half of the maximum flow depth has been recognized as acceptable professional practice (Seybert, 2006). These recommendations for the freeboard, F, can be collectively expressed as ⎧ ⎪ 0.15 m, y < 0.30 m ⎪ ⎪ ⎪ ⎪ ⎨ 0.30 m, y Ú 0.30 m, V … 1.72 m/s F= (5.34) ⎪ 2 ⎪ V ⎪ ⎪ ⎪ ⎩0.15 + 2g m, y Ú 0.30 m, V > 1.72 m/s For channels on steep slopes, it is also recommended that the freeboard be at least equal to the flow depth (USFHWA, 2005). In general, channel linings should extend to at least the freeboard elevation. At channel bends, additional freeboard must be provided to accommodate the superelevation of the water surface. The superelevation, hs [L], of the water surface is given by hs =

V2T grc

(5.35)

where V is the average velocity in the channel [LT−1 ], T is the top width of the channel [L], and rc is the radius of curvature of the centerline of the channel [L]. Equation 5.35 is valid only for subcritical-flow conditions, in which case the elevation of the water surface at the outer channel bank will be hs /2 higher than the centerline water-surface elevation, and the elevation of the water surface at the inner channel bank will be hs /2 lower than the centerline water elevation. Equation 5.35 is a theoretical relation derived from the momentum equation (normal to the flow direction) and assumes a uniform velocity and constant curvature across the stream. If the effects of nonuniform velocity distribution and variation in curvature across the stream are taken into account, the superelevation, hs , may be as much as 20% higher than given by Equation 5.35 (Finnemore and Franzini, 2002). The additional freeboard to accommodate the superelevation of the water surface around bends is hs /2, and this additional freeboard need only be provided in the vicinity of the bend. In order to minimize flow disturbances around bends, it is also recommended that the radius of curvature be at least three times the channel top width (USACE, 1995). EXAMPLE 5.7 A trapezoidal drainage channel has a bottom width of 2.5 m, side slopes of 3:1 (H:V), longitudinal slope of 0.8%, and is lined with angular stones for protection against erosion. At the design flow rate, the flow is subcritical, the velocity in the channel is 60 cm/s, and the flow depth is 0.75 m. How far above the normal flow depth should the lining be extended in straight segments of the channel? How would this change in a bend segment with a radius of curvature of 30 m? Solution From the given data: b = 2.5 m, m = 3, S0 = 0.008, V = 0.60 m/s, and y = 0.75 m. Since y Ú 0.30 m and V … 1.72 m/s, the freeboard, F, estimated by Equation 5.34 is 0.30 m. In the straight segments of the channel, the lining should extend a minimum of 30 cm above the design water surface.

198

Chapter 5

Design of Drainage Channels For bend segments with rc = 30 m, the superelevation is given by Equation 5.35 as hs =

V2T V 2 [b + 2my] 0.602 [2.5 + 2(3)(0.75)] = = = 0.01 m grc grc (9.81)(30)

Therefore, an additional freeboard of 0.01/2 m = 0.005 m = 0.5 cm will be required in bend segments. However, given the small increase in depth due to superelevation and the fact that a conservative freeboard is already being used in the straight segments, the design engineer may choose not to adjust the lining around bends.

5.3 Design of Channels with Rigid Linings The objective in designing channels with rigid linings is to determine the channel dimensions required to safely convey a design flow rate in a channel of given shape, lining material, and longitudinal slope. A recommended design procedure is as follows: Step 1: Estimate the roughness coefficient, n [dimensionless], for the specified lining material and design flow rate, Q [m3 /s]. Guidance for estimating Manning’s n for rigidboundary channels are listed in Table 5.4. ASCE (1992) has recommended that openchannel designs not use a roughness coefficient lower than 0.013 for well-troweled concrete, and other finishes should have proportionally higher n values assigned to them. TABLE 5.4: Manning Roughness Coefficients in Open Channels with Rigid Linings

Range of Manning’s n Type Cement Concrete

Concrete bottom float finished with sides of

Grouted riprap Stone masonry Soil cement Asphalt Brick Masonry Dressed ashlar

Characteristics

Minimum

Normal

Maximum

Neat surface Mortar Trowel finish Float finish Finished, with gravel on bottom Unfinished Gunite, good section Gunite, wavy section On good excavated rock On irregular excavated rock

0.010 0.011 0.011 0.013 0.015

0.011 0.013 0.013 0.015 0.017

0.013 0.015 0.015 0.016 0.020

0.014 0.016 0.018 0.017 0.022

0.017 0.019 0.022 0.020 0.027

0.020 0.023 0.025 — —

0.015 0.017 0.016

0.017 0.020 0.020

0.020 0.024 0.024

0.020 0.020 0.028 0.030 0.020 0.016 0.013 0.011 0.012 0.017 0.023 0.013

0.025 0.030 0.030 0.032 0.022 0.016 0.013 0.013 0.015 0.025 0.032 0.015

0.030 0.035 0.040 0.042 0.025 0.018 — 0.015 0.018 0.030 0.035 0.017

Dressed stone in mortar Random stone in mortar Cement rubble masonry, plastered Cement rubble masonry Dry rubble or riprap — — — — Smooth Glazed In cement mortar Cemented rubble Dry rubble —

Sources: USFHWA (2005); Chow (1959).

Section 5.3

Design of Channels with Rigid Linings

199

Step 2: Compute the normal depth of flow, y [m], using the Manning equation Q=

2 1 1 AR 3 S02 n

(5.36)

where A is the flow area [m2 ], R is the hydraulic radius [m], and S0 is the longitudinal slope of the channel [dimensionless]. If appropriate, the relative dimensions of the best hydraulic section may be specified. Step 3: Estimate the required freeboard and increase the freeboard in channel bends as appropriate to account for superelevation. As an additional constraint in designing concrete-lined channels, ASCE (1992) recommends that flow velocities not exceed 2.1 m/s (7 ft/s) or result in a Froude number greater than 0.8 for nonreinforced linings, and that flow velocities not exceed 5.5 m/s (18 ft/s) for reinforced linings. EXAMPLE 5.8 Design a lined trapezoidal channel to carry 20 m3 /s on a longitudinal slope of 0.0015. The lining of the channel is to be float-finished concrete. Consider: (a) the best hydraulic section, and (b) a section with side slopes of 1.5 : 1 (H : V). Solution (a) According to Table 5.4, n = 0.015. Using the best trapezoidal hydraulic section, the bottom width, b, and side slope, m, are given by Equations 5.9 and 5.10 as b = 1.15y,

m = 0.58

(= 60◦ angle)

and, according to Table 5.1, A = 1.73y2 ,

P = 3.46y,

T = 2.31y,

R=

A = 0.5y P

Substituting the geometric characteristics of the channel into the Manning equation yields 20 =

2 1 1 (1.73y2 )(0.5y) 3 (0.0015) 2 0.015

or

8

y 3 = 7.12 which leads to y = 2.09 m and hence the bottom width, b, of the channel is given by b = 1.15y = 1.15(2.09) = 2.40 m The flow area is therefore given by A = 1.73y2 = 1.73(2.09)2 = 7.6 m2 and the average velocity, V, is 20 Q = = 2.6 m/s A 7.6 Since the velocity is greater than 2.1 m/s, the lining should be reinforced. Since y Ú 0.30 m and V > 1.72 m/s, the freeboard, F, estimated by Equation 5.34 is given by V=

F = 0.15 +

V2 2.62 = 0.15 + = 0.49 m 2g 2(9.81)

The minimum depth of the channel to be excavated and lined is equal to the normal depth plus the freeboard, 2.09 m + 0.49 m = 2.58 m. The channel is to have a bottom width of 2.40 m and side slopes of 0.58:1 (H:V).

200

Chapter 5

Design of Drainage Channels (b) If the channel side slope is 1.5:1, then m = 1.5 and Equation 5.11 gives the bottom width, b, of the best hydraulic section (given that m = 1.5) as



 1 + m2 − m y = 2 1 + 1.52 − 1.5 y = 0.60y b=2 The top width, T, flow area, A, and hydraulic radius, R, are given by T = b + 2my = 0.60y + 2(1.5)y = 3.6y A = (b + my)y = (0.60y + 1.5y)y = 2.10y2 R=

A 2.10y2 = = 0.499y P 4.21y

Substituting the geometric characteristics of the channel into the Manning equation gives 20 =

2 1 1 (2.10y2 )(0.499y) 3 (0.0015) 2 0.015

or 8

y 3 = 5.86 which leads to y = 1.94 m and hence the bottom width, b, of the channel is given by b = 0.60y = 0.60(1.94) = 1.16 m The flow area is therefore given by A = 2.10y2 = 2.10(1.94)2 = 7.90 m2 and the average velocity, V, is 20 Q = = 2.53 m/s A 7.90 Since the velocity is greater than 2.1 m/s, the lining should be reinforced. Since y Ú 0.30 m and V > 1.72 m/s, the freeboard, F, estimated by Equation 5.34 is given by V=

F = 0.15 +

2.532 V2 = 0.15 + = 0.48 m 2g 2(9.81)

The minimum depth of the channel to be excavated and lined is equal to the normal depth plus the freeboard, 1.94 m + 0.48 m = 2.42 m. The channel is to have a bottom width of 1.16 m and side slopes of 1.5:1 (H:V).

Channels with rigid linings are frequently used to drain roads with steep slopes where erosion in the channel might be a problem. A typical roadside rigid-lining (concrete) drainage channel on a relatively steep slope is shown in Figure 5.8. From the viewpoint of practical construction, the dimensions of lined channels should usually be specified to the nearest 5 cm (2 in.) (Kay, 1998). 5.4 Design of Channels with Flexible Linings Several types of flexible-lining materials are used in practice, such as rolled erosion control products (RECPs), vegetative lining, riprap, and gabions. A brief description of these linings are as follows: Rolled erosion control products (RECPs). RECPs are temporary degradable or long-term nondegradable materials manufactured or fabricated into rolls designed to reduce soil erosion and assist in the growth, establishment, and protection of vegetation. Vegetative lining. Vegetative lining consists of seeded or sodded grass placed in and along the channel. There is usually a transition period between seeding and vegetation establishment, and temporary flexible linings (e.g., RECPs) provide erosion protection during the establishment period.

Section 5.4

Design of Channels with Flexible Linings

201

FIGURE 5.8: Roadside channel with rigid lining

Gravel and cobble linings. Gravel is characterized by stone sizes in the range of 2–64 mm (0.08–2.5 in.), and cobble is characterized by stone sizes in the range 64–256 mm (2.5–10 in.). Gravel lining is sometimes referred to as gravel mulch, and cobble linings are often used when a decorative channel design is needed. Riprap lining. Riprap is a loose assemblage of large angular broken stones that is commonly used to armor streambeds, bridge abutments, pilings, and other shoreline structures against scour. Gabions. Gabions consist of wire containers containing large stones. The wire container is usually rectangular and made of steel wire woven in a uniform pattern, with reinforced corners and edges made of heavier wire. Gabions are typically anchored to the channel side slope. Wire-enclosed riprap is typically used when rock riprap is either not available or not large enough to be stable. 5.4.1

General Design Procedure

Parameters that influence the design of channels with flexible linings are: (1) Manning’s n of the lining material [dimensionless], (2) the effective shear stress exerted by the flowing fluid on the lining material, τe [FL−2 ], and (3) the permissible shear stress of the lining material, τp [FL−2 ]. The approaches used to estimate n, τe , and τp can vary significantly between linings and flow conditions; however, the design procedure for all flexible linings is similar. Consider a typical design problem in which the design flow rate, Q, the longitudinal slope, S0 , and the channel shape are pre specified. The following design procedure is recommended for determining the channel dimensions and identifying an acceptable lining material: Step 1: Select the lining type. Step 2: Calculate the normal flow depth, y [m], using the Manning equation, taking into account any relationship between the Manning’s n and y that might be a property of the selected lining. Typical values of Manning’s n for unlined and RECP-lined channels are given in Table 5.5. In the case of gravel-mulch, cobble, and riprap linings, n is usually dependent on the flow depth and roughness height, and typical relationships are shown in Table 5.6. Step 3: Calculate the maximum shear stress on the perimeter of the channel. In cases where the lining consists of cohesive materials, the design is governed by the maximum shear stress on the bottom of the channel, τb [FL−2 ], which is estimated by τb = γ yS0

(5.37)

202

Chapter 5

Design of Drainage Channels TABLE 5.5: Typical Manning’s n for RECP Linings

Range of Manning’s n Lining category

Lining type

Minimum

Normal

Maximum

Unlined

Bare soil

0.016

0.020

0.025

RECP

Open-weave textile e.g., jute net Erosion control blankets e.g., curled wood mat Turf reinforcement mat e.g., synthetic mat

0.028

0.025

0.022

0.028

0.035

0.045

0.024

0.030

0.036

Source: USFHWA (2005). TABLE 5.6: Typical Manning’s n for Riprap, Cobble, and Gravel Linings

Flow depth Lining category

Lining type

0.15 m (0.5 ft)

0.50 m (1.6 ft)

1.0 m (3.3 ft)

Gravel mulch

d50 = 25 mm (1 in.) d50 = 50 mm (2 in.)

0.040 0.056

0.033 0.042

0.031 0.038

Cobbles

d50 = 100 mm (4 in.)



0.055

0.047

Rock riprap

d50 = 150 mm (6 in.) d50 = 300 mm (12 in.)

∗ ∗

0.069 ∗

0.056 0.080

Source: USFHWA (2005). Note: ∗ depends on channel slope

In cases where the lining consists of noncohesive materials, the design is usually governed by the maximum shear stress on the sides of the channel, τs [FL−2 ], which is estimated by τs = Ks τb (5.38) where Ks is the side-shear-stress factor [dimensionless]. In trapezoidal channels, Ks depends on the side slopes and can be estimated using Equation 5.18. In some cases, a safety factor (SF) might be applied (as a multiplicative factor) to τb and τs in order to account for uncertainties in design. Typically 1 … SF … 1.5, with SF = 1 being most common. Step 4: Estimate the permissible shear stress on the perimeter of the channel. In cases where the lining consists of cohesive materials, the permissible shear stress, τp [FL−2 ], is the same for both the bottom and sides of the channel. Linings of cohesive materials include bare-soil linings with PI (plasticity index) Ú 10. Typical permissible shear stresses for such materials are shown in Table 5.7. For noncohesive lining materials, which includes bare soil with PI < 10, gravel mulch, and riprap, the permissible shear stress on the sides of the channel, τps , is reduced relative to the permissible shear stress on the bottom of the channel, τp , according to Equation 5.29 which gives τps = Kτp

(5.39)

where K is the tractive force ratio [dimensionless] that can be estimated using Equation 5.29. Step 5: Verify the adequacy of the lining. An adequate lining requires that τb … τp and τs … τps . If the lining is inadequate, repeat Steps 1 to 5 for different linings until an adequate lining is found. Several adequate linings might be identified. The preferred lining is selected by considering factors such as cost and practicality.

Section 5.4

Design of Channels with Flexible Linings

203

TABLE 5.7: Typical Permissible Shear Stresses for Bare Soil and Stone Linings

Permissible shear stress Lining category

Lining type

Pa

lb/ft2

Bare soil cohesive (PI = 10)

Clayey sands Inorganic silts Silty sands

1.8–4.5 1.1–4.0 1.1–3.4

0.037–0.095 0.027–0.11 0.024–0.072

Bare soil cohesive (PI Ú 20)

Clayey sands Inorganic silts Silty sands Inorganic clays

4.5 4.0 3.5 6.6

0.094 0.083 0.072 0.14

Bare soil noncohesive (PI < 10)

d75 < 1.3 mm (0.05 in.) d75 = 7.5 mm (0.3 in.) d75 = 15 mm (0.6 in.)

1.0 5.6 11

0.02 0.12 0.24

Gravel mulch

d50 = 25 mm (1 in.) d50 = 50 mm (2 in.)

19 38

0.4 0.8

Rock riprap

d50 = 0.15 m (0.5 ft) d50 = 0.30 m (1 ft)

113 227

2.4 4.8

Source: USFHWA (2005).

EXAMPLE 5.9 Design a trapezoidal drainage channel to accommodate a peak flow rate of 0.5 m3 /s on a slope of 0.5%. Local regulations require that the side slope of the channel be no greater than 3:1 (H:V). A field investigation has indicated that the native soil is noncohesive and has a 75-percentile grain size of 10 mm. Solution From the given data: Q = 0.5 m3 /s, S0 = 0.005, m = 3, and d75 = 10 mm. Use the most efficient trapezoidal section with m = 3, in which case Equation 5.11 gives the bottom width, b, as



 1 + m2 − m y = 2 1 + 32 − 3 y = 0.325y b=2 The corresponding flow area, A, and wetted perimeter, P, are given by A = by + my2 = (0.325y)y + (3)y2 = 3.325y2   P = b + 2y 1 + m2 = (0.325y) + 2y 1 + 32 = 6.650y Step 1: Consider the case where no lining is used. In this case, the perimeter of the channel consists of bare soil. Step 2: A typical Manning’s n for channels in bare soil is given in Table 5.5 as n = 0.020. The Manning equation gives 5

1 A 3 21 S Q= n P 32 0 5

0.5 =

1 1 (3.325y2 ) 3 (0.005) 2 2 0.020 (6.650y) 3

which yields y = 0.364 m and b = 0.325y = 0.118 m. Step 3: The maximum shear stress on the bottom of the channel, τb , is given by (assuming γ = 9790 N/m3 ) τb = γ yS0 = (9790)(0.364)(0.005) = 17.8 Pa

204

Chapter 5

Design of Drainage Channels The side-shear-stress factor, Ks , is given by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 and so the maximum shear stress exerted on the side of the channel, τs , is given by τs = Ks τb = (0.868)(17.8) = 15.4 Pa Assume a safety factor (SF) equal to one. Step 4: The permissible shear stress, τp , on the noncohesive bare-soil lining can be estimated (interpolated) from Table 5.7 for d75 = 10 mm as 7.4 Pa. Since the soil is noncohesive, the permissible shear stress on the side of the channel, τps (=Kτp ), will be less than 7.4 Pa. Step 5: The maximum shear stress on bottom of the channel (17.8 Pa) is greater than the permissible shear stress on the bare-soil lining (7.4 Pa), and so the bare-soil lining is inadequate. Try another lining. Step 1: Try a gravel-mulch lining with d50 = 25 mm. Step 2: For typical gravel-mulch linings, the Manning’s n depends on the flow depth as shown in Table 5.6. This functional relationship can be expressed as n(y) and the Manning equation gives 5

Q=

1 A 3 12 S n P 23 0 5

1 1 (3.325y2 ) 3 0.5 = (0.005) 2 n(y) (6.650y) 32

which simplifies to 1 8 y 3 = 3.377 (5.40) n(y) Solving Equation 5.40 simultaneously with the n-versus-y relationship in Table 5.6 (with linear interpolation) yields y = 0.445 m and b = 0.325y = 0.145 m. Step 3: The maximum shear stress on the bottom of the channel, τb , is given by τb = γ yS0 = (9790)(0.445)(0.005) = 21.8 Pa and the maximum shear stress exerted on the side of the channel, τs , is given by τs = Ks τb = (0.868)(21.8) = 18.9 Pa Step 4: The permissible shear stress, τp , on the gravel-mulch lining is estimated from Table 5.7 for d50 = 25 mm as 19 Pa. Since the lining is noncohesive, the permissible shear stress on the side of the channel, τps (=Kτp ), will be less than 19 Pa. Step 5: The maximum shear stress on bottom of the channel (21.8 Pa) is greater than the permissible shear stress on the gravel-mulch lining with d50 = 25 mm (19 Pa), and so the gravel-mulch lining with d50 = 25 mm is inadequate. Try another lining. Step 1: Try gravel-mulch lining with d50 = 50 mm. Step 2: Solving the Manning equation (Equation 5.40) simultaneously with the n-versus-y relationship in Table 5.6 yields y = 0.484 m and b = 0.325y = 0.157 m. Step 3: The maximum shear stress on the bottom of the channel, τb , is given by τb = γ yS0 = (9790)(0.484)(0.005) = 23.7 Pa and the maximum shear stress exerted on the side of the channel, τs , is given by τs = Ks τb = (0.868)(23.7) = 20.6 Pa

Section 5.4

Design of Channels with Flexible Linings

205

Step 4: The permissible shear stress, τp , on the gravel-mulch lining can be estimated from Table 5.7 for d50 = 50 mm as 38 Pa. The angle of repose, α, of the gravel mulch can be estimated from Table 5.2 (Equation 2) for subangular-shaped stones as α = α0 (d50 )0.00778 = (34.3)(50)0.00778 = 35.4◦ The side-slope angle θ is given by     1 1 = tan−1 = 18.4◦ θ = tan−1 m 3 The tractive-force ratio, K, is given by Equation 5.29 as     sin2 θ sin2 (18.4) K= 1 − = 1 − = 0.839 sin2 α sin2 (35.4) and hence the permissible shear stress on the side of the channel, τps , is given by τps = Kτp = (0.839)(38) = 31.9 Pa Step 5: Since the maximum shear stress on bottom of the channel (23.7 Pa) is less than the permissible shear stress on the bottom of the channel (38 Pa), and the maximum shear stress on the side of the channel (20.6 Pa) is less permissible shear stress on the side of the channel (31.9 Pa), the lining is adequate. For a depth of flow of 0.484 m and bottom width of 0.157 m, the flow area is 0.779 m2 and the flow velocity (for Q = 0.5 m3 /s) is 0.64 m/s and hence the recommended freeboard (according to Equation 5.34) is 0.30 m. The height of the lining above the bottom of the channel should be at least 0.484 m + 0.30 m = 0.784 m. These dimensions can be rounded to the nearest centimeter for final specification. Final specification: The recommended channel design has a bottom width of 0.16 m, side slopes of 3:1, a gravel-mulch lining with d50 = 50 mm, and the lining should extend at least 0.78 m above the bottom of the channel. The specified channel is the hydraulically most efficient trapezoidal channel having side slopes of 3:1.

5.4.2

Vegetative Linings and Bare Soil

Vegetative linings mostly consist of grass and are used primarily as long-term linings on mild slopes in humid areas. A typical grass-lined roadside channel is shown in Figure 5.9. Grass linings are sometimes grouped into classes according to their retardance, and the five FIGURE 5.9: Grass-lined roadside channel

206

Chapter 5

Design of Drainage Channels TABLE 5.8: Retardance in Grass-Lined Channels

Retardance

Cover

Condition

A

Reed canary grass Yellow bluestem Ischaemum Weeping lovegrass

Excellent stand, tall, average 91 cm (36 in.) Excellent stand, tall, average 91 cm (36 in.) Excellent stand, tall, average 76 cm (30 in.)

B

Smooth bromegrass Bermuda grass Native grass mixture (little bluestem, blue grama, and other long and short Midwest grasses) Tall fescue Lespedeza sericea Tall fescue, with bird’s foot Trefoil or lodino Blue gamma Kudzu

Good stand, mowed, average 30–40 cm (12–16 in.) Good stand, tall, average 30 cm (12 in.) Good stand, unmowed

Good stand, unmowed, average 45 cm (18 in.) Good stand, not woody, tall, 48 cm (19 in.) Good stand, uncut, average 45 cm (18 in.)

Alfalfa

Good stand, uncut, average 28 cm (11 in.) Very dense growth, uncut Dense growth, uncut Good stand, tall, average 61 cm (24 in.) Good stand, unmowed, average 33 cm (13 in.) Good stand, uncut, average 28 cm (11 in.)

C

Bahia Bermuda grass Redtop Grass-legume mixture–summer Centipede grass Kentucky bluegrass Crabgrass Common Lespedeza

Good stand, uncut, 15–18 cm (6–7 in.) Good stand, mowed, average 15 cm (6 in.) Good stand, headed, 40–60 cm (16–24 in.) Good stand, uncut, 15–20 cm (6–8 in.) Very dense cover, average 15 cm (6 in.) Good stand, headed, 15–30 cm (6–12 in.) Fair stand, uncut, average 25 cm to 120 cm (10–48 in.) Good stand, uncut, average 28 cm (11 in.)

D

Bermuda grass Red fescue Buffalo grass Grass-legume mixture–fall, spring Lespedeza sericea

Good stand, cut to 6 cm (2.5 in.) Good stand, headed 30–45 cm (12–18 in.) Good stand, uncut, 8–15 cm (3–6 in.) Good stand, uncut, 10–13 cm (4–5 in.)

Weeping lovegrass

E

Common Lespedeza

After cutting to 5 cm (2 in.) height; very good stand before cutting Excellent stand, uncut, average 11 cm (4.5 in.)

Bermuda grass Bermuda grass

Good stand, cut to 4 cm (1.5 in.) Burned stubble

Sources: USFHWA (2005); Coyle (1975).

retardance classifications (A-E) are shown in Table 5.8. Grass lining is usually initiated by sodding, with sods laid parallel to the flow direction and secured to the ground by pins and staples. If adequate protection can be established during growth, seeding can also be used. On steep slopes or in arid areas, other linings such as riprap are preferable to grass, since grass covers under these conditions are usually not sustainable. Bare-soil linings occur when the channel is excavated in the native soil and no additional lining is provided. Channel shapes. Channel shapes commonly used for grass-lined and bare-soil channels are trapezoidal and triangular; however, both triangular and trapezoidal sections usually transform into parabolic sections over time if the channel is poorly maintained or simply left

Section 5.4

Design of Channels with Flexible Linings

207

alone (Seybert, 2006). In designing grass-lined and bare-soil channels, the ability of tractors, or other farm-type machinery, to cross the channels during periods of no flow is an important consideration, and this may require that side slopes of a channel be designed to allow tractors to cross, rather than for hydraulic efficiency or stability. Side slopes of 4:1 (H:V) or 5:1 (H:V) are typically adequate for this purpose. Manning’s n. The density, stiffness, and height of grass stems are the main properties of grass that relate to flow resistance and erosion control. The density and stiffness properties of grass are defined by the density–stiffness coefficient, Cs [dimensionless], for conditions ranging from excellent to poor as follows: ⎧ ⎪ ⎪ 580, excellent condition ⎪ ⎪ ⎪ ⎪ 290, very good condition ⎨ (5.41) Cs = 106, good condition ⎪ ⎪ ⎪ 24, fair condition ⎪ ⎪ ⎪ ⎩ 8.6, poor condition where “good condition” indicates a stem density of 2000–4000 stems/m2 (200–400 stems/ft2 ), “poor condition” indicates a stem density around one-third of that density, and “excellent condition” indicates a stem density around five-thirds of that density. The combined effect of density, stiffness, and grass height is defined by the grass–roughness coefficient, Cn [dimensionless], as Cn = 0.35Cs0.10 h0.528 (5.42) where h is the grass height [m]. For agricultural ditches, grass heights can reach 30–100 cm (12–39 in.); however, near a roadway grass heights are kept much lower for safety reasons and are typically in the range of 8–23 cm (3–9 in.). Manning’s n for grass linings depends on the grass properties and the shear force exerted by the flow and can be estimated by the relation n = Cn τ0−0.4

(5.43)

where τ0 is the mean boundary shear stress [Pa] given by Equation 5.15 and repeated here for convenience as (5.44) τ0 = γ RS0 where γ is the specific weight of water [N/m3 ], R is the hydraulic radius [m], and S0 is the channel slope [dimensionless]. For channels in bare soil (i.e., without lining) the Manning roughness, commonly designated by ns , depends on the 75-percentile particle size, d75 [mm], and can be estimated by (USFHWA, 2005) ⎧ ⎨0.016, d75 < 1.3 mm (0.05 in.) 1 ns = (5.45) ⎩0.015d 6 , d75 Ú 1.3 mm (0.05 in.) 75 Effective shear stress. Grass lining reduces the shear stress at the soil surface, and the remaining shear at the soil surface is termed the effective shear stress. When the effective shear stress is less than the allowable shear stress for the soil surface, then erosion of the soil surface will be controlled. The effective stress, τe [FL−2 ], on the soil surface is estimated by 

ns τe = τb (1 − Cf ) n

2 (5.46)

where τb is the shear stress exerted by the flowing water on the bottom of the channel [FL−2 ], Cf is the grass cover factor [dimensionless], ns is the soil–grain roughness given by

208

Chapter 5

Design of Drainage Channels TABLE 5.9: Cover Factors, Cf , for Uniform Stands of Grass

Condition Growth form Sod Bunch Mixed

Excellent

Very good

Good

Fair

Poor

0.98 0.55 0.82

0.95 0.53 0.79

0.90 0.50 0.75

0.84 0.47 0.70

0.75 0.41 0.62

Source: USFHWA (2005).

Equation 5.45 [dimensionless], and n is the lining roughness given by Equation 5.43 [dimensionless]. The grass cover factor, Cf , varies with cover density and grass growth form (sod or bunch) and can be estimated using the values in Table 5.9. Bunch grasses tend to grow in “bunches” leaving open-soil areas, while sod grasses are more dispersed and tend to cover most of the soil surface. Permissible soil shear stress. Erosion of the grass lining occurs when the effective shear stress on the underlying soil exceeds the permissible soil shear stress. For noncohesive soils, the permissible shear stress, τp,n [Pa], can be estimated by the relation  τp,n =

1 Pa, 0.75d75 Pa,

d75 < 1.3 mm (0.05 in.) 1.3 mm (0.05 in.) < d75 < 50 mm (2 in.)

(5.47)

where d75 is in millimeters. In contrast to noncohesive soils, the permissible shear stress for cohesive soils depends on cohesive strength and soil density. Cohesive strength is a function of the plasticity index, PI [dimensionless], and the soil density is a function of the void ratio, e [dimensionless]. The permissible shear stress on cohesive soils, τp,c [Pa] can be estimated by (USFHWA, 2005) τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6

(5.48)

where c1 –c6 are coefficients that can be estimated using Table 5.10. A simplified approach for estimating τp,c based on Equation 5.48 is to use Figure 5.10, where fine-grained soils are grouped together (GM, CL, SC, ML, SM, and MH), coarse-grained soils (GC) are grouped separately, and clays (CH) fall between the two groups. Figure 5.10 is applicable for soils that are within 5% of a typical unit weight for a soil class. For sands and gravels (SM, SC, GM, GC) typical soil unit weight is approximately 1.6 ton/m3 (100 lb/ft3 ), for silts and lean clays (ML, CL) 1.4 ton/m3 (90 lb/ft3 ), and fat clays (CH, MH) 1.3 ton/m3 (80 lb/ft3 ). EXAMPLE 5.10 A trapezoidal channel with a bottom width of 1 m, side slopes of 4:1 (H:V), and a longitudinal slope of 2% is excavated in very fine sand. The sand is classified as ML soil, with a plasticity index of 12, a void ratio of 0.4, and a 75-percentile grain size of approximately 0.1 mm. The design flow rate to be accommodated by the channel is 0.8 m3 /s and it is proposed to protect the channel from erosion using a sod-grass lining with a height of 10 cm in good condition. Assess the adequacy of the proposed lining. Solution From the given data: b = 1 m, m = 4, S0 = 0.02, soil classification ML, PI = 12, e = 0.4, d75 = 0.1 mm, Q = 0.8 m3 /s, sod-grass lining in good condition, and h = 0.10 m. From the given channel dimensions, the area, A, wetted perimeter, P, and hydraulic radius, R, are given by A = by + my2 = (1)y + (4)y2 = y + 4y2   P = b + 2y 1 + m2 = (1) + 2y 1 + 42 = 1 + 8.246y R=

y + 4y2 A = P 1 + 8.246y

Section 5.4

Design of Channels with Flexible Linings

209

TABLE 5.10: Coefficients for Permissible Soil Shear Stress

ASTM class∗

Typical names

Applicable range

c1

c2

1.07

14.3

c3

c4

c5

c6

GM

Silty gravels Gravel-sand-silt mixtures

10 … PI … 20 PI Ú 20

47.7 1.42 0.076 1.42

−0.61 −0.61

4.8 * 10−3 48

GC

Clayey gravels Gravel-sand-clay mixtures

10 … PI … 20 0.0477 2.86 42.9 1.42 PI Ú 20 0.119 1.42

−0.61 −0.61

4.8 * 10−2 48

SM

Silty sands Sand-silt mixtures

10 … PI … 20 PI Ú 20

1.07

7.15

11.9 1.42 0.058 1.42

−0.61 −0.61

4.8 * 10−3 48

SC

Clayey sands Sand-clay mixtures

10 … PI … 20 PI Ú 20

1.07

14.3

47.7 1.42 0.076 1.42

−0.61 −0.61

4.8 * 10−3 48

ML

Inorganic silts Very fine sands Rock flour Silty or clayey fine sands

10 … PI … 20 PI Ú 20

1.07

7.15

11.9 1.48 0.058 1.48

−0.57 −0.57

4.8 * 10−3 48

CL

Inorganic clays, low or medium PI Gravelly clays Sandy clays Silty clays Lean clays

10 … PI … 20 PI Ú 20

1.07

14.3

47.7 1.48 0.076 1.48

−0.57 −0.57

4.8 * 10−3 48

MH

Inorganic silts 10 … PI … 20 0.0477 1.43 10.7 1.38 −0.373 4.8 * 10−2 Micaceous or diatomaceous PI Ú 20 0.058 1.38 −0.373 48 fine sands or silts Elastic silts

CH

Inorganic clays of high plasticity Fat clays

PI Ú 20

0.097 1.38 −0.373

48

Note: ∗ ASTM Classification of soils (ASTM D 2487) is also known as the Unified Soil Classification System and is given in Appendix F. FIGURE 5.10: Cohesive soil permissible shear stress Source: USFHWA (2005).

Stress range, Pa (lb/ft2) 1062

12 11–28 >25

Low density Medium density High density Source: ASCE (2007).

TABLE 6.2: Average Per-Capita Wastewater Residential Flow Rates

Flow rate City Winnipeg, Manitoba Seattle, WA San Diego, CA Milwaukee, WI Tampa, FL Boulder, CO U.S. Average Denver, CO Phoenix, AZ Phoenix, AR Eugene, OR Los Angeles, CA

(L/d/person)

(gal/d/person)

210 220 220 240 250 250 250 260 270 300 320 340

55 58 58 63 66 66 66 69 71 79 85 90

Source: ASCE (2007).

where Qr is the average flow rate from residential sources [L3 T−1 ], d1 is the residential dwelling unit (DU) density [DU · L−2 ], d2 is the number of persons per dwelling unit [persons · DU−1 ], qr is the per-capita flow rate from residential sources [L3 T−1 person−1 ], and Ar is the residential service area [L2 ]. Zoning regulations typically divide residential areas into different dwelling-unit densities, and an example of such a division is shown in Table 6.1. Population densities in residential areas at the end of the design period are commonly taken as the saturation densities based on the community master plan. The average per-capita residential wastewater flows, qr , vary considerably within the United States, and several measured per-capita flow rates from residential areas are listed in Table 6.2. These data show average per-capita flow rates in the range of 210–340 L/d/person (55–90 gal/d/person), with a U.S. average of 250 L/d/person (66 gal/day/person). Local regulatory agencies usually specify the per-capita residential wastewater flow rates to be used in their jurisdiction. The average flows from residential sources calculated using Equation 6.1 can be adjusted to account for vacancy rates by reducing the number of dwelling units per unit area. Other residential sources such as hotels, motels, prisons, jails, nursing homes, and seasonal-use facilities can be converted into equivalent full-time residents, and their average-flow contributions calculated using Equation 6.1. Values of Qr given by Equation 6.1 must be calculated at both the beginning and the end of the design period so that minimum and maximum flow rates to be accommodated by the sewer system can be determined. 6.2.2

Nonresidential Sources

Nonresidential sources of wastewater flow are usually associated with commercial and industrial sources. For commercial sources, the wastewater flow rate is related to the type and size of the business, and the number of employees. For each type of commercial operation, the wastewater flow contributed to the sanitary-sewer system can be estimated using the relationship

Section 6.2

Quantity of Wastewater

Qc = A * FAR * e * qc

231

(6.2)

where Qc is the average wastewater flow rate from the commercial source [L3 T−1 ], A is the land area occupied [L2 ], FAR is the floor-area–land-area ratio [dimensionless], e is the number of employees per unit floor-area [employees · L−2 ], and qc is the per-capita wastewater flow rate [L3 T−1 employee−1 ]. Industrial operations typically use water to manufacture products, and wastewater generation from industrial sources must usually be developed on a case-by-case basis. The type of industry, the size of the industry, operational techniques, and method of onsite wastewater treatment are all factors to be considered in estimating wastewater flows from industrial sources. 6.2.3

Inflow and Infiltration (I/I)

Inflow is defined as the component of sewer flow that enters the sewer conduit from surfacewater sources. The modes of inflow include flooded sewer vents, leaky manholes, illicitly connected storm drains, basement drains, roof drains, and/or sources other than groundwater. Inflow usually originates from rainfall and/or snowmelt. In many existing sewer systems, inflow is the largest flow component on rainy days and is often responsible for the backup of wastewater into basements and homes, or the bypassing of untreated wastewater to streams and other watercourses. Infiltration is defined as water that enters the sewer conduit from groundwater, and sources of infiltration flows include broken pipes, cracks, loose pipe joints, faulty connections, and manhole walls. In areas where the water table is far below the sanitary-sewer conduit, infiltration is not usually a problem. However, under these circumstances, leakage from the pipe, called exfiltration, can cause subsurface pollution, thereby providing another reason to control leakage in sewer pipes. Inflow is difficult to separate from infiltration, and so these two contributions are usually lumped together and called “infiltration and inflow,” which is usually abbreviated as “I/I” and enunciated as “I and I.” Regulatory agencies in most states specify maximum allowances for I/I, and typical I/I allowances are shown in Table 6.3. It is apparent from Table 6.3 that I/I regulations can take two alternate forms, either expressing I/I as a flow rate per unit contributing area in (m3 /d)/ha or as a flow rate per unit diameter per unit length in (m3 /d)/100-mm diameter/km. In both approaches there is order-of-magnitude variability in the regulatory rates. Based on a survey of 128 cities, ASCE (1982) reported an average design infiltration allowance of 3.9 (m3 /d)/100-mm/km. TABLE 6.3: Infiltration/Inflow Allowances

Location Lethbridge, Alberta State of Maryland London, Ontario Kingston, Ontario State of Oregon Edmonton, Alberta Kingston, Ontario Edmonton, Alberta State of New Hampshire State of New Hampshire Union County, NC Antigonish, Nova Scotia Source: ASCE (2007).

Allowance

Remarks

2.2 (m3 /d)/ha 3.8 (m3 /d)/ha 8.6 (m3 /d)/ha 12 (m3 /d)/ha 19 (m3 /d)/ha 24 (m3 /d)/ha 28 (m3 /d)/ha 34 (m3 /d)/ha 3 0.93 (m /d)/100-mm diameter/km 1.9 (m3 /d)/100-mm diameter/km 2.8 (m3 /d)/100-mm diameter/km 5.6 (m3 /d)/100-mm diameter/km

— — — Residential areas — Not in sag locations Industrial & commercial areas In sag locations Pipe diameters … 1220 mm Pipe diameters > 1220 mm — —

232

Chapter 6

Design of Sanitary Sewers

In spite of regulatory values being expressed in (m3 /d)/100-mm/km, McGhee (1991) reported that the diameter of sewer pipes has little effect on infiltration inflow, since larger diameter sewers tend to have better joint workmanship, offsetting the increased size of the potential infiltration area. The actual amount of infiltration depends on the quality of sewer installation, the height of the water table, and the properties of the surrounding soil. Expansive soils tend to pull joints apart, while granular soils permit water to move easily through joints and breaks. The use of pressure sewers, which are also called force mains, can reduce I/I to near zero. 6.2.4

Peaking Factors

A peaking factor represents the ratio of the peak flow rate to the average flow rate. In designing sanitary sewers, two peaking factors are commonly used: the maximum-flow peaking factor, PFmax , and the minimum-flow peaking factor, PFmin . The factor PFmax is defined as the ratio of the peak 1-hour flow rate to the average flow rate, and PFmin is defined as the ratio of the peak 1-hour flow rate during the week with the lowest average flow rate to the average flow rate (ASCE, 2007). These peaking factors are typically applied to the sum of the residential and commercial flows. The peaking factors are used to determine the design minimum flow rate at the beginning of the design period, Qmin , and to determine the design maximum flow rate at the end of the design period, Qmax , according to the relations Qmin = PFmin * Qavg1 + Qind1 + QI/I1

(6.3)

Qmax = PFmax * Qavg2 + Qind2 + QI/I2

(6.4)

where Qavg1 and Qavg2 are the average residential-plus-commercial flow at the beginning and end of the design period [L3 T−1 ], respectively; Qind1 and Qind2 are the industrial flows at the beginning and end of the design period [L3 T−1 ], respectively; and QI/I1 and QI/I2 are the I/I contributions to the sewer flow at the beginning and end of the design period [L3 T−1 ], respectively. Residential and commercial wastewater flows vary continuously, with typically very low flows between 2:00 A.M. and 6:00 A.M. and peak flows occurring at various times during the daylight hours. In cases where the industrial component is relatively small, Qind1 and Qind2 can be included in Qavg1 and Qavg2 , respectively, when calculating Qmin and Qmax using Equations 6.3 and 6.4. The I/I component typically remains fairly constant during the day, except during and immediately following periods of heavy rainfall, when significant increases usually occur. When the I/I flow is low and major industrial flows are absent, the I/I can be considered part of the lumped domestic-plus-commercial flow and not handled separately. When commercial, institutional, or industrial wastewaters make up a significant portion of the average flows (such as 25% or more of all flows, exclusive of infiltration), peaking factors for the various flow components should be estimated separately. The peaking factors PFmax and PFmin in Equations 6.3 and 6.4 both correspond to the ratio of the peak 1-hour flow rate to the average flow rate, with the only difference being that PFmax corresponds to an average flow rate at the end of the design period, Qavg2 , and PFmin corresponds to an average flow rate at the beginning of the design period, Qavg1 . The relationship of the peaking factor, PF [dimensionless], to the average flow rate, Qavg [m3 /s], can be estimated by (ASCE, 2007; Merritt, 2009) PF =

⎧ ⎨1.88Q−0.095 avg

Qavg Ú 0.00368 m3 /s

⎩0.281Q−0.44

Qavg < 0.00368 m3 /s

avg

(6.5)

Hence, using Qavg = Qavg1 in Equation 6.5 yields PF = PFmin , and using Qavg = Qavg2 yields PF = PFmax . The formula for Qavg Ú 0.00368 m3 /s (58 gpm∗ ) in Equation 6.5 was developed by the Bureau of Engineering, City of Los Angeles (LA), California, and is sometimes referred to as the LA formula. The formula for Qavg < 0.00368 m3 /s (58 gpm) in Equation 6.5 ∗ gpm = (U.S.) gallon per minute

Section 6.2

Quantity of Wastewater

233

was developed by Merritt (2009) based on work reported by Zhang and others (2005) and is sometimes called the Poisson rectangular pulse (PRP) model. For a typical per-capita waste water flow contribution of 300 L/d/person (80 gal/d/person), the LA formula is applicable for service populations greater than around 1100 people, while the PRP model is applicable to service populations less than around 1100 people. Peaking factors generally decrease with an increasing average flow rate and also decrease with an increasing service population. Based on the requirements of various sewer agencies in the United States, the following peaking factors are typical (ASCE, 2007): ⎧ ⎪ ⎨2.1 P > 500, 000 persons (6.6) PF = 2.6 100, 000 persons … P … 500, 000 persons ⎪ ⎩3.0 P < 100, 000 persons where P is the resident population in the service area. In general, peaking factors specified by regulatory agencies with jurisdiction over the project area must be used. In lieu of such guidance, local measurements in either the project area or similar areas are the next-best alternative, with generalized formulae such as Equations 6.5 and 6.6 being the last resort. EXAMPLE 6.1 A new trunk sewer∗ is to be designed for a 25-km2 city that when fully developed will include 60% residential, 30% commercial, and 10% industrial development. The residential area will consist of 40% large lots (6 persons/ha), 55% small single-family lots (75 persons/ha), and 5% multistory apartments (2500 persons/ha). The design average wastewater flow rates are: 300 L/d/person for large and singlefamily lots, 200 L/d/person for apartments, 25,000 L/d/ha for commercial areas, and 40,000 L/d/ha for industrial areas. Infiltration and inflow is 1000 L/d/ha for the entire area. It is anticipated that when the new trunk sewer is first installed the average flow will be 50% of the flow when the city is fully developed. Estimate the maximum and minimum flow rates to be handled by the trunk sewer. Solution From the given data, the total area of the city is 25 km2 = 2500 ha and the fully developed residential area is 60% of 2500 ha = 1500 ha. Taking the average per-capita flow rate as 300 L/d/person for large and single-family lots and 200 L/d/person for apartments gives the wastewater flows in the following table:

Type Large lots Small single-family lots Multistory apartments Total

Area (ha)

Density (persons/ha)

Population

Flow (m3 /s)

0.40 (1500) = 600 0.55 (1500) = 825 0.05 (1500) = 75

6 75 2500

3600 61,875 187,500

0.013 0.215 0.434

252,975

0.662

The commercial sector of the city covers 30% of 2500 ha = 750 ha, with a flow rate per unit area of 25,000 L/d/ha = 2.89 * 10−4 m3 /s/ha. Hence the average flow rate from the commercial sector is (2.89 * 10−4 )(750) = 0.217 m3 /s. The industrial sector of the city covers 10% of 2500 ha = 250 ha, with a flow rate per unit area of 40,000 L/d/ha = 4.63 * 10−4 m3 /s/ha. Hence the average flow rate from the commercial sector is (4.63 * 10−4 )(250) = 0.116 m3 /s. The infiltration and inflow from the entire area is 1000 L/ha * 2500 ha = 2.5 * 106 L/d = 0.029 m3 /s. On the basis of these calculations, the average daily wastewater flow rate (excluding I/I) at the end of the design period (when the city is fully developed) is 0.662 + 0.217 + 0.116 = 0.995 m3 /s. Since ∗ A trunk sewer is defined as a sewer conduit receiving sewage from many tributaries serving a large territory.

234

Chapter 6

Design of Sanitary Sewers the average flow rate at the beginning of the design period is 50% of the flow rate at the end of the design period, then the average flow rate when the trunk sewer is first installed is 0.50 (0.995 m3 /s) = 0.498 m3 /s. Using Equation 6.5, the peaking factors can be estimated as −0.095 = 1.88 PFmax = 1.88Q−0.095 avg2 = 1.88(0.995) −0.095 = 2.01 PFmin = 1.88Q−0.095 avg1 = 1.88(0.498)

The maximum and minimum flow rates are estimated by multiplying the corresponding average wastewater flow rates by these factors and adding the I/I. Thus Maximum flow rate = 1.88(0.995) + 0.029 = 1.90 m3 /s Minimum flow rate = 2.01(0.498) + 0.029 = 1.03 m3 /s The trunk sewer must be of sufficient size to accommodate the peak flow rate of 1.90 m3 /s, and must also generate sufficient shear stress on the bottom of the sewer to prevent solids accumulation under minimum-flow conditions of 1.03 m3 /s.

6.3 Hydraulics of Sewers Flows in sanitary sewers can be described by either the Manning or the Darcy–Weisbach equations, which can be expressed in the forms ⎧ 1 ⎪ ⎪ R 6 A√RS ⎪ ⎪ 0 ⎨ n Q=  ⎪ ⎪ 8g √ ⎪ ⎪ ⎩ f A RS0

Manning equation (6.7) Darcy–Weisbach equation

where Q is the volumetric flow rate [L3 T−1 ], R is the hydraulic radius [L], n is the Manning roughness, A is the flow area [L2 ], S0 is the slope of the sewer [dimensionless], g is gravity [LT−2 ], and f is the friction factor [dimensionless] that can be estimated by the Colebrook equation,   1 2.51   = −2 log (6.8) + 14.8R f 4Re f where  is the roughness height [L], and Re is the Reynolds number [dimensionless] defined as VR (6.9) Re = ν where V is the average velocity [LT−1 ], and ν is the kinematic viscosity [L2 T−1 ]. In cases where solution of the Colebrook equation for f is a computational inconvenience (because it is implicit in f ), the Colebrook equation can be approximated by the Swamee–Jain equation (Swamee and Jain, 1976) which is given by Equation 2.39 for pipe flows and is adapted for open-channel applications (by replacing the pipe diameter by 4 times the hydraulic radius) as f =

0.25

1.65  + log 14.8R Re0.9

2

(6.10)

This approximation is valid for 10−6 … /4R … 10−2 and 1250 … Re … 2.5 * 107 , where Re is defined by Equation 6.9. Regardless of which equation is used for estimating f , it is

Section 6.3

Hydraulics of Sewers

235

apparent from Equation 6.7 that both the Manning and Darcy–Weisbach equations yield the same flow rate (Q) for any given value of hydraulic radius (R) and slope (S0 ) provided that Manning’s n satisfies the relation  1 8g R6 = (6.11) n f Taking g = 9.807 m/s2 , Equation 6.11 can be put in the convenient form 1

1

n = 0.1129R 6 f 2

(6.12)

In modern engineering practice, the Darcy–Weisbach equation is considered to be a more accurate representation of the flow in sewer conduits, while the Manning equation remains widely used, particularly in the United States. To ensure that both approaches yield the same results, Manning’s n must be specified in accordance with Equation 6.12. Based on the combination of Equations 6.12 and 6.8, it has been demonstrated that for circular sewers when the depth of flow is more than 15% of the diameter, Manning’s n shows essentially no variation with depth of flow, and when the depth of flow is greater than 10% of the diameter, the value of n will not change within two significant digits (Merritt, 1998). Approximation of a constant n value when the depth of flow exceeds 15% of the diameter is therefore justified (ASCE, 2007). Both the Darcy–Weisbach and Manning equations give a relationship between the flow rate, Q, and the depth of flow, h, in the sewer, and calculation of this relationship requires specification of the roughness height, , or Manning roughness, n, the hydraulic radius as a function of the flow depth, R(h), the flow area as a function of the flow depth, A(h), and the kinematic viscosity of the fluid, ν. Typical values of  range from about 0.03 mm for smoothfinish concrete to 0.0015 mm for plastics (ASCE, 2007), and ν can be estimated directly from the temperature of the fluid and for water and wastewater is normally taken as 10−6 m2 /s at 20◦ C . The functions R(h) and A(h) are determined based on the shape of the sewer cross section. Consider the circular-pipe cross section in Figure 6.1, where h is the depth of flow and θ is the water-surface angle. The depth of flow, h, cross-sectional area, A, wetted perimeter, P, and hydraulic radius, R, can be expressed in terms of θ by the geometric relations

D θ h= 1 − cos (6.13) 2 2 A= P=

θ − sin θ 8

Dθ 2

D2

A D sin θ R= = 1 − P 4 θ

(6.14) (6.15) (6.16)

These relationships give the parametric relationship between h, R, and A, where θ is the parameter. FIGURE 6.1: Flow in partially filled pipe

D

θ h

236

Chapter 6

Design of Sanitary Sewers

EXAMPLE 6.2 A 455-mm diameter concrete sewer pipe is observed to have a flow depth of 100 mm when the flow rate is 5.30 L/s. Determine the area, wetted perimeter, and hydraulic radius of the flow section. Estimate the average velocity. Solution From the given data: D = 455 mm, h = 100 mm, and Q = 5.30 L/s = 5.30 * 10−3 m3 /s. Use Equation 6.13 to calculate θ , such that

θ D 1 − cos h= 2 2

θ 455 1 − cos 100 = 2 2 which yields θ = 1.952 radians. Substitute for θ in Equations 6.14 to 6.16 to determine A, P, and R, which yields

  1.952 − sin(1.952) θ − sin θ 2 D = (455)2 = 2.65 * 104 mm2 = 0.0265 m2 A= 8 8 P=

(455)(1.952) Dθ = = 444 mm 2 2

R=

A 2.65 * 104 = = 59.7 mm P 444

The average velocity, V, is given by V=

Q 5.30 * 10−3 = = 0.200 m/s A 0.0265

In describing the hydraulics of flow in sewer pipes, the actual inside diameter of the pipe should ideally be used in design calculations rather than the nominal diameter of the pipe, which is an approximate diameter that is used for easy reference. However, the slight difference between the actual inside diameter and the nominal diameter of sewer pipes is often neglected in calculations, which is justified by the much larger uncertainties that are usually associated with other design variables such as flow rate. The (inside) top of a sewer pipe is commonly called the crown, and the (inside) bottom of a sewer pipe is called the invert. 6.3.1

Manning Equation with Constant n

In cases where a constant Manning’s n is specified along with Q, D, and S0 , it is convenient to substitute Equations 6.14 and 6.16 directly into the Manning equation (Equation 6.7) which yields the convenient form, 2

5

8

− 12

θ − 3 (θ − sin θ ) 3 − 20.16nQD− 3 S0

=0

(6.17)

which can be solved for θ [radians], which is then substituted into Equation 6.13 to obtain the depth of flow in the sewer pipe. The average flow velocity is obtained by first calculating the flow area, A, from Equation 6.14, and then calculating the average velocity, V, by V=

Q A

(6.18)

The only obstacle to solving Equation 6.17 for θ , and hence obtaining h and V, is that Equation 6.17 is an implicit equation in θ that must be solved numerically.

Section 6.3

Hydraulics of Sewers

237

EXAMPLE 6.3 Water flows at a rate of 4.00 m3 /s in a circular concrete sewer of diameter 1500 mm and longitudinal slope 1.00%. Manning’s n of the sewer pipe can be taken as 0.013. Estimate the normal depth of flow and the average velocity in the sewer. Solution From the given data: Q = 4.00 m3 /s, D = 1500 mm, S0 = 0.0100, and n = 0.013. According to Equation 6.17, 2

5

8

− 12

θ − 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2 θ − 3 (θ



5 sin θ) 3



8 1 20.16(0.013)(4)(1.5)− 3 (0.0100)− 2

=0 =0

which yields θ = 3.30 radians. Therefore, the normal flow depth, h, and area, A, are given by Equations 6.13 and 6.14 as



D θ 1.5 3.30 h= 1 − cos = 1 − cos = 0.81 m 2 2 2 2 3.30 − sin 3.30 θ − sin θ D2 = (1.5)2 = 0.970 m2 A= 8 8 The average flow velocity, V, in the sewer is given by V=

4.00 Q = = 4.12 m/s A 0.970

Hence the normal flow depth is 0.81 m and the average velocity is 4.12 m/s.

The hydraulics of flows in circular conduits, including sanitary sewers, for a constant value of Manning’s n are summarized in Figure 6.2. The velocity, V, and flow rate, Q, are normalized by their full-flow values Vfull and Qfull , respectively, where the Manning equation gives FIGURE 6.2: Flow in circular conduits

1.0 0.9

Ratio of depth/diameter, h/D

0.8 0.7

Q Qfull

0.6 0.5 0.4 0.3

V Vfull

0.2 0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

V Q or Vfull Qfull

0.8

0.9

1.0

1.1

1.2

1.3

238

Chapter 6

Design of Sanitary Sewers

Vfull =

2 D 3 12 S0 4  π D2 = Vfull 4

1 23 12 1 R S = n full 0 n

Qfull = Vfull Afull



(6.19) (6.20)

and hence V = Vfull Q = Qfull



4R D 4R D

2 3

(6.21) 2 3

4A π D2

(6.22)

The curves in Figure 6.2 are useful for visualizing the variation of Q and V with flow depth, h. However, the actual values of Q and V for any given value of h should be calculated using Equations 6.13 to 6.18 rather than estimated visually from the curves plotted in Figure 6.2. The combination of Equations 6.17 and 6.13 gives a nonlinear relationship between the flow rate, Q, and the depth of flow, h, and it can be shown that the maximum flow rate occurs when h/D L 0.94. This condition manifests itself as a flow instability when the pipe is flowing almost full, and there is a consequent tendency for the pipe to run temporarily full at irregular intervals (Henderson, 1966; Hager, 1999). This condition, sometimes referred to as slugging (Sturm, 2010), results in streaming air pockets at the crown of the pipe and pulsations that could damage pipe joints and cause undesirable fluctuations in flow rate. This condition is typically avoided in practice by designing pipes such that h/D … 0.75 under maximum-flow conditions. A related consequence of the nonlinear variation of flow rate with depth in circular conduits is that the flow rate under full-flow condition, Qfull , occurs both under the full-flow condition (when h/D = 1) and at partially full condition (h/D L 0.82), which means that there are two normal depths of flow for Q = Q full ! In addition to the aforementioned nonlinearity in flow rate, it can be shown that the velocity is the same whether the pipe flows half full or completely full. According to the Manning equation (Equation 6.17), the velocity increases with depth of flow until it reaches a maximum at h/D L 0.81; the velocity then decreases with increasing depth and becomes equal to the half-full velocity when the pipe flows full. 6.3.2

Manning Equation with Variable n

The Manning equation with variable n is used to match the flow estimated by the Manning equation to the flow estimated by the Darcy–Weisbach equation, in which case Manning’s n must satisfy Equation 6.12. For any given sewer-pipe diameter, D, it has been shown that when h/D Ú 0.15 Manning’s n shows essentially no variation with depth, and when h/D Ú 0.10 the value of n will not change within two significant digits (Merritt, 1998). Therefore, assignment of a constant Manning’s n is justified under these flow conditions for a given diameter, D, and roughness height, . However, for any given value of , the appropriate (constant) value of Manning’s n will vary as a function of D, and this relationship can be determined by using Equation 6.12. For smooth-finish concrete, where  = 0.03 mm, the relationship between n and D has been calculated by ASCE (2007) and is given in Table 6.4. The n values shown for the “Extra Care” condition in Table 6.4 were derived for full flow at a velocity of 0.6 m/s (2 ft/s) in a smooth-finish concrete pipe of given diameter and roughness height equal to 0.03 mm (0.0001 ft); the temperature of the flowing sewage was taken as 15.6◦ C (60◦ F). Pipes in “Typical” condition are assumed to have n values 15% higher than the Extra Care values, and pipes in “Substandard” condition are assumed to have n values 30% higher than Extra Care values. ASCE (2007) recommends that the Typical condition be used in design. The n values given in Table 6.4 account for the variation of n with pipe diameter, while ignoring smaller variations caused by flow depth, velocity, and temperature variations. Design engineers should take into account that if the pipe roughness exceeds 0.03 mm, which

TABLE 6.4: Recommended Manning’s n versus Pipe Diameter

Pipe diameters Condition Extra care Typical Substandard

150 mm (6 in.)

205 mm (8 in.)

255 mm (10 in.)

305 mm (12 in.)

380 mm (15 in.)

455 mm (18 in.)

610 mm (24 in.)

760 mm (30 in.)

915 mm (36 in.)

1220 mm (48 in.)

1525 mm (60 in.)

0.0092 0.0106 0.0120

0.0093 0.0107 0.0121

0.0095 0.0109 0.0123

0.0096 0.0110 0.0125

0.0097 0.0112 0.0126

0.0098 0.0113 0.0127

0.0100 0.0115 0.0130

0.0102 0.0117 0.0133

0.0103 0.0118 0.0134

0.0105 0.0121 0.0137

0.0107 0.0123 0.0139

Source: ASCE (2007).

239

240

Chapter 6

Design of Sanitary Sewers

is characteristic of smooth-finish concrete, n values begin to increase significantly from the values shown in Table 6.4; however, the usual presence of a thin slime film moderates the increased roughness. Under low-flow conditions where h/D < 0.15, the flow depth is preferably calculated using the Darcy–Weisbach equation, since the constant-n assumption of the Manning equation is not valid under these circumstances. Historically, n = 0.013 has been taken as characteristic of concrete sewer pipes; however, this n value is based on very old data (early 1900s), and more recent measurements using modern sewer pipes and joints have shown n values in the range of 0.008–0.011, which are consistent with the values shown in Table 6.4 (ASCE, 2007). A rationale for using n = 0.013 in design is that the higher n value accounts for such conditions as pipe misalignment and joint irregularities, interior corrosion and coating buildup, cracks and breaks, protruding or interfering laterals, and sediment buildup. However, it has also been asserted that these effects seldom justify using an n value as high as 0.013 (ASCE, 2007). EXAMPLE 6.4 A sewer line is to consist of several 610-mm diameter concrete-pipe segments laid on an average slope of 0.1%. The pipe finish is estimated to have a roughness height of 0.1 mm. Assuming that the sewage flow has a temperature of 20◦ C and the sewer pipes are in typical condition, estimate Manning’s n when the pipe is 10%, 30%, 50%, 80%, and 100% full. Based on these results, do you anticipate that there will be significant variation of Manning’s n with flow depth? Solution From the given data: D = 610 mm = 0.610 m, S0 = 0.1% = 0.001,  = 0.1 mm, T = 20◦ C, and h/D = 0.10, 0.30, 0.50, 0.80, and 1.00. At 20◦ C, the kinematic viscosity of water is ν = 10−6 m2 /s. For any given values of D (= 0.610 m) and h/D, the central angle, θ , the flow area, A, and the hydraulic radius, R, are given by Equations 6.13, 6.14, and 6.16 as h θ = 2 cos−1 1 − 2 D θ − sin θ θ − sin θ θ − sin θ D2 = (0.610)2 = 0.3721 A= 8 8 8 sin θ 0.610 sin θ sin θ D 1 − = 1 − = 0.1525 1 − R= 4 θ 4 θ θ Taking the unknown variable as Q, the following system of equations must be satisfied by Q: (Q/A)R Re = ν 0.25 f =

2 1.65  + log 14.8R Re0.9 1

1

n = 0.1129R 6 f 2

(6.24)

(6.25)

2

Q=

(6.23)

5

θ − 3 (θ − sin θ) 3 8

(6.26)

− 12

20.16nD− 3 S0

Substituting known quantities in Equations 6.23 to 6.26 and solving for Q and n yields the following results: (1)

(3)

(4)

h/D

(2) Q (m3 /s)

n

Typical n

(5) n/nfull (%)

0.10 0.30 0.50 0.80 1.00

0.006 0.052 0.130 0.253 0.261

0.0099 0.0100 0.0101 0.0102 0.0101

0.0114 0.0115 0.0116 0.0117 0.0116

98 99 100 101 100

Section 6.3

Hydraulics of Sewers

241

The input values of h/D are shown in Column 1, and the calculated values of Q and n are shown in Columns 2 and 3, respectively. The “Typical” n value is obtained by increasing the calculated n value in Column 3 by 15% and is shown in Column 4. The ratio of Manning’s n for any given h/D to the full-flow Manning’s n is shown in Column 5. Based on these results, it is apparent that Manning’s n will not vary by more than 2% for the flow depths considered, and hence Manning’s n can be assumed to be constant. A typical value of n = 0.0116 could be assumed. This is close to the value of n = 0.0115 given in Table 6.4 for a roughness height of 0.03 mm.

An alternative to using the Manning equation with variable n is to simply use the Darcy–Weisbach equation directly, and this direct approach is common in the United Kingdom (Brown and Davies, 2000). However, in the United States, formulation of sewer hydraulics in terms of the Manning equation remains commonplace. 6.3.3

Self-Cleansing

Sewers must be designed to ensure that solids do not accumulate on the bottom of the conduit and affect their carrying capacity. The suspended-solids content of domestic sewage is typically in the range of 100–500 mg/L, and includes both mineral and organic material in a variety of sizes and densities. To prevent solids accumulation in sewers, the recommended design approach is to identify a “design particle” and then ensure that the minimum flow rate in the sewer (when it is first put into service) creates sufficient shear stress on the bottom of the sewer to move the design particle along the bottom of the sewer. Sewers designed to prevent solids accumulation are called self-cleansing. This recommended design approach to ensure self-cleansing is called the tractive force method. Using the tractive force method, the average shear stress, τ0 [FL−2 ], on the perimeter of the sewer conduit is calculated using the relation τ0 = γ RS0 (6.27) where γ is the specific weight of water [FL−3 ], R is the hydraulic radius of the flow section [L], and S0 is the slope of the sewer conduit [dimensionless]. The average shear stress, τ0 , is commonly referred to as the tractive force (although it is actually a stress), and the design value of τ0 is estimated under minimum-flow conditions when the sewer is first put into service. Estimation of the minimum flow rate, Qmin , is described in Section 6.2.4. The critical shear stress, τc [Pa], for a design particle can be estimated by the relation (Walski et al., 2004) τc = 0.867d0.277

(6.28)

where d is the nominal diameter [mm] of a discrete sand particle with a specific gravity of 2.7. Design particle sizes should typically be in the range of 1.0–1.5 mm depending on local conditions (Merritt, 2009), and these values substituted in Equation 6.28 yield τc of 0.87–0.97 Pa, respectively. For sewers transporting larger-than-usual particle sizes, design particle sizes of 2 mm or higher can be used in such extraordinary conditions (ASCE, 2007). Equation 6.28 is applicable to discrete granular particles and not particles embedded in a cohesive matrix, the formation of a cohesive matrix being prevented by the maintenance of shear stresses exceeding the critical shear stress. The design flow rate for self-cleansing is normally taken as Qmin as calculated in Section 6.2.4, which corresponds to the highest 1-hour flow rate in the lowest flow week when the sewer is first put into service. This means that the critical self-cleansing condition is mostly exceeded at other times of the year. Ideally, self-cleansing conditions should be achieved at least once per day (Butler and Davies, 2000); however, in cul-de-sacs served by short deadend sections, it will often be impossible to meet self-cleansing requirements, and such pipes may require periodic flushing to scour the accumulated sediments. Since self-cleansing of a sanitary sewer is assessed under minimum-flow conditions, it is very important that the design minimum flow rate, Qmin , be estimated as accurately as possible and not overestimated. If a sewer conduit is designed for an overestimated Qmin , then the conduit will not be selfcleansing under actual minimum-flow conditions.

242

Chapter 6

Design of Sanitary Sewers

It is still common practice to assure self-cleansing by specifying a minimum permissible velocity under full-flow conditions. This minimum full-flow velocity is sometimes called the self-cleansing velocity. A typical self-cleansing velocity is 0.60 m/s (2 ft/s), although selfcleansing velocities as high as 1.1 m/s (3.5 ft/s) have been used (e.g., Nalluri and Ghani, 1996). The fundamental limitation of using a specified self-cleansing velocity in the design of sewer pipes is that the tractive force under minimum-flow conditions will vary depending on the diameter of the pipe, so self-cleansing might not be assured for all pipe diameters, particularly for sewers with relatively small flows.

EXAMPLE 6.5 It is proposed to lay a sewer pipe of diameter 610 mm on a slope of 0.045%, and the minimum flow rate expected upon installation is 7 L/s. The pipe roughness is estimated as 0.03 mm. Local regulations require a minimum permissible full-flow velocity of 0.60 m/s to assure self-cleansing. If the minimum tractive force required for self-cleansing is 1 Pa (based on a particle size of 1.67 mm), determine whether the minimum permissible velocity criterion will assure self-cleansing. Solution From the given data: D = 610 mm = 0.610 m, S0 = 0.045% = 0.00045, Qmin = 7 L/s = 0.007 m3 /s,  = 0.03 mm, Vpm = 0.60 m/s, and τc = 1 Pa. Under full-flow conditions, the velocity of flow in the pipe, Vfull , is given by Manning’s equation as Vfull =

1

2 3

1 2

1

R S = nfull full 0 nfull



0.610 4

2 3

1

(0.00045) 2 =

0.00606 nfull

(6.29)

For a roughness height of 0.03 mm, nfull can be calculated using the Darcy–Weisbach equation and Equation 6.12 which yields nfull = 0.00992, and hence Equation 6.29 yields Vfull =

0.00606 = 0.61 m/s 0.00992

Since Vfull > Vpm (i.e., 0.61 m/s > 0.60 m/s) the pipe configuration is adequate to assure self-cleansing, as per the regulatory requirement. Under minimum-flow conditions (Qmin = 0.007 m3 /s), solution of the Manning equation (with variable n) yields n = 0.00986 and a hydraulic radius of R = 0.0514 m. Hence the minimum shear stress on the pipe boundary, τmin , is given by (taking γ = 9790 N/m3 ) τmin = γ RS0 = (9790)(0.0514)(0.00045) = 0.23 Pa Since the minimum shear stress (0.23 Pa) is less than the critical shear stress (1 Pa), the sewer pipe will not be self-cleansing for a particle size of 1.67 mm; the largest particle that a 0.23 Pa shear stress will move is 0.25 mm. Based on these results, if the sewer pipe is assessed for self-cleansing on the basis of the minimum permissible full-flow velocity, then in reality it will not be self-cleansing, and particle buildup within the sewer pipe will likely occur.

6.3.4

Scour Prevention

The specification of a maximum permissible velocity is important to prevent excessive scouring of sewer conduits and hazardous conditions for sewer workers. ASCE (2007) recommends that flow velocities in sanitary sewers should not be greater than 3.5–4.5 m/s (10–15 ft/s). 6.3.5

Design Computations for Diameter and Slope

The typical objective in the hydraulic design of a sewer conduit is to determine the appropriate diameter, D, and slope, S0 , for a given maximum flow rate, Qmax , minimum flow rate, Qmin , pipe material or roughness, , and design critical shear stress, τc , for self-cleansing. Constraints in the hydraulic design are as follows:

Section 6.3

Hydraulics of Sewers

243

 When Q = Qmax : The relative flow depth must not exceed (h/D)max , and the velocity must not exceed Vlim .  When Q = Qmin : The sewer must be self-cleansing, which requires that τ0 Ú τc . The relationship between flow rate, Q, and depth-of-flow, h, can be calculated using either the Manning equation or the Darcy–Weisbach equation. If the Manning equation is used, an n value that provides consistency with the Darcy–Weisbach equation should also be used. The typical sequence of design computations is as follows: Step 1: Determine the pipe slope required to match the given invert depth at the upstream end of the pipe and meet the minimum-cover requirement at the downstream end of the pipe; call this slope Sref1 . Step 2: Assume a commercial-size pipe diameter, D. On the first iteration, D should be equal to the diameter of the upstream pipe, or equal to the minimum allowable pipe diameter if there is no upstream pipe. Step 3: Specify Q = Qmax and h/D = (h/D)max , and use the Manning or Darcy–Weisbach equation to find the required slope, Sref2 . Step 4: Specify Q = Qmin , and use either the Manning or the Darcy–Weisbach equation to find the required slope, Sref3 , for the shear stress on the bottom of the pipe to equal the critical shear stress, τc . Step 5: Assign the pipe slope, S0 = max(Sref1 , Sref2 , Sref3 ). Step 6: Verify that V … Vlim when Q … Qmax . If not, adjust the pipe diameter and repeat Steps 1–5. This procedure yields corresponding values of D and S0 that satisfy the hydraulic design constraints, where S0 is the minimum slope that meets all of the design constraints for the given diameter, D. There can be multiple (D,S0 ) combinations that satisfy the design constraints, and the designer will usually consider how the entire sewer system fits together before making a final decision on the appropriate (D,S0 ) combination for any individual pipe. Usually the first preference is to use the same diameter as the upstream pipe. EXAMPLE 6.6 It is desired to use a 610-mm-diameter concrete pipe in a segment of a sewer line. The maximum and minimum flow rates in the pipe segment are 0.30 m3 /s and 0.015 m3 /s, respectively. Design specifications require that the roughness height of the concrete pipe be taken as 0.1 mm, the critical shear stress for self-cleansing as 0.87 Pa, and the sewage temperature as 20◦ C. Regulatory requirements state that the pipe must flow no more than 75% full under maximum-flow conditions, and the maximum velocity must be less than 4.0 m/s. Based on the invert elevation of the upstream pipe segment and minimum-cover requirement, any pipe slope greater than zero is feasible. Determine the minimum allowable pipe slope that could be used. Solution From the given data: D = 610 mm = 0.610 m, Qmax = 0.30 m3 /s, Qmin = 0.015 m3 /s,  = 0.1 mm = 0.0001 m, τc = 0.87 Pa, T = 20◦ C, (h/D)max = 0.75, and Vlim = 4.0 m/s. Step 1: From the given information, any pipe slope greater than zero meets the minimum-cover requirement. Therefore, Sref1 = 0.0. Step 2: The pipe diameter under consideration is D = 610 mm. This is likely equal to the pipe diameter of the upstream segment. Step 3: When Q = Qmax = 0.30 m3 /s and h/D = (h/D)max = 0.75, the following calculations are made:

  h −1 1 − 2 = 2 cos−1 1 − 2(0.75) = 4.189 radians θmax = 2 cos D max

  sin θmax 0.610 D sin(4.189) 1 − = Rmax = 1 − = 0.184 m 4 θmax 4 4.189

244

Chapter 6

Design of Sanitary Sewers

 θmax − sin θmax 4.189 − sin(4.189) 2 = D = (0.610)2 = 0.235 m2 8 8 

Amax

(Qmax /Amax )Rmax (0.30/0.235)(0.184) = 2.35 * 105 = ν 10−6 0.0001 = = 0.000543 0.184 ⎡ ⎡   ⎤−2 ⎤−2  1.65 0.000543 1.65 ⎦ = 0.25 ⎣log ⎦ = 0.25 ⎣log + + 14.8Rmax 14.8 (2.35 * 105 )0.9 Re0.9 max

Remax =  Rmax fmax

= 0.0141 1

1

1

1

6 2 fmax = 0.1129(0.184) 6 (0.0141) 2 = 0.0101 nmax = 0.1129Rmax ⎤2 ⎡ ⎡

Smax

⎤2 −8 − 83 20.16(0.0101)(0.30)(0.610) ⎢ 20.16nmax Qmax D 3 ⎥ ⎦ = 0.00159 =⎣ 2 ⎦ =⎣ 2 5 5 −3 (4.189)− 3 [4.189 − sin(4.189)] 3 θmax [θmax − sin θmax ] 3

Based on these results, the minimum slope required for the sewer to flow no more than 75% full is 0.00159. Hence, Sref2 = 0.00159. Step 4: Taking Q = Qmin = 0.015 m3 /s and τmin = 0.87 Pa, the following equations are solved for Smin :



D 0.610 sin θmin sin θmin Rmin = = 1 − 1 − 4 θmin 4 θmin     − sin θmin − sin θmin θ θ D2 = min (0.610)2 Amin = min 8 8 (0.015/Amin )Rmin (Qmin /Amin )Rmin = ν 10−6  0.001 = Rmin Rmin ⎡ ⎛ ⎞⎤−2  1.65 ⎠⎥ ⎢ fmin = 0.25 ⎣log ⎝ + ⎦ 14.8Rmin Re0.9 min

Remin =

1

1

6 2 nmin = 0.1129Rmin fmin ⎡

⎤2 8 D− 3

⎤2

⎡ 8 (0.015)(0.610)− 3

⎥ ⎢ 20.16nmin Qmin ⎢ ⎥ = ⎢ 20.16nmin Smin = ⎢ ⎦ ⎣ −2 ⎣ −2 5 5 3 3 θmin [θmin − sin θmin ] 3 θmin [θmin − sin θmin ] 3

⎥ ⎥ ⎦

τmin = γ Rmin Smin Simultaneous solution of the above equations (with the given values of Qmin and τmin ) yields Smin = 0.00166. Hence, Sref3 = 0.00166. Step 5: To meet the physical constraints, the maximum flow depth limitation, and the minimum boundary shear stress requirements, the minimum required pipe slope, S0 , is given by S0 = max(Sref1 , Sref2 , Sref3 ) = max(0, 0.00159, 0.00166) = 0.00166 Therefore the minimum boundary shear stress requirement controls the minimum-slope requirement. Step 6: Determine the average velocity when Q = Qmax = 0.30 m3 /s and S0 = 0.00166. Application of the Manning equation yields Vmax = 1.66 m/s. Since the maximum velocity

Section 6.3

Hydraulics of Sewers

245

(1.66 m/s) is less than the given limit (4.0 m/s), the calculated minimum-allowable slope of 0.00166 is acceptable. The calculations presented here are easily performed using a spreadsheet with solver capability; the unknown variable can be taken as θ.

6.3.6

Hydraulics of Manholes

The head loss, hL [L], as sewage flows through a manhole is typically estimated using the relation hL = K

V22 2g

(6.30)

where K is a head-loss coefficient [dimensionless], V2 is the velocity in the pipe downstream of the manhole [LT−1 ], and g is gravity [LT−2 ]. For manholes with one inflow pipe and one outflow pipe, values of K depend on the angle, θ , between the inlet pipe and the outlet pipe, and typical values are as follows:  0.15, θ = 0◦ (straight through) K= (6.31) 1.00, θ = 90◦ For contoured flow-through manholes, typically hL … 0.6 cm (0.02 ft). The energy grade line across a manhole is illustrated in Figure 6.3, where it is apparent that in order to avoid backwater effects a drop of z [L] is required such that  z = (y2 − y1 ) +

V2 V22 − 1 2g 2g

+ hL

(6.32)

were y1 and y2 are the normal flow depths in the sewer conduits upstream and downstream of the manhole [L], respectively, and V1 and V2 are the velocities in the sewer conduits upstream and downstream of the manhole [LT−1 ], respectively. The maximum flow rate, Qmax , should be used in calculating z since this represents the worst-case scenario for sewer backwater effects. If a negative drop (i.e., a rise) is calculated from Equation 6.32, then zero drop is used, since a rise would be an obstacle to bedload sediment transport and self-cleansing. If two or more sewers enter the same manhole, then z in Equation 6.32 should be calculated for each entrance–exit pair and the maximum z should be used at the manhole. The largest manhole drops will be required at locations where the sewer transitions from steep to flat slopes. Under typical circumstances z … 3 cm (0.10 ft). If manhole drops are not needed and head losses are negligible, it might be prudent to specify a drop of z = 1.5 cm (0.05 ft) to accommodate a small head loss and as a construction tolerance so that slight horizontal misalignment of the manhole will not result in an outlet at a higher elevation than the lowest inlet (ASCE, 2007). FIGURE 6.3: Energy grade line through manhole

Manhole

D1 Energy Grade Line hL

2

V1 2g

y1 2

Δz

y2

V2 2g

D2 Flow

246

Chapter 6

Design of Sanitary Sewers

EXAMPLE 6.7 A manhole is to be placed at a location where the slope of a 760-mm-diameter pipe changes from 0.5% to 0.2%. The roughness height of the concrete pipe is estimated as 1 mm and the design maximum flow rate is 0.485 m3 /s. If the inflow and outflow pipes are located directly opposite each other, determine the minimum drop in the pipe invert that must be provided at the manhole. Solution From the given data: D = 760 mm = 0.760 m, S1 = 0.5% = 0.005, S2 = 0.2% = 0.002,  = 1 mm = 0.001 m, and Q = 0.485 m3 /s. Assume a temperature of 20◦ C so that ν = 10−6 m2 /s. Step 1: Determine the upstream flow conditions. The following equations are solved for the value of the central flow angle, θ1 :



D 0.760 sin θ1 sin θ1 R1 = = 1 − 1 − 4 θ1 4 θ1     θ − sin θ1 θ − sin θ1 D2 = 1 (0.760)2 A1 = 1 8 8 (Q/A1 )R1 (0.485/A1 )R1 = ν 10−6 ⎡ ⎡ ⎛ ⎛ ⎞⎤−2 ⎞⎤−2  0.001 1.65 ⎠⎥ 1.65 ⎠⎥ ⎢ ⎢ f1 = 0.25 ⎣log ⎝ + + ⎦ = 0.25 ⎣log ⎝ 0.9 ⎦ 14.8R1 14.8R1 Re0.9 Re 1 1

Re1 =

1

1

n1 = 0.1129R16 f12 ⎡

⎤2

⎤2



8 QD− 3

8 (0.485)(0.760)− 3

⎥ ⎢ 20.16n1 ⎢ ⎥ = ⎢ 20.16n1 S1 = ⎢ ⎦ ⎣ −2 ⎣ 5 5 −2 θ1 3 [θ1 − sin θ1 ] 3 θ1 3 [θ1 − sin θ1 ] 3

⎥ ⎥ = 0.005 ⎦

(6.33)

The solution of these equations is θ1 = 3.310 radians, which corresponds to a flow depth y1 = 0.412 m and an average velocity V1 = 1.93 m/s. Step 2: Determine the downstream flow conditions. Use the same equations as in Step 1 with the exception that the slope in Equation 6.33 is taken as 0.002 instead of 0.005. This yields a flow depth y2 = 0.570 m and an average velocity V2 = 1.33 m/s. Step 3: Determine the head loss at the manhole. Since the flow is straight through the manhole, the head-loss coefficient can be estimated as K = 0.15 and the head loss, hL , at the manhole is given by Equation 6.30 as V2 (1.33)2 = 0.014 m hL = K 2 = (0.15) 2g 2(9.81) Step 4: Determine the required drop at the manhole. The required drop, z, is given by Equation 6.32 as ⎡ ⎤ V12 V22 ⎦ + hL − z = (y2 − y1 ) + ⎣ 2g 2g

= (0.570 − 0.412) +

1.332 1.932 − 2(9.81) 2(9.81)

+ 0.014

= 0.072 m Therefore a minimum drop of 7.2 cm should be provided at the manhole to avoid backwater conditions.

Section 6.4

System Design Criteria

247

6.4 System Design Criteria Sanitary-sewer systems have a variety of components, which typically include pipes, manholes, and pump stations. Criteria governing the design of several key system components are given in the following sections. 6.4.1

System Layout

To layout a sanitary-sewer system requires selection of the system outlet; determination of the tributary area (the “sewershed”); location of the main sewer; determination of whether there is a need for, and the location of, pumping stations and force mains; location of underground rock formations; and location of water and gas lines, electrical, telephone, and television wires, and other underground utilities. The main sewer is the principal sewer to which branch sewers are tributary. In larger systems, the main sewer is also called the trunk sewer. The selected system outlet depends on the scope and objectives of the particular project and may consist of a pumping station, an existing trunk or main sewer, or a treatment plant. Since the flows in sewer systems are typically driven by gravity, preliminary layouts are usually made using topographic maps. Trunk and main sewers are located at the lower elevations of the service area, although existing roadways and the availability of rights-of-way may affect the exact locations. In developed areas, sanitary sewers are commonly located at or near the centers of roadways and alleys. In very wide streets, however, it might be more economical to install sanitary sewers on both sides of the street. When sanitary sewers are in close proximity to public water supplies, it is common practice to use pressure-type sewer pipe (i.e., a force main), concrete encasement of the sewer pipe, sewer pipe with joints that meet stringent infiltration/exfiltration requirements, or at least to put water pipes and sewer pipes on opposite sides of the street. Most building codes prohibit sanitary-sewer installation in the same trench as water mains, and require that sewers be at least 3 m (10 ft) horizontally from water mains and, where they cross, at least 450 mm (18 in.) vertical separation between sewers and water mains (Lagvankar and Velon, 2000). 6.4.2

Pipe Material

A variety of pipe materials are used in practice, including concrete, vitrified clay, cast iron, ductile iron, and various thermoplastic materials including PVC. Pipes are broadly classified as either rigid pipes or flexible pipes. Rigid pipes derive a substantial part of their loadcarrying capacity from the structural strength inherent in the pipe wall, while flexible pipes derive their load-carrying capacity from the interaction of the pipe and the embedment soils affected by the deflection of the pipe to the equilibrium point under load. Pipe materials classified as rigid and flexible are listed in Table 6.5, and the advantages and disadvantages to using various pipe materials, along with typical Manning’s n values, are given in Table 6.6. The type of pipe material to be used in any particular case is dictated by several factors, including the type of wastewater to be transported (residential, industrial, or combination), scour and abrasion conditions, installation requirements, type of soil, trench-load conditions, bedding and initial backfill material available, infiltration/exfiltration requirements, and cost effectiveness. Concrete, vitrified clay, and plastic pipe are the most common materials used in gravity-driven sanitary sewers, with concrete being the most common, and clay and plastic pipe used in preference to concrete when corrosion of concrete is a concern. The commercially available diameters of concrete pipe are given in Appendix E.3. It is useful to keep in

TABLE 6.5: Rigid and Flexible Pipe Materials

Rigid pipe Concrete Cast iron Vitrified clay

Flexible pipe Ductile iron Steel Thermoplastic (e.g., PVC)

248 TABLE 6.6: Sewer-Pipe Materials

Material

Advantages

Disadvantages

Manning’s n

Available diameters

Concrete

(1) Readily available in most localities (2) Wide range of structural and pressure strengths (3) Wide range of nominal diameters (4) Wide range of laying lengths

(1) High weight (2) Subject to corrosion on interior if atmosphere over wastewater contains hydrogen sulfide (3) Subject to corrosion on exterior if buried in an acid or highsulfate environment

0.011–0.015

300 mm–3050 mm

Vitrified clay

(1) High resistance to chemical corrosion (2) Not susceptible to damage from hydrogen sulfide (3) High resistance to abrasion (4) Wide range of fittings available (5) Manufactured in standard and extra-strength classifications

(1) Limited range of sizes available (2) High weight (3) Subject to shear and beam breakage when improperly bedded (4) Brittle

0.010–0.015

100 mm–610 mm

Ductile iron

(1) Long laying lengths (2) High pressure and loadbearing capacity (3) High impact strength (4) High beam strength

(1) Subject to corrosion where acids are present (2) Susceptible to hydrogen sulfide attack (3) Subject to chemical attack in corrosive soils (4) High weight

0.011–0.015

100 mm–910 mm

Plastic

(1) Light weight (2) Long laying lengths (3) High impact strength (4) Ease in field cutting and tapping (5) Corrosion resistant (6) Low friction

(1) Subject to attack by certain organic chemicals (2) Subject to excessive deflection when improperly bedded and haunched (3) Limited range of sizes available (4) Subject to surface changes (5) Effected by long-term ultraviolet exposure

0.010–0.015

100 mm–1220 mm

Section 6.4

System Design Criteria

249

mind that the size of reinforced concrete pipe varies in 75-mm (3-in.) increments for diameters in the range of 305–915 mm (12–36 in.), and in 150-mm (6-in.) increments for diameters in the range of 915–2745 mm (36–108 in.). Ductile-iron pipe is often used for river crossings, where the pipe must support unusually high load, where an unusually leak proof sewer is required, or where unusual root problems are likely to develop. Ductile-iron pipe should not be used where the groundwater is brackish, unless suitable protective measures are taken. Ductile-iron pipe is employed in sewerage primarily for force mains and for piping in and around buildings. 6.4.3

Depth of Sanitary Sewer

Sanitary sewers should be buried to a sufficient depth that they can receive the contributed flow from the tributary area by gravity flow. Deep basements and buildings on land substantially below street level may require individual pumping facilities. In the northern United States, a minimum cover of 3 m (10 ft) is typically required to prevent freezing, while in the southern United States the minimum cover is dictated by traffic loads, and ranges upward from 0.75 m (2.5 ft), depending on the pipe size and anticipated loads (McGhee, 1991). The depth of sanitary sewers is such that they pass under all other utilities, with the possible exception of storm sewers. 6.4.4

Diameter and Slope of Pipes

Sanitary sewers typically have a minimum diameter of 205 mm (8 in.). Service connections typically have diameters in the range of 100–150 mm (4–6 in.) and slopes in the range of 1%–2%. In some developments where the houses are set far back from the street, the length and slope of the house service connections may determine the minimum sewer depths. The diameters of sanitary sewers should normally not decrease in the downstream direction so as to prevent sediment accumulation and blockage where this reduction occurs. However, some jurisdictions allow reductions for diameters greater than about 610 mm (24 in.) if the smaller diameter can be sustained for a considerable distance. 6.4.5

Hydraulic Criteria

It is common practice in the United States that sanitary sewers be designed to flow threefourths full at the maximum design flow rate. This practice ensures proper ventilation in sewers. Head losses in bends with pipe diameters greater than 915 mm (36 in.) are difficult to quantify and are usually accounted for by increasing the Manning’s n in the bend by 25%– 40% (McGhee, 1991). 6.4.6

Manholes

Manholes provide access to the sewer system for preventive maintenance and emergency service. Manholes are generally located at the junctions of sanitary sewers; at changes of grade, size, or alignment; and at the beginning of the sewer system. A typical manhole is illustrated in Figure 6.4(a), showing that manhole covers are typically 0.61 m (2 ft) in diameter, with a working space in the manhole typically 1.2 m (4 ft) in diameter. More modern manholes have eccentric cones to provide a vertical side for the steps. Manholes are typically precast and delivered to the site ready for installation, and a typical precast manhole is shown in Figure 6.5. The invert of the manhole should conform to the shape and slope of the incoming and outgoing sewer lines. In some cases, a section of pipe is laid through the manhole and the upper portion is sawed or broken off. More commonly, U-shaped channels are constructed within manholes, and in these cases the elevated area surrounding these channels are called benches. The spacing between manholes varies with local conditions, regulations, and methods of sewer maintenance. In the absence of other requirements, manholes should be spaced at distances not greater than 120 m (400 ft) for sewers 380 mm (15 in.) or less, and 150 m (500 ft) for sewers 455–760 mm (18–30 in.), except that distances up to 185 m (600 ft) may be acceptable where adequate modern cleaning equipment for such spacing is available. Greater spacing

250

Chapter 6

Design of Sanitary Sewers

FIGURE 6.4: Typical manholes Source: ASCE (2007).

0.61 m

1.2 m 1.2 m

(a) Typical manhole

(b) Drop manhole

FIGURE 6.5: Precast manhole

up to 300 m (1000 ft) may be acceptable for larger sewers that a person can walk through. A commonly used design criterion in some jurisdictions is that sewers with diameters of 610 mm (24 in.) or less should be straight between manholes (Lagvankar and Velon, 2000). In cases where a sewer pipe enters a manhole at an elevation considerably higher than the outgoing pipe, it is generally not acceptable to let the incoming wastewater simply pour into the manhole, since this does not provide an acceptable workspace for maintenance and repair, and can also cause excessive aeration and scour within the drainage system. Under these conditions, a drop manhole, illustrated in Figure 6.4(b), is used. Drop manholes are typically specified when the invert of the inflow pipe is more than 0.6 m (2 ft) above the elevation that would be obtained by matching the crowns of the inflow and outflow pipes. Drop manholes are also commonly used in the sewer systems of steep urban areas so as to avoid using steep slopes in sewer pipes. In some drop-manhole configurations, the upstream flow enters the manhole directly without using the vertical feeder pipe shown in Figure 6.4(b). For these types of drop manholes, caution should be taken since under some flow conditions only limited energy dissipation will occur in the manhole (Granata et al., 2011). In unusual cases where large drops are to be accommodated, a directly connected series of drop manholes can be considered, and design guidelines for such systems can be found in the work of Camino et al., (2011).

Section 6.4

System Design Criteria

251

The center of streets and street intersections are common locations for manholes. Sanitary-sewer manholes should not be located or constructed in a way that allows surface water to enter through the manhole cover. Manholes that are not in the pavement, especially in open country, should have their rims set above grade (i.e., above ground elevation) to avoid the inflow of stormwater and to facilitate field location. 6.4.7

Pump Stations

Pump stations, also known as lift stations, are frequently necessary in flat terrain to raise the wastewater to a higher elevation so that gravity flow can continue at reasonable slopes and depths. Pump stations typically have a wet well and a dry well; the wet well receives and temporarily stores the wastewater flow and is sized for a 10–30-minute detention time, and the dry well contains the pump and motor. Ventilation, humidity control, and standby power supply are important design considerations. For smaller pump stations with capacities less than 40 L/s (700 gpm), at least two pumps should be provided; for larger pump stations, three or more pumps should be provided. In both cases, the pump capacities should be sufficient to pump at the maximum wastewater flow rate if any one pump is out of service. To avoid clogging, the associated suction and discharge piping should be at least 100 mm (4 in.) in diameter and the pump should be capable of passing 75-mm (3-in.) diameter solids. In cases where pumping directly into a gravity-flow sanitary sewer is not practical, wastewater can be pumped from the wet well through a force main to another gravity sewer, or directly to a wastewater-treatment facility. Some pump stations consist principally of above-ground piping that contains the inflow and outflow pipes to and from the wet well, respectively, and accumulated sewage in the wet well is discharged using submersible pumps that are directly connected to the outflow piping. This type of pump station is shown in Figure 6.6, where two submersible pumps are connected to the outflow pipes in the wet well. If large increases in flow rates are expected over the design life of a pump station and the intermittent selfcleansing associated with having small flows in large-capacity pump stations is not feasible, then two parallel lines through pump stations might be preferable. In these cases, one line is used until its capacity is exceeded. 6.4.8

Force Mains

Where pumping of sewage is required, such as in flat terrains, force mains (pressure conduits) must be designed to carry the flows. In these cases, the costs of pumping and associated equipment are important considerations. Force mains must be able to transport sewage at velocities sufficient to avoid deposition and yet not so high as to create pipe-erosion problems. Force mains are generally 200 mm (8 in.) in diameter or greater, but for small pumping stations smaller pipes may sometimes be acceptable. Velocities encountered in force-main operations are in the range of 0.6–3 m/s (2–10 ft/s). In designing force mains, high points in lines should be avoided if possible, since this eliminates the need for air-relief valves. Good design practice dictates that the hydraulic gradient should lie above the force main at all points during periods of minimum-flow pumping. Service connections to force mains are FIGURE 6.6: Pump station

Outflow Wet well Outflow pipes with Submersible pumps Inflow (a) Pump station

(b) Wet well

252

Chapter 6

Design of Sanitary Sewers

called pressure sewers, and customers discharging to pressure sewers must have their own pump. Head losses in force mains are best calculated using the Darcy–Weisbach equation with roughness heights depending on the type and state of the pipe material; typical roughness heights for pipe materials used in force mains can be found in Table 2.1. In the United States, head losses in force mains are frequently calculated using the Hazen–Williams formula, hL =

10.7L 1.852 D4.87 CH

Q1.852

(6.34)

where hL is the head loss [m], L is the pipe length [m], CH is the Hazen–Williams “C-factor” [dimensionless], D is the pipe diameter [m], and Q is the flow rate [m3 /s]. Typical values of CH for pipe materials used in force mains can be found in Table 2.2. Caution has to be taken in using the Hazen–Williams formula since it is an empirical equation that is only valid under flow conditions that are not fully turbulent. Local head losses at fixtures such as valves, tees, bends, and reducers are typically calculated using relations of the form h0 = K

V2 2g

(6.35)

where h0 is the local loss [L], K is the local loss coefficient [dimensionless], V is the average flow velocity [LT−1 ], and g is the acceleration due to gravity [LT−2 ]. Loss coefficients at fixtures commonly found in force mains are given in Figure 2.7. 6.4.9

Hydrogen-Sulfide Control

Hydrogen-sulfide (H2 S) generation is a common problem in sanitary sewers. Among the problems associated with H2 S generation are odor, health hazard to maintenance crews, and corrosion of unprotected sewer pipes produced from cementitious materials and metals. The design of sanitary sewers seeks to avoid septic conditions and to provide an environment that is relatively free of H2 S. Generation of H2 S in sanitary sewers results primarily from the action of sulfate-reducing bacteria (SRB) on the pipe floor which convert sulfates to hydrogen sulfide under anaerobic (reducing) conditions. Compounds such as sulfite, thiosulfate, free sulfur, and other inorganic sulfur compounds occasionally found in wastewater can also be reduced to sulfide. Corrosive conditions occur when sulfuric acid (H2 SO4 ) is derived through the oxidation of hydrogen sulfide by aerobic bacteria (Thiobacillus thiooxidans) and fungi that reside on the moist exposed sewer pipe wall. This process can only take place where there is an adequate supply of H2 S (>2 ppm), high relative humidity, and atmospheric oxygen. The effect of H2 SO4 on concrete surfaces exposed to the sewer environment can be devastating. Concrete pipes, asbestos-cement pipes, and mortar linings on ferrous pipes experience surface reactions in which the surface material is converted to an expanding pasty mass which may fall away and expose new surfaces to corrosive attack. The color of corroded concrete surfaces can be various shades of yellow caused by the direct oxidation of H2 S to elemental sulfur. Entire pump stations have been known to collapse due to loss of structural stability from corrosion. Hydrogen sulfide gas is extremely toxic and can cause death at concentrations as low as 300 ppm (0.03%) in air. A person who ignores the first odor of the gas quickly loses the ability to smell the gas, eliminating further warning and leading to deadly consequences. Significant factors that contribute to H2 S generation are high wastewater temperatures and low flow velocities. The potential for sulfide generation in sewers with diameters less than 600 mm (24 in.) can be assessed using the value of the Z variable, where (Pomeroy and Parkhurst, 1977) Z = 0.308

EBOD 1 2

S0 Q

1 3

*

P B

(6.36)

Section 6.4

System Design Criteria

253

TABLE 6.7: Sulfide Generation Based on Z Values

Z values

Sulfide condition

Z < 5000 Sulfide rarely generated 5000 … Z … 10,000 Marginal condition for sulfide generation. At Z L 7500, low H2 S concentrations are likely. At Z L 10,000 odor and corrosion problems can occur. Z > 10,000 Sulfide generation common. At Z L 15,000 frequent problems with odor and significant corrosion problems can be expected. Sources: ASCE (1982); Butler and Davies (2011).

where EBOD is the effective biochemical oxygen demand [mg/L], defined by the relation EBOD = BOD5 * 1.07T−20

(6.37)

where BOD5 is the average 5-day biochemical oxygen demand [mg/L] at 20◦ C during the highest 6-hour flow period of the day, T is the temperature of the wastewater in the sewer [◦ C], S0 is the slope of the sewer [dimensionless], Q is the flow rate in the sewer [m3 /s], P is the wetted perimeter [m], and B is the top width [m] of the sewer flow. The wastewater pH is assumed to be in the range 7–8. The relationship given by Equation 6.36 is commonly referred to as the Z formula, and the relationship between the calculated Z value and the potential for sulfide generation is given in Table 6.7. The Z formula given by Equation 6.36 is widely used in practice (Butler and Davies, 2011). However, it has been suggested that the biodegradability of the wastewater in sanitary sewers is an important factor influencing sulfide generation that is not incorporated in the Z formula. Vollertsen (2006) has suggested that the value of Z calculated using Equation 6.36 be modified by the factor, fZ , where fZ = 1 + 10

BOD5 − 0.47 COD

(6.38)

where COD is the chemical oxygen demand [mg/L], BOD5 is the 5-day biochemical oxygen demand [mg/L], and biodegradability is measured by the ratio of BOD5 /COD. If Equation 6.38 is used, then the modified Z value is equal to fZ * Z and the same criteria in Table 6.7 can be used to evaluate the likelihood of sulfide generation. It is usually impractical or impossible to design a sulfide-free sewer system, and engineers endeavor to minimize sulfide generation and use corrosion-resistant materials to the maximum extent possible. Increased turbulence within sewers will increase the rate at which H2 S is released from wastewater, and structures causing avoidable turbulence should be identified and retrofitted to produce a more streamlined flow. Wastewater-treatment plants are usually located at the terminus of sewer systems, and chlorination with either elemental chlorine or hypochlorite quickly destroys sulfide and odorous organic sulfur compounds. Chlorination in sanitary sewers is generally considered impractical. The dissolving of air or oxygen in the wastewater as it moves through the sewer system, addition of chemicals such as iron and nitrate salts, and periodic sewer flushing are effective sulfide-control measures.

EXAMPLE 6.8 A 915-mm-diameter concrete sewer is laid on a slope of 0.9% and is to carry 1.7 m3 /s of domestic wastewater. Manning’s n of the sewer pipe is estimated as 0.013. If the 5-day BOD of the wastewater at 20◦ C is expected to be 300 mg/L, determine the potential for sulfide generation when the wastewater temperature is 25◦ C.

254

Chapter 6

Design of Sanitary Sewers Solution From the given data, Q = 1.7 m3 /s, D = 915 mm = 0.915 m, S0 = 0.009, n = 0.013, and Equation 6.17 gives the flow angle, θ, by the relation 2

5

8

− 12

θ − 3 (θ − sin θ ) 3 − 20.16nQD− 3 S0 2

5

8

=0

1

θ − 3 (θ − sin θ ) 3 − 20.16(0.013)(1.7)(0.915)− 3 (0.009)− 2 = 0 which yields θ = 4.31 radians. The flow perimeter, P, top width, B, and the ratio P/B are given by Dθ 2 θ θ D sin = D sin B=2 2 2 2 P=

P Dθ/2 θ 4.31 = = = = 2.57 B D sin (θ/2) 2 sin (θ/2) 2 sin (4.31/2) The effective BOD, EBOD at T = 25◦ C, is given by Equation 6.37 as EBOD = BOD5 * 1.07T−20 = 300 * 1.0725−20 = 421 mg/L According to the Z formula, Equation 6.36, Z = 0.308

EBOD 1 2

S0

1 Q3

*

421 P = 0.308 * 2.57 = 2948 1 1 B (0.009) 2 (1.7) 3

Therefore, according to Table 6.7, hydrogen sulfide will be rarely generated.

6.4.10

Combined Sewers

Combined sewers carry both stormwater runoff and sewage, and are found in many older cities in the United States. Typically, urban combined-sewer systems are designed to carry flow that is about four to eight times the average dry-weather flow (sewage flows), while treatment plants serving these systems are typically designed to handle mixed flows that are four to six times the average dry-weather flow. Overflows from combined systems in most urban areas occur on average 10–60 times per year (Novotny, 2003). Under normal circumstances, combined sewers transport water through partially full pipes in an open-channel regime. However, flow conditions within sewers are significantly altered during intense rain events as the pipes transition to a pressurized full-pipe condition with the air at the pipe crown often expelled by a hydraulic bore propagating on the freesurface flow. Associated problems include geysering through vertical shafts and structural damages resulting from surges and waterhammer pressures, such as described by Guo and Song (1990) in the Chicago TARP system and Zhou et al. (2002) in the drainage system of Edmonton, Canada. Historically, different numerical techniques have been employed to predict flow in the open-channel and pressurized regimes, and no existing technique is fully satisfactory when both conditions exist. A proposed technique that can be used to simulate flow in both regimes is given by Vasconcelos and et al. (2006). 6.5 Design Computations Hydraulic design computations for sanitary sewers seek to determine the slope and diameter of each pipe segment such that the pipe has adequate capacity under maximum-flow conditions and is self-cleansing under the minimum-flow conditions. The procedure for determining acceptable diameter-slope combinations in each pipe segment is described in detail in Section 6.3.5. However, this procedure can become tedious when repeated manually for a large number of pipe segments, and design aids are useful in such circumstances.

Section 6.5

6.5.1

Design Computations

255

Design Aids

Design aids have been developed to facilitate the computation of Manning’s n and for determining the minimum slope for self-cleansing. These design aids provide close approximations to the results obtained when calculations are done manually; however, when computations are automated using a computer program, the exact methods presented previously should be used. 6.5.1.1

Manning’s n

Manning’s n can be calculated exactly for any given pipe diameter, flow, and roughness height using the Colebrook equation (to estimate the friction factor) along with the relationship between Manning’s n and the friction factor given by Equation 6.12. However, it has also been shown (ASCE, 2007) that for depths of flow greater than 15% of the diameter Manning’s n is relatively insensitive to the depth of flow, velocity of flow, and temperature of the sewage. Manning’s n is much more sensitive to the pipe diameter and the roughness height. For most concrete pipes of a given diameter, it is recommended to specify a roughness height of 0.03 mm (0.0001 ft) and take the design value of Manning’s n to be a so-called “Typical” value that is 15% higher than the n value calculated using Equation 6.12. These “Typical” values of Manning’s n are shown in Table 6.4 for several common pipe diameters. The n values corresponding directly to a roughness height of 0.03 mm are used to characterize concrete pipes in “Extra Care” condition, and n values 30% higher characterize pipes in “Substandard” condition. Pipes in “Typical” condition usually reflect operating conditions, and the corresponding n values shown in Table 6.4 are appropriate for most designs. Some regulations, particularly in the United States, require that n values in concrete sewers be taken as 0.013. Although this requirement is typically an overestimate of n and does not take into account the reality that n is a function of pipe diameter, local regulatory requirements should be followed when they exist. Using an overestimate of n to assess pipe capacity under maximum-flow conditions will generally lead to an overly conservative sewerpipe design. 6.5.1.2

Minimum slope for self-cleansing

The minimum slope required for self-cleansing depends on the minimum flow rate, pipe diameter, pipe roughness, and the design particle size for bedload transport. A typical design particle size is 1 mm, although 1.5 mm is also used and 1.67 mm is also convenient in that it corresponds to a critical boundary stress of exactly 1 Pa, according to Equation 6.28. The exact procedure for calculating the minimum slope required for self-cleansing is given in Section 6.3.5; however, in the usual cases where the roughness height is taken as 0.03 mm and the pipe is in “Typical” condition, the slopes required for self-cleansing are closely approximated by the empirical relations shown in Table 6.8. The predictions of these equations have correlation coefficients greater than 0.999 when compared with the exact values for relative flow depths, h/D, in the range of 0.1–0.4, and can be extended to the range of 0.05–0.50 with very little loss of accuracy (Merritt, 2009). For relative flow depths outside this range, which is uncommon for minimum-flow conditions, and for design particle sizes outside the range of 1.0–1.67 mm, the minimum self-cleansing slope should be calculated using the exact approach described in Section 6.3.5. The self-cleansing slope values derived from Table 6.8 would normally be rounded to three significant digits. In cases where local regulatory requirements state that a Manning’s n of 0.013 must be used in design, this will typically lead to an underdesign of the pipe for self-cleansing, since 0.013 is typically higher than the actual n value as computed via Equation 6.12. For any given pipe slope, this leads to an overestimate of the flow depth and associated boundary shear stress under minimum-flow conditions. Therefore, the professional recommendation of the design engineer should be to determine Manning’s n using Equation 6.12 when designing pipes for self-cleansing. As minimum flows decrease, the required slope for self-cleansing increases. For very small minimum flows the required minimum slopes might become unreasonably steep and for economical reasons the design engineer might want to set the slope at a lower value

256

Chapter 6

Design of Sanitary Sewers TABLE 6.8: Self-Cleansing Slopes as Functions of Design Minimum Flow Rates

Pipe diameter (mm)

1.0 mm

1.5 mm

1.67 mm

150

0.00539Q−0.5692 min

0.00624Q−0.5697 min

0.00650Q−0.5699 min

225

0.00589Q−0.5720 min

0.00683Q−0.5723 min

0.00711Q−0.5724 min

300

0.00629Q−0.5736 min

0.00729Q−0.5739 min

0.00759Q−0.5740 min

375

0.00662Q−0.5748 min

0.00768Q−0.5751 min

0.00799Q−0.5751 min

450

0.00786Q−0.5758 min

0.00801Q−0.5780 min

0.00833Q−0.5761 min

500

0.00707Q−0.5763 min

0.00820Q−0.5764 min

0.00854Q−0.5764 min

525

0.00716Q−0.5766 min

0.00830Q−0.5767 min

0.00864Q−0.5768 min

600

0.00738Q−0.5772 min

0.00856Q−0.5774 min

0.00892Q−0.5775 min

675

0.00759Q−0.5778 min

0.00880Q−0.5779 min

0.00917Q−0.5781 min

750

0.00778Q−0.5783 min

0.00902Q−0.5783 min

0.00940Q−0.5784 min

825

0.00796Q−0.5786 min

0.00923Q−0.5788 min

0.00961Q−0.5789 min

900

0.00813Q−0.5790 min

0.00942Q−0.5791 min

0.00981Q−0.5792 min

975

0.00828Q−0.5793 min

0.00960Q−0.5795 min

0.01000Q−0.5797 min

1050

0.00846Q−0.5804 min

0.00978Q−0.5799 min

0.01018Q−0.5799 min

1125

0.00870Q−0.5837 min

0.00993Q−0.5800 min

0.01035Q−0.5802 min

1200

0.00893Q−0.5864 min

0.01017Q−0.5821 min

0.01052Q−0.5807 min

1500

0.00971Q−0.5932 min

0.01114Q−0.5908 min

0.01155Q−0.5898 min

2000

0.01077Q−0.5996 min

0.01239Q−0.5979 min

0.01288Q−0.5976 min

Design particle size

Note: Q in liters per second. Values of self-cleansing slopes are determined using variable Manning’s n derived from the Darcy–Weisbach equation at 20◦ C and  = 0.03 mm (0.0001 ft); the n value is increased by 15% above that derived from the Darcy–Weisbach equation.

than required, relying on manual cleaning until the minimum flow rate increases to the self-cleansing flow rate that corresponds to the selected pipe slope. This course of action is particularly appropriate in cases where the estimated minimum flow rate cannot be estimated with certainty. A possible approach might be to limit the relative flow depth, h/D, to 10% or 15% and determine the corresponding minimum slope, Smin , and minimum flow rate, Qmin , for self-cleansing. The pipe would then be laid on the appropriate minimum slope and selfcleansing would be attained when the minimum flow rate in the pipe equals or exceeds the corresponding design minimum flow rate. To facilitate this approach, values of Smin and Qmin for design particle sizes of 1.0 mm, 1.5 mm, and 1.67 mm are shown in Table 6.9. ASCE (2007) recommends using a 1-mm particle unless local conditions indicate that granular sediments in local sewage are larger than found in typical domestic sewage.

TABLE 6.9: Minimum Slopes at 15% and 10% Full

1 mm Diameter (mm) 150 mm 200 mm 250 mm 300 mm 375 mm 450 mm 525 mm 600 mm 750 mm 900 mm 1050 mm 1200 mm 1500 mm

15% Full Smin Qmin (–) (L/s) 0.00625 0.00468 0.00375 0.00312 0.00250 0.00208 0.00179 0.00156 0.00125 0.00104 0.00089 0.00078 0.00063

0.74 1.39 2.21 3.23 5.15 7.59 10.5 13.9 22.2 32.6 44.7 59.5 95.1

10% Full Smin Qmin (–) (L/s) 0.00914 0.00685 0.00548 0.00457 0.00365 0.00305 0.00261 0.00228 0.00183 0.00152 0.00131 0.00114 0.00092

0.40 0.71 1.13 1.67 2.69 3.94 5.46 6.37 11.5 16.9 23.2 30.9 49.3

Design particle size 1.5 mm 15% Full 10% Full Smin Qmin Smin Qmin (–) (L/s) (–) (L/s) 0.00685 0.00514 0.00410 0.00342 0.00274 0.00228 0.00195 0.00171 0.00137 0.00114 0.00098 0.00086 0.00068

0.79 1.44 2.29 3.40 5.41 7.93 11.0 14.6 23.2 34.3 47.0 62.3 99.1

0.01000 0.00751 0.00600 0.00500 0.00400 0.00334 0.00286 0.00250 0.00200 0.00167 0.00143 0.00125 0.00100

0.40 0.76 1.19 1.76 2.80 4.13 5.72 7.56 12.1 17.8 23.2 32.3 51.5

1.67 mm 15% Full 10% Full Smin Qmin Smin Qmin (–) (L/s) (–) (L/s) 0.00702 0.00526 0.00420 0.00350 0.00281 0.00234 0.00201 0.00175 0.00140 0.00117 0.00100 0.00088 0.00070

0.79 1.47 2.32 3.43 5.46 8.04 11.2 14.7 23.5 34.5 47.6 63.1 100

0.01026 0.00770 0.00615 0.00512 0.00410 0.00342 0.00293 0.00257 0.00205 0.00171 0.00146 0.00128 0.00103

0.42 0.76 1.22 1.78 2.83 4.19 5.78 7.67 12.2 18.0 24.6 32.6 52.1

257

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Design of Sanitary Sewers

6.5.2

Procedure for System Design

Basic information required prior to computing the sizes and slopes of sewer pipes includes: (1) a topographic map showing the proposed locations of the sewer lines, (2) the tributary areas to each line, (3) the (final) ground-surface elevations along each line, (4) the elevations of the basements of low-lying houses and other buildings that contribute flows to the sewer, and (5) the elevations of existing sanitary sewers which must be intercepted. After the sewer layout has been developed, design computations can be performed using the spreadsheet shown in Figure 6.7. Design computations begin with the characteristics of the sewer-pipe configuration in Columns 1 to 5 and lead to the computation of the sewer slopes in Column 14, diameters in Column 15, and sewer invert elevations in Columns 21 and 22. The steps to be followed in using Figure 6.7 in the design of sanitary sewers are as follows: Step 1: Computations begin with the uppermost pipe in the sewer system. List the pipeline number in Column 1 (usually starting from the Number 1), list the street location of the pipe in Column 2, list the beginning and ending manhole numbers in Columns 3 and 4 (the manhole numbers usually start from 1 at the uppermost manhole), and list the length of the sewer pipe in Column 5. The ground-surface elevations at the upstream and downstream manhole locations are listed in Columns 23 and 24. These ground-surface elevations are used as reference elevations to ensure that the cover depth is acceptable and to compare the pipe slope with the ground slope. Step 2: In Column 6, list the land area that will contribute wastewater flow to the sewer line. The contributing land area can be estimated using a topographic map and the proposed development plan. Step 3: In Column 7, list the total area contributing wastewater flow to the sewer pipe. This total contributing area is the sum of the area that contributes directly to the pipe (listed in Column 6) and the area that contributes flow to the upstream pipes that feed the sewer pipe. Step 4: In Column 8, the contribution of infiltration and inflow (I/I) to the pipe flow is listed. I/I is usually calculated by multiplying the length of the pipe by a design inflow rate in m3 /d/km and adding this inflow to the I/I contribution from all upstreamconnected pipes. Step 5: In Column 9, the maximum sewage flow rate is calculated by multiplying the contributing area listed in Column 7 by the average wastewater flow rate at the end of the design period (usually given in m3 /d/ha) and the peaking factor (derived using a relation similar to Equation 6.5). Step 6: In Column 10, the peak design flow rate is calculated as the sum of I/I (Column 8) and the peak wastewater flow rate (Column 9). Step 7: In Columns 11, 12, and 13, the minimum design flow rate is calculated using a similar procedure to that used in calculating the peak design flow rate. The I/I contribution (Column 11) is typically the same as calculated in Step 5 (Column 8); the minimum wastewater flow rate (Column 12) is the average wastewater flow rate when the sewer system first becomes operational multiplied by the peaking factor (derived using a relation similar to Equation 6.5); and the minimum design flow rate (Column 13) is the sum of I/I (Column 11) and the minimum wastewater flow rate (Column 12). Step 8: For a given pipe diameter, the slope of the sewer (Column 14) is equal to the steeper of the ground slope and the minimum slope required for self-cleansing. The minimum self-cleansing slope can be estimated from the minimum design flow rate (in Column 13) using Table 6.8. Once the slope is determined, the flow depth and velocity at the maximum flow rate (Column 10) are calculated. If the flow depth under maximum-flow conditions exceeds the acceptable limit (e.g., 0.75D), or the maximum-flow velocity exceeds an acceptable limit (3.5 m/s), then either the pipe

Manhole no.

Line no. Location From (2) (3) (1)

Area

Length Increment Total (ha) (m) To (ha) (6) (5) (7) (4)

Maximum flow

I/I (L/s) (8)

Sewage (L/s) (9)

Total (L/s) (10)

Sewer invert Ground surface elevation elevation

Minimum flow

I/I (L/s) (11)

Sewage (L/s) (12)

Total (L/s) (13)

Min Manhole tractive Max Max Slope invert Fall in of Diam force depth velocity drop sewer sewer (mm) (Pa) (mm) (m/s) (m) (m) (18) (14) (15) (16) (17) (20) (19)

FIGURE 6.7: Typical spreadsheet for design of sanitary sewers

Upper end (m) (21)

Lower Upper Lower end end end (m) (m) (m) (22) (23) (24)

259

260

Chapter 6

Design of Sanitary Sewers

Step 9:

Step 10: Step 11:

Step 12:

Step 13:

diameter is increased to the next commercial size (e.g., see Appendix E.3) or the pipe slope (Column 14) is increased until self-cleansing, maximum flow depth, and maximum-velocity criteria are all met. After this iteration, the diameter of the sewer pipe is listed in Column 15, the minimum tractive force (= γ Rmin S0 ) in Column 16, the maximum flow depth in Column 17, and the maximum velocity in Column 18. The minimum tractive force in Column 16 should generally be less than that corresponding to the design particle size, except in cases where a minimum relative flow depth is used as an overriding criterion; in such cases the pipe segment should be flagged for manual cleaning, and the minimum flow rate required for selfcleansing should also be noted. In cul-de-sacs and other dead-end street sections, it is often impossible to satisfy the self-cleansing requirement. Such pipes will usually require periodic flushing to scour accumulated sediments. The minimum practicable slope for construction is usually around 0.08%. When good estimates of the maximum design flow rate are available, it is recommended that the required drop at each manhole be calculated as described in Section 6.3.6. In cases where manhole losses cannot be estimated accurately, the following guidelines are sometimes followed: (1) If the inflow and outflow pipes are in different directions, the pipe invert is dropped by 30 mm (1.2 in.) to compensate for the energy losses; and (2) If the diameter of the sewer pipe leaving a manhole is larger than the diameter of the sewer pipe entering the manhole, head losses are accounted for by matching the crown elevations of the entering and leaving pipes. Using arbitrary fixed drops at each manhole is not recommended. In the calculation spreadsheet, the drop in the sewer invert is listed in Column 19. The fall in the sewer line is equal to the product of the slope (Column 14) and the length of the sewer (Column 5) and is listed in Column 20. The invert elevations of the upper and lower ends of the sewer line are listed in Columns 21 and 22. The difference in these elevations is equal to the fall in the sewer calculated in Column 20, and the invert elevation at the upper end of the sewer differs from the invert elevation at the lower end of the upstream pipe by the manhole invert drop listed in Column 19. Repeat Steps 2 to 11 for all connected pipes, proceeding downstream until the sewer main is reached. Then repeat Steps 2 to 11 for the sewer main until the connection to the next lateral is reached. Repeat Steps 2 to 12, beginning with the outermost pipe in the next sewer lateral. Continue designing the sewer main until the outlet point of the sewer system is reached.

The design procedure described here is usually automated to some degree and is most easily implemented using a spreadsheet program. The design procedure is illustrated by the following example.

EXAMPLE 6.9 A sewer system is to be designed to service the development of multistory apartments shown in Figure 6.8. The average per-capita wastewater flow rate is estimated to be 300 L/d/person, and the infiltration and inflow (I/I) is estimated to be 70 m3 /d/km. The sewer system is to join an existing main sewer at manhole (MH) 5, where the average wastewater flow rate at the end of the design period is 0.370 m3 /s. The I/I contribution to the flow in the main sewer at MH 5 is negligible. The existing sewer main at MH 5 is 1065 mm in diameter, has an invert elevation of 55.35 m, and is laid on a slope of 0.9%. The layout of the proposed sewer system shown in Figure 6.8 is based on the topography of the area. Pipe lengths, contributing areas, and ground-surface elevations are given in the following table:

Section 6.5 FIGURE 6.8: Residential sewer project

64.0 m

Main sewer (existing) 3

5

7

9

8

10

60.0 m

16

15

17 Q Avenue

14

21

B Street

11

19

18

C Street

23 24

26

25

D Street

Main street

22

A Street

4

12

6

P Avenue

62.0 m

62.0 m

2

20

261

60.0 m

1

13

Design Computations

Manhole no. From To (3) (4)

Length (m) (5)

Contributing area (ha) (6)

Ground-surface elevation Upper end Lower end (m) (m) (7) (8)

Line no. (1)

Location (2)

0

Main Street



5







60.04

1 2 3 4 5

A Street A Street A Street A Street Main Street

1 2 3 4 5

2 3 5 5 12

53 91 100 89 69

0.47 0.50 0.44 0.90 0.17

65.00 63.80 62.40 61.88 60.04

63.80 62.40 60.04 60.04 60.04

6 7 8 9 10 11 12

B Street P Avenue B Street Q Avenue B Street B Street Main Street

6 7 8 9 10 11 12

8 8 10 10 12 12 19

58 50 91 56 97 125 75

0.43 0.48 0.39 0.88 0.45 0.90 0.28

65.08 63.60 63.20 62.72 62.04 61.88 60.04

63.20 63.20 62.04 62.04 60.04 60.04 60.20

13 14 15 16 17 18 19

C Street P Avenue C Street Q Avenue C Street C Street Main Street

13 14 15 16 17 18 19

15 15 17 17 19 19 26

57 53 97 63 100 138 78

0.60 0.76 0.51 0.94 0.46 1.41 0.30

64.40 63.24 62.84 62.12 61.60 61.92 60.20

62.84 62.84 61.60 61.60 60.20 60.20 60.08

When the system is first installed, it is estimated that the average flow rates will be 30% of the average flows when the area is fully developed. Design the sewer system along A Street and the first extension segment of the sewer main between manholes 5 and 12. The saturation density of the area being served is 347 persons/ha. Local municipal guidelines require that the sewer pipes have a minimum cover of 2 m, a minimum slope of 0.08%, a minimum allowable pipe diameter of 150 mm, and be designed for self-cleansing based on a 1.5-mm particle with a specific gravity of 2.7.

262

Chapter 6

Design of Sanitary Sewers Solution Preliminary calculations and specifications: From the given data, the average wastewater flow rates at the beginning and end of design period, Qavg2 and Qavg2 , respectively, and the I/I flow can be expressed in convenient units as follows: Qavg2 = 300 L/d/person * 347 persons/ha = 104,000 L/d/ha = 1.20 L/s/ha Qavg1 = 0.30Qavg2 = 0.30(1.20) = 0.360 L/s/ha QI/I2 = 70 m3 /d/km = 8.10 * 10−4 L/s/m QI/I1 = QI/I2 = 8.10 * 10−4 L/s/m Since the design particle diameter is 1.5 mm, the design boundary shear stress (i.e., tractive force), τc , is given by Equation 6.28 as τc = 0.867d0.277 = 0.867(1.5)0.277 = 0.97 Pa Therefore, sewers in which the boundary shear stress is greater than or equal to 0.97 Pa under minimum-flow conditions will be taken as self-cleansing. In accordance with conventional practice, the sewer system will be designed to meet the following additional constraints: Vmax = 3.5 m/s h … 0.75 D Hydraulics of existing sewer: The results of the design computations are shown in Figure 6.9. The computations begin with Line 0, which is the existing sewer main that must be extended to accommodate the sewer lines in the proposed residential development. The average flow rate in the sewer main at the end of the design period is 0.370 m3 /s = 370 L/s, and the average flow rate at the beginning of the design period is 0.3(0.370) = 0.111 m3 /s = 111 L/s. Using Equation 6.5, the peaking factors and corresponding flow rates are given by −0.095 = 2.07 PFmax = 1.88Q−0.095 avg2 = 1.88(0.370) −0.095 = 2.32 PFmin = 1.88Q−0.095 avg1 = 1.88(0.111)

Qmax = PFmax Qavg2 = (2.07)(0.370) = 0.766 m3 /s = 766 L/s Qmin = PFmin Qavg1 = (2.32)(0.111) = 0.258 m3 /s = 258 L/s Hence, the maximum flow rate is 766 L/s (Column 10) and the minimum flow rate is 258 L/s (Column 13). The n value for a 1065-mm concrete pipe in typical condition is 0.0120 (ASCE, 2007). With a slope of 0.009 (Column 14) and a diameter of 1065 mm (Column 15), the depth of flow at the maximum flow rate is 373 mm (Column 17) and the corresponding maximum velocity is 2.76 m/s (Column 18). Under minimum-flow conditions, the depth of flow is 214 mm with a corresponding boundary shear stress of 11.4 Pa (Column 16). The invert elevation of the main sewer at MH 5 is 55.35 m (Column 22) and the ground-surface elevation at MH 5 is 60.04 m (Column 24). Sewer Line 1: The design of the sewer system begins with Line 1 on A Street, which goes from MH 1 to MH 2 and is 53 m long. The area contributing wastewater flow is 0.47 ha (Column 7). The maximum and minimum wastewater flow rates are calculated as follows:

Manhole no.

Line no. Location From (1) (2) (3) 0 Main Street − 1 2 3 4 5

A Street A Street A Street A Street Main Street

1 2 3 4 5

Area

Maximum flow

Sewer invert Ground surface elevation elevation

Minimum flow

Length Increment Total (ha) (ha) To (m) (4) (5) (6) (7) − − − 5

I/I (L/s) (8) −

Sewage (L/s) (9) −

Total (L/s) (10) 766

I/I (L/s) (11) −

Sewage (L/s) (12) −

Total (L/s) (13) 258

Slope of sewer (14) 0.009

2 3 5 5 12

0.043 0.117 0.200 0.072 0.328

4.27 6.40 7.90 6.13 770

4.31 6.25 8.10 6.22 770

0.043 0.117 0.200 0.072 0.328

2.18 3.27 4.02 3.13 258

2.22 3.38 4.22 3.20 258

0.0226 0.0154 0.0236 0.0207 0.0008

53 91 100 89 69

0.47 0.50 0.44 0.90 0.17

0.47 0.97 1.41 0.90 309.96

FIGURE 6.9: Sewer design calculations

Diam (mm) (15) 1065 150 150 150 150 1065

Manhole Min tractive Max Max invert force depth velocity drop (m) (Pa) (mm) (m/s) (16) (17) (18) (19) − 11.4 373 2.76 1.53 3.40 5.24 4.19 1.07

40 54 54 49 780

1.15 1.13 1.4 1.23 0.23

− − − − 0.03

Fall in Upper sewer end (m) (m) (20) (21) − − 1.20 1.40 2.36 1.84 0.06

62.85 61.65 60.25 59.73 55.35

Lower Upper Lower end end end (m) (m) (m) (22) (23) (24) − 55.35 60.04 61.65 60.25 57.89 57.89 55.27

65.00 63.80 62.40 61.88 60.04

63.80 62.40 60.04 60.04 60.04

263

264

Chapter 6

Design of Sanitary Sewers QI/I2 = 8.10 * 10−4 L/s/m * 53 m = 0.043 L/s = 0.000043 m3 /s QI/I1 = QI/I2 = 0.000043 m3 /s Qavg2 = 1.20 L/s/ha * 0.47 ha = 0.565 L/s = 0.000565 m3 /s Qavg1 = 0.3Qavg2 = 0.3(0.000565) = 0.000170 m3 /s −0.44 = 7.55 PFmax = 0.281Q−0.44 avg2 = 0.281(0.000565) −0.44 = 12.8 PFmin = 0.281Q−0.44 avg1 = 0.281(0.000170)

Qmax,sewage = PFmax Qavg2 = (7.55)(0.000565) = 0.00427 m3 /s = 4.27 L/s Qmin,sewage = PFmin Qavg1 = (12.8)(0.000170) = 0.00218 m3 /s = 2.18 L/s Qmax = Qmax,sewage + QI/I2 = 4.27 + 0.043 = 4.31 L/s Qmin = Qmin,sewage + QI/I1 = 2.18 + 0.043 = 2.22 L/s

Hence, the maximum flow rate is 4.31 L/s (Column 10) and the minimum flow rate is 2.22 L/s (Column 13). The n value for a 150-mm concrete pipe in typical condition is 0.0106. The minimum slope for self-cleansing, Smin calculated from the appropriate equation in Table 6.8 along with the ground slope, Sground , is as follows: Smin = 0.00624Q−0.5697 = 0.00624(2.22)−0.5697 = 0.00396 min Sground =

65.00 − 63.80 = 0.0226 53

Hence, to maintain a minimum cover of 2 m, specify the pipe slope, S0 , to equal the ground slope (= 0.0226). With a pipe slope of 0.0226 (Column 14) and a diameter of 150 mm (Column 15), the depth of flow at the maximum flow rate is 40 mm (Column 17) and the corresponding maximum velocity is 1.15 m/s (Column 18). Under minimum-flow conditions, the depth of flow is 28 mm with a corresponding boundary shear stress of 3.82 Pa (Column 16). The pipe has adequate capacity, is self-cleansing, and meets all design constraints. The drop in the sewer, z, is given by z = LS0 = (53)(0.0226) = 1.20 m The sewer invert at the upper end to have a 2.00-m cover is 65.00 m − 2.00 m − 0.150 m = 62.85 m (Column 21), and the sewer invert at the lower end is 62.85 m − 1.20 m = 61.65 m. Sewer Lines 2 to 4: The design of Lines 2 and 3 follows the same sequence as for Line 1, with the exception that the wastewater flows in each pipe are derived from the sum of the contributing areas of all upstream pipes plus the pipe being designed, and that the I/I flow in each pipe is the sum of all upstream I/I flows plus the I/I contribution to the pipe being designed. Using this approach, the invert elevation at the end of Lines 3 and 4 is 57.89 m (Column 22), where the sewer laterals join the main sewer. The invert elevation of the sewer main is 55.35 m, which is 57.89 m − 55.35 m = 3.54 m below the invert of the laterals. A special drop-manhole structure will be required at this intersection. Sewer Line 5: The main sewer leaving MH 5 (Line 5) is designed next. The tributary area to Line 5 is the sum of the contributing areas of all contributing sewer laterals (Lines 1 to 4, 1.41 + 0.9 = 2.31 ha) plus the equivalent area of the average flow in the main sewer upstream of MH 5 (0.37 m3 /s ÷ 1.20 L/s/ha = 307.48 ha) plus the area that contributes directly to Line 5 (0.17 ha). Hence the total contributing area is 2.31 + 307.48 + 0.17 = 309.96 ha (Column 7). The I/I

Problems

265

contribution to Line 5 is the sum of the I/I contributions to all upstream laterals (0.200 + 0.072 = 0.272 L/s) plus the I/I contribution directly to Line 5 (69 m * 8.10 * 10−4 L/s/m = 0.056 L/s) for a total I/I contribution of 0.272 + 0.056 = 0.328 L/s (Column 8). The maximum and minimum flow rates are calculated using the peaking flow factors as previously described. Since the required minimum slope is less than the practical slope of 0.08%, a slope of 0.08% is used. A manhole drop at the end of the pipe of 0.03 m is used to account for energy losses at the manhole, where laterals intersect.

Problems 6.1. Estimate the maximum and minimum design wastewater flow rates from a 65-ha residential development that when fully developed will consist of 10% large lots (6 persons/ha), 75% small single-family lots (75 persons/ha), and 15% small two-family lots (125 persons/ha). The average wastewater flow rate when the sewers are first installed is expected to be 30% of the average wastewater flow rate when the area is fully developed. Assume an average per-capita flow rate of 350 L/d/person. 6.2. A fully developed 45-km2 city will have land uses that are 65% residential, 25% commercial, and 10% industrial. The residential development will be 15% large lots (6 persons/ha), 75% small single-family lots (75 persons/ha), and 10% multistory apartments (2500 persons/ha). The average domestic wastewater flow rate can be taken as 500 L/d/person, the average commercial flow rate as 50,000 L/d/ha, and the average industrial flow rate as 90,000 L/d/ha. When the sewer is first installed, the average wastewater flow rate will be 35% of the average flow rate expected when the area is fully developed. Infiltration and inflow is estimated as 1500 L/d/ha for the entire area. Estimate the maximum and minimum flow rates to be handled by the main sewer. 6.3. A sewer pipe has a diameter of 760 mm and flows threefourths full when the flow rate is 260 L/s. Estimate the average velocity of flow under this condition. 6.4. Show that the Manning equation can be written in the form given by Equation 6.17. 6.5. Water flows at a rate of 7 m3 /s in a circular concrete sewer of diameter 1600 mm. If the slope of the sewer is 0.01, estimate the depth of flow and velocity in the sewer. What diameter of pipe would be required for the pipe to flow three-quarters full? 6.6. A circular sanitary sewer is designed to carry the maximum flow of sewage while flowing 70% full at a velocity of 0.9 m/s. If the ratio of maximum to average and average to minimum flows are 2.5 and 2.0 respectively, estimate the proportionate depth of flow. 6.7. Water flows at 3.5 m3 /s in a 1400-mm-diameter concrete sewer in which Manning’s n is 0.015. Find the slope at which the sewer flows half full. 6.8. A sewer 450 mm in diameter with an invert slope of 1 in 350 is flowing at two-thirds of the full depth. Calculate

the velocity and the rate of flow in the sewer if n equals 0.013. Is the velocity self-cleansing? 6.9. Show that the flow velocity in a circular pipe is the same whether the pipe is flowing half full or completely full. 6.10. A concrete sanitary sewer is to be designed to flow onehalf full when the flow rate is 30 L/s. If the sewer is on a slope of 0.005 and the Manning roughness is 0.015, what commercial size of pipe is required? Compare your results using the Manning and Darcy–Weisbach equations, taking into consideration that for fully turbulent flow Manning’s n and the equivalent sand roughness, ks , are approximately related by 1

n = 0.039ks6 Determine whether the same Manning’s n should be used for the pipe flowing full and half full. 6.11. Determine the relative depth of flow in a partially full circular pipe that yields to the same flow rate as when the pipe is flowing full. 6.12. A sanitary sewer is to serve a uniformly distributed population of 5,000 along a 1 km road. The average ground slope for the first 500 m is 1 in 350 and 1 in 1,000 for the remaining. Design the sewer. 6.13. A sewer line is laid to serve a community of 250 persons/ha in an area of 50 ha. The average water supply is 250 L/d/person. The available ground slope is 1 in 1,000. Estimate the diameter of a sewer to carry the peak discharge when flowing half-full. Check the velocity for self-cleansing. Use Manning’s formula (n equals 0.013). 6.14. The main sewer of a town was designed to serve an area of 40 km2 . The town has an average population of 200 persons/ha. The average rate of sewage flow is 270 L/d/person. The maximum flow of run-off is 50% in excess of the average, which is the same as the rainfall equivalent of 20 mm in 24 hours. What should the capacity of the town sewer be? 6.15. The design maximum and minimum flow rates in a concrete sewer pipe are 0.60 m3 /s and 0.030 m3 /s, respectively. The roughness height of the concrete pipe is 1.5 mm, the critical shear stress for self-cleansing is

266

6.16.

6.17.

6.18.

6.19.

6.20.

Chapter 6

Design of Sanitary Sewers

2.0 Pa, and the sewage temperature as 20◦ C. The pipe must flow no more than 75% full under maximumflow conditions, and the maximum velocity must be less than 4.0 m/s. Any pipe slope greater than 0.1% is feasible. Determine the minimum allowable pipe slope for a 915-mm-diameter pipe. A sanitary-sewer system is to be designed for a newly developed 1700-ha area that is expected to have a population of 10,000 people when the sewers are first installed and a population of 50,000 people when the area is fully developed. Local regulations require that the sewer system be designed for an average per-capita flow rate of 250 L/d/person, accommodate an infiltration and inflow of 2 m3 /d/ha, not flow more than 75% full, be selfcleansing for 1.5-mm particles, have a minimum diameter of 150 mm, and have a maximum velocity less than 3 m/s. Field surveys indicate that the ground slope is 1% along the planned route of the sewer system. If the main pipe segment draining the area is to be made of concrete and have the same slope as the ground, determine the required (commercial-size) pipe diameter. A town has a population of 5,000 and area of 50 ha. The average runoff coefficient for this area is 0.65. The inlet time and the time of flow in the sewer are 10 and 20 minutes respectively. Estimate the discharge for a proposed combined system of the sewers. A 1220-mm-diameter concrete sewer is to carry 0.50 m3 /s of domestic wastewater at a temperature of 23◦ C on a slope of 0.009. Manning’s n has a design value of 0.013. Estimate the maximum 5-day BOD for which hydrogen sulfide generation will not be a problem. Find the minimum value of gradient required to transport coarse sand through a sewer of 200 mm diameter with sand particles of 0.2 mm diameter and specific gravity of 2.65. Assume the Darcy coefficient of friction as 0.03 and noticeable quality of solids flowing in the sewage in suspension as 0.04. A minimum wastewater flow rate of 0.15 m3 /s is to be transported in a 915-mm-diameter concrete pipe. If the design particle size is 1.5 mm, estimate the minimum slope of the pipe to assure self-cleansing. If the maximum flow rate is 0.29 m3 /s, and regulatory requirements are that the pipe not flow more than 75% full and have a maximum velocity less than 3.5 m/s, determine the minimum allowable slope to achieve all design objectives. Assuming the pipe is laid on the selected slope and carries an average flow of 0.25 m3 /s, assess the potential for hydrogen sulfide generation if the temperature of the wastewater is 25◦ C and the 5-day BOD is 250 mg/L at 20◦ C.

6.21. A previously designed sewer line is made of 535-mmdiameter reinforced concrete pipe and enters a manhole with a crown elevation of 2.50 m. The maximum design flow rate in the sewer line is 0.30 m3 /s, and the minimum design flow rate is 0.07 m3 /s. The downstream sewer line is to have a maximum allowable flow depth equal to 75% of the pipe diameter, a minimum cover of 2.00 m, and a design particle size of 1.5 mm. The ground-surface elevation upstream of the new sewer line is 5.00 m, the ground-surface elevation at the manhole downstream of the new sewer line is 4.50 m, and the new sewer line is 150 m long. Determine: (a) the diameter and slope of the new sewer line, and (b) the crown and invert elevations at the upstream and downstream ends of the new sewer line. 6.22. A separate sewerage system has to be designed for a city having an area of 20 ha with a population of 10,000 for a rainfall frequency of 2 years. The average water supply for the city is 250 L/d/person. The maximum onehour rainfall having 2 years frequency is 40 mm. Using a runoff coefficient of 0.55 and the dispersion factor for an area of 20 ha equaling 1, compute the maximum storm drainage discharge and also compute the peak design discharge for which the sewers of this city will be designed. The time of concentration is 20 minutes. 6.23. A sewer system is to be designed to service the area shown in Figure 6.8. The average per-capita wastewater flow rate is estimated to be 250 L/d/person, and the infiltration and inflow (I/I) is estimated to be 100 m3 /d/km. The sewer system is to join an existing main sewer at manhole (MH) 5, where the average wastewater flow rate at the end of the design period is 0.40 m3 /s. The I/I contribution to the flow in the main sewer at MH 5 is negligible, and the main sewer at MH 5 is 1220 mm in diameter, has an invert elevation of 55.00 m, and is laid on a slope of 0.8%. The layout of the sewer system shown in Figure 6.8 is based on the topography of the area, and the pipe lengths, contributing areas, and ground-surface elevations are shown in Table 6.10. It is estimated that when the new sewers are first installed the average flows will be 20% of the average flows when the area is fully developed. Design the sewer system along A Street and the first extension segment of the sewer main between manholes 5 and 12. The saturation density of the area being served is 600 persons/ha. Local municipal guidelines require that the sewer pipes have a minimum cover of 1 m, a minimum slope of 0.08%, a minimum allowable pipe diameter of 150 mm, and be designed for self-cleansing based on a 1.0-mm particle with a specific gravity of 2.7.

Problems TABLE 6.10 Ground-surface elevation Manhole no. From To (3) (4)

Length (m) (5)

Contributing area (ha) (6)

Upper end (m) (7)

Lower end (m) (8)

Line no. (1)

Location (2)

0

Main Street



5







60.04

1 2 3 4 5

A Street A Street A Street A Street Main Street

1 2 3 4 5

2 3 5 5 12

55 90 100 90 70

0.47 0.50 0.44 0.90 0.17

65.00 63.80 62.40 61.88 60.04

63.80 62.40 60.04 60.04 60.04

267

C H A P T E R

7

Design of Hydraulic Structures 7.1 Introduction The adequate design of hydraulic structures is usually essential to ensuring that waterresource systems function as intended. Most water-resource systems are designed with primary objectives related to flood control and/or water supply, and the common types of structures used in these systems are covered in this chapter. Culverts are used to pass small drainage channels under roadways, gates are used to control the flow of water, weirs are used as discharge structures in stormwater-management systems and also for flow measurement, spillways are used to discharge excess water from storage reservoirs, stilling basins are used to dissipate energy downstream of spillways, and dams are used to regulate the flow of water and to support the generation of hydroelectric power. 7.2 Culverts Culverts are short conduits that are designed to pass peak flood discharges under roadways or other embankments. Because of the function they perform, culverts are commonly included in a class of structures called cross-drainage structures, which also include bridges. Culverts perform a similar function to that of bridges, but, unlike bridges, they have small spans that are typically less than 6 m (20 ft) and can be designed to have a submerged inlet. Typical cross sections of culverts include circular, arched, rectangular, and oval shapes. Culverts can have either a single barrel or multiple barrels. A typical single-barrel circular culvert is shown in Figure 7.1(a), and a multibarrel rectangular culvert is shown in Figure 7.1(b); rectangular culverts are sometimes called box culverts. Culvert pipes with bottoms buried in the flow way are called embedded culverts or buried-invert culverts, and culverts without below-ground bottoms are called or open-bottom culverts or bottomless culverts. Embedded or open-bottom culverts are commonly used to facilitate the passage of fish and other aquatic organisms since they do not produce high velocities at low flow rates as in the case of circular (non-embedded) culverts, and the natural streambed is maintained. Embedded and open-bottom culverts provide a natural channel bottom that might be preferable in channels with high sediment transport, particularly with coarse materials (gravels and cobbles) where abrasion caused by moving sediment can destroy the culvert invert (USFHWA, 2012). Multibarrel culverts are frequently necessary to pass wide shallow streams under roadways. Culvert design typically requires the calculation of the dimensions of a barrel cross section that passes a given flow rate when the water is ponded to a maximum-allowable height at the culvert entrance. 7.2.1

Hydraulics

In analyzing culvert flows, the normal and critical depths are both useful reference depths. The normal depth of flow is determined by solving the Manning equation, and the critical depth of flow is determined by setting the Froude number equal to unity (i.e., Fr = 1). In cases where the cross section is circular, it is convenient to put the Manning equation in the form  5 θ − sin θ 3 nQ = 20.16 8 √ (7.1) 2 θ3 D 3 S0 where n is the Manning roughness [dimensionless], Q is the flow rate [L3 T−1 ], S0 is the slope of the culvert [dimensionless], D is the culvert diameter [L], and θ [radians] is the central angle that is related to the depth of flow, y [L], by 268

Section 7.2

Culverts

269

FIGURE 7.1: Typical culverts: (a) single-barrel circular; (b) multibarrel rectangular (box)

(b)

(a)

   D θ y= 1 − cos 2 2

(7.2)

The critical depth can be conveniently obtained by solution of the following (Fr = 1) relation 

θ − sin θ  sin θ2

3 = 512

Q2 gD5

(7.3)

where g is gravity [LT−2 ]. For given values of Q, n, S0 , and D, both Equations 7.1 and 7.3 can be independently solved for θ , and these two values of θ are used in Equation 7.2 to find the normal depth (yn ) and the critical depth (yc ), respectively. The solution of Equations 7.3 and 7.2 for yc requires numerical solution since Equation 7.3 is implicit in θ . An approximate explicit estimate of yc [m] is (French, 1985)  yc =

1.01 D0.26



Q2 g

0.25 (7.4)

for 0.2D … yc … 0.8D, where D is in m, Q is in m3 /s, and g is 9.81 m/s2 . The exact solution for the critical depth given by Equations 7.3 and 7.2 is generally preferable to using Equation 7.4. EXAMPLE 7.1 A 760-mm-diameter concrete culvert has a design flow rate of 1.0 m3 /s and is to be laid on a slope of 0.8%. Determine the normal and critical depths of flow. Solution From the given data: D = 0.760 m, Q = 1.0 m3 /s, and S0 = 0.008. Assume that n = 0.013 for concrete pipe. Substituting in Equation 7.1 gives  5 θ − sin θ 3 θ

2 3

= 20.16

nQ

8 D3 S

= 20.16 0

(0.013)(1.0) = 6.091 8√ (0.760) 3 0.008

which yields θ = 4.396 radians. Substituting θ into Equation 7.2 gives       D θ 0.760 4.396 y= 1 − cos = 1 − cos = 0.603 m 2 2 2 2

270

Chapter 7

Design of Hydraulic Structures Therefore, the normal depth of flow is 0.603 m. Substituting the given data into Equation 7.3 gives  3 θ − sin θ (1.0)2 Q2  = 512 = 205.8 = 512 gD5 (9.81)(0.760)5 sin θ2 which yields θ = 4.476 radians. Substituting θ into Equation 7.2 gives       θ 4.476 D 0.760 y= 1 − cos = 1 − cos = 0.615 m 2 2 2 2 Therefore, the critical depth of flow is 0.615 m. An approximation of the critical depth can be obtained by using Equation 7.4, which gives  0.25   2 0.25 

1.01 (1.0)2 Q 1.01 = = 0.613 m yc = g 9.81 D0.26 (0.760)0.26 In this case, the approximate value of the critical depth (0.613 m) is very close to the exact value (0.615 m).

The hydraulic analysis of flow in culverts is complicated by the fact that there are several possible flow regimes, with the governing flow equation being dependent on the flow regime. Culvert flow regimes can be classified into six types, depending primarily on the headwater and tailwater elevations relative to the crown of the culvert, and whether the slope is hydraulically mild or steep. These six flow regimes can be grouped into either submerged-entrance conditions or free-entrance conditions, illustrated in Figures 7.2 and 7.3, respectively. The entrance to a culvert is regarded as submerged when the depth, H, of water upstream of the culvert exceeds 1.2D, where D is the diameter or height of the culvert. Some engineers take this limit as 1.5D (e.g., French, 1985; Sturm, 2010); however, the water surface will impinge on the headwall when the headwater depth is about 1.2D if the critical depth, yc , in the culvert occurs at the culvert entrance and yc Ú 0.8D (Finnemore and Franzini, 2002). There is some experimental evidence indicating the transition is at H = D (e.g., Haderlie et al., 2008). Fundamentally, the minimum headwater depth to cause inlet submergence depends on the shape of the culvert entrance (Jain, 2001). The depth of water at the culvert entrance is called the headwater depth, and the depth of water at the culvert exit is the tailwater depth. 7.2.1.1

Submerged entrances

In the case of a submerged entrance, Figure 7.2 shows three possible flow regimes: (a) The outlet is submerged (Type 1 flow); (b) the outlet is not submerged and the normal depth of flow in the culvert is larger than the culvert height, D (Type 2 flow); and (c) the outlet is not submerged and the normal depth of flow in the culvert is less than the culvert height (Type 3 flow). Types 1 and 2 flow. In Type 1 flow, applying the energy equation between Sections 1 (headwater) and 3 (tailwater) leads to h = hi + hf + ho

(7.5)

where h is the difference between the headwater and tailwater elevations, hi is the entrance loss, hf is the head loss due to friction in the culvert, and ho is the exit loss. Equation 7.5 neglects the velocity heads at Sections 1 and 3, which are usually small compared with the other terms, and also neglects the friction losses between Section 1 and the culvert entrance, and the friction loss between the culvert exit and Section 3. Using the Manning equation to calculate hf [m] within the culvert, then hf =

n2 V 2 L 4

R3

(7.6)

Section 7.2 FIGURE 7.2: Flow through culvert with submerged entrance

HW

TW

Culverts

271

Δh

H V

D

(a) Submerged outlet (Type 1)

1

3

HW Δh TW

H

yn

V

L

1

3 2

(b) Normal depth > Barrel height (Type 2)

HW

h H TW

(c) Entrance control, Normal depth < Barrel height (Type 3)

where n is the Manning roughness coefficient [dimensionless], V is the velocity of flow [m/s], L is the length [m], and R is the hydraulic radius of the culvert [m]. Head loss due to friction within the culvert is usually minor, except in long rough barrels on flat slopes (Bodhaine, 1976). The entrance loss, hi , is given by hi = ke

V2 2g

(7.7)

where ke is the entrance loss coefficient. Entrance losses are caused by the sudden contraction and subsequent expansion of the stream within the culvert barrel, and the entrance geometry has an important influence on this loss. The exit loss, ho , can be expressed in the form

Vd2 V2 − αd ho = ko α (7.8) 2g 2g where ko is the exit loss coefficient, α is the energy coefficient inside the culvert at the exit, αd is the energy coefficient at the cross section just downstream of the culvert, and Vd is the average velocity at the cross section just downstream of the culvert. For a sudden expansion of flow, such as in a typical culvert, the exit loss coefficient, ko , is normally set to 1.0, while in general exit loss coefficients can vary between 0.3 and 1.0. The exit loss coefficient should

272

Chapter 7

Design of Hydraulic Structures

FIGURE 7.3: Flow through culvert with free entrance

HW Δh H

yn

yc

(a) Mild slope, Low tailwater (Type 4)

1

HW Δh H

yc

yn

1.4yc

1

(b) Steep slope, Low tailwater (Type 5)

HW Δh H yc

1

(c) Mild slope, Tailwater submerges yc (Type 6)

be reduced as the transition becomes less abrupt. It is commonplace for culverts to exit into large open bodies of water, in which case it is assumed that ko = 1.0, α = αd = 1.0, and Vd = 0, which yields the commonly used relation ho =

V2 2g

(7.9)

In cases where a culvert discharges into a downstream channel and the flow rate in the downstream channel is equal to the culvert discharge, such as in driveway cross-drains and fish-passage culverts, experiments (Tullis et al., 2008) have indicated that exit losses in such culverts are more accurately estimated using the relation   Ap 2 V2 ho = ko where ko = 1 − (7.10) 2g Ac where Ap and Ac are the area of the culvert pipe and downstream channel [L2 ], respectively. Combining Equations 7.5 to 7.7 with Equation 7.9 yields the following form of the energy equation between Sections 1 and 3: h =

n2 V 2 L R

4 3

+ ke

V2 V2 + 2g 2g

(7.11)

Equation 7.11 can also be applied between Section 1 (headwater) and Section 2 (culvert exit) in Type 2 flow, illustrated in Figure 7.2(b), where the velocity head at the exit, V 2 /2g, is equal

Section 7.2

Culverts

273

to the exit loss in Type 1 flow. Equation 7.11 reduces to the following relationship between the difference in the water-surface elevations on both sides of the culvert, h, and the discharge through the culvert, Q:   2gh Q = A  2  2gn L + ke + 1 4 R3

(7.12)

where A is the cross-sectional area of the culvert. It should be noted that h is equal to the difference between the headwater and tailwater elevations for a submerged outlet (Type 1 flow), and h is equal to the difference between the headwater and the crown of the culvert exit for an unsubmerged outlet when the normal depth of flow in the culvert exceeds the height of the culvert (Type 2 flow). If the Darcy–Weisbach equation is used to calculate the head loss in the culvert, then Equation 7.12 takes the form   2gh  Q = A  fL + ke + 1 4R

(7.13)

where f is the Darcy–Weisbach friction factor. Under both Type 1 and Type 2 conditions, the flow is said to be under outlet control, since either the water depth at the outlet or the elevation of the crown of the culvert at the outlet influences the discharge through the culvert. Equation 7.12 is the basis for the culvert-design nomographs developed by the U.S. Federal Highway Administration (2012); however, it is seldom justified to use these nomographs in lieu of Equation 7.12. Type 3 flow. In Type 3 flow, the inlet is submerged and the culvert entrance will not admit water fast enough to fill the culvert. In other words, the culvert capacity is limited by the inlet capacity. In this case, the culvert inlet behaves like an orifice and the discharge through the culvert, Q, is related to the head on the center of the orifice, h, by the relation Q = Cd A 2gh

(7.14)

where Cd is the coefficient of discharge and h is equal to the vertical distance from the centroid of the culvert entrance to the water surface at the entrance. The coefficient of discharge, Cd , is frequently taken as 0.62 for square-edged entrances and 1.0 for well-rounded entrances. However, original data from USGS (Bodhaine, 1976) indicate that, for squareedged entrances, Cd depends on H/D, with Cd = 0.44 when H/D = 1.4 and Cd = 0.59 when H/D = 5. Beveling or rounding culvert entrances generally reduces the flow contraction at the inlet and results in higher values of Cd than for square entrances, and a 45◦ bevel is recommended for ease of construction (U.S. Federal Highway Administration, 2012). In cases where the culvert entrance acts like an orifice, downstream conditions do not influence the flow through the culvert and the flow is said to be under inlet control. In general, when the inlet to a structure controls the capacity of the structure to transmit water, the structure is said to be under inlet control. According to ASCE (1992), Equation 7.14 is applicable only when H/D Ú 2, but the error in Equation 7.14 is less than 2% when H/D Ú 1.2 (Finnemore and Franzini, 2002). The occurrence of Type 3 flow requires a relatively square entrance that will cause contraction of the flow area to less than the area of the culvert barrel. In addition, the combination of barrel length, roughness, and bed slope must be such that the contracted flow will not expand to the full area of the barrel. If the water surface of the expanded flow comes into contact with the top of the culvert, Type 2 flow will occur because the passage of air to the culvert will be sealed off, causing the culvert to flow full throughout its length. In such cases, the culvert is called hydraulically long; otherwise, the culvert is hydraulically short. Bodhaine (1976) reported that a culvert has to meet the criterion that L … 10D to allow Type 3 flow

274

Chapter 7

Design of Hydraulic Structures

on a mild slope, and this rule can be used to differentiate between hydraulically short and hydraulically long culverts. Based on experiments conducted by the U.S. National Bureau of Standards, now the National Institute of Science and Technology (NIST), the following nondimensional best-fit power relationship has been developed to facilitate the description of Type 3 flow through culverts: H = 32.2 c Fr2 + Y − 0.5S (7.15) D where H is the headwater specific energy, D is the height of the culvert entrance, and Fr is the Froude number at the culvert entrance defined by Fr =

Q A gD

(7.16)

where Q is the flow rate through the culvert, A is the full cross-sectional area of the culvert, S is the slope of the culvert, and c and Y are empirical constants that depend on the culvert shape, material, and inlet configuration and are given in Table 7.1. The factor of 32.2 in Equation 7.15 is equal to the acceleration due to gravity in U.S. Customary units (ft/s2 ), which is necessary to be able to use the values of c reported by NIST. Equation 7.15 is dimensionally homogeneous and is applicable for Fr ≥ 0.7. Equation 7.15 is predicated on the observation that a critical section (where critical flow occurs) exists in the culvert approximately one-half pipe diameter downstream of the inlet, hence the factor of 0.5 in Equation 7.15 represents the ratio of distance to critical section divided by the diameter, and S is the slope of the culvert between the entrance and the critical section. If the culvert is hydraulically long and the slope is mild, then a hydraulic jump can be expected to occur within the culvert, assuming that the normal flow depth is less than the culvert diameter. When applying Equation 7.15 to mitered inlets, a slope correction factor of +0.7S should be used instead of −0.5S. In specialized cases of embedded or bottomless culverts, more appropriate values of c and Y for use in Equation 7.15 can be found in USFHWA (2012). Of all the equations proposed for describing Type 3 flow, Equation 7.15 is most widely used through hand calculation, nomographs (USFHWA, 2012), or computer models such as HEC-RAS (USACE, 2010) and HY-8 (USFHWA, 2005a). A close approximation to the NIST equation (Equation 7.15) for Type 3 flow conditions can be obtained by applying the energy equation at the entrance to a culvert. Considering a box culvert and neglecting the entrance loss gives H=

V2 + Cc D 2g

(7.17)

where H is the headwater specific energy, V is the velocity within the culvert entrance, Cc is the contraction coefficient associated with flow passing the crown of the culvert entrance, and D is the height of the culvert. Using Equation 7.17 gives the culvert discharge, Q, as (7.18) Q = (Cb B)(Cc D)V = Cb Cc A 2g(H − Cc D) where Cb is a width contraction coefficient associated with the culvert entrance edge conditions, and B is the culvert span (width). Equation 7.18 can be expressed in the nondimensional form 1 H = Fr2 + Cc (7.19) D 2(Cb Cc )2 where Fr is the Froude number at the culvert entrance, defined by Equation 7.16 and A (= B * D) is the full culvert cross-sectional area. Values of Cb and Cc that give the closest agreement between Equation 7.19 and Equation 7.15 have been estimated by Charbeneau et al. (2006) and are given in Table 7.2. In applying Equation 7.19 to circular culverts, D is taken as the diameter of the culvert. The primary advantage to using Equation 7.19 rather

TABLE 7.1: Constants for Empirical Culvert-Design Equations

Shape and material Circular concrete

Circular CMP*

Circular Rectangular box, concrete Rectangular box, concrete

Box CM

Inlet shape Square edge with headwall Groove end with headwall Groove end projecting Headwall Mitered to slope Projecting Beveled ring, 45◦ Beveled ring, 33.7◦ 30◦ –75◦ wingwall flares 30◦ and 75◦ wingwall flares 0◦ wingwall flares 45◦ wingwall flare, w/D = 0.043 18◦ to 33.7◦ wingwall flare, w/D = 0.083 90◦ headwall, 19-mm chamfers 90◦ headwall, 45◦ bevels 90◦ headwall, 33.7◦ bevels 19-mm chamfers, 45◦ skewed headwall 19-mm chamfers, 30◦ skewed headwall 19-mm chamfers, 15◦ skewed headwall 45◦ bevels, 10◦ –45◦ skewed headwall 19-mm chamfers, 45◦ wingwall flare, nonoffset 19-mm chamfers, 18.4◦ wingwall flare, nonoffset 19-mm chamfers, 18.4◦ wingwall flare, nonoffset, 30◦ skew Top bevels, 45◦ wingwall flare, offset Top bevels, 33.7◦ wingwall flare, offset Top bevels, 18.4◦ wingwall flare, offset 90◦ headwall Thick wall projecting Thin wall projecting

c

Y

0.0398 0.0292 0.0317 0.0379 0.0463 0.0553 0.0300 0.0243 0.0347 0.0400 0.0423 0.0309 0.0249 0.0375 0.0314 0.0252 0.04505 0.0425 0.0402 0.0327 0.0339 0.0361 0.0386 0.0302 0.0252 0.0227 0.0379 0.0419 0.0496

0.67 0.74 0.69 0.69 0.75 0.54 0.74 0.83 0.81 0.80 0.82 0.80 0.83 0.79 0.82 0.865 0.73 0.705 0.68 0.75 0.803 0.806 0.71 0.835 0.881 0.887 0.69 0.64 0.57

Form (Type 5) 1

1

1 1

2

1

K or K

M or M

0.0098 0.0018 0.0045 0.0078 0.0210 0.0340 0.0018 0.0018 0.026 0.061 0.061 0.510 0.486 0.515 0.495 0.486 0.545 0.533 0.522 0.498 0.497 0.493 0.495 0.497 0.495 0.493 0.0083 0.0145 0.0340

2.0 2.0 2.0 2.0 1.33 1.5 2.5 2.5 1.0 0.75 0.75 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667 2.0 1.75 1.5

275

276 TABLE 7.1: (Continued)

Shape and material Ellipse concrete

Arch CM†

Circular Elliptical inlet face

Rectangular concrete Rectangular concrete Rectangular concrete

Inlet shape Horizontal ellipse, square edge with headwall Horizontal ellipse, groove end with headwall Horizontal ellipse, groove end projecting Vertical ellipse, square edge with headwall Vertical ellipse, groove end with headwall Vertical ellipse, groove end projecting 46-cm corner radius, 90◦ headwall 46-cm corner radius, mitered to slope 46-cm corner radius, projecting 46-cm corner radius, projecting 46-cm corner radius, no bevels 46-cm corner radius, 33.7◦ bevels 79-cm corner radius, projecting 79-cm corner radius, no bevels 79-cm corner radius, 33.7◦ bevels 90◦ headwall Mitered to slope Thin-wall projecting Smooth-tapered inlet throat Rough-tapered inlet throat Tapered inlet, beveled edges Tapered inlet, square edges Tapered inlet, thin edge projecting Tapered inlet throat Side tapered, less favorable edge Side tapered, more favorable edge Side tapered, less favorable edge Side tapered, more favorable edge

Source: U.S. Federal Highway Administration (2012). Notes: *CMP = corrugated metal pipe; † CM = corrugated metal.

c

Y

0.0398 0.0292 0.0317 0.0398 0.0292 0.0317 0.0379 0.0463 0.0496 0.0496 0.0368 0.0269 0.0496 0.0368 0.0269 0.0379 0.0473 0.0496 0.0196 0.0210 0.0368 0.0478 0.0598 0.0179 0.0446 0.0378 0.0446 0.0378

0.67 0.74 0.69 0.67 0.74 0.69 0.69 0.75 0.57 0.57 0.68 0.77 0.57 0.68 0.77 0.69 0.75 0.57 0.90 0.90 0.83 0.80 0.75 0.97 0.85 0.87 0.65 0.71

Form (Type 5) 1

1

2 2

2 2 2

K or K

M or M

0.0100 0.0018 0.0045 0.010 0.0018 0.0095 0.0083 0.0300 0.0340 0.0300 0.0088 0.0030 0.0300 0.0088 0.0030 0.0083 0.0300 0.0340 0.534 0.519 0.536 0.5035 0.547 0.475 0.56 0.56 0.50 0.50

2.0 2.5 2.0 2.0 2.5 2.0 2.0 1.0 1.5 1.5 2.0 2.0 1.5 2.0 2.0 2.0 1.0 1.5 0.555 0.64 0.622 0.719 0.80 0.667 0.667 0.667 0.667 0.667

Section 7.2

Culverts

277

TABLE 7.2: Contraction Coefficients

Culvert type Circular concrete

Rectangular box

Rectangular Multibarrel box FIGURE 7.4: Variation of Type 2 flow (Type 2A)

Description

Cb

Cc

Square edge with headwall Groove end with headwall Groove end projecting 30◦ –75◦ wingwall flares 90◦ and 15◦ wingwall flares 0◦ wingwall flares Tapered inlet throat Vertical headwall

0.944 1.000 0.998 0.854 0.815 0.792 0.982 1.000

0.662 0.729 0.712 0.752 0.754 0.749 0.910 0.667

HW Δh H

yn

D

TW

1 (D+y ) c 2

yc L

1

3 2

than Equation 7.15 is that Equation 7.19 contains less parameters and is capable of estimating flows with comparable accuracy to Equation 7.15. Whereas Types 1 to 3 flows include most design circumstances when the culvert entrance is submerged, it is important to keep in mind that other submerged-entrance flow regimes are possible. For example, a scenario in which the culvert entrance is fully submerged, the tailwater is low, and the culvert flows partially full at the exit is certainly possible, even in cases where the normal depth of flow in the culvert exceeds the culvert diameter. This scenario is illustrated in Figure 7.4. In these circumstances, the culvert flow can be approximated by assuming that the hydraulic grade line at the culvert exit is at a point halfway between the critical depth, yc , and the height of the culvert, D, and then using Equation 7.12 with h equal to the difference between the assumed hydraulic grade line elevation at the exit and the headwater elevation (USFHWA, 2012). In cases where the actual tailwater elevation exceeds the assumed hydraulic grade line elevation at the exit, the actual tailwater elevation should be used in this approximation. The flow regime shown in Figure 7.4 can be designated as Type 2A flow and provides a reasonable approximation to reality as long as the culvert flows full over at least part of its length. If the culvert does not flow full over most of its length, the actual backwater profile in the culvert should be calculated. Type 2 and Type 2A flow regimes are both possible under submerged-entrance and low-tailwater conditions; however, a more conservative estimate of the culvert capacity is achieved by assuming the Type 2 flow shown in Figure 7.2 versus Type 2A flow shown in Figure 7.4. 7.2.1.2

Unsubmerged entrances

Submerged entrances generally lead to greater flows through culverts than unsubmerged entrances, which are also called free entrances. In some cases, however, culverts must be designed so that the entrances are not submerged. Such cases include those in which the top of the culvert forms the base of a roadway. In the case of an unsubmerged entrance, Figure 7.3 shows three possible flow regimes: (a) The culvert has a mild slope and a low tailwater, in which case the critical depth occurs somewhere near the exit of the culvert (Type 4 flow); (b) the culvert has a steep slope and a low tailwater, in which case the critical depth occurs somewhere near the entrance of the culvert, at approximately 1.4yc downstream from the entrance, and the flow approaches normal depth at the outlet end (Type 5 flow); and (c) the culvert has a mild slope and the tailwater submerges yc (Type 6 flow).

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Type 4 flow. For Type 4 flow (mild slope, low tailwater), the critical flow depth occurs at the exit of the culvert. Applying the energy equation between the headwater and the culvert exit gives V2 V2 h + 1 − = hi + hf (7.20) 2g 2g where h is the difference between the headwater elevation and the elevation of the (critical) water surface at the exit of the culvert, V1 is the headwater velocity, hi is the entrance loss given by Equation 7.7, and hf is the friction loss in the culvert given by Equation 7.6. Equation 7.20 assumes that the culvert slope is mild, and the velocity of the headwater is not neglected, as in the case of a ponded headwater where H/D>1.2. Equation 7.20 yields the following expression for the discharge, Q, through the culvert: 

  V12  − hi − hf Q = Ac 2g h + 2g

(7.21)

where Ac is the flow area at the critical-flow section at the exit of the culvert. Equation 7.21 is an implicit expression for the discharge, since the critical depth (a component of h), headwater velocity, V1 , entrance loss, hi , and the friction loss, hf , all depend on the discharge, Q. Type 5 flow. For Type 5 flow (steep slope, low tailwater), the critical flow depth occurs at the entrance of the culvert. Applying the energy equation between the headwater and the culvert entrance gives V2 V2 = hi h + 1 − (7.22) 2g 2g where h is the difference between the headwater elevation and the elevation of the (critical) water surface at the entrance of the culvert. Equation 7.22 leads to the following expression for the discharge, Q, through the culvert: 

  V12  − hi Q = Ac 2g h + 2g

(7.23)

Equation 7.23 is an implicit expression for the discharge, since the critical depth (a component of h), headwater velocity, V1 , and entrance loss, hi , all depend on the discharge, Q. In some cases, the approach velocity head, V12 /2g, is neglected; the entrance loss is given by Equation 7.7; and Equation 7.23 is put in the form (Sturm, 2010) Q = Cd Ac 2g(H − yc )

(7.24)

where Cd is defined by the relation 1 Cd = √ 1 + ke

(7.25)

and H is the headwater depth. The USGS (Bodhaine, 1976) developed values for Cd as a function of the headwater depth-to-diameter ratio, H/D. For circular culverts set flush in a vertical headwall, Cd = 0.93 for H/D < 0.4, and Cd decreases to 0.80 at H/D = 1.5, where the entrance is submerged. For box culverts set flush in a vertical headwall, Cd can be taken as 0.95. Based on experiments conducted by the U.S. National Bureau of Standards, now the National Institute of Science and Technology (NIST), the following nondimensional best-fit power relationships have been developed to facilitate the description of Type 5 flow through culverts:

Section 7.2

⎧ ⎪ M E ⎪ ⎨ c + K · (32.2) 2 · FrM − 0.5S H D =  D ⎪ ⎪ ⎩K · (32.2) M2 · FrM

Culverts

279

(Form 1) (7.26) (Form 2)

where H is the headwater specific energy [L]; D is the height of the culvert entrance [L]; Fr is the Froude number at the culvert entrance defined by Equation 7.16 [dimensionless]; Ec is the specific energy under critical-flow conditions at the culvert entrance [L]; S is the slope of the culvert between the culvert entrance and critical section [dimensionless]; and K, K , M, and M are dimensionless empirical constants that depend on the culvert configuration and are given in Table 7.1. The factor of 32.2 in Equation 7.26 is equal to the acceleration due to gravity in U.S. Customary units (ft/s2 ), which is necessary to be able to use the values of K and K reported by NIST. Equation 7.26 is dimensionally homogeneous and is applicable for Fr … 0.6. In specialized cases of embedded culverts, more appropriate values of K, K , M, and M can be found in USFHWA (2012). Form 1 of Equation 7.26 is preferred, while Form 2 is easier to use since calculation of the critical flow depth is not required. However, some experimental data have indicated that Form 2 provided better correlation with observations that Form 1 (e.g., Haderlie et al., 2008). When applying Form 1 of Equation 7.26 to mitered inlets, a slope correction factor of +0.7S should be used instead of −0.5S. Of all the equations proposed for describing Type 5 flow, Equation 7.26 is the most widely used. A close approximation to the NIST relationships for Type 5 flow (Equation 7.26) can be obtained by applying the energy equation at the entrance to a culvert. For the specific case of box culverts, assuming that critical flow is established within the culvert barrel near the entrance and that head losses between the headwater and critical section are negligible, the energy equation can be expressed in the form 2  Q Cb Byc H = Ec = yc + (7.27) 2g where H is the headwater specific energy, Ec is the specific energy at the critical section within the culvert entrance, yc is the critical depth, Cb is a width contraction coefficient associated with the culvert entrance edge conditions, and B is the culvert width. For critical flow in a rectangular box culvert, yc = 2/3Ec = 2/3H and Equation 7.27 can be expressed as (Charbeneau et al., 2006) 3 H = D 2



1 Cb

2 3

2

Fr 3

(7.28)

where Fr is the Froude number at the culvert entrance, defined by Equation 7.16, D is the culvert height, and A (= B * D) is the full culvert cross-section area. Values of Cb that give the closest agreement between Equation 7.28 and the NIST relationships for Type 5 flow (Equation 7.26) have been estimated by Charbeneau et al. (2006) and are given in Table 7.2. In applying Equation 7.28 to circular culverts, D is taken as the diameter of the culvert. The primary advantage to using Equation 7.28 rather than Equation 7.26 is that Equation 7.28 contains less parameters and is capable of estimating flows with comparable accuracy to NIST equations. As the culvert entrance becomes submerged, Type 5 flow becomes Type 3 flow and curve-matching equations are typically used to describe the transitional behavior of the culvert flow. For example, the HY-8 computer program (USFHWA, 2005a) uses a fifth-order polynomial to approximate the H versus Q relationship in the transition from Type 5 to Type 3 flow. Type 6 flow. For Type 6 flow (mild slope, tailwater submerges yc ), the water surface at the culvert exit is approximately equal to the tailwater elevation. Applying the energy equation between the headwater and the culvert exit gives

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h +

V12 V2 − = hi + hf 2g 2g

(7.29)

where h is the difference between the headwater elevation and the tailwater elevation at the exit of the culvert. Equation 7.29 leads to the following expression for the discharge, Q, through the culvert: 

 2  V Q = A2g h + 1 − hi − hf (7.30) 2g where A is the flow area at the exit of the culvert. Equation 7.30 is an implicit expression for the discharge, since the headwater velocity, V1 , entrance loss, hi , and the friction loss, hf , all depend on the discharge, Q. 7.2.2

Design Constraints

It is generally necessary to take into account regulatory requirements in designing culverts. Typical design criteria found in regulations are as follows: Flow. Several flow rates are typically considered in culvert design. A minimum flow rate is considered to ensure a self-cleansing velocity, a design flow rate is considered to ensure that flooding does not occur with unacceptable regularity, and a maximum flow rate is considered to ensure that the culvert does not cause major flooding during extreme runoff events. Minimum (self-cleansing) flow rates typically are taken as 2-year flow events,∗ design flow rates as 10-year or 25-year flow events, and maximum flow rates as 100-year flow events. Roadway overtopping may be allowed only for maximum-flow events. Headwater Elevation. The allowable headwater elevation is usually the primary basis for sizing a culvert. The allowable headwater elevation is taken as the elevation above which damage may be caused to adjacent property and/or the roadway, and is determined from an evaluation of land-use upstream of the culvert and the proposed or existing roadway elevation. The higher the allowable headwater elevation, the smaller the minimum-required culvert size and therefore the more economical the design. Typical HW/D ratios in the United States are in the range of 1.0-1.5 (USFHWA, 2012). Typically a 45-cm (18-in.) freeboard is required. Slope. The culvert slope should approximate the existing topography, and the maximum slope is typically 10% for concrete pipe and 15% for corrugated metal pipe (CMP) without using a pipe restraint. Size. Local drainage regulations often require a minimum culvert diameter, usually 30–60 cm (12–24 in.). Debris potential is an important consideration in determining the minimum acceptable size of a culvert, and some localities require that the engineer assume 25% debris blockage in computing the required size of any culvert. Shape. Circular culverts are the most common shape. They are generally reasonably priced, can support high structural loads, and are hydraulically efficient. Limited fill height might require the use of a pipe arch or ellipse, which are more expensive than circular pipes. Arches require special attention to their foundations, where failure due to scour is a common concern. Allowable Velocities. Both minimum and maximum velocities must be considered in designing a culvert. A minimum velocity in a culvert of 0.6–0.9 m/s (2–3 ft/s) at the 2-year flow rate is frequently required to assure self-cleansing. The maximum-allowable velocity for corrugated metal pipe is 3–5 m/s (10–15 ft/s), and usually there is no specified maximum-allowable velocity for reinforced concrete pipe, although velocities greater than 4–5 m/s (12–15 ft/s) are rarely used because of potential problems with scour. ∗ An n-year event is equalled or exceeded once every n years on average.

Section 7.2

Culverts

281

Material. The most common culvert materials are concrete (reinforced or nonreinforced), corrugated metal (aluminum or steel), and plastic (high-density polyethylene, HDPE; or polyvinyl chloride, PVC). The selection of a culvert material depends on the required structural strength, hydraulic roughness, durability, corrosion and abrasion resistance, and cost. In general, corrugated culverts have significantly higher frictional resistance than concrete culverts, and most cities require the use of concrete pipe for culverts placed in critical areas or within the public right-of-way. Corrosion is generally a concern with corrugated metal culverts. Culverts may also be lined with other materials to inhibit corrosion and abrasion, or to reduce hydraulic resistance. For example, corrugated metal culverts may be lined with asphaltic concrete or a polymer material. Recommended Manning’s n values for culvert design are given in Table 7.3. Inlet. The geometry of a culvert entrance is an important aspect of culvert design, since the culvert entrance exerts a significant influence on the hydraulic characteristics, size, and cost of the culvert. The four standard inlet types are: (1) flush setting in a vertical headwall, (2) wingwall entrance, (3) projecting entrance, and (4) mitered entrance set flush with a sloping embankment. Structural stability, aesthetics, and erosion control are among the factors that influence the selection of the inlet configuration. Headwalls increase the efficiency of an inlet, provide embankment stability and protection against erosion, and shorten the length of the required structure. Headwalls are usually required for all metal culverts and where buoyancy protection is necessary. Wingwalls are used where the side slopes of the channel adjacent to the inlet are unstable or where the culvert is skewed to the normal channel flow. The entrance loss coefficient, ke , used to describe the entrance losses in most discharge formulae depends on the pipe material, shape, and entrance type, and can be estimated using the guidelines in Table 7.4. In some cases, the invert of the culvert entrance is depressed below the bottom of the incoming stream bed to increase the headwater depth and the capacity of the culvert. This design will only be effective for flows under inlet control, such as Type 3 TABLE 7.3: Manning’s n in Culverts

Type of conduit

Wall and joint description

n

Concrete pipe

Good joints, smooth walls Good joints, rough walls Poor joints, rough walls Badly spalled

0.011–0.013 0.014–0.016 0.016–0.017 0.015–0.020

Concrete box

Good joints, smooth, finished walls Poor joints, rough, unfinished walls

0.012–0.015 0.014–0.018

Spiral rib metal pipe

19-mm * 19-mm recesses at 30-cm spacing, good joints

0.012–0.013

Corrugated metal pipe, pipe arch, and box

68-mm * 13-mm annular corrugations 68-mm * 13-mm helical corrugations 150-mm * 25-mm helical corrugations 125-mm * 25-mm corrugations 75-mm * 25-mm corrugations 150-mm * 50-mm structural plate 230-mm * 64-mm structural plate

0.022–0.027 0.011–0.023 0.022–0.025 0.025–0.026 0.027–0.028 0.033–0.035 0.033–0.037

Polyethylene

Corrugated Smooth

0.018–0.025 0.009–0.015

PVC

Smooth

0.009–0.011

Source: U.S. Federal Highway Administration (2005a; 2012).

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Culvert type and entrance conditions

ke

Pipe, concrete: Projecting from fill, socket end (groove end) Projecting from fill, square-cut end Headwall or headwall and wingwalls Socket end of pipe (groove end) Square edge Rounded (radius = D/12) Mitered to conform to fill slope End section conforming to fill slope Beveled edges, 33.7◦ or 45◦ bevels Side- or slope-tapered inlet

0.2 0.5 0.2 0.7 0.5 0.2 0.2

Pipe, or pipe arch, corrugated metal: Projecting from fill (no headwall) Headwall or headwall and wingwalls, square edge Mitered to conform to fill slope, paved or unpaved slope End section conforming to fill slope Beveled edges, 33.7◦ or 45◦ bevels Side- or slope-tapered inlet

0.9 0.5 0.7 0.5 0.2 0.2

Box, reinforced concrete: Headwall parallel to embankment (no wingwalls) Square edged on 3 edges Rounded on 3 edges Wingwalls at 30◦ to 75◦ to barrel Square edged at crown Crown edge rounded Wingwalls at 10◦ to 25◦ to barrel Square edged at crown Wingwalls parallel (extension of sides) Square edged at crown Side or slope-tapered inlet

0.2 0.5

0.5 0.2 0.4 0.2 0.5 0.7 0.2

Source: U.S. Federal Highway Administration (2012).

and Type 5 flows. The difference in elevation between the stream bed and the invert of the culvert entrance is called the fall. If high headwater depths are to be encountered, or the approach velocity in the channel will cause scour, a short channel apron should be provided at the toe of the headwall. In cases of embedded or bottomless culverts, entrance-loss coefficients 10%–65% higher than those shown in Table 7.4 should be expected (Tullis et al., 2008). Outlet. Outlet protection should be provided where discharge velocities will cause erosion problems. Protection against erosion at culvert outlets varies from limited riprap placement to complex and expensive energy dissipation devices such as stilling basins, impact basins, and drop structures. Outlet protection is typically designed for a 25-year flow rate. Debris Control. Usage of smooth well-designed inlets, avoidance of multiple barrels, and alignment of culverts with natural drainage channels will help pass most floating debris. In cases where debris blockage is unavoidable, debris control structures may be required. 7.2.3

Sizing Calculations

Culverts must be sized to accommodate a given design flow rate, Qdesign , under the constraint of a maximum-allowable headwater depth, Hmax . For circular culverts, sizing of the culvert

Section 7.2

Culverts

283

requires finding a commercially available pipe diameter, D, such that Q Ú Qdesign and H … Hmax . To accomplish the sizing objective, two alternative approaches can be used. In one alternative, the headwater depth, H, is set equal to Hmax and Q is calculated for incremental values of D (by commercial size) until Q Ú Qdesign . In an alternative approach, the flow rate through the culvert, Q, is set equal to Qdesign and H is calculated for incremental values of D until H … Hmax . Both alternative approaches lead to the same result by different paths, with different information provided along the paths. If the minimum size of a culvert to pass the design flow rate is too large to meet the headwater constraint, multiple barrels may be used. In this situation, the design flow rate is divided by the number of barrels and each barrel is designed for the per-barrel flow rate. This approach assumes that the culvert barrels perform the same whether they are in a single-barrel or multibarrel configuration. Although this is not exactly true, inner barrels tend to perform less efficiently than outer barrels, the error in estimating the multibarrel culvert capacity as the sum of identical single-barrel capacities is relatively small and within acceptable bounds (Haderlie et al., 2008). Once the size of the culvert barrel(s) is determined, it is useful to calculate the headwater elevation versus the flow rate through the culvert, and such relations are called culvert performance curves. 7.2.3.1

Fixed-headwater method

In calculating Q for a given H, the value of H is typically set to the maximum-allowable value of H (= Hmax ), which is usually determined by the elevation of the roadway under which the culvert passes. The tailwater elevation is either set by normal-flow conditions in the downstream channel, a constant (ponded) elevation in the receiving water body, or derived from a given depth-flow rate relation for the downstream channel (called a rating curve). Based on assumed values of H and D, it is first determined whether the culvert entrance is submerged: if H/D > 1.2, the culvert entrance is submerged; otherwise, the entrance is not submerged. Once the submergence condition is determined, the discharge capacity of the culvert can be calculated as follows: Submerged entrance. When the culvert entrance is submerged, the flow is Type 1, 2, or 3. The flow type and the associated flow rate are determined by the following procedure: Step 1: Determine whether the culvert exit is submerged. • If the culvert exit is submerged, then the flow is Type 1 and the culvert capacity is given by Equation 7.12. • If the culvert exit is not submerged, then the flow is either Type 2 or 3. Continue to Step 2. Step 2: Assume that the flow is Type 2 and calculate the discharge using Equation 7.12. Determine the corresponding normal depth of flow in the culvert. • If the normal depth of flow is greater than the height of the culvert, then Type 2 flow is confirmed and the calculated discharge is the culvert capacity. • If the normal depth of flow is less than the height of the culvert, then the flow is probably Type 3. Continue to Step 3. Step 3: Assume that the flow is Type 3 and calculate the discharge using Equation 7.14 or 7.15. Determine the corresponding normal depth of flow in the culvert. • If the normal depth of flow is less than the culvert height, then Type 3 flow is confirmed and the calculated discharge is the culvert capacity. • If the normal depth of flow is greater than the culvert height, then neither Type 2 nor Type 3 flow can be confirmed and some intermediate flow regime is probably occurring. The culvert capacity should be taken as the lesser of the discharges calculated using the Type 2 and Type 3 discharge equations. It is useful to note that circular culverts always flow full when the discharge exceeds 1.07 Qfull , where Qfull is the full-flow discharge calculated using the Manning equation.

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Unsubmerged entrance. When the culvert entrance is unsubmerged, the flow is possibly Type 4, 5, or 6. The flow type and associated flow rate are determined by the following procedure: Step 1: Assume that the flow is Type 4; use Equation 7.21, along with the critical-flow equation (Equation 4.100), to calculate the discharge, Q, and the critical depth, yc . Use Q in the Manning equation to calculate the normal depth, yn . • If yn Ú yc and the tailwater depth is less than yc , then Type 4 flow is confirmed and the calculated discharge is the culvert capacity. • Otherwise, continue to Step 2. Step 2: Assume that the flow is Type 5; use Equation 7.23, along with the critical-flow equation (Equation 4.100), to calculate the discharge, Q, and the critical depth, yc . Alternatively, Equation 7.26 can be used to determine Q. Use Q in the Manning equation to calculate the normal depth, yn . • If yn yc and the tailwater depth is greater than yc , then Type 6 flow is confirmed and the calculated discharge is the culvert capacity. • Otherwise, the culvert capacity should be taken as the lesser of the discharges calculated using the Type 4, Type 5, and Type 6 discharge equations. In the above-described approach, the submergence criterion is taken as H/D>1.2. If the Charbeneau et al. (2006) equations are used to describe Type 3 (submerged) and Type 5 (unsubmerged) flow, then the appropriate submergence criterion is H/D>1.5Cc where Cc is given in Table 7.2. In calculating the capacities of culverts, it is important to keep in mind that there are flow regimes other than the Type 1 to Type 6 flows identified here, and that under unsteadyflow conditions the flow regime can depend on the time sequence of flows (e.g., Meselhe and Hebert, 2007). Type 1 to Type 6 flow conditions are representative of most culvert flows under (steady-state) design conditions, and they can generally be confirmed by using the calculated flow rates to verify the assumed flow conditions. EXAMPLE 7.2 What is the capacity of a 1.22-m by 1.22-m concrete box culvert (n = 0.013) with a rounded entrance (ke = 0.05, Cd = 0.95) if the culvert slope is 0.5%, the length is 36.6 m, and the maximum-allowable headwater level is 1.83 m above the culvert invert? Consider the following cases: (a) free-outlet conditions, and (b) tailwater elevation 0.304 m above the crown of the culvert at the outlet. What must the headwater elevation be in case (b) for the culvert to pass the flow rate that exists in case (a)? Solution Since the headwater depth exceeds 1.2 times the height of the culvert opening, the culvert entrance is submerged. An elevation view of the culvert is shown in Figure 7.5. FIGURE 7.5: Elevation view of culvert

Q

L⫽

1.83 m

36.

6m

1.2

2m S⫽

0.5 %

Q

Section 7.2

Culverts

285

(a) For free-outlet conditions, two types of flow are possible: the normal depth of flow is greater than the culvert height (Type 2) or the normal depth of flow is less than the culvert height (Type 3). To determine the flow type, assume a certain type of flow, calculate the discharge and depth of flow, and see if the assumption is confirmed. If the assumption is not confirmed, then the assumed flow type is incorrect. Assuming Type 2 flow, then the flow-rate equation, Equation 7.12, is given by   2g h  (7.31) Q = A  2gn2 L + ke + 1 4 R3

where h is the difference in water levels between the entrance and exit of the culvert [= 1.83 + 0.005(36.6) − 1.22 = 0.793 m], n is the roughness coefficient (= 0.013), L is the length of the culvert (= 36.6 m), and R is the hydraulic radius given by R=

A P

where A is the cross-sectional area of the culvert and P is the wetted perimeter of the culvert, A = (1.22)(1.22) = 1.49 m2 P = 4(1.22) = 4.88 m and therefore R=

1.49 = 0.305 m 4.88

Substituting into Equation 7.31 gives   2(9.81)(0.793)  Q = (1.49)  2(9.81)(0.013)2 (36.6) + 0.05 + 1 4 (0.305) 3

which simplifies to Q = 4.59 m3 /s The next step is to calculate the normal depth of flow at a discharge of 4.59 m3 /s using the Manning equation, where 2 1 1 Q = AR 3 S02 n and S0 is the slope of the culvert (= 0.005). If the normal depth of flow is yn , then the flow area, A, wetted perimeter, P, and hydraulic radius, R, are given by A = byn = 1.22yn P = b + 2yn = 1.22 + 2yn R=

1.22yn A = P 1.22 + 2yn

The Manning equation gives 5

4.59 =

1 (1.22yn ) 3 1 (0.005) 2 0.013 (1.22 + 2y ) 23 n

which yields yn = 1.25 m Therefore, the initial assumption that the normal depth of flow is greater than the height of the culvert (= 1.22 m) is verified, and Type 2 flow is confirmed. The flow rate through the culvert is equal to 4.59 m3 /s. A variation of Type 2 flow could exist where flow in the culvert transitions from full flow to the tailwater depth before exiting the culvert, this was referred to previously as Type 2A flow. Under Type 2A flow conditions, the flow depth at the culvert exit is taken as equal to (yc + D)/2

286

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Design of Hydraulic Structures or the tailwater depth, whichever is greater, and full-flow conditions are assumed in calculating the head loss between the entrance and exit of the culvert. In the present case, the tailwater depth under the free-flow condition is unknown and will be assumed to be less than (yc + D)/2. Under this circumstance,   1 1 3 2 Q2 3 Q2 = = 0.409Q 3 yc = gb2 (9.81)(1.22)2 2

2 0.409Q 3 + 1.22 yc + D = = 0.205Q 3 + 0.61 2 2   2 yc + D = 1.83 + 0.005(36.6) − (0.205Q 3 + 0.61) h = HW + S0 L − 2 2

= 1.403 − 0.205Q 3 Substituting into Equation 7.12 using the previously calculated full-flow parameters gives      2(9.81)(1.403 − 0.205Q 23 ) 2gh   Q = A = (1.49)   2gn2 L  2(9.81)(0.013)2 (36.6) + 0.05 + 1 + ke + 1 4 4 R3

(0.305) 3

which yields Q = 4.69 m3 /s. This flow rate is about 2% higher than the flow rate determined under Type 2 conditions (i.e., 4.59 m3 /s). These results confirm that assuming Type 2 conditions rather than Type 2A conditions will generally result in conservative estimates of culvert capacity. Furthermore, the computations for Type 2 flow are much less demanding than for Type 2A flow, particularly if computation of the critical flow depth in a circular conduit is required. In the present problem the culvert was rectangular and so this latter complication did not occur. (b) In this case, the tailwater is 0.304 m above the crown of the culvert at the outlet, and therefore the difference in water levels between the inlet and outlet, h, is 1.83 m − 1.22 m + 0.005(36.6) m − 0.304 m = 0.489 m. The flow equation in this case, Type 1 flow, is given by Equation 7.31 with h = 0.489 m, which gives   2(9.81)(0.489)  Q = (1.49)  2(9.81)(0.013)2 (36.6) + 0.05 + 1 4

(0.305) 3

which simplifies to Q = 3.60 m3 /s Therefore, when the tailwater depth rises to 0.305 m above the crown of the culvert exit, the discharge decreases from 4.59 m3 /s to 3.60 m3 /s. When the headwater is at a height x above the crown of the culvert at the inlet and the tailwater is 0.305 m above the crown of the culvert at the outlet, the flow rate through the culvert is 4.59 m3 /s. The difference between the headwater and tailwater elevations, h, is given by h = x + 0.005(36.6) − 0.305 = x − 0.122 The flow equation for Type 1 flow (Equation 7.31) requires that   2(9.81)(x − 0.122)  4.59 = (1.49)  2(9.81)(0.013)2 (36.6) + 0.05 + 1 4 (0.305) 3

which leads to x = 0.915 m Therefore, the headwater depth at the entrance of the culvert for a flow rate of 4.59 m3 /s is 1.22 m + 0.915 m = 2.14 m.

Section 7.2

Culverts

287

TABLE 7.5: Calculation of Headwater Depths for Various Flow Types

7.2.3.2

Type

Equation number

Notes

1&2 3 4 5 6

7.11 7.14 or 7.15 7.20 7.22 or 7.26 7.29

h has different meaning for Type 1 and Type 2 flows Equation 7.15 is preferred — Form 1 of Equation 7.26 is preferred; Form 2 is second —

Fixed-flow method

Some approaches to culvert design calculate the required headwater depth, H, to pass a given design flow rate, Q, through a culvert of diameter, D, for given tailwater conditions. In this approach, the headwater depth, H, required for each type of flow is calculated and then the available headwater depth must be sufficient to accommodate the maximum H under design flow conditions. The equations used to calculate H for each flow type are given in Table 7.5. This design approach of finding H for given Q and D is in widespread use, and the Type 3 and Type 5 culvert equations expressed in terms of H/D are particularly suited to this approach. It is also useful to recall that Type 3 and Type 5 flows are under inlet control, which means that inlet conditions alone control the flow rate, Q, through the culvert. Conversely, Types 1, 2, 4, and 6 flows are under outlet control, which means that both inlet and outlet conditions determine Q. The procedure used in the fixed-flow method of calculating H for a given Q and D is similar to that used in the fixed headwater method and can be summarized as follows: Step 1: Assume a possible flow type and calculate H. If the calculated H is consistent with the assumed flow type, then the calculated H is the actual headwater depth. If not, go to Step 2. Step 2: Repeat Step 1 for each possible flow type until the calculated H is consistent with the assumed flow type. If none of the possible flow types are confirmed, then take H as the maximum of all calculated values of H.

EXAMPLE 7.3 A 915-mm-diameter concrete culvert is 20 m long and is laid on a horizontal slope. The culvert entrance is flush with the headwall with a grooved end and the estimated entrance loss coefficient is 0.2. The design flow rate is 1.70 m3 /s and under design conditions the tailwater depth is 0.75 m. Estimate the headwater depth required for the culvert to accommodate the design flow rate. Solution From the given data: D = 0.915 m, Q = 1.70 m3 /s, L = 20 m, S0 = 0, ke = 0.2, and TW = 0.75 m. For a concrete culvert it can be assumed that n = 0.013. Since the culvert is horizontal, yn = q and since the exit is not submerged the only possible flow regimes are Types 2, 3, and 6. These are considered sequentially as follows: Type 2 Flow: For Type 2 flow, the difference between the headwater elevation and the crown of the culvert exit, h, is given by Equation 7.11. From the given data, π 2 D 4 Q = V= A D = R= 4 A=

π (0.915)2 = 0.6576 m2 4 1.70 = 2.585 m/s 0.6576 0.915 = 0.2288 m 4

=

288

Chapter 7

Design of Hydraulic Structures and substituting into Equation 7.11 gives h = H − D=

n2 V 2 L 4 R3

+ ke

V2 V2 + 2g 2g

(0.013)2 (2.585)2 (20) 4 (0.2288) 3

+ 0.2

(2.585)2 (2.585)2 + = 0.570 m 2(9.81) 2(9.81)

where H is the headwater depth. The calculated result that H − D = 0.570 m validates the assumption of Type 2 flow, and gives H = D + 0.570 m = 0.915 m + 0.570 m = 1.485 m It is noteworthy that the calculated value of H − D will always be positive; therefore if the culvert is hydraulically long (L > 10D) a horizontal slope and an unsubmerged outlet will always support Type 2 flow. Type 2 is not the only possible type of flow, since Type 3 flow might be supported in cases where the culvert is hydraulically short (L < 10D), and Type 6 flow might also be possible in cases where the entrance is not submerged. If the tailwater elevation was very low (not in this case), Type 5 flow would also be a possibility. The other possible flow types are considered below. Type 3 Flow: For Type 3 flow, the headwater depth can be calculated using Equation 7.15, which requires that Fr > 0.7. In this case 2.585 V = 0.863 = Fr = gD (9.81)(0.915) Therefore, application of Equation 7.15 is validated. For a culvert entrance flush with the headwall and with a grooved end, Table 7.1 gives c = 0.0292 and Y = 0.74. Substituting into Equation 7.15 gives H = 32.2 c Fr2 + Y − 0.5S0 D H = 32.2(0.0292)(0.863)2 + 0.74 − 0.5(0) 0.915 which yields H = 1.318 m. This result indicates that Type 3 flow will require a headwater depth of 1.318 m. However, Type 3 flow is very unlikely because the culvert is hydraulically long (L > 10D) and horizontal (yn = q), so the flow will most likely expand and fill the culvert before reaching the exit, thus attaining Type 2 flow. Type 6 Flow: For Type 6 flow, the headwater depth is calculated using Equation 7.29. Neglecting the headwater velocity, Equation 7.29 can be expressed as h +

V12 2g



V2 = h i + hf 2g ⎛

⎞2 nQ ⎠ L (H − TW) + 0 − = ke + ⎝ 2 2 2 2gA 2gA AR3 Q2

Q2

(7.32)

where A and R represent the average flow area and hydraulic radius at the entrance and exit of the culvert. Equation 7.32 is an implicit equation for the headwater depth, H, since A and R will also depend on H. Any attempt to solve this equation numerically will show that there is no solution for H … D and so Type 6 flow is not possible. Collectively, the results presented here have demonstrated that Type 2 flow is the likely regime, and this will require a headwater depth of 1.485 m when the culvert is passing the design flow rate.

Section 7.2

7.2.3.3

Culverts

289

Minimum-performance method

Both the fixed-headwater and the fixed-flow methods involve the two-step procedure of: (1) assume a flow type; and (2) validate the flow type. A commonly used approximate method that is widely advocated (e.g., ASCE, 2006; USFHWA, 2012) is to skip the validation step and simply use the most conservative value of the calculated flow rate or headwater as the design value. This approach has the potential to over design the culvert structure. EXAMPLE 7.4 How would the required headwater depth in Example 7.3 change if the minimum-performance method were used? Solution The possible flow regimes are Types 2, 3, and 6, and the calculated headwater depths for these cases are as follows:

Type

Headwater depth (m)

2 3 6

1.485 1.318 N/A

Based on these results, the minimum-performance method would require a headwater depth of 1.485 m corresponding to Type 2 flow. In this case the minimum-performance method yields the same result as the exact method where the validity of the assumed flow type was also determined.

7.2.4

Roadway Overtopping

In cases where the culvert headwater elevation exceeds the roadway crest elevation (i.e., the roadway is overtopped), the flow must be partitioned between flow through the culvert and flow over the roadway. Under overtopping conditions, the roadway is typically assumed to perform like a rectangular weir, in which case the flow rate over the roadway, Qr , is given by 3

Qr = Cd LHr2

(7.33)

where Cd is the discharge coefficient, L is the roadway length over the culvert, and Hr is the head of water over the crest of the roadway. The discharge coefficient, Cd , can be estimated from the head of water over the roadway (Hr ), the width of the roadway (Lr ), and the submergence depth downstream of the roadway (yd ), using the relations shown in Figure 7.6, where the discharge coefficient is expressed in the form Cd = kr Cr

(7.34)

where Cr is derived from Hr using either Figure 7.6(a) for Hr /Lr > 0.15 or Figure 7.6(b) for Hr /Lr … 0.15, and kr is derived from Figure 7.6(c) for a given value of yd /Hr . Values of Cr and kr derived from Figure 7.6 are used to calculate Cd using Equation 7.34, and this value of Cd is used in Equation 7.33 to calculate the flow rate over the roadway. Application of the weir equation (Equation 7.33) to describe the flow over a roadway does not take into account the effect of rails on the sides of the roadway. In cases of overflowing bridges, rails have been found to have a significant effect of the head-discharge relationship under overflow conditions (Klenzendorf and Charbeneau, 2009). An iterative approach is usually required to determine the division of flow between the culvert and the roadway. This requires that different headwater elevations be assumed until the sum of the flow rates through the culvert and over the roadway is equal to the given total flow rate to be accommodated by the system.

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FIGURE 7.6: Discharge coefficient for roadway overtopping Source: USFHWA (2012).

Hr

Flow

yd Lr

1.70 1.65

1.75

el

v Gra

1.65

el

av

Gr

1.60 Cr 1.55

Paved

1.70 Cr

Paved

1.50 1.45 1.40

1.60 5

6

7 8 Hr (cm)

9

0

10

0.2 0.4

0.6 0.8 1.0 1.2 Hr (m)

(b) Discharge coefficient for Hr ≤ 0.15 Lr

(a) Discharge coefficient for Hr > 0.15 Lr 1.00

Paved

0.90 Gravel

0.80 kr 0.70 0.60 0.50 0.6

0.7

0.8 yd Hr

0.9

1.0

(c) Submergence factor

EXAMPLE 7.5 A culvert under a roadway is to be designed to accommodate a 100-year peak flow rate of 2.49 m3 /s. The invert elevation at the culvert inlet is 289.56 m, the invert elevation at the outlet is 288.65 m, and the length of the culvert is to be 22.9 m. The channel downstream of the culvert has a rectangular cross section with a bottom width of 1.5 m, a slope of 4%, and a Manning’s n of 0.045. The paved roadway crossing the culvert has a length of 15.2 m, an elevation of 291.08 m, and a width of 18.3 m. Considering a circular reinforced concrete pipe (RCP) culvert with a diameter of 610 mm and a conventional square-edge inlet and headwall, determine the depth of water flowing over the roadway, the flow rate over the roadway, and the flow rate through the culvert. Solution For the given design flow rate, the tailwater elevation can be derived from the normal-flow condition in the downstream channel. Characteristics of the rectangular downstream channel are given as: b = 1.5 m, S0 = 0.04, and n = 0.045. Taking Q = 2.49 m3 /s, the Manning equation gives 2 1 1 Q = AR 3 S02 n 2  3 1 1 1.5yn (1.5yn ) 2.49 = (0.04) 2 0.045 1.5 + 2yn

Section 7.2

Culverts

291

which yields a normal flow depth, yn = 0.73 m. Since the invert elevation of the downstream channel at the culvert outlet is 288.65 m, the tailwater elevation, TW, under the design condition is given by TW = 288.65 m + 0.73 m = 289.38 m Since the diameter of the culvert is 0.61 m and the tailwater depth is 0.73 m, the culvert outlet is submerged; and since the roadway elevation is 291.08 m and the tailwater elevation is 289.38 m, the tailwater is below the roadway. Assuming that roadway overtopping (by the headwater) occurs under the design condition, the design flow rate is equal to the sum of the flow rate through the culvert and the flow rate over the roadway such that   3 2gh  2 L H (7.35) + C Q = A r R d  2gn2 L + k + 1 e 4 R3

where Type 1 flow through the culvert exists (see Equation 7.12). From the given data: Q = 2.49 m3 /s, D = 0.61 m, A = π D2 /4 = 0.292 m2 , n = 0.012 (Table 7.3 for concrete pipe, good joints, smooth walls), L = 22.9 m, R = D/4 = 0.153 m, ke = 0.5 (Table 7.4 for headwall, square edge), LR = 15.2 m, and h = (Roadway elevation + Hr ) − Tailwater elevation = (291.08 + Hr ) − 289.38 = 1.70 + Hr

(7.36)

Combining Equations 7.35 and 7.36 with the given data yields   3 2(9.81)(1.70 + Hr )  2.49 = 0.292 + Cd (15.2)Hr2 2  2(9.81)(0.012) (22.9) + 0.5 + 1 4 (0.153) 3

which simplifies to

3 2.49 = 0.855 1.70 + Hr + 15.2Cd Hr2

(7.37)

The discharge coefficient, Cd , depends on the head over the roadway, Hr , via the graphical relations in Figure 7.6. Taking Lr = 18.3 m and yd = 0 (since the tailwater is below the roadway), the simultaneous solution of Equation 7.37 and the graphical relations in Figure 7.6 is done by iteration in the following table: (1) Hr (m)

(2) Hr /Lr

(3) Cr

(4) yd /Hr

(5) kr

(6) Cd = kr Cr

(7) Hr (m)

1.00 0.14

0.055 0.008

1.68 1.66

0.00 0.00

1.00 1.00

1.68 1.66

0.14 0.14

Column 1 is the assumed Hr in meters, Column 2 is Hr /Lr , Column 3 is Cr derived from Hr /Lr and Hr using Figure 7.6, Column 4 is yd /Hr , Column 5 is kr derived from yd /Hr using Figure 7.6, Column 6 is Cd obtained by multiplying Columns 3 and 5 and Column 7 is obtained by substituting Cd in Column 6 into Equation 7.37 and solving for Hr . The iterations indicate that Cd = 1.66, Hr = 0.14 m, and the flow rate over the roadway, Qr , is given by 3

3

Qr = Cd LR Hr2 = (1.66)(15.2)(0.14) 2 = 1.32 m3 /s The corresponding flow rate through the culvert is equal to 2.49 m3 /s − 1.32 m3 /s = 1.17 m3 /s. Therefore, a culvert diameter of 610 mm will result in roadway overtopping, with a flow rate of 1.17 m3 /s passing through the culvert, 1.32 m3 /s passing over the roadway, and a depth of flow over the roadway equal to 14 cm. A larger culvert diameter could be explored if less roadway overtopping at the design flow rate is desired.

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Design of Hydraulic Structures

7.2.5

Riprap/Outlet Protection

Outlet protection is required when there is a possibility of the native soil being eroded by the water exiting the culvert. This possibility is typically assessed by comparing the flow velocity at the culvert exit to the scour velocity of the native soil at the outlet. The velocity in the culvert barrel is assessed under design conditions and typical scour velocities of various soils are given in Table 7.6. In cases where the flow velocity at the culvert exit exceeds the scour velocity of the native soil, the native soil is usually overlain by riprap. Riprap consists of broken rock, cobbles, or boulders placed on the perimeter of a channel to protect against the erosive action of water. Riprap is a common erosion-control lining used at culvert outlets, storm-sewer outfalls, and around bridge abutments, especially in areas where suitable rock materials are readily available. A ground lining using riprap or any other material is commonly called an apron. A culvert exit with a riprap apron is shown in Figure 7.7. In this particular case, there is a small concrete apron between the culvert exit and the beginning of the riprap apron. Design of riprap outlet protection includes specifying: (1) the type and size of stone, (2) the thickness of the stone lining, and (3) the length and width of the apron. Several design standards have been developed, and local regulatory requirements should always be followed if they exist. In lieu of regulatory requirements, the following design guidelines can be followed (Gribbin, 2007): Type of Stone. Stones used for riprap should be hard, durable, and angular. Angularity, a feature of crushed stone from a quarry, helps to keep the stones locked together when subjected to the force of moving water. TABLE 7.6: Scour Velocities of Various Soils

Permissible Velocity Soil Sand Sandy loam Silt loam Sandy clay loam Clay loam Clay, fine gravel Cobbles Shale

FIGURE 7.7: Culvert exit with riprap apron

(m/s)

(ft/s)

0.5 0.8 0.9 1.1 1.2 1.5 1.7 1.8

1.6 2.6 3.0 3.6 3.9 4.9 5.6 5.9

Section 7.3

Gates

293

Size of Stone. Size is normally measured by the median diameter (by weight), d50 [m], and can be selected using the following formula (originally developed by USEPA, 1976): d50

0.044 = TW



Q D

4 3

(7.38)

where Q is the design flow rate [m3 /s], D is the culvert width or pipe diameter [m], and TW is the tailwater depth [m]. In cases where the pipe does not discharge into a well-defined channel, TW is taken as the flow depth in the pipe at the outlet. Thickness of Stone Lining. The thickness of the blanket of stones should be three times the median stone size if no filter fabric liner between the stones and the ground is used. If a filter fabric liner is used, the thickness should be twice the median stone size. Length of Apron. The length of the apron, La , depends on the relative magnitudes of TW and D/2 according to the relations (originally developed by USEPA, 1976) ⎧ 5.43Q ⎪ ⎪ ⎪ TW Ú D/2 ⎪ ⎨ D 32 (7.39) La = ⎪ 3.26Q ⎪ ⎪ ⎪ + 7D TW < D/2 ⎩ 3 D2 where La and D are in meters, and Q is in m3 /s. Width of Apron. If a channel exists downstream of the culvert outlet, the riprap width is dictated by the width of the channel. Riprap should cover at least the width of the channel and extend up the sides of the channel to at least 0.3 m (1 ft) above the design tailwater depth but not lower than two thirds of the vertical conduit dimension above the culvert invert. Additional requirements are: (1) the side slopes of the riprap-lined channel section should be less than or equal to 2:1 (H:V); (2) the bottom grade should be level (0% slope); and (3) there should be no drop at the end of the apron or at the outlet of the culvert. If a well-defined channel does not exist downstream of the culvert outlet, the width, W, of the apron at the culvert outlet should be at least three times the culvert width, and at the end of the apron the width should be at least ⎧ ⎪ ⎨3D + 0.4La TW Ú D/2 W= (7.40) ⎪ ⎩3D + La TW < D/2 where W, D, and La are in meters, or any consistent length units. In cases where a wingwall surrounds the culvert outlet, the width of the riprap lining starts with the width of the wingwall and ends with a width W at a distance La from the culvert outlet. 7.3 Gates Gates are used to regulate the flow in open channels. They are designed for either overflow or underflow operation, with underflow operation appropriate for channels in which there is a significant amount of floating debris. Two common types of gates are vertical gates and radial gates (also called Tainter∗ gates), which are illustrated in Figure 7.8. Vertical gates are supported by vertical guides with roller wheels, and large hydrostatic forces usually induce significant frictional resistance to raising and lowering the gates. An example of a vertical gate structure containing two gates and a close-up view of a gate lift are shown in Figure 7.9. Vertical gates are sometimes referred to as vertical lift gates, sluice gates, or vertical sluice ∗ Tainter gates were patented in the United States by Jeremiah B. Tainter in 1886.

294

Chapter 7

Design of Hydraulic Structures

FIGURE 7.8: Types of gates

Flow

Flow

(a) Vertical gate

(b) Radial (Tainter) gate

FIGURE 7.9: Vertical gate structure (with two gates) and close-up view of gate lift

FIGURE 7.10: Tainter gates Source: U.S. Army Corps of Engineers (2011).

gates, where a sluice is an artificial channel for conducting water, with a gate to regulate the flow. The conventional radial (Tainter) gate consists of an arc-shaped face plate supported by radial struts that are attached to a central horizontal shaft called a trunnion, which transmits the hydrostatic force to the supporting structure. Since the vector of the resultant hydrostatic force passes through the axis of the horizontal shaft, only the weight of the gate needs to be lifted to open the gate. Radial gates are economical to install and are widely used in both underflow and overflow applications. An example of radial gates in operation over a spillway is shown in Figure 7.10. 7.3.1

Free Discharge

Applying the energy equation to both vertical and radial gates as shown in Figure 7.11 yields y1 +

V12 V2 = y2 + 2 2g 2g

(7.41)

Section 7.3 FIGURE 7.11: Flow-through gates

Gate

Gates

295

Energy grade line

V12 2g

Gate

y1

V22 2g

V1 Flow

y1 yg

θ

y2 = Cc yg

y2 = Cc yg

yg

(a) Vertical gate

(b) Radial gate

where Sections 1 and 2 are upstream and downstream of the gate, respectively, and energy losses are neglected. Expressing Equation 7.41 in terms of the flow rate, Q, leads to

y1 +

Q2 2gb2 y21

= y2 +

Q2 2gb2 y22

and solving for Q gives  Q = by1 y2

2g y1 + y2

(7.42)

where b is the width of the gate. The depth of flow downstream of the gate, y2 , is less than the gate opening, yg , since the streamlines of the flow contract as they move past the gate (see Figure 7.11). The downstream location where the depth of flow is most contracted is called the vena contracta, and y2 denotes the depth at the vena contracta. Denoting the ratio of the downstream depth, y2 , to the gate opening, yg , by the coefficient of contraction, Cc , where Cc =

y2 yg

(7.43)

then Equations 7.42 and 7.43 can be combined to yield the following expression for the discharge through a gate: Q = Cd byg 2gy1

(7.44)

where Cd is the discharge coefficient or sluice coefficient given by Cd = 

Cc 1 + Cc

yg y1

(7.45)

The form of the discharge equation given by Equation 7.44 expresses the discharge in terms of an “orifice-flow” velocity, 2gy1 , times the flow area through the gate, byg , times a discharge coefficient, Cd , to account for deviations from the orifice-flow assumption. On the basis of Equation 7.45, the discharge coefficient depends on the amount of flow contraction as measured by Cc and yg /y1 . In the case of a vertical gate with a sharp edge, it has been reported that (Rajaratnam and Subramanya, 1967b)

296

Chapter 7

Design of Hydraulic Structures

Cc = 0.61

(7.46)

whenever 0 15, the weir is called a sill, the discharge can be computed from the critical-flow equation by assuming yc = H + Hw (Chaudhry, 1993), and the discharge coefficient is given by (Jain, 2001) 3  Hw 2 Cd = 1.06 1 + (7.71) H A general expression for Cd that is valid for any value of H/Hw has been proposed by Swamee (1988) and is given by  Cd = 1.06

14.14 8.15 + H/Hw

10

 +

H/Hw H/Hw + 1

15 −0.01 (7.72)

Equations 7.70 to 7.72 are valid only if the pressure under the nappe is atmospheric, and so for these equations to be valid the underside of the nappe must be aerated. Full aeration under the nappe is attained when the tailwater is more than 5 cm (2 in.) below the weir crest (Bos, 1988). It is convenient to express the discharge formula, Equation 7.67, as 3

Q = Cw bH 2

(7.73)

where Cw is called the weir coefficient and is related to the discharge coefficient by Cw =

2 Cd 2g 3

(7.74)

Taking Cd = 0.62 in Equation 7.74 yields Cw = 1.83, and Equation 7.73 becomes 3

Q = 1.83bH 2

(7.75)

which gives good results if H/Hw < 0.4, which is within the usual operating range of most weirs. Equation 7.75 is applicable in SI units, where Q is in m3 /s, and b and H are in meters. In accordance with Equation 7.73, for a rectangular weir of a given width, the flow rate over the weir depends only on the height of water, H, above the crest of the weir. Consequently, measurements of H can be used in Equation 7.73 to determine the flow rate over the weir, and hence the weir can be used as a flow-measuring device. The accuracy of the weir-discharge formula, Equation 7.73, depends significantly on the location of the gaging station for measuring the upstream head, H, and it is recommended that measurements of H be taken between 4H and 5H upstream of the weir (Ackers et al., 1978). In addition, the behavior of uncontracted weirs is complicated by the fact that air is trapped beneath the

304

Chapter 7

Design of Hydraulic Structures

nappe, which tends to be entrained into the jet, thereby reducing the air pressure beneath the nappe and drawing the nappe toward the face of the weir. To avoid this effect, a vent is sometimes placed beneath the weir to maintain atmospheric pressure. In the case of unsuppressed (contracted) weirs, the air beneath the nappe is in contact with the atmosphere and venting is not necessary. Experiments have shown that the effect of side contractions is to reduce the effective width of the nappe by 0.1H, and that flow rate over the weir, Q, can be estimated by 3

Q = Cw (b − 0.1nH)H 2

(7.76)

where Cw is the weir coefficient calculated using Equation 7.74, b is the width of the contracted weir, and n is the number of sides of the weir that are contracted, usually equal to 2. Equation 7.76 gives acceptable results as long as b > 3H. A type of contracted weir that is related to the rectangular sharp-crested weir is the Cipolletti weir, which has a trapezoidal cross section with side slopes 1 : 4 (H : V) and is illustrated in Figure 7.16. The advantage of using a Cipolletti weir is that corrections for end contractions are not necessary, and the discharge over a Cipolletti weir is greater than the discharge over a rectangular weir with the same crest width, b. The discharge formula for a Cipolletti weir can be written simply as 3

Q = Cw bH 2

(7.77)

where b is the bottom width of the Cipolletti weir. The minimum head on standard rectangular and Cipolletti weirs is 6 mm (0.2 in.), and at heads less than 6 mm (0.2 in.) the nappe does not spring free of the crest (Aisenbrey et al., 1974). The sharp-crested weir is a control structure, since the flow rate over the weir is determined by the stage just upstream of the weir. This control relationship assumes that the water downstream of the weir, called the tailwater, does not interfere with the operation of the weir. If the tailwater elevation rises above the crest of the weir, then the flow rate becomes influenced by the downstream flow conditions, and the weir is submerged. The discharge over a submerged weir, Qs , can be estimated in terms of the upstream and downstream heads on the weir using Villemonte’s formula (Villemonte, 1947) ⎡ ⎤   3 0.385 2 y Qs d ⎦ = ⎣1 − Q H

(7.78)

where Q is the calculated flow rate assuming the weir is not submerged, yd is the head downstream of the weir, and H is the head upstream of the weir; both yd and H are measured relative to the crest of the weir. The head downstream of the weir, yd , is approximately equal to the difference between the downstream water-surface elevation and the crest of the FIGURE 7.16: Cipolletti weir

4

H

1 b

Section 7.4

Weirs

305

weir. Consistent with these definitions, when yd = 0, Equation 7.78 gives Qs = Q. In using Equation 7.78, it is recommended that H be measured at least 2.5H upstream of the weir and that yd be measured beyond the turbulence caused by the nappe (Brater et al., 1996). A more recent formula for calculating the discharge over a submerged weir is given by Abu-Seida and Quraishi (1976) as   Qs yd yd (7.79) 1 − = 1 + Q 2H H When used for flow measurement, weirs should be designed to discharge freely rather than submerged because of greater accuracy in flow estimation. EXAMPLE 7.8 A weir is to be installed to measure flow rates in the range of 0.5–1.0 m3 /s. If the maximum (total) depth of water that can be accommodated at the weir is 1 m and the width of the channel is 4 m, determine the crest height of a suppressed weir that should be used to measure the flow rate. Solution The flow over the weir is illustrated in Figure 7.17, where the crest height of the weir is Hw and the flow rate is Q. The height of the water over the crest of the weir, H, is given by H = 1 − Hw Assuming that H/Hw < 0.4, then Q is related to H by Equation 7.75, where 3

Q = 1.83 bH 2 Taking b = 4 m and Q = 1 m3 /s (the maximum flow rate will give the maximum head, H), then

Q H= 1.83b



2 3

=

1 1.83(4)

2 3

= 0.265 m

The crest height, Hw , is therefore given by Hw = 1 − 0.265 = 0.735 m and

H 0.265 = 0.36 = Hw 0.735 The initial assumption that H/Hw < 0.4 is therefore validated, and the height of the weir should be 0.735 m.

FIGURE 7.17: Weir flow

H Q

1m Hw Q

Rectangular sharp-crested weirs with very small widths are called slit weirs, and are characterized by the dependence of Cd on the Reynolds number. Slit weirs are used for measuring very small flow rates, on the order of 0.01 L/s (0.2 gpm), and values of Cd for these types of weirs can be found in the work of Aydin et al. (2006). Multislit weirs are also used to measure small flow rates, and values of Cd as a function of Reynolds number for these types of weirs can be found in the work of Ramamurthy et al. (2007).

306

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Design of Hydraulic Structures

7.4.1.2

V-notch weirs

A V-notch weir is a sharp-crested weir that has a V-shaped opening instead of a rectangularshaped opening. These weirs, also called triangular weirs, are typically used instead of rectangular weirs when lower discharges are desired for a given head, H, or where lower flow rates are to be measured with greater accuracy than can be achieved with rectangular weirs. V-notch weirs are usually limited to flow rates of 0.28 m3 /s (10 cfs) or less, and are frequently found in small irrigation canals. An operating V-notch weir is shown in Figure 7.18. The basic theory of V-notch weirs is the same as that for rectangular weirs, where the theoretical flow rate over the weir, Q, is given by ! H Q= v2 b dz2 0

! = 0

⎡ H

⎤ 12 V12 ⎣2g H − z2 + ⎦ b dz2 2g

where b is the width of the V-notch weir at elevation z2 and is given by   θ b = 2z2 tan 2

(7.80)

(7.81)

where θ is the angle at the apex of the V-notch, as illustrated in Figure 7.19. Combining Equations 7.80 and 7.81 leads to ! Q= 0

FIGURE 7.18: Operating V-notch weir

FIGURE 7.19: Schematic diagram of V-notch weir

Channel

b dz2 H

θ Weir

Hw

z2

⎡ H

⎤ 12   V12 ⎣2g H − z2 + ⎦ 2z2 tan θ dz2 2g 2

(7.82)

Section 7.4

Weirs

307

The approach velocity, V1 , is usually negligible for the low velocities that are typically handled by V-notch weirs, and therefore Equation 7.82 can be approximated by   $  % 12 θ 2g H − z2 Q= 2z2 tan dz2 2 0   5 8 θ H2 = 2g tan 15 2 !

H

(7.83)

As in the case of a rectangular weir, the theoretical discharge given by Equation 7.83 is corrected by a discharge coefficient, Cd , to account for discrepancies in the assumptions leading to Equation 7.83. The actual flow rate, Q, over a V-notch weir is therefore given by Q=

  5 θ 8 Cd 2g tan H2 15 2

(7.84)

where Cd generally depends on Re, We, θ , and H. The vertex angles used in V-notch weirs are usually between 10◦ and 120◦ , with the most commonly encountered notch angles being 120◦ , 90◦ , 60◦ , and 45◦ (Cruise et al., 2007; Davie, 2008). Weirs with notch angles of 90◦ are sometimes called Thomson weirs. Values of Cd for a variety of notch angles, θ , and heads, H, are plotted in Figure 7.20. It is apparent from Figure 7.20 that the minimum discharge coefficient corresponds to a notch angle of 90◦ and the minimum value of Cd for all angles is 0.581; the rise in Cd at heads less than 15 cm (6 in.) is due to incomplete contraction. Using Cd = 0.58 for engineering calculations is usually acceptable, provided that 20◦ < θ < 100◦ and H > 5 cm (2 in.) (Potter and Wiggert, 1991; White, 1994). For H < 5 cm (2 in.), both viscous and surface-tension effects may be important and a recommended value of Cd is given by (White, 1994) 1.19

Cd = 0.583 +

1

(7.85)

(ReWe) 6

where Re is the Reynolds number defined by 1

1

g2 H 2 Re = ν FIGURE 7.20: Discharge coefficient in V-notch weirs

0.70

Discharge coefficient, Cd

0.68 θ = 10o 0.66 0.64 20o 0.62 45o

0.60 Cd = 0.581

0.58 5

10

90o

60o

15

20

Head, H (cm)

25

30

(7.86)

308

Chapter 7

Design of Hydraulic Structures

and We is the Weber number defined by We =

ρgH 2 σ

(7.87)

where ν is the kinematic viscosity and σ is the surface tension of water. EXAMPLE 7.9 A V-notch weir is to be used to measure channel flow rates in the range 0.1–0.2 m3 /s. What is the maximum head of water on the weir for a vertex angle of 45◦ ? Solution The maximum head of water occurs at the maximum flow rate, so Q = 0.2 m3 /s will be used to calculate the maximum head. The relationship between the head and flow rate is given by Equation 7.84, which can be put in the form  H=





2 5

15Q

8Cd 2g tan (θ/2)

=

8Cd



2 5

15(0.2) 2(9.81) tan (45◦ /2)

=

0.530 2

m

Cd5

The discharge coefficient as a function of H for θ = 45◦ is given in Figure 7.20, and some iteration is necessary to find H. These iterations are summarized in the following table: Assumed H (cm)

Cd (Figure 7.20)

Calculated H (cm)

12 64.9 65.8

0.6 0.581 0.581

64.9 65.8 65.8

Therefore, the maximum head expected at the V-notch weir is approximately 66 cm.

An alternative flow equation for V-notch weirs has been proposed by Kindsvater and Shen (USBR, 1997), where Q=

8 Cd 2g tan 15

  5 θ (H + k) 2 2

(7.88)

where Cd and k are both functions of the notch angle, θ . LMNO Engineering and Research Software (LMNO, 1999) developed the following analytic functions for Cd and k to match the graphical functions recommended by Kindsvater and Shen (USBR, 1997): Cd = 0.6072 − 0.000874θ + 6.1 * 10−6 θ 2 k = 4.42 − 0.1035θ + 1.005 * 10−3 θ 2 − 3.24 * 10−6 θ 3

(7.89) (7.90)

where θ is in degrees and k is in millimeters. The Kindsvater and Shen equation, Equation 7.88, has been recommended by several reputable organizations and researchers (International Organization for Standardization, 1980; American Society for Testing and Materials, 1993; U.S. Bureau of Reclamation, 1997; Martinez et al., 2005).

Section 7.4

Weirs

309

EXAMPLE 7.10 Repeat the previous example using the Kindsvater–Shen equation. Compare the results. Solution The maximum head of water occurs at the maximum flow rate, so Q = 0.2 m3 /s will be used to calculate the maximum head. Using θ = 45◦ , Equations 7.89 and 7.90 yield Cd = 0.6072 − 0.000874θ + 6.1 * 10−6 θ 2 = 0.6072 − 0.000874(45◦ ) + 6.1 * 10−6 (45◦ )2 = 0.580 k = 4.42 − 0.1035θ + 1.005 * 10−3 θ 2 − 3.24 * 10−6 θ 3 = 4.42 − 0.1035(45◦ ) + 1.005 * 10−3 (45◦ )2 − 3.24 * 10−6 (45◦ )3 = 1.50 mm = 0.0015 m The relationship between the head and flow rate is given by Equation 7.88, which yields  H=

15Q

8Cd 2g tan (θ/2)



2 5

− k=



15(0.2)

8(0.580) 2(9.81) tan (45◦ /2)

2 5

− 0.0015 = 0.657 m

Therefore, the maximum head expected at the V-notch weir is approximately 0.66 m. This is the same result that was obtained using the graphical method. Since the Kindsvater–Shen equation does not require iteration, it is clearly preferable in this case.

The following guidelines have been suggested by the U.S. Bureau of Reclamation (USBR, 1997) for using V-notch weirs to measure flow rates in open channels:  The head (H) should be measured at a distance of at least 4H upstream of the weir.  The weir should be 0.8–2 mm (0.03–0.08 in.) thick at the crest. If the bulk of the weir is thicker than 2 mm (0.08 in.), the downstream edge of the crest can be chamfered at an angle greater than 45◦ (60◦ is recommended) to achieve the desired thickness of the edges. This should avoid having water cling to the downstream face of the weir.  The water surface downstream of the weir should be at least 6 cm (2.5 in.) below the bottom of the crest to allow a free-flowing waterfall.  The measured head (H) should be greater than 6 cm (2.5 in.) due to potential measurement error at such small heads and the possibility that the nappe may cling to the weir. In cases where the tailwater elevation rises above the crest of the weir, the flow rate is influenced by downstream flow conditions. Under this submerged condition, the discharge over the weir, Qs , can be estimated in terms of the upstream and downstream heads on the weir using Villemonte’s formula (Equation 7.78), with the exceptions that the exponent of yd /H is taken as 52 instead of 32 , and the unsubmerged discharge, Q, is calculated using Equation 7.84 or Equation 7.88. In stormwater-management applications, V-notch weirs with invert angles less than 20◦ and heights less than 50 mm (2 in.) are usually undesirable, since such weirs are difficult to build and easy to clog. 7.4.1.3

Compound weirs

A compound weir is a combination of different types of weirs in a single structure. The most common compound weir consists of a rectangular weir with a V-notch in the middle as shown in Figure 7.21. These weirs are commonly used in stormwater-management systems, where the V-notch is used to regulate the release of an initial volume of runoff stored in a detention area (e.g., the water-quality volume), while the rectangular component of the weir provides for the controlled release of larger volumes of runoff. In these cases, the V-notch weir is

310

Chapter 7

Design of Hydraulic Structures

FIGURE 7.21: Compound weir

b

H1

θ

H2

frequently referred to as the bleeder. When the water level behind the compound weir is within the V-notch, the compound weir performs like a V-notch weir, with the discharge relation given by Equation 7.84 or 7.88. When the water level is above the V-notch, the weir discharge is taken as the sum of the discharge through the V-notch and the rectangular weir, with the V-notch discharge, Qv , given by the orifice-flow equation Qv = Cvd A 2gH (7.91) where Cvd is a discharge coefficient that accounts for head loss and flow contraction through the V-notch, A is the area of the V-notch, and H is the height of the water surface above the centroid of the V-notch. Combining the discharge over the rectangular weir, given by Equation 7.67, with the discharge through the V-notch, given by Equation 7.91, yields the following discharge equation for flow over a compound weir:    3 2 H2 2 Q = Cd 2gbH1 + Cvd A 2g H1 + 3 3

(7.92)

where Cd is the discharge coefficient for the rectangular weir, given by Equation 7.70; H1 and H2 are the heights shown in Figure 7.21; and Cvd is typically taken as 0.6 (McCuen, 1998). EXAMPLE 7.11 The outlet structure from a stormwater detention area is a compound weir consisting of a rectangular weir with a V-notch in the middle. The notch angle is 90◦ , the height of the notch is 20 cm, and the apex of the notch is placed at the bottom of the detention area. Determine the required crest length of the rectangular weir such that the discharge through the compound weir is 3.0 m3 /s when the water depth in the detention area is 1.00 m. Solution From the given data: θ = 90◦ , H2 = 20 cm = 0.20 m, and H1 = 1.00 − H2 = 1.00 − 0.20 = 0.80 m. The discharge, Q, over the weir is given by Equation 7.92 as    3 2 H2 2 (7.93) Q = Cd 2gbH1 + Cvd A 2g H1 + 3 3 The discharge coefficient, Cd , is given by Equation 7.70 as Cd = 0.611 + 0.075

H1 H2

(7.94)

Section 7.4

Weirs

311

which yields 0.80 = 0.911 0.20 Since H1 /H2 = 4, Equation 7.94 is valid, and Cd can be taken as 0.911 in the weir design. The area, A, of the V-notch is given by    ◦ θ 90 = 0.202 tan = 0.04 m2 A = H22 tan 2 2 Cd = 0.611 + 0.075

Taking Cvd = 0.6 and Q = 3 m3 /s, substituting into Equation 7.93 gives    3 2 0.20 3 = (0.911) 2(9.81)b(0.80) 2 + 0.6(0.04) 2(9.81) 0.80 + 3 3 which yields b = 1.51 m Therefore, a compound weir with a crest length of 1.51 m will yield a discharge of 3 m3 /s when the water depth in the detention area is 1 m.

A minor drawback in using the compound weir shown in Figure 7.21 is that the flowhead relations given by Equations 7.91 and 7.92 are discontinuous at H = H1 . There are other less widely used compound-weir geometries that do not have this discontinuity, such as a triangular weir with a V-notch in the middle (Martinez et al., 2005). 7.4.1.4

Other types of sharp-crested weirs

Sharp-crested weirs with cross-sectional shapes other than rectangular, V-notch, and compound shapes have been developed with the primary motive of matching a cross-sectional shape to a desired head-discharge relationship. Notable shapes are the Sutro∗ weir which has a linear head-discharge relationship, and the polynomial weir (Baddour, 2008) whose shape parameters can be varied to produce a range of head-discharge relationships. Sharp-crested weirs have been modified to include low-level openings that allow the passage of solids that would otherwise be deposited and accumulated behind the weir. A summary of the performance of such weirs can be found in Samani and Mazaheri (2009). A labyrinth weir has a crest length that is longer than the width of the channel in which the weir is located. The crests of labyrinth weirs follow a path that is not a straight line between the sides of the channel, which allows for increased flow rates for a given head compared to regular straight-line weirs. An example of a labyrinth weir is shown in Figure 7.22, where the weir crest follows a FIGURE 7.22: Triangular labyrinth weir

∗ Developed by Harry Sutro in the early 1900s.

312

Chapter 7

Design of Hydraulic Structures

FIGURE 7.23: Wooden weir in a coastal drainage channel

triangular (zig zag) path. A class of labyrinth weirs called piano key weirs have been shown to be practical and effective add-on structures that increase the capacity of existing spillways where width constraints prevent increasing the (straight-line) length of the spillway (Ribeiro et al., 2012). A variation on the sharp-crested weir occurs when the crest of the weir is rounded instead of being sharp. Such round-crested weirs are less frequently studied and used than their sharp-crested counterparts, but they are unique in that their discharge coefficient depends on the radius of curvature of the crest. A summary of the performance of such weirs can be found in the work of Castro-Orgaz et al. (2008). Weirs are sometimes used to control overland flows in coastal areas. In such applications, weirs are placed in drainage channels to retain freshwater runoff behind the weir and provide increased freshwater heads to limit saltwater intrusion. An example of such a weir is shown in Figure 7.23, where the weir is constructed of wood and is designed such that elevation of the crest can be adjusted by adding or removing planks of the wood. 7.4.2

Broad-Crested Weirs

Broad-crested weirs, also called long-based weirs, have significantly broader crests than sharpcrested weirs. These weirs are usually constructed of concrete, have rounded edges, and are capable of handling much larger discharges than sharp-crested weirs. There are several different designs of broad-crested weirs, of which the rectangular (broad-crested) weir can be considered representative. 7.4.2.1

Rectangular weirs

A typical rectangular broad-crested weir is illustrated in Figure 7.24. These weirs operate on the theory that the elevation of the weir above the channel bottom is sufficient to create critical-flow conditions over the weir. Under these circumstances, the estimated flow rate over the weir, Q, is given by Q = yc bVc

(7.95)

where yc is the critical depth of flow over the weir, b is the width of the weir, and Vc is the velocity at critical flow. If E1 is the specific energy of the flow at Section 1 just upstream of the weir, and the energy loss between this upstream section and the critical-flow section over the weir is negligible, then the energy equation requires that

Section 7.4 FIGURE 7.24: Flow over a rectangular broad-crested weir

Weirs

313

1 Energy grade line V 12 2g

H Q

h1

Critical flow

Q

Hw L Longitudinal section view

E1 = Hw + yc +

Vc2 2g

(7.96)

where Hw is the height of the weir crest above the upstream channel. Under critical-flow conditions, the Froude number is equal to 1, hence Vc =1 gyc

(7.97)

Combining Equations 7.96 and 7.97 yields the following expression for yc : yc =

2 2 (E1 − Hw ) = H 3 3

(7.98)

where H is the energy of the upstream flow measured relative to the weir-crest elevation. The upstream energy, H, can be written as H = h1 +

V12 2g

(7.99)

where h1 is the elevation of the upstream water surface above the weir crest, and V1 is the average velocity of flow upstream of the weir. Combining Equations 7.95, 7.97, and 7.98 leads to the following estimate of the flow rate over a rectangular broad-crested weir: Q=



3  2 2 gb H 3

(7.100)

In reality, the energy loss over the weir is not negligible and the estimated flow rate, Q, must be corrected to account for the energy loss. The correction factor is the discharge coefficient, Cd , and the actual flow rate, Q, over the weir is given by Q = Cd



3  2 2 gb H 3

(7.101)

where values of Cd can be estimated using the relation (Chow, 1959) Cd =

0.65 1

(1 + H/Hw ) 2

(7.102)

314

Chapter 7

Design of Hydraulic Structures

Alternatively, Swamee (1988) proposed the relation ⎡

  13 ⎤0.1 h1 5 h1 ⎢ ⎥ + 1500 ⎢ L ⎥ L ⎢ ⎥ Cd = 0.5 + 0.1 ⎢  3 ⎥ h ⎣ ⎦ 1 1 + 1000 L

(7.103)

To ensure proper operation of a broad-crested weir, flow conditions must be restricted to the operating range of 0.08 < H/L < 0.50 (Bos, 1988) or 0.08 < H/L < 0.33 (French, 1985) or 0.08 < h1 /L < 0.33 (Jain, 2001; Sturm, 2010). These requirements are very similar, particularly in the usual circumstance where the velocity head upstream of the weir is negligible, in which case h1 L H. Overall, it is recommended that broad-crested weirs be operated in the range 0.08 < h1 /L < 0.33. Flow conditions for various values of h1 /L are listed in Table 7.7. As an alternative to Equation 7.101, the flow rate over a broad-crested weir can be expressed in terms of the upstream depth, h1 , instead of the upstream head, H, in which case the discharge equation is given by 

Q=

√ Cd gb

2 h1 3

3 2

(7.104)

where the discharge coefficient Cd in Equation 7.104 is different (typically higher) from the Cd in Equation 7.101. Using Equation 7.104, values of 0.80 < Cd < 0.85 have been reported for properly operated broad-crested weirs. It has been further suggested that Cd can be approximated as a constant provided that 0.08 < h1 /L < 0.33 and 0.18 < h1 /(h1 + Hw ) < ¨¸ 0.36, with the constant Cd being taken as a fixed value within the range of 0.80–0.85 (Go¨ g˘ us et al., 2007). A more refined estimate of Cd within this range has been suggested by Azimi and Rajaratnam (2009) as 2    h1 h1  + 0.89 (7.105) − 0.38 Cd = 0.95 h1 + Hw h1 + Hw This estimate of Cd is appropriate for broad-crested weirs with squared upstream edges. For rounded upstream edges, the following expression for Cd is more appropriate (Azimi and Rajaratnam, 2009)   h1  Cd = 0.90 + 0.146 (7.106) h1 + Hw Broad-crested weirs with rounded upstream entrances will deliver higher flow rates than those with squared entrances. TABLE 7.7: Flow Conditions over Broad-Crested Weirs h1 /L

Weir classification

Flow condition

h1 /L < 0.08

Long crested

The critical-flow section is near the downstream end of the weir, the flow over the weir crest is subcritical, and the value of Cd depends on the resistance of the weir surface. This type of weir is of limited use for flow measurements.

0.08 < h1 /L < 0.33

Broad crested

The region of parallel flow occurs near the middle section of the crest. The variation of Cd with h1 /L is small.

0.33 < h1 /L < 1.5

Narrow crested

The streamlines are curved over the entire crest; there is no region of parallel flow over the crest.

Sharp crested

The flow separates at the upstream end and does not reattach to the crest. The flow is unstable.

h1 /L > 1.5

Section 7.4

Weirs

315

Broad-crested weirs that have sloped upstream and downstream faces are commonly referred to as embankment weirs. It has been shown that a more gradual 2:1 (H:V) slope on the upstream weir face provides a higher discharge coefficient than either a 1:1 or vertical slope (Sargison and Percy, 2009). Providing a vertical face at the downstream end of the weir produces leaping flow and prevents cavitation at high flow rates. The use of a vertical instead of a sloped downstream face has negligible effect on the discharge coefficient. An advantage of a broad-crested weir is that the tailwater can be above the crest of the weir without affecting the head-discharge relationship as long as the control section is unaffected. The limit of the head, yd , downstream of the weir so that the discharge does not decrease by more than 1% is called the modular limit, which is usually expressed in terms of yd /H. For rectangular broad-crested weirs, yd /H = 0.66 can usually be taken as the modular limit (Bos, 1988). EXAMPLE 7.12 A 20-cm-high broad-crested weir is placed in a 2-m-wide channel. Estimate the flow rate in the channel if the depth of water upstream of the weir is 50 cm. Solution Upstream of the weir, h1 = 0.5 m − 0.2 m = 0.30 m, and H = h1 +

V12 2g

= h1 +

Q2 2gA21

= 0.30 +

Q2 = 0.30 + 0.0510Q2 2(9.81)(0.5 * 2)2

The discharge coefficient, Cd , is given by Cd =

0.65 (1 +

1 H/Hw ) 2

=

0.65 1

[1 + (0.30 + 0.0510Q2 )/0.2] 2

=

0.65 1

[2.5 + 0.255Q2 ] 2

where Hw has been taken as 0.2 m. The discharge over the weir is therefore given by √



2 H Q = Cd gb 3

3 2

=

0.65 1 0.255Q2 ] 2



2 9.81(2) (0.30 + 0.0510Q2 ) 3

3 2

[2.5 +  1 (0.30 + 0.0510Q2 )3 2 = 2.22 2.5 + 0.255Q2

which yields Q = 0.23 m3 /s This solution assumes that the length of the weir is such that 0.08 < h1 /L < 0.33.

7.4.2.2

Compound weirs

Broad-crested weirs can also be used in compound weirs, where smaller flows go through a lower rectangular or V section and higher flows go through a wider rectangular section. A typical broad-crested compound weir with a rectangular low-flow section is shown in Figure 7.25. When the critical flow depth, yc , is in the low-flow section (yc … z), the discharge FIGURE 7.25: Broad-crested compound weir

Channel

B b

z

yc

Hw

Weir

316

Chapter 7

Design of Hydraulic Structures

through the broad-crested weir is given by Equation 7.101, and when the critical flow depth is in the high-flow section (yc > z), the discharge relationship is determined using the criticalflow relation Q2 A3 = c g B

(7.107)

where B is the width of the high-flow section, Ac is the flow area given by Ac = bz + (yc − z)B

(7.108)

and b and z are the width and depth, respectively, of the low-flow section. Combining Equations 7.107 and 7.108 with the energy and continuity equations, and applying a discharge coefficient to account for theoretical discrepancies gives the discharge through the compound weir under high-flow conditions as  1   32  g 2 bz 2z 2 H − − Q = Cd bz + B B 3 3B 3

(7.109)

where H is the specific energy upstream of the weir. Values of the discharge coefficient, Cd , depend on H/L and b/z, and a (graphical) relationship based on laboratory experiments can ¨ ¸ et al. (2006). be found in the work of Go¨ g˘ us As an alternative to using Equation 7.109, it has been shown that the discharge through compound broad-crested weirs can be accurately approximated as the sum of the discharges through subsections of the weir using the conventional head-discharge relationships for noncompound weirs (Jan et al., 2009). For example, the compound weir in Figure 7.25 could be modeled as three simple broad-crested weirs: one center weir and two “shoulder” weirs. In practical applications, it is desirable to provide a small transverse slope on the bottom of the high-flow rectangular section (called the “shoulder” of the weir). This feature in which the shoulder slopes toward the center of the weir provides a more predictable dischargeversus-head relationship as the flow depth transitions from the notch section to the high-flow section. Furthermore, the use of an inclined-ramp approach to the control section of a compound broad-crested weir is commonly used to increase the discharge coefficient relative to using the blunt approach illustrated in Figure 7.25. Computer programs are available to facilitate the design and calibration of compound broad-crested weirs, particularly with respect to estimating the discharge coefficient (e.g., Wahl et al., 2007). 7.4.2.3

Gabion weirs

Conventional broad-crested weirs are made of impermeable materials such as concrete, which under low-flow conditions prevent the longitudinal movement of small aquatic life, which can have a negative impact on the water environment. Gabion weirs have been suggested as an alternative that mitigates the aforementioned restriction. Gabions consist of stones encased in a wire mesh, and in gabion weirs concrete is replaced by gabions. Flow over broad-crested gabion weirs can be quantified using the relation 3 √ Q = Cd gbH 2

(7.110)

where Q is the flow rate, Cd is the discharge coefficient, b is the channel width (equal to the weir width), and H is the water head above the crest of the weir. Experimental results have shown that for unsubmerged gabion weirs the discharge coefficient, Cd , in Equation 7.110 can be estimated using the relation (Mohamed, 2010) Cd = −1.31 + 0.47 log(Re) − 0.84

dm H + 1.01 L Hw

(7.111)

Section 7.5

Spillways

317

where Re is the Reynolds number given by Re =

Qρ bμ

(7.112)

where ρ and μ are the density and dynamic viscosity of water, respectively; L is the length of the weir; dm is the mean stone size in the gabion; and Hw is the height of the weir. In cases where flow over the broad-crested weir is submerged, Equation 7.110 is also applicable, with the discharge coefficient, Cd , given by Cd = −1.77 + 0.55 log(Re) − 0.78

dm H y + 0.35 + 0.085 L Hw H

(7.113)

where y is the difference between the headwater and tailwater elevations. Application of Equations 7.111 and 7.113 generally lead to lower values of H for any given value of Q when comparing gabion weirs with those constructed of impermeable materials. 7.5 Spillways A spillway is a structure that is used to safely discharge excess water that is stored in a reservoir behind the spillway, where “excess water” refers to water stored above the crest elevation of the spillway. In reservoirs designed for water regulation and hydroelectric power generation, spillways typically function infrequently. Spillways are categorized as controlled or uncontrolled, depending on whether or not they are equipped with gates. 7.5.1

Uncontrolled Spillways

The shapes of spillways are generally designed for uncontrolled conditions. If control features such as gates are used on a spillway, then the performance of the gate is considered separately. The shape of an overflow spillway is usually made to conform to the profile of a fully ventilated nappe of water flowing over a sharp-crested weir that is coincident with the upstream face of the spillway. This shape depends on the head over the crest, the inclination of the upstream face, and the velocity of approach. Crest shapes of ogee∗ spillways have been studied extensively by the U.S. Bureau of Reclamation (USBR, 1977). Ogee spillways are sometimes referred to as ogee-crest weirs. The head over the crest of a spillway is measured above the apex of the spillway, which is point O in Figure 7.26. High spillways are defined as those in which the ratio of crest height, P, to design head, Hd , is greater than 1 (i.e., P/Hd > 1), and low spillways are those in which P/Hd < 1. When the actual head on the spillway is less than the design head, the trajectory of the nappe falls below the crest profile, creating positive pressures on the crest, and reducing the discharge coefficient. Conversely, when the actual head is higher than the design head, the nappe trajectory is higher than the crest profile, creating negative pressures on the crest and increasing the discharge coefficient. When negative pressures are generated on the crest, cavitation conditions are possible and should generally be avoided. Experiments have shown that cavitation on the spillway crest occurs whenever the pressure on the spillway is less than −7.6 m (−25 ft) of water, and it is recommended that spillways be designed such that the maximum expected head will result in a pressure on the spillway crest of not less than −4.6 m (−15 ft) of water (U.S. Department of the Army, 1986). It is generally desirable to design the crest shape of a high-overflow spillway for a design head, Hd , that is less than the head on the crest corresponding to the maximum reservoir level, He ; and, to avoid cavitation problems, the U.S. Bureau of Reclamation (1987) recommends that He /Hd not exceed 1.33. Experiments have shown that, in cases where the pressure on the spillway is equal to −4.6 m (−15 ft), the relationship between the design head, Hd [m], and the head at the maximum reservoir level, He [m], is given by (Reese and Maynord, 1987)

∗ An ogee refers to an S-shaped profile.

318

Chapter 7

Design of Hydraulic Structures

FIGURE 7.26: Ogee-crest spillway

Approach head, he Pool elevation Hd

Design head

He + x-axis

O B

PT (B − y) 2 x2 + =1 B2 A2

y

A

Hd + y-axis

=

1 K

x Hd

n

Point of tangency (PT)

Crest axis

1

P

a

1 Fs

& For He Ú 9.1 m (30 ft) :

Hd = &

For He < 9.1 m (30 ft) :

Hd =

0.33He1.22 0.30He1.26

0.7He 0.74He

(without piers) (with piers)

(7.114)

(without piers) (with piers)

(7.115)

When a crest with piers is designed for negative pressures, the piers must be extended downstream beyond the negative-pressure zone in order to prevent aeration of the nappe, nappe separation or undulation, and loss of the underdesign efficiency advantage (U.S. Army Corps of Engineers, 1986). An important part of spillway design is specification of the shape of the spillway. The U.S. Bureau of Reclamation (1977) separates the shape of the spillway into two quadrants, one upstream and one downstream of the crest axis (apex) shown in Figure 7.26. The equation describing the surface of the spillway in the downstream quadrant is given by y 1 = Hd K



x Hd

n (7.116)

where (x, y) are the coordinates of the crest profile relative to the apex (point O) as shown in Figure 7.26, and K and n are constants that depend on the upstream-face inclination and velocity of approach. According to Murphy (1973), n can be taken as 1.85 in all cases, and K varies with the ratio of the crest height, P, to the design head, Hd , as shown in Figure 7.27(a). Values of K range from 2.0 for high values of P/Hd to 2.2 for low values of P/Hd . The crest profile merges with the straight downstream section of slope α, at a location given by XDT = 0.485(Kα)1.176 Hd

(7.117)

where XDT is the horizontal distance from the apex to the downstream tangent point. Typically, only high spillways have a tangent point, since only this type of spillway tends to have

Section 7.5 FIGURE 7.27: Coordinate coefficients for spillway crests

Spillways

319

10.0 8.0 6.0

Crest height, P Design head, Hd

4.0

2.0

1.0 0.8 0.6 0.4

0.2 0.15 1.90

2.10 K (a)

2.30

0.21

0.23

0.25 A Hd (b)

0.27

0.29

0.12

0.14

0.16 B Hd (c)

0.18

a straight section between the crest-curve and the toe-curve portions of the spillway. In low spillways, there is usually a continuous curvature between the crest curve and the toe curve. The chute is that portion of the spillway that connects the crest curve to the terminal structure, and the flow in the spillway chute is generally supercritical. When the spillway is an integral part of a concrete gravity monolith, the chute is usually very steep, and slopes in the range from 0.6:1 to 0.8:1 (H:V) are common (Prakash, 2004). An (elliptical) equation describing the surface of an ogee spillway in the upstream quadrant is given by x2 (B − y)2 + =1 A2 B2

(7.118)

where (x, y) are the coordinates of the crest profile relative to the apex (point O) as shown in Figure 7.26, and A and B are equation parameters that can be estimated using Figures 7.27(b) and 7.27(c) for various values of P/Hd . For an inclined upstream face with a slope of Fs , the location of the point of tangency with the elliptical shape is given by the following equation: XUT =

A2 Fs 1

(A2 Fs2 + B2 ) 2

(7.119)

where XUT is the horizontal distance from the apex to the upstream tangent point. Since the crest of a spillway is approximately conformed to the profile of the lower nappe surface from a (ventilated) sharp-crested weir, the discharge, Q, over a weir corresponding to a head, He , on the spillway crest is given by the relation 3

Q = CLe He2

(7.120)

where C is the coefficient of discharge, and Le is the effective length of the spillway crest. In cases where crest piers and abutments cause side contractions of the overflow, the effective

320

Chapter 7

Design of Hydraulic Structures

length, Le , is less than the actual crest length, L, according to the relation Le = L − wN − 2(NKp + Ka )He

(7.121)

where N is the number of piers, w is the width of each pier, Kp is the pier contraction coefficient, and Ka is the abutment contraction coefficient. Recommended values of Kp and Ka are given below (U.S. Department of the Army, 1986; U.S. Bureau of Reclamation, 1987): ⎧ ⎪ ⎨0.02 for square-nose piers (7.122) Kp = 0.01 for round-nose piers ⎪ ⎩0.0 for pointed-nose piers ⎧ ◦ ⎪ ⎨0.2 for square abutments with headwalls at 90 to the flow direction Ka = 0.1 for rounded abutments with 0.15Hd … r … 0.5Hd (7.123) ⎪ ⎩0.0 for rounded abutments with r > 0.5H d where r is the radius of curvature of the abutments. The discharge coefficient, C, of an overflow spillway is influenced by a number of factors, including: (1) the crest height-to-head ratio; (2) the difference between the actual head and the design head; (3) the upstream face slope; and (4) the downstream submergence. The discharge coefficients for spillways can be estimated using relationships developed by the U.S. Bureau of Reclamation (1987) and shown in Figure 7.28. The curve shown in Figure 7.28(a) gives the basic discharge coefficient for a vertical-faced spillway with atmospheric pressure on the crest (i.e., at the design head), the curve in Figure 7.28(b) gives the correction factor for a sloping upstream face, and the curve in Figure 7.28(c) gives the correction factor for a head other than the design head. The overall discharge coefficient for the spillway is equal to the product of the basic discharge coefficient in Figure 7.28(a) with the correction factors in Figures 7.28(b) and 7.28(c). The increase in discharge coefficient, C, for heads greater than the design head is the basic reason for underdesigning spillway crests to obtain greater efficiency. The basic discharge coefficient given in Figure 7.28(a) is the same as for a vertical sharp-crested weir forming the upstream face of the spillway, with an adjustment to account for the fact that the head on a sharp-crested weir is measured relative to the top of the vertical face of the weir, while the head on a spillway is measured relative to the crest of the spillway. For very high spillways (P/Hd W 1) the basic discharge coefficient is independent of P/Hd and is equal to 2.18, which is appropriate for P/Hd Ú 3. A sloping upstream face can be used to prevent separation eddies that might occur on the vertical face of low spillways, and a sloping upstream face sometimes increases the discharge coefficient. EXAMPLE 7.13 Design an overflow spillway with an effective crest length of 60 m that will discharge at a design flow rate of 1500 m3 /s at a maximum-allowable pool elevation of 400 m. The bottom elevation behind the spillway is 350 m, the upstream face of the spillway is vertical, and the spillway chute is to have a slope of 1:2 (H:V). Solution Assuming that the spillway is sufficiently high that the ratio of the spillway height, P, to the design head, Hd , is greater than 3, Figure 7.28(a) gives the basic discharge coefficient as C0 = 2.18. Assuming that the effective head, He , is less than 9.1 m, then for the limiting case where the water pressure on the spillway is equal to −4.6 m, Equation 7.115 requires that He 1 = 1.43 = Hd 0.7 and Figure 7.28(c) gives C/C0 = 1.05. The discharge coefficient, C, under maximum headwater conditions can now be calculated using the relation

C C0 = (1.05)(2.18) = 2.29 C= C0

Section 7.5 FIGURE 7.28: Discharge coefficient for overflow spillways

Spillways

Basic discharge coefficient, C0

2.20 2.10 2.00 1.90

Hd P

1.80 1.70 0

0.5

1.0

1.5 Value of

2.0

2.5

3.0

P Hd

(a) Basic discharge coefficient

Ratio of coefficients,

Cinclined Cvertical

1.04

Slope (H :V)

1:1

P

1.02

2:3

1:3

18°26

2:3

33°41

1:1

45°00

1:3

1.00

0.98

Angle with the vertical

Hd

0

0.5

1.0

1.5

P Value of Hd (b) Correction factor for sloping upstream face

C Ratio of coefficients, C0

1.1

1.0 h0 H

Hd

0.9

P

0.8

0

0.2

0.4

0.6

0.8

1.0

1.2

Ratio of head on crest to design head,

1.4 H Hd

(c) Correction factor for other than design head

1.6

321

322

Chapter 7

Design of Hydraulic Structures From the given data, Q = 1500 m3 /s, Le = 60 m, and therefore the effective head, He , on the spillway is given by Equation 7.120 as

Q He = CLe



2 3

=

1500 (2.29)(60)

2 3

= 4.92 m

Since this calculated value of He is less than 9.1 m, the initial assumption that He < 9.1 m used in estimating the discharge coefficient is validated. The design head, Hd , corresponding to He = 4.92 m is given by Equation 7.115 as Hd = 0.7He = 0.7(4.92) = 3.44 m The depth of water, d, upstream of the spillway is given by d = 400 m−350 m = 50 m and the approach velocity, v0 , can be estimated by v0 =

1500 Q = = 0.5 m/s Le d (60)(50)

with a velocity head, h0 , given by v2 0.52 = 0.01 m h0 = 0 = 2g 2(9.81) Therefore, at the maximum pool elevation, the height of water above the spillway is He − h0 = 4.92 m − 0.01 m = 4.91 m, and the required crest elevation of the spillway is 400 m − 4.91 m = 395.09 m. Having determined the required crest elevation of the spillway (= 395.09 m) to pass the design flow rate (= 1500 m3 /s), the next step is to determine the required shape of the spillway in the vicinity of the crest. Since the maximum pool elevation is 400 m and the bottom elevation behind the spillway is 350 m, the height of the spillway crest, P, is given by P = 400 m − 350 m − 4.91 m = 45.09 m and

45.09 P = 13.1 = Hd 3.44 which indicates a “high spillway” with negligible approach velocity. Figure 7.27(a) gives K = 2.0, and the profile of the downstream quadrant of the spillway, taking n = 1.85, is given by n

y x 1 = Hd K Hd 1 y = 3.44 2.0



 x 1.85 3.44

which simplifies to y = 0.175x1.85 Since the downstream slope of the spillway is 1 : 2 (H : V), then α = 2 and the horizontal distance from the apex to the downstream tangent point, XDT , is given by Equation 7.117 as XDT = 0.485(Kα)1.176 Hd XDT = 0.485(2.0 * 2)1.176 3.44 which gives XDT = 8.52 m Therefore, at a distance of 8.52 m downstream of the crest, the curved spillway profile merges into the linear profile of the spillway chute, which has a slope of 1 : 2 (H : V). The corresponding y value is 1.85 = 0.175(8.52)1.85 = 9.21 m. y = 0.175XDT

Section 7.5

Spillways

323

Taking P/Hd = 13.1 in Figure 7.27(b) and (c) gives A/Hd = 0.28 and B/Hd = 0.165, which yields A = 0.28(3.44) = 0.963 m, B = 0.165(3.44) = 0.568 m, and the profile of the upstream quadrant of the spillway is given by Equation 7.118 as (B − y)2 x2 + =1 2 A B2 x2 (0.568 − y)2 + =1 2 0.963 0.5682 which simplifies to x2 + 2.87(0.568 − y)2 = 0.927 The upstream crest profile given by this equation intersects the vertical (upstream) spillway wall at x = −A = −0.963 m and y = B = 0.568 m.

Submerged flow. A submerged-flow condition occurs when the tailwater elevation is above the crest of the spillway as shown in Figure 7.29, where va is the approach velocity, P is the crest height, He is the upstream head relative to the crest, hd is the difference between the upstream head and the tailwater elevation, and d is the tailwater depth on the downstream apron of the spillway. The discharge, Q, over a submerged spillway is calculated using the free-flow discharge equation (Equation 7.120), with the free-flow discharge coefficient, C, reduced by an amount that depends on the ratios hd /He and (hd + d)/He as derived from Figure 7.30 (USACE, 1990b). Application of Figure 7.30 is illustrated in the following example. EXAMPLE 7.14 Consider an ogee spillway under the conditions shown in Figure 7.31, where the headwater is 1.0 m above the crest and the tailwater is 0.5 m above the crest. The height of the crest relative to the bottom of the upstream reservoir is 3.5 m; the height of the crest relative to the downstream apron is 4.0 m. The effective length of the spillway crest is 3.0 m and the design head is 1.0 m. Estimate the spillway discharge under the given conditions. What percentage reduction in free-flow discharge is caused by submergence of the spillway? Solution From the given data: Le = 3.0 m, P = 3.5 m, Hd = 1.0 m, and hd = 1.0 m − 0.5 m = 0.5 m. From these data, P/Hd = 3.5/1.0 = 3.5, and Figure 7.28(a) gives the basic discharge coefficient, C0 , as 2.18. Since the upstream face of the spillway is vertical, and assuming the upstream velocity head is small, then He L Hd = 1 m, and Figure 7.28(b) and (c) indicate that the free-flow discharge coefficient, C, is equal to C0 , and hence C = 2.18. The free-flow discharge, Q0 , is given by Equation 7.120 as

FIGURE 7.29: Submerged flow at a spillway

2

Va

2g

hd He

Va

d P Spillway

Apron

324

Chapter 7

Design of Hydraulic Structures

FIGURE 7.30: Discharge-coefficient reduction percentages for submerged flow

1.2

Source: U.S. Army Corps of Engineers (1990b).

1.1

1 0.5

0

2

The contour lines give the decrease in coefficient of discharge in percent

3 1.0 4 6

0.9

0 8

0.8

10

0.5

0.7 1

15 hd He

0.5 0.6 2

20 0.5

3

1 4

0.4

2 6

3

0.3

4

8

6

10

8

0.2

10 15

15 0.1

0.0 1.0

20

20

30 40 50 60 80

30 40 50 6 800 1.5

2.0

2.5

3.0 hd + d He

3.5

4.0

4.5

5.0 A

Section 7.5 FIGURE 7.31: Submerged flow

Spillways

325

0.5 m 1.0 m

3.5 m

4.0 m

Spillway

⎡ 3 2

Q0 = CLe He = (2.18)(3.0) ⎣1.0 +

Q20 2gA2

⎤3 2



⎦ = (2.18)(3.0) ⎣1.0 +

Q20 2(9.81)(4.5 * 3)2

⎤3 2



which yields Q0 = 6.67 m3 /s. This flow corresponds to an upstream velocity, va , and velocity head, v2a /2g, given by 6.67 v2 Q = 0.49 m/s ' a = 0.01 m va = 0 = A 4.5 * 3 2g Therefore He = 1.0 m + 0.01 m = 1.01 m, and hd 0.5 = 0.50 = He 1.01 hd + d 0.5 + 4.0 = 4.46 = He 1.01 Using these parameter values in Figure 7.30 gives a discharge-coefficient reduction of approximately 2.8% caused by submergence. Hence, the adjusted discharge coefficient is C = (1 − 0.028)(2.18) = 2.12 The corresponding discharge, Q, under the given submerged condition is   3 3 2 3 Q2 2 Q2 2 = (2.12)(3.0) 1.0 + Q = CLe He = (2.12)(3.0) 1.0 + 2gA2 2(9.81)(4.5 * 3)2 which yields Q = 6.47 m3 /s. This flow corresponds to an upstream velocity, va , and velocity head, v2a /2g, given by Q 6.47 v2 = = 0.48 m/s ' a = 0.01 m va = A 4.5 * 3 2g Since this approach velocity is approximately the same as calculated for the free-flow condition, no further iteration is necessary. The submerged discharge is 6.47 m3 /s and the reduction in discharge caused by submergence is 2.8%.

7.5.2

Controlled (Gated) Spillways

Gated spillways are used to control discharges from reservoirs and also to control flows in rivers and canals. The flow at a gated spillway is “controlled” when the gate opening influences the discharge, and is “uncontrolled” when the gate opening does not influence the discharge, such as when the gate is completely out of the water. Both vertical lift gates and Tainter gates are commonly used on spillway crests, with Tainter gates more common in cases where the discharge is free, and vertical gates more common in cases where the discharge is submerged. Vertical lift gates are more commonly found on low-ogee-crest dams and navigation dams with low sills where reservoir pool control normally requires gate operation at partial openings (USACE, 1990b). Gates can be seated either on the crest of the spillway as shown in Figure 7.32(a) or seated downstream of the spillway crest as shown in Figure 7.32(b). When the gate is seated on the crest of the spillway, the tangent to the spillway is horizontal and it is generally assumed that the gate-discharge equations for horizontal channels as described in detail in Section 7.3 are applicable. In cases where the gate is seated downstream of the spillway crest, modifications to these discharge equations are required.

326

Chapter 7

Design of Hydraulic Structures

FIGURE 7.32: Gate locations on spillway

Lip Lip Crest of spillway

Crest

Reservoir

Reservoir Gate seat

(b) Gate downstream of crest

7.5.2.1

Gate seat

(a) Gate at crest

Gates seated on the spillway crest

In cases where the gate is seated on the spillway crest, the flow rate, Q [L3 T−1 ], through the gate is typically estimated using equations of the following forms ⎧ ⎨C1 hL 2gH, controlled, free Q= ⎩C2 hL 2g(H − ht ), controlled, submerged

(7.124)

where C1 and C2 are dimensionless constants that are best determined by calibration, h is the height of the gate opening [L], L is the width of the gate opening [L], H is the height of the headwater above the spillway crest [L], g is gravity [LT−2 ], and ht is the height of the tailwater above the spillway crest [L]. Typical values of the dimensionless constants for ogee spillways with vertical lift gates are C1 = 1.0 and C2 = 0.71 (Ansar and Chen, 2009). The critical depth of flow is a useful reference depth in determining the flow regime through gated spillways. For rectangular openings, the critical depth, yc , through an opening of width L is given by

1 Q2 3 yc = (7.125) gL2 Gated spillways are controlled when the gate opening, h, is smaller than yc , and submerged when the tailwater depth, ht , is greater than yc . When the tailwater is above the weir crest but not of sufficient depth to affect the flow, this condition is referred to as modular submergence. The limiting condition at which the tailwater begins to influence the flow is called the modular submergence limit (Tullis, 2011). In cases where the gate opening does not affect the flow (i.e., h > yc ), the uncontrolled spillway equations described in the previous section are applicable. EXAMPLE 7.15 Water-level measurements are collected via telemetry from a 3-m-wide ogee spillway with a vertical lift gate seated on the crest of the spillway. These measurements indicate that the headwater and tailwater levels are 1.00 m and 0.50 m above the spillway crest, respectively, and the gate opening is 0.25 m. Estimate the discharge through the gate. Solution From the given data: H = 1.00 m, ht = 0.50 m, h = 0.25 m, L = 3 m, and these dimensions are illustrated in Figure 7.33. It is apparent from these data that the flow through the gate is likely to be controlled-submerged flow. Assuming that C2 = 0.71, Equation 7.124 gives Q = C2 hL 2g(H − ht ) = (0.71)(0.25)(3) 2(9.81)(1.00 − 0.50) = 1.67 m3 /s

Section 7.5 FIGURE 7.33: Gated spillway

Spillways

327

Vertical-lift gate

1.00 m 0.50 m

0.25 m

Spillway

Validation that the gate is both controlled and submerged can be determined by comparing the gate opening and tailwater depth to the critical dept, yc , given by  yc =

Q2 gL2



1 3

=

(1.67)2 (9.81)(3)2

1 3

= 0.32 m

Since h < yc (0.25 m < 0.32 m) the flow is controlled, and since hd > yc (0.50 m > 0.32 m) the flow is submerged. Therefore the assumed discharge equation is confirmed, and the estimated discharge is 1.67 m3 /s.

7.5.2.2

Gates seated downstream of the spillway crest

In cases where the gate is seated downstream of the spillway crest, variations of the discharge relationships given in Equation 7.124 are sometimes used, and some common cases are described below. Tainter gate with free discharge. For a Tainter gate located downstream of the spillway crest, the flow rate, Q [L3 T−1 ], through the gate is given by the orifice equation Q = Cd Lh 2gH

(7.126)

where Cd is the discharge coefficient [dimensionless], L is the width of the gate opening [L], h is the height of the gate opening [L], g is gravity [LT−2 ], and H is the head measured as the vertical distance between the upstream water surface and the center of the gate opening [L]. The discharge coefficient, Cd , is related to the distance of the gate seat downstream of the spillway crest and the opening of the gate according to the empirical relation shown in Figure 7.34 (U.S. Army Corps of Engineers, 1990b), where X is the horizontal distance of the seat from the crest, Hd is the design head of the spillway, and θ is the angle that the face of the gate makes with the tangent to the spillway. The angle θ is related to the gate opening by the relation θ = cos−1



a − h R

 (7.127)

where a is the height of the gate trunnion above the spillway and R is the radius of the gate.

328

Chapter 7

Design of Hydraulic Structures

FIGURE 7.34: Tainter gate on spillway

0.95 0.90

θ

H

Lip h

0.85

R a

0.80 Cd

Strut Trunnion

X/Hd

Gate seat

0.75

X/Hd = 0.1−0.3

0.70

X/Hd = 0.0

0.65 0.60 0.55 50

55

60

65

70

75

80

85

90

95

100

105

110

θ Degrees

EXAMPLE 7.16 A Tainter gate is located on a spillway as shown in Figure 7.35. The gate seat is located 2.00 m horizontally and 0.87 m vertically below the crest of the spillway, the gate has a radius of 2.75 m, and the trunnion is 2.10 m above the spillway tangent, which makes an angle of 30◦ with the horizontal. The design head on the spillway crest is 7.00 m and the crest length is 5.00 m. Determine the flow rate under the gate when the water is ponded to a height of 1.00 m above the spillway crest and the gate is opened 1 m. Solution From the given data: X = 2.00 m, Hd = 7.00 m, a = 2.10 m, h = 1.00 m, R = 2.75 m, and L = 5.00 m. The gate angle, θ, is given by Equation 7.127 as     a − h 2.10 − 1.00 = cos−1 = 66◦ θ = cos−1 R 2.75 In this case, X/Hd = 2.00/7.00 = 0.29 and θ = 66◦ , and Figure 7.34 gives the discharge coefficient as Cd = 0.67. When the ponded water is 1.00 m above the crest, the head on the gate opening, H, can be approximated by (neglecting the velocity head) 1.00 h cos (30◦ ) = 0.87 m + 1.00 m − cos (30◦ ) = 1.44 m 2 2 and the corresponding flow rate under the gate is given by Equation 7.126 as Q = Cd Lh 2gH = (0.67)(5.00)(1.00) 2(9.81)(1.44) = 17.8 m3 /s H = 0.87 m + 1.00 m −

FIGURE 7.35: Operational Tainter gate

Tainter gate

2.75 m 1.00 m 1m

0.87 m

2.10 m Spillway

2.00 m

Spillway 30°

tangent

Section 7.5 FIGURE 7.36: Vertical lift gate downstream of spillway crest

Spillways

329

Vertical lift gate

H1 H0

H2

Spillway Flow

Vertical gate with free discharge. For a vertical gate seated downstream of the spillway crest and with a free discharge, the flow rate through the gate, Q [L3 T−1 ], can be estimated using the relation (U.S. Army Corps of Engineers, 1990b) 3

3

H2 − H2 Q = 2 3 1 Q0 H02

(7.128)

where Q0 is the free-flow discharge without the gate [L3 T−1 ], H0 is the reservoir head on the spillway crest used to calculate Q0 [L], H2 is the reservoir head on the gate seat [L], and H1 is the reservoir head on the lip of the gate. These dimensions are shown in Figure 7.36. EXAMPLE 7.17 A vertical gate is located downstream of a spillway crest as shown in Figure 7.37. When the gate is out of the water, the free-flow discharge is 10 m3 /s. Estimate the discharge when the gate opening is 0.50 m and the pool elevation is 1.40 m above the lip of the gate. Solution From the given data: Q0 = 10 m3 /s, H1 = 1.40 m, and h = 0.50 m. Under the given operating conditions the reservoir head on the gate seat, H2 , is given by H2 = H1 + h = 1.40 m + 0.50 m = 1.90 m and the head on the crest of the spillway is 0.50 m + 1.40 m − 0.25 m = 1.65 m. Assuming that the pool elevation remains as shown in Figure 7.37 when the gate is fully open (and out of the water), then H0 = 1.65 m. Substituting the given and derived parameters into Equation 7.128 gives FIGURE 7.37: Operational vertical gate

Gate

1.40 m

Crest 0.50 m 0.25 m

Spillway

Flow

330

Chapter 7

Design of Hydraulic Structures 3

3

H2 − H2 Q = 2 3 1 Q0 H02 3

3

(1.90) 2 − (1.40) 2 Q = 3 10 (1.65) 2 which yields a gate discharge of Q = 4.5 m3 /s.

7.6 Stilling Basins The terminal structure of a spillway is usually designed to dissipate the energy associated high-flow velocities at the base of the spillway. A stilling basin forces the occurrence of a hydraulic jump within the basin, stabilizes the jump, and reduces the length required for the jump to occur. An example of a stilling basin immediately downstream of a gated spillway (under construction) is shown in Figure 7.38. In this case, the stilling basin consists of a concrete apron with baffle blocks that create the necessary energy loss to cause a hydraulic jump to occur. Stilling basins are usually constructed of concrete, and low-velocity subcritical flow exiting a stilling basin is necessary to prevent erosion and the undermining of the associated dam and spillway structures. 7.6.1

Type Selection

The U.S. Bureau of Reclamation has developed several standard stilling-basin designs (U.S. Bureau of Reclamation, 1987), and the selection of the appropriate stilling-basin design is based on the Froude number (Fr) of the incoming flow. The selection guidelines are given below. 1.7 < Fr 4.5: For Froude numbers greater than 4.5, either Type III or Type II basins are recommended. The Type III basin, shown in Figure 7.40, includes baffle blocks, and so is limited to applications where the incoming velocity does not exceed 18 m/s (60 ft/s). For inflow velocities exceeding 18 m/s (60 ft/s), the Type II basin, shown in Figure 7.41, is recommended. The Type II basin is slightly longer than the Type III basin, has no baffle blocks, has a dentated end sill, and the tailwater is recommended to be 5% greater than the conjugate depth (of the hydraulic jump) to help prevent sweepout. The stilling basins shown in Figures 7.39 to 7.41 have several features in common, specifically: (1) chute blocks at the entrance to the stilling basin, which block a portion of the incoming flow; this stabilizes the jump and reduces the required length of the basin; (2) an end sill, which is a gradual rise at the end of the basin that shortens the jump and prevents scour downstream; and (3) baffle blocks on the floor of the basin, which dampen velocities. Baffle blocks are used only in stilling basins with relatively low inflow velocities; otherwise cavitation damage may result. Vertical sidewalls are preferred over sloped walls for stilling basin structures, since vertical sidewalls promote stable flow and are more likely to contain hydraulic jump turbulence (ASCE, 2006). 7.6.2

Design Procedure

The dimensions of the structural features within the standard stilling basins are determined by the inflow Froude number as shown in Figures 7.39 to 7.41. Once the inflow Froude number is determined and the appropriate stilling basin is selected, the required elevation of the bottom of the stilling basin must be determined. A recommended design procedure is given below. Step 1: Determine inflow conditions. Selecting the type of stilling basin requires predicting the flow rate and velocity at the toe of the spillway. According to the U.S. Bureau of Reclamation (USBR, 1987), if the stilling basin is located immediately downstream of the crest of an overflow spillway, or if the spillway chute is no longer than the hydraulic head, energy losses can be neglected. In this case, the hydraulic head is defined as the difference in elevation between the reservoir water surface and the water surface downstream of the stilling basin. If the spillway chute length is between one and five times the hydraulic head, an energy loss of 10% of the hydraulic head is recommended. For spillway chute lengths in excess of five times the hydraulic head, an energy loss of 20% of the hydraulic head is recommended. For more accurate estimates of head loss, the gradually varied flow equation can be solved along the spillway chute; however, it should be recognized that this equation is not valid in the vicinity of the spillway crest, where the flow is not gradually varied and the boundary layer is not fully developed. Step 2: Determine the floor elevation of the stilling basin. The elevation of the floor of the stilling basin is determined by matching the tailwater and conjugate elevation curves over a range of operating discharges. The tailwater elevation is the elevation of the water surface immediately downstream of the stilling basin, which is normally determined by backwater computation from a control section in the downstream channel or can be approximated using the normal depth of flow in the downstream channel. The conjugate elevation is the sequent elevation of the hydraulic jump that occurs in the stilling basin. If the tailwater elevation is lower than the conjugate elevation of the jump, then the jump may be swept out of the basin. On the other hand, a tailwater elevation that is higher than the conjugate elevation causes the jump to back up against the spillway chute and be drowned, no longer dissipating as much energy. The ideal situation is one in which the conjugate elevation of the hydraulic

Section 7.6

Stilling Basins

FIGURE 7.40: Type III stilling basin characteristics, Fr > 4.5 and V1 … 18 m/s Source: U.S. Bureau of Reclamation (1987).

w1 = y1 S1= y1

End sill

0.2h3

Chute blocks

Baffle blocks 0.375h3 w3 = 0.75h3

0.5y1

S3 = 0.75h3 h1 = y1 2:1 Slope h3

1:1 Slope

h4

0.8y2 L (a) Type III basin dimensions

24

Froude number 10 12

8

14

16

18 24

Note: y1 and y2 are the initial and sequent depths of the hydraulic jump in the stilling basin.

20

T.W. depth y1

6

20

0

. = T.W y2

16

1.

y2 1 ⫽ ( y1 2

1 + 8Fr2 − 1 )

16

12

12

8

8

T.W. depth y1

4

(b) Minimum tailwater depths 4 Baffle block height h3 End sill height h4

2

0

2

h3 or h4 y1 y1

h3 or h4 y1 y1

4

0

(c) Height of baffle blocks and end sill

L y2

3

L y2

3

2

4

6

8

10 12 Froude number (d) Length of jump

14

16

2 18

333

Design of Hydraulic Structures

FIGURE 7.41: Type II stilling basin characteristics, Fr > 4.5 Source: U.S. Bureau of Reclamation (1987).

Dentated sill Chute blocks

S1 = y1 w1 = y1

0.02y2 y1 2

S2 = 0.15y2 w2 = 0.15y2

Slope = 2:1

h2 = 0.2y2

h1 = y1 L

(a) Type II basin dimensions Froude number 4

6

8

10

12

14

16

18

Note: y1 and y2 are the initial and sequent depths of the hydraulic jump in the stilling basin.

24

24

T.W. depth y1

20

20

05

. = T.W y2

16

y2 1 ⫽ ( y1 2

1 + 8Fr2 − 1 )

16

12

12

8

8

4

L y2

1.

4

(b) Minimum tailwater depths

5

5

4

4

3

4

6

8

10 12 Froude number (c) Length of jump

T.W. depth y1

Chapter 7

14

16

18

3

L y2

334

Section 7.6

Stilling Basins

335

jump perfectly matches the tailwater elevation over the full range of operating discharges, but this is unlikely to occur. Instead, the basin floor elevation is normally set such that the conjugate elevation matches the tailwater elevation at the maximum design discharge, assuming that the conjugate elevation is less than the tailwater elevation when the discharge is less than the maximum discharge. In some cases, the tailwater elevation equals the conjugate elevation at a discharge less than the maximum discharge, and in these cases the stilling basin should be designed for the lower discharge. In no case should the tailwater elevation be less than the conjugate elevation within the operating range of the stilling basin. The basin floor elevation need not be the same as the bottom elevation of the downstream channel. EXAMPLE 7.18 The maximum design discharge over a spillway is 280 m3 /s, and the spillway and stilling basin are 12 m wide. The reservoir behind the spillway has a water-surface elevation of 60.00 m, and the river watersurface elevation downstream of the stilling basin is 30.00 m. Assuming a 10% loss of hydraulic head in the flow down the spillway, find the elevation of the floor of the stilling basin so that the hydraulic jump forms in the basin. Design the stilling basin. Solution From the given data: Q = 280 m3 /s and b = 12 m. A schematic diagram showing the spillway, stilling basin, and downstream river is given in Figure 7.42. Since the water loses 10% of its hydraulic head flowing down the spillway, and the water-elevation behind the spillway is 60 m, then, taking the elevation of the bottom of the stilling basin as X, the specific energy, E1 , at entrance to the stilling basin (i.e., base of the spillway) is given by E1 = (60.00 − X) − 0.10(60.00 − 30.00) = 57.00 − X Taking y1 as the depth of flow at the entrance to the stilling basin, E1 = y1 + 57 − X = y1 +

Q2 2g(by1 )2 2802 2(9.81)(12y1 )2

which simplifies to X = 57 − y1 −

FIGURE 7.42: Stilling basin

27.75

(7.129)

y21

El.60.00 m

Spillway Tailwater (river) El.30.00 m

y2 y1

El. X L Stilling basin

336

Chapter 7

Design of Hydraulic Structures Since the elevation of the water-surface downstream of the stilling basin is 30.00 m, the conjugate depth of the hydraulic jump, y2 , is given by (7.130) y2 = 30 − X and the relationship between y1 and y2 is given by the hydraulic jump equation ⎛ ⎛ ⎞ ⎞     2 2   y ⎜ 8q ⎟ y1 ⎜ 8(280/12) ⎟ y2 = 1 ⎝−1 + 1 + ⎝−1 + 1 + ⎠= ⎠ 2 2 gy31 9.81y31 which simplifies to

⎞ ⎛  y1 ⎝ 444.0 ⎠ y2 = −1 + 1 + 2 y31

(7.131)

Combining Equations 7.129 to 7.131 yields 30 − 57 + y1 +

27.75 y21

⎞ ⎛  y1 ⎝ 444.0 ⎠ = −1 + 1 + 2 y31

which simplifies to 1.5y1 +

27.75 y21



y1 2

 1 +

444.0 y31

− 27 = 0

The solutions of this equation are y1 = 27.1 m and y1 = 0.87 m. These depths correspond to subcritical and supercritical flows, respectively (Fr = 0.1 and Fr = 9.2). Since the upstream depth of flow must be supercritical, take y1 = 0.87 m, and Equation 7.129 gives X = 57 − 0.87 −

27.75 = 19.5 m 0.872

Therefore, the stilling basin must have an elevation of 19.5 m for the hydraulic jump to occur within the stilling basin. Since the Froude number of the flow entering the stilling basin is 9.2 and the entrance velocity is 280/(0.87 * 12) = 26.8 m/s, a Type II basin is required. Using this type of stilling basin, it is recommended that the tailwater depth in the stilling basin be 5% greater than the sequent depth of the hydraulic jump to prevent sweepout. Since the elevation of the stilling basin was calculated assuming that the sequent depth and tailwater depth in the stilling basin were the same, the invert elevation of the stilling basin should be recalculated taking the tailwater depth to be 5% greater than the sequent depth. In this case, Equation 7.130 becomes 1.05y2 = 30 − X and the required elevation of the stilling basin becomes 18.6 m, with an entering flow depth of 0.86 m, Froude number of 9.3, and velocity of 27.1 m/s. This indicates a Type II basin, for which the appropriate invert has now been calculated. The dimensions of the chute blocks and dentated sill in the Type II stilling basin are given in Figure 7.41 in terms of the entering depth (y1 ) of 0.86 m and sequent depth (y2 ) given by Equation 7.131 as 10.9 m. The length, L, of the stilling basin is derived from Figure 7.41(c), where Fr = 9.3, L/y2 = 4.3, and hence L = 4.3(10.9) = 47 m.

In cases where flow rates are small, energy dissipation can be accomplished within the spillway itself by using steps, with additional energy dissipation accomplished by placing gabions on the steps. The former structure is sometimes referred to as a horizontal-stepped weir, and the latter structure as a gabion-stepped weir. Design guidelines for these structures can be found in Boes and Hager (2003) and Chinnarasri et al. (2008). 7.7 Dams and Reservoirs Reservoir impoundments are created by building dams, and these impoundments typically serve multiple purposes, including water supply, flood control, hydroelectric power generation, navigation, and water-based recreation activities. Dams and reservoirs have enhanced

Section 7.7

Dams and Reservoirs

337

the health and economic prosperity of people around the world for centuries; however, dam construction can also have negative effects such as reduced streamflows, degraded water quality, and impacts on migrating fish. The principal parts of a dam are illustrated in Figure 7.43. The face of a dam is the exposed surface of the structure, which contains materials such as rockfill, concrete, or earth. There are both upstream and downstream faces. Abutments are the sides of the dam structure that tie into canyon walls or contact the far sides of a river valley. Left and right abutments are identified by facing downstream from the reservoir. The top of the dam is called the crest, while the point of intersection between the downstream face and the natural ground surface is called the toe. The crest usually has a road or walkway along the top that follows the entire length of the dam, where the length is measured from abutment to abutment. A parapet wall is often built along the dam crest for ornamental, safety, and wave-control purposes. All dams have a spillway to allow excess water to flow past the dam in a safe manner. Dams on rivers used for navigation often include locks, which are rectangular box-like structures with gates at both ends that allow vessels to move upstream and downstream through the dam. The highest lock in the United States is the John Day lock on the Columbia River with a lift of 34.5 m (114 ft). 7.7.1

Types of Dams

There are four types of dams: gravity concrete, buttress, concrete arch, and earthen embankment dams. These types of dams are described and illustrated below. Gravity Concrete Dam. A gravity concrete dam is a solid concrete structure that uses its mass (weight) to hold back water. It requires massive amounts of concrete, especially at its base, to provide the weight necessary to withstand the hydrostatic force exerted by the water impounded behind the dam. An example of a gravity concrete dam is the Grand Coulee Dam on the Columbia River, which is shown in Figure 7.44(a). The Grand Coulee Dam has a height of 168 m (551 ft), a crest length of 1730 m (1.08 mi), a spillway discharge capacity of 28,300 m3 /s (106 ft3 /s), and a hydroelectric power generation capacity of 6.48 GW. It is operated by the U.S. Bureau of Reclamation and impounds the Franklin D. Roosevelt reservoir (Lake Roosevelt), with a storage capacity of 6.40 * 109 m3 (5.19 * 106 ac·ft), which extends 240 km (150 mi) upstream to the Canadian border. The Grand Coulee Dam is the largest concrete structure and hydroelectric facility in the United States, and is designated as a National Historic Civil Engineering Landmark by the American Society of Civil Engineers. FIGURE 7.43: Principal parts of a dam Source: Wisconsin Department of Natural Resources, 2012.

Right abutment Downstream slope Upstream slope Principal chute spillway Spillway training walls

Berm

Top of dam Riprap Toe drain outlet

Toe of embank m

ent

Emerg ency s

pillway

Left abutment

338

Chapter 7

Design of Hydraulic Structures

FIGURE 7.44: Major dams in the United States Source: U.S. Bureau of Reclamation (2012b; 2012c).

(a) Grand Coulee Dam

(b) Hoover Dam

Concrete Arch Dam. A concrete arch dam has a curvature design that arches across a canyon and has abutments embedded into solid rock walls. Hydrostatic forces push against the curve of the concrete arch and compress the material inside the structure. This pressure makes the dam more solid and dissipates water pressure into the canyon walls. Arch dams require less concrete than gravity concrete dams but must have solid rock walls as anchors for the abutments. Arch dams have thinner sections than comparable gravity dams, and are feasible only in canyons with walls capable of withstanding the thrust produced by the arch action. An example of a concrete arch dam is the Hoover Dam on the Colorado River, which is shown in Figure 7.44(b). Unlike the Grand Coulee Dam, the spillway at the Hoover Dam is located on the side of the upstream reservoir, and the water that flows over the side spillway is routed through a tunnel and is discharged via two “flip bucket” energy-dissipating structures in front of the dam as shown in Figure 7.44(b). The Hoover Dam has a height of 221 m (725 ft), a crest length of 379 m (1240 ft), a spillway discharge capacity of 11,300 m3 /s (399,000 ft3 /s), and a hydroelectric power generation capacity of 1.34 GW. It is operated by the U.S. Bureau of Reclamation and impounds Lake Mead, which is the largest reservoir in the United States, with a storage capacity of 3.67 * 1010 m3 (29.8 * 106 ac·ft) and a surface area of 658 km2 (254 mi2 ). The Hoover Dam is the highest concrete dam in the Western Hemisphere and is designated as a National Historic Civil Engineering Landmark by the American Society of Civil Engineers. Buttress Dam. A gravity concrete dam may have triangular supports called buttresses on the downstream side to strengthen it and to distribute water pressure to the foundation. Such dams are commonly placed in a separate category from concrete gravity dams and are called buttress dams. Typical configurations for buttress dams include the flatslab configuration and multiple-arch configuration. The face of a flat-slab buttress dam is a series of flat reinforced concrete slabs, while the face of a multiple-arch buttress dam consists of a series of arches that permit wider spacing of the buttresses. Buttress dams usually require only one-third to one-half as much concrete as gravity dams of similar height. Consequently, they may be used on foundations that are too weak to support a gravity dam. An example of a buttress dam with a multiple-arch configuration is the Bartlett Dam on the Verde River (near Phoenix, Arizona), which is shown

Section 7.7

Dams and Reservoirs

339

FIGURE 7.45: Bartlett Dam Source: U.S. Bureau of Reclamation (2012).

in Figure 7.45. The Bartlett Dam has a height of 94 m (308 ft), a crest length of 251 m (823 ft), and a spillway discharge capacity of 8140 m3 /s (287,500 ft3 /s). It impounds Bartlett Lake, which has a surface area of 11 km2 (2700 ac). Earthen Embankment Dam. More than 50% of the total volume of an earthen embankment dam or earthfill dam consists of compacted earth material. Large earthen dams have an impervious core of clay, or other material of low permeability, that prevents reservoir water from rapidly seeping through or beneath the foundation of the structure. Most large earthen dams also have drains installed along the downstream toe of the dam. These small parallel conduits transport any seepage water inside the earth dam to locations downstream and away from the toe of the dam. This prevents water from building up inside the structure and eventually eroding the material within the dam. The base of an earthen dam is very broad when compared to the crest of the dam. A 2 : 1 (H : V) downstream slope and a 3 : 1 (H : V) upstream slope are fairly common. The upstream face of the dam is often covered with riprap to prevent erosion by wave action. Earthen embankment dams are usually the most economical to build in small watersheds or across broad valleys. Almost 80% of all dams in the world are earthen dams (Cech, 2002). An example of an earthen embankment dam is the Prosser Creek Dam on Prosser Creek (near Truckee, California), which is shown in Figure 7.46. The FIGURE 7.46: Prosser Creek Dam Source: U.S. Bureau of Reclamation (2012a).

340

Chapter 7

Design of Hydraulic Structures

Prosser Creek Dam has a height of 50 m (163 ft) and a crest length of 558 m (1830 ft). It impounds Prosser Creek Reservoir, which has a storage capacity of 3.7 * 107 m3 (3 * 104 ac·ft). In addition to the above classifications, dams may also be categorized as overflow or nonoverflow dams. Overflow dams are concrete structures designed for water to flow over their crests, while nonoverflow dams have spillways to prevent overtopping. Earthen embankment dams are damaged by the erosive action of overflowing water and are generally designed as nonoverflow dams. The geology, topography, and streamflow at a site generally dictate the type of dam that is appropriate for a particular location. Commonly used types of dam spillways are overflow spillways, chute spillways, shaft spillways, side-channel spillways, and limited-service spillways. An overflow spillway is a section of dam designed to permit water to pass over its crest, and overflow spillways are widely used on gravity, arch, and buttress dams, and are generally an integral part of the dam structure. The Grand Coulee Dam shown in Figure 7.44(a) has an overflow spillway. Chute spillways, also called trough spillways, are normally used with earth- or rock-filled embankment dams, and are normally designed to minimize excavation by setting the invert profile to approximate the profile of the natural ground. Shaft spillways include various configurations of crest designs, with or without gates, all of which transition into a closed conduit (tunnel) system immediately downstream from the crest. Side-channel spillways consist of an overflow weir discharging into a narrow channel in which the direction of flow is approximately parallel to the weir crest. Limited-service spillways are designed to operate very infrequently, and with the knowledge that some degree of damage or erosion will occur during operation of the spillway. 7.7.2

Reservoir Storage

The three major components of reservoir storage are: (1) dead storage for sediment collection, (2) active storage or conservation storage for firm and secondary yields, and (3) flood storage for reduction of downstream flood damages. These storage components are illustrated in Figure 7.47. Firm yield, which is also called safe yield, can be defined as the maximum amount of water that will be consistently available from a reservoir based on historical streamflow records, whereas secondary yield is any amount greater than firm yield. Dead/Inactive Storage. Dead or inactive storage is equal to the capacity at the bottom of the reservoir that is reserved for sediment accumulation during the anticipated life of the reservoir. The top of the inactive pool elevation may be fixed by the invert of the lowest outlet, or by conditions of operating efficiency for hydroelectric turbines. Typically, the dead-storage capacity is lost to sedimentation gradually over many decades. However, in extreme cases, small reservoirs have been almost completely filled with sediment during a single major flood event. Active Storage. Active storage is the storage between the top of the dead storage pool and the normal reservoir water-surface elevation. The active storage capacity is sometimes referred to as the active conservation pool or conservation storage. This storage is FIGURE 7.47: Reservoir storage zones

Dam Maximum pool elevation Inflow

Spillway Flood storage

Normal pool elevation

Hydroelectric turbine

Active storage T

Minimum pool elevation

Low-level outlet Tailwater

Dead storage

Section 7.7

Dams and Reservoirs

341

available for various uses such as water supply, irrigation, navigation, recreation, and hydropower generation. The reservoir water surface is maintained at or near the designated top of the conservation pool level as streamflows and water demands allow. Drawdowns are made as required to meet the various needs for water. Flood Storage. Flood storage is the storage between the top of the conservation pool and the maximum permissible water level in the reservoir. This storage is reserved for flood control and is usually kept unused or empty most of the time to be used for temporary storage of floodwater during storm events. The flood-storage volume includes the flood-control pool and surcharge capacity. Under normal operating conditions surface runoff is handled within the flood-control pool. There is an additional surcharge capacity, which is the storage available between the normal flood-control pool and the maximum permissible water level in the reservoir. The surcharge capacity is not used except during abnormal conditions, and is reserved for extreme flood events that are larger than the design flood. The crest of the emergency spillway is typically located at the top of the flood-control pool, with normal flood releases being made through other outlet structures. The top of the surcharge storage is established during project design from the perspective of dam safety. Freeboard. The freeboard is the difference in elevation between the dam crest and the maximum permissible water level in the reservoir. This allowance provides for wave action and an additional factor of safety against overtopping. For many reservoir projects, a full range of outflow rates are discharged through a single spillway. However, some reservoirs have more than one spillway, with a service spillway conveying smaller, frequently occurring releases, and an emergency spillway that is used only rarely during extreme floods. The crest elevation of the service spillway is typically located at the top of the conservation pool, and the crest of the emergency spillway is located at the top of the flood-control pool. Outlet works are used for releases from storage both below and above the spillway crest; however, discharge capacities for outlet works are typically much smaller than for spillways. The components of an outlet-works facility include an intake structure in the reservoir, one or more conduits or sluices through the dam, gates located either in the intake structure or conduits, and a stilling basin or other energy-dissipation structure at the downstream end. 7.7.2.1

Sediment accumulation

The storage zones illustrated in Figure 7.47 present a simplified view of reservoir capacity, since sedimentation storage capacity is generally provided in all storage zones. Typically, sediment reserve storage capacity is provided to accommodate sediment deposition expected to occur over a specified design life which, for large projects, is typically on the order of 50–100 years. Reservoir sedimentation amounts are predicted as the sediment yield entering the reservoir multiplied by a trap efficiency. The reservoir trap efficiency (TE) is a measure of the proportion of the inflowing sediment that is deposited, where TE (%) =

sediment amount deposited * 100 sediment amount entering

(7.132)

Analyses of sediment measurements from many reservoirs resulted in Figure 7.48, which may be used to estimate the TE as a function of the ratio of the reservoir storage capacity to the average annual inflow volume for normal-ponded reservoirs (USBR, 1987). Determining the TE of a reservoir allows designers and managers to plan the daily operation of the reservoir, and to estimate when dredging might become necessary. The reservoir TE curve shown in Figure 7.48 was originally developed by Brune (1953) and has been widely used and revised. Several empirical equations have been developed and used to estimate TE, and the most prominent of these equations are given in Table 7.8. In using any of the empirical equations listed in Table 7.8, several ancillary considerations should be taken into account. The Brown (1949) equation was one of the first to be developed

342

Chapter 7

Design of Hydraulic Structures

FIGURE 7.48: Reservoir trap efficiency Source: USBR (1987).

100 90 Sediment trapped (%)

80 70 60 50 40 30 20 10 0 0.001

0.01 0.1 1.0 10 Ratio of reservoir storage capacity to average annual inflow volume (years)

100

TABLE 7.8: Trap Efficiency Empirical Equations

Equation

Nomenclature 

TE = 100 1 −  TE = 100 1 −

1 1 + 2.1 * 10−5 (C/W) 1 1 + 0.1(C/I)

TE = −22 +

C = reservoir capacity (m3 )

Brown (1949)

W = watershed area (km2 )



C = reservoir capacity [L3 ]

Brune (1953)

I = annual inflow volume [L3 ]

  log(C/I) TE = 100 0.970.19 

Reference



C = reservoir capacity [L3 ]

119.6(C/I) 0.012 + 1.02(C/I)



Dendy and Cooper (1974)

I = annual inflow volume [L3 ] C = reservoir capacity [L3 ]

Heinemann (1981)

I = annual inflow volume [L3 ] 

(C/I)2 TE = 100 − 1000 L

−0.2 + 12

C = reservoir capacity (m3 )

Churchill (1947);

I = inflow rate (m3 /s) L = reservoir length (m)

USACE (1995)

for TE estimation, and this equation is still useful when no other information is available other than the reservoir capacity (C) and the watershed area (W). The main limitation of the Brown (1949) equation is that it does not take into account reduced inflows that occur in drought years. The other empirical equations account for variations in the reservoir inflow rate (I), with C/I providing a measure of the detention time in the reservoir. An advantage of the Brune (1953) equation is that it allows for negative trap efficiencies, which indicates the occurrence of scour. The Dendy and Cooper (1974) equation was developed from a study of 17 small reservoirs with catchment areas less than 60 km2 (23 mi2 ). The Heinemann (1981) equation was developed for small agricultural ponds with catchment areas less than 40 km2 (15 mi2 ). It is generally recognized that empirical equations used to estimate TE can give only a general estimate of TE since these equations do not take into account reservoir flow

Section 7.7

Dams and Reservoirs

343

dynamics. Reductions in TE with time as the reservoir fills can be a significant factor that should be taken into account (e.g., Minear and Kondolf, 2009). Comparative studies on the relative performance of the TE equations are rare; however, application of these equations to the Coralville reservoir in Iowa, with a drainage area of 8100 km2 (3100 mi2 ), showed that the Churchill (1947) equation provided the best agreement with observations (Espinosa-Villegas and Schnoor, 2009). EXAMPLE 7.19 A reservoir covers an area of 400 km2 and has an average depth of 24.8 m. The inflow to the reservoir is from a river with an average flow rate of 3000 m3 /s and a typical suspended-sediment concentration of 150 mg/L. Estimate the rate at which sediment is accumulating in the reservoir, the rate at which the depth of the reservoir is decreasing due to sediment accumulation, and the average suspended sediment concentration in the water released from the reservoir. Assume that the accumulated sediment has a bulk density of 1600 kg/m3 . Solution From the given data, the average flow rate into the reservoir is 3000 m3 /s = 9.46 * 1010 m3 /year, and the typical suspended-sediment concentration is 150 mg/L = 0.150 kg/m3 , therefore the average sediment load entering the reservoir is given by sediment load = inflow rate * suspended-sediment concentration = 9.46 * 1010

kg m3 * 0.150 = 1.42 * 1010 kg/year year m3

The area of the reservoir is 400 km2 = 4 * 108 m2 , and the average depth of the reservoir is 24.8 m, therefore the reservoir storage capacity is given by storage capacity = area of reservoir * average depth = 4 * 108 m2 * 28.4 m = 9.92 * 109 m3 and

9.92 * 109 m3 storage capacity = = 0.10 year annual inflow 9.46 * 1010 m3 /year Based on this ratio of storage capacity to annual inflow (= 0.10 year), the percent of sediment trapped in the reservoir is estimated from Figure 7.48 as 87%. Since the average sediment load delivered by the river to the reservoir is 1.42 * 1010 kg/year, the rate at which sediment is accumulating in the reservoir is 0.87 * 1.42 * 1010 kg/year = 1.24 * 1010 kg/year. Since the bulk density of the sediment accumulating at the bottom of the reservoir is 1600 kg/m3 and the area of the reservoir is 400 km2 = 4 * 108 m2 , the rate at which sediment volume is accumulating is given by sediment volume accumulation rate = =

sediment trap rate sediment bulk density 1.24 * 1010 kg/year 1600 kg/m3

= 7.75 * 106 m3 /year

Since the plan area of the reservoir is 400 km2 = 4 * 108 m2 , the rate of sediment accumulation on the bottom of the reservoir is given by rate of sediment accumulation = =

sediment volume accumulation rate reservoir area 7.75 * 106 m3 /year 4 * 108 m2

= 0.019 m/year = 1.9 cm/year

At this rate, it will take approximately 130 years for the reservoir capacity to decrease by 10% due to sediment accumulation. Since 87% of the incoming sediment is trapped by the reservoir and the sediment load delivered by the river to the reservoir is 1.42 * 1010 kg/year, the sediment load released from the reservoir is

344

Chapter 7

Design of Hydraulic Structures (1 − 0.87)(1.42 * 1010 ) kg/year = 5.36 * 109 kg/year. Assuming that the average flow rate of water released from the reservoir is equal to the average flow rate entering the reservoir (= 9.46 * 1010 m3 /year), then sediment concentration downstream = =

sediment load release from reservoir flow rate from reservoir 5.36 * 109 kg/year 9.46 * 1010 m3 /year

= 0.057 kg/m3 = 57 mg/L

Therefore, the reservoir reduces the suspended-sediment concentration from 150 mg/L to 57 mg/L, a reduction of 62%. It is noteworthy that the reservoir trap efficiency (= 87%) is not equal to the reduction in suspended-solids concentration (= 62%).

7.7.2.2

Determination of storage requirements

A relatively simple method for determining reservoir storage requirements is based on a mass diagram, which shows the cumulative reservoir inflows and demands plotted against time. Water demand includes water withdrawn directly from the reservoir, required downstream releases, and evaporation of water stored in the reservoir. When the cumulative differences between inflow and demand are plotted on a mass diagram, the maximum vertical distance between the highest point of the cumulative difference curve and the lowest subsequent point represents the required capacity. This method, sometimes referred to as a mass-curve analysis or Rippl analysis (Rippl, 1883), assumes that future inflows to a reservoir will be a duplicate of the historical record. Commonly, storage calculations are based on inflows during a critical low-flow period such as the most severe drought of record. Once the critical period is chosen, the required storage is determined using the mass-curve analysis. Reservoirs in humid regions tend to refill annually and function principally to smooth out intraannual (seasonal) fluctuations in flows. In more arid regions, additional carry over storage is required to smooth out interannual variations, and reservoirs in these regions fill only rarely. EXAMPLE 7.20 A reservoir is to be sized to stabilize the flows in a river and provide a dependable source of irrigation water. During a critical 24-month drought, the average monthly inflows to the reservoir and the average water demand for downstream flow, irrigation, and evaporation are given in the following table:

Month∗

Average inflow (m3 /s)

Average demand (m3 /s)

1 2 3 4 5 6 7 8

1450 1730 1910 1600 1770 1860 1480 1530

1600 1700 1800 1900 2000 2000 2000 2000

Month

Average inflow (m3 /s)

Average demand (m3 /s)

9 10 11 12 13 14 15 16

2150 2200 1970 1810 1540 1420 1640 1890

1900 1800 1700 1600 1600 1700 1800 1900

Month

Average inflow (m3 /s)

Average demand (m3 /s)

17 18 19 20 21 22 23 24

1950 1870 1720 1830 1920 1850 1840 1680

2000 2000 2000 2000 1900 1800 1700 1600

Note: ∗ Month “1” is January.

Use these data to estimate the required storage volume of the reservoir. Solution From the given data, the cumulative inflow, demand, and difference between inflow and demand are tabulated below, and a plot of cumulative inflow minus demand versus time is shown in Figure 7.49.

Section 7.7

Dams and Reservoirs

Month

Cumulative inflow (* 1010 m3 )

Cumulative demand (* 1010 m3 )

Cumulative inflow − demand (* 1010 m3 )

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

0.39 0.81 1.32 1.73 2.21 2.69 3.09 3.50 4.05 4.64 5.15 5.64 6.05 6.39 6.83 7.32 7.84 8.33 8.79 9.28 9.78 10.27 10.75 11.20

0.43 0.84 1.32 1.81 2.35 2.87 3.40 3.94 4.43 4.91 5.36 5.78 6.21 6.62 7.11 7.60 8.13 8.65 9.19 9.72 10.22 10.70 11.14 11.57

−0.04 −0.03 0.00 −0.08 −0.14 −0.18 −0.32 −0.44 −0.38 −0.27 −0.20 −0.15 −0.16 −0.23 −0.27 −0.28 −0.29 −0.32 −0.40 −0.44 −0.44 −0.42 −0.39 −0.37

345

The plotted data in Figure 7.49 indicate that the reservoir storage must be sufficient to accommodate the cumulative deficit between demand and inflow that occurs between Month 3 and Month 21, and this deficit is equal to 0.44 * 1010 m3 . Assuming that the reservoir is full at the beginning of this critical interval, then an active reservoir storage of 0.44 * 1010 m3 will be sufficient. If the reservoir is assumed to be partially full at the beginning of the critical interval between Months 3 and 21, then the unfilled volume must be added to the deficit volume (0.44 * 1010 m3 ) to determine the required storage volume of the reservoir. 0.05 Cumulative inflow minus demand (× 1010 m3)

FIGURE 7.49: Cumulative difference between inflow and outflow volumes

0.00 −0.05 −0.10 −0.15 −0.20 −0.25 Required storage

−0.30 −0.35 −0.40 −0.45 −0.50

2

4

6

8

10

12 14 Month

16

18

20

22

24

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7.7.3

Hydropower

Hydroelectric power is an important source of energy, with hydropower responsible for about 15% of the electric energy generated in the United States and over 70% of electric energy generated in Brazil and Norway. There are two types of hydropower plants: (1) runof-river plants that use direct streamflow, and where energy output is directly related to the instantaneous flow rate in the river; and (2) storage plants that use a reservoir to store water, and where energy is produced using controlled water releases. Run-of-river plants use the sustained flow of a stream or river to turn the turbines for electricity generation. This type of hydropower plant usually has limited storage capacity and provides a continuous output of electricity. Storage plants use a reservoir of sufficient size to increase the amount of water available for power generation and to carry over water through dry periods. A pumpedstorage plant uses external power during low periods of demand to pump water back up the system to a headwater reservoir, and this water is subsequently used for hydropower generation during peak demand periods. This recycling of water represents an economic efficiency between peak and off-peak demand requirements. 7.7.3.1

Turbines

Turbines are the central components of all hydropower facilities. Their role is to extract energy from flowing water and convert it to mechanical energy to drive electric generators. There are two basic types of hydraulic turbines: impulse turbines and reaction turbines. Impulse Turbines. In an impulse turbine, a free jet of water impinges on a rotating element, called a runner, which is exposed to atmospheric pressure. The runner in an impulse turbine is sometimes called an impulse wheel or Pelton wheel, in honor of Lester A. Pelton (1829–1908), who contributed much to its development in the early gold-mining days in California. A typical impulse turbine is shown in Figure 7.50, where the runner has a series of split buckets located around its periphery. When the water-jet strikes the dividing ridge of the bucket, it is split into two parts that discharge at both sides of the bucket. Only one jet is used on small turbines, but two or more jets impinging at different points around the runner are often used on large units. The jet flows are usually controlled by a needle nozzle, and some jet velocities exceed 150 m/s (490 ft/s). The generator rotor is usually mounted on a horizontal shaft between two bearings with the runner installed on the projecting end of the shaft; this is called a single-overhung installation. In some cases, runners are installed on both sides of the generator to equalize bearing loads, and this is called a double-overhung installation. The diameters of runners range up to about 5 m (16 ft). The schematic layout of an impulse-turbine installation is shown in Figure 7.51. Water is delivered from the upstream reservoir to the turbine through a large-diameter pipe called a penstock. The head loss in the penstock between the reservoir and the nozzle entrance is hL , and the static head, H, is equal to the difference between the FIGURE 7.50: Impulse turbine Source: Capture 3D, Inc., Costa Mesa, CA, and GOM mbH, Braunschweig, Germany (2012).

Section 7.7 FIGURE 7.51: Impulse-turbine system

Water surface

Dams and Reservoirs

347

Energy grade line hL

Reservoir Q Impulse turbine

he

H

Penstock

Tailwater

water elevation in the reservoir and the elevation of the nozzle. The nozzle is considered to be an integral part of the impulse turbine, and so the net head or effective head, he , on the turbine is equal to the static head minus the head loss, hL , and is given by he = H − hL

(7.133)

The head at the entrance to the nozzle is expended in four ways: (1) energy is lost in fluid friction in the nozzle, known as nozzle loss; (2) energy is lost in fluid friction over the buckets of the turbine runner; (3) kinetic energy is carried away in the water discharged from the buckets; and (4) energy is used in rotating the buckets. Therefore, the head directly available to the runner, hT , is given by hT = he − kj

Vj2 2g

− k

v22 V2 − 2 2g 2g

(7.134)

where kj is a nozzle head-loss coefficient that depends on the geometry of the nozzle, Vj is the jet velocity, k is the bucket friction loss coefficient, v2 is the velocity of water relative to the bucket at the exit from the bucket, and V2 is the absolute velocity of the water leaving the bucket. Typical values of k are in the range of 0.2–0.6. The (theoretical) power extracted by the turbine, PT , is therefore given by PT = γ QhT

(7.135)

where γ is the specific weight of water. Interestingly, for a given pipeline, there is a unique jet diameter that will deliver maximum power to a jet. This is apparent by noting that the power of the jet, Pjet , issuing from the nozzle is given by Pjet = γ Q

Vj2 2g

(7.136)

As the size of the nozzle opening is increased, the flow rate, Q, increases while the jet velocity, Vj , decreases; hence there is some intermediate size of nozzle opening that will provide maximum power to the jet. The hydraulic efficiency, ηT , of an impulse turbine is the ratio of the power delivered to the turbine buckets to the power in the flow at the entrance to the nozzle. Thus, for impulse turbines, ηT =

γ QhT hT = γ Qhe he

(7.137)

The overall efficiency, η, of an impulse turbine is less than the hydraulic efficiency, ηT , because of that part of the energy delivered to the buckets that is lost in the mechanical

348

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friction in the bearings and the air resistance associated with the spinning runner. The overall efficiency of an impulse turbine, η, is given by η=

Tω output (shaft) power = input power γ Qhe

(7.138)

where T is the torque delivered to the shaft by the turbine, and ω is the angular speed of the runner. EXAMPLE 7.21 It is proposed to install a small Pelton-wheel turbine system where the static head above the nozzle is 50 m, and the delivery line is 455-mm-diameter concrete pipe with a length of 150 m and a roughness height of 1 mm. The discharge nozzle has a diameter of 100 mm and a head-loss coefficient of 0.5. Model tests on the Pelton wheel show that the bucket friction loss coefficient is 0.3, the velocity of water relative to the bucket at the exit from the bucket is 1 m/s, and the absolute velocity of the water leaving the bucket is 5 m/s. Estimate the power that could be derived from the system and the hydraulic efficiency of the turbine. Solution From the given data: H = 50 m, D = 455 mm, L = 150 m, ks = 1 mm, Dj = 100 mm, kj = 0.5, k = 0.3, v2 = 1 m/s, and V2 = 5 m/s. The flow rate, Q, is determined by application of the energy equation between the upstream reservoir and the exit of the nozzle, which requires that H −

Q2 Q2 fL Q2 − kj = D 2gA2 2gA2j 2gA2j

(7.139)

where f is the friction factor, and A and Aj are the cross-sectional areas of the pipe and jet, respectively. Using the Swamee–Jain equation (Equation 2.39) to relate f to Q, Equation 7.139 yields Q = 0.200 m3 /s. The corresponding velocities in the delivery pipeline and nozzle jet are V = 1.23 m/s and Vj = 25.4 m/s, and the friction factor is f = 0.0244. Using these derived data yields the following results: fL V 2 (0.0244)(150) 1.232 he = H − h L = H − = 50 − = 49.4 m D 2g (0.455) 2(9.81) h T = h e − kj

Vj2

v2 V2 25.42 12 52 − k 2 − 2 = 49.4 − 0.5 − 0.3 − = 31.6 m 2g 2g 2g 2(9.81) 2(9.81) 2(9.81)

PT = γ QhT = (9.79)(0.200)(31.6) = 61.9 kW h 31.6 = 0.64 ηT = T = he 49.4 For the given configuration, the expected power from the system is 61.9 kW with a hydraulic efficiency of 64%.

Reaction Turbines. A reaction turbine is one in which flow takes place in a closed chamber under pressure, and the flow through a reaction turbine may be radially inward, axial, or mixed. There are two types of reaction turbines in common use, the Francis turbine and the axial-flow turbine, also called a propeller turbine. The Francis turbine is named after the American engineer James B. Francis (1815– 1892), who designed, built, and tested the first efficient inward-flow turbine in 1849. All inward-flow turbines are called Francis turbines. In the conventional Francis turbine, water enters a scroll case and moves into the runner through a series of guide vanes with contracting passages that convert pressure head to velocity head. These vanes, called wicket gates, are adjustable so that the quantity and direction of flow can be controlled. A Francis turbine runner is shown in Figure 7.52. When installed, flow will be in the radial direction toward the top of the runner shown in Figure 7.52 and the flow will exit the runner through the underside; such turbines are therefore called mixed-flow turbines. A scroll case that surrounds the runner is designed to decrease the

Section 7.7

Dams and Reservoirs

349

FIGURE 7.52: Francis turbine Source: ANDRITZ (2012).

cross-sectional area in proportion to the decreasing flow rate passing a given section of the casing. Constant rotative speed of the runner under varying load is achieved by a governor that actuates a mechanism that regulates the gate openings. A relief valve or surge tank is generally necessary to prevent serious water-hammer pressures. The propeller turbine is an axial-flow machine with its runner confined in a closed conduit. The usual runner has four to eight blades mounted on a hub, with very little clearance between the blades and the conduit wall. A Kaplan turbine, named in honor of the Austrian engineer Viktor Kaplan (1876–1934), is a propeller turbine with movable blades whose pitch can be adjusted to suit existing operating conditions. A typical Kaplan turbine runner is shown in Figure 7.53. Other types of propeller turbines include the Deriaz turbine, an adjustable-blade, diagonal-flow turbine where the flow is directed inward as it passes through the blades, and the tube turbine, an inclined-axis type that is particularly well adapted to low-head installations, since the water passages can be formed directly in the concrete structure of the dam. To operate properly, reaction turbines must have a submerged discharge. After passing through the runner, the water enters a draft tube, which directs the water to the discharge location in the downstream channel called the tailrace. Care must be taken to ensure that the pressure at the entrance to the draft tube is greater than the vapor pressure of water so that cavitation does not occur; this will generally require that the turbine be less than 10 m (33 ft) above the tailrace. A schematic diagram of a reaction-turbine system is shown in Figure 7.54. Applying the energy equation between the upstream reservoir and the tailrace, the head, hT , that can be extracted from the water by the runner is given by

hT = H − hL − hDT −

V2 2g

(7.140)

where H is the height of the water surface in the upstream reservoir above the water surface in the tailrace, hL is the head loss between the reservoir and the turbine inlet, hDT is the sum of the head losses in the turbine and the draft tube, and V is the velocity in the tailrace. In most cases, V 2 /2g is relatively small and may be neglected. The (theoretical) power that can be extracted by the turbine, PT , is given by Equation 7.135.

350

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Design of Hydraulic Structures

FIGURE 7.53: Kaplan axial flow turbine Source: U.S. Army Corps of Engineers (2005g).

FIGURE 7.54: Reaction-turbine system

Water surface

Energy grade line hL

Reservoir Q

H Turbine

Penstock

Runner V2 2g

Draft tube

Tailrace

V

Pump turbines used at pumped-storage facilities are very similar in design and construction to the Francis turbine. When water enters the rotor at the periphery and flows inward, the machine acts as a turbine. When water enters the center (or eye) and flows outward, the machine acts as a pump. The pump turbine is connected to a motor generator, which acts as either a motor or generator, depending on whether the pump or turbine mode is being used. EXAMPLE 7.22 A hydropower facility is being planned in which the water surface in the tailrace is 75 m below the water surface of a reservoir. The 1.70-m-diameter concrete-lined tunnel leading from the reservoir to the Francis turbine intake (i.e., the penstock) is 300 m long and has an estimated roughness height of 10 mm. Model studies indicate that when the flow rate through the system is 16 m3 /s, the combined head loss in the turbine and draft tube is 2.0 m, and the average velocity in the tailrace is 0.50 m/s. Estimate the power that can be extracted from the system. Solution From the given data: H = 75 m, D = 1.70 m, L = 300 m, ks = 10 mm, Q = 16 m3 /s, hDT = 2.0 m, and V = 0.5 m/s. The velocity (= Q/A) in the penstock can be calculated as Vp = 7.05 m/s. Using the given values of Q, D, and ks , and assuming that the kinematic viscosity of water, ν, is 10−6 m2 /s (at 20◦ C), the friction factor, f , of the penstock can be calculated using the Swamee–Jain equation (Equation 2.39) as f = 0.0319. The head, hT , extracted by the turbine is given by Equation 7.140 as

Section 7.7

Dams and Reservoirs

351

2

hT = H −

(0.0319)(300) 7.052 0.52 fL Vp V2 − h DT − = 75 − − 2.0 − = 58.72 m D 2g 2g 1.70 2(9.81) 2(9.81)

which gives PT = γ QhT = (9.79)(16)(58.72) = 9200 kW = 9.2 MW Therefore the system will extract 9.2 MW of power from the water flowing through the turbine. Because of transmission inefficiencies, the actual amount of electrical power available for distribution will be less.

7.7.3.2

Turbine performance

Similarity laws derived in Chapter 2 for pumps also apply to reaction turbines. Therefore, the performance of a homologous series of reaction turbines can be described by the following functional relation:   gh Q =f (7.141) ω2 D2 ωD3 where h is the head extracted by the turbine, ω is the angular speed of the runner, D is the characteristic size of the runner, Q is the volumetric flow rate through the turbine, and g is the acceleration due to gravity. In a similar fashion to pumps, the operating point of greatest efficiency is defined by the dimensionless specific speed, ns , where 1

ns =

ωQ 2 3

(7.142)

(gh) 4

and any consistent set of units can be used. In the United States, it is common to define the specific speed of a turbine in U.S. Customary units, in which case the specific speed, Ns , is expressed in the form ne bhp Ns = (7.143) 5 h4 where ne is the speed of the runner in revolutions per minute, bhp is the shaft or brake horsepower (= Tω) in horsepower, and h is the extracted head in feet, where all performance variables are at the point of maximum efficiency. Comparing Equations 7.142 and 7.143 gives the approximate relation Ns = 43.4ns

(7.144)

For ns < 0.2, impulse turbines are typically most efficient; for 0.2 … ns < 2, Francis turbines are most efficient, and for 2 … ns … 5, propeller turbines are most efficient. These efficiencies indicate that high-head installations work best with impulse turbines and low-head installations work best with propeller turbines. Impulse turbines are commonly used for heads greater than 150–300 m (500–1000 ft), while the upper limit for using a Francis turbine is on the order of 450 m (1500 ft) because of possible cavitation and the difficulty of building casings to withstand such high pressures. Turbines are generally operated at constant speed. In the United States, 60 Hz electric current is used, and under such conditions the rotative speed, n [rpm], of a turbine is related to the number of pairs of poles, N, by the relation n=

3600 N

(7.145)

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Chapter 7

Design of Hydraulic Structures

The capacity of a hydroelectric generating plant is defined as the maximum rate at which the plant can produce electricity. The installed capacity or hydropower generation potential for a given site is estimated by the relation P = γ Qhη

(7.146)

where P is the hydropower generation potential [kW], γ is the specific weight of water [= 9.79 kN/m3 ], Q is the volumetric flow rate [m3 /s], h is the available head [m], and η is the turbine efficiency [dimensionless], which is usually in the range of 0.80–0.90. To maximize use of available water, a value of Q which is exceeded 10%–30% of the time may be selected to estimate the installed capacity of the facility. Several values of Q and the corresponding h may be examined to estimate the values that result in optimum installed capacity based on water-use and economic considerations. For preliminary planning, optimum installed capacity may be that beyond which relatively large increases in Q are required to obtain relatively small increases in P. It is considered good practice to have at least two turbines at an installation so that the plant can continue operation while one of the turbines is shut down for repairs, maintenance, or inspection. To estimate the annual hydropower generation potential of a facility, a reservoir and power-plant operation study has to be conducted with sequences of available flows and corresponding heads for relatively long periods of time—for example, 10–50 years or more. For these estimates, the overall water-to-wire efficiency should be used, which varies in the range of 0.70–0.86 and includes efficiencies of the turbine, generator, transformers, and other equipment. In addition, adjustments to the efficiency may be made for tailwater fluctuations and unscheduled down time. EXAMPLE 7.23 A hydroelectric project is to be developed along a river where the 90-percentile flow rate is 2240 m3 /s, and a Rippl analysis indicates that storage to a height of 30 m above the downstream stage is required to meet all the demands. Estimate the installed capacity that would be appropriate for these conditions. Solution Assuming that the capacity of the turbines will be sufficient to pass a flow rate, Q, of 2240 m3 /s at a head of 30 m, and taking the turbine efficiency, η, as 0.85 and γ = 9.79 kN/m3 yields P = γ Qhη = (9.79)(2240)(30)(0.85) = 5.59 * 105 kW = 559 MW Therefore, to fully utilize the available head and anticipated flow rates, an installed capacity of 559 MW is required. This analysis neglects all hydraulic head losses in the system, so the estimated capacity is an upper bound.

7.7.3.3

Feasibility of hydropower

The extraction of hydropower is potentially feasible in cases where water supplies are in excess of demand requirements, and in cases where the total quantity of water available over a period of time is equal to or in excess of the demand requirements, although the demand requirements may at times exceed the availability of water. In both of these cases and where the sites are within reasonable transmission distance to a power market, there is a potential for hydropower development. The design and construction of a hydroelectric plant is very complex and can easily take more than a decade from the beginning of the design phase to the actual generation of electricity. Depending on the installed capacity, hydroelectric power plants can be classified as shown in Table 7.9 (Prakash, 2004). The largest hydroelectric facility in the world is currently the Three Gorges Dam in China, designed to produce 22.5 GW of electrical energy. Of fundamental importance in assessing the feasibility of hydropower development is the maximum amount of power that can be extracted from any given site at which the general condition is illustrated in Figure 7.55. In cases where the downstream channel can be approximated as rectangular, it can be shown that the maximum power per unit width of tailwater channel that can be extracted by any hydraulic device, pmax , is given by (Pelz, 2011)

Problems

353

TABLE 7.9: Classification of Hydropower Capacity

FIGURE 7.55: Extraction of hydropower

Classification

Installed capacity (MW)

Conventional Small-scale Mini Micro

>15 1–15 0.1–1 xn ) = 1 − P(xi … xn )

(8.6)

The function P(xi … xn ) is called the cumulative distribution function and is commonly written as F(xn ), where F(xn ) = P(xi … xn ) (8.7) These definitions of the probability distribution function, f (xn ), and cumulative distribution function, F(xn ), can be expanded to describe several random variables simultaneously. As an illustration, consider the case of two discrete random variables, X and Y, with elements xp and yq , where p ∈ [1, N] and q ∈ [1, M]. The joint probability distribution function of X and Y, f (xp , yq ), is the probability of xp and yq occurring simultaneously. Such distributions are called bivariate probability distributions. Joint probability distributions involving more than one variable are generally referred to as multivariate probability distributions. Based

364

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Probability and Statistics in Water-Resources Engineering

on the definition of the bivariate probability distribution, f (xp , yq ), the following conditions must necessarily be satisfied f (xp , yq ) Ú 0, M N  

5 xp , yq

f (xp , yq ) = 1

(8.8) (8.9)

p=1 q=1

As in the case of a single random variable, the cumulative distribution function, F(xp , yq ), is defined as the probability that the outcomes are simultaneously less than or equal to xp and yq , respectively. Hence F(xp , yq ) is related to the joint probability distribution as follows: F(xp , yq ) =

 

f (xi , yj )

(8.10)

xi …xp yj …yq

In multivariate analysis, the probability distribution of any single random variate is called the marginal probability distribution function. In the case of a bivariate probability function, f (xp , yq ), the marginal probability of xp , g(xp ), is given by the relation g(xp ) =

M 

f (xp , yq )

(8.11)

q=1

Several probability functions can be derived from the basic function describing the probability of occurrence of discrete events. The particular function of interest in any given case (e.g., cumulative or marginal) depends on the question being asked. 8.2.2

Continuous Probability Distributions

If X is a random variable with a continuous sample space, then there are an infinite number of possible outcomes and the probability of occurrence of any single outcome is zero. This problem is addressed by defining the probability of an outcome being in the range [x, x + x] as f (x) x, where f (x) is the probability density function. Based on this definition, any valid probability density function must satisfy the following conditions: f (x) Ú 0, 

+q −q

f (x ) dx = 1

5x

(8.12) (8.13)

The cumulative distribution function, F(x), describes the probability that the outcome of a random process will be less than or equal to x and is related to the probability density function by the equation  x F(x) = f (x ) dx (8.14) −q

which can also be written as f (x) =

dF(x) dx

(8.15)

In describing the probability distribution of more than one random variable, the joint probability density function is used. In the case of two variables, X and Y, the probability that x will be in the range [x, x + x] and y will be in the range [y, y + y] is approximated by f (x, y) x y, where f (x, y) is the joint probability density function of x and y. As in the case of discrete random variables, the bivariate cumulative distribution function, F(x, y), and marginal probability distributions, g(x) and h(y), are defined as

Section 8.2

 F(x, y) =

y

−q −q

 g(x) =

q

−q

 h(y) = 8.2.3



x

q

−q

Probability Distributions

f (x , y ) dx dy

365

(8.16)

f (x, y ) dy

(8.17)

f (x , y) dx

(8.18)

Mathematical Expectation and Moments

Assuming that xi is an outcome of the discrete random variable X, f (xi ) is the probability distribution function of X, and g(xi ) is an arbitrary function of xi , then the expected value of the function g, represented by g, is defined by the equation g =

N 

g(xi )f (xi )

(8.19)

i=1

where N is the number of possible outcomes in the sample space, X. The expected value of a random function is equal to the arithmetic average of the function calculated from an infinite number of random outcomes, and the analogous result for a function of a continuous random variable is given by  +q g = g(x )f (x ) dx (8.20) −q

where g(x) is a continuous random function and f (x) is the probability density function of x. Several random functions are particularly useful in characterizing the distribution of random variables. The first is simply g(x) = x

(8.21)

In this case g = x corresponds to the arithmetic average of the outcomes over an infinite number of realizations. The quantity x is commonly called the mean of the random variable and is usually denoted by μx . According to Equations 8.19 and 8.20, μx is defined for both discrete and continuous random variables by

μx =

⎧ N ⎪  ⎪ ⎪ ⎪ ⎪ xi f (xi ) ⎪ ⎨

(discrete)

i=1

⎪  ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

+q −q

(8.22) x f (x ) dx

(continuous)

A second random function that is frequently used is g(x) = (x − μx )2

(8.23)

which equals the square of the deviation of a random outcome from its mean. The expected value of this quantity is referred to as the variance of the random variable and is usually denoted by σx2 . According to Equations 8.19 and 8.20, the variance of discrete and continuous random variables are given by the relations

366

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Probability and Statistics in Water-Resources Engineering

σx2

=

⎧ N ⎪  ⎪ ⎪ ⎪ ⎪ (xi − μx )2 f (xi ) ⎪ ⎨

(discrete)

i=1

⎪  ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

+q −q

(8.24) (x − μx )2 f (x ) dx

(continuous)

The square root of the variance, σx , is called the standard deviation of x and measures the average magnitude of the deviation of the random variable from its mean. Random outcomes occur that are either less than or greater than the mean, μx , and the symmetry of these outcomes about μx is measured by the skewness or skewness coefficient, which is the expected value of the function (x − μx )3 g(x) = (8.25) σx3 If the random outcomes are symmetrical about the mean, the skewness is equal to zero; otherwise, a nonsymmetric distribution will have a positive or negative skew. There is no universal symbol that is used to represent the skewness, but in this text skewness will be represented by gx . For discrete and continuous random variables,

gx =

⎧ N ⎪ ⎪ 1  ⎪ ⎪ (xi − μx )3 f (xi ) ⎪ ⎪ ⎨ σx3

(discrete)

i=1

⎪  +q ⎪ ⎪ 1 ⎪ ⎪ (x − μx )3 f (x ) dx ⎪ ⎩σ3 −q x

(8.26) (continuous)

The variables μx , σx , and gx are measures of the average, variability about the average, and the symmetry about the average, respectively. EXAMPLE 8.1 A water-resource system is designed such that the probability, f (xi ), that the system capacity is exceeded xi times during the 50-year design life is given by the following discrete probability distribution: xi

f (xi )

0 1 2 3 4 5 6 >6

0.13 0.27 0.28 0.18 0.09 0.03 0.02 0.00

What is the mean number of system failures expected in 50 years? What are the variance and skewness of the number of failures? Solution The mean number of failures, μx , is defined by Equation 8.22, where μx =

N  i=1

xi f (xi ) = (0)(0.13) + (1)(0.27) + (2)(0.28) + (3)(0.18) + (4)(0.09) + (5)(0.03) + (6)(0.02) = 2

Section 8.2

Probability Distributions

367

The variance of the number of failures, σx2 , is defined by Equation 8.24 as σx2 =

N  (xi − μx )2 f (xi ) = (0 − 2)2 (0.13) + (1 − 2)2 (0.27) + (2 − 2)2 (0.28) + (3 − 2)2 (0.18) i=1

+ (4 − 2)2 (0.09) + (5 − 2)2 (0.03) + (6 − 2)2 (0.02) = 1.92 The skewness coefficient, gx , is defined by Equation 8.26 as gx =

N  1  1 (0 − 2)3 (0.13) + (1 − 2)3 (0.27) + (2 − 2)3 (0.28) + (3 − 2)3 (0.18) (xi − μx )3 f (xi ) = 3 σx3 (1.92) 2 i=1 + (4 − 2)3 (0.09) + (5 − 2)3 (0.03) + (6 − 2)3 (0.02) = 0.631

EXAMPLE 8.2 The probability density function, f (t), of the time between storms during the summer in Miami is estimated as ⎧ ⎨0.014e−0.014t , t > 0 f (t) = ⎩ 0, t … 0 where t is the time interval between storms in hours. Estimate the mean, standard deviation, and skewness of t. Solution The mean interstorm time, μt , is given by Equation 8.22 as

 q  q  q

  μt = t f (t ) dt = t 0.014e−0.014t dt = 0.014 t e−0.014t dt 0

0

0

The quantity in parentheses can be conveniently integrated by using the result (Dwight, 1961)  q 1 xe−ax dx = 2 a 0 Hence, the expression for μt can be written as

μt = 0.014

1 0.0142

= 71 h

The variance of t, σt2 , is given by Equation 8.24 as  q  q  q   (t − μt )2 f (t ) dt = (t − 71)2 0.014e−0.014t dt = 0.014 (t2 − 142t + 5041)e−0.014t dt σt2 = 0

0

0

Using the integration results (Dwight, 1961),  q  q 2 1 2 −ax x e dx = ; xe−ax dx = ; 3 2 a a 0 0

 q 0

e−ax dx =

1 a

the expression for σt2 can be written as 

σt2 = 0.014

 2 142 5041 = 5102 h2 − + 0.014 0.0143 0.0142

√ The standard deviation, σt , of the storm interevent time is therefore equal to 5102 = 71 h. The skewness of t, gt , is given by Equation 8.26 as  q  q  1 1 (t − μt )3 f (t ) dt = (t − 71)3 0.014e−0.014t dt gt = 3 3 (71) σt 0 0 =

  0.014 q 3 (t − 213t2 + 15,123t − 357,911)e−0.014t dt (71)3 0

368

Chapter 8

Probability and Statistics in Water-Resources Engineering Using the previously cited integration formulae plus (Dwight, 1961)  q 0

gives gt = 3.91 * 10−8



x3 e−ax dx =

6 a4

6 2 1 1 − 213 + 15,123 − 357,911 4 3 2 0.014 0.014 0.014 0.014

= 2.1

The positive skewness (= 2.1) indicates that the probability distribution of t has a long tail to the right of μt (toward higher values of t).

8.2.4

Return Period

Consider a random variable X, with the outcome having a return period T given by xT . Let p be the probability that X Ú xT in any observation, or p = P(X Ú xT ). For each observation, there are two possible outcomes, either X Ú xT with probability p or X < xT with probability 1 − p. Assuming that all observations are independent, then the probability of a return period τ is the probability of τ − 1 observations where X < xT followed by an observation where X Ú xT . The expected value of τ , τ , is therefore given by τ  =

q 

τ (1 − p)τ −1 p

τ =1

= p + 2(1 − p)p + 3(1 − p)2 p + 4(1 − p)3 p + · · · = p[1 + 2(1 − p) + 3(1 − p)2 + 4(1 − p)3 + · · · ]

(8.27)

The expression within the brackets has the form of the power series expansion (1 + x)n = 1 + nx +

n(n − 1) 2 n(n − 1)(n − 2) 3 x + x + ··· 2 6

(8.28)

with x = −(1 − p) and n = −2. Equation 8.27 can therefore be written in the form τ  =

p 1 = 2 p [1 − (1 − p)]

(8.29)

Since T = τ , Equation 8.29 can be expressed as 1 p

(8.30)

1 P(X Ú xT )

(8.31)

1 T

(8.32)

T= or T= which can also be written as

P(X Ú xT ) =

In civil engineering practice it is more common to describe an event by its return period than its exceedance probability. For example, floodplains are usually delineated for the “100-year flood,” which has an exceedance probability of 1% in any given year. However, because of the common misconception that the 100-year flood occurs once every 100 years, ASCE (1996a) recommends that the reporting of return periods should be avoided, and annual exceedance probabilities reported instead.

Section 8.2

Probability Distributions

369

EXAMPLE 8.3 Analyses of the maximum-annual floods over the past 150 years in a small river indicate the following cumulative distribution:

n

Flow, xn (m3 /s)

P(X < xn )

1 2 3 4 5 6 7 8 9 10 11

0 25 50 75 100 125 150 175 200 225 250

0 0.19 0.35 0.52 0.62 0.69 0.88 0.92 0.95 0.98 1.00

Estimate the magnitudes of the floods with return periods of 10, 50, and 100 years. Solution According to Equation 8.32, floods with return periods, T, of 10, 50, and 100 years have 1 = 0.10, 1 = 0.02, and 1 = 0.01. These exceedance probabilities corexceedance probabilities of 10 50 100 respond to cumulative probabilities of 1 − 0.1 = 0.9, 1 − 0.02 = 0.98, and 1 − 0.01 = 0.99, respectively. Interpolating from the given cumulative probability distribution gives the following results:

8.2.5

Return period (years)

Flow (m3 /s)

10 50 100

163 225 238

Common Probability Functions

There are an infinite number of valid probability distributions. Engineering analyses, however, are typically confined to well-defined distributions that are relatively simple, associated with identifiable processes, and can be fully characterized by a relatively small number of parameters. Probability functions that can be expressed analytically are the functions of choice in engineering applications. In using theoretical probability distribution functions to describe observed phenomena, it is important to understand the fundamental processes that lead to these distributions, since there is an implicit assumption that the theoretical and observed processes are the same. Probability distribution functions are either discrete or continuous. Commonly encountered discrete and continuous probability distribution functions, and their associated processes, are described in the following sections. 8.2.5.1

Binomial distribution

The binomial distribution, also called the Bernoulli distribution, is a discrete probability distribution that describes the probability of outcomes as either successes or failures. For example, in studying the probability of the annual maximum flow in a river exceeding a certain value, the maximum flow in any year may be deemed either a success (the flow exceeds the specified value) or a failure (the flow does not exceed the specified value). Since many engineered systems are designed to operate below certain threshold (design) conditions, the analysis in terms of success and failure is fundamental to the assessment of system reliability. Specifically, the binomial distribution describes the probability of n successes in N trials, given

370

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Probability and Statistics in Water-Resources Engineering

FIGURE 8.1: Binomial probability distribution

0.30 N = 10 p = 0.5

0.25

f(n)

0.20 0.15 0.10 0.05 0.00

0

1

2

3

4

5

6

7

8

9

10

n

that the outcome in any one trial is independent of the outcome in any other trial and the probability of a success in any one trial is p. Such trials are called Bernoulli trials, and the process generating Bernoulli trials is called a Bernoulli process. The binomial (Bernoulli) probability distribution, f (n), is given by f (n) =

N! pn (1 − p)N−n n!(N − n)!

(8.33)

This discrete probability distribution follows directly from permutation and combination analysis and can also be written in the form N n (8.34) f (n) = p (1 − p)N−n n N N! = n n!(N − n)!

where

(8.35)

is commonly referred to as the binomial coefficient. The parameters of the binomial distribution are the total number of outcomes, N, and the probability of success, p, in each outcome, and this distribution is illustrated in Figure 8.1. The mean, variance, and skewness coefficient of the binomial distribution are given by μn = Np,

σn2 = Np(1 − p),

g= 

1 − 2p Np(1 − p)

(8.36)

The Bernoulli probability distribution, f (n), is symmetric for p = 0.5, skewed to the right for p < 0.5, and skewed to the left for p > 0.5. EXAMPLE 8.4 The capacity of a stormwater-management system is designed to accommodate a storm with a return period of 10 years. What is the probability that the stormwater system will fail once in 20 years? What is the probability that the system fails at least once in 20 years? Solution The stormwater-management system fails when the magnitude of the design storm is 1 = 0.1. The exceeded. The probability, p, of the 10-year storm being exceeded in any year is 10 probability of the 10-year storm being exceeded once in 20 years is given by the binomial distribution, Equation 8.33, as N! f (n) = pn (1 − p)N−n n!(N − n)!

Section 8.2

Probability Distributions

371

where n = 1, N = 20, and p = 0.1. Substituting these values gives f (1) =

20! (0.1)1 (1 − 0.1)20−1 = 0.27 1!(20 − 1)!

Therefore, the probability that the stormwater-management system fails once in 20 years is 27%. The probability, P, of the 10-year storm being equaled or exceeded at least once in 20 years is given by P=

20 

f (i) = 1 − f (0) = 1 −

i=1

20! (0.1)0 (1 − 0.1)20−0 = 1 − 0.12 = 0.88 0!(20 − 0)!

The probability that the design event will be exceeded (at least once) is referred to as the risk of failure, and the probability that the design event will not be exceeded is called the reliability of the system. In this example, the risk of failure is 88% over 20 years and the reliability of the system over the same period is 12%.

8.2.5.2

Geometric distribution

The geometric distribution is a discrete probability distribution function that describes the number of Bernoulli trials, n, up to and including the one in which the first success occurs. The probability that the first success in a Bernoulli trial occurs on the nth trial is found by noting that for the first exceedance to be on the nth trial, there must be n − 1 preceding trials without success followed by a success. The probability of this sequence of events is (1 − p)n−1 p, and therefore the (geometric) probability distribution of the first success being in the nth trial, f (n), is given by f (n) = p(1 − p)n−1

(8.37)

The parameter of the geometric distribution is the probability of success, p, and this distribution is illustrated in Figure 8.2. The mean and variance of the geometric distribution are given by 1 1 − p μn = , σn2 = (8.38) p p2 The probability distribution of the number of Bernoulli trials between successes can be found from the geometric distribution by noting that the probability that n trials elapse between successes is the same as the probability that the first success is in the n + 1st trial. Hence the probability of n trials between successes is equal to p(1 − p)n . FIGURE 8.2: Geometric probability distribution

0.6 p = 0.5 0.5

f(n)

0.4 0.3 0.2 0.1 0.0 0

1

2

3

4

5 n

6

7

8

9

10

372

Chapter 8

Probability and Statistics in Water-Resources Engineering

EXAMPLE 8.5 A stormwater-management system is designed for a 10-year storm. Assuming that exceedance of the 10-year storm is a Bernoulli process, what is the probability that the capacity of the stormwatermanagement system will be exceeded for the first time in the fifth year after the system is constructed? What is the probability that the system capacity will be exceeded within the first 5 years? 1 = 0.1, and the probability that the 10-year storm will be Solution From the given data, n = 5, p = 10 exceeded for the first time in the fifth year is given by the geometric distribution (Equation 8.37) as

f (5) = (0.1)(1 − 0.1)4 = 0.066 = 6.6% The probability, P, that the 10-year storm will be exceeded within the first 5 years is given by P=

5 

p(1 − p)n−1 =

n=1

8.2.5.3

5 

(0.1)(1 − 0.1)n−1 =

n=1

5 

(0.1)(0.9)n−1 = 0.41 = 41%

n=1

Poisson distribution

The Poisson process is a limiting case of an infinite number of Bernoulli trials, where the number of trials, N, becomes large and the probability of success in each trial, p, becomes small in such a way that the expected number of successes, Np, remains constant. Under ´ these circumstances it can be shown that (Benjamin and Cornell, 1970; Thiebaux, 1994) lim

N→q, p→0

N! (Np)n e−Np pn (1 − p)N−n = n!(N − n)! n!

(8.39)

This approximation can be safely applied when N Ú 100, p … 0.01, and Np … 20 (Devore, 2000). Denoting the expected number of successes, Np, by λ, the probability distribution of the number of successes, n, in an interval in which the expected number of successes is given by λ is described by the Poisson distribution: f (n) =

λn e−λ n!

(8.40)

Haan (1977) describes the Poisson process as corresponding to the case where, for a given interval of time, the measurement increments decrease, resulting in a corresponding decrease in the event probability within the measurement increment, in such a way that the total number of events expected in the overall time interval remains constant and equal to λ. The only parameter in the Poisson distribution is λ. The mean, variance, and skewness of the discrete Poisson distribution are given by μn = λ,

σn2 = λ,

1

gn = λ− 2

(8.41)

Poisson distributions for several values of λ are illustrated in Figure 8.3. The Poisson process for a continuous time scale is defined in a similar way to the Poisson process on a discrete time scale. The assumptions underlying the Poisson process on a continuous time scale are: (1) the probability of an event in a short interval between t and t + t is λ t (proportional to the length of the interval) for all values of t; (2) the probability of more than one event in any short interval t to t + t is negligible in comparison to λt; and (3) the number of events in any interval of time is independent of the number of events in any other nonoverlapping interval of time. Extending the result for the discrete Poisson process given by Equation 8.40 to the continuous Poisson process, the probability distribution of the number of events, n, in time, t, for a Poisson process is given by f (n) =

(λt)n e−λt n!

(8.42)

Section 8.2 FIGURE 8.3: Poisson probability distribution

0.7

0.4

0.4

f(n)

f(n)

0.5

0.3

0.2

0.1

0.1

1

2

3

4

5 n

6

7

8

9

0.0

10

0.7

0

1

2

3

4

5 n

6

7

8

9

10

0.7 λ = 4.5

0.6 0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1 0

1

2

3

4

5 n

6

7

8

9

λ = 6.5

0.6

f(n)

f(n)

0.3

0.2

0

λ = 2.5

0.6

0.5

0.0

373

0.7

λ = 0.5

0.6

0.0

Probability Distributions

10

0.0

0

1

2

3

4

5 n

6

7

8

9

10

The parameter of the Poisson distribution for a continuous process is λt, where λ is the average rate of occurrence of the event. The mean, variance, and skewness of the continuous Poisson distribution are given by μn = λt,

σn2 = λt,

1

gn = (λt)− 2

(8.43)

The occurrences of storms and major floods have both been successfully described as Poisson processes (Borgman, 1963; Shane and Lynn, 1964). EXAMPLE 8.6 A flood-control system is designed for a runoff event with a 50-year return period. Assuming that exceedance of the 50-year runoff event is a Poisson process, what is the probability that the design event will be exceeded twice in the first 10 years of the system? What is the probability that the design event will be exceeded more than twice in the first 10 years? Solution The probability distribution of the number of exceedances is given by Equation 8.40 (since the events are discrete). Over a period of 10 years, the expected number of exceedances, λ, of the 50-year event is given by 1 = 0.2 λ = Np = (10) 50 and the probability of the design event being exceeded twice is given by Equation 8.40 as 0.22 e−0.2 = 0.016 = 1.6% 2! The probability, P, that the design event is exceeded more than twice in 10 years can be calculated using the relation   0.21 e−0.2 0.22 e−0.2 0.20 e−0.2 + + = 0.001 P = 1 − [f (0) + f (1) + f (2)] = 1 − 0! 1! 2! f (2) =

Therefore, there is only a 0.1% chance that the 50-year design event will be exceeded more than twice in any 10-year interval.

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8.2.5.4

Exponential distribution

The exponential distribution describes the probability distribution of the time between occurrences of an event in a continuous Poisson process. For any time duration, t, the probability of zero events during time t is given by Equation 8.42 as e−λt . Therefore, the probability that the time between events is less than t is equal to 1 − e−λt . This is equal to the cumulative probability distribution, F(t), of the time, t, between occurrences, which is equal to the integral of the probability density function, f (t). The probability density function can therefore be derived by differentiating the cumulative distribution function; hence, f (t) =

d d F(t) = (1 − e−λt ) dt dt

(8.44)

which simplifies to f (t) = λe−λt

(8.45)

This distribution is called the exponential probability distribution. The mean, variance, and skewness coefficient of the exponential distribution are given by μt =

1 , λ

σt2 =

1 , λ2

gt = 2

(8.46)

The positive and constant value of the skewness indicates that the distribution is skewed to the right for all values of λ. The exponential probability distribution is illustrated in Figure 8.4. The cumulative distribution function, F(t), of the exponential distribution function is given by  F(t) =

t

λe−λτ dτ

(8.47)

0

which leads to F(t) = 1 − e−λt

(8.48)

In hydrologic applications, the exponential distribution is sometimes used to describe the interarrival times of random shocks to hydrologic systems, such as slugs of polluted runoff entering streams (Chow et al., 1988) and the arrival of storm events (Bonta, 2003). In more empirical applications, the exponential distribution has also been shown to adequately describe intensities and durations of rainfall measured at fixed locations (Kao and Govindaraju, 2007).

FIGURE 8.4: Exponential probability distribution

1.0λ 0.8λ f(t)

0.6λ 0.4λ 0.2λ 0

0

1

2

3

λt

4

5

Section 8.2

Probability Distributions

375

EXAMPLE 8.7 In the 86-year period between 1903 and 1988, Florida was hit by 56 hurricanes (Winsberg, 1990). Assuming that the time interval between hurricanes is given by the exponential distribution, what is the probability that there will be less than 1 year between hurricanes? What is the probability that there will be less than 2 years between hurricanes? Solution The mean time between hurricanes, μt , is estimated as 86 = 1.54 y μt L 56 The mean, μt , is related to the parameter, λ, of the exponential distribution by Equation 8.46, which gives 1 1 = 0.649 y−1 = λ= μt 1.54 The cumulative probability distribution, F(t), of the time between hurricanes is given by Equation 8.48 as F(t) = 1 − e−λt = 1 − e−0.649t The probability that there will be less than 1 year between hurricanes is given by F(1) = 1 − e−0.649(1) = 0.48 = 48% and the probability that there will be less than 2 years between hurricanes is F(2) = 1 − e−0.649(2) = 0.73 = 73%

8.2.5.5

Gamma/Pearson Type III distribution

The gamma distribution describes the probability of the time to the nth occurrence of an event in a Poisson process. This probability distribution is derived by finding the probability distribution of t = t1 + t2 + · · · + tn , where ti is the time from the i − 1th to the ith event. This is simply equal to the sum of time intervals between events. The probability distribution of the time, t, to the nth event can be derived from the exponential distribution to yield f (t) =

λn tn−1 e−λt (n − 1)!

(8.49)

where f (t) is the gamma probability distribution, illustrated in Figure 8.5 for several values of n. The mean, variance, and skewness coefficient of the gamma distribution are given by μt =

n , λ

σt2 =

n , λ2

gt =

2 n

(8.50)

In cases where n is not an integer, the gamma distribution, Equation 8.49, can be written in the more general form f (t) = FIGURE 8.5: Gamma probability distribution

λn tn−1 e−λt (n)

(8.51)

1.0λ

1.0λ

n=3

n=0

0.8λ

0.6λ

0.6λ

f(t)

f(t)

0.8λ

0.4λ

0.4λ

0.2λ

0.2λ

0

0

1

2

3

λt

4

5

0

0

1

2

3 λt

4

5

376

Chapter 8

Probability and Statistics in Water-Resources Engineering

where (n) is the gamma function defined by the equation  q (n) = tn−1 e−t dt

(8.52)

0

which has the property that (n) = (n − 1)!,

n = 1, 2, . . .

(8.53)

The gamma distribution has been widely used as an empirical probability distribution in hydrology, and is particularly useful for describing skewed variables without the need for transformation. The gamma distribution has been used to describe the distribution of precipitation depth for fixed durations and, correspondingly, rainfall intensities for fixed durations (e.g., Nadarajah and Kotz, 2007). The gamma distribution has been shown to adequately represent daily-averaged stream flows at many locations in the United States, with different gamma distributions appropriate for wet and dry seasons (Botter et al., 2007). The gamma distribution has also been shown to adequately describe annual flood peaks in the United States (e.g., Villarini et al., 2009). EXAMPLE 8.8 A commercial area experiences significant flooding whenever a storm yields more than 13 cm of rainfall in 24 hours. In a typical year, there are four such storms. Assuming that these rainfall events are a Poisson process, what is the probability that it will take more than 2 years to have 4 flood events? Solution The probability density function of the time for four flood events is given by the gamma distribution, Equation 8.49. From the given data, n = 4, μt = 365 d (= 1 year), and 4 n = 0.01096 d−1 = λ= μt 365 The probability of up to t0 days elapsing before the fourth flood event is given by the cumulative distribution function  t0  t0 n n−1 −λt  (0.01096)4 t0 4−1 −0.01096t λ t e dt = f (t) dt = t e dt F(t0 ) = (n − 1)! (4 − 1)! 0 0 0  t0 = 2.405 * 10−9 t3 e−0.01096t dt 0

According to Dwight (1961),



 t3 eat dt = eat

Using this relation to evaluate F(t0 ) gives ⎡  F(t0 ) = −2.405 * 10−9 ⎣e−0.01096t

t3 3t2 6t 6 − + − 2 3 a a a a4



⎤t0 3t2 6t 6 t3 ⎦ + + + 0.01096 0.010962 0.010963 0.010964 0

Taking t0 = 730 d (= 2 years) gives ⎡

⎤730 3 2 t 3t 6t 6 ⎦ + F(730) = −2.405 * 10−9 ⎣e−0.01096t + + = 0.958 0.01096 0.010962 0.010963 0.010964 

0

Hence, the probability that four flood events will occur in less than 2 years is 0.958, and the probability that it will take more than 2 years to have four flood events is 1 − 0.958 = 0.042 or 4.2%.

On the basis of the fundamental process leading to the gamma distribution, it should be clear why the gamma distribution is equal to the exponential distribution when n = 1. The gamma distribution given by Equation 8.51 can also be written in the form f (t) =

λ(λt)n−1 e−λt (n)

(8.54)

Section 8.2

Using the transformation

Probability Distributions

x = λt

377

(8.55)

the probability density function of x, p(x), is related to f (t) by the relation    dt  1 p(x) = f (t)   = f (t) dx λ

(8.56)

Combining Equations 8.54 to 8.56 and replacing n by α yields p(x) =

1 xα−1 e−x , (α)

x Ú 0

(8.57)

This distribution is commonly referred to as the one-parameter gamma distribution (Yevjevich, 1972). The mean, variance, and skewness coefficient of the one-parameter gamma distribution are given by 2 μx = α, σx2 = α, gx = √ (8.58) α Application of the one-parameter gamma distribution in hydrology is quite limited, primarily because of the difficulty in fitting a one-parameter distribution to observed hydrologic events (Yevjevich, 1972). A more general gamma distribution that does not have a lower bound of zero can be obtained by replacing x in Equation 8.57 by (x − γ )/β to yield the following probability density function: p(x) =

1 β α (α)

(x − γ )α−1 e

−(x−γ ) β

,

x Ú γ

(8.59)

where the parameter, γ , is the lower bound of the distribution, β is a scale parameter, and α is a shape parameter. The distribution function given by Equation 8.59 is called the threeparameter gamma distribution. The mean, variance, and skewness coefficient of the threeparameter (α, β, and γ ) gamma distribution are given by μx = γ + αβ,

σx2 = α,

2 gx = √ α

(8.60)

The three-parameter gamma distribution has the desirable properties (from a hydrologic viewpoint) of being bounded on the left and a positive skewness. Pearson distributions. Pearson (1930) proposed a general equation for a distribution that fits many distributions, including the gamma distribution, by choosing appropriate parameter values. A form of the Pearson function similar to the gamma distribution is known as the Pearson Type III distribution and is commonly expressed in the form p(x) =

λβ (x − )β−1 e−λ(x− ) (β)

(8.61)

where λ, β, and are parameters of the distribution which can be expressed in terms of the mean (μx ), variance (σx2 ), and skewness coefficient (gx ) as √ 2 β β 2 (8.62) , λ= , = μx − β= gx σx λ The Pearson Type III distribution was first applied in hydrology to describe the probability distribution of annual-maximum flood peaks (Foster, 1924), and is the preferred method in the Chinese Hydraulic Design Code for estimating annual-maximum flood peaks in China

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(Liu et al., 2011). When the measured data are very positively skewed, the data are usually log transformed (i.e., x is the logarithm of a variable), and the distribution is called the log-Pearson Type III distribution; this distribution is commonly referred to in practice as the LP3 distribution. The LP3 distribution is widely used in hydrology, primarily because it has been (officially) recommended for application to annual peak flood flows by the U.S. Interagency Advisory Committee on Water Data (1982). More recent comparisons with U.S. flood data reveal that the LP3 distribution provides a reasonable model of the distribution of annual flood series from unregulated watersheds for log-space skews gx … 1.414 (Griffis and Stedinger, 2007b). The LP3 distribution is also recommended for application to low-flow estimates by the American Society of Civil Engineers (ASCE, 1980). An example of frequency analysis using the LP3 distribution is given in Section 8.3.3. If a LP3 distribution is assumed for annual flood peaks, then the usual problem is to estimate the flow for a given return period. The confidence limits of the estimated flow will depend on the length of the flow record used to estimate the LP3 parameters, and these confidence limits can be estimated using graphical relationships found in McCuen and Galloway (2010). 8.2.5.6

Normal distribution

The normal distribution, also called the Gaussian distribution, is a symmetrical bell-shaped curve describing the probability density of a continuous random variable. The functional form of the normal distribution is given by 

 1 x − μx 2 1 f (x) = √ exp − 2 σx σx 2π

(8.63)

where the parameters μx and σx are equal to the mean and standard deviation of x, respectively. Normally distributed random variables are commonly described by the shorthand notation N(μ, σ 2 ), and the shape of the normal distribution is illustrated in Figure 8.6. Most hydrologic variables cannot (theoretically) be normally distributed, since the range of normally distributed random variables is from −q to q, and negative values of many hydrologic variables do not make sense. However, if the mean of a random variable is more than three or four times greater than its standard deviation, errors in the normal-distribution assumption can, in many cases, be neglected (Haan, 1977). As an application example, annual rainfall on the Greek island of Crete has been shown to mostly follow a normal distribution (Naoum and Tsanis, 2003). FIGURE 8.6: Normal probability distribution

0.40 µx = 0, σx = 1

f(x)

0.30 µx = 1, σx = 2

0.20

0.10

0.00

–6

–4

–2

0 x

2

4

6

Section 8.2

Probability Distributions

379

It is sometimes convenient to work with the standard normal deviate, z, which is defined by x − μx (8.64) z= σx where x is normally distributed. The probability density function of z is therefore given by z2 1 f (z) = √ e− 2 2π

(8.65)

Equation 8.64 guarantees that z is normally distributed with a mean of zero and a variance of unity, and is therefore an N(0,1) variate. The cumulative distribution, F(z), of the standard normal deviate is given by  z  z z 2 1   f (z ) dz = √ e− 2 dz (8.66) F(z) = 2π −q −q where F(z) is sometimes referred to as the area under the standard normal curve, and these values are tabulated in Appendix C.1. Values of F(z) can be approximated by the analytic relation (Abramowitz and Stegun, 1965)  F(z) L

B, 1 − B,

z≤0 z≥0

(8.67)

where B=

−4 1 1 + 0.196854|z| + 0.115194|z|2 + 0.000344|z|3 + 0.019527|z|4 2

(8.68)

and the error in F(z) using Equation 8.68 is less than 0.00025. Conditions under which any random variable can be expected to follow a normal distribution are specified by the central limit theorem: Central limit theorem. If Sn is the sum of n independently and identically distributed random variables Xi , each having a mean μ and variance σ 2 , then in the limit as n approaches infinity, the distribution of Sn approaches a normal distribution with mean nμ and variance nσ 2 . Unfortunately, not many hydrologic variables are the sum of independent and identically distributed random variables. Under some very general conditions, however, it can be shown that if X1 , X2 , . . . , Xn are n random variables with μXi = μi and σX2 i = σi2 , then the sum, Sn , defined by Sn = X1 + X2 + · · · + Xn (8.69) is a random variable whose probability distribution approaches a normal distribution with mean, μ, and variance, σ 2 , given by μ=

n 

μi

(8.70)

σi2

(8.71)

i=1 2

σ =

n  i=1

The application of this result generally requires a large number of independent variables to be included in the sum Sn , and the probability distribution of each Xi has negligible influence on the distribution of Sn . In hydrologic applications, annual averages of variables such as precipitation and evaporation are assumed to be calculated from the sum of independent identically distributed measurements, and they tend to follow normal distributions (Chow et al., 1988).

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Chapter 8

Probability and Statistics in Water-Resources Engineering

EXAMPLE 8.9 The annual rainfall in the Upper Kissimmee River basin has been estimated to have a mean of 130.0 cm and a standard deviation of 15.6 cm (Chin, 1993b). Assuming that the annual rainfall is normally distributed, what is the probability of having an annual rainfall of less than 101.6 cm? Solution From the given data: μx = 130.0 cm and σx = 15.6 cm. For x = 101.6 cm, the standard normal deviate, z, is given by Equation 8.64 as 101.6 − 130.0 x − μx = −1.82 = z= σx 15.6 The probability that z … −1.82, F(−1.82), is given by Equation 8.67, where 1 2 1 = 2

B=

 

1 + 0.196854|z| + 0.115194|z|2 + 0.000344|z|3 + 0.019527|z|4

−4

1 + 0.196854| − 1.82| + 0.115194| − 1.82|2 + 0.000344| − 1.82|3 + 0.019527| − 1.82|4

−4

= 0.034

and therefore F(−1.82) = 0.034 = 3.4% There is a 3.4% probability of having an annual rainfall less than 101.6 cm.

8.2.5.7

Log-normal distribution

In cases where the random variable, X, is equal to the product of n random variables X1 , X2 , . . . , Xn , the logarithm of X is equal to the sum of n random variables, where ln X = ln X1 + ln X2 + · · · + ln Xn

(8.72)

Therefore, according to the central limit theorem, ln X will be asymptotically normally distributed and X is said to have a log-normal distribution. Defining the random variable, Y, by the relation Y = ln X (8.73) then if Y is normally distributed, the theory of random functions can be used to show that the probability distribution of X, the log-normal distribution, is given by   (ln x − μy )2 1 f (x) = (8.74) exp − , x > 0 √ 2σy2 xσy 2π where μy and σy2 are the mean and variance of Y, respectively. The mean, variance, and skewness of a log-normally distributed variable, X, are given by ⎛ ⎞ σy2  ⎠ , σx2 = μ2x exp (σy2 ) − 1 , gx = 3Cv + Cv3 μx = exp ⎝μy + (8.75) 2 where Cv is the coefficient of variation defined as σx μx

Cv = If Y is defined by the relation

(8.76)

Y = log10 X

(8.77)

then Equation 8.74 still describes the probability density of X, with ln x replaced by log10 x, and the moments of X are related to the moments of Y by 

μx = 10

μy +

σy2 2



 ,

σx2

=

μ2x

(σy2 )

10

 − 1 ,

gx = 3Cv + Cv3

(8.78)

Section 8.2

Probability Distributions

381

Many hydrologic variables exhibit a marked skewness, largely because they cannot be negative. Whereas the normal distribution allows the random variable to range without limit from negative infinity to positive infinity, the log-normal distribution has a lower limit of zero. The log-normal distribution has been found to reasonably describe such variables as daily rainfall depths, daily peak discharge rates, and the distribution of hydraulic conductivity in porous media. The log-normal distribution is routinely used in China to describe annual flood peaks (Singh and Strupczewski, 2002b) and has been shown to adequately describe annual flood peaks in the United States (e.g., Villarini et al., 2009). In water-quality analyses, the log-normal distribution has been found to describe the occurrence of bacteria (coliforms) in public water-distribution systems (Christian and Pipes, 1983). EXAMPLE 8.10 Annual-maximum discharges in the Guadalupe River near Victoria, Texas, show a mean of 801 m3 /s and a standard deviation of 851 m3 /s. If the capacity of the river channel is 900 m3 /s, and the flow is assumed to follow a log-normal distribution, what is the probability that the maximum discharge will exceed the channel capacity? Solution From the given data: μx = 801 m3 /s and σx = 851 m3 /s. Equation 8.75 gives ⎛ ⎞ σy2 ⎠ = 801 exp ⎝μy + (= μx ) 2 and



(801)2 exp σy2 − 1 = (851)2 (= σx2 )

which are solved to yield μy = 6.31 and σy = 0.870 3 When x = 900 m /s, the log-transformed variable, y, is given by y = ln 900 = 6.80 and the normalized random variate, z, is given by z=

y − μy 6.80 − 6.31 = 0.563 = σy 0.870

The probability that z … 0.563, F(0.563), is given by Equation 8.67, where −4 1 B = 1 + 0.196854|z| + 0.115194|z|2 + 0.000344|z|3 + 0.019527|z|4 2 1 1 + 0.196854|0.563| + 0.115194|0.563|2 + 0.000344|0.563|3 = 2 −4 + 0.019527|0.563|4 = 0.286 which yields F(0.563) = 1 − 0.286 = 0.714 = 71.4% The probability that the maximum discharge in the river is greater than the channel capacity of 900 m3 /s is therefore equal to 1 − 0.714 = 0.286 = 28.6%. This is the probability of flooding in any given year.

8.2.5.8

Uniform distribution

The uniform distribution describes the behavior of a random variable in which all possible outcomes are equally likely within the range [a, b]. The uniform distribution can be applied to either discrete or continuous random variables. For a continuous random variable, x, the uniform probability density function, f (x), is given by f (x) =

1 , b − a

a … x … b

(8.79)

382

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Probability and Statistics in Water-Resources Engineering

where the parameters a and b define the range of the random variable. The mean, μx , and variance, σx2 , of a uniformly distributed random variable are given by μx =

1 (a + b), 2

σx2 =

1 (b − a)2 12

(8.80)

EXAMPLE 8.11 Conflicting data from several remote rain gages in a region of the Amazon basin indicate that the annual rainfall in 1997 was between 810 mm and 1080 mm. Assuming that the uncertainty can be described by a uniform probability distribution, what is the probability that the rainfall is greater than 1000 mm? Solution In this case, a = 810 mm, b = 1080 mm, and the uniform probability distribution of the 1997 rainfall is given by Equation 8.79 as f (x) =

1 = 0.00370 1080 − 810

for

810 mm … x … 1080 mm

The probability, P, that x Ú 1000 mm is therefore given by P = 0.00370 * (1080 − 1000) = 0.296 = 29.6%

8.2.5.9

Extreme-value distributions

Extreme values are either maxima or minima of sets of random variables. Consider the sample set X1 , X2 , . . . , Xn , and let Y be the largest of the sample values. If F(y) is the probability that Y < y and PXi (xi ) is the probability that Xi < xi , then the probability that Y < y is equal to the probability that all the xi s are less than y, which means that F(y) = PX1 (y)PX2 (y) . . . PXn (y) = [PX (y)]n

(8.81)

This equation gives the cumulative probability distribution of the extreme value, Y, in terms of the cumulative probability distribution of each outcome, which is called the parent distribution. The probability density function, f (y), of the extreme value, Y, is therefore given by f (y) =

d F(y) = n[PX (y)]n−1 pX (y) dy

(8.82)

where pX (y) is the probability density function of each outcome. Equation 8.82 indicates that the probability distribution of extreme values, f (y), depends on both the sample size and the parent distribution. Equation 8.82 was derived for maximum values, and a similar result can be derived for minimum values (see Problem 8.19). Distributions of extreme values selected from large samples of many probability distributions have been shown to converge to one of three types of extreme-value distribution (Fisher and Tippett, 1928): Type I, Type II, or Type III. In Type I distributions, the parent distribution is unbounded in the direction of the desired extreme, and all moments of the distribution exist. In Type II distributions, also called Frechet distributions, the parent distribution is unbounded in the direction of the desired extreme and all moments of the distribution do not exist. In Type III distributions, the parent distribution is bounded in the direction of the desired extreme. The distributions of maxima of hydrologic variables are typically of Type I, since most hydrologic variables are (quasi-) unbounded to the right; Type II distributions are seldom used in hydrologic applications; and the distributions of minima are typically of Type III, since many hydrologic variables are bounded on the left by zero. Extreme-value Type I (Gumbel) distribution. The extreme-value Type I distribution requires that the parent distribution be unbounded in the direction of the extreme value. Specifically, Type I distributions require that the parent distribution falls off in an exponential manner,

Section 8.2

Probability Distributions

383

such that the upper tail of the cumulative distribution function, PX (x), can be expressed in the form PX (x) = 1 − e−g(x) with g(x) an increasing function of x (Benjamin and Cornell, 1970). In using the Type I distribution to estimate maxima, parent distributions with the unbounded property include the normal, log-normal, exponential, and gamma distributions. Using the normal distribution as the parent distribution, the probability density function for the Type I extreme-value distribution is given by (Gumbel, 1958) 1 f (x) = exp a

!

 " x − b x − b ; − exp ; , a a

−q < x < q,

−q < b < q,

a > 0 (8.83)

where a and b are scale and location parameters, b is the mode of the distribution and the minus of the ; used for maximum values. The plus of the ; is used for minimum values. The Type I distribution for maximum values is illustrated in Figure 8.7. The Type I extremevalue distribution is sometimes referred to as the Gumbel extreme-value distribution, the Fisher–Tippett Type I distribution, or the double-exponential distribution. The mean, variance, and skewness coefficient for the extreme-value Type I distribution applied to maximum values are μx = b + 0.577a,

σx2 = 1.645a2 ,

gx = 1.1396

(8.84)

gx = −1.1396

(8.85)

and for the Type I distribution applied to minimum values, μx = b − 0.577a,

σx2 = 1.645a2 ,

By using the transformation y=

x − b a

(8.86)

the extreme-value Type I distribution can be written in the form f (y) = exp [;y − exp (;y)] which yields the following cumulative distribution functions: ⎧ ⎨exp [−exp (−y)] (maxima) F(y) = ⎩1 − exp [−exp (y)] (minima) FIGURE 8.7: Type I extreme-value probability distribution

0.4

0.3

f(x)

a=1 0.2

0.1

0.0 –5

a=2

–4

–3

–2

–1

0 (x–b)/a

1

2

3

4

5

(8.87)

(8.88)

384

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Probability and Statistics in Water-Resources Engineering

These cumulative distributions are most useful in determining the return periods of extreme events, such as flood flows (annual-maximum flows), maximum rainfall, and maximum wind speed (Gumbel, 1954). The extreme-value Type I (Gumbel) distribution is used extensively in flood studies in the United Kingdom and in many other parts of the world (Cunnane, 1988), and has been applied by the U.S. National Weather Service in analyzing annual-maximum precipitation amounts for given durations. EXAMPLE 8.12 The annual-maximum discharges in the Guadalupe River near Victoria, Texas, between 1935 and 1978 show a mean of 811 m3 /s and a standard deviation of 851 m3 /s. Assuming that the annual-maximum flows are described by an extreme-value Type I (Gumbel) distribution, estimate the annual-maximum flow rate with a return period of 100 years. Solution From the given data: μx = 811 m3 /s and σx = 851 m3 /s. According to Equation 8.84, the scale and location parameters, a and b, are derived from μx and σx by a= √

σx 1.645

851 = √ = 664 m3 /s 1.645

b = μx − 0.577a = 811 − 0.577(664) = 428 m3 /s These parameters are used to transform the annual maxima, X, to the normalized variable, Y, where Y=

X − 428 X − b = a 664

1 = 0.01 and the cumulative For a return period of 100 years, the exceedance probability is 100 probability, F(y), is 1 − 0.01 = 0.99. The extreme-value Type I probability distribution given by Equation 8.88 yields 0.99 = exp [−exp (−y100 )]

where y100 is the event with a return period of 100 years. Solving for y100 gives y100 = −ln (−ln 0.99) = 4.60 The annual-maximum flow with a return period of 100 years, x100 , is related to y100 by − 428 x y100 = 100 664 which leads to x100 = 428 + 664y100 = 428 + 664(4.60) = 3482 m3 /s Therefore, a flow of 3482 m3 /s has a return period of 100 years.

It has been suggested that the Gumbel distribution tends to underestimate the magnitude of the rarest rainfall events, and that the log-Gumbel distribution might perform better in predicting hydrological variables (Brooks and Carruthers, 1953; Bernier, 1959). The log´ Gumbel distribution is sometimes called the Frechet distribution. Extreme-value Type III (Weibull) distribution. The extreme-value Type III distribution requires that the parent distribution be bounded in the direction of the extreme value. Specifically, Type III distributions (for minima) require that the left-hand tail of the cumulative distribution function, PX (x), rises from zero for values of x Ú c such that PX (x) = α(x − c)a ,

x Ú c

where c is the lower limit of x (Benjamin and Cornell, 1970). The extreme-value Type III distribution has mostly been used in hydrology to estimate low streamflows, which are bounded

Section 8.2

Probability Distributions

385

on the left by zero. The extreme-value Type III distribution for minimum values is commonly called the Weibull distribution (after Weibull, 1939) and for c = 0 is given by a−1 −a

f (x) = ax

b

  x a exp − , b

x Ú 0,

a, b > 0

(8.89)

The Weibull distribution is illustrated in Figure 8.8. The mean and variance of the Weibull distribution are given by 

 1 1 2 2 2 2 μx = b 1 + , σx = b  1 + −  1 + (8.90) a a a and the skewness coefficient is







 1 + 3a − 3 1 + 2a  1 + 1a + 2 3 1 + gx = 



 32  1 + 2a −  2 1 + 1a

1 a

(8.91)

where (n) is the gamma function defined by Equation 8.52. The cumulative Weibull distribution is given by   x a F(x) = 1 − exp − (8.92) b which is useful in determining the return period of minimum events. If the lower bound of the parent distribution is nonzero, then a displacement parameter, c, must be added to the Type III extreme-value distributions for minima, and the probability density function becomes (Haan, 1977) f (x) = a(x − c)

a−1

(b − c)

−a



 x − c a exp − b − c

(8.93)

and the cumulative distribution then becomes 

 x − c a F(x) = 1 − exp − b − c FIGURE 8.8: Type III extreme-value (Weibull) probability distribution

0.6 b=2

0.5 a=3

f(x)

0.4 0.3 a=2

0.2

a=1

0.1 0.0

0

2

4

6 x

8

10

(8.94)

386

Chapter 8

Probability and Statistics in Water-Resources Engineering

Equation 8.93 is sometimes called the three-parameter Weibull distribution or the bounded exponential distribution. The mean and variance of the three-parameter Weibull distribution are given by 

 1 2 1 2 2 2 μx = c + (b − c) 1 + , σx = (b − c)  1 + −  1 + (8.95) a a a and the skewness coefficient of the three-parameter Weibull distribution is given by Equation 8.91. The two-parameter Weibull distribution (c = 0) is commonly used to describe the distribution of wind speeds, and is appropriate for averaging periods less than about 10 minutes (Ro and Hunt, 2007). EXAMPLE 8.13 The annual-minimum flows in a river at the location of a water-supply intake have a mean of 123 m3 /s and a standard deviation of 37 m3 /s. Assuming that the annual low flows have a lower bound of 0 m3 /s and are described by an extreme-value Type III distribution, what is the probability that an annual-low flow will be less than 80 m3 /s? Solution From the given data: μx = 123 m3 /s and σx = 37 m3 /s. According to Equation 8.90, the parameters of the probability distribution, a and b, must satisfy the relations

1 = 123 (= μx ) (8.96) b 1 + a and



2 b2  1 +

a



1 − 2 1 +

a

 = (37)2

(= σx2 )

Combining these equations to eliminate b yields 

 1 2 1232 2

 1 + −  1 + = 372 a a  2 1 + 1a or





 1 + 2a −  2 1 + 1a

= 0.0905  2 1 + 1a

This equation can be solved for a by using the Solver capability of an electronic spreadsheet with a built-in gamma function (or any other numerical means) which yields a = 3.65 and Equation 8.96 gives b=



123

 1 + 1a

=

123 123 =

= 136 1 0.903  1 + 3.65

The cumulative (Weibull) distribution of the annual-low flows, X, is given by Equation 8.92 as    

x a x 3.65 F(x) = 1 − exp − = 1 − exp − b 136 and the probability that X … 80 m3 /s is given by  F(80) = 1 − exp −



80 3.65 = 0.134 = 13.4% 136

There is a 13.4% probability that the annual-low flow is less than 80 m3 /s.

Section 8.2

Probability Distributions

387

Generalized extreme-value distribution. The generalized extreme-value (GEV) distribution incorporates the extreme-value Type I, II, and III distributions for maxima. The GEV distribution was first proposed by Jenkinson (1955) to describe the distribution of the largest values of meteorological data when the limiting form of the extreme-value distribution is unknown. The GEV distribution has to date been used to model a wide variety of natural extremes, including annual flood peaks, rainfall, wind speeds, wave heights, and other maxima (Martins and Stedinger, 2000). The GEV cumulative distribution function is given by

F(x) =

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ exp ⎪ ⎪ ⎪ ⎪ ⎨

⎧ ⎫  1 ⎪ ⎪ c⎬ ⎨ c(x − a) − 1 − , ⎪ ⎪ b ⎩ ⎭

c Z 0 (8.97)

⎧ ⎪  ⎫ ⎪ ⎪ ⎨ ⎪ ⎪ (x − a) ⎬ ⎪ ⎪ exp − exp − , ⎪ ⎪ ⎩ ⎭ ⎩ b

c=0

where a, b, and c are location, scale, and shape parameters, respectively. For c = 0, the GEV distribution is the same as the extreme-value Type I distribution; for c < 0 the GEV distribution corresponds to the Type II distribution for maxima that have a finite lower bound of a + b/c; and for c > 0 the GEV distribution corresponds to the Type III distribution for maxima that have a finite upper bound at a + b/c. The mean, variance, and skewness coefficient for the GEV distribution are given by (Kottegoda and Rosso, 1997) μx = a + σx2 =

b [1 − (1 + c)], c

(for c > −1)

2 b [(1 + 2c) −  2 (1 + c)], c

gx = sign (c)

(8.98)

(for c > −0.5)

−(1 + 3c) + 3(1 + c)(1 + 2c) − 2 3 (1 + c) 3

[(1 + 2c) −  2 (1 + c)] 2

(8.99) ,

for c > −

1 3

(8.100)

where it is noted that the moments only exist for certain values of c. In many applications, the GEV parameters (a, b, and c) are estimated from the mean, standard deviation, and skewness (μx , σx , and gx ) of the observed data; this approach to estimating the parameters of a probability distribution is called the method of moments (see Section 8.3.2.1). Using Equations 8.98 to 8.100 to estimate the GEV parameters can be cumbersome since a numerical solution is required. Whereas these calculations are fairly manageable, they can be facilitated by the following approximate relations (Bhunya et al., 2007b) ⎧ ⎨ 0.0087gx3 + 0.0582gx2 − 0.32gx + 0.2778, c= & ' ⎩ −0.31158 1 − exp[−0.4556(gx − 0.97134)] + 1.13828,

−0.7 … gx … 1.15 gx Ú 1.15 (8.101)

b = −0.1429c3 − 0.7631c2 + 1.0145c + 0.7795, σx ⎧ ⎨ 0.514075c1.33199 − 0.44901,

a − μx = ⎩ σx 19.357c4 + 13.749c3 + 4.484c2 + 0.5212c − 0.4427,

−0.5 … c … 0.5 (8.102) 0.01 … c … 0.5 −0.5 … c … 0.01 (8.103)

388

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Probability and Statistics in Water-Resources Engineering

Estimation of the GEV parameters from the moments of the observed data (i.e., μx , σx , and gx ) is appropriate for modest sample sizes; however, for small sample sizes, parameter estimation using the method of L-moments (see Section 8.3.2.3) are likely to perform better (Hosking, 1985; Madsen et al., 1997; Bhunya et al., 2007b). The generalized extreme-value distribution has been used in the United States to describe annual flood peaks (e.g., Villarini and Smith, 2010), is commonly used in Great Britain to describe annual flood extremes (Singh and Strupczewski, 2002b), has been found to perform well in describing flood flows some river basins in Turkey (Saf, 2009), and has been used describe the distribution of annual-maximum rainfall amounts for a given duration (Overeem et al., 2007; Mailhot and Duchesne, 2010; Villarini et al., 2011).

EXAMPLE 8.14 The annual-maximum 24-hour rainfall amounts in a subtropical area have a mean of 15.4 cm, a standard deviation of 3.42 cm, and a skewness coefficient of 1.24. Assuming that the annual-maximum 24-hour rainfall is described by a general extreme-value (GEV) distribution, estimate the annual-maximum 24-hour rainfall with a return period of 100 years. Solution From the given data: μx = 15.4 cm, σx = 3.42 cm, and gx = 1.24. In order to find the GEV cumulative distribution function, the parameters a, b, and c must first be determined. The shape parameter, c, can be determined directly from the skewness coefficient, gx , using Equation 8.100, where gx = 1.24 = sign (c)

−(1 + 3c) + 3(1 + c)(1 + 2c) − 2 3 (1 + c) [(1 + 2c) −  2 (1 + c)]3/2

which can be solved to give c = −0.0163. Rearranging Equation 8.99 gives b as ( b=

c2 σx2 = (1 + 2c) −  2 (1 + c)

(

(−0.0163)2 (3.42)2 = 2.61 cm 1.020 − 1.0102

and rearranging Equation 8.98 gives a as a = μx −

2.61 b [1 − (1 + c)] = 15.4 − [1 − 1.010] = 13.9 cm c −0.0163

Hence the cumulative distribution function, F(x), is given by Equation 8.97 as ⎫ ⎧  1 ⎪ ⎪ ⎨ c⎬ c(x − a) = exp F(x) = exp − 1 − ⎪ ⎪ b ⎭ ⎩ which yields

⎧ ⎫   1 ⎪ ⎪ ⎨ −0.0163 ⎬ −0.0163(x − 13.9) − 1 − ⎪ ⎪ 2.61 ⎩ ⎭

)  * −61.3 F(x) = exp − 1 + 0.00625(x − 13.9)

For a return period of 100 years, 1 = 0.99 and 100 * )  −61.3 F(x100 ) = 0.99 = exp − 1 + 0.00625(x100 − 13.9)

F(x100 ) = 1 −

which yields x100 = 26.4 cm Therefore, an annual-maximum 24-hour rainfall of 26.4 cm has a return period of 100 years.

Section 8.2 FIGURE 8.9: Chi-Square probability distribution

Probability Distributions

389

0.5 0.4 ν=2

f(x)

0.3 0.2

ν=4

0.1 0.0

0

2

4

6

8

10

12

14

16

x

8.2.5.10

Chi-square distribution

Unlike the previously described probability distributions, the chi-square distribution is not used to describe hydrologic processes but is more commonly used in testing how well observed outcomes of hydrologic processes are described by theoretical probability distributions. The chi-square distribution is the probability distribution of a variable obtained by adding the squares of ν normally distributed random variables, all of which have a mean of zero and a variance of 1. That is, if X1 , X2 , . . . , Xν are normally distributed random variables with mean of zero and variance of 1 and a new random variable χ 2 is defined such that χ2 =

ν 

Xi2

(8.104)

i=1

then the probability density function of χ 2 is defined as the chi-square distribution and is given by f (x) =

− 1− ν2

x

−x

e2 , ν ν 22  2

x, ν > 0

(8.105)

where x = χ 2 and ν is called the number of degrees of freedom. The chi-square distribution is illustrated in Figure 8.9. The mean and variance of the chi-square distribution are given by μχ 2 = ν,

σχ22 = 2ν

(8.106)

The chi-square distribution is commonly used in determining the confidence intervals of various sample statistics. The cumulative chi-square distribution is given in Appendix C.3. EXAMPLE 8.15 A random variable is calculated as the sum of the squares of 10 normally distributed variables with a mean of zero and a standard deviation of 1. What is the probability that the sum is greater than 20? What is the expected value of the sum? Solution The random variable cited here is a χ 2 variate with 10 degrees of freedom. The cumulative distribution of the χ 2 variate is given in Appendix C.3 as a function of the number of degrees of freedom, ν, and the exceedance probability, α. In this case, for ν = 10 and χ 2 = 20, Appendix C.3 gives (by interpolation) α = 0.031

390

Chapter 8

Probability and Statistics in Water-Resources Engineering Therefore, the probability that the sum of the squares of 10 N(0,1) variables exceeds 20 is 0.031 or 3.1%. The expected value of the sum is, by definition, equal to the mean, μχ 2 , which is given by Equation 8.106 as μχ 2 = ν = 10

8.3 Analysis of Hydrologic Data 8.3.1

Estimation of Population Distribution

Utilization of hydrologic data in engineering practice is typically a two-step procedure. In the first step, the cumulative probability distribution of the measured data is compared with a variety of theoretical distribution functions, and the theoretical distribution function that best fits the probability distribution of the measured data is taken as the population distribution. In the second step, the exceedance probabilities of selected events are estimated using the theoretical population distribution. Some federal agencies require the use of specific probability distributions for describing certain hydrologic events, such as peak river flows, and in some cases they also specify the minimum size of the data set to be used in estimating the parameters of the distribution. In estimating probability distributions from measured data, the measurements are usually assumed to be independent and drawn from identical probability distributions. In cases where the observations are correlated, alternative methods such as time-series analysis are used in lieu of probability theory. The most common methods of estimating population distributions from measured data are: (1) visually comparing the probability distribution of the measured data with various theoretical distributions, and picking the closest distribution that is consistent with the underlying process generating the sample data, and (2) using hypothesis-testing methods to quantitatively assess whether various probability distributions are consistent with the probability distribution of the measured data. In estimating the occurrence of rare events from a fitted probability distribution, particular care should be taken to ensure a good match in the tail of the fitted distribution (e.g. El Adlouni et al., 2008). 8.3.1.1

Probability distribution of observed data

To assist in identifying a theoretical probability distribution that adequately describes observed outcomes, it is generally useful to graphically compare the probability distribution of observed outcomes with various theoretical probability distributions. The first step in plotting the probability distribution of observed data is to rank the data, such that for N observations a rank of 1 is assigned to the observation with the largest magnitude and a rank of N is assigned to the observation with the lowest magnitude. The exceedance probability of the m-ranked observation, xm , denoted by PX (X > xm ), is commonly estimated by the relation PX (X > xm ) =

m , N + 1

m = 1, . . . , N

(8.107)

or as a cumulative distribution function PX (X < xm ) = 1 −

m , N + 1

m = 1, . . . , N

(8.108)

Equation 8.107 is called the Weibull formula (Weibull, 1939) and is widely used in practice. The main drawback of the Weibull formula for estimating the cumulative probability distribution from measured data is that it is asymptotically exact (as the number of observations approaches infinity) only for a population with an underlying uniform distribution, which is relatively rare in nature. To address this shortcoming, Gringorten (1963) proposed that the exceedance probability of observed data be estimated using the relation PX (X > xm ) =

m − a , N + 1 − 2a

m = 1, . . . , N

(8.109)

Section 8.3

Analysis of Hydrologic Data

391

where a is a parameter that depends on the population distribution. For the normal, lognormal, and gamma distributions, a = 0.375; for the Gumbel and Weibull distributions, a = 0.44; and for the GEV distribution, a = 0.40 (Cunnane, 1978; Stedinger et al., 1993; Castellarin, 2007). Bedient and Huber (2002) suggest that a = 0.40 is a good compromise for the usual situation in which the exact distribution is unknown. EXAMPLE 8.16 The annual peak flows in the Guadalupe River near Victoria, Texas, between 1965 and 1978 are as follows: Peak flow

Peak flow

Year

(m3 /s)

Year

(m3 /s)

1965 1966 1967 1968 1969 1970 1971

425 260 1982 1254 430 277 276

1972 1973 1974 1975 1976 1977 1978

1657 937 714 855 399 1543 360

Use the Weibull and Gringorten formulae to estimate the cumulative probability distribution of annual peak flow and compare the results. Solution The rankings of flows between 1965 and 1978 are shown in Columns 1 and 2 of the following table: (1) Rank, m

(2) Flow, qm (m3 /s)

(3) Weibull P(Q > qm )

(4) Gringorten P(Q > qm )

1 2 3 4 5 6 7 8 9 10 11 12 13 14

1982 1657 1543 1254 937 855 714 430 425 399 360 277 276 260

0.067 0.133 0.200 0.267 0.333 0.400 0.467 0.533 0.600 0.667 0.733 0.800 0.867 0.933

0.042 0.113 0.183 0.253 0.324 0.394 0.465 0.535 0.606 0.676 0.746 0.817 0.887 0.958

In this case, N = 14 and the Weibull exceedance probabilities are given by Equation 8.107 as m m = P(Q > qm ) = 14 + 1 15 These probabilities are tabulated in Column 3. The Gringorten exceedance probabilities are given by (assuming a = 0.40) m − a m − 0.40 m − 0.40 P(Q > qm ) = = = N + 1 − 2a 14 + 1 − 2(0.40) 14.2 and these probabilities are tabulated in Column 4. The probability functions estimated using the Weibull and Gringorten formulae are compared in Figure 8.10, where the distributions appear very similar. At high flows, the Gringorten formula assigns a lower exceedance probability and a correspondingly higher return period.

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Probability and Statistics in Water-Resources Engineering

FIGURE 8.10: Comparison between Weibull and Gringorten probability distributions

0.0

1.0

0.1

0.9

Gringorten 0.2

Cumulative probability

0.8 0.7

Weibull

0.3

0.6

0.4

0.5

0.5

0.4

0.6

0.3

0.7

0.2

0.8

0.1

0.9

0.0 0

200

400

600

800

Exceedance probability

392

1.0 1000 1200 1400 1600 1800 2000 2200

Flow (m3/s)

Visual comparison of the probability distribution calculated from observed data with various theoretical population distributions gives a preliminary idea of which theoretical distributions might provide the best fit to the observed data. A quantitative comparison between the observed and theoretical distributions can be obtained using the probability plot correlation coefficient (PPCC) originally developed by Filliben (1975). The PPCC test uses the correlation coefficient, r, between the ordered observations, xi , and the corresponding fitted quantiles, Mi , determined from the calculated cumulative probability, Pi , for each xi . It is inferred that the observations are described by the fitted distribution if the value of r is close to 1.0. The correlation coefficient, r, is given by

+N i=1 (xi − x) Mi − M r= , (8.110) 2 +N +N

2 i=1 (xi − x) i=1 Mi − M where x and M denote the mean values of the observations, xi , and the fitted quantiles, Mi , respectively, and N is the sample size. The fitted quantile, Mi , is related to the calculated cumulative probability, Pi , for each xi by the assumed theoretical probability distribution, according to the relation Mi = −1 (Pi ) (8.111) where −1 is the inverse of theoretical cumulative distribution function. Correlation between the observed and theoretical probability distribution cannot be rejected at the q significance level when (8.112) r > rq (N) where rq (N) is the PPCC test statistic for a given sample size, N, and significance level, q. Values of rq for an assumed normal distribution can be estimated by the relation (Heo et al., 2008) ⎧ ⎪ 1.29 + 0.283 · ln(q) + (0.887 − 0.751q + 3.21q2 ) · ln(N) ⎪ ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ ⎨0.505 + 2.13q + (0.857 − 0.133q) · ln(N) 1 ln = ⎪ 1 − rq  

⎪ ⎪ ⎪ 1 −2.42+5.63q−3.11q2 ⎪ ⎪ ⎪ N ⎪ ⎩ −5.22 + 12.3q − 6.81q2

0.005 … q < 0.1 0.1 … q < 0.9 0.9 … q … 0.995 (8.113)

Values of rq for Gumbel, gamma (Pearson Type III), and GEV distributions can be found in Heo et al. (2008).

Section 8.3

Analysis of Hydrologic Data

393

EXAMPLE 8.17 The annual peak flows in the Guadalupe River cited in the previous example appear to fit a log-normal distribution. Use the probability plot correlation coefficient to evaluate whether this approximation is justified at the 95% confidence level. Solution The rank-ordered annual peak flows in the Guadalupe River are tabulated in Columns 1 and 2 in Table 8.1, and the logs (to base 10) of the flows are shown as log Qi in Column 3. The mean, X, and standard deviation, SX , of the log flows in Column 3 are calculated as X = 2.806 and SX = 0.313. The cumulative probabilities from the ranked measurements using the Gringorten formula with a = 0.375 (for a log-normal distribution) are shown as Pi in Column 4. The standard normal deviate, zi , corresponding to Pi is shown in Column 5, and the theoretical value of log Qi (= Mi ) is derived from zi according to the relation Mi = X + zi SX = 2.806 + 0.313zi The calculated values of Mi are shown in Column 6. The correlation, r, between log Qi and Mi is calculated using Equation 8.110 as 0.97, and the 95% significance value of r given by Equation 8.113 is r0.95 = 0.99. Therefore, although the correlation, r, is high, it is not sufficiently high to justify the assertion, with 95% confidence, that the probability distribution of annual peak flows in the Guadalupe River is lognormal. The relationship between the theoretical log-normal distribution and the frequency distribution of observations is shown in Figure 8.11, where it is apparent that the match is less than ideal. TABLE 8.1: Calculation of Probability Plot Correlation Coefficient

(1)

FIGURE 8.11: Observed and theoretical probability distributions

(3) log Qi (log m3 /s)

(4)

(5)

i

(2) Qi (m3 /s)

Pi

zi

(6) Mi (log m3 /s)

1 2 3 4 5 6 7 8 9 10 11 12 13 14

1982 1657 1543 1254 937 855 714 430 425 399 360 277 276 260

3.30 3.22 3.19 3.10 2.97 2.93 2.85 2.63 2.63 2.60 2.56 2.44 2.44 2.41

0.96 0.89 0.82 0.75 0.68 0.61 0.54 0.46 0.39 0.32 0.25 0.18 0.11 0.04

1.71 1.21 0.90 0.66 0.45 0.27 0.09 −0.09 −0.27 −0.45 −0.66 −0.90 −1.21 −1.71

3.34 3.18 3.09 3.01 2.95 2.89 2.83 2.78 2.72 2.66 2.60 2.52 2.43 2.27

3.4

Flow (m3/s)

3.2

Observations Log-normal distribution (theory)

3.0 2.8 2.6 2.4 2.2 0.0

0.2

0.6 0.4 Cumulative probability

0.8

1.0

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8.3.1.2

Hypothesis tests

As an alternative to the direct comparison of observed probability distributions with various theoretical probability distributions, quantitative comparisons can also be made using hypothesis tests. The two most common hypothesis tests for quantitatively assessing whether an observed probability distribution can be approximated by a given (theoretical) population distribution are the chi-square test and the Kolmogorov–Smirnov test. The chi-square test. Based on sampling theory, it is known that if N outcomes are divided into M classes, with Xm being the number of outcomes in class m, and pm being the theoretical probability of an outcome being in class m, then the random variable χ2 =

M  (Xm − Npm )2 Npm

(8.114)

m=1

has a chi-square distribution. The number of degrees of freedom is M − 1 if the expected frequencies can be computed without having to estimate the population parameters from the sample statistics, while the number of degrees of freedom is M − 1− n if the expected frequencies are computed by estimating n population parameters from sample statistics. In applying the chi-square goodness of fit test, the null hypothesis is taken as H0 : The samples are drawn from the proposed probability distribution. The null hypothesis is accepted at the α significance level if χ 2 ∈ [0, χα2 ] and rejected otherwise. EXAMPLE 8.18 Analysis of a 47-year record of annual rainfall indicates the following frequency distribution: Range (mm)

Number of outcomes

Range (mm)

Number of outcomes

1500

7 5 3 2 2 1

The measured data also indicate a mean of 1225 mm and a standard deviation of 151 mm. Using a 5% significance level, assess the hypothesis that the annual rainfall is drawn from a normal distribution. Solution The first step in the analysis is to derive the theoretical frequency distribution. Appendix C.1 gives the cumulative probability distribution of the standard normal deviate, z, which is defined as x − 1225 x − μx = z= σx 151 where x is the annual rainfall. Converting the annual rainfall amounts into standard normal deviates, z, yields: Rainfall (mm) z P(Z < z) 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500

−1.49 −1.16 −0.83 −0.50 −0.17 0.17 0.50 0.83 1.16 1.49 1.82

0.07 0.12 0.20 0.31 0.43 0.57 0.69 0.80 0.88 0.93 0.97

Section 8.3

Analysis of Hydrologic Data

395

and therefore the theoretical frequencies are given by Range (mm)

Theoretical probability, pm

Theoretical outcomes, Npm

1500

0.07 0.05 0.08 0.11 0.12 0.14 0.12 0.11 0.08 0.05 0.04 0.03

3.29 2.35 3.76 5.17 5.64 6.58 5.64 5.17 3.76 2.35 1.88 1.41

where the total number of observations, N, is equal to 47. Based on the observed and theoretical frequency distributions, the chi-square statistic is given by Equation 8.114 as χ2 =

(3 − 2.35)2 (4 − 3.76)2 (5 − 5.17)2 (6 − 5.64)2 (2 − 3.29)2 + + + + 3.29 2.35 3.76 5.17 5.64 +

(7 − 5.64)2 (5 − 5.17)2 (3 − 3.76)2 (7 − 6.58)2 + + + 6.58 5.64 5.17 3.76

(2 − 1.88)2 (1 − 1.41)2 (2 − 2.35)2 + + 2.35 1.88 1.41 = 1.42 +

Since both the mean and standard deviation were estimated from the measured data, the χ 2 statistic has M − 1 − n degrees of freedom, where M = 12 (= number of intervals), and n = 2 (= number of population parameters estimated from measured data), hence 12 − 1 − 2 = 9 degrees of freedom. Using a 5% significance level, the hypothesis that the observations are drawn from a normal distribution is accepted if 2 0 … 1.42 … χ0.05 2 Appendix C.3 gives that for 9 degrees of freedom, χ0.05 = 16.919. Since 0 … 1.42 … 16.919, the hypothesis that the annual rainfall is drawn from a normal distribution is accepted at the 5% significance level.

The effectiveness of the chi-square test is diminished if both the number of data intervals, called cells, is less than 5 and the expected number of outcomes in any cell is less than 5 (Haldar and Mahadevan, 2000; McCuen, 2002a). Kolmogorov–Smirnov test. This test differs from the chi-square test in that no parameters from the theoretical probability distribution need to be estimated from the observed data. For this reason, the Kolmogorov–Smirnov test is classified as a nonparametric test. The procedure for implementing the Kolmogorov–Smirnov test is as follows (Haan, 1977): Step 1: Let PX (x) be the specified theoretical cumulative distribution function under the null hypothesis. Step 2: Let SN (x) be the sample cumulative distribution function based on N observations. For any observed x, SN (x) = k/N, where k is the number of observations less than or equal to x. Step 3: Determine the maximum deviation, D, defined by   D = max PX (x) − SN (x) (8.115)

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Step 4: If, for the chosen significance level, the observed value of D is greater than or equal to the critical value of the Kolmogorov–Smirnov statistic tabulated in Appendix C.4, the hypothesis is rejected. An advantage of the Kolmogorov–Smirnov test over the chi-square test is that it is not necessary to divide the data into intervals; thus any error of judgment associated with the number or size of the intervals is avoided (Haldar and Mahadevan, 2000). EXAMPLE 8.19 Use the Kolmogorov–Smirnov test at the 10% significance level to assess the hypothesis that the data in Example 8.18 are drawn from a normal distribution. Solution Based on the given data, the measured and theoretical probability distributions of the annual rainfall are as follows: (1) Rainfall x (mm) 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500

(2) Normalized rainfall (x − μx )/σx −1.490 −1.159 −0.828 −0.497 −0.166 0.166 0.497 0.828 1.159 1.490 1.821

(3)

(4)

(5)

SN (x)

PX (x)

  PX (x) − SN (x)

2/47 = 0.043 5/47 = 0.106 9/47 = 0.191 14/47 = 0.298 20/47 = 0.426 27/47 = 0.574 34/47 = 0.723 39/47 = 0.830 42/47 = 0.894 44/47 = 0.936 46/47 = 0.979

0.068 0.123 0.204 0.310 0.434 0.566 0.690 0.796 0.877 0.932 0.966

0.025 0.017 0.013 0.012 0.008 0.008 0.033 0.034 0.017 0.004 0.013

The normalized rainfall amounts in Column 2 are calculated using the rainfall amount, x, in Column 1; mean, μx , of 1225 mm; and standard deviation, σx , of 151 mm. The sample cumulative distribution function, SN (x), in Column 3 is calculated using the relation k N where k is the number of measurements less than or equal to the given rainfall amount, x, and N = 47 is the total number of measurements. The theoretical cumulative distribution function, PX (x), in Column 4 is calculated using the normalized rainfall (Column 2) and the standard normal-distribution function given in Appendix C.1. The absolute value of the difference between the theoretical and sample distribution functions is given in Column 5. The maximum difference, D, between the theoretical and sample distribution functions is equal to 0.034 and occurs at an annual rainfall of 1350 mm. For a sample size of 11 and a significance level of 10%, Appendix C.4 gives the critical value of the Kolmogorov–Smirnov statistic as 0.352. Since 0.034 < 0.352, the hypothesis that the measured data are from a normal distribution is accepted at the 10% significance level. SN (x) =

The Kolmogorov–Smirnov test is generally more efficient than the chi-square test when the sample size is small (McCuen, 2002a). However, neither of these tests is very powerful in the sense that the probability of accepting a false hypothesis is quite high, especially for small samples (Haan, 1977). Furthermore, it should be kept in mind that probability distributions that demonstrate a good fit with observed data might perform poorly in estimating the frequencies of outcomes in the tails of the fitted distributions (e.g., Mitosek et al., 2006). In cases where there are several competing distributions, goodness-of-fit statistics can be used to rank the effectiveness of the competing distributions. Both the chi-square and the Kolmogorov–Smirnov test statistics are widely used in engineering applications, although other goodness-of-fit statistics have also been proposed (e.g., Mitosek et al., 2006).

Section 8.3

8.3.1.3

Analysis of Hydrologic Data

397

Model selection criteria

Limitations of using hypothesis testing as the sole basis for selecting a probability model to describe hydrologic observations are: (1) multiple distributions might be found suitable to describe a given set of observations; and (2) hypothesis testing does not account for the tradeoff between increasing the number of distribution parameters and improved predictability. These limitations can be addressed by using various model selection criteria. The most commonly used model selection criteria are the Akaike information criterion (Akaike, 1973) and the Bayesian information criterion (Schwarz, 1978). Suppose that a candidate probability density function, f , for an observed set of data is given by f (x|θ ), where x is an individual observation and θ is the vector of model parameters, then the likelihood, L(θ), of the observed data set, xi , i = [1, N], for a best-estimate modelparameter set, θˆ , is given by L(θˆ ) =

N -

ˆ f (xi |θ)

(8.116)

i=1

ˆ is defined by the The Akaike information criterion (AIC) for the probability model, f (x|θ), relation AIC = −2 ln(L) + 2np

(8.117)

where np is the number of parameters in the model. The Bayesian information criterion (BIC) is defined by the relation BIC = −2 ln(L) + ln(N)np

(8.118)

When all candidate models are considered, the model with the lowest AIC or BIC is considered to be the best (most parsimonious) model. The AIC and BIC formulations given by Equations 8.117 and 8.118 are quite similar, and in most cases these criteria give similar results and a similar ranking of preferred models. Other model selection criteria, such as the Anderson–Darling criterion, have also been proposed as bases for identifying the optimal model (Laio et al., 2009). 8.3.2

Estimation of Population Parameters

The probability distribution of a random variable X is typically written in the form fX (x). However, this distribution can also be written in the expanded form fX (x | θ1 , θ2 , . . . , θm ) to indicate that the probability distribution of the random variable also depends on the values of the m parameters θ1 , θ2 , . . . , θm . This explicit expression for the probability distribution is particularly relevant when the population parameters are estimated using sample statistics, which are themselves random variables. The most common methods for estimating the parameters from measured data are the method of moments, the maximum-likelihood method, and the method of L-moments. 8.3.2.1

Method of moments

The method of moments is based on the observation that the parameters of a probability distribution can usually be expressed in terms of the first few moments of the distribution. These moments can be estimated using sample statistics, and then the parameters of the distribution can be calculated using the relationship between the population parameters and the moments. The three moments most often used in hydrology are the mean, standard deviation, and skewness, defined by Equations 8.22, 8.24, and 8.26, respectively, for both discrete and continuous probability distributions. In practical applications, these moments must be estimated from finite samples, and the following equations usually provide the best estimates: μˆ x =

N 1  xi N i=1

(8.119)

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 1 (xi − μˆ x )2 N − 1 N

σˆ x2 =

i=1

gˆ x =

N (N − 1)(N − 2)

(8.120)

+N

i=1 (xi

− μˆx )3

(8.121)

σˆx3

where μˆ x , σˆ x2 , and gˆ x are unbiased estimates of the mean (μx ), variance (σx2 ), and skewness (gx ) of the population, respectively, from which the N samples of x, denoted by xi , are drawn (Benjamin and Cornell, 1970). Unbiased estimates of the variance, σˆ x2 , and skewness, gˆ x , are generally dependent on the underlying population distribution. The accuracy of the estimated skewness, gˆ x , is usually of most concern, since it involves the summation of the cubes of deviations from the mean and is therefore subject to larger errors in its computation. Although Equation 8.121 is most often used to estimate the skewness, other estimates have ´ and Robitaille, 1975; Tasker and Stedinger, 1986). also been proposed (Bobee EXAMPLE 8.20 The number of hurricanes hitting Florida each year between 1903 and 1988 is given in the following table (Winsberg, 1990):

1900 1910 1920 1930 1940 1950 1960 1970 1980

00

01

02

03

04

05

06

07

08

09

1 0 0 0 2 1 0 0

1 1 0 1 0 0 0 0

0 0 1 0 0 0 1 0

1 0 0 2 0 1 3 0 0

1 0 2 0 1 0 0 0 0

0 1 1 2 2 0 1 1 2

2 2 2 1 1 1 2 0 0

0 1 0 0 2 0 0 0 1

0 0 2 0 2 0 1 0 0

1 1 1 1 1 0 0 1

If the number of hurricanes per year is described by a Poisson distribution, estimate the parameters of the probability distribution. What is the probability of three hurricanes hitting Florida in 1 year? Solution The Poisson probability distribution, f (n), is given by (λt)n e−λt n! where n is the number of occurrences per year and the parameter, λt, is related to the mean, μn , and standard deviation, σn , of n by √ μn = λt, σn = λt f (n) =

Based on the 86 years of data (1903–1988), the mean number of occurrences per year, μn , is estimated by the first-order moment, N, as 1  1 (56) = 0.65 μn L N = ni = 86 86 Hence λt L 0.65 and the probability distribution of the number of hurricanes per year in Florida is given by 0.65n e−0.65 f (n) = n! Putting n = 3 gives the probability of three hurricanes in 1 year as 0.653 e−0.65 = 0.02 = 2% 3! It is interesting to note that the probability of at least one hurricane hitting Florida in any year is given by 1 − f (0) = 1 − e−0.65 = 0.48 = 48%. f (3) =

Section 8.3

Analysis of Hydrologic Data

399

Estimates of the mean, variance, and skewness given by Equations 8.119 to 8.121 are based on samples of a random variable and are therefore random variables themselves. The standard deviation of an estimated parameter is commonly called the standard error, and the standard errors of the estimated mean, variance, and skewness are given by the following relations σˆ x Sμˆ x = √ N , Sσˆ x = σˆ x

(8.122) 1 + 0.75gˆ x 2N . N / 0.5

 Sgˆ x = 10A−B log10

10

(8.123) (8.124)

where A and B are given by A=

! −0.33 + 0.08|gˆ x |

if |gˆ x | … 0.90

−0.52 + 0.30|gˆ x |

if |gˆ x | > 0.90

and

! B=

0.94 − 0.26|gˆ x |

if |gˆ x | … 1.50

0.55

if |gˆ x | > 1.50

(8.125)

(8.126)

where Sμˆ x , Sσˆ x , and Sgˆ x are the standard errors of μˆ x , σˆ x , and gˆ x , respectively. Combining Equations 8.119 to 8.126 shows that, for a given sample size, the relative accuracy of the skewness is much less than the relative accuracies of both the mean and standard deviation, especially for small sample sizes.

EXAMPLE 8.21 Annual-maximum peak flows in the Rocky River from 1960 to 1980 show a mean of 52 m3 /s, standard deviation of 21 m3 /s, and skewness of 0.8. Calculate the standard errors and assess the relative accuracies of the estimated parameters. Solution From the given data: N = 21 (1960–1980), μˆ = 52 m3 /s, σˆ = 21 m3 /s, and gˆ = 0.8. Substituting these values into Equations 8.122 to 8.124 yields 21 σˆ Sμˆ = √ = √ = 4.6 m3 /s N 21 ( , 1 + 0.75gˆ 1 + 0.75(0.8) = 21 = 4.1 m3 /s Sσˆ = σˆ 2N 2(21) 0.5  Sgˆ = 10A−B log10 (N/10) where Equations 8.125 and 8.126 give A = −0.33 + 0.08|0.8| = −0.27 B = 0.94 − 0.26|0.8| = 0.73 and therefore

0.5  Sgˆ = 10−0.27−0.73 log10 (21/10) = 0.56

Hence, the standard errors of the mean, standard deviation, and skewness are 4.6 m3 /s, 4.1 m3 /s, and 0.56, respectively. Using the ratio of the standard error to the estimated value as a measure of the accuracy of the estimated value, then

400

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Sμˆ 4.6 * 100 = * 100 = 8.8% μˆ 52

S 4.1 accuracy of standard deviation = σˆ * 100 = * 100 = 20% σˆ 21 Sgˆ 0.56 accuracy of skewness = * 100 = * 100 = 70% gˆ 0.8 Based on these results, the accuracies of the estimated parameters deteriorate as the order of the moments associated with the parameters increases.

8.3.2.2

Maximum-likelihood method

The maximum-likelihood method selects the population parameters that maximize the likelihood of the observed outcomes. Consider the case of n independent outcomes x1 , x2 , . . . , xn , where the probability of any outcome, xi , is given by pX (xi | θ1 , θ2 , . . . , θm ), where θ1 , θ2 , . . . , θm are the population parameters. The probability of the n observed (independent) outcomes is then given by the product of the probabilities of each of the outcomes. This product is called the likelihood function, L(θ1 , θ2 , . . . , θm ), where L(θ1 , θ2 , . . . , θm ) =

n -

pX (xi | θ1 , θ2 , . . . , θm )

(8.127)

i=1

The values of the parameters that maximize the value of L are called the maximumlikelihood estimates of the parameters. Since the form of the probability function, pX (x|θ1 , θ2 , . . . , θm ), is assumed to be known, the maximum-likelihood estimates can be derived from Equation 8.127 by equating the partial derivatives of L with respect to each of the parameters, θi , to zero. This leads to the following m equations ⭸L = 0, ⭸θi

i = 1, . . . , m

(8.128)

This set of m equations can be solved simultaneously to yield the m maximum-likelihood parameters θˆ1 , θˆ2 , . . . , θˆm . In some cases, it is more convenient to maximize the natural logarithm of the likelihood function than the likelihood function itself. This approach is particularly convenient when the probability distribution function involves an exponential term. It should be noted that since the logarithmic function is monotonic, values of the estimated parameters that maximize the logarithm of the likelihood function also maximize the likelihood function. EXAMPLE 8.22 Use the maximum-likelihood method to estimate the parameter in the Poisson distribution of hurricane hits described in the previous example. Compare this result with the parameter estimate obtained using the method of moments. Solution The Poisson probability distribution can be written as f (n) =

θ n e−θ n!

where θ = λt is the population parameter of the Poisson distribution. Define the likelihood function, L , for the 86 years of data as

Section 8.3 86 -

L =

Analysis of Hydrologic Data

401

f (ni )

i=1

where ni is the number of occurrences in year i. It is more convenient to work with the log-likelihood function, L, which is defined as   86 86 86 86     1 1  + ni ln θ − θ = + ln θ ln f (ni ) = ln ni − 86θ L = ln L = ln ni ! ni ! i=1

i=1

i=1

i=1

Taking the derivative with respect to θ and putting ⭸L/⭸θ = 0 gives

⭸L 1  ni − 86 = 0 = ⭸θ θ 86

i=1

which leads to

1  ni 86 86

θ=

i=1

This is the same estimate of θ that was derived in the previous example using the method of moments.

The method of moments and the maximum-likelihood method do not always yield the same estimates of the population parameters. The maximum-likelihood method is generally preferred over the method of moments, particularly for large samples (Haan, 1977). The method of moments is severely affected if the data contain errors in the tails of the distribution, where the moment arms are long (Chow, 1954), and is particularly severe in highly skewed distributions (Haan, 1977). In contrast, the relative asymptotic bias of upper quartiles (i.e., high-return-period events) is smallest for the method of moments and is largest for the maximum-likelihood method when the true distribution is either the log-normal distribution or the gamma distribution and another distribution is fitted to it (Strupczewski et al., 2002). Several useful relationships between measured data and maximum-likelihood parameter estimates for a variety of distributions can be found in Bury (1999). 8.3.2.3

Method of L-moments

The typically small sample sizes available for characterizing hydrologic time series yield estimates of the third and higher moments that are usually very uncertain. This has led to the use of an alternative approach for estimating the parameters of probability distributions called L-moments. The rth probability-weighted moment, βr , is defined by the relation  βr =

+q

−q

x[FX (x)]r fX (x)dx

(8.129)

where FX and fX are the cumulative distribution function and probability density function of x, respectively. The L-moments, λr , are linear combinations of the probability-weighted moments, βr , and the first four L-moments are computed as λ1 = β0

(8.130)

λ2 = 2β1 − β0

(8.131)

λ3 = 6β2 − 6β1 + β0

(8.132)

λ4 = 20β3 − 30β2 + 12β1 − β0

(8.133)

Since the probability-weighted moments involve raising values of FX rather than x to powers, and because FX … 1, estimates of probability-weighted moments and L-moments are

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much less susceptible to the influences of a few large or small values in the sample. Hence, L-moments are generally preferable to product moments for estimating the parameters of probability distributions of hydrologic variables. Consider a sample of N measured values of a random variable, X. To estimate the L-moments of the probability distribution from which the sample was taken, first rank the values as x1 … x2 … x3 … · · · … xN , and estimate the probability-weighted moments as follows (Hosking and Wallis, 1997): N 1  b0 = xi N

(8.134)

i=1

 1 (i − 1)xi N(N − 1)

(8.135)

 1 (i − 1)(i − 2)xi N(N − 1)(N − 2)

(8.136)

N

b1 =

i=2

N

b2 =

i=3

 1 (i − 1)(i − 2)(i − 3)xi b3 = N(N − 1)(N − 2)(N − 3) N

(8.137)

i=4

The sample estimates of the first four L-moments, denoted by L1 to L4 , are then calculated by substituting b0 to b3 for β0 to β3 , respectively, in Equations 8.130 to 8.133. The L-moments of various probability distributions are given in terms of the parameters of the distributions in Table 8.2. Equating the sample L-moments to the theoretical L-moments of a distribution and then solving for the distribution parameters constitutes the method of L-moments.

EXAMPLE 8.23 The annual-maximum flows for the period 1946–1970 in Dry-Gulch Creek are as follows:

Year

Maximum flow (m3 /s)

Year

Maximum flow (m3 /s)

1946 1947 1948 1949 1950 1951 1952 1953 1954

126 178 251 35 71 501 891 18 2239

1955 1956 1957 1958 1959 1960 1961 1962 1963

3 2 141 282 112 40 63 398 708

Year

Maximum flow (m3 /s)

1964 1965 1966 1967 1968 1969 1970

11 1122 2 1259 158 126 35

A histogram of these data indicates that they are likely drawn from a log-normal distribution. Use the method of L-moments to estimate the mean and standard deviation of the log-normal distribution, and compare these parameters with those estimated using the method of moments.

TABLE 8.2: Moments and L-Moments of Common Probability Distributions

Distribution Normal

Parameters

Moments

L-Moments

μX , σX

μX = μX

λ1 = μX σ λ2 = X π 1/2

σX = σX ⎛ Log-normal (Y = ln X)

μY = μY

μY , σY

ξ, η

μX = ξ + σX =

Gumbel

Generalized extreme value

ξ, α

ξ , α, κ

1 η

λ1 = ξ +

1 η2

λ2 =

1 2η

λ1 = ξ + 0.5772α

2 = 1.645α 2 σX

λ2 = 0.6931α

α [1 − (1 + κ)], κ > −1 κ 2 ) * 2 = α σX (1 + 2κ) − [(1 + κ)]2 , κ

λ1 = ξ +

κ > −

1 3

α [1 − (1 + κ)] κ

α (1 − 2−κ )(1 + κ) κ 2(1 − 3−κ ) λ3 − 3 = λ2 (1 − 2−κ ) λ2 =

κ > −0.5

−(1 + 3κ) + 3(1 + κ)(1 + 2κ) − 2 3 (1 + κ) − 1, [(1 + 2κ) −  2 (1 + κ)]3/2

Sources: Dingman (2002), Hosking and Wallis (1997), Kottegoda and Rosso (1997).

1 η

μX = ξ + 0.5772α

μX = ξ +

γX = sign(κ)



⎠ λ1 = exp ⎝μY + 2 ⎛ ⎞

σY2 ⎠ erf σY λ2 = exp ⎝μY + 2 2

σY = σY

Exponential

σY2

403

404

Chapter 8

Probability and Statistics in Water-Resources Engineering Solution For the N = 25 years of data, the rank (i) and corresponding flow (Qi ) are given in Columns 1 and 2, respectively, in the following table: (1)

(2)

(3)

(4)

(5)

i

Qi (m3 /s)

(1 − i)Qi

ln Qi

(ln Qi − μY )2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

2 2 3 11 18 35 35 40 63 71 112 126 126 141 158 178 251 282 398 501 708 891 1122 1259 2239

2 6 33 72 175 210 280 504 639 1120 1386 1512 1833 2212 2670 4016 4794 7164 9519 14,160 18,711 24,684 28,957 53,736

0.693 0.693 1.099 2.398 2.890 3.555 3.555 3.689 4.143 4.263 4.718 4.836 4.836 4.949 5.063 5.182 5.525 5.642 5.986 6.217 6.562 6.792 7.023 7.138 7.714

15.314 15.314 12.305 4.878 2.945 1.105 1.105 0.842 0.215 0.118 0.013 0.053 0.053 0.117 0.208 0.331 0.844 1.072 1.904 2.592 3.826 4.778 5.839 6.409 9.655

Sum

8772

178,395

115.163

91.836

The first probability-weighted moment, b0 , is given by Equation 8.134 as b0 =

N 1  1 (8772) = 350.9 Qi = N 25 i=1

The product (1 − i)Qi is shown in Column 3, and the second probability-weighted moment, b1 , is given by Equation 8.135 as b1 =

N  1 1 (178, 395) = 297.3 (i − 1)Qi = N(N − 1) 25(25 − 1) i=2

The first L-moment, λ1 , is given by Equation 8.130 as λ1 = b0 = 350.9 and the second L-moment, λ2 , is given by Equation 8.131 as λ2 = 2b1 − b0 = 2(297.3) − 350.9 = 243.8 The relationship between the L-moments, λ1 and λ2 , and the parameters of the log-normal distribution, μY and σY (where Y = ln Q), is given in Table 8.2, which yields

Section 8.3 ⎛

σY2

Analysis of Hydrologic Data

405



⎠ λ1 = 350.9 = exp ⎝μY + 2 ⎛



⎠ erf σY λ2 = 243.8 = exp ⎝μY + 2 2 σY2

and solving for μY and σY gives μY = 4.81 σY = 1.45 Hence, using the method of L-moments, the estimated mean and standard deviation of the log-normal distribution are 4.81 and 1.45, respectively, where the flows are given in m3 /s. Column 4 in the preceding table gives ln Qi , and Column 5 gives (ln Qi − μY )2 , and the method of moments gives the mean (μY ) and standard deviation (σY ) of the log-normal distribution as 1  1 ln Qi = (115.163) = 4.61 N 25 i=1 0 ( 1 N 1  . /2 1 1 1 ln Qi − μY = σY = 2 91.836 = 1.96 N − 1 25 − 1 N

μY =

i=1

Hence, using the method of moments, the estimated mean and standard deviation of the log-normal distribution are 4.61 and 1.96, respectively. The difference between the mean and standard deviation (μY , σY ) of the log-normal distribution estimated using the method of moments (4.61, 1.96) compared with the mean and standard deviation estimated using the method of L-moments (4.81, 1.45) is accounted for by the few outliers in the short period of record, which arguably tend to make the method of L-moments parameter estimates more reliable than the method of moments parameter estimates.

8.3.3

Frequency Analysis

Frequency analysis is concerned with estimating the relationship between an event, x, and the return period, T, of that event. Recalling that T is related to the exceedance probability of x, PX (X Ú x), by the relation 1 (8.138) T= PX (X Ú x) then T is related to the cumulative distribution function of x, PX (X < x), by the relation T=

1 1 − PX (X < x)

(8.139)

A fundamental component of frequency analysis is the assumption that the underlying probability distribution of an event is known. However, in reality, the underlying probability distribution of an observed event can only be hypothesized, and the adequacy of this hypothesis must be assessed using tests such as the Chi-Square and Kolmogorov–Smirnov tests before proceeding with a frequency analysis based on an assumed probability distribution. The uncertainty of sample estimates of parameters in probability distributions requires caution when using data records shorter than 10 years, and caution should also be exercised when estimating the values of variables with recurrence intervals longer than twice the record length (Viessman and Lewis, 1966; Davie, 2002). The conventional frequency analyses presented here are based on an assumed population probability distribution, however,

406

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Probability and Statistics in Water-Resources Engineering

alternative, and sometimes more accurate, nonparametric methods using artificial neural networks have also been proposed for use in frequency analysis (e.g., He and Valeo, 2009). Many cumulative distribution functions cannot be expressed analytically and are tabulated as functions of normalized variables, X  , where X =

X − μx σx

(8.140)

and μx and σx are the mean and standard deviation of the population, respectively. The cumulative probability distribution of X  , PX  (X  < x ), for many distributions depends only on x and the skewness coefficient, gx , of the population and can be readily obtained from statistical tables. Denoting the value of X  with return period T by xT , for many distributions xT = KT (T, gx )

(8.141)

where KT (T, gx ) is derived from the cumulative distribution function of X  and is commonly called the frequency factor. Combining Equations 8.140 and 8.141 leads to xT = μx + KT σx

(8.142)

where xT is the realization of X with return period T. The frequency factor is applicable to many, but not all, probability distributions. The frequency factors for a few probability distributions that are commonly used in hydrologic practice are described here. 8.3.3.1

Normal distribution

In the case of a normal distribution, the variable X  is equal to the standard normal deviate and has a N(0, 1) distribution. The cumulative distribution function of the standard normal deviate is given in Appendix C.1, and the frequency factor is equal to the standard normal deviate corresponding to a given exceedance probability or return period. The frequency factor, KT , can also be approximated by the empirical relation (Abramowitz and Stegun, 1965) KT = w − where

2.515517 + 0.802853w + 0.010328w2 1 + 1.432788w + 0.189269w2 + 0.001308w3 ⎡  ⎤ 1 2 1 ⎦ ⎣ w = ln 2 , p

0 < p … 0.5

(8.143)

(8.144)

and p is the exceedance probability (= 1/T). When p > 0.5, 1 − p is substituted for p in Equation 8.144 and the value of KT computed using Equation 8.143 is given a negative sign. According to Abramowitz and Stegun (1965), the error in using Equation 8.143 to estimate the frequency factor is less than 0.00045. Alternative expressions for the normal-distribution frequency factor (Equation 8.143) having comparable or lesser accuracy have also been proposed (e.g., Swamee and Rathie, 2007b). EXAMPLE 8.24 Annual rainfall at a given location can be described by a normal distribution with a mean of 127 cm and a standard deviation of 19 cm. Estimate the magnitude of the 50-year annual rainfall. Solution The 50-year rainfall, x50 , can be written in terms of the frequency factor, K50 , as x50 = μx + K50 σx From the given data, μx = 127 cm, σx = 19 cm, and the exceedance probability, p, is given by p=

1 1 = = 0.02 T 50

Section 8.3

Analysis of Hydrologic Data

407

The intermediate variable, w, is given by Equation 8.144 as  ⎤12

 12 1 1 ⎦ = ln = 2.797 w = ⎣ln p2 0.022 ⎡ 

Equation 8.143 gives the frequency factor, K50 , as K50 = w −

2.515517 + 0.802853w + 0.010328w2 1 + 1.432788w + 0.189269w2 + 0.001308w3

= 2.797 −

2.515517 + 0.802853(2.797) + 0.010328(2.797)2 1 + 1.432788(2.797) + 0.189269(2.797)2 + 0.001308(2.797)3

= 2.054 The 50-year rainfall is therefore given by x50 = μx + K50 σx = 127 + 2.054(19) = 166 cm

8.3.3.2

Log-normal distribution

In the case of a log-normal distribution, the random variable is first transformed using the relation Y = ln X (8.145) and the value of Y with return period T, yT , is given by yT = μy + KT σy

(8.146)

where μy and σy are the mean and standard deviation, respectively, of Y, and KT is the frequency factor of the standard normal deviate with return period T. The value of the original variable, X, with return period T, xT , is then given by xT = ln−1 yT = eyT

(8.147)

EXAMPLE 8.25 The annual rainfall at a given location has a log-normal distribution with a mean of 127 cm and a standard deviation of 19 cm. Estimate the magnitude of the 50-year annual rainfall. Solution From the given data, μx = 127 cm, σx = 19 cm, and the mean, μy , and standard deviation, σy , of the log-transformed variable, Y = ln X, are related to μx and σx by Equation 8.75, where ⎛ ⎞ σy2  ⎠ , σx2 = μ2x exp (σy2 ) − 1 μx = exp ⎝μy + 2 These equations can be put in the form   μ4x 1 , μy = ln 2 μ2x + σx2 which lead to μy =

σy =

σy =

σ 2 + μ2x ln x μ2x

  1274 1 ln = 4.83 2 1272 + 192 (

and

(

ln

192 + 1272 = 0.149 1272

408

Chapter 8

Probability and Statistics in Water-Resources Engineering Since the rainfall, X, is log-normally distributed, Y (= ln X) is normally distributed and the value of Y corresponding to a 50-year return period, y50 , is given by y50 = μy + K50 σy where K50 is the frequency factor for a normal distribution and a return period of 50 years. In the previous example it was shown that K50 = 2.054, therefore y50 = 4.83 + 2.054(0.149) = 5.14 and the corresponding 50-year rainfall, x50 , is given by x50 = ey50 = e5.14 = 171 cm

8.3.3.3

Gamma/Pearson Type III distribution

In the case of the Pearson Type III distribution, also called the three-parameter gamma distribution, the frequency factor depends on both the return period, T, and the skewness coefficient, gx . If the skewness coefficient falls between −1 and +1, approximate values of the frequency factor for the gamma/Pearson Type III distribution, KT , can be estimated using the relation (Kite, 1977; Viessman and Lewis, 2003) KT =

* 1 )  [(xT − k)k + 1]3 − 1 3k

(8.148)

where xT is the standard normal deviate corresponding to the return period T and k is related to the skewness coefficient by gx k= (8.149) 6 When the skewness, gx , is equal to zero, the expanded form of Equation 8.148 indicates that KT = xT , and the gamma/Pearson Type III distribution is identical to the normal distribution. In cases where the skewness coefficient falls outside the −1 to +1 range, the frequency factors given in Appendix C.2 should be used. Methods for calculating the confidence limits of the Pearson Type III distribution can be found in Ames (2006). EXAMPLE 8.26 The annual-maximum discharges in a river show a mean of 811 m3 /s, a standard deviation of 851 m3 /s, and a skewness of 0.94. Assuming a Pearson Type III distribution, estimate the 100-year discharge. If subsequent measurements indicate that the skewness coefficient is actually equal to 1.52, how does this affect the estimated 100-year discharge? Solution From the given data: μx = 811 m3 /s, σx = 851 m3 /s, and gx = 0.94. The standard normal deviate corresponding to a 100-year return period, x100 , is 2.33 (Appendix C.1), k = gx /6 = 0.94/6 = 0.157, and the frequency factor K100 is given by Equation 8.148 as ) * * 1 )  1 [(x100 − k)k + 1]3 − 1 = [(2.33 − 0.157)(0.157) + 1]3 − 1 = 3.00 K100 = 3k 3(0.157) The 100-year discharge, x100 , is therefore given by x100 = μx + K100 σx = 811 + (3.00)(851) = 3364 m3 /s If the skewness is equal to 1.52, this is outside the [−1, +1] range of applicability of Equation 8.148, and the frequency factor must be obtained from the tabulated frequency factors in Appendix C.2. For a return period of 100 years and a skewness coefficient of 1.52, Appendix C.2 gives K100 = 3.342, and the 100-year discharge is given by x100 = μx + K100 σx = 811 + (3.342)(851) = 3655 m3 /s Therefore, the 62% change in estimated skewness from 0.94 to 1.52 causes a 9% change in the 100-year discharge from 3364 m3 /s to 3655 m3 /s.

Section 8.3

8.3.3.4

Analysis of Hydrologic Data

409

Log-Pearson Type III distribution

In the case of a log-Pearson Type III distribution, the random variable is first transformed using the relation Y = ln X (8.150) The value of Y with return period T, yT , is given by yT = μy + KT σy

(8.151)

where μy and σy are the mean and standard deviation, respectively, of Y, and KT is the frequency factor of the Pearson Type III distribution with return period T and skewness coefficient gy . The value of the original variable, X, with return period T, xT , is then given by xT = ln−1 yT = eyT

(8.152)

Whenever the skewness of Y (= ln X) is equal to zero, the log-Pearson Type III distribution is identical to the log-normal distribution. EXAMPLE 8.27 The natural logarithms of annual-maximum discharges (in m3 /s) in a river show a mean of 6.33, a standard deviation of 0.862, and a skewness of −0.833. Assuming a log-Pearson Type III distribution, estimate the 100-year discharge. Solution From the given data, μy = 6.33, σy = 0.862, and gy = −0.833, which yields k = gy /6 = −0.833/6 = −0.139. The standard normal deviate corresponding to a 100-year return period, x100 , is 2.33 (Appendix C.1), and the frequency factor for a 100-year return period, K100 , is given by Equation 8.148 as ) * * 1 )  1 [(x100 − k)k + 1]3 − 1 = [(2.33 + 0.139)(−0.139) + 1]3 − 1 = 1.72 K100 = 3k 3(−0.139) The 100-year log-discharge, y100 , is therefore given by y100 = μy + K100 σy = 6.33 + (1.72)(0.862) = 7.81 and the 100-year discharge, x100 , is x100 = ey100 = e7.81 = 2465 m3 /s

The U.S. Interagency Advisory Committee on Water Data published an important document called Bulletin 17B (USIAC, 1982), which provides detailed guidance on the application of the log-Pearson Type III distribution to annual-maximum flood flows. Of particular concern in applying the log-Pearson Type III distribution to observed data is that skew coefficients estimated from small samples might be highly inaccurate. In cases where this is a concern, Bulletin 17B outlines a procedure for developing regional skew coefficients based on at least 40 gaging stations located within a 160-km (100-mi) radius of the site of concern, with each station having at least 25 years of data. If this procedure is not feasible, the generalized skew map shown in Figure 8.12 is provided as an easier but less accurate alternative. This map of generalized logarithmic skew coefficients for peak annual flows was developed from skew coefficients computed at 2972 gaging stations, all having at least 25 years of record, following the procedures outlined in Bulletin 17B. Depending on the number of years of record, regionalized skew coefficients are used either in lieu of or in combination with values computed from observed flows at the particular stream gage of concern. Equation 8.153 allows a weighted skew coefficient GW to be computed by combining a regionalized skew coefficient GR and station skew coefficient G, GW =

(MSER )(G) + (MSES )(GR ) MSER + MSES

(8.153)

410

Chapter 8

Probability and Statistics in Water-Resources Engineering

FIGURE 8.12: Generalized skew coefficients of the logarithms of annual-maximum streamflow Source: Interagency Advisory Committee on Water Data (Bulletin 17B, 1982).

0.2

−0.1

−0.3

−0.4

0.6

0 20.3

−0.4

−0.1

0.3 0

0 0.3

−0.2 −0.1 −0.4 −0.3

0.5 0.7

0.1

0 −0.3

0 0 0.6 0.3

−0.3

−0.1 −0.3 0.2 −0.2

0.7 0.5 0.3

−0.2

−0.1

0

−0.4

0 −0.1

20.1

−0.2

0.2 0

−0.1 0 −0.2

0.2 0 −0.3

0.2 −0.1

where the station skew G is computed from observed flows at the station of interest; the regional skew, GR , is either developed from multiple stations following procedures outlined in Bulletin 17B or read from Figure 8.12; MSES denotes the mean-square error of the station skew, which can be estimated for N years of data using the relation (Wallis et al., 1974) MSES = 10[A−B(log10 (N/10))]

(8.154)

where

A=

B=

⎧ ⎨ −0.33 + 0.08|G|

if |G| … 0.90

⎩ −0.52 + 0.30|G|

if |G| > 0.90

⎧ ⎨

0.94 − 0.26|G|

if |G| … 1.50



0.55

if |G| > 1.50

and MSER is the mean-square error of the regional skew. If GR is taken from Figure 8.12, then MSER is equal to 0.302. Recent studies have further validated using this method to estimate the skewness of log-transformed annual peak flows in the United States (Griffis and Stedinger, 2007c; 2009). However, issues still remain as to the fundamental validity of using regional skews that neglect local storage effects (McCuen and Smith, 2008), and the need for more formal criteria in selecting the most appropriate regional data to supplement local data (Chebana and Ouarda, 2008). EXAMPLE 8.28 Observations of annual-maximum flood flows in the Ocmulgee River near Macon, Georgia, from 1980– 2000 yield a skew coefficient of the log-transformed data equal to −0.309. Determine the (weighted) skew coefficient that should be used for flood-frequency analysis. Solution In central Georgia, where Macon is located, Figure 8.12 gives a generalized skew coefficient GR of −0.1, and MSER is equal to 0.302. From the observed annual-maximum flood flows, G = −0.309, and for the 21-year period of record (1980–2000), N = 21. By definition

Section 8.3

Analysis of Hydrologic Data

411

A = −0.33 + 0.08|G| = −0.33 + 0.08|−0.309| = −0.305 B = 0.94 − 0.26|G| = 0.94 − 0.26|−0.309| = 0.860 . . N //  . . 21 //  = 10 −0.305−0.860 log10 10 = 0.262 MSES = 10 A−B log10 10 and the weighted skew coefficient is given by GW =

(MSER )(G) + (MSES )(GR ) (0.302)(−0.309) + (0.262)(−0.1) = = −0.212 MSER + MSES 0.302 + 0.262

Therefore, a skew coefficient of −0.212 is more appropriate than the measured skew coefficient of −0.309 for frequency analyses of flood flows in the Ocmulgee River at Macon, Georgia. As the period of record increases, it is expected that the measured skew coefficient will approach the generalized skew coefficient derived from Figure 8.12.

8.3.3.5

Extreme-value Type I distribution

In the case of an extreme-value Type I (Gumbel) distribution, the frequency factor, KT , can be written as (Chow, 1953) ⎫  √ ⎧

⎬ 6⎨ T KT = − 0.5772 + ln ln π ⎩ T − 1 ⎭

(8.155)

This expression for KT is valid only in the limit as the number of samples approaches infinity. For a finite sample size KT varies with the sample size as shown in Table 8.3. TABLE 8.3: Extreme-Value Type I (Gumbel) Frequency Factors

Return period Sample size

2.33

5

10

20

25

50

75

100

1000

15 20 25 30 40 50 60 70 75 100 q

0.065 0.052 0.044 0.038 0.031 0.026 0.023 0.020 0.019 0.015 −0.067

0.967 0.919 0.888 0.866 0.838 0.820 0.807 0.797 0.794 0.779 0.720

1.703 1.625 1.575 1.541 1.495 1.466 1.446 1.430 1.423 1.401 1.305

2.410 2.302 2.235 2.188 2.126 2.086 2.059 2.038 2.029 1.998 1.866

2.632 2.517 2.444 2.393 2.326 2.283 2.253 2.230 2.220 2.187 2.044

3.321 3.179 3.088 3.026 2.943 2.889 2.852 2.824 2.812 2.770 2.592

3.721 3.563 3.463 3.393 3.301 3.241 3.200 3.169 3.155 3.109 2.911

4.005 3.386 3.729 3.653 3.554 3.491 3.446 3.413 3.400 3.340 3.137

6.265 6.006 5.842 5.727 5.476 5.478 5.410 5.359 5.338 5.261 4.900

Source: Viessman and Lewis (2005)

EXAMPLE 8.29 Based on 30 years of data, the annual-maximum discharges in a river have a mean of 811 m3 /s and a standard deviation of 851 m3 /s. Assuming that the annual maxima are described by an extremevalue Type I distribution, estimate the 100-year discharge. What would be the 100-year discharge if the statistics were based on 60 years of data? Compare your results with the prediction made using Equation 8.155.

412

Chapter 8

Probability and Statistics in Water-Resources Engineering Solution From the given data: μx = 811 m3 /s, σx = 851 m3 /s, and the 100-year discharge, x100 , is given by x100 = μx + K100 σx = 811 + K100 851 For 30 years of data, the 100-year frequency factor is given by Table 8.3 as K100 = 3.653, and the 100-year discharge is given by x100 = 811 + (3.653)851 = 3920 m3 /s For 60 years of data, the 100-year frequency factor is given by Table 8.3 as K100 = 3.446, and the 100-year discharge is given by x100 = 811 + (3.446)851 = 3740 m3 /s For an infinite number of years of data, the 100-year frequency factor is given by Equation 8.155 as ⎫  √ ⎧

⎬ 6⎨ 100 K100 = − 0.5772 + ln ln = 3.137 π ⎩ 100 − 1 ⎭ which gives x100 = 811 + (3.137)851 = 3480 m3 /s Hence, for 30, 60, and an infinite number of years of data, the estimated 100-year discharge is 3920 m3 /s, 3740 m3 /s, and 3480 m3 /s, respectively. Based on these results, for the same statistics, the estimated 100-year discharge decreases as the number of years of data increases. There is approximately 13% difference between the 100-year discharge based on 30 years of data and the 100-year discharge based on an infinite number of years of data.

8.3.3.6

General extreme-value (GEV) distribution

In the case of the general extreme-value (GEV) distribution, the value of the random variable, xT , with a return period T is given by (Kottegoda and Rosso, 1997; Overeem et al., 2007). ⎧ 

c  ⎪ b T ⎪ ⎪ a + ⎪ c Z 0 1 − ln ⎪ ⎨ c T − 1 (8.156) xT =

⎪ ⎪ ⎪ T ⎪ ⎪ c=0 ⎩a + b ln ln T − 1 where a, b, and c, are the location, scale, and shape parameters, respectively, of the GEV distribution.

EXAMPLE 8.30 The annual-maximum 24-hour rainfall in a subtropical area is described by a general extreme-value distribution with parameters: a = 13.8 cm, b = 2.17 cm, and c = −0.129. Determine the annual-maximum 24-hour rainfall with a return period of 20 years. Solution From the given data: T = 20 years, a = 13.8 cm, b = 2.17 cm, and c = −0.129. The annualmaximum 24-hour rainfall with a return period of 20 years is given by Equation 8.156 as  

c 

−0.129  b 2.17 T 20 = 13.8 + = 21.7 cm 1 − ln 1 − ln x20 = a + c T − 1 −0.129 20 − 1 Therefore, an annual-maximum 24-hour rainfall of 21.7 cm has a return period of 20 years.

Section 8.4

Uncertainty Analysis

413

8.4 Uncertainty Analysis The error in a measured variable is defined as the difference between the measured value of the variable and the true value of the variable. The error can usually be assumed to be random, and characterized by a probability distribution. Also, in cases where measured data are used to calculate quantities that are not directly measurable, calculated quantities deviate from their true values, with errors whose probability distributions depend on the probability distributions of the errors in the measured data. The objective of uncertainty analysis is to estimate the statistics of quantities that are calculated from measured (uncertain) data. Whenever mathematical models are used to estimate quantities from measured data, uncertainty in the calculated quantities may occur from four possible sources: (1) uncertainty in the values of parameters and constants in the model equations; (2) uncertainty in the measured or estimated data; (3) errors introduced by the numerical method that solves the model equations; and (4) errors in using the selected equations to describe the process being modeled. This section addresses the first two sources of uncertainty. The third source of error can be identified using numerical experiments in which the model’s numerical solution is compared with results from accurate analytical solutions. The fourth source of error is related to the validity of the model itself, and such errors can be estimated only by comparing model predictions with direct measurements of those same quantities in simple cases where all other uncertainties are negligible. Consider the typical case in which the calculated quantity Y is being estimated using N measured variables Xi , i = 1, N. Functionally, this may be written as Y = φ(X)

(8.157)

where X is a vector having elements Xi . If the true values of the measured quantities are denoted by Xi , then, assuming the model is valid, the true value of Y, Y, is given by Y = φ(X)

(8.158)

where X is a vector having elements Xi . The error in Y, Y, is then given by Y = Y − Y = φ(X) − φ(X)

(8.159)

Using a multidimensional Taylor expansion, φ(X) can be expressed in the following form:  N  ⭸φ  (Xi − Xi ) φ(X) = φ(X) + ⭸Xi X=X i=1  N N 1   ⭸2 φ  + (Xi − Xi )(Xj − Xj ) + R  2! ⭸Xi ⭸Xj  i=1 j=1

(8.160)

X=X

where the remainder, R, involves higher-order products of the errors, (Xi − Xi ). If the measurement errors are small, then higher-order terms are much smaller than lower-order terms, and retaining only first-order terms in Equation 8.160 results in the following approximate relationship  N  ⭸φ  φ(X) = φ(X) + (Xi − Xi ) (8.161) ⭸Xi X=X i=1

Combining Equations 8.159 and 8.161 gives the error in the calculated quantity, Y, in terms of the errors in the measured quantities, Xi = (Xi − Xi ), hence Y =

N  i=1

 φX Xi i

(8.162)

414

Chapter 8

Probability and Statistics in Water-Resources Engineering  where φX is given by i  φX = i

 ⭸φ  ⭸Xi X=X

(8.163)

Equation 8.162 is a stochastic equation relating the error in the derived quantity, Y, to the errors in measured quantities, Xi , provided these latter errors are small. Taking the ensemble average of Equation 8.162 yields Y =

N 

 φX Xi  i

(8.164)

i=1

indicating that the average error in the estimated quantity is the sum of the average errors  . Typically, measurements are in the measured quantities, weighted by the sensitivities φX i assumed to be without bias, in which case the average errors, Xi , are equal to zero and, according to Equation 8.164, the calculated result would then also be without bias. Another statistical parameter of interest is the variance of the calculated quantity, Y, which is denoted by σY2 and defined by σY2 = (Y)2  (8.165) Combining Equations 8.162 and 8.165 yields σY2 =

N N  

 φX φ  Xi Xj  i Xj

(8.166)

i=1 j=1

where Xi Xj  is the covariance between the errors in Xi and Xj . In many cases the errors in different measured quantities are uncorrelated, leading to error covariances of zero. Under these circumstances, Equation 8.166 becomes σY2 =

N 

 (φX )2 σX2 i i

(8.167)

i=1

EXAMPLE 8.31 The flow rate, Q, in a trapezoidal open channel can be expressed as Q = V(b + my)y where V is the average velocity, b is the bottom width, m is the side slope, and y is the depth of flow. The variables V, b, y, and m are subject to measurement errors, and the means and standard deviations of these variables are given by: Variable

Mean

V b m y

0.9 m/s 10 m 2.0 2.4 m

Standard Deviation 0.10 m/s 0.30 m 0.2 0.3 m

Use a first-order uncertainty analysis to estimate the mean and standard deviation of Q. Solution According to Equation 8.158, the expected value of Q, Q, is given by Q = V[b + my]y = 0.9[10 + (2.0)(2.4)](2.4) = 32.0 m3 /s

Problems

415

2 , can be estimated using Equation 8.167, which can be written as The variance of Q, σQ

 2 = σQ



⭸Q  ⭸V X=X

where  ⭸Q  ⭸V X=X  ⭸Q  ⭸b X=X  ⭸Q  ⭸m X=X  ⭸Q  ⭸y X=X



2 σV2 +



⭸Q  ⭸b X=X



2 σb2 +



⭸Q  ⭸m X=X



2 2 + σm



⭸Q  ⭸y X=X

2 σy2 (8.168)

 = (b + my)yX=X = [b + my]y = [10 + (2.0)(2.4)](2.4) = 35.5 (m3 /s)/(m/s)  = VyX=X = Vy = (0.9)(2.4) = 2.16 (m3 /s)/m   = Vy2 

X=X

= Vy2 = (0.9)(2.4)2 = 5.18 m3 /s

 = V(b + 2my)X=X = V[b + 2my] = (0.9)[10 + 2(2.0)(2.4)] = 17.64 (m3 /s)/m

Substituting these derivatives into Equation 8.168 with the given standard errors of V, b, m, and y yields 2 = (35.5)2 (0.10)2 + (2.16)2 (0.30)2 + (5.18)2 (0.20)2 + (17.64)2 (0.30)2 = 42.10 (m3 /s)2 σQ

which gives the standard error of Q, σQ , as σQ =



42.10 = 6.5 m3 /s

First-order uncertainty analysis is sometimes referred to as first-order error analysis (Yan et al., 2006) or first-order second moment analysis. The major advantage of first-order uncertainty analysis (FOUA) is its simplicity, requiring knowledge of only the first two statistical moments of the basic variables and a simple sensitivity analysis to calculate the incremental change in model output corresponding to an incremental change in model input. The applicability of FOUA, however, is inherently restricted by the degree of nonlinearity in the model. For nonlinear systems, this analysis becomes increasingly inaccurate as the basic variables depart from their central values (Melching and Yoon, 1997). The FOUA approximation deteriorates when the coefficient of variation is greater than 10%–20%, which is quite common. Problems 8.1. A flood-control system is designed such that the probability that the system capacity is exceeded X times in 30 years is given by the following discrete probability distribution: xi 0 1 2 3 4

f (xi )

xi

f (xi )

0.04 0.14 0.23 0.24 0.18

5 6 7 8 >9

0.10 0.05 0.02 0.01 0.00

What is the mean number of system failures expected in 30 years? What are the variance and skewness of the number of failures? 8.2. The probability distribution of the time between flooding in a residential area is given by

t2 1 f (t) = √ e− 2 , t > 0 2π =0 , t … 0

where t is the time interval between flood events in days. Estimate the mean, standard deviation, coefficient of variation, and skewedness of t. 8.3. The annual-maximum flow rates in a river are known to vary between 4 m3 /s and 10 m3 /s and have a probability distribution of the form 2 f (x) = α * e−4x

Where x is the annual-maximum flow rate and α is a constant. Determine: (a) the value of α, (b) the return period of a maximum flow rate of 7 m3 /s, and (c) the mean and standard deviation of the annual-maximum flow rates.

416

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8.4. The flow rates during the summer in a regulated channel vary between 1 m3 /s and 2 m3 /s, and all flow rates between these values are equally likely. Derive an expression for the probability density function of the flow rates in the river, and use this distribution to calculate the mean, variance, and skewness of the streamflows. Identify the flow rate that has a return period of 50 years. [Hint: Your derived 3 q expression must satisfy the f (Q) dQ = 1.] relations: f (Q) Ú 0 and −q 8.5. Analysis of data on maximum one-day rainfall depth at a city indicated that a depth of 320 mm had a return period of 50 years. Determine the probability of a one-day rainfall depth equal to or greater than 320 mm at the city occurring (a) once in 20 successive years, (b) twice in 15 successive years, and (c) at least once in 20 successive years. 8.6. Use the theory of combinations to show that if the probability of an event occurring in any trial is p, and if all trials are independent, then the probability of the event occurring n times in N trials, f (n), is given by f (n) =

N! pn (1 − p)N−n n!(N − n)!

8.7. A flood-control system is designed for a rainfall event with a return period of 25 years. What is the probability that the design rainfall will be exceeded five times in 10 years? What is the probability that the design rainfall will be equalled or exceeded at least once in a 10-year period? 8.8. A water-resource system is designed for a hydrologic event with a 25-year return period. Assuming that the hydrologic events can be taken as a Bernoulli process, what is the probability that the design event will be exceeded once in the first 5 years of system operation? More than once in 5 years? 8.9. A stormwater-management system is being designed for a (design) life of 10 years, and a (design) runoff event is to be selected such that there is only a 1% chance that the system capacity will be exceeded during the design life. Determine the return period of the design runoff event. If the cumulative distribution function of the peak runoff rate, Q (in m3 /s), in any runoff event is given by F(Q) = exp [−exp (−Q)] what is the design Q? 8.10. A water-resource system is designed for a 50-year storm. Assuming that the occurrence of the 50-year storm is a Bernoulli process, what is the probability that the 50year storm will be equalled or exceeded 1 year after the system is constructed? Within the first 6 years? 8.11. Determine the mean capacity of a storm sewer pipe, the coefficient of variation of the pipe capacity, and the standard deviation of the pipe capacity using Manning’s equation for full pipe flow. The values of the mean and coefficient of variation of roughness coefficient (n) are 0.014 and 0.011, pipe diameter (D in m) are 1.5 and 0, and bed slope (S) are and 0.0012 and 0.052, respectively.

8.12. A drainage system is designed for a storm with a return period of 10 years. What is the average interval between floods? What is the probability that there will be more than 6 months between floods? 8.13. Explain why a gamma distribution with n = 1 is the same as the exponential distribution. 8.14. A residential development in the southern United States floods whenever more than 10.2 cm of rain falls in 24 hours. In a typical year, there are three such storms. Assuming that these rainfall events are a Poisson process, what is the probability that it will take more than 1 year to have three flood events? 8.15. Two pumps operate in parallel to provide water to a village located in a tourist area. The water demand is subject to considerable weekly and seasonal fluctuations. Each unit has the capacity to supply 80% of demand when the other pump fails. The probability of failure of each pump is 10% and that of both pumps is 3%. What is the probability that the village demand will be satisfied? 8.16. Traces of toxic waste from an unknown source are found in a stream from tests on the water. The mean concentration is found to be 1 mg/L. What is the probability that the concentration of pollutant will be in the range 0.5 to 2.5 mg/L? Assume that the distribution is exponential. 8.17. The annual rainfall in a rural town is shown to be lognormally distributed with a mean of 114 cm and a standard deviation of 22 cm. Estimate the rainfall having a return period of 100 years. How would your estimate differ if the rainfall were normally distributed? 8.18. Streamflow data collected from several sources indicate that the peak flow rate in a drainage channel at a given location is somewhere in the range between 95 and 115 m3 /s. Assuming that the uncertainty can be described by a uniform distribution, what is the probability that the peak flow rate is greater than 100 m3 /s? 8.19. The probability distribution for the maximum values in a sample set is given by Equation 8.82. Derive a similar expression for the minimum values in a sample set. 8.20. The annual maximum flood magnitude in a river shows a mean of 170 m3 /s and a standard deviation of 113 m3 /s. Assuming that the uncertainty can be described by a normal distribution, what is the probability that the annual maximum flood magnitude would exceed 284 m3 /s? Determine the flood magnitude with a return period of 100 years. 8.21. The annual-minimum rainfall amounts have been tabulated for a region and show a mean of 710 mm and a standard deviation of 112 mm. Assuming that these minima are described by an extreme-value Type I distribution, estimate the average interval between years where the minimum rainfall exceeds 600 mm. 8.22. Determine the mean coefficient of variation and standard deviation of runoff using the rational equation with

Problems

8.23.

8.24.

8.25.

8.26.

the parameters coefficient of runoff (C), rainfall intensity (I in mm/hr), and area (A in km2 ). The values of mean and coefficient of variation of C, I, and A are 0.9 and 0.08, 100 and 0.5, and 0.2 and 0.005, respectively. Two types of plants are used to treat sewage effluent from two similar areas of a city. Out of 90 tests made on the output from plant X, 33 tests showed that the pollution has been reduced significantly, whereas 44 tests out of 100 from plant Y show that pollution has been reduced to the same or lower level. Are the effects of the plants in reducing pollution different? The annual-maximum streamflows in a river have a mean of 63.2 m3 /s, a standard deviation of 13.7 m3 /s, and a skewness coefficient of 1.86. Assuming that the annual-maximum streamflows are described by a general extreme-value (GEV) distribution, estimate the annual-maximum streamflow with a return period of 50 years. A random variable, X, is equal to the sum of the squares of 15 normally distributed variables with a mean of zero and a standard deviation of 1. What is the probability that X is greater than 25? What is the expected value of X? Exceedance probabilities of measured data can be estimated using the Gringorten formula PX (X > xm ) =

m − a N + 1 − 2a

where xm is a measurement with rank m, N is the total number of measurements, and a is a constant that depends on the population probability distribution. Values of a are typically in the range of 0.375 to 0.44. Determine the formula for the return period of xm . What is the maximum error in the return period that can result from uncertainty in a? 8.27. The annual peak flow rates in a river between 1980 and 1996 are as follows:

Year

Peak flow (ft3 /s)

Year

Peak flow (ft3 /s)

1980 1981 1982 1983 1984 1985 1986 1987 1988

8000 5550 3390 6390 5889 7182 10,584 11,586 8293

1989 1990 1991 1992 1993 1994 1995 1996 −

9193 5142 7884 4132 12136 5129 7236 6222 −

8.29.

8.30.

8.31.

8.32.

8.33.

8.34.

Use the Weibull and Gringorten formulae to estimate the cumulative probability distribution of peak flow rate and compare the results. 8.28. Analysis of a 50-year record of annual rainfall indicates the following frequency distribution:

8.35.

Range (mm)

Number of outcomes

1560

4 5 6 6 5 7 4 3 6 4

417

The measured data also indicate a mean of 1260 mm and a standard deviation of 234 mm. Using the chi-square test with a 5% significance level, assess the hypothesis that the annual rainfall is drawn from a normal distribution. Use the Kolmogorov–Smirnov test at the 10% significance level to assess the hypothesis that the data in Problem 8.28 are drawn from a normal distribution. A sample of a random variable has a mean of 1.6, and the random variable is assumed to have an exponential distribution. Use the method of moments to determine the parameter (λ) in the exponential distribution. A sample of a random variable has a mean of 1.6 and a standard deviation of 1.2. If the random variable is assumed to have a log-normal distribution, apply the method of moments to estimate the parameters in the log-normal distribution. Annual-maximum peak flow rates in the Davidian River from 1940 to 1980 show a mean of 35 m3 /s, standard deviation of 15 m3 /s, and skewness of 0.4. Calculate the standard errors and assess the relative accuracies of the estimated parameters. A sample of random data is to be fitted to the exponential probability distribution given by Equation 8.45. Derive the maximum-likelihood estimate of the distribution parameter λ. Compare your result with the method of moments estimate. Derive the maximum-likelihood estimators of the parameters in the normal distribution. The annual-maximum flow rates for the period 1970 to 1996 in the Bucking Bronco River are as follows:

Year

Maximum flow (m3 /s)

Year

Maximum flow (m3 /s)

Year

Maximum flow (m3 /s)

1970 1971 1972 1973 1974 1975 1976 1977 1978

189 267 377 53 107 752 1337 27 3359

1979 1980 1981 1982 1983 1984 1985 1986 1987

5 3 212 423 168 60 95 597 1062

1988 1989 1990 1991 1992 1993 1994 1995 1996

17 1683 4 1889 237 189 53 550 320

418

8.36.

8.37.

8.38.

8.39.

8.40.

8.41.

8.42.

Chapter 8

Probability and Statistics in Water-Resources Engineering

A histogram of these data indicate that they are likely drawn from a log-normal distribution. Use the method of L-moments to estimate the mean and standard deviation of the log-normal distribution, and compare these parameters with those estimated using the method of moments. Determine the return period that should be used in a small dam so that the design flood is exceeded with a probability of not more than the 0.05 during a 50-year design period. The annual rainfall at a given location has a log-normal distribution with a mean of 152 cm and a standard deviation of 30 cm. Estimate the magnitude of the 10-year and 100-year annual rainfall amounts. Compare your results with those obtained in Problem 8.36. The annual-maximum discharges in a river show a mean of 480 m3 /s, a standard deviation of 320 m3 /s, and a skewness of 0.87. Assuming a Pearson Type III distribution, use the frequency factor to estimate the 50-year and 100-year annual-maximum discharges. Compare the Pearson Type III frequency factor for a 100-year event estimated using Equation 8.148 with the same frequency factor estimated using Appendix C.2, for skewness coefficients of 3.0, 2.0, 1.0, 0.5, and 0.0. How do these results support the assertion that Equation 8.148 should be used only when the skewness coefficient is in the range [−1, +1]? The annual-maximum discharges in a river show a mean of 480 m3 /s, a standard deviation of 320 m3 /s, and a logtransformed skewness of 0.1. Assuming a log-Pearson Type III distribution, estimate the 100-year annualmaximum discharge. Observations of annual-maximum flood flows in the Mississippi River near Baton Rouge, Louisiana, from 1985 to 2000 yield a skew coefficient of the logtransformed data equal to −0.210. Determine the (weighted) skew coefficient that should be used for flood-frequency analysis. Based on 25 years of data, the annual-maximum discharges in a river have a mean of 480 m3 /s and a standard deviation of 320 m3 /s. Assuming that the annual maxima are described by an extreme-value Type I

distribution, estimate the 100-year discharge. Compare your result with the prediction made using Equation 8.155, and with the result obtained in Problem 8.20. 8.43. Consider the series of annual-maximum streamflow measurements given in the table below. (a) Estimate the return period of a 30-m3 /s (annual-maximum) flow using the Weibull formula; and (b) estimate the return period of a 30-m3 /s (annual-maximum) flow using the Gringorten formula. It is postulated that the observed data follow the extreme-value Type I distribution. (c) Use the method of moments to estimate the parameters (a and b) of the extreme-value Type I distribution; (d) use the chi-square test with six classes to assess whether the data fit an extreme-value Type I distribution; and (e) assuming that the data are adequately described by an extreme-value Type I distribution, use the frequency factor to determine the annual-maximum streamflow with a 100-year return period.

Year

Flow (m3 /s)

Year

Flow (m3 /s)

1970 1971 1972 1973 1974 1975 1976 1977

20 23 13 28 35 19 14 10

1978 1979 1980 1981 1982 1983 1984 −

31 25 9 40 22 19 17 −

8.44. The annual-maximum streamflows in a river are described by a general extreme-value distribution with parameters: a = 55.2 m3 /s, b = 8.68 m3 /s, and c = −0.163. Determine the annual-maximum streamflow with a return period of 100 years. 8.45. A trapezoidal channel is estimated to have a bottom width of 2 m, side slopes of 3:1 (H:V), and a flow depth of 1 m. If the coefficient of variation of each of these measurements is 10%, estimate the coefficient of variation of the flow area calculated from these measurements.

C H A P T E R

9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions 9.1 Introduction Surface-water hydrology is the science that encompasses the distribution, movement, and properties of natural water above the surface of the earth. Applications of surface-water hydrology in engineering practice are mostly encompassed within the field of engineering hydrology, which includes modeling rainfall events and predicting the quantity and quality of the resulting surface runoff. Designing systems to control the quantity and quality of surface runoff is the responsibility of a water-resources engineer. The land area that can contribute surface runoff to any particular location is determined by the shape and topography of the land area surrounding the given location. The potential contributing area is called the watershed, and the area within a watershed over which any particular rainfall event occurs is called the catchment area. In most engineering applications, the watershed and catchment area are taken to be the same, and are sometimes referred to as the drainage basin or the drainage area. The runoff from a watershed or catchment is concentrated at the catchment outlet, which is sometimes called the pour point of the catchment. The characteristics of the catchment determine the quantity, quality, and timing of the surface runoff for a given rainfall event. The types of pollutants contained in surface runoff generally depend on the land uses within the catchment area, and pollutants in surface-water runoff that are typically of concern include suspended solids, heavy metals, nutrients, organics, oxygen-demanding substances, and pathogenic microorganisms. These pollutants may be in solution, in suspension, or attached to particles of sediment. To facilitate water management in the United States, the country is divided into a hierarchy of watersheds called hydrologic units with the following four levels: region, subregion, accounting unit, and cataloging unit. Regions represent the largest scale of watershed and are shown in Figure 9.1, and these 21 regions are further divided by lower-level topography into 222 subregions, 352 accounting units, and 2262 cataloging units. Hydrologic unit codes are used to identify hydrologic units, with regions identified by a 2-digit code, subregions by a 4-digit code, accounting units by a 6-digit code, and cataloging units by an 8-digit code. As an example, Lake Winnebago (Wisconsin) is located within region 04, subregion 0403, accounting unit 040302, and cataloging unit 04030204; note the hierarchy embedded in the numbering system. In engineering practice, it is common to refer to the cataloging unit by its “HUC8” number (referring to the 8-digit hydrologic unit code), which for the Lake Winnebago watershed is 04030204. 9.2 Rainfall The precipitation of water vapor from the atmosphere occurs in many forms, the most important of which are rain and snow. Hail and sleet are less-frequent forms of precipitation. Engineered drainage systems in most urban communities are designed primarily to control the runoff from rainfall. The formation of precipitation usually results from the lifting of moist air masses within the atmosphere, which results in the cooling of the air mass and condensation of moisture. The four conditions that must be present for the production of precipitation are: (1) cooling of the air mass, (2) condensation of water droplets onto nuclei, (3) growth of water droplets, and (4) mechanisms to cause a sufficient density of the droplets for the precipitation to fall to the ground. 419

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions

(17)

NO RT HW

(01) RED-RAINYA

(09) EST

G GRR EEAATT LLA AK KEESS MISSISSIPPI (04)

MISSOURI

GR EAT BAS

UPPER

(10)

(07)

IN

NIA FOR CALI

(16)

MID–-A TLA NT IC

IC

UPPE COLO R RADO

(02) OHIO (05)

(14)

(18)

E ND

(13)

ISSI PPI

AARRKKAA D N NSSAASS–-WHITE-–RREED (11)

RA RIO G

LOW ER COL ORA DO (15)

TE XA S–-G ULF (12)

SS (06) NE TE N

EE

G UL F

CI F

NEW E NGL AN D

SOU RIS-

PA

FIGURE 9.1: Hydrologic regions in the United States Source: Seaber et al. (1987).

ER MIS S

Chapter 9

LO W

420

C– TI AN L AT SOUTH (03)

(02) Water Resources Regions

(08)

0

400 MILES

ALASKA (19) Kauai Oahu

CARIBBEAN (21)

(20) Molokai Maui Moui

HAWAII 0

400 MILES

0

Hawaii 150 MILES

Puerto Rico

0

80 MILES

Cloud droplets generally form on condensation nuclei, which are usually less than 1 µm in diameter and typically consist of sea salts, dust, or combustion by-products. These particles are called atmospheric aerosols, which also play an important role in regulating the amount of solar radiation that reaches the surface of the earth (Hendriks, 2010). In pure air, condensation of water vapor to form liquid water droplets occurs only when the air is supersaturated with water molecules. In cold clouds, where the temperature is below freezing, water droplets and ice crystals are both formed and the greater saturation vapor pressure over water compared to over ice causes the growth of ice crystals at the expense of water droplets. This is called the Bergeron–Findeisen process or ice-crystal process. In warm clouds, the primary mechanism for the growth of water droplets is collision and coalescence of water droplets. When the moisture droplets and/or ice crystals are large enough, the precipitation falls to the ground. Moisture droplets larger than about 0.1 mm are large enough to fall, and these droplets grow as they collide and coalesce to form larger droplets. Rain drops falling to the ground are typically in the size range of 0.5–3 mm. Drizzle is a subcategory of rain with droplet sizes less than 0.5 mm. Most of the clouds and atmospheric moisture are contained within 5500 m (18,000 ft) of the surface of the earth. The main mechanisms of air-mass lifting are frontal lifting, orographic lifting, and convective lifting. In frontal lifting, warm air is lifted over cooler air by frontal passage. The resulting precipitation events are called cyclonic or frontal storms, and the zone where the warm and cold air masses meet is called a front. Frontal precipitation is the dominant type of precipitation in the north-central United States and other continental areas (Elliot, 1995). In a warm front, warm air advances over a colder air mass with a relatively slow rate of ascent (0.3%–1% slope) causing a large area of precipitation in advance of the front, typically 300–500 km (200–300 mi) ahead of the front. In a cold front, warm air is pushed upward at a relatively steep slope (0.7%–2%) by advancing cold air, leading to smaller precipitation areas

Section 9.2

Rainfall

421

in advance of the cold front. Precipitation rates are generally higher in advance of cold fronts than in advance of warm fronts. In orographic lifting, warm air rises as it flows over hills or mountains, and the resulting precipitation events are called orographic storms. An example of the orographic effect can be seen in the northwestern United States, where the westerly airflow results in higher precipitation and cooler temperatures to the west of the Cascade mountains (e.g., Seattle, Washington) than to the east (e.g., Boise, Idaho). Orographic precipitation is a major factor in most mountainous areas, and the amount of precipitation typically shows a strong correlation with elevation (e.g., Naoum and Tsanis, 2003). The lee side of mountainous areas is usually dry, since most of the moisture is precipitated on the windward side, and this is sometimes called the rain shadow effect (Davie, 2008). In convective lifting, air rises by virtue of being warmer and less dense than the surrounding air, and the resulting precipitation events are called convective storms or, more commonly, thunderstorms. A typical thunderstorm is illustrated in Figure 9.2, where the localized nature of such storms is clearly apparent. Convective precipitation is common during the summer months in the central United States and other continental climates with moist summers. Convective storms are typically of short duration, and usually occur on hot midsummer days as late-afternoon storms. 9.2.1

Measurement of Rainfall

Records of rainfall have been collected for more than 2000 years (Ward and Robinson, 1999). Rainfall is typically measured using rain gages operated by government agencies such as the National Weather Service and local drainage districts. Rainfall amounts are described by the volume of rain falling per unit area and are given as a depth of water. Gages for measuring rainfall are categorized as either nonrecording (manual) or recording. In the United States, most of the rainfall data reported by the National Weather Service (NWS) are collected manually using a standard rain gage consisting of a 20.3-cm (8-in.) diameter funnel that passes water into a cylindrical measuring tube, the whole assembly being placed within an overflow can. The measuring tube has a cross-sectional area one-tenth that of the collector funnel; therefore, a 2.5-mm (0.1-in.) rainfall will occupy a depth of 25 mm (1 in.) in the collector tube. The capacity of the standard collector tube is 50 mm (2 in.), and rainfall in excess of this amount collects in the overflow can. The manual NWS gage is primarily used for collecting daily rainfall amounts. Automatic-recording gages are usually used for measuring rainfall at intervals less than one day, and for collecting data in remote locations. Recording gages use either a tipping bucket, weighing mechanism, or float device. The original tipping-bucket rain gage was invented by Sir Christopher Wren in about 1662, and accounts for nearly half of all recording rain gages worldwide (Upton and Rahimi, 2003). A typical tipping-bucket rain gage is FIGURE 9.2: Typical thunderstorm

422

Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions

FIGURE 9.3: Tipping-bucket rain gage

(a) Rain gage

(b) Support structure

shown in Figure 9.3, where Figure 9.3(a) shows a close-up top view of the rain gage with projections that help prevent the splattering of incident rainfall. The structure that usually supports the rain gage is shown in Figure 9.3(b), where a solar panel (front) provides power for the data recorder (back), and the rain gage is on top of the structure. The tipping-bucket gage is based on funneling the collected rain to a small bucket that tilts and empties each time it fills, generating an electronic pulse with each tilt. The number of bucket-tilts per time interval provides a basis for determining the precipitation depth over time. Typical problems with tipping-bucket rain gages include blockages, wetting and evaporation losses of typically 0.05 mm (0.002 in.) per rainfall event (Niemczynowicz, 1986), rain missed during the tipping process which typically takes about half a second (Marsalek, 1981), wind effects, position, shelter, and dynamic rainfall effects (Habib et al., 2008). A weighing-type gage is based on continuously recording the weight of the accumulated precipitation, and float-type recording gages operate by catching rainfall in a tube containing a float whose rise is recorded with time. Older weighing-type gages are mechanical-weighing and analog-recording devices, while modern weighing-type gages are electronic-weighing and digital-recording devices; significant discrepancies in measured rainfall intensities between analog and digital gages have been found for rainfall durations less than 5 min (Keefer et al., 2008). The accuracy of both manual and automatic-recording rain gages is typically on the order of 0.25 mm (0.01 in.) per rainfall event. Rain-gage measurements are point measurements of rainfall and are only representative of a small area surrounding the rain gage. Areas on the order of 25 km2 (10 mi2 ) have been taken as characteristic of rain-gage measurements (Gupta, 1989; Ponce, 1989), although considerably smaller characteristic areas can be expected in regions where convection storms are common. In the United States, the National Weather Service makes extensive use of Weather Surveillance Radar 1988 Doppler (WSR-88D), commonly known as next-generation weather radar (NEXRAD). A typical WSR-88D tower is shown in Figure 9.4, and each WSR-88D station measures weather activity over a 230-km (143-mi) radius area with a 10-cm (3.9-in) wavelength signal. This radar system provides estimates of hourly rainfall from the reflectivity of the S-band signal (1.55–3.9 GHz) within cells that are approximately 4 km * 4 km (2.5 mi * 2.5 mi). Since radar systems measure only the droplet size and movement of water in the atmosphere, and not the volume of water falling to the ground, radar-estimated rainfall should generally be corrected to correlate with field measurements of rainfall on the ground. Comparisons of rainfall predictions using WSR-88D with rainfall measurements using dense networks of rain gages have shown that WSR-88D estimates tend to be about 5%–10% lower than long-term rain-gage measurements (Johnson et al., 1999), and comparisons of WSR-88D with daily rain-gage measurements have shown significant spatial variability in radar accuracy, with typical annual bias on the order of −15%, correlation between radar

Section 9.2

Rainfall

423

FIGURE 9.4: Weather radar station (WSR-88D) Source of close-up view: NOAA (2012a).

(a) Far view

(b) Near view

and rain-gage measurements on the order of 0.7, and typical agreement in detecting the occurrence of rainfall on the order of 85% (Young and Brunsell, 2008). Errors in WSR88D estimates of rainfall have been shown to depend on the type of rainfall, with opposite biases for frontal and convective rainfall (Skinner et al., 2009). In spite of measurement inaccuracies, radar provides the advantage of covering large areas with high spatial and temporal resolution. Spatially distributed rainfall measurements provided by NEXRAD and other high-resolution weather radar systems have been incorporated directly into rainfallrunoff models to provide improved estimates of runoff distribution (e.g., Gires et al., 2012, Vieux et al., 2009). For example, real-time high-resolution NEXRAD measurements within 1 km * 1 km (0.6 mi * 0.6 mi) cells have been combined with hydrologic models to provide near-real-time flood alerts in the Brays Bayou Watershed in Houston, Texas (Fang et al., 2008). Care must be taken in using radar measurements to estimate short-term (e.g., 10-min) rainfall, since radars sample precipitation aloft and by the time the precipitation reaches the ground, wind-induced drift may have caused a horizontal shift of several kilometers (Islam and Rasmussen, 2008). Such conditions occur during intense convective storms, which typically have strong spatial and temporal gradients of intensity. Errors in short-term radar-measured rainfall can sometimes be reduced by increased spatial or temporal averaging (e.g., Knox and Anagnostou, 2009). 9.2.2

Statistics of Rainfall Data

Rainfall measurements are seldom used directly in engineering design, but rather the statistics of rainfall measurements are typically used. Rainfall statistics are most commonly presented in the form of intensity-duration-frequency (IDF) curves, which express the relationship between the average intensity in a rainstorm and the averaging time (= duration), with the average intensity having a given probability of occurrence. Such curves are also called intensity-frequency duration (IFD) curves (e.g., Gyasi-Agyei, 2005). A typical IDF curve (for Miami, Florida) for return periods ranging from 2 to 100 years is shown in Figure 9.5. To fully understand the meaning and application of the IDF (or IFD) curve, it is best to review how this curve is derived from rainfall measurements. The data required to calculate the IDF curve are a record of rainfall measurements in the form of the depth of rainfall during fixed intervals of time, t, typically on the order of 5 min. For a rainfall record containing several years of data, the following computations lead to the IDF curve:

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Rainfall intensity in inches per hour

FIGURE 9.5: Intensityduration-frequency (IDF) curve Source: Florida Department of Transportation (2000).

15

15

10 9 8 7 6 5

10 9 8 7 6 5

2 Years 3 Years 5 Years 10 Years 25 Years 50 Years 100 Years

4 3

4 3 2

2

1.0 0.9 0.8 0.7 0.6 0.5

1.0 0.9 0.8 0.7 0.6 0.5

Miami, FL Zone 10

0.4

0.4 0.3

0.3 0.2 0.8 10

15

20 Min

30

40 50 60

2

3

4

5

10

15

20 24

0.2

H Duration

Step 1: For a given duration of time (= averaging period), starting with t, determine the annual-maximum rainfall (AMR) for this duration in each year. Step 2: The AMR values, one for each year, are rank-ordered, and the return period, T, for each AMR value is estimated using the Weibull formula T=

n + 1 m

(9.1)

where n is the number of years of data and m is the rank of the data corresponding to the event with return period T. As an alternative to using Equation 9.1 to estimate the return periods of the AMR values, an extreme-value distribution (typically Type I or Type II) may be fitted to the AMR values for the given duration, t (e.g., Koutsoyiannis et al., 1998). Step 3: Steps 1 and 2 are repeated, with the duration increased by t. An upper-limit duration of interest needs to be specified, and for urban-drainage applications the upper-limit duration of interest is typically on the order of 1–2 h. Step 4: For each return period, T, the AMR versus duration can be plotted, and this relationship is called the depth-duration-frequency curve. Dividing each AMR value by the corresponding duration yields the average intensity, which is plotted versus the duration, for each return period, to yield the IDF curve. This procedure is illustrated in the following example.

Section 9.2

Rainfall

425

EXAMPLE 9.1 A rainfall record contains 32 years of rainfall measurements at 5-min intervals. The annual-maximum rainfall amounts for intervals of 5 min, 10 min, 15 min, 20 min, 25 min, and 30 min have been calculated and ranked. The top three annual-maximum rainfall amounts, in millimeters, for each time increment are given in the following table:

Rank

5

10

1 2 3

12.1 11.0 10.7

18.5 17.9 17.5

t in min 15 20 24.2 22.1 21.9

28.3 26.0 25.2

25

30

29.5 28.4 27.6

31.5 30.2 29.9

Calculate the IDF curve for a return period of 20 years. Solution For each time interval, t, there are n = 32 ranked rainfall amounts of annual maxima. The relationship between the rank, m, and the return period, T, is given by Equation 9.1 as

T=

32 + 1 33 n + 1 = = m m m

The return period can therefore be used in lieu of the rankings, and the given data can be put in the form: Return period, T (years)

t in min 10 15 20

5

33 16.5 11

12.1 11.0 10.7

18.5 17.9 17.5

24.2 22.1 21.9

28.3 26.0 25.2

25

30

29.5 28.4 27.6

31.5 30.2 29.9

The rainfall increments with a return period, T, of 20 years can be linearly interpolated between the rainfall increments corresponding to T = 33 years and T = 16.5 years to yield:

Duration (min) Rainfall (mm)

5 11.2

10 18.0

15 22.5

20 26.5

25 28.6

30 30.5

and the average intensities for each duration are obtained by dividing the rainfall increments by the corresponding duration to yield:

Duration (min) Intensity (mm/h)

5 134

10 108

15 90

20 79

25 69

30 61

These points define the IDF curve for a return period of 20 years, and this IDF curve is plotted in Figure 9.6.

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FIGURE 9.6: IDF curve for 20-year return period

140 130 Return period ⫽ 20 years 120

Intensity (mm/h)

110 100 90 80 70 60 50

0

5

10

15 20 Duration (min)

25

30

35

Many of the data used in deriving IDF curves are derived from tipping-bucket rain-gage measurements. These gages, the most popular type of rain gage employed worldwide, are known to underestimate rainfall at high intensities because of the rain missed during the tipping of the bucket. Unless the rain gage used in deriving the IDF curve is dynamically calibrated, or the rain gage is self-calibrating, the derived IDF curve can lead to significant underdesign of urban drainage systems (Molini et al., 2005). EXAMPLE 9.2 How many years of rainfall data are required to derive the IDF curve for a return period of 10 years? Solution The return period, T, is related to the number of years of data, n, and the ranking, m, by T=

n + 1 m

For T = 10 years and m = 1, n = mT − 1 = (1)(10) − 1 = 9 years Therefore, a minimum of 9 years of data are required to estimate the IDF curve for a 10-year return period.

Partial-duration series. The previous example has illustrated the derivation of IDF curves from n annual maxima of rainfall measurements, where the series of annual maxima is called the annual series. As an alternative to using an n-year annual series to derive IDF curves, a partial-duration series is sometimes used in which the largest n rainfall amounts in an n-year record are selected for each duration, regardless of the year in which the rainfall amounts occur. In this case, the return period, T, assigned to each rainfall amount is still calculated using Equation 9.1. Analysis of the annual-maximum series produces estimates of the average period between years when a particular value is exceeded, while analysis of the partial-duration series produces estimates of the average period between values of a particular magnitude. The frequency distribution of rainfall amounts derived using the

Section 9.2

Rainfall

427

TABLE 9.1: Factors for Converting Partial-Duration Series to Annual Series

Return period (years) 2 5 10 25 >25

Factor 0.88 0.96 0.99 1.00 1.00

Source: Frederick et al. (1977).

partial-duration series differs from that derived using the annual series. The relationship between the return period of an event derived from an annual series, Ta , and the return period of that same event derived from a partial-duration series, Tp , can be estimated by (Chow, 1964) 1  Tp = −  (9.2) ln 1 − Ta−1 In general, Tp < Ta , with the difference between Tp and Ta decreasing as Ta increases. A rainfall amount with a return period derived from a partial-duration series can be converted to a corresponding rainfall amount for an annual series by using the empirical factors given in Table 9.1. These factors are applicable for return periods greater than 2 years. Partialduration and annual series are approximately the same at the larger recurrence intervals, but for smaller recurrence intervals the partial-duration series will normally indicate events of greater magnitude. For rainfall records shorter than 20–25 years it is recommended that return periods be estimated using partial-duration series (NRC, 2009).

EXAMPLE 9.3 A 10-year rainfall record measures rainfall increments at 5-min intervals. The top six rainfall increments derived from the partial-duration series are as follows: Rank 5-min rainfall (mm)

1 22.1

2 21.9

3 21.4

4 20.7

5 20.3

6 19.8

Estimate the frequency distribution of the annual maxima with return periods greater than 2 years. Solution The ranked data were derived from a 10-year partial-duration series (n = 10), where the return period, T, is given by 10 + 1 11 n + 1 = = T= m m m Using this (Weibull) relation, the freq