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WATER-RESOURCES ENGINEERING
Third Edition
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David A. Chin
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Library of Congress Cataloging-in-Publication Data
Chin, David A.
Water-resources engineering / David A. Chin. – 3rd ed.
p. cm.
ISBN-13: 978-0-13-283321-9 (alk. paper)
ISBN-10: 0-13-283321-2 (alk. paper)
1. Hydraulics. 2. Hydrology. 3. Waterworks. 4. Water resources
development. I. Title.
TC160.C52 2014
627–dc23
2012018911
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
Production Project Manager / Editorial Production Manager: Greg Dulles
Cover Photo: United States Bureau of Reclamation
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© 2013, 2010, 2006, 2000 Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved. No part of this book may be reproduced in any form or by any means, without
permission in writing from the publisher.
Pearson Prentice Hall™ is a trademark of Pearson Education, Inc.
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Printed in the United States of America.
10 9 8 7 6 5 4 3 2 1
ISBN: 0-13-283321-2
Pearson Education Ltd., London
Pearson Education Australia Pty. Ltd., Sydney
Pearson Education Singapore, Pte. Ltd.
Pearson Education North Asia Ltd., Hong Kong
Pearson Education Canada, Inc., Toronto
Pearson Educación de Mexico, S.A. de C.V.
Pearson Education—Japan, Tokyo
Pearson Education Malaysia, Pte. Ltd.
Pearson Education, Upper Saddle River, New Jersey
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To Andrew and Stephanie.
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“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
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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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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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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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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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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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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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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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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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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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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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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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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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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 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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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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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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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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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
700
700
700
700
702
706
709
713
714
718
718
728
736
741
741
742
744
746
746
750
752
752
761
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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
771
771
771
774
774
775
777
777
778
779
779
783
784
785
788
793
794
794
795
796
798
803
808
809
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 . . . . . . . . . . . . . . . . . . . . .
815
815
815
818
819
819
820
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xiii
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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
820
821
823
823
825
825
828
829
A Units and Conversion Factors
A.1 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.2 Conversion Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
831
831
832
B Fluid Properties
B.1 Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
B.2 Organic Compounds Found in Water . . . . . . . . . . . . . . . . . . . . . .
B.3 Air at Standard Atmospheric Pressure . . . . . . . . . . . . . . . . . . . . .
834
834
834
836
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 . . . . . . . . . .
837
837
839
841
842
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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
843
843
844
844
844
844
845
845
845
845
848
849
E Pipe Specifications
E.1 PVC Pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
E.2 Ductile-Iron Pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
E.3 Concrete Pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
E.4 Physical Properties of Common Pipe Materials . . . . . . . . . . . . . . . .
850
850
850
851
851
F Unified Soil Classification System
F.1 Definition of Soil Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
852
852
853
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Bibliography
854
Index
912
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Preface
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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,
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Preface
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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.
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What’s New in the Third Edition
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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.
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xvii
• 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.
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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
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C H A P T E R
1
Introduction
1.1 Water-Resources Engineering
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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.
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1.2 The Hydrologic Cycle
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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
1
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2
Chapter 1
Introduction
FIGURE 1.1: Hydrologic
cycle
Sun
Hydrologic Cycle
Snowfall
Condensation
and cloud
formation
Snowcap
Rainfall
Evapotranspiration
Snowmelt
Mountain
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Evaporation
Surface
water
Surface
runoff
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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
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Section 1.2
The Hydrologic Cycle
3
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
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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
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TABLE 1.2: Estimated World Water Quantities
Volume
(*103 km3 )
Percent
total water
(%)
Oceans
Groundwater
Fresh
Saline
Soil moisture
Polar ice
Other ice and snow
Lakes
Fresh
Saline
Marshes
Rivers
Biological water
Atmospheric water
1,338,000
96.5
Total water
Freshwater
1,385,984.61
35,029.21
Item
10,530
12,870
16.5
24,023.5
340.6
0.76
0.93
0.0012
1.7
0.025
91
85.4
11.47
2.12
1.12
12.9
0.007
0.006
0.0008
0.0002
0.0001
0.001
100
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Percent
freshwater
(%)
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).
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4
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).
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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
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TABLE 1.4: Climate Spectrum
Climate
Mean annual
precipitation
(mm)
Mean annual
evapotranspiration
(mm)
Length of
rainy season
(months)
Superarid
Hyperarid
Arid
Semiarid
Subhumid
Humid
Hyperhumid
Superhumid
<100
100–200
200–400
400–800
800–1600
1600–3200
3200–6400
Ú 6400
<3000
2400–3600
2000–2400
1600–2000
1200–1600
1200
1200
1200
<1
1–2
2–3
3–4
4–6
6–9
9–12
12
Source: Ponce et al. (2000).
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Section 1.3
Humid climate
Percentage of rainfall
80%
60%
40%
20%
0%
5
100%
100%
Percentage of rainfall
FIGURE 1.3: Typical water
budgets in different
climates
Design of Water-Resource Systems
ET
80%
60%
40%
20%
0%
Runoff Recharge
Semiarid climate
ET
Runoff Recharge
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Percentage of rainfall
100%
Arid climate
80%
60%
40%
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20%
0%
ET
Runoff Recharge
withdrawal, and return flows within watersheds can have significant impacts on regional
ecologies and should generally be taken into account in designing water-resource systems.
1.3 Design of Water-Resource Systems
The uncertainty and natural variability of hydrologic processes require that most waterresource systems be designed with some degree of risk (of failure). Approaches to designing
such systems can be classified as either frequency-based design, risk-based design, or criticalevent design. In frequency-based design, the exceedance probability of the design event is
selected a priori and the water-resource system is designed to accommodate all lesser events
up to and including an event with the selected exceedance probability. The water-resource
system will then be expected to fail with a probability equal to the exceedance probability
of the design event. The frequency-based design approach is commonly used in designing
the minor structures of urban drainage systems. For example, urban storm-drainage systems are typically designed for precipitation events with return periods of 10 years or less,
where the return period of an event is defined as the reciprocal of the (annual) exceedance
probability of the event. In risk-based design, systems are designed such that the sum of the
capital cost and the cost of failure is minimized. Capital costs tend to increase and the cost of
failure tends to decrease with increasing system capacity. Because any threats to human life
are generally assigned extremely high failure costs, structures such as large dams are usually
designed for rare hydrologic events with long return periods and commensurate small failure
risks. In some extreme cases, where the consequences of failure are truly catastrophic, waterresource systems are designed for the largest possible magnitude of a hydrologic event. This
approach is called critical-event design, and the value of the design (hydrologic) variable in
this case is referred to as the estimated limiting value (ELV).
Water-resource systems can be broadly categorized as water-control systems or wateruse systems, with design objectives as shown in Table 1.5; however, these systems are not
mutually exclusive. The following sections present a brief overview of the design objectives
in water-control and water-use systems.
1.3.1
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Water-Control Systems
Water-control systems are primarily designed to control the spatial and temporal distribution of surface runoff resulting from rainfall events. Flood-control structures and storage
impoundments reduce the peak flows in streams, rivers, and drainage channels, thereby
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6
Chapter 1
Introduction
TABLE 1.5: Design Objectives of Water-Resource Systems
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Water-control systems
Water-use systems
Drainage
Flood control
Salinity control
Sediment control
Pollution abatement
Domestic and industrial water supply
Wastewater treatment
Irrigation
Hydropower generation
reducing the occurrence of floods. A flood is defined as a high flow that exceeds the capacity
of a stream or drainage channel, and the elevation at which water overflows the embankments of a stream or drainage channel is called the flood stage. A floodplain is the normally
dry land adjoining rivers, streams, lakes, bays, or oceans that is inundated during flood events.
Typically, flows with return periods from 1.5 to 3 years represent bankfull conditions in natural streams, with larger flows causing inundation of the floodplain. The 100-year flood has
been adopted by the U.S. Federal Emergency Management Agency (FEMA) as the base
flood for delineating floodplains, and the area inundated by the 500-year flood is sometimes
also delineated to indicate additional areas at a lessened risk. Encroachment onto floodplains
reduces the capacity of the watercourse and increases the extent of the floodplain. Approximately 7%–10% of the land in the United States is in a floodplain. The largest floodplain
areas in the United States are in the South; the most populated floodplains are along the
north Atlantic coast, the Great Lakes region, and in California.
In urban settings, water-control systems include storm-sewer systems for collecting
and transporting surface runoff, and storage reservoirs that attenuate peak runoff rates and
reduce pollutant loads in drainage channels. Urban stormwater control systems are typically
designed to prevent flooding from runoff events with return periods of 10 years or less. For
runoff events that exceed the design event, surface (street) flooding usually results.
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1.3.2
Water-Use Systems
Water-use systems are designed to support human habitation and include water-treatment
systems, water-distribution systems, wastewater-collection systems, wastewater-treatment systems, and irrigation systems. Water for these systems are generally derived from rivers, lakes,
and groundwater; and great care must be taken to ensure that water withdrawals from these
sources are sustainable. The design capacity of water-use systems is generally dictated by the
population and characteristics of the service area, commercial and industrial requirements,
crop requirements, and/or the economic design life of the system. Some key design aspects
of these systems are given below.
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Domestic water-supply systems. These systems typically include water-extraction facilities,
such as wellfields, that must extract water from the source at rates that do not cause adverse
effects on the source water and those that depend on it; water-treatment plants that must
produce water of sufficient quality to meet drinking water standards; and water-distribution
systems that must deliver peak demands while sustaining adequate water pressures.
Domestic wastewater-collection systems. Sanitary sewers and associated appurtenances
must have sufficient capacity to transport wastes without overflowing into the streets. These
systems typically terminate in a wastewater-treatment plant that must provide a sufficient
level of treatment for the intended use of the effluent, and effluent discharge systems that do
not degrade the receiving waters.
Irrigation systems. In agricultural areas, water requirements of crops are met by a combination of rainfall and irrigation. The design of irrigation systems requires the estimation of crop
evapotranspiration rates and leaching requirements. Usually, irrigation water is obtained
from groundwater and care must be taken to ensure that the extraction rates are sustainable
and long-term declines in groundwater resources will not occur.
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Section 1.3
Design of Water-Resource Systems
7
Hydroelectric power systems. In rivers where there are large seasonal variations in flow
volumes and droughts occur frequently, construction of dams and the generation of hydroelectric power might be economically feasible. In such cases, water reservoirs are sized to
provide sufficient capacity to store excess water in years and seasons of surplus, and provide
water to support human habitation in years and seasons of water deficit. Water released
from the storage reservoirs behind dams can be used to supply hydroelectric power, which is
a clean and nonpolluting source of electricity.
1.3.3
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Supporting Federal Agencies in the United States
The design of water-resource systems usually involves interaction with government agencies. Collection of hydrologic and geologic data, granting of development permits, specification of design criteria, and use of government-developed computer codes for developing
models of water-resource systems are some of the many areas in which water-resources
engineers interact with government agencies. The following are some of the key federal
water-resources agencies in the United States:
National Climatic Data Center (NCDC). The world’s largest active source of weather data;
it produces numerous climate publications and responds to data requests from all over
the world. Most of the data available from NCDC are collected and analyzed by the
National Weather Service (NWS), Military Services, Coast Guard, Federal Aviation
Administration, and cooperative observers.
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U.S. Army Corps of Engineers (USACE). USACE is responsible for the planning, construction, operation, and maintenance of a variety of water-resource facilities whose objectives include navigation, flood control, water supply, recreation, hydroelectric power
generation, water-quality control, and other purposes.
U.S. Bureau of Reclamation (USBR). USBR is responsible for planning, construction, operation, and maintenance of a variety of water-resource facilities whose objectives include
irrigation, power generation, recreation, fish and wildlife preservation, and municipal
water supply. Most activities are confined to the 17 (arid) states west of the Mississippi
River. Besides being the largest wholesale supplier of water in the United States, USBR
is the sixth largest hydroelectric supplier in the United States.
U.S. Environmental Protection Agency (USEPA). USEPA is responsible for the implementation and enforcement of federal environmental laws. The agency’s mission is to protect public health and to safeguard and improve the natural environment—air, water,
and land—upon which human life depends.
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U.S. Geological Survey (USGS). The Water Resources Division of the USGS has primary
federal responsibility for collection and dissemination of measurements of stream discharge and stage, reservoir and lake stage and storage, groundwater levels, well and
spring discharge, and the quality of surface and groundwater in the United States.
USGS maintains a network of thousands of stream gages and groundwater monitoring
wells.
U.S. National Weather Service (NWS). NWS, under the direction of the National Oceanic
and Atmospheric Administration (NOAA), has a mandate to collect hydrologic data
and provide weather, hydrologic, and climate forecasts. Data collected by NWS include
rainfall, temperature, and evaporation measurements at over 10,000 locations in the
United States. The NWS uses its River Forecast System at 13 River Forecast Centers
to provide daily river-stage forecasts at approximately 4000 locations.
U.S. Natural Resources Conservation Service (NRCS). NRCS works with landowners on
private lands to conserve natural resources. NRCS provides technical and financial
assistance to farmers and ranchers for flood protection, recreation, and water-supply
development in watersheds with areas less than 100,000 ha (400 mi2 ). The NRCS
publishes general soil maps for each state and detailed soil maps for each county in
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8
Chapter 1
Introduction
TABLE 1.6: Federal Agencies Relevant to Water-Resources Engineering in the United States
Organization
Web address
National Climatic Data Center (NCDC)
U.S. Army Corps of Engineers (USACE)
U.S. Bureau of Reclamation (USBR)
U.S. Environmental Protection Agency (USEPA)
U.S. Geological Survey (USGS)
U.S. National Weather Service (NWS)
U.S. Natural Resources Conservation Service (NRCS)
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Problem
www.ncdc.noaa.gov
www.usace.army.mil
www.usbr.gov
www.epa.gov
www.usgs.gov
www.nws.noaa.gov
www.nrcs.usda.gov
the United States. The Snowpack Telemetry (SNOTEL) system operated by NRCS
provides year-round temperature and precipitation data in remote, mountainous areas
primarily in the western United States.
These federal agencies provide a wealth of data and information on water resources,
relevant government regulations, and useful computer software that can be found on the
Internet. Several of the more useful Web sites currently in use (and likely to be around for
the foreseeable future) are listed in Table 1.6.
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1.1. Search the Internet to determine the mean annual rainfall and evapotranspiration of Boston, Massachusetts,
and Santa Fe, New Mexico. Classify the climate in these
cities.
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C H A P T E R
2
Fundamentals of Flow
in Closed Conduits
2.1 Introduction
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Flow in closed conduits includes all cases where the flowing fluid completely fills the conduit.
The cross sections of closed conduits can be of any shape or size and the conduits can be
made of a variety of materials. Engineering applications of the principles of flow in closed
conduits include the design of municipal water-supply systems and transmission lines. The
basic equations governing the flow of fluids in closed conduits are the continuity, momentum,
and energy equations, and the most useful forms of these equations for application to pipeflow problems are derived in this chapter. The governing equations are presented in forms
that are applicable to any fluid flowing in a closed conduit, but particular attention is given
to the flow of water.
Flows in pipe networks are a natural extension of flows in single pipelines, and methods of calculating flows and pressure distributions in pipeline systems are also covered here.
These methods are particularly applicable to the analysis and design of municipal waterdistribution systems, where the engineer is frequently interested in assessing the effects of
various modifications to the system. Because transmission of water in closed conduits is typically accomplished using pumps, the fundamentals of pump operation and performance are
also presented in this chapter. A sound understanding of pumps is important in selecting the
appropriate pump to achieve the desired operational characteristics in water-transmission
systems.
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2.2 Single Pipelines
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The governing equations for flows in pipelines are derived from the conservation laws of
mass, momentum, and energy, and the forms of these equations that are most useful for
application to closed-conduit flow are derived in the following sections.
2.2.1
Steady-State Continuity Equation
Consider the application of the continuity equation to the control volume illustrated in
Figure 2.1. Fluid enters and leaves the control volume normal to the control surfaces, with the
inflow velocity denoted by v1 (r) and the outflow velocity by v2 (r), where r is the radial position vector originating at the centerline of the conduit. Both the inflow and outflow velocities
vary across the control surface. The steady-state continuity equation for an incompressible
fluid can be written as
!
!
v1 dA =
v2 dA
(2.1)
A1
A2
Defining V1 and V2 as the average velocities across A1 and A2 , respectively, where
!
1
V1 =
v1 dA
A1 A1
and
V2 =
1
A2
!
A2
v2 dA
(2.2)
(2.3)
9
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10
Chapter 2
Fundamentals of Flow in Closed Conduits
FIGURE 2.1: Flow
through closed conduit
Control
volume
r
v2
v1
Velocity
distribution
Boundary of
control volume
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Flow
the steady-state continuity equation becomes
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V1 A1 = V2 A2 (= Q)
(2.4)
The terms on each side of Equation 2.4 are equal to the volumetric flow rate, Q. The steadystate continuity equation simply states that the volumetric flow rate across any surface normal to the flow is a constant.
EXAMPLE 2.1
Water enters a pump through a 150-mm diameter intake pipe and leaves through a 200-mm diameter
discharge pipe. If the average velocity in the intake pipeline is 1 m/s, calculate the average velocity in
the discharge pipeline. What is the flow rate through the pump?
Solution In the intake pipeline, V1 = 1 m/s, D1 = 0.15 m and
π
π
A1 = D21 = (0.15)2 = 0.0177 m2
4
4
In the discharge pipeline, D2 = 0.20 m and
π
π
A2 = D22 = (0.20)2 = 0.0314 m2
4
4
According to the continuity equation,
V1 A1 = V2 A2
Therefore,
V2 = V1
The flow rate, Q, is given by
"
A1
A2
#
= (1)
"
0.0177
0.0314
#
= 0.56 m/s
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Q = A1 V1 = (0.0177)(1) = 0.0177 m3 /s
The average velocity in the discharge pipeline is 0.56 m/s, and the flow rate through the pump is
0.0177 m3 /s.
2.2.2
Steady-State Momentum Equation
Consider the application of the momentum equation to the control volume illustrated in
Figure 2.1. Under steady-state conditions, the component of the momentum equation in the
direction of flow (x-direction) can be written as
!
$
Fx =
ρvx v · n dA
(2.5)
A
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Section 2.2
Single Pipelines
11
%
where
Fx is the sum of the x-components of the forces acting on the fluid in the control volume, ρ is the density of the fluid, vx is the component of the flow velocity in the
x-direction, and v · n is the component of the flow velocity normal to the control surface.
Since the unit normal vector, n, in Equation 2.5 is directed outward from the control volume,
the momentum equation for an incompressible fluid (ρ = constant) can be written as
!
!
$
v22 dA − ρ
v21 dA
(2.6)
Fx = ρ
A2
A1
where the integral terms depend on the velocity distributions across the inflow and outflow
control surfaces. The velocity distribution across each control surface is generally accounted
for by the momentum correction coefficient, β, defined by the relation
ww
1
β=
AV 2
!
A
v2 dA
(2.7)
where A is the area of the control surface and V is the average velocity over the control
surface. The momentum coefficients for the inflow and outflow control surfaces, A1 and A2 ,
are given by β1 and β2 , where
!
1
β1 =
v2 dA
(2.8)
A1 V12 A1 1
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β2 =
1
A2 V22
!
A2
v22 dA
(2.9)
Substituting Equations 2.8 and 2.9 into Equation 2.6 leads to the following form of the
momentum equation
$
Fx = ρβ2 V22 A2 − ρβ1 V12 A1
(2.10)
Recalling that the continuity equation states that the volumetric flow rate, Q, is the same
across both the inflow and outflow control surfaces, where
Q = V1 A1 = V2 A2
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(2.11)
then combining Equations 2.10 and 2.11 leads to the following form of the steady-state
momentum equation
$
(2.12)
Fx = ρβ2 QV2 − ρβ1 QV1
or
$
Fx = ρQ(β2 V2 − β1 V1 )
(2.13)
In many cases of practical interest, the velocity distribution across the cross section of the
closed conduit is approximately uniform, in which case the momentum coefficients, β1 and
β2 , are approximately equal to unity and the steady-state momentum equation becomes
$
Fx = ρQ(V2 − V1 )
(2.14)
Consider the common case of flow in a straight pipe with a uniform circular cross section
illustrated in Figure 2.2, where the average velocity remains constant at each cross section,
V1 = V2 = V
(2.15)
then the steady-state momentum equation becomes
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12
Chapter 2
Fundamentals of Flow in Closed Conduits
FIGURE 2.2: Forces on
flow in closed conduit
2
τ0 PL
1
A
p2
z2
A
p 1A
θ
Flow
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L
z1
γAL
$
Fx = 0
(2.16)
The forces that act on the fluid in a control volume of uniform cross section are illustrated in Figure 2.2. At Section 1, the average pressure over the control surface is equal to
p1 and the elevation of the midpoint of the section relative to a defined datum is equal to z1 ,
at Section 2, located a distance L downstream from Section 1, the pressure is p2 , and the
elevation of the midpoint of the section is z2 . The average shear stress exerted on the fluid
by the pipe surface is equal to τ0 , and the total shear force opposing flow is τ0 PL, where P
is the perimeter of the pipe. The fluid weight acts vertically downward and is equal to γ AL,
where γ is the specific weight of the fluid and A is the cross-sectional area of the pipe. The
forces acting on the fluid system that have components in the direction of flow are the shear
force, τ0 PL; the weight of the fluid in the control volume, γ AL; and the pressure forces on
the upstream and downstream faces, p1 A and p2 A, respectively. Substituting the expressions
for the forces into the momentum equation, Equation 2.16, yields
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p1 A − p2 A − τ0 PL − γ AL sin θ = 0
(2.17)
where θ is the angle that the pipe makes with the horizontal and is given by the relation
sin θ =
z2 − z1
L
Combining Equations 2.17 and 2.18 yields
p1
p2
τ0 PL
−
− z2 + z1 =
γ
γ
γA
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(2.18)
(2.19)
Defining the total head, or mechanical energy per unit weight, at Sections 1 and 2 as h1 and
h2 , where
h1 =
V2
p1
+
+ z1
γ
2g
(2.20)
h2 =
p2
V2
+
+ z2
γ
2g
(2.21)
and
then the head loss between Sections 1 and 2, 'h, is given by
#
"
#
"
p2
p1
'h = h1 − h2 =
+ z1 −
+ z2
γ
γ
(2.22)
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Section 2.2
Single Pipelines
13
Combining Equations 2.19 and 2.22 leads to the following expression for head loss:
'h =
τ0 PL
γA
(2.23)
In this case, the head loss, 'h, is entirely due to pipe friction and is commonly denoted by hf .
In the case of pipes with circular cross sections, Equation 2.23 can be written as
hf =
τ0 (π D)L
4τ0 L
=
2
γD
γ (π D /4)
(2.24)
where D is the diameter of the pipe. The ratio of the cross-sectional area, A, to the perimeter,
P, is defined as the hydraulic radius, R, where
ww
R=
A
P
(2.25)
and the head loss can be written in terms of the hydraulic radius as
hf =
τ0 L
γR
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(2.26)
The form of the momentum equation given by Equation 2.26 is of limited utility in that
the head loss, hf , is expressed in terms of the boundary shear stress, τ0 , which is not an easily
measurable quantity. However, the boundary shear stress, τ0 , can be expressed in terms of
measurable flow variables using dimensional analysis, where τ0 can be taken as a function of
the mean flow velocity, V; density of the fluid, ρ; dynamic viscosity of the fluid, µ; diameter
of the pipe, D; characteristic size of roughness projections, ϵ; characteristic spacing of the
roughness projections, ϵ ′ ; and a (dimensionless) form factor, m, that depends on the shape of
the roughness elements on the surface of the conduit. This functional relationship can be
expressed as
τ0 = f1 (V, ρ, µ, D, ϵ, ϵ ′ , m)
(2.27)
According to the Buckingham pi theorem, this relationship between eight variables in three
fundamental dimensions can also be expressed as a relationship between five nondimensional
groups. The following relation is proposed:
'
&
τ0
ϵ ϵ′
(2.28)
= f2 Re, , , m
D D
ρV 2
where Re is the Reynolds number defined by
Re =
ρVD
µ
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(2.29)
The relationship given by Equation 2.28 is as far as dimensional analysis goes, and experiments are necessary to determine an empirical relationship between the nondimensional
groups. Nikuradse (1932; 1933) conducted a series of experiments on the turbulent flow of
water in pipes in which the inner surfaces were roughened with sand grains of uniform diameter, ϵ. In these experiments, the spacing, ϵ ′ , and shape, m, of the roughness elements (sand
grains) were constant and Nikuradse’s experimental data fitted to the following functional
relation:
"
#
τ0
ϵ
Re,
(2.30)
=
f
3
D
ρV 2
It is convenient for subsequent analysis to introduce a factor of 8 into this relationship, which
can then be written as
"
#
τ0
ϵ
1
f
Re,
(2.31)
=
8
D
ρV 2
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14
Chapter 2
Fundamentals of Flow in Closed Conduits
or simply
τ0
f
=
2
8
ρV
(2.32)
where the dependence of the friction factor, f , on the Reynolds number, Re, and relative
roughness, ϵ/D, is understood. Combining Equations 2.32 and 2.24 leads to the following
form of the momentum equation for flows in circular pipes:
hf =
ww
fL V 2
D 2g
(2.33)
This equation, called the Darcy–Weisbach equation,∗ expresses the frictional head loss, hf , of
the fluid over a length L of pipe in terms of measurable parameters, including the pipe diameter (D), average flow velocity (V), and the friction factor (f ) that characterizes the shear
stress of the fluid on the pipe. Equation 2.33 is sometimes referred to simply as the Darcy
equation; however, this is inappropriate, since it was Weisbach who first proposed the exact
form of Equation 2.33 in 1845, with Darcy’s contribution on the functional dependence of f
on V and D in 1857 (Rouse and Ince, 1957; Brown, 2002). The differences between laminar
and turbulent flow were later quantified by Reynolds† in 1883 (Reynolds, 1883).
Nikuradse (1932; 1933) performed landmark experiments to study the head losses in
pipes that were artificially roughened by sticking sand grains of uniform size onto the surface of smooth pipes. Based on Nikuradse’s experimental results, Prandtl and von Kármán
developed the following empirical formulae for estimating the friction factor in turbulent
pipe flows:
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eer
in
⎧
&
'
"
#
⎪
⎪
2.51
ks
⎪
⎪
(
−2
log
,
Prandtl
equation
for
smooth
pipe
L
0
⎪
⎪
⎨
D
Re f
1
( =
⎪
#
"
f
⎪
⎪
ks /D
⎪
⎪
⎪
,
−2
log
⎩
3.7
"
#
ks
von Kármán equation for rough pipe
W0
D
(2.34)
g.n
et
where ks is the size of the sand grains on the surface of the pipe. Turbulent flow in pipes is generally present when Re > 4000, and transition to turbulent flow begins at about Re = 2300. A
pipe behaves like a smooth pipe when the friction factor does not depend on the height of the
roughness projections on the wall of the pipe and therefore depends only on the Reynolds
number. In rough pipes, the friction factor is determined solely by the relative roughness,
ks /D, and is independent of the Reynolds number. The smooth-pipe case generally occurs
at lower Reynolds numbers, when the roughness projections are submerged within the viscous boundary layer. At higher values of the Reynolds number, the thickness of the viscous
boundary layer decreases and eventually the roughness projections protrude sufficiently far
outside the viscous boundary layer that the shear stress of the pipe boundary is dominated by
the hydrodynamic drag associated with the roughness projections into the main body of the
flow. Under these circumstances, the flow in the pipe becomes fully turbulent, the friction factor is independent of the Reynolds number, and the pipe is considered to be (hydraulically)
rough. The flow is actually turbulent under both smooth-pipe and rough-pipe conditions,
but the flow is termed fully turbulent when the friction factor is independent of the Reynolds
number. Between the smooth- and rough-pipe conditions, there is a transition region in which
the friction factor depends on both the Reynolds number and the relative roughness.
∗ Henry Darcy (1803–1858) was a nineteenth-century French engineer; Julius Weisbach (1806–1871) was a
German engineer of the same era. Weisbach proposed the use of a dimensionless resistance coefficient, and Darcy
carried out the tests on water pipes.
† Osbourne Reynolds (1842–1912) was a British engineer who made seminal contributions in the area of fluid
dynamics.
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Section 2.2
ww
Single Pipelines
15
A limitation of the Nikuradse experiments on sand-roughened pipes was that the
roughness heights were of uniform size, whereas in commercial pipes the roughness heights
are generally nonuniform in size. Colebrook and White (1939) performed experiments on
commercial pipes and showed that head losses in commercial pipes could still be characterized asymptotically by the Prandtl and von Kármán equations (Equation 2.34); however,
the transition region where f depends on both Re and ks /D has a different behavior in
commercial pipes than in Nikuradse’s sand-roughened pipes. Colebrook (1939) identified
the equivalent sand roughness of several commercial pipe materials that would give the same
head losses as sand-roughened pipes under fully turbulent conditions and developed the following relationship that adequately describes the friction factor of commercial pipes under
turbulent-flow conditions:
&
'
ks /D
2.51
1
(
( = −2 log
+
(2.35)
3.7
f
Re f
This equation is commonly referred to as the Colebrook equation or Colebrook–White equation. Values of friction factor, f , predicted by the Colebrook equation (Equation 2.35) are
generally accurate to within 10%–15% of experimental data (Granger, 1985; Alexandrou,
2001; Finnemore and Franzini, 2002), and the Colebrook equation asymptotes to the Prandtl
and von Kármán equations given by Equation 2.34. The accuracy of the Colebrook equation
deteriorates significantly for small pipe diameters, and it is recommended that this equation
not be used for pipes with diameters smaller than 2.5 mm (0.1 in.) (Yoo and Singh, 2005). The
Colebrook equation is valid only for turbulent flow (rough, smooth, and transition between
rough and smooth). In cases where the flow is in the laminar-turbulent transition region,
alternative approximations should be used (e.g., Cheng, 2008); however, such conditions are
rare in water engineering.
The equivalent sand roughness, ks , of several commercial pipe materials is given in
Table 2.1. These values of ks apply to clean new pipe only; pipe that has been in service
for a long time usually experiences corrosion or scale buildup that results in values of ks
orders of magnitude larger than the values given in Table 2.1 (Echávez, 1997; Butler and
Davies, 2011). The rate of increase of ks with time depends primarily on the quality of the
water being transported, and the roughness coefficients for older water mains are usually
determined through field testing. The expression for the friction factor derived by Colebrook
(Equation 2.35) was plotted by Moody (1944) in what is commonly referred to as the Moody
diagram,∗ reproduced in Figure 2.3. The Moody diagram indicates that for Re … 2000, the
flow is laminar and the friction factor is given by
w.E
asy
En
gin
eer
in
64
f =
Re
g.n
et
(2.36)
which is commonly known as the Hagen–Poiseuille formula, which can be derived theoretically based on the assumption of laminar flow of a Newtonian fluid. For 2000 < Re … 4000
there is no fixed relationship between the friction factor and the Reynolds number or relative
roughness, and flow conditions are generally uncertain. Beyond a Reynolds number of 4000,
the flow is turbulent and the friction factor is controlled by the thickness of the laminar
boundary layer relative to the height of the roughness projections on the surface of the pipe.
The dashed line in Figure 2.3 indicates the boundary between the fully turbulent flow regime,
where f is independent of Re, and the transition regime, where f depends on both Re and
the relative roughness, ks /D. The equation of this dashed line is given by (Mott, 1994)
1
Re
( =
200(D/ks )
f
(2.37)
∗ This type of diagram was originally suggested by Blasius in 1913 and Stanton in 1914 (Stanton and Pannell,
1914). The Moody diagram is sometimes called the Stanton diagram.
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16
Chapter 2
Fundamentals of Flow in Closed Conduits
TABLE 2.1: Typical Equivalent Sand Roughness for Various Materials
Equivalent sand
roughness, ks
(mm)
Material
ww
Asbestos cement:
Coated
Uncoated
Brass
Brick
Concrete:
General
Steel forms
Wooden forms
Centrifugally spun
Clay
Copper
Corrugated metal
Glass
Iron:
Cast iron
General
Lined with asphalt
Uncoated
Coated
Ductile iron
Lined with bitumen
Lined with spun concrete
Galvanized iron
Wrought iron
Lead
Plastic
PVC
Steel
Coal-tar enamel
New unlined
Riveted
Wood stave
0.038
0.076
0.0015–0.003
0.6–6.0
0.3–3.0
0.18
0.6
0.13–0.36
0.03–0.15
0.0015–0.003
45
0.0015–0.003
w.E
asy
En
gin
eer
in
0.2–5.5
0.1–2.1
0.226
0.102
0.26
0.12
0.030–0.038
0.102–4.6
0.046–2.4
0.0015
0.0015–0.06
0.0015
g.n
et
0.0048
0.028–0.076
0.9–9.0
0.18
Sources: Butler and Davies (2011); Haestad et al., (2004); Haestad
Methods, Inc. (2002); Moody (1944); Sanks (1998).
The line in the Moody diagram corresponding to a relative roughness of zero describes the
friction factor for pipes that are hydraulically smooth.
Although the Colebrook equation (Equation 2.35) can be used to calculate the friction
factor, this equation has the drawback that it is an implicit equation for the friction factor
and must be solved iteratively. This minor inconvenience was circumvented by Swamee and
Jain (1976), who suggested the following explicit equation for the friction factor:
#
"
1
5.74
ks
ks /D
( = −2 log
, 10−6 …
+
… 10−2 , 5000 … Re … 108
(2.38)
0.9
3.7
D
f
Re
where, according to Swamee and Jain (1976), Equation 2.38 deviates by less than 1% from
the Colebrook equation within the entire turbulent-flow regime, provided that the restrictions on ks /D and Re are honored. The Swamee–Jain equation (Equation 2.38) can be more
conveniently written as
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Section 2.2
Transitional zone
0.038
0.036
=
0.034
Rough turbulent zone
0.01
64
Re
0.008
0.032
0.030
0.006
Laminar flow
0.004
Friction factor, f
0.028
ww
0.026
0.024
0.002
0.022
0.020
0.001
0.018
0.0006
0.016
0.0004
Smooth pipes
0.014
0.0002
w.E
asy
En
gin
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in
0.0001
0.012
0.00005
0.010
0.008
17
1
3
10
Relative roughness, ks /D
FIGURE 2.3: Moody
diagram
Source: Moody (1944).
Single Pipelines
2
3 4 5 6 7891
104
2
3 4 5 6 7891
105
2
3 4 5 6 7891
106
2
0.00001
3 4 5 6 7891
107
2
3 4
Reynolds number, Re
f =-
0.25
"
5.74
ks
+
log
3.7D
Re0.9
(2.39)
#.2
In addition to the Swamee–Jain equation, several other explicit equations have been proposed that approximate, to within about 1%, the friction factor given by the Colebrook
equation (e.g., Haaland, 1983; Sonnad and Goudar, 2006). Uncertainties in the relative
roughness and in the data used to produce the Colebrook equation make the use of severalplace accuracy in pipe-flow problems unjustified. As a rule of thumb, an accuracy of 10% in
calculating the friction factor from either the Colebrook equation or close approximations is
to be expected (Gerhart et al., 1992; Munson et al., 1994).
EXAMPLE 2.2
g.n
et
Water from a treatment plant is pumped into a distribution system at a rate of 4.38 m3 /s, a pressure of
480 kPa, and a temperature of 20◦ C. The pipe has a diameter of 750 mm and is made of ductile iron.
Estimate the pressure 200 m downstream of the treatment plant if the pipeline remains horizontal.
Compare the friction factor estimated using the Colebrook equation to the friction factor estimated
using the Swamee–Jain equation. After 20 years in operation, scale buildup is expected to cause the
equivalent sand roughness of the pipe to increase by a factor of 10. Determine the effect on the water
pressure 200 m downstream of the treatment plant.
Solution According to the Darcy–Weisbach equation, the difference in total head, 'h, between the
upstream section (at exit from treatment plant) and the downstream section (200 m downstream from
the upstream section) is given by
fL V 2
'h =
D 2g
where f is the friction factor, L is the pipe length between the upstream and downstream sections
(= 200 m), D is the pipe diameter (= 750 mm), and V is the velocity in the pipe. The velocity, V, is
given by
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18
Chapter 2
Fundamentals of Flow in Closed Conduits
Q
A
where Q is the flow rate in the pipe (= 4.38 m3 /s) and A is the area of the pipe cross section given by
V=
A=
π 2
π
D = (0.75)2 = 0.442 m2
4
4
The pipeline velocity is therefore
V=
ww
4.38
Q
=
= 9.91 m/s
A
0.442
The friction factor, f , in the Darcy–Weisbach equation is calculated using the Colebrook equation:
.
1
ks
2.51
(
( = −2 log
+
3.7D
f
Re f
Here Re is the Reynolds number and ks is the equivalent sand roughness of ductile iron (= 0.26 mm).
The Reynolds number is given by
VD
Re =
ν
where ν is the kinematic viscosity of water at 20◦ C, which is equal to 1.00 * 10−6 m2 /s. Therefore
w.E
asy
En
gin
eer
in
Re =
(9.91)(0.75)
VD
=
= 7.43 * 106
ν
1.00 * 10−6
Substituting into the Colebrook equation leads to
.
2.51
1
0.26
( = −2 log
(
+
(3.7)(750)
f
7.43 * 106 f
which yields
f = 0.016
The head loss, 'h, between the upstream and downstream sections can now be calculated using the
Darcy–Weisbach equation as
'h =
(0.016)(200) (9.91)2
fL V 2
=
= 21.4 m
D 2g
0.75
(2)(9.81)
Using the definition of head loss, 'h,
'h =
"
p1
+ z1
γ
#
−
"
p2
+ z2
γ
#
g.n
et
where p1 and p2 are the upstream and downstream pressures, γ is the specific weight of water, and z1
and z2 are the upstream and downstream pipe elevations. Since the pipe is horizontal, z1 = z2 and 'h
can be written in terms of the pressures at the upstream and downstream sections as
p
p
'h = 1 − 2
γ
γ
In this case, p1 = 480 kPa, γ = 9.79 kN/m3 , and therefore
21.4 =
p2
480
−
9.79
9.79
which yields
p2 = 270 kPa
Therefore, the pressure 200 m downstream of the treatment plant is 270 kPa. The Colebrook equation
required that f be determined from an implicit equation, but the explicit Swamee–Jain approximation
for f is given by
0
/
5.74
ks
1
( = −2 log
+
3.7D
f
Re0.9
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Section 2.2
Single Pipelines
19
Substituting for ks , D, and Re gives
which yields
.
0.26
5.74
1
( = −2 log
+
(3.7)(750)
(7.43 * 106 )0.9
f
f = 0.016
ww
This is the same friction factor obtained using the Colebrook equation within an accuracy of two significant digits.
After 20 years, the equivalent sand roughness, ks , of the pipe is 2.6 mm, the (previously calculated) Reynolds number is 7.43 * 106 , and the Colebrook equation gives
.
2.6
1
2.51
( = −2 log
(
+
(3.7)(750)
f
7.43 * 106 f
which yields
f = 0.027
w.E
asy
En
gin
eer
in
The head loss, 'h, between the upstream and downstream sections is given by the Darcy–Weisbach
equation as
(0.027)(200) (9.91)2
fL V 2
=
= 36.0 m
'h =
D 2g
0.75
(2)(9.81)
Hence the pressure, p2 , 200 m downstream of the treatment plant is given by the relation
p
p
'h = 1 − 2
γ
γ
where p1 = 480 kPa, γ = 9.79 kN/m3 , and therefore
36.0 =
p2
480
−
9.79
9.79
which yields
p2 = 128 kPa
g.n
et
Therefore, pipe aging over 20 years will cause the pressure 200 m downstream of the treatment plant
to decrease from 270 kPa to 128 kPa. This is quite a significant drop and shows why velocities as high
as 9.91 m/s are not used in these types of pipelines, even for short lengths of pipe.
The previous example (Example 2.2) illustrates the case where the flow rate through a
pipe is known and the objective is to calculate the pressure drop over a given length of the
pipe. The approach is summarized as follows: (1) calculate the Reynolds number, Re, and
the relative roughness, ks /D, from the given data; (2) use the Colebrook equation (Equation 2.35) or Swamee–Jain equation (Equation 2.38) to calculate f ; and (3) use the calculated
value of f to calculate the head loss from the Darcy–Weisbach equation (Equation 2.33), and
the corresponding pressure drop from Equation 2.22.
Flow rate for a given head loss. In many cases, the flow rate through a pipe is not controlled but attains a level that matches the pressure drop available. For example, the flow
rate through a faucet in home plumbing is determined by the pressure drop in the service
line between the pressurized water main at one end and the atmosphere at the other end.
A useful approach to this problem that uses the Colebrook
equation has been suggested by
(
Fay (1994), where the first step is to calculate Re f using the rearranged Darcy–Weisbach
equation
&
'1
(
2ghf D3 2
Re f =
(2.40)
ν2L
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20
Chapter 2
Fundamentals of Flow in Closed Conduits
(
Using this value of Re f , solve for Re using the rearranged Colebrook equation
&
'
(
ks /D
2.51
(
Re = −2.0(Re f ) log
+
3.7
Re f
(2.41)
Using this value of Re, the flow rate, Q, can then be calculated by
Q=
1
1
π D2 V = π DνRe
4
4
(2.42)
This approach must necessarily be validated by verifying that Re > 4000, which is required
for application of the Colebrook equation. Equations 2.40 to 2.42 can be combined to yield
the following combined form of the Darcy–Weisbach and Colebrook equations
ww
Q = −0.965D2
1
&
'
gDhf
ks /D
1.784ν
ln
+ (
L
3.7
D gDhf /L
w.E
asy
En
gin
eer
in
(2.43)
This equation is dimensionally homogeneous, and is particularly useful since the flow rate,
Q, can be explicitly calculated for given values of D, hf , L, and ks .
EXAMPLE 2.3
A 50-mm diameter galvanized iron service pipe is connected to a water main in which the pressure is
450 kPa gage. If the length of the service pipe to a faucet is 40 m and the faucet is 1.2 m above the main,
estimate the flow rate when the faucet is fully open.
Solution The head loss, hf , in the pipe is estimated by
#
"
#
"
poutlet
pmain
+ zmain −
+ zoutlet
hf =
γ
γ
where pmain = 450 kPa, zmain = 0 m, poutlet = 0 kPa, and zoutlet = 1.2 m. Therefore, taking γ =
9.79 kN/m3 (at 20◦ C) gives
#
"
2
3
450
+ 0 − 0 + 1.2 = 44.8 m
hf =
9.79
g.n
et
Also, since D = 50 mm, L = 40 m, ks = 0.15 mm (from Table 2.1), and ν = 1.00 * 10−6 m2 /s (at 20◦ C),
the combined Darcy–Weisbach and Colebrook equation (Equation 2.43) yields
⎛
⎞
4
gDh
k
1.784ν
/D
f
⎜ s
⎟
ln ⎝
+ 8
Q = −0.965D2
⎠
L
3.7
D gDh /L
f
1
= −0.965(0.05)2
-
0.15/50
1.784(1.00 * 10−6 )
(9.81)(0.05)(44.8)
(
ln
+
40
3.7
(0.05) (9.81)(0.05)(44.8)/40
.
= 0.0126 m3 /s = 12.6 L/s
The faucet can therefore be expected to deliver 12.6 L/s when fully open.
Diameter for a given flow rate and head loss. In many cases, the design engineer must select
a size of pipe to provide a given level of service. For example, the minimum required flow rate
for an available head loss might be specified for a water delivery pipe, and the design engineer
desires to calculate the minimum diameter pipe that will satisfy these design constraints. This
is the usual design problem in gravity-driven pipe networks (e.g., Jones, 2011). Solution of this
design problem necessarily requires a numerical procedure and can be easily accomplished
by using either a programmable calculator or a spreadsheet with solver capability.
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Section 2.2
Single Pipelines
21
EXAMPLE 2.4
A galvanized iron service pipe from a water main is required to deliver 200 L/s during a fire. If the
length of the service pipe is 35 m and the head loss in the pipe is not to exceed 50 m, calculate the
minimum pipe diameter that can be used.
Solution From the given data: Q = 200 L/s = 0.200 m3 /s, L = 35 m, and hf = 50 m. Assume that
ν = 10−6 m2 /s (at 20◦ C) and take ks = 0.15 mm (from Table 2.1). Substituting these data into Equation
2.43 gives
ww
Q = −0.965D2
4
0.2 = −0.965D2
4
gDhf
L
⎡
⎤
1.784ν ⎥
⎢ ks /D
ln ⎣
+ 8
⎦
3.7
D gDh /L
f
.
(9.81)D(50)
0.00015
1.784(10−6 )
ln
+ (
(35)
3.7D
D (9.81)D(50)/(35)
This is an implicit equation in D that can be solved numerically to yield D = 136 mm.
w.E
asy
En
gin
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in
Most numerical procedures that could be used for obtaining D in the previous example
converge fairly quickly, and do not pose any computational difficulties. In lieu of determining
an exact numerical solution, Swamee and Jain (1976) have suggested the following explicit
formula for calculating the pipe diameter, D:
⎡
D = 0.66 ⎣k1.25
s
&
LQ2
ghf
'4.75
+ νQ9.4
&
L
ghf
'5.2 ⎤0.04
⎦
(2.44)
Equation 2.44 is dimensionally homogeneous, is valid for 4000 … Re … 3 * 108 and 10−6 …
ks /D … 2 * 10−2 , and will yield a D within 5% of the value obtained by an exact solution to
the Darcy–Weisbach and Colebrook equations. Use of Equation 2.44 is illustrated by repeating the previous example.
EXAMPLE 2.5
g.n
et
A galvanized iron service pipe from a water main is required to deliver 200 L/s during a fire. If the length
of the service pipe is 35 m, and the head loss in the pipe is not to exceed 50 m, use the Swamee–Jain
approximation given by Equation 2.44 to calculate the minimum pipe diameter that can be used.
Solution Since ks = 0.15 mm, L = 35 m, Q = 0.2 m3 /s, hf = 50 m, ν = 1.00 * 10−6 m2 /s, the
Swamee–Jain approximation (Equation 2.44) gives
⎡
D = 0.66 ⎣k1.25
s
&
LQ2
ghf
'4.75
+ νQ9.4
&
L
ghf
'5.2 ⎤0.04
⎦
⎧
.4.75
.5.2⎫0.04
⎨
⎬
(35)(0.2)2
35
1.25
−6
9.4
= 0.66 (0.00015)
+ (1.00 * 10 )(0.2)
⎩
⎭
(9.81)(50)
(9.81)(50)
= 0.140 m
The calculated pipe diameter (140 mm) is about 3% higher than calculated by simultaneous solution of
the Darcy–Weisbach and Colebrook equations (i.e., 136 mm).
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22
Chapter 2
Fundamentals of Flow in Closed Conduits
2.2.3
Steady-State Energy Equation
General conditions between two sections of a closed conduit are illustrated in Figure 2.4.
Between the upstream and downstream sections of the conduit (Sections 1 and 2) there can
be an exchange of heat between the fluid and the conduit, and there can also be a machine
such as a pump or turbine that inputs or extracts energy from the flowing fluid. The steadystate energy equation for the control volume illustrated in Figure 2.4 is given by the first law
of thermodynamics as
dW
dQh
−
=
dt
dt
ww
!
A
ρe v · n dA
(2.45)
where Qh is the heat added to the fluid in the control volume [E∗ ], W is the work done by
the fluid in the control volume [E], A is the surface area of the control volume [L2 ], ρ is the
density of the fluid in the control volume [ML−3 ], and e is the internal energy per unit mass
of fluid in the control volume [EM−1 ] given by
e = gz +
v2
+ u
2
w.E
asy
En
gin
eer
in
(2.46)
where z [L] is the elevation of the fluid mass having a velocity v [LT−1 ] and internal energy
u [E]. By convention, the heat added to a system and the work done by a system are positive
quantities. 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 boundaries of the control
volume. As the fluid moves across the control surface with velocity v, the power (= rate of
doing work) expended by the fluid against the external pressure forces is given by
!
dWp
=
pv · n dA
(2.47)
dt
A
where Wp is the work done against external pressure forces [E]. The work done by a fluid
in the control volume is typically separated into work done against external pressure forces,
Wp , plus work done against rotating surfaces, Ws , commonly referred to as the shaft work.
The rotating element is called a rotor in a gas or steam turbine, an impeller in a pump, and a
runner in a hydraulic turbine. The rate at which work is done by a fluid system, dW/dt, can
therefore be written as
!
dWp
dW
dWs
dWs
=
+
=
(2.48)
pv · n dA +
dt
dt
dt
dt
A
FIGURE 2.4: Energy
balance in closed conduit
Heat flux, Qh
1
Shaft work, Ws
g.n
et
2
Inflow, Q
Outflow, Q
Control
volume
∗ E = ML2 T−2 denotes unit of energy, e.g., joule.
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Section 2.2
Single Pipelines
23
Combining Equation 2.48 with the steady-state energy equation (Equation 2.45) leads to
"
#
!
dQh
p
dWs
ρ
−
=
+ e v · n dA
(2.49)
dt
dt
ρ
A
Substituting the definition of the internal energy, e, given by Equation 2.46 into Equation 2.49
yields
&
'
!
dWs
dQh
v2
−
=
ρ h + gz +
v · n dA
(2.50)
dt
dt
2
A
where h is the enthalpy of the fluid defined by
h=
ww
p
+ u
ρ
(2.51)
# and the rate at
Denoting the rate at which heat is being added to the fluid system by Q,
# s , the
which work is being done against moving impervious boundaries (shaft work) by W
energy equation can be written in the form
&
'
!
v2
#Q − W
# s=
ρ h + gz +
v · n dA
(2.52)
2
A
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Consider the term h + gz, where
"
#
p
p
+ u + gz = g
+ z + u
h + gz =
ρ
γ
(2.53)
and γ is the specific weight of the fluid. Equation 2.53 indicates that h + gz can be assumed
to be constant across the inflow and outflow openings illustrated in Figure 2.4, since a hydrostatic pressure distribution across the inflow/outflow boundaries guarantees that p/γ + z is
constant across the inflow/outflow boundaries normal to the flow direction, 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 2.52 can be integrated to yield
!
!
!
v2
# − W
# s = (h1 + gz1 )
Q
ρv · n dA +
ρ v · n dA + (h2 + gz2 )
ρv · n dA
2
A1
A1
A2
!
v2
+
ρ v · n dA
2
A2
!
!
!
v3
= −(h1 + gz1 )
ρv1 dA −
ρ 1 dA + (h2 + gz2 )
ρv2 dA
2
A1
A1
A2
!
v3
(2.54)
+
ρ 2 dA
2
A2
g.n
et
where the subscripts 1 and 2 refer to the inflow and outflow boundaries, respectively, and
the negative signs result from the fact that the unit normal points out of the control volume,
causing v · n to be negative on the inflow boundary and positive on the outflow boundary.
Equation 2.54 can be further simplified by noting that the assumption of steady flow
requires that the rate of mass inflow to the control volume is equal to the rate of mass outflow,
and denoting the mass flow rate by m,
# the continuity equation requires that
!
!
m
# =
ρv1 dA =
ρv2 dA
(2.55)
A1
A2
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24
Chapter 2
Fundamentals of Flow in Closed Conduits
Furthermore, the constants α1 and α2 can be defined by the equations
!
V3
v3
ρ dA = α1 ρ 1 A1
2
2
A1
!
V3
v3
ρ dA = α2 ρ 2 A2
2
2
A2
ww
(2.56)
(2.57)
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 constants α1 and α2
are determined by the velocity distributions across the flow boundaries, and these constants
are called kinetic energy correction factors. If the velocity is constant across a flow boundary,
then it is clear from Equation 2.56 that the kinetic energy correction factor for that boundary
is equal to unity; for any other velocity distribution, the kinetic energy factor is greater than
unity. Combining Equations 2.54 to 2.57 leads to
# − W
# s = −(h1 + gz1 )m
Q
# − α1 ρ
V13
V3
A1 + (h2 + gz2 )m
# + α2 ρ 2 A2
2
2
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(2.58)
Invoking the continuity equation requires that
#
ρV1 A1 = ρV2 A2 = m
(2.59)
and combining Equations 2.58 and 2.59 leads to
⎡&
'
&
'⎤
2
2
V
V
# − W
# s=m
# ⎣ h2 + gz2 + α2 2 − h1 + gz1 + α1 1 ⎦
Q
2
2
which can be put in the form
&
'
&
'
#
# s
V12
V22
p2
u2
p1
u1
W
Q
+
+ z2 + α2
+
+ z1 + α1
−
=
−
γ
g
2g
γ
g
2g
mg
#
mg
#
(2.60)
(2.61)
and can be further rearranged into the useful form
⎤
⎡
.
' &
'
&
#
# s
V12
V22
W
p2
1
p1
Q
⎦ +
(2.62)
+ α1
+ z1 =
+ α2
+ z2 + ⎣ (u2 − u1 ) −
γ
2g
γ
2g
g
mg
#
mg
#
g.n
et
Two key terms can be identified in Equation 2.62: the (shaft) work done by the fluid
per unit weight, hs , defined by the relation
hs =
# s
W
mg
#
(2.63)
and the energy loss per unit weight, commonly called the head loss, hL , defined by the
relation
#
1
Q
(2.64)
hL = (u2 − u1 ) −
g
mg
#
Combining Equations 2.62 to 2.64 leads to the most common form of the steady-state energy
equation for flow in closed conduits
&
V2
p1
+ α1 1 + z1
γ
2g
'
=
&
V2
p2
+ α2 2 + z2
γ
2g
'
+ hL + hs
(2.65)
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Section 2.2
Single Pipelines
25
where a positive head loss indicates an increase in internal energy (manifested by an increase
in temperature) and/or a loss of heat, and a positive value of hs is associated with work being
done by the fluid, such as in moving a turbine runner. Some practitioners incorrectly refer
to Equation 2.65 as the Bernoulli equation, which bears some resemblance to Equation 2.65
but is different in several important respects. Fundamental differences between the energy
equation and the Bernoulli equation are that the Bernoulli equation is derived from the
momentum equation, which is independent of the energy equation, the Bernoulli equation
does not account for fluid friction, and the Bernoulli equation is applicable only for flow
along a streamline.
2.2.3.1
Energy and hydraulic grade lines
The total head, h, [L] of a fluid at any cross section of a pipe is defined by
ww
h=
p
V2
+ α
+ z
γ
2g
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(2.66)
where p is the pressure in the fluid at the centroid of the cross section [FL−2 ], γ is the specific
weight of the fluid [FL−3 ], α is the kinetic energy correction factor [dimensionless], V is the
average velocity across the pipe cross section [LT−1 ], and z is the elevation of the centroid of
the pipe cross section [L]. The total head measures the average mechanical energy per unit
weight of the fluid flowing across a pipe cross section. The energy equation, Equation 2.65,
states that changes in the total head along the pipe are described by
h(x + 'x) = h(x) − (hL + hs )
(2.67)
where x is the coordinate measured along the pipe centerline, 'x is the distance between
two cross sections in the pipe, hL is the head loss, and hs is the shaft work done by the fluid
over the distance 'x. The practical application of Equation 2.67 is illustrated in Figure 2.5,
where the head loss, hL , between two sections a distance 'x apart is indicated. At each
FIGURE 2.5: Head loss
along pipe
Head loss, hL
α1
V 12
2g
α2
fluid level
2g
p
1
p1
2
2
γ
g.n
et
Energy grade line (EGL)
V 22
Q
Hydraulic grade line (HGL)
γ
∆x
Q
z1
z2
Datum
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26
Chapter 2
Fundamentals of Flow in Closed Conduits
ww
FIGURE 2.6: Pump effect
on flow in pipeline
cross section, the total (mechanical) energy, h, is plotted relative to a defined datum, and
the locus of these points is called the energy grade line. The energy grade line at each pipe
cross section is located a distance p/γ + αV 2 /2g vertically above the centroid of the cross
section, and between any two cross sections the elevation of the energy grade line falls by
a vertical distance equal to the head loss, hL , plus the shaft work, hs , done by the fluid. The
hydraulic grade line measures the hydraulic head p/γ + z at each pipe cross section. It
is located a distance p/γ above the pipe centroid and indicates the elevation to which the
fluid would rise in an open tube (piezometer) connected to the wall of the pipe section. The
hydraulic grade line is therefore located a distance αV 2 /2g below the energy grade line. In
most water-supply applications, the velocity heads are negligible and the hydraulic grade line
closely approximates the energy grade line.
Both the hydraulic grade line and the energy grade line are useful in visualizing the
state of the fluid as it flows along the pipe and are frequently used in assessing the performance of fluid-delivery systems. Most fluid-delivery systems, for example, require that the
fluid pressure remain positive, in which case the hydraulic grade line must remain above the
pipe. In circumstances where additional energy is required to maintain acceptable pressures
in pipelines, a pump is installed along the pipeline to elevate the energy grade line by an
amount hs , which also elevates the hydraulic grade line by the same amount. This condition is illustrated in Figure 2.6. In cases where the pipeline upstream and downstream of the
pump are of the same diameter, the velocity heads αV 2 /2g both upstream and downstream
of the pump are the same, and the head added by the pump, hs , goes entirely to increase the
pressure head, p/γ , of the fluid. It should also be clear from Figure 2.5 that the pressure head
in a pipeline can be increased by simply increasing the pipeline diameter, which reduces the
velocity head, αV 2 /2g, and thereby increases the pressure head, p/γ , to maintain the same
approximately total energy at the pipe section. Expansion losses will cause some reduction
in total energy.
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hs
α1
α2
V 22
Energy grade line (EGL)
2g
2g
p
2
2
1
γ
p1
γ
g.n
et
Hydraulic grade line (HGL)
V 12
Pump
X
∆x
z1
z2
Datum
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Section 2.2
2.2.3.2
Single Pipelines
27
Velocity profile
The momentum and energy correction factors, α and β, depend on the cross-sectional velocity distribution. The velocity profile in both smooth and rough pipes of circular cross section
can be estimated by the semi-empirical equation
-
(
(
v(r) = (1 + 1.326 f ) − 2.04 f log
ww
"
R
R − r
#.
V
(2.68)
where v(r) is the velocity at a radial distance r from the centerline of the pipe, R is the radius
of the pipe, f is the friction factor, and V is the average velocity across the pipe.
The velocity distribution given by Equation 2.68 agrees well with velocity measurements in both smooth and rough pipes. This equation, however, is not applicable within the
small region close to the centerline of the pipe and is also not applicable in the small region
close to the pipe boundary. This is apparent since at the axis of the pipe dv/dr must be equal
to zero, but Equation 2.68 does not have a zero slope at r = 0. At the pipe boundary v must
be equal to zero, but Equation 2.68 gives a velocity of zero at a small distance from the wall,
with a velocity of −q at r = R. The energy and momentum correction factors, α and β,
derived from the velocity profile given by Equation 2.68 are (Moody, 1950)
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α = 1 + 2.7f
(2.69)
β = 1 + 0.98f
(2.70)
Another commonly used equation to describe the velocity distribution in turbulent pipe
flow is the empirical power law equation given by
"
#1
r n
v(r) = V0 1 −
R
(2.71)
where V0 is the centerline velocity and n is a function of the Reynolds number, Re. Values of
n are typically in the range of 6–10 and can be approximated by (Schlichting, 1979; Fox and
McDonald, 1992)
n = 1.83 log Re − 1.86
g.n
et
(2.72)
The power law is not applicable within 0.04R of the wall, since the power law gives an infinite
velocity gradient at the wall. Although the power-law profile fits measured velocities close
to the centerline of the pipe, it does not give zero slope at the centerline. The kinetic energy
coefficient, α, derived from the power law equation is given by
α=
(1 + n)3 (1 + 2n)3
4n4 (3 + n)(3 + 2n)
(2.73)
For n between 6 and 10, α varies from 1.08 to 1.03.
For turbulent flow in pipes, α is usually in the range of 1.03–1.06 and β in the range of
1.005–1.05. In most engineering applications, α and β are taken as unity since errors introduced by these assumptions are usually negligible.
2.2.3.3
Head losses in transitions and fittings
The head losses in straight pipes of constant diameter are primarily caused by friction
between the moving fluid and the pipe boundary and are estimated using the Darcy–
Weisbach equation. Flow through pipe fittings, around bends, and through changes in pipeline
geometry causes additional head losses, h0 , that are quantified by an equation of the form
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28
Chapter 2
Fundamentals of Flow in Closed Conduits
FIGURE 2.7: Loss
coefficients for
transitions and fittings
Description
Sketch
Pipe entrance
D
Additional Data
K
r/D
K
0.0
0.1
>0.2
0.50
0.12
0.03
V
r
K
Reentrant flow
V
0.8 –1
K
Pipe exit
ww
V
1
V
1
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D2/D1
Contraction
D1
D2
θ
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
90° smooth 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
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
g.n
et
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
0.4 – 0.9
1.7 – 2.0
0.9 –1.5
0.3 – 0.4
h0 =
$
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
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Section 2.2
ww
Single Pipelines
29
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.
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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
P
3m
100-mm PVC (pump to reservoir)
50-mm PVC (well to pump)
g.n
et
5m
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
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30
Chapter 2
Fundamentals of Flow in Closed Conduits
π 2
π
D = (0.05)2 = 0.001963 m2
4 1
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 =
ww
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 = /
0
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
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and for the 100-mm pipe
Re2 =
which leads to
(0.637)(0.10)
V2 D2
=
= 6.37 * 104
ν
1.00 * 10−6
f2 = -
0.25
5.74
log10
(6.37 * 104 )0.9
.2 = 0.0197
g.n
et
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.
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Section 2.2
2.2.3.4
Single Pipelines
31
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
ww
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
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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).
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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)
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32
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
ww
f = -
0.25
"
5.74
ks /4R
+
log
3.7
Re0.9
= ⎡
#.2
which yields
0.25
'⎤2
0.00120
5.74
⎣log
⎦
+
3.7
(4.00 * 106 )0.9
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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
g.n
et
(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
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Section 2.2
Single Pipelines
33
TABLE 2.2: Pipe Roughness Coefficients
CH
Pipe material
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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
—
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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 Garcı́a, 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
g.n
et
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)
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34
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
ww
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
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(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
D3
= 6.35
(0.013)2 (500)(1)2
4
(0.15) 3
= 6.73 m
(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
g.n
et
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
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Section 2.2
Single Pipelines
35
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.
ww
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
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$
!
!
!
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 =
g.n
et
!
[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
Flow velocity V
po
ρ
c
po + ∆ p
ρ + ∆ρ
Rapidly
closed value
x
V=0
V
Control
volume
Pressure
wave
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36
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
(2.89)
'p = ρ0 cV
ww
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
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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
Recalling the definition of the bulk modulus of elasticity, Ev , as
Ev =
'p
'ρ/ρ0
then Equations 2.92 and 2.93 combine to give
c=
VEv
'p
(2.91)
(2.92)
g.n
et
(2.93)
(2.94)
and substituting Equation 2.89 into Equation 2.94 yields wave speed, c, in terms of the fluid
properties as
4
Ev
c=
(2.95)
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).
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Section 2.2
Single Pipelines
37
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=
ww
FIGURE 2.10:
Propagation of pressure
wave
4
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
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Hydraulic grade line
po
(a)
Source Pipeline
V
L
c
∆p
po
V
(b)
V=o
Pressure wave
c
∆p
po
V
(c)
po
V
(d)
V=o
c
Closed valve
g.n
et
∆p
V=o
po
V
(e)
∆p
c
V=o
po
(f)
V
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38
Chapter 2
Fundamentals of Flow in Closed Conduits
ww
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
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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
g.n
et
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
4
4
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).
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Section 2.3
Pipe Networks
39
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
ww
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
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NP(j)
$
i=1
Qij − Fj = 0,
j = 1, . . . , NJ
(2.98)
g.n
et
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
B
Qa1
A
Qb2
Q5
B
Qb1
C
Loop
Node
(a)
Qd1
Q2
D
Qd2
C
(b)
Qc1
Qc2
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40
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 −
ww
"
$
fL
+
Km
D
#
Q|Q|
Q
+
hp
2
|Q|
2gA
(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.
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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
g.n
et
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
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Section 2.3
Pipe Networks
41
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)
Pipe 2: hC = hB − 2 2
= 87.3 −
D2 2gA2
0.40
(2)(9.81)(0.126)2
2
ww
= 87.3 − 85.2Q22
(2.101)
Q23
f L Q23
(0.0168)(650)
= 87.3 −
Pipe 3: hC = hB − 3 3
D3 2gA2
0.50
(2)(9.81)(0.196)2
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3
= 87.3 − 29.0Q23
(2.102)
Q24
f L Q24
(0.0156)(400)
Pipe 4: hD = hC − 4 4
= hC −
D4 2gA2
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
g.n
et
(2.106)
Q3 = 1.26 m3 /s
and from Equation 2.106
According to Equation 2.102
Q2 = 0.74 m3 /s
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)
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42
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)
ww
$
j=1
(hL,ij − hp,ij ) = 0,
i = 1, . . . , NL
w.E
asy
En
gin
eer
in
(2.107)
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
g.n
et
(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)
and n = 2. If the flow in each pipe is approximated as Q̂, and 'Q is the error in this estimate,
then the actual flow rate, Q, is related to Q̂ and 'Q by
(2.110)
Q = Q̂ + 'Q
and the head loss in each pipe is given by
hL = rQn
= r(Q̂ + 'Q)n
n
n−1
= r Q̂ + nQ̂
n(n − 1) n−2
'Q +
Q̂ ('Q)2 + · · · + ('Q)n
2
.
(2.111)
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Section 2.3
Pipe Networks
43
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
hL L rQ̂n + rnQ̂n−1 'Q
ww
(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
hL = rQ̂|Q̂|n−1 + rn|Q̂|n−1 'Q
w.E
asy
En
gin
eer
in
(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
NP(i)
NP(i)
$
j=1
rij Qj |Qj |n−1 + 'Qi
$
j=1
rij n|Qj |n−1 = 0,
i = 1, . . . , NL
(2.114)
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)
rij Qj |Qj |n−1
j=1
'Qi = − $NP(i)
j=1
nrij |Qj |n−1
g.n
et
(2.115)
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.
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44
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
20
r=6
ww
50
r=1
r=3
20
r=2
15
1
35
70
50
2
II
35
I
100
r=5
4
100
30
30
w.E
asy
En
gin
eer
in
(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)
r Qj |Qj |
ij
j=1
NP(i)
'Qi = − $
j=1
2rij |Qj |
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 = −
g.n
et
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
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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
45
The flow correction for loop II, 'QII , is therefore given by
'QII = −
−2799
= 11.1
253
and the corrected flows are
ww
Loop
Pipe
Q
II
1–2
2–3
3–1
26.1
−23.9
−2.7
w.E
asy
En
gin
eer
in
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
rQ|Q|
2r|Q|
2
I
4–1
1–3
3–4
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
47.7
1.4
−52.3
13,663
6
−13,666
3
573
8
523
1104
29.1
−20.9
1.4
847
−874
6
−21
58
84
8
150
47.7
1.5
−52.3
13,662
7
−13,668
1
573
9
523
1104
29.2
−20.8
1.5
853
−865
7
−5
58
83
9
150
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
'Q
−1.1
3.0
0.0
Corrected Q
47.7
1.6
−52.3
29.1
−20.9
1.4
g.n
et
47.7
1.4
−52.3
29.2
−20.8
1.5
0.1
0.0
47.7
1.5
−52.3
29.2
−20.8
1.5
0.0
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46
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.
ww
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.
w.E
asy
En
gin
eer
in
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
g.n
et
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
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Section 2.4
FIGURE 2.14: Types of
pumps
47
A
Impeller
Flow
Flow
Eye of
impeller
Eye of
impeller
Casing
A
ww
Pumps
View A–A
(a) Centrifugal pump
w.E
asy
En
gin
eer
in
ω
Flow
Outlet
vane
Inlet
vane
Input
shaft
Impeller
(b) Axial-flow pump
g.n
et
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)
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48
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
Gate valve (service)
Check valve
Foot valve
Flow
ww
Strainer
(a) Centrifugal pump
w.E
asy
En
gin
eer
in
Motor
Flow
Guide vanes
Impeller
Flow
Intake
(b) Axial-flow pump
FIGURE 2.16:
(a) Centrifugal pump;
and (b) axial-flow pump
(a)
g.n
et
(b)
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Section 2.4
Pumps
49
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
ghp
ww
or
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
w.E
asy
En
gin
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 = f3
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,
eer
in
g.n
et
FIGURE 2.17:
Performance curves of a
homologous series of
pumps
max
ghp
K1
P
ω2D 2
Efficiency
curve
K2
Q
ωD 3
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50
Chapter 2
Fundamentals of Flow in Closed Conduits
ghp
= K1
ω2 D2
Q
= K2
ωD3
and
(2.120)
Eliminating D from these equations yields
E
FK
F 2
=
G 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
ww
ns =
ωQ 2
3
(ghp ) 4
(2.122)
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
w.E
asy
En
gin
eer
in
1
Ns =
ωQ 2
3
4
(2.123)
hp
g.n
et
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
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Section 2.4
Pumps
51
TABLE 2.3: Pump Selection Guidelines
Type of pump
Range of
specific speeds, ns
(dimensionless)
Centrifugal
Mixed flow
Axial flow
0.15–1.5
1.5–3.7
3.7–5.5
Typical
flow rates
(L/s)
(gpm)
<60
60–300
>300
Typical
efficiencies
(%)
70–94
90–94
84–90
<950
950–4750
>4750
FIGURE 2.18: Axial-flow
pump operating in a
canal
Source: South Florida
Water Management
District.
ww
w.E
asy
En
gin
eer
in
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)
g.n
et
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)
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52
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
ww
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
w.E
asy
En
gin
eer
in
1 − η2
=
1 − η1
D1
D2
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.
g.n
et
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
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Section 2.4
1 − η2
=
1 − η1
1 − η2
=
1 − 0.85
"
"
D1
D2
Pumps
53
#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
ww
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
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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
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
30
73% (Efficiency)
(Pump curve)
10
7ft
10 ft
17.5 in
(Power curve)
15.7 in
50
25
0
1000
2000
0
250
500
3000
750
4000
1000
10
5
5000
6000
1250
kW
60
40
13.9 in
12.1 in
0
20
16 ft (Required NPSH)
hp
75
m
25
15
0ft
20
g.n
et
RPM 885
Model: 3409
Size: 14 16 -17
Imp. Dwg: 08217815
Pattern: P-2599
Eye Area: 171.0 in2
ft
17.5 in
A-7874-8
CDS
CENTRIFUGAL PUMP
CHARACTERISTICS
20
7000
1500
8000 gpm(US)
0
1750 m3/hr
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54
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 =
ww
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.
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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
hp = 'z + Q2
$
$ Km
fL
+
2gA2 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.
g.n
et
FIGURE 2.20: Pipeline
system
Q
∆z
P
Pump
Q
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Section 2.4
System curve
Pump
characteristic
curve
∆z
0
2.4.2.3
ww
55
Operating
point
Pump head, hp
FIGURE 2.21: Operating
point in pipeline system
Pumps
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
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NPSHA =
&
ps
V2
+ s + zs
γ
2g
'
−
"
pv
+ zs
γ
#
=
g.n
et
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 ,
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56
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
ww
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.
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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 = -
log
0.25
"
ks
5.74
3.7D + Re0.9
#.2
where Re is the Reynolds number given by
Re =
QD
VD
=
ν
Aν
(2.138)
g.n
et
(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
⎛
⎢
⎜ 4.6 * 10−5
⎢
⎢log ⎜
⎝ 3.7(0.2) + "
⎣
which simplifies to
f =
5.74
Q(0.2)
(0.03142)(1.00 * 10−6 )
(2.140)
⎞⎤ 2
⎟⎥
⎥
#0.9 ⎟
⎠⎥
⎦
1
(2.141)
4[log(6.216 * 10−5 + 4.32 * 10−6 Q−0.9 )]2
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)
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Section 2.4
Pumps
57
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
H
I2
Q
1375 15,850
hp
= 5.2 + .
"
3.281
H
I−0.9 # 2
Q
log 6.216 * 10−5 + 4.32 * 10−6 15,850
which simplifies to
hp = 17.1 +
ww
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:
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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
3
1
=
(885)(5000) 2
3
= 3232
g.n
et
(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 =
−5
4[log(6.216 * 10
+ 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
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58
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
16.6 in
70
15.7 in
60
14.8 in
73%
79%
83%
85%
86%
85%
83%
79%
73%
13.9 in
13 in
12 in
50
40
30
ww
55% 65%
A-7874-8
CDS
Centrifugal pump
characteristics
10
20
System curve
5
10 ft 16 ft
7ft
hp
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17.5 in
75
15.7 in
50
25
0
20
15
0ft
10
m
25
0
1000
2000
0
250
500
3000
4000
750
5000
1000
40
13.9 in
12.1 in
6000
1250
7000
1500
kW
60
20
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)
g.n
et
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
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Section 2.4
FIGURE 2.23: Pumps in
series
hp
Q
ww
Q
P
P
59
EGL
Two pumps in series
hp
Q
Pumps
hp
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.
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FIGURE 2.24: Pumps in
parallel
hp
g.n
et
EGL
Q
P
2Q
Q
P
2Q
hp
Two pumps in parallel
One pump
Pump system
Q
(a)
(b)
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60
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
ww
and the characteristic curve of the pump system is
Hp = 36 − 0.3Q2
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(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
g.n
et
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)
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Section 2.4
61
ω2 > ω1
Pump head, hp
FIGURE 2.25: Pump
curves for variable-speed
motor
Pumps
h2
h1
P2
P1
@ ω2
@ ω1
Q1 Q2
Flow rate, Q
then
ww
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
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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
g.n
et
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
which leads to
0.25h2 = 12 − 0.1(0.5Q2 )2
h2 = 48 − 0.1Q22
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62
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 at 20◦ C is flowing in a 100-mm diameter pipe at
an average velocity of 2 m/s. If the diameter of the pipe is
suddenly expanded to 150 mm, what is the new velocity
in the pipe? What are the volumetric and mass flow rates
in the pipe?
2.2. A 200-mm diameter pipe divides into two smaller pipes,
each of diameter 100 mm. If the flow divides equally
between the two smaller pipes and the velocity in the
200-mm pipe is 1 m/s, calculate the velocity and flow rate
in each of the smaller pipes.
2.3. The velocity distribution in a pipe is given by the equation
!
" #2 $
r
(2.149)
v(r) = V0 1 −
R
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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 =
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asy
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gin
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 momentum correction coefficient, β, for
the velocity distribution given in Equation 2.149.
2.5. Water is flowing in a horizontal 200-mm diameter pipe
at a rate of 0.06 m3 /s, and the pressures at sections 100 m
apart are equal to 500 kPa at the upstream section and
400 kPa at the downstream section. Estimate the average shear stress on the pipe and the friction factor, f .
2.6. Water at 20◦ C flows at a velocity of 2 m/s in a 250-mm
diameter horizontal ductile-iron pipe. Compare the friction factors derived from the Moody diagram, the Colebrook equation, and the Swamee–Jain equation. State
whether the flow is fully turbulent or not. 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
0.25
%
{log[(ks /D)/3.7 + 2.51/(Re f )]}2
Why is this equation termed “(slightly) more convenient”?
2.9. 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.
2.10. Water leaves a treatment plant in a 500-mm diameter
ductile-iron pipeline at a pressure of 600 kPa and at a
flow rate of 0.50 m3 /s. If the elevation of the pipeline at
the treatment plant is 120 m, estimate the pressure in the
pipeline 1 km downstream where the elevation is 100 m.
Assess whether the pressure in the pipeline would be
sufficient to serve the top floor of a 10-story building
(approximately 30 m high).
2.11. A 25-mm diameter galvanized iron service pipe is connected to a water main in which the pressure is 400 kPa.
If the length of the service pipe to a faucet is 20 m and
the faucet is 2.0 m above the main, estimate the flow rate
when the faucet is fully open.
2.12. A galvanized iron service pipe from a water main is
required to deliver 300 L/s during a fire. If the length of
the service pipe is 40 m and the head loss in the pipe is
not to exceed 45 m, calculate the minimum pipe diameter that can be used. Use the Colebrook equation in your
calculations.
2.13. Repeat Problem 2.12 using the Swamee–Jain approximation (Equation 2.44).
2.14. Use the velocity distribution given in Problem 2.3 to
estimate the kinetic energy correction factor, α, for turbulent pipe flow.
2.15. The velocity profile, v(r), for turbulent flow in smooth
pipes is sometimes estimated by the seventh-root law,
originally proposed by Blasius (1913):
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in
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et
"
r
v(r) = V0 1 −
R
#1
7
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Problems
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.
2.16. Show that the kinetic energy correction factor, α, corresponding to the power-law velocity profile is given by
Equation 2.73. Use this result to confirm your answer to
Problem 2.15.
2.17. Water enters and leaves a pump in pipelines of the
same diameter and approximately the same elevation. If
the pressure on the inlet side of the pump is 30 kPa and
a pressure of 500 kPa is desired for the water leaving the
pump, what is the head that must be added by the pump,
and what is the power delivered to the fluid?
2.18. Water leaves a reservoir at 0.06 m3 /s through a 200-mm
riveted steel pipeline that protrudes into the reservoir
and then immediately turns a 90◦ bend with a local
(minor) loss coefficient equal to 0.3. Estimate the length
of pipeline required for the friction losses to account
for 90% of the total losses, which includes both friction
losses and so-called “minor losses.” Would it be fair to
say that for pipe lengths shorter than the length calculated in this problem, the word “minor” should not be
used?
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A
2.20. 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
Transmission
line
P
Pump
Elev. 4 m
B
FIGURE 2.26: Water pumped into transmission line
head that must be added by the pump; (b) estimate 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.
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in
2.19. Water in a household plumbing system originates at the
neighborhood water main where the pressure is 480 kPa,
the velocity is 5 m/s, and the elevation is 2.44 m. A
19-mm (3/4-in.) copper service line supplies water to a
two-story residence where the faucet in the master bedroom is 40 m (of pipe) away from the main and at an elevation of 7.62 m. If the sum of the minor-loss coefficients
is 3.5, estimate the maximum (open faucet) flow. Is this
flow rate typical of a bathroom faucet? How would this
flow be affected by the operation of other faucets in the
house?
Elev. 10 m
Elev. 7 m
Supply
reservoir
63
g.n
et
Turbine
Open globe valve
40 m
T
125 m
FIGURE 2.27: Flow out of a reservoir
2.24. Water flows at 5 m3 /s in a 1 m * 2 m rectangular concrete
pipe. Calculate the head loss over a length of 100 m.
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?
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64
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 flows at a velocity of 2 m/s in a 300-mm
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. Calculate the Hazen–Williams roughness coefficient and the Manning coefficient that should
be used to obtain the same head loss as the Darcy–
Weisbach equation.
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 100-m long pipe at 3 m/s. If the water
temperature is 20◦ C, determine the minimum valveclosure time to avoid creating water-hammer pressures.
What is the maximum water-hammer pressure that can
occur? How is this pressure affected if the water temperature drops to 10◦ C?
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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
Elev.
80 m
LAJ = 900 m
B
J
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in
2.35. Based on your result in Problem 2.34, do you think that
water hammer can be a serious problem in household
plumbing?
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?
LBJ = 800 m
Elev.
60 m
C
LJC = 700 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
g.n
et
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
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Problems
Flow
A
C
P
D
B
Flow
E
Flow
F
Flow
G
Flow
65
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:
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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
0.5 m3/s
A
w.E
asy
En
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in
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
F
Pipe
L
(m)
D
(mm)
AB
BC
CD
DE
EF
FA
BE
1000
750
800
700
900
900
950
300
325
200
250
300
250
350
0.5 m3/s
E
D
FIGURE 2.31: Two-loop pipe network
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
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
P
150 m
100 m
0.06 m3/s
g.n
et
0.1 m3/s
150 m
0.07 m3/s
100 m
150 m
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.
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66
Chapter 2
Fundamentals of Flow in Closed Conduits
48,000 m3/d
8000 m /d
8000 m /d
3
8000 m /d
3
C
H
3
D
8000 m3/d
8000 m3/d
8000 m3/d
F
G
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E
FIGURE 2.33: Water-supply system
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.
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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 homologous series of centrifugal pumps is driven by
1200-rpm motor. For a 500-mm size, the best efficiency
of 81% occurs when the flow rate is 250 L/s and the total
dynamic head is 63.7 m. What is the best-efficiency operating point for a 250-mm size? Estimate the efficiency?
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
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.
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et
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
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Problems
%
59
%
%
55
57
%
%
53
%
50
45
%
55
% 5
3%
5.5 in.
120
5.0 in.
100
45
%
0
30
40
NPSH
20
0
20
REQ
20
40
60
80
100
10
5
10
w.E
asy
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in
0
NPSH in meters
140
60
ww
10
%
6.0 in.
80
20
59
57
160
NPSH in feet
Total head in feet
Total head in meters
%
6.5 in.
%
50
30
1½ × 1½ × 7B
3500 rpm
59.5%
200
180
40
%
220
60
50
35
7.0 in.
40
240
70
Bell & Gossett
Series 80-SC
25
%
260
67
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
100 m
P
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
g.n
et
hp = 20 − 4713Q2
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
10 m
E
3m
B
1m
Lower
reservoir
FIGURE 2.36: Water pumped from lower to upper reservoir
F
C
P
D
3m
A
FIGURE 2.37: Water-delivery system
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10 m
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68
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.
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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
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(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
50
43% 50%
55%
60%
4 in.
64%
3.75 in.
40
10
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
64%
3.5 in.
60%
55%
30
3.25 in.
3.0 in.
5
NPSHR - ft
10
0
50%
43%
20
0
g.n
et
5 ft
16 ft
6 ft
7 ft
0
0
20
40
10
60
80
100
20
Capacity
9 ft
120
140
30
160
180 gpm
40 m /h
3
FIGURE 2.38: Goulds Model 3656 pump curves
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Problems
C Elev. 2.00 m
p ≥ 350 kPa
C
P
P
B
A
P
69
Elev. 3.00 m
B
A
P
D Elev. 5.00 m
p ≥ 350 kPa
P
FIGURE 2.39: Parallel-pump 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:
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FIGURE 2.40: Water-supply system
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
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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
g.n
et
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.
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C H A P T E R
3
Design of Water-Distribution
Systems
3.1 Introduction
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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.
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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.
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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.
70
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Section 3.2
Water Demand
71
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
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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:
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N
!
"
#
qi * Pi
Q=
(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)
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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.
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72
Chapter 3
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TABLE 3.1: Typical Distribution of Per-Capita Water Demand
Category
Average use
(L/d) (gal/d)
Residential
Commercial
Industrial
Public
Loss
380
115
85
65
40
100
30
22
17
11
56
17
12
9
6
Total
685
180
100
Percent of total
Source: Solley (1998).
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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.
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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
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(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,
dP
dt
dP
dt
dP
dt
! k3 (Psat − P )
! k2
! k1P
Time, t
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(3.4)
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Section 3.2
Water Demand
73
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:
(3.6)
P(t) = P0 + k2 t
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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:
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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.
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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 ).
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74
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).
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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.
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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.
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(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
Population trend
350
(a)
(b)
(e)
(d)
C
300
B
A
250
200
Census data
150
100
1940
1960
1980
2000
2020
2040
Year
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Section 3.2
Water Demand
75
(b) Arithmetic growth is described by Equation 3.6 as
(3.13)
P(t) = P0 + k2 t
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
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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
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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
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(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 )
P0 P2 − P 21
=
2(225)(270)(330) − (270)2 (225 + 330)
(225)(330) − (270)2
= −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
=
1 + aebt
1 + 1.67e−0.0310t
(3.19)
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Chapter 3
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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.
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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.
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FIGURE 3.3: Typical daily
cycles in water demand
2.0
g.n
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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
14
16
18
20
22
24
Hour
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Section 3.2
Water Demand
77
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
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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)
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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
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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
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Chapter 3
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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
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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
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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.
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TABLE 3.5: Needed Fire Flow for One- and Two-Family Dwellings
Distance between buildings
(m)
(ft)
Needed fire flow
(L/min) (gpm)
>30
9.5–30
3.5–9.5
<3.5
2000
3000
4000
6000
>100
30–100
10–30
<10
500
750
1000
1500
Source: AWWA (1992).
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79
TABLE 3.6: Required Fire-Flow Durations
Required fire flow
(L/min)
(gpm)
<9000
11,000–13,000
15,000–17,000
19,000–21,000
23,000–26,000
26,000–30,000
30,000–34,000
34,000–38,000
38,000–45,000
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<2500
3000–3500
4000–4500
5000–5500
6000–7000
7000–8000
8000–9000
9000–10,000
10,000–12,000
Duration
(h)
2
3
4
5
6
7
8
9
10
Source: AWWA (1992).
EXAMPLE 3.2
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Estimate the flow rate and volume of water required to provide adequate fire protection to a 10-story
noncombustible building with an effective floor area of 8000 m2 .
Solution The NFF can be estimated by Equation 3.20 as
NFFi = Ci Oi (X + P)i
where the construction factor, Ci , is given by
&
Ci = 220 F Ai
For the 10-story building, F = 0.8 (Table 3.3, noncombustible, Class 3 construction), and Ai = 8000 m2 ,
hence
√
Ci = 220(0.8) 8000 = 16, 000 L/min
where Ci has been rounded to the nearest 1000 L/min. The occupancy factor, Oi , is given by Table 3.4
as 0.75 (C-1 noncombustible); (X + P)i can be estimated by the median value of 1.4; and hence the
needed fire flow, NFF, is given by
NFFi = (16, 000)(0.75)(1.4) = 17, 000 L/min
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This flow must be maintained for a duration of 4 hours (Table 3.6). Hence the required volume, V, of
water is given by
V = 17, 000 * 4 * 60 = 4.08 * 106 L = 4080 m3
Fire hydrants are placed throughout the service area to provide either direct hose
connections for firefighting or connections to pumper trucks, also known as fire engines. A
single-hose stream is generally taken as 1000 L/min (250 gpm), called the standard fire stream,
and hydrants are typically located at street intersections or spaced 60–150 m (200–600 ft)
apart. In high-value districts, additional hydrants may be necessary in the middle of long
blocks to supply the required fire flows.
3.2.4
Design Flows
Typical design capacities of various components of a water-supply system are given in
Table 3.7, where low-lift pumps refer to low-head, high-rate units that convey the raw-water
supply to the treatment facility and high-lift pumps deliver finished water from the treatment
facility into the distribution network at suitable pressures. The required capacities shown in
Table 3.7 consist of various combinations of the maximum daily demand, maximum hourly
demand, and the fire demand. Typically, the delivery pipelines from the water source to the
treatment plant, as well as the treatment plant itself, are designed with a capacity equal to
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TABLE 3.7: Typical Design Periods and Capacities in Water-Supply Systems
Design period
(years)
Component
1. Source of supply:
River
Wellfield
Reservoir
2. Conveyance:
Intake conduit
Conduit to treatment plant
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3. Pumps:
Low-lift
High-lift
Design capacity
Indefinite
10–25
25–50
Maximum daily demand
Maximum daily demand
Average annual demand
25–50
25–50
Maximum daily demand
Maximum daily demand
10
10
Maximum daily demand, one reserve unit
Maximum hourly demand, one reserve unit
4. Treatment plant
10–15
Maximum daily demand
5. Service reservoir
20–25
Working storage plus fire demand plus
emergency storage
6. Distribution system:
(a) Supply pipe or conduit
25–50
Greater of (1) maximum daily demand plus
fire demand, or (2) maximum hourly
demand
Same as for supply pipes
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(b) Distribution grid
Full development
Source: Gupta (2008).
the maximum daily demand. However, because of the high cost of providing treatment, some
utilities have trended toward using peak flows averaged over a longer period than 1 day to
design water-treatment plants (such as 2–5 days), and relying on system storage to meet peak
demands above treatment capacity. This approach has serious water-quality implications and
should be avoided if possible (AWWA, 2004). The flow rates and pressures in the distribution
system are analyzed under both maximum daily plus fire demand and the maximum hourly
demand, and the larger flow rate governs the design. Pumps are sized for a variety of conditions from maximum daily to maximum hourly demand, depending on their function in the
distribution system. Additional reserve capacity is usually installed in water-supply systems
to allow for redundancy and maintenance requirements.
EXAMPLE 3.3
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A metropolitan area has a population of 130,000 people with an average daily demand of 600 L/d/person.
If the needed fire flow is 20,000 L/min, estimate: (a) the design capacities for the wellfield and the
water-treatment plant; (b) the duration that the fire flow must be sustained and the volume of water
that must be kept in the service reservoir in case of a fire; and (c) the design capacity of the main supply
pipeline to the distribution system.
Solution
(a) The design capacity of the wellfield should be equal to the maximum daily demand (Table 3.7).
With a demand factor of 1.8 (Table 3.2), the per-capita demand on the maximum day is equal to
1.8 * 600 = 1080 L/day/person. Since the population served is 130,000 people, the design capacity
of the wellfield, Qwell , is given by
Qwell = 1080 * 130, 000 = 1.4 * 108 L/d = 1.62 m3 /s
The design capacity of the water-treatment plant is also equal to the maximum daily demand,
and therefore should also be taken as 1.62 m3 /s.
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81
(b) The needed fire flow, Qfire , is 20,000 L/min = 0.33 m3 /s. According to Table 3.6, the fire flow must
be sustained for 5 hours. The volume, Vfire , required for the fire flow will be stored in the service
reservoir and is given by
Vfire = 0.33 * 5 * 3600 = 5940 m3
(c) The required flow rate in the main supply pipeline is equal to the maximum daily demand plus
fire demand or the maximum hourly demand, whichever is greater.
Maximum daily demand + fire demand = 1.62 + 0.33 = 1.95 m3 /s
Maximum hourly demand =
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3.25
* 1.62 = 2.92 m3 /s
1.80
where a demand factor of 3.25 has been assumed for the maximum hourly demand. The main
supply pipe to the distribution system should therefore be designed with a capacity of 2.92 m3 /s.
The water pressure within the distribution system must be above acceptable levels when the
system demand is 2.92 m3 /s.
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3.3 Components of Water-Distribution Systems
Water-distribution systems consist of a network of pipelines and appurtenances designed
to deliver drinking-quality water at adequate flow rates and pressures to customers. The
methodology for calculating the distribution of flows and pressures in pipe networks was
covered in Section 2.3. The design and specification of key components of water-distribution
systems are covered in the following sections.
3.3.1
Pipelines
Water-distribution systems typically consist of connected pipe loops throughout the service
area. Pipelines in water-distribution systems include transmission lines, arterial mains, and
distribution mains. Transmission lines carry flow from the water-treatment plant to the service
area, typically have diameters greater than 600 mm (24 in.), and are usually on the order of
3 km (2 mi) apart. Arterial mains are connected to transmission mains and are laid out in
interlocking loops with the pipelines not more than 1 km (0.6 mi) apart and diameters in
the range of 400–500 mm (16–20 in.). Distribution mains form a grid over the entire service
area, with diameters in the range of 150–300 mm (6–12 in.), and supply water to every user.
A typical distribution main in a trench before backfilling is shown in Figure 3.4. Pipelines in
distribution systems are collectively called water mains, and a pipe that carries water from a
main to a building or property is called a service line.
FIGURE 3.4:
Water-distribution main
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FIGURE 3.5: Distribution
pipes crossing bridge
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Water mains are normally installed within the rights-of-way of streets, and dead ends
in water-distribution systems should be avoided whenever possible, since the lack of flow in
such lines can contribute to water-quality problems. At bridge-over-water crossings, water
mains are typically located above ground and alongside the bridge as shown in Figure 3.5. In
the particular case shown in Figure 3.5, there are two parallel water pipes, one for potable
water and the other for (recycled) non-potable water. Wherever such dual systems exist,
these systems are generally differentiated by the pipes being painted different colors; in the
present case the potable pipeline was green and the nonpotable pipeline was purple. Also
apparent in Figure 3.5 are the air-relief valves (located just above the pipe bends) that are
used to expel any air that might accumulate in the pipes.
Pipelines in water-distribution systems should be buried to a depth below the frost line
in northern climates and at a depth sufficient to cushion the pipe against traffic loads in
warmer climates. In warmer climates, a cover of 1.2–1.5 m (4–5 ft) is used for large mains
and 0.75–1.0 m (2.5–3 ft) for smaller mains. In areas where frost penetration is a significant
factor, mains can have as much as 2.5 m (8 ft) of cover. Trenches for water mains should be
as narrow as possible while still being wide enough to allow for proper joining and compaction around the pipe. The suggested trench width is the nominal pipe diameter plus 0.6 m
(2 ft); in deep trenches, sloping may be necessary to keep the trench wall from caving in.
Trench bottoms should be undercut 15–25 cm (6–10 in.), and sand, clean fill, or crushed
stone installed to provide a cushion against the bottom of the excavation, which is usually
rock. Standards for pipe construction, installation, and performance are published by the
American Water Works Association in its C-series standards, which are continuously being
updated.
Pipelines in water-distribution systems are typically designed with constraints relating
to the minimum pipe size and commercially available materials that will perform adequately
under operating conditions.
3.3.1.1
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Minimum size
The size of a water main determines its carrying capacity. Main sizes must be selected to
provide the capacity to meet peak domestic, commercial, and industrial demands in the service area, and must also provide for fire flow at the necessary pressure. For fire protection,
insurance underwriters typically require a minimum main size of 150 mm (6 in.) for residential areas and 200 mm (8 in.) for high-value districts (such as sports stadiums, shopping
centers, and libraries) if cross-connecting mains are not more than 180 m (600 ft) apart. On
principal streets, and for all long lines not connected at frequent intervals, 300-mm (12 in.)
and larger mains are required. Connections for fire hydrants require a minimum pipe size of
150 mm (6 in.).
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3.3.1.2
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Components of Water-Distribution Systems
83
Service lines
Service lines are pipes, including accessories, that carry water from the main to the point of
service, which is normally a meter setting or curb stop located at the property line. Service
lines can be any size, depending on how much water is required to serve a particular customer. Single-family residences are most commonly served with 20-mm (3/4-in.) diameter
service lines, while larger residences and buildings located far from the main connection
should have a 25-mm (1-in.) or larger service line. To properly size service lines it is essential
to know the peak demands than any service tap will be called on to serve. Materials used
for service-line pipe and tubing are typically either copper (tubing) or plastic, which includes
polyvinyl chloride (PVC) and polyethylene (PE). Type K copper is the most commonly used
material for copper service lines. Older service lines used lead and galvanized iron, which
are no longer recommended. The valve used to connect a small-diameter service line to
a water main is called a corporation stop, sometimes loosely referred to as a corporation
cock, corporation tap, or simply “corp” or “stop.” Tapping a water main and inserting a
corporation stop directly into the pipe wall requires a tapping machine, and taps are typically installed at the 10 or 2 o’clock position on the pipe. Guidelines for designing water
service lines and meters are given in AWWA Manual M22 (latest edition). Good construction
practices must be used when installing service lines to avoid costly repairs in the future. This
includes burying the pipe below frost lines, maintaining proper ditch conditions, proper backfill, trench compaction, and protection from underground structures that may cause damage
to the pipe.
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3.3.1.3
Pipe materials
Pipeline materials should be selected based on a consideration of service conditions, availability, properties of the pipe, and economics. In selecting pipe materials the following considerations should be taken into account:
◃ Most water-distribution mains in older cities in the United States are made of (gray)
cast-iron pipe (CIP), with many cities having CIP over 100 years old and still providing satisfactory service. Cast-iron pipe (CIP) is no longer manufactured in the United
States.
◃ For new distribution mains, ductile-iron pipe (DIP) is most widely used for pipe diameters up to 760 mm (30 in.). DIP has all the good qualities of CIP plus additional strength
and ductility.
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Before making the final selection of a pipe material, alternative pipe materials should be
considered and the rationale for selecting a particular pipe material should be clearly articulated. The key features of several pipe materials that are used in water-distribution systems
are given below.
Ductile iron. Ductile-iron pipe (DIP) is manufactured with diameters in the range of
76–1625 mm (3–64 in.). For diameters in the range of 100–500 mm (4–20 in.), standard
commercial sizes are available in 50-mm (2-in.) increments, while for diameters in the
range of 600–1200 mm (24–48 in.), the size increments are 150 mm (6 in.). The standard
lengths of DIP are 5.5 m (18 ft) and 6.1 m (20 ft). DIP is usually coated (outside and
inside) with a bituminous coating to minimize corrosion. An internal cement-mortar
lining 1.5–3 mm (0.06–0.12 in.) thick is common, and external polyethylene wraps are
used to reduce corrosion in corrosive soil environments. DIP used in water systems in
the United States is provided with a cement-mortar lining unless otherwise specified by
the purchaser. The design of DIP and fittings is covered in AWWA Manual M41 (latest
edition) and guidance for DIP lining is covered in AWWA Standard C105 (latest edition). Tests conducted by the Ductile Iron Pipe Research Association (DIPRA) suggest
that a Hazen–Williams C-value of 140 is appropriate for the design of cement-mortar
lined DIP. A variety of joints are available for use with DIP, which includes push-on
(the most common), mechanical, flanged, ball-and-socket, and numerous other joint
designs. A stack of DIP is shown in Figure 3.6, where the bell and spigot pipe ends
that facilitate push-on connection are apparent. A rubber gasket, to ensure a tight fit,
is contained in the bell side of the pipe.
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FIGURE 3.6: Ductile-iron
pipe
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Steel. Steel pipe usually compares favorably with DIP for diameters larger than 400 mm
(16 in.). As a consequence, steel pipe is primarily used for transmission lines in waterdistribution systems. Steel pipe is available in diameters in the range of 100–3600 mm
(4–144 in.). The standard length of steel pipe is 12.2 m (40 ft). The interior of steel pipe
is usually protected with either cement mortar or epoxy, and the exterior is protected by
a variety of plastic coatings, bituminous materials, and polyethylene tapes, depending
on the degree of protection required. Guidance for the design of steel pipe is covered
in AWWA Manual M11 (latest edition), and linings for steel pipe are covered under
AWWA Standard C205 (latest edition) and AWWA Standard C210 (latest edition).
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Plastic. Plastic materials used for fabricating water-main pipe include polyvinyl chloride
(PVC), polyethylene (PE), and polybutylene (PB). PVC pipe is by far the most widely
used type of plastic pipe material for small-diameter water mains and gravity-driven
water-distribution systems (e.g., Jones, 2011). The American Water Works Association
Standard (C900) for PVC pipe with diameters in the range of 100–300 mm (4–12 in.)
and laying lengths of 6.1 m (20 ft) is based on the same outside diameter as for DIP.
In this way, standard DIP fittings can be used with PVC pipe. PVC pipe is commonly
available with diameters in the range of 100–900 mm (4–36 in.). Extruded PE and PB
pipe is primarily used for water service pipe in small sizes; however, the use of PB
has decreased significantly because of structural difficulties caused by premature pipe
failures. The impact strength of PVC pipe is very low, so PVC pipes should be buried
where possible in a stabilized or backfilled trench with a depth of around 1 m (3 ft).
Research has documented that pipe materials such as PVC, PE, and PB may be subject to permeation by lower-molecular-weight organic solvents or petroleum products
(AWWA, 2002b). If a water pipe must pass through an area subject to contamination,
caution should be used in selecting PVC, PE, and PB pipes. In the hydraulic design of
PVC pipes, a roughness height of 0.0015 mm (= 1.5 µm) or a Hazen–Williams C-value
of 150 is appropriate for design (AWWA, 2002b). Note that the roughness height is
approximately the size of a bacterium! Details of large-diameter PE pipe are found
in AWWA Standard C906 (latest edition) and information on PVC water-main pipe is
available in AWWA Manual M23 (latest edition).
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Asbestos cement. Asbestos-cement (A-C) pipe has been widely installed in waterdistribution systems, especially in areas where metallic pipe is subject to corrosion,
such as in coastal areas. It has also been installed in remote areas, where its light weight
makes it much easier to transport and install than DIP. Common diameters are in the
range of 100–900 mm (4–36 in.). The U.S. Environmental Protection Agency banned
most uses of asbestos in 1989 and, due to the manufacturing ban, new A-C pipe is no
longer being installed in the United States.
Fiberglass. Fiberglass pipe is available for potable water in diameters in the range of
25–3600 mm (1–144 in.). Advantages of fiberglass pipe include corrosion resistance,
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light weight, low installation cost, ease of repair, and hydraulic smoothness. Disadvantages include susceptibility to mechanical damage, low modulus of elasticity, and
lack of a standard joining system. Fiberglass pipe is covered in AWWA Standard C950
(latest edition).
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Concrete. Concrete pressure pipe provides a combination of the high tensile strength of steel
and the high compressive strength and corrosion resistance of concrete. The pipe is
available in diameters in the range of 250–6400 mm (10–256 in.) and in standard lengths
in the range of 3.7–12.2 m (12–40 ft). The design of concrete pressure pipe is covered in
AWWA Manual M9 (latest edition). Concrete pipe is available with various types of liners and reinforcement; the four types in common use in the United States and Canada
are prestressed concrete cylinder pipe, bar-wrapped concrete cylinder pipe, reinforced
concrete cylinder pipe, and reinforced concrete noncylinder pipe. The manufacture of
prestressed concrete cylinder pipe is covered under AWWA Standard C301 (latest edition) and AWWA Standard C304 (latest edition), bar-wrapped concrete cylinder pipe
is covered under AWWA Standard C303 (latest edition), reinforced concrete cylinder
pipe is covered under AWWA Standard C300 (latest edition), and reinforced concrete
noncylinder pipe is covered under AWWA Standard C302 (latest edition).
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3.3.2
Pumps
Service pressures are typically maintained by pumps, with head losses and increases in
pipeline elevations acting to reduce pressures, and decreases in pipeline elevations acting to
increase pressures. When portions of the distribution system are separated by long distances
or significant changes in elevation, booster pumps are sometimes used to maintain acceptable
service pressures. In some cases, fire-service pumps are used to provide additional capacity
for emergency fire protection. Pumps operate at the intersection of the pump performance
curve and the system curve, and since the system curve is significantly affected by variations
in water demand, there is a significant variation in pump operating conditions. In most cases,
the range of operating conditions is too wide to be met by a single pump, and multiple-pump
installations or variable-speed pumps are required.
3.3.3
Valves
Valves in water-distribution systems are designed to perform several different functions.
Their primary functions are to isolate pipelines for maintenance and repair, regulate pressure
and throttle flow, prevent backflow, and relieve pressure. Shutoff valves or gate valves are
typically provided at 350-m (1100-ft) intervals so that areas within the system can be isolated
for repair or maintenance; air-relief valves or air-and-vacuum relief valves are required at high
points to release trapped air; blowoff valves or drain valves may be required at low points;
and backflow-prevention devices are required by applicable regulations to prevent contamination from backflows of nonpotable water into the distribution system from system outlets.
A typical backflow preventer that is commonly found outside a tall office buildings is shown
in Figure 3.7. To maintain the performance of water-distribution systems it is recommended
that all valves in distribution systems be operated through a full cycle and then returned to
their normal positions on a regular schedule. The time interval between operations should
be determined by the manufacturer’s recommendations, size of the valve, severity of the
operating conditions, and the importance of the installation (AWWA, 2003d).
3.3.4
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Meters
The water meter is a changeable component of a customer’s water system. Unlike the service
line and water tap, which when incorrectly sized will generally require expensive excavation and retapping, water meters can usually be changed less expensively. Selection of the
type and size of a meter should be based primarily on the range of flow, and the pressure
loss through the meter should also be a consideration. For many single-family residences,
a 20-mm (3/4-in.) service line with a 15-mm (1/2-in.) meter is typical, while in areas where
irrigation is prevalent, 20–25-mm (3/4–1-in.) meters may be more common. Undersizing the
meter can cause pressure-related problems, and oversizing can result in reduced revenue and
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FIGURE 3.7: Backflow
preventer
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inaccurate meter recordings (since the flows do not register). Some customers, such as hospitals, schools, and factories with processes requiring uninterrupted water service, should have
bypasses installed around the meter so that meter test and repair activities can be performed
at scheduled intervals without inconvenience to either the customer or the utility. The bypass
should be locked and valved appropriately.
3.3.5
Fire Hydrants
Fire hydrants are one of the few parts of a water-distribution system that are visible to
the public, so keeping them well maintained can help a water utility present a good public
image. Fire hydrants are direct connections to the water mains and, in addition to providing
an outlet for fire protection, they are used for flushing water mains, flushing sewers, filling
tank trucks for street washing, tree spraying, and providing a temporary water source for
construction jobs. A typical fire hydrant is shown in Figure 3.8, along with the pipe connection between the water main and the fire hydrant. The vertical pipe connecting the water
main to the fire hydrant is commonly called a riser. The water utility is usually responsible
for keeping hydrants in working order, although fire departments sometimes assume this
responsibility. Standard practice is to install hydrants only on mains 150 mm (6 in.) or larger;
however, larger mains are often necessary to ensure that the residual pressure during fire
flow remains greater than 140 kPa (20 psi). Guidelines for the placement of hydrants are as
follows (AWWA, 2003c):
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◃ Not too close to buildings, since firefighters will not position their fire (pumper) trucks
where a building wall could fall on them.
◃ Preferably located near street intersections, where the hose can be strung to fight a fire
in any of several directions.
◃ Far enough from a roadway to minimize the danger of their being struck by vehicles.
FIGURE 3.8: Fire hydrant
and connection to water
main
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◃ Close enough to the pavement to ensure a secure connection with the pumper and
hydrant without the risk of the truck getting stuck in mud or snow.
◃ In areas of heavy snow, hydrants must be located where they are least liable to be
covered by plowed snow or struck by snow-removal equipment.
◃ Hydrants should be high enough off the ground that valve caps can be removed with a
standard wrench, without the wrench hitting the ground.
Fire hydrants should be inspected and operated through a full cycle on a regular schedule,
and the hydrant should be flushed to prevent sediment buildup in the hydrant or connecting
piping.
3.3.6
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Water-Storage Reservoirs
Water usually enters the water-distribution system at a fairly constant rate from the treatment plant. To accommodate fluctuations in demand, a storage reservoir is typically located
at the head of the system to store the excess water during periods of low demand and provide water during periods of high demand. In addition to the operational storage required
to accommodate diurnal (24-hour cycle) variations in water demand, storage facilities are
also used to provide storage to fight fires, to provide storage for emergency conditions, and
to equalize pressures in water-distribution systems.
Storage facilities are classified as either elevated storage, ground storage, or standpipes.
The function and relative advantages of these types of systems are given below.
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Elevated Storage Tanks. Elevated storage tanks are constructed above ground such that the
height of the water in the tank is sufficient to deliver water to the distribution system
at the required pressure. Elevated storage tanks are usually made of steel. The storage
tank is generally supported by a steel or concrete tower, the tank is directly connected
to the distribution system through a riser (i.e., vertical pipe), the water level in the tank
is equal to the elevation of the hydraulic grade line in the distribution system (at the
outlet of the storage tank), and the elevated storage tank is said to float on the system.
A typical elevated storage tank is illustrated in Figure 3.9. Elevated storage is useful
in the case of fires and emergency conditions, since pumping of water from elevated
tanks is not necessary, although the water must generally be pumped into the tanks.
Occasionally, system pressure could become so high that the tank would overflow; to
prevent this, altitude valves must be installed on the tank fill line.
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Ground Storage Reservoirs. Ground storage reservoirs are constructed at or below ground
level and usually discharge water to the distribution system through pumps. These
systems, which are sometimes referred to simply as distribution-system reservoirs or
ground-level tanks, are usually used where very large quantities of water must be stored
FIGURE 3.9: Elevated
storage tank
Access
manhole
Roof vent
Overflow
pipe
Normal operating range
Fire/Emergency range
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or when an elevated tank is objectionable to the public. When a ground-level or buried
reservoir is located at a low elevation in the distribution system, water is admitted
through a remotely operated valve, and a pump station is provided to transfer water
into the distribution system. In some cases, the land over a buried reservoir can be used
for recreational facilities. Ground storage reservoirs are typically constructed of steel
or concrete.
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Standpipes. A storage tank that rests on the ground with a height that is greater than its
diameter is generally referred to as a standpipe. In most installations, only water in the
upper portion of the tank will furnish usable system pressure, so most larger standpipes
are equipped with an adjacent pumping system that can be used in an emergency to
pump water to the system from the lower portion of the tank. A standpipe is illustrated in Figure 3.10. Standpipes combine the advantages of elevated storage with
the ability to store large quantities of water, and they are usually constructed of steel.
Standpipes taller than 15 m (50 ft) are usually uneconomical, since it tends to be more
economical to build an elevated tank than to accommodate the dead storage that must
be pumped into the system.
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Storage facilities in a distribution system are required to have sufficient volume to meet the
following design criteria:
◃ Supply peak demands in excess of the maximum daily demand using no more than 50%
of the available storage capacity.
◃ Supply the critical fire demand.
◃ Supply the average daily demand of the system for the estimated duration of a possible
emergency.
Conventional design practice is to rely on pumping to meet the daily operational demands
up to the maximum daily demand; where detailed demand data are not available, the storage available to supply the peak demands should equal 20%–25% of the maximum daily
demand volume. Sizing the storage volume for fire protection is based on the product of
the critical fire flow and duration for the service area. In extremely large systems, where fire
demands may be only a small fraction of the maximum daily demand, fire storage may not be
necessary (Walski, 2000). Emergency storage is generally necessary to provide water during
power outages, breaks in water mains, problems at treatment plants, unexpected shutdowns
of water-supply facilities, and other sporadic events. Emergency volumes for most municipal
water-supply systems are in the range of 1–2 days of supply capacity at the average daily
demand. The recommended standards for water works developed by the Great Lakes-Upper
Mississippi River Board of State and Provincial Public Health and Environmental Managers
FIGURE 3.10: Standpipe
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Useful storage
Over
flow
Supporting storage
Drain
Inflow/Outflow
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Section 3.3
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Components of Water-Distribution Systems
89
(GLUMRB, 2007) suggest a minimum emergency storage capacity equal to the average daily
system demand.
In cases where elevated storage tanks are used, the minimum acceptable height of
water in a tank is determined by computing the minimum acceptable head in the service
area and then adding to that figure an estimate of the head losses between the critical service
location and the location of the elevated service tank, under the condition of average daily
demand. The maximum height of water in the elevated tank is then determined by adding
the minimum acceptable head to the head loss between the tank location and the critical
service location under the condition of maximum hourly demand. The difference between
the calculated minimum and maximum heights of water in the elevated storage tank is then
specified as the normal operating range within the tank. The normal operating range for
water in elevated tanks is usually limited to 4.5–6 m (15–20 ft), so that fluctuations in pressure are limited to 35–70 kPa (5–10 psi). In most cases, the operating range is located in
the upper half of the storage tank, with storage in the lower half of the tank reserved for
firefighting and emergency storage. Any water stored in elevated tanks less than 14 m (46 ft)
above ground is referred to as ineffective storage, since the pressure in connected distribution
pipes will be less than the usual minimum acceptable pressure during emergency conditions
of 140 kPa (20 psi). Operational storage in elevated tanks is normally at elevations of more
than 25 m (80 ft) above the ground, since under these conditions the pressure in connected
distribution pipes will exceed the usual minimum acceptable pressure during normal operations of 240 kPa (35 psi). Elevated tanks generally have only a single pipe connection to the
distribution system, and this single pipe handles both inflows and outflows from the storage
tank. This piping arrangement is in contrast to ground storage reservoirs, which have separate
inflow and outflow piping. The inflow piping delivers the outflow from the water-treatment
facility to the reservoir, while the outflow piping delivers the water from the reservoir to the
pumps that input water into the distribution system. The largest elevated storage tank in the
United States has a volume of 15,520 m3 (550,000 ft3 ). Elevated storage tanks are best placed
on the downstream side of the location with the largest demand from the water source, with
the advantages being that: (1) if a pipe breaks near the source, the break will not result
in disconnecting all the storage from the customers; and (2) if flow reaches the center of
demand from more than one direction, the flow carried by any individual pipe will be lower
and pipe sizes will generally be smaller, with associated cost savings (Walski, 2000). If there
are multiple storage tanks in the distribution system, the tanks should be placed at roughly
the same distance from the source or sources, and all tanks should have approximately the
same overflow elevation (otherwise, it may be impossible to fill the highest tank without
overflowing or shutting off the lower tanks).
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EXAMPLE 3.4
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A service reservoir is to be designed for a water-supply system serving 250,000 people with an average
demand of 600 L/d/capita, and a needed fire flow of 37,000 L/min. Estimate the required volume of
storage.
Solution The required storage is the sum of three components: (1) volume to supply the demand in
excess of the maximum daily demand, (2) fire storage, and (3) emergency storage.
The volume to supply the peak demand can be taken as 25% of the maximum daily demand
volume. Taking the maximum daily demand factor as 1.8 (Table 3.2), then the maximum daily flow rate,
Qmax , is given by
Qmax = (1.8)(600)(250, 000) = 2.7 * 108 L/d = 2.7 * 105 m3 /d
The storage volume to supply the peak demand, Vpeak , is therefore given by
Vpeak = (0.25)(2.7 * 105 ) = 67, 500 m3
According to Table 3.6, the 37,000 L/min (= 0.62 m3 /s) fire flow must be maintained for at least
9 hours. The volume to supply the fire demand, Vfire , is therefore given by
Vfire = 0.62 * 9 * 3600 = 20,100 m3
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The emergency storage, Vemer , can be taken as the average daily demand, in which case
Vemer = 250,000 * 600 = 150 * 106 L = 150,000 m3
The required volume, V, of the service reservoir is therefore given by
V = Vpeak + Vfire + Vemer = 67, 500 + 20, 100 + 150, 000 = 237,600 m3
The service reservoir should be designed to store approximately 238,000 m3 of water. This large volume
will require a ground storage tank (recall that the largest elevated-tank volume in the United States is
15,520 m3 ), and it is interesting to note that most of the storage in the service reservoir is reserved for
emergencies.
EXAMPLE 3.5
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A water-supply system is to be designed in an area where the minimum allowable pressure in the distribution system is 300 kPa. A hydraulic analysis of the distribution network under average-daily-demand
conditions indicates that the head loss between the low-pressure service location, which has a pipeline
elevation of 5.40 m, and the location of the elevated storage tank is 10.0 m. Under maximum-hourlydemand conditions, the head loss between the low-pressure service location and the elevated storage
tank is 12.0 m. Determine the normal operating range for the water stored in the elevated tank.
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Solution Under average-daily-demand conditions, the elevation z0 of the hydraulic grade line (HGL)
at the reservoir location is given by
p
z0 = min + zmin + hL
γ
where pmin = 300 kPa, γ = 9.79 kN/m3 , zmin = 5.40 m, and hL = 10.0 m, which yields
z0 =
300
+ 5.4 + 10.0 = 46.0 m
9.79
Under maximum-hourly-demand conditions, the elevation z1 of the HGL at the service reservoir is
given by
300
z1 =
+ 5.4 + 12.0 = 48.0 m
9.79
Therefore, the operating range in the storage tank should be between elevations 46.0 m and 48.0 m.
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It is important to keep in mind that the best hydraulic location and most economical
design are not always the deciding factors in the location of an elevated tank. In some cases,
the only acceptable location will be in an industrial area or public park. In cases where public
opinion is very strong, a water utility may have to construct ground-level storage, which is
more aesthetically acceptable.
3.4
Performance Criteria for Water-Distribution Systems
The primary functions of water-distribution systems are to: (1) meet the water demands of
users while maintaining acceptable pressures in the system; (2) supply water for fire protection at specific locations within the system, while maintaining acceptable pressures for
normal service throughout the remainder of the system; and (3) provide a sufficient level
of redundancy to support a minimum level of reliable service during emergency conditions, such as an extended loss of power or a major water-main failure. Real-time operation
of water-distribution systems is typically based on remote measurements of pressures and
storage-tank water levels within the distribution system. The pressure and water-level data
are typically transmitted to a central control facility via telemetry, and adjustments to the
operation of the distribution system are made from the central facility by remote control of
pumps and valves within the distribution system. These electronic control systems are generally called supervisory control and data acquisition (SCADA) systems. Operating criteria for
service pressures and storage facilities are described below.
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Section 3.4
3.4.1
Performance Criteria for Water-Distribution Systems
91
Service Pressures
Maintaining adequate pressures in the distribution system while supplying the service
demands requires that the system be analyzed on the basis of allowable pressures. Minimum
acceptable pressures are necessary to prevent contamination of the water supply from cross
connections. Criteria for minimum acceptable service pressures such as those recommended
by the Great Lakes-Upper Mississippi River Board of State and Provincial Public Health
and Environmental Managers (GLUMRB, 2007) and those endorsed by the American Water
Works Association (AWWA, 2003c) are typical of most water-distribution systems, and are
listed in Table 3.8. During main breaks, when the pressure in water-supply pipelines can drop
below 140 kPa (20 psi), it is not uncommon for a water utility to issue a “boil water” advisory
because of the possibility of system contamination from cross connections. There are several
considerations in assessing the adequacy of service pressures, including:
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◃ Flow is adequate for residential areas if the pressure is not reduced below 240 kPa
(35 psi).
◃ The pressure required at street level for excellent flow to a 3-story building is about
290 kPa (40 psi).
◃ It is usually desirable to maintain normal pressures of 410–520 kPa (60–75 psi), since
these pressures are adequate to supply ordinary consumption for buildings up to 10
stories and to provide adequate sprinkler service in buildings of 4–5 stories.
◃ Pressures higher than 650 kPa (95 psi) should be avoided if possible because of
excessive leakage and water use, and the added burden of installing and maintaining
pressure-reducing valves and other specialized equipment.
◃ The pressure required for adequate flow to a 20-story building is about 830 kPa
(120 psi), which is not desirable because of the associated leakage and waste.
◃ Very tall buildings are usually served with their own pumping equipment.
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Customers do not generally like high pressure because water comes out of an open faucet
with too much force (AWWA, 2003c). In addition, excessive pressures decrease the life of
water heaters and other plumbing fixtures.
3.4.2
Allowable Velocities
Maximum allowable velocities in pipeline systems are imposed to control friction losses and
hydraulic transients. Maximum allowable velocities of 0.9–1.8 m/s (3–6 ft/s) are common in
water-distribution pipes, and the American Water Works Association recommends a limit
of about 1.5 m/s (5 ft/s) under normal operating conditions, but velocities may exceed this
guideline under fire-flow conditions (AWWA, 2003c). The importance of controlling the maximum velocities in water-distribution systems is supported by the fact that a sudden change
in velocity of 0.3 m/s (1 ft/s) in water transmission and distribution systems can increase
the pressure in a pipe by approximately 345 kPa (50 psi), while the standard design for ductile iron pipe allows a total pressure of 690 kPa (100 psi) (AWWA, 2003d).
3.4.3
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Water Quality
The quality of water delivered to consumers can be significantly influenced by various
components of a water-distribution system. The principal factors affecting water quality in
TABLE 3.8: Minimum Acceptable Pressures in Distribution Systems
Demand condition
Minimum acceptable pressure
(kPa)
(psi)
Average daily demand
Maximum daily demand
Maximum hourly demand
Fire situation
Emergency conditions
240–410
240–410
240–410
>140
>140
35–60
35–60
35–60
>20
>20
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Chapter 3
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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.
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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.
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Section 3.5
Building Water-Supply Systems
93
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.
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(a)
(b)
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(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:
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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.
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Chapter 3
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FIGURE 3.12: Isometric
view of residential
water-supply system
A
B
Hot water
Cold water
C
z
Service line
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Water heater
y
x
Backflow preventer
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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
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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
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Section 3.5
Building Water-Supply Systems
95
TABLE 3.9: Water-Supply Fixture Units (WSFU)
Load values in WSFU
Fixture
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Occupancy
Supply control
Flush tank
Flush valve
Faucet
Faucet
Faucet
Faucet
Cold
Hot
Total
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
Bathroom group
Bathroom group
Bathtub
Bathtub
Bidet
Combination fixture
Private
Private
Private
Public
Private
Private
Dishwashing machine
Drinking fountain
Kitchen sink
Kitchen sink
Laundry trays (1 to 3)
Private
Automatic
Offices, etc.
9.5-mm (3/8-in.) valve
Private
Faucet
Hotel, restaurant Faucet
Private
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
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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
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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
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Chapter 3
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TABLE 3.10: Typical Minimum Flow Rates and Pressures in Building Fixtures
Flow rate
(L/min)
Pressure
(kPa)
Flow rate
(gpm)
Pressure
(psi)
Bathtub
Bidet
Combination fixture
Dishwasher, residential
Drinking fountain
Laundry tray
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
Lavatory, private
Lavatory, public
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
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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
$
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%
!
!
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
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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
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97
(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
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TABLE 3.11: Head Loss in Standard Fittings in Terms of Equivalent Pipe Lengths
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Fitting Diameter
90◦ Ell
45◦ Ell
90◦ Tee
Gate
valve
Globe
valve
Angle
valve
(mm)
(in.)
(m)
(m)
(m)
(m)
(m)
(m)
9.5
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 14
1.219
0.732
1.829
0.244
10.668
5.486
38
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
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
12.7
Source: Uniform Plumbing Code (2012).
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Chapter 3
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
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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:
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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.
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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
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Section 3.5
Building Water-Supply Systems
99
FIGURE 3.14: Factory
water-supply system
Building
46 m
D
E
Legend:
Floor 2
Valve
Tee
Ell
D1
46 m
Cold water pipe
Hot water pipe
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E’
4.0 m
4.0 m
46 m
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C
F
Floor 1
46 m
2.5 m
C’
F’
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
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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
Re =
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
f = $
0.25
,
ks
+ 5.74
log 3.7D
Re0.9
- %2
= $
0.25
,
0.0015 +
5.74
log 3.7(64)
135,6000.9
= 0.0173
-%2
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100
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)
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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,
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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 =
and the friction factor, f , is
f = $
log
(2.05)(0.064)
VD
= 13,100
=
ν
10−6
0.25
,
ks
5.74
3.7D + Re0.9
= $
-%2
0.25
log
,
0.0015 +
5.74
3.7(64)
13,1000.9
= 0.0174
-%2
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
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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:
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Problems
101
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.
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Problems
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.
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3.1. Derive an expression for the population, P, versus time,
t, where the growth rate is: (a) geometric, (b) arithmetic,
and (c) declining.
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. A city founded in 1950 had a population of 13,000 in
1960; 125,000 in 1975; and 300,000 in 1990. Assuming
that the population growth follows a logistic curve, estimate the saturation population of the city, if possible.
3.4. The average demand of a population served by a waterdistribution system is 580 L/d/person, and the population at the end of the design life is estimated to be
100,000 people. Estimate the maximum daily demand
and maximum hourly demand at the end of the design
life.
3.5. Estimate the flow rate and volume of water required to
provide adequate fire protection to a five-story office
building constructed of joisted masonry. The effective
floor area of the building is 5000 m2 .
3.6. What is the maximum fire flow and corresponding duration that can be used for any building?
3.7. A water-supply system is being designed to serve a population of 200,000 people, with an average per-capita
demand of 600 L/d/person and a needed fire flow of
28,000 L/min. If the water supply is to be drawn from
a river, what should be the design capacity of the supply pumps and water-treatment plant? For what duration must the fire flow be sustained, and what volume of
water must be kept in the service reservoir to accommodate a fire? What should be the design capacity of the
distribution pipes?
3.8. What is the minimum acceptable water pressure in a
distribution system under average-daily-demand conditions?
3.9. A water-supply utility is expecting a service-area population of 200,000 people at the end of the design period.
The expected per-capita demand is 500 L/d/person and
the design fire flow is 30,000 L/min. (a) What should
be the design capacity and minimum diameter of the
main distribution pipeline? What pipe material would
you specify? How deep below the ground surface would
you bury the pipe? (b) If the water main leaving the
treatment plant has a design flow rate of 3.80 m3 /s, a
diameter of 1800 mm, and a pressure of 550 kPa, how
many kilometers of pipeline is required for the pressure to drop to 350 kPa? Assume that it is an old pipe
and that there is no change in elevation along the pipe.
Contrast your results using the Darcy–Weisbach and
Hazen–Williams formulae. Which approach is preferable and why?
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102
Chapter 3
Design of Water-Distribution Systems
C
20 m
B
20 m
3m
A
P
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. For a particular office building,
the total fixture units is determined to be 120 and the
corresponding flow rate is 4.67 L/s. The pressure at the
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water main is estimated to be 380 kPa, and the maximum
velocity in the plumbing is not to exceed 2.4 m/s, copper
lines are to be used, the length of the line to the most
remote fixture is 110 m, the minimum allowable pressure
in the pipe (after accounting for head losses) is 240 kPa,
and the elevation difference between the curb and the
most remote fixture in the building is 3 m. Copper pipe
is available in diameters starting at 12.5 mm (1/2 in.) and
increasing in increments of 6.25 mm (1/4 in.). Determine
the minimum diameter that could be used for the plumbing line. Neglect local losses.
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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?
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C H A P T E R
4
Fundamentals of Flow
in Open Channels
4.1 Introduction
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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.
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4.2 Basic Principles
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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
103
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104
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
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1
V1 =
A1
!
1
A2
!
and
A1
(4.2)
v1 dA
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V2 =
A2
(4.3)
v2 dA
then, for an incompressible fluid (ρ = constant) such as water, the steady-state continuity
equation (Equation 4.1) can be written as
(4.4)
V1 A1 = V2 A2
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
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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
γA∆x
τ P∆
0
z1
Datum
Water surface
y2
x
Q
∆x
z2
θ
x
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Section 4.2
"
Fx =
!
A2
ρv2x dA −
!
A1
Basic Principles
ρv2x dA
105
(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)
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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)
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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
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(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 A$x sin θ − τ0 P$x + 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
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106
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
$z = z2 − z1 ,
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$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
w.E
asy
En
gin
eer
in
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)
g.n
et
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, ϵ, ϵ ′ , m, s)
(4.22)
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Section 4.2
Basic Principles
107
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], ϵ is the characteristic
size of the roughness projections on the channel boundary [L], ϵ ′ 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
ϵ ϵ′
= f1 Re,
,
, m, s
(4.23)
4R 4R
ρV 2
where Re is the Reynolds number defined by the relation
ww
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:
)
*
ϵ
τ0
1
= f Re,
(4.25)
8
4R
ρV 2
w.E
asy
En
gin
eer
in
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 ϵ/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
⎪
⎪
Re < 105
(4.26)
⎪ 1.78Re 8 ,
⎨
'
(
1
+ =
2.51
⎪
f
⎪
+ ,
(4.27)
−2.0 log10
Re > 105
⎪
⎩
Re f
g.n
et
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108
Chapter 4
Fundamentals of Flow in Open Channels
These relations are the same as the Blasius and Prandtl–von Kármán equations for flow in
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
ww
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
Prandtl–von Kármán equation that is used in pipe flow. In the transition region between
(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
w.E
asy
En
gin
eer
in
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 asymptotes to the Prandtl–von Kármán equation (Equation 4.27), which is known to deviate by
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
0
1
τ0
= gRSf
(4.30)
u∗ =
ρ
g.n
et
and ν is the kinematic viscosity of the fluid. The approximate ranges for the three types of
turbulent flow are as follows:
⎧
⎪
smooth
⎨<5,
u∗ ks
= 5 to 70, transition
(4.31)
⎪
ν
⎩>70,
rough
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
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Section 4.2
1
+ = 1.2 + 2.03 log10
f
ww
)
R
d84
Basic Principles
*
109
(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
by taking ks equal to 2.4d90 , 2.8d84 , or 6.1d50 (López and Barragán, 2008).
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
w.E
asy
En
gin
eer
in
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:
2
8g 1
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
g.n
et
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)
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110
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
ww
FIGURE 4.3: Flow in a
trapezoidal channel
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.
T = 3 + 4y
w.E
asy
En
gin
eer
in
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 8gS
0
g.n
et
⎤
−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
and other large open channels began much earlier. In 1775, Chézy∗ proposed the following
∗ Antoine de Chézy (1718–1798) was a French engineer.
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Section 4.2
Basic Principles
expression for the mean velocity in an open channel
1
V = C RSf
111
(4.39)
where C was referred to as the Chézy 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
Chézy coefficient on the Reynolds number and the relative roughness was not considered.
Comparing Equations 4.35 and 4.39, the Chézy coefficient is related to the friction factor by
2
8g
C=
(4.40)
f
ww
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
w.E
asy
En
gin
eer
in
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 4.41 or 4.42 is combined with the Chézy equation, the resulting expression is called
the Manning equation or Strickler equation (in Europe) and is given by
⎧
1 2 1
⎪
⎪
⎪
⎨nR3 S2
V=
⎪
1.486 2 1
⎪
⎪
⎩
R3 S2
n
(SI units)
g.n
et
(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
2
1
R6
8g
=
(4.44)
n
f
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112
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
ww
n
1
6
ks
= 0.039
w.E
asy
En
gin
eer
in
(4.46)
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
1
ks RSf > 2.2 * 10−5
(4.48)
g.n
et
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
R/ks = 500
1
n/ks6
R/ks = 4
1
n/ks 6 = 0.039
0.01
1
10
100
1000
R/ks
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Section 4.2
Basic Principles
113
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
1
0.039ks6 .
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
ww
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
w.E
asy
En
gin
√
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.
eer
in
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
g.n
et
(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.
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114
Chapter 4
Fundamentals of Flow in Open Channels
TABLE 4.1: Semiempirical Expressions for Manning’s n
Expression for n
1
6
0.047d50
1
6
0.0025d90
1
6
0.0039d90
1
6
0.047d75
1
6
0.046d50
ww
Conditions
Reference
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
R6
[7.69 ln(R/d84 ) + 63.4]
1
R6
w.E
asy
En
gin
eer
in
[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.
g.n
et
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
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Section 4.2
Basic Principles
115
TABLE 4.2: Manning Coefficient for Open Channels
Channel type
ww
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
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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
g.n
et
(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
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116
Chapter 4
Fundamentals of Flow in Open Channels
TABLE 4.3: Composite-Roughness Formulae
Formula∗
⎛
Assumption
⎞2
N
"
3
2 ⎟
3
⎜
P i ni ⎟
⎜
⎟
⎜ i=1
⎟
⎜
ne = ⎜
⎟
⎟
⎜
P
⎠
⎝
⎛
ww
⎞1
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)
2
N
"
⎜
Pi n2i ⎟
⎟
⎜
⎟
⎜ i=1
⎟
ne = ⎜
⎟
⎜
⎟
⎜
P
⎠
⎝
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⎛
Reference
⎞2
N
"
3
2 ⎟
3
⎜
P i ni ⎟
⎜
⎟
⎜ i=1
⎟
⎜
ne = ⎜
⎟
⎟
⎜
P
⎠
⎝
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)
5
ne =
PR 3
5
n
"
Pi R 3
i
ni
i=1
N
"
ln ne = i=1
N
"
i=1
ne = 1
A
ne =
3
Pi yi2 ln ni
3
2
Pi yi
=N
i=1 Ai ni
1
1 =N P R 3 n
1
i=1 i i i
PR 3
Total shear velocity equal to weighted sum
of subsection shear velocities
g.n
et
Yen (1991)
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
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Section 4.2
ww
Basic Principles
117
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
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TABLE 4.4: Manning Roughness Coefficient for Floodplains
Surface
Minimum
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
g.n
et
Maximum
Typical
0.035
0.050
0.030
0.035
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
0.025
0.030
0.020
0.025
0.030
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118
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).
ww
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:
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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
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
295
633
yi
(m)
g.n
et
1.00
2.00
3.50
5.00
3.50
2.00
1.00
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
4
150 m
2
3
1
4
2
2m
5
2
1
3m
6
7
4
1
15 m
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Section 4.2
Basic Principles
119
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:
ww
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
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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.
g.n
et
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
can possibly be linked to the Chézy assumption that the surface resistance is proportional to
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,
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120
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] (López 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
ww
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),
)
*
y
1+
v(y) = V +
gdS0 1 + 2.3 log
κ
d
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(4.56)
where V is the depth-averaged velocity [LT−1 ], κ is the von Kármán constant (L 0.4) [dimensionless], 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
v(y) = Vmax
) *1
y 7
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)
g.n
et
(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
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Section 4.2
Basic Principles
121
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
ww
Q=
N
"
(4.60)
V i Ai
i=1
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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
g.n
et
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
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122
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
ww
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
'
(
!
v2
#Q =
ρ h + gz +
v · n dA
(4.67)
2
A
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Considering the term h + gz, then
)
*
p
p
+ u + gz = g
+ z + u
h + gz =
ρ
γ
(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
#Q = (h1 + gz1 )
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
g.n
et
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
!
A2
ρ
V3
v3
dA = α2 ρ 2 A2
2
2
(4.72)
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Section 4.2
Basic Principles
123
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
ww
ρV1 A1 = ρV2 A2 = m
#
(4.74)
and combining Equations 4.73 and 4.74 leads to
⎡'
(
'
(⎤
V12
V22
# =m
⎦
Q
# ⎣ h2 + gz2 + α2
− h1 + gz1 + α1
2
2
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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
#
(4.75)
(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
#
g.n
et
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)
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124
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
ww
(
'
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
(4.84)
h1 = h2 + hL
w.E
asy
En
gin
eer
in
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
2
= (S − S0 cos θ )L
(4.86)
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
S=
hL
L
and S0 is the slope of the channel defined as
S0 =
z1 − z2
L cos θ
g.n
et
(4.87)
(4.88)
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
cos θ
0.001
0.01
0.1
1
0.9999995
0.99995
0.995
0.707
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Section 4.2
%
V2
y + α
2g
&1
= (S − S0 )L,
2
Basic Principles
S0 < 0.1
125
(4.89)
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
%
ww
V2
y + α
2g
&1
= (Sf − S0 )L,
2
S0 < 0.1
(4.90)
where Sf is the frictional head loss per unit length. Equation 4.90 is sometimes written in the
expanded form
w.E
asy
En
gin
eer
in
'
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)
g.n
et
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
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126
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
(4.95)
E1 − E2 = hL + $z
ww
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
(4.96)
y=E
w.E
asy
En
gin
eer
in
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
g.n
et
(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
Depth of flow, y
FIGURE 4.6: Typical
specific energy diagram
y
y2
=
(4.99)
E
yc
y1
Q = constant
Emin
Specific energy, E
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Section 4.2
FIGURE 4.7: Typical
channel cross section
Basic Principles
127
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
ww
(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
w.E
asy
En
gin
eer
in
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 =
g.n
et
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).
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128
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
Under critical-flow conditions
ww
FIGURE 4.8: Trapezoidal
cross section
A = 3y + 2y2
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 )
w.E
asy
En
gin
eer
in
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.
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
g.n
et
where b is the width of the channel. The flow area, A, and top width, T, are given by
(4.107)
A = by
(4.108)
T=b
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
(4.110)
q2
2gy2c
(4.111)
3
and corresponding critical energy
Ec = yc +
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Section 4.2
Basic Principles
129
Combining Equations 4.110 and 4.111 leads to the following simplified form of the minimum
specific energy in rectangular channels:
Ec =
ww
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
w.E
asy
En
gin
eer
in
FIGURE 4.9: Specific
energy diagram for
rectangular channel
Depth of flow, y
y
!
E
q increases
1'
y1'
y2'
g.n
et
2'
yc
y2
y1
2
1
Ec
q ! q3
q ! q2
q ! q1
E1 (! E2)
Specific energy, E
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130
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?
ww
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,
w.E
asy
En
gin
eer
in
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
E1 = y1 +
V12
2g
= 0.85 +
1.002
= 0.901 m
2(9.81)
Equating the specific energies at Sections 1 and 2 yields
0.901 = y2 +
Q2
2gA2
1.102
= y2 +
2(9.81)[(1.30 − 0.30)y2 ]2
which simplifies to
y2 +
FIGURE 4.10: Constriction
in rectangular channel
0.0617
y22
g.n
et
= 0.901
0.3 m
Pier
y2
0.85 m
1.30 m
1.30 m
Upstream of constriction
At constriction
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Section 4.2
Basic Principles
131
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
ww
√
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
w.E
asy
En
gin
eer
in
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
0.901 =
%
&1
3
3
1.102
2 b2 (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.”
g.n
et
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)
⎧
⎫
⎡
⎡
⎤⎤⎪
⎪
⎡
⎤
⎪
⎪
0
3
⎪
⎪
*
)
⎨
⎥⎬
⎢ 2
ysub
⎢1
−1 ⎣− Ec 2 ⎦ − 2 π ⎥⎥
(4.113)
= cos 2 sin−1 ⎢
cos
cos
⎣
⎦⎦
⎣ 3
⎪
⎪
E
3
E
3
⎪
⎪
⎪
⎪
⎩
⎭
⎧
⎪
⎪
⎪
⎨
⎡
⎤⎫
⎡
⎡
⎤⎤ ⎪
⎪
0
⎪
) *3
⎬
⎢ 2
ysup
Ec 2 ⎦⎥⎥
⎢1
−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,
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132
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
ww
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)
w.E
asy
En
gin
eer
in
C
C
C V2
2C
V
C
C
he = C Cα2 2 − α1 1 C
C 2g
2g C
(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
g.n
et
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 =
'
$ y + α
$x
V2
2g
(
(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
D 2E
$ 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
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Section 4.3
S0 − Sf = lim
133
D 2E
$ α V2g
$y
+ lim
$x
$x→0
'
(
dy
d
V2
=
+
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
$x→0 $x
ww
Water-Surface Profiles
(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
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where T is the top width of the channel and the hydraulic depth, D, of the channel is
defined as
A
D=
(4.120)
T
then the Froude number, Fr, of the flow can be written as
2
√
V
dA
Q T
Q
Fr = +
= +
=+
3
dy
gD
A gA
gA
or
Fr2 =
Q2 dA
gA3 dy
Combining Equations 4.118 and 4.122 and rearranging yields
S0 − Sf
dy
=
dx
1 − αFr2
(4.121)
g.n
et
(4.122)
(4.123)
This differential equation describes the water-surface profile in open channels, and its original derivation has been attributed to Bélanger (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)
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134
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 .
ww
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.
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TABLE 4.7: Hydraulic Classification of Slopes
Name
Type
Condition
Mild
Steep
Critical
Horizontal
Adverse
M
S
C
H
A
yn > yc
yn < yc
yn = yc
yn = q
S0 < 0
g.n
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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
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Section 4.3
FIGURE 4.11: Slope
classifications
1
dy
Horizontal
2
dx
dy
yn
3
dx
yc
dy
Water-Surface Profiles
135
! ", M1 profile
! #, M2 profile
! ", M3 profile
dx
Mild slope
1
Horizontal
2
3
ww
yn
dy
dx !
yc
dy
dx
dy
dx
Steep slope
",
S1
!#
, S2
!"
, S3
pro
file
pro
file
pro
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1
file
Horizontal
3
yn ! y
c
dy
dx ! ", C1 pro
file
dy
dx ! ", C3
profile
Critical slope
2
3
yn ! ∞
dy
dx
dy
yc
dx
! #, H2 profile
! ", H3 profile
Horizontal slope
file
2 pro
dy ! #, A
dx
yc
2
3
file
3 pro
dy ! ", A
dx
Adverse slope
g.n
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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
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136
Chapter 4
Fundamentals of Flow in Open Channels
Since
and
An = (b + myn )yn = (5 + 1.5yn )yn
+
+
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
When the flow conditions are critical, then
ww
yn = 2.25 m
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
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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
g.n
et
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
Sf =
⎡
⎤2
(0.04)(60)
⎣
⎦
2
(44)(2.27) 3
= 0.000997
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Section 4.3
Water-Surface Profiles
137
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 =
ww
(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.
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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
Q2 Tc
=1
gA3c
(4.129)
g.n
et
(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
yn
0
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138
Chapter 4
Fundamentals of Flow in Open Channels
Q2
2gA2n
(4.131)
2 1
1
An Rn3 S02
n
(4.132)
H = yn +
and
Q=
ww
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.
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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
g.n
et
Eliminating Q from these equations yields the more convenient combined form as
'
(
gA3c
1
Ac
H = yc +
' H = yc +
2
Tc
2Tc
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=
2
F
G
E3
D
G
G (9.81) 2yc + 3y2c
3
gAc
H
=
= 78.3 m3 /s
Tc
2 + 2(3)yc
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Section 4.3
Water-Surface Profiles
139
Determine the normal depth of flow, yn , corresponding to Q = 78.3 m3 /s by applying the Manning
equation which requires that
E5
D
5
3
byn + my2n
1
1
1 An3 12
S0 = D
S02
Q=
2
2
E
+
n 3
n
3
Pn
b + 2yn 1 + m2
1
78.3 =
0.020 D
D
2yn + 3y2n
+
E5
3
2 + 2yn 1 + 32
E 2 (0.01)
1
2
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.
ww
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
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P1 − P2 = ρQ(V2 − V1 )
g.n
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(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)
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140
Chapter 4
Fundamentals of Flow in Open Channels
)
Q
Q
γ y1 A1 − γ y2 A2 = ρQ
−
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
ww
(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
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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
y2 =
⎛
y1 ⎝
−1 +
2
2
1 +
8q2
gy31
⎞
⎠
which can also be written in the following nondimensional form:
*
)
1
y2
1
2
−1 + 1 + 8Fr1
=
y1
2
g.n
et
(4.139)
(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 equation, although some have argued that this equation was originally derived by Bélanger (1841)
and should be called Bélanger’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
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Section 4.3
Water-Surface Profiles
141
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
ww
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)
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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
g.n
et
(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
<5%
5%–15%
15%–45%
45%–70%
70%–85%
Standing waves
Smooth rise
Unstable; avoid
Best design range
Choppy, intermittent
Name
Undular jump
Weak jump
Oscillating jump
Stable jump
Strong jump
Source: USBR (1987).
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142
Chapter 4
Fundamentals of Flow in Open Channels
Outside of the 4.5 < Fr1 < 9.0 range, disturbances caused by the hydraulic jump tend to
propagate downstream, which is usually undesirable. The energy loss in the hydraulic jump,
$E, is given by
'
(
'
(
V12
V22
$E = y1 +
− y2 +
(4.145)
2g
2g
Combining Equation 4.145 with the equation relating the conjugate depths of the hydraulic
jump, Equation 4.140, leads to the following expression for the energy loss in a hydraulic
jump in a rectangular channel:
$E =
ww
(y2 − y1 )3
4y1 y2
(4.146)
Hydraulic jumps are sometimes used as self-aerators to enhance the dissolved oxygen concentration in water. Measurements in hydraulic jumps have shown that oxygen concentration
increases across a hydraulic jump in direct proportion to energy dissipation rate per unit
width, γ q$E, according to the relation (Kucukali and Cokgor, 2009)
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Cd − Cu
= 0.0015(γ q$E) + 0.01
Cs − Cu
(4.147)
where Cd , Cu , and Cs are the downstream, upstream, and saturation concentrations of dissolved oxygen, respectively, at 20◦ C. Equation 4.147 was derived from experimental results
where γ q$E < 180 Watt/m and should not be extrapolated beyond this range. The expression on the left-hand side of Equation 4.147 is called the aeration efficiency and varies
between 0 and 1.
The air-entrainment characteristics at any distance, x, from the beginning of a hydraulic
jump can be expressed in the form (Chanson, 2011)
$
#
J x
I
Qair
(4.148)
= 0.3387Fr0.202
exp
−0.103
+
0.0073Fr
1
1
Q + Qair
y1
where Qair is the air flow rate, Q is the water flow rate, Fr1 is the upstream Froude number,
and y1 is the upstream flow depth. Equation 4.148 can be used to estimate the rate of air
entrainment with distance from the start of the hydraulic jump, and indicates decreasing
aeration (deaeration) with distance downstream from the toe of the jump.
EXAMPLE 4.8
g.n
et
Water flows down a spillway at the rate of 12 m3 /s per meter of width into a horizontal channel, where
the velocity at the channel entrance is 20 m/s. Determine the (downstream) depth of flow in the channel
that will cause a hydraulic jump to occur at the toe of the spillway (i.e., where the spillway meets the
channel), and determine the power loss in the jump per meter of width.
Solution In this case, q = 12 m2 /s, and V1 = 20 m/s. Therefore the initial depth of flow, y1 , is given by
y1 =
q
12
= 0.60 m
=
V1
20
and the corresponding Froude number, Fr1 , is given by
Fr1 =
q
12
+
+
= 8.24
=
y1 gy1
0.60 (9.81)(0.60)
which confirms that the flow is supercritical. The conjugate depth is given by Equation 4.140, where
'
(
1
)
*
1
y2
1
1
−1 + 1 + 8(8.24)2 = 11.2
=
−1 + 1 + 8Fr21 =
y1
2
2
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Section 4.3
Water-Surface Profiles
143
Therefore the conjugate downstream depth, y2 , is
y2 = 11.2y1 = 11.2(0.60) = 6.70 m
Hence, a downstream flow depth of 6.70 m will cause a hydraulic jump to occur at the toe of the spillway.
The energy loss in the hydraulic jump, $E, is given by Equation 4.146 as
$E =
(6.70 − 0.60)3
(y2 − y1 )3
=
= 14.1 m
4y1 y2
4(0.60)(6.70)
and therefore the power loss, P, per unit width in the jump is given by
P = γ q $E = 9790(12)(14.1) = 1.66 * 106 W = 1.7 MW
ww
The location of a hydraulic jump is important in determining channel wall heights
as well as the flow conditions (subcritical or supercritical) in the channel. The mean location of a hydraulic jump is usually estimated by computing the upstream and downstream
water-surface profiles, and the jump is located where the upstream and downstream water
depths are equal to the conjugate depths of the hydraulic jump. In many cases, the location
of a hydraulic jump is controlled by installing baffle blocks, sills, drops, or rises in the bottom
of the channel to create sufficient energy loss that the hydraulic jump forms at the location of
these structures. Hydraulic structures that are specifically designed to induce the formation
of hydraulic jumps are called stilling basins.
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4.3.4
Computation of Water-Surface Profiles
The differential equation describing the water-surface profile in an open channel is given by
Equation 4.123, which can be written in the form
dy
= F(x, y)
dx
(4.149)
where y is the depth of flow in the channel, x is the distance along the channel, and F(x, y) is
a function defined by the relation
F(x, y) =
S0 − Sf
1 − αFr2
(4.150)
g.n
et
where S0 is the channel slope, Sf is the slope of the energy grade line, Fr is the Froude
number of the flow, and α is the kinetic energy correction factor. The function F(x, y) can
be calculated using given values of y, Q, S0 , α, and channel geometry (which can be a function of x), and by using Equation 4.121 to estimate the Froude number and the Manning or
Darcy–Weisbach equation to estimate the slope of the energy grade line, Sf .
A basic assumption in calculating water-surface profiles is that the head loss between
upstream and downstream sections can be estimated using either the Manning or Darcy–
Weisbach equation without regard to trends in depth. This approximation requires that flow
conditions change gradually, and such flow conditions are called gradually varied flow (GVF).
Conversely, flows that are not gradually varied are called rapidly varied flow (RVF). Important differences between RVF and GVF are: (1) in RVF, there is significant acceleration
normal to the streamlines, causing a nonhydrostatic pressure distribution; (2) in RVF, significant depth variations occur over short distances and so boundary friction is relatively small;
and (3) in RVF, the kinetic energy correction factor, α, and the momentum correction factor,
β, are much greater than unity. Examples of RVF include hydraulic jumps and flow over
spillways. This section is concerned with the computation of water-surface profiles for GVF
conditions.
Under GVF conditions, the Manning approximation (in SI units) to the friction slope,
Sf , is given by
'
(2
nQ
Sf =
(4.151)
2
AR 3
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144
Chapter 4
Fundamentals of Flow in Open Channels
where n is the Manning roughness coefficient, Q is the flow rate, and A and R are the area
and hydraulic radius of the cross section, respectively. It is sometimes convenient to define
the conveyance, K, by the relation
2
1
K = AR 3
(4.152)
n
in which case the friction slope, Sf , can be put in the form
Sf =
ww
)
Q
K
*2
(4.153)
The solution of the water-surface profile equation (Equation 4.149) is usually obtained
using numerical integration, since an analytic solution is not possible and field information
for calculating the function F(x, y) is typically available only at discrete intervals along the
channel. As an alternative to direct numerical integration of Equation 4.149, water-surface
profiles can also be calculated directly from the energy equation. Rearranging the form of
the energy equation given by Equation 4.90 leads to
%
&1
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$L =
V2
y + α
2g
2
Sf − S0
(4.154)
where $L is the distance between the upstream section (Section 1) and the downstream
section (Section 2), and Sf is the mean slope of the energy grade line between Sections 1
and 2. The following alternative formulations have been used to estimate Sf :
Method 1: Average conveyance method.
Sf = #
Q2
K1 + K2
2
$2
Method 2: Average friction slope method.
Sf =
Sf 1 + Sf 2
2
Method 3: Geometric mean friction slope method.
1
Sf = Sf 1 Sf 2
(4.155)
g.n
et
(4.156)
(4.157)
Method 4: Harmonic mean friction slope method.
Sf =
2Sf 1 Sf 2
Sf 1 + Sf 2
(4.158)
where K1 and K2 are the conveyances at the upstream and downstream sections, and Sf 1 and
Sf 2 are the friction slopes at the upstream and downstream sections, respectively. Method 2
has been found to be most accurate for M1 profiles, while Method 4 is best for M2 profiles
(U.S. Army Corps of Engineers, 2008). Differences between methods become smaller as the
spacing between cross sections is reduced, and differences are typically minimal for crosssection spacings less than 150 m (500 ft) (U.S. Army Corps of Engineers, 1986; French, 2001).
In applying Equation 4.154 to calculate the water-surface profile, it is assumed that the
contraction or expansion loss is negligible in comparison to the friction loss. Contraction and
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Section 4.3
Water-Surface Profiles
145
expansion losses can be accounted for by Equation 4.115, and the friction loss accounted for
by Sf . In the general case where friction and expansion/contraction losses are both accounted
for, the appropriate energy equation is given by
%
&1
V2
y + α
2g
2
C
C
$L =
C
2
V12 CC
C C V2
− α1 C − S0
Sf +
Cα2
$L C 2g
2g C
ww
(4.159)
where C is the expansion or contraction coefficient between adjacent channel sections. For
gradual expansions and contractions, values of C are typically 0.3 and 0.1, respectively, while
for abrupt expansions and contractions, values of C are typically 0.8 and 0.6, respectively.
When the channel cross sections do not vary significantly, contraction and expansion losses
between sections are much less than friction losses and, given the uncertainty in estimating
friction losses, Equation 4.154 is an adequate representation of the energy equation in these
cases.
Whenever the water surface passes through critical depth the energy equation is not
applicable. It is applicable only to gradually varied flows, and the transition from subcritical
to supercritical or supercritical to subcritical is a rapidly varying flow. Such rapidly varying
flows occur at significant changes in channel slope, some bridge constrictions, drop-structures
and weirs, and some stream junctions. In many of these cases, empirical equations can be used
(e.g., weirs), while at others it is necessary to apply the momentum equation to determine the
changes in water-surface elevation.
Methods commonly used to determine the water-surface profile are the directintegration method, direct-step method, and standard-step method. These methods are all
based on the energy equation and yield essentially the same results in cases where they are
all applicable; their differences are related to the ease and efficiency of the computations.
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4.3.4.1
Direct-integration method
In applying the direct-integration method, the water-surface profile described by Equation 4.149 can be expressed in the finite difference form
y2 − y1
= F(x, y)
x2 − x1
where
x=
x1 + x2
2
and
y=
y1 + y2
2
g.n
et
(4.160)
(4.161)
and the subscripts refer to (adjacent) cross sections of the channel, where Section 1 is
upstream of Section 2. A convenient form of Equation 4.160 is
y2 = y1 + F(x, y)(x2 − x1 )
(4.162)
This equation is appropriate for computing the water-surface profile in the downstream direction. However, in most cases water-surface profiles are computed in the upstream direction,
in which case the following form of Equation 4.160 is more useful
y1 = y2 − F(x, y)(x2 − x1 )
(4.163)
In subcritical flow, calculations generally proceed in an upstream direction, while in supercritical flow calculations proceed in a downstream direction. In applying Equation 4.163, the
following procedure is suggested:
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146
Chapter 4
Fundamentals of Flow in Open Channels
Step 1: Starting with known flow conditions at Section 2, assume a depth, y1 , at location x1 ,
and then calculate y1 using Equation 4.163. On the first calculation, it is reasonable
to assume that y1 = y2 .
Step 2: Repeat Step 1 until the calculated value of y1 is equal to the assumed value of y1 .
This is then the depth of flow at x1 .
A similar procedure is used to apply Equation 4.162, with iterations on y2 rather than y1 .
EXAMPLE 4.9
ww
Water flows at 10 m3 /s in a rectangular concrete channel of width 5 m and longitudinal slope 0.001.
The Manning roughness coefficient, n, of the channel lining is 0.015, and the water depth is measured as
0.80 m at a gaging station. Use the direct-integration method to estimate the flow depth 100 m upstream
of the gaging station.
Solution From the given data: Q = 10 m3 /s, b = 5 m, S0 = 0.001, n = 0.015, y2 = 0.80 m, x1 = 0 m,
and x2 = 100 m. Assuming y1 = y2 = 0.80 m, the hydraulic parameters of the flow are:
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0 + 100
x + x2
=
= 50 m
x= 1
2
2
0.80 + 0.80
y + y2
=
= 0.80 m
y= 1
2
2
A = by = (5)(0.80) = 4.0 m2
P = b + 2y = 5 + 2(0.80) = 6.60 m
4.0
= 0.606 m
6.60
⎤2 ⎡
⎤2
nQ
(0.015)(10)
⎦ =⎣
⎦ = 0.00274
Sf = ⎣
2
2
3
3
(4.0)(0.606)
AR
R=
A
P
⎡
=
by
A
=
= y = 0.80 m
T
b
10
Q
= 2.5 m/s
=
V=
4
A
D=
2
Fr =
V
2
gD
=
(2.5)2
(9.81)(0.80)
= 0.80
Using these results, and assuming α = 1, Equation 4.150 gives
F(x, y) =
S0 − S f
2
1 − Fr
=
g.n
et
0.001 − 0.00274
= −0.0087
1 − 0.80
and the estimated depth 100 m upstream of the gaging station is given by Equation 4.163 as
y1 = y2 − F(x, y)(x2 − x1 ) = 0.80 − (−0.0087)(100 − 0) = 1.67 m
Since this calculated flow depth at Section 1 is significantly different from the assumed flow depth
(= 0.80 m), the calculations must be repeated, starting with the assumption that y1 = 1.67 m. These calculations are summarized in the following table, where the assumed values of y1 are given in Column 1
and the calculated values of y1 (using Equation 4.163) in Column 4:
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Section 4.3
ww
(1)
(2)
(3)
(4)
y1
(m)
Sf
Fr
y1
(m)
1.67
1.5
1.3
1.1
·
·
·
1.01
0.000761
0.000936
0.00122
0.00164
·
·
·
0.00193
0.216
0.268
0.352
0.476
·
·
·
0.559
0.77
0.79
0.83
0.92
·
·
·
1.01
2
Water-Surface Profiles
147
These results indicate that the depth 100 m upstream from the gaging station is approximately equal to
1.01 m.
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4.3.4.2
Direct-step method
In applying the direct-step method, the flow depth at one section is known, the flow depth
at a second section is specified, and the objective is to find the distance between these two
sections. With the flow conditions at two channel sections known, the terms on the right-hand
side of Equation 4.154 are evaluated to determine the distance, $L, between these sections.
These computations are then repeated to find the incremental distances between all adjacent
sections with specified flow depths, thereby yielding the water-surface profile.
The main limitations of the direct-step method are: (1) the water-surface profile is not
computed at predetermined locations, and (2) the method is suitable only for prismatic channels, where the shape of the channel cross section is independent of the interval, $L. In cases
where the flow conditions at specific locations in a prismatic or nonprismatic channel are
required, the standard-step method should be used.
EXAMPLE 4.10
g.n
et
Water flows at 12 m3 /s in a trapezoidal concrete channel (n = 0.015) of bottom width 4 m, side slopes
2 : 1 (H : V), and longitudinal slope 0.0009. If depth of flow at a gaging station is measured as 0.80 m,
use the direct-step method to find the location where the depth is 1.00 m.
Solution At the location where the depth is 1.00 m:
y1 = 1.00 m
A1 = [4 + 2y1 ]y1 = [4 + 2(1.00)](1.00) = 6.00 m2
√
√
P1 = 4 + 2 5y1 = 4 + 2 5(1.00) = 8.47 m
R1 =
A1
6.00
= 0.708 m
=
P1
8.47
Q
12
=
= 2.00 m/s
A1
6.00
⎤2
⎡
⎡
⎤2
⎢ nQ ⎥
(0.015)(12)
⎥ =⎣
⎦ = 0.00143
Sf 1 = ⎢
2 ⎦
2
⎣
3
3
(6.00)(0.708)
A1 R1
V1 =
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148
Chapter 4
Fundamentals of Flow in Open Channels
and where the depth is 0.80 m:
y2 = 0.80 m
A2 = [4 + 2y2 ]y2 = [4 + 2(0.80)](0.80) = 4.48 m2
√
√
P2 = 4 + 2 5 y2 = 4 + 2 5 (0.80) = 7.58 m
R2 =
V2 =
ww
A2
4.48
= 0.591 m
=
P2
7.58
Q
12
= 2.68 m/s
=
A2
4.48
⎡
⎤2 ⎡
⎤2
(0.015)(12)
⎢ nQ ⎥
⎦ = 0.00325
Sf 2 = ⎣
=⎣
2 ⎦
2
3
3
(4.48)(0.591)
A2 R2
Substituting the hydraulic parameters at Sections 1 and 2 into Equation 4.154, taking the velocity coefficients α1 and α2 equal to unity, and using Equation 4.156 to estimate the average friction slope, Sf ,
gives
#
$1
*
)
*
)
2
2
V
2.682
y + α 2g
−
0.80
+
1.00 + 2 2.00
* 9.81
2 * 9.81
2 =
D
E
= 26.4 m
$L =
0.00143
+
0.00325
S f − S0
− 0.0009
2
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Hence the depth in the channel increases to 1.00 m at a location that is approximately 26.4 m upstream
of the section where the depth is 0.80 m.
4.3.4.3
Standard-step method
In applying the standard-step method, the flow depth is known at one section and the objective is to find the flow depth at a second section a given distance away. The standard-step
method is similar to the direct-step method in being based on the solution of Equation 4.154;
in the standard-step method, however, $L is given and the flow depth at the second section
is unknown. The standard-step method is particularly useful in natural channels, where the
dimensions of the cross sections are typically measured at locations that are easily accessible.
EXAMPLE 4.11
g.n
et
Water flows in an open channel whose slope is 0.04%. The Manning roughness coefficient of the channel
lining is estimated to be 0.035, and the flow rate is 200 m3 /s. At a given section of the channel, the cross
section is trapezoidal, with a bottom width of 10 m, side slopes of 2:1 (H:V), and a depth of flow of 7 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 15 m and side slopes of 3:1 (H:V).
Solution A longitudinal view of the flow along with the upstream and downstream channel sections
are illustrated in Figure 4.14.
In this case, Q = 200 m3 /s, n = 0.035, and the flow conditions at Section 2 (downstream section) are
given as y2 = 7 m, b2 = 10 m, and m2 = 2 (side slope is m2 :1). Therefore,
A2 = [b2 + m2 y2 ]y2 = [10 + (2)(7)](7) = 168 m2
1
+
P2 = b2 + 2y2 1 + m22 = 10 + 2(7) 1 + 22 = 41.3 m
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Section 4.3
FIGURE 4.14: Calculation
of water-surface profile
Water-Surface Profiles
149
1
2
200 m 3
/s
y1
y2
Slope
100 m
ww
! 0.0
4%
1
7m
2
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10 m
Section 2
1
y1
3
15 m
Section 1
R2 =
V2 =
A2
168
= 4.07 m
=
P2
41.3
Q
200
=
= 1.19 m/s
A2
168
⎤2 ⎡
⎡
g.n
et
⎤2
(0.035)(200)
nQ
⎥
⎢
⎦ = 0.000267
=⎣
Sf 2 = ⎣
2 ⎦
2
3
3
(168)(4.07)
A2 R2
At the upstream section: b1 = 15 m, m1 = 3, and denoting the depth of flow at the upstream section as
y1 gives
A1 = (b1 + m1 y1 )y1 = (15 + 3y1 )y1
1
+
√
P1 = b1 + 2y1 1 + m21 = 15 + 2y1 1 + 32 = 15 + 2 10y1
R1 =
V1 =
A1
P1
Q
A1
⎡
(4.165)
(4.166)
(4.167)
⎤2
⎢ nQ ⎥
⎥
Sf 1 = ⎢
2 ⎦
⎣
3
A1 R1
Sf =
(4.164)
Sf 1 + S f 2
2
(4.168)
=
Sf 1 + 0.000267
2
= 0.5Sf 1 + 0.000134
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150
Chapter 4
Fundamentals of Flow in Open Channels
Applying the energy equation (Equation 4.154) between Sections 1 and 2 with $L = 100 m and taking
the velocity coefficients, α1 and α2 , to be equal to unity yields
$L =
100 =
#
$1
2
y + αV
2g
2
#
S f − S0
V2
1
y1 + 2(9.81)
$
#
$
1.192
− 7.00 + 2(9.81)
(4.169)
(0.5Sf 1 + 0.000134) − 0.0004
Substituting Equations 4.164 to 4.168 into Equation 4.169 and solving for y1 yields y1 = 7.02 m. Therefore, the depth of the flow 100 m upstream is 7.02 m.
ww
In computing water-surface profiles using the standard-step method, significant errors
can result if adjacent cross sections used in the computations are too far apart and the
hydraulic properties of the flow change too radically from one cross section to the next.
A common guideline is that the slope of the energy grade line should not decrease by more
than 50% or increase by more than 100% between sections (Dodson, 1999). Selecting cross
sections too far apart can result in there being no solution to the energy equation. In general,
cross sections should be located at changes in channel geometry and slope; above and below
major tributaries; and at structures such as bridges, submerged roads, and transitions.
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4.3.4.4
Practical considerations
In performing backwater computations there are several factors that should be considered to
ensure that the results reflect reality and are technically sound. Such considerations include
honoring the discharge-depth relation at control sections, and recognizing that the normal
and critical flow depths can vary between channel reaches. In the case of 100-year flood
flows, computed water-surface profiles are typically used to establish minimum ground-floor
elevations for buildings located in the floodplain of the channel. Computers are typically used
to perform backwater computations.
Control sections. The computation of a water-surface profile generally begins at a section
where the depth of flow is known. In most cases, the depth of flow is known at a section
where there is a unique relationship between the depth and the flow rate. Such sections are
called control sections. A typical control section requires that critical-flow conditions occur at
that section, in which case the depth of flow and the flow rate are related by Equation 4.100.
Examples of control sections include channel constrictions that choke the flow to create critical conditions at the control section and rectangular free overfalls where the depth-discharge
relationship can be estimated using the relation (Tiǧrek et al., 2008)
g.n
et
3
q = Cd yb2
where q is the discharge per unit width; Cd is a discharge coefficient given by
⎧
⎪
5.55
Fr … 1
⎪
⎪
⎪
⎨
√ (− 32
Cd = '
S0
⎪
⎪
⎪
0.361 − 0.00841
Fr > 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
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Section 4.3
Water-Surface Profiles
151
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.
ww
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.
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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
C
C
$L =
V12 CC
C CC V22
− α1 C − S0
Sf +
Cα2
$L C 2g
2g C
(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
g.n
et
(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
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152
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).
ww
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
w.E
asy
En
gin
#"
N
i=1 Ai
α=#
"N
i=1 Ki
$2
$3
N
%
K3
i
(4.177)
A2i
i=1
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
eer
in
g.n
et
This relationship can be conveniently represented by defining the compound-channel Froude
number, Fr∗ , as
αQ2 T
Q2 dα
Fr∗ =
(4.179)
−
gA3
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 250 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 = 250 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
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Section 4.3
FIGURE 4.15: Compound
channel
Water-Surface Profiles
153
23 m
180 m
130 m
n = 0.025
n = 0.090
n = 0.070
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.
ww
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:
w.E
asy
En
gin
A = w2 y2 = 23y2
P = w2 + 2y2 = 23 + 2y2
5
5
(23y2 ) 3
1 A3
1
K=
=
n2 P 23
0.025 (23 + 2y ) 23
2
α=1
Fr∗ =
αQ2 T
(1)(250)2 (23)
12.04
=
=
3
gA
(9.81)(23y2 )3
y32
(4.180)
eer
in
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 2.29 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
3
5
3
1 A1
1 A1
K1 =
=
n1 32
0.090 23
P1
P1
g.n
et
A2 = w2 y2 = 23y2
P2 = w2 + 2"z = 23 + 2(2.52) = 28.04 m
5
5
1 A23
1 A23
=
K2 =
n2 32
0.025 23
P2
P2
A3 = w3 y3 = 130(y2 − 2.52)
P3 = w3 + y3 = 130 + (y2 − 2.52)
5
3
5
3
1 A3
1 A3
=
K3 =
2
n3 3
0.070 23
P3
P3
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Fundamentals of Flow in Open Channels
2
Depth in main channel, y (m)
FIGURE 4.16: Flow
properties in compound
channel
4.0
4.0
3.5
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0
1
2
3
4
5
6
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0.0
7
Kinetic energy correction factor,
ww
Fr* = 1
Chapter 4
Depth in main channel, y2 (m)
154
0.5
1.0
1.5
2.0
2.5
3.0
Compound-channel Froude number, Fr*
A = A1 + A2 + A3
K = K1 + K2 + K3
⎛
⎞
3
K33
K23
A2 ⎝ K1
⎠
α=
+
+
K3 A21
A22
A23
w.E
asy
En
gin
(4.181)
"α
dα
L
dy
"y2
(4.182)
T = w1 + w2 + w3 = 180 + 23 + 130 = 333 m
Fr∗ =
αQ2 T
Q2 dα
−
gA3
2gA2 dy
(4.183)
eer
in
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
g.n
et
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
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Section 4.3
FIGURE 4.17: Cross
sections at bridge
Source: U.S. Army Corps
of Engineers (2010).
CR
Contraction reach
3
BU
BD
A
Le
ww
155
4
1
Lc
Water-Surface Profiles
B
C
D
2
Expansion reach
Idealized flow
transition
ER
1
w.E
asy
En
gin
eer
in
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.
g.n
et
Solution The energy equation to be satisfied between adjacent channel sections is given by Equation
4.173 and can be expressed as
$i
#
2
V
y + α 2g
i+1C
C
(4.185)
$L =
C
D
E
2
2C
V
V
C
C
zi −zi+1
C
i+1
i
Sf + $L Cαi+1 2g − αi 2g C −
$L
C
C
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
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156
Chapter 4
Fundamentals of Flow in Open Channels
FIGURE 4.18:
Flow-through bridge
2.5 m
60 m
150 m
170 m
2.5 m
Sections 1 and 4
10 m
10 m
ww
Sections 2 and 3
L
R
w.E
asy
En
gin
eer
in
5m
10 m
5m
Sections BD and BU
R
L
Flow
4
200 m
3
4m
BU
20 m
BD
2
4m
g.n
et
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.
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Section 4.3
Water-Surface Profiles
157
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
ww
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,
w.E
asy
En
gin
eer
in
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
3
5
1 A11
1 373.5 3
=
= 13573 m3 /s
K11 =
2
n11 3
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 + 2$z = 60 + 2(2.5) = 65 m
5
3
5
1 A12
1 299.4 3
=
= 27628 m3 /s
K12 =
n12 32
0.030 65 23
P12
g.n
et
y13 = y11 = 2.49 m
A13 = w13 y13 = (170)(2.49) = 423.3 m2
P13 = w13 + y13 = 170 + 2.49 = 172.49 m
5
3
5
1 A13
1 423.3 3
=
= 15403 m3 /s
K13 =
n13 32
0.050 172.49 23
P13
A1 = A11 + A12 + A13 = 373.5 + 299.4 + 423.3 = 1096.2 m2
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158
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
ww
Sf 1 =
'
V1 =
Q
300
= 0.27 m/s
=
A1
1096.2
Q
K1
(2
=
)
*
300 2
= 0.0000281
56604
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
w.E
asy
En
gin
eer
in
A21 = w21 y21 = (10)y21
P21 = w21 = 10 m
5
3
1 A21
K21 =
n21 23
P21
y22 = y22
A22 = w22 y22 = (60)y22
P22 = w22 + 2$z = 60 + 2(2.5) = 65 m
5
3
1 A22
K22 =
2
n22 3
P22
y23 = y21
A23 = w23 y23 = 10y23
P23 = w23 = 10 m
5
3
1 A23
K23 =
n23 23
P23
g.n
et
A2 = A21 + A22 + A23
K2 = K21 + K22 + K23
⎛
⎞
3
3
K2i
A22 "
⎝
⎠
α2 =
K23 i=1 A22i
*
Q 2
K2
Sf 2 =
)
V2 =
Q
A2
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Problems
159
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
=
S0 = 2
= 0.0005
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
ww
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
w.E
asy
En
gin
eer
in
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. An open channel has a trapezoidal cross section with a
bottom width of 5 m and side slopes of 2 : 1 (H : V). If
the depth of flow is 2 m and the average velocity in the
channel is 1 m/s, calculate the discharge in the channel.
4.2. Water flows at 8 m3 /s through a rectangular channel 4 m
wide and 3 m deep. If the flow velocity is 1 m/s, calculate
the depth of flow in the channel. If this channel expands
(downstream) to a width of 5 m and the depth decreases
by 0.5 m from the upstream depth, then what is the flow
velocity in the expanded section?
4.3. Show that for circular pipes of diameter, D, the hydraulic
radius, R, is related to the pipe diameter by 4R = D.
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. Water flows in a 8-m-wide rectangular channel that has a
longitudinal slope of 0.0001. The channel has an equiva-
lent sand roughness of 2 mm. Calculate the uniform flow
depth in the channel when the flow rate is 15 m3 /s. Use
the Darcy–Weisbach equation.
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
g.n
et
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
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160
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
n=
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. Water flows at 20 m3 /s in a trapezoidal channel that has
a bottom width of 2.8 m, side slope of 2:1 (H:V), longitudinal slope of 0.01, and a Manning’s n of 0.015. (a) Use
the Manning equation to find the normal depth of flow,
and (b) determine the equivalent sand roughness of the
channel. Assume the flow is fully turbulent.
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
ww
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. Water flows at a depth of 4.00 m in a trapezoidal,
concrete-lined section with a bottom-width of 4 m and
side slopes of 3:1 (H:V). The longitudinal slope of the
channel is 0.0001 and the water temperature is 20◦ C.
Assess the validity of using the Manning equation,
assuming n = 0.013.
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.]
w.E
asy
En
gin
eer
in
1
n = 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.
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
ks = 1 mm
15 cm
Slope = 2%
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
g.n
et
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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:
ww
Section
n
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. If the velocity profile in a channel is described by
the one-seventh power-law relationship, Equation 4.57,
determine: (a) the ratio of the average velocity to the
maximum velocity, and (b) the distance from the bottom
of the channel to where the average velocity occurs.
4.27. Water flows at 8.4 m3 /s in a trapezoidal channel with a
bottom width of 2 m and side slopes of 2:1 (H:V). Over
a distance of 100 m, the bottom width expands to 2.5 m,
with the side slopes remaining constant at 2:1. If the
depth of flow at both of these sections is 1 m and the
channel slope is 0.001, calculate the head loss between
the sections. What is the power in kilowatts that is dissipated?
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
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1
2
3
4
5
6
7
0.040
0.030
0.015
0.013
0.017
0.035
0.060
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
Left floodplain
Main channel
Right floodplain
n
0.040
0.016
0.050
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.
2
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.
4.29. Determine the critical depth for a flow of 30 m3 /s in a
rectangular channel with width 5 m. If the actual depth
of flow is equal to 3 m, is the flow supercritical or subcritical?
4.30. Determine the critical depth for a flow of 50 m3 /s in
a trapezoidal channel with bottom-width 4 m and side
slopes of 1.5 : 1 (H : V). If the actual depth of flow is 3 m,
calculate the Froude number and state whether the flow
is subcritical or supercritical.
4.31. A rectangular channel 2 m wide carries 3 m3 /s of water
at a depth of 1.2 m. If an obstruction 40 cm wide is placed
in the middle of the channel, find the elevation of the
water surface at the constriction. What is the minimum
width of the constriction that will not cause a rise in the
water surface upstream?
4.32. Water flows at 1 m3 /s in a rectangular channel of width
1 m and depth 1 m. What is the maximum contraction of
the channel that will not choke the flow?
Left floodplain
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Right floodplain
S ! 1%
S ! 4%
Main channel
3m
1
3
161
1
2
30 m
FIGURE 4.20: Flow in an open channel
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162
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
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3m
3
10 m
raised by 0.25 m, what is the depth of flow in the constriction?
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. Water flows at 20 m3 /s with a uniform depth of 3 m in
a trapezoidal channel of base width 3 m and side slopes
1 : 1. If a channel transition restricts the flow locally by
raising the side walls to the vertical position, calculate
the depth of water in the rectangular constriction. What
is the minimum allowable width of the constriction to
prevent choking?
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
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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 rectangular channel 3 m wide carries 4 m3 /s of water
at a depth of 1.5 m. If an obstruction 15 cm high is
placed across the channel, calculate the elevation of the
water surface over the obstruction. What is the maximum height of the obstruction that will not cause a rise
in the water surface upstream?
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 at 4.3 m3 /s in a rectangular channel of width
3 m and the depth of flow 1 m. If the channel width is
decreased by 0.75 m and the bottom of the channel is
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Problems
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?
4.44. 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?
4.45. Water flows at 30 m3 /s in a rectangular channel of width
8 m. Manning’s n of the channel is 0.035. Determine the
range of channel slopes that would be classified as steep
and the range that would be classified as mild.
4.46. 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.
4.47. Water discharges from a large storage reservoir via a
trapezoidal channel of bottom width 3.00 m, side slopes
3:1, and longitudinal slope of 0.5%. Manning n in the
channel is estimated as 0.025. Estimate the discharge
from the reservoir when the depth of the reservoir at the
discharge location is 2.00 m.
4.48. 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.]
4.49. If 100 m3 /s of water flows in a channel 8 m wide at a
depth of 0.9 m, calculate the downstream depth required
to form a hydraulic jump and the fraction of the initial
energy lost in the jump.
4.50. 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 for a rectangular section the dimensionless
head loss, hL /y1 , is given by
!
" #2 $
Fr21
hL
y2
y1
1 −
=1 −
+
y1
y1
2
y2
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163
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
my31
by21
Q2
+
+
=
2
3
gy1 (b + my1 )
by22
my32
Q2
2
3
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.
4.53. 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?
4.54. A flume with a triangular cross section and side slopes
of 2 : 1 (H : V) contains water flowing at 0.30 m3 /s at a
depth of 15 cm. Verify that the flow is supercritical and
calculate the conjugate depth.
4.55. 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?
4.56. 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?
4.57. If the depth of flow in the channel described in Problem
4.10 is measured as 1.4 m, find the location where the
depth is 1.6 m.
4.58. 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:
+
+
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where Fr1 is the upstream Froude number.
(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?
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164
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. A rectangular channel 6 m wide carries a discharge of
0.8 m3 /s. At a certain section the channel roughness
changes suddenly. The normal depths in the reaches
upstream and downstream of sudden change are 0.9 m
and 0.7 m, respectively. The channel slope is 0.5%. Using
a single step of the direct-step method, estimate the
length of the reach of nonuniform flow.
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?
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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
et
FIGURE 4.22: Flow in upstream and downstream sections
100 m
3m
Elevation = 50.00 m NAVD88
10 m
FIGURE 4.23: Floodplain
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Problems
!
V2
Z + α
2g
"B
A
165
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
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 250 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.
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Area (m2 )
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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
—
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C H A P T E R
5
Design of Drainage Channels
5.1 Introduction
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FIGURE 5.1:
Concrete-lined canal
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
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Section 5.2
Basic Principles
167
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
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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.
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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
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(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
y
A
1
Freeboard
m
b
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(5.3)
(5.4)
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168
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
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⎛
(5.6)
⎞3
⎜ Qn ⎟
c=⎝ 1 ⎠
S02
5
(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)
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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
g.n
et
(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).
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Section 5.2
Basic Principles
169
TABLE 5.1: Geometric Characteristics of Best Hydraulic Sections
ww
Shape
Best geometry
Trapezoid
Rectangle
Half of a hexagon
A
√ 2
3y
P
√
2 3y
T
√
4 3y
3
Half of a square
2y2
Triangle
Half of a square
4y
√
2 2y
2y
y2
Semicircle
—
Parabola
—
π y2
2
√
4 2y2
3
πy
√
8 2y
3
2y
√
2 2y
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)
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α=
)
'
1 (
m + 1 − 1 + m2
m
(5.12)
*
(5.14)
b = 2(1 − αm)y
1
y=
24
3
(2 − α 2 m) 8
nQ
√
S0
(5.13)
+3
8
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:
g.n
et
◃ 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
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170
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).
ww
EXAMPLE 5.1
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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
g.n
et
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
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Section 5.2
FIGURE 5.4: Shear-stress
distribution in a
trapezoidal channel
1
Basic Principles
171
1
m
m
y
b
τs
τs
τb
ww
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)
(5.16)
τb = γ yS0
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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
(5.17)
τs = Ks τb
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
g.n
et
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
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172
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
ww
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)
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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
0
u∗c ds
ν
1
g.n
et
(5.20)
where f2 denotes a functional relationship, u∗c is the friction velocity corresponding to τc and
2
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.
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Section 5.2
Basic Principles
FIGURE 5.5: Sheilds
diagram
γs
(g/cm3)
Material
Source: ASCE (2006b).
*
Dimensionless shear stress, τ =
τc
(γs – γ)d s
Amber
Lignite
Barite
ww
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
0.1
0.08
4
6 8 1
ds
ν
173
(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
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0.06
0.05
0.04
0.03
0.02
0.2
×
0.4 0.6
1.0
2
×
4
+
++
+×
6
Shields curve
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 N/m3 and
ν = 1.00 * 10−6 m2 /s. Hence,
# &
'
! "$
"( )
! "
!
,
*
+
ds $
γs
0.001
%0.1
− 1 gds =
0.1 2.65 − 1 (9.81)(0.001) = 40
ν
γ
1.00 * 10−6
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
Ap τp = µp wp
(5.23)
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174
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
ww
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
3
(5.26)
Fp = (τs Ap )2 + (wp sin θ )2
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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
3
wp cos θ tan α = (τps Ap )2 + (wp sin θ )2
which can be put in the form
4
wp
tan2 θ
τps =
cos θ tan α 1 −
Ap
tan2 α
g.n
et
(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
4
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
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Section 5.2
Basic Principles
175
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◦
†d
50 is in centimeters
ww
0
5
1.
7
1.
6
0.
0.
3
0.
2
15
0.
0.
01
07
0.
06
0.
0.
03
0.
0.
43
02
Median stone size, d 50 (ft)
39
k
ed
ro
c
41
Cr
us
h
ang
ula
r
37
Very
nded
35
33
Very ro
u
Source: USFHWA (2005).
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Angle of repose, α (degrees)
FIGURE 5.6: Angle of
repose of gravel and
riprap linings
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 .
31
00
0 10
20
30 40 50 60 100
200 300 400 600
Median stone size, d 50 (mm)
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176
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
ww
Angle of repose, α (degrees)
38
36
r
ula
ng
a
ery
34
V
te
era
32
d
Mo
30
28
26
Sli
24
M
r
ula
g
an
y
htl
g
Sli
y
htl
g
d
de
n
ou
r
y
el
t
ra
e
od
lar
gu
n
ly a
ry
Ve
d
de
n
ou
r
ed
nd
u
ro
22
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20
3
4
5
6
7 8 9 10
20
30
Particle size, d75 (mm)
40 50 60 70
100
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
0 1
0 1
1
1
= tan−1
= 26.6◦
θ = tan−1
m
2
g.n
et
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
1
0
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
5
6
6
sin2 (26.6)
= 0.698
K = 71 −
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
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Section 5.2
5.2.4
Basic Principles
177
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
(5.30)
τr = Kr τb
ww
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
⎪
⎪
⎪
0 12 T
0 1
⎨
r
r
r
Kr = 2.38 − 0.206 c + 0.0073 c , 2 < c < 10
(5.31)
⎪
T
T
T
⎪
⎪
r
⎪
c
⎪
⎩1.05,
Ú 10
T
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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.
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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=
and hence
(2.5)(1.00) + (3)(1.00)2
by + my2
'
'
=
= 0.623 m
b + 2y 1 + m2
2.5 + 2(1.00) 1 + 32
20
rc
=
= 2.35
T
8.5
Using Equation 5.31, the bend factor, Kr , can be estimated by
0 12
0 1
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
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178
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
R6 ⎠
0.623 6 ⎠
Lp = 0.604R ⎝
= 0.604(0.623) ⎝
= 10.8 m
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
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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.
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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
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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
Side slope (H:V)
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
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
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Section 5.2
Basic Principles
179
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
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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
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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).
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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.
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180
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
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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.
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TABLE 5.4: Manning Roughness Coefficients in Open Channels with Rigid Linings
Range of Manning’s n
Type
Characteristics
Minimum
Normal
Maximum
Cement
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
—
—
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
—
0.015
0.017
0.016
0.017
0.020
0.020
0.020
0.024
0.024
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
Concrete
Concrete bottom float
finished with sides of
Grouted riprap
Stone masonry
Soil cement
Asphalt
Brick
Masonry
Dressed ashlar
0.020
0.020
0.028
0.030
0.020
0.016
0.013
0.011
0.012
0.017
0.023
0.013
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Sources: USFHWA (2005); Chow (1959).
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Section 5.3
Design of Channels with Rigid Linings
181
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.
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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).
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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
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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).
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182
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 =
ww
2
1
1
(2.10y2 )(0.499y) 3 (0.0015) 2
0.015
or
8
y 3 = 5.86
which leads to
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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)
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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.
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Section 5.4
Design of Channels with Flexible Linings
183
FIGURE 5.8: Roadside
channel with rigid lining
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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
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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)
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184
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).
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TABLE 5.6: Typical Manning’s n for Riprap, Cobble, and Gravel Linings
Flow depth
0.15 m
(0.5 ft)
0.50 m
(1.6 ft)
1.0 m
(3.3 ft)
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
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Lining category
Lining type
Gravel mulch
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)
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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.
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Section 5.4
Design of Channels with Flexible Linings
185
TABLE 5.7: Typical Permissible Shear Stresses for Bare Soil and Stone Linings
Permissible shear stress
Lining category
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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
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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
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'
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
Q=
1 A 3 21
S
n P 32 0
5
1
1 (3.325y2 ) 3
0.5 =
(0.005) 2
0.020 (6.650y) 23
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
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186
Chapter 5
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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
ww
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
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5
1 A 3 12
S
Q=
n P 23 0
5
0.5 =
1
1 (3.325y2 ) 3
(0.005) 2
2
n(y) (6.650y) 3
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
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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
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Section 5.4
Design of Channels with Flexible Linings
187
Step 4: The permissible shear stress, τp , on the gravel-mulch lining can be estimated from Table 5.7
for d50 = 50 mm = 5 cm 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)(5)0.00778 = 34.7◦
The side-slope angle θ is given by
! "
! "
1
1
−1
−1
= tan
= 18.4◦
θ = tan
m
3
ww
The tractive-force ratio, K, is given by Equation 5.29 as
#
/
$
2
$
sin θ
sin2 (18.4)
K= 1 −
= %1 −
= 0.832
2
sin α
sin2 (34.7)
and hence the permissible shear stress on the side of the channel, τps , is given by
τps = Kτp = (0.832)(38) = 31.6 Pa
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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
(31.6 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
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188
Chapter 5
Design of Drainage Channels
TABLE 5.8: Retardance in Grass-Lined Channels
ww
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.)
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Weeping lovegrass
E
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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
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Section 5.4
Design of Channels with Flexible Linings
189
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.
ww
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
⎪
⎪
⎪
⎪
very good condition
⎨290,
(5.41)
Cs = 106,
good condition
⎪
⎪
⎪
24,
fair
condition
⎪
⎪
⎪
⎩ 8.6, poor condition
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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
τ0 = γ RS0
(5.44)
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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
τe = τb (1 − Cf )
0
ns
n
12
(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
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Chapter 5
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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).
ww
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
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τp,n =
8
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
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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
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Design of Channels with Flexible Linings
191
TABLE 5.10: Coefficients for Permissible Soil Shear Stress
ASTM
class∗
ww
Typical names
Applicable
range
GM
Silty gravels
Gravel-sand-silt mixtures
10 … PI … 20
PI Ú 20
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
SM
Silty sands
Sand-silt mixtures
10 … PI … 20
PI Ú 20
1.07
7.15
11.9 1.42
0.058 1.42
SC
Clayey sands
Sand-clay mixtures
10 … PI … 20
PI Ú 20
1.07
14.3
47.7 1.42
0.076 1.42
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
CL
Inorganic clays, low or
medium PI
Gravelly clays
Sandy clays
Silty clays
Lean clays
10 … PI … 20
PI Ú 20
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
c1
c2
c3
1.07
14.3
47.7 1.42
0.076 1.42
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PI Ú 20
1.07
14.3
c4
47.7 1.48
0.076 1.48
c5
c6
−0.61
−0.61
4.8 * 10−3
48
−0.61
−0.61
−0.61
−0.61
−0.61
−0.61
4.8 * 10−2
48
4.8 * 10−3
48
4.8 * 10−3
48
−0.57
−0.57
4.8 * 10−3
48
−0.57
−0.57
4.8 * 10−3
48
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).
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Stress range, Pa (lb/ft2)
10<PI<20
1.3 (0.03) to 4.5 (0.09)
20<PI
3.9 (0.08) to 4.5 (0.09)
20<PI
5.7 (0.12)
Fine grained
Cohesive
Clay
10<PI<20
4.6 (0.10) to 7.1 (0.15)
Coarse grained
20<PI
7.1 (0.15)
Step 1: Determine the flow depth in the channel. Equation 5.41 gives Cs = 106, and Equation 5.42
gives
Cn = 0.35Cs0.10 h0.528 = 0.35(106)0.10 (0.10)0.528 = 0.165
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The average shear stress on the channel boundary, τ0 , is given by Equation 5.44 as (assuming γ = 9790 N/m3 )
*
+
y + 4y2
195.8(y + 4y2 )
τ0 = γ RS0 = (9790)
(0.02) =
1 + 8.246y
1 + 8.246y
and Manning’s n is given by Equation 5.43 as
n = Cn τ0−0.4 = (0.165)
9
195.8(y + 4y2 )
1 + 8.246y
The Manning equation gives
ww
:−0.4
*
1 + 8.246y
= 0.0200
y + 4y2
+0.4
(5.49)
5
1 A 3 21
Q=
S
n P 32 0
5
0.8 =
1
1 (y + 4y2 ) 3
(0.02) 2
2
n (1 + 8.246y) 3
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(5.50)
Solving Equations 5.49 and 5.50 simultaneously yields y = 0.301 m and n = 0.039.
Step 2: Determine the effective stress on the underlying soil. The maximum shear stress on the
bottom of the channel, τb , is given by
τb = γ yS0 = (9790)(0.301)(0.02) = 58.9 Pa
For sod grass in good condition, Table 5.9 gives Cf = 0.90, for d75 = 0.1 mm Equation 5.45
gives ns = 0.016, and the effective shear stress on the soil underlying the grass lining is
given by Equation 5.46 as
τe = τb (1 − Cf )
0
1
1
0
ns 2
0.016 2
= (58.9)(1 − 0.9)
= 1.0 Pa
n
0.039
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Step 3: Determine the permissible shear stress on the soil underlying the vegetative lining and
assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters
to estimate the permissible shear stress are given in Table 5.10 as: c1 = 1.07, c2 = 7.15,
c3 = 11.9, c4 = 1.48, c5 = −0.57, and c6 = 4.8 * 10−3 . The permissible shear stress on the
underlying soil, τp,c , is then given by Equation 5.48 as
τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6
(
)
= [1.07(12)2 + 7.15(12) + 11.9][1.48 + (−0.57)(0.4)]2 4.8 * 10−3
= 1.9 Pa
Since the shear stress on the underlying soil (1.0 Pa) is less than the permissible shear stress
on the underlying soil (1.9 Pa), the proposed grass lining is adequate.
Drainage channels in bare soils are seldom used as permanent channels in practice since
under these conditions the perimeter shear stress acts directly on the soil, which severely
limits the capacity of the channel relative to what it would be with a lining. A unique feature
to be considered when designing channels with bare-soil linings is that while for noncohesive
soils the critical-shear-stress conditions usually occur on the sides of the channel, for cohesive
soils they occur on the bottom of the channel. The design of a bare-soil channel is illustrated
in the following example.
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Section 5.4
Design of Channels with Flexible Linings
193
EXAMPLE 5.11
Design a trapezoidal channel in bare soil to carry 1 m3 /s on a slope of 0.2%. The channel will be
excavated in noncohesive coarse alluvium with a 75-percentile diameter of 1 cm. The soil particles
are moderately rounded.
Solution From the given data: Q = 1 m3 /s, S0 = 0.002, and d75 = 10 mm (moderately rounded). For
stability and functionality of the channel, specify the side slopes as 3:1 (H:V), so m = 3.
Step 1: Determine the maximum depth of flow for stability of the channel. The side-slope angle,
θ , and side-shear-stress factor, Ks , are given by
0 1
0 1
1
1
= tan−1
= 18.4◦
θ = tan−1
m
3
Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868
ww
The angle of repose, α, for d75 = 10 mm (moderately rounded) is given by Figure 5.7 as
α = 27.0◦ , and so the tractive force ratio, K, is given by
5
4
6
6
sin2 θ
sin2 (18.4)
K= 1 −
= 71 −
= 0.719
sin2 α
sin2 (27.0)
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It is worth noting that since K < Ks it is guaranteed that the side shear stress controls the
stability of the channel. Equation 5.47 gives the permissible shear stress on the soil as
τp = 0.75d75 = 0.75(10) = 7.5 Pa
and so the permissible shear stress on the side of the channel, τps , is given by
τps = Kτp = (0.719)(7.5) = 5.4 Pa
(5.51)
The maximum shear stress exerted by flowing water on the sides of the channel, τs , is given
by (for γ = 9790 N/m3 )
τs = Ks γ yS0 = (0.868)(9790)y(0.002) = 17.0y
(5.52)
At the limit of stability, τs = τps and combining Equations 5.51 and 5.52 gives
17.0y = 5.4 Pa
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which yields y = 0.317 m. Therefore, as long as the depth of flow is less than 31.7 cm the
channel will be stable.
Step 2: Determine the minimum width of the channel to accommodate the design flow rate. For a
trapezoidal channel, the area, A, and wetted perimeter, P, are given by
A = by + my2 = b(0.317) + 3(0.317)2 = 0.317b + 0.301
'
'
P = b + 2y 1 + m2 = b + 2(0.317) 1 + 32 = b + 2.005
Since d75 Ú 1.3 mm, Equation 5.45 estimates Manning’s n as
1
1
6
n = ns = 0.015d75
= 0.015(10) 6 = 0.022
and the Manning equation requires that
5
1 A 3 12
S
Q=
n P 23 0
5
1=
1
1 (0.317b + 0.301) 3
(0.002) 2
2
0.022 (b + 2.005) 3
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Chapter 5
Design of Drainage Channels
which yields b = 2.96 m. Therefore, as long as the bottom width of the channel is greater
than or equal to 2.96 m the channel will be stable. This bottom width should be checked
against the bottom width of the best hydraulic section, which is given by Equation 5.11 as
)
('
)
('
1 + m2 − m = 2(0.317)
1 + 32 − 3 = 0.103 m
b = 2y
Therefore in this case the best hydraulic section will not be stable, and so the required
bottom width of the channel is 2.96 m.
Step 3: Calculate the freeboard and finalize the channel design. For a flow rate of 1 m3 /s,
b = 2.96 m, y = 0.317 m, and m = 3, the flow area is 1.24 m2 and the average velocity is
V = 0.81 m/s. Therefore, in accordance with Equation 5.34, the minimum required freeboard is 0.30 m. Hence the final channel design has a bottom width of 2.96 m, side slopes
of 3:1 (H:V), and a minimum depth of 0.317 m + 0.30 m L 0.62 m.
ww
Retardance-based design of vegetative linings. The retardance-based approach derives
some of its parameters from the retardance of the vegetative lining in accordance with the
classifications shown in Table 5.8. Values of grass-roughness coefficient, Cn , and densitystiffness coefficient, Cs , are assigned based on typical vegetation characteristics within each
retardance classification, and these assigned values are given in Table 5.11. Retardance
classes A and B are not commonly found in roadway applications, since they reflect overgrown conditions, and also class E is not commonly found in roadside channels, since it
reflects a drainage channel in poor condition. Typical values of Cn in roadside drainage
channels are in the range of 0.1–0.3.
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TABLE 5.11: Typical Values of Stem Height, Cs , and Cn for Various Retardance Classes
Retardance class
Stem height (cm)
Density-stiffness coefficient, Cs
Grass-roughness coefficient, Cn
A
B
C
D
E
91
390
0.605
61
81
0.418
20
47
0.220
10
33
0.147
4
44
0.093
Source: USFHWA (2005).
EXAMPLE 5.12
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Consider a trapezoidal channel with a bottom width of 1 m, side slopes of 4:1 (H:V), and a longitudinal
slope of 2%. The channel is excavated in a cohesionless soil with a 75-percentile grain size of approximately 0.8 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 vegetation with class C retardance. Assess the adequacy of
the proposed lining.
Solution From the given data: b = 1 m, m = 4, S0 = 0.02, cohesionless soil, d75 = 0.8 mm,
Q = 0.8 m3 /s, and class C retardance. 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
Step 1: Determine the flow depth in the channel. Table 5.11 gives Cn = 0.220. The average shear
stress on the channel boundary, τ0 , is given by Equation 5.44 as (assuming γ = 9790 N/m3 )
*
+
y + 4y2
195.8(y + 4y2 )
τ0 = γ RS0 = (9790)
(0.02) =
1 + 8.246y
1 + 8.246y
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Section 5.4
Design of Channels with Flexible Linings
and Manning’s n is given by Equation 5.43 as
9
*
:−0.4
+0.4
195.8(y + 4y2 )
1 + 8.246y
= 0.0267
n = Cn τ0−0.4 = (0.220)
1 + 8.246y
y + 4y2
195
(5.53)
The Manning equation gives
5
Q=
1 A 3 21
S
n P 32 0
5
1
1 (y + 4y2 ) 3
0.8 =
(0.02) 2
n (1 + 8.246y) 23
(5.54)
Solving Equations 5.53 and 5.54 simultaneously yields y = 0.339 m and n = 0.050.
ww
Step 2: Determine the effective shear stress on the underlying soil. The maximum shear stress on
the bottom of the channel, τb , is given by
τb = γ yS0 = (9790)(0.339)(0.02) = 66.4 Pa
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Assuming a mixed cover in good condition, Table 5.9 gives Cf = 0.75, for d75 = 0.8 mm,
Equation 5.45 gives ns = 0.016, and the effective shear stress on the soil underlying the
grass lining is given by Equation 5.46 as
1
0 12
0
ns
0.016 2
= (66.4)(1 − 0.75)
= 1.7 Pa
τe = τb (1 − Cf )
n
0.050
Step 3: Determine the permissible shear stress on the soil underlying the vegetative lining and
assess the adequacy of the lining. Since the underlying soil is noncohesive, the permissible
shear stress is given by Equation 5.47 as 1 Pa. Since the shear stress on the underlying soil
(1.7 Pa) is greater than the permissible shear stress on the underlying soil (1.0 Pa), the
proposed grass lining is inadequate.
Alternative retardance-based design of vegetative linings. Many guidelines and regulatory
requirements for designing drainage channels with vegetative linings still use an alternative
retardance-based approach instead of the more state-of-the-art approaches described above.
The alternative retardance-based approach assigns a retardance to the vegetative lining in accordance with the classifications shown in Table 5.8, and Manning’s n is expressed
directly in terms of the flow depth, channel slope, and retardance according to the relation
1
n=
1.22R 6
a + 19.97 log(R1.4 S00.4 )
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(5.55)
where R is the hydraulic radius [m], S0 is the channel slope [dimensionless], and a depends
on the retardance class of the vegetation according to the relation
⎧
⎪
⎪
30.2, Class A
⎪
⎪
⎪
⎪
37.4,
Class B
⎨
(5.56)
a = 44.6, Class C
⎪
⎪
⎪49.0, Class D
⎪
⎪
⎪
⎩52.1, Class E
The effective shear stress, τb [FL−2 ], on the channel boundary is taken as
τb = γ yS0
(5.57)
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196
Chapter 5
Design of Drainage Channels
TABLE 5.12: Typical Permissible Shear Stresses of Vegetative Linings
Permissible shear stress
Lining category
Lining type
Pa
lb/ft2
Vegetative
Class A
Class B
Class C
Class D
Class E
177.2
100.6
47.9
28.7
16.8
3.70
2.10
1.00
0.60
0.35
Source: USFHWA (1988).
ww
where y is the maximum flow depth [L], and the permissible shear stress, τp [FL−2 ], is taken
to be a function of the retardance of the lining material as shown in Table 5.12. A limitation of
the alternative retardance-based design is that it does not take into consideration the ability
of the underlying soil to support the vegetation.
EXAMPLE 5.13
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Consider a trapezoidal channel with a bottom width of 1 m, side slopes of 4:1 (H:V), and a longitudinal
slope of 2%. The design flow rate to be accommodated by the channel is 0.8 m3 /s. It is proposed that
the channel be lined with a good stand of Buffalo grass having a height of 10 cm during the season in
which the design flow rate is likely to occur. Using the alternative retardance-based approach, assess
the adequacy of the proposed lining.
Solution From the given data: b = 1 m, m = 4, S0 = 0.02, and Q = 0.8 m3 /s. For a good stand of Buffalo grass of height 10 cm, Table 5.8 indicates a retardance classification of D. 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
(5.58)
Step 1: Determine the flow depth in the channel. For a retardance of D, Equation 5.56 gives
a = 49.0, and Equation 5.55 gives Manning’s n as
1
n=
g.n
et
1
1.22R 6
1.22R 6
< =
<
;
;
1.4 (0.02)0.4
49.0 + 19.97 log R1.4 S0.4
49.0
+
19.97
log
R
0
The Manning equation requires that
5
Q=
1 A 3 21
S
n P 32 0
0.8 =
1
1 (y + 4y2 ) 3
(0.02) 2
2
n (1 + 8.246y) 3
(5.59)
5
(5.60)
Solving Equations 5.58, 5.59, and 5.60 simultaneously yields y = 0.357 m and n = 0.056.
Step 2: Determine the maximum shear stress on the bottom of the channel. The maximum shear
stress on the bottom of the channel, τb , is given by (assuming γ = 9790 N/m3 )
τb = γ yS0 = (9790)(0.357)(0.02) = 69.9 Pa
Step 3: Determine the permissible shear stress on the channel lining. The permissible shear stress
on a (retardance) class D lining, τp , is given in Table 5.12 as 28.7 Pa. Since the maximum
shear stress exerted on the bottom of the channel (69.9 Pa) is greater than the permissible
shear stress on the grass lining (28.7 Pa), the proposed lining is inadequate.
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Section 5.4
5.4.3
Design of Channels with Flexible Linings
197
RECP Linings
A variety of rolled erosion control products are used for erosion protection, and these products generally consist of materials that are stitched or bound into a fabric. Proper installation
of RECPs is critical to their performance, with installation requirements typically including
stapling the lining to the channel perimeter and lapping of adjacent fabric edges.
Manning’s n. The Manning’s n of a RECP is assumed to depend on the boundary shear
stress exerted by the flowing water. Consequently, manufacturers provide functional relationships between n values and the boundary shear stress, τ . Based on this functional relationship,
three points on this curve are identified: (τlower ,n lower ), (τmid ,nmid ), and (τupper ,nupper ), where
(τmid ,nmid ) is a midrange point, τupper = 2τmid , and τlower = τmid /2. Based on these values,
Manning’s n can be estimated by the relation
ww
n = aτ0b
(5.61)
where τ0 is the average shear stress on the channel boundary as given by Equation 5.44 [Pa],
and a and b are determined from the n versus τ relationship as
4 0
1 0
1
nupper
nmid
ln
(5.62)
b = −1.44 ln
nlower
n mid
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−b
a = nmid · τmid
(5.63)
Effective shear stress. RECPs provide shear-stress reduction primarily by providing cover
for the soil surface, and the effective stress, τe [Pa], on the soil surface can be estimated by
the relation
1
10
0
6.5
τI
τe = τb −
(5.64)
4.3
τI
where τb is the shear stress exerted by the flowing water on the bottom of the channel [Pa],
and τI is the shear stress on the RECP that results in 12.5 mm (0.5 in.) of soil erosion [Pa].
The value of τI is determined based on a standard soil specified in a standard testing protocol
(USFHWA, 2005).
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Permissible soil shear stress. The permissible shear stress on the underlying soil can be
estimated using Equation 5.47 for noncohesive soils and Equation 5.48 for cohesive soils.
The RECP lining is adequate when the effective soil shear stress is less than the permissible
soil shear stress.
EXAMPLE 5.14
Consider a trapezoidal channel with a bottom width of 1.5 m, side slopes of 3:1 (H:V), and a longitudinal
slope of 2%. The channel is excavated in very fine sand (ML classification) that has a plasticity index
of 15 and a void ratio of 0.4. The design flow rate in the channel is 1.2 m3 /s. Assess the adequacy of
a temporary RECP lining for which the shear stress that causes 12.5 mm of soil loss is 60 Pa, and the
roughness rating characteristics of the lining are as follows:
Applied shear
(Pa)
Manning’s n
35
70
140
0.038
0.034
0.031
Solution From the given data: b = 1.5 m, m = 3, S0 = 0.02, Q = 1.2 m3 /s, soil classification ML,
PI = 15, and e = 0.4. For the proposed RECP lining, τI = 60 Pa. For any flow depth, y, the flow area,
A, wetted perimeter, P, and hydraulic radius, R, are given by
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198
Chapter 5
Design of Drainage Channels
A = by + my2 = 1.5y + 3y2
'
'
P = b + 2y 1 + m2 = 1.5 + 2y 1 + 32 = 1.5 + 6.325y
R=
1.5y + 3y2
A
=
P
1.5 + 6.325y
Step 1: Determine the flow depth in the channel. For the RECP lining, the parameters a and b are
given by Equations 5.62 and 5.63 as
5 *
4 0
+ *
+
6
1 0
1
6
nupper
nmid
0.031
0.034
7
ln
= −0.146
b = −1.44 ln
ln
= −1.44 ln
nlower
nmid
0.038
0.034
−b
a = nmid · τmid
= (0.034) · (70)0.146 = 0.0632
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The average shear stress on the channel boundary, τ0 , is given by (assuming γ = 9790 N/m3 )
*
+
*
+
1.5y + 3y2
1.5y + 3y2
τ0 = γ RS0 = (9790)
(0.02) = 195.8
1.5 + 6.325y
1.5 + 6.325y
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Manning’s n is given by Equation 5.61 as
⎡
*
*
+⎤−0.146
+−0.146
2
1.5y
+
3y
1.5y + 3y2
b
⎣
⎦
= 0.0292
n = aτ0 = (0.0632) 195.8
1.5 + 6.325y
1.5 + 6.325y
(5.65)
The Manning equation requires that
5
1 A 3 21
S
Q=
n P 32 0
5
1.2 =
1
1 (1.5y + 3y2 ) 3
(0.02) 2
2
n (1.5 + 6.325y) 3
(5.66)
Solving Equations 5.65 and 5.66 simultaneously gives y = 0.330 m and n = 0.036.
Step 2: Determine the effective stress on the underlying soil. The maximum shear stress on the
bottom of the channel, τb , is given by
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τb = γ yS0 = (9790)(0.330)(0.02) = 64.6 Pa
and the effective shear stress on the underlying soil, τe , is given by Equation 5.64 as
10
1 0
10
1
0
6.5
60
6.5
τ
= 64.6 −
= 5.5 Pa
τe = τb − I
4.3
τI
4.3
60
Step 3: Determine the permissible shear stress on the underlying soil and assess the adequacy of
the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible
shear stress are given in Table 5.10 as: c1 = 1.07, c2 = 7.15, c3 = 11.9, c4 = 1.48, c5 = −0.57,
and c6 = 4.8 * 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given
by Equation 5.48 as
τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6
= [1.07(15)2 + 7.15(15) + 11.9][1.48 + (−0.57)(0.4)]2 (4.8 * 10−3 )
= 2.2 Pa
Since the shear stress on the underlying soil (5.5 Pa) is greater than the permissible shear
stress on the underlying soil (2.2 Pa), the proposed RECP lining is inadequate. A more
effective temporary lining should be sought.
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Section 5.4
5.4.4
Design of Channels with Flexible Linings
199
Riprap, Cobble, and Gravel Linings
Riprap, cobble, and gravel linings are typically used as permanent linings. These linings are
characterized by their 50-percentile (median) sizes, d50 , with typical ranges as follows:
Gravel: d50 = 15–64 mm (0.6–2.5 in.)
Cobble: d50 = 64–130 mm (2.5–5.0 in.)
Riprap: d50 = 130–550 mm (5.0–22.0 in.)
ww
Other differences between gravel, cobble, and riprap linings include gradation and angularity. The smallest stone size usually considered for rock riprap is about 50 mm (2 in.), and
stone sizes greater than about 300 mm (12 in.) are uncommon in small-to-moderately sized
drainage channels on all but the steepest of slopes (Froehlich, 2011). A typical riprap-lined
drainage channel is shown in Figure 5.11, where the size and angularity of the stones are
clearly apparent.
Manning’s n. For gravel and riprap linings, Manning’s n has been shown to be a function of
a variety of factors including the flow depth, stone sizes on the perimeter of the channel, and
the friction slope. A variety of relationships between Manning’s n and these parameters have
been suggested; however, for the conditions commonly encountered in roadside and other
small channels, USFHWA (2005) recommends relationships developed by Blodgett (1986)
and Bathurst (1985) which give
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⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
1
0.319y 6
0
y
2.25 + 5.23 log
n=
d50
⎪
1
⎪
⎪
6
y
⎪
⎪
⎪
⎩ √gf f f
1 2 3
1
1.5 …
0.3 <
y
… 185 (Blodgett, 1986)
d50
(5.67)
y
< 1.5 (Bathurst, 1985)
d50
where y is the average flow depth [m]; d50 is the 50-percentile gravel or riprap size on the
channel boundary [m]; g is gravity [= 9.81 m/s2 ], and f1 , f2 , and f3 are defined as
f1 =
FIGURE 5.11:
Riprap-lined drainage
channel
0
0.28Fr
β
1log(0.755/β)
g.n
et
(5.68)
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200
Chapter 5
Design of Drainage Channels
f2 = 13.434
f3 =
0
0 1−β
T
y
T
d50
10.492
0.118
β 1.025(T/d50 )
(5.69)
(5.70)
where T is the top width of the flow [m] and
Q
'
A gy
0
1
1
0
d50 0.453 y 0.814
β = 1.14
T
d50
Fr =
ww
(5.71)
(5.72)
Exerted shear stress. In riprap, cobble, and gravel linings the maximum shear stress on the
bottom of the channel, τb [FL−2 ], is calculated using the relation
τb = SF · γ yS0
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(5.73)
where SF is a safety factor. The maximum shear stress on the side of the channel, τs , is
given by
τs = Ks τb
(5.74)
where Ks is the side-shear-stress factor which, for trapezoidal channels, depends on the side
slope and can be estimated by Equation 5.18.
Permissible shear stress. The permissible shear stress, τp [Pa], on the bottom of riprap-,
cobble-, and gravel-lined channels is estimated by (USFHWA, 2005)
τp =
⎧
⎪
⎪
⎨
τ∗ (γs − γ )d50 ,
1
⎪
⎪
⎩ τ∗ (γs − γ )d50 ,
)
for S0 … 5%
for S0 > 10%
(5.75)
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where τ∗ is Shield’s parameter [dimensionless], γs is the specific weight of the stone [N/m3 ],
γ is the specific weight of water [= 9790 N/m3 at 20◦ C], d50 is the median stone size [m],
and ) is an adjustment factor to account for the effect of the longitudinal component of
the stone weight on reducing the permissible stress. Typically γs = 25, 900 N/m3 ; however
the site-specific value for γs should be used if different. The Shield’s parameter, τ∗ , can be
assumed to depend on the Reynolds number, Re, of the flow as follows
8
0.047 Re … 4 * 104 , SF = 1.0
τ∗ =
(5.76)
0.15
Re Ú 2 * 105 , SF = 1.5
where SF is the safety factor, and for 4 * 104 < Re < 2 * 105 values of τ∗ and SF should
be interpolated between those given in Equation 5.76. The Reynolds number, Re, is defined
by the relation
'
gyS0 d50
u∗ d50
Re =
=
(5.77)
ν
ν
where u∗ is the friction velocity (= τ0 /ρ), g is gravity [m/s2 ], and ν is the kinematic viscosity
of water [m2 /s]. The adjustment factor, ), is given by
)=
Ks [1 + sin(φ + β)] tan α
2(cos θ tan α − SF sin θ cos β)
(5.78)
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Section 5.4
Design of Channels with Flexible Linings
201
where φ is the longitudinal-slope angle [degrees], θ is the side-slope angle [degrees], α is the
angle of repose of the lining material, and β is defined as
⎛
⎞
⎜
⎟
cos φ
⎜
⎟
β = tan−1 ⎜
⎟
2
sin
θ
⎝
⎠
+ sin φ
η tan θ
(5.79)
where η is the stability number that is calculated using the relation
η=
ww
τs
τ∗ (γs − γ )d50
(5.80)
The permissible shear stress on the bottom of the channel, τp , given by Equation 5.75 covers
cases in which S0 … 5% and S0 > 10%; for intermediate cases it is recommended that the
lower value of τp be used in design (USFHWA, 2005).
The permissible shear stress on the side of the channel, τps [FL2 ], is related to the permissible shear stress on the bottom of the channel, τp [FL2 ], by the tractive force ratio, K
[dimensionless], such that
τps = Kτp
(5.81)
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where K is a function of the side-slope angle, θ [degrees], and angle of repose, α [degrees],
as given by Equation 5.29 and repeated here for convenience as
4
sin2 θ
K= 1 −
(5.82)
sin2 α
The angle of repose as a function of the median stone size, d50 , as recommended by the U.S.
Federal Highway Administration (USFHWA, 2005) is given in Figure 5.6.
Stability of the channel lining requires that τb < τp and τs < τps . In using stone linings,
it is quite common to use different stone sizes on the bottom and sides of the channel to meet
these requirements.
EXAMPLE 5.15
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A riprap lining with a median stone size of 200 mm composed mostly of subangular stones with specific
weight 25.9 kN/m3 is proposed for a trapezoidal channel having a bottom width of 1.00 m, side slopes
of 3:1 (H:V), and a longitudinal slope of 3%. The design flow rate is 2.5 m3 /s. Assess the adequacy of
the proposed lining.
Solution From the given data: d50 = 0.200 m, γs = 25.9 kN/m3 , b = 1.00 m, m = 3, S0 = 0.03, and
Q = 2.5 m3 /s. For any flow depth, y, the wetted perimeter, P, the flow area, A, the top width, T, and the
average depth, y, are given by
'
'
P = b + 2y 1 + m2 = 1.00 + 2y 1 + 32 = 1 + 6.325y
A = by + my2 = (1.00)y + 3y2 = y + 3y2
T = b + 2my = (1.00) + 2(3)y = 1 + 6y
y=
y + 3y2
A
=
T
1 + 6y
(5.83)
Step 1: Determine the flow depth in the channel. Assume that 1.5 … y/d50 … 185, then Manning’s
n is given by Equation 5.67 as
1
n=
0.319y 6
( )
y
2.25 + 5.23 log d
50
(5.84)
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202
Chapter 5
Design of Drainage Channels
and the Manning equation requires that
5
Q=
1 A 3 21
S
n P 32 0
5
1
1 (y + 3y2 ) 3
2.5 =
(0.03) 2
n (1 + 6.325y) 23
(5.85)
Solving Equations 5.83, 5.84, and 5.85 simultaneously yields y = 0.661 m, y = 0.397 m, and
y/d50 = 1.99. Since 1.5 … y/d50 … 185, the assumed expression for n (Equation 5.84) is
validated.
Step 2: Determine the maximum shear stresses exerted on the bottom and sides of the channel.
The maximum shear stress on the bottom of the channel, τb , is given by Equation 5.73 as
(assuming γ = 9790 N/m3 )
ww
τb = SF · γ yS0 = SF · (9790)(0.661)(0.03) = 194 · SF Pa
(5.86)
The side-shear-stress factor, Ks , is given by Equation 5.18 as
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Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868
and so the maximum shear stress on the sides of the channel, τs , is given by
τs = Ks τb = (0.868)(194 · SF) = 168 · SF Pa
(5.87)
Step 3: Determine the permissible shear stress on the bottom and sides of the channel and assess
the adequacy of the lining. The shear-velocity Reynolds number, Re, under design-flow
conditions is given by Equation 5.77 as (assuming ν = 10−6 m2 /s)
'
'
gyS0 d50
(9.81)(0.397)(0.03)(0.2)
=
= 6.8 * 104
Re =
ν
10−6
Using this value of Re, the corresponding values of τ∗ and SF are interpolated from
Equation 5.76 to yield
τ∗ = 0.065 and SF = 1.09
and hence the permissible shear stress on the bottom of the channel, τp , is given by
Equation 5.75 as
g.n
et
τp = τ∗ (γs − γ )d50 = (0.065)(25,900 − 9790)(0.2) = 209 Pa
The side-slope angle, θ, is given by
0 1
1
= 18.4◦
θ = tan−1
3
and the angle of repose, α, can be estimated from Figure 5.6, which yields
8
41.5◦ , very angular
α=
very rounded
39◦ ,
Since the riprap is subangular, taking the average of α for “very angular” and “very
rounded” stones gives α = 40.3◦ and hence the tractive force ratio, K, is given by Equation
5.82 as
5
4
6
2
6
sin θ
sin2 (18.4◦ )
K= 1 −
= 71 −
= 0.873
2
sin α
sin2 (40.3◦ )
The permissible shear stress on the side of the channel, τps , is given by 5.81 as
τps = Kτp = (0.873)(209) = 182 Pa
Using the calculated safety factor of SF = 1.09, the maximum shear stress on the bottom of
the channel is given by Equation 5.86 as τb = 194(1.09) = 211 Pa and the maximum shear
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Section 5.4
Design of Channels with Flexible Linings
203
stress on the side of the channel is given by Equation 5.87 as τs = 168(1.09) = 183 Pa.
Since the maximum shear stress on the bottom of the channel (211 Pa) is greater than the
permissible shear stress on the bottom of the channel (209 Pa), and the maximum shear
stress on the sides of the channel (183 Pa) is greater than the permissible shear stress on
the sides of the channel (182 Pa), the proposed lining is inadequate. A riprap lining with
larger median diameter should be considered.
Addition design considerations. The two most important considerations in designing riprap
linings are: (1) riprap gradation and thickness, and (2) use of filter material under riprap. The
following guidelines are suggested by the U.S. Federal Highway Administration (USFHWA,
2005):
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◃ Riprap gradation should follow a smooth-size distribution curve so that interstices
formed by larger stones are filled with smaller sizes. Most riprap gradations will fall
into the range of d100 /d50 and d50 /d20 between 1.5 and 3.0, which is acceptable.
◃ The thickness of a riprap lining should equal the diameter of the largest rock size in the
gradation. For most gradations, this will mean a thickness of from 1.5 to 3.0 times the
mean riprap diameter.
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5.4.5
Gabions
Gabions are rock-filled wire containers that utilize smaller rock sizes than would be required
by riprap. Gabion baskets are individual rectangular wire mesh containers filled with rock
and frequently used in grade-control structures and retaining walls. Gabion mattresses are
also rock-filled wire mesh containers; however, gabion mattresses are composed of a series
of integrated cells that hold the rock, allowing for a greater spatial extent in each unit. A
gabion basket is shown in Figure 5.12(a) and a channel lined with gabion baskets is shown in
Figure 5.12(b). Limestone rocks were used in this case.
Manning’s n. The effect of the mesh on Manning’s n can usually be neglected and so n
values can be determined using the d50 of enclosed rock along with Equation 5.67.
Exerted shear stress. The maximum shear stress on the bottom of the channel, τb , is calculated using Equation 5.73, and the maximum shear stress on the side of the channel is
calculated using Equation 5.74.
g.n
et
Permissible shear stress. The permissible shear stress for gabions can be estimated based on
either the size of the rock fill or the thickness of the gabion mattress. Typically, the permissible
shear stress is estimated using both methods, and the largest value is taken as the permissible
shear stress. If the size of the rock is characterized by d50 [m] and 0.076 m … d50 … 0.457 m
FIGURE 5.12: Gabions as
a Channel Lining
(a) Gabion basket
(b) Gabion-lined channel
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Chapter 5
Design of Drainage Channels
(0.25 ft … d50 … 1.5 ft), then the permissible shear stress, τp [Pa], is given by (USFHWA,
2005)
τp = τ∗ (γs − γ )d50
(5.88)
where τ∗ is the Shield’s parameter [dimensionless], γs is the specific weight of the rock [N/m3 ],
and γ is the specific weight of water [N/m3 ]. In utilizing Equation 5.88, τ∗ is typically taken
as 0.10. If the permissible shear stress, τp , is determined based on the thickness of the gabion
mattress, MT [m], and 0.152 m … MT … 0.457 m (0.5 ft … MT … 1.5 ft), then τp [Pa] is
given by
τp = 0.0091(γs − γ )(MT + 1.24)
(5.89)
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Rock sizes in gabion mattresses typically range from 8–15 cm (3–6 in.) in the 15 cm (0.5 ft)
thick mattresses to 12–30 cm (4–12 in.) in 46 cm (1.5 ft) thick mattresses. When comparing
the shear stress exerted by the flowing water to the permissible shear stress, it is recommended to use a safety factor, SF. In accordance with Equation 5.76, SF = 1.25 corresponds
to τ∗ = 0.10. Alternatively, SF can be calculated based on the SF versus Reynolds number
given in Equation 5.76. The permissible shear stress on the side of the channel is the same
as the permissible shear stress on the bottom of the channel. Since the exerted shear stress
is less on the sides of the channel, as long as the exerted shear stress on the bottom of the
channel is less than the permissible shear stress the gabion lining will be stable.
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EXAMPLE 5.16
It is proposed to line a trapezoidal channel with gabion mattresses having a thickness of 0.30 m and
containing stones with a median size of 150 mm and specific weight of 25.9 kN/m3 . The channel has a
bottom width of 1.00 m, side slopes of 3:1 (H:V), and a longitudinal slope of 6%. The design flow rate
is 0.7 m3 /s. Is the proposed gabion lining adequate?
Solution From the given data: MT = 0.30 m, d50 = 0.15 m, γs = 25.9 kN/m3 , b = 1.00 m, m = 3,
S0 = 6%, and Q = 0.7 m3 /s. For any flow depth, y, the wetted perimeter, P, the flow area, A, the top
width, T, and the average depth, y, are given by
'
'
P = b + 2y 1 + m2 = 1.00 + 2y 1 + 32 = 1 + 6.325y
A = by + my2 = (1.00)y + 3y2 = y + 3y2
T = b + 2my = (1.00) + 2(3)y = 1 + 6y
y=
A
y + 3y2
=
T
1 + 6y
g.n
et
(5.90)
Step 1: Determine the flow depth in the channel. Assume that 1.5 … y/d50 … 185, then Manning’s
n is given by Equation 5.67 as
1
n=
0.319y 6
( )
y
2.25 + 5.23 log d
50
(5.91)
and the Manning equation requires that
5
1 A 3 21
S
Q=
n P 32 0
5
0.7 =
1
1 (y + 3y2 ) 3
(0.06) 2
2
n (1 + 6.325y) 3
(5.92)
Solving Equations 5.90 to 5.92 simultaneously yields y = 0.326 m, y = 0.218 m, and y/d50
= 1.45. Since y/d50 < 1.5, the assumed expression for n (Equation 5.91) is not validated.
Assume that 0.3 < y/d50 < 1.5, then Manning’s n is given by Equation 5.67 as
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Section 5.5
Composite Linings
205
1
n= √
y6
gf1 f2 f3
(5.93)
where
0.7
'
(y + 3y2 ) (9.81)y
*
+0.814
0
0
1
10.453 0
1
d50 0.453
0.15
y
y 0.814
β = 1.14
= 1.14
T
d50
1 + 6y
0.15
Fr =
ww
0
1
0.28Fr log(0.755/β)
β
1
0
1 + 6y 0.492 1.025((1+6y)/0.15)0.118
β
f2 = 13.434
0.15
0
1
1 + 6y −β
f3 =
y
f1 =
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(5.94)
(5.95)
(5.96)
(5.97)
(5.98)
Solving Equations 5.92 to 5.98 simultaneously yields y = 0.262 m, y = 0.182 m, and
y/d50 = 1.21. Since 0.3 < y/d50 < 1.5 the assumed expression for n (Equation 5.93)
is validated. It is also noted that Equation 5.94 gives Fr = 1.12, indicating that the flow is
supercritical.
Step 2: Determine the maximum shear stress exerted on the bottom of the channel. The default
safety factor, SF, for gabions is 1.25, which is based on an assumed τ∗ = 0.10. The safety
factor can also be calculated based on the friction-velocity Reynolds number, Re, where
(assuming ν = 10−6 m2 /s)
'
'
gyS0 d50
(9.81)(0.182)(0.06)(0.15)
=
Re =
= 4.9 * 104
ν
10−6
and interpolating with Equation 5.76 gives SF = 1.03. Use the more conservative
SF = 1.25. The maximum shear stress on the bottom of the channel, τb is therefore given
by (assuming γ = 9790 N/m3 )
g.n
et
τb = SF · γ yS0 = (1.25) · (9790)(0.262)(0.06) = 192 Pa
Step 3: Determine the permissible shear stress on the bottom of the channel and assess the adequacy of the lining. Taking τ∗ = 0.10, Equation 5.88 gives
τp = τ∗ (γs − γ )d50 = (0.10)(25,900 − 9790)(0.15) = 242 Pa
and taking Equation 5.89 gives
τp = 0.0091(γs − γ )(MT + 1.24) = 0.0091(25,900 − 9790)(0.30 + 1.24) = 225 Pa
Therefore, taking the maximum of the two estimations gives τp = 242 Pa. Since the maximum shear stress on the bottom of the channel (192 Pa) is less than the permissible shear
stress (242 Pa), the proposed lining is adequate.
5.5 Composite Linings
In cases where different lining materials are used for different parts of the channel perimeter,
the lining is called a composite lining. Two cases in which composite linings are used are:
(1) when vegetative linings experience frequent low flows for which they are not adapted;
in such cases, concrete or riprap can be used to form a low-flow channel; and (2) when it is
cost effective to use a lining with lower permissible shear stress on the side of the channel
to match the lower shear stress exerted by the flowing water. In composite channels, the
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206
Chapter 5
Design of Drainage Channels
effective Manning’s n is obtained by dividing the channel boundary into two parts: the lowflow bottom and the sides. Assigning a Manning’s n to each of these parts of the boundary,
and assuming that the flow overlying each part of the boundary has the same mean velocity,
then the effective Manning’s n for the composite section is given by
⎡
PL
+
n=⎣
P
0
PL
1 −
P
10
ns
nL
13
2
⎤2
3
⎦ nL
(5.99)
where nL is Manning’s n of the low-flow (bottom) lining, ns is Manning’s n of the highflow (side) lining, PL is the wetted perimeter of the low-flow channel, and P is the wetted
perimeter of the entire channel.
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EXAMPLE 5.17
A roadside channel is to be constructed with a low-flow concrete bottom and grass sides. To further
channelize low flows, the concrete bottom is made triangular with side slopes 12:1 (H:V). The top
width of the concrete-bottom section is 1.00 m, the longitudinal slope of the channel is 2.5%, and the
sides of the channel have a slope of 3:1 (H:V) and are lined with sod grass with a height of 10 cm. The
channel is excavated in noncohesive soil with a 75-percentile grain size of approximately 1 mm. For a
design flow rate of 4.0 m3 /s, assess the adequacy of the channel lining and determine the depth of flow
in the channel.
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Solution From the given data: m1 = 12, m2 = 3, Tb = 1.00 m, S0 = 0.025, h = 0.10 m, d75 = 1 mm,
and Q = 4.0 m3 /s. For a concrete lining, Table 5.4 gives nL = 0.013 (trowel finish). For the grass lining,
Equation 5.41 gives Cs = 106 and Equation 5.42 gives
Cn = 0.35Cs0.10 h0.528 = 0.35(106)0.10 (0.10)0.528 = 0.165
For a depth of flow, y, expressions for the flow area, A, wetted perimeter, P, and low-flow wetted
perimeter, PL , are given by
T
A= b
2
*
1.00
=
2
Tb
2m1
*
+
1.00
2(12)
⎡
+ ⎣Tb +
+
⎡
*
Tb
y −
2m1
+ ⎣1.00 +
*
+
⎤*
g.n
et
Tb
m2 ⎦ y −
2m1
⎤*
+
+
+
1.00
1.00
⎦
y −
(3)
y −
2(12)
2(12)
= 3y2 + 0.75y − 0.0157
*
+
*
+
3
3
Tb
Tb
2
2
P = 2 1 + m1
+ 2 1 + m2 y −
2m1
2m1
*
*
+
+
'
'
1.00
1.00
= 2 1 + 122
+ 2 1 + 32 y −
2(12)
2(12)
= 0.739 + 6.325y
*
+
*
+
3
'
Tb
1
2
2
P L = 2 1 + m1
= 2 1 + 12
= 1.003 m
2m1
2(12)
Note: Since PL (= 1.003 m) is very close to Tb = (1.00 m) it is apparent that the channel could be
approximated as a trapezoidal channel.
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Section 5.5
Composite Linings
207
Step 1: Determine the flow depth in the channel. Manning’s n is given by Equation 5.99 as
⎡
P
n=⎣ L +
P
⎡
⎤2
0
10 13 3
PL
ns 2 ⎦
1 −
nL
P
nL
1.003
=⎣
+
0.739 + 6.325y
0
1.003
1 −
0.739 + 6.325y
10
ns
0.013
13
2
For grassed sides ns is given by Equation 5.43 as
ww
⎤2
3
⎦ (0.013)
(5.100)
ns = Cn τ0−0.4 = Cn [γ RS0 ]−0.4
⎡
⎤−0.4
*
+
2 + 0.75y − 0.0157
3y
= (0.165) ⎣9790
(0.025)⎦
0.739 + 6.325y
9
3y2 + 0.75y − 0.0157
= 0.0183
0.739 + 6.325y
:−0.4
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(5.101)
From the Manning equation
5
Q=
1 A 3 12
S
n P 23 0
4.0 =
1
1 (3y2 + 0.75y − 0.0157) 3
(0.025) 2
2
n
(0.739 + 6.325y) 3
5
(5.102)
Solving Equations 5.100 to 5.102 simultaneously yields y = 0.571 m and n = 0.026.
Step 2: Determine the effective stress on the soil underlying the side of the channel. There is no
need to calculate the maximum shear stress on the bottom of the channel, since with a
concrete bottom erosion will not be an issue. For a side slope of 3:1 (H:V), the side-shearstress factor, Ks , can be estimated by Equation 5.18 as
g.n
et
Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868
and so the maximum shear stress on the sides of the channel, τs , can be estimated as
(assuming γ = 9790 N/m3 )
τs = Ks τb = Ks [γ yS0 ] = (0.868)[(9790)(0.571)(0.025)] = 121 Pa
For sod grass in good condition, Table 5.9 gives Cf = 0.90, for d75 = 1 mm Equation 5.45
gives ns = 0.016, and the maximum effective shear stress, τe , on the soil underlying the
grass lining on the side of the channel is given by Equation 5.46 as
1
0 12
0
ns
0.016 2
= (121)(1 − 0.90)
= 4.6 Pa
τe = τs (1 − Cf )
n
0.026
Step 3: Determine the permissible shear stress on the side of the channel and assess the adequacy
of the lining. For noncohesive soil, the permissible shear stress, τp,n , is given by Equation
5.47 as
τp,n = 1 Pa
Since the effective shear stress (4.6 Pa) is greater than the permissible stress (1 Pa), the proposed grass lining is inadequate. An alternative lining or alternative channel dimensions
should be considered.
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Chapter 5
Design of Drainage Channels
Problems
5.1. Show that the best hydraulic section for a rectangularshaped section is one in which the bottom width is equal
to twice the flow depth.
5.2. Show that the best hydraulic section for a triangularshaped section is one in which the top width is equal to
twice the flow depth.
5.3. How do the side slopes in the best trapezoidal section
compare with the side slopes in the best triangular
section?
5.4. A highway median channel is to be sized to accommodate a peak runoff rate of 2.4 m3 /s. Regulations require
that the channel side slopes be 3:1 (H:V), and site conditions require a lining with an n value of 0.020 and a
longitudinal slope of 0.8%. Determine the dimensions
of the most efficient trapezoidal section.
5.5. An existing trapezoidal channel has a bottom width of
3 m, side slopes of 2:1 (H:V), a longitudinal slope of
0.5%, and a lining material with a Manning’s n of 0.020
and a permissible shear stress of 25 Pa. The channel is
lined up to an elevation 1 m above the bottom of the
channel. Determine the maximum flow rate for which
the lining material will be stable.
5.6. Consider a channel with flexible lining in which the lining material has a permissible shear stress of τp . Use the
Manning equation to show that the permissible velocity,
vp , in the channel is given by
* +1
0.010 R4 6 12
τp
vp =
n
y3
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the stone lining is estimated to be 200 Pa. Determine
the maximum flow depth in the channel for the lining
to remain stable.
5.11. A trapezoidal drainage channel has a lining material
with a permissible shear stress of 60 Pa, a Manning’s n
of 0.023, a bottom width of 2 m, side slopes of 4:1 (H:V),
a longitudinal slope of 0.5%, and a design flow depth
of 0.80 m. Determine the maximum radius of curvature
that can be used without having to provide additional
armoring to the channel.
5.12. What type of material can withstand vertical side slopes?
What is a typical side slope used in roadside channels?
5.13. Under what conditions would you use a freeboard
greater than 30 cm in a drainage channel?
5.14. A trapezoidal channel has been designed with a bottom
width of 10 m, side slopes of 2:1 (H:V), and a longitudinal slope of 0.053%. The Manning’s n of the channel
material is estimated to be 0.030. The design flow rate
is 28 m3 /s and the radius of curvature of the channel is
100 m. Determine the design depth of flow in the channel and the minimum required freeboard.
5.15. Design the most efficient (straight) lined trapezoidal
channel to carry 30 m3 /s on a longitudinal slope of 0.002.
The lining of the channel is to be float-finished concrete.
5.16. A lined rectangular channel is to be constructed on a
slope of 0.042% to handle a design flow rate of 4.5 m3 /s.
The lining of the channel is to be unfinished concrete,
and the maximum radius of curvature is 150 m. Determine the minimum depth and width of the channel to be
excavated.
5.17. A lined (straight) triangular channel is to be constructed
on a slope of 0.032% to handle a design flow rate of
4.4 m3 /s. The lining of the channel is to be smooth
asphalt. Determine the dimensions of the most efficient
channel.
5.18. A parabolic-shaped channel is to be excavated and lined
with mortar. If the channel is to carry 6 m3 /s on a slope
of 0.05%, determine the minimum depth of channel to
be excavated. Give the equation of the channel sides.
5.19. A trapezoidal channel has a design peak flow rate of
0.42 m3 /s, a bottom width of 0.4 m, side slopes of 3:1, and
a longitudinal slope of 0.8%. If an open-graded gravelmulch lining is to be used, estimate an adequate median
stone size for the lining.
5.20. A triangular channel is to be used to provide drainage
for a peak flow rate of 0.2 m3 /s, and the channel is to
be located adjacent to a roadway that has a longitudinal slope of 0.05%. Informal proposals from two engineering design/construction firms have been received. A
Brazilian firm proposes to use an unfinished concrete
lining with the best hydraulic section and will charge
$100/m3 of channel volume. A Spanish firm proposes to
use a locally available gravel-mulch lining, with round
stones having a median size of 50 mm. The Spanish
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where R is the hydraulic radius and y is the depth of flow.
Explain why it is better to design a channel based on
maximum permissible shear stress rather than maximum
permissible velocity.
5.7. The bottom sediments of an open channel consist of
cohesionless particles with a median grain size of 2.5 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 sediments.
5.8. Derive Equation 5.29 from Equations 5.25 and 5.28.
5.9. The theoretical tractive-force ratio, K, in trapezoidal
channels is given by Equation 5.29, and the bottom and
side shear stresses are given by Equations 5.16 and 5.17.
Use these relations to identify the side slopes under
which the critical shear stress will be on the side of an
unlined (trapezoidal) channel. If the side slopes of a particular channel are 25◦ and the angle of repose is 30◦ , will
the critical shear stress on the sides be reached before
the critical shear stress on the bottom is exceeded?
5.10. A trapezoidal channel is to be laid on a slope of 1%,
have a bottom width of 4 m, side slopes of 2.5:1 (H:V),
and lined with angular stones of median size 250 mm and
85-percentile size 300 mm. The stone lining is derived
from an open gradation. The permissible shear stress of
g.n
et
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Problems
firm does not guarantee that they will use the best
hydraulic section and will charge $70/m3 of channel volume. Design both channel alternatives, determine the
likely best alternative, and estimate the difference in cost
per km of roadway.
5.21. Evaluate the adequacy of using a sod-grass lining of
characteristic height 7.5 cm in good condition in a trapezoidal channel having a bottom width of 0.90 m, side
slopes of 3:1 (H:V), and a longitudinal slope of 3%. The
design flow rate is 0.5 m3 /s. The channel is underlain by
clayey sand (SC classification) with a plasticity index of
16 and a void ratio of 0.5.
5.22. A trapezoidal channel is to be excavated on a slope of
0.1% and have a design discharge of 35 m3 /s. Right-ofway constraints require the top width of the channel to
be 17 m, and the channel is expected to consist of clean
gravel (moderately angular) with a 75-percentile diameter of 3 cm. If freeboard is taken to be zero, determine
the required bottom width and side slopes of the channel. Is the required channel practical?
5.23. A roadway median channel in Georgia is to be excavated
in compact clay and is expected to convey a peak flow
rate of 0.5 m3 /s on a slope of 0.2%. The clay has a plasticity index of 25, a void ratio of 0.4, and has the ASTM
classification CH. Design the unlined median channel.
Compare the designed channel to the best hydraulic
section.
5.24. An existing channel is on a 0.3% slope in moderately
angular coarse gravel that has a 75-percentile diameter
of 3 cm. The channel is 3 m deep with a bottom width
of 6 m and side slopes of 2.5:1 (H:V). Estimate the safe
discharge capacity of the channel.
5.25. Evaluate the adequacy of a vegetation lining with class
D retardance in a trapezoidal channel having a bottom
width of 0.90 m, side slopes of 3:1 (H:V), and a longitudinal slope 3%. The design flow rate is 0.5 m3 /s. The
channel is underlain by clayey sand (SC classification)
with a plasticity index of 16 and a void ratio of 0.5.
5.26. A large housing development has an existing trapezoidal
swale with a bottom width of 5 m, side slopes of 3:1
(H:V), and a total depth of 1 m. The swale has a longitudinal slope of 0.1% and is lined with a dense (sodded)
cover of centipede grass in good condition with a retardance classification of C. The channel is underlain by
native compacted sandy-clay soil with a plasticity index
of 14, a void ratio of 0.4, and an ASTM classification
of CL. (a) Determine the flow rate that can be safely
accommodated by the swale. (b) If construction activity degrades the grass in the swale such that the lining
is transformed back to the native compacted sandy-clay
soil, determine the adjusted flow rate that could be safely
accommodated by the channel.
5.27. A roadside channel is to be lined with Bermuda grass
and regularly mowed to a height of 4 cm. The channel
is to be designed for a peak flow rate of 1.1 m3 /s and a
longitudinal slope of 1.4%, and it is to have a triangular
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209
cross section with side slopes of 3:1 (H:V). Using the
alternative retardance-based approach, determine the
depth of flow in the channel and assess the adequacy of
the channel lining.
5.28. A trapezoidal drainage channel on a golf course is lined
with Kentucky bluegrass and is expected to convey
10 m3 /s under design conditions. The longitudinal slope
of the drainage channel is 0.1%. If the channel has a
bottom width of 3 m, side slopes of 2:1 (H:V), and a
total depth of 2 m, use the alternative retardance-based
approach to assess the adequacy of the existing channel.
5.29. An existing triangular channel has a depth of 2 m and
side slopes of 4:1 (H:V), a l o n g i t u d i n a l s l o p e o f
1%, and is lined with a new type of Bermuda grass
which has a retardance classificationof C and maximum
permissible shear stress of 98.9 Pa. Use the alternative
retardance-based approach to determine the maximum
flow rate that can be handled by this channel?
5.30. A trapezoidal channel is to be designed to accommodate
a flow rate of 4 m3 /s on a longitudinal slope of 0.8%.
Excavation constraints limit the flow depth to 1 m and
it is proposed that the lining of the channel be centipede
grass with an average height of 15 cm, which is classified as sod grass in good condition. The grass lining is
underlain by a sandy soil with a void ratio of 0.45, a 75percentile grain size of 0.8 mm, and a plasticity index
of 8. (a) Design the bottom width and side slopes of
the channel; (b) assess the adequacy of the grass lining using the conventional approach; and (c) assess the
adequacy of the lining using the alternative retardancebased approach.
5.31. A temporary RECP lining is being considered for a
trapezoidal roadside channel with a bottom width of
0.9 m, side slopes of 3:1 (H:V), and a longitudinal slope
of 3%. The channel is excavated in clayey sand (SC classification) with a plasticity index of 16 and a void ratio
of 0.5. The design flow rate in the channel is 0.3 m3 /s.
Assess the adequacy of a lining for which the shear stress
at 12.5 mm soil loss is 100 Pa, and the roughness rating is
as follows:
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ing
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t
Applied shear
(Pa)
Manning’s n
50
100
200
0.040
0.036
0.033
5.32. A riprap lining is being proposed for a trapezoidal
drainage channel having a bottom width of 0.6 m, side
slopes of 3:1 (H:V), and a longitudinal slope of 2%.
The design flow rate is 1.13 m3 /s. Assess the adequacy
of riprap that has a median diameter of 150 mm and the
stone is mostly subangular.
5.33. Design a trapezoidal channel to carry 0.2 m3 /s on a
longitudinal slope of 0.2%. The channel is to be excavated in coarse alluvium with a 50-percentile diameter of
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210
Chapter 5
Design of Drainage Channels
25 mm, with the grains on the perimeter of the channel
moderately rounded.
5.34. Gabion baskets are proposed as a lining for a trapezoidal
channel with a bottom width of 0.60 m, side slopes of
3:1 (H:V), and longitudinal slope of 9%. The design flow
rate is 0.28 m3 /s. The gabion baskets have a thickness of
0.23 m and contain stones with a median size of 150 mm
and specific weight of 25.9 kN/m3 . Is the proposed lining
adequate?
5.35. A trapezoidal channel is to be constructed with a composite lining in which the bottom of the channel is made
of concrete to accommodate low flows, and the sides of
the channel are lined with grass. The grass lining is to
be a mixture of sod and bunch grasses, with good coverage and average height of 20 cm. The soil underlying the
grass is a clayey sand (SC classification) with a plasticity
index of 16 and a void ratio of 0.5. The bottom width
of the channel is 0.90 m, the side slopes are 3:1 (H:V),
and the longitudinal slope is 2%. The design flow rate is
0.28 m3 /s. Assess the adequacy of the proposed lining.
5.36. A median channel is to have a low-flow concrete bottom and grass sides. The bottom width of the channel is
1.2 m, the longitudinal slope of the channel is 1.5%, and
the sides of the channel have a slope of 4:1 (H:V). The
sides of the channel are lined with Class B mixed vegetation in good condition underlain by very fine sand with
a plasticity index of 12, a void ratio of 0.3, and an ASTM
classification of ML. For a design flow rate of 0.7 m3 /s,
assess the adequacy of the channel lining.
5.37. Show that the flow area, A, and wetted perimeter, P, of
a channel with a triangular bottom section are given by
.
Tb Tb
A=
2 2m1
⎡
. ⎤.
T
T
b
b
m2 ⎦ y −
+ ⎣Tb + y −
2m1
2m1
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(5.103)
.
4
Tb
2
P = 2 1 + m1
2m1
4
+ 2 1 + m22
Tb
y −
2m1
.
(5.104)
where Tb is the top width of the bottom section, m1 is
the side slope of the bottom section, y is the depth of
flow, and m2 is the side slope of the channel.
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C H A P T E R
6
Design of Sanitary Sewers
6.1 Introduction
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Sanitary sewers are underground conduits that are used to transport domestic, commercial,
and industrial wastewaters to disposal or treatment locations. Sanitary sewers must also be
designed to accommodate groundwater infiltration as well as illicit and extraneous inflows
from sources such as roof leaders, basement drains, and submerged manhole covers. Sanitary
sewers are also called foul sewers in the United Kingdom (Butler and Davies, 2011). The
term “sewer” generically refers to an underground pipeline system that transports drainage
in urban areas. In cases where the drainage is wastewater, the sewers are called sanitary
sewers, when the drainage is stormwater runoff they are called storm sewers, and when the
sewers transport both wastewater and stormwater runoff they are called combined sewers.
The design period of a sanitary-sewer system is typically determined by the planning
horizon of the local planning department and the expected useful life of the sewer pipe.
Planning horizons are typically in the range of 10-50 years, however, in some cases the saturation population or build-out population of the service area is used in lieu of the population
at the end of the planning horizon in determining the design capacity of sanitary-sewer pipes.
The useful life of sanitary-sewer pipes is usually on the order of 50 years (ASCE, 2007). In
cases where there might be significant additional development beyond the design period,
it is generally prudent to secure easements and rights-of-way for future expansions. Pump
stations in sanitary-sewer systems can be designed with shorter design periods since they are
relatively short lived and easier to update. Sewage-treatment plants, which are also relatively
easy to update, are generally designed for periods of about 20 years.
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6.2 Quantity of Wastewater
Flows in sanitary sewers broadly consist of two components: flows contributed through service connections and flows contributed by infiltration and inflow. Service connections include
connections to residential, commercial, and industrial users, and these sources of wastewater flow are typically considered separately since their usage characteristics are significantly
different from each other. Community master plans typically lay out the current and future
land uses in an area, and these master plans are usually the bases for estimating the locations of service connections and the magnitudes of wastewater flows contributed at these
connections. Infiltration and inflow (I/I) contributions are largely the result of pipe defects
and illicit connections, and are difficult to predict. To estimate the quantity of wastewater to
be accommodated by sewer pipes, all sources of flow must be estimated and aggregated.
Sewer flows must be estimated at both the beginning and the end of the design period.
Flows at the beginning of the design period, which is when the sewer system is first put into
service, must be adequate to prevent the buildup of solids in the sewers (i.e., the sewers must
be self-cleansing), and the sewer system must also have sufficient capacity to accommodate
the flows that will exist at the end of the design period.
6.2.1
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Residential Sources
Average wastewater flows from residential sources are typically estimated using the following
relationship,
Qr = d1 * d2 * qr * Ar
(6.1)
211
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212
Chapter 6
Design of Sanitary Sewers
TABLE 6.1: Typical Residential Densities
Unit density
Classification
(DU/ha)
(DU/ac)
Occupancy
(persons/DU)
Low density
Medium density
High density
…10
10–25
>25
4
4–10
>10
3.0
2.8
2.5
Density
(persons/ha)
(persons/ac)
30
28–70
>62
12
11–28
>25
Source: ASCE (2007).
TABLE 6.2: Average Per-Capita Wastewater Residential Flow Rates
Flow rate
City
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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
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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
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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
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Section 6.2
Quantity of Wastewater
Qc = A * FAR * e * qc
213
(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
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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.
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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
Allowance
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
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Remarks
—
—
—
Residential areas
—
Not in sag locations
Industrial & commercial areas
In sag locations
Pipe diameters … 1220 mm
Pipe diameters > 1220 mm
—
—
Source: ASCE (2007).
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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
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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
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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
⎩0.281Q−0.44
avg
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Qavg Ú 0.00368 m3 /s
Qavg < 0.00368 m3 /s
(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
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Section 6.2
Quantity of Wastewater
215
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
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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.
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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
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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.
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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
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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.
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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,
'
(
ϵ
1
2.51
&
& = −2 log
(6.8)
+
14.8R
f
4Re f
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where ϵ 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
ϵ
+
log
14.8R
Re0.9
+,2
(6.10)
This approximation is valid for 10−6 … ϵ/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
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Section 6.3
Hydraulics of Sewers
217
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
R6
8g
=
(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
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(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, ϵ, 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 ϵ 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
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+
A=
*
P=
Dθ
2
R=
*
+
A
D
sin θ
=
1 −
P
4
θ
θ − sin θ
8
D2
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(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
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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
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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
D2 =
(455)2 = 2.65 * 104 mm2 = 0.0265 m2
A=
8
8
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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
g.n
et
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
−1
θ − 3 (θ − sin θ ) 3 − 20.16nQD− 3 S0 2 = 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.
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Section 6.3
Hydraulics of Sewers
219
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
−1
θ − 3 (θ − sin θ ) 3 − 20.16nQD− 3 S0 2 = 0
2
5
8
1
θ − 3 (θ − sin θ ) 3 − 20.16(0.013)(4)(1.5)− 3 (0.0100)− 2 = 0
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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
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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
1.0
0.9
0.8
Ratio of depth/diameter, h/D
FIGURE 6.2: Flow
in circular conduits
0.7
Q
Qfull
0.6
g.n
et
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
0.8
0.9
1.0
1.1
1.2
1.3
V
Q
or
Vfull
Qfull
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220
Chapter 6
Design of Sanitary Sewers
+2
D 3 12
S0
4
'
(
π D2
Qfull = Vfull Afull = Vfull
4
1 23 12
1
Vfull = Rfull
S0 =
n
n
*
(6.19)
(6.20)
and hence
V
=
Vfull
Q
=
Qfull
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*
*
4R
D
4R
D
+2
3
+2 *
3
(6.21)
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.
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6.3.2
Manning Equation with Variable n
g.n
et
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, ϵ. However, for any given value of ϵ, 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 ϵ = 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
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.n
TABLE 6.4: Recommended Manning’s n versus Pipe Diameter
Pipe diameters
Condition
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.)
Extra care
Typical
Substandard
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).
et
221
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222
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).
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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?
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Solution From the given data: D = 610 mm = 0.610 m, S0 = 0.1% = 0.001, ϵ = 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
ϵ
+
log
14.8R
Re0.9
1
1
n = 0.1129R 6 f 2
Q=
2
θ − 3 (θ
5
− sin θ ) 3
8
g.n
et
(6.23)
(6.24)
(6.25)
(6.26)
−1
20.16nD− 3 S0 2
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
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Section 6.3
Hydraulics of Sewers
223
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.
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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)
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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
g.n
et
(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.
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224
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
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It is proposed to lay a sewer pipe of diameter 915 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 for a Manning’s n of 0.013 to assure selfcleansing. 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 = 915 mm = 0.915 m, S0 = 0.045% = 0.00045, Qmin = 7 L/s =
0.007 m3 /s, ϵ = 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
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Vfull =
1
nfull
2
1
3
Rfull
S02 =
1
nfull
!
0.915
4
"2
3
1
(0.00045) 2 =
0.007934
nfull
(6.29)
Assessment of the minimum flow velocity for self cleansing regulations requires the assumption that
nfull = 0.013, and hence Equation 6.29 yields
Vfull =
0.007934
= 0.61 m/s
0.013
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 = 0.00987 and a hydraulic radius of R = 0.0473 m. Hence the minimum shear stress
on the pipe boundary, τmin , is given by (taking γ = 9790 N/m3 )
τmin = γ RS0 = (9790)(0.0473)(0.00045) = 0.21 Pa
Since the minimum shear stress (0.21 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 0.21 Pa shear stress will
move is 0.006 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, ϵ, and design critical shear stress, τc , for self-cleansing.
Constraints in the hydraulic design are as follows:
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Section 6.3
Hydraulics of Sewers
225
◃ 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:
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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.
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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.
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et
Solution From the given data: D = 610 mm = 0.610 m, Qmax = 0.30 m3 /s, Qmin = 0.015 m3 /s, ϵ =
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:
)
,
* +
0
1
h
−1
θmax = 2 cos
1 − 2
= 2 cos−1 1 − 2(0.75) = 4.189 radians
D max
)
,
.
/
sin θmax
0.610
D
sin(4.189)
1 −
=
Rmax =
1 −
= 0.184 m
4
θmax
4
4.189
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Chapter 6
Design of Sanitary Sewers
)
,
/
θmax − sin θmax
4.189 − sin(4.189)
2
Amax =
D =
(0.610)2 = 0.235 m2
8
8
.
(Qmax /Amax )Rmax
(0.30/0.235)(0.184)
= 2.35 * 105
=
ν
10−6
0.0001
ϵ
=
= 0.000543
Rmax
0.184
⎡
⎡ '
'
(⎤−2
(⎤−2
ϵ
1.65
0.000543
1.65
⎦ = 0.25 ⎣log
⎦
fmax = 0.25 ⎣log
+
+
14.8Rmax
14.8
(2.35 * 105 )0.9
Re0.9
max
Remax =
= 0.0141
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1
1
1
1
6
2
fmax
= 0.1129(0.184) 6 (0.0141) 2 = 0.0101
nmax = 0.1129Rmax
⎤2 ⎡
⎡
⎤2
−8
− 83
20.16(0.0101)(0.30)(0.610)
⎢ 20.16nmax Qmax D 3 ⎥
⎦ = 0.00159
Smax = ⎣ 2
⎦ =⎣
5
5
−3
− 23
3
(4.189)
[4.189
−
sin(4.189)]
3
θmax [θmax − sin θmax ]
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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
ϵ
0.001
=
Rmin
Rmin
⎡
⎛
⎞⎤−2
Remin =
⎢
fmin = 0.25 ⎣log ⎝
1
6
ϵ
1.65 ⎠⎥
+
⎦
14.8Rmin
Re0.9
min
1
2
nmin = 0.1129Rmin fmin
⎡
− 83
⎤2
⎡
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et
⎤2
− 38 ⎥
⎥
⎢ 20.16nmin Qmin D
⎢
⎥ = ⎢ 20.16nmin (0.015)(0.610)
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
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Section 6.3
Hydraulics of Sewers
227
(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
ww
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◦
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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
(
(6.32)
+ hL
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
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Manhole
D1
Energy Grade Line
hL
2
V1
2g
y1
2
∆z
y2
V2
2g
Flow
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228
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, ϵ =
1 mm = 0.001 m, and Q = 0.485 m3 /s. Assume a temperature of 20◦ C so that ν = 10−6 m2 /s.
ww
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
ϵ
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 =
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1
1
n1 = 0.1129R16 f12
⎡
⎢
S1 = ⎢
⎣
− 83
⎤2
⎡
⎥
⎢
20.16n1 QD
⎥ = ⎢ 20.16n1 (0.485)(0.760)
⎣
5 ⎦
5
− 23
−2
θ1 [θ1 − sin θ1 ] 3
θ1 3 [θ1 − sin θ1 ] 3
⎤2
− 83 ⎥
⎥ = 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)
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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.
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Section 6.4
System Design Criteria
229
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
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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).
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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
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TABLE 6.5: Rigid and Flexible Pipe Materials
Rigid pipe
Flexible pipe
Concrete
Cast iron
Vitrified clay
Ductile iron
Steel
Thermoplastic (e.g., PVC)
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230
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.n
TABLE 6.6: Sewer-Pipe Materials
Material
Concrete
Advantages
Disadvantages
Manning’s n
Available diameters
(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
et
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Section 6.4
System Design Criteria
231
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
ww
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.
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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
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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
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232
Chapter 6
Design of Sanitary Sewers
FIGURE 6.4: Typical
manholes
Source: ASCE (2007).
0.61 m
1.2 m
1.2 m
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FIGURE 6.5: Precast
manhole
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(a) Typical manhole
(b) Drop manhole
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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).
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Section 6.4
System Design Criteria
233
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
ww
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.
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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
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FIGURE 6.6: Pump station
Outflow
Wet well
Outflow pipes with
Submersible pumps
Inflow
(a) Pump station
(b) Wet well
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234
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 =
ww
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
V2
2g
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h0 = K
(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
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(6.36)
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Section 6.4
System Design Criteria
235
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
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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
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+
BOD5
− 0.47
fZ = 1 + 10
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.
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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.
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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
2
5
5
8
−1
θ − 3 (θ − sin θ ) 3 − 20.16nQD− 3 S0 2 = 0
8
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=
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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
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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).
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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.
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Section 6.5
6.5.1
Design Computations
237
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
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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.
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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
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Chapter 6
Design of Sanitary Sewers
TABLE 6.8: Self-Cleansing Slopes as Functions of Design Minimum Flow Rates
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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
2000
0.01077Q−0.5996
min
0.01239Q−0.5979
min
Design particle size
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0.01155Q−0.5898
min
0.01288Q−0.5976
min
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 ϵ = 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.
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TABLE 6.9: Minimum Slopes at 15% and 10% Full
Diameter
(mm)
15% Full
Smin
Qmin
(–)
(L/s)
10% Full
Smin
Qmin
(–)
(L/s)
Design particle size
1.5 mm
15% Full
10% Full
Smin
Qmin
Smin
Qmin
(–)
(L/s)
(–)
(L/s)
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
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
0.00914
0.00685
0.00548
0.00457
0.00365
0.00305
0.00261
0.00228
0.00183
0.00152
0.00131
0.00114
0.00092
0.00685
0.00514
0.00410
0.00342
0.00274
0.00228
0.00195
0.00171
0.00137
0.00114
0.00098
0.00086
0.00068
1 mm
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
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
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
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Chapter 6
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:
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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
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Manhole no.
Line
no. Location From
(2)
(3)
(1)
To
(4)
Area
Length Increment Total
(ha)
(m)
(ha)
(6)
(5)
(7)
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 Upper Lower Upper Lower
of Diam force depth velocity drop sewer end end
end
end
sewer (mm) (Pa) (mm) (m/s)
(m)
(m)
(m)
(m)
(m)
(m)
(18)
(14) (15) (16) (17)
(20) (21) (22)
(23)
(24)
(19)
FIGURE 6.7: Typical spreadsheet for design of sanitary sewers
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242
Chapter 6
Design of Sanitary Sewers
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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%.
Step 9: 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.
Step 10: 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.
Step 11: 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.
Step 12: 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.
Step 13: 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.
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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:
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Section 6.5
FIGURE 6.8: Residential
sewer project
64.0 m
20
P Avenue
13
3
62.0 m
5
7
9
8
10
14
16
15
17
Q Avenue
6
60.0 m
62.0 m
2
1
21
22
60.0 m
A Street
4
12
243
B Street
11
19
18
C Street
23
24
26
25
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D Street
Main street
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Main sewer (existing)
Design Computations
Length
(m)
(5)
Contributing
area (ha)
(6)
Ground-surface elevation
Upper end
Lower end
(m)
(m)
(7)
(8)
Line no.
(1)
Location
(2)
Manhole no.
From
To
(3)
(4)
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
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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.
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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
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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
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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)
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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:
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Manhole no.
Line
no. Location From
(1)
(2)
(3)
0 Main Street −
To
(4)
5
1
2
3
4
5
2
3
5
5
12
A Street
A Street
A Street
A Street
Main Street
1
2
3
4
5
Area
Length Increment Total
(ha)
(ha)
(m)
(5)
(6)
(7)
−
−
−
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
Maximum flow
Sewer invert Ground surface
elevation
elevation
Minimum flow
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
Manhole
Min
tractive Max Max
invert Fall in Upper Lower Upper Lower
Slope
end
of Diam force depth velocity drop sewer end end end
(m)
(m)
(m)
(m) (m)
(m)
sewer (mm) (Pa) (mm) (m/s)
(14) (15) (16) (17) (18)
(19)
(20) (21) (22) (23)
(24)
−
−
−
−
55.35
0.009 1065 11.4 373 2.76
60.04
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 150
0.0154 150
0.0236 150
0.0207 150
0.0008 1065
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
1.20
1.40
2.36
1.84
0.06
62.85 61.65
61.65 60.25
60.25 57.89
59.73 57.89
55.35 55.27
65.00
63.80
62.40
61.88
60.04
63.80
62.40
60.04
60.04
60.04
FIGURE 6.9: Sewer design calculations
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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
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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
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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
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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
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Problems
247
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. Neglect I/I.
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. Water flows at 3 m3 /s in a circular concrete sewer laid on
a slope of 0.005. If a velocity of 2 m/s is desired in the
sewer, calculate the diameter of sewer pipe that should
be used. What would the depth of flow be in the sewer
pipe?
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. Show that the maximum flow rate in a circular pipe
occurs when h/D L 0.94, where h is the depth of flow and
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D is the diameter of the pipe. Assume that Manning’s n
is constant with depth.
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
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t
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. Consider a 915-mm-diameter concrete pipe with a characteristic roughness height of 1 mm, in typical condition,
and laid on a slope of 0.4%. Estimate Manning’s n when
the pipe is 10%, 50%, and 100% full. Assume a temperature of 20◦ C.
6.13. A minimum flow of 7 L/s is to be transported in a concrete sewer pipe laid on a slope of 0.5%. The estimated
Manning’s n of the sewer is 0.013, and the critical shear
stress for self cleansing is 1.5 Pa. Determine the range of
commercial-size pipe diameters that would assure selfcleansing.
6.14. A sewer pipe of diameter 915 mm is to be laid on a
slope of 0.027%, and the minimum flow rate expected
upon installation is 15 L/s. The pipe roughness is
estimated as 0.03 mm and local regulations require a
minimum permissible velocity of 0.60 m/s for a Manning’s n of 0.013 to assure self-cleansing. The minimum
tractive force required for self-cleansing is 0.9 Pa based
on a particle size of 1 mm. Will the minimum permissible
velocity criterion assure self-cleansing?
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
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248
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.
6.16. 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.
6.17. A 455-mm-diameter concrete pipe changes slope from
0.4% to 0.1% at a manhole. The roughness height of
the pipe is 1.5 mm and the design maximum flow rate
is 84.2 L/s. The inflow and outflow pipes are located
directly opposite each other. Determine the required
drop at the manhole.
6.18. 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.
6.19. A sanitary sewer is to transport wastewater with a
5-day BOD of 300 mg/L at 20◦ C. The flow rate of the
wastewater is 10 L/s, the diameter of the pipe is 205 mm,
Manning’s n is 0.013, and the slope is 0.1%. Assess the
potential for hydrogen sulfide generation.
6.20. 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.
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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. ASCE recommends that when designing a 610-mm
(24-in.) sanitary sewer an n value of 0.0115 should be
used. (a) State the hydraulic and pipe conditions and
any other assumptions used in determining this n value.
Using these stated conditions, calculate the n value yourself. Do you agree with the ASCE claim? (b) If the flow
rate in a 610-mm sewer is 5 L/s and the design particle
has a diameter of 1 mm, calculate the minimum pipe
slope required for self-cleansing. Compare this slope
with that estimated using the ASCE empirical formulae
for self-cleansing slope.
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.
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Problems
TABLE 6.10
Ground-surface elevation
Line no.
(1)
Location
(2)
Manhole no.
From
To
(3)
(4)
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
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Length
(m)
(5)
Contributing
area (ha)
(6)
Upper end
(m)
(7)
Lower end
(m)
(8)
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C H A P T E R
7
Design of Hydraulic Structures
7.1 Introduction
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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.
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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
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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
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Section 7.2
Culverts
251
FIGURE 7.1: Typical
culverts: (a) single-barrel
circular; (b) multibarrel
rectangular (box)
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(b)
(a)
#
$ %&
D
θ
y=
1 − cos
2
2
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(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
g.n
et
(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
2
θ3
= 20.16
nQ
8+
D3 S
0
= 20.16
(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
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252
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
5
gD
(9.81)(0.760)5
sin θ
2
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
ww
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).
w.E
asy
En
gin
eer
in
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
g.n
et
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
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(7.6)
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Section 7.2
FIGURE 7.2: Flow
through culvert with
submerged entrance
HW
TW
Culverts
253
∆h
H
V
D
(a) Submerged outlet (Type 1)
1
3
HW
∆h
ww
H
TW
yn
V
w.E
asy
En
gin
eer
in
L
1
3
2
(b) Normal depth > Barrel height (Type 2)
HW
H
h
(c) Entrance control, Normal depth < Barrel height (Type 3)
TW
g.n
et
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
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254
Chapter 7
Design of Hydraulic Structures
FIGURE 7.3: Flow
through culvert with free
entrance
HW
∆h
H
yn
yc
(a) Mild slope, Low tailwater (Type 4)
1
HW
ww
∆h
H
yc
yn
w.E
asy
En
gin
eer
in
1.4yc
1
(b) Steep slope, Low tailwater (Type 5)
HW
∆h
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
g.n
et
(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
4
R3
+ 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
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Section 7.2
Culverts
255
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:
.
/
2g"h
Q = A/
/
0 2gn2 L
+ ke + 1
4
R3
ww
(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
.
/
2g"h
/
Q = A/
0 fL
+ ke + 1
4R
w.E
asy
En
gin
eer
in
(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
g.n
et
(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
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256
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 =
ww
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
w.E
asy
En
gin
eer
in
H=
V2
+ Cc D
2g
g.n
et
(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
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TABLE 7.1: Constants for Empirical Culvert-Design Equations
ww
w.E
asy
En
gin
eer
ing
.n
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
et
257
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258
ww
w.E
asy
En
gin
eer
ing
.n
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
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
et
Source: U.S. Federal Highway Administration (2012).
Notes: *CMP = corrugated metal pipe; † CM = corrugated metal.
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Section 7.2
Culverts
259
TABLE 7.2: Contraction Coefficients
Culvert type
Circular concrete
Rectangular box
Rectangular
Multibarrel box
FIGURE 7.4: Variation of
Type 2 flow (Type 2A)
ww
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
∆h
H
yn
w.E
asy
En
gin
eer
in
D
L
1
TW
1 (D+y )
c
2
yc
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
g.n
et
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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260
Chapter 7
Design of Hydraulic Structures
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:
ww
. )
*
/
2
/
V
Q = Ac 02g "h + 1 − hi − hf
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.
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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:
. )
*
/
2
/
V
Q = Ac 02g "h + 1 − hi
2g
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(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:
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Section 7.2
⎧
⎪
M
E
⎪
⎨ c + K · (32.2) 2 · FrM − 0.5S
H
= D
′
D ⎪
⎪
⎩K′ · (32.2) M2 · FrM′
ww
Culverts
(Form 1)
261
(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
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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
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(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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262
Chapter 7
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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:
. )
*
/
/
V12
0
− hi − hf
Q = A 2g "h +
(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.
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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:
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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.
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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.
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Section 7.2
Culverts
263
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.
ww
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
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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
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
g.n
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0.012–0.015
0.014–0.018
Source: U.S. Federal Highway Administration (2005a; 2012).
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TABLE 7.4: Culvert Entrance Loss Coefficients
Culvert type and entrance conditions
ww
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
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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
Source: U.S. Federal Highway Administration (2012).
0.2
0.5
0.5
0.2
0.4
0.2
0.5
0.7
0.2
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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
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Section 7.2
ww
Culverts
265
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.
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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.
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• 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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Chapter 7
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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.
ww
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 less than yc , then Type 5 flow is confirmed and
the calculated discharge is the culvert capacity.
• Otherwise, continue to Step 3.
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Step 3: Assume that the flow is Type 6; use Equation 7.30 to calculate the discharge, Q, and
use Q to calculate the normal depth, yn , and critical depth, yc .
• 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
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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.
1.2
6m
2m
S!
0.5
%
Q
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Section 7.2
Culverts
267
(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/
0 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
ww
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
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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)/
0 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=
The Manning equation gives
1.22yn
A
=
P
1.22 + 2yn
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5
4.59 =
which yields
1
(1.22yn ) 3
1
(0.005) 2
0.013 (1.22 + 2y ) 23
n
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
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Chapter 7
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 =
2
2
gb
(9.81)(1.22)
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
ww
= 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 )
2g"h
/
/
Q=A
= (1.49)/
/
0 2gn2 L
0 2(9.81)(0.013)2 (36.6) + 0.05 + 1
+ ke + 1
4
4
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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)/
0 2(9.81)(0.013)2 (36.6) + 0.05 + 1
4
(0.305) 3
which simplifies to
Q = 3.60 m3 /s
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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)/
0 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.
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Section 7.2
Culverts
269
TABLE 7.5: Calculation of Headwater Depths for Various Flow Types
7.2.3.2
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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:
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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
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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
π 2
D = (0.915)2 = 0.6576 m2
4
4
1.70
Q
=
= 2.585 m/s
V=
A
0.6576
0.915
D
=
= 0.2288 m
R=
4
4
A=
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270
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
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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.
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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
g.n
et
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
(H − TW) + 0 −
−
V2
= h i + hf
2g
Q2
2
2gA
= ke
Q2
2
2gA
+
⎛
⎞2
nQ
⎝
⎠
2
L
(7.32)
AR3
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.
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Section 7.2
7.2.3.3
Culverts
271
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:
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Type
Headwater depth
(m)
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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
g.n
et
(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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272
Chapter 7
Design of Hydraulic Structures
FIGURE 7.6: Discharge
coefficient for roadway
overtopping
Source: USFHWA (2012).
Hr
Flow
yd
Lr
1.70
1.65
1.75
ww
1.60
Cr 1.55
vel
Gra
1.65
el
av
Gr
1.60
Paved
1.70
Cr
Paved
1.50
1.45
1.40
5
6
7
8
Hr (cm)
9
0
10
0.2 0.4
0.6 0.8 1.0 1.2
Hr (m)
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(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
(c) Submergence factor
EXAMPLE 7.5
1.0
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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
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Section 7.2
Culverts
273
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
2g"h
/
(7.35)
+ Cd LR Hr2
Q = A/
0 2gn2 L
+ ke + 1
4
R3
ww
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
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"h = (Roadway elevation + Hr ) − Tailwater elevation
= (291.08 + Hr ) − 289.38
(7.36)
= 1.70 + Hr
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
+ 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
1.00
0.14
0.055
0.008
1.68
1.66
0.00
0.00
1.00
1.00
1.68
1.66
g.n
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(7)
Hr
(m)
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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274
Chapter 7
Design of Hydraulic Structures
7.2.5
ww
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):
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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
FIGURE 7.7: Culvert exit
with riprap apron
Soil
(m/s)
(ft/s)
Sand
Sandy loam
Silt loam
Sandy clay loam
Clay loam
Clay, fine gravel
Cobbles
Shale
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
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Section 7.3
Gates
275
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):
0.044
d50 =
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.
ww
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
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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
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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.
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276
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
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FIGURE 7.10: Tainter
gates
Source: U.S. Army Corps
of Engineers (2011).
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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)
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Section 7.3
FIGURE 7.11:
Flow-through gates
Gate
277
Energy grade line
V12
2g
Gate
y1
V22
2g
V1
Flow
y1
yg
θ
y2 = Cc yg
yg
(a) Vertical gate
ww
Gates
y2 = Cc yg
(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
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y1 +
Q2
Q2
=
y
+
2
2gb2 y21
2gb2 y22
and solving for Q gives
Q = by1 y2
9
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
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(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
yg
1 + Cc
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)
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Chapter 7
Design of Hydraulic Structures
Cc = 0.61
(7.46)
whenever 0 <yg /E1 < 0.5, where E1 is the specific energy of the flow upstream of the gate,
defined by E1 = y1 + (V12 /2g). Experimental results by Lin et al. (2002) found values of Cc
for sharp-edged vertical gates in range of 0.59–0.61, and these results are consistent with the
reported range of 0.58–0.63 by Rajaratnam and Subramanya (1967a), the suggested value of
0.60 by Henry (1960), and the suggested value of 0.61 by Henderson (1966) and Rajaratnam
and Subramanya (1967b). For vertical gates with rounded edges, values of Cc in the range
0.65–0.75 have been reported (Lin et al., 2002). In the case of radial gates, the contraction
coefficient, Cc , is generally greater than 0.61 and is commonly expressed as a function of the
angle θ (shown in Figure 7.11) as
ww
$ %2
$ %
θ
θ
+ 0.36
Cc = 1 − 0.75
90
90
(7.47)
where θ is measured in degrees. Equation 7.47 gives results that are accurate to within ; 5%
provided that θ < 90◦ (Henderson, 1966). Analyses of field data reported by Shahrokhnia
and Javan (2006) indicate that the discharge coefficient, Cd , for radial gates is approximately
equal to 0.57.
Flows through gates may be free or submerged depending on the tailwater depth. If the
tailwater depth is less than the conjugate depth of the vena contracta, the flow through the
gate will be free, and if the tailwater depth is greater than the conjugate depth of the vena
contracta, the flow through the gate will be submerged. The flow condition in which the tailwater depth is equal to the conjugate depth of the vena contracta is called the distinguishing
condition. Under this condition, the flow through the vertical gate is given by Equation 7.44,
and the tailwater depth, y3 , is related to the depth at the vena contracta, y2 , by the hydraulic
jump equation (Equation 4.139) given by
w.E
asy
En
gin
eer
in
⎞
⎛
9
y2 ⎝
8Q2 ⎠
y3 =
−1 + 1 +
2
gb2 y32
(7.48)
g.n
et
Combining Equations 7.44, 7.45, and 7.48 yields the following expression for the tailwater
depth, y3 , under the distinguishing condition,
y3 =
⎤
⎡9
Cc yg
16
⎣ 1 +
− 1⎦
2
η(1 + η)
where
η = Cc
yg
y1
(7.49)
(7.50)
For a known gate opening, yg , contraction coefficient, Cc , and upstream water depth, y1 , the
maximum-allowable downstream depth for free flow can be determined from Equation 7.49,
which represents the distinguishing condition between free flow and submerged flow through
the gate. For a given upstream depth and gate opening, the maximum-allowable downstream
tailwater depth under the free-flow condition increases with increasing contraction coefficient, implying that a gate with a large contraction coefficient, such as a radial gate, is more
suitable for flow control under the free-flow condition (Lin et al., 2002).
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Section 7.3
Gates
279
EXAMPLE 7.6
A vertical sluice gate is opened 10 cm and the depth of water behind the gate is 1 m. If the width of
the rectangular sluice is 1.5 m, estimate the free-flow discharge through the gate and the distinguishing
condition for the tailwater depth.
Solution From the given data: yg = 0.10 m, y1 = 1 m, and b = 1.5 m. Assuming Cc = 0.61,
Equation 7.50 gives
yg
0.1
= 0.61
η = Cc
= 0.061
y1
1
and, for free-flow conditions, Equation 7.45 gives the discharge coefficient, Cd , as
ww
Cd = +
Cc
1 + η
0.61
= √
= 0.592
1 + 0.061
The free-flow discharge through the gate is given by Equation 7.44 as
+
+
Q = Cd byg 2gy1 = 0.592(1.5)(0.1) 2(9.81)(1) = 0.393 m3 /s
w.E
asy
En
gin
eer
in
The distinguishing condition for the tailwater depth is given by Equation 7.49 as
⎡9
⎤
⎡9
⎤
Cc yg
16
16
0.61(0.10)
⎣ 1 +
⎣ 1 +
y3 =
− 1⎦ =
− 1⎦ = 0.45 m
2
η(1 + η)
2
0.061(1 + 0.061)
Therefore, the flow rate through the gate will be 0.393 m3 /s as long as the tailwater depth is less than or
equal to 0.45 m. If the tailwater depth exceeds 0.45 m, then the flow through the gate will be submerged
and the flow rate will be less than 0.393 m3 /s.
7.3.2
Submerged Discharge
In cases where the discharge through the gate opening is supercritical and the depth of flow
downstream of the gate exceeds the conjugate depth of the vena contracta (i.e., exceeds the
distinguishing condition), the outflow will be submerged and the discharge equation given by
Equation 7.44 will not be applicable. This condition is illustrated in Figure 7.12. An approximate analysis of the submerged-flow condition assumes that all head losses occur in the flow
downstream of the gate, between Sections 2 and 3, in which case the energy equation can be
written as
y1 + α1
Q2
2gb2 y21
= y + α2
Q2
2gb2 y22
g.n
et
(7.51)
where y is the depth of flow immediately downstream of the gate, and α1 and α2 are the
energy or Coriolis coefficients that account for nonuniform velocity distributions at Sections
FIGURE 7.12: Submerged
flow through gates
Energy grade line
Gate
y1
V1
Flow
y
yg
y3
y2
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280
Chapter 7
Design of Hydraulic Structures
1 and 2, respectively. Between Sections 2 and 3, the momentum equation (Equation 4.136)
can be written as
y2
y2
Q2
Q2
+ β2 2 = 3 + β3 2
2
2
gb y2
gb y3
(7.52)
where β2 and β3 are the momentum or Boussinesq coefficients that account for non uniform
velocity distributions at Sections 2 and 3, respectively. The flow rate, Q, can be estimated
by simultaneous solution of Equations 7.51 and 7.52, where y1 and y3 are usually known
and y2 is estimated by Cc yg . Typical values of the energy and momentum coefficients are:
α1 = α2 = 1.05 and β2 = β3 = 1.02 (Lozano et al., 2009). Taking α1 = α2 = β2 = β3 = 1.0,
Lin et al. (2002) provided the following analytic solution to Equations 7.51 and 7.52:
ww
⎡
Q = Cc
⎣ξ −
9
ξ2 −
$
⎤1
% 2
%2 $
1 ⎦
1
1 − 2
− 1
η2
λ
1
− η
η
+
byg 2gy1
w.E
asy
En
gin
eer
in
(7.53)
where
y1
y3
%2
$
1
− 1
+ 2(λ − 1)
ξ=
η
(7.54)
λ=
(7.55)
and η is given by Equation 7.50; the parameter λ is sometimes called the inlet depth factor
(Habibzadeh et al., 2011b). To estimate the flow rate through a gate under submerged-flow
conditions, Equation 7.53 should be used.
In the case of vertical sluice gates, it has been shown that for submerged flow Cc
remains close to its free-flow value when the gate opening is small. However, for large openings of vertical sluice gates under submerged conditions, Cc can be much higher than the
free-flow value (Belaud et al., 2009; Cassan and Belaud, 2012). Experiments on sluice gates
under submerged-flow conditions have shown that using Equation 7.53 with Cc = 0.611 yields
discharge errors on the order of 6% (Sepúlveda et al., 2009). These errors can be reduced
by using alternate discharge methods that require field calibration data; for example, the
calibrated model proposed by Ferro (2001) yields discharge errors on the order of 3%. Also,
locally calibrated values of α and β in Equations 7.51 and 7.52 can be effective in reducing
discharge errors (e.g., Castro-Orgaz et al., 2010). It has also been shown that improved discharge estimates for sluice gates under both submerged and unsubmerged conditions can
be achieved by introducing an additional term to account for the head loss between the
upstream reservoir and the vena contracta (e.g., Habibzadeh et al., 2011a), where the magnitude of this additional term would need to be determined by calibration.
g.n
et
EXAMPLE 7.7
Water is ponded behind a vertical gate to a height of 4 m in a rectangular channel of width 7 m. If the
tailwater depth is 3.5 m and the gate opening is 1.15 m, what is the discharge through the gate?
Solution From the given data: y1 = 4 m, b = 7 m, yg = 1.15 m, and y3 = 3.5 m. It must first be
determined whether free-flow or submerged-flow conditions exist. The distinguishing condition is given
by Equation 7.49, where the contraction coefficient, Cc , can be assumed equal to 0.61, and η is given by
Equation 7.50 as
yg
1.15
= 0.61
= 0.175
η = Cc
y1
4
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Section 7.3
Gates
281
Substituting into Equation 7.49 gives the distinguishing condition as
⎤
⎤
⎡9
⎡9
Cc yg
0.61(1.15)
16
16
⎣ 1 +
⎣ 1 +
− 1⎦ =
− 1⎦ = 2.76 m
y3 =
2
η(1 + η)
2
0.175(1 + 0.175)
Since the tailwater elevation (= 3.5 m) exceeds 2.76 m, then the flow is submerged and Equation 7.53
must be used to calculate the flow rate through the gate. From the given data, Equations 7.54 and 7.55
give
4
y
λ= 1 =
= 1.14
y3
3.5
%2
%2
$
$
1
1
− 1
− 1
+ 2(λ − 1) =
+ 2(1.14 − 1) = 22.5
ξ=
η
0.175
ww
Substituting into Equation 7.53 gives
⎡
⎣ξ −
9
ξ2 −
$
⎤1
%2 '
( 2
1 − 1
1 − 12 ⎦
2
η
λ
w.E
asy
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gin
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in
Q = Cc
= 0.61
#
22.5 −
1
η − η
:
22.52 −
= 19.1 m3 /s
'
( '
+
byg 2gy1
2
1
1 − 12
− 1
0.1752
1.14
1
− 0.175
0.175
(
&1
2
+
(7)(1.15) 2(9.81)(4)
Therefore, under the given headwater, tailwater, and gate-opening conditions, the (submerged) discharge through the gate is estimated as 19.1 m3 /s.
A semi-empirical discharge equation for a sluice gate under submerged-flow conditions
can be derived by applying the energy equation to Sections 1 and 2, neglecting the energy
loss and the velocity term at Section 1, and taking α1 = α2 = 1.00 to give
+
(7.56)
Q = Cd yg 2g(y1 − y)
g.n
et
where Cd is a coefficient of discharge that is generally considered to depend on yg and y1 − y.
Several empirical equations have been proposed for estimating Cd (e.g., Lozano et al., 2009).
With proper calibration, Equation 7.53 or 7.56 can be used to estimate submerged sluice-gate
discharges with comparable accuracy.
7.3.3
Empirical Equations
The accuracies of the theoretical gate-discharge equations described previously depend on
the accuracy with which the contraction coefficient, Cc , and discharge coefficient, Cd , are
estimated. Empirical equations have been shown to have comparable or better accuracies
in many cases. Empirical equations for describing the flow through radial gates proposed by
Shahrokhnia and Javan (2006) based on dimensional analysis and calibration with numerous
laboratory and field data are given by
⎧
)
*0.40
⎪
⎪
"H
⎪
⎪
0.88
(radial gate, free flow)
⎪
)
* ⎪
⎪
yg
⎨
2
Q
(7.57)
=
)
*0.33
⎪
b2 gy3g
⎪
⎪
⎪
"H
⎪
⎪
(radial gate, submerged flow)
⎪
⎩1.14 yg
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282
Chapter 7
Design of Hydraulic Structures
where "H is differential energy head given by
"H =
⎧
⎪
V12
⎪
⎪
⎪
y
+
1
⎪
⎨
2g
⎪
⎪
V12
⎪
⎪
⎪
− y3
+
y
⎩ 1
2g
(radial gate, free flow)
(7.58)
(radial gate, submerged flow)
where V1 is the upstream flow velocity. Equation 7.57 was derived from data reported by
Buyalski (1983), Safarinezhad (1991), Webby (1999), and Shahrokhnia and Javan (2006).
7.4 Weirs
ww
Weirs are elevated structures in open channels that are used to control outflow and/or measure flow from basins and drainage channels. In these structures, water flows through an
elevated opening of regular shape (typically rectangular, triangular, or trapezoidal) in which
the relationship between flow rate and depth of flow above the crest of the structure is fixed
and known.
w.E
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in
7.4.1
Sharp-Crested Weirs
Sharp-crested, or thin-plate, weirs consist of a plastic or metal plate that is set vertically and
across the width of a channel. For a weir to be considered sharp crested, the usual criterion
is that the ratio of the height of water above the crest of the weir to the thickness of the weir
must be greater than 1.5. Under this condition, the flow separates from the upstream edge of
the weir, which is a characteristic of sharp-crested weirs. In flow-measurement applications,
typical thicknesses of sharp-crested weirs are 1–2 mm (0.04–0.08 in.) (Martinez et al., 2005).
Sharp-crested weirs are classified according to their cross-sectional shape, and the main types
of sharp-crested weirs are rectangular and triangular (V-notch) weirs.
7.4.1.1
Rectangular weirs
Rectangular weirs have a rectangular cross sections. In suppressed (uncontracted) rectangular weirs, the rectangular opening spans the entire width of the channel; in unsuppressed
(contracted) rectangular weirs, the rectangular opening spans only a portion of the width
of the channel. Suppressed rectangular weirs are sometimes called Bazin weirs. Both suppressed and unsuppressed rectangular weirs are illustrated in Figure 7.13, and pictures of
operating rectangular sharp-crested weirs are shown in Figure 7.14.
An elevation view of the flow over a sharp-crested weir is illustrated in Figure 7.15,
where at Section 1, just upstream of the weir, the flow is approximately horizontal, the pressure distribution is approximately hydrostatic, and the head (or mechanical energy per unit
weight) of the fluid, E1 , is given by
FIGURE 7.13: Schematic
diagrams of rectangular
sharp-crested weirs
b
H
Hw
g.n
et
b
Channel
H
Weir
(a) Suppressed weir
Hw
Weir
(b) Unsuppressed weir
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Section 7.4
Weirs
283
FIGURE 7.14: Operating
rectangular sharp-crested
rectangular weirs
FIGURE 7.15: Flow over a
sharp-crested weir
2
V1
1
2
2g
ww
Energy grade line
p1
H
γ
z1
z2
w.E
asy
En
gin
eer
in
Datum
Nappe
Weir
E1 = H +
V12
2g
(7.59)
where H is the elevation of the water surface above the crest of the weir and V1 is the average
velocity at Section 1. At Section 2, over the crest of the weir, the flow can be approximated
as horizontal, with a head, E2 , at any elevation, z2 , given by
E2 =
v2
p2
+ 2 + z2
γ
2g
g.n
et
(7.60)
where p2 /γ and v2 are the pressure head and velocity, respectively, at elevation z2 . The jet
of water that flows over the weir is commonly referred to as the nappe. The distribution
of fluid pressure at Section 2 is such that it is equal to atmospheric pressure both at the top
and bottom water surfaces, and increases above atmospheric pressure between the two water
surfaces. The bottom water surface is where the water springs clear of the crest at Section 2. If
the flow depth at Section 2 is small, then the pressure may be assumed equal to atmospheric
pressure throughout the depth, and E2 becomes
E2 =
v22
+ z2
2g
(7.61)
Assuming that the head loss is negligible along a streamline crossing Section 1, at elevation
z1 , and leaving Section 2 at elevation z2 , then
E1 = E2
(7.62)
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284
Chapter 7
Design of Hydraulic Structures
Combining Equations 7.59, 7.61, and 7.62 leads to
⎡
*⎤ 12
V12
⎦
v2 = ⎣2g H − z2 +
2g
)
(7.63)
? across Section 2 can be calculated by integrating the flow rates
The estimated flow rate, Q,
across elements of area b dz2 , where b is the width of the rectangular weir. Therefore
@ H
?=
Q
v2 b dz2
0
ww
⎡
*⎤ 12
V12
⎣2g H − z2 +
⎦ b dz2
=
2g
0
⎤
⎡
) *3
*3
)
2
2
2
2
+
V1
V
2
⎥
⎢
= b 2g ⎣ H + 1
−
⎦
3
2g
2g
@ H
)
w.E
asy
En
gin
eer
in
(7.64)
This equation assumes that the elevation of the water surface at Section 2 is equal to the
elevation of the water surface at Section 1. This condition is physically impossible but does
not necessarily lead to a significant error in the estimated flow rate at Section 2. If V12 /2g is
negligible compared with H, then Equation 7.64 reduces to
+
? = 2 2gbH 32
Q
3
(7.65)
? over the weir in terms of measurable quantities, b and
This expression gives the flow rate, Q,
H, and illustrates why the weir is a hydraulic structure that is used for measuring flow rates
in open channels based on the upstream water stage, H.
The weir equation given by Equation 7.65 was derived with the following theoretical discrepancies: (1) the pressure distribution in the water over the crest of the weir is
not uniformly atmospheric; (2) the water surface does not remain horizontal as the water
approaches the weir; (3) viscous effects that cause a nonuniform velocity and a loss of energy
between Sections 1 and 2 have been neglected; and (4) the approach velocity head, V12 /2g,
might not be negligible. The error in the flow rate resulting from these theoretical discrepancies is accounted for by a discharge coefficient, Cd , defined by the relation
Cd =
Q
?
Q
g.n
et
(7.66)
where Q is the actual flow rate over the weir. Combining Equations 7.65 and 7.66 leads to
the following expression for the flow rate over a weir in terms of the discharge coefficient:
Q=
3
2 +
Cd 2gbH 2
3
(7.67)
The original derivation of Equation 7.67 has been credited to Giovanni Poleni (1683–1761)
(Rouse and Ince, 1963). It can be shown by dimensional analysis that
%
$
H
Cd = f Re, We,
(7.68)
Hw
where Re is the Reynolds number, We is the Weber number, and Hw is the height of the weir
crest above the bottom of the channel, commonly referred to as the crest height. The Weber
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Section 7.4
Weirs
285
number is a nondimensional measure of the relative magnitude of the inertial force to the
surface tension force, and is commonly denoted by
We =
LV 2 ρ
gσ
(7.69)
where L is the characteristic length scale, V is the velocity, ρ is the density, and σ is the
surface tension. Experiments have shown that H/Hw is the most important variable affecting
Cd , with We important only at low heads; Re is usually sufficiently high that viscous effects
can be neglected. An empirical formula for Cd is (Rouse, 1946; Blevins, 1984)
Cd = 0.611 + 0.075
ww
H
Hw
(7.70)
which is valid for H/Hw < 5, and is approximate up to H/Hw = 10. For H/Hw > 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
w.E
asy
En
gin
eer
in
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
g.n
et
(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
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286
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
ww
(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
w.E
asy
En
gin
eer
in
3
(7.77)
Q = Cw bH 2
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)
⎡
Qs
= ⎣1 −
Q
$
yd
H
%3
2
⎤0.385
⎦
g.n
et
(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
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Section 7.4
Weirs
287
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
1 −
(7.79)
= 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
ww
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
w.E
asy
En
gin
eer
in
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
g.n
et
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).
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288
Chapter 7
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
? is given by
theoretical flow rate over the weir, Q,
@ H
?=
Q
v2 b dz2
0
ww
=
@ H
0
⎡
*⎤ 12
V12
⎣2g H − z2 +
⎦ b dz2
2g
)
(7.80)
where b is the width of the V-notch weir at elevation z2 and is given by
$ %
θ
b = 2z2 tan
2
w.E
asy
En
gin
eer
in
(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
FIGURE 7.18: Operating
V-notch weir
FIGURE 7.19: Schematic
diagram of V-notch weir
@ H
0
⎡
*⎤ 12
$ %
V12
⎣2g H − z2 +
⎦ 2z2 tan θ dz2
2g
2
)
(7.82)
g.n
et
Channel
b
H
θ
z2
dz2
Weir
Hw
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Section 7.4
Weirs
289
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
$ %
@ HC
!
"D 12
θ
2g H − z2
2z2 tan
dz2
2
0
$ %
5
8+
θ
H2
=
2g tan
15
2
?=
Q
(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
ww
$ %
+
5
θ
8
Q=
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)
w.E
asy
En
gin
eer
in
1.19
Cd = 0.583 +
where Re is the Reynolds number defined by
1
FIGURE 7.20: Discharge
coefficient in V-notch
weirs
1
Re =
g2 H 2
ν
25
30
0.70
0.68
Discharge coefficient, Cd
(7.85)
1
(ReWe) 6
g.n
et
(7.86)
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)
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290
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◦ ?
ww
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
#
15Q
&2
5
#
15(0.2)
&2
w.E
asy
En
gin
eer
in
H=
+
8Cd 2g tan (θ/2)
=
8Cd
+
2(9.81) tan (45◦ /2)
5
=
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.
g.n
et
An alternative flow equation for V-notch weirs has been proposed by Kindsvater and
Shen (USBR, 1997), where
+
8
Cd 2g tan
Q=
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
(7.89)
k = 4.42 − 0.1035θ + 1.005 * 10−3 θ 2 − 3.24 * 10−6 θ 3
(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).
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Section 7.4
Weirs
291
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
ww
= 1.50 mm = 0.0015 m
The relationship between the head and flow rate is given by Equation 7.88, which yields
#
&2
#
&2
w.E
asy
En
gin
eer
in
H=
15Q
+
8Cd 2g tan (θ/2)
5
− k=
+
15(0.2)
8(0.580) 2(9.81) tan (45◦ /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.
g.n
et
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
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292
Chapter 7
Design of Hydraulic Structures
FIGURE 7.21: Compound
weir
b
H1
θ
ww
H2
w.E
asy
En
gin
eer
in
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:
9 $
%
3
2 +
H2
2
Q = Cd 2gbH1 + Cvd A 2g H1 +
3
3
g.n
et
(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
9 $
%
3
+
2
H
2
(7.93)
Q = Cd 2gbH1 + Cvd A 2g H1 + 2
3
3
The discharge coefficient, Cd , is given by Equation 7.70 as
Cd = 0.611 + 0.075
H1
H2
(7.94)
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Section 7.4
Weirs
293
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
9
%
$
+
3
2
0.20
2
3 = (0.911) 2(9.81)b(0.80) + 0.6(0.04) 2(9.81) 0.80 +
3
3
which yields
ww
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.
w.E
asy
En
gin
eer
in
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
g.n
et
∗ Developed by Harry Sutro in the early 1900s.
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294
Chapter 7
Design of Hydraulic Structures
FIGURE 7.23: Wooden
weir in a coastal drainage
channel
ww
w.E
asy
En
gin
eer
in
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
g.n
et
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
? is given by
over the weir, Q,
? = y bV
Q
c
c
(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
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Section 7.4
FIGURE 7.24: Flow over a
rectangular
broad-crested weir
Weirs
295
1
Energy grade line
V 12
2g
H
Q
h1
Critical
flow
Q
Hw
L
Longitudinal section view
ww
E1 = Hw + yc +
Vc2
2g
w.E
asy
En
gin
eer
in
(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
g.n
et
(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:
%3
$
2
? = √gb 2 H
Q
3
(7.100)
? must
In reality, the energy loss over the weir is not negligible and the estimated flow rate, Q,
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)
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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
⎥
h1
⎣
⎦
1 + 1000
L
ww
(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
w.E
asy
En
gin
eer
in
$
√
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 (Göğüş
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)
Cd′ = 0.95
− 0.38
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
g.n
et
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.
h1 /L > 1.5
Sharp crested
The flow separates at the upstream end and does not reattach to
the crest. The flow is unstable.
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Section 7.4
Weirs
297
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
ww
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
w.E
asy
En
gin
eer
in
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
%3
$
,
-3
√
2
2
0.65
2
√
2) 2
H
(0.30
+
0.0510Q
Q = Cd gb
=
9.81(2)
1
3
3
[2.5 + 0.255Q2 ] 2
#
(0.30 + 0.0510Q2 )3
= 2.22
2.5 + 0.255Q2
which yields
Q = 0.23 m3 /s
&1
2
g.n
et
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
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298
Chapter 7
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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
ww
(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
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(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
be found in the work of Göğüş et al. (2006).
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
g.n
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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)
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Section 7.5
Spillways
299
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
ww
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
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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)
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∗ An ogee refers to an S-shaped profile.
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300
Chapter 7
Design of Hydraulic Structures
FIGURE 7.26: Ogee-crest
spillway
Approach
head, he
Pool
elevation
Hd
Design
head
He
PT
(B − y) 2
x2
+
=1
2
A
B2
+ x-axis
O
A
B
y
Hd
+ y-axis
=
x n
Hd
Point of
tangency
(PT)
Crest
axis
1
P
ww
1
K
a
1
Fs
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E
For He Ú 9.1 m (30 ft) :
Hd =
For He < 9.1 m (30 ft) :
E
Hd =
0.33He1.22
0.30He1.26
(without piers)
(with piers)
(7.114)
0.7He
0.74He
(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
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(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
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Section 7.5
FIGURE 7.27: Coordinate
coefficients for spillway
crests
Spillways
301
10.0
8.0
6.0
Crest height, P
Design head, Hd
4.0
2.0
1.0
0.8
0.6
ww
0.4
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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
g.n
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(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
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302
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
ww
(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.
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EXAMPLE 7.13
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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
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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
0.5
1.0
1.5
2.0
2.5
3.0
P
Value of
Hd
(a) Basic discharge coefficient
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1.04
Ratio of coefficients,
Cinclined
Cvertical
1:1
Slope (H :V)
Angle with
the vertical
1:3
18°26
2:3
33°41
1:1
45°00
Hd
P
1.02
2:3
1:3
1.00
0.98
0
0.5
1.0
1.5
P
Value of
Hd
(b) Correction factor for sloping upstream face
1.1
C
Ratio of coefficients,
C0
ww
0
1.0
h0
H
Hd
0.9
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P
0.8
0
0.2
0.4
0.6
0.8
1.0
1.2
H
Ratio of head on crest to design head,
Hd
1.4
1.6
(c) Correction factor for other than design head
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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
ww
and the approach velocity, v0 , can be estimated by
v0 =
1500
Q
=
= 0.5 m/s
Le d
(60)(50)
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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
which simplifies to
$
%
x 1.85
3.44
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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
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Section 7.5
Spillways
305
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
A2
B2
x2
(0.568 − y)2
+
=1
0.9632
0.5682
which simplifies to
x2 + 2.87(0.568 − y)2 = 0.927
ww
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.
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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?
g.n
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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
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306
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
10
0.5
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0.7
1
15
hd
He
ww
8
0.8
0.5
0.6
2
20
0.5
3
1
4
0.4
2
6
3
0.3
4
8
6
0.2
10
15
15
0.1
0.0
1.0
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10
8
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
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Section 7.5
FIGURE 7.31: Submerged
flow
307
0.5 m
1.0 m
3.5 m
4.0 m
Spillway
3
2
⎡
Q0 = CLe He = (2.18)(3.0) ⎣1.0 +
ww
Spillways
Q20
⎤3
2
2gA2
⎡
⎦ = (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
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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 +
2
2
2gA
2(9.81)(4.5 * 3)
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
g.n
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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.
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308
Chapter 7
Design of Hydraulic Structures
FIGURE 7.32: Gate
locations on spillway
Lip
Lip
Crest
of spillway
Crest
Reservoir
Gate seat
(b) Gate downstream of crest
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7.5.2.1
Reservoir
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
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⎧
+
⎨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
g.n
et
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
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Section 7.5
FIGURE 7.33: Gated
spillway
Spillways
309
Vertical-lift
gate
1.00 m
0.50 m
0.25 m
Spillway
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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
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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.
g.n
et
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.
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310
Chapter 7
Design of Hydraulic Structures
FIGURE 7.34: Tainter gate
on spillway
0.95
0.90
θ
H
h
0.85
Cd
0.80
Strut
Trunnion
Lip
R
a
X/Hd
Gate seat
0.75
X/Hd = 0.1−0.3
0.70
X/Hd = 0.0
0.65
0.60
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0.55
50
55
60
65
70
75
80
85
90
95
100
105
110
θ Degrees
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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)
FIGURE 7.35: Operational
Tainter gate
g.n
et
h
1.00
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 −
Tainter gate
2.75 m
1.00 m
1m
0.87 m
2.10 m
Spillway
2.00 m
Spillway
30°
tangent
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Section 7.5
FIGURE 7.36: Vertical lift
gate downstream of
spillway crest
Spillways
311
Vertical lift gate
H1
H0
H2
Spillway
Flow
ww
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)
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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.
g.n
et
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
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312
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
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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.
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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 <2.5: For incoming Froude numbers in the range 1.7–2.5, the hydraulic jump is
weak and no special appurtenances are necessary. The key design parameter is the basin
length. The length of the stilling basin should be at least five times the conjugate depth
of the hydraulic jump contained within the stilling basin. This is called a Type I stilling
basin.
g.n
et
2.5 < Fr < 4.5: For Froude numbers in the range 2.5–4.5, a transition jump forms with considerable wave action. A Type IV stilling basin, shown in Figure 7.39, is recommended for
this jump. This basin includes chute blocks and a solid end sill, which tends to hold the
jump in the stilling basin. There are no baffle blocks within the basin. The recommended
FIGURE 7.38: Gated
spillway with stilling
basin
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Section 7.6
Stilling Basins
FIGURE 7.39: Type IV
stilling basin
characteristics,
2.5 < Fr < 4.5
Source: U.S. Bureau of
Reclamation (1987).
Chute blocks
End sill
Fractional space
w ! Max. tooth width ! y1
2y1, min.
Space ! 2.5w
Top surface on 5° slope
1.25y1
L1
(a) Type IV basin dimensions
4
5
Note:
y1 and y2 are the initial and
sequent depths of the hydraulic
jump in the stilling basin.
7
T.W. depth
y1
6
5
. !
T.W
y2
4
5
1
1.
y2
1
! (
y1
2
3
2
6
1 + 8Fr2 − 1 )
4
g.n
et
3
2
(b) Minimum tailwater depths
6
6
5
4
5
2
3
Froude number
4
5
L1
y2
7
Froude number
3
T.W. depth
y1
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2
L1
y2
ww
2y1
Sill optional
4
(c) Length of jump
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314
Chapter 7
Design of Hydraulic Structures
tailwater depth is 10% greater than the conjugate depth to help prevent sweepout of
the jump. Because considerable wave action can remain downstream of the basin, this
jump and basin are sometimes avoided altogether by widening the basin to increase the
Froude number.
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.
ww
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).
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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
g.n
et
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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).
0.2h3
Chute blocks
w1 = y1
S1= y1
End sill
Baffle blocks
0.375h3
w3 = 0.75h3
0.5y1
S3 = 0.75h3
h1 = y1
1:1 Slope
0.8y2
h4
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L
(a) Type III basin dimensions
24
Froude number
10
12
8
14
16
Note:
y1 and y2 are the initial and
sequent depths of the hydraulic
jump in the stilling basin.
20
. =
T.W
y2
16
0
1.
y2
1
! (
y1
2
1 + 8Fr2 − 1 )
12
8
(b) Minimum tailwater depths
h3 or h4
y1
y1
4
18
24
20
16
12
g.n
et
8
4
Baffle block height h3
End sill height h4
2
0
2
h3 or h4
y1
y1
T.W. depth
y1
6
T.W. depth
y1
4
0
(c) Height of baffle blocks and end sill
3
L
y2
3
L
y2
ww
2:1 Slope
h3
2
4
6
8
10
12
Froude number
14
16
2
18
(d) Length of jump
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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
0.02y2
y1
S1 = y1 2
w1 = y1
S2 = 0.15y2
w2 = 0.15y2
L
(a) Type II basin dimensions
Froude number
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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
. =
T.W
y2
16
05
1.
y2
1
! (
y1
2
20
1 + 8Fr2 − 1 )
16
12
8
4
L
y2
ww
Slope = 2:1
h2 = 0.2y2
h1 = y1
12
g.n
et
8
4
(b) Minimum tailwater depths
5
5
4
4
3
4
6
8
10
12
Froude number
T.W. depth
y1
Chapter 7
14
16
18
L
y2
316
3
(c) Length of jump
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Section 7.6
Stilling Basins
317
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
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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.
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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
y21
El.60.00 m
g.n
et
(7.129)
Spillway
Tailwater (river)
El.30.00 m
y2
y1
El. X
L
Stilling basin
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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
y1 ⎜
8q
8(280/12)
⎜
⎟
⎟
1
y2 =
⎝−1 + 01 +
⎝−1 + 01 +
⎠=
⎠
3
3
2
2
gy1
9.81y1
which simplifies to
⎞
⎛
9
y1 ⎝
444.0 ⎠
y2 =
−1 + 1 +
2
y3
(7.131)
1
Combining Equations 7.129 to 7.131 yields
ww
30 − 57 + y1 +
27.75
y21
⎞
⎛
9
y1 ⎝
444.0 ⎠
=
−1 + 1 +
2
y3
1
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which simplifies to
1.5y1 +
27.75
y21
y
− 1
2
9
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
g.n
et
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
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Dams and Reservoirs
319
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).
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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
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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
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FIGURE 7.44: Major dams
in the United States
Source: U.S. Bureau of
Reclamation (2012b;
2012c).
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(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.
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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
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321
FIGURE 7.45: Bartlett
Dam
Source: U.S. Bureau of
Reclamation (2012).
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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
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FIGURE 7.46: Prosser
Creek Dam
Source: U.S. Bureau of
Reclamation (2012a).
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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).
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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.
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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.
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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
Flood storage
Normal pool elevation
Spillway
Hydroelectric turbine
Active storage
T
Minimum pool elevation
Low-level outlet
Tailwater
Dead storage
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Section 7.7
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323
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.
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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.
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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
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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
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FIGURE 7.48: Reservoir
trap efficiency
Source: USBR (1987).
100
90
Sediment trapped (%)
80
70
60
50
40
30
20
10
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0
0.001
0.01
0.1
1.0
10
Ratio of reservoir storage capacity to average annual inflow volume (years)
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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 = 100 − 1000
(C/I)2
L
Brown (1949)
W = watershed area (km2 )
C = reservoir capacity [L3 ]
I = annual inflow volume [L3 ]
&−0.2
Reference
Brune (1953)
I = annual inflow volume [L3 ]
119.6(C/I)
0.012 + 1.02(C/I)
#
C = reservoir capacity (m3 )
C = reservoir capacity [L3 ]
%
$
log(C/I)
TE = 100 0.970.19
TE = −22 +
&
&
+ 12
C = reservoir capacity [L3 ]
I = annual inflow volume [L3 ]
C = reservoir capacity (m3 )
I = inflow rate (m3 /s)
L = reservoir length (m)
Dendy and Cooper (1974)
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Heinemann (1981)
Churchill (1947);
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
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Dams and Reservoirs
325
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 .
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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
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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
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= 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
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(1 − 0.87)(1.42 * 1010 ) kg/year = 1.85 * 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
1.85 * 109 kg/year
9.46 * 1010 m3 /year
= 0.0196 kg/m3 = 19.6 mg/L
Therefore, the reservoir reduces the suspended-sediment concentration from 150 mg/L to 19.6 mg/L, a
reduction of 87%. It is noteworthy that the reservoir trap efficiency (= 87%) is equal to the reduction
in suspended-solids concentration.
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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.
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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
g.n
Month
Average
inflow
(m3 /s)
Average
demand
(m3 /s)
Month
Average
inflow
(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
17
18
19
20
21
22
23
24
1950
1870
1720
1830
1920
1850
1840
1680
et
Average
demand
(m3 /s)
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.
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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
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327
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
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−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
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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
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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
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FIGURE 7.50: Impulse
turbine
Source: Capture 3D, Inc.,
Costa Mesa, CA, and
GOM mbH,
Braunschweig, Germany
(2012).
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Section 7.7
FIGURE 7.51:
Impulse-turbine system
Water surface
Dams and Reservoirs
329
Energy grade line
hL
Reservoir
Q
Impulse turbine
he
H
Penstock
Tailwater
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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
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(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
v2
V2
− k 2 − 2
2g
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
g.n
et
(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
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330
Chapter 7
Design of Hydraulic Structures
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
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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.
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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
=
2
2
D 2gA
2gA
2gA2
j
(7.139)
j
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:
he = H − h L = H −
h T = h e − kj
fL V 2
(0.0244)(150) 1.232
= 50 −
= 49.4 m
D 2g
(0.455)
2(9.81)
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
ηT = T =
= 0.64
he
49.4
g.n
et
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
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Section 7.7
Dams and Reservoirs
331
FIGURE 7.52: Francis
turbine
Source: ANDRITZ (2012).
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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
g.n
et
(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.
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332
Chapter 7
Design of Hydraulic Structures
FIGURE 7.53: Kaplan
axial flow turbine
Source: U.S. Army Corps
of Engineers (2005g).
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FIGURE 7.54:
Reaction-turbine system
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Water surface
Energy grade line
hL
Reservoir
Q
Penstock
Turbine
H
Runner
V2
2g
Draft tube
Tailrace
V
g.n
et
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
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Section 7.7
Dams and Reservoirs
333
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
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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
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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
(gh) 4
(7.142)
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
g.n
et
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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334
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
(7.146)
P = γ Qhη
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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.
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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
g.n
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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)
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Problems
335
TABLE 7.9: Classification of Hydropower Capacity
FIGURE 7.55: Extraction
of hydropower
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Classification
Installed capacity
(MW)
Conventional
Small-scale
Mini
Micro
>15
1–15
0.1–1
<0.1
2
V1
2g
y1
H
Any hydraulic
machine
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eff
2
V2
2g
∆z
y2
Flow
$ %5
5
3
2 2
2
pmax =
ρg 2 Heff
5
(7.147)
where Heff is equal to the upstream specific energy, y1 + V12 /2g plus the drop in ground level,
"z, ρ is the density of water, and g is the acceleration due to gravity.
EXAMPLE 7.24
g.n
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It is proposed to dam a 30-m-wide stream such that the depth of water on the upstream side of the dam
is 7 m. Estimate the maximum power that could be extracted by a turbine at this location.
Solution From the given data, b = 30 m, "z = 0 m, and Heff = 7 m. These approximations assume
that the downstream channel is rectangular and the velocity head behind the dam is negligible. Taking
ρ = 998 kg/m3 and using Equation 7.147 gives the maximum power, Pmax , as
Pmax = bpmax = b
$ %5
$ %5
5
3
3
5
2 2
2 2
2
ρg 2 Heff
= (30)
(998)(9.81) 2 (7) 2 = 12.1 * 106 W = 12.1 MW
5
5
Therefore, the theoretical maximum power that could be extracted at this location is 12.1 MW.
Problems
7.1. Determine the normal and critical depths of flow in
a concrete culvert with a diameter of 1220 mm and a
design flow rate of 2.5 m3 /s. The culvert has a slope of
0.5%.
7.2. A 500-mm-diameter concrete drainage culvert (n =
0.013) is to be placed under a roadway. During the
design storm, it is expected that water will pond behind
the culvert to a height 20 cm above the crown of the
culvert. If the culvert entrance is to be well rounded
(ke = 0.05, Cd = 0.95), the slope of the culvert is
2%, the length of the culvert is 20 m, and the exit
is not submerged, estimate the discharge through the
culvert.
7.3. An existing drainage channel has a trapezoidal cross
section with a bottom width of 3.5 m, side slopes of 2:1
(H:V), a longitudinal slope of 0.5%, and a Manning’sn of
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Chapter 7
Design of Hydraulic Structures
0.025. A roadway is to be constructed across the channel
and a culvert structure is to be used to pass a design flow
rate of 3 m3 /s under the roadway. The culvert barrel(s)
is to be approximately horizontal, 10 m long, and, under
design conditions, ponding to a depth of 3 m will be
allowed at the entrance to the culvert structure. Design
the culvert.
7.4. A hydraulically short horizontal culvert is to be placed
at the end of a drainage channel to pass a peak flow
rate of 1 m3 /s under a roadway. The allowable ponding
depth in the drainage channel leading to the culvert is
1 m. The culvert is to be made of (commercial size) circular concrete pipe that is square to the headwall, and the
minimum allowable culvert diameter is 150 mm. Use the
applicable NIST equation to determine the required culvert diameter. How would the required diameter change
if you used a grooved-end culvert entrance instead of a
square-edged entrance? How would this design compare
to using the appropriate non-NIST equation. What conclusion can you draw from your analysis?
7.5. A culvert is to be designed to pass water under a roadway into a recreational lake, where the design elevation
of the lake surface is 10.00 m and the bottom elevation of
the lake is 9.50 m. The culvert is to be 380-mm-diameter
corrugated metal pipe (CMP) with 75-mm * 25-mm corrugations, 8 m long, and have a horizontal slope and
a mitered entrance. Calculate the culvert performance
curve from the no-flow condition up to a headwater elevation of 12 m. If the culvert is required to pass 0.30 m3 /s
under design conditions, what is the minimum safe elevation of the roadway? If the lake elevation drops to
9.75 m, determine the percentage change in the culvert
capacity.
7.6. What is the capacity of a 1.5-m by 1.5-m concrete box
culvert (n = 0.013) with a rounded entrance (ke = 0.05,
Cd = 0.95) if the culvert slope is 0.7%, the length is 40 m,
and the headwater level is 2 m above the culvert invert?
Consider the following cases: (a) free-outlet conditions,
and (b) tailwater elevation 0.5 m above the top of the
box 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)?
7.7. A concrete pipe culvert (n = 0.013) is to be sized to pass
0.80 m3 /s when the headwater depth is 1 m. Site conditions require that the longitudinal slope of the culvert be
2% and the length of the culvert be 10 m. If the culvert
entrance conditions are such that ke = 0.5, Cd L 1, and
the discharge is free, determine the required diameter of
the culvert.
7.8. A circular concrete culvert is designed to pass water
under a roadway. Under design conditions, the headwater depth is 2 m, the tailwater depth is 1 m, the flow rate
through the culvert is 1 m3 /s, the length of the culvert
is 15 m, and the slope is 1.5%. Determine the required
diameter of the culvert.
7.9. A trapezoidal drainage ditch has a base width of 2 m,
side slopes of 3:1 (H:V), a longitudinal slope of 2%, and
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a total depth of 3 m. The design flow rate for the channel is 8 m3 /s. A driveway is to be constructed across the
channel and a culvert is to be used to pass flow in the
drainage ditch under the driveway. Under design conditions, ponding will be allowed to within 30 cm of the
top of the drainage ditch. What commercial-size culvert
would you recommend? What material would you use
for the culvert pipe?
7.10. A 2-m * 2-m concrete box culvert is to handle a design
flow rate of 4 m3 /s. The culvert is to be laid on a slope
of 0.1, be 25 m long, and have a low tailwater elevation.
Assuming an entrance loss coefficient of 0.1, determine
the headwater depth under design conditions.
7.11. A culvert consists of two 1830-mm-diameter RCPs with
lengths of 6.1 m. Estimate the flow rate through the culvert when the headwater depth is 1.170 m and the tailwater depth is 0.920 m. The entrance has a vertical headwall
and the velocity upstream of the culvert entrance is negligible.
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7.12. A 915-mm-diameter concrete culvert is 15 m long and is
laid on a horizontal slope, and entrance conditions are
such that the entrance-loss coefficient is 0.2. The design
flow rate and tailwater depth are 1.5 m3 /s and 0.4 m,
respectively. If Type 2 flow conditions are assumed, estimate and compare the headwater depth for the following cases: (a) the culvert flows full for the entire length,
and (b) the culvert flows full for only part of the length,
as shown in Figure 7.4.
7.13. A drainage canal has a trapezoidal cross section with a
bottom width of 5 m, side slopes of 2:1 (H:V), a longitudinal slope of 0.5%, And a Manning roughness coefficient of 0.022. A 10-m long, 2-m * 2-m box culvert
is to be placed in the canal to permit drainage under
a roadway that crosses the canal. When the flow rate
through the culvert is 10 m3 /s, what is the depth of flow
at the entrance and exit of the culvert? Does the culvert have an adequate capacity to pass the flow without
the entrance being submerged? [Hint: (1) Assume that
the culvert tailwater depth is equal to the normal depth
of flow in the downstream channel; and (2) calculate
the water-surface profile in the culvert to determine the
depth of flow at the culvert entrance.]
7.14. Design a culvert to pass a 25-year peak runoff rate of
0.20 m3 /s under a roadway. The elevation of the roadway surface is 50.17 m; the width of the roadway is 15 m;
the bottom elevation of the inlet and outlet channels is
48.01 m; the inlet and outlet channels are both trapezoidal with bottom widths of 2 m, side slopes of 2:1
(H:V), lined with riprap, and have a longitudinal slope
of 0.1%. Local drainage regulations require a minimum
culvert diameter of 455 mm.
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7.15. A drainage channel has a bottom width of 2 m, side
slopes of 3:1 (H:V), a total depth of 2 m, a longitudinal
slope of 0.5%, and an estimated Manning’s n of 0.018.
A 15-m-long multibarrel culvert is to be used to pass the
flow in the channel under a roadway, with eachbarrel of
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Problems
the culvert consisting of a 455-mm-diameter reinforced
concrete pipe. The design flow rate in the channel is 1.5
m3 /s and the culvert is on the same slope as the channel. The culvert entrance consists of a headwall with
a square-edged entrance to each barrel. (a) Determine
the number of barrels required in the culvert. (b) Under
design conditions, what is the water depth just upstream
of the culvert (i.e., the headwater depth) and how far
upstream of the culvert until the water depth is reduced
by 50%.
7.16. Is it possible to have a condition in which a culvert
entrance is not submerged while its exit is submerged? If
so, sketch the flow regime within the culvert and describe
how you would calculate the capacity of the culvert.
7.17. A 450-mm * 450-mm concrete box culvert is being
proposed to provide drainage across a rural driveway.
The culvert is to be placed in a vertical headwall with
45◦ wingwall flares, 4 m long, and have a longitudinal
slope of 3%. Use the NIST equations (Equations 7.15
and 7.26) to calculate the culvert performance curve for
flow rates up to 0.6 m3 /s. State your assumptions in using
these equations, and confirm your results for a flow rate
of 0.6 m3 /s using the USFHWA nomograph shown in
Figure 7.56.
7.18. A stream is to pass under a rural roadway via a 1220mm-diameter concrete culvert that is 12-m long and
laid on a horizontal slope. The culvert entrance is flush
with the headwall with a grooved end and the estimated
entrance loss coefficient is 0.3. The design flow rate is
2.3 m3 /s. Under design conditions the tailwater depth is
0.60 m. Estimate the headwater depth required for the
culvert to accommodate the design flow rate.
7.19. How would the required headwater depth in Problem
7.18 change if the minimum-performance method were
used?
7.20. A culvert under a roadway is to be designed for a 100year peak surface runoff rate of 3.00 m3 /s. The invert
elevation at the culvert inlet is 12.11 m, the invert elevation at the outlet is 11.71 m, and the length of the culvert
is to be 20 m. The channel downstream of the culvert
has a trapezoidal cross section with a bottom width of
1 m, side slopes of 2:1 (H:V), a longitudinal slope of 2%,
and a Manning’s n of 0.040. The gravel roadway crossing the culvert has a length of 15.0 m, an elevation of
13.50 m, and a width of 16.2 m. Considering a circular
reinforced concrete pipe (RCP) culvert with a diameter
of 760 mm and an entrance that is mitered to conform to
the fill slope, determine the depth of water flowing over
the roadway, the flow rate over the roadway, and the flow
rate through the culvert.
7.21. The 450-mm-diameter concrete culvert shown in
Figure 7.57 passes surface runoff from a drainage channel to an open area. The culvert has an allowable headwater depth of 0.6 m, a headwall with square edges at
the entrance, a longitudinal slope of 1%, and a length
of 3.6 m. (a) Determine the flow capacity of the culvert;
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337
and (b) design the riprap apron at the exit of the culvert. Consider all applicable equations in determining
the flow capacity of the culvert, and the apron design
should include the length, width, thickness, and stone
size.
7.22. A vertical gate is installed at the end of a canal that is 7 m
wide; the depth of flow is 3.5 m. If the gate has a width
of 3 m, estimate the flow rate through the gate when the
gate is raised 0.5 m. Neglect the effect of the downstream
water depth.
7.23. Explain why the contraction coefficient of a vertical gate
with a rounded edge is greater than that of a vertical gate
with a sharp edge.
7.24. Derive Equation 7.49.
7.25. If a vertical gate has an opening of 70 cm and discharges
16 m3 /s into a 5-m-wide channel, estimate the maximum
depth of water downstream of the gate for which the discharge under the gate will not be “drowned.” [Hint: As
long as a hydraulic jump can form, the discharge will not
be drowned.]
7.26. Water flows under a vertical gate and into a 3-m-wide
rectangular channel where the flow rate and depth of
flow are 4 m3 /s and 1.1 m, respectively. Estimate the
minimum gate opening to prevent the formation of a
hydraulic jump. What energy loss (in kilowatts) can be
expected in a hydraulic jump?
7.27. Water is released from under a vertical gate into a 2-mwide lined rectangular channel. The gate opening is 50
cm, and the flow rate into the channel is 10 m3 /s. The
channel is lined with reinforced concrete and has a Manning roughness coefficient of 0.015, a horizontal slope,
and merges with a river where the depth of flow is 3.5 m.
Does a hydraulic jump occur in the lined channel? If so,
how far downstream of the gate does the jump occur,
and what could be done to ensure that a hydraulic jump
does not occur?
7.28. A vertical sluice gate is opened 20 cm and the depth of
water behind the gate is 1.7 m. If the width of the channel is 1.5 m, estimate the free-flow discharge through the
gate and the distinguishing condition for the tailwater
depth.
7.29. Derive Equation 7.53.
7.30. Water is ponded behind a vertical gate to a height of 5 m
in a rectangular channel of width 8 m. If the tailwater
depth is 4.0 m and the gate opening is 1.40 m, what is the
discharge through the gate?
7.31. Water is ponded behind a vertical gate to a height of 5 m
in a rectangular channel of width 8 m. Calculate the gate
opening that will release 50 m3 /s through the gate. How
would this discharge be affected by a downstream flow
depth of 4 m?
7.32. A vertical gate is to be designed to release water from an
upstream reservoir into a downstream canal. The design
elevation of the water surface in the upstream reservoir
is 13.00 m, the downstream water-surface elevation is
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Chapter 7
Design of Hydraulic Structures
12
600
11
500
10
400
9
300
8
200
7
HW
D
1.75
1.90
2.05
Inlet
(1)
(2)
(3)
(2)
8
7
6
HW
feet
5
3.5
3.8
4.1
4
3
80
2
(3)
9
8
7
6
5
4
3
2
60
50
1.5
Headwater depth in terms of height (HW/D)
Ratio of discharge to width (Q/B) in cfs per foot
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5
(1)
Example
5 ft x 2 ft Box Q = 75 cfs
Q/B = 15 cfs/ft
100
6
Height of box (D) in feet
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40
1.5
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3
2
30
ple
am
Ex
20
10
Angle of
wingwall
flare
8
6
5
4
3
HW
D
(1)
(2)
(3)
Wingwall
flare
30o to 75o
90o and 15o
0o
0.8
1
0.6
0.5
5
4
3
2
1.5
0.9
1.0
1.0
0.8
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
0.4
0.4
0.35
0.35
0.7
0.6
2
1
8
7
6
1.0
0.5
To use scale (2) or (3),
project horizintally to
scale (1) then use straight
inclined line through D
and Q scales, or reverse
as illustrated.
10
0.4
0.30
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Headwater depth
for box culverts
with inlet control
FIGURE 7.56: USFHWA culvert nomograph
Source: USFHWA (2012).
10.00 m, and the elevation of the bottom of the channel
is 8.00 m. The gate is 3 m wide. Determine the range
of gate openings for which a hydraulic jump will occur
downstream of the gate. Considering all possible gate
openings, determine the discharge through the gate as
a function of the gate opening.
7.33. The contraction coefficient for a radial gate is given by
$ %2
$ %
θ
θ
+ 0.36
Cc = 1 − 0.75
90
90
Find the value of θ that gives the minimum value of Cc .
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Problems
339
Culvert
Flow
600 mm
Riprap outlet
protection
450 mm
Low tailwater
Slope = 1
%
Open area
3.6 m
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FIGURE 7.57: Drainage culvert with outlet armoring
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7.34. Water flows at 10 m3 /s in a 5-m-wide channel. What is
the height of a suppressed rectangular (sharp-crested)
weir that will cause the depth of flow in the channel to
be 2 m?
7.35. How does a sill differ from a weir? Show that Equation 7.72 gives a reasonable approximation to the discharge coefficient for both weirs and sills.
7.36. A sharp-crested rectangular weir is to be installed to
measure flow rates that are expected to be on the order
of 1 to 3 m3 /s. If the depth of the channel is such that
the maximum depth of water that can be accommodated
at the weir is 3.0 m and the width of the channel is 5 m,
determine the height of a suppressed weir that can be
used to measure the flow rate.
7.37. The stage in a river is to be maintained at an elevation
of 20.00 m behind a sharp-crested rectangular weir. If
the river is 10 m wide and 1 m deep and the flow rate
in the river is 5 m3 /s, what is the crest elevation of the
suppressed weir that must be used?
7.38. A weir is to be placed downstream of a gate to control submergence of the gate discharge. Consider the
condition shown in Figure 7.58, where the water depth
upstream of the gate is 3 m and the gate opening is 0.5 m.
The width of the channel is 4 m and the weir is at a sufficient distance downstream of the gate such that flow
conditions immediately upstream of the weir are subcritical. Determine the minimum height of the weir required
to ensure that the gate discharge is submerged.
7.39. A contracted sharp-crested rectangular weir is to be
used to measure the flow rate in an open channel that
is 5 m wide and 2 m deep. It is desired that the depth of
water over the crest of the weir be no less than 0.5 m
when the flow rate is 1 m3 /s and that the water level be
at least 30 cm below the top of the channel. What should
the crest length of the weir be, and how far below the top
of the channel should the crest of the weir be located?
Sketch the weir.
7.40. A recreational lake is to be created along a stream by
installing a rectangular sharp-crested weir across the
stream. The design flow rate through the planned lake
is 55 m3 /s and the lake is to be 100 m long and 15 m
wide, have vertical (walled) sides, and have a longitudinal slope of 0.5%. The Manning’s n of the lake section of
the stream is estimated to be 0.018. (a) If the water depth
at the downstream end of the lake (where the weir is
located) is to be maintained at 5 m and the weir crest is to
be 10 m long, determine the required height of the weir
crest. (b) Estimate the depth of water at the upstream
end of the lake.
7.41. A sharp-crested weir is to be placed 100 m downstream
of a water park on a river such that the river stage at the
park is 5.50 m when the flow rate in the river is 25 m3 /s.
The longitudinal slope of the river is 0.1%, the elevation
of the bottom of the river at the park location is 2.00 m,
the side slopes are 2:1 (H:V), and the bottom width is
6.00 m. It can be assumed that the Manning’s n of the
river is 0.05. Specify a crest length and elevation of the
weir crest that will have the desired effect.
7.42. Water flows at a rate of 5 m3 /s in a rectangular channel
of width 10 m. The normal depth of flow in the channel is 5 m. If a sharp-crested (suppressed) weir of height
5 m is installed across the channel, determine the new
flow depth in the channel for the same flow rate. What
would the flow depth be if the rectangular weir were contracted (on both sides) to a width of 7 m? What would
the flow depth be if a 7-m-wide Cipolletti weir were
used?
7.43. A Cipolletti weir is to be designed to measure the flow
rate in a channel. Under design conditions, the normal
depth of flow is 2 m and the flow rate is 5 m3 /s. Design
the weir with a top width of 10 m.
7.44. A 0.8-m-high suppressed rectangular weir is placed in a
1-m-wide rectangular channel. If the water depth over
the weir is observed to be 20 cm, estimate the flow rate
in the channel. If the downstream water depth rises to
5 cm above the crest of the weir and the upstream water
depth remains at 20 cm above the crest, estimate the new
flow rate in the channel.
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340
Chapter 7
Design of Hydraulic Structures
Gate
Weir
Flow
3m
Flow
0.5 m
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FIGURE 7.58: Gate/weir combination
7.45. A suppressed rectangular weir is being used to measure
flow rates in an irrigation canal. The weir is 5 m wide and
2 m high, and, under flood conditions, the upstream and
downstream depths are measured as 2.5 m and 2.3 m,
respectively. Use all applicable formulae to estimate the
range of possible flow rates over the weir under these
conditions. Assess the reliability of your flow-rate estimate. How could a more precise estimate of the flow rate
be obtained?
7.46. An outlet weir is to be used to form a lake at a park. The
lake is to have a surface elevation of 101.40 m, the elevation of the bottom of the lake is 96.00 m, and the flow
rate through the lake under design conditions is 4 m3 /s.
The crest length of the outlet weir is constrained to be
less than or equal to 6 m. (a) Design the crest elevation
and height of the weir; and (b) Determine the maximum
downstream stage that will not reduce the design weir
flow rate by more that 10%.
7.47. A small irrigation channel has a trapezoidal cross section
with a longitudinal slope of 0.1%, bottom width of
2 m, side slopes of 2 : 1 (H : V), and a total channel
depth of 2 m. The maximum flow rate expected in
the channel is 3 m3 /s, and the maximum flow depth at
a flow-measurement structure is not to exceed 1.5 m.
(a) Design an unsuppressed rectangular weir of width
2 m that can be placed in the channel to measure the
flow rate. (b) If the elevation of the water surface downstream of the weir is 20 cm below the water surface
upstream of the weir, by how much does this affect
the free-flow weir capacity (in m3 /s)? (c) What range of
elevation differences between the headwater and tailwater will limit the reduction in weir capacity to less
than 10%?
7.48. Equations 7.78 and 7.79 have been proposed for estimating the flow rates over submerged rectangular weirs.
Compare the predicted discharges given by these formulae and assess whether either one is preferable.
7.49. A V-notch weir is to be used to measure a maximum flow
rate of 12 L/s. If a notch angle of 30◦ is selected, what
should the depth of the notch be?
7.50. If a V-notch weir has a notch angle of 50◦ and a height
of 50 cm, estimate the capacity of the weir.
7.51. A 90◦ V-notch weir is to be used to measure flow rates
as low as 1 L/min. Estimate the corresponding depth of
flow over the weir, accounting for viscous and surfacetension effects if necessary.
7.52. A 70◦ V-notch weir is observed to have a 40-mm head.
Estimate the flow rate over the weir.
7.53. A V-notch weir is to be designed to measure the flow
rate in an irrigation channel. Under design conditions,
the flow rate in the channel is 0.30 m3 /s and the depth of
flow is 1 m. Design the weir with a top width of 1 m.
7.54. The flow through a V-notch weir is commonly expressed
in the form
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$ %
+
5
θ
8
Cd 2g tan
H2
15
2
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(a) Use the Kindsvater and Shen equation to derive an
analytic expression for Cd as a function of H and θ. Give
the units of H and θ in your expression. (b) A V-notch
weir is to be designed as part of the discharge structure
of a detention pond. The weir is to be sized such that it
releases 20,000 m3 of water from the detention pond in
no less than 24 hours. If a 90◦ notch angle must be used,
what height of weir is required?
7.55. Determine the crest length of a compound weir that has
a notch angle of 90◦ , a notch height of 20 cm, and a discharge of 1.5 m3 /s when the water depth is 1.20 m. The
apex of the notch is placed at ground level.
7.56. A 25-cm-high broad-crested weir is placed in a 1.5m-wide rectangular channel. If the maximum depth of
water that can be measured upstream of the weir is equal
to 75 cm, what is the maximum flow rate that can be
measured by the weir?
7.57. A broad-crested rectangular weir of length 1 m, width
1 m, and height 30 cm is being considered to measure the
flow rate in a canal. For what range of flow rates would
this weir length be adequate?
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Problems
7.58. A broad-crested weir is to be used to measure the
flow rate in an irrigation channel. The design section
upstream of the weir is rectangular with a width of 3 m,
and the depth of flow is 4 m at a flow rate of 5 m3 /s.
Design the height and length of the weir.
7.59. A broad-crested weir is to be constructed to measure
the flow rate in a 3-m-wide rectangular channel. If the
flow rate in the channel is expected to be in the range of
0.8–1.5 m3 /s and the depth of the channel is 2.5 m, select
a height and length of weir such that the water level in
the channel does not rise any higher than 2 m above the
bottom of the channel.
7.60. Excess water from surface runoff is to be drained
from a pond with a bottom elevation of 9.10 m. The
minimum-allowable water elevation in the pond is
11.20 m, the maximum-allowable elevation is 11.70 m,
and the maximum-allowable discharge from the pond is
0.75 m3 /s. The pond outlet structure is to be a broadcrested weir. Determine the length and width of the
weir. Take into account that there is more than one credible expression for estimating the discharge coefficient
of the weir, and assume that the velocity head upstream
of the weir is negligible.
7.61. The broad-crested compound weir shown in Figure 7.59
is to be designed to regulate the outflow from a drainage
channel. When the depth of water in the channel is less
than 0.25 m no outflow will be allowed, when the depth
is 0.50 m the outflow rate is to be 0.08 m3 /s, and when
the depth is 0.75 m the outflow rate is to be 0.50 m3 /s.
Determine the required dimensions b, B, and L of the
weir. [Hint: Use the subsection approach.]
7.62. Show that Equation 7.117 can be derived from Equation 7.116. Explain your approach. [Hint: dy/dx = α at
x = XDT .]
7.63. Design a 40-m-long overflow spillway that will discharge
a design flow rate of 1600 m3 /s at a maximum-allowable
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341
pool elevation of 200 m. The bottom elevation behind
the spillway is 170 m, the upstream face of the spillway
has a 1:1 slope, and the spillway chute is to have a slope
of 1:1.5 (H:V).
7.64. An overflow spillway is to be designed to discharge
500 m3 /s when the maximum pool elevation is at 200 m.
The bottom elevation of the reservoir behind the dam
is 150 m, and the spillway is to have an upstream face
with a 1:1 slope. To accommodate three Tainter gates,
the spillway will need two 1-m-wide round-nosed piers
between two square abutments. Determine the required
spillway lengths for crest elevations of 195, 190, 185, and
180 m.
7.65. If a spillway were used instead of a broad-crested weir
as the discharge structure in Problem 7.60, specify the
dimensions of the spillway and assess the size efficiency
gained, if any, in using a spillway instead of a weir.
7.66. A spillway design is being considered in which the headwater is 2.0 m above the crest and the tailwater is 0.7 m
above the crest. The height of the crest relative to the
bottom of the upstream reservoir is 4.2 m; the height of
the crest relative to the downstream apron is 4.8 m. The
effective length of the spillway crest is 5.0 m, the design
head is 2.0 m, and the upstream face is vertical. Estimate
the spillway discharge under the given conditions.
7.67. Gated ogee spillways are commonly used to regulate
flows in canals, and when the gate is completely out of
water the structure behaves just like an ungated ogee
spillway. An existing spillway is being investigated where
the bottom elevation of the upstream and downstream
canals is 0.50 m, the crest elevation of the spillway is
2.5 m, the crest length is 10 m, and the shape of the spillway is based on a design head of 0.70 m. (a) Estimate
the flow capacity of the spillway under design conditions.
(b) What is the maximum-allowable stage upstream of
the spillway and what is the discharge capacity of the
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b
0.25 m
0.25 m
0.25 m
FIGURE 7.59: Compound-weir discharge structure
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342
Chapter 7
Design of Hydraulic Structures
structure under this condition? (c) What would happen
if the maximum stage is exceeded?
7.68. You have been charged with calculating the performance of a gate that controls the discharge from a spillway. When the spillway gate is completely open, the
spillway behaves like a regular spillway, with critical
depth occurring over the crest of the spillway. A vertical
gate is mounted above the spillway crest, and when the
gate is lowered to control the flow, the flow through the
gate opening behaves like a sluice gate on a horizontal
floor. The spillway has a crest length of 10 m, the discharge coefficient is 2.2, and the contraction coefficient
for the sluice gate is 0.61. For the case in which upstream
water is ponded to a height of 2.0 m above the crest of
the spillway, determine the spillway discharge as a function of the gate opening for gate openings between 0.0
and 2.0 m in increments of 0.20 m.
7.69. An ogee spillway is 5 m wide with a vertical lift gate
seated on the crest of the spillway. If the headwater and
tailwater levels are 2.00 m and 0.70 m above the spillway crest, respectively, and the gate opening is 0.40 m,
estimate the discharge through the gate.
7.70. Consider the Tainter gate described in Example 7.16.
Compare the discharge when the gate opening is 0.50 m
to the discharge when the gate opening is 1.00 m?
7.71. Consider the vertical gate described in Example 7.17.
Compare the discharge when the gate opening is 1.00 m
to the discharge when the gate opening is 0.50 m?
7.72. A spillway is 10 m wide and discharges 120 m3 /s under
design conditions. The reservoir level behind the spillway is at elevation 100.00 m, and the water-surface elevation of the receiving water downstream of the stilling
basin is 80.00 m. Assuming a 15% loss of hydraulic head
in the flow down the spillway, find the required elevation
of the floor of the stilling basin and design the stilling
basin.
7.73. A 10-m-wide emergency spillway with well-rounded
abutments has a crest height 5.00 m above the bottom of a storage reservoir in which the depth at
the maximum-pool elevation is 6.00 m. The emergency
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spillway terminates with a stilling basing in a 10-m-wide
rectangular concrete channel that has an invert elevation
equal to the elevation of the bottom of the storage reservoir. If energy losses down the spillway are neglected,
what type of stilling basin should be used?
7.74. A spillway is to be used to control discharges from a
reservoir such that when the water level in the reservoir
is at elevation 80.00 m, the free-flow discharge from the
reservoir is 100 m3 /s. The spillway discharges are to be
controlled by three vertical-lift gates with each gate having a width of 3 m, and the piers between the gates each
having a width of 1 m. The elevation of the bottom of
the reservoir is 67.00 m, and the water-surface elevation
downstream of the spillway under design conditions is
70.00 m. A 10% loss in available energy is expected as
water flows down the spillway. Determine the required
height and crest length of the spillway, the bottom elevation of the stilling basin, and the type of stilling basin
that should be used.
7.75. A reservoir covers an area of 850 km2 and has an average depth of 18.7 m. The inflow to the reservoir is from
a river with an average flow rate of 2500 m3 /s and an
average suspended-sediment concentration of 250 mg/L.
Estimate the rate at which the depth of the reservoir is
decreasing due to sediment accumulation, and the time
it will take for the reservoir storage to decrease by 10%.
Assume that the accumulated sediment has a bulk density of 1600 kg/m3 .
7.76. During a critical 24-month period, the average monthly
inflows to a reservoir and the average water demand are
given in Table 7.10. Use these data to estimate the minimum required storage volume of the reservoir.
7.77. At a Pelton-wheel installation the water-surface elevation of the supply reservoir is 85 m above the nozzle; the
delivery line has a diameter of 0.60 m, a length of 300 m,
and a roughness height of 8 mm. The discharge nozzle
has a diameter of 50 mm and a head-loss coefficient of
0.8. The bucket friction loss coefficient is 0.5, the velocity
of water relative to the bucket at the exit from the bucket
is 2 m/s, and the absolute velocity of the water leaving
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TABLE 7.10
Average Average
Average Average
Average Average
inflow demand
inflow demand
inflow demand
Month∗ (m3 /s)
(m3 /s) Month (m3 /s)
(m3 /s) Month (m3 /s)
(m3 /s)
1
2
3
4
5
6
7
8
850
1130
1310
1000
1170
1260
880
930
1000
1100
1200
1300
1400
1400
1400
1400
9
10
11
12
13
14
15
16
1550
1600
1370
1210
940
820
1040
1290
1300
1200
1100
1000
1000
1100
1200
1300
17
18
19
20
21
22
23
24
1350
1270
1120
1230
1320
1250
1240
1080
1400
1400
1400
1400
1300
1200
1100
1000
Note: ∗ Month “1” is January.
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Problems
the bucket is 6 m/s. Determine the power that could be
derived from the system and the hydraulic efficiency of
the turbine.
7.78. A hydropower installation is to be located where the
downstream water-surface elevation is 100 m below
the water-surface elevation in the reservoir. The 2.0m-diameter concrete-lined penstock is 500 m long and
has an estimated roughness height of 15 mm. When
the flow rate through the system is 20 m3 /s, the combined head loss in the turbine and draft tube is 5.0
m, and the average velocity in the tailrace is 0.80 m/s.
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Estimate the power that can be extracted from the
system.
7.79. A hydroelectric project is to be developed where the 90percentile flow rate is 1630 m3 /s, and storage to a height
of 15 m above the downstream stage is required to meet
all the demands. Estimate the maximum installed capacity that would be appropriate for these conditions.
7.80. A 10-m-wide stream is to be dammed such that the depth
of water on the upstream side of the dam is 5 m. Estimate the maximum amount of hydroelectric power that
could be extracted at this location.
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C H A P T E R
8
Probability and Statistics
in Water-Resources Engineering
8.1 Introduction
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All hydrologic phenomena obey the laws of nature and can, in theory, be described by
solutions of the relevant fundamental equations. In most cases, however, the uncertainty
and amount of spatial and temporal detail that needs to be known about the physical system,
combined with the limitations of analytical and numerical techniques for solving systems
of equations, preclude an exact description of most natural phenomena. Processes that can
be specified with certainty are called deterministic processes, while those that cannot are
called stochastic processes. The two types of stochastic processes commonly encountered in
engineering practice are: (1) processes where the outcome is uncertain because of the lack
of information on the process or parameters affecting the process, called epistemic uncertainty, and (2) processes where the outcome is uncertain because the process generating
the outcome is fundamentally random, called natural uncertainty. In the first case, epistemic
uncertainty can be reduced, or even eliminated, by additional measurements of the variables
affecting the process or by the use of more accurate models. In the second case, natural
randomness is intrinsic to the process and uncertainty in the outcome might not be reduced
by additional measurements. In many cases, recorded data include both epistemic and natural uncertainty, and separation of these components is a challenging task. An example of
this is in the analysis of climatic data, where separation of random climate variability from
more predictable long-term climate changes is a common challenge. Whether uncertainty is
epistemic, natural, or a combination of both, an uncertain outcome is generally defined by
a probability distribution that describes the relative likelihood of all possible outcomes. In
practice, uncertain variables are collectively referred to as random variables.
A common application of probability theory in water-resources engineering involves
the assignment of an exceedance probability, Pe , to hydrologic events. A system designed
to handle a particular hydrologic event can be expected to fail with a probability equal to
the probability of exceedance, Pe , of the design event. This probability of system failure is
called the risk of failure, and the probability that the system will not fail, 1 − Pe , is called the
reliability of the system. Exceedance probabilities of hydrologic events are usually stated in
terms of the probability that a given event is exceeded in any given year. The average number
of years between exceedances is called the return period, T, which is related to the annual
exceedance probability, Pe , by
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T=
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1
Pe
(8.1)
The return period, T, is sometimes called the average recurrence interval (ARI) and is usually a criterion for selecting design events in water-resource systems. In cases where waterresource systems are to be in place for only a short period of time (less than a year), it might
be appropriate to calculate return periods of design events based on their occurrence within
a specific season of the year (e.g., McCuen and Beighley, 2003).
The determination of flows corresponding to a given return period is also of central
importance in the restoration of natural channels, since it is commonly assumed that natural channels evolve to contain surface runoff having approximately a 2-year return period
(e.g., He and Wilkerson, 2011).
344
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Section 8.2
Probability Distributions
345
If a water-resource system is designed for a hydrologic event with exceedance probability, Pe , and the system has a design life of N years, then the risk that the system will fail
within its design life, Pf , is given by
Pf = 1 − (1 − Pe )N
(8.2)
This formulation is particularly useful in assessing the cost of failure, since the expected
damage cost is equal to the probability of failure multiplied by the cost associated with the
failure.
8.2
ww
Probability Distributions
A probability function defines the relationship between the outcome of a random process
and the probability of occurrence of that outcome. If the sample space, S, contains discrete
elements, then the sample space is called a discrete sample space; if S is a continuum, the
sample space is called a continuous sample space. The sample space of a random variable is
commonly denoted by an upper-case letter (e.g., X), and the corresponding lower-case letter
denotes an element of the sample space (e.g., x). Discrete probability distributions describe
the probability of outcomes in discrete sample spaces, while continuous probability distributions describe the probability of outcomes in continuous sample spaces.
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8.2.1
Discrete Probability Distributions
If X is the sample space of a discrete random variable, with outcomes x1 , x2 , . . . , xN , then the
probability of occurrence of each element, xn , can be written as a probability function, f (xn ),
sometimes referred to as the probability distribution function. If f (xn ) is a valid probability
function, then it must necessarily have the following properties:
f (xn ) Ú 0,
N
!
n=1
5 xn
(8.3)
f (xn ) = 1
(8.4)
where N is the number of elements in X. In many engineering applications, the quantity of
interest is the probability that a random process will generate an outcome that is greater than
or equal to some outcome xn , where xn is a design variable and exceedance of xn will result
in system failure.
Based on the definition of the probability distribution function, the probability that an
outcome is less than or equal to xn is given by
!
P(xi … xn ) =
f (xi )
(8.5)
xi …xn
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and the probability that the outcome is greater than xn is given by
P(xi > 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
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346
Chapter 8
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on the definition of the bivariate probability distribution, f (xp , yq ), the following conditions
must necessarily be satisfied
f (xp , yq ) Ú 0,
N !
M
!
p=1 q=1
5 xp , yq
(8.8)
f (xp , yq ) = 1
(8.9)
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 ) =
ww
! !
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
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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
g.n
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(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
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Section 8.2
F(x, y) =
g(x) =
h(y) =
8.2.3
" x " y
−q −q
" q
−q
" q
−q
Probability Distributions
f (x′ , y′ ) dx′ dy′
347
(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
ww
⟨g⟩ =
N
!
g(xi )f (xi )
(8.19)
i=1
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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
(8.21)
g(x) = x
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
⎪
⎪
⎪
⎪
⎪
x′ f (x′ ) dx′
⎩
g.n
et
(8.22)
(continuous)
−q
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
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348
Chapter 8
Probability and Statistics in Water-Resources Engineering
σx2 =
⎧
N
⎪
⎪
⎪!
⎪
⎪
(xi − µx )2 f (xi )
⎪
⎨
(discrete)
i=1
⎪
" +q
⎪
⎪
⎪
⎪
⎪
(x′ − µx )2 f (x′ ) dx′
⎩
(8.24)
(continuous)
−q
ww
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,
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gx =
⎧
N
⎪
⎪
1 !
⎪
⎪
(xi − µx )3 f (xi )
⎪
⎪
⎨ σx3
(discrete)
i=1
⎪
" +q
⎪
⎪
1
⎪
⎪
(x′ − µx )3 f (x′ ) dx′
⎪ 3
⎩
σx −q
(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
g.n
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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
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Section 8.2
Probability Distributions
349
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
N
gx =
ww
'
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
3
σx
2
(1.92)
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
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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 =
a2
0
Hence, the expression for µt can be written as
+
µt = 0.014
1
0.0142
,
= 71 h
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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
(t′2 − 142t′ + 5041)e−0.014t dt′
σt2 =
0
0
0
Using the integration results (Dwight, 1961),
" q
" q
2
1
x2 e−ax dx =
;
xe−ax dx =
;
3
a
a2
0
0
the expression for σt2 can be written as
σt2 = 0.014
-
" q
0
e−ax dx =
1
a
.
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) 0
σt 0
=
"
′
0.014 q ′3
(t − 213t′2 + 15,123t′ − 357,911)e−0.014t dt′
3
(71) 0
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350
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
0.014
0.0144
0.0143
0.0142
,
= 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
ww
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
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⟨τ ⟩ =
q
!
τ =1
τ (1 − p)τ −1 p
= 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
=
p
[1 − (1 − p)]2
Since T = ⟨τ ⟩, Equation 8.29 can be expressed as
T=
1
p
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(8.29)
(8.30)
or
T=
1
P(X Ú xT )
(8.31)
1
T
(8.32)
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.
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Section 8.2
Probability Distributions
351
EXAMPLE 8.3
Analyses of the maximum-annual floods over the past 150 years in a small river indicate the following
cumulative distribution:
ww
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
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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
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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
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352
Chapter 8
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
ww
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
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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
where
+ ,
N
N!
=
n
n!(N − n)!
(8.35)
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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)!
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Section 8.2
Probability Distributions
353
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=
ww
20
!
i=1
f (i) = 1 − f (0) = 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
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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
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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
6
7
8
9
10
n
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354
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=
ww
8.2.5.3
5
!
n=1
p(1 − p)n−1 =
5
!
n=1
(0.1)(1 − 0.1)n−1 =
5
!
(0.1)(0.9)n−1 = 0.41 = 41%
n=1
Poisson distribution
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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; Thiébaux, 1994)
N!
(Np)n e−Np
pn (1 − p)N−n =
n!
N→q, p→0 n!(N − n)!
lim
(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)
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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 λ!t; 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)
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Section 8.2
FIGURE 8.3: Poisson
probability distribution
0.7
0.5
0.4
0.4
f(n)
f(n)
0.5
0.3
0.3
0.2
0.2
0.1
0.1
0
1
2
3
4
5
n
6
7
8
9
λ = 2.5
0.6
0.0
10
0.7
0
1
2
3
5
n
6
7
8
9
10
λ = 6.5
0.6
0.5
0.5
0.4
0.4
f(n)
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0.3
0.3
0.2
0.2
0.1
0.1
0.0
4
0.7
λ = 4.5
0.6
f(n)
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355
0.7
λ = 0.5
0.6
0.0
Probability Distributions
0
1
2
3
4
5
n
6
7
8
9
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
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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
1
0
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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Chapter 8
Probability and Statistics in Water-Resources Engineering
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
ww
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
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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
0
λe−λτ dτ
(8.47)
which leads to
F(t) = 1 − e−λt
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(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
4
5
λt
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Section 8.2
Probability Distributions
357
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
ww
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
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t
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
0.8λ
0.8λ
0.6λ
0.6λ
0.4λ
0.2λ
0
(8.51)
1.0λ
n=1
f(t)
f(t)
1.0λ
λn tn−1 e−λt
%(n)
n =3
0.4λ
0.2λ
0
1
2
3
λt
4
5
0
0
1
2
3
4
λt
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358
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)!,
ww
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).
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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
n
4
= 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
2
t3
3t2
6t
6
−
+
−
a
a2
a3
a4
Using this relation to evaluate F(t0 ) gives
⎡
2
3
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3
2
3t
6t
6
t
⎦
+
F(t0 ) = −2.405 * 10−9 ⎣e−0.01096t
+
+
0.01096
0.010962
0.010963
0.010964
Taking t0 = 730 d (= 2 years) gives
⎡
0
3⎤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
2
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)
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Section 8.2
Using the transformation
Probability Distributions
359
(8.55)
x = λt
the probability density function of x, p(x), is related to f (t) by the relation
8 8
8 dt 8
1
p(x) = f (t) 88 88 = f (t)
dx
λ
(8.56)
Combining Equations 8.54 to 8.56 and replacing n by α yields
p(x) =
ww
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
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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
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(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 − ϵ)β−1 e−λ(x−ϵ)
%(β)
(8.61)
where λ, β, and ϵ 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=
, ϵ = 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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360
Chapter 8
Probability and Statistics in Water-Resources Engineering
ww
(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 watershe