Water-Resources Engineering

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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 ´ de Mexico, S.A. de C.V. Pearson Educacion Pearson Education—Japan, Tokyo Pearson Education Malaysia, Pte. Ltd. Pearson Education, Upper Saddle River, New Jersey

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To Andrew and Stephanie.

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

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Contents Preface

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

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

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

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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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545 545 545 549 550 554 560 565 566 567 568 571 576 577 578 583 584

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

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

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

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

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

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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 .

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815 815 815 818 819 819 820

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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

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

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

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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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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

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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

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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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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.

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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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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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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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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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The Hydrologic Cycle

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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

Item

Volume (*103 km3 )

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

1,338,000

Total water Freshwater

1,385,984.61 35,029.21

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

Percent total water (%) 96.5 0.76 0.93 0.0012 1.7 0.025 0.007 0.006 0.0008 0.0002 0.0001 0.001 100 2.5

Percent freshwater (%)

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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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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 Superarid Hyperarid Arid Semiarid Subhumid Humid Hyperhumid Superhumid

Mean annual precipitation (mm)

Mean annual evapotranspiration (mm)

0.2

r

K Reentrant flow

V

0.8–1

K Pipe exit

V 1

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V

1

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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

r

K θ = 20°

K θ = 180° 1.00 0.87 0.70 0.41 0.15

0.30 0.25 0.15 0.10

Without vanes

K = 1.1

With vanes

K = 0.2

r/D

D 90° smooth bend

K θ = 180° 0.50 0.49 0.42 0.27 0.20 0.10

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90° miter bend

K θ = 60° 0.08 0.08 0.07 0.06 0.06 0.06

1 2 4 6

g .n K θ = 90° 0.35 0.19 0.16 0.21 K

Threaded pipe fittings

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

4–10 5.0 0– 0.2 5.6 2.2

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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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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.

w .E asy En g 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

ine eri n 15 m

Reservoir

5m

5m

2m

P

3m

100-mm PVC (pump to reservoir)

g .n 5m

et

50-mm PVC (well to pump) Well

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

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

V12

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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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

0.005 Q = = 0.637 m/s A2 0.007854

PVC pipe can be considered smooth (ks L 0) and the friction factor, f , can be estimated using the Swamee–Jain equation as 0.25 f =   5.74 2 log10 Re0.9

where Re is the Reynolds number. At 20◦ C, the kinematic viscosity, ν, is equal to 1.00 * 10−6 m2 /s and for the 50-mm pipe (2.54)(0.05) V D Re1 = 1 1 = = 1.27 * 105 ν 1.00 * 10−6 which leads to 0.25 f1 =  2 = 0.0170 5.74 log10 (1.27 * 105 )0.9

w .E asy En g and for the 100-mm pipe

Re2 =

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

which leads to

f2 = 

ine eri n 0.25

5.74 log10 (6.37 * 104 )0.9

2 = 0.0197

g .n

Substituting the values of these parameters into the energy equation yields     (0.0197)(22) (0.0170)(8) 2.542 0.6372 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.

et

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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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

w .E asy En g

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).

ine eri n

g .n

et

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=

(2)(1) A = = 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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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

w .E asy En g

0.25

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

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

ine eri n

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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Single Pipelines

33

TABLE 2.2: Pipe Roughness Coefficients

CH Pipe material

ww

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

n

Range

Typical

Range

Typical

120–140 40–100 100–140

130 80 120

— — 0.011–0.015

0.013 0.025 0.013

80–150 100–140 120–145

120 110 130 140 120 110 140 —

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

0.012 0.015 — 0.011 0.012 — 0.009 0.025

100–140 110–140 135–150 —

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

w .E asy En g

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

ine eri n

g .n

ks = D(3.320 − 0.021CH D0.01 )2.173 exp(−0.04125CH D0.01 )

et

(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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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

w .E asy En g

(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

= 6.73 m

ine eri n 4

(0.15) 3

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

g .n

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     2.51 2.51 0.26 ks 1 = −2 log = −2 log + + 3.7D 3.7(150) f Re f 1.5 * 105 f which yields

et

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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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

w .E asy En g 

ine eri n

  ⭸ 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

⭸ [ρ0 V(L − ct)A] + ρ0 V(−VA) ⭸t

et

(2.87)

where A is the cross-sectional area of the pipe and L is the length of the control volume. Simplifying Equation 2.87 gives FIGURE 2.9: Pressure wave

L ct

po Flow velocity V

ρ

c

Rapidly closed value

po + ∆p

ρ + ∆ρ

x

V=0

V Control volume

Pressure wave

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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

w .E asy En g 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

ine eri n

(2.91)

ρ V = ρ0 c

(2.92)

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

g .n

(2.93)

et

(2.94)

and substituting Equation 2.89 into Equation 2.94 yields wave speed, c, in terms of the fluid properties as  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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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



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

w .E asy En g

Hydraulic grade line

po

Source Pipeline V

(a)

L

c

∆p

po

V=o

(b)

Pressure wave c

∆p

po V

V=o

(c)

po V

ine eri n Closed valve

V

c

g .n

et

∆p

(d) V=o

po V

(e)

∆p

c

V=o

po V (f)

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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

w .E asy En g

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

ine eri n

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   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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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

w .E asy En g NP(j)



ine eri n

Qij − Fj = 0,

i=1

(2.98)

j = 1, . . . , NJ

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

Qa1

A

g .n

et

Qb2

Q5

B

Qb1

B C

D Loop Node

(a)

Qd1 Q2

D

Qd2

C

(b)

Qc1

Qc2

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Chapter 2

Fundamentals of Flow in Closed Conduits

2.3.1

Nodal Method

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



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

ww

EXAMPLE 2.10

w .E asy En g

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

ine eri n

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

B

C

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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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 (0.0168)(650) f L Q23 = 87.3 − Pipe 3: hC = hB − 3 3 D3 2gA2 0.50 (2)(9.81)(0.196)2 3

w .E asy En g = 87.3 − 29.0Q23

(2.102)

Q24 (0.0156)(400) f L Q24 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

ine eri n

(2.104) (2.105)

Junction C: Q2 + Q3 = Q4

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

g .n

2 = (0.583 + 1)Q3 or

et

(2.106)

Q3 = 1.26 m3 /s

and from Equation 2.106

Q2 = 0.74 m3 /s

According to Equation 2.102 hC = 87.3 − 29.0Q23 = 87.3 − 29.0(1.26)2 = 41.3 m and Equation 2.103 gives hD = hC − 3.07Q24 = 41.3 − 3.07(2)2 = 29.0 m Therefore, since the total head at D, hD , is equal to 29.0 m, then V2 p p 5.202 + 3.5 29.0 = D + 4 + zD = D + γ 2g 9.79 (2)(9.81)

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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



(hL,ij − hp,ij ) = 0,

i = 1, . . . , NL

(2.107)

j=1

where NP(i) is the number of pipes in loop i, hL,ij is the head loss in pipe j of loop i, hp,ij is the head added by any pumps that may exist in line ij, and NL is the number of loops in the network. Combining Equations 2.98 and 2.107 with an expression for calculating the head losses in pipes, such as the Darcy–Weisbach equation, and the pump characteristic curves, which relate the head added by the pump to the flow rate through the pump, yields a complete mathematical description of the flow problem. Solution of this system of flow equations is complicated by the fact that the equations are nonlinear, and numerical methods must be used to solve for the flow distribution in the pipe network.

w .E asy En g

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

ine eri n hL = rQn

g .n

(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

et

(2.109)

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

(2.110)

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

(2.111)

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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 ˆ n + rnQ ˆ n−1 Q hL L rQ

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 ˆ Q| ˆ n−1 + rn|Q| ˆ n−1 Q hL = rQ|

w .E asy En g

(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)

 j=1

NP(i)

rij Qj |Qj |n−1 + Qi



rij n|Qj |n−1 = 0,

i = 1, . . . , NL

(2.114)

j=1

ine eri n

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

Qi = −

NP(i)

rij Qj |Qj |n−1

j=1 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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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

50

r=1

r=6

r=3

20

15

1

35 r=2

50

2 II

70

35 I

ww

100

r=5

4

100

30

30

(a)

w .E asy En g

3

30

(b)

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

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

ine eri n j=1

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

Pipe

I

4–1 1–3 3–4

Q

rQ|Q|

2r|Q|

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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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 g

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

Loop

Pipe

Q

2

I

4–1 1–3 3–4

48.8 2.7 −51.2

II

3

I

II

4

I

II

1–2 2–3 3–1

4–1 1–3 3–4

1–2 2–3 3–1

4–1 1–3 3–4

1–2 2–3 3–1

rQ|Q|

2r|Q|

14,289 22 −13,107 1204

586 16 512 1114

681 −1142 −8 −469

52 96 10 157

13,663 6 −13,666 3

573 8 523 1104

847 −874 6 −21

58 84 8 150

13,662 7 −13,668 1

573 9 523 1104

853 −865 7 −5

58 83 9 150

Q

ine eri n

26.1 −23.9 −1.6

47.7 1.4 −52.3

29.1 −20.9 1.4

47.7 1.5 −52.3

29.2 −20.8 1.5

Corrected Q 47.7 1.6 −52.3

−1.1

3.0

29.1 −20.9 1.4

g .n 47.7 1.4 −52.3

0.0

et

29.2 −20.8 1.5 0.1 47.7 1.5 −52.3 0.0 29.2 −20.8 1.5 0.0

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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. 2.3.3

Application of Computer Programs

w .E asy En g

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

ine eri n

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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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 g

ω

Flow

Input shaft

Outlet vane

Inlet vane

Impeller

ine eri n

(b) Axial-flow pump

g .n

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

et

(2.116)

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Chapter 2

Fundamentals of Flow in Closed Conduits

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

Reducer (if necessary)

P U M P

Flow Check valve

Gate valve (service)

Foot valve Flow

Strainer (a) Centrifugal pump

ww

Motor

w .E asy En g

Flow Guide vanes

Impeller

Flow

ine eri n

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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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 g

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

ine eri n

This relationship describes the performance of all (geometrically similar) pumps in which viscous effects are negligible, but the exact form of the function in Equation 2.119 depends on the geometry of the pump. A series of pumps having the same shape (but different sizes) are expected to have the same functional relationships between ghp /(ω2 D2 ) and Q/(ωD3 ) as well as η and Q/(ωD3 ). A class of pumps that have the same shape (i.e., are geometrically similar) is called a homologous series, and the performance characteristics of a homologous series of pumps are described by curves such as those in Figure 2.17. Pumps are selected to meet specific design conditions and, since the efficiency of a pump varies with the operating condition, it is usually desirable to select a pump that operates at or near the point of maximum efficiency, indicated by the point P in Figure 2.17. The point of maximum efficiency of a pump is commonly called the best-efficiency point (BEP), and sometimes the nameplate or design point. Maintaining operation near the BEP will allow a pump to function for years with little maintenance, and as the operating point moves away from the BEP, pump thrust and radial loads increase, which increases the wear on the pump bearings and shaft. For these reasons, it is generally recommended that pump operation should be maintained between 70% and 130% of the BEP flow rate (Lansey and El-Shorbagy, 2001). At the BEP in Figure 2.17,

FIGURE 2.17: Performance curves of a homologous series of pumps

g .n

et

max

ghp

K1

P

Efficiency curve

ω2D 2

K2 Q ωD 3

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Chapter 2

Fundamentals of Flow in Closed Conduits

ghp = K1 ω2 D2

Q = K2 ωD3

and

(2.120)

Eliminating D from these equations yields 1

& 'K ' 2 = ( 3 3 (ghp ) 4 K12 ωQ 2

(2.121)

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

ns =

ww

ωQ 2

(2.122)

3

(ghp ) 4

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

w .E asy En g

ine eri n 1

Ns =

ωQ 2

(2.123)

3 4

hp

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

g .n

et

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Pumps

51

TABLE 2.3: Pump Selection Guidelines

Range of specific speeds, ns (dimensionless)

Type of pump Centrifugal Mixed flow Axial flow

0.15–1.5 1.5–3.7 3.7–5.5

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

Typical efficiencies (%) 70–94 90–94 84–90

4750

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

ww

w .E asy En g

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

ine eri n

Synchronous speed (rpm) =

3600 no. of pairs of poles

(2.124)

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

g .n

et

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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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 − η2 = 1 − η1

w .E asy En g



D1 D2

1 4

(2.129)

where η1 and η2 are the efficiencies of homologous pumps of diameters D1 and D2 , respectively. The effect of changes in flow rate on efficiency is sometimes estimated using the relation   0.94 − η2 Q1 0.32 = (2.130) 0.94 − η1 Q2 where Q1 and Q2 are corresponding homologous flow rates.

EXAMPLE 2.12

ine eri n

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     500 Q2 = (2400)(400)3 (2400)(300)3 which yields Q2 = 210 L/s. Also, (ω1 )2 (D1 )2



=



89.5 (2400)2 (400)2



=



 

hp1

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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Pumps

53

1

1 − η2 = 1 − η1



D1 D2

1 − η2 = 1 − 0.85



400 300

4

1

4

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

ww

2.4.2

Pump Selection

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

w .E asy En g 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

17.5 in 55% 65%

16.6 in

70

15.7 in

60

14.8 in

73% 79%

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

83% 85% 86%

40

g .n

85% 83% 79%

13.9 in 13 in 12 in

50

A-7874-8

CDS

CENTRIFUGAL PUMP CHARACTERISTICS

ft 80

ine eri n

et 20

15

73% (Efficiency) 10

30 0ft

(Pump curve)

20 10

7ft

10 ft

17.5 in

(Power curve)

75

15.7 in

50 25 1000

0

250

2000 500

3000 750

4000 1000

5000

6000

1250

kW 60 40

13.9 in

12.1 in 0

5

16 ft (Required NPSH)

hp

0

m 25

20 7000 1500

8000 gpm(US)

0

1750 m3/hr

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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.

w .E asy En g 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



ine eri n

(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

Δz P Pump

Q

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55

Operating point

Pump head, hp

FIGURE 2.21: Operating point in pipeline system

Pumps

System curve Pump characteristic curve

Δz 0

2.4.2.3

ww

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

w .E asy En g NPSHA =



ps V2 + s + zs γ 2g



ine eri n −



pv + zs γ



=

ps V2 pv + s − γ 2g γ

g .n

(2.133)

et

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 =

pv p0 − 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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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.

w .E asy En g

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 = 

ine eri n

log

where Re is the Reynolds number given by

0.25



ks 5.74 3.7D + Re0.9

Re =

(2.138)

2

QD VD = ν Aν

g .n

(2.139)

et

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 ⎟⎥ ⎥ 0.9 ⎟ ⎠⎥ ⎦

(2.140)

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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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 ) *2 Q 1375 15,850 hp = 5.2 +    3.281 *−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:

w .E asy En g

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

ine eri n

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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Chapter 2

Fundamentals of Flow in Closed Conduits

FIGURE 2.22: Pump operating points

Goulds Pumps ITT Industries

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

ft 17.5 in

80

55% 65% 73%

16.6 in

70

15.7 in

60

14.8 in

79%

83%

85% 86%

85% 83% 79% 73%

13.9 in 13 in 12 in

50 40 30

A-7874-8

CDS

Centrifugal pump characteristics

10

20

ww

System curve

5

10 ft 16 ft

7ft

hp

w .E asy En g

17.5 in

75

15.7 in

50 25

0

1000

0

250

2000

500

3000

4000

750

5000

1000

6000

1250

kW 60 40

13.9 in

12.1 in

0

20

15

0ft

10

m 25

20 7000 1500

8000 gpm(US)

0

1750 m3/hr

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

ine eri n

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

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

et

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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hp

Pumps

59

EGL Two pumps in series

hp hp Q

ww

Q

Q

P

P

One pump

Q

Pump system

Q

(a)

(b)

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

w .E asy En g

FIGURE 2.24: Pumps in parallel

hp

ine eri n

g .n

et

EGL

Q P 2Q

Q P

2Q

Two pumps in parallel hp One pump

Pump system

Q

(a)

(b)

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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

w .E asy En g

(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

ine eri n

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 f (Q/ωD3 )

g .n

et

where 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   Q1 h1 = 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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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

w .E asy En g 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

ine eri n

A pump with a 1200-rpm motor has a performance curve of

hp = 12 − 0.1Q2

g .n

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.

et

Solution From the given data: ω1 = 1200 rpm, ω2 = 2400 rpm, and the affinity laws (Equation 2.148) give that Q1 =

h1 =

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

h2 =

12002 h2 = 0.25h2 24002

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

h2 = 48 − 0.1Q22

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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

ww

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 =

w .E asy En g

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

2.9.

2.10.

2.11.

2.12.

2.13. 2.14.

2.15.

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

Why is this equation termed “(slightly) more convenient”? If you had your choice of estimating the friction factor either from the Moody diagram or from the Colebrook equation, which one would you pick? Explain your reasons. 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). 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. 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. Repeat Problem 2.12 using the Swamee–Jain approximation (Equation 2.44). Use the velocity distribution given in Problem 2.3 to estimate the kinetic energy correction factor, α, for turbulent pipe flow. The velocity profile, v(r), for turbulent flow in smooth pipes is sometimes estimated by the seventh-root law, originally proposed by Blasius (1913):

ine eri n

g .n



r v(r) = V0 1 − R

et

1 7

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 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?

ww

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.

w .E asy En g

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

ine eri n

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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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?

ww

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

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?

=

Elev. 80 m

900 m B

LBJ = 800 m J

w .E asy En g

2.35. Based on your result in Problem 2.34, do you think that water hammer can be a serious problem in household plumbing?

LAJ

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:

ine eri n 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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A C

P

Flow

F

Flow

G

Flow

D

B

Flow

E

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:

Pipe

ww 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

hf = rQn

F

E

You can assume that the flow in each pipe is hydraulically rough.

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

ine eri n 0.05 m3/s

150 m

P

150 m 100 m 0.06 m3/s

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

D

FIGURE 2.31: Two-loop pipe network

100 m

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

C

0.5 m3/s

w .E asy En g

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

B

A

0.1 m3/s

150 m

g .n 0.07 m3/s 150 m

100 m

et

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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Chapter 2

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

3

8000 m /d

3

3

8000 m /d

C

H

8000 m /d

D

8000 m3/d

8000 m3/d

8000 m3/d

F

G

ww

E

FIGURE 2.33: Water-supply system

2.44. What is the constant that can be used to convert the specific speed in SI units (Equation 2.122) to the specific speed in U.S. Customary units (Equation 2.123)?

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.

w .E asy En g

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.

ine eri n

g .n

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

et

Upper reservoir

Pipeline 61 m

Pump/turbine

Lower reservoir FIGURE 2.34: Pumped storage system

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Bell & Gossett % 59

%

%

55

1½ × 1½ × 7B 3500 rpm 57

%

%

220

53

%

50

45

%

%

70

40

7.0 in.

35

240

%

Series 80-SC

25

%

59.5%

200

60

59

260

6.5 in.

%

140

55 % 5 3%

5.5 in.

120 5.0 in.

100

45 %

80

10

20

60

ww 10 0

NPSH in meters

57 6.0 in.

NPSH in feet

30

Total head in feet

40

160

% 50

Total head in meters

180 50

67

30

40 NPSH

20

20

REQ

0 0

20

40

60

80

100

w .E asy En g

5

10

120

140

0 180

160

0

Capacity in U.S. gallons per minute

0

5

10

15 20 25 Capacity in cubic meters/hr

35

30

40

FIGURE 2.35: Pump performance curves

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

hp = 15 − 0.1Q2

ine eri n

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

hp = 20 − 4713Q2 100 m

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

Upper reservoir

P

et

10 m E

3m C

B P

1m Lower reservoir

FIGURE 2.36: Water pumped from lower to upper reservoir

F 10 m D

3m

A

FIGURE 2.37: Water-delivery system

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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.

ww

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

w .E asy En g

(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

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

ft

30%

Total dynamic head

15

43% 50% 55%

60%

4 in.

50

ine eri n

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

64%

3.75 in.

40

64% 3.5 in.

10

g .n

60% 55%

30

et

50%

3.25 in.

43%

20

3.0 in.

5 NPSHR - ft

10

5 ft

16 ft 6 ft 7 ft

0

0

0 0

20

40 10

60

80

100

20 Capacity

9 ft 120

140 30

160

180 gpm 3 40 m /h

FIGURE 2.38: Goulds Model 3656 pump curves

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C Elev. 2.00 m p ≥ 350 kPa

C

P B

A

69

P

P

Elev. 3.00 m

B

A

P

D Elev. 5.00 m p ≥ 350 kPa

P

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

A and B, and the performance curve for each pump in the system is given by

of at least 350 kPa at C and D. All pipes are to be made of ductile iron. (a) Determine the minimum power that must be delivered by the pump. (b) A pump manufacturer will be able to match the minimum-power operating condition by providing several “micro-pumps” in series where each micro-pump has a performance characteristic given by

hp = 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:

ww

w .E asy En g

Line

Length (km)

Diameter (mm)

Flow (L/s)

AB BC BD

1.05 2.80 2.50

200 150 150

27 12 15

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

hp = 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

ine eri n

hp = 6 − 0.05Q2

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

g .n

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.

et

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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.

w .E asy En g 3.2 Water Demand

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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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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:

w .E asy En g

N    Q= qi * Pi

(3.1)

i=1

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

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Q=q * P

(3.2)

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

Estimation of per-capita demand

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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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Chapter 3

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

Category

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

Percent of total

Residential Commercial Industrial Public Loss

380 115 85 65 40

100 30 22 17 11

56 17 12 9 6

Total

685

180

100

Source: Solley (1998).

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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)

et

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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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

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

EXAMPLE 3.1

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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.

w .E asy En g 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

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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

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

et

Population trend

350

C

300

B A

250 200

Census data

150 100 1940

1960

2000

1980

2020

2040

Year

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(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 P = P0 ek1 t

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(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

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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 =

et

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

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

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

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

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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

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2.0

et

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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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

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

Class of construction

Description

F

1 2 3 4 5 6

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

1.5 1.0 0.8 0.8 0.6 0.6

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

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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

w .E asy En g 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) >30 9.5–30 3.5–9.5 100 30–100 10–30 20 >20

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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.

w .E asy En g 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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Building Water-Supply Systems

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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)

w .E asy En g (c)

(b)

(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

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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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et

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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FIGURE 3.12: Isometric view of residential water-supply system

A

B

Hot water C Cold water

z Service line Water heater

ww

y x

Backflow preventer Meter

w .E asy En g 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

ine eri n

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

g .n

et

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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Building Water-Supply Systems

95

TABLE 3.9: Water-Supply Fixture Units (WSFU)

Load values in WSFU Fixture

ww

Occupancy

Supply control

Cold

Hot

Total

Bathroom group Bathroom group Bathtub Bathtub Bidet Combination fixture

Private Private Private Public Private Private

Flush tank Flush valve Faucet Faucet Faucet Faucet

2.7 6 1 3 1.5 2.25

1.5 3 1 3 1.5 2.25

3.6 8 1.4 4 2 3

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

Private Offices, etc. Private Hotel, restaurant Private

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

— 0.25 1 3 1

1.4 — 1 3 1

1.4 0.25 1.4 4 1.4

Lavatory Lavatory Service sink Shower head Shower head

Private Public Offices, etc. Public Private

Faucet Faucet Faucet Mixing valve Mixing valve

0.5 1.5 2.25 3 1

0.5 1.5 2.25 3 1

0.7 2 3 4 1.4

Urinal Urinal Urinal

Public Public Public

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

— — —

10 5 3

w .E asy En g Washing machine, 3.6 kg (8 lb) Private Washing machine, 3.6 kg (8 lb) Public Washing machine, 6.8 kg (15 lb) Public Water closet Water closet Water closet Water closet Water closet

FIGURE 3.13: Relationship between demand and fixture units

1 2.25 3

1 2.25 3

1.4 3 4

Flush valve Flush tank Flush valve Flush tank Flushometer tank

6 2.2 10 5 2

— — — — —

6 2.2 10 5 2

ine eri n

Private Private Public Public Public or private

Source: International Plumbing Code (2012).

Automatic Automatic Automatic

500

Demand (L /min)

400

g .n

et

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

Flow rate (L/min)

Pressure (kPa)

Flow rate (gpm)

Pressure (psi)

15 8 15 10.4 2.8 15

138 138 55 55 55 55

4 2 4 2.75 0.75 4

20 20 8 8 8 8

8 8

55 55

2 2

8 8

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

11 11 9.5 11

55 138 55 55

3 3 2.5 3

8 20 8 8

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

45 95 95

172 310 241

12 25 25

25 45 35

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

11 23

138 138

3 6

20 20

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

ww

w .E asy En g 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

ine eri n

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



g .n

   V12 p1 + + z = hf + hl γ 2g

et

(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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⎧

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

ww

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

w .E asy En g

ine eri n

TABLE 3.11: Head Loss in Standard Fittings in Terms of Equivalent Pipe Lengths

90◦ Ell

45◦ Ell

90◦ Tee

Gate valve

Globe valve

Angle valve

(in.)

(m)

(m)

(m)

(m)

(m)

(m)

0.305

0.183

0.457

0.061

2.438

1.219

0.610

0.366

0.914

0.122

4.572

2.438

19.1

3 8 1 2 3 4

0.762

0.457

1.219

0.152

6.096

3.658

25.4

1

0.914

0.549

1.524

0.183

7.620

4.572

32

1 41

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

Fitting Diameter (mm) 9.5 12.7

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et

Source: Uniform Plumbing Code (2012).

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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:

w .E asy En g 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

ine eri n

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.

g .n

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.

et

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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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

ww

E’

4.0 m 4.0 m 46 m

w .E asy En g C

F Floor 1

46 m C’

F’

2.5 m

Backflow preventer

Main

A

Meter

2.5 m

B

1.7 m

17 m

ine eri n 1m

B’

Water heater

TABLE 3.12: Results of Design Calculations

g .n

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

et

22.33 21.97 17.05 21.26 16.34

216 188 164 166 157

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

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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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)

ww

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,

w .E asy En g

Fitting Length = tee loss = 3.658 m

and so the total equivalent pipe length of BC is 2.5 m + 3.658 m = 6.158 m L 6.16 m. The Reynolds number, Re, in pipe BC is Re =

(2.05)(0.064) VD = 13,100 = ν 10−6

and the friction factor, f , is f = 

log

0.25

ks 5.74 3.7D + Re0.9

=  2

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

0.25

ine eri n log

5.74 0.0015 + 3.7(64) 13,1000.9

= 0.0174 2

(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

g .n

Terminal Head = starting head − friction loss

= 22.33 m − 0.36 m = 21.97 m

et

Since the change in elevation, z, between B and C is 2.5 m, the terminal pressure, p, is line BC is given by     V2 2.052 p = γ terminal head − z − = (9.79) 21.97 − 2.5 − = 188 kPa 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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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.

ww

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.

w .E asy En g

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?

ine eri n

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Chapter 3

Design of Water-Distribution Systems C 20 m

B P 20 m 3m

A

P

FIGURE 3.15: Water-distribution system

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

ww

w .E asy En g

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.

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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et

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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.

w .E asy En g

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4.2 Basic Principles

g .n

et

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   ρv2 dA (4.1) ρv1 dA = 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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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

ww

1 V1 = A1



1 A2



and V2 =

w .E asy En g

A1

v1 dA

(4.2)

v2 dA

(4.3)

A2

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

ine eri n

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

g .n

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

et

1 Hydrostatic pressure distribution

Control volume 2

y1

γAΔx Water surface

y2

τ PΔ 0 x z1

Q

Δx z2

Datum

θ x

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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)

ww

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:  (4.9) Fx = ρβ2 V22 A2 − ρβ1 V12 A1

w .E asy En g

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)

ine eri n

and the momentum equation (Equation 4.9) can be written as  Fx = ρβ2 QV2 − ρβ1 QV1

g .n

(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 )

et

(4.12)

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

(4.13)

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

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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 ,

ww

y = y2 − y1 ,

(4.16)

V = V2 − V1

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

w .E asy En g

ine eri n

(4.18)

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

V2 2g

(4.19)

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

g .n

et

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, ǫ, ǫ ′ , m, s)

(4.22)

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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], ǫ is the characteristic size of the roughness projections on the channel boundary [L], ǫ ′ 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 ǫ ǫ′ = 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: 

ǫ τ0 1 (4.25) = f Re, 8 4R ρV 2

w .E asy En g

ine eri n

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 ǫ/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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Chapter 4

Fundamentals of Flow in Open Channels

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

1 ks = −2 log10 (4.28) 12R f

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 ´ an ´ equation that is used in pipe flow. In the transition region between Prandtl–von Karm (hydraulically) smooth and rough flow, the friction factor depends on both the Reynolds number and the relative roughness and can be approximated by (ASCE, 1963)  2.5 1 ks = −2 log10 + (4.29) 12R f Re f

w .E asy En g

This relation differs slightly from the Colebrook equation for transition flow in closed conduits, but is still called the Colebrook equation and is commonly applied in both smooth and rough flow in open channels. Caution should be exercised in applying Equation 4.29 to smooth-flow conditions in open channels since under these conditions Equation 4.29 asymp´ an ´ equation (Equation 4.27), which is known to deviate by totes to the Prandtl–von Karm up to 20% from measurements (Cheng et al., 2011). Equation 4.29 was originally suggested by Henderson (1966) for wide open channels (width/depth Ú 10), and others have suggested similar formulations with different constants (e.g., Yen, 1991). The three types of flow (smooth, transition, rough) can be delineated using the shear velocity Reynolds number, also known as the roughness Reynolds number, ks u∗ /ν, where u∗ is the shear velocity defined by   τ0 = gRSf u∗ = (4.30) ρ

ine eri n

and ν is the kinematic viscosity of the fluid. The turbulent flow are as follows: ⎧ ⎪ ⎨70,

g .n

approximate ranges for the three types of smooth transition rough

et

(4.31)

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

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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 ´ ´ 2008). by taking ks equal to 2.4d90 , 2.8d84 , or 6.1d50 (Lopez and Barragan, For channels lined with submerged vegetation, the sources of drag within the channel are fundamentally different from the boundary drag in channels lined with sediment, gravel, or rocks. In the case of submerged vegetation, the friction factor, f , can be estimated by (Cheng, 2011)

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

w .E asy En g

kv =

π λ D 41 − λ

(4.34)

where λ is the fraction of the channel bottom occupied by vegetation stems [dimensionless], D is the average stem diameter [L], and h is the depth of water above the vegetation. Equation 4.33 provides a good fit to the observed data for 1 * 10−5 … kv /h … 3 * 10−2 . Regardless of the empirical equation used to estimate the friction factor, f , combining Equations 4.21 and 4.25 leads to the following form of the momentum equation, which is most commonly used in practice:  8g  V= RSf (4.35) f

ine eri n

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 1 2.5 0.625ν ⎜ s ⎟ = −2 log10 ⎜ + + ⎟ = −2 log10 ρV(4R) ⎠ ⎝ 12R 12R f RV f f μ

(4.37)

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Chapter 4

Fundamentals of Flow in Open Channels

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

(4.38)

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

ww

FIGURE 4.3: Flow in a trapezoidal channel

Solution The flow in the channel is illustrated in Figure 4.3. T = 3 + 4y

w .E asy En g

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

ine eri n 0

g .n

⎤ −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

et

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

4.2.2.2

Manning equation

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

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Basic Principles

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

111

(4.39)

where C was referred to as the Ch´ezy coefficient. Equation 4.39 has exactly the same form as Equation 4.35 and was derived in the same way, except that the functional dependence of the ´ coefficient on the Reynolds number and the relative roughness was not considered. Chezy ´ coefficient is related to the friction factor by Comparing Equations 4.35 and 4.39, the Chezy  8g C= (4.40) f

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 g

and n is a coefficient that is characteristic of the surface roughness alone. Since C is not a dimensionless quantity, values of n were specified to be consistent with length units measured in meters and time in seconds. If the length units are measured in feet, then Equation 4.41 becomes 1 R6 C = 1.486 (4.42) n where 1.486 is the cube root of 3.281, the number of feet in a meter. When either ´ equation, the resulting expression is called Equation 4.41 or 4.42 is combined with the Chezy the Manning equation or Strickler equation (in Europe) and is given by

ine eri n

⎧ 1 2 1 ⎪ ⎪ ⎪ ⎨nR3 S2 V= ⎪ 1.486 2 1 ⎪ ⎪ ⎩ R3 S2 n

(SI units)

(4.43)

g .n

(U.S. Customary units)

et

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

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Chapter 4

Fundamentals of Flow in Open Channels

and substituting the expression for the friction factor under fully turbulent conditions (Equation 4.28) into Equation 4.44 and rearranging yields 1 R 6 ks n 

= 1 R ks6 2.0 log 12 ks 1 8g



(4.45)

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

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

ww

1 6

ks

(4.46)

= 0.039

1

w .E asy En g

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

ine eri n

g .n

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

et

0.10

1

n/ks6 in fully turbulent flow

1

n/ks6

R/ks = 4

R/ks = 500

1

n/ks 6 = 0.039

0.01

1

10

100

1000

R/ks

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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

1.64 R = 781 = ks 0.0021   ks RS0 = 0.0021 (1.64)(0.0005) = 6.01 * 10−5

w .E asy En g

√ 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.

ine eri n

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

(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

et

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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Chapter 4

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

Expression for n

Conditions

Reference

1

6 0.047d50

Uniform sand

Strickler (1923)

—

Keulegan (1938)

Sand mixtures

Meyer–Peter and Muller (1948)

Gravel-lined canals

Lane and Carlson (1953)

Riprap-lined channels

Maynord (1991)

—

Limerinos (1970)

Gravel streams with S0 > 0.004

Bathurst (1985)

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

Wide channels

Dingman (1984)

R6 [7.64 ln(R/d50 ) + 15.5]

Riprap-lined channels

Froehlich (2012)

1 6

0.0025d90 1

6 0.0039d90 1

6 0.047d75 1

6 0.046d50 1

ww

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

[7.64 ln(R/d84 ) + 65.3]

w .E asy En g 1

1

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.

ine eri n

g .n

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

et

(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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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

w .E asy En g

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

ine eri n

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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Chapter 4

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

Formula∗ ⎛

Assumption

N 

⎛

ww

⎞2 3

3 2

⎜ P i ni ⎜ ⎜ i=1 ⎜ ne = ⎜ ⎜ P ⎝

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

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)

⎞1 2

N 

⎜ Pi n2i ⎟ ⎟ ⎜ ⎟ ⎜ i=1 ⎟ ne = ⎜ ⎟ ⎜ ⎟ ⎜ P ⎠ ⎝

⎞2

w .E asy En g ⎛

N 

3 2

⎜ P i ni ⎜ ⎜ i=1 ⎜ ne = ⎜ ⎜ P ⎝ 5

ne =

PR 3

3

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

5

n  Pi Ri3

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)

ni

i=1

N 

3

Pi yi2 ln ni

ln ne = i=1 N 

3 2

Pi yi

 ne = 1 N An A i=1 i i 1 1 N P R 3 n i 1 i=1 i i PR 3

ine eri n

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)

i=1

ne =

Reference

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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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

w .E asy En g

ine eri n

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

0.025 0.030

g .n Maximum 0.035 0.050

Typical

et

0.030 0.035

0.020 0.025 0.030

0.040 0.045 0.050

0.030 0.035 0.040

0.035 0.035 0.040 0.045 0.070

0.070 0.060 0.080 0.110 0.160

0.050 0.050 0.060 0.070 0.100

0.110 0.030 0.030

0.200 0.050 0.050

0.150 0.040 0.040

0.080

0.120

0.100

0.100

0.160

0.120

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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:

w .E asy En g

Section

n

1 2 3 4 5 6 7

0.040 0.030 0.015 0.013 0.017 0.035 0.060

ine eri n

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)

1 2 3 4 5 6 7

8.25 100 6.71 15.0 6.71 150 8.25 295

Ai (m2 )

Ri (m)

ni

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

yi (m)

g .n 1.00 2.00 3.50 5.00 3.50 2.00 1.00

et

633

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

100 m 1

1

150 m 2m

2 3

4

4

5

1

1 2

3m

6

7

1 4

2

15 m

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

ww

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

w .E asy En g

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.

ine eri n

g .n

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

et

Other equations

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

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Fundamentals of Flow in Open Channels

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

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

w .E asy En g

(4.56)

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

1 y 7 d

ine eri n

(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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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=

w .E asy En g

N 

(4.60)

V i Ai

i=1

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

Steady-State Energy Equation

ine eri n

g .n

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

et

(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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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

w .E asy En g 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) ρ ρv · n dA + 2 A2 A2

ine eri n

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) A1

A2

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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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

w .E asy En g

(4.75)

which can be put in the form   # V12 V22 p2 p1 u2 u1 Q + + z2 + α2 + + z1 + α1 = − γ g 2g γ g 2g mg #

(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 1 p1 p2 Q ⎦ + α1 + z1 = + α2 + z2 + ⎣ (u2 − u1 ) − (4.77) γ 2g γ 2g g mg #

ine eri n

The energy loss per unit weight or head loss, hL , is defined by the relation hL =

# 1 Q (u2 − u1 ) − g mg #

g .n

(4.78)

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

(4.79)

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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Chapter 4

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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



(4.83)

+ hL

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 g

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

ine eri n 2

(4.86)

= (S − S0 cos θ )L

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

(4.87)

et

(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 0.001 0.01 0.1 1

cos θ 0.9999995 0.99995 0.995 0.707

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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 g 

V2 y1 + α1 1 + z1 2g





V2 = y2 + α2 2 + z2 2g



+ hf ,

(4.91)

S0 < 0.1

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

ine eri n

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

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

et

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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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

w .E asy En g

(4.96)

y=E

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

ine eri n

(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

(4.98)

et

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

Depth of flow, y

y

=

(4.99)

E

y2

yc y1 Q = constant Emin Specific energy, E

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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 g

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=

ine eri n

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

g .n

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)

et

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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Chapter 4

Fundamentals of Flow in Open Channels

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

Q2 A3 = c g Tc 3 2 and since Q = 10 m /s and g = 9.81 m/s , under critical-flow conditions

ww FIGURE 4.8: Trapezoidal cross section

(3yc + 2y2c )3 102 = 9.81 (3 + 4yc )

Solving for yc yields

w .E asy En g

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

y

3m

1 2

ine eri n

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

where b is the width of the channel. The flow area, A, and top width, T, are given by A = by

et

(4.107) (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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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 g

FIGURE 4.9: Specific energy diagram for rectangular channel

Depth of flow, y

y

⫽

g .n

et

E

q increases

1'

y1'

ine eri n

y2' 2' yc y2 y1

2 1 Ec

q ⫽ q3 q ⫽ q2 q ⫽ q1

E1 (⫽ E2)

Specific energy, E

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Chapter 4

Fundamentals of Flow in Open Channels

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

ww

(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 g

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 +

ine eri n

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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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 g

If b is the width of the constriction that causes critical flow, then  1 3 (Q/b)2 3 E1 = 2 g

From the given data: E1 = 0.901 m, and Q = 1.10 m3 /s. Therefore 0.901 =

which gives

1  3 3 1.102 2 b2 (9.81)

ine eri n b = 0.75 m

If the constricted channel width is any less than 0.75 m, the flow will be choked and the upstream flow depth will increase—in other words, the flow will “back up.” Note: Part (a) of this problem required finding alternate flow depths as the cubic roots of the specific energy equation, which can be easily done using a programmable calculator. Alternatively, the following equations could be used to provide a direct solution of the specific energy equation (Abdulrahman, 2008) ⎫ ⎧ ⎡ ⎡ ⎤⎤⎪ ⎪ ⎤ ⎡ ⎪ ⎪  3 ⎪ ⎪ 

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

g .n

⎡ ⎤⎫ ⎡ ⎡ ⎤⎤ ⎪ ⎪  ⎪

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

et

(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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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 g

$ $ $ V2 2$ V $ $ he = C $α2 2 − α1 1 $ $ 2g 2g $

(4.115)

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

Profile Equation

ine eri n

g .n

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



et

(4.116)

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

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%  α

x→0

= =

ww

=

V2 2g

y + lim x x x→0  dy V2 d α + dx dx 2g  dy V 2 dy d α + dx dy 2g dx ⎡ ⎤  dy ⎣ d Q2 ⎦ 1 + α dx dy 2gA2   dy Q2 dA 1 − α dx gA3 dy

S0 − Sf = lim =

Water-Surface Profiles

133

&

(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

w .E asy En g

where T is the top width of the channel and the hydraulic depth, D, of the channel is defined as A (4.120) D= T

then the Froude number, Fr, of the flow can be written as  √ V dA Q T Q Fr = = = 3 dy gD A gA gA

or

ine eri n

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

(4.122)

et

(4.123)

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

2 3

⎞2

Sf ⎜ An Rn ⎟ =⎝ 2 ⎠ S0 AR 3

(4.124)

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

2

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

when

y < yn ,

and

Sf < S0

when

y > yn

(4.125)

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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.

w .E asy En g

ine eri n

TABLE 4.7: Hydraulic Classification of Slopes

Name Mild Steep Critical Horizontal Adverse

Type

Condition

M S C H A

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

g .n

et

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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1

dy

Horizontal

dx

2

dy

yn 3

dx

yc

dy

Water-Surface Profiles

135

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

dx

Mild slope

1 Horizontal

2 3

yn

ww

dy dx ⫽

yc

dy

⫽⫺ , S2

dx dy dx

⫹, S1

⫽⫹

Steep slope

, S3

w .E asy En g 1

pro

file

pro

file

pro

file

Horizontal

3

yn ⫽ y c

dy dx

⫽ ⫹, C1

dy

dx

⫽ ⫹, C3

profile

profile

Critical slope

dy

2 3

yn ⫽ ∞

dx dy

yc

dx

⫽ ⫹, H3 profile

Horizontal slope dy dx

yc

2 3

⫽ ⫺,

dy dx

ine eri n

⫽ ⫺, H2 profile

rofile

A2 p

rofile

A3 p ⫽ ⫹,

Adverse slope

g .n

et

5

2 1 1 An3 21 1 S Q = An Rn3 S02 = 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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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

yn = 2.25 m A3c Q2 = Tc g

ww

where Ac and Tc are the area and top width, respectively. The left-hand side of this equation is a function of the critical flow depth, yc , and the right-hand side is in terms of the given data. Thus A3c 602 = 367 = Tc 9.81

w .E asy En g 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

ine eri n

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

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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 dy 0.01 − 0.000997 = = 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.

w .E asy En g

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

ine eri n

H = yc +

and

Q2 2gA2c

Q2 Tc =1 gA3c

(4.129)

g .n

(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

et

B Lake or Reservoir H d

C S0 = Mild slope

yc

S0 = Steep slope

Slop

e, S

yn

0

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Chapter 4

Fundamentals of Flow in Open Channels

H = yn +

Q2 2gA2n

(4.131)

and Q=

ww

2 1 1 An Rn3 S02 n

(4.132)

where Equation 4.131 is required for conservation of energy and Equation 4.132 is the condition for normal flow. To determine the actual flow rate, Equations 4.129 and 4.130 are first solved simultaneously for Q and yc , and then the calculated value of Q is used in the Manning equation to calculate yn . If yn < yc , a steep slope is confirmed and the calculated Q is the actual Q. If yn > yc , then the slope is mild and the actual Q is determined by simultaneous solution of Equations 4.131 and 4.132. It is relevant to note that when the slope is mild, a M1 profile is required between the upstream reservoir and the downstream channel. However, this would require that the flow depth increase in the downstream direction (as per Figure 4.11), which would not be possible in the present case. As a consequence, the depth of flow goes immediately to the normal flow depth. The problem illustrated in Figure 4.12 is commonly referred to as the lake discharge problem, and the aforementioned solution works in most cases. In cases where the downstream channel is mild and the channel is not sufficiently long, uniform flow might not be established in the channel, and the uniform-flow equation, Equation 4.132, must be replaced by the (energy) equation for gradually varied flow. In this case, the channel is called hydraulically short. Conversely, if the channel is mild and sufficiently long for uniform flow to be established, then the channel is hydraulically long. For steep slopes, the attainment of normal flow in the downstream channel is not required for Equations 4.129 and 4.130 to be valid, only that the downstream flow be supercritical.

w .E asy En g 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.

ine eri n

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

Eliminating Q from these equations yields the more convenient combined form as  gA3c 1 Ac H = yc + ' H = yc + 2 Tc 2Tc 2gAc

et

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=



gA3c Tc

' ( % &3 ( ( (9.81) 2yc + 3y2c ) = = 78.3 m3 /s 2 + 2(3)yc

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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 % &5 5 3 byn + my2n 1 1 1 An3 12 S02 S0 = % Q= 2 2 & n 3 n 3 Pn b + 2yn 1 + m2 1 78.3 = 0.020 %

%

2yn + 3y2n

&5 3

2 + 2yn 1 + 32

1

2 & 2 (0.01) 3

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

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

w .E asy En g

ine eri n

P1 − P2 = ρQ(V2 − V1 )

g .n

(4.133)

et

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 E1 Specific energy, E (b)

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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

w .E asy En g

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

ine eri n

(4.138)

where q is the flow per unit width. Equation 4.138 can be solved for y2 to yield y2 =

⎛

y1 ⎝ −1 + 2



1 +

8q2 gy31

⎞ ⎠

which can also be written in the following nondimensional form: 

 y2 1 2 −1 + 1 + 8Fr1 = y1 2

g .n

(4.139)

et

(4.140)

where y2 /y1 is commonly called the sequent depth ratio, and Fr1 is the upstream Froude number defined by q V1 Fr1 = = (4.141) gy1 y1 gy1

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

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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 (4.142) − 1 = 2 exp − 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)

w .E asy En g

ine eri n

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

(4.144)

et

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

Fr1

Energy dissipation

Characteristics

1.0–1.7 1.7–2.5 2.5–4.5 4.5–9.0 >9.0

1 ⎪ ⎩ n

(4.170)

(4.171)

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

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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.

w .E asy En g

Natural channels. In natural channels the standard-step method is used with the energy equation given by Equation 4.159, repeated here as 1 V2 y + α 2g 2 $ $ L = V12 $$ C $$ V22 − α1 $ − S0 Sf + $α2 L $ 2g 2g $ 

ine eri n

(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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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= (4.176) v dA L  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 g

 N

i=1 Ai

α= N

i=1 Ki

2

3

N  K3 i

i=1

(4.177)

A2i

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

ine eri n

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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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: A = w2 y2 = 23y2

w .E asy En g

P = w2 + 2y2 = 23 + 2y2 5

5

(23y2 ) 3 1 A3 1 K= = n2 P 32 0.025 (23 + 2y ) 23 2 α=1

Fr∗ =

(1)(250)2 (23) 12.04 αQ2 T = = 3 gA (9.81)(23y2 )3 y32

ine eri n

(4.180)

Since the velocity in each part of the compound channel is taken as being equal to the mean velocity in that part of the channel, this requires that α = 1 when the flow is confined to the main channel. To determine the critical flow depth when the flow is confined to the main channel, set Fr∗ = 1 in Equation 4.180 and solve for y2 . This yields a critical flow depth, yc , of 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

K1 =

g .n

5 3

1 A1 1 A1 = n1 32 0.090 32 P1 P1

et

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

5

1 A23 1 A23 = K2 = n2 32 0.025 32 P2 P2

A3 = w3 y3 = 130(y2 − 2.52) P3 = w3 + y3 = 130 + (y2 − 2.52) 5

5

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

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4.0

3.5

3.5

3.0 2.5 2.0 1.5 1.0 0.5 0.0

3.0 2.5 2.0 1.5 1.0 0.5

0

1

2

3

4

5

6

7

0.0 0.0

Kinetic energy correction factor,

ww

Fr* = 1

2

FIGURE 4.16: Flow properties in compound channel

Depth in main channel, y2 (m)

Chapter 4

Depth in main channel, y (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 g

(4.181)

α dα L dy y2

(4.182)

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

Fr∗ =

Q2 dα αQ2 T − gA3 2gA2 dy

ine eri n

(4.183)

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

4.3.4.5

Profiles across bridges

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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155

4

1 CR

Contraction reach Lc

3

BU BD A

Le

ww

Water-Surface Profiles

B

C

D

2

Expansion reach Idealized flow transition

ER 1

w .E asy En g

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

ine eri n

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+1$ $ (4.185) L = $ % & 2 2$ V V $ $ zi −zi+1 C i+1 i Sf + L $αi+1 2g − αi 2g $ − L $ $

where i and i + 1 denote the upstream and downstream sections, respectively, L is the distance between the sections, y is the depth of flow (in the main channel), α is the kinetic energy correction factor calculated using Equation 4.177, V is the average velocity, Sf is the average friction slope between the upstream and downstream sections, C is the energy loss factor taken as 0.5 for expanding flow and

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Chapter 4

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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

5m

10 m

R

5m

w .E asy En g

Sections BD and BU

R

L

Flow

200 m

ine eri n 4

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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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 =

Le 1440 = = 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 200 Lc = 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 g

y11 = 88.0 − 83.01 − 2.5 = 2.49 m

A11 = w11 y11 = (150)(2.49) = 373.5 m2

ine eri n

P11 = w11 + y11 = 150 + 2.49 = 152.49 m 5

5

3 1 373.5 3 1 A11 = = 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 + 2z = 60 + 2(2.5) = 65 m 5

g .n

5

et

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

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

5

3 1 A13 1 423.3 3 K13 = = = 15403 m3 /s 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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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

2

 300 2 = 0.0000281 56604

Sf 1 =



V1 =

300 Q = 0.27 m/s = A1 1096.2

Q K1

=



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

w .E asy En g

P21 = w21 = 10 m 5

3 1 A21 K21 = n21 23 P21

y22 = y22

A22 = w22 y22 = (60)y22

ine eri n

P22 = w22 + 2z = 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

K23 =

1 A23 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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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 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:

ww

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 g

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-

ine eri n

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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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 g 1

ks = 1 mm

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.

15 cm Slope = 2%

1

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

ine eri n

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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Downloaded From : www.EasyEngineering.net 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

1 2 3 4 5 6 7

0.040 0.030 0.015 0.013 0.017 0.035 0.060

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

w .E asy En g

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

4.29.

4.30.

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.

4.31.

4.32.

2

S=

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. 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? 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. 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? 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?

ine eri n

Left floodplain

g .n

et

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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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

ww

4.38.

4.39.

3m

3 10 m

4.40.

w .E asy En g

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

4.41.

4.42.

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

4.43.

raised by 0.25 m, what is the depth of flow in the constriction? 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? 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? 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? 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. 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. 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

ine eri n

g .n

et

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4.44.

4.45.

4.46.

4.47.

4.48.

4.49.

4.50.

the channel is 20 m3 /s and the flow depth in the approach channel is 2 m. (a) Taking into account contraction and expansion head losses, what is the flow depth at the bridge constriction and the flow depth downstream of the bridge? (b) What is the percentage error in the calculated flow depths if energy losses are neglected? Water flows at 36 m3 /s in a rectangular channel of width 10 m and Manning’s n of 0.030. If the depth of flow at a channel section is 3 m and the slope of the channel is 0.001, classify the water-surface profile. What is the slope of the water surface at the observed section? Would the water-surface profile be much different if the depth of flow were equal to 2 m? Water flows at 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. Water flows in a trapezoidal channel where the bottom width is 6 m and side slopes are 2:1 (H: V). The channel lining has an estimated Manning’s n of 0.045, and the slope of the channel is 1.5%. When the flow rate is 80 m3 /s, the depth of flow at a gaging station is 5 m. Classify the water-surface profile, state whether the depth increases or decreases in the downstream direction, and calculate the slope of the water surface at the gaging station. On the basis of this water-surface slope, estimate the depths of flow 100 m downstream and 100 m upstream of the gaging station. Water 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. 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.] 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. 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 y2 hL y1 1 − =1 − + y1 y1 2 y2

ww

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

4.53.

w .E asy En g

where Fr1 is the upstream Froude number.

163

4.54.

4.55.

4.56.

4.57.

4.58.

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. Water flows in a horizontal trapezoidal channel at 21 m3 /s, where the bottom width of the channel is 2 m, side slopes are 1 : 1, and the depth of flow is 1 m. Calculate the downstream depth required for a hydraulic jump to form at this location. What would be the energy loss in the jump? A 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. Water flows at 10 m3 /s in a rectangular channel of width 5.5 m. The slope of the channel is 0.15% and the Manning roughness coefficient is 0.038. Estimate the depth 100 m upstream of a section where the flow depth is 2.2 m using: (a) the direct-integration method, and (b) the standard-step method. Approximately how far upstream of this section would you expect to find uniform flow? Water flows at 5 m3 /s in a 4-m-wide rectangular channel that is laid on a slope of 4%. If the channel has a Manning’s n of 0.05 and the depth at a given section is 1.5 m, how far upstream or downstream is the depth equal to 1 m? If the depth of flow in the channel described in Problem 4.10 is measured as 1.4 m, find the location where the depth is 1.6 m. A trapezoidal canal has a longitudinal slope of 1%, side slopes of 3 : 1 (H : V), a bottom width of 3.00 m, a Manning’s n of 0.015, and carries a flow of 20 m3 /s. The depth of flow at a gaging station is observed to be 1.00 m. Answer the following questions: +

+

ine eri n

g .n

et

(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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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).

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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?

w .E asy En g

ine eri n

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

g .n

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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V2 Z + α 2g

B

A

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

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165

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.

w .E asy En g TABLE 4.9

Area (m2 )

Water surface elevation (m)

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 —

ine eri n Wetted perimeter (m)

Section A

Section B

— 52.21 50.84 48.77 47.40 46.02

36.27 35.05 33.83 32.77 31.70 —

g .n

et

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C H A P T E R

5

Design of Drainage Channels 5.1 Introduction

ww

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

w .E asy En g

ine eri n

g .n

et

166

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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. 5.2.1

Best Hydraulic Section

w .E asy En g

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

ine eri n ⎞3

⎛

5

2 ⎜ Qn ⎟ A = ⎝ 1 ⎠ P5 S02

g .n

(5.2)

et

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

T = b + 2 my Freeboard y

A

1 m

b

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(5.3) (5.4)

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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

(5.6)

where c is a constant given by

⎛

⎞3

5

⎜ Qn ⎟ c=⎝ 1 ⎠ S02

ww

(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)

w .E asy En g

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 P = 2 3y, b=2 (5.10) 3

ine eri n

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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169

TABLE 5.1: Geometric Characteristics of Best Hydraulic Sections

Shape

A √ 2 3y

P √ 2 3y

T √ 4 3 3y 2y

Trapezoid

Half of a hexagon

Rectangle

Half of a square

2y2

Half of a square

y2

4y √ 2 2y

2y

π y2 2 √ 4 2y2 3

πy √ 8 3 2y

2y √ 2 2y

Triangle

ww

Best geometry

Semicircle

—

Parabola

—

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)

w .E asy En g α=

 1 m + 1 − 1 + m2 m

b = 2(1 − αm)y y=

1

24 3

(2 − α 2 m) 8



nQ √ S0

(5.12) (5.13)

3 8

(5.14)

where y is the normal depth of flow in the channel derived using the Manning equation. In cases where it is not practical to use m = 1 in the compound section, Equations 5.12 to 5.14 give the most efficient compound section for any given value of m. Although the best hydraulic section appears to be the most economical in terms of excavation and channel lining, it is important to note that this section might not always be the most economical section for one or more of the following reasons:

ine eri n

g .n

⊲ 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

et

b + 2mαy

(1−α)y y

αy m:1 (H:V)

b

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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).

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EXAMPLE 5.1 A triangular concrete-lined drainage channel is to be designed on a slope of 0.5% and accommodate a peak runoff rate of 0.15 m3 /s. Determine the optimal dimensions of the channel. Assume n = 0.013 for the concrete lining.

w .E asy En g

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 Q=

ine eri n 5

2 1 1 A 3 21 1 AR 3 S02 = S n n P 23 0 5

0.15 =

1 (y2 ) 3 1 (0.005) 2 0.013 (2√2y) 32

g .n

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.

et

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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Basic Principles

1

171

1 m

m y b

τs

τs τb

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

ww

(5.16)

τb = γ yS0

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

w .E asy En g

(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

ine eri n

g .n

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.

et

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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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

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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)

w .E asy En g

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

ine eri n

τc = f2 (γs − γ )ds



u∗c ds ν



(5.20)

g .n

where f2 denotes a functional relationship, u∗c is the friction velocity corresponding to τc and  defined as τc (5.21) u∗c = ρ

et

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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Basic Principles

FIGURE 5.5: Sheilds diagram

γs

(g/cm3)

Material

Source: ASCE (2006b).

Amber Lignite Barite

τc (γs – γ)d s *

Dimensionless shear stress, τ =

ww

173

Fully developed turbulent velocity profile 1.0 0.8 0.6 0.5 0.4

Turbulent boundary layer

0.3 0.2

Value of 2

4

6 8 1

ds ν

(Shields)

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

1.06 1.27 2.7 4.25 2.65 2.65 2.65 2.65 2.61 2.10 7.9

γs 0.1 ( γ _ 1) gds

2

4

6 100

2

4

6

1000

0.1 0.08 0.06 0.05 0.04 0.03

w .E asy En g 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

ine eri n

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

g .n

et

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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Chapter 5

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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  (5.26) Fp = (τs Ap )2 + (wp sin θ )2

w .E asy En g

where τs is the shear stress exerted by the flowing fluid on the side of the channel, and θ is the angle that the side of the channel makes with the horizontal [degrees]. When motion is incipient on the side of the channel, the force tending to move the particle is equal to the frictional force, Ff [F], which is equal to the component of the particle weight normal to the side of the channel, multiplied by the coefficient of friction (tan α), in which case Ff = wp cos θ tan α

(5.27)

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

τps

ine eri n

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

g .n

(5.28)

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



1 −

sin2 θ sin2 α

et

(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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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 is in centimeters 50

ine eri n

Source: USFHWA (2005).

39

5

1.

6 0. 7 1. 0

k

ed

et

ang ula r

37

g .n

Very

nded

35

33

Very ro u

Angle of repose, α (degrees)

ro c

41

0.

3

0.

2

15

0.

0.

01

07

0.

06

0.

03

0.

43

0.

02

Median stone size, d 50 (ft)

0.

FIGURE 5.6: Angle of repose of gravel and riprap linings

w .E asy En g Cr us h

ww

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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Chapter 5

Design of Drainage Channels

FIGURE 5.7: Angle of repose of noncohesive material

0.1 42

Source: USBR (1953).

40

Particle size, d75 (in.) 0.2

0.3

0.6

0.4

0.8 1.0

2.0

3.0

4.0

40 50 60 70

100

ww

Angle of repose, α (degrees)

38 36

r

ula

ng

a ery

34

te

era

32

d Mo

30 28 26

Sli

y

y htl

g

M

d

de

n ou

r

y

el

t ra

e od

24

r

ula

g an

htl

g Sli

lar

gu

n ly a

V

ry

d

de

n ou

r

ed

nd

u ro

Ve

22

w .E asy En g 20

3

4

5

6

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

EXAMPLE 5.4

A trapezoidal channel with bottom width 3 m and side slopes 2:1 (H:V) is lined with stones of median size 200 mm and 85-percentile size 280 mm. The stones are angular and the permissible shear stress of the stones are estimated to be 170 Pa. Determine the permissible shear stress on the bottom and sides of the channel.

ine eri n

Solution From the given data: b = 3 m, m = 2, d50 = 200 mm, and d85 = 280 mm. The permissible shear stress on the bottom of the channel, τp , is 170 Pa. The side-slope angle, θ, is given by     1 1 = tan−1 = 26.6◦ θ = tan−1 m 2

g .n

Since no information is given about the grading of the stone lining (open versus narrow), the angle of repose, α, will be estimated using both Table 5.2 and Figure 5.6. The smaller value of α will be used in design, since this gives the more conservative (i.e., lower) estimate of the tractive force ratio. The value of α estimated using Table 5.2 is  0.125   d85 280 0.125 = (37.1) = 38.7◦ α = α0 d50 200

et

For d50 = 200 mm, Figure 5.6 gives α = 41.5◦ for very angular stone and α = 39◦ for very rounded stone. Since the stone in this case is characterized as “angular,” we can estimate the angle of repose by the average value of α = 40◦ . Comparing the estimates of α derived from Table 5.2 (38.7◦ ) and Figure 5.6 (40◦ ), the more conservative estimate of α = 38.7◦ will be used. The tractive force ratio, K, is therefore given by Equation 5.29 as    sin2 (26.6) = 0.698 K = 1 − sin2 (38.7)

Since the permissible shear stress on bottom of the channel, τp , is 170 Pa, the permissible shear stress on the side of the channel, τps , is given by τps = Kτp = (0.698)(170) = 119 Pa

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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 ⎪ ⎪ ⎪  2 T   ⎨ 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

w .E asy En g

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

ine eri n

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.

g .n

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

et

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  2   rc rc + 0.0073 = 2.38 − 0.206(2.35) + 0.0073(2.35)2 = 1.93 Kr = 2.38 − 0.206 T T Therefore, the maximum shear stress expected in the bend, τr , is given by Equation 5.30 as τr = Kr τb = (1.93)(78.3) = 151 Pa

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Chapter 5

Design of Drainage Channels This increased shear stress persists for a distance Lp downstream of the bend, where Lp is given by Equation 5.32 as ⎛ ⎞ ⎛ ⎞ 7 7 6 6 R 0.623 ⎠ = 0.604(0.623) ⎝ ⎠ = 10.8 m Lp = 0.604R ⎝ nb 0.020

It is apparent from the results of this example that the maximum shear stress in a bend can be significantly higher than in the straight segment of the channel; the factor is 1.93 in this example. Also, the increased shear stress can persist for a significant distance downstream, approximately 4.3 channel bottom-widths in the present example.

5.2.5

ww

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.

w .E asy En g EXAMPLE 5.6

ine eri n

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

g .n

et

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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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

ww

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

w .E asy En g

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 V2T grc

ine eri n hs =

(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).

g .n

et

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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Chapter 5

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

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

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

5.3 Design of Channels with Rigid Linings The objective in designing channels with rigid linings is to determine the channel dimensions required to safely convey a design flow rate in a channel of given shape, lining material, and longitudinal slope. A recommended design procedure is as follows:

ww

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.

w .E asy En g

TABLE 5.4: Manning Roughness Coefficients in Open Channels with Rigid Linings

Type

Cement

Concrete

Concrete bottom float finished with sides of

Grouted riprap Stone masonry Soil cement Asphalt Brick Masonry Dressed ashlar

Range of Manning’s n

Characteristics

Minimum

Normal

Maximum

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

0.010 0.011 0.011 0.013 0.015

0.011 0.013 0.013 0.015 0.017

0.013 0.015 0.015 0.016 0.020

0.014 0.016 0.018 0.017 0.022

0.017 0.019 0.022 0.020 0.027

0.020 0.023 0.025 — —

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

ine eri n 0.020 0.020 0.028 0.030 0.020 0.016 0.013 0.011 0.012 0.017 0.023 0.013

g .n 0.017 0.020 0.020

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

et

0.020 0.024 0.024 0.030 0.035 0.040 0.042 0.025 0.018 — 0.015 0.018 0.030 0.035 0.017

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

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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.

ww

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).

w .E asy En g 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=

ine eri n

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

g .n

b = 1.15y = 1.15(2.09) = 2.40 m The flow area is therefore given by

et

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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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 



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

A = (b + my)y = (0.60y + 1.5y)y = 2.10y2 R=

A 2.10y2 = = 0.499y P 4.21y

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

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

or

ww

8

y 3 = 5.86

which leads to

w .E asy En g

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=

ine eri n

2.532 V2 = 0.15 + = 0.48 m F = 0.15 + 2g 2(9.81) The minimum depth of the channel to be excavated and lined is equal to the normal depth plus the freeboard, 1.94 m + 0.48 m = 2.42 m. The channel is to have a bottom width of 1.16 m and side slopes of 1.5:1 (H:V).

g .n

et

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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Design of Channels with Flexible Linings

183

FIGURE 5.8: Roadside channel with rigid lining

ww

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.

w .E asy En g

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

ine eri n

g .n

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:

et

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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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).

ww

TABLE 5.6: Typical Manning’s n for Riprap, Cobble, and Gravel Linings

Flow depth

w .E asy En g Lining category

Lining type

0.15 m (0.5 ft)

0.50 m (1.6 ft)

1.0 m (3.3 ft)

Gravel mulch

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

0.040 0.056

0.033 0.042

0.031 0.038

Cobbles

d50 = 100 mm (4 in.)

∗

0.055

0.047

Rock riprap

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

∗ ∗

0.069 ∗

0.056 0.080

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

ine eri n

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)

g .n

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

et

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

Permissible shear stress Lining category

ww

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

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

113 227

2.4 4.8

w .E asy En g Rock riprap

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.

ine eri n

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

g .n

  P = b + 2y 1 + m2 = (0.325y) + 2y 1 + 32 = 6.650y

et

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 (0.005) 2 0.5 = 0.020 (6.650y) 32

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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Chapter 5

Design of Drainage Channels The side-shear-stress factor, Ks , is given by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 and so the maximum shear stress exerted on the side of the channel, τs , is given by τs = Ks τb = (0.868)(17.8) = 15.4 Pa Assume a safety factor (SF) equal to one. Step 4: The permissible shear stress, τp , on the noncohesive bare-soil lining can be estimated (interpolated) from Table 5.7 for d75 = 10 mm as 7.4 Pa. Since the soil is noncohesive, the permissible shear stress on the side of the channel, τps (=Kτp ), will be less than 7.4 Pa. Step 5: The maximum shear stress on bottom of the channel (17.8 Pa) is greater than the permissible shear stress on the bare-soil lining (7.4 Pa), and so the bare-soil lining is inadequate. Try another lining.

ww

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

w .E asy En g

5

1 A 3 21 S Q= n P 23 0

0.5 =

which simplifies to

5

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

ine eri n

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

g .n

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

et

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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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 The tractive-force ratio, K, is given by Equation 5.29 as    2  sin θ sin2 (18.4) = 1 − = 0.832 K= 1 − 2 sin α sin2 (34.7)

ww

and hence the permissible shear stress on the side of the channel, τps , is given by τps = Kτp = (0.832)(38) = 31.6 Pa

w .E asy En g

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

ine eri n

Vegetative Linings and Bare Soil

g .n

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

et

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Chapter 5

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

Retardance

ww

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

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.)

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.)

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.)

w .E asy En g Weeping lovegrass

C

D

E

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.)

ine eri n

g .n

et

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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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

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)

w .E asy En g

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

ine eri n

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)

g .n

et

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 )



ns n

2

(5.46)

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

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Chapter 5

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

Condition Growth form Sod Bunch Mixed

Excellent

Very good

Good

Fair

Poor

0.98 0.55 0.82

0.95 0.53 0.79

0.90 0.50 0.75

0.84 0.47 0.70

0.75 0.41 0.62

Source: USFHWA (2005).

Equation 5.45 [dimensionless], and n is the lining roughness given by Equation 5.43 [dimensionless]. The grass cover factor, Cf , varies with cover density and grass growth form (sod or bunch) and can be estimated using the values in Table 5.9. Bunch grasses tend to grow in “bunches” leaving open-soil areas, while sod grasses are more dispersed and tend to cover most of the soil surface.

ww

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

w .E asy En g τp,n =



1 Pa, 0.75d75 Pa,

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

(5.47)

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

ine eri n

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 ).

g .n

EXAMPLE 5.10

et

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=

A y + 4y2 = P 1 + 8.246y

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191

TABLE 5.10: Coefficients for Permissible Soil Shear Stress

ASTM class∗

ww

Typical names

Applicable range

c1

c2

1.07

14.3

c3

c4

c5

c6 4.8 * 10−3 48

GM

Silty gravels Gravel-sand-silt mixtures

10 … PI … 20 PI Ú 20

47.7 1.42 0.076 1.42

−0.61 −0.61

GC

Clayey gravels Gravel-sand-clay mixtures

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

−0.61 −0.61

SM

Silty sands Sand-silt mixtures

10 … PI … 20 PI Ú 20

1.07

7.15

11.9 1.42 0.058 1.42

−0.61 −0.61

SC

Clayey sands Sand-clay mixtures

10 … PI … 20 PI Ú 20

1.07

14.3

47.7 1.42 0.076 1.42

−0.61 −0.61

ML

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

10 … PI … 20 PI Ú 20

1.07

7.15

11.9 1.48 0.058 1.48

−0.57 −0.57

4.8 * 10−3 48

1.07

14.3

47.7 1.48 0.076 1.48

−0.57 −0.57

4.8 * 10−3 48

w .E asy En g

10 … PI … 20 PI Ú 20

4.8 * 10−2 48 4.8 * 10−3 48 4.8 * 10−3 48

CL

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

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

ine eri n

PI Ú 20

0.097 1.38 −0.373

48

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

g .n

Stress range, Pa (lb/ft2)

1062

12 11–28 >25

Source: ASCE (2007). TABLE 6.2: Average Per-Capita Wastewater Residential Flow Rates

Flow rate City

ww

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

w .E asy En g Source: ASCE (2007).

(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

ine eri n

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

g .n

et

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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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

ww

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.

w .E asy En g

ine eri n

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

g .n

et

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

w .E asy En g

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 =

ine eri n

⎧ −0.095 ⎨1.88Qavg

⎩0.281Q−0.44 avg

g .n

et

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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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

ww

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.

w .E asy En g 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.

ine eri n

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

g .n

Area (ha)

Density (persons/ha)

Population

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

6 75 2500

3600 61,875 187,500

252,975

Flow (m3 /s)

et 0.013 0.215 0.434 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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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 PFmax = 1.88Qavg2 = 1.88(0.995)−0.095 = 1.88 −0.095 = 1.88(0.498)−0.095 = 2.01 PFmin = 1.88Qavg1

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.

w .E asy En g 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

ine eri n

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,  ǫ 1 2.51   = −2 log (6.8) + 14.8R f 4Re f

g .n

et

where ǫ 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 ǫ + log 14.8R Re0.9 

2

(6.10)

This approximation is valid for 10−6 … ǫ/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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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

ww

(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, ǫ, 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 ǫ 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

w .E asy En g

ine eri n

θ − sin θ 8



A=



P=

Dθ 2

R=

 sin θ A D 1 − = P 4 θ

D2

g .n

et

(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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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

 D θ 1 − cos h= 2 2

 θ 455 100 = 1 − cos 2 2

ww

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

w .E asy En g 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

ine eri n

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

− 12

θ − 3 (θ − sin θ ) 3 − 20.16nQD− 3 S0

=0

(6.17)

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

Q A

(6.18)

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

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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, 8

5

2

− 21

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

8

5

=0

1

θ − 3 (θ − sin θ) 3 − 20.16(0.013)(4)(1.5)− 3 (0.0100)− 2 = 0 which yields θ = 3.30 radians. Therefore, the normal flow depth, h, and area, A, are given by Equations 6.13 and 6.14 as

ww



  D 1.5 θ 3.30 h= = = 0.81 m 1 − cos 1 − cos 2 2 2 2   3.30 − sin 3.30 θ − sin θ D2 = (1.5)2 = 0.970 m2 A= 8 8

w .E asy En g

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.

ine eri n

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

FIGURE 6.2: Flow in circular conduits

1.0 0.9

Ratio of depth/diameter, h/D

0.8 0.7

g .n

Q Qfull

0.6

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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Chapter 6

Design of Sanitary Sewers

Vfull

2 D 3 12 S0 4  π D2 = Vfull 4

1 23 21 1 = Rfull S0 = n n

Qfull = Vfull Afull



(6.19) (6.20)

and hence

ww

V = Vfull



4R D

Q = Qfull



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.

w .E asy En g 6.3.2

ine eri n

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, ǫ. However, for any given value of ǫ, 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 ǫ = 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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ww

w .E asy En g

TABLE 6.4: Recommended Manning’s n versus Pipe Diameter

Pipe diameters

Condition Extra care Typical Substandard

150 mm (6 in.) 0.0092 0.0106 0.0120

Source: ASCE (2007).

205 mm (8 in.)

255 mm (10 in.)

305 mm (12 in.)

380 mm (15 in.)

455 mm (18 in.)

610 mm (24 in.)

760 mm (30 in.)

915 mm (36 in.)

1220 mm (48 in.)

1525 mm (60 in.)

0.0093 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

ine eri n

g .n

et

221

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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?

w .E asy En g

Solution From the given data: D = 610 mm = 0.610 m, S0 = 0.1% = 0.001, ǫ = 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

ine eri n

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 ǫ + 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)

− 12

20.16nD− 3 S0

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

(3)

(4)

h/D

(2) Q (m3 /s)

n

Typical n

(5) n/nfull (%)

0.10 0.30 0.50 0.80 1.00

0.006 0.052 0.130 0.253 0.261

0.0099 0.0100 0.0101 0.0102 0.0101

0.0114 0.0115 0.0116 0.0117 0.0116

98 99 100 101 100

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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. 6.3.3

ww

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)

w .E asy En g

ine eri n

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

(6.28)

et

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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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, ǫ = 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

w .E asy En g Vfull =

1

nfull

2

1

3 S02 = Rfull

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

ine eri n

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 )

g .n

τmin = γ RS0 = (9790)(0.0473)(0.00045) = 0.21 Pa

et

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, ǫ, and design critical shear stress, τc , for self-cleansing. Constraints in the hydraulic design are as follows:

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⊲ 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.

w .E asy En g

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

ine eri n

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.

g .n

et

Solution From the given data: D = 610 mm = 0.610 m, Qmax = 0.30 m3 /s, Qmin = 0.015 m3 /s, ǫ = 0.1 mm = 0.0001 m, τc = 0.87 Pa, T = 20◦ C, (h/D)max = 0.75, and Vlim = 4.0 m/s. Step 1: From the given information, any pipe slope greater than zero meets the minimum-cover requirement. Therefore, Sref1 = 0.0. Step 2: The pipe diameter under consideration is D = 610 mm. This is likely equal to the pipe diameter of the upstream segment. Step 3: When Q = Qmax = 0.30 m3 /s and h/D = (h/D)max = 0.75, the following calculations are made:

   h −1 1 − 2 θmax = 2 cos = 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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 θmax − sin θmax 4.189 − sin(4.189) 2 = 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 = = 0.000543 0.184 ⎡  ⎡  ⎤−2 ⎤−2 1.65 ǫ 0.000543 1.65 ⎦ = 0.25 ⎣log ⎦ = 0.25 ⎣log + + 14.8Rmax 14.8 (2.35 * 105 )0.9 Re0.9 max

Remax = ǫ Rmax fmax

= 0.0141

1

ww

1

1

1

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

Smax

⎤2 − 38 −8 20.16(0.0101)(0.30)(0.610) ⎢ 20.16nmax Qmax D 3 ⎥ ⎦ = 0.00159 =⎣ 2 ⎦ =⎣ 5 5 −3 − 23 3 (4.189) [4.189 − sin(4.189)] 3 θmax [θmax − sin θmax ]

w .E asy En g

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     θ − sin θmin θ − sin θmin D2 = min (0.610)2 Amin = min 8 8

ine eri n

(0.015/Amin )Rmin (Qmin /Amin )Rmin = ν 10−6 ǫ 0.001 = Rmin Rmin ⎡ ⎛ ⎞⎤−2 ǫ 1.65 ⎢ ⎠⎥ fmin = 0.25 ⎣log ⎝ + ⎦ 0.9 14.8Rmin Remin

Remin =

1 6

1 2

nmin = 0.1129Rmin fmin ⎡

− 38

⎤2

g .n

et

⎤2 8 − ⎢ 20.16nmin (0.015)(0.610) 3 ⎥ ⎡

⎢ 20.16nmin Qmin D ⎥ ⎥ =⎢ 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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(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◦

w .E asy En g

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

ine eri n

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

g .n

et

Manhole

D1 Energy Grade Line hL

2

V1 2g

y1 2

Δz

y2

V2 2g

D2 Flow

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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, ǫ = 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 ⎛ 0.001 ǫ 1.65 1.65 ⎢ ⎢ ⎠⎥ ⎠⎥ f1 = 0.25 ⎣log ⎝ + + ⎦ = 0.25 ⎣log ⎝ 0.9 ⎦ 14.8R1 14.8R1 Re0.9 Re 1 1

Re1 =

w .E asy En g 1

1

n1 = 0.1129R16 f12 ⎡

− 83

⎤2

⎤2 8 − ⎢ 20.16n1 (0.485)(0.760) 3 ⎥ ⎡

⎥ ⎢ 20.16n1 QD ⎥ =⎢ S1 = ⎢ ⎣ ⎣ −2 5 ⎦ 3 θ1 [θ1 − sin θ1 ] 3

ine eri n 5 −2 θ1 3 [θ1 − sin θ1 ] 3

⎥ = 0.005 ⎦

(6.33)

The solution of these equations is θ1 = 3.310 radians, which corresponds to a flow depth y1 = 0.412 m and an average velocity V1 = 1.93 m/s. Step 2: Determine the downstream flow conditions. Use the same equations as in Step 1 with the exception that the slope in Equation 6.33 is taken as 0.002 instead of 0.005. This yields a flow depth y2 = 0.570 m and an average velocity V2 = 1.33 m/s. Step 3: Determine the head loss at the manhole. Since the flow is straight through the manhole, the head-loss coefficient can be estimated as K = 0.15 and the head loss, hL , at the manhole is given by Equation 6.30 as

g .n

V2 (1.33)2 = 0.014 m hL = K 2 = (0.15) 2g 2(9.81)

et

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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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).

w .E asy En g 6.4.2

Pipe Material

ine eri n

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

g .n

et

TABLE 6.5: Rigid and Flexible Pipe Materials

Rigid pipe Concrete Cast iron Vitrified clay

Flexible pipe Ductile iron Steel Thermoplastic (e.g., PVC)

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TABLE 6.6: Sewer-Pipe Materials

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

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

w .E asy En g

ine eri n

100 mm–910 mm

g .n

100 mm–1220 mm

et

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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

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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.

w .E asy En g 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

ine eri n

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

g .n

et

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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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

(a) Typical manhole

w .E asy En g

(b) Drop manhole

ine eri n

g .n

et

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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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

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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.

w .E asy En g 6.4.8

Force Mains

ine eri n

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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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

w .E asy En g

h0 = K

V2 2g

(6.35)

where h0 is the local loss [L], K is the local loss coefficient [dimensionless], V is the average flow velocity [LT−1 ], and g is the acceleration due to gravity [LT−2 ]. Loss coefficients at fixtures commonly found in force mains are given in Figure 2.7. 6.4.9

Hydrogen-Sulfide Control

Hydrogen-sulfide (H2 S) generation is a common problem in sanitary sewers. Among the problems associated with H2 S generation are odor, health hazard to maintenance crews, and corrosion of unprotected sewer pipes produced from cementitious materials and metals. The design of sanitary sewers seeks to avoid septic conditions and to provide an environment that is relatively free of H2 S. Generation of H2 S in sanitary sewers results primarily from the action of sulfate-reducing bacteria (SRB) on the pipe floor which convert sulfates to hydrogen sulfide under anaerobic (reducing) conditions. Compounds such as sulfite, thiosulfate, free sulfur, and other inorganic sulfur compounds occasionally found in wastewater can also be reduced to sulfide. Corrosive conditions occur when sulfuric acid (H2 SO4 ) is derived through the oxidation of hydrogen sulfide by aerobic bacteria (Thiobacillus thiooxidans) and fungi that reside on the moist exposed sewer pipe wall. This process can only take place where there is an adequate supply of H2 S (>2 ppm), high relative humidity, and atmospheric oxygen. The effect of H2 SO4 on concrete surfaces exposed to the sewer environment can be devastating. Concrete pipes, asbestos-cement pipes, and mortar linings on ferrous pipes experience surface reactions in which the surface material is converted to an expanding pasty mass which may fall away and expose new surfaces to corrosive attack. The color of corroded concrete surfaces can be various shades of yellow caused by the direct oxidation of H2 S to elemental sulfur. Entire pump stations have been known to collapse due to loss of structural stability from corrosion. Hydrogen sulfide gas is extremely toxic and can cause death at concentrations as low as 300 ppm (0.03%) in air. A person who ignores the first odor of the gas quickly loses the ability to smell the gas, eliminating further warning and leading to deadly consequences. Significant factors that contribute to H2 S generation are high wastewater temperatures and low flow velocities. The potential for sulfide generation in sewers with diameters less than 600 mm (24 in.) can be assessed using the value of the Z variable, where (Pomeroy and Parkhurst, 1977)

ine eri n

Z = 0.308

EBOD 1 2

S0 Q

1 3

*

P B

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(6.36)

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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

ww

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

w .E asy En g

ine eri n

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

5

8

− 12

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

−

5 sin θ) 3

−

8 1 20.16(0.013)(1.7)(0.915)− 3 (0.009)− 2

=0 =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=

ww

Dθ/2 θ 4.31 P = = = = 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

ine eri n

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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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.

w .E asy En g 6.5.1.2

ine eri n

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

−0.5692 0.00539Qmin

−0.5697 0.00624Qmin

−0.5699 0.00650Qmin

225

−0.5720 0.00589Qmin

−0.5723 0.00683Qmin

−0.5724 0.00711Qmin

300

−0.5736 0.00629Qmin

−0.5739 0.00729Qmin

−0.5740 0.00759Qmin

375

−0.5748 0.00662Qmin

−0.5751 0.00768Qmin

−0.5751 0.00799Qmin

450

−0.5758 0.00786Qmin

−0.5780 0.00801Qmin

−0.5761 0.00833Qmin

500

−0.5763 0.00707Qmin

−0.5764 0.00820Qmin

−0.5764 0.00854Qmin

525

−0.5766 0.00716Qmin

−0.5767 0.00830Qmin

−0.5768 0.00864Qmin

600

−0.5772 0.00738Qmin

−0.5774 0.00856Qmin

−0.5775 0.00892Qmin

675

−0.5778 0.00759Qmin

−0.5779 0.00880Qmin

−0.5781 0.00917Qmin

750

−0.5783 0.00778Qmin

−0.5783 0.00902Qmin

−0.5784 0.00940Qmin

825

−0.5786 0.00796Qmin

−0.5788 0.00923Qmin

−0.5789 0.00961Qmin

900

−0.5790 0.00813Qmin

−0.5791 0.00942Qmin

−0.5792 0.00981Qmin

975

−0.5793 0.00828Qmin

−0.5795 0.00960Qmin

−0.5797 0.01000Qmin

1050

−0.5804 0.00846Qmin

−0.5799 0.00978Qmin

−0.5799 0.01018Qmin

1125

−0.5837 0.00870Qmin

−0.5800 0.00993Qmin

−0.5802 0.01035Qmin

1200

−0.5864 0.00893Qmin

−0.5821 0.01017Qmin

−0.5807 0.01052Qmin

1500

−0.5932 0.00971Qmin

2000

−0.5996 0.01077Qmin

Design particle size

w .E asy En g

ine eri n −0.5908 0.01114Qmin

−0.5979 0.01239Qmin

g .n

−0.5898 0.01155Qmin −0.5976 0.01288Qmin

et

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 ǫ = 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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ww

w .E asy En g

TABLE 6.9: Minimum Slopes at 15% and 10% Full

1 mm

Diameter (mm) 150 mm 200 mm 250 mm 300 mm 375 mm 450 mm 525 mm 600 mm 750 mm 900 mm 1050 mm 1200 mm 1500 mm

15% Full Smin Qmin (–) (L/s)

0.00625 0.00468 0.00375 0.00312 0.00250 0.00208 0.00179 0.00156 0.00125 0.00104 0.00089 0.00078 0.00063

0.74 1.39 2.21 3.23 5.15 7.59 10.5 13.9 22.2 32.6 44.7 59.5 95.1

10% Full Smin Qmin (–) (L/s)

0.00914 0.00685 0.00548 0.00457 0.00365 0.00305 0.00261 0.00228 0.00183 0.00152 0.00131 0.00114 0.00092

0.40 0.71 1.13 1.67 2.69 3.94 5.46 6.37 11.5 16.9 23.2 30.9 49.3

Design particle size 1.5 mm 15% Full 10% Full Smin Qmin Smin Qmin (–) (L/s) (–) (L/s)

0.00685 0.00514 0.00410 0.00342 0.00274 0.00228 0.00195 0.00171 0.00137 0.00114 0.00098 0.00086 0.00068

0.79 1.44 2.29 3.40 5.41 7.93 11.0 14.6 23.2 34.3 47.0 62.3 99.1

0.01000 0.00751 0.00600 0.00500 0.00400 0.00334 0.00286 0.00250 0.00200 0.00167 0.00143 0.00125 0.00100

0.40 0.76 1.19 1.76 2.80 4.13 5.72 7.56 12.1 17.8 23.2 32.3 51.5

1.67 mm 15% Full 10% Full Smin Qmin Smin Qmin (–) (L/s) (–) (L/s) 0.00702 0.00526 0.00420 0.00350 0.00281 0.00234 0.00201 0.00175 0.00140 0.00117 0.00100 0.00088 0.00070

0.79 1.47 2.32 3.43 5.46 8.04 11.2 14.7 23.5 34.5 47.6 63.1 100

0.01026 0.00770 0.00615 0.00512 0.00410 0.00342 0.00293 0.00257 0.00205 0.00171 0.00146 0.00128 0.00103

ine eri n

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

w .E asy En g

ine eri n

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ww Manhole no.

Line no. Location From (1) (2) (3)

w .E asy En g Area

Length Increment Total (ha) (m) To (ha) (6) (7) (5) (4)

Maximum flow

I/I (L/s) (8)

Sewage (L/s) (9)

Total (L/s) (10)

Sewer invert Ground surface elevation elevation

Minimum flow

I/I (L/s) (11)

Sewage (L/s) (12)

Total (L/s) (13)

Min Manhole tractive Max Max Slope invert Fall in of Diam force depth velocity drop sewer sewer (mm) (Pa) (mm) (m/s) (m) (m) (18) (14) (15) (16) (17) (20) (19)

ine eri n

Upper end (m) (21)

Lower Upper Lower end end end (m) (m) (m) (22) (23) (24)

FIGURE 6.7: Typical spreadsheet for design of sanitary sewers

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241

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Chapter 6

Design of Sanitary Sewers

Step 9:

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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%. When good estimates of the maximum design flow rate are available, it is recommended that the required drop at each manhole be calculated as described in Section 6.3.6. In cases where manhole losses cannot be estimated accurately, the following guidelines are sometimes followed: (1) If the inflow and outflow pipes are in different directions, the pipe invert is dropped by 30 mm (1.2 in.) to compensate for the energy losses; and (2) If the diameter of the sewer pipe leaving a manhole is larger than the diameter of the sewer pipe entering the manhole, head losses are accounted for by matching the crown elevations of the entering and leaving pipes. Using arbitrary fixed drops at each manhole is not recommended. In the calculation spreadsheet, the drop in the sewer invert is listed in Column 19. The fall in the sewer line is equal to the product of the slope (Column 14) and the length of the sewer (Column 5) and is listed in Column 20. The invert elevations of the upper and lower ends of the sewer line are listed in Columns 21 and 22. The difference in these elevations is equal to the fall in the sewer calculated in Column 20, and the invert elevation at the upper end of the sewer differs from the invert elevation at the lower end of the upstream pipe by the manhole invert drop listed in Column 19. Repeat Steps 2 to 11 for all connected pipes, proceeding downstream until the sewer main is reached. Then repeat Steps 2 to 11 for the sewer main until the connection to the next lateral is reached. Repeat Steps 2 to 12, beginning with the outermost pipe in the next sewer lateral. Continue designing the sewer main until the outlet point of the sewer system is reached.

w .E asy En g Step 10:

Step 11:

Step 12:

Step 13:

ine eri n

g .n

et

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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64.0 m

Main sewer (existing) 3

5

7

9

8

10

14

16

15

17

21 22

A Street

B Street

11

19

18

C Street

23 24

w .E asy En g Manhole no. From To (3) (4)

Line no. (1)

Location (2)

0

Main Street

—

5

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

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

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

26

25

D Street

Main street

Q Avenue

P Avenue

60.0 m

4

12

6

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62.0 m

62.0 m

2

20

243

60.0 m

1

13

Design Computations

Ground-surface elevation Upper end Lower end (m) (m) (7) (8)

Length (m) (5)

Contributing area (ha) (6)

—

—

—

60.04

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

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

57 53 97 63 100 138 78

0.60 0.76 0.51 0.94 0.46 1.41 0.30

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g .n 64.40 63.24 62.84 62.12 61.60 61.92 60.20

et

62.84 62.84 61.60 61.60 60.20 60.20 60.08

When the system is first installed, it is estimated that the average flow rates will be 30% of the average flows when the area is fully developed. Design the sewer system along A Street and the first extension segment of the sewer main between manholes 5 and 12. The saturation density of the area being served is 347 persons/ha. Local municipal guidelines require that the sewer pipes have a minimum cover of 2 m, a minimum slope of 0.08%, a minimum allowable pipe diameter of 150 mm, and be designed for self-cleansing based on a 1.5-mm particle with a specific gravity of 2.7.

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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

w .E asy En g

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

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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 = 1.88(0.370)−0.095 = 2.07 PFmax = 1.88Qavg2 −0.095 PFmin = 1.88Qavg1 = 1.88(0.111)−0.095 = 2.32

g .n

Qmax = PFmax Qavg2 = (2.07)(0.370) = 0.766 m3 /s = 766 L/s

et

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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ww Manhole no.

Line no. Location From (1) (2) (3) 0 Main Street − 1 2 3 4 5

A Street A Street A Street A Street Main Street

1 2 3 4 5

w .E asy En g Area

Maximum flow

Sewer invert Ground surface elevation elevation

Minimum flow

Length Increment Total (ha) (ha) To (m) (4) (5) (6) (7) − − − 5

I/I (L/s) (8) −

Sewage (L/s) (9) −

Total (L/s) (10) 766

I/I (L/s) (11) −

Sewage (L/s) (12) −

Total (L/s) (13) 258

Slope of sewer (14) 0.009

2 3 5 5 12

0.043 0.117 0.200 0.072 0.328

4.27 6.40 7.90 6.13 770

4.31 6.25 8.10 6.22 770

0.043 0.117 0.200 0.072 0.328

2.18 3.27 4.02 3.13 258

2.22 3.38 4.22 3.20 258

0.0226 0.0154 0.0236 0.0207 0.0008

53 91 100 89 69

0.47 0.50 0.44 0.90 0.17

0.47 0.97 1.41 0.90 309.96

Diam (mm) (15) 1065 150 150 150 150 1065

Manhole Min tractive Max Max invert force depth velocity drop (m) (Pa) (mm) (m/s) (16) (17) (18) (19) − 11.4 373 2.76 1.53 3.40 5.24 4.19 1.07

40 54 54 49 780

1.15 1.13 1.4 1.23 0.23

− − − − 0.03

ine eri n

Fall in Upper sewer end (m) (m) (20) (21) − − 1.20 1.40 2.36 1.84 0.06

62.85 61.65 60.25 59.73 55.35

Lower Upper Lower end end end (m) (m) (m) (22) (23) (24) − 55.35 60.04 61.65 60.25 57.89 57.89 55.27

65.00 63.80 62.40 61.88 60.04

63.80 62.40 60.04 60.04 60.04

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 PFmax = 0.281Qavg2 = 0.281(0.000565)−0.44 = 7.55 −0.44 PFmin = 0.281Qavg1 = 0.281(0.000170)−0.44 = 12.8

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

w .E asy En g

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: −0.5697 Smin = 0.00624Qmin = 0.00624(2.22)−0.5697 = 0.00396

Sground =

ine eri n

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

g .n

et

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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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

w .E asy En g 6.11.

6.12.

6.13.

6.14.

6.15.

1

n = 0.039ks6 Determine whether the same Manning’s n should be used for the pipe flowing full and half full. 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. 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. 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. 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? 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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6.16.

Chapter 6

Design of Sanitary Sewers

2.0 Pa, and the sewage temperature as 20◦ C. The pipe must flow no more than 75% full under maximumflow conditions, and the maximum velocity must be less than 4.0 m/s. Any pipe slope greater than 0.1% is feasible. Determine the minimum allowable pipe slope for a 915-mm-diameter pipe. A sanitary-sewer system is to be designed for a newly developed 1700-ha area that is expected to have a population of 10,000 people when the sewers are first installed and a population of 50,000 people when the area is fully developed. Local regulations require that the sewer system be designed for an average per-capita flow rate of 250 L/d/person, accommodate an infiltration and inflow of 2 m3 /d/ha, not flow more than 75% full, be selfcleansing for 1.5-mm particles, have a minimum diameter of 150 mm, and have a maximum velocity less than 3 m/s. Field surveys indicate that the ground slope is 1% along the planned route of the sewer system. If the main pipe segment draining the area is to be made of concrete and have the same slope as the ground, determine the required (commercial-size) pipe diameter. A 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. 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. 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. 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.

ww

6.17.

6.18.

6.19.

6.20.

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.

w .E asy En g

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Length (m) (5)

Contributing area (ha) (6)

Upper end (m) (7)

Lower end (m) (8)

Line no. (1)

Location (2)

0

Main Street

—

5

—

—

—

60.04

1 2 3 4 5

A Street A Street A Street A Street Main Street

1 2 3 4 5

2 3 5 5 12

55 90 100 90 70

0.47 0.50 0.44 0.90 0.17

65.00 63.80 62.40 61.88 60.04

63.80 62.40 60.04 60.04 60.04

ww

w .E asy En g

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249

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C H A P T E R

7

Design of Hydraulic Structures 7.1 Introduction

ww

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.

w .E asy En g 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 250

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Culverts

251

FIGURE 7.1: Typical culverts: (a) single-barrel circular; (b) multibarrel rectangular (box)

(b)

(a)

ww

   D θ y= 1 − cos 2 2

w .E asy En g

(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 =



ine eri n 1.01 D0.26



Q2 g

0.25

g .n

(7.4)

et

for 0.2D … yc … 0.8D, where D is in m, Q is in m3 /s, and g is 9.81 m/s2 . The exact solution for the critical depth given by Equations 7.3 and 7.2 is generally preferable to using Equation 7.4. EXAMPLE 7.1

A 760-mm-diameter concrete culvert has a design flow rate of 1.0 m3 /s and is to be laid on a slope of 0.8%. Determine the normal and critical depths of flow. Solution From the given data: D = 0.760 m, Q = 1.0 m3 /s, and S0 = 0.008. Assume that n = 0.013 for concrete pipe. Substituting in Equation 7.1 gives  5 θ − sin θ 3 θ

2 3

= 20.16

nQ 8 D3 S

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 0.760 θ 4.396 = = 0.603 m 1 − cos 1 − cos y= 2 2 2 2

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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= = = 0.615 m 1 − cos 1 − cos 2 2 2 2

Therefore, the critical depth of flow is 0.615 m. An approximation of the critical depth can be obtained by using Equation 7.4, which gives  0.25   2 0.25 

1.01 (1.0)2 Q 1.01 = = 0.613 m yc = g 9.81 D0.26 (0.760)0.26

ww

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 g

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

ine eri n

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)

Downloaded From : www.EasyEngineering.net Section 7.2 FIGURE 7.2: Flow through culvert with submerged entrance

HW

Culverts

253

TW Δh

H V

D

(a) Submerged outlet (Type 1)

1

3

HW Δh

ww

TW

H

yn

V

w .E asy En g L

1

3 2

(b) Normal depth > Barrel height (Type 2)

HW

h

H

ine eri n

TW

(c) Entrance control, Normal depth < Barrel height (Type 3)

g .n

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

et

(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 ho = ko α − αd (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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Chapter 7

Design of Hydraulic Structures

FIGURE 7.3: Flow through culvert with free entrance

HW Δh H

yn

yc

(a) Mild slope, Low tailwater (Type 4)

1

HW Δh

ww

H

yc

yn

1.4yc

w .E asy En g 1

(b) Steep slope, Low tailwater (Type 5)

HW

Δh

H

1

yc

ine eri n

(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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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:   2gh Q = A   2gn2 L + ke + 1 4 R3

(7.12)

where A is the cross-sectional area of the culvert. It should be noted that h is equal to the difference between the headwater and tailwater elevations for a submerged outlet (Type 1 flow), and h is equal to the difference between the headwater and the crown of the culvert exit for an unsubmerged outlet when the normal depth of flow in the culvert exceeds the height of the culvert (Type 2 flow). If the Darcy–Weisbach equation is used to calculate the head loss in the culvert, then Equation 7.12 takes the form

ww

  2gh  Q = A  fL + ke + 1 4R

w .E asy En g

(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.

ine eri n

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

(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

et

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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 g

ine eri n

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 = (7.19) Fr2 + Cc 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

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

w .E asy En g

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

ine eri n 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

g .n

et

257

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ww

TABLE 7.1: (Continued)

Shape and material Ellipse concrete

Arch CM†

Inlet shape

c

w .E asy En g

Circular Elliptical inlet face

Rectangular concrete Rectangular concrete Rectangular concrete

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

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

Y 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

ine eri n 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

g .n

et

Source: U.S. Federal Highway Administration (2012). Notes: *CMP = corrugated metal pipe; † CM = corrugated metal.

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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 Δh H

yn

D

w .E asy En g yc

TW

1 (D+y ) c 2

L

1

3 2

than Equation 7.15 is that Equation 7.19 contains less parameters and is capable of estimating flows with comparable accuracy to Equation 7.15. Whereas Types 1 to 3 flows include most design circumstances when the culvert entrance is submerged, it is important to keep in mind that other submerged-entrance flow regimes are possible. For example, a scenario in which the culvert entrance is fully submerged, the tailwater is low, and the culvert flows partially full at the exit is certainly possible, even in cases where the normal depth of flow in the culvert exceeds the culvert diameter. This scenario is illustrated in Figure 7.4. In these circumstances, the culvert flow can be approximated by assuming that the hydraulic grade line at the culvert exit is at a point halfway between the critical depth, yc , and the height of the culvert, D, and then using Equation 7.12 with h equal to the difference between the assumed hydraulic grade line elevation at the exit and the headwater elevation (USFHWA, 2012). In cases where the actual tailwater elevation exceeds the assumed hydraulic grade line elevation at the exit, the actual tailwater elevation should be used in this approximation. The flow regime shown in Figure 7.4 can be designated as Type 2A flow and provides a reasonable approximation to reality as long as the culvert flows full over at least part of its length. If the culvert does not flow full over most of its length, the actual backwater profile in the culvert should be calculated. Type 2 and Type 2A flow regimes are both possible under submerged-entrance and low-tailwater conditions; however, a more conservative estimate of the culvert capacity is achieved by assuming the Type 2 flow shown in Figure 7.2 versus Type 2A flow shown in Figure 7.4. 7.2.1.2

ine eri n

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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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 2g 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.

w .E asy En g

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 (7.22) h + 1 − 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:

ine eri n



  2  V Q = Ac 2g h + 1 − hi 2g

(7.23)

g .n

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 )

et

(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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⎧ ⎪ 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

w .E asy En g

ine eri n

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

g .n

et

(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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Chapter 7

Design of Hydraulic Structures

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  − 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.

et

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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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.

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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 Concrete pipe

Wall and joint description Good joints, smooth walls Good joints, rough walls Poor joints, rough walls Badly spalled

n 0.011–0.013 0.014–0.016 0.016–0.017 0.015–0.020

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Concrete box

Good joints, smooth, finished walls Poor joints, rough, unfinished walls

0.012–0.015 0.014–0.018

Spiral rib metal pipe

19-mm * 19-mm recesses at 30-cm spacing, good joints

0.012–0.013

Corrugated metal pipe, pipe arch, and box

68-mm * 13-mm annular corrugations 68-mm * 13-mm helical corrugations 150-mm * 25-mm helical corrugations 125-mm * 25-mm corrugations 75-mm * 25-mm corrugations 150-mm * 50-mm structural plate 230-mm * 64-mm structural plate

0.022–0.027 0.011–0.023 0.022–0.025 0.025–0.026 0.027–0.028 0.033–0.035 0.033–0.037

Polyethylene

Corrugated Smooth

0.018–0.025 0.009–0.015

PVC

Smooth

0.009–0.011

Source: U.S. Federal Highway Administration (2005a; 2012).

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Design of Hydraulic Structures 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

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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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et

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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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.

w .E asy En g 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:

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Submerged entrance. When the culvert entrance is submerged, the flow is Type 1, 2, or 3. The flow type and the associated flow rate are determined by the following procedure:

Step 1: Determine whether the culvert exit is submerged.

• If the culvert exit is submerged, then the flow is Type 1 and the culvert capacity is given by Equation 7.12. • If the culvert exit is not submerged, then the flow is either Type 2 or 3. Continue to Step 2.

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Step 2: Assume that the flow is Type 2 and calculate the discharge using Equation 7.12. Determine the corresponding normal depth of flow in the culvert. • If the normal depth of flow is greater than the height of the culvert, then Type 2 flow is confirmed and the calculated discharge is the culvert capacity. • If the normal depth of flow is less than the height of the culvert, then the flow is probably Type 3. Continue to Step 3. Step 3: Assume that the flow is Type 3 and calculate the discharge using Equation 7.14 or 7.15. Determine the corresponding normal depth of flow in the culvert. • If the normal depth of flow is less than the culvert height, then Type 3 flow is confirmed and the calculated discharge is the culvert capacity. • If the normal depth of flow is greater than the culvert height, then neither Type 2 nor Type 3 flow can be confirmed and some intermediate flow regime is probably occurring. The culvert capacity should be taken as the lesser of the discharges calculated using the Type 2 and Type 3 discharge equations. It is useful to note that circular culverts always flow full when the discharge exceeds 1.07 Qfull , where Qfull is the full-flow discharge calculated using the Manning equation.

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Unsubmerged entrance. When the culvert entrance is unsubmerged, the flow is possibly Type 4, 5, or 6. The flow type and associated flow rate are determined by the following procedure: Step 1: Assume that the flow is Type 4; use Equation 7.21, along with the critical-flow equation (Equation 4.100), to calculate the discharge, Q, and the critical depth, yc . Use Q in the Manning equation to calculate the normal depth, yn . • If yn Ú yc and the tailwater depth is less than yc , then Type 4 flow is confirmed and the calculated discharge is the culvert capacity. • Otherwise, continue to Step 2. Step 2: Assume that the flow is Type 5; use Equation 7.23, along with the critical-flow equation (Equation 4.100), to calculate the discharge, Q, and the critical depth, yc . Alternatively, Equation 7.26 can be used to determine Q. Use Q in the Manning equation to calculate the normal depth, yn .

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• 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.

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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.

6m

1.2

2m S⫽

0.5 %

Q

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(a) For free-outlet conditions, two types of flow are possible: the normal depth of flow is greater than the culvert height (Type 2) or the normal depth of flow is less than the culvert height (Type 3). To determine the flow type, assume a certain type of flow, calculate the discharge and depth of flow, and see if the assumption is confirmed. If the assumption is not confirmed, then the assumed flow type is incorrect. Assuming Type 2 flow, then the flow-rate equation, Equation 7.12, is given by   2g h  (7.31) Q = A  2gn2 L + ke + 1 4 R3

where h is the difference in water levels between the entrance and exit of the culvert [= 1.83 + 0.005(36.6) − 1.22 = 0.793 m], n is the roughness coefficient (= 0.013), L is the length of the culvert (= 36.6 m), and R is the hydraulic radius given by R=

ww

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)  2(9.81)(0.013)2 (36.6) + 0.05 + 1 4 which simplifies to

(0.305) 3

ine eri n Q = 4.59 m3 /s

The next step is to calculate the normal depth of flow at a discharge of 4.59 m3 /s using the Manning equation, where 2 1 1 Q = AR 3 S02 n and S0 is the slope of the culvert (= 0.005). If the normal depth of flow is yn , then the flow area, A, wetted perimeter, P, and hydraulic radius, R, are given by A = byn = 1.22yn P = b + 2yn = 1.22 + 2yn R=

1.22yn A = P 1.22 + 2yn

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The Manning equation gives 5

4.59 =

1 (1.22yn ) 3 1 (0.005) 2 0.013 (1.22 + 2y ) 23 n

which yields yn = 1.25 m

Therefore, the initial assumption that the normal depth of flow is greater than the height of the culvert (= 1.22 m) is verified, and Type 2 flow is confirmed. The flow rate through the culvert is equal to 4.59 m3 /s. A variation of Type 2 flow could exist where flow in the culvert transitions from full flow to the tailwater depth before exiting the culvert, this was referred to previously as Type 2A flow. Under Type 2A flow conditions, the flow depth at the culvert exit is taken as equal to (yc + D)/2

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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 yc + D 0.409Q 3 + 1.22 = = 0.205Q 3 + 0.61 2 2   2 yc + D = 1.83 + 0.005(36.6) − (0.205Q 3 + 0.61) h = HW + S0 L − 2 2

= 1.403 − 0.205Q 3

ww

Substituting into Equation 7.12 using the previously calculated full-flow parameters gives      2(9.81)(1.403 − 0.205Q 32 ) 2gh   = (1.49) Q=A   2gn2 L  2(9.81)(0.013)2 (36.6) + 0.05 + 1 + ke + 1 4 4

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(0.305) 3

R3

which yields Q = 4.69 m3 /s. This flow rate is about 2% higher than the flow rate determined under Type 2 conditions (i.e., 4.59 m3 /s). These results confirm that assuming Type 2 conditions rather than Type 2A conditions will generally result in conservative estimates of culvert capacity. Furthermore, the computations for Type 2 flow are much less demanding than for Type 2A flow, particularly if computation of the critical flow depth in a circular conduit is required. In the present problem the culvert was rectangular and so this latter complication did not occur. (b) In this case, the tailwater is 0.304 m above the crown of the culvert at the outlet, and therefore the difference in water levels between the inlet and outlet, h, is 1.83 m − 1.22 m + 0.005(36.6) m − 0.304 m = 0.489 m. The flow equation in this case, Type 1 flow, is given by Equation 7.31 with h = 0.489 m, which gives   2(9.81)(0.489)  Q = (1.49)  2(9.81)(0.013)2 (36.6) + 0.05 + 1

ine eri n 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)  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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TABLE 7.5: Calculation of Headwater Depths for Various Flow Types

7.2.3.2

ww

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.

et

Solution From the given data: D = 0.915 m, Q = 1.70 m3 /s, L = 20 m, S0 = 0, ke = 0.2, and TW = 0.75 m. For a concrete culvert it can be assumed that n = 0.013. Since the culvert is horizontal, yn = q and since the exit is not submerged the only possible flow regimes are Types 2, 3, and 6. These are considered sequentially as follows: Type 2 Flow: For Type 2 flow, the difference between the headwater elevation and the crown of the culvert exit, h, is given by Equation 7.11. From the given data, π 2 D 4 Q = V= A D = R= 4 A=

π (0.915)2 = 0.6576 m2 4 1.70 = 2.585 m/s 0.6576 0.915 = 0.2288 m 4

=

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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

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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

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.

et

Type 6 Flow: For Type 6 flow, the headwater depth is calculated using Equation 7.29. Neglecting the headwater velocity, Equation 7.29 can be expressed as h +

V12 2g

−

V2 = h i + hf 2g ⎛

⎞2 nQ ⎠ L = ke + ⎝ (H − TW) + 0 − 2 2 2 3 2gA 2gA AR Q2

Q2

(7.32)

where A and R represent the average flow area and hydraulic radius at the entrance and exit of the culvert. Equation 7.32 is an implicit equation for the headwater depth, H, since A and R will also depend on H. Any attempt to solve this equation numerically will show that there is no solution for H … D and so Type 6 flow is not possible. Collectively, the results presented here have demonstrated that Type 2 flow is the likely regime, and this will require a headwater depth of 1.485 m when the culvert is passing the design flow rate.

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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:

ww

Type

Headwater depth (m)

2 3 6

1.485 1.318 N/A

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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

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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

(7.33)

et

where Cd is the discharge coefficient, L is the roadway length over the culvert, and Hr is the head of water over the crest of the roadway. The discharge coefficient, Cd , can be estimated from the head of water over the roadway (Hr ), the width of the roadway (Lr ), and the submergence depth downstream of the roadway (yd ), using the relations shown in Figure 7.6, where the discharge coefficient is expressed in the form Cd = kr Cr

(7.34)

where Cr is derived from Hr using either Figure 7.6(a) for Hr /Lr > 0.15 or Figure 7.6(b) for Hr /Lr … 0.15, and kr is derived from Figure 7.6(c) for a given value of yd /Hr . Values of Cr and kr derived from Figure 7.6 are used to calculate Cd using Equation 7.34, and this value of Cd is used in Equation 7.33 to calculate the flow rate over the roadway. Application of the weir equation (Equation 7.33) to describe the flow over a roadway does not take into account the effect of rails on the sides of the roadway. In cases of overflowing bridges, rails have been found to have a significant effect of the head-discharge relationship under overflow conditions (Klenzendorf and Charbeneau, 2009). An iterative approach is usually required to determine the division of flow between the culvert and the roadway. This requires that different headwater elevations be assumed until the sum of the flow rates through the culvert and over the roadway is equal to the given total flow rate to be accommodated by the system.

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FIGURE 7.6: Discharge coefficient for roadway overtopping Source: USFHWA (2012).

Hr

Flow

yd Lr

1.70 1.65

1.75

Cr 1.55

vel

Gra

1.65

el

av

Gr

1.60

Paved

1.70 Cr

Paved

1.50 1.45

ww

1.40

1.60 5

6

8 7 Hr (cm)

9

0

10

0.2 0.4

0.6 0.8 1.0 1.2 Hr (m)

w .E asy En g

(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

ine eri n 0.9

(c) Submergence factor

EXAMPLE 7.5

1.0

g .n

et

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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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 2gh  (7.35) + Cd LR Hr2 Q = A  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 h = (Roadway elevation + Hr ) − Tailwater elevation

w .E asy En g

= (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)  + 0.5 + 1 4 (0.153) 3

which simplifies to

ine eri n

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 (7) Hr (m)

0.14 0.14

et

Column 1 is the assumed Hr in meters, Column 2 is Hr /Lr , Column 3 is Cr derived from Hr /Lr and Hr using Figure 7.6, Column 4 is yd /Hr , Column 5 is kr derived from yd /Hr using Figure 7.6, Column 6 is Cd obtained by multiplying Columns 3 and 5 and Column 7 is obtained by substituting Cd in Column 6 into Equation 7.37 and solving for Hr . The iterations indicate that Cd = 1.66, Hr = 0.14 m, and the flow rate over the roadway, Qr , is given by 3

3

Qr = Cd LR Hr2 = (1.66)(15.2)(0.14) 2 = 1.32 m3 /s The corresponding flow rate through the culvert is equal to 2.49 m3 /s − 1.32 m3 /s = 1.17 m3 /s. Therefore, a culvert diameter of 610 mm will result in roadway overtopping, with a flow rate of 1.17 m3 /s passing through the culvert, 1.32 m3 /s passing over the roadway, and a depth of flow over the roadway equal to 14 cm. A larger culvert diameter could be explored if less roadway overtopping at the design flow rate is desired.

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Design of Hydraulic Structures

7.2.5

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):

w .E asy En g

Type of Stone. Stones used for riprap should be hard, durable, and angular. Angularity, a feature of crushed stone from a quarry, helps to keep the stones locked together when subjected to the force of moving water. TABLE 7.6: Scour Velocities of Various Soils

Permissible Velocity

Soil

(m/s)

ine eri n

Sand Sandy loam Silt loam Sandy clay loam Clay loam Clay, fine gravel Cobbles Shale

FIGURE 7.7: Culvert exit with riprap apron

(ft/s)

0.5 0.8 0.9 1.1 1.2 1.5 1.7 1.8

1.6 2.6 3.0 3.6 3.9 4.9 5.6 5.9

g .n

et

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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): d50

0.044 = TW



Q D

4 3

(7.38)

where Q is the design flow rate [m3 /s], D is the culvert width or pipe diameter [m], and TW is the tailwater depth [m]. In cases where the pipe does not discharge into a well-defined channel, TW is taken as the flow depth in the pipe at the outlet. Thickness of Stone Lining. The thickness of the blanket of stones should be three times the median stone size if no filter fabric liner between the stones and the ground is used. If a filter fabric liner is used, the thickness should be twice the median stone size.

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 La = (7.39) ⎪ 3.26Q ⎪ ⎪ ⎪ + 7D TW < D/2 ⎩ 3 D2

w .E asy En g

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

ine eri n

g .n

et

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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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

ww FIGURE 7.10: Tainter gates Source: U.S. Army Corps of Engineers (2011).

w .E asy En g

ine eri n

g .n

et

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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Gate

277

Energy grade line

V12 2g

Gate

y1

V22 2g

V1 Flow

y1 yg

θ

y2 = Cc yg

y2 = Cc yg

yg

(a) Vertical gate

ww

Gates

(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 Q2

w .E asy En g y1 +

2gb2 y21

= y2 +

Q2 2gb2 y22

and solving for Q gives

Q = by1 y2



2g y1 + y2

(7.42)

where b is the width of the gate. The depth of flow downstream of the gate, y2 , is less than the gate opening, yg , since the streamlines of the flow contract as they move past the gate (see Figure 7.11). The downstream location where the depth of flow is most contracted is called the vena contracta, and y2 denotes the depth at the vena contracta. Denoting the ratio of the downstream depth, y2 , to the gate opening, yg , by the coefficient of contraction, Cc , where

ine eri n Cc =

y2 yg

g .n

(7.43)

et

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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Cc = 0.61

(7.46)

whenever 0 15, the weir is called a sill, the discharge can be computed from the critical-flow equation by assuming yc = H + Hw (Chaudhry, 1993), and the discharge coefficient is given by (Jain, 2001) 3  Hw 2 Cd = 1.06 1 + (7.71) H

w .E asy En g

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

ine eri n

(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

(7.73)

et

where Cw is called the weir coefficient and is related to the discharge coefficient by 2 Cw = 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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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 g

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)

ine eri n

⎡

Qs = ⎣1 − Q



yd H

3 2

⎤0.385 ⎦

g .n

(7.78)

et

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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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 g

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

The crest height, Hw , is therefore given by

2

3



1 1.83(4)

2 3

ine eri n =

= 0.265 m

Hw = 1 − 0.265 = 0.735 m and

g .n

0.265 H = 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

et

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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7.4.1.2

V-notch weirs

A V-notch weir is a sharp-crested weir that has a V-shaped opening instead of a rectangularshaped opening. These weirs, also called triangular weirs, are typically used instead of rectangular weirs when lower discharges are desired for a given head, H, or where lower flow rates are to be measured with greater accuracy than can be achieved with rectangular weirs. V-notch weirs are usually limited to flow rates of 0.28 m3 /s (10 cfs) or less, and are frequently found in small irrigation canals. An operating V-notch weir is shown in Figure 7.18. The basic theory of V-notch weirs is the same as that for rectangular weirs, where the theoretical flow rate over the weir, Q, is given by ! H Q= v2 b dz2 0

ww

=

!

0

H

⎡

⎤ 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 g

(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

!

0

H

⎡

⎤ 21   V12 ⎣2g H − z2 + ⎦ 2z2 tan θ dz2 2g 2

ine eri n

g .n

(7.82)

et

Channel

b dz2 H

θ

z2

Weir Hw

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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   $  % 21 θ 2g H − z2 2z2 tan dz2 2 0   5 8 θ H2 = 2g tan 15 2

Q=

!

H

(7.83)

As in the case of a rectangular weir, the theoretical discharge given by Equation 7.83 is corrected by a discharge coefficient, Cd , to account for discrepancies in the assumptions leading to Equation 7.83. The actual flow rate, Q, over a V-notch weir is therefore given by

ww

  5 θ 8 Q= H2 Cd 2g tan 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 g

ine eri n 1.19

Cd = 0.583 +

(7.85)

1

(ReWe) 6

where Re is the Reynolds number defined by

1

Re = FIGURE 7.20: Discharge coefficient in V-notch weirs

1

g2 H 2 ν

0.70

Discharge coefficient, Cd

0.68

g .n

(7.86)

et

θ = 10o 0.66 0.64 20o 0.62 45o

0.60 Cd = 0.581

0.58 5

10

90o

60o

15

20

25

30

Head, H (cm)

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Chapter 7

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and We is the Weber number defined by We =

ρgH 2 σ

(7.87)

where ν is the kinematic viscosity and σ is the surface tension of water. EXAMPLE 7.9 A V-notch weir is to be used to measure channel flow rates in the range 0.1–0.2 m3 /s. What is the maximum head of water on the weir for a vertex angle of 45◦ ? Solution The maximum head of water occurs at the maximum flow rate, so Q = 0.2 m3 /s will be used to calculate the maximum head. The relationship between the head and flow rate is given by Equation 7.84, which can be put in the form

ww

H=





15Q

2

5

=

w .E asy En g 8Cd 2g tan (θ/2)



8Cd



2 5

15(0.2) 2(9.81) tan (45◦ /2)

=

0.530 2

m

Cd5

The discharge coefficient as a function of H for θ = 45◦ is given in Figure 7.20, and some iteration is necessary to find H. These iterations are summarized in the following table: Assumed H (cm)

Cd (Figure 7.20)

Calculated H (cm)

12 64.9 65.8

0.6 0.581 0.581

64.9 65.8 65.8

ine eri n

Therefore, the maximum head expected at the V-notch weir is approximately 66 cm.

g .n

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

et

(7.88)

where Cd and k are both functions of the notch angle, θ . LMNO Engineering and Research Software (LMNO, 1999) developed the following analytic functions for Cd and k to match the graphical functions recommended by Kindsvater and Shen (USBR, 1997): Cd = 0.6072 − 0.000874θ + 6.1 * 10−6 θ 2 k = 4.42 − 0.1035θ + 1.005 * 10−3 θ 2 − 3.24 * 10−6 θ 3

(7.89) (7.90)

where θ is in degrees and k is in millimeters. The Kindsvater and Shen equation, Equation 7.88, has been recommended by several reputable organizations and researchers (International Organization for Standardization, 1980; American Society for Testing and Materials, 1993; U.S. Bureau of Reclamation, 1997; Martinez et al., 2005).

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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 

15Q

2

5



w .E asy En g H=

8Cd 2g tan (θ/2)

− k=



15(0.2)

8(0.580) 2(9.81) tan (45◦ /2)

2 5

− 0.0015 = 0.657 m

Therefore, the maximum head expected at the V-notch weir is approximately 0.66 m. This is the same result that was obtained using the graphical method. Since the Kindsvater–Shen equation does not require iteration, it is clearly preferable in this case.

The following guidelines have been suggested by the U.S. Bureau of Reclamation (USBR, 1997) for using V-notch weirs to measure flow rates in open channels:

ine eri n

⊲ 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 25 instead of 23 , 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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FIGURE 7.21: Compound weir

b

H1

θ

ww

H2

w .E asy En g

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:

ine eri n

   3 2 H2 2 Q = Cd 2gbH1 + Cvd A 2g H1 + 3 3

g .n

(7.92)

et

where Cd is the discharge coefficient for the rectangular weir, given by Equation 7.70; H1 and H2 are the heights shown in Figure 7.21; and Cvd is typically taken as 0.6 (McCuen, 1998). EXAMPLE 7.11

The outlet structure from a stormwater detention area is a compound weir consisting of a rectangular weir with a V-notch in the middle. The notch angle is 90◦ , the height of the notch is 20 cm, and the apex of the notch is placed at the bottom of the detention area. Determine the required crest length of the rectangular weir such that the discharge through the compound weir is 3.0 m3 /s when the water depth in the detention area is 1.00 m. Solution From the given data: θ = 90◦ , H2 = 20 cm = 0.20 m, and H1 = 1.00 − H2 = 1.00 − 0.20 = 0.80 m. The discharge, Q, over the weir is given by Equation 7.92 as    3 2 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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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    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 g

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

ine eri n

g .n

et

∗ Developed by Harry Sutro in the early 1900s.

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FIGURE 7.23: Wooden weir in a coastal drainage channel

ww

w .E asy En g

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

ine eri n

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 over the weir, Q, is given by Q = yc bVc

(7.95)

where yc is the critical depth of flow over the weir, b is the width of the weir, and Vc is the velocity at critical flow. If E1 is the specific energy of the flow at Section 1 just upstream of the weir, and the energy loss between this upstream section and the critical-flow section over the weir is negligible, then the energy equation requires that

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Weirs

295

1 Energy grade line 2

V1

2g

H Q

h1

Critical flow

Q

Hw L Longitudinal section view

ww

E1 = Hw + yc +

w .E asy En g

Vc2 2g

(7.96)

where Hw is the height of the weir crest above the upstream channel. Under critical-flow conditions, the Froude number is equal to 1, hence Vc =1 gyc

(7.97)

Combining Equations 7.96 and 7.97 yields the following expression for yc : yc =

2 2 (E1 − Hw ) = H 3 3

ine eri n

(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

(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 2 √ Q = gb H 3

et

(7.100)

In reality, the energy loss over the weir is not negligible and the estimated flow rate, Q, must be corrected to account for the energy loss. The correction factor is the discharge coefficient, Cd , and the actual flow rate, Q, over the weir is given by

Q = Cd

√

3  2 2 gb H 3

(7.101)

where values of Cd can be estimated using the relation (Chow, 1959) Cd =

0.65 1

(7.102)

(1 + H/Hw ) 2

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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 g Q=



√ Cd′ gb

2 h1 3

3

2

(7.104)

where the discharge coefficient Cd′ in Equation 7.104 is different (typically higher) from the Cd in Equation 7.101. Using Equation 7.104, values of 0.80 < Cd′ < 0.85 have been reported for properly operated broad-crested weirs. It has been further suggested that Cd′ can be approximated as a constant provided that 0.08 < h1 /L < 0.33 and 0.18 < h1 /(h1 + Hw ) < ¨¸ 0.36, with the constant Cd′ being taken as a fixed value within the range of 0.80–0.85 (Go¨ g˘ us et al., 2007). A more refined estimate of Cd′ within this range has been suggested by Azimi and Rajaratnam (2009) as 2    h1 h1 + 0.89 (7.105) Cd′ = 0.95 − 0.38 h1 + Hw h1 + Hw

ine eri n

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.

Sharp crested

The flow separates at the upstream end and does not reattach to the crest. The flow is unstable.

h1 /L > 1.5

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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 V2 Q2 Q2 = 0.30 + 0.0510Q2 = 0.30 + H = h 1 + 1 = h1 + 2g 2(9.81)(0.5 * 2)2 2gA21

w .E asy En g The discharge coefficient, Cd , is given by Cd =

0.65

(1 +

1 H/Hw ) 2

=

0.65

[1 + (0.30 +

1 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 0.65 2 2 √ 2) 2 H (0.30 + 0.0510Q = Q = Cd gb 9.81(2) 1 3 3 [2.5 + 0.255Q2 ] 2  1 (0.30 + 0.0510Q2 )3 2 = 2.22 2.5 + 0.255Q2

which yields

ine eri n

Q = 0.23 m3 /s

g .n

This solution assumes that the length of the weir is such that 0.08 < h1 /L < 0.33.

7.4.2.2

Compound weirs

et

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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through the broad-crested weir is given by Equation 7.101, and when the critical flow depth is in the high-flow section (yc > z), the discharge relationship is determined using the criticalflow relation Q2 A3 = c g B

(7.107)

where B is the width of the high-flow section, Ac is the flow area given by Ac = bz + (yc − z)B

(7.108)

and b and z are the width and depth, respectively, of the low-flow section. Combining Equations 7.107 and 7.108 with the energy and continuity equations, and applying a discharge coefficient to account for theoretical discrepancies gives the discharge through the compound weir under high-flow conditions as

ww

 23  1   g 2 bz 2z 2 H − − Q = Cd bz + B B 3 3B 3

w .E asy En g

(7.109)

where H is the specific energy upstream of the weir. Values of the discharge coefficient, Cd , depend on H/L and b/z, and a (graphical) relationship based on laboratory experiments can ¨ ¸ et al. (2006). be found in the work of Go¨ g˘ us As an alternative to using Equation 7.109, it has been shown that the discharge through compound broad-crested weirs can be accurately approximated as the sum of the discharges through subsections of the weir using the conventional head-discharge relationships for noncompound weirs (Jan et al., 2009). For example, the compound weir in Figure 7.25 could be modeled as three simple broad-crested weirs: one center weir and two “shoulder” weirs. In practical applications, it is desirable to provide a small transverse slope on the bottom of the high-flow rectangular section (called the “shoulder” of the weir). This feature in which the shoulder slopes toward the center of the weir provides a more predictable dischargeversus-head relationship as the flow depth transitions from the notch section to the high-flow section. Furthermore, the use of an inclined-ramp approach to the control section of a compound broad-crested weir is commonly used to increase the discharge coefficient relative to using the blunt approach illustrated in Figure 7.25. Computer programs are available to facilitate the design and calibration of compound broad-crested weirs, particularly with respect to estimating the discharge coefficient (e.g., Wahl et al., 2007). 7.4.2.3

ine eri n

Gabion weirs

g .n

et

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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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 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.

w .E asy En g 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)

ine eri n

g .n

et

∗ An ogee refers to an S-shaped profile.

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FIGURE 7.26: Ogee-crest spillway

Approach head, he Pool elevation Hd

Design head

He + x-axis

O B

PT (B − y) 2 x2 + =1 2 A B2

y

A

Hd + y -axis

=

x Hd

n

Point of tangency (PT)

Crest axis

1

P

a

1

ww

1 K

Fs

w .E asy En g For He Ú 9.1 m (30 ft) :

For He < 9.1 m (30 ft) :

Hd =

&

0.33He1.22 0.30He1.26

(without piers) (with piers)

ine eri n Hd =

&

0.7He 0.74He

(without piers) (with piers)

(7.114)

(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

g .n

et

(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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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

0.2

w .E asy En g 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

ine eri n

x2 (B − y)2 + =1 A2 B2

g .n

(7.118)

et

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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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.

w .E asy En g

ine eri n

EXAMPLE 7.13

g .n

et

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 1 He = 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 C= C0 = (1.05)(2.18) = 2.29 C0

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Downloaded From : www.EasyEngineering.net 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

1.0

1.5

2.0

2.5

3.0

P Value of Hd (a) Basic discharge coefficient

w .E asy En g Ratio of coefficients,

Cinclined Cvertical

1.04

1:1

Slope (H :V)

1.02

2:3

1:3

18°26

2:3

33°41

1:1

45°00

1:3

1.00

0.98

Angle with the vertical

Hd P

0

0.5

ine eri n 1.0

1.5

P Value of Hd

(b) Correction factor for sloping upstream face 1.1 C Ratio of coefficients, C0

ww

0.5

1.0 h0 H

Hd

0.9

g .n

et

P

0.8

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

H Ratio of head on crest to design head, Hd (c) Correction factor for other than design head

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303

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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 and the approach velocity, v0 , can be estimated by

ww

v0 =

1500 Q = = 0.5 m/s Le d (60)(50)

with a velocity head, h0 , given by

w .E asy En g

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

ine eri n

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 x y 1 = Hd K Hd 1 y = 3.44 2.0



 x 1.85 3.44

g .n

et

which simplifies to y = 0.175x1.85

Since the downstream slope of the spillway is 1 : 2 (H : V), then α = 2 and the horizontal distance from the apex to the downstream tangent point, XDT , is given by Equation 7.117 as XDT = 0.485(Kα)1.176 Hd XDT = 0.485(2.0 * 2)1.176 3.44 which gives XDT = 8.52 m

Therefore, at a distance of 8.52 m downstream of the crest, the curved spillway profile merges into the linear profile of the spillway chute, which has a slope of 1 : 2 (H : V). The corresponding y value is 1.85 = 0.175(8.52)1.85 = 9.21 m. y = 0.175XDT

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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.

w .E asy En g EXAMPLE 7.14

ine eri n

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

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

et

2

Va

2g

hd He

Va

d P Spillway

Apron

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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

w .E asy En g 0.7

1

15

0.5

hd He

ww

8

0.8

0.6

2

20

0.5

1

0.4

2

ine eri n 4

6

3

0.3

3

4

8

6

10

15

15 0.1

20

20 30

30 40 50 60 80

40 50 60 80 0.0 1.0

g .n

10

8

0.2

1.5

2.0

2.5

3.0 hd + d He

3.5

4.0

et

4.5

5.0 A

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Spillways

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

⎤3

2

Q20 2gA2

⎡

⎦ = (2.18)(3.0) ⎣1.0 +

⎤3 2

Q20 2(9.81)(4.5 * 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

w .E asy En g

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 Q2 2 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)

ine eri n

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

et

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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FIGURE 7.32: Gate locations on spillway

Lip Lip Crest of spillway

Crest

Reservoir

Reservoir Gate seat

Gate seat

(b) Gate downstream of crest

ww

7.5.2.1

(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

w .E asy En g

⎧ ⎨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

ine eri n

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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Downloaded From : www.EasyEngineering.net Section 7.5 FIGURE 7.33: Gated spillway

Spillways

309

Vertical-lift gate

1.00 m 0.50 m

0.25 m

Spillway

ww

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

w .E asy En g 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

ine eri n

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

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

et

(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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FIGURE 7.34: Tainter gate on spillway

0.95 0.90

θ

H

h

0.85

R a

0.80 Cd

Strut Trunnion

Lip

X/Hd

Gate seat

0.75

X/Hd = 0.1−0.3

0.70

X/Hd = 0.0

0.65 0.60 0.55

ww

50

55

60

65

70

75

80

85

90

95

100

105

110

θ Degrees

w .E asy En g 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     2.10 − 1.00 a − h = cos−1 = 66◦ θ = cos−1 R 2.75

ine eri n

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

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

et

2.75 m 1.00 m 1m

0.87 m

2.10 m Spillway

2.00 m

Spillway 30°

tangent

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Downloaded From : www.EasyEngineering.net 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)

w .E asy En g

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

ine eri n

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

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

et

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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Chapter 7

Design of Hydraulic Structures 3

3

H2 − H2 Q = 2 3 1 Q0 H02 3

3

(1.90) 2 − (1.40) 2 Q = 3 10 (1.65) 2 which yields a gate discharge of Q = 4.5 m3 /s.

7.6 Stilling Basins

ww

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.

w .E asy En g 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.

ine eri n

1.7 < Fr 4.5: For Froude numbers greater than 4.5, either Type III or Type II basins are recommended. The Type III basin, shown in Figure 7.40, includes baffle blocks, and so is limited to applications where the incoming velocity does not exceed 18 m/s (60 ft/s). For inflow velocities exceeding 18 m/s (60 ft/s), the Type II basin, shown in Figure 7.41, is recommended. The Type II basin is slightly longer than the Type III basin, has no baffle blocks, has a dentated end sill, and the tailwater is recommended to be 5% greater than the conjugate depth (of the hydraulic jump) to help prevent sweepout.

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).

w .E asy En g 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.

ine eri n

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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Stilling Basins

FIGURE 7.40: Type III stilling basin characteristics, Fr > 4.5 and V1 … 18 m/s Source: U.S. Bureau of Reclamation (1987).

w1 = y1 S1= y1

End sill

0.2h3

Chute blocks

Baffle blocks 0.375h3 w3 = 0.75h3

0.5y1

S3 = 0.75h3 h1 = y1 2:1 Slope 1:1 Slope

h4

0.8y2 L

w .E asy En g

(a) Type III basin dimensions

24

Froude number 10 12

8

14

16

. = T.W y2

16

12

8

20

0

1.

ine eri n y2 1 ⫽ ( y1 2

1 + 8Fr2 − 1 )

(b) Minimum tailwater depths

16

12

g .n 8

Baffle block height h3 End sill height h4

2

0

2

et

h3 or h4 y1 y1

4

4 h3 or h4 y1 y1

18 24

Note: y1 and y2 are the initial and sequent depths of the hydraulic jump in the stilling basin.

20

T.W. depth y1

6

T.W. depth y1

4

0

(c) Height of baffle blocks and end sill

3 L y2

3 L y2

ww

h3

2

4

6

8

10 12 Froude number

14

16

2 18

(d) Length of jump

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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

Slope = 2:1

h2 = 0.2y2

h1 = y1 L

Froude number 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

20

T.W. depth y1

20

05

. = T.W y2

16

12

8

4

1.

y2 1 ⫽ ( y1 2

ine eri n 1 + 8Fr2 − 1 )

16

12

g .n 8

4

(b) Minimum tailwater depths

5

5

4

4

3

4

6

8

10 12 Froude number

T.W. depth y1

w .E asy En g 4

L y2

ww

(a) Type II basin dimensions

14

16

18

et

L y2

316

3

(c) Length of jump

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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

ww

The maximum design discharge over a spillway is 280 m3 /s, and the spillway and stilling basin are 12 m wide. The reservoir behind the spillway has a water-surface elevation of 60.00 m, and the river watersurface elevation downstream of the stilling basin is 30.00 m. Assuming a 10% loss of hydraulic head in the flow down the spillway, find the elevation of the floor of the stilling basin so that the hydraulic jump forms in the basin. Design the stilling basin. Solution From the given data: Q = 280 m3 /s and b = 12 m. A schematic diagram showing the spillway, stilling basin, and downstream river is given in Figure 7.42. Since the water loses 10% of its hydraulic head flowing down the spillway, and the water-elevation behind the spillway is 60 m, then, taking the elevation of the bottom of the stilling basin as X, the specific energy, E1 , at entrance to the stilling basin (i.e., base of the spillway) is given by

w .E asy En g

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 +

ine eri n

57 − X = y1 +

which simplifies to

Q2 2g(by1 )2

2802 2(9.81)(12y1 )2

X = 57 − y1 − FIGURE 7.42: Stilling basin

27.75 y21

El.60.00 m

g .n

(7.129)

et

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 + 1 + ⎝−1 + 1 + ⎠= ⎠ 3 3 2 2 gy1 9.81y1 which simplifies to

⎞ ⎛  444.0 ⎠ y1 ⎝ −1 + 1 + y2 = 2 y31

(7.131)

Combining Equations 7.129 to 7.131 yields

ww

30 − 57 + y1 +

27.75 y21

which simplifies to

w .E asy En g 1.5y1 +

27.75 y21

⎞ ⎛  y1 ⎝ 444.0 ⎠ = −1 + 1 + 2 y31

y − 1 2



1 +

444.0 y31

− 27 = 0

The solutions of this equation are y1 = 27.1 m and y1 = 0.87 m. These depths correspond to subcritical and supercritical flows, respectively (Fr = 0.1 and Fr = 9.2). Since the upstream depth of flow must be supercritical, take y1 = 0.87 m, and Equation 7.129 gives X = 57 − 0.87 −

27.75 = 19.5 m 0.872

ine eri n

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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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). 7.7.1

Types of Dams

w .E asy En g

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.

ine eri n

Right abutment Downstream slope Upstream slope

g .n

et

Principal chute spillway Spillway training walls

Berm

Top of dam Riprap Toe drain outlet

Toe of embank m Emerg

ent

ency s

Left abutment pillway

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FIGURE 7.44: Major dams in the United States Source: U.S. Bureau of Reclamation (2012b; 2012c).

ww

w .E asy En g (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.

ine eri n

g .n

et

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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FIGURE 7.45: Bartlett Dam Source: U.S. Bureau of Reclamation (2012).

ww

w .E asy En g

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

ine eri n

g .n

et

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).

ww

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.

w .E asy En g 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.

ine eri n

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.

g .n

et

Active Storage. Active storage is the storage between the top of the dead storage pool and the normal reservoir water-surface elevation. The active storage capacity is sometimes referred to as the active conservation pool or conservation storage. This storage is FIGURE 7.47: Reservoir storage zones

Dam Maximum pool elevation Inflow

Spillway Flood storage

Normal pool elevation

Hydroelectric turbine

Active storage T

Minimum pool elevation

Low-level outlet Tailwater

Dead storage

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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.

ww

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.

w .E asy En g

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

ine eri n

g .n

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

et

(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

ww

0 0.001

0.01 0.1 1.0 10 Ratio of reservoir storage capacity to average annual inflow volume (years)

w .E asy En g

100

TABLE 7.8: Trap Efficiency Empirical Equations

Equation



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

C = reservoir capacity (m3 )

Reference Brown (1949)

W = watershed area (km2 )

ine eri n

Brune (1953)

I = annual inflow volume [L3 ] C = reservoir capacity [L3 ]

Dendy and Cooper (1974)

I = annual inflow volume [L3 ]

119.6(C/I) 0.012 + 1.02(C/I) 

Nomenclature

C = reservoir capacity [L3 ]

  log(C/I) TE = 100 0.970.19 TE = −22 +



−0.2



+ 12

C = reservoir capacity [L3 ]

I = annual inflow volume [L3 ] C = reservoir capacity (m3 ) I = inflow rate (m3 /s) L = reservoir length (m)

g .n

Heinemann (1981)

et

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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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 .

ww

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

w .E asy En g

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

ine eri n

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

g .n

sediment trap rate sediment volume accumulation rate = sediment bulk density =

1.24 * 1010 kg/year 1600 kg/m3

et

= 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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Chapter 7

Design of Hydraulic Structures (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.

7.7.2.2

ww

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.

w .E asy En g EXAMPLE 7.20

ine eri n

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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ww

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

w .E asy En g

ine eri n

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

g .n

et

−0.10 −0.15 −0.20 −0.25 Required storage

−0.30 −0.35 −0.40 −0.45 −0.50

2

4

6

8

10

12 14 Month

16

18

20

22

24

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7.7.3

Hydropower

Hydroelectric power is an important source of energy, with hydropower responsible for about 15% of the electric energy generated in the United States and over 70% of electric energy generated in Brazil and Norway. There are two types of hydropower plants: (1) runof-river plants that use direct streamflow, and where energy output is directly related to the instantaneous flow rate in the river; and (2) storage plants that use a reservoir to store water, and where energy is produced using controlled water releases. Run-of-river plants use the sustained flow of a stream or river to turn the turbines for electricity generation. This type of hydropower plant usually has limited storage capacity and provides a continuous output of electricity. Storage plants use a reservoir of sufficient size to increase the amount of water available for power generation and to carry over water through dry periods. A pumpedstorage plant uses external power during low periods of demand to pump water back up the system to a headwater reservoir, and this water is subsequently used for hydropower generation during peak demand periods. This recycling of water represents an economic efficiency between peak and off-peak demand requirements.

ww

7.7.3.1

Turbines

Turbines are the central components of all hydropower facilities. Their role is to extract energy from flowing water and convert it to mechanical energy to drive electric generators. There are two basic types of hydraulic turbines: impulse turbines and reaction turbines.

w .E asy En g

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

ine eri n

g .n

et

FIGURE 7.50: Impulse turbine Source: Capture 3D, Inc., Costa Mesa, CA, and GOM mbH, Braunschweig, Germany (2012).

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Water surface

Dams and Reservoirs

329

Energy grade line hL

Reservoir Q Impulse turbine

he

H

Penstock

Tailwater

water elevation in the reservoir and the elevation of the nozzle. The nozzle is considered to be an integral part of the impulse turbine, and so the net head or effective head, he , on the turbine is equal to the static head minus the head loss, hL , and is given by

ww

(7.133)

he = H − hL

w .E asy En g

The head at the entrance to the nozzle is expended in four ways: (1) energy is lost in fluid friction in the nozzle, known as nozzle loss; (2) energy is lost in fluid friction over the buckets of the turbine runner; (3) kinetic energy is carried away in the water discharged from the buckets; and (4) energy is used in rotating the buckets. Therefore, the head directly available to the runner, hT , is given by

hT = he − kj

Vj2 2g

− k

v22 V2 − 2 2g 2g

ine eri n

(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

(7.135)

et

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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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

ww

It is proposed to install a small Pelton-wheel turbine system where the static head above the nozzle is 50 m, and the delivery line is 455-mm-diameter concrete pipe with a length of 150 m and a roughness height of 1 mm. The discharge nozzle has a diameter of 100 mm and a head-loss coefficient of 0.5. Model tests on the Pelton wheel show that the bucket friction loss coefficient is 0.3, the velocity of water relative to the bucket at the exit from the bucket is 1 m/s, and the absolute velocity of the water leaving the bucket is 5 m/s. Estimate the power that could be derived from the system and the hydraulic efficiency of the turbine. Solution From the given data: H = 50 m, D = 455 mm, L = 150 m, ks = 1 mm, Dj = 100 mm, kj = 0.5, k = 0.3, v2 = 1 m/s, and V2 = 5 m/s. The flow rate, Q, is determined by application of the energy equation between the upstream reservoir and the exit of the nozzle, which requires that

w .E asy En g H −

Q2 Q2 fL Q2 − kj = 2 2 D 2gA 2gAj 2gA2j

(7.139)

where f is the friction factor, and A and Aj are the cross-sectional areas of the pipe and jet, respectively. Using the Swamee–Jain equation (Equation 2.39) to relate f to Q, Equation 7.139 yields Q = 0.200 m3 /s. The corresponding velocities in the delivery pipeline and nozzle jet are V = 1.23 m/s and Vj = 25.4 m/s, and the friction factor is f = 0.0244. Using these derived data yields the following results: he = H − h L = H −

h T = h e − kj

ine eri n

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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331

FIGURE 7.52: Francis turbine Source: ANDRITZ (2012).

ww

w .E asy En g

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

ine eri n

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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FIGURE 7.53: Kaplan axial flow turbine Source: U.S. Army Corps of Engineers (2005g).

ww FIGURE 7.54: Reaction-turbine system

w .E asy En g Water surface

Energy grade line hL

Reservoir

Q

Penstock

H Turbine

ine eri n Runner V2 2g

Draft tube

Tailrace

V

g .n

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.

et

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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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

ww

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

w .E asy En g

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

(7.142)

3

(gh) 4

ine eri n

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

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

et

(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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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

ww

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.

w .E asy En g 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.

ine eri n

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

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

et

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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335

TABLE 7.9: Classification of Hydropower Capacity

FIGURE 7.55: Extraction of hydropower

Classification

Installed capacity (MW)

Conventional Small-scale Mini Micro

>15 1–15 0.1–1 xn ) = 1 − P(xi … xn )

(8.6)

The function P(xi … xn ) is called the cumulative distribution function and is commonly written as F(xn ), where F(xn ) = P(xi … xn ) (8.7) These definitions of the probability distribution function, f (xn ), and cumulative distribution function, F(xn ), can be expanded to describe several random variables simultaneously. As an illustration, consider the case of two discrete random variables, X and Y, with elements xp and yq , where p ∈ [1, N] and q ∈ [1, M]. The joint probability distribution function of X and Y, f (xp , yq ), is the probability of xp and yq occurring simultaneously. Such distributions are called bivariate probability distributions. Joint probability distributions involving more than one variable are generally referred to as multivariate probability distributions. Based

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Chapter 8

Probability and Statistics in Water-Resources Engineering

on the definition of the bivariate probability distribution, f (xp , yq ), the following conditions must necessarily be satisfied f (xp , yq ) Ú 0, 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

w .E asy En g

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

ine eri n

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

(8.12)

et

(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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8.2.3

F(x, y) =



g(x) =



h(y) =



x



y

Probability Distributions

f (x′ , y′ ) dx′ dy′

347

(8.16)

−q −q q

f (x, y′ ) dy′

(8.17)

f (x′ , y) dx′

(8.18)

−q q

−q

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

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)

w .E asy En g

−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

ine eri n

(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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Probability and Statistics in Water-Resources Engineering

σx2

ww

=

⎧ N ⎪ ⎪ ⎪ ⎪ ⎪ (xi − μx )2 f (xi ) ⎪ ⎨

(discrete)

i=1

⎪  ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

+q

−q

(x′ − μx )2 f (x′ ) dx′

(8.24) (continuous)

The square root of the variance, σx , is called the standard deviation of x and measures the average magnitude of the deviation of the random variable from its mean. Random outcomes occur that are either less than or greater than the mean, μx , and the symmetry of these outcomes about μx is measured by the skewness or skewness coefficient, which is the expected value of the function (x − μx )3 g(x) = (8.25) σx3 If the random outcomes are symmetrical about the mean, the skewness is equal to zero; otherwise, a nonsymmetric distribution will have a positive or negative skew. There is no universal symbol that is used to represent the skewness, but in this text skewness will be represented by gx . For discrete and continuous random variables,

w .E asy En g gx =

⎧ N ⎪ ⎪ 1  ⎪ ⎪ (xi − μx )3 f (xi ) ⎪ ⎪ ⎨ σx3

(discrete)

i=1

⎪  +q ⎪ ⎪ 1 ⎪ ⎪ (x′ − μx )3 f (x′ ) dx′ ⎪ ⎩σ3 x −q

(8.26) (continuous)

ine eri n

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

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

et

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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The variance of the number of failures, σx2 , is defined by Equation 8.24 as σx2 =

N  (xi − μx )2 f (xi ) = (0 − 2)2 (0.13) + (1 − 2)2 (0.27) + (2 − 2)2 (0.28) + (3 − 2)2 (0.18) i=1

+ (4 − 2)2 (0.09) + (5 − 2)2 (0.03) + (6 − 2)2 (0.02) = 1.92

The skewness coefficient, gx , is defined by Equation 8.26 as gx =

ww

N  1  1 (0 − 2)3 (0.13) + (1 − 2)3 (0.27) + (2 − 2)3 (0.28) + (3 − 2)3 (0.18) (xi − μx )3 f (xi ) = 3 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

w .E asy En g

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

ine eri n

Hence, the expression for μt can be written as

μt = 0.014

1 0.0142

= 71 h

g .n

The variance of t, σt2 , is given by Equation 8.24 as  q  q  q ′ ′ (t′ − 71)2 0.014e−0.014t dt′ = 0.014 (t′2 − 142t′ + 5041)e−0.014t dt′ (t′ − μt )2 f (t′ ) 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

 q 0

e−ax dx =

et

1 a

the expression for σt2 can be written as σt2 = 0.014



 142 5041 2 = 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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Chapter 8

Probability and Statistics in Water-Resources Engineering Using the previously cited integration formulae plus (Dwight, 1961)  q 0

x3 e−ax dx =

6 a4

gives gt = 3.91 * 10−8



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

w .E asy En g 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 +

ine eri n

n(n − 1)(n − 2) 3 n(n − 1) 2 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

g .n

(8.29)

et

(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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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

w .E asy En g

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

ine eri n

g .n

et

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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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

w .E asy En g 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) p (1 − p)N−n f (n) = n

where

ine eri n

N! N = n n!(N − n)!

(8.35)

is commonly referred to as the binomial coefficient. The parameters of the binomial distribution are the total number of outcomes, N, and the probability of success, p, in each outcome, and this distribution is illustrated in Figure 8.1. The mean, variance, and skewness coefficient of the binomial distribution are given by μn = Np,

σn2 = Np(1 − p),

g= 

g .n

1 − 2p

Np(1 − p)

et

(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! pn (1 − p)N−n f (n) = n!(N − n)!

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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=

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%.

ww

8.2.5.2

Geometric distribution

w .E asy En g

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

ine eri n

(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

g .n

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

et

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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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

The Poisson process is a limiting case of an infinite number of Bernoulli trials, where the number of trials, N, becomes large and the probability of success in each trial, p, becomes small in such a way that the expected number of successes, Np, remains constant. Under ´ these circumstances it can be shown that (Benjamin and Cornell, 1970; Thiebaux, 1994)

w .E asy En g lim

N→q, p→0

(Np)n e−Np N! pn (1 − p)N−n = n!(N − n)! n!

(8.39)

This approximation can be safely applied when N Ú 100, p … 0.01, and Np … 20 (Devore, 2000). Denoting the expected number of successes, Np, by λ, the probability distribution of the number of successes, n, in an interval in which the expected number of successes is given by λ is described by the Poisson distribution:

ine eri n

f (n) =

λn e−λ n!

(8.40)

Haan (1977) describes the Poisson process as corresponding to the case where, for a given interval of time, the measurement increments decrease, resulting in a corresponding decrease in the event probability within the measurement increment, in such a way that the total number of events expected in the overall time interval remains constant and equal to λ. The only parameter in the Poisson distribution is λ. The mean, variance, and skewness of the discrete Poisson distribution are given by μn = λ,

σn2 = λ,

gn = λ

− 21

g .n

et

(8.41)

Poisson distributions for several values of λ are illustrated in Figure 8.3. The Poisson process for a continuous time scale is defined in a similar way to the Poisson process on a discrete time scale. The assumptions underlying the Poisson process on a continuous time scale are: (1) the probability of an event in a short interval between t and t + t is λ t (proportional to the length of the interval) for all values of t; (2) the probability of more than one event in any short interval t to t + t is negligible in comparison to λt; and (3) the number of events in any interval of time is independent of the number of events in any other nonoverlapping interval of time. Extending the result for the discrete Poisson process given by Equation 8.40 to the continuous Poisson process, the probability distribution of the number of events, n, in time, t, for a Poisson process is given by f (n) =

(λt)n e−λt n!

(8.42)

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0.7

0.5

0.4

0.4

f(n)

f(n)

0.5

0.3

0.2

0.1

0.1

1

2

3

4

5 n

6

7

8

9

0.0

10

1

2

3

4

5 n

6

7

8

9

10

0.4

0.4

f(n)

0.5

w .E asy En g 0.3

0.3

0.2

0.2

0.1

0.1

0

1

2

3

4

5 n

6

7

8

9

λ = 6.5

0.6

0.5

0.0

0

0.7 λ = 4.5

0.6

f(n)

0.3

0.2

0

λ = 2.5

0.6

0.7

ww

355

0.7

λ = 0.5

0.6

0.0

Probability Distributions

10

0.0

0

1

2

3

4

5 n

6

7

8

9

10

The parameter of the Poisson distribution for a continuous process is λt, where λ is the average rate of occurrence of the event. The mean, variance, and skewness of the continuous Poisson distribution are given by μn = λt,

ine eri n

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

g .n

et

A flood-control system is designed for a runoff event with a 50-year return period. Assuming that exceedance of the 50-year runoff event is a Poisson process, what is the probability that the design event will be exceeded twice in the first 10 years of the system? What is the probability that the design event will be exceeded more than twice in the first 10 years? Solution The probability distribution of the number of exceedances is given by Equation 8.40 (since the events are discrete). Over a period of 10 years, the expected number of exceedances, λ, of the 50-year event is given by 1 = 0.2 λ = Np = (10) 50 and the probability of the design event being exceeded twice is given by Equation 8.40 as 0.22 e−0.2 = 0.016 = 1.6% 2! The probability, P, that the design event is exceeded more than twice in 10 years can be calculated using the relation   0.21 e−0.2 0.22 e−0.2 0.20 e−0.2 + + = 0.001 P = 1 − [f (0) + f (1) + f (2)] = 1 − 0! 1! 2! f (2) =

Therefore, there is only a 0.1% chance that the 50-year design event will be exceeded more than twice in any 10-year interval.

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8.2.5.4

Exponential distribution

The exponential distribution describes the probability distribution of the time between occurrences of an event in a continuous Poisson process. For any time duration, t, the probability of zero events during time t is given by Equation 8.42 as e−λt . Therefore, the probability that the time between events is less than t is equal to 1 − e−λt . This is equal to the cumulative probability distribution, F(t), of the time, t, between occurrences, which is equal to the integral of the probability density function, f (t). The probability density function can therefore be derived by differentiating the cumulative distribution function; hence, f (t) =

d d F(t) = (1 − e−λt ) dt dt

(8.44)

which simplifies to

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

w .E asy En g 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

ine eri n

F(t) =

which leads to



t

λe−λτ dτ

(8.47)

0

F(t) = 1 − e−λt

g .n

(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

et

1.0λ 0.8λ f(t)

0.6λ 0.4λ 0.2λ 0

0

1

2

3

4

5

λt

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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 F(1) = 1 − e−0.649(1) = 0.48 = 48%

w .E asy En g

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

ine eri n

f (t) =

λn tn−1 e−λt (n − 1)!

(8.49)

g .n

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 n μt = , λ

σt2

n = 2, λ

2 gt = n

et

(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

1.0λ

λn tn−1 e−λt Ŵ(n)

(8.51)

1.0λ

n=1

n =3

0.8λ

0.6λ

0.6λ

f(t)

f(t)

0.8λ

0.4λ

0.4λ

0.2λ

0.2λ

0

0

1

2

3

λt

4

5

0

0

1

2

3

4

λt

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where Ŵ(n) is the gamma function defined by the equation  q Ŵ(n) = tn−1 e−t dt

(8.52)

0

which has the property that Ŵ(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). EXAMPLE 8.8

w .E asy En g

A commercial area experiences significant flooding whenever a storm yields more than 13 cm of rainfall in 24 hours. In a typical year, there are four such storms. Assuming that these rainfall events are a Poisson process, what is the probability that it will take more than 2 years to have 4 flood events? Solution The probability density function of the time for four flood events is given by the gamma distribution, Equation 8.49. From the given data, n = 4, μt = 365 d (= 1 year), and 4 n = 0.01096 d−1 = λ= μt 365 The probability of up to t0 days elapsing before the fourth flood event is given by the cumulative distribution function  t0 n n−1 −λt  t0  (0.01096)4 t0 4−1 −0.01096t λ t e dt = f (t) dt = F(t0 ) = t e dt (n − 1)! (4 − 1)! 0 0 0  t0 t3 e−0.01096t dt = 2.405 * 10−9

ine eri n 0

According to Dwight (1961),



t3 eat dt = eat



t3 3t2 6t 6 − + − a a2 a3 a4

Using this relation to evaluate F(t0 ) gives ⎡ 



g .n

et

⎤t0 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

⎤730 2 3 3t t 6t 6 ⎦ + F(730) = −2.405 * 10−9 ⎣e−0.01096t = 0.958 + + 0.01096 0.010962 0.010963 0.010964 

0

Hence, the probability that four flood events will occur in less than 2 years is 0.958, and the probability that it will take more than 2 years to have four flood events is 1 − 0.958 = 0.042 or 4.2%.

On the basis of the fundamental process leading to the gamma distribution, it should be clear why the gamma distribution is equal to the exponential distribution when n = 1. The gamma distribution given by Equation 8.51 can also be written in the form f (t) =

λ(λt)n−1 e−λt Ŵ(n)

(8.54)

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Using the transformation (8.55)

x = λt the probability density function of x, p(x), is related to f (t) by the relation    dt  1 p(x) = f (t)   = f (t) dx λ

(8.56)

Combining Equations 8.54 to 8.56 and replacing n by α yields p(x) =

ww

1 xα−1 e−x , Ŵ(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

w .E asy En g

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 Ŵ(α)

(x − γ )α−1 e

−(x−γ ) β

,

x Ú γ

ine eri n

(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

g .n

(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.

et

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 − ǫ)β−1 e−λ(x−ǫ) Ŵ(β)

(8.61)

where λ, β, and ǫ 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) β= , λ= , ǫ = μx − 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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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 watersheds for log-space skews gx … 1.414 (Griffis and Stedinger, 2007b). The LP3 distribution is also recommended for application to low-flow estimates by the American Society of Civil Engineers (ASCE, 1980). An example of frequency analysis using the LP3 distribution is given in Section 8.3.3. If a LP3 distribution is assumed for annual flood peaks, then the usual problem is to estimate the flow for a given return period. The confidence limits of the estimated flow will depend on the length of the flow record used to estimate the LP3 parameters, and these confidence limits can be estimated using graphical relationships found in McCuen and Galloway (2010). 8.2.5.6

Normal distribution

The normal distribution, also called the Gaussian distribution, is a symmetrical bell-shaped curve describing the probability density of a continuous random variable. The functional form of the normal distribution is given by

w .E asy En g



 1 x − μx 2 1 f (x) = √ exp − 2 σx σx 2π

(8.63)

where the parameters μx and σx are equal to the mean and standard deviation of x, respectively. Normally distributed random variables are commonly described by the shorthand notation N(μ, σ 2 ), and the shape of the normal distribution is illustrated in Figure 8.6. Most hydrologic variables cannot (theoretically) be normally distributed, since the range of normally distributed random variables is from −q to q, and negative values of many hydrologic variables do not make sense. However, if the mean of a random variable is more than three or four times greater than its standard deviation, errors in the normal-distribution assumption can, in many cases, be neglected (Haan, 1977). As an application example, annual rainfall on the Greek island of Crete has been shown to mostly follow a normal distribution (Naoum and Tsanis, 2003).

FIGURE 8.6: Normal probability distribution

ine eri n

0.40

g .n

et

µx = 0, σx = 1

f(x)

0.30 µx = 1, σx = 2

0.20

0.10

0.00

–6

–4

–2

0

2

4

6

x

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It is sometimes convenient to work with the standard normal deviate, z, which is defined by x − μx z= (8.64) σx where x is normally distributed. The probability density function of z is therefore given by z2 1 f (z) = √ e− 2 2π

ww

(8.65)

Equation 8.64 guarantees that z is normally distributed with a mean of zero and a variance of unity, and is therefore an N(0,1) variate. The cumulative distribution, F(z), of the standard normal deviate is given by  z  z z′ 2 1 f (z′ ) dz′ = √ e− 2 dz′ (8.66) F(z) = 2π −q −q where F(z) is sometimes referred to as the area under the standard normal curve, and these values are tabulated in Appendix C.1. Values of F(z) can be approximated by the analytic relation (Abramowitz and Stegun, 1965)

w .E asy En g F(z) L



B, 1 − B,

z≤0 z≥0

(8.67)

where

B=

−4 1 1 + 0.196854|z| + 0.115194|z|2 + 0.000344|z|3 + 0.019527|z|4 2

(8.68)

and the error in F(z) using Equation 8.68 is less than 0.00025. Conditions under which any random variable can be expected to follow a normal distribution are specified by the central limit theorem:

ine eri n

Central limit theorem. If Sn is the sum of n independently and identically distributed random variables Xi , each having a mean μ and variance σ 2 , then in the limit as n approaches infinity, the distribution of Sn approaches a normal distribution with mean nμ and variance nσ 2 .

g .n

Unfortunately, not many hydrologic variables are the sum of independent and identically distributed random variables. Under some very general conditions, however, it can be shown that if X1 , X2 , . . . , Xn are n random variables with μXi = μi and σX2 i = σi2 , then the sum, Sn , defined by Sn = X1 + X2 + · · · + Xn (8.69)

et

is a random variable whose probability distribution approaches a normal distribution with mean, μ, and variance, σ 2 , given by μ= σ2 =

n 

μi

(8.70)

σi2

(8.71)

i=1

n  i=1

The application of this result generally requires a large number of independent variables to be included in the sum Sn , and the probability distribution of each Xi has negligible influence on the distribution of Sn . In hydrologic applications, annual averages of variables such as precipitation and evaporation are assumed to be calculated from the sum of independent identically distributed measurements, and they tend to follow normal distributions (Chow et al., 1988).

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EXAMPLE 8.9 The annual rainfall in the Upper Kissimmee River basin has been estimated to have a mean of 130.0 cm and a standard deviation of 15.6 cm (Chin, 1993b). Assuming that the annual rainfall is normally distributed, what is the probability of having an annual rainfall of less than 101.6 cm? Solution From the given data: μx = 130.0 cm and σx = 15.6 cm. For x = 101.6 cm, the standard normal deviate, z, is given by Equation 8.64 as x − μx 101.6 − 130.0 = −1.82 = z= σx 15.6 The probability that z … −1.82, F(−1.82), is given by Equation 8.67, where 1 2 1 = 2

B=

ww

 

1 + 0.196854|z| + 0.115194|z|2 + 0.000344|z|3 + 0.019527|z|4

−4

1 + 0.196854| − 1.82| + 0.115194| − 1.82|2 + 0.000344| − 1.82|3 + 0.019527| − 1.82|4

−4

= 0.034

and therefore F(−1.82) = 0.034 = 3.4% There is a 3.4% probability of having an annual rainfall less than 101.6 cm.

w .E asy En g 8.2.5.7

Log-normal distribution

In cases where the random variable, X, is equal to the product of n random variables X1 , X2 , . . . , Xn , the logarithm of X is equal to the sum of n random variables, where ln X = ln X1 + ln X2 + · · · + ln Xn

(8.72)

Therefore, according to the central limit theorem, ln X will be asymptotically normally distributed and X is said to have a log-normal distribution. Defining the random variable, Y, by the relation Y = ln X (8.73)

ine eri n

then if Y is normally distributed, the theory of random functions can be used to show that the probability distribution of X, the log-normal distribution, is given by   (ln x − μy )2 1 exp − f (x) = (8.74) , x > 0 √ 2σy2 xσy 2π

g .n

et

where μy and σy2 are the mean and variance of Y, respectively. The mean, variance, and skewness of a log-normally distributed variable, X, are given by ⎛ ⎞ σy2  ⎠ , σx2 = μ2x exp (σy2 ) − 1 , gx = 3Cv + Cv3 (8.75) μx = exp ⎝μy + 2 where Cv is the coefficient of variation defined as

σx μx

Cv =

(8.76)

If Y is defined by the relation Y = log10 X

(8.77)

then Equation 8.74 still describes the probability density of X, with ln x replaced by log10 x, and the moments of X are related to the moments of Y by 

μx = 10

μy +

σy2 2



,

σx2

=

μ2x



(σy2 )

10

 − 1 ,

gx = 3Cv + Cv3

(8.78)

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Many hydrologic variables exhibit a marked skewness, largely because they cannot be negative. Whereas the normal distribution allows the random variable to range without limit from negative infinity to positive infinity, the log-normal distribution has a lower limit of zero. The log-normal distribution has been found to reasonably describe such variables as daily rainfall depths, daily peak discharge rates, and the distribution of hydraulic conductivity in porous media. The log-normal distribution is routinely used in China to describe annual flood peaks (Singh and Strupczewski, 2002b) and has been shown to adequately describe annual flood peaks in the United States (e.g., Villarini et al., 2009). In water-quality analyses, the log-normal distribution has been found to describe the occurrence of bacteria (coliforms) in public water-distribution systems (Christian and Pipes, 1983). EXAMPLE 8.10

ww

Annual-maximum discharges in the Guadalupe River near Victoria, Texas, show a mean of 801 m3 /s and a standard deviation of 851 m3 /s. If the capacity of the river channel is 900 m3 /s, and the flow is assumed to follow a log-normal distribution, what is the probability that the maximum discharge will exceed the channel capacity? Solution From the given data: μx = 801 m3 /s and σx = 851 m3 /s. Equation 8.75 gives ⎞ ⎛ σy2 ⎠ = 801 (= μx ) exp ⎝μy + 2

w .E asy En g and

which are solved to yield



 (801)2 exp σy2 − 1 = (851)2 (= σx2 )

μy = 6.31 and σy = 0.870 3 When x = 900 m /s, the log-transformed variable, y, is given by y = ln 900 = 6.80 and the normalized random variate, z, is given by

z=

ine eri n

y − μy 6.80 − 6.31 = 0.563 = σy 0.870

The probability that z … 0.563, F(0.563), is given by Equation 8.67, where −4 1 B = 1 + 0.196854|z| + 0.115194|z|2 + 0.000344|z|3 + 0.019527|z|4 2 1 1 + 0.196854|0.563| + 0.115194|0.563|2 + 0.000344|0.563|3 = 2 −4 + 0.019527|0.563|4 = 0.286

g .n

et

which yields

F(0.563) = 1 − 0.286 = 0.714 = 71.4%

The probability that the maximum discharge in the river is greater than the channel capacity of 900 m3 /s is therefore equal to 1 − 0.714 = 0.286 = 28.6%. This is the probability of flooding in any given year.

8.2.5.8

Uniform distribution

The uniform distribution describes the behavior of a random variable in which all possible outcomes are equally likely within the range [a, b]. The uniform distribution can be applied to either discrete or continuous random variables. For a continuous random variable, x, the uniform probability density function, f (x), is given by f (x) =

1 , b − a

a … x … b

(8.79)

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where the parameters a and b define the range of the random variable. The mean, μx , and variance, σx2 , of a uniformly distributed random variable are given by μx =

1 (a + b), 2

σx2 =

1 (b − a)2 12

(8.80)

EXAMPLE 8.11 Conflicting data from several remote rain gages in a region of the Amazon basin indicate that the annual rainfall in 1997 was between 810 mm and 1080 mm. Assuming that the uncertainty can be described by a uniform probability distribution, what is the probability that the rainfall is greater than 1000 mm? Solution In this case, a = 810 mm, b = 1080 mm, and the uniform probability distribution of the 1997 rainfall is given by Equation 8.79 as

ww

f (x) =

1 = 0.00370 1080 − 810

for 810 mm … x … 1080 mm

The probability, P, that x Ú 1000 mm is therefore given by

w .E asy En g

P = 0.00370 * (1080 − 1000) = 0.296 = 29.6%

8.2.5.9

Extreme-value distributions

Extreme values are either maxima or minima of sets of random variables. Consider the sample set X1 , X2 , . . . , Xn , and let Y be the largest of the sample values. If F(y) is the probability that Y < y and PXi (xi ) is the probability that Xi < xi , then the probability that Y < y is equal to the probability that all the xi s are less than y, which means that

ine eri n

F(y) = PX1 (y)PX2 (y) . . . PXn (y) = [PX (y)]n

(8.81)

This equation gives the cumulative probability distribution of the extreme value, Y, in terms of the cumulative probability distribution of each outcome, which is called the parent distribution. The probability density function, f (y), of the extreme value, Y, is therefore given by f (y) =

d F(y) = n[PX (y)]n−1 pX (y) dy

g .n

(8.82)

et

where pX (y) is the probability density function of each outcome. Equation 8.82 indicates that the probability distribution of extreme values, f (y), depends on both the sample size and the parent distribution. Equation 8.82 was derived for maximum values, and a similar result can be derived for minimum values (see Problem 8.19). Distributions of extreme values selected from large samples of many probability distributions have been shown to converge to one of three types of extreme-value distribution (Fisher and Tippett, 1928): Type I, Type II, or Type III. In Type I distributions, the parent distribution is unbounded in the direction of the desired extreme, and all moments of the distribution exist. In Type II distributions, also called Frechet distributions, the parent distribution is unbounded in the direction of the desired extreme and all moments of the distribution do not exist. In Type III distributions, the parent distribution is bounded in the direction of the desired extreme. The distributions of maxima of hydrologic variables are typically of Type I, since most hydrologic variables are (quasi-) unbounded to the right; Type II distributions are seldom used in hydrologic applications; and the distributions of minima are typically of Type III, since many hydrologic variables are bounded on the left by zero. Extreme-value Type I (Gumbel) distribution. The extreme-value Type I distribution requires that the parent distribution be unbounded in the direction of the extreme value. Specifically, Type I distributions require that the parent distribution falls off in an exponential manner,

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such that the upper tail of the cumulative distribution function, PX (x), can be expressed in the form PX (x) = 1 − e−g(x) with g(x) an increasing function of x (Benjamin and Cornell, 1970). In using the Type I distribution to estimate maxima, parent distributions with the unbounded property include the normal, log-normal, exponential, and gamma distributions. Using the normal distribution as the parent distribution, the probability density function for the Type I extreme-value distribution is given by (Gumbel, 1958) 1 f (x) = exp a

!

"  x − b x − b , ; − exp ; a a

−q < x < q,

−q < b < q,

a > 0 (8.83)

ww

where a and b are scale and location parameters, b is the mode of the distribution and the minus of the ; used for maximum values. The plus of the ; is used for minimum values. The Type I distribution for maximum values is illustrated in Figure 8.7. The Type I extremevalue distribution is sometimes referred to as the Gumbel extreme-value distribution, the Fisher–Tippett Type I distribution, or the double-exponential distribution. The mean, variance, and skewness coefficient for the extreme-value Type I distribution applied to maximum values are

w .E asy En g μx = b + 0.577a,

σx2 = 1.645a2 ,

gx = 1.1396

(8.84)

gx = −1.1396

(8.85)

and for the Type I distribution applied to minimum values, μx = b − 0.577a,

By using the transformation

σx2 = 1.645a2 ,

ine eri n

y=

x − b a

(8.86)

the extreme-value Type I distribution can be written in the form f (y) = exp [;y − exp (;y)]

g .n

et

(8.88)

0.4

0.3 a=1

f(x)

FIGURE 8.7: Type I extreme-value probability distribution

which yields the following cumulative distribution functions: ⎧ ⎨exp [−exp (−y)] (maxima) F(y) = ⎩1 − exp [−exp (y)] (minima)

(8.87)

0.2

0.1

0.0 –5

a=2

–4

–3

–2

–1

0

1

2

3

4

5

(x–b)/a

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These cumulative distributions are most useful in determining the return periods of extreme events, such as flood flows (annual-maximum flows), maximum rainfall, and maximum wind speed (Gumbel, 1954). The extreme-value Type I (Gumbel) distribution is used extensively in flood studies in the United Kingdom and in many other parts of the world (Cunnane, 1988), and has been applied by the U.S. National Weather Service in analyzing annual-maximum precipitation amounts for given durations. EXAMPLE 8.12 The annual-maximum discharges in the Guadalupe River near Victoria, Texas, between 1935 and 1978 show a mean of 811 m3 /s and a standard deviation of 851 m3 /s. Assuming that the annual-maximum flows are described by an extreme-value Type I (Gumbel) distribution, estimate the annual-maximum flow rate with a return period of 100 years. Solution From the given data: μx = 811 m3 /s and σx = 851 m3 /s. According to Equation 8.84, the scale and location parameters, a and b, are derived from μx and σx by

ww

a= √

851 = √ = 664 m3 /s 1.645 1.645 σx

b = μx − 0.577a = 811 − 0.577(664) = 428 m3 /s

w .E asy En g

These parameters are used to transform the annual maxima, X, to the normalized variable, Y, where Y=

X − 428 X − b = a 664

1 = 0.01 and the cumulative For a return period of 100 years, the exceedance probability is 100 probability, F(y), is 1 − 0.01 = 0.99. The extreme-value Type I probability distribution given by Equation 8.88 yields 0.99 = exp [−exp (−y100 )]

ine eri n

where y100 is the event with a return period of 100 years. Solving for y100 gives y100 = −ln (−ln 0.99) = 4.60

The annual-maximum flow with a return period of 100 years, x100 , is related to y100 by − 428 x y100 = 100 664 which leads to

g .n

x100 = 428 + 664y100 = 428 + 664(4.60) = 3482 m3 /s

Therefore, a flow of 3482 m3 /s has a return period of 100 years.

et

It has been suggested that the Gumbel distribution tends to underestimate the magnitude of the rarest rainfall events, and that the log-Gumbel distribution might perform better in predicting hydrological variables (Brooks and Carruthers, 1953; Bernier, 1959). The log´ Gumbel distribution is sometimes called the Frechet distribution. Extreme-value Type III (Weibull) distribution. The extreme-value Type III distribution requires that the parent distribution be bounded in the direction of the extreme value. Specifically, Type III distributions (for minima) require that the left-hand tail of the cumulative distribution function, PX (x), rises from zero for values of x Ú c such that PX (x) = α(x − c)a ,

x Ú c

where c is the lower limit of x (Benjamin and Cornell, 1970). The extreme-value Type III distribution has mostly been used in hydrology to estimate low streamflows, which are bounded

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on the left by zero. The extreme-value Type III distribution for minimum values is commonly called the Weibull distribution (after Weibull, 1939) and for c = 0 is given by   x a f (x) = axa−1 b−a exp − , b

x Ú 0,

(8.89)

a, b > 0

The Weibull distribution is illustrated in Figure 8.8. The mean and variance of the Weibull distribution are given by 

 1 1 2 2 2 2 μx = bŴ 1 + , σx = b Ŵ 1 + − Ŵ 1 + (8.90) a a a

ww

and the skewness coefficient is







Ŵ 1 + 3a − 3Ŵ 1 + 2a Ŵ 1 + a1 + 2Ŵ 3 1 + gx = 

  23

2 1 2 Ŵ 1 + a − Ŵ 1 + a

w .E asy En g

1 a



(8.91)

where Ŵ(n) is the gamma function defined by Equation 8.52. The cumulative Weibull distribution is given by   x a F(x) = 1 − exp − (8.92) b

which is useful in determining the return period of minimum events. If the lower bound of the parent distribution is nonzero, then a displacement parameter, c, must be added to the Type III extreme-value distributions for minima, and the probability density function becomes (Haan, 1977)

ine eri n



a  x − c f (x) = a(x − c)a−1 (b − c)−a exp − b − c and the cumulative distribution then becomes



 x − c a F(x) = 1 − exp − b − c FIGURE 8.8: Type III extreme-value (Weibull) probability distribution

g .n

(8.93)

et

(8.94)

0.6 b=2

0.5 a=3

f(x)

0.4 0.3 a=2

0.2

a=1

0.1 0.0

0

2

4

6

8

10

x

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Equation 8.93 is sometimes called the three-parameter Weibull distribution or the bounded exponential distribution. The mean and variance of the three-parameter Weibull distribution are given by 

 1 2 1 μx = c + (b − c)Ŵ 1 + , σx2 = (b − c)2 Ŵ 1 + − Ŵ2 1 + (8.95) a a a and the skewness coefficient of the three-parameter Weibull distribution is given by Equation 8.91. The two-parameter Weibull distribution (c = 0) is commonly used to describe the distribution of wind speeds, and is appropriate for averaging periods less than about 10 minutes (Ro and Hunt, 2007). EXAMPLE 8.13

ww

The annual-minimum flows in a river at the location of a water-supply intake have a mean of 123 m3 /s and a standard deviation of 37 m3 /s. Assuming that the annual low flows have a lower bound of 0 m3 /s and are described by an extreme-value Type III distribution, what is the probability that an annual-low flow will be less than 80 m3 /s?

w .E asy En g

Solution From the given data: μx = 123 m3 /s and σx = 37 m3 /s. According to Equation 8.90, the parameters of the probability distribution, a and b, must satisfy the relations

1 = 123 (= μx ) (8.96) bŴ 1 + a

and



2 b2 Ŵ 1 +

a

1 − Ŵ2 1 +

a



= (37)2

(= σx2 )

Combining these equations to eliminate b yields 

 1 2 1232 2

 Ŵ 1 + − Ŵ 1 + = 372 a a Ŵ 2 1 + a1 or

ine eri n



 Ŵ 1 + 2a − Ŵ 2 1 + 1a 

= 0.0905 Ŵ 2 1 + a1

g .n

et

This equation can be solved for a by using the Solver capability of an electronic spreadsheet with a built-in gamma function (or any other numerical means) which yields a = 3.65 and Equation 8.96 gives b=

123

Ŵ 1 + 1a

 =

123

1 Ŵ 1 + 3.65

 =

123 = 136 0.903

The cumulative (Weibull) distribution of the annual-low flows, X, is given by Equation 8.92 as    

x a x 3.65 = 1 − exp − F(x) = 1 − exp − b 136 and the probability that X … 80 m3 /s is given by 



80 3.65 F(80) = 1 − exp − = 0.134 = 13.4% 136

There is a 13.4% probability that the annual-low flow is less than 80 m3 /s.

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Generalized extreme-value distribution. The generalized extreme-value (GEV) distribution incorporates the extreme-value Type I, II, and III distributions for maxima. The GEV distribution was first proposed by Jenkinson (1955) to describe the distribution of the largest values of meteorological data when the limiting form of the extreme-value distribution is unknown. The GEV distribution has to date been used to model a wide variety of natural extremes, including annual flood peaks, rainfall, wind speeds, wave heights, and other maxima (Martins and Stedinger, 2000). The GEV cumulative distribution function is given by

F(x) =

ww

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ exp ⎪ ⎪ ⎪ ⎪ ⎨

⎧ ⎫  1 ⎪ ⎪ c⎬ ⎨ c(x − a) − 1 − , ⎪ ⎪ b ⎩ ⎭

c Z 0 (8.97)

⎧ ⎪  ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ (x − a) ⎪ ⎪ − exp − exp , ⎪ ⎪ ⎩ ⎭ ⎩ b

c=0

where a, b, and c are location, scale, and shape parameters, respectively. For c = 0, the GEV distribution is the same as the extreme-value Type I distribution; for c < 0 the GEV distribution corresponds to the Type II distribution for maxima that have a finite lower bound of a + b/c; and for c > 0 the GEV distribution corresponds to the Type III distribution for maxima that have a finite upper bound at a + b/c. The mean, variance, and skewness coefficient for the GEV distribution are given by (Kottegoda and Rosso, 1997)

w .E asy En g μx = a + σx2

b [1 − Ŵ(1 + c)], c

(for c > −1)

2 b = [Ŵ(1 + 2c) − Ŵ 2 (1 + c)], c

gx = sign (c)

(8.98)

ine eri n (for c > −0.5)

−Ŵ(1 + 3c) + 3Ŵ(1 + c)Ŵ(1 + 2c) − 2Ŵ 3 (1 + c) 3

[Ŵ(1 + 2c) − Ŵ 2 (1 + c)] 2

,



(8.99)

1 for c > − 3

g .n

(8.100)

where it is noted that the moments only exist for certain values of c. In many applications, the GEV parameters (a, b, and c) are estimated from the mean, standard deviation, and skewness (μx , σx , and gx ) of the observed data; this approach to estimating the parameters of a probability distribution is called the method of moments (see Section 8.3.2.1). Using Equations 8.98 to 8.100 to estimate the GEV parameters can be cumbersome since a numerical solution is required. Whereas these calculations are fairly manageable, they can be facilitated by the following approximate relations (Bhunya et al., 2007b) ⎧ ⎨ 0.0087gx3 + 0.0582gx2 − 0.32gx + 0.2778, c= & ' ⎩ −0.31158 1 − exp[−0.4556(gx − 0.97134)] + 1.13828, b = −0.1429c3 − 0.7631c2 + 1.0145c + 0.7795, σx ⎧ ⎨ 0.514075c1.33199 − 0.44901,

a − μx = ⎩ σx 19.357c4 + 13.749c3 + 4.484c2 + 0.5212c − 0.4427,

et

−0.7 … gx … 1.15 gx Ú 1.15 (8.101) −0.5 … c … 0.5

(8.102)

0.01 … c … 0.5 −0.5 … c … 0.01 (8.103)

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Estimation of the GEV parameters from the moments of the observed data (i.e., μx , σx , and gx ) is appropriate for modest sample sizes; however, for small sample sizes, parameter estimation using the method of L-moments (see Section 8.3.2.3) are likely to perform better (Hosking, 1985; Madsen et al., 1997; Bhunya et al., 2007b). The generalized extreme-value distribution has been used in the United States to describe annual flood peaks (e.g., Villarini and Smith, 2010), is commonly used in Great Britain to describe annual flood extremes (Singh and Strupczewski, 2002b), has been found to perform well in describing flood flows some river basins in Turkey (Saf, 2009), and has been used describe the distribution of annual-maximum rainfall amounts for a given duration (Overeem et al., 2007; Mailhot and Duchesne, 2010; Villarini et al., 2011).

EXAMPLE 8.14

ww

The annual-maximum 24-hour rainfall amounts in a subtropical area have a mean of 15.4 cm, a standard deviation of 3.42 cm, and a skewness coefficient of 1.24. Assuming that the annual-maximum 24-hour rainfall is described by a general extreme-value (GEV) distribution, estimate the annual-maximum 24-hour rainfall with a return period of 100 years. Solution From the given data: μx = 15.4 cm, σx = 3.42 cm, and gx = 1.24. In order to find the GEV cumulative distribution function, the parameters a, b, and c must first be determined. The shape parameter, c, can be determined directly from the skewness coefficient, gx , using Equation 8.100, where

w .E asy En g gx = 1.24 = sign (c)

−Ŵ(1 + 3c) + 3Ŵ(1 + c)Ŵ(1 + 2c) − 2Ŵ 3 (1 + c) [Ŵ(1 + 2c) − Ŵ 2 (1 + c)]3/2

which can be solved to give c = −0.0163. Rearranging Equation 8.99 gives b as b=

(

c2 σx2 = Ŵ(1 + 2c) − Ŵ 2 (1 + c)

and rearranging Equation 8.98 gives a as a = μx −

(

(−0.0163)2 (3.42)2 = 2.61 cm 1.020 − 1.0102

ine eri n

2.61 b [1 − Ŵ(1 + c)] = 15.4 − [1 − 1.010] = 13.9 cm c −0.0163

g .n

Hence the cumulative distribution function, F(x), is given by Equation 8.97 as ⎫ ⎧  1 ⎪ ⎪ ⎨ c(x − a) c ⎬ F(x) = exp − 1 − = exp ⎪ ⎪ b ⎭ ⎩

which yields

et

⎫ ⎧   1 ⎪ ⎪ ⎨ −0.0163(x − 13.9) −0.0163 ⎬ − 1 − ⎪ ⎪ 2.61 ⎭ ⎩

)  * −61.3 F(x) = exp − 1 + 0.00625(x − 13.9)

For a return period of 100 years,

1 = 0.99 and 100 )  * −61.3 F(x100 ) = 0.99 = exp − 1 + 0.00625(x100 − 13.9) F(x100 ) = 1 −

which yields

x100 = 26.4 cm Therefore, an annual-maximum 24-hour rainfall of 26.4 cm has a return period of 100 years.

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Probability Distributions

371

0.5 0.4 ν=2

f(x)

0.3 0.2

ν=4

0.1 0.0

0

2

4

6

8

10

12

14

16

x

ww

8.2.5.10 Chi-square distribution Unlike the previously described probability distributions, the chi-square distribution is not used to describe hydrologic processes but is more commonly used in testing how well observed outcomes of hydrologic processes are described by theoretical probability distributions. The chi-square distribution is the probability distribution of a variable obtained by adding the squares of ν normally distributed random variables, all of which have a mean of zero and a variance of 1. That is, if X1 , X2 , . . . , Xν are normally distributed random variables with mean of zero and variance of 1 and a new random variable χ 2 is defined such that

w .E asy En g

χ2 =

ν 

Xi2

(8.104)

i=1

ine eri n

then the probability density function of χ 2 is defined as the chi-square distribution and is given by

 − 1− ν2

x

−x

e2 , f (x) = ν ν 22 Ŵ 2

x, ν > 0

g .n

(8.105)

et

where x = χ 2 and ν is called the number of degrees of freedom. The chi-square distribution is illustrated in Figure 8.9. The mean and variance of the chi-square distribution are given by μχ 2 = ν,

σχ22 = 2ν

(8.106)

The chi-square distribution is commonly used in determining the confidence intervals of various sample statistics. The cumulative chi-square distribution is given in Appendix C.3. EXAMPLE 8.15 A random variable is calculated as the sum of the squares of 10 normally distributed variables with a mean of zero and a standard deviation of 1. What is the probability that the sum is greater than 20? What is the expected value of the sum? Solution The random variable cited here is a χ 2 variate with 10 degrees of freedom. The cumulative distribution of the χ 2 variate is given in Appendix C.3 as a function of the number of degrees of freedom, ν, and the exceedance probability, α. In this case, for ν = 10 and χ 2 = 20, Appendix C.3 gives (by interpolation) α = 0.031

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Probability and Statistics in Water-Resources Engineering Therefore, the probability that the sum of the squares of 10 N(0,1) variables exceeds 20 is 0.031 or 3.1%. The expected value of the sum is, by definition, equal to the mean, μχ 2 , which is given by Equation 8.106 as μχ 2 = ν = 10

8.3 Analysis of Hydrologic Data 8.3.1

ww

Estimation of Population Distribution

Utilization of hydrologic data in engineering practice is typically a two-step procedure. In the first step, the cumulative probability distribution of the measured data is compared with a variety of theoretical distribution functions, and the theoretical distribution function that best fits the probability distribution of the measured data is taken as the population distribution. In the second step, the exceedance probabilities of selected events are estimated using the theoretical population distribution. Some federal agencies require the use of specific probability distributions for describing certain hydrologic events, such as peak river flows, and in some cases they also specify the minimum size of the data set to be used in estimating the parameters of the distribution. In estimating probability distributions from measured data, the measurements are usually assumed to be independent and drawn from identical probability distributions. In cases where the observations are correlated, alternative methods such as time-series analysis are used in lieu of probability theory. The most common methods of estimating population distributions from measured data are: (1) visually comparing the probability distribution of the measured data with various theoretical distributions, and picking the closest distribution that is consistent with the underlying process generating the sample data, and (2) using hypothesis-testing methods to quantitatively assess whether various probability distributions are consistent with the probability distribution of the measured data. In estimating the occurrence of rare events from a fitted probability distribution, particular care should be taken to ensure a good match in the tail of the fitted distribution (e.g. El Adlouni et al., 2008).

w .E asy En g 8.3.1.1

ine eri n

Probability distribution of observed data

To assist in identifying a theoretical probability distribution that adequately describes observed outcomes, it is generally useful to graphically compare the probability distribution of observed outcomes with various theoretical probability distributions. The first step in plotting the probability distribution of observed data is to rank the data, such that for N observations a rank of 1 is assigned to the observation with the largest magnitude and a rank of N is assigned to the observation with the lowest magnitude. The exceedance probability of the m-ranked observation, xm , denoted by PX (X > xm ), is commonly estimated by the relation PX (X > xm ) =

m , N + 1

g .n

m = 1, . . . , N

et

(8.107)

or as a cumulative distribution function PX (X < xm ) = 1 −

m , N + 1

m = 1, . . . , N

(8.108)

Equation 8.107 is called the Weibull formula (Weibull, 1939) and is widely used in practice. The main drawback of the Weibull formula for estimating the cumulative probability distribution from measured data is that it is asymptotically exact (as the number of observations approaches infinity) only for a population with an underlying uniform distribution, which is relatively rare in nature. To address this shortcoming, Gringorten (1963) proposed that the exceedance probability of observed data be estimated using the relation PX (X > xm ) =

m − a , N + 1 − 2a

m = 1, . . . , N

(8.109)

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373

where a is a parameter that depends on the population distribution. For the normal, lognormal, and gamma distributions, a = 0.375; for the Gumbel and Weibull distributions, a = 0.44; and for the GEV distribution, a = 0.40 (Cunnane, 1978; Stedinger et al., 1993; Castellarin, 2007). Bedient and Huber (2002) suggest that a = 0.40 is a good compromise for the usual situation in which the exact distribution is unknown. EXAMPLE 8.16 The annual peak flows in the Guadalupe River near Victoria, Texas, between 1965 and 1978 are as follows: Peak flow

ww

Peak flow

Year

(m3 /s)

Year

(m3 /s)

1965 1966 1967 1968 1969 1970 1971

425 260 1982 1254 430 277 276

1972 1973 1974 1975 1976 1977 1978

1657 937 714 855 399 1543 360

w .E asy En g

Use the Weibull and Gringorten formulae to estimate the cumulative probability distribution of annual peak flow and compare the results. Solution The rankings of flows between 1965 and 1978 are shown in Columns 1 and 2 of the following table: (1)

Rank, m

(2) Flow, qm (m3 /s)

(3) Weibull P(Q > qm )

(4) Gringorten P(Q > qm )

1 2 3 4 5 6 7 8 9 10 11 12 13 14

1982 1657 1543 1254 937 855 714 430 425 399 360 277 276 260

0.067 0.133 0.200 0.267 0.333 0.400 0.467 0.533 0.600 0.667 0.733 0.800 0.867 0.933

0.042 0.113 0.183 0.253 0.324 0.394 0.465 0.535 0.606 0.676 0.746 0.817 0.887 0.958

ine eri n

g .n

et

In this case, N = 14 and the Weibull exceedance probabilities are given by Equation 8.107 as m m = P(Q > qm ) = 14 + 1 15 These probabilities are tabulated in Column 3. The Gringorten exceedance probabilities are given by (assuming a = 0.40) m − a m − 0.40 m − 0.40 P(Q > qm ) = = = N + 1 − 2a 14 + 1 − 2(0.40) 14.2 and these probabilities are tabulated in Column 4. The probability functions estimated using the Weibull and Gringorten formulae are compared in Figure 8.10, where the distributions appear very similar. At high flows, the Gringorten formula assigns a lower exceedance probability and a correspondingly higher return period.

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FIGURE 8.10: Comparison between Weibull and Gringorten probability distributions

1.0

0.0

0.9

0.1

Gringorten 0.2

Cumulative probability

0.8 0.7

Weibull

0.6

0.4

0.5

0.5

0.4

0.6

0.3

0.7

0.2

0.8

0.1

0.9

0.0 0

ww

0.3

200

400

600

800

Exceedance probability

374

1.0 1000 1200 1400 1600 1800 2000 2200

Flow (m3/s)

w .E asy En g

Visual comparison of the probability distribution calculated from observed data with various theoretical population distributions gives a preliminary idea of which theoretical distributions might provide the best fit to the observed data. A quantitative comparison between the observed and theoretical distributions can be obtained using the probability plot correlation coefficient (PPCC) originally developed by Filliben (1975). The PPCC test uses the correlation coefficient, r, between the ordered observations, xi , and the corresponding fitted quantiles, Mi , determined from the calculated cumulative probability, Pi , for each xi . It is inferred that the observations are described by the fitted distribution if the value of r is close to 1.0. The correlation coefficient, r, is given by

 +N x) M − M (x − i i i=1 r= , (8.110)

2 +N + N 2 i=1 (xi − x) i=1 Mi − M

ine eri n

g .n

where x and M denote the mean values of the observations, xi , and the fitted quantiles, Mi , respectively, and N is the sample size. The fitted quantile, Mi , is related to the calculated cumulative probability, Pi , for each xi by the assumed theoretical probability distribution, according to the relation Mi = −1 (Pi ) (8.111)

et

where −1 is the inverse of theoretical cumulative distribution function. Correlation between the observed and theoretical probability distribution cannot be rejected at the q significance level when r > rq (N) (8.112)

where rq (N) is the PPCC test statistic for a given sample size, N, and significance level, q. Values of rq for an assumed normal distribution can be estimated by the relation (Heo et al., 2008) ⎧ ⎪ 1.29 + 0.283 · ln(q) + (0.887 − 0.751q + 3.21q2 ) · ln(N) ⎪ ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ ⎨0.505 + 2.13q + (0.857 − 0.133q) · ln(N) 1 ln = ⎪ 1 − rq 

 ⎪ ⎪ ⎪ 1 −2.42+5.63q−3.11q2 ⎪ ⎪ ⎪ N ⎪ ⎩ −5.22 + 12.3q − 6.81q2

0.005 … q < 0.1 0.1 … q < 0.9 0.9 … q … 0.995 (8.113)

Values of rq for Gumbel, gamma (Pearson Type III), and GEV distributions can be found in Heo et al. (2008).

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EXAMPLE 8.17 The annual peak flows in the Guadalupe River cited in the previous example appear to fit a log-normal distribution. Use the probability plot correlation coefficient to evaluate whether this approximation is justified at the 95% confidence level.

ww

Solution The rank-ordered annual peak flows in the Guadalupe River are tabulated in Columns 1 and 2 in Table 8.1, and the logs (to base 10) of the flows are shown as log Qi in Column 3. The mean, X, and standard deviation, SX , of the log flows in Column 3 are calculated as X = 2.806 and SX = 0.313. The cumulative probabilities from the ranked measurements using the Gringorten formula with a = 0.375 (for a log-normal distribution) are shown as Pi in Column 4. The standard normal deviate, zi , corresponding to Pi is shown in Column 5, and the theoretical value of log Qi (= Mi ) is derived from zi according to the relation Mi = X + zi SX = 2.806 + 0.313zi The calculated values of Mi are shown in Column 6. The correlation, r, between log Qi and Mi is calculated using Equation 8.110 as 0.97, and the 95% significance value of r given by Equation 8.113 is r0.95 = 0.99. Therefore, although the correlation, r, is high, it is not sufficiently high to justify the assertion, with 95% confidence, that the probability distribution of annual peak flows in the Guadalupe River is lognormal. The relationship between the theoretical log-normal distribution and the frequency distribution of observations is shown in Figure 8.11, where it is apparent that the match is less than ideal.

w .E asy En g

TABLE 8.1: Calculation of Probability Plot Correlation Coefficient

(1)

FIGURE 8.11: Observed and theoretical probability distributions

(3) log Qi (log m3 /s)

(4)

(5)

i

(2) Qi (m3 /s)

Pi

zi

1 2 3 4 5 6 7 8 9 10 11 12 13 14

1982 1657 1543 1254 937 855 714 430 425 399 360 277 276 260

3.30 3.22 3.19 3.10 2.97 2.93 2.85 2.63 2.63 2.60 2.56 2.44 2.44 2.41

0.96 0.89 0.82 0.75 0.68 0.61 0.54 0.46 0.39 0.32 0.25 0.18 0.11 0.04

1.71 1.21 0.90 0.66 0.45 0.27 0.09 −0.09 −0.27 −0.45 −0.66 −0.90 −1.21 −1.71

(6) Mi (log m3 /s) 3.34 3.18 3.09 3.01 2.95 2.89 2.83 2.78 2.72 2.66 2.60 2.52 2.43 2.27

ine eri n

g .n

et

3.4

Flow (m3/s)

3.2

Observations Log-normal distribution (theory)

3.0 2.8 2.6 2.4 2.2 0.0

0.2

0.6 0.4 Cumulative probability

0.8

1.0

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8.3.1.2

Hypothesis tests

As an alternative to the direct comparison of observed probability distributions with various theoretical probability distributions, quantitative comparisons can also be made using hypothesis tests. The two most common hypothesis tests for quantitatively assessing whether an observed probability distribution can be approximated by a given (theoretical) population distribution are the chi-square test and the Kolmogorov–Smirnov test. The chi-square test. Based on sampling theory, it is known that if N outcomes are divided into M classes, with Xm being the number of outcomes in class m, and pm being the theoretical probability of an outcome being in class m, then the random variable χ2 =

ww

M  (Xm − Npm )2 Npm

(8.114)

m=1

has a chi-square distribution. The number of degrees of freedom is M − 1 if the expected frequencies can be computed without having to estimate the population parameters from the sample statistics, while the number of degrees of freedom is M − 1− n if the expected frequencies are computed by estimating n population parameters from sample statistics. In applying the chi-square goodness of fit test, the null hypothesis is taken as H0 : The samples are drawn from the proposed probability distribution. The null hypothesis is accepted at the α significance level if χ 2 ∈ [0, χα2 ] and rejected otherwise.

w .E asy En g EXAMPLE 8.18

Analysis of a 47-year record of annual rainfall indicates the following frequency distribution: Range (mm)

Number of outcomes

1500

7 5 3 2 2 1

g .n

The measured data also indicate a mean of 1225 mm and a standard deviation of 151 mm. Using a 5% significance level, assess the hypothesis that the annual rainfall is drawn from a normal distribution.

et

Solution The first step in the analysis is to derive the theoretical frequency distribution. Appendix C.1 gives the cumulative probability distribution of the standard normal deviate, z, which is defined as x − μx x − 1225 z= = σx 151 where x is the annual rainfall. Converting the annual rainfall amounts into standard normal deviates, z, yields: Rainfall (mm) z P(Z < z) 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500

−1.49 −1.16 −0.83 −0.50 −0.17 0.17 0.50 0.83 1.16 1.49 1.82

0.07 0.12 0.20 0.31 0.43 0.57 0.69 0.80 0.88 0.93 0.97

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and therefore the theoretical frequencies are given by

ww

Range (mm)

Theoretical probability, pm

Theoretical outcomes, Npm

1500

0.07 0.05 0.08 0.11 0.12 0.14 0.12 0.11 0.08 0.05 0.04 0.03

3.29 2.35 3.76 5.17 5.64 6.58 5.64 5.17 3.76 2.35 1.88 1.41

where the total number of observations, N, is equal to 47. Based on the observed and theoretical frequency distributions, the chi-square statistic is given by Equation 8.114 as

w .E asy En g χ2 =

(3 − 2.35)2 (4 − 3.76)2 (5 − 5.17)2 (6 − 5.64)2 (2 − 3.29)2 + + + + 3.29 2.35 3.76 5.17 5.64

+

(7 − 5.64)2 (5 − 5.17)2 (3 − 3.76)2 (7 − 6.58)2 + + + 6.58 5.64 5.17 3.76

(2 − 1.88)2 (1 − 1.41)2 (2 − 2.35)2 + + 2.35 1.88 1.41 = 1.42 +

ine eri n

Since both the mean and standard deviation were estimated from the measured data, the χ 2 statistic has M − 1 − n degrees of freedom, where M = 12 (= number of intervals), and n = 2 (= number of population parameters estimated from measured data), hence 12 − 1 − 2 = 9 degrees of freedom. Using a 5% significance level, the hypothesis that the observations are drawn from a normal distribution is accepted if 2 0 … 1.42 … χ0.05

g .n

2 Appendix C.3 gives that for 9 degrees of freedom, χ0.05 = 16.919. Since 0 … 1.42 … 16.919, the hypothesis that the annual rainfall is drawn from a normal distribution is accepted at the 5% significance level.

et

The effectiveness of the chi-square test is diminished if both the number of data intervals, called cells, is less than 5 and the expected number of outcomes in any cell is less than 5 (Haldar and Mahadevan, 2000; McCuen, 2002a). Kolmogorov–Smirnov test. This test differs from the chi-square test in that no parameters from the theoretical probability distribution need to be estimated from the observed data. For this reason, the Kolmogorov–Smirnov test is classified as a nonparametric test. The procedure for implementing the Kolmogorov–Smirnov test is as follows (Haan, 1977): Step 1: Let PX (x) be the specified theoretical cumulative distribution function under the null hypothesis. Step 2: Let SN (x) be the sample cumulative distribution function based on N observations. For any observed x, SN (x) = k/N, where k is the number of observations less than or equal to x. Step 3: Determine the maximum deviation, D, defined by   D = max PX (x) − SN (x) (8.115)

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Step 4: If, for the chosen significance level, the observed value of D is greater than or equal to the critical value of the Kolmogorov–Smirnov statistic tabulated in Appendix C.4, the hypothesis is rejected. An advantage of the Kolmogorov–Smirnov test over the chi-square test is that it is not necessary to divide the data into intervals; thus any error of judgment associated with the number or size of the intervals is avoided (Haldar and Mahadevan, 2000). EXAMPLE 8.19 Use the Kolmogorov–Smirnov test at the 10% significance level to assess the hypothesis that the data in Example 8.18 are drawn from a normal distribution. Solution Based on the given data, the measured and theoretical probability distributions of the annual rainfall are as follows:

ww

(1) Rainfall x (mm)

(2) Normalized rainfall (x − μx )/σx

(3)

(4)

(5)

SN (x)

PX (x)

2/47 = 0.043 5/47 = 0.106 9/47 = 0.191 14/47 = 0.298 20/47 = 0.426 27/47 = 0.574 34/47 = 0.723 39/47 = 0.830 42/47 = 0.894 44/47 = 0.936 46/47 = 0.979

0.068 0.123 0.204 0.310 0.434 0.566 0.690 0.796 0.877 0.932 0.966

  PX (x) − SN (x)

w .E asy En g 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500

−1.490 −1.159 −0.828 −0.497 −0.166 0.166 0.497 0.828 1.159 1.490 1.821

ine eri n

0.025 0.017 0.013 0.012 0.008 0.008 0.033 0.034 0.017 0.004 0.013

The normalized rainfall amounts in Column 2 are calculated using the rainfall amount, x, in Column 1; mean, μx , of 1225 mm; and standard deviation, σx , of 151 mm. The sample cumulative distribution function, SN (x), in Column 3 is calculated using the relation

k N where k is the number of measurements less than or equal to the given rainfall amount, x, and N = 47 is the total number of measurements. The theoretical cumulative distribution function, PX (x), in Column 4 is calculated using the normalized rainfall (Column 2) and the standard normal-distribution function given in Appendix C.1. The absolute value of the difference between the theoretical and sample distribution functions is given in Column 5. The maximum difference, D, between the theoretical and sample distribution functions is equal to 0.034 and occurs at an annual rainfall of 1350 mm. For a sample size of 11 and a significance level of 10%, Appendix C.4 gives the critical value of the Kolmogorov–Smirnov statistic as 0.352. Since 0.034 < 0.352, the hypothesis that the measured data are from a normal distribution is accepted at the 10% significance level. SN (x) =

g .n

et

The Kolmogorov–Smirnov test is generally more efficient than the chi-square test when the sample size is small (McCuen, 2002a). However, neither of these tests is very powerful in the sense that the probability of accepting a false hypothesis is quite high, especially for small samples (Haan, 1977). Furthermore, it should be kept in mind that probability distributions that demonstrate a good fit with observed data might perform poorly in estimating the frequencies of outcomes in the tails of the fitted distributions (e.g., Mitosek et al., 2006). In cases where there are several competing distributions, goodness-of-fit statistics can be used to rank the effectiveness of the competing distributions. Both the chi-square and the Kolmogorov–Smirnov test statistics are widely used in engineering applications, although other goodness-of-fit statistics have also been proposed (e.g., Mitosek et al., 2006).

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Analysis of Hydrologic Data

379

Model selection criteria

Limitations of using hypothesis testing as the sole basis for selecting a probability model to describe hydrologic observations are: (1) multiple distributions might be found suitable to describe a given set of observations; and (2) hypothesis testing does not account for the tradeoff between increasing the number of distribution parameters and improved predictability. These limitations can be addressed by using various model selection criteria. The most commonly used model selection criteria are the Akaike information criterion (Akaike, 1973) and the Bayesian information criterion (Schwarz, 1978). Suppose that a candidate probability density function, f , for an observed set of data is given by f (x|θ ), where x is an individual observation and θ is the vector of model parameters, then the likelihood, L(θ), of the observed data set, xi , i = [1, N], for a best-estimate modelparameter set, θˆ , is given by L(θˆ ) =

ww

N i=1

ˆ f (xi |θ)

(8.116)

ˆ is defined by the The Akaike information criterion (AIC) for the probability model, f (x|θ), relation

w .E asy En g

AIC = −2 ln(L) + 2np

(8.117)

where np is the number of parameters in the model. The Bayesian information criterion (BIC) is defined by the relation BIC = −2 ln(L) + ln(N)np

(8.118)

When all candidate models are considered, the model with the lowest AIC or BIC is considered to be the best (most parsimonious) model. The AIC and BIC formulations given by Equations 8.117 and 8.118 are quite similar, and in most cases these criteria give similar results and a similar ranking of preferred models. Other model selection criteria, such as the Anderson–Darling criterion, have also been proposed as bases for identifying the optimal model (Laio et al., 2009). 8.3.2

ine eri n

Estimation of Population Parameters

g .n

The probability distribution of a random variable X is typically written in the form fX (x). However, this distribution can also be written in the expanded form fX (x | θ1 , θ2 , . . . , θm ) to indicate that the probability distribution of the random variable also depends on the values of the m parameters θ1 , θ2 , . . . , θm . This explicit expression for the probability distribution is particularly relevant when the population parameters are estimated using sample statistics, which are themselves random variables. The most common methods for estimating the parameters from measured data are the method of moments, the maximum-likelihood method, and the method of L-moments. 8.3.2.1

et

Method of moments

The method of moments is based on the observation that the parameters of a probability distribution can usually be expressed in terms of the first few moments of the distribution. These moments can be estimated using sample statistics, and then the parameters of the distribution can be calculated using the relationship between the population parameters and the moments. The three moments most often used in hydrology are the mean, standard deviation, and skewness, defined by Equations 8.22, 8.24, and 8.26, respectively, for both discrete and continuous probability distributions. In practical applications, these moments must be estimated from finite samples, and the following equations usually provide the best estimates: μˆ x =

N 1  xi N

(8.119)

i=1

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Chapter 8

Probability and Statistics in Water-Resources Engineering N

σˆ x2 =

 1 (xi − μˆ x )2 N − 1

(8.120)

i=1

N gˆ x = (N − 1)(N − 2)

+N

i=1 (xi

− μˆx )3

(8.121)

σˆx3

where μˆ x , σˆ x2 , and gˆ x are unbiased estimates of the mean (μx ), variance (σx2 ), and skewness (gx ) of the population, respectively, from which the N samples of x, denoted by xi , are drawn (Benjamin and Cornell, 1970). Unbiased estimates of the variance, σˆ x2 , and skewness, gˆ x , are generally dependent on the underlying population distribution. The accuracy of the estimated skewness, gˆ x , is usually of most concern, since it involves the summation of the cubes of deviations from the mean and is therefore subject to larger errors in its computation. Although Equation 8.121 is most often used to estimate the skewness, other estimates have ´ and Robitaille, 1975; Tasker and Stedinger, 1986). also been proposed (Bobee

ww

EXAMPLE 8.20 The number of hurricanes hitting Florida each year between 1903 and 1988 is given in the following table (Winsberg, 1990):

w .E asy En g 1900 1910 1920 1930 1940 1950 1960 1970 1980

00

01

02

03

04

05

06

07

08

09

1 0 0 0 2 1 0 0

1 1 0 1 0 0 0 0

0 0 1 0 0 0 1 0

1 0 0 2 0 1 3 0 0

1 0 2 0 1 0 0 0 0

0 1 1 2 2 0 1 1 2

2 2 2 1 1 1 2 0 0

0 1 0 0 2 0 0 0 1

0 0 2 0 2 0 1 0 0

1 1 1 1 1 0 0 1

ine eri n

If the number of hurricanes per year is described by a Poisson distribution, estimate the parameters of the probability distribution. What is the probability of three hurricanes hitting Florida in 1 year? Solution The Poisson probability distribution, f (n), is given by

g .n

(λt)n e−λt n! where n is the number of occurrences per year and the parameter, λt, is related to the mean, μn , and standard deviation, σn , of n by √ μn = λt, σn = λt f (n) =

et

Based on the 86 years of data (1903–1988), the mean number of occurrences per year, μn , is estimated by the first-order moment, N, as 1 1  (56) = 0.65 μn L N = ni = 86 86 Hence λt L 0.65 and the probability distribution of the number of hurricanes per year in Florida is given by 0.65n e−0.65 f (n) = n! Putting n = 3 gives the probability of three hurricanes in 1 year as

0.653 e−0.65 = 0.02 = 2% 3! It is interesting to note that the probability of at least one hurricane hitting Florida in any year is given by 1 − f (0) = 1 − e−0.65 = 0.48 = 48%. f (3) =

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Analysis of Hydrologic Data

381

Estimates of the mean, variance, and skewness given by Equations 8.119 to 8.121 are based on samples of a random variable and are therefore random variables themselves. The standard deviation of an estimated parameter is commonly called the standard error, and the standard errors of the estimated mean, variance, and skewness are given by the following relations σˆ x Sμˆ x = √ N , Sσˆ x = σˆ x

ww

1 + 0.75gˆ x 2N . N / 0.5

 Sgˆ x = 10A−B log10

where A and B are given by A=

(8.122)

! −0.33 + 0.08|gˆ x | −0.52 + 0.30|gˆ x |

w .E asy En g and

B=

!

0.94 − 0.26|gˆ x | 0.55

(8.123) (8.124)

10

if |gˆ x | … 0.90

(8.125)

if |gˆ x | > 0.90 if |gˆ x | … 1.50

(8.126)

if |gˆ x | > 1.50

where Sμˆ x , Sσˆ x , and Sgˆ x are the standard errors of μˆ x , σˆ x , and gˆ x , respectively. Combining Equations 8.119 to 8.126 shows that, for a given sample size, the relative accuracy of the skewness is much less than the relative accuracies of both the mean and standard deviation, especially for small sample sizes.

EXAMPLE 8.21

ine eri n

Annual-maximum peak flows in the Rocky River from 1960 to 1980 show a mean of 52 m3 /s, standard deviation of 21 m3 /s, and skewness of 0.8. Calculate the standard errors and assess the relative accuracies of the estimated parameters.

g .n

Solution From the given data: N = 21 (1960–1980), μˆ = 52 m3 /s, σˆ = 21 m3 /s, and gˆ = 0.8. Substituting these values into Equations 8.122 to 8.124 yields Sμˆ Sσˆ Sgˆ

21 σˆ = √ = √ = 4.6 m3 /s N 21 ( , 1 + 0.75gˆ 1 + 0.75(0.8) = 21 = 4.1 m3 /s = σˆ 2N 2(21)  0.5 = 10A−B log10 (N/10)

et

where Equations 8.125 and 8.126 give

A = −0.33 + 0.08|0.8| = −0.27 B = 0.94 − 0.26|0.8| = 0.73 and therefore

 0.5 Sgˆ = 10−0.27−0.73 log10 (21/10) = 0.56

Hence, the standard errors of the mean, standard deviation, and skewness are 4.6 m3 /s, 4.1 m3 /s, and 0.56, respectively. Using the ratio of the standard error to the estimated value as a measure of the accuracy of the estimated value, then

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Chapter 8

Probability and Statistics in Water-Resources Engineering accuracy of mean =

Sμˆ 4.6 * 100 = * 100 = 8.8% μˆ 52

S 4.1 accuracy of standard deviation = σˆ * 100 = * 100 = 20% σˆ 21 Sgˆ 0.56 accuracy of skewness = * 100 = * 100 = 70% gˆ 0.8 Based on these results, the accuracies of the estimated parameters deteriorate as the order of the moments associated with the parameters increases.

8.3.2.2

ww

Maximum-likelihood method

The maximum-likelihood method selects the population parameters that maximize the likelihood of the observed outcomes. Consider the case of n independent outcomes x1 , x2 , . . . , xn , where the probability of any outcome, xi , is given by pX (xi | θ1 , θ2 , . . . , θm ), where θ1 , θ2 , . . . , θm are the population parameters. The probability of the n observed (independent) outcomes is then given by the product of the probabilities of each of the outcomes. This product is called the likelihood function, L(θ1 , θ2 , . . . , θm ), where

w .E asy En g

L(θ1 , θ2 , . . . , θm ) =

n i=1

pX (xi | θ1 , θ2 , . . . , θm )

(8.127)

The values of the parameters that maximize the value of L are called the maximumlikelihood estimates of the parameters. Since the form of the probability function, pX (x|θ1 , θ2 , . . . , θm ), is assumed to be known, the maximum-likelihood estimates can be derived from Equation 8.127 by equating the partial derivatives of L with respect to each of the parameters, θi , to zero. This leads to the following m equations

ine eri n

⭸L = 0, ⭸θi

(8.128)

i = 1, . . . , m

g .n

This set of m equations can be solved simultaneously to yield the m maximum-likelihood parameters θˆ1 , θˆ2 , . . . , θˆm . In some cases, it is more convenient to maximize the natural logarithm of the likelihood function than the likelihood function itself. This approach is particularly convenient when the probability distribution function involves an exponential term. It should be noted that since the logarithmic function is monotonic, values of the estimated parameters that maximize the logarithm of the likelihood function also maximize the likelihood function.

et

EXAMPLE 8.22 Use the maximum-likelihood method to estimate the parameter in the Poisson distribution of hurricane hits described in the previous example. Compare this result with the parameter estimate obtained using the method of moments. Solution The Poisson probability distribution can be written as f (n) =

θ n e−θ n!

where θ = λt is the population parameter of the Poisson distribution. Define the likelihood function, L′ , for the 86 years of data as

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L′ =

Analysis of Hydrologic Data

383

f (ni )

i=1

where ni is the number of occurrences in year i. It is more convenient to work with the log-likelihood function, L, which is defined as   86 86 86 86     1 1 ′ + ni ln θ − θ = + ln θ ln f (ni ) = ln ni − 86θ ln L = ln L = ni ! ni ! i=1

i=1

i=1

i=1

Taking the derivative with respect to θ and putting ⭸L/⭸θ = 0 gives 86

⭸L 1  ni − 86 = 0 = ⭸θ θ i=1

which leads to

ww

86

θ=

1  ni 86 i=1

This is the same estimate of θ that was derived in the previous example using the method of moments.

w .E asy En g

The method of moments and the maximum-likelihood method do not always yield the same estimates of the population parameters. The maximum-likelihood method is generally preferred over the method of moments, particularly for large samples (Haan, 1977). The method of moments is severely affected if the data contain errors in the tails of the distribution, where the moment arms are long (Chow, 1954), and is particularly severe in highly skewed distributions (Haan, 1977). In contrast, the relative asymptotic bias of upper quartiles (i.e., high-return-period events) is smallest for the method of moments and is largest for the maximum-likelihood method when the true distribution is either the log-normal distribution or the gamma distribution and another distribution is fitted to it (Strupczewski et al., 2002). Several useful relationships between measured data and maximum-likelihood parameter estimates for a variety of distributions can be found in Bury (1999). 8.3.2.3

Method of L-moments

ine eri n

g .n

The typically small sample sizes available for characterizing hydrologic time series yield estimates of the third and higher moments that are usually very uncertain. This has led to the use of an alternative approach for estimating the parameters of probability distributions called L-moments. The rth probability-weighted moment, βr , is defined by the relation βr =



+q

x[FX (x)]r fX (x)dx

−q

et

(8.129)

where FX and fX are the cumulative distribution function and probability density function of x, respectively. The L-moments, λr , are linear combinations of the probability-weighted moments, βr , and the first four L-moments are computed as λ1 = β0

(8.130)

λ3 = 6β2 − 6β1 + β0

(8.132)

λ2 = 2β1 − β0

(8.131)

λ4 = 20β3 − 30β2 + 12β1 − β0

(8.133)

Since the probability-weighted moments involve raising values of FX rather than x to powers, and because FX … 1, estimates of probability-weighted moments and L-moments are

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much less susceptible to the influences of a few large or small values in the sample. Hence, L-moments are generally preferable to product moments for estimating the parameters of probability distributions of hydrologic variables. Consider a sample of N measured values of a random variable, X. To estimate the L-moments of the probability distribution from which the sample was taken, first rank the values as x1 … x2 … x3 … · · · … xN , and estimate the probability-weighted moments as follows (Hosking and Wallis, 1997):

b0 =

N 1  xi N

b1 =

 1 (i − 1)xi N(N − 1)

b2 =

 1 (i − 1)(i − 2)xi N(N − 1)(N − 2)

(8.134)

i=1

N

ww

(8.135)

i=2

N

w .E asy En g b3 =

(8.136)

i=3

N

 1 (i − 1)(i − 2)(i − 3)xi N(N − 1)(N − 2)(N − 3)

(8.137)

i=4

The sample estimates of the first four L-moments, denoted by L1 to L4 , are then calculated by substituting b0 to b3 for β0 to β3 , respectively, in Equations 8.130 to 8.133. The L-moments of various probability distributions are given in terms of the parameters of the distributions in Table 8.2. Equating the sample L-moments to the theoretical L-moments of a distribution and then solving for the distribution parameters constitutes the method of L-moments.

EXAMPLE 8.23

ine eri n

The annual-maximum flows for the period 1946–1970 in Dry-Gulch Creek are as follows:

Year

Maximum flow (m3 /s)

Year

Maximum flow (m3 /s)

1946 1947 1948 1949 1950 1951 1952 1953 1954

126 178 251 35 71 501 891 18 2239

1955 1956 1957 1958 1959 1960 1961 1962 1963

3 2 141 282 112 40 63 398 708

g .n

Year

Maximum flow (m3 /s)

1964 1965 1966 1967 1968 1969 1970

11 1122 2 1259 158 126 35

et

A histogram of these data indicates that they are likely drawn from a log-normal distribution. Use the method of L-moments to estimate the mean and standard deviation of the log-normal distribution, and compare these parameters with those estimated using the method of moments.

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ww

TABLE 8.2: Moments and L-Moments of Common Probability Distributions

Distribution

Parameters

Normal

w .E asy En g

Log-normal (Y = ln X)

Moments

L-Moments λ1 = μX σ λ2 = X π 1/2

μX = μX

μX , σX

σX = σX

μY , σY

⎛

μY = μY

ξ, η

μX = ξ + σX =

Gumbel

ξ, α

1 η2

Generalized extreme value

ξ , α, κ

1 η

ine eri n

λ1 = ξ + λ2 =

μX = ξ + 0.5772α 2 = 1.645α 2 σX

α [1 − Ŵ(1 + κ)], κ > −1 κ 2 ) * 2 = α σX Ŵ(1 + 2κ) − [Ŵ(1 + κ)]2 , κ

μX = ξ +

γX = sign(κ)

κ > −0.5

−Ŵ(1 + 3κ) + 3Ŵ(1 + κ)Ŵ(1 + 2κ) − 2Ŵ 3 (1 + κ) − 1, [Ŵ(1 + 2κ) − Ŵ 2 (1 + κ)]3/2

Sources: Dingman (2002), Hosking and Wallis (1997), Kottegoda and Rosso (1997).

⎞

⎠ λ1 = exp ⎝μY + 2 ⎛ ⎞

σY2 ⎠ erf σY λ2 = exp ⎝μY + 2 2

σY = σ Y

Exponential

σY2

κ > −

1 3

1 η

1 2η

λ1 = ξ + 0.5772α λ2 = 0.6931α λ1 = ξ +

α [1 − Ŵ(1 + κ)] κ

g .n

α (1 − 2−κ )Ŵ(1 + κ) κ 2(1 − 3−κ ) λ3 − 3 = λ2 (1 − 2−κ ) λ2 =

et

385

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Chapter 8

Probability and Statistics in Water-Resources Engineering Solution For the N = 25 years of data, the rank (i) and corresponding flow (Qi ) are given in Columns 1 and 2, respectively, in the following table:

ww

(1)

(2)

(3)

(4)

(5)

i

Qi (m3 /s)

(1 − i)Qi

ln Qi

(ln Qi − μY )2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

2 2 3 11 18 35 35 40 63 71 112 126 126 141 158 178 251 282 398 501 708 891 1122 1259 2239

2 6 33 72 175 210 280 504 639 1120 1386 1512 1833 2212 2670 4016 4794 7164 9519 14,160 18,711 24,684 28,957 53,736

0.693 0.693 1.099 2.398 2.890 3.555 3.555 3.689 4.143 4.263 4.718 4.836 4.836 4.949 5.063 5.182 5.525 5.642 5.986 6.217 6.562 6.792 7.023 7.138 7.714

15.314 15.314 12.305 4.878 2.945 1.105 1.105 0.842 0.215 0.118 0.013 0.053 0.053 0.117 0.208 0.331 0.844 1.072 1.904 2.592 3.826 4.778 5.839 6.409 9.655

Sum

8772

178,395

115.163

91.836

w .E asy En g

ine eri n

The first probability-weighted moment, b0 , is given by Equation 8.134 as b0 =

N 1 1  (8772) = 350.9 Qi = N 25

g .n

i=1

et

The product (1 − i)Qi is shown in Column 3, and the second probability-weighted moment, b1 , is given by Equation 8.135 as

b1 =

N  1 1 (178, 395) = 297.3 (i − 1)Qi = N(N − 1) 25(25 − 1) i=2

The first L-moment, λ1 , is given by Equation 8.130 as λ1 = b0 = 350.9 and the second L-moment, λ2 , is given by Equation 8.131 as λ2 = 2b1 − b0 = 2(297.3) − 350.9 = 243.8 The relationship between the L-moments, λ1 and λ2 , and the parameters of the log-normal distribution, μY and σY (where Y = ln Q), is given in Table 8.2, which yields

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σY2

⎛

σY2

λ1 = 350.9 = exp ⎝μY +

2

Analysis of Hydrologic Data

387

⎞ ⎠ ⎞

⎠ erf σY λ2 = 243.8 = exp ⎝μY + 2 2

and solving for μY and σY gives

μY = 4.81 σY = 1.45 Hence, using the method of L-moments, the estimated mean and standard deviation of the log-normal distribution are 4.81 and 1.45, respectively, where the flows are given in m3 /s.

ww

Column 4 in the preceding table gives ln Qi , and Column 5 gives (ln Qi − μY )2 , and the method of moments gives the mean (μY ) and standard deviation (σY ) of the log-normal distribution as N

1 1  ln Qi = (115.163) = 4.61 N 25 i=1 0 ( 1 N 1  . /2 1 1 1 ln Qi − μY = σY = 2 91.836 = 1.96 N − 1 25 − 1

w .E asy En g μY =

i=1

Hence, using the method of moments, the estimated mean and standard deviation of the log-normal distribution are 4.61 and 1.96, respectively. The difference between the mean and standard deviation (μY , σY ) of the log-normal distribution estimated using the method of moments (4.61, 1.96) compared with the mean and standard deviation estimated using the method of L-moments (4.81, 1.45) is accounted for by the few outliers in the short period of record, which arguably tend to make the method of L-moments parameter estimates more reliable than the method of moments parameter estimates.

8.3.3

Frequency Analysis

ine eri n

g .n

Frequency analysis is concerned with estimating the relationship between an event, x, and the return period, T, of that event. Recalling that T is related to the exceedance probability of x, PX (X Ú x), by the relation 1 (8.138) T= PX (X Ú x)

et

then T is related to the cumulative distribution function of x, PX (X < x), by the relation T=

1 1 − PX (X < x)

(8.139)

A fundamental component of frequency analysis is the assumption that the underlying probability distribution of an event is known. However, in reality, the underlying probability distribution of an observed event can only be hypothesized, and the adequacy of this hypothesis must be assessed using tests such as the Chi-Square and Kolmogorov–Smirnov tests before proceeding with a frequency analysis based on an assumed probability distribution. The uncertainty of sample estimates of parameters in probability distributions requires caution when using data records shorter than 10 years, and caution should also be exercised when estimating the values of variables with recurrence intervals longer than twice the record length (Viessman and Lewis, 1966; Davie, 2002). The conventional frequency analyses presented here are based on an assumed population probability distribution, however,

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alternative, and sometimes more accurate, nonparametric methods using artificial neural networks have also been proposed for use in frequency analysis (e.g., He and Valeo, 2009). Many cumulative distribution functions cannot be expressed analytically and are tabulated as functions of normalized variables, X ′ , where X′ =

X − μx σx

(8.140)

and μx and σx are the mean and standard deviation of the population, respectively. The cumulative probability distribution of X ′ , PX ′ (X ′ < x′ ), for many distributions depends only on x′ and the skewness coefficient, gx , of the population and can be readily obtained from statistical tables. Denoting the value of X ′ with return period T by x′T , for many distributions x′T = KT (T, gx )

ww

(8.141)

where KT (T, gx ) is derived from the cumulative distribution function of X ′ and is commonly called the frequency factor. Combining Equations 8.140 and 8.141 leads to (8.142)

xT = μx + KT σx

w .E asy En g

where xT is the realization of X with return period T. The frequency factor is applicable to many, but not all, probability distributions. The frequency factors for a few probability distributions that are commonly used in hydrologic practice are described here. 8.3.3.1

Normal distribution

In the case of a normal distribution, the variable X ′ is equal to the standard normal deviate and has a N(0, 1) distribution. The cumulative distribution function of the standard normal deviate is given in Appendix C.1, and the frequency factor is equal to the standard normal deviate corresponding to a given exceedance probability or return period. The frequency factor, KT , can also be approximated by the empirical relation (Abramowitz and Stegun, 1965) KT = w − where

ine eri n

2.515517 + 0.802853w + 0.010328w2 1 + 1.432788w + 0.189269w2 + 0.001308w3

⎡  ⎤ 1 2 1 w = ⎣ln 2 ⎦ , p

0 < p … 0.5

g .n

(8.143)

(8.144)

et

and p is the exceedance probability (= 1/T). When p > 0.5, 1 − p is substituted for p in Equation 8.144 and the value of KT computed using Equation 8.143 is given a negative sign. According to Abramowitz and Stegun (1965), the error in using Equation 8.143 to estimate the frequency factor is less than 0.00045. Alternative expressions for the normal-distribution frequency factor (Equation 8.143) having comparable or lesser accuracy have also been proposed (e.g., Swamee and Rathie, 2007b). EXAMPLE 8.24 Annual rainfall at a given location can be described by a normal distribution with a mean of 127 cm and a standard deviation of 19 cm. Estimate the magnitude of the 50-year annual rainfall. Solution The 50-year rainfall, x50 , can be written in terms of the frequency factor, K50 , as x50 = μx + K50 σx From the given data, μx = 127 cm, σx = 19 cm, and the exceedance probability, p, is given by p=

1 1 = = 0.02 T 50

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Analysis of Hydrologic Data

389

The intermediate variable, w, is given by Equation 8.144 as  ⎤21

 12 1 1 ⎣ ⎦ = ln = 2.797 w = ln p2 0.022 ⎡ 

Equation 8.143 gives the frequency factor, K50 , as K50 = w −

2.515517 + 0.802853w + 0.010328w2 1 + 1.432788w + 0.189269w2 + 0.001308w3

= 2.797 − = 2.054

2.515517 + 0.802853(2.797) + 0.010328(2.797)2 1 + 1.432788(2.797) + 0.189269(2.797)2 + 0.001308(2.797)3

The 50-year rainfall is therefore given by

ww

x50 = μx + K50 σx = 127 + 2.054(19) = 166 cm

w .E asy En g 8.3.3.2

Log-normal distribution

In the case of a log-normal distribution, the random variable is first transformed using the relation Y = ln X (8.145) and the value of Y with return period T, yT , is given by (8.146)

yT = μy + KT σy

ine eri n

where μy and σy are the mean and standard deviation, respectively, of Y, and KT is the frequency factor of the standard normal deviate with return period T. The value of the original variable, X, with return period T, xT , is then given by xT = ln−1 yT = eyT

EXAMPLE 8.25

(8.147)

g .n

The annual rainfall at a given location has a log-normal distribution with a mean of 127 cm and a standard deviation of 19 cm. Estimate the magnitude of the 50-year annual rainfall.

et

Solution From the given data, μx = 127 cm, σx = 19 cm, and the mean, μy , and standard deviation, σy , of the log-transformed variable, Y = ln X, are related to μx and σx by Equation 8.75, where ⎛ ⎞ σy2  ⎠ , σx2 = μ2x exp (σy2 ) − 1 μx = exp ⎝μy + 2 These equations can be put in the form   μ4x 1 , μy = ln 2 μ2x + σx2 which lead to

and

σy =

(

σ 2 + μ2x ln x μ2x

  1274 1 μy = ln = 4.83 2 1272 + 192 σy =

(

ln

192 + 1272 = 0.149 1272

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Chapter 8

Probability and Statistics in Water-Resources Engineering Since the rainfall, X, is log-normally distributed, Y (= ln X) is normally distributed and the value of Y corresponding to a 50-year return period, y50 , is given by y50 = μy + K50 σy where K50 is the frequency factor for a normal distribution and a return period of 50 years. In the previous example it was shown that K50 = 2.054, therefore y50 = 4.83 + 2.054(0.149) = 5.14 and the corresponding 50-year rainfall, x50 , is given by x50 = ey50 = e5.14 = 171 cm

8.3.3.3

ww

Gamma/Pearson Type III distribution

In the case of the Pearson Type III distribution, also called the three-parameter gamma distribution, the frequency factor depends on both the return period, T, and the skewness coefficient, gx . If the skewness coefficient falls between −1 and +1, approximate values of the frequency factor for the gamma/Pearson Type III distribution, KT , can be estimated using the relation (Kite, 1977; Viessman and Lewis, 2003)

w .E asy En g KT =

* 1 ) ′ [(xT − k)k + 1]3 − 1 3k

(8.148)

where x′T is the standard normal deviate corresponding to the return period T and k is related to the skewness coefficient by gx k= (8.149) 6

When the skewness, gx , is equal to zero, the expanded form of Equation 8.148 indicates that KT = x′T , and the gamma/Pearson Type III distribution is identical to the normal distribution. In cases where the skewness coefficient falls outside the −1 to +1 range, the frequency factors given in Appendix C.2 should be used. Methods for calculating the confidence limits of the Pearson Type III distribution can be found in Ames (2006). EXAMPLE 8.26

ine eri n

g .n

The annual-maximum discharges in a river show a mean of 811 m3 /s, a standard deviation of 851 m3 /s, and a skewness of 0.94. Assuming a Pearson Type III distribution, estimate the 100-year discharge. If subsequent measurements indicate that the skewness coefficient is actually equal to 1.52, how does this affect the estimated 100-year discharge?

et

Solution From the given data: μx = 811 m3 /s, σx = 851 m3 /s, and gx = 0.94. The standard normal deviate corresponding to a 100-year return period, x′100 , is 2.33 (Appendix C.1), k = gx /6 = 0.94/6 = 0.157, and the frequency factor K100 is given by Equation 8.148 as ) * * 1 ) ′ 1 [(x100 − k)k + 1]3 − 1 = [(2.33 − 0.157)(0.157) + 1]3 − 1 = 3.00 K100 = 3k 3(0.157) The 100-year discharge, x100 , is therefore given by

x100 = μx + K100 σx = 811 + (3.00)(851) = 3364 m3 /s If the skewness is equal to 1.52, this is outside the [−1, +1] range of applicability of Equation 8.148, and the frequency factor must be obtained from the tabulated frequency factors in Appendix C.2. For a return period of 100 years and a skewness coefficient of 1.52, Appendix C.2 gives K100 = 3.342, and the 100-year discharge is given by x100 = μx + K100 σx = 811 + (3.342)(851) = 3655 m3 /s Therefore, the 62% change in estimated skewness from 0.94 to 1.52 causes a 9% change in the 100-year discharge from 3364 m3 /s to 3655 m3 /s.

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8.3.3.4

Analysis of Hydrologic Data

391

Log-Pearson Type III distribution

In the case of a log-Pearson Type III distribution, the random variable is first transformed using the relation Y = ln X (8.150) The value of Y with return period T, yT , is given by (8.151)

yT = μy + KT σy

where μy and σy are the mean and standard deviation, respectively, of Y, and KT is the frequency factor of the Pearson Type III distribution with return period T and skewness coefficient gy . The value of the original variable, X, with return period T, xT , is then given by xT = ln−1 yT = eyT

ww

(8.152)

Whenever the skewness of Y (= ln X) is equal to zero, the log-Pearson Type III distribution is identical to the log-normal distribution. EXAMPLE 8.27

w .E asy En g

The natural logarithms of annual-maximum discharges (in m3 /s) in a river show a mean of 6.33, a standard deviation of 0.862, and a skewness of −0.833. Assuming a log-Pearson Type III distribution, estimate the 100-year discharge.

Solution From the given data, μy = 6.33, σy = 0.862, and gy = −0.833, which yields k = gy /6 = −0.833/6 = −0.139. The standard normal deviate corresponding to a 100-year return period, x′100 , is 2.33 (Appendix C.1), and the frequency factor for a 100-year return period, K100 , is given by Equation 8.148 as ) * * 1 1 ) ′ [(x100 − k)k + 1]3 − 1 = [(2.33 + 0.139)(−0.139) + 1]3 − 1 = 1.72 K100 = 3k 3(−0.139)

ine eri n

The 100-year log-discharge, y100 , is therefore given by

y100 = μy + K100 σy = 6.33 + (1.72)(0.862) = 7.81

and the 100-year discharge, x100 , is

x100 = ey100 = e7.81 = 2465 m3 /s

g .n

The U.S. Interagency Advisory Committee on Water Data published an important document called Bulletin 17B (USIAC, 1982), which provides detailed guidance on the application of the log-Pearson Type III distribution to annual-maximum flood flows. Of particular concern in applying the log-Pearson Type III distribution to observed data is that skew coefficients estimated from small samples might be highly inaccurate. In cases where this is a concern, Bulletin 17B outlines a procedure for developing regional skew coefficients based on at least 40 gaging stations located within a 160-km (100-mi) radius of the site of concern, with each station having at least 25 years of data. If this procedure is not feasible, the generalized skew map shown in Figure 8.12 is provided as an easier but less accurate alternative. This map of generalized logarithmic skew coefficients for peak annual flows was developed from skew coefficients computed at 2972 gaging stations, all having at least 25 years of record, following the procedures outlined in Bulletin 17B. Depending on the number of years of record, regionalized skew coefficients are used either in lieu of or in combination with values computed from observed flows at the particular stream gage of concern. Equation 8.153 allows a weighted skew coefficient GW to be computed by combining a regionalized skew coefficient GR and station skew coefficient G, GW =

(MSER )(G) + (MSES )(GR ) MSER + MSES

et

(8.153)

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Probability and Statistics in Water-Resources Engineering

FIGURE 8.12: Generalized skew coefficients of the logarithms of annual-maximum streamflow Source: Interagency Advisory Committee on Water Data (Bulletin 17B, 1982).

0.2

−0.1

−0.3

0.6

0 20.3

−0.4

−0.4

−0.1

0.3 0

0.5 0.7

0.1

0 −0.3

0 0 0.6 0.3

−0.4

−0.3

−0.1 −0.3 0.2 −0.2

0.7 0.5 0.3

−0.2

−0.1

0

ww

0 0.3

−0.2 −0.1 −0.4 −0.3

0 −0.1

20.1

−0.2

0.2 0

−0.1 0 −0.2

w .E asy En g

0.2 0 −0.3

0.2 −0.1

where the station skew G is computed from observed flows at the station of interest; the regional skew, GR , is either developed from multiple stations following procedures outlined in Bulletin 17B or read from Figure 8.12; MSES denotes the mean-square error of the station skew, which can be estimated for N years of data using the relation (Wallis et al., 1974) MSES = 10[A−B(log10 (N/10))]

where

ine eri n

A=

⎧ ⎨ −0.33 + 0.08|G|

B=

⎧ ⎨

(8.154)

if |G| … 0.90

⎩ −0.52 + 0.30|G|

if |G| > 0.90

0.94 − 0.26|G|

if |G| … 1.50

0.55

if |G| > 1.50

⎩

g .n

et

and MSER is the mean-square error of the regional skew. If GR is taken from Figure 8.12, then MSER is equal to 0.302. Recent studies have further validated using this method to estimate the skewness of log-transformed annual peak flows in the United States (Griffis and Stedinger, 2007c; 2009). However, issues still remain as to the fundamental validity of using regional skews that neglect local storage effects (McCuen and Smith, 2008), and the need for more formal criteria in selecting the most appropriate regional data to supplement local data (Chebana and Ouarda, 2008). EXAMPLE 8.28 Observations of annual-maximum flood flows in the Ocmulgee River near Macon, Georgia, from 1980– 2000 yield a skew coefficient of the log-transformed data equal to −0.309. Determine the (weighted) skew coefficient that should be used for flood-frequency analysis. Solution In central Georgia, where Macon is located, Figure 8.12 gives a generalized skew coefficient GR of −0.1, and MSER is equal to 0.302. From the observed annual-maximum flood flows, G = −0.309, and for the 21-year period of record (1980–2000), N = 21. By definition

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Analysis of Hydrologic Data

393

A = −0.33 + 0.08|G| = −0.33 + 0.08|−0.309| = −0.305 B = 0.94 − 0.26|G| = 0.94 − 0.26|−0.309| = 0.860 . 21 //  . . N //  . = 10 −0.305−0.860 log10 10 = 0.262 MSES = 10 A−B log10 10

and the weighted skew coefficient is given by GW =

(MSER )(G) + (MSES )(GR ) (0.302)(−0.309) + (0.262)(−0.1) = = −0.212 MSER + MSES 0.302 + 0.262

Therefore, a skew coefficient of −0.212 is more appropriate than the measured skew coefficient of −0.309 for frequency analyses of flood flows in the Ocmulgee River at Macon, Georgia. As the period of record increases, it is expected that the measured skew coefficient will approach the generalized skew coefficient derived from Figure 8.12.

ww

8.3.3.5

Extreme-value Type I distribution

In the case of an extreme-value Type I (Gumbel) distribution, the frequency factor, KT , can be written as (Chow, 1953)

w .E asy En g

⎫  √ ⎧

⎬ ⎨ 6 T KT = − 0.5772 + ln ln π ⎩ T − 1 ⎭

(8.155)

This expression for KT is valid only in the limit as the number of samples approaches infinity. For a finite sample size KT varies with the sample size as shown in Table 8.3.

ine eri n

TABLE 8.3: Extreme-Value Type I (Gumbel) Frequency Factors

Sample size

2.33

5

10

15 20 25 30 40 50 60 70 75 100 q

0.065 0.052 0.044 0.038 0.031 0.026 0.023 0.020 0.019 0.015 −0.067

0.967 0.919 0.888 0.866 0.838 0.820 0.807 0.797 0.794 0.779 0.720

1.703 1.625 1.575 1.541 1.495 1.466 1.446 1.430 1.423 1.401 1.305

Return period

20

25

50

75

100

1000

2.410 2.302 2.235 2.188 2.126 2.086 2.059 2.038 2.029 1.998 1.866

2.632 2.517 2.444 2.393 2.326 2.283 2.253 2.230 2.220 2.187 2.044

3.321 3.179 3.088 3.026 2.943 2.889 2.852 2.824 2.812 2.770 2.592

3.721 3.563 3.463 3.393 3.301 3.241 3.200 3.169 3.155 3.109 2.911

4.005 3.386 3.729 3.653 3.554 3.491 3.446 3.413 3.400 3.340 3.137

6.265 6.006 5.842 5.727 5.476 5.478 5.410 5.359 5.338 5.261 4.900

g .n

et

Source: Viessman and Lewis (2005)

EXAMPLE 8.29 Based on 30 years of data, the annual-maximum discharges in a river have a mean of 811 m3 /s and a standard deviation of 851 m3 /s. Assuming that the annual maxima are described by an extremevalue Type I distribution, estimate the 100-year discharge. What would be the 100-year discharge if the statistics were based on 60 years of data? Compare your results with the prediction made using Equation 8.155.

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Chapter 8

Probability and Statistics in Water-Resources Engineering Solution From the given data: μx = 811 m3 /s, σx = 851 m3 /s, and the 100-year discharge, x100 , is given by x100 = μx + K100 σx = 811 + K100 851 For 30 years of data, the 100-year frequency factor is given by Table 8.3 as K100 = 3.653, and the 100-year discharge is given by x100 = 811 + (3.653)851 = 3920 m3 /s For 60 years of data, the 100-year frequency factor is given by Table 8.3 as K100 = 3.446, and the 100-year discharge is given by x100 = 811 + (3.446)851 = 3740 m3 /s

ww

For an infinite number of years of data, the 100-year frequency factor is given by Equation 8.155 as ⎫  √ ⎧

⎬ 6⎨ 100 K100 = − = 3.137 0.5772 + ln ln π ⎩ 100 − 1 ⎭

w .E asy En g which gives

x100 = 811 + (3.137)851 = 3480 m3 /s

Hence, for 30, 60, and an infinite number of years of data, the estimated 100-year discharge is 3920 m3 /s, 3740 m3 /s, and 3480 m3 /s, respectively. Based on these results, for the same statistics, the estimated 100-year discharge decreases as the number of years of data increases. There is approximately 13% difference between the 100-year discharge based on 30 years of data and the 100-year discharge based on an infinite number of years of data.

8.3.3.6

ine eri n

General extreme-value (GEV) distribution

In the case of the general extreme-value (GEV) distribution, the value of the random variable, xT , with a return period T is given by (Kottegoda and Rosso, 1997; Overeem et al., 2007). ⎧ 

c  ⎪ b T ⎪ ⎪ a + ⎪ c Z 0 1 − ln ⎪ ⎨ c T − 1 (8.156) xT =

⎪ ⎪ ⎪ T ⎪ ⎪ c=0 ⎩a + b ln ln T − 1

g .n

et

where a, b, and c, are the location, scale, and shape parameters, respectively, of the GEV distribution.

EXAMPLE 8.30 The annual-maximum 24-hour rainfall in a subtropical area is described by a general extreme-value distribution with parameters: a = 13.8 cm, b = 2.17 cm, and c = −0.129. Determine the annual-maximum 24-hour rainfall with a return period of 20 years. Solution From the given data: T = 20 years, a = 13.8 cm, b = 2.17 cm, and c = −0.129. The annualmaximum 24-hour rainfall with a return period of 20 years is given by Equation 8.156 as  

c 

−0.129  b 2.17 T 20 = 13.8 + = 21.7 cm 1 − ln 1 − ln x20 = a + c T − 1 −0.129 20 − 1 Therefore, an annual-maximum 24-hour rainfall of 21.7 cm has a return period of 20 years.

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Uncertainty Analysis

395

8.4 Uncertainty Analysis

ww

The error in a measured variable is defined as the difference between the measured value of the variable and the true value of the variable. The error can usually be assumed to be random, and characterized by a probability distribution. Also, in cases where measured data are used to calculate quantities that are not directly measurable, calculated quantities deviate from their true values, with errors whose probability distributions depend on the probability distributions of the errors in the measured data. The objective of uncertainty analysis is to estimate the statistics of quantities that are calculated from measured (uncertain) data. Whenever mathematical models are used to estimate quantities from measured data, uncertainty in the calculated quantities may occur from four possible sources: (1) uncertainty in the values of parameters and constants in the model equations; (2) uncertainty in the measured or estimated data; (3) errors introduced by the numerical method that solves the model equations; and (4) errors in using the selected equations to describe the process being modeled. This section addresses the first two sources of uncertainty. The third source of error can be identified using numerical experiments in which the model’s numerical solution is compared with results from accurate analytical solutions. The fourth source of error is related to the validity of the model itself, and such errors can be estimated only by comparing model predictions with direct measurements of those same quantities in simple cases where all other uncertainties are negligible. Consider the typical case in which the calculated quantity Y is being estimated using N measured variables Xi , i = 1, N. Functionally, this may be written as

w .E asy En g

(8.157)

Y = φ(X)

where X is a vector having elements Xi . If the true values of the measured quantities are denoted by Xi , then, assuming the model is valid, the true value of Y, Y, is given by (8.158)

Y = φ(X)

ine eri n

where X is a vector having elements Xi . The error in Y, Y, is then given by (8.159)

Y = Y − Y = φ(X) − φ(X)

Using a multidimensional Taylor expansion, φ(X) can be expressed in the following form:

g .n

 N  ⭸φ  φ(X) = φ(X) + (Xi − Xi ) ⭸Xi X=X i=1  N N 1   ⭸2 φ  (Xi − Xi )(Xj − Xj ) + R +  2! ⭸Xi ⭸Xj  i=1 j=1

X=X

et

(8.160)

where the remainder, R, involves higher-order products of the errors, (Xi − Xi ). If the measurement errors are small, then higher-order terms are much smaller than lower-order terms, and retaining only first-order terms in Equation 8.160 results in the following approximate relationship  N  ⭸φ  φ(X) = φ(X) + (Xi − Xi ) (8.161) ⭸Xi X=X i=1

Combining Equations 8.159 and 8.161 gives the error in the calculated quantity, Y, in terms of the errors in the measured quantities, Xi = (Xi − Xi ), hence Y =

N  i=1

′ φX Xi i

(8.162)

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Chapter 8

Probability and Statistics in Water-Resources Engineering ′ where φX is given by i

 ⭸φ  = ⭸Xi X=X

′ φX i

(8.163)

Equation 8.162 is a stochastic equation relating the error in the derived quantity, Y, to the errors in measured quantities, Xi , provided these latter errors are small. Taking the ensemble average of Equation 8.162 yields

Y =

ww

N  i=1

′ φX Xi  i

(8.164)

indicating that the average error in the estimated quantity is the sum of the average errors ′ in the measured quantities, weighted by the sensitivities φX . Typically, measurements are i assumed to be without bias, in which case the average errors, Xi , are equal to zero and, according to Equation 8.164, the calculated result would then also be without bias. Another statistical parameter of interest is the variance of the calculated quantity, Y, which is denoted by σY2 and defined by σY2 = (Y)2  (8.165)

w .E asy En g

Combining Equations 8.162 and 8.165 yields σY2 =

N N   i=1 j=1

′ φX φ ′ Xi Xj  i Xj

(8.166)

where Xi Xj  is the covariance between the errors in Xi and Xj . In many cases the errors in different measured quantities are uncorrelated, leading to error covariances of zero. Under these circumstances, Equation 8.166 becomes σY2

EXAMPLE 8.31

ine eri n

=

N  i=1

′ (φX )2 σX2 i i

The flow rate, Q, in a trapezoidal open channel can be expressed as Q = V(b + my)y

(8.167)

g .n

et

where V is the average velocity, b is the bottom width, m is the side slope, and y is the depth of flow. The variables V, b, y, and m are subject to measurement errors, and the means and standard deviations of these variables are given by: Variable

Mean

Standard Deviation

V b m y

0.9 m/s 10 m 2.0 2.4 m

0.10 m/s 0.30 m 0.2 0.3 m

Use a first-order uncertainty analysis to estimate the mean and standard deviation of Q. Solution According to Equation 8.158, the expected value of Q, Q, is given by Q = V[b + my]y = 0.9[10 + (2.0)(2.4)](2.4) = 32.0 m3 /s

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397

2 , can be estimated using Equation 8.167, which can be written as The variance of Q, σQ 2 = σQ

ww



⭸Q  ⭸V X=X

where  ⭸Q  ⭸V X=X  ⭸Q  ⭸b X=X  ⭸Q  ⭸m X=X  ⭸Q  ⭸y X=X



2

σV2 +



⭸Q  ⭸b X=X 

2

σb2 +



⭸Q  ⭸m X=X 

2

2 + σm



⭸Q  ⭸y X=X 

2

σy2 (8.168)

 = (b + my)yX=X = [b + my]y = [10 + (2.0)(2.4)](2.4) = 35.5 (m3 /s)/(m/s)  = VyX=X = Vy = (0.9)(2.4) = 2.16 (m3 /s)/m

  = Vy2 

X=X

= Vy2 = (0.9)(2.4)2 = 5.18 m3 /s

 = V(b + 2my)X=X = V[b + 2my] = (0.9)[10 + 2(2.0)(2.4)] = 17.64 (m3 /s)/m

Substituting these derivatives into Equation 8.168 with the given standard errors of V, b, m, and y yields

w .E asy En g

2 = (35.5)2 (0.10)2 + (2.16)2 (0.30)2 + (5.18)2 (0.20)2 + (17.64)2 (0.30)2 = 42.10 (m3 /s)2 σQ

which gives the standard error of Q, σQ , as

σQ =

√

42.10 = 6.5 m3 /s

First-order uncertainty analysis is sometimes referred to as first-order error analysis (Yan et al., 2006) or first-order second moment analysis. The major advantage of first-order uncertainty analysis (FOUA) is its simplicity, requiring knowledge of only the first two statistical moments of the basic variables and a simple sensitivity analysis to calculate the incremental change in model output corresponding to an incremental change in model input. The applicability of FOUA, however, is inherently restricted by the degree of nonlinearity in the model. For nonlinear systems, this analysis becomes increasingly inaccurate as the basic variables depart from their central values (Melching and Yoon, 1997). The FOUA approximation deteriorates when the coefficient of variation is greater than 10%–20%, which is quite common.

Problems 8.1. A flood-control system is designed such that the probability that the system capacity is exceeded X times in 30 years is given by the following discrete probability distribution: xi 0 1 2 3 4

f (xi )

xi

f (xi )

0.04 0.14 0.23 0.24 0.18

5 6 7 8 >9

0.10 0.05 0.02 0.01 0.00

What is the mean number of system failures expected in 30 years? What are the variance and skewness of the number of failures? 8.2. The probability distribution of the time between flooding in a residential area is given by

ine eri n

g .n

! 0.143e−0.143t , f (t) = 0,

et

t > 0 t … 0

where t is the time interval between flood events in days. Estimate the mean, standard deviation, and skewness of t. 8.3. The annual-maximum flow rates in a river are known to vary between 4 m3 /s and 10 m3 /s and have a probability distribution of the form α f (x) = x2 where x is the annual-maximum flow rate and α is a constant. Determine: (a) the value of α, (b) the return period of a maximum flow rate of 7 m3 /s, and (c) the mean and standard deviation of the annual-maximum flow rates.

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Probability and Statistics in Water-Resources Engineering

8.4. The flow rates during the summer in a regulated channel vary between 1 m3 /s and 2 m3 /s, and all flow rates between these values are equally likely. Derive an expression for the probability density function of the flow rates in the river, and use this distribution to calculate the mean, variance, and skewness of the streamflows. Identify the flow rate that has a return period of 50 years. [Hint: Your derived 3 q expression must satisfy the f (Q) dQ = 1.] relations: f (Q) Ú 0 and −q 8.5. The average annual rainfall over a certain area is estimated to be 1524 mm, and the rainfall is as likely to be above average in any year (wet year) as it is to be below average (dry year). For any 20-year interval, what is the probability of experiencing the following events: (a) one wet year with 19 dry years, and (b) 10 wet years with 10 dry years? 8.6. Use the theory of combinations to show that if the probability of an event occurring in any trial is p, and if all trials are independent, then the probability of the event occurring n times in N trials, f (n), is given by

ww f (n) =

8.12. A drainage system is designed for a storm with a return period of 10 years. What is the average interval between floods? What is the probability that there will be more than 6 months between floods? 8.13. Explain why a gamma distribution with n = 1 is the same as the exponential distribution. 8.14. A residential development in the southern United States floods whenever more than 10.2 cm of rain falls in 24 hours. In a typical year, there are three such storms. Assuming that these rainfall events are a Poisson process, what is the probability that it will take more than 1 year to have three flood events? 8.15. The annual rainfall in Everglades National Park has been estimated to have a mean of 141.2 cm and a standard deviation of 28.2 cm (Chin, 1993a). Assuming that the annual rainfall is normally distributed, what is the probability of having: (a) an annual rainfall of less than 127.0 cm, (b) an annual rainfall of less than 152.4 cm, and (c) an annual rainfall between 127.0 and 152.4 cm? 8.16. Annual-maximum discharges in a river show a mean of 620 m3 /s and a standard deviation of 311 m3 /s. If the capacity of the river is 780 m3 /s and the annualmaximum discharge can be assumed to follow a lognormal distribution, what is the probability that the annual-maximum discharge will exceed the channel capacity? 8.17. The annual rainfall in a rural town is shown to be lognormally distributed with a mean of 114 cm and a standard deviation of 22 cm. Estimate the rainfall having a return period of 100 years. How would your estimate differ if the rainfall were normally distributed? 8.18. Streamflow data collected from several sources indicate that the peak flow rate in a drainage channel at a given location is somewhere in the range between 95 and 115 m3 /s. Assuming that the uncertainty can be described by a uniform distribution, what is the probability that the peak flow rate is greater than 100 m3 /s? 8.19. The probability distribution for the maximum values in a sample set is given by Equation 8.82. Derive a similar expression for the minimum values in a sample set. 8.20. The annual-maximum discharges in a river show a mean of 480 m3 /s and a standard deviation of 320 m3 /s. Assuming that the annual-maximum discharges are described by an extreme-value Type I (Gumbel) distribution, use Equation 8.88 to estimate the annualmaximum discharges corresponding to return periods of 50 and 100 years. 8.21. The annual-minimum rainfall amounts have been tabulated for a region and show a mean of 710 mm and a standard deviation of 112 mm. Assuming that these minima are described by an extreme-value Type I distribution, estimate the average interval between years where the minimum rainfall exceeds 600 mm. 8.22. The annual-low flow rates in a drainage channel have a mean of 43 m3 /s and a standard deviation of 12 m3 /s.

w .E asy En g

N! pn (1 − p)N−n n!(N − n)!

8.7. A flood-control system is designed for a rainfall event with a return period of 25 years. What is the probability that the design rainfall will be exceeded five times in 10 years? What is the probability that the design rainfall will be equalled or exceeded at least once in a 10-year period? 8.8. A water-resource system is designed for a hydrologic event with a 25-year return period. Assuming that the hydrologic events can be taken as a Bernoulli process, what is the probability that the design event will be exceeded once in the first 5 years of system operation? More than once in 5 years? 8.9. A stormwater-management system is being designed for a (design) life of 10 years, and a (design) runoff event is to be selected such that there is only a 1% chance that the system capacity will be exceeded during the design life. Determine the return period of the design runoff event. If the cumulative distribution function of the peak runoff rate, Q (in m3 /s), in any runoff event is given by F(Q) = exp [−exp (−Q)]

what is the design Q? 8.10. A water-resource system is designed for a 50-year storm. Assuming that the occurrence of the 50-year storm is a Bernoulli process, what is the probability that the 50year storm will be equalled or exceeded 1 year after the system is constructed? Within the first 6 years? 8.11. South Florida experiences a 6-month wet season between May and October. Within the wet season there are, on average, 35 days with more than 2.5 cm of rainfall. Assuming that the time between 2.5-cm rainfall events is exponentially distributed, what is the probability that there is less than 1 week between 2.5-cm rainfall events during the wet season?

ine eri n

g .n

et

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8.23.

8.24.

8.25.

8.26.

Assuming that the annual-low flow rates have a lower bound of 0 m3 /s and are described by an extreme-value Type III distribution, what is the probability that an annual-low flow rate will be less than 10 m3 /s? Measurements of annual-minimum flow rates in a river indicate that the minimum annual-minimum is 9 m3 /s, an annual-minimum flow rate of 34 m3 /s has a return of 10 years, and an annual-minimum flow rate of 76 m3 /s has a return period of 50 years. If these data follow an extreme-value Type III distribution, what would the 100year annual-minimum flow rate be? The annual-maximum streamflows in a river have a mean of 63.2 m3 /s, a standard deviation of 13.7 m3 /s, and a skewness coefficient of 1.86. Assuming that the annual-maximum streamflows are described by a general extreme-value (GEV) distribution, estimate the annual-maximum streamflow with a return period of 50 years. A random variable, X, is equal to the sum of the squares of 15 normally distributed variables with a mean of zero and a standard deviation of 1. What is the probability that X is greater than 25? What is the expected value of X? Exceedance probabilities of measured data can be estimated using the Gringorten formula

ww

w .E asy En g

PX (X > xm ) =

m − a N + 1 − 2a

where xm is a measurement with rank m, N is the total number of measurements, and a is a constant that depends on the population probability distribution. Values of a are typically in the range of 0.375 to 0.44. Determine the formula for the return period of xm . What is the maximum error in the return period that can result from uncertainty in a? 8.27. The annual peak flow rates in a river between 1980 and 1996 are as follows:

Year

Peak flow (ft3 /s)

1980 1981 1982 1983 1984 1985 1986 1987 1988

8000 5550 3390 6390 5889 7182 10,584 11,586 8293

Year

Peak flow (ft3 /s)

1989 1990 1991 1992 1993 1994 1995 1996 −

9193 5142 7884 4132 12136 5129 7236 6222 −

8.29.

8.30.

8.31.

8.32.

8.33.

8.34.

Use the Weibull and Gringorten formulae to estimate the cumulative probability distribution of peak flow rate and compare the results. 8.28. Analysis of a 50-year record of annual rainfall indicates the following frequency distribution:

8.35.

Range (mm)

Number of outcomes

1560

4 5 6 6 5 7 4 3 6 4

399

The measured data also indicate a mean of 1260 mm and a standard deviation of 234 mm. Using the chi-square test with a 5% significance level, assess the hypothesis that the annual rainfall is drawn from a normal distribution. Use the Kolmogorov–Smirnov test at the 10% significance level to assess the hypothesis that the data in Problem 8.28 are drawn from a normal distribution. A sample of a random variable has a mean of 1.6, and the random variable is assumed to have an exponential distribution. Use the method of moments to determine the parameter (λ) in the exponential distribution. A sample of a random variable has a mean of 1.6 and a standard deviation of 1.2. If the random variable is assumed to have a log-normal distribution, apply the method of moments to estimate the parameters in the log-normal distribution. Annual-maximum peak flow rates in the Davidian River from 1940 to 1980 show a mean of 35 m3 /s, standard deviation of 15 m3 /s, and skewness of 0.4. Calculate the standard errors and assess the relative accuracies of the estimated parameters. A sample of random data is to be fitted to the exponential probability distribution given by Equation 8.45. Derive the maximum-likelihood estimate of the distribution parameter λ. Compare your result with the method of moments estimate. Derive the maximum-likelihood estimators of the parameters in the normal distribution. The annual-maximum flow rates for the period 1970 to 1996 in the Bucking Bronco River are as follows:

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Year

Maximum flow (m3 /s)

Year

Maximum flow (m3 /s)

Year

Maximum flow (m3 /s)

1970 1971 1972 1973 1974 1975 1976 1977 1978

189 267 377 53 107 752 1337 27 3359

1979 1980 1981 1982 1983 1984 1985 1986 1987

5 3 212 423 168 60 95 597 1062

1988 1989 1990 1991 1992 1993 1994 1995 1996

17 1683 4 1889 237 189 53 550 320

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8.36.

8.37.

8.38.

8.39.

8.40.

8.41.

8.42.

Chapter 8

Probability and Statistics in Water-Resources Engineering

A histogram of these data indicate that they are likely drawn from a log-normal distribution. Use the method of L-moments to estimate the mean and standard deviation of the log-normal distribution, and compare these parameters with those estimated using the method of moments. The annual rainfall at a given location is normally distributed with a mean of 152 cm and a standard deviation of 30 cm. Estimate the magnitude of the 10-year and 100-year annual rainfall amounts. The annual rainfall at a given location has a log-normal distribution with a mean of 152 cm and a standard deviation of 30 cm. Estimate the magnitude of the 10-year and 100-year annual rainfall amounts. Compare your results with those obtained in Problem 8.36. The annual-maximum discharges in a river show a mean of 480 m3 /s, a standard deviation of 320 m3 /s, and a skewness of 0.87. Assuming a Pearson Type III distribution, use the frequency factor to estimate the 50-year and 100-year annual-maximum discharges. Compare the Pearson Type III frequency factor for a 100-year event estimated using Equation 8.148 with the same frequency factor estimated using Appendix C.2, for skewness coefficients of 3.0, 2.0, 1.0, 0.5, and 0.0. How do these results support the assertion that Equation 8.148 should be used only when the skewness coefficient is in the range [−1, +1]? The annual-maximum discharges in a river show a mean of 480 m3 /s, a standard deviation of 320 m3 /s, and a logtransformed skewness of 0.1. Assuming a log-Pearson Type III distribution, estimate the 100-year annualmaximum discharge. Observations of annual-maximum flood flows in the Mississippi River near Baton Rouge, Louisiana, from 1985 to 2000 yield a skew coefficient of the logtransformed data equal to −0.210. Determine the (weighted) skew coefficient that should be used for flood-frequency analysis. Based on 25 years of data, the annual-maximum discharges in a river have a mean of 480 m3 /s and a standard deviation of 320 m3 /s. Assuming that the annual maxima are described by an extreme-value Type I

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distribution, estimate the 100-year discharge. Compare your result with the prediction made using Equation 8.155, and with the result obtained in Problem 8.20. 8.43. Consider the series of annual-maximum streamflow measurements given in the table below. (a) Estimate the return period of a 30-m3 /s (annual-maximum) flow using the Weibull formula; and (b) estimate the return period of a 30-m3 /s (annual-maximum) flow using the Gringorten formula. It is postulated that the observed data follow the extreme-value Type I distribution. (c) Use the method of moments to estimate the parameters (a and b) of the extreme-value Type I distribution; (d) use the chi-square test with six classes to assess whether the data fit an extreme-value Type I distribution; and (e) assuming that the data are adequately described by an extreme-value Type I distribution, use the frequency factor to determine the annual-maximum streamflow with a 100-year return period.

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Year

Flow (m3 /s)

1970 1971 1972 1973 1974 1975 1976 1977

20 23 13 28 35 19 14 10

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Year

Flow (m3 /s)

1978 1979 1980 1981 1982 1983 1984 −

31 25 9 40 22 19 17 −

8.44. The annual-maximum streamflows in a river are described by a general extreme-value distribution with parameters: a = 55.2 m3 /s, b = 8.68 m3 /s, and c = −0.163. Determine the annual-maximum streamflow with a return period of 100 years. 8.45. A trapezoidal channel is estimated to have a bottom width of 2 m, side slopes of 3:1 (H:V), and a flow depth of 1 m. If the coefficient of variation of each of these measurements is 10%, estimate the coefficient of variation of the flow area calculated from these measurements.

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C H A P T E R

9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions 9.1 Introduction

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Surface-water hydrology is the science that encompasses the distribution, movement, and properties of natural water above the surface of the earth. Applications of surface-water hydrology in engineering practice are mostly encompassed within the field of engineering hydrology, which includes modeling rainfall events and predicting the quantity and quality of the resulting surface runoff. Designing systems to control the quantity and quality of surface runoff is the responsibility of a water-resources engineer. The land area that can contribute surface runoff to any particular location is determined by the shape and topography of the land area surrounding the given location. The potential contributing area is called the watershed, and the area within a watershed over which any particular rainfall event occurs is called the catchment area. In most engineering applications, the watershed and catchment area are taken to be the same, and are sometimes referred to as the drainage basin or the drainage area. The runoff from a watershed or catchment is concentrated at the catchment outlet, which is sometimes called the pour point of the catchment. The characteristics of the catchment determine the quantity, quality, and timing of the surface runoff for a given rainfall event. The types of pollutants contained in surface runoff generally depend on the land uses within the catchment area, and pollutants in surface-water runoff that are typically of concern include suspended solids, heavy metals, nutrients, organics, oxygen-demanding substances, and pathogenic microorganisms. These pollutants may be in solution, in suspension, or attached to particles of sediment. To facilitate water management in the United States, the country is divided into a hierarchy of watersheds called hydrologic units with the following four levels: region, subregion, accounting unit, and cataloging unit. Regions represent the largest scale of watershed and are shown in Figure 9.1, and these 21 regions are further divided by lower-level topography into 222 subregions, 352 accounting units, and 2262 cataloging units. Hydrologic unit codes are used to identify hydrologic units, with regions identified by a 2-digit code, subregions by a 4-digit code, accounting units by a 6-digit code, and cataloging units by an 8-digit code. As an example, Lake Winnebago (Wisconsin) is located within region 04, subregion 0403, accounting unit 040302, and cataloging unit 04030204; note the hierarchy embedded in the numbering system. In engineering practice, it is common to refer to the cataloging unit by its “HUC8” number (referring to the 8-digit hydrologic unit code), which for the Lake Winnebago watershed is 04030204.

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9.2 Rainfall The precipitation of water vapor from the atmosphere occurs in many forms, the most important of which are rain and snow. Hail and sleet are less-frequent forms of precipitation. Engineered drainage systems in most urban communities are designed primarily to control the runoff from rainfall. The formation of precipitation usually results from the lifting of moist air masses within the atmosphere, which results in the cooling of the air mass and condensation of moisture. The four conditions that must be present for the production of precipitation are: (1) cooling of the air mass, (2) condensation of water droplets onto nuclei, (3) growth of water droplets, and (4) mechanisms to cause a sufficient density of the droplets for the precipitation to fall to the ground. 401

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IC

(17)

NO RT HW

(01) RED-RAINYA

(09) EST

G GRR EEAATT LLA AK KEESS MISSISSIPPI (04)

MISSOURI

GR EAT BAS

UPPER

(10)

(07)

IN

NIA FOR CALI

(16)

UPPE COLO R RADO

(02) OHIO (05)

(14)

(18)

(13)

ER MIS S

TE XA S–-G ULF (12)

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ISSI PPI

AARRKKAA D N NSSAASS–-WHITE-–RREED (11)

E ND

LOW ER COL ORA DO (15)

RA RIO G

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MID–-A TLA NT IC

CI F

NEW E NGL AN D

SOU RIS-

PA

FIGURE 9.1: Hydrologic regions in the United States Source: Seaber et al. (1987).

EE SS (06) NE TE N

G UL F

Chapter 9

LO W

402

C– TI AN L T A SOUTH (03)

(02) Water Resources Regions

(08)

0

400 MILES

ALASKA (19)

0

400 MILES

0

Kauai Oahu

CARIBBEAN (21)

(20) Molokai Maui Moui

HAWAII

Hawaii 150 MILES

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Puerto Rico

0

80 MILES

Cloud droplets generally form on condensation nuclei, which are usually less than 1 µm in diameter and typically consist of sea salts, dust, or combustion by-products. These particles are called atmospheric aerosols, which also play an important role in regulating the amount of solar radiation that reaches the surface of the earth (Hendriks, 2010). In pure air, condensation of water vapor to form liquid water droplets occurs only when the air is supersaturated with water molecules. In cold clouds, where the temperature is below freezing, water droplets and ice crystals are both formed and the greater saturation vapor pressure over water compared to over ice causes the growth of ice crystals at the expense of water droplets. This is called the Bergeron–Findeisen process or ice-crystal process. In warm clouds, the primary mechanism for the growth of water droplets is collision and coalescence of water droplets. When the moisture droplets and/or ice crystals are large enough, the precipitation falls to the ground. Moisture droplets larger than about 0.1 mm are large enough to fall, and these droplets grow as they collide and coalesce to form larger droplets. Rain drops falling to the ground are typically in the size range of 0.5–3 mm. Drizzle is a subcategory of rain with droplet sizes less than 0.5 mm. Most of the clouds and atmospheric moisture are contained within 5500 m (18,000 ft) of the surface of the earth. The main mechanisms of air-mass lifting are frontal lifting, orographic lifting, and convective lifting. In frontal lifting, warm air is lifted over cooler air by frontal passage. The resulting precipitation events are called cyclonic or frontal storms, and the zone where the warm and cold air masses meet is called a front. Frontal precipitation is the dominant type of precipitation in the north-central United States and other continental areas (Elliot, 1995). In a warm front, warm air advances over a colder air mass with a relatively slow rate of ascent (0.3%–1% slope) causing a large area of precipitation in advance of the front, typically 300–500 km (200–300 mi) ahead of the front. In a cold front, warm air is pushed upward at a relatively steep slope (0.7%–2%) by advancing cold air, leading to smaller precipitation areas

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FIGURE 9.2: Typical thunderstorm

Rainfall

403

in advance of the cold front. Precipitation rates are generally higher in advance of cold fronts than in advance of warm fronts. In orographic lifting, warm air rises as it flows over hills or mountains, and the resulting precipitation events are called orographic storms. An example of the orographic effect can be seen in the northwestern United States, where the westerly airflow results in higher precipitation and cooler temperatures to the west of the Cascade mountains (e.g., Seattle, Washington) than to the east (e.g., Boise, Idaho). Orographic precipitation is a major factor in most mountainous areas, and the amount of precipitation typically shows a strong correlation with elevation (e.g., Naoum and Tsanis, 2003). The lee side of mountainous areas is usually dry, since most of the moisture is precipitated on the windward side, and this is sometimes called the rain shadow effect (Davie, 2008). In convective lifting, air rises by virtue of being warmer and less dense than the surrounding air, and the resulting precipitation events are called convective storms or, more commonly, thunderstorms. A typical thunderstorm is illustrated in Figure 9.2, where the localized nature of such storms is clearly apparent. Convective precipitation is common during the summer months in the central United States and other continental climates with moist summers. Convective storms are typically of short duration, and usually occur on hot midsummer days as late-afternoon storms. 9.2.1

Measurement of Rainfall

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Records of rainfall have been collected for more than 2000 years (Ward and Robinson, 1999). Rainfall is typically measured using rain gages operated by government agencies such as the National Weather Service and local drainage districts. Rainfall amounts are described by the volume of rain falling per unit area and are given as a depth of water. Gages for measuring rainfall are categorized as either nonrecording (manual) or recording. In the United States, most of the rainfall data reported by the National Weather Service (NWS) are collected manually using a standard rain gage consisting of a 20.3-cm (8-in.) diameter funnel that passes water into a cylindrical measuring tube, the whole assembly being placed within an overflow can. The measuring tube has a cross-sectional area one-tenth that of the collector funnel; therefore, a 2.5-mm (0.1-in.) rainfall will occupy a depth of 25 mm (1 in.) in the collector tube. The capacity of the standard collector tube is 50 mm (2 in.), and rainfall in excess of this amount collects in the overflow can. The manual NWS gage is primarily used for collecting daily rainfall amounts. Automatic-recording gages are usually used for measuring rainfall at intervals less than one day, and for collecting data in remote locations. Recording gages use either a tipping bucket, weighing mechanism, or float device. The original tipping-bucket rain gage was invented by Sir Christopher Wren in about 1662, and accounts for nearly half of all recording rain gages worldwide (Upton and Rahimi, 2003). A typical tipping-bucket rain gage is

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions

FIGURE 9.3: Tipping-bucket rain gage

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(a) Rain gage

(b) Support structure

shown in Figure 9.3, where Figure 9.3(a) shows a close-up top view of the rain gage with projections that help prevent the splattering of incident rainfall. The structure that usually supports the rain gage is shown in Figure 9.3(b), where a solar panel (front) provides power for the data recorder (back), and the rain gage is on top of the structure. The tipping-bucket gage is based on funneling the collected rain to a small bucket that tilts and empties each time it fills, generating an electronic pulse with each tilt. The number of bucket-tilts per time interval provides a basis for determining the precipitation depth over time. Typical problems with tipping-bucket rain gages include blockages, wetting and evaporation losses of typically 0.05 mm (0.002 in.) per rainfall event (Niemczynowicz, 1986), rain missed during the tipping process which typically takes about half a second (Marsalek, 1981), wind effects, position, shelter, and dynamic rainfall effects (Habib et al., 2008). A weighing-type gage is based on continuously recording the weight of the accumulated precipitation, and float-type recording gages operate by catching rainfall in a tube containing a float whose rise is recorded with time. Older weighing-type gages are mechanical-weighing and analog-recording devices, while modern weighing-type gages are electronic-weighing and digital-recording devices; significant discrepancies in measured rainfall intensities between analog and digital gages have been found for rainfall durations less than 5 min (Keefer et al., 2008). The accuracy of both manual and automatic-recording rain gages is typically on the order of 0.25 mm (0.01 in.) per rainfall event. Rain-gage measurements are point measurements of rainfall and are only representative of a small area surrounding the rain gage. Areas on the order of 25 km2 (10 mi2 ) have been taken as characteristic of rain-gage measurements (Gupta, 1989; Ponce, 1989), although considerably smaller characteristic areas can be expected in regions where convection storms are common. In the United States, the National Weather Service makes extensive use of Weather Surveillance Radar 1988 Doppler (WSR-88D), commonly known as next-generation weather radar (NEXRAD). A typical WSR-88D tower is shown in Figure 9.4, and each WSR-88D station measures weather activity over a 230-km (143-mi) radius area with a 10-cm (3.9-in) wavelength signal. This radar system provides estimates of hourly rainfall from the reflectivity of the S-band signal (1.55–3.9 GHz) within cells that are approximately 4 km * 4 km (2.5 mi * 2.5 mi). Since radar systems measure only the droplet size and movement of water in the atmosphere, and not the volume of water falling to the ground, radar-estimated rainfall should generally be corrected to correlate with field measurements of rainfall on the ground. Comparisons of rainfall predictions using WSR-88D with rainfall measurements using dense networks of rain gages have shown that WSR-88D estimates tend to be about 5%–10% lower than long-term rain-gage measurements (Johnson et al., 1999), and comparisons of WSR-88D with daily rain-gage measurements have shown significant spatial variability in radar accuracy, with typical annual bias on the order of −15%, correlation between radar

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Rainfall

405

FIGURE 9.4: Weather radar station (WSR-88D) Source of close-up view: NOAA (2012a).

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(a) Far view

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(b) Near view

and rain-gage measurements on the order of 0.7, and typical agreement in detecting the occurrence of rainfall on the order of 85% (Young and Brunsell, 2008). Errors in WSR88D estimates of rainfall have been shown to depend on the type of rainfall, with opposite biases for frontal and convective rainfall (Skinner et al., 2009). In spite of measurement inaccuracies, radar provides the advantage of covering large areas with high spatial and temporal resolution. Spatially distributed rainfall measurements provided by NEXRAD and other high-resolution weather radar systems have been incorporated directly into rainfallrunoff models to provide improved estimates of runoff distribution (e.g., Gires et al., 2012, Vieux et al., 2009). For example, real-time high-resolution NEXRAD measurements within 1 km * 1 km (0.6 mi * 0.6 mi) cells have been combined with hydrologic models to provide near-real-time flood alerts in the Brays Bayou Watershed in Houston, Texas (Fang et al., 2008). Care must be taken in using radar measurements to estimate short-term (e.g., 10-min) rainfall, since radars sample precipitation aloft and by the time the precipitation reaches the ground, wind-induced drift may have caused a horizontal shift of several kilometers (Islam and Rasmussen, 2008). Such conditions occur during intense convective storms, which typically have strong spatial and temporal gradients of intensity. Errors in short-term radar-measured rainfall can sometimes be reduced by increased spatial or temporal averaging (e.g., Knox and Anagnostou, 2009). 9.2.2

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Statistics of Rainfall Data

Rainfall measurements are seldom used directly in engineering design, but rather the statistics of rainfall measurements are typically used. Rainfall statistics are most commonly presented in the form of intensity-duration-frequency (IDF) curves, which express the relationship between the average intensity in a rainstorm and the averaging time (= duration), with the average intensity having a given probability of occurrence. Such curves are also called intensity-frequency duration (IFD) curves (e.g., Gyasi-Agyei, 2005). A typical IDF curve (for Miami, Florida) for return periods ranging from 2 to 100 years is shown in Figure 9.5. To fully understand the meaning and application of the IDF (or IFD) curve, it is best to review how this curve is derived from rainfall measurements. The data required to calculate the IDF curve are a record of rainfall measurements in the form of the depth of rainfall during fixed intervals of time, t, typically on the order of 5 min. For a rainfall record containing several years of data, the following computations lead to the IDF curve:

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions

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Rainfall intensity in inches per hour

FIGURE 9.5: Intensityduration-frequency (IDF) curve Source: Florida Department of Transportation (2000).

15

15

10 9 8 7 6 5

10 9 8 7 6 5

2 Years 3 Years 5 Years 10 Years 25 Years 50 Years 100 Years

4 3

4 3 2

2

1.0 0.9 0.8 0.7 0.6 0.5

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1.0 0.9 0.8 0.7 0.6 0.5

Miami, FL Zone 10

0.4

0.4 0.3

0.3

0.2 0.8 10

15

20 Min

30

40 50 60

2

3

4

5

10

15

20 24

0.2

H

Duration

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Step 1: For a given duration of time (= averaging period), starting with t, determine the annual-maximum rainfall (AMR) for this duration in each year. Step 2: The AMR values, one for each year, are rank-ordered, and the return period, T, for each AMR value is estimated using the Weibull formula n + 1 T= m

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(9.1)

where n is the number of years of data and m is the rank of the data corresponding to the event with return period T. As an alternative to using Equation 9.1 to estimate the return periods of the AMR values, an extreme-value distribution (typically Type I or Type II) may be fitted to the AMR values for the given duration, t (e.g., Koutsoyiannis et al., 1998). Step 3: Steps 1 and 2 are repeated, with the duration increased by t. An upper-limit duration of interest needs to be specified, and for urban-drainage applications the upper-limit duration of interest is typically on the order of 1–2 h. Step 4: For each return period, T, the AMR versus duration can be plotted, and this relationship is called the depth-duration-frequency curve. Dividing each AMR value by the corresponding duration yields the average intensity, which is plotted versus the duration, for each return period, to yield the IDF curve. This procedure is illustrated in the following example.

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407

EXAMPLE 9.1 A rainfall record contains 32 years of rainfall measurements at 5-min intervals. The annual-maximum rainfall amounts for intervals of 5 min, 10 min, 15 min, 20 min, 25 min, and 30 min have been calculated and ranked. The top three annual-maximum rainfall amounts, in millimeters, for each time increment are given in the following table:

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t in min 15 20

Rank

5

10

1 2 3

12.1 11.0 10.7

18.5 17.9 17.5

24.2 22.1 21.9

28.3 26.0 25.2

25

30

29.5 28.4 27.6

31.5 30.2 29.9

Calculate the IDF curve for a return period of 20 years. Solution For each time interval, t, there are n = 32 ranked rainfall amounts of annual maxima. The relationship between the rank, m, and the return period, T, is given by Equation 9.1 as

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32 + 1 33 n + 1 = = m m m

The return period can therefore be used in lieu of the rankings, and the given data can be put in the form: Return period, T (years)

5

33 16.5 11

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10

12.1 11.0 10.7

18.5 17.9 17.5

24.2 22.1 21.9

28.3 26.0 25.2

25

30

29.5 28.4 27.6

31.5 30.2 29.9

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The rainfall increments with a return period, T, of 20 years can be linearly interpolated between the rainfall increments corresponding to T = 33 years and T = 16.5 years to yield: Duration (min) Rainfall (mm)

5 11.2

10 18.0

15 22.5

20 26.5

25 28.6

30 30.5

and the average intensities for each duration are obtained by dividing the rainfall increments by the corresponding duration to yield:

Duration (min) Intensity (mm/h)

5 134

10 108

15 90

20 79

25 69

30 61

These points define the IDF curve for a return period of 20 years, and this IDF curve is plotted in Figure 9.6.

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FIGURE 9.6: IDF curve for 20-year return period

140 130 Return period ⫽ 20 years 120

Intensity (mm/h)

110 100 90 80

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70 60

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0

5

10

15 20 Duration (min)

25

30

35

Many of the data used in deriving IDF curves are derived from tipping-bucket rain-gage measurements. These gages, the most popular type of rain gage employed worldwide, are known to underestimate rainfall at high intensities because of the rain missed during the tipping of the bucket. Unless the rain gage used in deriving the IDF curve is dynamically calibrated, or the rain gage is self-calibrating, the derived IDF curve can lead to significant underdesign of urban drainage systems (Molini et al., 2005). EXAMPLE 9.2

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How many years of rainfall data are required to derive the IDF curve for a return period of 10 years? Solution The return period, T, is related to the number of years of data, n, and the ranking, m, by n + 1 T= m For T = 10 years and m = 1, n = mT − 1 = (1)(10) − 1 = 9 years

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Therefore, a minimum of 9 years of data are required to estimate the IDF curve for a 10-year return period.

Partial-duration series. The previous example has illustrated the derivation of IDF curves from n annual maxima of rainfall measurements, where the series of annual maxima is called the annual series. As an alternative to using an n-year annual series to derive IDF curves, a partial-duration series is sometimes used in which the largest n rainfall amounts in an n-year record are selected for each duration, regardless of the year in which the rainfall amounts occur. In this case, the return period, T, assigned to each rainfall amount is still calculated using Equation 9.1. Analysis of the annual-maximum series produces estimates of the average period between years when a particular value is exceeded, while analysis of the partial-duration series produces estimates of the average period between values of a particular magnitude. The frequency distribution of rainfall amounts derived using the

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TABLE 9.1: Factors for Converting Partial-Duration Series to Annual Series

Return period (years) 2 5 10 25 >25

Factor 0.88 0.96 0.99 1.00 1.00

Source: Frederick et al. (1977).

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partial-duration series differs from that derived using the annual series. The relationship between the return period of an event derived from an annual series, Ta , and the return period of that same event derived from a partial-duration series, Tp , can be estimated by (Chow, 1964) 1  Tp = −  (9.2) ln 1 − Ta−1

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In general, Tp < Ta , with the difference between Tp and Ta decreasing as Ta increases. A rainfall amount with a return period derived from a partial-duration series can be converted to a corresponding rainfall amount for an annual series by using the empirical factors given in Table 9.1. These factors are applicable for return periods greater than 2 years. Partialduration and annual series are approximately the same at the larger recurrence intervals, but for smaller recurrence intervals the partial-duration series will normally indicate events of greater magnitude. For rainfall records shorter than 20–25 years it is recommended that return periods be estimated using partial-duration series (NRC, 2009).

EXAMPLE 9.3

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A 10-year rainfall record measures rainfall increments at 5-min intervals. The top six rainfall increments derived from the partial-duration series are as follows: Rank 5-min rainfall (mm)

1 22.1

2 21.9

3 21.4

4 20.7

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6 19.8

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Estimate the frequency distribution of the annual maxima with return periods greater than 2 years. Solution The ranked data were derived from a 10-year partial-duration series (n = 10), where the return period, T, is given by 10 + 1 11 n + 1 = = T= m m m Using this (Weibull) relation, the frequency distribution of the 5-min rainfall increments is given by Return period (y) 5-min rainfall (mm)

11 22.1

5.5 21.9

3.7 21.4

2.8 20.7

2.2 20.3

1.8 19.8

The calculated rainfall increments for the partial-duration series can be converted to corresponding rainfall increments for the annual series of maxima by using the factors in Table 9.1. Applying linear interpolation in Table 9.1 leads to the following conversion factors: Return period (y) Factor

11 0.99

5.5 0.96

3.7 0.93

2.8 0.90

2.2 0.89

1.8 —

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions Applying these factors to the partial-duration frequency distribution leads to the following frequency distribution of annual-maxima 5-min rainfall increments: Return period (y) 5-min rainfall (mm)

11 21.9

5.5 21.0

3.7 19.9

2.8 18.6

2.2 18.1

This analysis can be repeated for different durations in order to determine the annual series IDF curves from a partial-duration series.

ww

The previous problems demonstrate a basic approach in which IDF curves are constructed from measured rainfall data. The main drawback of this approach is that it does not produce smooth IDF curves for short rainfall records. Alternative approaches to constructing IDF curves include: (1) fitting extreme-type probability distributions (such as the Gumbel and GEV distributions) to observed rainfall maxima (e.g., Koutsoyiannis et al., 1998; Overeem et al., 2007), (2) fitting a time-series model of rainfall to continuous rainfall records and then constructing the IDF curve from the simulated rainfall time series (e.g., Langousis and Veneziano, 2007), and (3) using marginal and hybrid methods (e.g., Veneziano et al., 2007). Particular care should be taken in using IDF curves estimated for short-duration (e.g., 5 min) storms where available data are limited (Di Baldassarre et al., 2006). At locations where a long time series of daily rainfall measurements exists and there is only a short record of subdaily rainfall measurements (used to derive the IDF curve), effective methods have been proposed for disaggregating the daily measurements to supplement the derivation of the local IDF curve (e.g., Wu et al., 2009).

w .E asy En g 9.2.2.1

Rainfall statistics in the United States

The frequency distributions of local rainfall in the United States have been published by Hershfield (1961) for storm durations from 30 min to 24 h, and return periods from 1 to 100 years. Hershfield’s paper is commonly referred to as TP-40 (an acronym for the U.S. Weather Bureau∗ Technical Paper Number 40, published by Hershfield in 1961) or simply the Rainfall Frequency Atlas. The frequency distributions in TP-40 were derived from data at approximately 4000 stations by assuming a Gumbel distribution. The data in TP-40 for the 11 western states have been updated by Miller et al. (1973). These 11 western states are Montana, Wyoming, Colorado, New Mexico, Idaho, Utah, Nevada, Arizona, Washington, Oregon, and California. Frequency distributions for local rainfall in Alaska can be found in TP-47 (Miller, 1963), for Hawaii in TP-43 (U.S. Department of Commerce, 1962), and for Puerto Rico and the Virgin Islands in TP-42 (U.S. Department of Commerce, 1961). The U.S. Weather Bureau has also published rainfall frequency maps for storm durations from 2 to 10 d (USWB, 1964) in TP-49. In designing urban drainage systems, rainfall durations of less than 60 min and sometimes as brief as 5 min are commonly used. The rainfall statistics for durations of 5–60 min have been published by Frederick et al. (1977) for the eastern and central United States. This paper is commonly referred to as HYDRO-35 (an acronym for the NOAA Technical Memorandum Number NWS HYDRO-35), and these results partially supersede those in TP-40. In most of the aforementioned publications, rainfall frequency distributions for durations less than 1 d were derived by assuming a constant ratio between subdaily rainfall amounts and daily rainfall amounts in the annual maxima—an assumption that has been shown to be of limited accuracy (Faiers and Keim, 2008). However, this does not necessarily compromise the utility of the derived rainfall frequency distributions in engineering applications. Revised rainfall frequency atlases in the United States are published aperiodically by the National Weather Service, and the latest documents for any area of interest should generally be consulted. 9.2.2.2

ine eri n

g .n

et

Secondary estimation of IDF curves

Although IDF curves are available for many locations within the United States, usually from a local water-management or drainage district, there are some locations for which IDF curves are not available. In cases where local IDF curves are not available, they can be estimated from the published National Weather Service frequency distributions in TP-40 using a ∗ The United States Weather Bureau is now called the National Weather Service.

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.8

.6 .6

1

.6 .6 .6

1 .8

ww

.6 .8

1

1.2

1.4 1.4

.6

1

1.2

1.4

.8

1 1.2 1.2 1.2 1.2 1 1.6 1.6

1

1

.6 .8 1.2 2

1.6

1.2

1 1 1

Pacific Ocean

w .E asy En g 1

1.2 1.4 1.6 1.8 2.22 2.2 2 1.8 1.6 1.4

.6

1.2 1 1.4

1.2

1.4 1.4

1.4 1.6 1.8 2 2.2

2.4

2.6 2.8

3.4

1.6 1.8 2

2.2 2.4 2.6

2.8

ine eri n 3.6 3.8

3.6 3.4 3.8

3.4 3.4 3.2 3

3

3

Gulf of Mexico

3.2

FIGURE 9.7: Ten-year 1-h rainfall (inches) Source: Hershfield (1961).

3

3 2.8 6 2. 2.6 2.8 3 3.2 3.4 3.6 3.8 3.8 3.6 3.4 3 3 3.2

Atlantic Ocean

g .n

et

411

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ww 6 4

4 3

6

2 2 2.5

8

1.5

4

6

6

8

3.5

3

2.5 3

w .E asy En g

4 3

2 3

6

6

1.5

8

6 6 8 6 6 8 6

5

8 6

1.5

8 3

6 6

6

4 2 4 1.5 6 4 4 6

6

1.5

3

8

6

4

4

3

6

3 4 4 322 2.5 3

Pacific Ocean

5

6

2.5

2.5

6

3

4 3.5 3.5

3.5 3 3 3.5

6

ine eri n 10

9

4

Atlantic Ocean

g .n 8

9

8

6

7

7 6

9 10

5

6

Gulf of Mexico 7

8

7

9

8

et

FIGURE 9.8: Ten-year 24-h rainfall (inches) Source: Hershfield (1961).

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1

1

1

ww

1

1

1

1.5 2 2

2

1.5

1.5

1 1.5 2 2.5 2.5 2

1

1.5

2

w .E asy En g

1.5 1.5 1.5

1.5

2

2.5

Pacific Ocean

3

3 2.5 2

2 2.5

2 2.5 3

3.5

4

ine eri n 4.5

3 3.5

5

4

5

4.5 4.5 4

4

Gulf of Mexico 4.5

FIGURE 9.9: One-hundred-year 1-h rainfall (inches) Source: Hershfield (1961).

Atlantic Ocean

4

3.5 3.5 4 4.5 5

g .n 4

4

5 4.5

et

413

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions

methodology proposed by Chen (1983). This method is based on three rainfall depths derived 10 from TP-40: the 10-year 1-h rainfall (R10 1 ) shown in Figure 9.7, the 10-year 24-h rainfall (R24 ) 100 shown in Figure 9.8, and the 100-year 1-h rainfall (R1 ) shown in Figure 9.9. The IDF curve can then be expressed as a relationship between the average intensity, i (in./h), for a rainfall duration, t (min), by a i= (9.3) (t + b1 )c1 where a is a constant given by a=

a1 R10 1





Tp (x − 1) log 10





+ 1

(9.4)

10 a1 , b1 , and c1 are empirical functions of R10 1 /R24 derived from Figure 9.10, x is defined by

FIGURE 9.10: Constants in Chen IDF curve Source: Chen (1983).

w .E asy En g

R100 1

(9.5)

R10 1

and Tp is the return period derived from a partial-duration series, which is assumed to be related to return period, Ta , for the corresponding annual-maximum series by Equation 9.2. An IDF curve can be developed for any location using the Chen (1983) method, which is appropriate for storm durations ranging from 5 min to 24 h and return periods greater than or equal to 1 year. In applying the Chen method within the United States, it is important to 10 note that values of R10 1 , R24 , interpolated from Figures 9.7 and 9.8, respectively, have standard errors of at least 10%, and values of R100 1 interpolated from Figure 9.9 have standard errors of at least 20% (Hershfield, 1961). a1 40

b1 16

c1 1.0

15 35

14 13 12

30

0.8

11 10

25

0.9

0.7

9 8

0.6

7 20

6

0.5

5 15

4 3 2

10

0.3

1 0

5

0.4

Values of standard storm parameters

ww

x=

ine eri n

c1

g .n

et

b1

a1

0.2

–1 –2

0.1

–3 0

–4

0

0

10 20 30 40 50 60 Ratio of 1-h to corresponding 10 24-h rainfall depth (R 10 1 / R 24 ), in percent

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415

The Chen method has been widely applied both within and outside the United States; however, in countries such as Canada, India, and Italy, alternative generic IDF curves have proven to be more appropriate than the Chen method (Alila, 2000). Other formulations for the IDF curve have also been proposed for use within the United States (e.g., Froehlich, 2010). Urbanization can have a significant effect on local rainfall characteristics (e.g., Ntelekos et al., 2008), and the stationarity of IDF curves derived from historical data is sometimes questionable. For example, predevelopment rainfall characteristics in the cities of St. Louis, Phoenix, and Houston are demonstrably different from current rainfall characteristics (Huff and Vogel, 1978; Balling and Brazel, 1987; Burian and Shepherd, 2005). Some climate-model simulations indicate that global warming will lead to increased rainfall intensities and hence alter IDFs based on historical data (Guo and Senior, 2006; Mailhot and Duchesne, 2010). The extent of this effect will vary by location and could potentially have a significant effect on drainage design.

ww

EXAMPLE 9.4 Estimate the IDF curve for 50-year storms in Miami using the Chen method. What is the average intensity in a 50-year 1-h storm?

w .E asy En g

10 100 = 4.7 in. Solution For Miami, R10 1 = 3.6 in. (Figure 9.7), R24 = 9 in. (Figure 9.8), and R1 (Figure 9.9). For an annual return period, Ta , equal to 50 years,

Tp = − x=

1

ln(1 − Ta−1 )

R100 1 R10 1

=

=−

1 ln(1 − 50−1 )

= 49.5 years

4.7 = 1.31 3.6

10 Since R10 1 /R24 = 3.6/9 = 0.4 = 40%, then Figure 9.10 gives a1 = 22.8, b1 = 7.5, and c1 = 0.74. Equation 9.4 gives

ine eri n

a = a1 R10 1 [(x − 1) log(Tp /10) + 1] = (22.8)(3.6)[(1.31 − 1) log(49.5/10) + 1] = 99.8

and therefore the IDF curve is given by i=

99.8 a = (t + b1 )c1 (t + 7.5)0.74

For a storm of duration, t, equal to 1 h (= 60 min), i=

99.8 = 4.4 in./h (60 + 7.5)0.74

g .n

et

A 50-year 1-h storm in Miami therefore has an average intensity of 4.4 in./h (110 mm/h).

Wenzel (1982) developed empirical IDF curves for several large cities in the United States using the Chen (1983) form of the IDF curves given by Equation 9.3. Values of a, b1 , and c1 for the IDF curves corresponding to a 10-year return period are given in Table 9.2. These IDF curves are particularly useful to engineers, since most drainage systems are designed for rainfall events with return periods on the order of 10 years. Functional forms of the IDF curve other than that given by Equation 9.3 have been proposed and are widely used in practice. Three common examples are i=

aT m , (b + t)n

i = a + b(ln t) + c(ln t)2 + d(ln t)3 ,

i=

a tb

(9.6)

where i is the average rainfall intensity, t is the duration, T is the return period, and a, b, c, d, m, and n are locally calibrated constants. For the latter two expressions in Equation 9.6, the constants are functions of the return period, T.

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions TABLE 9.2: Ten-Year IDF Constants for Major U.S. Cities

City Atlanta, Georgia Chicago, Illinois Cleveland, Ohio Denver, Colorado Helena, Montana Houston, Texas Los Angeles, California Miami, Florida New York, New York Olympia, Washington Santa Fe, New Mexico St. Louis, Missouri

ww

a

b1

c1

64.1 60.9 47.6 50.8 30.8 98.3 10.9 79.9 51.4 6.3 32.2 61.0

8.16 9.56 8.86 10.50 9.56 9.30 1.15 7.24 7.85 0.60 8.54 8.96

0.76 0.81 0.79 0.84 0.81 0.80 0.51 0.73 0.75 0.40 0.76 0.78

Source: Wenzel (1982).

w .E asy En g 9.2.3

Spatial Averaging and Interpolation of Rainfall

Surface runoff from catchment areas is typically estimated using rainfall amounts that are spatially averaged over the catchment. Spatially averaged rainfall is sometimes referred to as mean areal precipitation (U.S. Army Corps of Engineers, 2000). Rainfall is never uniformly distributed in space, and spatially averaged rainfall tends to be scale dependent (i.e., it depends on the size of the averaging area). Consider the averaging area and distribution of rain gages illustrated in Figure 9.11. The objective is to estimate the average rainfall over the averaging area based on rainfall measurements at the individual rain gages. The basis of the estimation scheme is an interpolation function that estimates the rainfall at any point in the given area in terms of rainfall measurements at individual rain gages. A linear interpolation function has the form

ine eri n

ˆ P(x) =

N 

λi P(xi )

(9.7)

i=1

g .n

ˆ where P(x) is the interpolated rainfall at location x, P(xi ) is the measured precipitation at rain gage i that is located at xi , λi is the weight given to the measurement at station i, and these weights generally satisfy the relation N  i=1

FIGURE 9.11: Estimation of spatially averaged rainfall

Rain gage

λi = 1

et

Averaging grid

Boundary of averaging area

Δy O Δx

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(9.8)

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Rainfall

417

There are a variety of ways to assign the weights, λi , depending on the underlying assumptions about the spatial distribution of the rainfall. Some of the more common approaches are as follows: Uniform-Weight Method. This method assumes that the rainfall is uniformly distributed within a given area. Under this assumption, equal weight is assigned to each station within a predefined area such that 1 λi = (9.9) N The estimated rainfall at any point is simply equal to the arithmetic average of the measured data within the given area.

ww

Geostatistical Methods. The weights, λi , are calculated using geostatistical methods of which kriging is the most common. In the widely used ordinary kriging approach, the weights, λi , are given by the solutions to the following set of linear equations, N 

λj γ (sij ) − μ = γ (si0 ),

w .E asy En g j=1

N  i=1

i = 1, . . . , N

(9.10)

λi = 1

(9.11)

where μ is a constant (called the Lagrange multiplier), sij is the distance between measurement locations xi and xj , si0 is the distance between the measurement location xi and the prediction location x, and γ (s) is the semivariogram which measures the degree of dissimilarity between observations. The experimental semivariogram, γˆ (s), is calculated from measured data using the relation γˆ (s) =

ine eri n

Ns 1  [P(xi ) − P(xi + s)]2 2Ns

(9.12)

i=1

g .n

where Ns is the number of pairs of measurement locations that are a distance s apart. In the general case, γ (s) might be a function of the direction of s, but in most cases this dependence is neglected. Experimental semivariograms as calculated using Equation 9.12 are typically erratic and are commonly approximated by smoothed curves such as the exponential semivariogram given by 

 s γ (s) = a 1 − exp − b



et

(9.13)

where a and b are constants called the “scale” and “range,” respectively. Several other semivariogram models in addition to Equation 9.13 are also used in practice (e.g., Goovaerts, 1997). Equations 9.10 and 9.11 yield N + 1 equations in N + 1 unknowns (λi , i = 1, . . . , N; μ) which give the station weights necessary to interpolate the rainfall at location x given measured rainfall at N nearby stations. Nearest-Neighbor Method. The rainfall at any point is estimated by the rainfall at the nearest station. Under this assumption, λi = 1 for the nearest station, and λi = 0 for all other stations. This approach is equivalent to the Thiessen polygon approach (Thiessen, 1911; Thiessen and Alter, 1911) using the grid method of implementation (Han and Bray, 2006). This binary allocation of weights (λi = 0 or 1) should not be used to

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions

estimate rainfall depths in mountainous watersheds, since elevation is also a strong factor influencing the areal distribution. Inverse-Distance Method. The weight assigned to each measurement station is inversely proportional to the distance from the estimation point to the measurement station. This approach is frequently referred to as the reciprocal-distance approach (Wei and McGuinness, 1973; Simanton and Osborn, 1980) or inverse-distance-weighted approach (Zhang and Srinivasan, 2009). A particular example of the reciprocal-distance approach is the inverse-distance-squared method in which the station weights are given by 1 λi =

ww

d2i

(9.14)

N  1 i=1

d2i

where di is the distance from the estimation location to station i, and N is the number of stations within some defined radius of where the rainfall is to be estimated. In general, the distance need not be squared but could be raised to some other power. In applying the inverse-distance-squared method, the U.S. Army Corps of Engineers (USACE, 2000) recommends centering North/South and East/West reference axes at the location where the rainfall is to be estimated, and then using the nearest measurement within each of the four quadrants to estimate the rainfall.

w .E asy En g

Other Methods. Other, less common, weighting schemes that have been shown to provide superior rainfall interpolation under particular regional (e.g., mountainous) conditions are the multiquadric-biharmonic method (Garcia et al., 2008), thin plate smoothing splines (Saghafian and Bondarabadi, 2008), and multiple linear regression (Hrachowitz and Weiler, 2011).

ine eri n

There is no generally superior rainfall-interpolation method, since the accuracy of interpolation methods generally depend on the characteristics of the study area, data-point density, and the spatial pattern of rainfall. Existing rainfall studies and regulatory requirements within a particular area might provide sufficient guidance for selecting an appropriate interpolation method. After specifying the station weights in the rainfall interpolation formula, Equation 9.7, the next step is to numerically discretize the averaging area by an averaging grid as indicated in Figure 9.11. The definition of the averaging grid requires specification of the origin, O; discretizations in the x- and y-directions, x and y; and specification of the number of cells ˆ j ), at the center, xj , of each cell is then in each of the coordinate directions. The rainfall, P(x calculated using the interpolation formula with specified weights, and the average rainfall over the entire area, P, is given by

g .n

et

J

P=

1 ˆ P(xj )Aj A

(9.15)

j=1

where A is the averaging area, J is the number of cells that contain a portion of the averaging area, and Aj is the amount of the averaging area contained in cell j. This method is well suited to spreadsheet calculation or to computer programs written specifically for this task.

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419

EXAMPLE 9.5 The spatially averaged rainfall is to be calculated for the catchment area shown in Figure 9.12. There are five rain gages in close proximity to the catchment area, and the Cartesian coordinates (x,y) and rainfall, P, measured at these gages during a particular 1-h interval are as follows:

ww

Gage

x (km)

A B C D E

1.3 1.0 4.2 3.5 2.1

y (km) 7.0 3.7 4.9 1.4 −1.0

P (mm) 60 90 65 35 20

Use the Thiessen grid method with the 1-km * 1-km grid shown in Figure 9.12 to estimate the average rainfall over the catchment area during the 1-h interval.

w .E asy En g

Solution Using the Thiessen grid method, the weights, λi , are assigned to each cell in Figure 9.12 using the convention that the nearest station has a weight of 1, and all other stations have weights of zero. Computations are summarized in Table 9.3, where the row numbers increase from the bottom to the top and the column numbers increase from left to right in Figure 9.12. The station weights assigned to each cell, based on the rain gage closest to the center of the cell, are shown in Columns 3 to 7 in Table 9.3, and the area, Ai , of the catchment contained in each cell, i, is shown in Column 8. The rainfall amount, Pi , assigned to each cell is equal to the rainfall at the nearest station (according to the Thiessen grid method) and is given in Column 9. The total area, A, of the catchment is obtained by summing the values in Column 8 and is equal to 28.91 km2 . The average rainfall over the catchment, P, is given by

P=

ine eri n

1  1 (1707.2) = 59 mm Pi Ai = A 28.91

The average rainfall over the entire catchment is therefore equal to 59 mm.

FIGURE 9.12: Catchment area

y Column 6 Row 7

1 km A 1 km

g .n

et

C Catchment B

7 km

D

Column 1 Row 1

x

E 6 km

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions TABLE 9.3: Calculation of Catchment-Averaged Rainfall

(1)

(2)

(3)

(4)

Row

Column

A

B

1

1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6

(5)

(6)

(7)

D

E

Weights, λi

2

ww

C

1 1 1 1 1 1 1

w .E asy En g 3

4

5

6

7

Total

1 1 1 1 1

1 1

1 1 1

1 1 1

1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1

(8)

(9)

Area, Ai

Rainfall, Pi

P * A

(km2 )

(mm)

(km2 · mm)

0.01 0.7 0.95 0.9 0.8 0.1 0.15 1 1 1 1 0.3 0.7 1 1 1 1 0.2 0.95 1 1 1 1 0.4 0.6 1 1 1 1 0.75 0 0.7 1 1 1 0.35 0 0.15 0.55 0.45 0.2 0

ine eri n

28.91

20 20 35 35 35 35 20 35 35 35 35 35 90 90 35 35 35 35 90 90 90 65 65 65 90 90 90 65 65 65 60 60 65 65 65 65 60 60 60 65 65 65

g .n

(10)

et

0.2 14 33.25 31.5 28 3.5 3 35 35 35 35 10.5 63 90 35 35 35 7 85.5 90 90 65 65 26 54 90 90 65 65 48.75 0 42 65 65 65 22.75 0 9 33 29.25 13 0

1707.2

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ww

Rainfall

421

Station weights are used in an operational setting where it is not practical to perform more detailed analysis of precipitation fields. For example, the National Weather Service River Forecast System (NWSRFS), which is the tool employed to generate hydrologic forecasts in real time throughout the United States, uses station weights to generate mean areal precipitation (MAP) time series for model calibration and to compute MAP for operational forecasting on large rivers at (typically) 6-h intervals (NWS, 2002). The two primary methods used by the National Weather Service to compute station weights are the Thiessen and inverse-distance-squared weighting methods (NWS, 2002); however, it is recognized that alternative approaches might be superior under certain circumstances (Fiedler, 2003; Zhang and Srinivasan, 2009). In cases where there are significant differences between the elevations of rain gages and the locations where rainfall amounts are to be estimated, kriging (Mair and Fares, 2011), cokriging (Sarangi et al., 2005), and nonstandard techniques (Chang et al., 2005) have been shown to provide improved accuracy compared to the Thiessen and inverse-distance-squared weighting methods. The accuracy of estimated mean areal precipitation over a catchment is directly related to the density of rain gages within the catchment and to the spatial characteristics of storms occurring within the catchment. Accordingly, the National Weather Service recommends that the minimum number of rain gages, N, for a local flood-warning network within a catchment of area A is given by

w .E asy En g

N = 0.73A0.33

(9.16)

where A is in square kilometers. However, even if more than the minimum number of gages are used, not all storms will be adequately gaged. It is interesting to note that rain gages are typically 20–30 cm (8–12 in.) in diameter, and these measurements are routinely used to estimate the average rainfall over areas exceeding 1 km2 (0.4 mi2 ). Clearly, isolated storms may not be measured well if storm cells are located over areas that are not gaged or if the distribution of rainfall within storm cells is very nonuniform over the catchment area. 9.2.4

Design Rainfall

ine eri n

A hypothetical rainfall event corresponding to a specified return period is usually the basis for the design of stormwater-management systems. In contrast to the single-event designstorm approach, a continuous-simulation approach is sometimes used where a historical rainfall record is used as input to a rainfall-runoff model; the resulting runoff is analyzed to determine the peak runoff rate or hydrograph corresponding to a given return period. The design-storm approach is more widely used in engineering practice than the continuoussimulation approach. Design storms can be either synthetic or actual (historic) design storms, with synthetic storms defined from historical rainfall statistics. Synthetic design storms are characterized by their return period, duration, depth, temporal distribution, and spatial distribution. The selection of these quantities for design purposes is described in the following sections. 9.2.4.1

g .n

et

Return period

The return period of a design rainfall should be selected on the basis of economic efficiency. Typical return periods are (ASCE, 2006): ⊲ Two to 15 years, with 10 years most common, for storm sewers in residential areas. ⊲ Ten to 100 years, for storm sewers in commercial and high-value districts. In selecting the return period for a particular project, local drainage regulations should be followed if they exist. An implicit assumption in designing drainage systems for a given return period of rainfall is that the return period of the resulting runoff is equal to the return period of the design rainfall. Based on this assumption, the risk of failure of the drainage system is taken as being equal to the exceedance probability of the design rainfall.

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9.2.4.2

ww

Rainfall duration

The duration of a design storm is usually selected on the basis of the time-response characteristics of the catchment. The time response of a catchment is measured by the travel time of surface runoff from the most remote point of the catchment to the catchment outlet and is called the time of concentration. On small urban catchments with areas less than 40 ha (100 ac), current practice is to select the duration of the design storm as equal to the time of concentration. This approach usually leads to the maximum peak runoff rate for a given return period. For the design of detention basins, however, the duration requiring the largest detention volume is most critical, and several different storm durations may need to be tried to identify the most critical design-storm duration. For catchments with high infiltration losses, the duration of the critical storm associated with the maximum peak runoff rate may be shorter than the time of concentration (Chen and Wong, 1993). Many drainage districts require that the performance of drainage systems be analyzed for a standard 24-h storm with a specified return period, typically on the order of 25 years. It has been shown that 24 h is an adequate design-storm duration for watersheds in Maryland with areas in the range of 5–130 km2 (2–50 mi2 ) (Levy and McCuen, 1999). 9.2.4.3

Rainfall depth

w .E asy En g

The design-rainfall depth for a selected return period and duration is obtained directly from the intensity-duration-frequency (IDF) curve of the catchment. The IDF curve can be obtained from regulatory manuals that govern local drainage designs, estimated from local rainfall measurements, or (in the United States) derived from National Weather Service (NWS) publications such as TP-40 (Hershfield, 1961). 9.2.4.4

Temporal distribution

Realistic temporal distributions of rainfall within design storms are best determined from historical rainfall measurements. In many cases, however, either the data are not available or such a detailed analysis cannot be justified. Under these conditions, the designer must resort to empirical distributions. Frequently used methods for estimating the rainfall distribution in storms are the triangular method, alternating-block method, and the NRCS 24-h hyetograph.

ine eri n

FIGURE 9.13: Triangular rainfall distribution

Normalized intensity, i/iave

Triangular method. A common approximation for the temporal rainfall distribution within a given storm is the triangular hyetograph shown in Figure 9.13. In this approximation, the instantaneous rainfall intensity, i [LT−1 ], is normalized relative to the average rainfall intensity over the rainfall event, iave [LT−1 ], the time, t [T], is normalized relative to the duration of the event, td [T], and the time at which the peak rainfall occurs is denoted by tp [T]. The triangular hyetograph assumes that the peak rainfall intensity is 2.0 times the average rainfall intensity within the storm. Yen and Chow (1980) and El-Jabi and Sarraf (1991) studied a wide variety of storms and found that tp /td was typically in the range of 0.32–0.51, while McEnroe (1986) found much shorter time to peak by considering only 1-h storms with return periods of 2 years and greater.

g .n

et

2.0

tp/td

1.0

t Normalized time, t d

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Determination of the actual rainfall hyetograph from the nondimensional form shown in Figure 9.13 requires that the storm duration, td , and the average intensity, iave , be specified, where the relationship between td and iave is derived from the IDF curve. EXAMPLE 9.6 The IDF curve for 10-year storms in Houston, Texas, is given by i=

2497 (td + 9.30)0.80

where i is the average intensity in mm/h and td is the storm duration in minutes. Assuming that the peak rainfall rate occurs at 40% of the rainfall duration, estimate the triangular hyetograph for a 1-h storm.

ww

Solution From the given data, td = 1 h = 60 min, tp = 0.4td = 0.4(60) = 24 min, and the average intensity, iave , is given by the IDF curve as iave =

2497 2497 = = 84.1 mm/h (td + 9.30)0.80 (60 + 9.30)0.80

w .E asy En g

The peak rainfall intensity, ipeak , is therefore given by ipeak = 2.0iave = 2.0(84.1) = 168 mm/h

FIGURE 9.14: Estimated 10-year 1-h rainfall hyetograph in Houston, Texas

Rainfall intensity, mm/h

The triangular hyetograph for this 10-year 1-h storm is illustrated in Figure 9.14.

200

168 mm/h

150

100 50

24 Time, min

60

ine eri n

g .n

et

Alternating-block method. A hyetograph can be derived from the IDF curve for a selected storm duration and return period using the alternating-block method. The hyetograph generated by the alternating-block method divides a rainfall event into n time intervals of duration t, for a total storm duration of nt. The procedure for determining the alternating-block hyetograph from the IDF curve is as follows: Step 1: Select a return period and duration for the storm. Step 2: Read the average intensity from the IDF curve for storms of duration t, 2t, . . . , nt. Determine the corresponding rainfall depths by multiplying each intensity by the corresponding duration. Step 3: Calculate the difference between successive rainfall depth values. These differences are equal to the amount of rainfall during each additional unit of time t. Step 4: Reorder the rainfall increments into a time sequence with the maximum intensity occurring at the center of the storm, and the remaining rainfall increments arranged in descending order alternately to the right and left of the central block.

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The alternating-block method assumes that the maximum rainfall for any duration less than or equal to the total storm duration has the same return period. Based on this assumption, the IDF curve is said to be “nested” within the hyetograph. Field data confirm that this assumption is very conservative, particularly for longer storms.

EXAMPLE 9.7 The IDF curve for 10-year storms in Houston, Texas, is given by 2497 (t + 9.30)0.80

i=

where i is the average intensity in mm/h, and t is the duration in minutes. Use the alternating-block method to calculate the hyetograph for a 10-year 1-h storm using 9 time intervals. Compare this result with the triangular hyetograph determined in the previous example.

ww

Solution For a 60-min storm with 9 time intervals, the time increment, t, is given by t =

w .E asy En g

60 = 6.67 min 9

The average intensities for storm durations equal to multiples of t are derived from the IDF curve using t = t, 2t, . . . , 9t and the results are given in Column 3 of the following table: (1)

(2)

(3)

(4)

Increment

t (min)

i (mm/h)

it (mm)

(5) Rainfall amount (mm)

1 2 3 4 5 6 7 8 9

6.67 13.33 20.00 26.67 33.33 40.00 46.67 53.33 60.00

272 206 167 142 124 110 99.8 91.2 84.1

30.2 45.8 55.7 63.1 68.9 73.3 77.6 81.1 84.1

30.2 15.6 9.90 7.40 5.80 4.40 4.30 3.50 3.00

(6) Intensity (mm/h)

ine eri n

272 140 89.1 66.6 52.2 39.6 38.7 31.5 27.0

g .n

et

The rainfall for each duration, t, is given in Column 4 (= Col. 2 * Col. 3), the rainfall increments corresponding to the duration increments are given in Column 5, and the corresponding intensities are given in Column 6. In accordance with the alternating-block method, the maximum intensity (= 272 mm/h) is placed at the center of the storm, and the other intensities are arranged in descending order alternately to the right and left of the center block. The alternating-block hyetograph is therefore given by: Time (min)

Average intensity (mm/h)

0–6.67 6.67–13.33 13.43–20.00 20.00–26.67 26.67–33.33 33.33–40.00 40.00–46.67 46.67–53.33 53.33–60.00

27.0 38.7 52.2 89.1 272 140 66.6 39.6 31.5

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In the triangular hyetograph derived in the previous example, the location of the peak intensity can be varied, while in the alternating-block method the peak intensity always occurs at 50% of the storm duration. Also, the triangular hyetograph derived in the previous example resulted in a maximum intensity of 168 mm/h, while the alternating-block method resulted in a much higher maximum intensity of 272 mm/h. This difference is a result of the short duration used to calculate the peak intensity and reflects the conservative nature of the alternating-block method.

The alternating-block method is the recommended approach for constructing frequencybased hypothetical storms; however, the durations of storms constructed using this method should generally be less than 10 d (U.S. Army Corps of Engineers, 2000).

ww

FIGURE 9.15: Geographic boundaries of NRCS rainfall distributions

NRCS 24-hour hyetographs. The Natural Resources Conservation Service (formerly the Soil Conservation Service) has developed 24-h rainfall distributions for four geographic regions in the United States (SCS, 1986). These rainfall distributions are approximately consistent with local IDF curves (constructed in a way similar to the alternating-block method using 6-min time increments), and the geographic boundaries corresponding to these rainfall distributions are shown in Figure 9.15. Types I and IA rainfall distributions are characteristic of a Pacific maritime climate with wet winters and dry summers. The Type I distribution is applicable to Hawaii, the coastal side of the Sierra Nevada in southern California, and the interior regions of Alaska. Type IA storms are representative of low-intensity rainfall associated with frontal storms on the coastal side of the Sierra Nevada and the Cascade Mountains in Oregon, Washington, northern California, and the coastal regions of Alaska. Type II storms are representative of high-intensity thunderstorms in most regions of the continental United States, with the exceptions of the Gulf Coast regions of Texas, Louisiana, Alabama, south Florida, and most of the Atlantic coastline, which are characterized by Type III rainfall, where tropical storms are prevalent and produce large 24-h rainfall amounts. Puerto Rico and the Virgin Islands are characterized by Type II rainfall. The 24-h cumulative rainfall distributions are illustrated in Figure 9.16, where the abscissa is the time in hours, and the ordinate gives the dimensionless precipitation, P/PT , where P is the cumulative rainfall (a function of time) and PT is the total 24-h rainfall amount. The coordinates of the rainfall distributions are given in Table 9.4. The peak rainfall intensity occurs at the time when the slope of the

w .E asy En g

ine eri n

g .n

et

Rainfall distribution Type I Type IA Type II Type III

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FIGURE 9.16: NRCS 24-h rainfall distributions

Fraction of 24-h rainfall

1.0

0.5 IA

III I II

0 0

ww

3

6

9

15 12 Time (h)

18

w .E asy En g

21

24

TABLE 9.4: NRCS 24-h Rainfall Distributions

Time (h)

Type I P/PT

Type IA P/PT

Type II P/PT

Type III P/PT

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.5 13.0 13.5 14.0 14.5 15.0 15.5 16.0

0.000 0.008 0.017 0.026 0.035 0.045 0.055 0.065 0.076 0.087 0.099 0.112 0.126 0.140 0.156 0.174 0.194 0.219 0.254 0.303 0.515 0.583 0.624 0.655 0.682 0.706 0.728 0.748 0.766 0.783 0.799 0.815 0.830

0.000 0.010 0.022 0.036 0.051 0.067 0.083 0.099 0.116 0.135 0.156 0.179 0.204 0.233 0.268 0.310 0.425 0.480 0.520 0.550 0.577 0.601 0.623 0.644 0.664 0.683 0.701 0.719 0.736 0.753 0.769 0.785 0.800

0.000 0.005 0.011 0.017 0.023 0.029 0.035 0.041 0.048 0.056 0.064 0.072 0.080 0.090 0.100 0.110 0.120 0.133 0.147 0.163 0.181 0.203 0.236 0.283 0.663 0.735 0.776 0.804 0.825 0.842 0.856 0.869 0.881

0.000 0.005 0.010 0.015 0.020 0.026 0.032 0.037 0.043 0.050 0.057 0.065 0.072 0.081 0.089 0.102 0.115 0.130 0.148 0.167 0.189 0.216 0.250 0.298 0.500 0.702 0.751 0.785 0.811 0.830 0.848 0.867 0.886

ine eri n

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TABLE 9.4: (Continued)

ww

Time (h)

Type I P/PT

Type IA P/PT

Type II P/PT

Type III P/PT

16.5 17.0 17.5 18.0 18.5 19.0 19.5 20.0 20.5 21.0 21.5 22.0 22.5 23.0 23.5 24.0

0.844 0.857 0.870 0.882 0.893 0.905 0.916 0.926 0.936 0.946 0.956 0.965 0.974 0.983 0.992 1.000

0.815 0.830 0.844 0.858 0.871 0.884 0.896 0.908 0.920 0.932 0.944 0.956 0.967 0.978 0.989 1.000

0.893 0.903 0.913 0.922 0.930 0.938 0.946 0.953 0.959 0.965 0.971 0.977 0.983 0.989 0.995 1.000

0.895 0.904 0.913 0.922 0.930 0.939 0.948 0.957 0.962 0.968 0.973 0.979 0.984 0.989 0.995 1.000

w .E asy En g Source: SCS (1986).

cumulative rainfall distribution is steepest, which for the NRCS 24-h hyetographs are 8.2 h (Type IA), 10.0 h (Type I), 12.0 h (Type II), and 12.2 h (Type III). Comparing the NRCS 24-h rainfall distributions indicates that Type IA yields the least intense storms, Type II the most intense storms, and Type II and Type III distributions are very similar to each other. The NRCS 24-h rainfall distributions apply for all storm frequencies, with each frequency having a different 24-h rainfall depth, PT . Since NRCS 24-h hyetographs are derived to be approximately consistent with local IDF curves for any given return period, they represent an unlikely distribution of rainfall in which annual-maximum rainfall amounts with the same return period occur over all durations within the same storm. Consequently, actual 24-h hyetographs contain lower average rainfall intensities (e.g., Powell et al., 2008; Guo and Hargadin, 2009; Kimoto et al., 2011), and runoff control structures designed using NRCS hyetographs tend to be more conservative than if these designs are based on observed hyetographs. Several empirical expressions have been derived to match the cumulative rainfall dis˜ tributions shown in Figure 9.16 (e.g., Haan et al., 1994; Munoz-Carpena and Parsons, 2004; Froehlich, 2009; Froehlich, 2010a), and these empirical distributions are particularly useful in extracting the rainfall hyetographs with smaller time intervals. It should be kept in mind that the NRCS cumulative rainfall distributions were derived using rainfall increments of 6 min, and therefore extracting hyetographs with 6-min time increments is appropriate.

ine eri n

g .n

et

EXAMPLE 9.8 The precipitation resulting from a 10-year 24-h storm on the Gulf Coast of Texas is estimated to be 180 mm. Calculate the NRCS 24-h hyetograph. Solution On the Gulf Coast of Texas, 24-h storms are characterized by Type III rainfall. The hyetograph is determined by multiplying the ordinates of the Type III hyetograph in Table 9.4 by PT = 180 mm to yield the following select points (this example is for illustrative purposes only and not all hyetograph points are shown):

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ww

9.2.4.5

Time (h)

P/PT

Cumulative precipitation, P (mm)

0 3 6 9 12 15 18 21 24

0 0.032 0.072 0.148 0.500 0.848 0.922 0.968 1.000

0 6 13 27 90 153 166 174 180

Spatial distribution

The spatial distribution of storms is usually important in calculating the runoff from large catchments. For any given return period and duration, the spatially averaged rainfall depth over an area is typically less than the maximum point-rainfall depth measured at any location within the area. The areal-reduction factor (ARF) is defined as the ratio of the areal-average rainfall to the point-rainfall depth and, in the absence of local data, can be estimated using Figure 9.17 (U.S. Weather Bureau, 1957; Hershfield, 1961). In applying the relationships shown in Figure 9.17, it should be understood that the ARF values were estimated as the ratio of the average annual-maximum areal precipitation depth for the analysis period to the average annual-maximum point precipitation depth in the area for the analysis period. ARFs tend to be less when the ratio of point measurements to areal averages in individual storms are considered (Olivera et al., 2008). Areal-average rainfall is usually assumed to be uniformly distributed over the catchment, and for catchment areas less than 25 km2 (10 mi2 ), areal-reduction factors are not recommended (Jens, 1979; World Meteorological Organization, 1983). The areal-reduction factor, F, given in Figure 9.17 can be approximated by the relation (Eagleson, 1972; Leclerc and Schaake, 1972)

w .E asy En g F = 1 − exp

Source: Hershfield (1961).

    1 1 −1.1td4 + exp −1.1td4 − 0.01A

100 Areal-average rainfall as percentage of point rainfall for same return period

FIGURE 9.17: Areal-reduction factor vs. catchment area and storm duration

ine eri n 24 h

g .n

(9.17)

et

90 6h 80 3h 70 1h 60

30 min

50 0

50

100

150

200 250 Area (mi 2 )

300

350

400

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where td is the rainfall duration in hours, and A is the catchment area in square miles. Although Equation 9.17 was based on areal-reduction factors (ARFs) derived from a limited data set many years ago, inclusion of more recent data and re-estimation of the ARFs leads to approximately the same relation (Allen and DeGaetano, 2005). Caution should be exercised in using generalized areal-reduction factors, since local and regional effects may lead to significantly different reduction factors (Huff, 1995; Olivera et al., 2008).

EXAMPLE 9.9 The local rainfall resulting from a 10-year 24-h storm on the Gulf Coast of Texas is 180 mm. Estimate the average rainfall on a 100-km2 catchment. Solution From the given data: td = 24 h, A = 100 km2 = 40 mi2 , and the areal-reduction factor is given by Equation 9.17 as     1

1

ww

F = 1 − exp −1.1td4

+ exp −1.1td4 − 0.01A





1 1 = 1 − exp −1.1(24) 4 + exp −1.1(24) 4 − 0.01(40)

w .E asy En g = 0.97

Therefore, for a local-average rainfall of 180 mm, the average precipitation over a 100-km2 catchment is expected to be 0.97 * 180 mm = 175 mm.

Areal-reduction factors are particularly useful in converting temporal rainfall distributions derived from point measurements to area-averaged temporal rainfall distributions. The conversion process is to simply multiply the ordinates of the point-rainfall distribution by the areal-reduction factor corresponding to the area of the catchment under consideration. The exact pattern of spatial variation in storm depth is normally disregarded. It has been shown that the spatial distribution of rainfall does not significantly affect the runoff hydrograph from a catchment provided that the rainfall correlation length scale exceeds the hillslope length scale and the catchment is not too large, typically smaller than around ´ 1000 km2 (400 mi2 ) (Nicotina et al., 2008). However, in such cases, total rainfall amounts and the temporal variation of rainfall remain as important influences on the runoff hydrograph. In some cases, the effect of the averaging area on spatially averaged rainfall intensity is incorporated directly into the constants of the IDF curve. In these cases, the IDF curves are called intensity-duration-area-frequency (IDAF) curves. For example, in Italy, IDAF curves are commonly expressed in the form (De Michele et al., 2011)

ine eri n

i=

a(T, A) tb(T,A)

g .n

et

(9.18)

where i is the average rainfall intensity, t is the duration, and a and b are functions of the return period, T, and averaging area, A. 9.2.5

Extreme Rainfall

In large water-resource projects it is frequently necessary to consider the consequences of extreme rainfall events. The probable maximum precipitation (PMP) is the theoretically greatest depth of precipitation for a given duration that is physically possible over a given area at a particular geographical location at a certain time of the year (U.S. National Weather Service, 1982). Estimates of PMP are required for design situations in which failure could result in catastrophic consequences. For example, a flood overtopping an embankment dam could breach the dam and result in severe downstream flooding that is much worse than if the dam did not exist. The most common methods used to estimate the PMP are the rational estimation method and the statistical estimation method.

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9.2.5.1

Rational estimation method

The rational estimation method for determining the PMP is based on the relation PMP =

ww



precipitation moisture



* (moisture supply)max

(9.19)

max

where the maximum ratio of precipitation to atmospheric moisture (first term) is obtained from historical rainfall and meteorological records, and the maximum moisture supply (second term) is obtained from meteorological tables of effective precipitable water based on the maximum persisting dew point for a given drainage basin. Studies of PMPs in the United States using the rational estimation method have been conducted by the U.S. National Weather Service (NWS) for the entire country. The results of these studies are contained in several hydrometeorological reports (HMRs) that can be downloaded from the NWS at www.nws.noaa.gov, and the PMP at any location for any storm area and storm duration can be obtained from these reports. The World Meteorological Organization (1986) addresses PMP estimates for regions throughout the world. 9.2.5.2

Statistical estimation method

w .E asy En g

The most widely used statistical method for estimating the PMP was proposed by Hershfield (1961; 1965), and has become one of the standard methods suggested by the World Meteorological Organization (WMO, 1986) for estimating the PMP at any location. The procedure is based on the equation (9.20)

Pm = P + km σP

where Pm is the annual-maximum rainfall amount in a given duration, P and σP are the mean and standard deviation of the annual-maximum rainfall amounts for a given duration, and km is a factor that depends on the storm duration and P. Hershfield (1961) analyzed 24-h rainfall records from 2600 stations with a total of 95,000 station-years of data and recommended taking km as 15 when estimating the PMP; some later investigations have recommended taking km as 20 for 24-h PMPs (National Research Council, 1985). EXAMPLE 9.10

ine eri n

g .n

The annual-maximum 24-h rainfall amounts in a tropical city have a mean of 15.2 cm and a standard deviation of 4.8 cm. Estimate the probable maximum precipitation (PMP) at this location.

et

Solution From the given data: P = 15.2 cm, and σP = 4.8 cm. Taking km = 20 gives the probable maximum precipitation, Pm , as Pm = P + km σP = 15.2 + (20)(4.8) = 111 cm

Therefore, the PMP at this location is 111 cm. In this case, the PMP is 7.3 times higher than the mean annual-maximum rainfall.

An important difference between the statistical estimation method and the rational estimation method, from an engineering viewpoint, is that the rational estimation method implies zero risk of exceeding the PMP, while the statistical method implies that any rainfall amount has a finite risk of being exceeded. A study by Koutsoyiannis (1999) indicates that the PMP amounts calculated using the rational method have return periods on the order of 60,000 years, and the factor km in Equation 9.20 should be taken as a random variable with a generalized extreme value (GEV) distribution given by ⎧  −7.69 ⎫ ⎨ ⎬ 0.13(km − 0.44) F(km ) = exp − 1 + ⎩ ⎭ 0.60

(9.21)

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where F(km ) is the cumulative probability distribution function of the frequency factor km in Equation 9.20. The value of km given by Equation 9.21, for a specified exceedance probability, is used in Equation 9.20 to calculate the PMP depth for a 24-h duration. Estimation of the PMP depth for durations other than 24 h, with the same return period, can be derived from the 24-h rainfall amount using the intensity-duration-frequency (IDF) curve for the particular geographic area. IDF curves can generally be expressed in the form i=

f (T) g(t)

(9.22)

where i is the average rainfall intensity in a storm of duration t and return period T, f (T) is a function of the return period, T, and g(t) is a function of the storm duration, t. For storms with a return period T, Equation 9.22 gives the following relation between the rainfall amount in 24 h and the rainfall in duration t:

ww

f (T) =

P24 Pt g(24) = g(t) 24 t

(9.23)

where Pt and P24 are the rainfall amounts in duration t h and 24 h, respectively. Equation 9.23 simplifies to the more useful form

w .E asy En g

Pt =

t g(24) P24 24 g(t)

(9.24)

The utility of this equation in estimating the rainfall in extreme storms of duration t h is illustrated in the following example.

EXAMPLE 9.11

ine eri n

The annual-maximum 24-h rainfall amounts on a Pacific island have a mean of 32.0 cm and a standard deviation of 15.6 cm. The IDF curve for the 10-year storm on the island is given by the Chen method as i=

2030

(t + 7.25)0.75

g .n

where i is the intensity in mm/h and t is the duration in minutes. Find the annual-maximum 6-h rainfall with a return period of 100,000 years. Solution From the given data, the annual-maximum 24-h rainfall series gives: P = 32.0 cm, and σP = 15.6 cm. For an extreme storm with a return period, T, of 100,000 years, Equation 9.21 gives ⎧  −7.69 ⎫ ⎨ ⎬ 0.13(km − 0.44) 1 = exp − 1 + F(km ) = 1 − (9.25) ⎩ ⎭ 100, 000 0.60

et

where km is the frequency factor with an exceedance probability of 100,000 years. Solving Equation 9.25 gives km = 16.45, and therefore the 24-h rainfall extreme, P24 , with a return period of 100,000 years is given by Equation 9.20 as P24 = P + km σP = 32.0 + 16.45(15.6) = 288.6 cm

(9.26)

The IDF curve for the island is given by the Chen method in the form i=

f (T) f (T) = g(t) (t + 7.25)0.75

If t is expressed in hours, then g(t) is given by g(t) = (60t + 7.25)0.75

(9.27)

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6 g(24) 6 (60 * 24 + 7.25)0.75 (288.6) = 202 cm P = 24 g(6) 24 24 (60 * 6 + 7.25)0.75

Therefore, the 6-h annual-maximum rainfall with a return period of 100,000 years is approximately 202 cm.

It is important to keep in mind that return periods associated with PMP events are generally large, on the order of tens of thousands of years, and cannot be verified in practice. In fact, given that the time scale of climate change is also on the order of tens of thousands of years or even less, estimates of PMP return periods must be considered highly uncertain. In cases where climate change is involved, PMPs derived from numerical atmospheric models are preferable to statistical models (e.g., Ohara et al., 2011).

ww

9.2.5.3

World-record precipitation amounts

The maximum observed rainfall amounts on the Earth for given durations can be approximated by the following relation (World Meteorological Organization, 1983)

w .E asy En g

Pm,obs = 422td0.475

(9.28)

where Pm,obs is the maximum observed point rainfall in millimeters and td is the storm duration in hours. Several of the world’s maximum observed point rainfalls as a function of duration are listed in Table 9.5. As a point of reference, the maximum 1-d rainfall measured in the United States was 109.2 cm (43 in.) in Alvin near Houston, Texas (Bedient et al., 2013).

EXAMPLE 9.12

ine eri n

Estimate the maximum amount of rainfall that has been observed in 24 h.

Solution According to Equation 9.28, for a 24-h storm (td = 24 h) the maximum observed rainfall amount, Pm,obs , is given by Pm,obs = 422td0.475 = 422(24)0.475 = 1910 mm = 191 cm

g .n

This is slightly higher than the actual observed 24-h maximum of 187 cm given in Table 9.5. It is noteworthy that this 24-h rainfall is significantly greater than the average annual rainfall in Miami, Florida (152 cm).

9.2.5.4

Probable maximum storm

et

The spatial and temporal distribution of the probable maximum precipitation (PMP) that generates the most severe runoff conditions is the probable maximum storm (PMS), and the most severe runoff condition, corresponding to the PMS, is called the probable maximum TABLE 9.5: Maximum Observed Point Rainfalls

Duration 1 min 1h 1d 1 week 1 month 1 year

Precipitation (cm) (in.) 3.8 40.1 187.0 465.3 930.0 2646.1

1.5 15.8 73.6 183.2 366.1 1041.8

Location

Date

Barot, Guadeloupe (West Indies) Shangdi, Inner Mongolia, China ´ Cilaos, La Reunion (Indian Ocean) ´ Commerson, La Reunion Cherrapunji, India Cherrapunji, India

November 26, 1970 July 3, 1975 March 15–16, 1952 January 21–27, 1980 July 1–31, 1861 August 1, 1860–July 31, 1861

Sources: National Weather Service, www.nws.noaa.gov; Ward and Trimble (2004); Dingman (2002).

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flood (PMF). The duration of the PMS that causes the largest flood at a site of interest is the critical duration for that drainage basin. The critical duration is determined by routing runoff hydrographs resulting from PMP rainfall of various durations and selecting the duration causing the maximum flood. In general, the critical duration should never be less than the time of concentration of the drainage basin. 9.3 Rainfall Abstractions The processes of interception, infiltration, and depression storage are commonly referred to as rainfall abstractions. These processes must generally be accounted for in estimating the surface runoff resulting from a given rainfall event. 9.3.1

ww

Interception

Interception is the process by which rainfall is abstracted prior to reaching the ground. The wetting of above-ground vegetation is typically the primary form of interception, although rainfall is also intercepted by buildings and other above-ground structures. In urban areas, the density of vegetation varies considerably and interception by vegetation can be significant. An experimental study in Dayton, Ohio, demonstrated that 22% canopy cover reduced runoff by 7%, and 29% canopy cover reduced runoff by 19% (Nowak et al., 1997). In an experimental urban area without vegetation, a rainfall interception rate on the order of 6% over a 5-month period has been reported (Nakayoshi et al., 2009). In areas where there is a significant amount of vegetation, such as forested areas, interception can dramatically reduce the amount of rainfall that reaches the ground. Therefore, in developments that involve the clearing of wooded areas, engineers must be prepared to account for the increased runoff that will occur as a result of reduced interception. Methods used for estimating interception are mostly empirical, where the amount of interception is expressed either as a fraction of the amount of precipitation or as an empirical function of the rainfall amount. Observed interception percentages over seasonal and annual time scales for several types of vegetation are summarized in Table 9.6. These data indicate that, on an (average) annual basis, tree interception can abstract as much as 48% of rainfall (Picea abies) and grasses on the order of 13% of rainfall. Extreme caution should be exercised expressing interception simply as a percentage of rainfall, since individual storm characteristics (intensity and duration), local climate, age and density of the vegetation, and rainfall distribution within the averaging period can all influence the interception percentage, particularly in forested areas. The location and density of rain gages below the canopy can also affect the accuracy of reported interception percentages (Shachnovich et al., 2008). Where possible, local measurements should be used to estimate interception losses. Many interception functions are similar to that originally suggested by Horton (1919), where the interception, I [mm], for a single storm is related to the rainfall amount, P [mm], by an equation of the form I = a + bPn (9.29)

w .E asy En g

ine eri n

g .n

et

where a, b, and n are empirical constants given in Table 9.7 for various types of trees. Application of Equation 9.29 can be problematic since the predicted interception can exceed the precipitation for small values of P, and for tree types where n equals 1.0 the interception approaches a constant rate as the precipitation amount becomes large, which is unrealistic. To address these limitations, a re-analysis of the Horton (1919) data for summer storms indicated that the following equation might also be appropriate I = cPm

(9.30)

where I and P are in millimeters, and best-fit values of the parameters c and m for various types of trees are given in Table 9.7 (ASCE, 1996). Surface vegetation generally has a finite interception capacity, which should not be exceeded by the estimated interception amount. The interception storage capacity of surface vegetation can range from less than 0.3 mm to 13 mm, with a typical value for turf grass of 1.3 mm.

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Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions TABLE 9.6: Interception Percentages in Selected Studies

Cover type Conifers Chamaecyparis obtusa Picea abies Pinus caribaea Pinus nigra Pinus pinaster Pinus radiata

ww

Pinus resinosa Pinus strobus Pinus taeda Pseudotsuga

Season/Period

Interception (%)

2 years Year 50 days Year Year 10 months 30 months Year Year Summer Year Year Year Summer Winter 6 months 2 years 18 months Summer

19 48 26 18 35 12 17 26 19 19 16 14 36 24 14 21 24 39 33

Year 66 days 30 months Year Summer Year Year NA Summer 4 years Summer Year Year Summer Winter

19 52 11 19 60 10 16 12–17 16 39 19 11 33 30 21

w .E asy En g Pseudotsuga menziesii

Tsuga canadensis

Evergreen hardwoods Acacia Dacryodes excelsa Eucalyptus gobulus Eucalyptus regnans Ficus benjamina Forest—Heath (Indonesia) Forest—Rain (Lowland) Forest—Rain (Tropical) Forest—Tropical Lauraceae Melaleuca quinquenervia Mixed eucalypts Notofagus sp. Notofagus/Podocarpus

Deciduous hardwoods Carpinus sp. Fagus grandifolia Fagus silvatica Fraxinus ornus Liriodendron Quercus ilex Quercus serrara

Year Summer Summer Winter Year Year Year 32 months

Reference Murakami (2007) Leyton et al. (1967) Lankreijer et al. (1999) Waterloo et al. (1999) Rutter et al. (1975) Gash et al. (1995) Valente et al. (1997) Feller (1981) Smith (1974) Voigt (1960) Helvey (1967) Swank et al. (1972) Aussenac and Boulangeat (1980) Rothacher (1963) Pypker et al. (2005) Link et al. (2004) Rutter et al. (1975) Voigt (1960)

Beard (1962) Schellekens et al. (1999) Valente et al. (1997) Feller (1981) Guevara-Escobar et al. (2007) Vernimmen et al. (2007) Vernimmen et al. (2007) Marin et al. (2000) Jackson (1971) Fleischbein et al. (2005) Woodall (1984) Smith (1974) Aldridge and Jackson (1973) Rowe (1979)

ine eri n 36 25 21 6 28 10 31 14

g .n

et

Leyton et al. (1967) Voigt (1960) Aussenac and Boulangeat (1980) ˇ Sraj et al. (2008) Helvey (1964) Limousin et al. (2008) Deguchi et al. (2006)

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TABLE 9.6: (Continued)

Season/Period

Interception (%)

Grasses Themeda sp. Cymbopogon sp.

Year Year

13 13

Soil cover Hardwood litter Pinus strobus litter Oak litter Pinus caribaea litter Pinus taeda litter

Year Year Year Year Year

3 3 2 8 4

Cover type

ww

Reference Beard (1962) Beard (1962)

Helvey (1964) Helvey (1967) Blow (1955) Waterloo et al. (1999) Swank et al. (1972)

TABLE 9.7: Parameters in Interpolation Equations

I = a + bPn b 0.18 0.23 0.23 0.20 1.26 1.01 0.23 0.22 1.01 0.40

w .E asy En g Tree type Apple Ash Beech Chestnut Elm Hemlock Maple Oak Pine Willow

a 1.00 0.38 0.51 1.00 0.00 0.00 0.76 0.76 0.00 0.51

I = cPm n 1.0 1.0 1.0 1.0 0.5 0.5 1.0 1.0 0.5 1.0

c 0.60 0.38 0.65 0.63 0.81 0.74 0.61 0.72 0.79 0.70

m 0.73 0.88 0.65 0.77 0.48 0.74 0.77 0.66 0.70 0.85

ine eri n

Note: In the interception formulas I and P are in millimeters.

More sophisticated interception functions have been suggested to account for the limited storage capacity of surface vegetation as well as evaporation of intercepted water during a storm (e.g., Meriam, 1960; Gray, 1973; Brooks et al., 1991). In these interception functions, the interception, I [L], is typically expressed in the form P

I = S(1 − e− S ) + KEt

g .n

et

(9.31)

where S is the available storage [L], P is the amount of rainfall during the storm [L], K is the ratio of the surface area of one side of the leaves to the projection of the vegetation on the ground (called the leaf area index) [dimensionless], E is the evaporation rate during the storm [LT−1 ], and t is the duration of the storm [T]. Determination of S can be an important source of error in estimating interception, and values of S are typically in the range of 3–4 mm for fully developed pine trees; 7 mm for spruce, fir, and hemlock; 3 mm for leafed-out hardwoods; and 1 mm for bare hardwoods (Helvey, 1971). A more recent review reported S values in the range of 0.3–3.0 mm in coniferous forests (Llorens and Gallart, 2000), and values generally less than 1.8 mm in broad-leaved deciduous forests (Deguchi et al., 2006). For P/S W 1 the exponential term in Equation 9.31 is negligible, which would almost certainly be the case for P > 25 mm (1 in.). More sophisticated models of canopy interception, particularly in sparsely forested areas, are still being developed and validated (e.g., Limousin et al., 2008), and models that take into account storage and evaporation from tree trunks have also been proposed (e.g., Gash et al., 1995). A comparative assessment of interception models can be found in the work of Muzylo et al. (2009). It is commonly recommended to use storm-event interception

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models, such as Equation 9.31, in formulating interception models for longer time intervals such as weeks and months (e.g., de Groen and Savenije, 2006). Rainfall that is not intercepted by tree canopies reaches the ground by either stemflow or throughfall. As the names imply, stemflow reaches the ground by flowing down the stems of plants or the trunks of trees, while throughfall reaches the ground directly after passing through the canopies. The distinction between stemflow and throughfall is primarily relevant to the ground ecology via their influence on the distribution of soil moisture and tree roots in the soil below the canopy (e.g., Andre´ et al., 2011). Interception by forest litter is much smaller than canopy interception. The amount of litter interception is largely dependent on the thickness of the litter, water-holding capacity, frequency of wetting, and evaporation rate. Studies have shown that storage in forest litter is only a few millimeters in most cases (Chang, 2002). Typically, about 1%–5% of annual precipitation and less than 50 mm/yr (2 in./yr) are lost to litter interception (Helvey and Patric, 1965).

ww

EXAMPLE 9.13 A pine forest is to be cleared for a commercial development in which all the trees on the site will be removed. The IDF curve for a 20-year rainfall is given by the relation

w .E asy En g

i=

2819 t + 16

where i is the rainfall intensity in mm/h and t is the duration in minutes. The storage capacity of the trees in the forest is estimated as 6 mm, the leaf area index is 7, and the evaporation rate during the storm is estimated as 0.2 mm/h. (a) Determine the increase in precipitation reaching the ground during a 20-min storm that will result from clearing the site; and (b) compare your result with the interception predicted by the Horton-type empirical equation of the form I = a + bPn , where a and b are constants and P is the precipitation amount.

Solution

ine eri n

(a) For a 20-min storm, the average intensity, i, is given by the IDF equation as i=

2819 2819 = = 78 mm/h t + 16 20 + 16

and the precipitation amount, P, is given by

P = it = (78)



20 60



= 26 mm

g .n

et

The interception, I, of the wooded area can be estimated by Equation 9.31, where S = 6 mm, P = 26 mm, K = 7, E = 0.2 mm/h, and t = 20/60 h = 0.33 h. Hence     26 P I = S 1 − e− S + KEt = 6 1 − e− 6 + (7)(0.2)(0.33) = 5.9 + 0.5 = 6.4 mm The wooded area intercepts approximately 6.4 mm of the 26 mm that falls on the wooded area. Prior to clearing the wooded area, the rainfall reaching the ground in a 20-min storm is 26−6.4 = 19.6 mm. After clearing the wooded area, the rainfall reaching the ground is expected to be 26 mm, an increase of 33% over the incident rainfall prior to clearing the wooded area. These calculations also show that evaporation contributes only 0.5 mm of the 6.4 mm intercepted, indicating that evaporation during a storm contributes relatively little to interception. (b) Using the interception formula I = a + bPn for pine woods, it can be assumed from the data in Table 9.7 that n = 0.5, a = 0 mm, and b = 1.01. Hence I = a + bPn = 0 + 1.01(26)0.5 = 5 mm The pine woods are estimated to intercept 5 mm of rainfall, in which case the predevelopment rainfall reaching the ground is 26 mm−5 mm = 21 mm and the postdevelopment rainfall reaching the ground is 26 mm, an increase of 24% over predevelopment conditions.

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There is a common perception that forest harvesting increases the frequency of “lowmagnitude” floods downstream of a harvested area, while the effect of harvesting on the frequency of “high-magnitude” floods is minimal. However, this perception should not be accepted as fact since its scientific bases, particularly as it relates to “high-magnitude” floods, are still being debated (e.g., Alila et al., 2009). 9.3.2

Depression Storage

Rainfall that accumulates in surface depressions during a storm is called depression storage. This portion of rainfall does not contribute to surface runoff; it either infiltrates or evaporates. Depression storage is generally expressed as an average depth over the catchment area, and typical depths of depression storage are given in Table 9.8. These typical values are for moderate ground slopes; the values would be larger for flat slopes, and smaller for steep slopes (Butler and Davies, 2011). In estimating surface runoff from rainfall, depression storage is usually deducted from the initial rainfall.

ww

EXAMPLE 9.14 A 10-min storm produces 12 mm of rainfall on an impervious parking lot. Estimate the fraction of this rainfall that becomes surface runoff.

w .E asy En g

Solution On the impervious parking lot, water trapped in depression storage forms puddles and does not contribute to runoff. Depression storage on impervious areas is typically in the range of 1.3–2.5 mm, with an average value of 1.9 mm. The fraction of runoff, C, is therefore estimated by

12 − 1.9 = 0.84 12 This indicates an abstraction of 16% for a 12-mm storm, which corresponds to 84% runoff. Clearly, the fraction of runoff will increase for higher precipitation amounts. C=

9.3.3

Infiltration

ine eri n

The process by which water seeps into the ground through the soil surface is called infiltration and is usually the dominant rainfall abstraction process. Infiltration capacity is determined primarily by the surface cover and the properties of the underlying soil. Soils covered with grass or other vegetation tend to have significantly higher infiltration capacities than bare soils. Bare-soil infiltration rates are considered high when they are greater than 25 mm/h (1 in./h) and low when they are less than 2.5 mm/h (0.1 in./h); grass cover tends to increase these values by a factor between 3 and 7.5 (Viessman and Lewis, 2003). Soils are typically classified by their particle-size distribution, and the U.S. Department of Agriculture (USDA) soil-classification system based on particle size is shown in Table 9.9.

g .n

et

TABLE 9.8: Typical Values of Depression Storage

Surface type

Depression storage (mm) (in.)

Pavement: Steep Flat Very flat Impervious areas Lawns Pasture Flat roofs Forest litter Cultivated land

0.5 1.5–3.5 3.4 1.3–2.5 2.5–5.1 5.1 2.5–7.5 7.5 1.3–2.5

0.02 0.06–0.14 0.13 0.05–0.1 0.1–0.2 0.2 0.1–0.3 0.3 0.05–0.1

Reference Pecher (1969); Viessman et al. (1977) Pecher (1969); Viessman et al. (1977) ASCE (2006) Tholin and Keifer (1960) Hicks (1944); ASCE (2006) ASCE (1992) Butler and Davies (2011) ASCE (2006) ASCE (2006)

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Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions TABLE 9.9: USDA Soil Classification System

Soil

Particle sizes

Clay Silt Sand Gravel FIGURE 9.18: USDA soil texture triangle

2 mm

10

100

20

90 80

30

60

Sandy clay

Silty clay loam

Clay loam

Loam

Sandy loam

10

Loamy sand

ine eri n Silt

10

20

30

40

50

60

70

80

90

100

Percent sand

Silt loam

90

20

80

Sandy clay loam

0

30

70

40

Sand

50

Silty clay

t

50

sil

Pe

nt

rce

w .E asy En g

10

nt

60

e rc

40

cla

y

Clay Pe

ww

70

g .n

Soil texture is determined by the proportions (by weight) of clay, silt, and sand, after particles larger than sand (>2 mm) are removed. The USDA classification system for defining soil texture is typically expressed in the ternary diagram shown in Figure 9.18, which is commonly called the USDA soil texture triangle. The USDA system classifies the soil into 12 textural classes. If a significant proportion of the soil (>15%) is larger than sand, an adjective such as “gravelly” or “stony” is added to the soil texture specification. The vertical soil profile is generally categorized into horizons that are distinguished by the proportion of organic material and the degree to which material has been removed (eluviated) or deposited (illuviated) by chemical and physical processes. Soil scientists identify soil horizons on the basis of color, texture, and structure, and these horizons are generally designated by letters as follows:

et

O-Horizon. Surface litter consisting primarily of organic matter. A-Horizon. Topsoil consisting of decaying organic matter and inorganic minerals. E-Horizon. The zone of leaching where percolating water dissolves water-soluble matter. B-Horizon. The subsoil below the A- and E-horizons that contain minerals and humic compounds. Usually contains more clay than the A-horizon.

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C-Horizon. A zone consisting primarily of undecomposed mineral particles and rock fragments. This is the soil parent material from which the A- and B-horizons were formed. R-Horizon. Bedrock, an impenetrable layer.

ww

Not all of the soil horizons listed above are present in all soils, and the boundaries between layers are commonly gradational. In engineering practice, it is common to assume that soils are uniform with depth; however, layered soil is much more common (Yang et al., 2006). Bare-soil infiltration rates can be significantly affected by the formation over time of a seal or crust at the soil surface. A surface seal is a very thin layer, typically of thickness 1–5 mm (0.04–0.2 in.), at or just below the soil surface that forms due to the breakdown of soil aggregates and dispersion of clay particles under raindrop impact. The clay particles fill the pores and create a layer with a saturated hydraulic conductivity several orders of magnitude less than the undisturbed soil (Miller and Radcliffe, 1992). Seals can be prevented by protecting the soil surface from raindrop impact with a mulch or crop residue, and by preventing chemical dispersion of clay particles through the use of soluble soil amendments (e.g., Miller and Baharuddin, 1986). Once a seal dries out, it develops a high soil strength due to the increased density of the layer and is called a crust. Quantitative estimation of runoff from rainfall generally requires an analytic description of the infiltration process. A variety of models are used to describe the infiltration process, with no one model being best for all cases. To appreciate the assumptions and limitations of the various models, it is important to look first at the fundamental physics of infiltration.

w .E asy En g 9.3.3.1

The infiltration process

Infiltration describes the entry of water into the soil through the soil surface, and percolation describes the movement of water within the soil. The infiltration rate is equal to the percolation rate just below the ground surface, and the percolation rate, q0 [LT−1 ], is given by the Buckingham–Darcy equation as: ⭸h q0 = −K(θ ) (9.32) ⭸z

ine eri n

where K(θ ) is the vertical hydraulic conductivity [LT−1 ] expressed as a function of the moisture content (= the volume of water per unit volume of the soil), θ [dimensionless]; and h is the piezometric head of the pore water [L] defined by h=

p + z γ

g .n

(9.33)

et

where p is the pore-water pressure [FL−2 ], γ is the specific weight of water [FL−3 ], and z is the vertical coordinate (positive upward) [L]. Since the pore-water pressure, p, is normally expressed as a gage pressure, a negative value of p indicates that the pressure in the pore water is below atmospheric pressure. The pore-water pressure, p, is commonly referred to simply as the pore pressure. In the unsaturated soil immediately beneath the ground surface, the soil moisture is usually under tension, with negative pore pressures. Laboratory and field experiments indicate that there is a fairly stable relationship between the pore pressure, p, and moisture content, θ , that is unique to each soil. The relationship between −p/γ and θ is called the retention curve, and a typical retention curve is illustrated in Figure 9.19(a). The capillary potential, ψ(θ ) [L], defined by ψ(θ ) = −

p γ

(9.34)

is commonly used in lieu of the pressure head and is closely related to the matric potential, which is defined as p/γ . The moisture retention curve, Figure 9.19(a), indicates that when the pores are completely filled with water at atmospheric pressure, the moisture content is equal to the saturated moisture content, θs , and the pore pressure and capillary potential are both equal to zero. The saturated moisture content, θs , is numerically equal to the porosity

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∂ψ(θ)

C(θ) ⫽ − ∂θ

FIGURE 9.19: Typical moisture retention curve and C(θ )

p

Chapter 9

Capillary potential, ψ(θ) ⫽ − γ

440

0

ww

θ0

0

θs

θs

θ0

Moisture content, θ

Moisture content, θ

(a) Moisture retention curve

(b) Derived function, C(θ)

of the soil. As the moisture content is reduced, the pore pressure decreases and the capillary potential increases in accordance with Equation 9.34. This trend continues until the moisture content is equal to θ0 , at which point the pore water becomes discontinuous and further reductions in the pore pressure do not result in a decreased moisture content. Combining the Buckingham–Darcy equation, Equation 9.32, with the definition of the piezometric head, Equation 9.33, yields the following expression for the vertical percolation rate (just below the ground surface) in terms of the capillary potential:

w .E asy En g

⭸h ⭸z    ⭸ p ⭸  −ψ(θ ) + z = −K(θ ) + z = −K(θ ) ⭸z γ ⭸z   ⭸ψ(θ ) = K(θ ) − 1 ⭸z

q0 = −K(θ )

ine eri n

(9.35)

The chain rule of differentiation guarantees that the suction gradient, ⭸ψ/⭸z, is given by ⭸ψ(θ ) ⭸ψ(θ ) ⭸θ = ⭸z ⭸θ ⭸z

g .n

(9.36)

et

where ⭸ψ/⭸θ is a soil property derived from the moisture retention curve, Figure 9.19(a), and ⭸θ/⭸z is called the wetness gradient. Combining Equations 9.36 and 9.35 yields the following relationship between the percolation rate, q0 , and the moisture content, θ :   ⭸ψ(θ ) ⭸θ − 1 q0 = K(θ ) ⭸θ ⭸z

⭸θ = −K(θ ) C(θ ) + 1 (9.37) ⭸z where C(θ ) is a soil property defined by C(θ ) = −

⭸ψ(θ ) ⭸θ

(9.38)

and is derived directly from the moisture retention curve for the soil. The inverse of C(θ ) is sometimes called the specific moisture capacity (e.g., Kemblowki and Urroz, 1999) or the specific water capacity (e.g., Mays, 2001). The functional form of C(θ ) corresponding to a typical moisture retention curve is illustrated in Figure 9.19(b), where it should be noted that C(θ ) is always positive. Although the percolation rate, q0 , just below the ground surface

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given by Equation 9.37 is numerically equal to the infiltration rate, f , Equation 9.37 requires that q0 is positive in the upward direction, while f is conventionally taken as positive in the downward direction. Consequently, the infiltration rate, f , is simply equal to negative q0 , in which case Equation 9.37 yields the following theoretical expression for the infiltration rate:

⭸θ f = K(θ ) C(θ ) + 1 (9.39) ⭸z Practical application of Equation 9.39 requires specification of the hydraulic conductivity function K(θ ). Field experiments indicate that K(θ ) can be adequately described by (Bear, 1979)   θ − θ0 3 K(θ ) = Ks (9.40) θs − θ0

ww

where Ks is the hydraulic conductivity at saturation. According to Equation 9.40, the hydraulic conductivity, K(θ ), increases monotonically from zero to Ks as θ increases from θ0 to θs . Combining Equations 9.39 and 9.40 yields the theoretical infiltration equation 

θ − θ0 θs − θ0

w .E asy En g f = Ks

3

⭸θ C(θ ) + 1 ⭸z

(9.41)

The fundamental infiltration process is illustrated in Figure 9.20 for the case where water is ponded above the ground surface. The initial moisture distribution between the ground surface and the water table (before ponding) approximates an equilibrium distribution where the conditions at the ground surface are described by θ = θ0 and ⭸θ/⭸z = 0. Under these conditions, Equation 9.41 indicates that f =0

ine eri n

(9.42)

Immediately after infiltration begins, the soil just below the ground surface becomes saturated, but it is still unsaturated further down in the soil column, leading to a sharp moisture gradient near the surface. Under these circumstances, θ = θs and ⭸θ/⭸z > 0 at the ground surface. The infiltration rate is given by Equation 9.41 as

⭸θ f = Ks C(θs ) + 1 (9.43) ⭸z

g .n

et

As infiltration proceeds, Ks and C(θs ) remain constant, and the moisture content gradient, ⭸θ/⭸z, gradually decreases to zero as the wetting front penetrates the soil column (see Figure 9.20). Therefore, the infiltration rate gradually decreases from its maximum value Ponded water Ground surface Initial moisture distribution Elevation, z

FIGURE 9.20: Infiltration process

Wetting front

Water table 0

θ0

θs

Moisture content, θ

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Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions

given by Equation 9.43 (⭸θ/⭸z > 0), at the beginning of the infiltration process, to the asymptotic minimum infiltration rate, fmin , when ⭸θ/⭸z = 0 and (9.44)

fmin = Ks

ww

The minimum infiltration rate is therefore equal to the (vertical) saturated hydraulic conductivity of the soil and this condition is called gravity drainage. As the ponded water continues to infiltrate into the soil, conditions near the ground surface remain approximately constant, with infiltration proceeding at the minimum rate. Eventually, the entire soil column becomes saturated and the recharge rate at the water table equals the infiltration rate. In reality, the soil column is typically stratified, with the infiltration rate being eventually limited by the rate of percolation through the least pervious subsoil layer. In homogeneous soils, experiments have shown that the asymptotic minimum infiltration rate is less than the saturated hydraulic conductivity that would be measured on a carefully saturated, intact core in the laboratory. This has been attributed to the effect of entrapped air under field conditions that reduces the cross-sectional area available for water flow (Radcliffe and Rasmussen, 2002). Bouwer (1978; 1966) recommended using a field-saturated hydraulic conductivity equal to 50% of the true saturated hydraulic conductivity for predicting the steady-state (minimum) infiltration rate of sandy soils, and 25% of the saturated hydraulic conductivity for predicting the steady-state infiltration rate of clays and loams. Several models are commonly used to estimate infiltration, and the validity of each of these models should be viewed relative to their consistency with the theoretical infiltration process described here. The models most frequently used to account for infiltration in engineering practice are the Horton, Green-Ampt, and Natural Resources Conservation Service (NRCS) models. No single approach works best for all situations, and in most cases the methods are limited by knowledge of their site-specific parameters. Many of these models distinguish between the actual infiltration rate, f , and the potential infiltration rate, fp , which is equal to the infiltration rate when water is ponded at the ground surface.

w .E asy En g 9.3.3.2

Horton model

ine eri n

Horton (1939; 1940) proposed the following empirical equation to describe the decline in the potential infiltration rate, fp [LT−1 ], as a function of time fp = fc + (f0 − fc )e−kt

[LT−1 ],

g .n

(9.45)

et

where f0 is the initial (maximum) infiltration rate fc is the asymptotic (minimum) infiltration rate (t → q) [LT−1 ], and k is a decay constant [T−1 ]. The potential infiltration rate, fp , is also commonly referred to as the infiltration capacity. It has been shown previously that the asymptotic minimum infiltration rate must be approximately equal to the saturated hydraulic conductivity of the soil. Equation 9.45 describes an infiltration capacity that decreases exponentially with time, ultimately approaching a constant value, and assumes an infiltration process described by the relation dfp = −k(fp − fc ) dt

(9.46)

The Horton model fits well with experimental data (Singh, 1989). Typical values of f0 , fc , and k are given in Table 9.10 (rank-ordered by fc ), and Singh (1992) recommends that f0 /fc be on the order of 5. The variability of the infiltration parameters in the Horton model reflects the condition that infiltration depends on several factors that are not explicitly accounted for in Table 9.10, such as the initial moisture content and organic content of the soil, vegetative cover, season, and degree of compaction (Linsley et al., 1982; Pitt et al., 2008a).

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TABLE 9.10: Typical Values of Horton Infiltration Parameters

fo (mm/h)

Soil type

fc (mm/h)

k (min−1 )

25 12 6 3

0.03 0.03 0.03 0.03

41§ –410§ 21§ –61† 13§ –26† 8§ –13† 7† 4† 2† 1† 1† 1† fp at the end of this interval and i < fp at the beginning of this interval, ponding begins

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Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions within this interval. Ponding occurs when fp = 80 mm/h, and substituting this value into Equation 9.52 yields the corresponding cumulative infiltration as F = 14.3 mm. If ponding begins at time tp when F = 14.3 mm, then   tp − 20 14.3 = 5.01 + 80 60 which yields tp = 27.0 min = 0.450 h. Had ponding occurred continuously from t = 0, then the time t′ to achieve a cumulative infiltration of 14.3 mm can be determined from Equation 9.48 since for continuous ponding f − fc (1 − e−kt ) F(t) = fc t + 0 k ′ 100 − 20 (1 − e−1.8t ) 14.3 = 20t′ + 1.8

ww

which yields t′ = 0.160 h. Using the calculated values of tp and t′ , values of F and fp as a function of time, t, after ponding can be derived from Equations 9.48 and 9.45 as ′ f − fc F(t) = fc (t − tp + t′ ) + 0 [1 − e−k(t−tp +t ) ] k 100 − 20 [1 − e−1.8(t−0.450+0.160) ] = 20(t − 0.450 + 0.160) + 1.8

w .E asy En g

= 20(t − 0.290) + 44.4[1 − e−1.8(t−0.290) ]

(9.53)

′ fp = fc + (f0 − fc )e−k(t−tp +t )

= 20 + (100 − 20)e−1.8(t−0.450+0.160)

= 20 + 80e−1.8(t−0.290)

ine eri n

(9.54)

At the end of this time interval, t = 0.5 h and Equations 9.53 and 9.54 yield F = 18.2 mm and fp = 74.8 mm/h. Since the rainfall amount in this time interval is 80 mm/h * 0.167 h = 13.3 mm and the infiltration amount in this time interval is 18.2 mm − 5.01 mm = 13.2 mm, then the rainfall excess is 13.3 mm − 13.2 mm = 0.1 mm which is stored in depression storage.

t = 30–40 min: Ponding continues at the beginning of this time interval since i > fp (100 mm/h > 74.8 mm/h). At the end of this time interval, t = 40 min = 0.667 h and Equations 9.53 and 9.54 give F = 29.4 mm and fp = 60.6 mm/h. So ponding continues at the end of this time interval, since i > fp (100 mm/h > 60.6 mm/h). Since the rainfall amount is 100 mm/h * 0.167 h = 16.7 mm and the infiltration amount is 29.4 mm − 18.2 mm = 11.2 mm, the rainfall excess is 16.7 mm − 11.2 mm = 5.5 mm, of which 1.9 mm goes to filling the depression storage (which has a given capacity of 2 mm) and 5.5 mm − 1.9 mm = 3.6 mm goes toward runoff.

g .n

et

t = 40–50 min: Ponding continues at the beginning of this time interval since i > fp (80 mm/h > 60.6 mm/h). At the end of this time interval, t = 50 min = 0.833 h and Equations 9.53 and 9.54 give F = 33.9 mm and fp = 50.1 mm/h. So ponding continues at the end of this time interval, since i > fp (80 mm/h > 50.1 mm/h). Since the rainfall amount is 80 mm/h * 0.167 h = 13.3 mm and the infiltration amount is 33.9 mm − 29.4 mm = 4.5 mm, the rainfall excess is 13.3 mm − 4.5 mm = 8.8 mm, all of which goes toward runoff since the depression storage is full. t = 50–60 min: Ponding does not occur at the beginning of this time interval since i < fp (10 mm/h < 60.6 mm/h); note also that i < fc . The rainfall amount in this interval is 10 mm/h * 0.167 h = 1.7 mm and the amount in depression storage is 2 mm. If both the rainfall and the depression storage infiltrate, then the value of F at the end of this time interval is F = 33.9 mm + 1.7 mm + 2 mm = 37.6 mm. Using F = 37.6 mm in Equation 9.52 yields fp = 51.1 mm/h. Since i < fp (10 mm/h < 51.1 mm/h) at the end of this interval, all rainfall and depression storage infiltrates and there is no runoff. The summary results shown in Table 9.11 indicate that the total runoff is 12.4 mm and the total rainfall is 50.0 mm. Ponding begins 27 min after the start of the storm, and the rainfall-runoff ratio is 12.4/50 * 100 = 24.8%.

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TABLE 9.11: Summary of Runoff Calculations Using Horton Method

Time (min) 0

ww

F (mm)

F (mm)

i (mm/h)

it (mm)

S (mm)

0.167

1.67

10

1.67

0

0

0

0.167

3.33

20

3.33

0

0

0

0

10

1.67

20

5.01

30

18.2

40

29.4

50

33.9

60

37.6

0.167

13.2

80

13.3

0.1

0.1

0

0.167

11.2

100

16.7

1.9

2.0

3.6

0.167

4.5

80

13.3

0

2.0

8.8

0.167

3.7

10

1.7

−2.0

0

0

Total:

w .E asy En g 9.3.3.3

S (mm)

Runoff (mm)

t (h)

50.0

12.4

Green–Ampt model

This physically based semiempirical model was first proposed by the Australian scientists W. H. Green and G. A. Ampt (1911) and was put on a firm physical basis by Philip (1954). The Green–Ampt model, sometimes called the delta-function model (Philip, 1993; Salvucci and Entekhabi, 1994), is considered to be one of the most realistic models of infiltration available to the engineer. A typical vertical section of soil is shown in Figure 9.23, where it is assumed that water is ponded to a depth H above the ground surface and that there is a sharp interface between the wet soil and the dry soil at a distance L below the ground surface. This interface is called the wetting front, and as the ponded water infiltrates into the soil, the wetting front moves downward. Flow through saturated porous media is described by Darcy’s law, which can be written in the form

ine eri n

qs = −Ks

dh ds

g .n

(9.55)

et

where qs is the flow per unit area or specific discharge in the s-direction [LT−1 ], Ks is the hydraulic conductivity of the saturated soil [LT−1 ], h is the piezometric head [L], and dh/ds is the gradient of the piezometric head in the s-direction [dimensionless]. In reality, Ks is usually less than the hydraulic conductivity at saturation, because of entrapped air which prevents complete saturation. Using a finite-difference approximation to the Darcy equation (Equation 9.55) over the depth of the wet soil, L, leads to FIGURE 9.23: Green–Ampt soil column

Ponded water Ground surface

H

L

θ = θs

Wet soil

Wetting front

θ = θi

Dry soil

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fp = −Ks

(− f ) − (H + L) L

(9.56)

where fp is the potential infiltration rate (equal to the specific discharge when water is ponded above the ground surface) [LT−1 ], − f is the head at the wetting front [L], and H + L is the head at the top of the wet soil [L]. The suction head, f , is defined as −p/γ , where p is the pore pressure in the water at the wetting front [FL−2 ] and p is generally negative (i.e., below atmospheric pressure), and γ is the specific weight of water [FL−3 ]. The suction head, f , is equal to the capillary potential under initial soil moisture conditions. Assuming that the soil was initially dry, the total depth of water infiltrated, F [L], is given by F = (n − θi )L

ww

(9.57)

where n is the porosity of the soil (equal to volumetric water content at saturation) [dimensionless], and θi is the initial volumetric water content of the dry soil [dimensionless]. The volumetric water content of a soil is defined as the volume of water in a soil sample divided by the volume of the sample. In reality, the value of n used in Equation 9.57 should be reduced to account for the entrapment of air. Assuming that H V (L + f ), then Equations 9.56 and 9.57 can be combined to yield

w .E asy En g

fp = Ks +

Ks (n − θi ) f F

(9.58)

Recognizing that the potential infiltration rate, fp , and the cumulative infiltrated amount, F, are related by fp =

dF dt

(9.59)

ine eri n

then Equations 9.58 and 9.59 can be combined to yield the following differential equation for F: Ks (n − θi ) f dF = Ks + (9.60) dt F

Separation of variables allows this equation to be written as 

F 0

F′ dF ′ = Ks F ′ + Ks (n − θi ) f



t

dt′

0

g .n

et

(9.61)

which incorporates the initial condition that F = 0 when t = 0. Integrating Equation 9.61 leads to the following equation for the cumulative infiltration versus time: 

F Ks t = F − (n − θi ) f ln 1 + (n − θi ) f



(9.62)

This equation assumes that water has been continuously ponded above the soil column and that infiltration has been at the potential rate from time t = 0, which will happen only if the rainfall intensity exceeds the infiltration capacity from the beginning of the storm. If the rainfall intensity, i, is initially less than the infiltration capacity, then the actual infiltration rate, f , will be equal to the rainfall intensity until the infiltration capacity, fp (which continually decreases), becomes less than the rainfall intensity. Assuming that this happens at time tp , then  i, t < tp (9.63) f = fp = i, t = tp

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Combining Equations 9.63 and 9.58 leads to the following expression for the total infiltrated depth, Fp , at t = tp : f (n − θi ) Fp = (9.64) i/Ks − 1 Since the infiltration rate has been equal to i up to this point, the time, tp , at which the infiltrated depth becomes equal to Fp is given by tp =

Fp i

(9.65)

For t > tp , the rainfall intensity exceeds the potential infiltration rate, and therefore infiltration continues at the potential rate given by Ks (n − θi ) f , F

f = fp = Ks +

ww

(9.66)

t > tp

and the cumulative infiltration as a function of time can be written in the form (Mein and Larson, 1973)

w .E asy En g



F Ks (t − tp + t ) = F − (n − θi ) f ln 1 + (n − θi ) f ′



,

(9.67)

t > tp

where t′ is the equivalent time to infiltrate Fp under the condition of surface ponding from t = 0. The formulation given by Equation 9.67 is sometimes called the Green–Ampt Mein Larson model (King et al., 1999). Care should be taken in applying Equation 9.64 since it gives negative or infinite values of Fp when i … Ks , which is a result of the fact that the Green–Ampt model can only be used to calculate the infiltration capacity when the rainfall rate exceeds the saturated hydraulic conductivity (i.e., i > Ks ). Typical values of the hydraulic properties for several soil textures are given in Table 9.12. These values should not be used for bare soils with crusted surfaces, and for finer-textured soils the hydraulic properties derived solely on the basis of soil texture should be regarded as highly uncertain (Twarakavi et al., 2010). A variety of empirical equations have been proposed for relating the saturated hydraulic conductivity, Ks , to the textural characteristics of soils (e.g., Smettem and Bristow, 1999). However, empirical relationships only provide rough estimates of Ks , and significant deviations from these estimates can occur depending on the type of vegetation covering the soil (e.g., Duan et al., 2012). The field capacity, θ0 , and

ine eri n

g .n

TABLE 9.12: Typical Values of Green–Ampt Parameters

USDA soil-texture class Sand Loamy sand Sandy loam Loam Silt loam Sandy clay loam Clay loam Silty clay loam Sandy clay Silty clay Clay

Hydraulic conductivity Ks (mm/h)

Wetting front suction head f (mm)

Porosity n

Field capacity θo

120 30 11 3 7 2 1 1 1 1 0.3

49–150∗ 61–250∗ 110–250∗ 89–350∗ 170 220 210 270 240 290 320–1000∗

0.437 0.437 0.453 0.463 0.501 0.398 0.464 0.471 0.430 0.479 0.475

0.062 0.105 0.190 0.232 0.284 0.244 0.310 0.342 0.321 0.371 0.378

et

Wilting point θw 0.024 0.047 0.085 0.116 0.135 0.136 0.187 0.210 0.221 0.251 0.265

Sources: Rawls et al. (1983) [data source unless otherwise noted]; ∗ Bouwer et al. (1999).

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the wilting point, θw , give some indication of the dry-soil moisture content, θi , where the field capacity is the residual water content after gravity drainage, and the wilting point is the limiting water content below which plants cannot extract water for transpiration. ASCE (1996b) recommends caution in applying the Green–Ampt method in forested areas, where the basic assumptions of the method may not be valid. The Green–Ampt method best describes the infiltration process in soils that exhibit a sharp wetting front, such as coarse-textured soils with uniform pore shapes (Ward and Dorsey, 1995). However, in many agricultural soils sharp wetting fronts are not the norm (Quisenberry and Phillips, 1976; Thomas and Phillips, 1979). In general, for coarse-textured soils with a narrow pore-size distribution, the wetting front will be more abrupt and in a fine-textured soil, the wetting front will be more diffuse.

EXAMPLE 9.17

ww

A catchment area consists almost entirely of loamy sand, which typically has a saturated hydraulic conductivity of 30 mm/h, average suction head of 61 mm, porosity of 0.44, field capacity of 0.105, and wilting point of 0.047. The catchment area is estimated to have a depression storage of 5 mm. You are in the process of designing a stormwater-management system to handle the runoff from this area, and have selected the following 1-h storm as the basis of your design:

w .E asy En g Interval (min)

Average rainfall (mm/h)

0–10 10–20 20–30 30–40 40–50 50–60

10 20 80 100 80 10

ine eri n

Determine the runoff versus time for average initial moisture conditions, and contrast the amount of rainfall with the amount of runoff. Use the Green–Ampt method for your calculations and assume that the initial moisture conditions are midway between the field capacity and wilting point.

Solution In the present case, Ks = 30 mm/h; θi = 21 (0.105 + 0.047) = 0.076; n = 0.44; and f = 61 mm. The infiltration capacity, fp , as a function of the cumulative infiltration, F, is given by Equation 9.58 as

g .n

Ks (n − θi ) f 30(0.44 − 0.076)(61) = 30 + F F 666.12 = 30 + F

fp = Ks +

et

(9.68)

The cumulative infiltration as a function of time (after ponding) is given by Equation 9.67 as   F ′ Ks (t − tp + t ) = F − (n − θi ) f ln 1 + (n − θi ) f   F ′ 30(t − tp + t ) = F − (0.44 − 0.076)(61) ln 1 + (0.44 − 0.076)(61) = F − 22.2 ln (1 + 0.0450F)

(9.69)

and if ponding occurs from t = 0, then 30t = F − 22.2 ln (1 + 0.0450F)

(9.70)

Each 10-min increment in the storm will now be taken sequentially, and the computation of the runoff is summarized in Table 9.13.

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TABLE 9.13: Computation of Rainfall Excess Using Green–Ampt Model Time (min)

ww

F (mm)

0

0.0

10

1.7

20

5.0

30

17.9

40

27.8

50

36.3

60

43.0

t (h)

F (mm)

i (mm/h)

it (mm)

Depression storage, S (mm)

0.167

1.7

10

1.7

0

0

0

0.167

3.3

20

3.3

0

0

0

0.167

12.9

80

13.4

0.5

0.5

0

0.167

9.9

100

16.7

4.5

5.0

2.3

0.167

8.5

80

13.4

0

5.0

4.9

0.167

6.7

10

1.7

−5.0

0

0

Total:

Total storage (mm)

Runoff (mm)

50.2

w .E asy En g

7.2

t = 0–10 min: During this period, the rainfall intensity (10 mm/h) is less than the saturated hydraulic conductivity, Ks (= 30 mm/h), and therefore no ponding occurs. The entire rainfall amount of 10 * (10/60) = 1.7 mm is infiltrated. t = 10–20 min: During this period, the rainfall intensity (20 mm/h) is still less than the minimum infiltration capacity (30 mm/h) and all the rainfall infiltrates. The rainfall amount during this period is 20 * (10/60) = 3.3 mm, and the cumulative infiltration after 20 min is 1.7 mm + 3.3 mm = 5.0 mm. t = 20–30 min: During this period the rainfall intensity (80 mm/h) exceeds the minimum infiltration capacity (= 30 mm/h), and therefore ponding is possible. Use Equation 9.68 to determine the cumulative infiltration, F80 , corresponding to a potential infiltration rate of 80 mm/h:

ine eri n

80 = 30 + which leads to

666.12 F80

F80 = 13.3 mm

g .n

After 20 min (0.334 h) the total infiltration was 5.0 mm, therefore if 0.334 + x is the time when the cumulative infiltration is 13.3 mm, then 5.0 + 80x = 13.3 which leads to x = 0.104 h (= 6.2 min)

et

Therefore, the time at which ponding occurs, tp , is tp = 0.334 h + 0.104 h = 0.438 h (= 26.3 min) The next step is to find the time, t′ , that it would take for 13.3 mm to infiltrate, if infiltration occurs at the potential rate from t = 0. Infiltration as a function of time is given by Equation 9.70, therefore 30t′ = 13.3 − 22.2 ln [1 + (0.0450)(13.3)] which leads to t′ = 0.096 h (= 5.8 min) Therefore, the equation for the cumulative infiltration as a function of time after t = 0.438 h is given by Equation 9.69, which can be written as 30(t − 0.438 + 0.096) = F − 22.2 ln (1 + 0.0450F)

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(9.71)

At the end of the current time period, t = 3(0.167) = 0.501 h, and substituting this value into Equation 9.71 leads to F = 17.9 mm Since the rainfall during this period is 13.4 mm, and the cumulative infiltration (= cumulative rainfall) up to the beginning of this period is 5 mm, the amount of rainfall that does not infiltrate is equal to 5 mm + 13.4 mm − 17.9 mm = 0.5 mm. This excess amount goes toward filling up the depression storage, which has a maximum capacity of 5 mm. The available depression storage at the end of this time period is 5 mm − 0.5 mm = 4.5 mm. Since the depression storage is not filled, there is no runoff. t = 30–40 min: Since the rainfall rate is now higher than during the previous time interval, ponding continues to occur. The time at the end of this period is 0.668 h. Substituting this value for t into Equation 9.71 gives the cumulative infiltration at the end of the time period as

ww

F = 27.8 mm Since the cumulative infiltration up to the beginning of this period is 17.9 mm, the infiltrated amount during this period is 27.8 mm − 17.9 mm = 9.9 mm. The rainfall during this period is 16.7 mm; therefore, the amount of rain that does not infiltrate is 16.7 mm − 9.9 mm = 6.8 mm. Since there is 4.5 mm in available depression storage, the amount of runoff is 6.8 mm − 4.5 mm = 2.3 mm. The depression storage is filled at the end of this time period.

w .E asy En g

t = 40–50 min: The rainfall rate (80 mm/h) is still higher than the infiltration capacity and ponding continues. The time at the end of this period is 0.835 h, and Equation 9.71 gives the cumulative infiltration at the end of this time period as F = 36.3 mm

The cumulative infiltration up to the beginning of this period is 27.8 mm; therefore, the infiltration during this time interval is 36.3 mm − 27.8 mm = 8.5 mm. The rainfall during this period is 13.4 mm, therefore the amount of rain that does not infiltrate is 13.4 mm − 8.5 mm = 4.9 mm. Since the depression storage is full, all 4.9 mm is contributed to runoff.

ine eri n

t = 50–60 min: The rainfall rate (10 mm/h) is below the minimum infiltration capacity (30 mm/h). Under these circumstances, the ponded water (i.e., the water in depression storage) will infiltrate and, if there is sufficient infiltration capacity, some of the rainfall may also infiltrate. The time at the end of this period is 1 h, and if infiltration continues at the potential rate, then the cumulative infiltration at the end of the period is given by Equation 9.71 as F = 44 mm

g .n

et

Because the cumulative infiltration up to the beginning of this period is 36.3 mm, the potential infiltration during this period is 44 mm − 36.3 mm = 7.7 mm. The rainfall during this period is 1.7 mm and there is 5 mm of depression storage to infiltrate. Since the rainfall plus depression storage is equal to 6.7 mm, then all the rainfall and depression storage is infiltrated during this time interval. The cumulative infiltration at the end of this period is 36.3 mm + 6.7 mm = 43 mm. The total rainfall during this storm is 50.2 mm and the total runoff is 7.2 mm. The runoff is therefore equal to 14% of the rainfall.

The runoff obtained by subtracting the Green–Ampt infiltration from the incident rainfall can be combined with a digital elevation model to predict areas of inundation resulting from various storm events (e.g., Chen et al., 2009). Although the Green–Ampt equation was derived for horizontal land surfaces, it is commonly applied to sloping surfaces without regard to the magnitude of the slope. Theoretical and numerical studies have shown that when the Green–Ampt formulation is modified to take into account the slope of the land surface, infiltration increases with increasing slope angle, for a given plan area (Chen and Young, 2006c). Conventional applications of the Green–Ampt equation also assume negligible ponding depth relative to the infiltration depth. This approximation might not be justified

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in low-permeability soils where there is lateral containment of ponded water. Under these circumstances, an extended Green–Ampt formulation in which the depth of ponded water is ´ taken into account might be more appropriate (e.g., Loaiciga and Huang, 2007). Nondimensionalization and approximation of the Green–Ampt equation. The fundamental form of the Green–Ampt equation given by Equation 9.62 can be expressed in nondimensional form as t∗ = F∗ − ln(1 + F∗ ) (9.72) where t∗ and F∗ are the nondimensional time and cumulative infiltration, respectively, that are normalized relative to the soil properties where

ww

t∗ =

Ks t f (n − θi )

(9.73)

F∗ =

F f (n − θi )

(9.74)

In a similar fashion, the Green–Ampt Mein Larson model given by Equation 9.67 can be expressed in nondimensional form by simply normalizing the ponding times, tp and t′ , by f (n − θi )/Ks . The Green–Ampt equation represented by Equation 9.72 expresses the time, t∗ , as a function of the cumulative infiltration, F∗ ; however, it is sometimes more desirable to express F∗ and a function of t∗ . This cannot be done algebraically, since Equation 9.72 is implicit in F∗ for any given t∗ . The relationship between F∗ and t∗ can be closely approximated by the following explicit equation (Swamee et al., 2012)

w .E asy En g

0.7  F∗ = 1.94t∗0.74 + t∗1.429

ine eri n

(9.75)

Using Equation 9.75 to express F∗ as a function of t∗ obviates the need to determine this relationship numerically and provides a useful expression that has an error of less than 1%. 9.3.3.4

NRCS curve-number model

The Natural Resources Conservation Service∗ (NRCS) curve-number model is the most widely used method for estimating rainfall excess (= rainfall minus abstractions) in the United States, and the popularity of this method can be attributed to its ease of application. The curve-number model was originally developed for calculating daily runoff as affected by land-use practices in small agricultural watersheds; then, because of its overwhelming success, it was subsequently adapted to urban catchments. The NRCS curve-number model separates the rainfall into three components: rainfall excess, Q, initial abstraction, Ia , and retention, F. These components are illustrated graphically in Figure 9.24. The initial abstraction includes the rainfall that is stored in the catchment area FIGURE 9.24: Components in the NRCS curve-number model

g .n

et

P, Q, F, Ia

P Q

Ia

F Time

∗ The Natural Resources Conservation Service (NRCS) was formerly named the Soil Conservation Service (SCS).

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before runoff begins. This includes interception, initial infiltration, and depression storage. If the amount of rainfall is less than the initial abstraction, then runoff does not occur. The retention, F, is the portion of the rainfall reaching the ground that is retained by the catchment and consists primarily of infiltrated water. The basic assumption of the NRCS model is that for any rainfall event the precipitation, P [L], runoff, Q [L], retention, F [L], and initial abstraction, Ia [L], are related by F Q (9.76) = S P − Ia where S is the potential maximum retention and measures the retention capacity of the catchment [L]. The maximum retention, S, does not include Ia . The rationale for Equation 9.76 is that for any rainfall event, the portion of available storage (= S) that is filled, F, is equal to the portion of available rainfall (= P − Ia ) that appears as runoff, Q. Equation 9.76 is applicable only when P > Ia . Conservation of mass requires that

ww

(9.77)

F = P − Q − Ia Eliminating F from Equations 9.76 and 9.77 yields (P − Ia )2 , (P − Ia ) + S

w .E asy En g Q=

(9.78)

P > Ia

Empirical data indicate that the initial abstraction, Ia , is directly related to the maximum retention, S, and the following relation is commonly assumed: Ia = 0.2 S

(9.79)

Some research has indicated that the factor of 0.2 is probably adequate for large storms in rural areas, but it is likely an overestimate for small to medium storms and is probably too high for urban areas (Singh, 1992; Schneider and McCuen, 2005). It has also been shown that the factor of 0.2 should be a function of the rainfall intensity (Chung et al., 2010). However, in spite of all these legitimate results, since the storage capacity, S, of many catchments has been determined based on Equation 9.79, it is recommended that the 0.2 factor be retained for consistency when using storage estimates calibrated from field measurements of rainfall and runoff. Combining Equations 9.78 and 9.79 leads to Q=

ine eri n

(P − 0.2S)2 , P + 0.8S

P > 0.2S

g .n

et

(9.80)

This equation is the basis for estimating the depth of runoff, Q, from a depth of rainfall, P, given the maximum retention, S. It should be noted, however, that the NRCS equation was originally developed as a runoff index for 24-h rainfall amounts and should be used with caution in attempting to analyze incremental runoff amounts during the course of a storm (Kibler, 1982) or runoff from durations other than 24 h. This latter limitation is elucidated by noting that the curve-number model predicts the same runoff for a given rainfall amount, regardless of the duration of the rainfall event. This is obviously not correct, since longduration low-intensity storms will have smaller runoff amounts than short-duration highintensity storms, when both storms have the same total rainfall. Modifications to the curvenumber model have been proposed to correct this deficiency (e.g., Jain et al., 2006a), but such modifications have not yet received widespread acceptance. For similar reasons, even if the duration of the rainfall event is fixed at 24 h, the appropriate value of S will depend on the distribution of rainfall within the 24-h interval, and therefore value of S can only represent a long-term rainfall-runoff relationship and not that within all 24-h rainfall-runoff events; this sensitivity has been demonstrated by Lamont et al. (2008). The curve-number model is less accurate when the runoff is less than 10 mm (0.4 in.), and in these cases another method should be used to determine the runoff (SCS, 1986). In general, low runoff amounts are particularly sensitive to the value of S, which can seldom be stated with certainty (Pandit

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and Heck, 2009). In applications, the curve-number model is quite satisfactory when used for its intended purpose, which is as an index to evaluate the effects of land-use changes and conservation practices on direct runoff. Since it was not developed to reproduce individual historical events, only limited success has been achieved in using it for that purpose. Instead of specifying S directly, a curve number, CN, is usually specified where CN is related to S by CN =

ww

1000 10 + 0.0394S

(9.81)

where S is given in millimeters. Equation 9.81 is modified from the original formula, which required S in inches. In the absence of available storage (S = 0 mm, impervious surface) the curve number is equal to 100, and for an infinite amount of storage (S = q) the curve number is equal to zero. The curve number therefore varies (theoretically) between 0 and 100, with a practical range between 40 and 98 (Seybert, 2006). A curve number of 40 corresponds to an available storage of 38 cm (15 in.) of rainfall, while a curve number of 98 corresponds to an available storage of 0.5 cm (0.2 in.) of rainfall. Since the curve number (CN) model (Equation 9.80) neglects rainfall intensities, values of CN can be expected to decrease with increasing storm depth, but ultimately approach a constant value (Hawkins, 1993). In practice this relationship between P and CN is usually neglected; however, this dependence should be taken into account when analyzing measured data (e.g., McCormick and Eshleman, 2011). The utilization of the curve number, CN, in the rainfall-runoff relation is the basis for the naming of the curve-number model. In fact, the NRCS adopted the term “curve number” because the rainfall-runoff equation may be expressed graphically with curves for different values of CN. In practical applications, the curve number is considered to be a function of several factors including hydrologic soil group, cover type, treatment (i.e., management practice), hydrologic condition, antecedent runoff condition, and impervious area in the catchment. Typical curve numbers corresponding to a variety of urban land uses are given in Table 9.14. Soils in the United States are classified into four hydrologic soil groups: A, B, C, and D; descriptions of these groups are given in Table 9.15. Soils are grouped based on profile characteristics that include depth, texture, organic matter content, structure, and degree of swelling when saturated. Minimum infiltration rates associated with the hydrologic soil groups are given in Table 9.15, where it should be noted that these rates are for bare soil after prolonged wetting, vegetated soils tend to have higher infiltration rates. The NRCS has classified more than 14,000 soils into these four groups. In the United States, local NRCS offices can usually provide information on local soils and their associated soil groups, but it should be noted that activities such as the operation of heavy equipment can substantially change the local soil characteristics. Soil compaction is of particular concern in lands reclaimed from surface mining, where soil compaction by heavy equipment can render most soils into the group D category (McCormick and Eshleman, 2011). There are a variety of methods for determining cover type, the most common ones being field reconnaissance, aerial photographs, and land-use maps. In agricultural practice, treatment is a cover-type modifier used to describe the management of cultivated lands. It includes mechanical practices such as contouring and terracing, and management practices such as crop rotations and reduced or no tillage. Hydrologic condition indicates the effects of cover type and treatment on infiltration and runoff, and is generally estimated from the density of plant and residue cover. Good hydrologic condition indicates that the soil has a low runoff potential for that specific hydrologic soil group, cover type, and treatment. The percentage of impervious area and the means of conveying runoff from impervious areas to the drainage systems should generally be considered in estimating curve numbers in urban areas. An impervious area is called “directly connected” if runoff from the area flows directly into the drainage system. The impervious area is called “not directly connected” if runoff from the impervious area flows over a pervious area and then into the drainage system. The antecedent runoff condition (ARC) is a measure of the actual available storage relative to the average available storage at the beginning of the rainfall event. The antecedent runoff condition is grouped into three categories: ARC I, ARC II, and ARC III. The average

w .E asy En g

ine eri n

g .n

et

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions TABLE 9.14: Curve Numbers for Various Urban Land Uses

Curve numbers for hydrologic soil group

ww

Cover type and hydrologic condition

A

B

C

D

Lawns, open spaces, parks, golf courses: Good condition: grass cover on 75% or more of the area Fair condition: grass cover on 50%–75% of the area Poor condition: grass cover on 50% or less of the area

39 49 68

61 69 79

74 79 86

80 84 89

Paved parking lots, roofs, driveways, etc.

98

98

98

98

Streets and roads: Paved with curbs and storm sewers Gravel Dirt Paved with open ditches

98 76 72 83

98 85 82 89

98 89 87 92

98 91 89 93

w .E asy En g Commercial and business areas

(85% impervious∗ )

89

92

94

95

Industrial districts

(72% impervious∗ )

81

88

91

93

Row houses, town houses, and residential with lot sizes … 0.05 ha (1/8 ac)

(65% impervious∗ )

77

85

90

92

Residential average lot size: 0.10 ha (1/4 ac) 0.14 ha (1/3 ac) 0.20 ha (1/2 ac) 0.40 ha (1 ac) 0.80 ha (2 ac)

(38% impervious∗ ) (30% impervious∗ ) (25% impervious∗ ) (20% impervious∗ ) (12% impervious∗ )

61 57 54 51 46

75 72 70 68 65

83 81 80 79 77

87 86 85 84 82

77

86

91

94

63 96

77 96

85 96

88 96

77 76 74

86 85 83

91 90 88

94 93 90

ine eri n

Developing urban areas (no vegetation established) Newly graded area Western desert urban areas: Natural desert landscaping (pervious area only) Artificial desert landscaping Cultivated agricultural land Fallow Straight row or bare soil Conservation tillage (Poor) Conservation tillage (Good)

g .n

et

Note: ∗ The impervious area is assumed to be directly connected to the drainage system, with the impervious area having a CN of 98, and the pervious area taken as equivalent to open space in good hydrologic condition.

curve numbers normally cited for a particular land area correspond to ARC II conditions, and these curve numbers can be adjusted for drier than normal conditions (ARC I) or wetter than normal conditions (ARC III) by using Table 9.16. The curve-number adjustments in Table 9.16 can be approximated by the relations (Hawkins et al., 1985; Chow et al., 1988) CNI =

CNII 2.3 − 0.013CNII

(9.82)

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457

TABLE 9.15: Description of NRCS Soil Groups

Minimum infiltration rate Group

ww

Description

(mm/h)

(in./h)

>7.6

>0.30

A

Deep sand; deep loess; aggregated silts

B

Shallow loess; sandy loam

3.8–7.6

0.15–0.30

C

Clay loams; shallow sandy loam; soils low in organic content; soils usually high in clay

1.3–3.8

0.05–0.15

D

Soils that swell significantly when wet; heavy plastic clays; certain saline soils

0.0–1.3

0.00–0.05

TABLE 9.16: Antecedent Runoff Condition Adjustments

CN for ARC II

w .E asy En g 100 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 0

and CNIII =

Corresponding CN for condition ARC I ARC III 100 87 78 70 63 57 51 45 40 35 31 27 23 19 15 12 9 7 4 2 0

100 99 98 97 94 91 87 83 79 75 70 65 60 55 50 45 39 33 26 17 0

ine eri n

CNII 0.43 + 0.0057CNII

g .n

et (9.83)

where CNI , CNII , and CNIII are the curve numbers under ARC I, ARC II, and ARC III conditions, respectively. The antecedent runoff condition is typically correlated with the 5-d antecedent rainfall (e.g., Bhuyan et al., 2003); however, this might not always be the case (e.g., Jain et al., 2006a). In continuous rainfall-runoff simulations, curve numbers can be adjusted daily based on the 5-d antecedent rainfall or, alternatively, curve numbers can be recalculated daily based on the variable daily available watershed storage, S. Current indications are that the latter approach yields more accurate results (Geetha et al., 2007; Brocca et al., 2009). Relating the watershed storage, S, to the baseflow in the receiving stream (which is presumably a function of the water-table elevation in the catchment) has also been shown to be generate fairly accurate runoff predictions (e.g., Shaw and Walter, 2009).

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The guidelines for selecting curve numbers given in Tables 9.14 and 9.16 are useful in cases where site-specific data on the maximum retention, S, either are not available or cannot be reasonably estimated. Most of the curve numbers given in Table 9.14 were estimated based on data from small agricultural watersheds with areas less than 4 ha (10 ac) located in the midwestern United States. Current empirical evidence suggests that hydrological systems are overdesigned when using curve numbers from Tables 9.14 and 9.16 (Schneider and McCuen, 2005). Whenever S is available for a catchment, then the curve number should be estimated directly from S using Equation 9.81, and, if necessary, adjusted using Table 9.16. For example, measurements in Florida have shown that estimating S from the depth to the water table in flat high-water-table watersheds is the most accurate way of estimating the curve number (Capece et al., 1987). As a precautionary note, the NRCS does not recommend the use of the curve-number model when CN is less than 40. EXAMPLE 9.18

ww

The drainage facilities of a catchment are to be designed for a rainfall event with a return period of 25 years and a duration of 2 h, where the IDF curve for 25-year storms is given by

w .E asy En g

i=

830 t + 33

where i is the rainfall intensity in cm/h, and t is the storm duration in minutes. A double-ring infiltrometer test on the soil shows that the minimum infiltration rate is on the order of 5 mm/h. The urban area being developed consists of mostly open space with less than 50% grass cover. Use the NRCS curve-number method to estimate the total amount of runoff (in cm), assuming the soil is in average condition at the beginning of the design storm. Estimate the percentage increase in runoff that would occur if heavy rainfall occurred within the previous 5 d and the soil was saturated.

Solution The amount of rainfall can be estimated using the IDF curve. For a 2-h 25-year storm, the average intensity is given by 830 = 5.42 cm/h i= 120 + 33 and the total amount of rainfall, P, in the 2-h storm is equal to (5.42)(2) = 10.8 cm. Since the minimum infiltration rate of the soil is 5 mm/h, according to Table 9.15 it can be inferred that the soil is in Group B. The description of the area, open space with less than 50% grass and Group B soil, is cited in Table 9.14 to have a typical curve number equal to 79. The curve number, CN, and soil storage, S (in mm), are related by Equation 9.81 as

ine eri n

CN =

1000 10 + 0.0394S

g .n

and therefore the storage, S, in this case is     1 1000 1000 1 − 10 = − 10 = 67.5 mm S= 0.0394 CN 0.0394 79

et

The runoff amount, Q, can be calculated from the rainfall amount, P (= 10.8 cm), and the maximum storage, S (= 6.75 cm) using Equation 9.80, where Q=

(P − 0.2S)2 , P + 0.8S

P > 0.2S

Since P (= 10.8 cm) > 0.2S (= 1.35 cm), this equation is valid and Q=

[10.8 − 0.2(6.75)]2 = 5.51 cm 10.8 + 0.8(6.75)

Hence, there is 5.51 cm of runoff from the storm with a rainfall amount of 10.8 cm. When the soil is saturated, Table 9.16 indicates that the curve number, CN, increases from 79 to 94 and the maximum available soil storage is given by     1000 1000 1 1 − 10 = − 10 = 16.2 mm S= 0.0394 CN 0.0394 94

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Rainfall Abstractions

459

and the runoff amount, Q, is given by Q=

[10.8 − 0.2(1.62)]2 = 9.07 cm 10.8 + 0.8(1.62)

Therefore, under saturated conditions, the runoff amount increases from 5.51 cm to 9.07 cm, which corresponds to a 64.6% increase.

ww

Soil type and soil properties at most locations in the United States can be obtained from the USDA-NRCS Soil Survey Geographic (SSURGO) database. The digital SSURGO maps duplicate the original soil-survey maps that were compiled to 1:24,0000 scale, 7.5-min orthophoto quadrangles. This database includes the hydrologic soil group and permeability range of each soil, and the permeability is commonly taken as being equal to the saturated hydraulic conductivity and the infiltration capacity of the soil. Some investigations have shown that the actual soil saturated hydraulic conductivity is often much higher than the soil permeability reported in soil surveys (Troch et al., 1993; Rossing, 1996); however, soil-survey permeabilities have sometimes been used as worst-case infiltration capacities in rainfall-runoff models (Walter et al., 2003). For larger-scale studies, the coarser State Soil Geographic (STATSGO) database with soil coverage at 1:250,000 scale is also available from USDA-NRCS. The smallest soil map unit represented in STATSGO is about 625 ha (1540 ac), whereas it is about 2 ha (5 ac) in the SSURGO database.

w .E asy En g

Physical interpretations of the CN model. The CN model is commonly regarded as a purely empirical model that fits the general relationship between total rainfall and total runoff. In support of this view, the total infiltration for a rainfall event is assumed to be given by the combination of Equations 9.77, 9.79, and 9.80, which yields F=

(P − 0.2S)S , P + 0.8S

ine eri n P > 0.2S

(9.84)

The infiltration rate, f , can be derived from Equation 9.84 by differentiation, where f = and i is the rainfall intensity given by

S2 i dF = dt (P + 0.8S)2

i=

dP dt

g .n

(9.85)

et

(9.86)

The functional form of the infiltration rate formula given by Equation 9.85 is not physically realistic, since it requires that the infiltration rate be generally dependent on the rainfall intensity (Morel-Seytoux and Verdin, 1981). The aforementioned derivation assumes that the CN model can be applied continuously during a rainfall event and that the difference between rainfall and runoff is infiltrated into the soil. It is apparent that these assumptions collectively yield unrealistic results regarding the infiltration process. An alternative physical interpretation of the CN model is that the difference between rainfall and runoff is accounted for by a variable catchment area contributing to runoff. This is referred to as the variable source area (VSA) mechanism of surface runoff. The VSA model assumes that runoff is caused by a saturation-excess mechanism in which shallow soils with low permeability cause surface saturation, resulting in varying fractions of the catchment contributing runoff. In accordance with the VSA model, the fraction of the catchment, fA , contributing to runoff can be expressed as fA =

dQ dPe

(9.87)

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Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions

where Q is the cumulative runoff [L] and Pe is the cumulative rainfall beyond the initial abstraction [L]. In the CN formulation, Pe is equal to P − Ia and hence Q=

Pe2 Pe + S

(9.88)

Differentiating Equation 9.88 with respect to Pe and using the definition of fA given by Equation 9.87 yields fA = 1 −

ww

S2 (Pe + S)2

(9.89)

This equation gives the fraction of the total catchment area that contributes to runoff as a function of the total rainfall excess and storage capacity of the catchment. There has been limited field validation of Equation 9.89; however, some field results support Equation 9.89 (e.g., Dahlke et al., 2012), thereby providing a physical (VSA) basis for the CN formulation. EXAMPLE 9.19

w .E asy En g

A catchment area is estimated to have a curve number of 79. Determine the fraction of the catchment area that yields runoff for a storm in which the total rainfall is 12 cm.

Solution From the given data: CN = 79 and P = 12 cm = 120 mm. Using the relationship between the storage capacity, S, and CN, along with the definition of the initial abstraction, Ia , yields     1 1000 1000 1 − 10 = − 10 = 67.5 mm S= 0.0394 CN 0.0394 79 Ia = 0.2S = 0.2(67.5) = 13.5 mm

Pe = P − Ia = 120 − 13.5 = 106.5 mm fA = 1 −

ine eri n

67.52 S2 =1 − = 0.850 (Pe + S)2 (106.5 + 67.5)2

Therefore, it is estimated that 85% of the catchment will contribute runoff when the total rainfall amount is 12 cm.

9.3.3.5

Comparison of infiltration models

g .n

et

The Horton, Green–Ampt, and curve-number models are all commonly used in engineering practice. These models are representative of the three classes of infiltration models: physically based, semi-empirical, and empirical models. The Green–Ampt model is a physically based model, the Horton model is a semi-empirical model, and the curve-number model is an empirical model. The scales at which infiltration models are applied include the local (soil column) scale, the agricultural field scale, and the catchment scale. An infiltration model that performs better at one scale might not necessarily perform better or even be applicable at another scale. The application of infiltration models such as the Horton and Green–Ampt models at catchment scales is physically unrealistic, where it is more common to lump all rainfall abstractions into a runoff coefficient, which is defined as the fraction of rainfall that becomes runoff. At catchment scales, the curve-number model can be used to estimate runoff coefficients, although this approach does not always yield reliable results (e.g., Merz et al., 2006). Some analyses have demonstrated that (field-scale) infiltration rates can increase with increasing rainfall intensity, a phenomenon that might be explained by an increase in contributing area to the field outlet as the rainfall intensity increases (Stone et al., 2008). In reality, local-scale infiltration is upscaled by interactive infiltration in which ponded rainfall generated at one location infiltrates at another location, thus dynamically coupling infiltration and overland flow. Studies have shown that area-averaged infiltration over statistically

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ww

Rainfall Abstractions

461

homogeneous areas can be as much as 10% higher due to interactive infiltration (Sheldon and Fiedler, 2008). To account for interactive infiltration at the catchment scale, field experiments have shown that it is sometimes appropriate to express the infiltration capacity of the catchment as a function of the rainfall intensity (Langhans et al., 2011). The effect of ground slope on infiltration is often overlooked, even though it is known that the infiltration rate is a function of the ground slope (e.g., Essig et al., 2009). Some comparative studies have indicated that the Green–Ampt model performs better than the curve-number model in predicting runoff volumes from catchments (Hjelmfelt, 1991; Van Mullem, 1991; Chahinian et al., 2005). This result is not unexpected, given the relatively poor performance of the curve-number model when applied to individual storms (Willeke, 1997), the general recognition that the curve-number method best represents a long-term expected-value relationship between rainfall and runoff (Smith, 1997), the recognition that the relation of soil and land-use conditions to curve number may vary regionally (Miller and Cronshey, 1989; Merz et al., 2006), and the recognition that the curve-number model does not properly account for soil moisture (Michel et al., 2005). Mishra et al. (2003) compared the performance of fourteen physically based, semi-empirical, and empirical infiltration models, including the Green–Ampt and Horton models (not including the curvenumber model), and concluded that semi empirical models tend to perform better than other physically based models, which are apparently better for laboratory-tested soils than for field soils. The results reported by Mishra et al. (2003) indicated that the Horton model performs significantly better than the Green–Ampt model. The measurable nature of the Green–Ampt parameters and the physical basis of the model are appealing features; however, both the Green–Ampt and Horton models suffer from being unbounded (Ponce and Hawkins, 1996), in that they allow an unlimited amount of infiltration and they do not account for the effects of spatial variability in parameters such as the saturated hydraulic conductivity (Woolhiser et al., 1996). A pragmatic reason for using the curve-number model is that it is a method supported by a U.S. government agency (the Department of Agriculture), which gives its users basic protection in litigation (Smith, 1997). The authoritative origin of the curve-number model apparently qualifies as a defense in legal proceedings that an engineer has performed a hydrologic analysis in accordance with generally accepted standards (Willeke, 1997).

w .E asy En g 9.3.4

ine eri n

Rainfall Excess on Composite Areas

In many cases, a catchment can be delineated into several subcatchments with each subcatchment having different abstraction characteristics. The runoff from such composite catchments depends on how the subcatchments are connected, and this is illustrated in Figure 9.25 for the case of pervious and impervious subcatchments within a larger catchment. The subcatchments consist of a pervious area and two separate impervious ones, with only one of the impervious areas directly connected to the catchment outlet. The runoff from the composite catchment is equal to the runoff from the impervious area, A1 , that is directly connected to the catchment outlet, plus the runoff from the pervious area, A2 , which assimilates a portion of the runoff from the impervious area, A3 , that is not directly connected to the catchment outlet. To calculate the rainfall excess on the composite catchment, the rainfall excesses on each of the subcatchments are first calculated using methods such as the NRCS curve-number method, and then these rainfall excesses are routed to the outlet of the composite catchment. Rainfall abstractions in each subcatchment can be accounted for by a functional relation such as Qi = fi (P), i = [1, 3] (9.90)

g .n

et

where Qi is the rainfall excess on subcatchment i, and fi is an abstraction function that relates Qi to the precipitation, P, on the subcatchment. If the NRCS curve-number method is used to calculate the rainfall excess on subcatchment i, then fi is given by ⎧ ⎪ ⎨ (P − 0.2Si )2 if P > 0.2Si fi (P) = (9.91) P + 0.8Si ⎪ ⎩0 if P … 0.2Si

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FIGURE 9.25: Runoff from composite catchment

Nondirectly connected impervious area, A3

Pervious area, A2

Directly connected impervious area, A1

ww

Runoff

w .E asy En g

where Si is the available storage in subcatchment i. It can usually be assumed that the precipitation depth, P, on each subcatchment is the same. Since the rainfall excess, Q3 , from the nondirectly connected impervious area, A3 , drains into the pervious area, A2 , the effective precipitation, Peff , on the pervious area is given by Peff = P + Q3

A3 A2

(9.92)

The total rainfall excess, Q, from the composite catchment is equal to the sum of the rainfall excess on the directly connected impervious area plus the rainfall excess on the (directly connected) pervious area. Therefore

ine eri n

A2 A1 + f2 (Peff ) A A   A1 A3 A2 + f2 P + Q3 = f1 (P) A A2 A

Q = f1 (P)

where A is the total area of the composite catchment given by A=

3  i=1

Ai

g .n

(9.93)

et

(9.94)

This routing methodology for estimating the rainfall excess on composite catchments can be extended to any arrangement of subcatchments, provided that the flow paths to the catchment outlet are clearly defined. It is widely accepted that urban-area runoff generally depends on both the amount of impervious area and the way this area is connected to outflow points by the drainage system (Meierdiercks et al., 2011). EXAMPLE 9.20 The commercial site illustrated in Figure 9.26 covers 25 ha, of which the parking lot covers 7.5 ha, the building covers 6.3 ha, open grass covers 10 ha, and the site is graded such that all of the onsite runoff is routed to a grass retention area that covers 1.2 ha. Both the roof of the building and the parking lot are impervious (CN = 98) and drain directly onto the grassed area, which contains Type B soil and is in good condition. For a storm with a precipitation of 200 mm, calculate the volume of surface runoff that enters the retention area.

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Rainfall Abstractions

463

Grass

Building Parking lot

Retention area

ww

Solution In this case, both the parking lot and the roof of the building are not directly connected to the retention area, and therefore the runoff must be routed to the retention area through the grass. For the parking lot and building, CN = 98, and the available storage, S1 , can be derived from Equation 9.81 as     1000 1000 1 1 − 10 = − 10 = 5 mm S1 = 0.0394 CN 0.0394 98

w .E asy En g

For a precipitation, P, of 200 mm, the rainfall excess, Q1 , from the parking lot and building is given by Q1 =

(P − 0.2S1 )2 [200 − (0.2)(5)]2 = = 194 mm P + 0.8S1 200 + (0.8)(5)

The total area of the parking lot and building is equal to 7.5 ha + 6.3 ha = 13.8 ha, and the grassed area (including the retention area) is 10 ha + 1.2 ha = 11.2 ha. The effective rainfall, Peff , on the grassed area is therefore given by 13.8 = 439 mm Peff = 200 + 194 11.2 The grassed area contains Type B soil and is in good condition, and Table 9.14 gives CN = 61. The available storage, S2 , is derived from Equation 9.81 as     1000 1000 1 1 − 10 = − 10 = 162 mm S2 = 0.0394 CN 0.0394 61

ine eri n

The rainfall excess from the 11.2-ha grassed area, Q2 , is given by Q2 =

g .n

(Peff − 0.2S2 )2 [439 − (0.2)(162)]2 = 291 mm = Peff + 0.8S2 439 + (0.8)(162)

and the rainfall excess, Q, from the entire 25-ha composite catchment is Q = 291

11.2 = 130 mm 25

et

The volume of runoff entering the retention area is (130 mm)(25 ha) = 32,500 m3 , which will be disposed of via infiltration in the retention area.

A commonly used approximation in estimating the rainfall excess from composite areas is to apply the NRCS curve-number method with an area-weighted curve number. In a composite catchment with n subcatchments, the area-weighted curve number, CNeff , is defined as n 1  CNeff = CNi Ai (9.95) A i=1

where CNi and Ai are the curve number and area, respectively, of subcatchment i, and A is the total area of the composite catchment. Curve number weighting is appropriate when the CN values within a catchment do not vary dramatically. In some cases, the catchment is

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simply broken down into pervious and impervious areas, the curve number of the impervious area is assumed to be 98, a fraction of impervious area is assumed to be directly connected to the catchment outlet, and the composite curve number of the catchment, CNeff , is taken as CNeff = CNp + f (98 − CNp )(1 − 0.5R)

(9.96)

where f is the impervious fraction, CNp is the curve number of the pervious area, and R is the fraction of the impervious area that is directly connected to the catchment outlet. Application of Equation 9.96 is not recommended for values of R greater than 0.3 (SCS, 1986). EXAMPLE 9.21

ww

The commercial site illustrated in Figure 9.26 covers 25 ha, of which the parking lot covers 7.5 ha, the building covers 6.3 ha, open grass covers 10 ha, and the site is graded such that all of the runoff is routed to a grass retention area that covers 1.2 ha. Both the roof of the building and the parking lot are impervious (CN = 98) and drain directly onto the grassed area (CN = 61). For a storm with a precipitation of 200 mm, use an area-weighted curve number to calculate the rainfall excess that enters the retention area. Contrast your result with that obtained using the routing method in Example 9.20. Solution The area-weighted curve number, CNeff , is given by Equation 9.95 as

w .E asy En g CNeff =

n 1 1  [(98)(7.5) + (98)(6.3) + (10)(61) + (1.2)(61)] = 81 CNi Ai = A 25 i=1

The available storage, S, corresponding to CNeff , is     1000 1 1 1000 − 10 = 60 mm − 10 = S= 0.0394 CNeff 0.0394 81

ine eri n

and the rainfall excess, Q, resulting from a rainfall, P, of 200 mm is given by Q=

[200 − (0.2)(60)]2 (P − 0.2S)2 = = 143 mm P + 0.8S 200 + (0.8)(60)

This estimated runoff of 143 mm is 10% higher than the 130 mm estimated by routing the rainfall excess from the subcatchments.

g .n

If the flow paths in the composite catchment are known, then it is generally preferable that the rainfall excesses be routed through the catchment, as described previously. 9.4 Baseflow

et

In cases where a drainage channel penetrates the saturated zone of an aquifer, the flow in the channel typically consists of both surface runoff and groundwater inflow. The flow resulting from surface runoff is called direct runoff, and the flow resulting from groundwater inflow is called baseflow. Baseflow is typically (quasi-) independent of the rainfall event, is equal to the flow of groundwater from the saturated zone into the channel, and depends on the difference between the water-table elevation in the surrounding aquifer and the water-surface elevation in the channel. In addition to direct runoff and baseflow, a third contribution to flow in drainage channels called interflow is sometimes present. Interflow is the inflow to the channel that occurs between the ground surface and the water table and is typically caused by low-permeability subsurface soil layers that impede the vertical infiltration of rainwater in combination with large lateral pores left by rotting tree roots and burrowing animals. Interflow contributions to river flows can be significant in forested areas. Several methods have been developed for separating the direct runoff and baseflow components in observed hydrographs (e.g., Singh, 1992; Tallaksen, 1995; McCuen, 2005). Methods of baseflow separation can be broadly divided into graphical separation methods and digital filtering methods (Chen et al., 2008). Graphical separation methods are more

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Baseflow

465

Peak flow

Discharge

Inflection point Runoff begins

Constant-slope method Concave method Constant-baseflow method

Time

ww

appropriate for individual storms, while digital filtering methods are more appropriate for automated baseflow separation in long time series of flows. The simplest graphical separation method is to assume a constant baseflow equal to the lowest discharge rate before the beginning of surface runoff; this method is called either the constant-baseflow method or the constant-discharge baseflow separation method (McCuen, 2005) and is illustrated in Figure 9.27. More sophisticated graphical separation methods are the constant-slope baseflow separation method and the concave baseflow separation method. In the constant-slope baseflow separation method, the baseflow is delineated by a straight line from the lowest prepeak flow to the inflection point on the recession limb of the hydrograph. In the concave baseflow separation method, the baseflow is delineated by a straight line extrapolated from the prepeak hydrograph recession to the time of the peak flow and connected to a straight line from the time of the peak flow to the inflection point on the recession limb of the hydrograph (McCuen, 2005). The inflection point is where the recession limb of the hydrograph goes from being concave to convex. These methods of baseflow separation are illustrated in Figure 9.27, and the concave and constant-slope methods are generally preferred. An empirical variation of the concave baseflow separation method is to assume that surface runoff ceases N d after the hydrograph peak, where N is estimated by

w .E asy En g

ine eri n

N = 0.8A0.2

and A is the catchment area in km2 . EXAMPLE 9.22

(9.97)

g .n

Streamflows at the outlet of a 3000-km2 watershed were recorded at 12-h intervals over a 9.5-d period and these data are tabulated as follows: Time (d)

Flow (m3 /s)

Time (d)

Flow (m3 /s)

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

500 481 463 925 1804 1526 1240 925 722 518

5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5

444 361 287 268 250 231 213 194 176 157

et

The rainfall causing the direct runoff began at a time of around 1.0 days and lasted approximately 1.3 d. Estimate the direct runoff hydrograph and the runoff depth using: (a) the constant-baseflow method, (b) the constant-slope method, and (c) the concave method. Which method agrees best with Equation 9.97?

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions Solution From the given data: A = 3000 km2 . The key points in the measured hydrograph are the peak flow and the inflection point. It is apparent from the measured streamflows that the peak streamflow occurs at t = 2.0 d. The inflection point on the recession side of the hydrograph occurs when the flows change from an increasing negative slope to a decreasing negative slope. The inflection point is located by plotting the incremental flows in Column 3 of Table 9.17, which shows that the inflection point is located at t = 3.5 d. Different baseflow separation techniques provide different estimates of the baseflow, Qb , and the corresponding direct runoff estimates, Qr , are given by TABLE 9.17: Baseflow Separation Results

Constant-Baseflow

ww

Constant-Slope

Concave

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

Time (d)

Flow (m3 /s)

 (m3 /s)

Baseflow (m3 /s)

Runoff (m3 /s)

Baseflow (m3 /s)

Runoff (m3 /s)

Baseflow (m3 /s)

Runoff (m3 /s)

0.0

500

500

0

500

0

500

0

481

0

481

0

481

0

463

0

463

0

463

0

463

462

555

370

445

480

463

1341

648

1156

426

1378

463

1063

740

786

592

934

463

777

833

407

759

481

−19 0.5

481

w .E asy En g −18

1.0

463

462

1.5

925

879

2.0

1804

−278

2.5

1526

−286

3.0

1240

−315

3.5

925

463

4.0

722

925

0

925

0

259

722

0

722

0

55

518

0

518

0

0

444

0

444

0

463

0

361

0

463

0

287

0

463

0

268

0

268

463

0

250

0

250

0

463

0

231

0

231

0

463

0

213

0

213

0

463

0

194

0

194

0

463

0

176

0

176

0

463

0

157

0

157

0

463 −204

4.5

518

463 −74

5.0

444

463 −83

5.5

361

ine eri n 462

−203

−74 6.0

287

g .n 361 287

−19 6.5

268

et 0 0 0

−18 7.0

250 −19

7.5

231 −18

8.0

213 −19

8.5

194 −18

9.0

176 −19

9.5

157

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Baseflow

467 (9.98)

Qr = Q − Qb

where Q is the measured streamflow. The runoff depth, h, is equal to the volume of runoff divided by the catchment area. In the present case, h can be estimated by ⎤ ⎡ N−1 N−1  1  1 ⎣1 (Q + Qr,N ) + h= Qr,i t = Qr,i ⎦ t (9.99) A A 2 r,1 i=2

i=1

where N is the number of runoff estimates, Qr,i is the average runoff rate between the ith and i+1st runoff estimates, t is the time interval between runoff estimates, and Qr,i is the ith runoff estimate. Values of Qb and h are calculated below for different baseflow separation methods and the results are shown in Table 9.17. Explanations are as follows:

ww

(a) For the constant-baseflow separation method, the baseflow is equal to the streamflow at t = 1 d. The baseflow is therefore taken as Qb = 463 m3 /s, which is shown in Column 4, and the corresponding direct runoff is calculated using Equation 9.98 and is shown in Column 5. The maximum rate of direct runoff is 1341 m3 /s. The depth of runoff, h, is given by Equation 9.99 (with necessary unit conversions) as ⎤ ⎡ N−1  1 ⎣1 (Qr,1 + Qr,N ) + Qr,i ⎦ t h= A 2

w .E asy En g i=2

1 = 3000 * 106



1 (0 + 0) + (462 + 1341 + 1063 + 777 + 462 + 259 + 55) (0.5 * 86400) 2

= 0.063 m = 6.3 cm

(b) For the constant-slope baseflow separation method, the baseflow is delineated by a straight line from the lowest prepeak flow at t = 1 d to the inflection point at t = 3.5 d. Since the corresponding flows at these times are 463 m3 /s and 925 m3 /s, the baseflow, Qb , for 1 d … t … 3.5 d is given by Qb = 463 +

ine eri n

925 − 463 (t − 1) = 463 + 184.8(t − 1) 3.5 − 1

The baseflow estimated using this equation is shown in Column 6, and the corresponding direct runoff is shown in Column 7. The maximum rate of direct runoff is 1156 m3 /s. The depth of runoff, h, is given by Equation 9.99 as

1 1 (0 + 0) + (370 + 1156 + 786 + 407) (0.5 * 86400) = 0.039 m = 3.9 cm h= 3000 * 106 2

g .n

et

(c) For the concave baseflow separation method, the baseflow is delineated by a straight line extrapolated from the prepeak hydrograph recession to the time of the peak flow and connected to a straight line from the time of the peak flow to the inflection point. It is apparent from the incremental flows shown in Column 3 that the prepeak hydrograph recession has a slope of (−19 − 18)/1 = 37 (m3 /s)/d, and so the baseflow between the beginning of the surface runoff at t = 1 d and the peak runoff at t = 2 d is given by Qb = 463 − 37(t − 1) which yields a baseflow at the time of the peak runoff (at t = 2 d) of Qb = 463 − 37(2 − 1) = 426 m3 /s. The baseflow between the peak at t = 2 d and the inflection point at t = 3.5 d (where Q = 925 m3 /s) is estimated by linear interpolation as Qb = 426 +

925 − 426 (t − 2) = 426 + 332.7(t − 2) 3.5 − 2

The estimated baseflows are shown Column 8, and the corresponding direct runoffs are shown in Column 9. The maximum rate of direct runoff is 1378 m3 /s. The depth of runoff, h, is given by Equation 9.99 as

1 1 (0 + 0) + (480 + 1378 + 934 + 481) (0.5 * 86400) = 0.047 m = 4.7 cm h= 3000 * 106 2

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions It is apparent from these results that the different baseflow separation methods can lead to much different results. For the three methods considered here, the estimated peak direct runoff rate is in the range of 1156–1378 m3 /s, and the estimated runoff depth is in the range of 3.9–6.3 cm. Equation 9.97 predicts that the surface runoff will end N d after the peak runoff rate, where N = 0.8A0.2 = 0.8(3000)0.2 = 4.0 d Based on the observed hydrograph, the constant-baseflow method estimates that surface runoff ends at (5.0 − 2.0) = 3.0 d after the peak, while both the constant-slope and concave methods estimate that surface runoff ends (3.5 − 2.0) = 1.5 d after the peak. Therefore, the constant-baseflow method provides the closest agreement with Equation 9.97.

ww Problems

The partitioning of total runoff into direct runoff and baseflow for any given storm event is usually dependent on the average intensity and duration of the storm as well as the total infiltration during the storm (Nejadhashemi et al., 2008). Baseflow contributed by infiltrating groundwater is a relatively slow process compared with overland flow and interflow and, as a consequence, overland flow and interflow are sometimes collectively referred to as quickflow (Fitts, 2002). The long-term ratio of baseflow volume to total streamflow volume is called the baseflow index (Eckhardt, 2008). In the Coastal Plain regions of the United States, baseflow is typically a significant portion of the flow in drainage channels (LaSage et al., 2008a).

w .E asy En g

9.1. Show that the maximum return period that can be investigated when using the Weibull formula to derive the IDF curve from an n-year annual rainfall series is equal to n + 1. 9.2. A rainfall record contains 50 years of measurements at 5-min intervals. The annual-maximum rainfall amounts for intervals of 5 min, 10 min, 15 min, 20 min, 25 min, and 30 min are ranked as follows:

Rank

5

10

1 2 3

26.2 25.3 24.2

45.8 44.0 42.2

t in min 15 20 60.5 58.1 55.8

72.4 69.6 66.8

25

30

81.8 78.6 75.5

89.7 86.3 82.8

where the rainfall amounts are in millimeters. Calculate the IDF curve for a return period of 40 years. 9.3. An 8-year rainfall record measures rainfall increments at 10-min intervals, and the top five annual-maximum rainfall increments are as follows: Rank 10-min rainfall (mm)

1

2

3

4

5

39.6

39.4

38.5

37.3

36.5

Estimate the frequency distribution of the annual maxima with return periods greater than 2 years. 9.4. Assuming that the ranked rainfall increments in Problem 9.2 were derived from a partial-duration series, calculate the annual-series IDF curve for a 40-year return period. 9.5. The rainfall data in Table 9.18 were compiled by determining the maximum amount of rainfall that occurred

for a given time interval during a given year. Use these data to develop IDF curves for 5-, 10-, and 25-yr return periods. If these IDF curves are to be fitted to a function of the form a i= t + b

ine eri n

estimate the values of a and b for a 5-year return period. 9.6. Use the Chen method to estimate the 10-year IDF curve for Atlanta, Georgia. 9.7. Use the Chen method to estimate the 10-year IDF curves for New York City and Los Angeles. Compare your results with those obtained by Wenzel (1982) and listed in Table 9.2. 9.8. The spatially averaged rainfall is to be calculated for the catchment area shown in Figure 9.12, where the coordinates of the nearby rain gages are as follows:

g .n

Gage

x (km)

A B C D E

1.3 1.0 4.2 3.5 2.1

y (km)

et

7.0 3.7 4.9 1.4 −1.0

and the coordinates of the grid origin are (0, 0). The measured rainfall at each of the gages during a 1-h interval is 60 mm at A, 90 mm at B, 65 mm at C, 35 mm at D, and 20 mm at E. Use the inverse-distance-squared method with the 1-km * 1-km grid shown in Figure 9.12 to estimate the average rainfall over the catchment during the 1-h interval.

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Downloaded From : www.EasyEngineering.net Problems TABLE 9.18: Rainfall Maxima (in cm) for Given Durations Year 1946 1947 1948 1949 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 1960 1961 1962 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990

1 1.7 1.9 1.9 1.9 1.9 2.1 1.7 1.7 1.9 2.4 1.0 2.7 1.3 2.8 2.0 1.8 1.5 1.4 1.6 1.2 1.4 1.7 1.9 1.8 1.3 1.8 1.6 1.9 3.1 1.5 2.9 1.4 1.6 2.2 2.5 2.1 1.6 1.2 1.5 1.7 3.7 2.7 2.0 1.9 1.2

2 2.9 2.6 2.3 2.4 1.9 2.4 2.5 3.6 2.1 3.7 1.3 2.8 1.5 3.2 2.1 2.7 2.6 2.5 1.8 1.7 1.5 1.8 2.6 3.2 1.6 2.1 2.0 2.8 3.4 1.6 3.4 2.5 2.1 2.3 3.0 3.0 2.3 1.8 4.2 2.7 6.7 4.8 6.9 2.9 1.4

ww

4 4.6 2.6 2.7 2.8 2.2 3.0 2.7 3.6 2.8 4.5 1.6 3.2 2.0 3.4 2.8 2.8 2.8 3.1 2.2 2.4 1.5 2.4 4.1 3.8 1.7 3.3 2.3 2.8 3.6 2.1 4.0 2.8 2.2 4.5 3.2 3.1 2.8 2.1 4.2 2.7 9.7 4.8 6.9 3.0 1.6

Duration (h) 6 10 4.8 4.8 3.0 3.7 2.9 3.1 4.0 4.7 2.5 2.5 3.5 4.0 2.8 3.0 3.6 4.3 3.0 3.8 4.5 4.6 1.7 1.7 3.7 3.7 2.6 3.5 4.1 5.2 4.1 5.6 2.8 2.8 3.1 3.8 3.7 4.2 2.6 3.6 2.7 3.0 2.3 2.5 2.7 2.9 5.1 5.1 3.9 3.9 1.7 1.7 3.4 3.7 2.6 2.7 2.9 2.9 3.6 3.6 2.9 3.6 4.1 4.4 2.8 2.8 2.2 2.2 5.2 7.6 3.2 4.3 3.1 3.1 3.2 3.8 2.5 2.5 4.2 4.2 2.7 2.8 10.4 10.4 4.8 6.2 6.9 6.9 3.0 3.0 1.7 2.1

12 4.8 3.7 3.1 4.9 2.5 4.2 3.0 4.3 3.8 4.6 1.7 3.8 3.8 5.7 6.0 2.8 4.5 4.3 3.6 3.1 2.7 2.9 5.2 3.9 1.7 4.2 2.7 2.9 3.6 3.6 4.5 2.8 2.2 7.7 4.6 3.1 4.1 2.5 4.2 2.9 10.7 6.2 6.9 3.0 2.1

24 5.1 4.8 3.2 5.7 3.0 5.1 4.0 4.7 3.8 5.4 1.8 4.4 4.1 6.0 7.9 2.8 6.4 4.5 4.1 3.3 2.8 4.4 5.4 3.9 2.7 4.9 2.7 2.9 3.8 4.2 4.9 2.9 2.7 7.8 4.7 5.0 5.0 2.6 4.2 3.5 11.7 6.7 10.1 3.2 2.5

9.11. The IDF curve for 10-year storms in Santa Fe, New Mexico, is given by i=

9.12.

9.13.

w .E asy En g

9.9. Repeat Problem 9.8 by assigning equal weight to each rain gage. Compare your result with that obtained using the inverse-distance-squared method. Which result do you think is more accurate? 9.10. The IDF curve for 10-year storms in Atlanta, Georgia, is given by i=

1628 (td + 8.16)0.76

where i is the average intensity in mm/h and td is the duration in minutes. Assuming that the maximum rainfall intensity occurs at 38% of the rainfall duration, estimate the triangular hyetograph for a 40-min storm.

469

9.14.

9.15.

9.16.

9.17.

9.18.

818 (td + 8.54)0.76

where i is the average intensity in mm/h and td is the duration in minutes. Assuming that the maximum rainfall intensity occurs at 44% of the rainfall duration, estimate the triangular hyetograph for a 50-min storm. Use the alternating-block method and the IDF curve given in Problem 9.11 to calculate the hyetograph for a 10-year 1-h storm using 11 time intervals. Derive the IDF curve for Boston, Massachusetts, using the Chen method. Use the derived IDF curve with the alternating-block method to determine the hyetograph for a storm with a return period of 10 years and a duration of 40 min. Use nine time intervals to construct the hyetograph. Use the 10-year IDF curve for Atlanta in Table 9.2 given by Wenzel (1982) and the alternating-block method to estimate the rainfall hyetograph for a 10-year 50-min storm in Atlanta. Use seven time intervals. The alternating-block method assumes that the maximum rainfall for any duration less than or equal to the total storm duration has the same return period. Discuss why this is a conservative assumption. The precipitation resulting from a 25-year 24-h storm in Miami, Florida, is estimated to be 260 mm. Calculate the NRCS 24-h hyetograph. The precipitation resulting from a 25-year 24-h storm in Atlanta, Georgia, is estimated to be 175 mm. Calculate the NRCS 24-h hyetograph, and compare it with the 25-year 24-h hyetograph for Miami that is calculated in Problem 9.16. The following function has been proposed as a good fit to the NRCS Type III rainfall distribution (Froehlich, 2009; 2010) ⎧ −0.504(12−t) + 0.151e−6.666(12−t) ⎪ ⎪ ⎪ 0.219e ⎪ ⎪ ⎨ + 0.0112t − 0.001, t … 12 h P = ⎪ P24 ⎪− 0.219e−0.504(t−12) − 0.151e−6.666(t−12) ⎪ ⎪ ⎪ ⎩ + 0.0112(t − 12) + 0.870, 12 < t … 24 h

ine eri n

g .n

et

where P24 is the 24-h total rainfall and P is the cumulative rainfall after time t. (a) Use the given function to determine the relationship between rainfall intensity and time. (b) Using the result in (a), and any other relevant calculations, estimate the maximum rainfall intensity in a 10-year 24-h storm in Miami. 9.19. The rainfall measured at a rain gage during a 10-h storm is 193 mm. Estimate the average rainfall over a catchment containing the rain gage, where the area of the catchment is 200 km2 . 9.20. Use Equation 9.17 to show that for long-duration storms the areal-reduction factor becomes independent of the

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions

storm duration. Under these asymptotic conditions, determine the relationship between the areal-reduction factor and the catchment area. 9.21. The annual-maximum 24-h rainfall amounts in a South American city have a mean of 202 mm and a standard deviation of 65 mm. Estimate the probable maximum precipitation (PMP) using the Hershfield method, and compare this value to the world’s largest observed 24-h rainfall. 9.22. Assess whether the Chen method always leads to an IDF curve of the form f (T) i= g(t) where i is the average rainfall intensity for a storm of duration t and return period T, f (T) is a term that depends only on the return period, T, of the storm, and g(t) is a term that depends only on the duration, t, of the storm. 9.23. The annual-maximum 24-h rainfall amounts on a tropical island have a mean of 320 mm and a standard deviation of 156 mm. The IDF curve for the 50-year storm on the island is given by

ww

9.28. For the pine forest described in Problem 9.26, estimate the interception using a Horton-type empirical equation of the form I = a + bPn , where a and b are constants and P is the precipitation amount. 9.29. A 25-min storm produces 30 mm of rainfall on the following surfaces: steep pavement, flat pavement, impervious surface, lawn, pasture, and forest litter. Assuming that depression storage is the dominant abstraction process, for each surface, estimate the fraction of rainfall that becomes surface runoff. On which surface is depression storage most significant, and on which surface is it least significant? 9.30. A soil has the Horton infiltration parameters: f0 = 200 mm/h, fc = 60 mm/h, and k = 4 min−1 . If rainfall in excess of 200 mm/h is maintained for 50 min, estimate the infiltration as a function of time. What is the infiltration rate at the end of 50 min? How would this rate be affected if the rainfall rate were less than 200 mm/h? 9.31. Show that the Horton infiltration model given by Equation 9.45 satisfies the differential equation

w .E asy En g

i=

6240

(t + 6.87)0.65

mm/h

where i is the intensity in mm/h and t is the duration in minutes. Find the annual-maximum 8-h rainfall with a return period of 50,000 years. 9.24. Estimate the maximum amount of rainfall that can be expected in 1 h. 9.25. Maximum precipitation amounts can be estimated using Equation 9.28, and observed precipitation maxima are listed in Table 9.5. Compare the predictions given by Equation 9.28 with the observations in Table 9.5. 9.26. A pine forest is to be cleared for a development in which all the trees on the site will be removed. The IDF curve for the area is given by 2819 i= t + 16

where i is the average rainfall intensity in mm/h and t is the duration in minutes. The storage capacity of the trees in the forest is estimated as 5 mm, the leaf area index is 6, and the evaporation rate during the storm is estimated as 0.3 mm/h. Determine the increase in precipitation reaching the ground during a 30-min storm that will result from clearing the site. 9.27. The storage capacity of a forest canopy covering a catchment is 9 mm, the leaf-area index is 8, and the evaporation rate during a storm is 0.5 mm/h. If a storm has the hyetograph calculated in Problem 9.14, estimate the amount of precipitation intercepted during each of the seven time intervals of the hyetograph. Determine the hyetograph of the rainfall reaching the ground.

dfp = −k(fp − fc ) dt

9.32. A catchment has the Horton infiltration parameters: f0 = 150 mm/h, fc = 50 mm/h, and k = 3 min−1 . The design storm is: Interval (min)

Average rainfall (mm/h)

0–10 10–20 20–30 30–40 40–50 50–60

20 40 80 170 90 20

ine eri n

g .n

et

Estimate the time when ponding begins. 9.33. An alternating-block analysis indicated the following rainfall distribution for a 70-min storm:

where each tration can parameters:

Interval

Rainfall (mm/h)

1 2 3 4 5 6 7

7 25 45 189 97 44 18

interval has a duration of 10 min. If infilbe described by the Horton model with f0 = 600 mm/h, fc = 30 mm/h, and

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Downloaded From : www.EasyEngineering.net Problems k = 0.5 min−1 , and the depression storage is 4 mm, use the Horton model to determine the distribution of runoff, and hence the total runoff. 9.34. An area consists almost entirely of sandy loam, which typically has a saturated hydraulic conductivity of 11 mm/h, average suction head of 110 mm, porosity of 0.45, field capacity of 0.190, wilting point of 0.085, and depression storage of 4 mm. The design rainfall is given as: Interval (min)

Average rainfall (mm/h)

0–10 10–20 20–30 30–40 40–50 50–60

20 40 60 110 60 20

ww

these time intervals. What is the runoff and associated curve number? 9.40. Use Equation 9.85 to calculate the average infiltration rate during the storm described in Problem 9.37. Compare this calculated infiltration rate with the given minimum infiltration rate of 10 mm/h. 9.41. Use Figure 9.8 and Table 9.4 to estimate the hourly rainfall intensity for a 10-year 24-h storm in Miami, Florida. (a) If this rain falls on a forested area with a storage capacity of 5 mm, a LAI of 7, and the area has an evaporation rate of 5 mm/d during the rainy season, estimate the fraction of rainfall that reaches the ground. (b) It is estimated that the infiltration capacity of any soil in nonforested areas can be taken as a constant, depending on the type of soil, and the depression storage is 5 mm. Estimate the runoff expected from the design storm in areas covered by sand and sandy loam. The infiltration capacities of these soils can be taken as 60 mm/h and 25 mm/h, respectively. (c) Estimate the curve numbers of the nonforested areas for the two types of soils.

w .E asy En g

Use the Green–Ampt method to determine the runoff versus time, and contrast the depth of rainfall with the depth of runoff. Assume that the initial moisture conditions are midway between the field capacity and wilting point. 9.35. Repeat Problem 9.34 for a sandy clay soil with a depression storage of 9 mm. [Hint: Use Table 9.12 to estimate the soil properties.] 9.36. Derive the NRCS curve-number model for the infiltration rate given by Equation 9.85. Explain why this infiltration model is unrealistic. 9.37. Drainage facilities are to be designed for a rainfall of return period 10 years and duration 1 h. The IDF curve is given by 203 i= (t + 7.24)0.73 where i is the average intensity in cm/h and t is the storm duration in minutes. The minimum soil infiltration capacity is 10 mm/h, and the area to be drained is primarily residential with lot sizes on the order of 0.2 ha (0.5 ac). Use the NRCS method to estimate the total amount of runoff (in cm), assuming the soil is in average condition at the beginning of the design storm. 9.38. Repeat Problem 9.37 for the case in which heavy rainfall occurs within the previous 5 d and the soil is saturated. 9.39. A drainage system in Atlanta, Georgia, is to be designed for a 4.5-h storm with a return period of 10 years. A site investigation has determined that the pervious areas on the site can be characterized by an average depression storage of 5 mm, and Hortonian infiltration characteristics with f0 = 80 mm/h, fc = 8 mm/h, and k = 0.5 h−1 . Construct the design rainfall distribution using 9 intervals in the alternating-block method, and then calculate the runoff by assuming Hortonian infiltration within

471

9.42. Data from a double-ring infiltrometer indicate that a soil in a catchment has the following Horton parameters: f0 = 250 mm/h, fc = 44 mm/h, and k = 0.13 min−1 . Observations also indicate that the catchment has an average depression storage of 6 mm. If the catchment is located where the 10-year 24-h rainfall is 229 mm and can be described by the NRCS Type II distribution, estimate the curve number for the site. Use hourly time increments in your analysis. Would the curve number be different for a 20-year 24-h rainfall? Is the dependence of the curve number on the rainfall amount physically reasonable?

ine eri n

9.43. An undeveloped parcel of land in south Florida has a water-table elevation 1.22 m below land surface, and an estimated cumulative water storage of 21 cm is recommended for use in the curve-number method. Consider the 1-d and 3-d storm events given in Tables 9.19 and 9.20, respectively, both of which have a total rainfall amount of 31.1 cm. Assuming that the infiltration capacity of the soil is a constant, plot the relationship between curve number and infiltration capacity for both the 1-d and 3-d storms. Use these results to determine the infiltration capacity corresponding to the 21 cm of available storage. If the actual infiltration capacity of the soil is 4 cm/h, estimate the curve numbers that should be used for the 1-d and 3-d storms. 9.44. A proposed 20-ha development includes 5 ha of parking lots, 10 ha of buildings, and 5 ha of grassed area. The runoff from the parking lots and buildings are both routed directly to grassed areas. If the grassed areas contain Type A soil in good condition, estimate the runoff from the site for a 180-mm rainfall event. 9.45. Repeat Problem 9.44 for the case where the runoff from the buildings, specifically the roofs, is discharged directly

g .n

et

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions TABLE 9.19: 1-Day Storm Event

Time (h)

Rainfall (cm)

Time (h)

Rainfall (cm)

Time (h)

Rainfall (cm)

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0

0.0 0.2 0.3 0.5 0.6 0.8 1.0 1.2 1.4 1.6 1.9 2.2 2.6 3.0 3.4 3.8 4.3

8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.5 13.0 13.5 14.0 14.5 15.0 15.5 16.0

4.8 5.3 5.9 6.6 7.4 8.4 9.9 20.4 22.7 23.9 24.7 25.4 26.0 26.4 26.9 27.4

16.5 17.0 17.5 18.0 18.5 19.0 19.5 20.0 20.5 21.0 21.5 22.0 22.5 23.0 23.5 24.0

27.6 27.9 28.2 28.5 28.8 29.0 29.3 29.6 29.8 30.0 30.2 30.4 30.5 30.7 30.9 31.1

ww

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Rainfall (cm) 0.0 0.2 0.27 0.41 0.55 0.69 0.82 0.98 1.1 1.3 1.4 1.5 1.7 1.8 1.9 2.1 2.2 2.35 2.51 2.65 2.79 2.93 3.06 3.20 3.34

CNc = CNp +

Time (h)

Rainfall (cm)

Time (h)

Rainfall (cm)

25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

3.54 3.75 3.95 4.16 4.34 4.55 4.75 4.96 5.17 5.37 5.58 5.76 5.97 6.17 6.38 6.58 6.79 7.00 7.20 7.41 7.59 7.80 8.00 8.21

49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72

8.44 8.66 8.93 9.24 9.62 10.1 10.7 11.3 12.1 13.1 14.4 23.2 25.7 26.9 27.6 28.3 28.7 29.1 29.6 30.0 30.2 30.5 30.8 31.1

onto the parking lots. Based on this result, what can you infer about the importance of directing roof drains to pervious areas?

I (98 − CNp ) 100

9.49. Consider the site described in Problem 9.48, with the exception that only a portion of the impervious area is directly connected to the drainage system. Show that the composite curve number of the site, CNc , is then given by

w .E asy En g

TABLE 9.20: 3-Day Storm Event

Time (h)

9.46. Repeat Problem 9.44 using an area-weighted curve number. How would your result change if the roof drains were directly connected to the parking lot? 9.47. Discuss why it is preferable to route rainfall excesses on composite areas rather than using weighted-average curve numbers. 9.48. Consider a site that is I percent impervious, the curve number of the impervious area is 98, and the curve number of the pervious area is CNp . If all of the impervious area is directly connected to the drainage system, show that the composite curve number of the site, CNc , is given by

CNc = CNp +

I (98 − CNp )(1 − 0.5R) 100

where R is the ratio of the unconnected impervious area to the total impervious area, and 50% of the unconnected impervious area runs onto the pervious surface. 9.50. The 10-year IDF curve for a catchment area is given by 6000 t + 20

ine eri n i=

mm/h

where t is the event duration in minutes. (a) If the soil can be characterized by an infiltration capacity of 25 mm/h, estimate the maximum 10-year runoff. (b) Field reconnaissance reveals that the infiltration capacity is distributed such that the upper half of the catchment is impervious and the lower half of the catchment has an infiltration capacity of 50 mm/h. Estimate the adjusted 10-year maximum runoff and assess the implication of your result. 9.51. Average daily streamflows on a river that drains a 2000 km2 watershed were recorded for 13 consecutive days and the measurements were as follows:

g .n

et

Time (d)

Flow (m3 /s)

Time (d)

Flow (m3 /s)

0 1 2 3 4 5 6

74 67 56 237 1920 607 168

7 8 9 10 11 12 —

108 82 69 63 57 51 —

Estimate the direct runoff hydrograph and the runoff depth using: (a) the constant-baseflow method, (b) the constant-slope method, and (c) the concave method.

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C H A P T E R

10

Fundamentals of Surface-Water Hydrology II: Runoff 10.1

ww

Introduction

Runoff models predict the temporal distribution of surface runoff at a catchment outlet based on the temporal distribution of effective rainfall and the catchment characteristics. The effective rainfall is defined as the incident rainfall minus the abstractions, and is sometimes referred to as the rainfall excess. The most important abstractions are usually infiltration and depression storage, and the catchment characteristics that are usually most important in translating the effective rainfall distribution to a runoff distribution at the catchment outlet are those related to the topography and surface cover of the catchment. Runoff models are classified as either distributed-parameter models or lumped-parameter models. Distributedparameter models account for runoff processes on scales smaller than the size of the catchment, such as accounting for the runoff from every roof, over every lawn, and in every street gutter, while lumped-parameter models consider the entire catchment as a single hydrologic element, with the runoff characteristics described by one or more (lumped) parameters. The process in which runoff from one part of a catchment infiltrates into another part of the catchment prior to reaching the catchment outlet is called the run-on process, and accounting for this process is particularly important in distributed hydrologic models (Morbidelli et al., 2008).

w .E asy En g 10.2

Mechanisms of Surface Runoff

ine eri n

A fundamental hypothesis that was originally made by Horton (1933b; 1945) is that overland flow occurs when the rainfall rate exceeds the infiltration capacity of the soil. This type of overland flow is commonly referred to as Hortonian overland flow or Hortonian runoff. In modern engineering practice, it is generally recognized that the two primary hydrological mechanisms that generate overland flow are infiltration excess and saturation excess. The saturation-excess mechanism is fundamentally different from infiltration-excess mechanism in that overland flow is generated at locations where the soil is saturated at the surface. Overland flow generated by saturation excess is sometimes called Hewlettian runoff after Hewlett (1961; 1972). Unlike Hortonian runoff, where infiltration capacity plays a controlling role in runoff generation, storage capacity is the primary control in Hewlettian runoff. Saturation excess is the basis of the concept of variable source-area hydrology that acknowledges that the spatial extent of ground saturation can vary between storms and hence cause variations in the portion of the catchment area that contributes to surface runoff. In the saturation-excess process, rainfall causes a thin layer of soil on some parts of the basin to saturate upward from some restricting boundary to the ground surface, especially in zones of shallow, wet, or less-permeable soil. Determining which process dominates, infiltration excess or saturation excess, is fundamental to identifying appropriate methods for describing the rainfall-runoff relation. In cases where significant surface runoff is observed while estimated infiltration capacities and rainfall intensities indicate that significant Hortonian runoff should not occur, saturation-excess runoff is likely to be the primary runoff mechanism. Identification of the appropriate runoff mechanism can be of great importance, since accurate partitioning of rainfall into infiltration and surface runoff affects the accuracy with which groundwater recharge and streamflow are predicted, and hence can have major implications on water management on seasonal and annual time scales (Alfa et al., 2011).

g .n

et

473

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff

ww

A wide variety of models are available for calculating runoff from rainfall, and their applicability must be assessed in light of the fundamental rainfall-runoff process. The applicability of various runoff models can be broadly associated with the scale of the catchment, which can be classified as small, midsize, or large (Ponce, 1989). In small catchments, the response of the catchment to rainfall events is sufficiently rapid and the catchment is sufficiently small that runoff during a relatively short time interval can be adequately modeled by assuming a constant rainfall in space and time. The rational method is the most widely used runoff model in small catchments. In midsize catchments, the slower response requires that the temporal distribution of rainfall be accounted for; however, the catchment is still smaller than the characteristic storm scale, and the rainfall can be assumed to be uniform over the catchment. Unit hydrograph models are the most widely used runoff models in midsize catchments. In large catchments, both the spatial and temporal variations in precipitation events must be incorporated in the runoff model, and models that explicitly incorporate routing methodologies are the most appropriate. Runoff regimes within a catchment vary from overland flow at the smallest scales to river flow at the largest scales, and runoff models must necessarily accommodate this scale effect. Small catchments have predominantly overland-flow runoff, while large catchments typically have a significant amount of runoff in identifiable rivers or drainage channels. As a consequence, the channel storage characteristics increase significantly from small catchments to large catchments.

w .E asy En g 10.3

Time of Concentration

The parameter most often used to characterize the response of a catchment to a rainfall event is the time of concentration, which is defined as the time to equilibrium of a catchment under a steady rainfall excess, or sometimes as the longest travel time that it takes for surface runoff to reach the discharge point of a catchment (Wanielista et al., 1997). Most equations for estimating the time of concentration, tc , express tc as a function of the effective rainfall intensity, ie , catchment length scale, L, average catchment slope, S0 , and a parameter that describes the catchment surface, C; hence the equations for tc typically have the functional form tc = f (ie , L, S0 , C) (10.1)

ine eri n

The time of concentration of a catchment includes the time of overland flow and the travel time in drainage channels leading to the catchment outlet. 10.3.1

Overland Flow

g .n

Several equations are commonly used to estimate the time of concentration for overland flow. The most popular ones are described here. 10.3.1.1

Kinematic-wave equation

et

A fundamental expression for the time of concentration in overland flow can be derived by considering the one-dimensional approximation of the surface-runoff process illustrated in Figure 10.1. The boundary of the catchment area is at x = 0, ie is the rainfall-excess rate [LT−1 ], y is the runoff flow depth [L], and q is the volumetric flow rate per unit width of the catchment area [L2 T−1 ]. Within the control volume of length x [L], the law of conservation of mass requires that the net mass inflow is equal to the rate of change of mass within the control volume. This law can be stated mathematically by the relation     ⭸(ρq) x ⭸(ρq) x ⭸y (ρq) − ρ x (10.2) + [ie ρ x] − (ρq) + = ⭸x 2 ⭸x 2 ⭸t where ρ is the density of the (runoff) water [ML−3 ], the first term in square brackets is the inflow rate to the control volume, the second term is the rate at which rainfall excess is entering the control volume, the third term is the outflow rate, and the righthand side of Equation 10.2 is equal to the rate of change of fluid mass within the control volume. Taking the density, ρ, as being constant and simplifying yields

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Rainf

Time of Concentration

475

all ex

cess

ie

Catch

ment

FIGURE 10.1: One-dimensional approximation of surface-runoff process

boun dary

Section 10.3

Control volume x ⫽0

y q Runo

ff

∆x

θ

x

ww

⭸q ⭸y + = ie ⭸t ⭸x

w .E asy En g

(10.3)

Equation 10.3 contains two unknowns, q and y, and a second relationship between these variables is needed to solve this equation. Normally, the second equation would be the momentum equation; however, a unique relationship between the flow rate, q, and the flow depth, y, can be assumed to have the form q = αym

(10.4)

where α and m are constants. The assumption of a relationship such as Equation 10.4 is justified in that equations describing steady-state flow in open channels, such as the Manning and Darcy–Weisbach equations, can be put in the form of Equation 10.4. Combining Equations 10.3 and 10.4 leads to the following differential equation for y:

ine eri n

⭸y ⭸y + αmym−1 = ie (10.5) ⭸t ⭸x This equation is known as the kinematic-wave model for flow over a sloped plane (Lighthill and Whitham, 1955). Solution of kinematic-wave equation (Equation 10.5) can be obtained by comparing it with the following equation,

g .n

dy ⭸y dx ⭸y = + = ie (10.6) dt ⭸t dt ⭸x which gives the rate of change of y with respect to t observed by moving at a velocity dx/dt. Consequently, Equation 10.5 is equivalent to the following pair of equations: dx = αmym−1 dt dy = ie dt

et

(10.7) (10.8)

where dx/dt is called the wave speed. Solution of Equation 10.8 subject to the boundary condition that y = 0 at t = 0 yields y = ie t (10.9) Substituting this result into Equation 10.7 and integrating subject to the boundary condition that x = 0 at t = 0 yields x = αiem−1 tm (10.10) Equations 10.9 and 10.10 are parametric equations describing the water surface illustrated in Figure 10.1, and the discharge at any location along the catchment area can be obtained by combining Equations 10.4 and 10.9 to yield

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff

q = α(ie t)m

(10.11)

Defining the time of concentration, tc , of a catchment as the time required for a kinematic wave to travel the distance, L, from the catchment boundary to the catchment outlet, then Equation 10.10 gives the time of concentration as

tc =



L αiem−1

1

m

(10.12)

Equation 10.12 was originally presented by Henderson and Wooding (1964) and Henderson (1966). If the Manning equation is used to relate the runoff rate to the depth of overland flow, then Equation 10.12 can be written in the form (ASCE, 1992)

ww

tc =

6.99 2 5

ie

w .E asy En g



nL √ S0

3 5

(10.13)

where tc is in minutes, ie in mm/h, and L in m; n is the Manning roughness coefficient for overland flow [dimensionless]; and S0 is the ground slope [dimensionless]. It is worthwhile to note that Equation 10.13 expresses tc as √ a function of both a rainfall characteristic (ie ) and a lumped catchment characteristic (nL/ S0 ). Equation 10.13 was originally presented by Woolhiser and Liggett (1967) and is recommended by ASCE for use in drainage design (ASCE, 1992). In some applications, the coefficient 6.99 is rounded to 7 (e.g., Wong, 2009). Estimates of Manning’s n for overland flow are given in Table 10.1, where the surface types are ordered by increasing roughness. Values of Manning’s n for overland flow are associated with very shallow flow depths and should not be confused with Manning’s n for open-channel flow; overland-flow values are typically much larger for sheet flow than for open-channel flow with the same surface description. Mean flow depths of 0.6 mm (0.02 in.) for paved areas and 6 mm (0.2 in.) for vegetated areas are typically associated with the Manning’s n for overland flow (Seybert, 2006). In reality, overland flow occurs as a discontinuous shallow sheet of water with threads of flow diverging and converging around microtopographic elevations, rocks, and vegetation; under these circumstances Manning’s n actually depends on both microtopographic parameters and the Reynolds number of the flow (Lopez-Sabater et al., 2002), and in overland-flow models Manning’s n might be more appropriately expressed as a function ¨ of flow depth (Mugler et al., 2011). The kinematic-wave model given by Equation 10.5 can also be solved for time-varying rainfall excess (e.g., Mizumura, 2006); however, the constant rainfall excess formulation presented here is much more commonly used in practice. The kinematic-wave model, which neglects the acceleration and pressure terms in the momentum equation, is only appropriate for slopes where S0 > 0.2% (Kazezyılmaz and Medina, 2007). The kinematic-wave time of concentration given by Equation 10.13 assumes that the surface runoff is described by the Manning equation, which is valid only for fully turbulent flows; however, at least a portion of the surface runoff will be in the laminar and transition regimes (Wong and Chen, 1997). This limitation associated with using the Manning equation can be addressed by using the Darcy–Weisbach equation, which yields α=



8gS0 Cν k



ine eri n

1 (2−k)

and

m=

3 2 − k

g .n

et

(10.14)

where ν is the kinematic viscosity of water, and C and k are parameters relating the Darcy–Weisbach friction factor, f , to the Reynolds number, Re: f =

C Rek

(10.15)

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Time of Concentration

477

TABLE 10.1: Manning’s n for Overland Flow

Surface type

ww

Smooth concrete Bare sand Asphalt Bare clay Smooth earth Graveled surface Bare clay-loam (eroded) Fallow (no residue) Cultivated soils (residue cover … 20%) Bare smooth soil Range (clipped) Range (natural) Sparse vegetation Grass (short prairie) Cultivated soils (residue cover > 20%) Light turf Woods, no underbrush Grass (dense) Lawns Dense turf Pasture Dense shrubbery and forest litter Woods (light underbrush) Grass (Bermuda) Grass (bluegrass sod) Woods (dense underbrush)

w .E asy En g

Manning’s n

Range

0.011 0.01 0.012 0.012 0.018 0.02 0.02 0.05 0.06 0.10 0.10 0.13 0.15 0.15 0.17 0.20 0.20 0.24 0.25 0.35 0.35 0.40 0.40 0.41 0.45 0.60

0.01–0.014 0.01–0.016 0.010–0.018 0.010–0.016 0.015–0.021 0.012–0.030 0.012–0.033 0.006–0.16 — — 0.02–0.24 0.01–0.32 — 0.10–0.25 — — 0.1–0.3 0.15–0.35 0.20–0.30 0.30–0.35 0.30–0.40 — 0.3–0.5 0.30–0.50 0.39–0.63 0.6–0.95

ine eri n

Sources: ASCE (1992; 2006a); Wurbs and James (2002); Crawford and Linsley (1966); Engman (1986); McCuen et al. (1996); McCuen (2005); Seybert (2006).

where k = 0 for fully turbulent flow, k = 1 for laminar flow, 0 < k < 1 for transitional flow, and the Reynolds number is defined by Re =

q ν

g .n

(10.16)

et

Laminar flow typically occurs where Re < 200, fully turbulent flow where Re > 2000, and transition flow where 200 < Re < 2000. Values of C for overland flow have not been widely published, but Radojkovic and Maksimovic (1987) and Wenzel (1970) indicate that for concrete surfaces C values of 41.8, 2, and 0.04 are appropriate for laminar, transition, and fully turbulent flow regimes, respectively. Combining the kinematic-wave expression for tc , Equation 10.12, with the Darcy–Weisbach equation, Equations 10.14 to 10.16, yields

tc =



0.21(3.6 * 106 ν)k CL2−k S0 i1+k e

1

3

(10.17)

where tc is in minutes, ν is in m2 /s, L is in meters, and ie is in mm/h. Equation 10.17, called the Chen and Wong formula (Wong, 2005), can be used to account for various flow regimes in overland flow, and it has been shown that assuming a single flow regime (laminar, transition, or fully turbulent) will tend to underestimate the time of concentration (Wong and Chen, 1997). Indications are that overland flow is predominantly in the transition regime and that Equation 10.17 may be most applicable using k L 0.5.

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff

10.3.1.2

ww

NRCS method

NRCS (SCS, 1986) proposed that overland flow consists of two sequential flow regimes: sheet flow and shallow concentrated flow. Sheet flow is characterized by runoff that occurs as a continuous sheet of water flowing over the land surface, while shallow concentrated flow is characterized by flow in isolated rills and then gullies of increasing proportions. Ultimately, most surface runoff enters open channels or drainage pipes, which is the third regime of surface runoff included in the time of concentration. In many cases, the time of concentration of a catchment is expressed as the sum of the travel time as sheet flow plus the travel time as shallow concentrated flow plus the travel time as open-channel flow. The flow characteristics of sheet flow are sufficiently different from shallow concentrated flow that separate equations are recommended. The NRCS equation for travel time in the sheet-flow regime was derived from the kinematic-wave time of concentration given by Equation 10.13. It has been shown that for rainfall-event durations longer than 15 min the IDF curves corresponding to the NRCS Type II and Type III rainfall distributions can be approximated by (Welle and Woodward, 1986) ⎧ 5.70P24 ⎪ ⎪ ⎪ Type II rainfall, t Ú 15 min ⎪ ⎨ t0.62 i= (10.18) ⎪ ⎪ 4.76P ⎪ 24 ⎪ Type III rainfall, t Ú 15 min ⎩ 0.63 t

w .E asy En g

where i is the average rainfall intensity [mm/h] over a duration t [min], and P24 is the 24-h rainfall depth [mm]. If the effective rainfall intensity, ie , is related to the actual rainfall intensity, i, by the runoff coefficient, C, such that (10.19)

ie = Ci

ine eri n

then combining Equations 10.13, 10.18 (averaged), and 10.19 gives the time of concentration in the sheet-flow regime as  4 5 nL 5.5 tf = √ (10.20) P24 C 32 √S0 The conventional NRCS equation for estimating the time of concentration in the sheet-flow regime is a special case of Equation 10.20 where C = 1, which gives 5.5 tf = √ P24



nL √ S0

4 5

g .n

et

(10.21)

where tf is in minutes, P24 is in mm, and L is in meters. Although the conventional form of the NRCS equation for tf is given by Equation 10.21 for all surfaces, some practitioners use the more general form given by Equation 10.20, in which case the estimated value of the runoff coefficient, C, of the catchment is used in preference to the assumption that C = 1. A common shortcut procedure in using Equation 10.21 is to use a return period of 2 years in calculating P24 ; however, it is generally preferable to use a return period for P24 that is consistent with the objectives of the analysis. Two limitations of Equations 10.20 and 10.21 are: (1) They were derived by assuming IDF relationships that are appropriate for Types II and III rainfall distributions, while for short-duration storms the rainfall intensities of Type I and IA rainfall are much less than for Type II and Type III rainfall; and (2) the IDF relations given by Equation 10.18 for Type II and Type III storms are inaccurate for durations less than about 15 min (Froehlich, 2011b). These limitations can both be addressed by using the following empirical relationship to estimate tf [min], ln tf = a + b ln N + c(ln N)2

(10.22)

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Time of Concentration

479

TABLE 10.2: Overland-Flow Coefficients

Rainfall I IA

a 1.75 − 0.227 ln P24 − 0.0210(ln P24 )2 1.54 − 0.0184 ln P24 − 0.0356(ln P24

)2

)2

II

1.94 − 0.409 ln P24 − 0.0049(ln P24

III

1.97 − 0.348 ln P24 − 0.00865(ln P24 )2

b

c

0.555

0.0188

0.573

0.0094

0.545

0.0216

0.555

0.0160

where N [m] is defined by N=

ww

nL √ C S0

(10.23)

2 3

and a, b, and c are constants given in Table 10.2. The flow length of the sheet-flow regime in unpaved areas should generally be less than 100 m (330 ft), while in paved areas longer lengths of sheet flow are acceptable with sheet flow usually terminating in gutters or low areas of parking lots. A maximum flow length of 130 m (400 ft) can be associated with sheet flow, although sheet flow should most often be associated with lengths less than or equal to 100 m (330 ft) (USFHWA, 2009).

w .E asy En g EXAMPLE 10.1

Surface runoff from a developed area is intercepted by a storm drain, where the catchment is characterized by a runoff coefficient of 0.70, an average slope of 0.03, a Manning’s n of 0.02, and a runoff length of 70 meters. The catchment is located in a region with Type I rainfall, and the 2-year 24-h rainfall is 90 mm. Estimate the time of concentration in the sheet-flow regime.

ine eri n

Solution From the given data: C = 0.70, S0 = 0.03, n = 0.02, L = 70 m, and P24 = 90 mm. Using the conventional NRCS formulation given by Equation 10.20 gives ⎡ ⎡ ⎤4 ⎤4 5 5 5.5 ⎣ nL ⎦ 5.5 ⎣ (0.02)(70) ⎦ tf = = = 3.7 min √ √ P24 C 32 S 90 (0.70) 32 0.03 0

g .n

et

If the imperviousness of the catchment is neglected, and the formulation given by Equation 10.21 is used, then 4   4 nL 5 5.5 (0.02)(70) 5 5.5 = √ = 3.1 min tf = √ P24 S0 90 0.03

The estimated values of tf derived from these two NRCS formulations are relatively close and differ by 16%. However, there remains a concern about the accuracy of these formulations on two counts: (1) These formulations were derived for Type II and Type III rainfall, whereas the area being considered has Type I rainfall, and (2) these formulations approximate the Type II and Type III rainfall intensities by expressions that are known to be inaccurate for durations less than 15 min and so a calculated value of tf = 3.7 min is inconsistent with the applicability of these equations. The aforementioned limitations in using the NRCS formulas are addressed by using Equation 10.22, where (0.02)(70) nL = 10.3 m N = 2 = 2√ C 3 S0 (0.70) 3 0.03 and according to Table 10.2 for Type I rainfall

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff a = 1.75 − 0.227 ln P24 − 0.0210(ln P24 )2 = 1.75 − 0.227 ln(90) − 0.0210(ln 90)2 = 0.303 b = 0.555 c = 0.0188 Substituting the values of N, a, b, and c into Equation 10.22 gives ln tf = a + b ln N + c[ln N]2 = 0.303 + (0.555) ln(10.3) + (0.0188)[ln(10.3)]2 = 1.70 which yields tf = e1.70 = 5.5 min. Based on these results, the most accurate estimate of the time of concentration in the sheet-flow regime is 5.5 min, which is approximately 49% higher than that given by Equation 10.20, and approximately 77% higher than that given by Equation 10.21. These results indicate that caution should be taken in applying the conventional NRCS formulations outside of the conditions for which they are applicable.

ww

After a maximum distance on the order of 100 m (330 ft), sheet flow becomes shallow concentrated flow, and the average velocity, Vsc , is taken to be a function of the slope of the flow path and the type of land surface in accordance with the Manning equation

w .E asy En g

Vsc =

1 2 21 R 3 S0 n

(10.24)

where n is the roughness coefficient, R is the hydraulic radius, and S0 is the slope of the flow path. For unpaved areas, it is commonly assumed that n = 0.05 and R = 12 cm (4.7 in.); and for paved areas, n = 0.025 and R = 6 cm (2.4 in). Equation 10.24 can also be expressed in the form

2

1

Vsc = kS02

ine eri n

(10.25)

where k (= R 3 /n) is called the intercept coefficient. Based on the aforementioned parameter values for n and R, k values of 4.91 m/s (16 ft/s) and 6.19 m/s (20 ft/s) for paved and unpaved areas, respectively, are typical. Several suggested values of k are given in Table 10.3. In addition to intercept coefficients for shallow concentrated flow, Table 10.3 also gives bulk intercept coefficients for the entire overland flow, including the sheet-flow and shallow concentrated flow regimes. The average velocity, Vsc , derived from Equation 10.25 is then combined with the flow length, Lsc , of shallow concentrated flow to yield the flow time, tsc , as tsc =

Lsc Vsc

g .n

et

(10.26)

The total time of concentration, tc , of overland flow is taken as the sum of the sheet-flow time, tf , (conventionally) given by Equation 10.21, and the shallow concentrated flow time, tsc , given by Equation 10.26. The overland-flow time of concentration is added to the channelflow time to obtain the time of concentration of the entire catchment. The NRCS approach of separately determining the travel times in the sheet flow, shallow concentrated flow, and channel-flow regimes is sometimes assumed a priori to be the best method for calculating the time of concentration (e.g., Sharifi and Hosseini, 2011). The U.S. Federal Highway Administration (USFHWA) has recommended using an approach that is similar to the NRCS three-segment method, with the only difference being that the kinematic-wave equation, Equation 10.13, is used to calculate the sheet-flow time of travel rather than the NRCS sheet-flow formula given by Equation 10.21 (USFHWA, 2008; 2009).

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Time of Concentration

481

TABLE 10.3: Intercept Coefficient for Overland-Flow Velocity versus Slope

Land cover/Flow regime

k (m/s)

Forest with heavy ground litter; hay meadow (overland flow) Trash fallow or minimum tillage cultivation; contour or strip cropped; woodland (overland flow) Short grass pasture (overland flow) Cultivated straight row (overland flow) Nearly bare and untilled (overland flow); alluvial fans in western mountain regions Grassed waterway (shallow concentrated flow) Unpaved (shallow concentrated flow) Paved area (shallow concentrated flow); small upland gullies

0.76 1.52 2.13 2.74 3.05 4.57 4.91 6.19

Source: U.S. Federal Highway Administration (2009).

ww

10.3.1.3

Kirpich equation

An empirical time of concentration formula that is especially popular is the Kirpich formula (Kirpich, 1940) given by

w .E asy En g

tc = 0.019

L0.77

(10.27)

S00.385

where tc is the time of concentration [min], L is the flow length [m], and S0 is the average slope along the flow path [dimensionless]. The Kirpich equation is applicable to mixed sheet and shallow-concentrated flow. Equation 10.27 was originally developed and calibrated from NRCS data reported by Ramser (1927) on seven partially wooded agricultural catchments in Tennessee, with catchment sizes in the range of 0.4–45 ha (1–112 ac) and with slopes in the range of 3%–10%; Equation 10.27 has found widespread use in urban applications to estimate both overland-flow and channel-flow travel times. Equation 10.27 is most applicable to natural basins with well-defined channels, bare-earth overland flow, and flow in mowed channels including roadside ditches (Debo and Reese, 1995). For surface conditions other than bare soil or roadside ditches, Rossmiller (1980) suggested that for concrete or asphalt surfaces, tc given by Equation 10.27 should be multiplied by 0.4; for concrete channels, multiply tc by 0.2; and for overland flow and flow in natural grass channels, multiply tc by 2. According to Prakash (2004), the Kirpich equation yields relatively low estimates of the time of concentration. The Kirpich formula is usually considered applicable to small agricultural watersheds with drainage areas less than 80 ha (200 ac). 10.3.1.4

ine eri n

g .n

et

Izzard equation

The Izzard equation (Izzard, 1944; 1946) was derived from laboratory experiments on pavements and turf where overland flow was dominant. The Izzard equation is given by 1

tc =

530KL 3 2 3

,

where ie L < 3.9 m2 /h

(10.28)

ie

where tc is the time of concentration [min], L is the overland-flow distance [m], ie is the effective rainfall intensity [mm/h], and K is a constant given by

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff TABLE 10.4: Values of cr in the Izzard Equation

Surface

cr

Very smooth asphalt Tar and sand pavement Crushed-slate roof Concrete Tar and gravel pavement Closely clipped sod Dense bluegrass

0.0070 0.0075 0.0082 0.012 0.017 0.016 0.060

Source: Izzard (1946).

ww

K=

2.8 * 10−6 ie + cr

(10.29)

1

S03 where cr is a retardance coefficient that is determined by the catchment surface as given in Table 10.4 and S0 is the catchment slope [dimensionless].

w .E asy En g 10.3.1.5

Kerby equation

The Kerby equation (Kerby, 1959) is given by 

Lr tc = 1.44 √ S0

0.467

,

(10.30)

where L < 365 m

ine eri n

where tc is the time of concentration [min], L is the length of flow in [m], r is a retardance roughness coefficient given in Table 10.5, and S0 is the slope of the catchment [dimensionless]. The Kerby equation is an empirical relation developed by Kerby (1959) using published research on airport drainage done by Hathaway (1945); consequently, it is sometimes referred to as the Kerby–Hathaway equation. Catchments with areas less than 4 ha (10 ac), slopes less than 1%, and retardance coefficients less than 0.8 were used in calibrating the TABLE 10.5: Values of r in the Kerby Equation

Surface

g .n r

Smooth pavements Asphalt/concrete Smooth bare packed soil, free of stones Light turf Poor grass on moderately rough ground Cultivated row crops Average grass Pasture Dense turf Dense grass Bermuda grass Deciduous timberland Conifer timberland, dense grass Deciduous timberland (w/ deep forest litter)

0.02 0.05–0.15 0.10 0.20 0.20 0.20 0.40 0.40 0.17–0.80 0.17–0.80 0.30–0.48 0.60 0.60–0.80 0.80

et

Sources: Kerby (1959); Westphal (2001); Seybert (2006).

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483

Kerby equation, and application of this equation should also be limited to this range. In addition, since the Kerby equation applies to the overland/sheet-flow regime, it is recommended that the length of flow, L, be less than 100 m (330 ft) (Westphal, 2001). EXAMPLE 10.2 An urban catchment with an asphalt surface has an average slope of 0.5%, and the distance from the catchment boundary to the outlet is 90 m. At the catchment location, the 2-year 24-hour rainfall is estimated as 50 mm. For a 20-min storm with an effective rainfall rate of 75 mm/h, estimate the time of concentration using: (a) the kinematic-wave equation, (b) the NRCS method, (c) the Kirpich equation, (d) the Izzard equation, and (e) the Kerby equation. Solution

ww

(a) Kinematic-wave equation: If the overland flow is assumed to be fully turbulent, the Manning form of the kinematic-wave equation (Equation 10.13) can be used. From the given data: L = 90 m, ie = 75 mm/h, S0 = 0.005, and for an asphalt surface Table 10.1 gives n = 0.012. According to Equation 10.13 

6.99

nL S0

3

w .E asy En g tc =

2

ie5

5

6.99

=

2

(75) 5



0.012 * 90 √ 0.005

3 5

= 6 min

If the overland flow is assumed to be in the transition range (which is more probable), then the Darcy–Weisbach form of the kinematic-wave equation (Equation 10.17) must be used. Substituting ν = 10−6 m2 /s, k = 0.5, and C = 2 into Equation 10.17 (Chen and Wong formula) gives tc =



1

0.21(3.6 * 106 ν)k CL2−k

3

So i1+k e

=



0.21(3.6 * 106 * 10−6 )0.5 (2)(90)2−0.5 (0.005)(75)1+0.5

ine eri n

1

3

= 6 min

Therefore, both forms of the kinematic-wave equation yield the same result (to the nearest minute). (b) NRCS method: Use n = 0.012, L = 90 m, S0 = 0.005, and P24 = 50 mm. According to the NRCS approach, assuming the sheet-flow regime (L … 100 m) yields the following estimate of the time of concentration: 4  4  nL 5 5.5 0.012 * 90 5 5.5 = √ = 7 min tc = √ P24 S0 50 0.005

g .n

et

(c) Kirpich equation: Use L = 90 m and S0 = 0.005. According to the Kirpich equation, with a factor of 0.4 to account for the asphalt surface, tc = 0.4(0.019)

L0.77 S00.385

= 0.4(0.019)

(90)0.77 (0.005)0.385

= 2 min

(d) Izzard equation: Use S0 = 0.005, L = 90 m, ie = 75 mm/h, and a retardance coefficient, cr , given by Table 10.4 as 0.007. The constant, K, is given by K=

2.8 * 10−6 ie + cr 1

S03

=

2.8 * 10−6 (75) + 0.007 1

(0.005) 3

= 0.0422

and the Izzard equation gives the time of concentration as 1

1

tc =

530KL 3 2

ie3

=

530(0.0422)(90) 3 2

(75) 3

= 6 min

In this case, ie L = (0.075)(90) = 6.75 m2 /h. Therefore, since ie L > 3.9 m2 /h, the Izzard equation is not strictly applicable.

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Fundamentals of Surface-Water Hydrology II: Runoff (e) Kerby equation: Use L = 90 m, S0 = 0.005, and a retardance coefficient, r, given by Table 10.5 as 0.02. According to the Kerby equation 

Lr tc = 1.44 S0

0.467



90 * 0.02 = 1.44 √ 0.005

0.467

= 7 min

The computed times of concentration are summarized in the following table: Equation Kinematic wave NRCS Kirpich Izzard Kerby

ww

tc (min) 6 7 2 6 7

Noting that the Kirpich equation would give tc = 5 min if the Rosmiller factor of 0.4 were not applied, the (overland flow) time of concentration of the catchment can be taken to be on the order of 6 min.

w .E asy En g 10.3.2

Channel Flow

Channel-flow elements in flow paths include street gutters, roadside swales, storm sewers, drainage channels, and small streams. In these cases, it is recommended that velocity-based equations such as the Manning and Darcy–Weisbach equations be used to estimate the flow time in each channelized segment (ASCE, 1992). Here the flow time, t0 , is estimated by the relation L t0 = (10.31) V0

ine eri n

where L is the length of the flow path in the channel and V0 is the estimated velocity of flow when the channel is flowing full, commonly referred to as the bankfull condition. The time of concentration of the entire catchment is equal to the sum of the time of concentration for overland flow and the time of travel in the channel.

EXAMPLE 10.3

g .n

et

A catchment consists of an asphalt pavement that drains into a rectangular concrete channel. The asphalt pavement has an average slope of 0.6%, and the distance from the catchment boundary to the drain is 50 m. The drainage channel is 30 m long, 25 cm wide, 20 cm deep, and has a slope of 0.8%. For an effective rainfall intensity of 60 mm/h, the flow rate in the channel is estimated to be 0.025 m3 /s. Estimate the time of concentration of the catchment. Solution The flow consists of both overland flow and channel flow. Use the kinematic-wave equation to estimate the time of concentration of overland flow, t1 , where L = 50 m, ie = 60 mm/h, S0 = 0.006, and for an asphalt surface Table 10.1 gives n = 0.012. The kinematic-wave equation (Equation 10.13) gives 3 3   6.99 0.012 * 50 5 6.99 nL 5 = = 5 min t1 = 2 √ 2 S0 0.006 (60) 5 ie5

The flow in the drainage channel is described by the Manning equation: Q=

5  2 1 2 1 1 A 3 2 1 1 A 3 21 AR 3 S02 = A S0 = S n n P n P 32 0

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485

where Q = 0.025 m3 /s, n = 0.013 (for concrete), and S0 = 0.008. Since the width of the channel is 0.25 m, the flow area, A, and wetted perimeter, P, can be written in terms of the depth of flow, d, in the drainage channel as A = 0.25d and P = 2d + 0.25. Therefore, the Manning equation can be put in the form 5 1 (0.25d) 3 1 (0.008) 2 0.025 = 0.013 (2d + 0.25) 32 which yields d = 0.10 m = 10 cm The flow area, A, in the drainage channel is given by A = 0.25d = (0.25)(0.10) = 0.025 m2 Therefore, the flow velocity, V0 , is given by 0.025 Q = = 1.0 m/s A 0.025 Since the length, L, of the drainage channel is 30 m, the flow time, t2 , in the channel is given by V0 =

ww

t2 =

30 L = = 30 s = 0.5 min V0 1.0

w .E asy En g

The time of concentration, tc , of the entire catchment area is equal to the overland-flow time, t1 , plus the channel-flow time, t2 . Hence tc = t1 + t2 = 5 + 0.5 = 5.5 min

Because of uncertainties in estimating tc , the time of concentration of the catchment can reasonably be taken as 6 min.

The time of travel, t0 [T], in a channel with negligible backwater effect and subject to a uniform lateral inflow per unit length, q [L2 T−1 ], and a constant upstream inflow, Qu [L3 T−1 ], can be estimated using the dimensionally homogeneous relation (Wong and Zhou, 2003)

t0 =



ine eri n

L αqβ−1

1   β 1 1 (λ + 1) β − λ β

(10.32)

g .n

where L is the length of channel [L], and α [L3−2β T−1 ] and β [dimensionless] are the kinematic-wave parameters relating the discharge, Q [L3 T−1 ], to the flow area, A [L2 ], according to the relation Q = αAβ (10.33) and λ is a dimensionless inflow ratio defined as λ=

Qu qL

et

(10.34)

Values of α and β for various channel shapes are shown in Table 10.6. Combining Equation 10.32 with the values of α and β in Table 10.6, it can be shown that a deep rectangular channel produces the longest time of travel and largest detention storage, while a parabolic channel has the shortest time of travel and the smallest detention storage. Across the spectrum of channel shapes, there can be a sixfold increase in time of travel depending on the shape. In estimating t0 for a natural or constructed channel, if sufficient data are available to divide the channel into many segments, total travel time can be expressed as the sum of the travel times in individual segments. In this circumstance, the overall travel time will likely be a function of the number of segments, the travel time will tend to increase with the number of segments, and the most accurate estimate of travel time will most likely be somewhere in between the coarsest and finest discretization (Pavlovic and Moglen, 2008). This outcome can be linked to the assumption of uniform flow where, in reality, the flow is gradually varied.

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff TABLE 10.6: Values of α and β for Channel Flow

Channel shape

α S0

Square

ww

β 4 3

2 33 n

Wide rectangular





Deep rectangular





S0 n S0 n



S0 n

Triangular



2

5 3

W 2

2

1

3

3

m 4(1 + m2 )

1 3

4 3

⎤1 ⎡ 3 S0 m ⎦ ⎣ n 2(1 + 1 + m2 )2



4 3

 2 0.493 S0 1 9 n W

13 9

Circular

0.501 S0 1 W6 n

5 4

Accuracy of Estimates

W = channel width [m] m = reciprocal of side slope [dimensionless]

Parabolic

10.3.3

S0 = channel slope [dimensionless] n = Manning roughness [dimensionless]

1 W

w .E asy En g Vertical curb

Notation

ine eri n

In applying any time of concentration formula it is essential that the flow regime and watershed characteristics associated with the formula be respected. Segments of the runoff path associated with sheet flow, shallow concentrated flow, and channel flow must first be identified and then the the appropriate equation(s) used, provided that the watershed characteristics are within the ranges for which the time of concentration formula was developed. Careful consideration should be given to the fact that the time of concentration might not necessarily correspond to the time of flow from the catchment boundary along the route of the principal flow path (i.e., the route of the main channel extended to the catchment boundary). In fact, it has been shown that in some cases the time of concentration derived from the principal flow path might have to be increased by 50% to accurately reflect the time of concentration of the entire catchment (McCuen, 2009). Wong (2005) compared 9 equations for estimating tc for overland flow in experimental plots having concrete and grass surfaces, and concluded that tc equations that do not account for rainfall intensity are valid for only a limited range of rainfall intensities. The equations used in the Wong (2005) study included, among others, the Izzard equation (Equation 10.28), the Kerby equation (Equation 10.30), and the Chen and Wong equation (Equation 10.17). The results of this study showed that best agreement with experimental data for both concrete and grass surfaces was obtained using the Chen and Wong equation. Fang et al. (2007) compared tc estimates by three independent research teams using data collected in 96 Texas watersheds, where all three teams used the NRCS method to estimate tc . Drainage areas of the 96 watersheds were in the range of 0.8–440 km2 (0.3–170 mi2 ). The resulting tc estimates varied considerably between teams, with the variability primarily caused by the differing analyst-derived parameters used to estimate travel times in the three flow regimes (i.e., sheet, shallow concentrated, and channel flow). Based on the results of

g .n

et

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ww

Peak-Runoff Models

487

this study, it is recommended that uncertainty be taken into account when estimating tc using the NRCS method. When conventional formulae are used to estimate tc , such as the Kirpich equation, it has been shown that the variability in tc estimates caused by using automated (DEM, GIS) versus manual methods are likely to be relatively minor compared to the variabilities between tc formulae (Fang et al., 2008). McCuen et al. (1984) compared 11 equations for estimating tc in 48 urban catchments in the United States. The catchments used in the study all had areas less than 16 km2 (6.2 mi2 ), average impervious areas of approximately 29%, and times of concentration in the range of 0.21– 6.14 h, with an average time of concentration of 1.5 h. The equations used in these studies included, among others, the kinematic-wave equation (Equation 10.13), the Kirpich equation (Equation 10.27), and the Kerby equation (Equation 10.30). The results of this study indicated that the error in the estimated value of tc exceeded 0.5 h for more than 50% of the catchments, with the standard deviation of the errors in the range of 0.37–2.27 h. These results indicate that relatively large errors can be expected in estimating tc from commonly used equations, and Singh (1989) has noted that this can lead to significant errors in design discharges derived from estimated values of tc . It is considered good practice to use at least three different methods to estimate the time of concentration and, within the range of these estimates, the final value should be selected using engineering judgment. Estimates of travel time in drainage channels are usually much more accurate than estimates of travel time in overland flow; therefore, in catchments where flow in drainage channels constitutes a significant portion of the travel time, the time of concentration can be estimated more accurately.

w .E asy En g 10.4

Peak-Runoff Models

Peak-runoff models estimate only the peak runoff rate, not the entire runoff hydrograph. Peak flow rates are required for the hydraulic design of bridges and culverts, for evaluation of flooding potential, and for the design of stormwater conveyance structures. 10.4.1

The Rational Method

ine eri n

The rational method is the most widely used peak-runoff method in urban hydrology and has been used by engineers since the nineteenth century (Mulvaney, 1851; Kuichling, 1889). European equivalents of the rational method are the Lloyd-Davies (1906) formula used in the United Kingdom, the Caquot (1941) formula used in France, and the Imhoff (1964) formula used in Central Europe. The rational method relates the peak-runoff rate, Qp [L3 T−1 ], to the rainfall intensity, i [LT−1 ], by the relation Qp = CiA

g .n

et

(10.35)

where C is the runoff coefficient [dimensionless], and A is the area of the catchment [L2 ]. Application of the rational method assumes that: (1) the entire catchment area is contributing to the runoff, in which case the duration of the storm must equal or exceed the time of concentration of the catchment; (2) the rainfall is distributed uniformly over the catchment area; and (3) all catchment losses are incorporated into the runoff coefficient, C. The runoff mechanism assumed by the rational method is illustrated in Figure 10.2 (for a constant rainfall intensity), where the runoff rate from a storm gradually increases until an equilibrium is reached in which the runoff rate, Q, is equal to the effective rainfall rate, Ci, over the entire catchment area, A. This condition occurs at the time of concentration, tc . Under these circumstances, the duration of the rainfall does not affect the peak runoff rate as long as the duration equals or exceeds tc . Since the storm duration must equal or exceed the time of concentration for the rational method to be applicable, and the average intensity of a storm is inversely proportional to the duration of the storm, then the rainfall intensity used in the rational method to generate the largest peak runoff rate must correspond to a storm whose duration is equal to the time of concentration. The rational method implicitly assumes that the rainfall excess, ie , is related to the rainfall rate, i, by

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Fundamentals of Surface-Water Hydrology II: Runoff

Runoff rate, Q

FIGURE 10.2: Runoff mechanism assumed by rational method

Rainfall intensity

488

i

Qp

tc

Time, t

(10.36)

ie = Ci

ww

which means that the rainfall-excess rate is a constant fraction, C, of the rainfall rate. The assumption that C is a constant that is independent of the rainfall rate and antecedent moisture conditions makes the application of the rational method questionable from a physical viewpoint, but this approximation improves as the imperviousness of the catchment increases. In addition to this flaw in the underlying theory of the rational method, the assumed uniformity of the rainfall distribution over the catchment area and the assumption of equilibrium conditions indicate that the rational method should be limited to areas smaller than about 80 ha (200 ac) (ASCE, 2006a), which typically have times of concentration less than 20 min. Debo and Reese (1995) recommend that the rational method be limited to areas smaller than 40 ha (100 ac), some jurisdictions limit the use of the rational method to areas smaller than 8 ha (20 ac) (South Carolina, 1992; New Jersey, 2004; North Carolina, 2007), and in some cases local characteristics may limit the application of the rational method to areas smaller than 4 ha (10 ac) (Haestad Methods, Inc., 2002). Use of the rational method is not recommended in any catchment where ponding of stormwater might affect the peak discharge or where the design and operation of large drainage facilities are to be undertaken. Application of the rational method requires a simultaneous solution of the IDF equation and the time-of-concentration expression to determine the rainfall intensity and time of concentration. Typically, the IDF curve can be expressed in the form

w .E asy En g

ine eri n i = f (tc )

and an expression for the time of concentration is of the form tc = g(i)

g .n

(10.37)

et

(10.38)

where f and g are known functions; Equation 10.37 implicitly assumes that the storm duration is equal to the time of concentration. Typical runoff coefficients for rainfall intensities with return periods of 2–10 years are shown in Table 10.7. Higher runoff coefficients should be used for less frequent storms having higher return periods, although the upscaling factor will depend on local climate and surface cover (e.g., Young et al., 2009). Runoff coefficients can be reasonably increased by 10% and 20% for 25- and 100-year storms, respectively (ASCE, 2006a), although an increase of 25% is sometimes applied to 100-year storms (USFHWA, 2009). The increased runoff coefficients must not be greater than 1.0. Under some circumstances, runoff coefficients have been reported to be negligibly small. For example, Valavala et al. (2006) reported that, for typical rainfall rates in the United States, there is little or no runoff from unclogged pervious concrete surfaces over sand subbases with concrete porosities in the range of 0.16–0.27 and surface slopes in the range of 2%–10%. However, this does not take into account the effects of pore clogging and subbase compaction. Typical values of the runoff coefficient, shown in Table 10.7, are considered appropriate for ground slopes less than around 5%; for higher slopes the runoff coefficient can be expected to depend on the ground slope and roughness. To support this assertion, ASCE

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489

TABLE 10.7: Typical Runoff Coefficients for Return Periods of 2–10 years

Runoff coefficient

Description of area

ww

Business: Downtown areas Neighborhood areas Residential: Single-family areas Multiunits, detached Multiunits, attached Residential, suburban Apartment dwelling areas Industrial: Light areas Heavy areas Parks, cemeteries Playgrounds Railroad yard areas Unimproved areas Pavement: Asphalt or concrete Brick Semipermeable blocks Roofs Lawns, sandy soil: Flat, 2% Average, 2%–7% Steep, Ú 7% Lawns, heavy soil: Flat, 2% Average, 2%–7% Steep, Ú 7%

w .E asy En g

0.70–0.95 0.50–0.70 0.30–0.50 0.40–0.75 0.60–0.75 0.25–0.40 0.50–0.70 0.50–0.80 0.60–0.90 0.10–0.25 0.20–0.35 0.20–0.35 0.10–0.30 0.70–0.95 0.70–0.85 0.60–0.70 0.75–0.95 0.05–0.10 0.10–0.15 0.15–0.20

ine eri n

0.13–0.17 0.18–0.22 0.25–0.35

Sources: ASCE (2006a); Hvitved-Jacobsen (2010).

g .n

(1969) indicated that C values for slopes of 7% might be twice the value for slopes of 2%, and McCuen (2005) provided C values for (forested) slopes greater than 6% that are 60% to more than 100% greater than the corresponding C values for slopes of 2% or less. The runoff coefficient as a function of the ground slope and roughness can be estimated using the following relation (Cutter and McCuen, 2007)   C = C0 + (0.96 − C0 )n0.5 1 − e−4.25(S−0.06)

et

(10.39)

where C0 is the runoff coefficient on a 6% ground slope, n is the Manning roughness coefficient, and S is the ground slope; Equation 10.39 is valid for slopes greater than 6%. It is common practice to use the impervious fraction of a catchment as a preliminary estimate of the runoff coefficient (e.g., Rodriguez et al., 2003; Dhakal et al., 2012); this approach is similar to the Lloyd-Davies variation of the rational method which is widely used in the United Kingdom and assumes that 100% of the runoff comes from impervious surfaces that are directly connected to the drainage system, and 0% of the runoff from pervious surfaces (Lloyd-Davies, 1906; Butler and Davies, 2011). This approach neglects the infiltration capacity and antecedent moisture conditions in the pervious fraction, which could have a significant effect on the rainfall-runoff process (Shuster et al., 2008).

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Fundamentals of Surface-Water Hydrology II: Runoff

EXAMPLE 10.4 A new 1.2-ha suburban residential development is to be drained by a storm sewer that connects to the municipal drainage system. The development is characterized by an average runoff coefficient of 0.4, a Manning’s n for overland-flow of 0.20, an average overland-flow length of 70 m, and an average slope of 0.7%. The time of concentration can be estimated by the kinematic-wave equation. Local drainage regulations require that the storm-sewer pipe be sized to accommodate the peak runoff rate resulting from a 10-year rainfall event. The 10-year IDF curve is given by i=

315.5 t0.81 + 6.19

where i is the rainfall intensity in cm/h and t is the duration in minutes. Local drainage regulations further require a minimum time of concentration of 5 min. Determine the peak runoff rate to be handled by the storm sewer.

ww

Solution The time of concentration, tc , is estimated by the kinematic-wave equation (Equation 10.13) as 3  6.99 nL 5 tc = 2 S0 5 ie where n = 0.20, L = 70 m, and S0 = 0.007, and therefore

w .E asy En g tc =

6.99 2 5

ie



0.20 * 70 √ 0.007

3 5

=

151 i0.4 e

min

(10.40)

where ie is in mm/h. Equating the storm duration to the time of concentration, the effective rainfall rate, ie , for a 10-year storm is given by the IDF relation as ie = Ci =

3155C

mm/h

ine eri n tc0.81 + 6.19

(10.41)

Combining Equations 10.40 and 10.41 with C = 0.4 yields

3155(0.4) 1262ie0.324 = ie =  0.81 58.2 + 6.19ie0.324 151 + 6.19 0.4 ie

g .n

Solving this equation gives ie = 58 mm/h and a corresponding time of concentration of 30 min, which exceeds the minimum allowable time of concentration of 5 min. The peak runoff rate, Qp , from the residential development is given by the rational formula as Qp = CiA = ie A

where ie = 58 mm/h = 1.61 * 10−5 m/s, and A = 1.2 ha = 1.2 * 104 m2 . Therefore

et

Qp = (1.61 * 10−5 )(1.2 * 104 ) = 0.19 m3 /s The storm sewer serving the residential area should be sized to accommodate 0.19 m3 /s.

It is important to keep in mind that the assumptions of the rational method are valid only for small catchments, with areas typically less than 80 ha (200 ac) and times of concentration typically less than 20 min. For larger catchments, the peak runoff rate will likely result from storms with durations considerably longer than the time of concentration of the catchment (Levy and McCuen, 1999). In using the rational method to calculate the peak runoff rate from a composite catchment, the rational formula should generally be applied in two ways (ASCE, 1992): (1) using the entire drainage area, and (2) using the most densely developed directly connected area. In using the entire drainage area, the rainfall duration is equated to the time of concentration of the entire catchment, the average intensity corresponding to a duration equal to the

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491

time of concentration is derived from the IDF curve, the average runoff coefficient is an area-weighted average, and the resulting peak runoff rate is calculated by substituting these values into the rational formula, Equation 10.35. Using only the portion of the catchment that is most densely developed and directly connected to the catchment outlet, the smaller (sub)catchment area results in a shorter time of concentration and higher average rainfall intensity, and a higher average runoff coefficient because of the more impervious nature of the subcatchment; the peak runoff rate from this area is obtained by substituting these values into Equation 10.35. If the increased average rainfall intensity and runoff coefficient are sufficient to offset the smaller catchment area, then the peak runoff rate from the densely developed directly connected area will be greater than the peak runoff rate from the entire catchment and will govern the design of the downstream drainage structures. EXAMPLE 10.5

ww

Consider the case where the residential development described in Example 10.4 contains 0.4 ha of impervious area that is directly connected to the storm sewer. If the runoff coefficient of the impervious area is 0.9, Manning’s n for overland flow on the impervious surface is 0.03, the average flow length is 20 m, and the average slope is 0.1%, estimate the design runoff rate to be handled by the storm sewer.

w .E asy En g

Solution The time of concentration, tc , of the impervious area is given by tc =

6.99 2

ie5



nL S0

3

5

where n = 0.03, L = 20 m, and S0 = 0.001. Therefore tc =

6.99 2 5

ie



0.03 * 20 √ 0.001

3 5

=

40.9 ie0.4

min

ine eri n

(10.42)

where ie is in mm/h. Taking the storm duration as tc , the 10-year effective rainfall rate, ie , is given by the IDF curve as 3155C ie = Ci = mm/h (10.43) tc0.81 + 6.19 Combining Equations 10.42 and 10.43 with C = 0.9 yields ie = 

3155(0.9) 2840ie0.324 = 0.81 20.2 + 6.19ie0.324 40.9 + 6.19 0.4 ie

g .n

et

Solving this equation gives ie = 303 mm/h and a corresponding time of concentration of 4.2 min, which is less than the minimum allowable time of concentration of 5 min. Taking tc = 5 min, the IDF relation yields 3155(0.9) = 288 mm/h ie = 0.81 5 + 6.19 The peak runoff rate, Qp , from the impervious area is given by the rational formula as Qp = CiA = ie A where ie = 288 mm/h = 8 * 10−5 m/s, and A = 0.4 ha = 4000 m2 ; therefore Qp = (8 * 10−5 )(4000) = 0.32 m3 /s Example 10.4 indicated that the peak runoff rate from the entire composite catchment is 0.19 m3 /s, and this example shows that the peak runoff rate from the directly connected impervious area is 0.32 m3 /s. The storm sewer serving this residential development should therefore be designed to accommodate 0.32 m3 /s.

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Low-impact developments (LIDs) attempt to maintain predevelopment hydrologic and ecological functions of their catchments. In assessing the pre- and postdevelopment peak runoff rates from LIDs, the path of surface runoff is particularly important and the effective runoff coefficient might not be accurately estimated by the weighted average of runoff coefficients from the pervious and impervious areas (Guo, 2008). Postdevelopment runoff volume generally increases above predevelopment runoff volume due to increased imperviousness of the catchment, and Guo and Cheng (2008) showed that this increased runoff can be entirely compensated for by increasing the onsite retention volume by D0 [L], according to the relation D0 = (C0 − 1)(I) ln(1 − Cv ) (10.44) P

ww

where P is the mean precipitation of individual rainfall events [L], C0 is the predevelopment runoff coefficient [dimensionless], I is the increase in imperviousness fraction between preand postdevelopment [dimensionless], and Cv is the capture rate of rainfall [dimensionless]. Values of P vary across the United States and are typically in the range of 10–20 mm (see Guo and Cheng, 2008, for map), C0 is typically in the range of 0–0.1, I is typically on the order of 0.1–0.5, and Cv is typically in the range of 0.75–0.85.

w .E asy En g 10.4.2

NRCS-TR55 Method

The Natural Resources Conservation Service (NRCS) computed the runoff from many small and midsize catchments using the NRCS regional 24-h hyetographs to describe the rainfall distribution, the NRCS curve-number model to calculate the rainfall excess, and the NRCS unit-hydrograph method to calculate the runoff hydrograph. These computations were performed using the TR-20 computer program. On the basis of these results, the NRCS proposed the graphical peak-discharge method or, more commonly, the TR-55 method, for estimating peak runoff rates from small and midsize catchments, with times of concentration in the range of 0.1–10 hours. In applying the TR-55 method, it is important to note that the NRCS regional 24-h hyetographs are designed to contain the (average) intensity of any duration of rainfall less than 24 h for the frequency of the event chosen (i.e., NRCS hyetographs are consistent with local IDF curves) and therefore peak runoff rates are generated from NRCS 24-h storms during a time interval of maximum rainfall excess roughly equal to the time of concentration of the catchment. The TR-55 method, named after the technical report in which it is described (SCS, 1986), expresses the peak runoff rate, qp [m3 /s] as

ine eri n

qp = qu AQFp

g .n

(10.45)

et

where qu is the unit peak discharge [m3 /s per cm of runoff per km2 of catchment area], A is the catchment area [km2 ], Q is the runoff [cm] from a 24-h storm with a given return period, and Fp is the pond and swamp adjustment factor [dimensionless]. The runoff, Q, is derived directly from the NRCS curve-number model, Equation 9.80, using the 24-h precipitation. Fp is derived from Table 10.8, assuming that the ponds and/or swampy areas are distributed throughout the catchment, and the unit-peak discharge, qu , is obtained using the empirical relation log (qu ) = C0 + C1 log tc + C2 (log tc )2 − 2.366 (10.46) where C0 , C1 , and C2 are constants obtained from Table 10.9, and tc is in hours. Values of C0 , C1 , and C2 are functions of Ia /P, where Ia is the initial abstraction of the catchment and P is the 24-hour rainfall that causes the runoff, Q. The relationship given by Equation 10.46 is illustrated for Type II rainfall in Figure 10.3, where it is apparent that the unit peak discharge, qu , decreases exponentially with increasing time of concentration, tc , and for any given value of tc the value of qu decreases with increasing values of Ia /P. Higher curve numbers correspond to lower values of Ia /P. Although the relationship between qu , tc , and Ia /P shown in Figure 10.3 is for Type II rainfall, the relationship for other rainfall types are similar. Values of tc in Equation 10.46 must be between 0.1 h and 10 h (calculated using the NRCS method described in Section 10.3.1). Values of Ia in Table 10.9 are derived using

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493

TABLE 10.8: Pond and Swamp Adjustment Factor, Fp

Percentage of pond and swamp areas

Fp

0 0.2 1.0 3.0 5.0∗

1.00 0.97 0.87 0.75 0.72

Note: ∗ If the percentage of pond and swamp areas exceeds 5%, then consideration should be given to routing the runoff through these areas.

ww

TABLE 10.9: Parameters Used to Estimate Unit Peak Discharge, qu

Rainfall type

Ia /P

w .E asy En g

Co

C1

C2

I

0.10 0.20 0.25 0.30 0.35 0.40 0.45 0.50

2.30550 2.23537 2.18219 2.10624 2.00303 1.87733 1.76312 1.67889

−0.51429 −0.50387 −0.48488 −0.45695 −0.40769 −0.32274 −0.15644 −0.06930

−0.11750 −0.08929 −0.06589 −0.02835 0.01983 0.05754 0.00453 0.0

IA

0.10 0.20 0.25 0.30 0.50

2.03250 1.91978 1.83842 1.72657 1.63417

−0.31583 −0.28215 −0.25543 −0.19826 −0.09100

−0.13748 −0.07020 −0.02597 0.02633 0.0

II

0.10 0.30 0.35 0.40 0.45 0.50

2.55323 2.46532 2.41896 2.36409 2.29238 2.20282

−0.61512 −0.62257 −0.61594 −0.59857 −0.57005 −0.51599

−0.16403 −0.11657 −0.08820 −0.05621 −0.02281 −0.01259

2.47317 2.39628 2.35477 2.30726 2.24876 2.17772

−0.51848 −0.51202 −0.49735 −0.46541 −0.41314 −0.36803

−0.17083 −0.13245 −0.11985 −0.11094 −0.11508 −0.09525

III

0.10 0.30 0.35 0.40 0.45 0.50

ine eri n

Ia = 0.2 S

g .n

et (10.47)

where S is obtained from the curve number in accordance with Equation 9.81. If Ia /P < 0.1, values of C0 , C1 , and C2 corresponding to Ia /P = 0.1 should be used, and if Ia /P > 0.5, values of C0 , C1 , and C2 corresponding to Ia /P = 0.5 should be used. These approximations result in reduced accuracy of the peak-discharge estimates (SCS, 1986), and McCuen and Okunola (2002) have noted that for times of concentration less than 0.3 h, the TR-55 method

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Fundamentals of Surface-Water Hydrology II: Runoff

FIGURE 10.3: Relationship between unit peak discharge and time of concentration

log10 (qu) (m3s–1cm–1km–2)

494

Type II Rainfall

3.0

Ia /P = 0.10

2.5

2.0 Ia /P = 0.50

1.5

0

2

4

6

8

10

Time of concentration, tc (h)

ww

may underestimate the peak discharge, relative to the TR-20 computer program, by as much as 15% for values of Ia /P near the lower limit of 0.1. The Federal Highway Administration (USFHWA, 1995) and the NRCS (SCS, 1986) have both recommended that the NRCS-TR55 method be used only with homogenous catchments, where the curve numbers vary within ; 5 between zones, CN of the catchment should be greater than 40, tc should be between 0.1 and 10 h, and tc should be approximately the same for all main channels. The NRCS has developed public-domain software for direct calculation of peak flows using the TR-55 method, and engineers are encouraged to use the latest version of such software.

w .E asy En g EXAMPLE 10.6

A 2.25-km2 catchment with 0.2% pond area is estimated to have a curve number of 85, a time of concentration of 2.4 h, and a 24-h Type III precipitation of 13 cm. Estimate the peak runoff rate from the catchment. Solution For CN = 85, the storage, S, is given by     1000 1000 1 1 − 10 = − 10 = 45 mm S= 0.0394 CN 0.0394 85 For P = 130 mm, the runoff, Q, is given by Q=

ine eri n

[P − 0.2S]2 [130 − 0.2(45)]2 = = 88 mm = 8.8 cm P + 0.8S 130 + 0.8(45)

From Table 10.8, Fp = 0.97. By definition

g .n

Ia 0.2S (0.2)(45) = = = 0.069 P P 130 Since Ia /P < 0.1, use Ia /P = 0.1 in Table 10.9, which gives C0 = 2.47317, C1 = −0.51848, C2 = −0.17083, and since tc = 2.4 h, Equation 10.46 gives log(qu ) = Co + C1 log tc + C2 (log tc )2 − 2.366

et

= 2.47317 − 0.51848 log(2.4) − 0.17083(log 2.4)2 − 2.366 = −0.116 which leads to

qu = 10−0.116 = 0.765 (m3 /s)/(cm·km2 )

Therefore, according to the TR-55 method, the peak discharge, qp , is given by Equation 10.45 as qp = qu AQFp = 0.765(2.25)(8.8)(0.97) = 14.7 m3 /s

It is important to recognize that hydrologic models used to estimate peak-runoff rates are not highly accurate. Calibrated peak-runoff models often have standard errors of 25% or more, with uncalibrated models having significantly higher standard errors (McCuen, 2001).

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10.5

ww

Continuous-Runoff Models

Continuous-Runoff Models

Continuous-runoff models estimate the entire runoff hydrograph from the rainfall excess remaining after initial abstraction, infiltration, and depression storage have been taken into account. The rainfall excess can be calculated for any given rainfall distribution using infiltration models such as the Horton and Green-Ampt models. Since surface conditions tend to be spatially variable within a catchment area, the rainfall excess will also be variable in both space and time, and the runoff hydrograph at the catchment outlet will depend on such factors as the topography and the surface roughness of the catchment. Four types of models that are commonly used in engineering practice to estimate the runoff hydrograph from the rainfall excess are: (1) unit-hydrograph models, (2) time-area models, (3) kinematic-wave models, and (4) nonlinear-reservoir models. These models are not normally used to estimate the peak runoff rate, since the accuracy of the estimated peak runoff rate using continuous runoff models depends on the temporal resolution of the rainfall excess (Aronica et al., 2005a). Other more comprehensive continuous-runoff models have been developed but have so far not become part of mainstream engineering practice (e.g., Jonsdottir et al., 2007). Continuous-runoff models are mostly used in designing storage reservoirs in stormwatermanagement systems. 10.5.1

Unit-Hydrograph Theory

w .E asy En g

The idea of using a unit hydrograph to describe the response of a catchment to rainfall excess was first introduced by Sherman (1932), and is currently the most widely used method of estimating runoff hydrographs. A unit hydrograph is defined as the temporal distribution of runoff rate resulting from a unit depth (1 cm or 1 in.) of rainfall excess occurring over a given duration, and distributed uniformly in time and space over the catchment area. To demonstrate the application of unit-hydrograph models, consider the unit hydrograph, ut (t), for a given catchment and rainfall-excess duration, t, illustrated in Figure 10.4(a). This is commonly referred to as the t-unit hydrograph. Assuming that the catchment response is linear, then the runoff hydrograph, Q(t), for a rainfall excess Pt occurring over a duration t is given by

ine eri n

Q(t) = Pt ut (t)

FIGURE 10.4: Applications of the unit hydrograph

495

Rainfall excess

Rainfall excess

P ∆t

1 ∆t ∆t

∆t

g .n

Runoff, Q (t)

Unit hydrograph, u(t)

u(t)

Pu(t) Time, t

Time, t

(a) Unit Hydrograph

(b) Runoff Hydrograph I

Rainfall excess P n∆t

(10.48)

1

2

3

∆t

∆t

∆t

n

et

Rainfall excess 1 ∆t ∆t

∆t

∆t

∆t

∆t

n∆t Runoff, Q (t) 1

2

3 n

Sum of runoff from individual storms

Runoff, S(t)

Time, t

Time, t

(c) Runoff Hydrograph II

(d) S—Hydrograph

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where the ordinates of Q(t) are equal to Pt times the ordinates of the unit hydrograph, ut (t). This is illustrated in Figure 10.4(b). If the rainfall excess, P, occurs over a duration equal to an integral multiple of t, say nt, and assuming the catchment response is linear, the response of the catchment is equal to that of n storms occurring sequentially, with the rainfall excess in each storm equal to Pnt /n. The runoff hydrograph, Q(t), is then given by

Q(t) =

n−1 Pnt  ut (t − i t) n

(10.49)

i=0

where the response of the catchment is equal to the summation of the responses to the incremental rainfall excesses of duration t. This is illustrated in Figure 10.4(c).

ww

EXAMPLE 10.7 The 10-min unit hydrograph for a 2.25-km2 urban catchment is given by

w .E asy En g Time (min)

0

30

60

90

120

150

180

210

240

270

300

330

360

390

Runoff (m3 /s)

0

1.2

2.8

1.7

1.4

1.2

1.1

0.91

0.74

0.61

0.50

0.28

0.17

0

(a) Verify that the unit hydrograph is consistent with a 1-cm rainfall excess, (b) estimate the runoff hydrograph for a 10-min rainfall excess of 3.5 cm, and (c) estimate the runoff hydrograph for a 20-min rainfall excess of 8.5 cm. Solution

ine eri n

(a) The area, Ah , under the unit hydrograph can be estimated by numerical integration as Ah = (1800)(1.2 + 2.8 + 1.7 + 1.4 + 1.2 + 1.1 + 0.91 + 0.74 + 0.61 + 0.50 + 0.28 + 0.17) = 22,698 m3

g .n

where the time increment between the hydrograph ordinates is 30 min = 1800 s. Since the area of the catchment is 2.25 km2 = 2.25 * 106 m2 , the depth, h, of rainfall excess is given by h=

22, 698 2.25 * 106

L 0.01 m = 1 cm

et

Since the depth of rainfall excess is 1 cm, the given hydrograph qualifies as a unit hydrograph. (b) For a 10-min rainfall excess of 3.5 cm, the runoff hydrograph is estimated by multiplying the ordinates of the unit hydrograph by 3.5. This yields the following runoff hydrograph:

Time (min)

0

30

60

90

Runoff (m3 /s) 0 4.2 9.8 6.0

120 150 180 210 240 270 300

330

4.9

0.98 0.60

4.2

3.9

3.2

2.6

2.1

1.8

360

390 0

(c) The runoff hydrograph for a 20-min rainfall excess of 8.5 cm is calculated by adding the runoff hydrographs from two consecutive 10-min events, with each event corresponding to a rainfall excess of 4.25 cm. The unit hydrograph is first interpolated for 10-min intervals and multiplied by 4.25 to give the runoff from a 10-min 4.25-cm event. This hydrograph is then added to the same hydrograph shifted forward by 10 min. The computations are summarized below, where Runoff 1 and Runoff 2 are the runoff hydrographs of the two 10-min 4.25-cm events, respectively.

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Continuous-Runoff Models

497

Time (min)

Runoff 1 (m3 /s)

Runoff 2 (m3 /s)

Total runoff (m3 /s)

Time (min)

Runoff 1 (m3 /s)

Runoff 2 (m3 /s)

Total runoff (m3 /s)

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200

0 1.70 3.40 5.10 7.35 9.65 11.90 10.33 8.80 7.23 6.80 6.38 5.95 5.65 5.40 5.10 4.97 4.80 4.68 4.42 4.12

0 0 1.70 3.40 5.10 7.35 9.65 11.90 10.33 8.80 7.23 6.80 6.38 5.95 5.65 5.40 5.10 4.97 4.80 4.68 4.42

0 1.70 5.10 8.50 12.45 17.00 21.55 22.23 19.13 16.03 14.03 13.18 12.33 11.60 11.05 10.50 10.07 9.77 9.48 9.10 8.54

210 220 230 240 250 260 270 280 290 300 310 320 330 340 350 360 370 380 390 400 —

3.87 3.61 3.40 3.15 2.98 2.76 2.59 2.42 2.30 2.13 1.83 1.49 1.19 1.02 0.89 0.72 0.47 0.26 0 0 —

4.12 3.87 3.61 3.40 3.15 2.98 2.76 2.59 2.42 2.30 2.13 1.83 1.49 1.19 1.02 0.89 0.72 0.47 0.26 0 —

7.99 7.48 7.01 6.55 6.13 5.74 5.35 5.01 4.72 4.43 3.96 3.32 2.68 2.21 1.91 1.61 1.19 0.73 0.26 0 —

w .E asy En g

For storms in which the duration of the rainfall excess is not an integral multiple of t, the runoff is derived from the S-hydrograph, St (t), which is defined as the response to a storm of infinite duration and intensity 1/t as illustrated in Figure 10.4(d). The S-hydrograph, which was originally proposed by Morgan and Hulinghorst (1939), is related to the t-unit hydrograph, ut (t), by

ine eri n

St (t) =

t/t 

ut (t − i t)

i=0

g .n

(10.50)

The S-hydrograph defined by Equation 10.50 converges quickly to a constant value, since the ordinates of the individual-storm unit hydrographs, ut (t), are only nonzero for a finite interval of time. The S-hydrograph is the response of a catchment to a rainfall excess of intensity 1/t beginning at t = 0 and continuing to t = q. If the rainfall excess were to begin at t = τ , the response of the catchment would be identical to the response to a rainfall excess beginning at t = 0, except that the S-hydrograph would begin at time t = τ instead of t = 0. In fact, linearity of the catchment response requires that the difference between the runoff hydrograph resulting from an infinite rainfall excess of intensity 1/t beginning at t = 0 and the runoff hydrograph resulting from an infinite rainfall excess of intensity 1/t beginning at t = τ must be equal to the runoff hydrograph resulting from a rainfall excess of intensity 1/t beginning at t = 0 and ending at t = τ . This is illustrated in Figure 10.5. Since the volume of rainfall is τ/t, the τ -unit hydrograph, denoted by uτ (t), is derived from the S-hydrograph by uτ (t) =

t [St (t) − St (t − τ )] τ

et

(10.51)

and the catchment response, Q(t), to a rainfall excess Pτ with duration τ is Q(t) = Pτ uτ =

Pτ t [St (t) − St (t − τ )] τ

(10.52)

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FIGURE 10.5: Estimation of unit hydrograph from S-hydrograph

Rainfall excess 1 ∆t

Runoff, S (t)

Rainfall excess 1 ∆t

␶ Runoff, S (t)

ww

Rainfall excess 1 ∆t

w .E asy En g ␶

Runoff, Q(t)

Time, t

The basic assumptions of the unit-hydrograph model are: (1) The rainfall excess has a constant intensity within the effective duration; (2) the rainfall excess is uniformly distributed throughout the catchment; (3) the base time (= duration) of the runoff hydrograph resulting from a rainfall excess of given duration is constant; (4) the ordinates of all the runoff hydrographs from the catchment are directly proportional to the amount of rainfall excess; and (5) for a given catchment, the hydrograph resulting from a given rainfall excess reflects the unchanging characteristics of the catchment. The last assumption is sometimes called the principle of invariance. The assumption of a uniformly distributed rainfall excess in space and time limits application of unit-hydrograph methods to relatively small catchment areas, typically in the range of 0.5 ha to 25 km2 (1 ac to 10 mi2 ), although even smaller catchment limitations, on the order of 2.5 km2 (1 mi2 ), are frequently used. If the catchment area is too large for the unit-hydrograph model to be applied over the entire catchment, then the area should be divided into smaller areas that are analyzed separately. The assumption of linearity has been found to be satisfactory in many practical cases (Chow et al., 1988); however, unit-hydrograph models are applicable only when channel conditions remain unchanged and catchments do not have appreciable storage.

ine eri n

g .n

et

EXAMPLE 10.8 The 30-min unit hydrograph for a 2.25-km2 catchment is given by

Time (min)

0

30

60

90

120

150

180

210

240

270

300

330

360

390

Runoff (m3 /s)

0

1.2

2.8

1.7

1.4

1.2

1.1

0.91

0.74

0.61

0.50

0.28

0.17

0

(a) Determine the S-hydrograph, (b) calculate the 40-min unit hydrograph for the catchment, and (c) verify that the 40-min unit hydrograph corresponds to a 1-cm rainfall excess.

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499

Solution (a) The S-hydrograph is calculated by summing the lagged unit hydrographs (Equation 10.50), and these computations are summarized in the following table:

Time (min)

Runoff* (m3 /s)

ww

ut (t) ut (t − ut (t − ut (t − ut (t − ut (t − ut (t − ut (t − ut (t − ut (t − ut (t − ut (t − ut (t − ut (t − ut (t −

t) 2t) 3t) 4t) 5t) 6t) 7t) 8t) 9t) 10t) 11t) 12t) 13t) 14t)

0

30

60

90

120

150

180

210

240

270

300

330

360

390

420

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2.8 1.2 0 0 0 0 0 0 0 0 0 0 0 0 0

1.7 2.8 1.2 0 0 0 0 0 0 0 0 0 0 0 0

1.4 1.7 2.8 1.2 0 0 0 0 0 0 0 0 0 0 0

1.2 1.4 1.7 2.8 1.2 0 0 0 0 0 0 0 0 0 0

1.1 1.2 1.4 1.7 2.8 1.2 0 0 0 0 0 0 0 0 0

0.91 1.1 1.2 1.4 1.7 2.8 1.2 0 0 0 0 0 0 0 0

0.74 0.91 1.1 1.2 1.4 1.7 2.8 1.2 0 0 0 0 0 0 0

0.61 0.74 0.91 1.1 1.2 1.4 1.7 2.8 1.2 0 0 0 0 0 0

0.50 0.61 0.74 0.91 1.1 1.2 1.4 1.7 2.8 1.2 0 0 0 0 0

0.28 0.50 0.61 0.74 0.91 1.1 1.2 1.4 1.7 2.8 1.2 0 0 0 0

0.17 0.28 0.50 0.61 0.74 0.91 1.1 1.2 1.4 1.7 2.8 1.2 0 0 0

0 0.17 0.28 0.50 0.61 0.74 0.91 1.1 1.2 1.4 1.7 2.8 1.2 0 0

0 0 0.17 0.28 0.50 0.61 0.74 0.91 1.1 1.2 1.4 1.7 2.8 1.2 0

0

1.2

4.0

5.7

7.1

8.3

9.4

10.3

11.1

11.7

12.2

12.4

12.6

12.6

12.6

w .E asy En g S-hydrograph

Note: *t = 30 min

(b) To estimate the 40-min unit hydrograph, the S-hydrograph must be shifted by 40 min and subtracted from the original S-hydrograph (Equation 10.51). These computations are summarized in Columns 2 to 4 in the following table.

(1) t (min) 0 30 60 90 120 150 180 210 240 270 300 330 360 390 420

ine eri n

(2) S(t) (m3 /s)

(3) S(t − 40) (m3 /s)

(4) S(t) − S(t − 40) (m3 /s)

0 1.2 4.0 5.7 7.1 8.3 9.4 10.3 11.1 11.7 12.2 12.4 12.6 12.6 12.6

0 0 0.8 3.1 5.1 6.6 7.9 9.0 10.0 10.8 11.5 12.0 12.3 12.5 12.6

0 1.2 3.2 2.6 2.0 1.7 1.5 1.3 1.1 0.9 0.7 0.4 0.3 0.1 0

(5) 40-min UH (m3 /s)

g .n 0 0.9 2.4 2.0 1.5 1.3 1.1 1.0 0.8 0.7 0.5 0.3 0.2 0.1 0

et

Note that values of S(t − 40) must be interpolated. From the given data, t = 30 min and τ = 40 min, therefore t/τ = 30/40 = 0.75, and the 40-min unit hydrograph is obtained by multiplying Column 4 by 0.75 (see Equation 10.51). The 40-min unit hydrograph is given in Column 5.

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Fundamentals of Surface-Water Hydrology II: Runoff (c) The area under the computed 40-min unit hydrograph is given by Area = (30)(60)(0.9 + 2.4 + 2.0 + 1.5 + 1.5 + 1.3 + 1.1 + 1.0 + 0.8 + 0.7 + 0.5 + 0.3 + 0.2 + 0.1) = 23, 040 m3 Since the catchment area is 2.25 km2 = 2.25 * 106 m2 , the depth of rainfall excess is given by Depth of rainfall excess =

23, 040 2.25 * 106

L 0.01 m = 1 cm

The depth of rainfall excess is equal to 1 cm, and therefore the 40-min unit hydrograph calculated in (b) is a valid unit hydrograph.

ww

Unit hydrographs can also be used to determine the runoff in cases where the rainfall excess is nonuniform in time. In these cases, the rainfall excess is discretized into a series of uniform-rainfall-excess events over time intervals, t, and the runoffs from each of these discrete events are added together to give the runoff from the entire event. Hence, if the rainfall excess is divided into N discrete events, with rainfall excesses Pi , i = 1, . . . , N, then the runoff, Q(t), from the entire event is given by

w .E asy En g Q(t) =

N 

Pi ut (t − i t + t)

(10.53)

i=1

where ut (t) is the t-unit hydrograph.

EXAMPLE 10.9

ine eri n

The 30-min unit hydrograph for a catchment is given by Time (min)

0

30

60

90

120

150

180

210

240

270

Runoff (m3 /s)

0

1.2

2.8

1.7

1.4

1.2

1.1

0.91

0.74

0.61

Estimate the runoff resulting from the following 90-min storm:

Time (min)

Rainfall excess (cm)

0–30 30–60 60–90

3.1 2.5 1.7

g .n 300

330

0.50

0.28

360

390

et 0.17

0

Solution The given 90-min storm can be viewed as three consecutive 30-min storms with rainfallexcess amounts of 3.1 cm, 2.5 cm, and 1.7 cm. The runoff from each storm is estimated by multiplying the 30-min unit hydrograph by 3.1, 2.5, and 1.7, respectively. The hydrographs are then lagged by 30 min and summed to estimate the total runoff from the storm (see Equation 10.53). These computations are summarized in the following table:

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ww

Time (min)

UH * 3.1 (m3 /s)

UH * 2.5 (m3 /s)

UH * 1.7 (m3 /s)

Total runoff (m3 /s)

0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450

0 3.7 8.7 5.3 4.3 3.7 3.4 2.8 2.3 1.9 1.6 0.9 0.5 0 0 0

0 0 3.0 7.0 4.3 3.5 3.0 2.8 2.3 1.9 1.5 1.3 0.7 0.4 0 0

0 0 0 2.0 4.8 2.9 2.4 2.0 1.9 1.5 1.3 1.0 0.9 0.5 0.3 0

0 3.7 11.7 14.3 13.4 10.1 8.8 7.6 6.4 5.3 4.4 3.2 2.1 0.9 0.3 0

w .E asy En g 10.5.2

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501

Instantaneous Unit Hydrograph

The instantaneous unit hydrograph (IUH) is the runoff hydrograph that would result if a unit depth of rainfall excess were instantaneously deposited uniformly over an area and then allowed to run off. To develop an IUH, the S-hydrograph derived from a t-unit hydrograph must first be obtained. The resulting S-hydrograph is then differenced by the time τ , and the difference scaled to develop the τ -unit hydrograph, which becomes the IUH in the limit as τ approaches zero. According to this definition, and using Equation 10.51, the ordinates of the IUH, QIUH (t), are given by

ine eri n

QIUH (t) = lim uτ (t) τ →0   t [St (t) − St (t − τ )] = lim τ τ →0   St (t) − St (t − τ ) = t lim τ τ →0 which yields QIUH (t) = t

dSt dt

(10.54)

g .n

(10.55)

et

(10.56)

(10.57)

This relation shows that the ordinates of the IUH can be determined directly from the slope of the S-hydrograph. The S-hydrograph can also be expressed in terms of the instantaneous unit hydrograph by integrating Equation 10.57, which yields St (t) =

1 t



t

QIUH dt

(10.58)

0

Substituting Equation 10.58 into Equation 10.51 gives the following relation between the τ -unit hydrograph, uτ (t), and the instantaneous unit hydrograph, QIUH (t):

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t [St (t) − St (t − τ )] τ    t  t−τ t 1 1 QIUH dt − QIUH dt = τ t 0 t 0

uτ (t) =

(10.59) (10.60)

which yields 1 uτ (t) = τ

ww



t

(10.61)

QIUH dt

t−τ

This relation shows that the instantaneous unit hydrograph, QIUH (t), is fundamental to determining the unit hydrograph for any duration, τ . For a finite time interval, τ , Equation 10.61 can be approximated by   1 QIUH (t) + QIUH (t − τ ) τ (10.62) uτ (t) L τ 2 which simplifies to

w .E asy En g uτ (t) L

 1 QIUH (t) + QIUH (t − τ ) 2

(10.63)

Use of this approximate equation is allowed for small values of τ and permits direct calculation of a unit hydrograph from an IUH. A major advantage of the IUH over the unit hydrograph is that the IUH does not require uniform excess precipitation for a specific duration. The IUH function is required to be greater than or equal to zero and integrate to unity, properties that are shared by probability distributions. This similarity is not coincidental, and one interpretation is that the IUH is equal to the probability density function of the overland-flow travel times within the ´ 1979). catchment (Rodr´ıguez-Iturbe and Valdes, The IUH of a catchment can be determined from field measurements by combining estimates of an effective-rainfall hyetograph with the corresponding direct-runoff hydrograph and using the relation 

ine eri n t

Q(t) =

0

ie (t − τ )QIUH (τ )dτ

g .n

(10.64)

where Q(t) is the direct runoff rate [LT−1 ], ie is effective rainfall rate [LT−1 ], and QIUH is the instantaneous unit hydrograph [T−1 ]. Determination of QIUH from measurements of Q(t) and estimates of ie (t) generally requires the assumption of a particular analytical form of the IUH and the subsequent optimization of the parameters of the analytical IUH (e.g., Cleveland et al., 2006; Singh, 2006; Singh, 2007e). The IUH is commonly approximated by the two-parameter gamma distribution (e.g., Singh, 2007e). In cases where runoff data are not available, methods have been proposed to derive the IUH from the geomorphologic characteristics of catchments (Sahoo et al., 2006; Bhunya et al., 2008) and directly from digital elevation models (Cleveland et al., 2008). 10.5.3

et

Unit-Hydrograph Models

Unit hydrographs are estimated in practice either from contemporaneous rainfall and runoff-rate measurements or by using empirical relationships. The estimation of unit hydrographs from measured rainfall and runoff hydrographs (e.g., Jain et al., 2006b) is seldom an option in urban hydrology, where runoff is to be estimated from planned developments. The most common method of estimating unit hydrographs in urban applications is to use empirical relationships to construct a synthetic unit hydrograph, where the shape of the unit hydrograph is related to the catchment characteristics and the duration of the rainfall excess. The shape parameters are usually the peak runoff rate, time to peak, and time base of the unit hydrograph. Using these shape parameters, a hydrograph with unit area must be

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503

constructed. Since probability distributions have unit areas, it is fairly common to use standard probability distributions to approximate unit hydrographs (e.g., Bhunya et al., 2007a). The gamma and beta distributions are the most common probability distribution functions used to approximate synthetic unit hydrographs; however, several other classical distributions can also be used (e.g., Nadarajah, 2007). A wide variety of methods to estimate synthetic unit hydrographs have been developed, and several of the more popular methods are described below. 10.5.3.1

ww

Snyder unit-hydrograph model

The Snyder unit-hydrograph model is an empirical model that was developed based on studies of 20 watersheds located primarily in the Appalachian Highlands in the United States (Snyder, 1938). Snyder collected rainfall and runoff data from gaged watersheds with areas in the range of 26–26,000 km2 (16–16,000 mi2 ), derived the unit hydrographs, and related the unit-hydrograph parameters to measurable watershed characteristics. Snyder’s observations indicated that the peak runoff rate per unit (1 cm) of rainfall excess, Qp [m3 /s], is given by Qp = 2.75

w .E asy En g

Cp A tl

(10.65)

where tl is the time lag [h], A is the catchment area [km2 ], and Cp is the peaking coefficient [dimensionless]. The time lag, tl , is defined as the time from the centroid of the rainfall excess to the occurrence of the peak runoff rate. The duration of rainfall excess, tr , is related to the time lag, tl , by tr =

tl 5.5

ine eri n

(10.66)

Based on Snyder’s observations, the time lag, tl [h], is related empirically to the catchment characteristics by tl = 0.75Ct (LLc )0.3

(10.67)

where Ct is the basin coefficient that accounts for the slope, land use, and associated storage characteristics of the river basin [dimensionless]; L is the length of the main stream from the outlet to the catchment boundary [km]; and Lc is the length along the main stream from the outlet to a point nearest to the catchment centroid [km] The term (LLc )0.3 is sometimes called the shape factor of the catchment (McCuen, 2005). The parameters Cp in Equation 10.65 and Ct in Equation 10.67 are best found via calibration, as they are not physically based. Snyder (1938) reported that Ct is typically in the range of 1.35–1.65, with a mean of 1.5; however, values of Ct have been found to vary significantly outside this range in very mountainous or very flat terrain. Bedient and Huber (1992) report that Cp is in the range of 0.4–0.8, where the larger values of Cp are associated with smaller values of Ct . The variability in Ct and Cp is evident from the range of values reported in Table 10.10. If the duration of the desired unit hydrograph for the watershed of interest is significantly different from that specified in Equation 10.66, the following relationship can be used to adjust the time lag, tl : tlR = tl + 0.25(tR − tr )

g .n

et

(10.68)

where tR is the desired duration of the rainfall excess, and tlR is the time lag of the desired unit hydrograph. For durations other than the standard duration, the peak runoff rate, QpR , is given by QpR = 2.75

Cp A tlR

(10.69)

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Fundamentals of Surface-Water Hydrology II: Runoff TABLE 10.10: Variability in Snyder Unit-Hydrograph Parameters

ww

Location

Reference

Ct

Cp

1.01–4.33

0.26–0.67

27 watersheds in Pennsylvania

Miller et al. (1983)

0.40–2.26

0.31–1.22

13 watersheds (1.28–192 km2 ) in central Texas

Hudlow and Clark (1969)

0.3–0.7

0.35–0.59

Northwestern United States (Ct ); Sacramento and lower San Joaquin rivers

Linsley (1943)

0.4–2.4

0.4–1.1

12 watersheds (0.05−635 km2 ) in eastern New South Wales, Australia

Cordery (1968)

The U.S. Army Corps of Engineers developed the following empirical equations to help define the shape of the Snyder unit hydrograph:

w .E asy En g

W50 =

2.14 (QpR /A)1.08

(10.70)

W75 =

1.22 (QpR /A)1.08

(10.71)

where W50 and W75 are the widths of the unit hydrograph [h] at 50% and 75% of QpR , where QpR is the peak discharge [m3 /s], and A is the catchment area [km2 ]. These time widths are proportioned such that one-third is before the peak and two-thirds after the peak. The time to peak discharge, Tp , is computed from the duration of the rainfall excess, tR , and time lag, tlR , using the relation 1 (10.72) Tp = tR + tlR 2

ine eri n

The time base of the unit hydrograph, TB , is computed such that the unit hydrograph represents 1 cm of runoff. A rough estimate of TB in small watersheds is three to five times Tp . EXAMPLE 10.10

g .n

et

Derive the 2-h Snyder unit hydrograph for a 60-km2 watershed where the main stream is 11 km long and the distance from the watershed outlet to the point on the stream nearest to the centroid of the watershed is 4 km. Previous investigations in similar terrain indicate that Cp = 0.6 and Ct = 1.5. Solution From the given data: A = 60 km2 , L = 11 km, Lc = 4 km, Cp = 0.6, and Ct = 1.5. According to Equation 10.67, the time lag, tl , is given by tl = 0.75Ct (LLc )0.3 = 0.75(1.5)(11 * 4)0.3 = 3.50 h This time lag corresponds to a rainfall-excess duration, tr , given by Equation 10.66 as 3.50 tl = = 0.636 h 5.5 5.5 Since a 2-h unit hydrograph is required, tR = 2 h, and the adjusted time lag, tlR , is given by Equation 10.68 as tlR = tl + 0.25(tR − tr ) = 3.50 + 0.25(2.00 − 0.636) = 3.84 h tr =

This gives a time to peak, Tp , as 2.00 t + 3.84 = 4.84 h Tp = R + tlR = 2 2

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505

and the peak flow rate, QpR , is given by Equation 10.69 as QpR = 2.75

Cp A (0.6)(60) = 25.8 m3 /s = 2.75 tlR 3.84

The hydrograph shape parameters, W50 and W75 , are given by Equations 10.70 and 10.71 as W50 = W75 =

2.14 (QpR /A)1.08 1.22 (QpR /A)1.08

= =

2.14 (25.8/60)1.08 1.22 (25.8/60)1.08

= 5.32 h = 3.04 h

Putting one-third of W50 and W75 before the peak and two-thirds of W50 and W75 after the peak leads to the following points on the unit hydrograph:

ww

t (h)

Q (m3 /s)

0.0 3.07 3.83 4.84 6.87 8.39 TB

0.0 12.9 19.4 25.8 19.4 12.9 0.0

w .E asy En g

Since the runoff depth is 1 cm and the area of the catchment is 60 km2 ,  1 1 (3.07)(12.9) + (12.9 + 19.4)(3.83 − 3.07) 1 cm = 2 2

ine eri n

1 1 (19.4 + 25.8)(4.84 − 3.83) + (25.8 + 19.4)(6.87 − 4.84) 2 2  1 (3600)(100) 1 + (19.4 + 12.9)(8.39 − 6.87) + (TB − 8.39)(12.9) 2 2 60 * 106

+

g .n

which yields TB = 14.80 h. As expected, TB is in the range of 3–5 times Tp . This completes the specification of the Snyder unit hydrograph, which is plotted in Figure 10.6. FIGURE 10.6: Snyder 2-h unit hydrograph

30

Runoff, Q (m3/s)

25

et

20 15 10 5 0

0

2

4

6

8 10 Time, t (h)

12

14

16

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10.5.3.2

NRCS dimensionless unit hydrograph

The Natural Resources Conservation Service (SCS, 1985; SCS, 1986) developed a dimensionless unit hydrograph that represents the average shape of a large number of unit hydrographs from small agricultural watersheds throughout the United States. This dimensionless unit hydrograph is a special case of the gamma function unit hydrograph that is frequently applied to urban catchments. The NRCS dimensionless unit hydrograph is illustrated in Figure 10.7(a) and expresses the normalized runoff rate, Q/Qp , as a function of the normalized time, t/Tp , where Qp is the peak runoff rate and Tp is the time to the peak of the hydrograph from the beginning of the rainfall excess. The coordinates of the NRCS dimensionless unit hydrograph are given in Table 10.11. The NRCS dimensionless unit hydrograph can be converted to an actual hydrograph by multiplying the abscissa by Tp and the ordinate by Qp . The time, Tp , is estimated using

ww FIGURE 10.7: NRCS unit hydrographs

Tp =

1 tr + tl 2

(10.73)

where tr is the duration of the rainfall excess and tl is the time lag from the centroid of the rainfall excess to the peak of the runoff hydrograph. NRCS recommends that the specified

w .E asy En g 1.0

Excess rainfall

0.8

Q

0.6

Qp

0.4 0.2 0

0

1

3

2

4

5

t Tp

tl

tr 2

ine eri n Qp

tr

Tp

1.67 Tp

TB

(a) Dimensionless unit hydrograph

Direct runoff

g .n

(b) Triangular unit hydrograph

TABLE 10.11: NRCS Dimensionless Unit Hydrograph

t/Tp

Q/Qp

t/Tp

Q/Qp

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4

0.000 0.100 0.310 0.660 0.930 1.000 0.930 0.780 0.560 0.390 0.280 0.207 0.147

2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0

0.107 0.077 0.055 0.040 0.029 0.021 0.015 0.011 0.008 0.006 0.004 0.002 0.000

et

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507

value of tr not exceed two-tenths of tc or three-tenths of Tp for the NRCS dimensionless unit hydrograph to be valid. The time lag, tl , can be estimated by tl = 0.6tc

(10.74)

for rural catchments and, if justified by available data, tl [h] can also be estimated using the more detailed relation tl =

ww

L0.8 (S + 2.54)0.7 1410Y 0.5

(10.75)

where L is the length to the catchment divide [m], and S is the potential maximum retention (cm) given by   1000 S = 2.54 − 10 (10.76) CN where CN is the curve number [dimensionless], and Y is the average catchment slope [%]. A historical account of the development of Equation 10.75 can be found in the work of Folmar et al. (2007). Care should be taken in estimating Y from digital elevation models (DEMs), since the value of Y can be sensitive to the DEM grid size (Hill and Neary, 2005). The curve number in Equation 10.76 should be based on antecedent runoff condition II (ARCII), since it is being used as a measure of the surface roughness and not runoff potential (Haan et al., 1994). Equation 10.75 is applicable to curve numbers in the range of 50–95, and catchment areas less than 8 km2 (3 mi2 ). Otherwise, Equation 10.74 should be used to estimate the basin lag, tl (Ponce, 1989). If the watershed is in an urban area, then tl given by Equation 10.75 is adjusted for imperviousness and/or improved water courses by the factor M, where

w .E asy En g

M = 1 − I(−6.8 * 10−3 + 3.4 * 10−4 CN − 4.3 * 10−7 CN2 − 2.2 * 10−8 CN3 )

ine eri n

(10.77)

and I is either the percentage impervious or the percentage of the main watercourse that is hydraulically improved from natural conditions. If part of the catchment area is impervious and part of the channel is improved, then two values of M are determined and both are multiplied by tl . Once tl is determined, the time to peak, Tp , is estimated using Equation 10.73. The time base of the unit hydrograph, TB , is equal to 5Tp , and both Tp and TB should be rounded to the nearest whole-number multiple of tr . A compilation of data from 10,000 runoff events in 52 nonurbanized watersheds in the United States (Folmar and Miller, 2008) showed that the NRCS time lag equation (Equation 10.75) underestimated measured time lags by around 60%, and a more accurate estimation of the time lag, tl [h], was found using the relation 0.65

tl = 0.0120L

g .n

et

(10.78)

where L is the longest flow distance [m]. Nonlinear analytic functions that closely approximate the NRCS curvilinear dimensionless unit hydrograph have been developed (e.g., Jeng, 2006) but are rarely used. The NRCS curvilinear dimensionless unit hydrograph is more commonly approximated by the triangular unit hydrograph illustrated in Figure 10.7(b). This triangular unit hydrograph incorporates the following key properties of the curvilinear dimensionless unit hydrograph: (1) The total volume under the hydrograph is the same, (2) the volume under the rising limb is the same, and (3) the peak discharge is the same. Taking Qp and Tp to be the same in both the NRCS curvilinear dimensionless unit hydrograph (Figure 10.7(a)) and the triangular approximation (Figure 10.7(b)), then the time base of the triangular unit hydrograph must be equal to 2.67Tp (= 8/3 Tp ). For a runoff depth, h, from a catchment area, A, the triangular unit hydrograph requires that A · h=

1 Qp (2.67Tp ) 2

(10.79)

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If the runoff depth, h, is 1 cm, A is in km2 , Qp is in (m3 /s)/cm, and Tp is in hours, then Equation 10.79 yields A Qp = 2.08 (10.80) Tp

ww

This estimate of Qp provides the peak runoff rate in both the curvilinear unit hydrograph (Figure 10.7(a)) and the triangular approximation (Figure 10.7(b)). The coefficient of 2.08 in Equation 10.80 is appropriate for the average rural experimental watersheds used in calibrating the formula, but it should be increased by about 20% for steep mountainous conditions and decreased by about 30% for flat swampy conditions. Flat swampy conditions are typically associated with areas having an average slope less than 0.5% (Lin and Perkins, 1989). Adjustment of the coefficient in Equation 10.80 must necessarily be accompanied by an adjustment to the time base of the unit hydrograph in order to maintain a runoff depth of 1 cm. According to Debo and Reese (1995), the triangular unit-hydrograph approximation produces results that are sufficiently accurate for most stormwater-management system designs, including curbs, gutters, storm drains, channels, ditches, and culverts.

w .E asy En g EXAMPLE 10.11

A 2.25-km2 urban catchment is estimated to have an average curve number of 70, an average slope of 0.5%, and a flow length from the catchment boundary to the outlet of 1680 m. If the imperviousness of the catchment is estimated at 40%, determine the NRCS unit hydrograph for a 30-min rainfall excess. Determine the approximate NRCS triangular unit hydrograph and verify that it corresponds to a rainfall excess of 1 cm. Solution The time lag of the catchment can be estimated using Equation 10.75, where L = 1680 m, Y = 0.5%, and S is derived from CN = 70 using Equation 10.76:     1000 1000 − 10 = 2.54 − 10 = 10.89 cm S = 2.54 CN 70

The time lag, tl , is therefore given by tl =

L0.8 (S + 2.54)0.7 1410Y 0.5

ine eri n

=

(1680)0.8 (10.89 + 2.54)0.7 1410(0.5)0.5

= 2.35 h

g .n

For a 40% impervious catchment, I = 40% and CN = 70, Equation 10.77 gives the correction factor, M, as

et

M = 1 − I[−6.8 * 10−3 + 3.4 * 10−4 CN − 4.3 * 10−7 CN2 − 2.2 * 10−8 CN3 ] = 1 − (40)[−6.8 * 10−3 + 3.4 * 10−4 (70) − 4.3 * 10−7 (70)2 − 2.2 * 10−8 (70)3 ] = 0.706 Applying the correction factor, M, to calculate the adjusted time lag yields tl = 0.706 * 2.35 h = 1.66 h For a 30-min rainfall excess, tr = 0.5 h and the time to peak, Tp , is given by Equation 10.73 as Tp =

1 1 tr + tl = (0.5) + 1.66 = 1.91 h 2 2

The NRCS unit hydrograph is limited to cases where tr /Tp … 0.3; in this case, tr /Tp = 0.5/1.91 = 0.26. The NRCS unit hydrograph is therefore applicable. The peak of the unit hydrograph, Qp , is estimated using Equation 10.80, where A = 2.25 km2 and Qp = 2.08

2.25 A = 2.45 (m3 /s)/cm = 2.08 Tp 1.91

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With Tp = 1.91 h and Qp = 2.45 (m3 /s)/cm, the NRCS 30-min unit hydrograph is obtained by multiplying t/Tp in Table 10.11 by 1.91 h and Q/Qp by 2.45 (m3 /s)/cm to yield

ww

t (h)

Q [(m3 /s)/cm]

t (h)

Q [(m3 /s)/cm]

0.0 0.38 0.76 1.15 1.53 1.91 2.29 2.67 3.06 3.44

0.0 0.25 0.76 1.62 2.28 2.45 2.28 1.91 1.37 0.96

3.82 4.20 4.58 4.97 5.35 5.73 6.49 8.02 8.79 9.55

0.69 0.51 0.36 0.26 0.19 0.14 0.07 0.03 0.01 0.00

For the approximate triangular unit hydrograph, Tp and Qp are the same as for the curvilinear unit hydrograph, but the base of the triangular unit hydrograph is 2.67Tp = 2.67 * 1.91 h = 5.10 h (= 18,360 s). The area under the triangular hydrograph (= volume of runoff) is given by

w .E asy En g Area =

1 1 base * height = 18, 360 * 2.45 = 22,500 m3 2 2

The catchment area is 2.25 km2 = 2.25 * 106 m2 and the depth of rainfall is equal to the volume of runoff divided by the area of the catchment. Hence Depth of rainfall =

22, 500 m3

2.25 * 106 m2

= 0.01 m = 1 cm

ine eri n

Since the runoff hydrograph corresponds to a unit depth (1 cm) of rainfall, it is a valid unit hydrograph.

10.5.3.3

Accuracy of unit-hydrograph models

According to Dunne and Leopold (1978), the unit-hydrograph method gives estimates of flood peaks that are usually within 25% of their true value; this is close enough for most planning purposes and about as close as can be expected, given the usual lack of detailed information about catchment processes and states. Errors larger than 25% can be expected if synthetic unit hydrographs are used without verification for the region of application (Dingman, 2002). Specialized unit hydrographs have been developed for some areas; for example, in the state of Colorado, the Colorado Unit Hydrograph Procedure (CUHP) was developed for metropolitan areas, and the major parameters used in the CUHP were calibrated by the area imperviousness ratio using urbanized watershed data. The CUHP has been widely accepted by metropolitan areas in the front range of the Rocky Mountains for predicting flood hydrographs. A comparison between the CUHP and other unit hydrograph methods can be found in Guo (2006a). 10.5.4

g .n

et

Time-Area Models

Time-area models describe the relationship between the travel time to the catchment outlet and the location within the catchment. This relationship is based on the estimated velocity of direct runoff and neglects storage effects. The time-area model can be illustrated by considering the runoff isochrones illustrated in Figure 10.8, and the corresponding histogram of incremental contributing area versus time shown in Figure 10.9. These figures indicate that the incremental area contributing to runoff varies throughout a storm, with full contribution from all areas for t Ú t5 . If the rainfall excess is expressed as the average rainfall-excess intensity over the time increments in the time-area histogram, then the runoff rate at the end of time increment j, Qj , can also be expressed in terms of the contributing areas by the relation

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FIGURE 10.8: Runoff isochrones

t4

t5

Catchment boundary t3

A5

t2

A4 A3

t1 A2 A1 Catchment outlet

Isochrone FIGURE 10.9: Time-area histogram

Area, A

ww

A3

A4

A2 A5

A1

w .E asy En g 0

t1

t2

t3

t4

t5

Time, t

Qj = ij A1 + ij−1 A2 + · · · + i1 Aj

or

j 

ine eri n

Qj =

(10.81)

ij−k+1 Ak

k=1

where ij and Aj are the average rainfall-excess intensity and incremental contributing area, respectively, during time increment j. The difficulty in estimating isochronal lines and the need to account for storage effects make the time-area model difficult to apply in practice. The time-area model uses a simple routing procedure that does not consider attenuation of the runoff hydrograph associated with storage effects, and is therefore limited to small watersheds with areas less than 200 ha (500 ac) (Ward and Trimble, 2004). As a direct result of neglecting attenuation, peak flows will likely be overestimated if the time-area model is applied to large watersheds.

g .n

et

EXAMPLE 10.12 A 100-ha catchment is estimated to have the following time-area relationship: Time (min)

Contributing area (ha)

0 5 10 15 20 25 30

0 3 9 25 51 91 100

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511

If the rainfall-excess distribution is given by

Time (min)

Average intensity (mm/h)

0–5 5–10 10–15 15–20

132 84 60 36

estimate the runoff hydrograph using the time-area model. Solution First, tabulate the contributing area for each of the given time intervals:

ww

Time interval (min)

Contributing area (ha)

0–5 5–10 10–15 15–20 20–25 25–30

3 6 16 26 40 9

w .E asy En g

The runoff hydrograph is calculated using Equation 10.81, and the computations are given in the following table (a factor of 0.002778 is used to convert ha·mm/h to m3 /s): Time (min) 0 5 10 15 20 25 30 35 40 45 50

ine eri n

Runoff hydrograph (m3 /s)

[132(3)](0.002778) = [84(3) + 132(6)](0.002778) = [60(3) + 84(6) + 132(16)](0.002778) = [36(3) + 60(6) + 84(16) + 132(26)](0.002778) = [36(6) + 60(16) + 84(26) + 132(40)](0.002778) = [36(16) + 60(26) + 84(40) + 132(9)](0.002778) = [36(26) + 60(40) + 84(9)](0.002778) = [36(40) + 60(9)](0.002778) = [36(9)](0.002778) = 0=

0 1.1 4.0 7.8 14.6 24.0 18.6 11.4 5.5 0.9 0

g .n

et

The expected runoff hydrograph in the above table has a time base of 50 min and a peak runoff rate of 24.0 m3 /s.

The Hydrologic Engineering Center (1998; 2000) has developed the following synthetic timearea relationship based on empirical analyses of several catchments ⎧  1.5 ⎪ t t ⎪ ⎪ 1.414 for 0 … … 0.5 ⎨ A t t c c = (10.82) 1.5  ⎪ Ac t t ⎪ ⎪ for 0.5 … … 1.0 ⎩1 − 1.414 1 − tc tc

where A is the contributing area at time t, Ac is the catchment area, and tc is the time of concentration of the catchment. For a rainfall excess of depth P occurring instantaneously

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over the catchment, the average runoff rate, I, at the catchment outlet between times t and t + t is (P)(A) (10.83) I= t where A is the incremental contributing area within the incremental time t. Taking P = 1 cm, and expressing A in km2 and t in hours, Equation 10.83 can be put in the form A (10.84) I = 2.78 t

ww

where I is in m3 /s. Equation 10.84 gives the coordinates of the translation hydrograph resulting from an instantaneous rainfall excess of 1 cm occurring over the entire catchment. To account for storage effects within the catchment, the Hydrologic Engineering Center (HEC) utilizes the continuity equation in the form   S2 − S1 O1 + O2 = (10.85) I − 2 t

w .E asy En g

where I is the mean inflow during time interval, t, O is the outflow, and S is the storage. The subscripts 1 and 2 refer to the beginning and end of the time interval, t. In applying Equation 10.85, I is the mean runoff rate during t accounting for translation effects only (as done in the previous example); and O1 and O2 are the outflows at the beginning and end of the time interval when storage effects are considered. Assuming that the storage, S, is linearly proportional to the outflow, O, commonly referred to as the linear-reservoir assumption, then S = RO (10.86)

ine eri n

where R is a proportionality constant that is sometimes called the storage coefficient and has units of time. A linear reservoir is defined as a reservoir/outlet combination in which the reservoir discharge is linearly proportional to the reservoir storage; a description of the physical shape of a linear reservoir can be found in Purcell (2006). Equations 10.85 and 10.86 combine to yield the relation O2 = CI + (1 − C)O1

where

2t C= 2R + t

g .n

(10.87)

et

(10.88)

Equation 10.87 is the basis for the Clark method of estimating the unit hydrograph from time-area curves (Clark, 1945), and this method is still widely used in current engineering practice (HEC, 2000; 2001). In the Clark method, the translation hydrograph is first calculated using Equations 10.82 and 10.84, and then the translation hydrograph is routed through a linear reservoir using Equation 10.87. In applying the Clark method, the resulting runoff hydrograph is, by definition, an instantaneous unit hydrograph (IUH). The unit hydrograph for any duration, τ , can then be determined from this IUH using Equation 10.63.

EXAMPLE 10.13 A catchment has an area of 95 km2 and a time of concentration of 7.5 h, and the Clark storage coefficient is estimated to be 3.4 h. Estimate the 2-h unit hydrograph for the catchment. Solution From the given data: Ac = 95 km2 , tc = 7.5 h, and R = 3.4 h. The computations are summarized in Table 10.12, and the sequence of calculations is described below.

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TABLE 10.12: Computation of IUH Using the Clark Method

(1) t (h)

(2) t/tc

(3) A/Ac

(4) A (km2 )

0.0

0.000

0.000

0.0

0.5 1.0 1.5 2.0

ww

2.5 3.0

0.067 0.133 0.200 0.267 0.333 0.400

0.024 0.069 0.126 0.195 0.272 0.358

4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5

8.0 8.5 9.0 9.5 · · · 29.5 30.0 30.5

0.467 0.533 0.600 0.667 0.733 0.800 0.867 0.933 1.000 — — — — · · · — — —

0.451 0.549 0.642 0.728 0.805 0.874 0.931 0.976 1.000 — — — — · · · — — —

(6) I (m3 /s)

2.3

12.8

4.2

23.5

5.5

30.4

6.5

36.0

7.4

40.9

8.1

45.2

8.8

49.1

9.4

52.0

8.8

49.1

8.1

45.2

7.4

40.9

2.3 6.5 12.0 18.5 25.9 34.0

w .E asy En g 3.5

(5) A (km2 )

42.8 52.2 61.0 69.1 76.5 83.0 88.5 92.7 95.0

95.0 — — — · · · — — —

(7) IUH (m3 /s) 0

0

1.8

0.9

4.7

2.4

8.3

4.1

12.1

6.0

16.0

8.9

20.0

12.4

24.0

16.1

27.8

19.9

30.7

23.4

32.7

26.4

33.8

28.9

ine eri n 6.5

36.0

5.5

30.4

4.2

23.5

2.3

12.8

0.0

0.0

(8) 2-h UH (m3 /s)

34.1

31.0

33.6

32.2

32.2

32.5

29.6

25.5 22.0 19.0 16.4 · · · 0.0 0.0 0.0

g .n 31.7

29.8 27.8 25.6 23.0 · · · 0.1 0.1 0.0

et

• Column 2 (= t/tc ) is derived from Column 1 (= t) by dividing Column 1 by tc (= 7.5 h); Column 3 (= A/Ac ) is derived from Column 2 using Equation 10.82; and Column 4 (= A) is derived from Column 3 by multiplying Column 3 by Ac (= 95 km2 ). • The incremental area, A, between isochrones is calculated by subtracting the areas included by adjacent isochrones in Column 4, and the results are shown in Column 5. • Assuming pure translation of the rainfall excess, the average runoff rate, I, during each time increment in response to an instantaneous rainfall excess of 1 cm (= 0.01 m) is given by

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Fundamentals of Surface-Water Hydrology II: Runoff Equation 10.84. The calculated runoff rate, I, is given in Column 6. The runoff hydrograph in Column 6 is the instantaneous unit hydrograph with storage effects neglected (i.e., pure translation). • Storage effects are accounted for by using the linear-reservoir routing relation given by Equation 10.87, where O2 = CI + (1 − C)O1 and C=

2t 2R + t

In this case, t = 0.5 h, R = 3.4 h, C=

2(0.5) = 0.137 2(3.4) + 0.5

and the linear-reservoir routing relation is given by

ww

O2 = 0.137I + (1 − 0.137)O1 = 0.137I + 0.863O1 The results of applying this routing relation are given in Column 7, where O1 and O2 are the routed runoff rates at the beginning and end of the time increment, respectively (in Column 7), and I is the average translated runoff rate during the time increment, given in Column 6. After applying the routing relation, the hydrograph shown in Column 7 is the instantaneous unit hydrograph with storage effects taken into account. • The 2-h unit hydrograph, u2 (t), is derived from the instantaneous unit hydrograph in Column 7 using Equation 10.63, 1 u2 (t) L [QIUH (t) + QIUH (t − 2)] 2 where QIUH (t) and QIUH (t − 2) are the instantaneous unit hydrograph runoff rates tabulated in Column 7. Applying this equation to the data in Column 7 yields the 2-h unit hydrograph given in Column 8, with the corresponding times given in Column 1.

w .E asy En g

ine eri n

The parameters of the Clark hydrograph are the time of concentration, tc , and the storage reservoir coefficient, R, which is the slope of the storage-outflow curve for the linear reservoir. In cases where rainfall and runoff measurements are available, tc can be estimated as the time from the end of a burst of rainfall excess to the inflection point on the receding limb of the resulting runoff hydrograph, and R can be estimated by dividing the direct-runoff discharge at the inflection point by the slope of the curve at that point (Bedient et al., 2008). 10.5.5

Kinematic-Wave Model

g .n

et

The kinematic-wave model describes runoff by solving the one-dimensional continuity equation ⭸y ⭸q + = ie (10.89) ⭸t ⭸x and a momentum equation of the form q = αym

(10.90)

where y is the flow depth [L], q is the flow per unit width of the catchment [L2 T−1 ], α and m are empirical constants, and ie is the effective rainfall [LT−1 ]. Combining Equations 10.89 and 10.90 yields the following kinematic-wave equation for overland flow: ⭸y ⭸y + αmym−1 = ie ⭸t ⭸x

(10.91)

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515

The kinematic-wave model solves Equation 10.91 for the flow depth, y, as a function of x and t for given initial and boundary conditions. This solution is then substituted into Equation 10.90 to determine the runoff rate, q, as a function of space and time. In cases where the rainfall excess is independent of x and t, the solution was derived previously and is given by Equation 10.11. In the more general case with variable effective rainfall, a numerical solution of Equation 10.91 is required. In most applications the kinematic-wave model is implemented using a constant Manning surface roughness; however, it has been shown that expressing the Manning roughness as a function of flow depth can yield more accurate runoff hydrographs (Rai et al., 2009). The kinematic-wave model is useful in developing surfacerunoff hydrographs for subwatersheds that contribute lateral flows to streams; for example, overland flow to a stream from the left and right sides of its bank may be computed using kinematic-wave models. In cases where overland flow enters a drainage channel, flow in the channel can be described by ⭸A ⭸Q + = q0 (10.92) ⭸t ⭸x and

w .E asy En g

Q = αAm

(10.93)

where A is the cross-sectional flow area [L2 ], Q is the flow rate in the channel [L3 T−1 ], α and m are empirical constants, and q0 is the overland inflow entering the channel per unit length of the channel [L2 T−1 ]. Combining Equations 10.92 and 10.93 yields the following kinematicwave equation for channel flow: ⭸A ⭸A + αmAm−1 = q0 ⭸t ⭸x

ine eri n

(10.94)

Because this equation does not allow for hydrograph diffusion, the application of the kinematic-wave equation is limited to flow conditions that do not demonstrate appreciable hydrograph attenuation. Accordingly, the kinematic-wave approximation works best when applied to short, well-defined channel reaches as found in urban drainage applications (ASCE, 1996). A typical kinematic-wave model divides the catchment into overland-flow planes that feed collector channels. The kinematic-wave model is used only in numerical models, and the documentation accompanying commercial software packages that implement the kinematic-wave model generally provides detailed descriptions of the numerical procedures used to solve the kinematic-wave equation. 10.5.6

Nonlinear-Reservoir Model

g .n

et

The nonlinear-reservoir model views a catchment as a very shallow reservoir, where the inflow is equal to the rainfall excess, the outflow is a (nonlinear) function of the depth of flow over the catchment, and the difference between the inflow and outflow is equal to the rate of change of storage within the catchment. The nonlinear-reservoir model can be expressed as A

dy = Aie − Q dt

(10.95)

where A is the surface area of the catchment [L2 ] and Q is the surface runoff rate at the catchment outlet [L3 T−1 ]. A particular application of the nonlinear-reservoir model is described by Huber and Dickinson (1988), where the runoff at the catchment outlet is given by the Manning equation as Q=

5 1 CW (y − yd ) 3 S02 n

(10.96)

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where C is a constant to account for the units (C = 1 for SI units, C = 1.49 for U.S. Customary units), W is a representative width of the catchment [L], n is an average value of the Manning roughness coefficient for the catchment, yd is the depression storage depth [L], and S0 is the average slope of the catchment [dimensionless]. Estimates of Manning’s n for overland flow are given in Table 10.1. The combination of Equations 10.95 and 10.96 can be put in the simple finite-difference form 1

CWS02 y2 − y1 = ie − t An

ww



y1 + y2 − yd 2

5

3

(10.97)

where t is the time step, y1 and y2 are the water depths at the beginning and end of the time step, and ie is the average effective rainfall over the time step. Equation 10.97 must be solved for y2 , which is then substituted into Equation 10.96 to determine the corresponding runoff rate, Q. EXAMPLE 10.14

w .E asy En g

A 1-ha catchment consists of mostly light turf with an average slope of 0.8%. The width of the catchment is approximately 100 m, and the depression storage is estimated to be 5 mm. Use the nonlinearreservoir model to calculate the runoff from the following 15-min rainfall excess:

Time (min)

Effective rainfall (mm/h)

0–5 5–10 10–15

120 70 50

ine eri n

Solution The nonlinear-reservoir model is given by Equations 10.96 and 10.97, where C = 1, W = 100 m, S0 = 0.008, A = 1 ha = 104 m2 , n = 0.20 (from Table 10.1), and yd = 5 mm = 0.005 m. Taking t = 5 min = 300 s and substituting the given parameters into Equation 10.97 gives 1

(1)(100)(0.008) 2 y2 − y1 = ie − 300 (104 )(0.20)



g .n

5 3 y1 + y2 − 0.005 2

which simplifies to ⎡

et

⎤ 5 3 + y y 2 y2 = y1 + 300 ⎣2.78 * 10−7 ie − 0.00447 1 − 0.005 ⎦ 2 

where the factor 2.78 * 10−7 is introduced to convert ie in mm/h to m/s. For given values of y1 , this equation is solved for y = y2 as a function of time. The corresponding runoff rate, Q, is given by Equation 10.96 as

Q=

5 5 1 5 1 (1)(100) CW (y − yd ) 3 S02 = (y − 0.005) 3 (0.008) 2 = 44.7(y − 0.005) 3 n 0.20

Starting with y1 = 0 m at t = 0, the runoff computations are tabulated for t = 0 to 60 min as follows:

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ie (mm/h)

0

Continuous-Runoff Models

y (m)

Q (m3 /s)

0.0100

0.00654

0.0154

0.0221

0.0187

0.0351

0.0177

0.0309

0.0168

0.0273

0.0160

0.0243

0.0153

0.0218

0.0147

0.0197

0.0141

0.0177

0.0136

0.0161

0.0131

0.0146

0.0127

0.0134

517

0 120

5 70 10 50 15 0 20 0 25 0

ww

30 0 35 0

w .E asy En g 40

0

45

0

50

0

55

0

60

ine eri n

Beyond t = 60 min, the runoff rate, Q, decreases gradually to zero.

10.5.7

Santa Barbara Urban Hydrograph Model

The Santa Barbara Urban Hydrograph (SBUH) model was developed for the Santa Barbara County Flood Control and Water Conservation District in California (Stubchaer, 1975) and has been adopted by other water-management agencies in the United States (e.g., the South Florida Water Management District, 2012). In the SBUH model, the impervious portion of the catchment is assumed to be directly connected to the drainage system, abstractions from rain falling on impervious surfaces are neglected, and abstractions from pervious areas are accounted for by using either the NRCS curve-number method or a similar technique. The SBUH combines the runoff from impervious and pervious areas to develop a runoff hydrograph that is routed through an imaginary reservoir that causes a time delay equal to the time of concentration of the catchment. The computations proceed through consecutive time intervals, t [T], during which the instantaneous runoff rate, I [L3 T−1 ], is calculated using the relation I = [ix + ie (1.0 − x)]A (10.98)

g .n

et

where i is the rainfall rate [LT−1 ], x is the fraction of the catchment that is impervious [dimensionless], ie is the effective rainfall rate [LT−1 ], and A is the area of the catchment [L2 ]. Over the finite interval, t, conservation of mass requires that the rate of runoff, I, minus the outflow rate, Q [LT−3 ], from the catchment is equal to the rate of change of catchment storage, S[L3 ]; and this requirement can be written in the finite-difference form     Ij + Ij−1 Qj + Qj−1 Sj − Sj−1 (10.99) − = 2 2 t

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where j is the time-step index. Approximating the catchment as a linear reservoir assumes that the catchment storage volume, S, is linearly proportional to the catchment outflow rate, Q, such that Sj = KQj and Sj−1 = KQj−1 (10.100) where K [T] is a proportionality constant. Also, since the time for the entire catchment to drain is, by definition, equal to the time of concentration, tc , the SBUH model takes K equal to tc , and Equation 10.100 becomes and

Sj = tc Qj

(10.101)

Sj−1 = tc Qj−1

Substituting Equation 10.101 into Equation 10.99 yields the following routing equation: Qj = Qj−1 + Kr (Ij−1 + Ij − 2Qj−1 )

ww

(10.102)

where Kr is a routing constant given by Kr =

w .E asy En g

t 2tc + t

(10.103)

An important limitation of the SBUH method is that the calculated peak discharge cannot occur after precipitation ceases. In reality, for short-duration storms over flat and large watersheds, the peak discharge can occur after rainfall ends. The SBUH method has been reported to greatly overpredict peak-flow rates for pasture conditions, and slightly overpredict peakflow rates for forest conditions (Jackson et al., 2001).

EXAMPLE 10.15

ine eri n

A 2.25-km2 catchment is estimated to have a time of concentration of 45 min, and 45% of the catchment is impervious. Estimate the runoff hydrograph for the following rainfall event:

Time (min)

Rainfall (mm/h)

Rainfall excess (mm/h)

0 10 20 30

0 210 126 78

0 150 102 66

g .n

Use a time increment of 10 min to calculate the runoff hydrograph.

et

Solution From the given data, t = 10 min and tc = 45 min, therefore Equation 10.103 gives Kr =

10 t = = 0.10 2tc + t 2(45) + 10

Also, from the given data, x = 0.45, A = 2.25 km2 , and Equation 10.98 gives the instantaneous runoff rate, I, as I = [ix + ie (1.0 − x)]A = [i(0.45) + ie (1.0 − 0.45)](2.25)(0.278) = 0.625[0.45i + 0.55ie ]

(10.104)

where the factor 0.278 is required to give I in m3 /s. The catchment runoff rate is given by Equation 10.102 as Qj = Qj−1 + Kr (Ij−1 + Ij − 2Qj−1 ) = Qj−1 + (0.10)(Ij−1 + Ij − 2Qj−1 )

(10.105)

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Beginning with I1 = Q1 = 0, Equation 10.104 is applied to calculate I at each time step, and Equation 10.105 is applied to calculate the runoff hydrograph. The results of these calculations are summarized in the following table:

ww

t (min)

i (mm/h)

ie (mm/h)

I (m3 /s)

Q (m3 /s)

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200

0 210 126 78 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 150 102 66 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 111 70.5 44.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 11.1 27.0 33.1 30.9 24.7 19.8 15.8 12.7 10.1 8.1 6.5 5.2 4.1 3.3 2.7 2.1 1.7 1.4 1.1 0.9

w .E asy En g

ine eri n

The runoff rate, Q, as a function of time reaches a peak at the time that the rainfall ends and decreases exponentially thereafter.

10.5.8

Extreme Runoff Events

The probable maximum flood (PMF) is the flood discharge that is expected from the most severe combination of meteorologic and hydrologic conditions that are possible in a region. Practical applications of such floods are usually confined to the determination of spillway requirements for high dams, but in unusual cases may constitute the design flood for local protection works where an exceptionally high degree of protection is advisable and economically obtainable. The standard project flood (SPF) is the most severe flood considered reasonably characteristic of a specific region. The SPF is intended as a practicable expression of the degree of protection that should be sought as a general rule in the design of flood-control works for communities where protection of human life and unusually high-valued property is involved. The standard project flood excludes extremely rare conditions and, typically, the peak discharge of an SPF is about 40%–60% of that of a PMF for the same basin and has a return period of about 500 years. The commonly used methodology for estimating the SPF in the United States has been documented by the U.S. Army Corps of Engineers (USACE, 1965). In most cases, the SPF is the runoff hydrograph from the standard project storm (SPS). In some cases, particularly in very large drainage basins, the SPF estimate may be based on a study of actual hydrographs or stages of record, or on other procedures not directly involving an SPS estimate. In cases where floods are predominantly the result of melting snow, the SPF estimate is based on estimates of the most critical combinations of snow, temperature, and water losses considered reasonably characteristic of the region. Since the statistical probability of SPF occurrence varies with the size of the drainage area and other hydrometeorological factors, it is not considered feasible to assign specific frequency estimates to SPF determinations in general.

g .n

et

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The standard project flood should not be confused with the design flood of a project. The term design flood refers to the flood hydrograph or peak-discharge value adopted as the basis for design of a particular project after full consideration has been given to flood characteristics, frequencies, and the economic and other practical considerations entering into selection of the design discharge criteria. The design flood for a particular project may be either greater or less than the standard project flood. 10.6

ww

Routing Models

Routing is the process of determining the spatial and temporal distribution of flow rate and flow depth along a watercourse such as a river or storm sewer. Routing is sometimes called flow routing or flood routing, and routing models are generally classified as either hydrologic-routing models or hydraulic-routing models. Hydrologic-routing models are based on the simultaneous solution of the continuity equation and a second equation which usually expresses the storage volume within a channel reach as a function of inflow and outflow. Hydraulic-routing models are based on the simultaneous solution of the continuity equation and momentum equation for open channels. Although hydraulic-routing models are generally considered to be more accurate than hydrologic-routing models, the simplicity and acceptable accuracy of hydrologic-routing models in some circumstances make them appealing, particularly in the design of storage ponds and reservoirs.

w .E asy En g 10.6.1

Hydrologic Routing

The basic equation used in hydrologic routing is the continuity equation dS = I(t) − O(t) dt

(10.106)

where S is the storage between upstream and downstream sections [L3 ], t is time [T], I(t) is the inflow rate at the upstream section [L3 T−1 ], and O(t) is the outflow rate at the downstream section [L3 T−1 ]. Hydrologic routing is frequently applied to storage reservoirs and stormwater-detention basins, where I(t) is the inflow rate to the storage reservoir, typically from rivers, streams, or drainage channels; O(t) is the outflow rate from the reservoir, typically over spillways, weirs, or orifice-type outlets; and S is the storage in the reservoir. Although the application of hydrologic routing to rivers and drainage channels is also well established, in many of these cases hydraulic routing is preferable. The procedure for hydrologic routing depends on the particular system being modeled. In the case of storage reservoirs, the storage, S, is typically a function of the outflow rate, O(t), and the modified Puls method (Puls, 1928)∗ is preferred. In the case of channel routing, S is typically related to both the upstream inflow rate, I(t), and downstream outflow rate, O(t), and other methods such as the Muskingum method are preferred. 10.6.1.1

ine eri n

g .n

Modified Puls method

et

Over the finite interval of time between t and t + t, Equation 10.106 can be written in the finite-difference form     S2 − S1 O1 + O2 I1 + I2 − (10.107) = t 2 2 where the subscripts 1 and 2 refer to the values of the variables at times t and t + t, respectively. For computational convenience, Equation 10.107 can be put in the form (I1 + I2 ) +



2S1 − O1 t



=



2S2 + O2 t



(10.108)

∗ The method proposed by Puls (1928) was modified by the U.S. Bureau of Reclamation (1949), hence the name modified Puls method.

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In using this form of the continuity equation to route the hydrograph through a storage reservoir, it is important to recognize what is known and what is to be determined. Since the inflow hydrograph is generally a given, then I1 and I2 are known at every time step. For ungated spillways, orifice-type outlets, and weir-type outlets, the discharge, O, can be expressed as function of the water-surface elevation (stage), Z, in the reservoir, and for any given reservoir topography the storage, S, in the reservoir can also be expressed as a function of Z. Hence the Z versus O relationship can be combined with the Z versus S relationship to yield a Z versus 2S/t + O relationship, as illustrated in Figure 10.10. This single relationship between Z and 2S/t + O is used to represent the performance of the combined reservoir and outlet structure. It is also apparent from Figure 10.10 that for any given value of 2S/t + O the corresponding values of Z, S, and O can be determined. Using the performance characteristics shown in Figure 10.10, the following procedure can be used to route the inflow hydrograph through a reservoir:

ww

Step 1: Substitute known values of I1 , I2 , and 2S1 /t − O1 into the left-hand side of Equation 10.108. This gives the value of 2S2 /t + O2 . Step 2: From the reservoir and outlet structure characteristics, determine the discharge, O2 , corresponding to the calculated value of 2S2 /t + O2 . Step 3: Subtract 2O2 from 2S2 /t + O2 to yield 2S2 /t − O2 at the end of this time step. Step 4: Repeat Steps 1 to 3 until the entire outflow hydrograph, O(t), is calculated.

w .E asy En g

In performing these computations, it is recommended to select a time step, t, such that there are a minimum of five or six points on the rising side of the inflow hydrograph, one of which coincides with the inflow peak. The storage-indication method is very similar to the modified Puls method, with the exception that the continuity equation is expressed in the form 

I1 + I 2 2



t +



O1 t S1 − 2



  O2 t = S2 + 2

ine eri n

(10.109)

rather than in the form given by Equation 10.108 (McCuen, 2005). The primary difference between the modified Puls method and the storage-indication method is that the modified Puls method is based on the relation between 2S/t + O and O, while the storage-indication method is based on the relation between S + Ot/2 and O; otherwise, the calculation procedures are almost identical. It is fairly common practice to refer to the modified Puls and storage-indication methods as being the same, although they are technically different. For irregular-shaped detention basins, the surface area, As , versus elevation, h, is determined from contour maps of the detention-basin site, and the storage-versus-elevation relation is subsequently calculated using the equation

g .n

Outflow, O

Reservoir characteristic

Storage, S

Stage, Z

Outlet structure characteristic

Stage, Z

FIGURE 10.10: Performance characteristics of reservoir and outlet structure

Stage, Z

As1 + As2 S2 = S1 + (h2 − h1 ) 2

et

(10.110)

Combined characteristic

2S +O ∆t

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Fundamentals of Surface-Water Hydrology II: Runoff

where S1 and As1 correspond to elevation h1 , and S2 and As2 correspond to elevation h2 . A more accurate relationship is the “frustrum-of-a-pyramid” formula proposed by Paine and Akan (2001) and given by h2 − h1 (As1 + As2 + As1 As2 ) 3

(10.111)

1

(10.112)

 1−N     N y − 1 2 exp exp ymax y max

(10.113)

S2 = S1 +

The difference between Equations 10.110 and 10.111 decreases as the height increment decreases. It has been shown (Mohammadzadeh-Habili et al., 2009) that many storageelevation and area-elevation curves can be closely approximated by the relations 

S = Smax 2 exp

ww

A = Amax







y ymax



N

− 1

y

where S and A are the storage and surface area of the reservoir at a vertical distance y above the zero-storage and zero-surface-area level, Smax and Amax are the maximum storage and maximum surface area of the reservoir corresponding to elevation ymax , and N is related to the characteristics of the storage reservoir according to the relation

w .E asy En g

N = 2(ln 2)

Smax Amax ymax

(10.114)

In cases where Equations 10.112 and 10.113 adequately describe the reservoir storage, which has been shown to commonly be the case, using these analytic relationships greatly facilitates the routing of runoff hydrographs through storage reservoirs.

EXAMPLE 10.16

ine eri n

A stormwater-detention basin is estimated to have the following storage characteristics: Stage (m)

Storage (m3 )

5.0 5.5 6.0 6.5 7.0 7.5 8.0

0 694 1525 2507 3652 4973 6484

g .n

et

The discharge weir from the detention basin has a crest elevation of 5.5 m, and the weir discharge, Q, is given by 3 Q = 1.83H 2 where Q is in m3 /s and H is the height of the water surface above the crest of the weir in meters. The catchment runoff hydrograph is given by Time (min)

0

30

60

90

120

150

180

210

240

270

300

330

360

390

Runoff (m3 /s)

0

2.4

5.6

3.4

2.8

2.4

2.2

1.8

1.5

1.2

1.0

0.56

0.34

0

If the prestorm stage in the detention basin is 5.0 m, estimate the discharge hydrograph from the detention basin.

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523

Solution This problem requires that the runoff hydrograph be routed through the detention basin. Using t = 30 min, the storage and outflow characteristics can be put in the convenient tabular form: Stage (m)

S (m3 )

O (m3 /s)

2S/t + O (m3 /s)

5.0 5.5 6.0 6.5 7.0 7.5 8.0

0 694 1525 2507 3652 4973 6484

0 0 0.647 1.83 3.36 5.18 7.23

0 0.771 2.34 4.62 7.42 10.7 14.4

and the computations in the routing procedure are summarized in the following table:

ww

(1) Time (min)

(2) I (m3 /s)

(3) 2S/t − O (m3 /s)

(4) 2S/t + O (m3 /s)

(5) O (m3 /s)

0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480

0 2.4 5.6 3.4 2.8 2.4 2.2 1.8 1.5 1.2 1.0 0.56 0.34 0 0 0 0

0 1.04 0.52 0.46 0.78 0.84 0.88 0.94 0.98 1.00 1.02 1.04 0.98 0.86 0.78 0.78 0.78

0 2.4 9.04 9.52 6.66 5.98 5.44 4.88 4.24 3.68 3.20 2.58 1.94 1.32 0.86 0.78 0.78

0 0.68 4.26 4.53 2.94 2.57 2.28 1.97 1.63 1.34 1.09 0.77 0.48 0.23 0.04 0.00 0.00

w .E asy En g

ine eri n

g .n

et

The sequence of computations begins with tabulating the inflow hydrograph in Columns 1 and 2, and the first row of the table (t = 0 min) is given by the initial conditions. At t = 30 min, I1 = 0 m3 /s, I2 = 2.4 m3 /s, 2S1 /t − O1 = 0 m3 /s, and Equation 10.108 gives   2S2 2S1 + O2 = (I1 + I2 ) + − O1 = 0 + 2.4 + 0 = 2.4 m3 /s t t This value of 2S/t + O is written in Column 4 (at t = 30 min), the corresponding outflow, O, from the detention basin is interpolated from the reservoir properties as 0.68 m3 /s, and this value is written in Column 5 (at t = 30 min), and the corresponding value of 2S/t − O is given by   2S 2S − O= + O − 2O = 2.4 − 2(0.68) = 1.04 m3 /s t t which is written in Column 3 (at t = 30 min). The variables at t = 30 min are now all known, and form a new set of initial conditions to repeat the computation procedure for subsequent time steps. The outflow hydrograph, given in Column 5, indicates a peak discharge of 4.53 m3 /s from the detention basin. This is a reduction of 19% from the peak catchment runoff rate (= basin inflow rate) of 5.6 m3 /s.

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It is interesting to note that the peak outflow rate from the reservoir must occur at the intersection of the inflow and outflow hydrographs, since before this point the inflow rate exceeds the outflow rate and the water level in the reservoir must be increasing; while beyond this point the outflow rate exceeds the inflow rate and the water level in the reservoir must be decreasing. The probability that the peak outflow rate from a detention reservoir exceeds the target value for preventing downstream flooding is sometimes called the hydrologic risk (Paik, 2008). This risk is typically associated with uncertainties in the inflow hydrograph, shape of the reservoir, and in models used for reservoir routing and calculating outlet discharge. 10.6.1.2

ww

Muskingum method

The Muskingum method for flood routing was originally developed by McCarthy (1938) for flood-control studies in the Muskingum River basin in eastern Ohio. It is primarily used for routing flows in drainage channels, including rivers and streams, and is the most widely used method of hydrologic stream-channel routing. The Muskingum method approximates the storage volume in a channel by a combination of prism storage and wedge storage, as illustrated in Figure 10.11 for the case in which the inflow rate exceeds the outflow rate. The prism storage is a volume of constant cross section corresponding to uniform flow in a prismatic channel; the wedge storage is generated by the passage of the flow hydrograph. A negative wedge is produced when the outflow rate exceeds the inflow rate, and it occurs as the water level recedes in the channel. Assuming that the flow area is directly proportional to the flow rate in the channel, the volume of prism storage can be expressed as KO, where K [T] is the travel time through the reach and O [L3 T−1 ] is the flow rate through the prism. The wedge storage can therefore be approximated by KX(I − O), where X is a weighting factor [dimensionless] in the range 0 … X … 0.5 and I is the upstream inflow rate. For reservoir-type storage, X = 0; for a full wedge, X = 0.5. The total storage, S [L3 ], between the inflow and outflow sections is therefore given by S = KO + KX(I − O) (10.115)

w .E asy En g or

ine eri n

S = K[XI + (1 − X)O]

(10.116)

Applying Equation 10.116 at time increments of t, the storage, S, in the channel between the inflow and outflow sections at times jt and (j + 1)t can be written as Sj = K[XIj + (1 − X)Oj ] and

g .n

Sj+1 = K[XIj+1 + (1 − X)Oj+1 ] respectively, and the change in storage over t is therefore given by   Sj+1 − Sj = K [XIj+1 + (1 − X)Oj+1 ] − [XIj + (1 − X)Oj ] FIGURE 10.11: Muskingum storage approximation

I –O Total flow = I

et

(10.117)

(10.118)

(10.119)

Wedge storage KX (I –O)

O

Channel

Prism storage KO Total flow = O

O

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525

The discretized form of the continuity equation, Equation 10.107, can be written as Sj+1 − Sj =

(Oj + Oj+1 ) (Ij + Ij+1 ) t − t 2 2

(10.120)

Combining Equations 10.119 and 10.120 yields the routing expression (10.121)

Oj+1 = C1 Ij+1 + C2 Ij + C3 Oj where C1 , C2 , and C3 are given by

ww

C1 =

t − 2KX 2K(1 − X) + t

(10.122)

C2 =

t + 2KX 2K(1 − X) + t

(10.123)

C3 =

2K(1 − X) − t 2K(1 − X) + t

(10.124)

w .E asy En g and it is apparent that

C1 + C2 + C3 = 1

(10.125)

The routing equation, Equation 10.121, is applied to a given inflow hydrograph, Ij (j = 1, J), and initial outflow, O1 , to calculate the outflow hydrograph, Oj (j = 2, J), at a downstream section. The constants in the routing equation, C1 , C2 , and C3 are expressed in terms of the routing time step, t, and the channel parameters K and X. Ideally, subreaches selected for channel routing should be such that the travel time, K, through each subreach is equal to the time step, t. If this is not possible, then to avoid negative flows, t should be selected such that (Hjelmfelt, 1985) 2KX … t … 2K(1 − X) (10.126)

ine eri n

where this condition ensures that the routing coefficients C1 to C3 are all positive. It is frequently recommended that t be assigned any convenient value between K/3 and K (ASCE, 1996b; Viessman and Lewis, 2003).

EXAMPLE 10.17

The flow hydrograph at a channel section is given by: Time (min)

Flow (m3 /s)

Time (min)

Flow (m3 /s)

0 30 60 90 120 150 180 210 240

10.0 10.0 25.0 45.0 31.3 27.5 25.0 23.8 21.3

270 300 330 360 390 420 450 480 —

19.4 17.5 16.3 13.5 12.1 10.0 10.0 10.0 —

g .n

et

Use the Muskingum method to estimate the hydrograph 1200 m downstream from the channel section. Assume that X = 0.2 and K = 40 min. Solution In accordance with Equation 10.126, select t such that 2KX … t … 2K(1 − X) ¡* 2(40)(0.2) … t … 2(40)(1 − 0.2) ¡* 16 min … t … 64 min and it is frequently recommended that

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff K … t … K 3 or

40 min … t … 40 min ¡* 13.3 min … t … 40 min 3 Taking t = 30 min, the Muskingum constants are given by Equations 10.122 to 10.124 as

ww

C1 =

t − 2KX 30 − 2(40)(0.2) = = 0.149 2K(1 − X) + t 2(40)(1 − 0.2) + 30

C2 =

30 + 2(40)(0.2) t + 2KX = = 0.489 2K(1 − X) + t 2(40)(1 − 0.2) + 30

C3 =

2(40)(1 − 0.2) − 30 2K(1 − X) − t = = 0.362 2K(1 − X) + t 2(40)(1 − 0.2) + 30

These results can be verified by taking C1 + C2 + C3 = 0.149 + 0.489 + 0.362 = 1. The Muskingum routing equation, Equation 10.121, is therefore given by Oj+1 = C1 Ij+1 + C2 Ij + C3 Oj = 0.149Ij+1 + 0.489Ij + 0.362Oj

w .E asy En g

This routing equation is applied repeatedly to the given inflow hydrograph, and the results are as follows: (1) Time

(2) I

(3) O

(1) Time

(2) I

(3) O

(min)

(m3 /s)

(m3 /s)

(min)

(m3 /s)

(m3 /s)

0 30 60 90 120 150 180 210 240 270 300

10.0 10.0 25.0 45.0 31.3 27.5 25.0 23.8 21.3 19.4 17.5

10.0 10.0 12.2 23.4 35.1 32.1 28.8 26.2 24.3 22.1 20.1

330 360 390 420 450 480 510 540 570 600 —

16.3 13.5 12.1 10.0 10.0 10.0 10.0 10.0 10.0 10.0 —

18.3 16.6 14.4 12.6 11.0 10.3 10.1 10.1 10.0 10.0 —

ine eri n

g .n

et

The computations begin with the inflow hydrograph in Columns 1 and 2 and the initial outflow rate, O, at t = 0 min. At t = 30 min, I1 = 10 m3 /s, I2 = 10 m3 /s, O1 = 10 m3 /s, and the Muskingum method (for j = 1) gives the outflow rate, O2 , at t = 30 min as O2 = 0.149I2 + 0.489I1 + 0.362O1 = 0.149(10) + 0.489(10) + 0.362(10) = 10 m3 /s This procedure is repeated for subsequent times, and the outflow hydrograph is shown in Column 3.

The proportionality factor, K, is a measure of the time of travel through the channel reach, and X in natural streams is typically in the range of 0.2–0.3, with mean values near 0.2 to 0.25 (Chow et al., 1988; Seybert, 2006). Experience has shown that for channels with mild slopes and over-bank flow, the parameter X will approach 0.0; while for steeper streams, with well-defined channels that do not have flows going out of the bank, X will be closer to 0.5 (Hydrologic Engineering Center, 2000). Values of X in excess of 0.5 produce hydrograph amplification, which is unrealistic. If measured inflow and outflow hydrographs are available for a channel reach, then K and X can be estimated by selecting the values of these parameters that given the best fit between the measured and predicted outflow hydrographs.

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Comparison between a computed and observed hydrograph is generally done using a goodness-of-fit index, which is typically called an objective function. Two of the most commonly used objective functions are the sum of absolute errors and the sum of squared residuals, defined by sum of absolute errors =

N     comp  − Oobs Oj j 

(10.127)

j=1

sum of squared residuals =

N 

comp

(Oj

2 − Oobs j )

(10.128)

j=1

comp

ww

where Oj and Oobs are the computed and observed outflows, respectively, at time step j j, and N is the number of time steps in the outflow hydrograph. The sum of absolute errors objective function gives equal weight to each difference between the computed and observed outflow, while the sum of squared residuals gives more weight to the larger differences. Since K is a measure of the travel time through the channel reach, K can be estimated by the observed time for the hydrograph peak to move through the channel reach. In the absence of measured data, K is usually taken as the estimated mean travel time between the inflow and outflow section and X is taken as 0.2. According to Chow et al. (1988), great accuracy in determining X may not be necessary, because the results of the Muskingum method are relatively insensitive to this parameter.

w .E asy En g EXAMPLE 10.18

Measured flows at an upstream and downstream section of a drainage channel are as follows:

Time (h) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

ine eri n

Upstream flow (m3 /s)

Downstream flow (m3 /s)

Time (h)

Upstream flow (m3 /s)

Downstream flow (m3 /s)

42.4 46.3 51.8 56.2 57.6 65.1 80.3 90.6 103.2

52.0 51.0 51.4 52.8 54.7 58.1 65.0 75.6 87.2

4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 —

81.5 78.3 65.8 59.6 56.3 50.8 53.1 50.7 —

93.3 85.3 77.5 68.8 61.7 56.1 52.8 51.1 —

g .n

et

Estimate the Muskingum coefficients X and K that should be used in routing flows through this section of the drainage channel. Solution From the given data: t = 0.5 h, and the Muskingum routing equation is given by the combination of Equations 10.121 to 10.124 as

Oj+1 =



     t + 2KX 2K(1 − X) − t t − 2KX Ij+1 + Ij + Oj (10.129) 2K(1 − X) + t 2K(1 − X) + t 2K(1 − X) + t

For given values of t, X, and K, Equation 10.129 is used to calculate the outflow hydrograph corresponding to the given inflow hydrograph, and the sum of absolute errors between the computed and

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff measured hydrograph is given by Equation 10.127. For various assumed values of X and K, the sum of absolute errors is given in the following table:

ww

(1)

(2)

X

K (h)

(3) Sum of absolute errors (m3 /s)

0.5 0.4 0.3 0.2 0.1 0.05 0.0 0.15 0.14 0.13

0.645 0.638 0.618 0.624 0.610 0.610 0.639 0.619 0.622 0.614

43.1 32.2 24.3 19.8 19.3 20.4 21.8 18.6 18.4 18.6

w .E asy En g

For each given value of X in Column 1, the value of K that minimizes the sum of absolute errors is given in Column 2, and the corresponding sum of absolute errors is given in Column 3. Based on the tabulated results, the values of X and K that minimize the sum of absolute errors are X = 0.14 and K = 0.622 h. The sum of squared residuals between a computed and measured hydrograph is given by Equation 10.128, and the sum of squared residuals for various values of X and K is given in the following table:

ine eri n

(1)

(2)

X

K (h)

0.5 0.4 0.3 0.2 0.1 0.05 0.0 0.07 0.08 0.06 0.15

0.597 0.619 0.631 0.636 0.636 0.635 0.633 0.636 0.636 0.636 0.636

(3) Sum of squared residuals (m6 /s2 ) 245.9 141.6 83.2 53.4 42.5 42.0 44.3 41.87 41.94 41.90 46.0

g .n

et

For each value of X in Column 1, the value of K that minimizes the sum of squared residuals is given in Column 2, and the corresponding sum of squared residuals is given in Column 3. Based on the tabulated results, the values of X and K that minimize the sum of squared residuals are X = 0.07 and K = 0.636 h. The optimal values of X and K based on the sum of absolute errors and the sum of squared residuals are different, and the computed outflow hydrographs for each optimal (X, K) combination are compared with the measured hydrograph in Figure 10.12. Clearly, both optimal (X, K) combinations give a good fit to the measured outflow hydrograph, with the value of X and K based on minimizing the sum of absolute errors (0.14, 0.622 h) giving a better fit to the hydrograph peak flow rate.

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100.0

Routing Models

529

Measured Hydrograph X ⫽ 0.07, K ⫽ 0.636 h X ⫽ 0.14, K ⫽ 0.622 h

90.0

Flow (m3/s)

80.0 70.0 60.0 50.0

ww

40.0 0

2

4

6

8

10

Time (h)

w .E asy En g

An alternative numerical procedure that is sometimes used for estimating K and X is to plot S versus XI + (1 − X)O for various values of X until this relationship is approximately linear in accordance with Equation 10.116, in which case K is estimated by the slope of the linear relationship. This approach, which is illustrated in Figure 10.13, identifies parameters that are most consistent with the basic assumptions of the Muskingum model and provides a direct indication of the validity of the Muskingum model. However, limitations of this approach are that it requires estimation of the channel storage between the inflow and outflow sections, and does not necessarily provide parameters that give the best fit between the measured and predicted outflow hydrographs.

ine eri n

Muskingum–Cunge method. In cases where flow measurements are not available, a method for estimating K and X was proposed by Cunge (1967), where K [T] is estimated by the relation x K= (10.130) c

g .n

where x is the distance between the inflow and outflow sections [L], and c is the wave celerity [LT−1 ], which can be taken as 5 c= v (10.131) 3

et

where v is the average velocity at the bankfull discharge [LT−1 ]. The coefficient 35 in Equation 10.131 is derived from the Manning equation applied to wide rectangular channels FIGURE 10.13: Estimation of parameters in the Muskingum model

Storage, S

X ≠ Xoptimal X = Xoptimal

S = K [XI + (1–X )O] slope = Koptimal

XI + (1–X )O

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Fundamentals of Surface-Water Hydrology II: Runoff

(Dingman, 1984); in triangular channels the coefficient is 34 , and in wide parabolic channels it is 11 9 (Viessman and Lewis, 2003). The value of X [dimensionless] suggested by Cunge (1967) is   q0 1 1 − (10.132) X= 2 S0 c x

ww

where q0 is the flow rate per unit top width [L2 T−1 ], calculated at the average flow rate (midway between the base flow and the peak flow of the inflow hydrograph) and S0 is the slope of the channel [dimensionless]. In cases where the Muskingum method is applied using values of K and X estimated from Equations 10.130 to 10.132, the approach is referred to as the Muskingum–Cunge method. The accuracy of this method is derived from the fact that taking K and X as given by Equations 10.130 and 10.132 makes the Muskingum equation equal to the finite-difference form of the diffusion-wave formulation of the momentum equation, which is in the class of more accurate hydraulic-routing methods. When using the Muskingum–Cunge method, the values of x and t should be selected to assure that the flood-wave details are properly routed. Nominally, the time to the peak inflow rate is broken into 5 or 10 time steps, t. To give both temporal and spatial resolution, the total reach length, L, can be divided into several increments of length x, and outflow from each is treated as inflow to the next.

w .E asy En g

Variable-parameter Muskingum methods. The parameters K and X of the Muskingum– Cunge method remain constant while routing a given flood wave and are therefore incapable of capturing the nonlinearity of the flood-wave propagation process. To overcome this limitation, the variable-parameter Muskingum–Cunge method (VPMC) is sometimes used in which the parameters K and X vary at every routing time step. In current practice, K and X are evaluated using the three-point and four-point averaging schemes given by Ponce and Chaganti (1994). Another variation of the Muskingum–Cunge method is the variable parameter Muskingum discharge hydrograph method (VMPD) in which K and X are given by

ine eri n

x c0   q0 1 1 − X= 2 S0 c0 x K=

(10.133)

g .n

(10.134)

where c0 and q0 are the wave celerity [LT−1 ] and flow rate per unit top width [L2 T−1 ] at a downstream section where the flow is relatively steady, and S0 is the slope of the channel [dimensionless]. A shortcoming of the VPMC method is that mass is not conserved and, in this respect, the VMPD method demonstrates better performance (Perumal and Sahoo, 2008).

et

Three-parameter Muskingum methods. The Muskingum model with parameters K and X is sometimes called the two-parameter Muskingum model. The less-frequently-used threeparameter Muskingum models were originally suggested by Gill (1978) and are based on one of the following storage functions S = K[XI m + (1 − X)Om ] S = K[XI + (1 − X)O]

m

(10.135) (10.136)

which have the additional parameter m [dimensionless] used as an exponent. The additional parameter presumably improves the prediction of the nonlinear relation between the accumulated storage and weighted flow rates. Calibration procedures for finding optimal values of K, X, and m are more difficult than in the two-parameter case, and several numerical methods have been proposed to estimate these parameters from observed hydrographs (e.g., Geem, 2006; Das, 2007c).

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10.6.2

Routing Models

531

Hydraulic Routing

In cases where there are significant backwater effects and where the channel is either very steep or very flat, dynamic effects may be significant and hydraulic routing is preferred over hydrologic routing. In hydraulic routing, the one-dimensional continuity and momentum equations are solved along the channel continuum. This approach is in contrast to hydrologic routing, which is based on the finite-difference approximation to the continuity equation plus a second equation that relates the channel storage to the flow rate. The one-dimensional continuity and momentum equations in open channels, first presented by Barre de Saint-Venant (1871), are commonly called the Saint-Venant equations, and can be written as ⭸Q ⭸A + =0 ⭸x ⭸t

ww

1 ⭸Q 1 ⭸ + A ⭸t A ⭸x



Q2 A



+ g

(10.137)

(1-D continuity)

⭸y − g(S0 − Sf ) = 0 ⭸x

(1-D momentum)

(10.138)

where Q is the volumetric flow rate [L3 T−1 ], x is the distance along the channel [L], t is time [T], A is the cross-sectional flow area [L2 ], g is gravity [LT−2 ], y is the depth of flow [L], S0 is the slope of the channel [dimensionless], and Sf is the slope of the energy grade line [dimensionless], which can be estimated using the Manning equation as

w .E asy En g

Sf =

n2 |Q|Q

(10.139)

4

R 3 A2

where n is the Manning roughness coefficient [dimensionless], and R is the hydraulic radius [L]. The assumptions inherent in the Saint-Venant equations are: (1) The flow is onedimensional (i.e., the depth and velocity vary only in the longitudinal direction of the channel); (2) the flow is gradually varied along the channel (i.e., the vertical pressure distribution is hydrostatic); (3) the longitudinal axis of the channel is straight; (4) the bottom slope of the channel is small and the channel bed is fixed (i.e., the effects of scour and deposition are negligible); and (5) the friction coefficients for steady-uniform flow are applicable. With the exception of a few simple cases, the Saint-Venant equations, given by Equations 10.137 and 10.138, cannot be solved analytically; they are usually solved with numerical models that use either implicit or explicit finite-difference algorithms or the method of characteristics (Amein, 1966). In many cases, some terms in the momentum equation are small compared with others in the same equation, and the momentum equation can be simplified by neglecting some of these terms. When the full momentum equation is used, this is commonly called the dynamic model and is given by Equation 10.138. The first two terms in the dynamic model (Equation 10.138) are the local and convective acceleration, also called the inertial terms, and neglecting these terms leads to the following diffusion model: g

ine eri n

⭸y − g(S0 − Sf ) = 0 ⭸x

g .n

(diffusion model)

et

(10.140)

The reason Equation 10.140 is called the “diffusion model” is made apparent when Equation 10.140 is combined with the continuity equation, Equation 10.137 (eliminating y), to yield ⭸Q ⭸Q ⭸2 Q (10.141) + C(Q) = D(Q) 2 ⭸t ⭸x ⭸x where C(Q) is an advective velocity, and D(Q) is a hydraulic diffusivity (Chow et al., 1988). Equation 10.141 is the classical form of the advection-diffusion equation, and describes a flow wave, Q(x, t), that gradually diffuses as it is advected along the channel. For wide rectangular channels in which the slope of the energy grade line is approximately equal to the slope of the channel, C(Q) and D(Q) can be estimated by the relations

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Fundamentals of Surface-Water Hydrology II: Runoff

C=

1 , mαQm−1

with

⎛

D=

2 3

Q 2BS0

(10.142)

⎞m

⎜ nP ⎟ α=⎝ 1 ⎠ S02

(10.143)

where m is the kinematic-wave ratio (= 3/5 for the Manning equation), B is the width of the channel, and P is the wetted perimeter. In cases where both the inertial terms and spatial variations in depth-of-flow are small relative to other terms, the momentum equation leads to the following kinematic model: (S0 − Sf ) = 0

ww

(kinematic model)

(10.144)

which includes only gravity and flow-resistance terms. The reason Equation 10.144 is called the “kinematic model” is made apparent when Equation 10.144 is combined with the continuity equation, Equation 10.137 (eliminating y), to yield

w .E asy En g

⭸Q ⭸Q + C(Q) =0 ⭸t ⭸x

(10.145)

where C(Q) is an advective velocity. Equation 10.145 describes a flow wave, Q(x, t), that is advected along the channel but does not diffuse. The kinematic model is frequently used for overland-flow routing. Both the diffusion model (Equation 10.141) and the kinematic model (Equation 10.145) are nonlinear equations in the flow rate, Q, and are commonly solved numerically. In applying these numerical solutions it is important to verify that mass and momentum are being conserved within acceptable bounds (Ga¸siorowski and Szymkiewicz, 2007).

EXAMPLE 10.19

ine eri n

Write a simple finite-difference model for solving the Saint-Venant equations.

g .n

Solution The dependent variables in the Saint-Venant equations are the flow rate, Q, and the depth of flow, y; and the independent variables are the distance along the channel, x, and the time, t. In a finite-difference approximation, the solution is calculated at discrete intervals, x, along the x-axis, and at discrete intervals, t, in time. Denoting the distance index by i and the time index by j, the following convention for specifying variables at index locations is adopted: Qi,j = Q(ix, jt) yi,j = y(ix, jt)

et

Ai,j = A(yi,j ) where Ai,j is the flow area corresponding to yi,j . The (explicit) finite-difference approximation to the one-dimensional continuity equation (Equation 10.137) can be written as Ai,j+1 − Ai,j Qi+1,j − Qi,j + =0 x t and the (explicit) finite-difference approximation to the one-dimensional momentum equation (Equation 10.138) can be written as ⎞ ⎛ 2 Q2i,j − yi,j y 1 1 ⎝ Qi+1,j 1 Qi,j+1 − Qi,j ⎠ + g i+1,j + − g(S0 − Si,j ) = 0 − Ai,j t Ai,j x Ai+1,j Ai,j x

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where Si,j is the friction slope calculated using the Manning equation with Qi,j and yi,j . These finitedifference equations, along with given initial and boundary conditions, can be used to solve for Q and y at the discrete points ix and jt along the x- and t-axes, respectively. This numerical scheme requires some constraints on the choice of the discrete intervals x and t in order to produce accurate and stable solutions (Ames, 1977).

10.7

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Water-Quality Models

Pollutants in stormwater runoff originate from a variety of sources, including oil from motor vehicles, suspended sediment from construction activities, chemicals from lawns, and fecal droppings from animals. Urban areas are of special concern with respect to their impact on water quality. While urban areas might be small compared with agricultural areas, they produce high concentrations of pollutants per unit area, and they lack the natural buffering that often exists in agricultural areas (McCuen, 2005). Field studies have demonstrated that stormwater runoff is frequently a major source of pollution in receiving waters, with pathogenic microorganisms and some heavy metals being mostly responsible for impairments of receiving streams in the United States (USEPA, 2012). Pretreatment of stormwater runoff prior to discharge into receiving waters is usually required by local regulations. The water quality in storm sewers and other urban drainage conduits is influenced by both runoff-related sources and nonstormwater sources. Nonstormwater sources of pollution in storm sewers include illicit connections, interactions with sanitary sewers, improper disposal of hazardous substances, spills, malfunctioning septic tanks, and infiltration of contaminated water into the storm sewers. The removal of nonstormwater sources of pollutants can result in a dramatic improvement in the quality of water discharged from storm sewers. Runoff-related sources of pollution in storm sewers mostly originate from land surfaces, where pollutants accumulate much more rapidly on impervious areas than they do on pervious areas, and the primary areas of accumulation are usually streets and gutters. Initial planning estimates of pollutant loads contained in stormwater runoff can be estimated using either event-mean concentrations or regression equations. The most widely used regression equations for pollutant loads are incorporated in the USGS and EPA models.

w .E asy En g 10.7.1

Event-Mean Concentrations

ine eri n

The majority of stormwater pollutants fall into the following categories: solids, oxygendemanding substances, nutrients, and heavy metals. Among the toxic heavy metals detected in stormwater runoff, lead, zinc, and copper are the most abundant and are detected most frequently. Typical concentrations of various pollutants in urban stormwater runoff are shown in Table 10.13. In many cases, a typical runoff concentration is assigned to a pollutant and this is called the event-mean concentration (EMC). Most pollutants in stormwater runoff have a strong affinity to suspended solids (SS), and the removal of SS often results in the removal of many other pollutants found in urban stormwater. According to ASCE (1992), for general-purpose planning, the concentrations of pollutants in runoff from large residential and commercial areas can be assumed to be in the range of the concentrations shown in Table 10.13, with central business districts usually having the highest pollutant loadings per unit area. Whalen and Cullum (1988) compared the quality of runoff from residential, commercial, light industrial, roadway, and mixed-urban land uses. Their results indicated that higher nutrient loads are generated by residential land uses, metal contamination is more widespread from commercial and roadway areas, and there are no discernible trends for suspended solids in runoff as a function of land use. Heavy metals found in urban runoff are typically 10–1000 times the concentration of metals found in sanitary sewage. The quality of urban runoff also depends on the stage of development. During the initial phase of urbanization, the dominant source of pollution will be sediments from bare-soil areas at construction sites. During the intermediate phase of urbanization, sediments from construction sites will decline, but sediments from stream-bank erosion will increase because of the increased runoff rate and volume. During the mature phase of urbanization (when the stream channels have stabilized and there is limited new construction), the primary source of pollution will be from washoff of accumulated deposits on impervious surfaces.

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ww

w .E asy En g

TABLE 10.13: Typical Water Quality of Runoff from Residential and Commercial Areas

Category Sediment

Sources Eroding rock, building sites, streets, lawns

Deleterious effects Clogs waterways, smothers bottom-living aquatic organisms and increases turbidity

Oxygen-demanding substrates

Decaying organic matter

Consume oxygen in water, sometimes creating an oxygen deficit that leads to fish deaths

Biochemical oxygen demand (BOD) Chemical oxygen demand (COD)

12–19 mg/L 82–178 mg/L

Nutrients

Nitrogen and phosphorus from landscape runoff, atmospheric deposition, and faulty septic tanks

Cause unwanted and uncontrolled growth of algae and aquatic weeds

Total phosphorous, as P (TP) Soluble phosphorous, as P (SP) Total Kjeldahl nitrogen, as N (TKN) Nitrite + Nitrate, as N (NO2+NO3)

0.42–0.88 mg/L 0.15–0.28 mg/L 1.90–4.18 mg/L 0.86–2.2 mg/L

Heavy metals

Vehicles, highway materials, atmospheric deposition, industry

Can disrupt the reproduction of fish and shellfish and accumulate in fish tissues

Total copper (Cu) Total lead (Pb) Total zinc (Zn)

43–118 μg/L 182–443 μg/L 202–633 μg/L

Source: USEPA (1983a).

Measure Total suspended solids (TSS)

ine eri n

Typical concentrations 180–548 mg/L

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EXAMPLE 10.20 Plans are being considered to rezone a 100-ha rural area for commercial and residential use. Annual rainfall in the area is 150 cm, the existing (rural) land is approximately 5% impervious, and the rezoned land is expected to be 40% impervious. Estimate the current and future amounts of suspended solids in the runoff. Solution Since the annual rainfall is 150 cm and the existing land is 5% impervious, existing annual runoff = 0.05(150 cm) = 7.5 cm and since the rezoned land is 40% impervious, expected annual runoff = 0.40(150 cm) = 60 cm

ww

According to Table 10.13, the EMC of suspended solids for residential and commercial areas is typically in the range of 180–548 mg/L with a midrange value of 364 mg/L. In the absence of any local EMC data, the EMC of suspended solids can be taken as 364 mg/L (= 0.364 kg/m3 ). Since the area of the catchment is 100 ha (= 106 m2 ), the total suspended solids in 7.5 cm (= 0.075 m) of annual runoff from the rural land is given by

w .E asy En g

existing suspended solids = (0.075 m)(106 m2 )(0.364 kg/m3 ) = 27,300 kg/year

and the suspended solids contained in 60 cm (0.60 m) of annual runoff from the rezoned/developed land is given by expected suspended solids = (0.60 m)(106 m2 )(0.364 kg/m3 ) = 218,400 kg/year

These results indicate that an order-of-magnitude increase in suspended solids contained in the surface runoff is to be expected, increasing from approximately 27,300 kg/year to 218,400 kg/year. An adequate stormwater-management system should be put in place to remove some of the anticipated suspended sediment load prior the runoff being discharged from the area.

10.7.2

Regression Equations

ine eri n

Regression equations are useful for estimating the annual pollutant loads on receiving waterbodies from stormwater runoff. The most widely used regression equations are those developed by USGS and EPA. 10.7.2.1

USGS model

g .n

et

The U.S. Geological Survey analyzed the pollutant loads resulting from 2813 storms at 173 urban stations in 30 metropolitan areas in the United States and developed empirical equations to estimate the annual pollutant loads in terms of the rainfall and catchment characteristics (Driver and Tasker, 1988; 1990). The USGS regression equations for annual load have the form Y = 0.454(N)(BCF)10[a+b

√

(DA)+c(IA)+d(MAR)+e(MJT)+f (X2)]

(10.146)

where Y is the annual pollutant load [kg] for the pollutants listed in Table 10.14, N is the average number of storms per year, BCF is a bias correction factor [dimensionless], DA is the total contributing drainage area [ha], IA is the impervious area as a percentage of the total contributing area [%], MAR is the mean annual rainfall [cm], MJT is the mean minimum January temperature [◦ C], and X2 is an indicator variable that is equal to 1.0 if commercial land use plus industrial land use exceeds 75% of the total contributing drainage area and is zero otherwise. The regression constants in Equation 10.146 depend on the type of pollutant and are given in Table 10.15. A storm is defined as a rainfall event with at least 1.3 mm (0.05 in.) of rain, and individual storms are separated by at least 6 consecutive hours

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff TABLE 10.14: Pollutants in USGS Formula

Pollutant

Y COD SS DS TN AN TP DP CU PB ZN

ww

Chemical oxygen demand Total suspended solids Dissolved solids Total nitrogen Total ammonia plus organic nitrogen Total phosphorus Dissolved phosphorus Total recoverable copper Total recoverable lead Total recoverable zinc

TABLE 10.15: Regression Constants for USGS Pollutant Load Equation

Y

a

b

c

d

e

f

BCF

COD SS DS TN AN TP DP CU PB ZN

1.1174 0.5926 1.1025 −0.2433 −1.4002 −2.0700 −1.3661 −1.9336 −1.9679 −1.6302

0.1427 0.0988 0.1583 0.1018 0.1002 0.1294 0.0867 0.1136 0.1183 0.1267

0.0051 — — 0.0061 0.0064 — — — 0.0070 0.0072

— 0.0104 — — 0.00890 0.00921 — — 0.00504 —

— −0.0535 −0.0418 — −0.0378 −0.0383 — −0.0254 — —

— — — −0.4442 −0.4345 — — — — —

1.298 1.521 1.251 1.345 1.277 1.314 1.469 1.403 1.365 1.322

w .E asy En g Source: Driver and Tasker (1990).

ine eri n

of no rainfall. In accordance with this definition, the mean number of storms per year in several cities within the United States is given in Table 10.16, where the cities have been rankordered by the number of storms. As in the case of most empirical relationships, application of Equation 10.146 is not recommended beyond the ranges of variables used in developing the equation, which are given in Table 10.17. TABLE 10.16: Mean Number of Storms per Year

City Miami, FL Cleveland, OH Bellevue, WA Seattle, WA Portland, OR Knoxville, TN Chicago, IL Boston, MA New York, NY New Orleans, LA Tampa, FL Winston-Salem, NC Philadelphia, PA St. Louis, MO

g .n

Mean number of storms

et

100 100 98 97 96 92 87 84 83 83 79 77 77 74

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537

TABLE 10.16: (Continued)

Houston, TX Denver, CO Austin, TX Dallas, TX San Francisco, CA Fresno, CA Los Angeles, CA Phoenix, AZ

70 57 54 53 46 41 32 31

Sources: Driver and Tasker (1988); McCuen (2005).

TABLE 10.17: Ranges of Variables Used in Developing USGS Equation

ww

Y

DA (ha)

IA (%)

MAR (cm)

MJT ◦C

COD SS DS TN AN TP DP CU PB ZN

4.9–183 4.9–183 5.2–117 4.9–215 4.9–183 4.9–215 5.2–183 3.6–215 4.9–215 4.9–215

4–100 4–100 19–99 4–100 4–100 4–100 5–99 6–99 4–100 13–100

21.3–157.5 21.3–125.4 26.0–95.5 30.0–157.5 21.3–157.5 21.3–157.5 21.3–117.3 21.3–157.5 21.3–157.5 21.3–157.5

−16.0–14.8 −16.0–10.1 −11.4–2.1 −16.0–14.8 −16.0–14.8 −16.0–14.8 −11.8–2.1 −9.3–14.8 −16.0–14.8 −11.4–14.8

w .E asy En g Source: Driver and Tasker (1990).

EXAMPLE 10.21

ine eri n

The phosphorus load from a 100-ha planned development in Fort Lauderdale, Florida, is to be assessed. The development will be 40% impervious, with 50% commercial and industrial use. There are typically 84 storms per year in Fort Lauderdale, the mean annual rainfall is 147 cm, and the mean minimum January temperature is 14.4◦ C. Estimate the annual load of total phosphorus contained in the runoff.

g .n

et

Solution From the given data: N = 84 storms, BCF = 1.314 (Table 10.15 for TP), DA = 100 ha, IA = 40%, MAR = 147 cm, MJT = 14.4◦ C, and X2 = 0 (since commercial plus industrial use is less than 75%). These variables are within the ranges given in Table 10.17, and therefore the USGS regression equation, Equation 10.146, can be used. The regression constants taken from Table 10.15 (for TP) are a = −2.0700, b = 0.1294, d = 0.00921, e = −0.0383, and c = f = 0. Substituting data into Equation 10.146 gives √ Y = 0.454(N)(BCF)10[a+b (DA)+c(IA)+d(MAR)+e(MJT)+f (X2)] √ = 0.454(84)(1.314)10[−2.0700+0.1294 (100)+0.00921(147)−0.0383(14.4)] = 53 kg/year The annual load of total phosphorus contained in the runoff from the planned development is expected to be on the order of 53 kg.

10.7.2.2

EPA model

The U.S. Environmental Protection Agency has also developed a set of empirical formulae that can be used to estimate the average annual pollutant loads in urban stormwater runoff

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff TABLE 10.18: Pollutant Loading Factor, α

Land use Residential Commercial Industrial Other

BOD5

SS

VS

PO4

N

0.799 3.200 1.210 0.113

16.3 22.2 29.1 2.7

9.4 14.0 14.3 2.6

0.0336 0.0757 0.0705 0.0099

0.131 0.296 0.277 0.060

Source: Heany et al. (1977).

(Heany et al., 1977). The empirical equation for urban areas having separate storm-sewer systems is given by Ms = 0.0442αPfs (10.147)

ww

where Ms is the amount of pollutant [kg] generated per hectare of land per year, α is a pollutant loading factor given in Table 10.18 for various pollutants (BOD5 = 5-day biochemical oxygen demand, SS = suspended solids, VS = volatile solids, PO4 = phosphate, and N = nitrogen), P is the precipitation [cm/year], f is a population density function, and s is a street-sweeping factor. The population density function, f , for residential areas is given by

w .E asy En g

f = 0.142 + 0.134D0.54

(10.148)

where D is the population density [persons per hectare]. For commercial and industrial areas, the population density function, f , is equal to 1.0; for other types of developed areas, such as parks, cemeteries, and schools, f is taken as equal to 0.142. The street-sweeping factor, s, depends on the sweeping interval, Ns (days); if Ns > 20 d, then s = 1.0, and if Ns … 20 d, then s is given by Ns s= (10.149) 20

ine eri n

Disposal of street-sweeping wastes may pose a problem because of possible high levels of lead, copper, zinc, and other wastes from automobile traffic. The average annual pollutant concentration can be derived from the annual pollutant load by dividing the annual pollutant load by the annual runoff. Heany et al. (1977) suggested that the annual runoff, R (cm), can be estimated using the formula 



I R = 0.15 + 0.75 100



P − 3.004d

g .n

0.5957

et

(10.150)

where I is the imperviousness of the catchment [%], P is the annual rainfall [cm], and d is the depression storage [cm], which can be estimated using the relation d = 0.64 − 0.476



I 100



(10.151)

EXAMPLE 10.22 A 100-ha residential development has a population density of 15 persons per hectare, streets are swept every 2 weeks, and the area is 40% impervious. The average annual rainfall is 147 cm. Estimate the annual load of phosphate (PO4 ) expected in the runoff. Solution From the given data: α = 0.0336 (Table 10.18: PO4 , Residential), P = 147 cm, D = 15 persons/ha, Ns = 14 d, Equation 10.148 gives f = 0.142 + 0.134D0.54 = 0.142 + 0.134(15)0.54 = 0.72

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539

and Equation 10.149 gives

14 Ns = = 0.7 20 20 Substituting these data into the EPA model, Equation 10.147, gives s=

Ms = 0.0442αPfs = 0.0442(0.0336)(147)(0.72)(0.7) = 0.11 kg/ha For a 100-ha development, the annual phosphate load is 100 ha * 0.11 kg/ha = 11 kg. The average concentration in the runoff is obtained by dividing the annual load (= 11 kg) by the annual runoff. Equation 10.151 gives the depression storage, d, as     40 I = 0.64 − 0.476 = 0.45 cm d = 0.64 − 0.476 100 100

ww

and Equation 10.150 gives the runoff, R, as  

 I R = 0.15 + 0.75 P − 3.004d 0.5957 100    40 = 0.15 + 0.75 (147) − 3.004(0.45)0.5957 100

w .E asy En g = 64 cm

Since the catchment area is 100 ha = 106 m2 , the volume, V, of annual runoff is 0.64 * 106 m3 and the average concentration, c, of PO4 is given by c=

Problems

11 kg

0.64 * 106 m3

10.1. Consider a catchment that has a runoff coefficient of 0.82, an average slope of 0.02, a Manning’s n of 0.03, and a runoff length of 50 m. If the catchment is located in a region with Type IA rainfall, and the 2-year 24-h rainfall is 80 mm, estimate the time of concentration in the sheet-flow regime. 10.2. A catchment with a grass surface has an average slope of 0.8%, and the distance from the catchment boundary to the outlet is 80 m. At the catchment location, the 2-year 24-hour rainfall is estimated as 85 mm. For a 30-min storm with an effective rainfall rate of 70 mm/h, estimate the time of concentration using: (a) the kinematicwave equation, (b) the NRCS method, (c) the Kirpich equation, (d) the Izzard equation, and (e) the Kerby equation. 10.3. What is the maximum flow distance that should be described by sheet flow? 10.4. Find α and m in the kinematic-wave model (Equation 10.4) corresponding to: (a) the Manning equation, and (b) the Darcy–Weisbach equation. 10.5. An asphalt pavement drains into a rectangular concrete channel. The catchment surface has an average slope of 1.0%, and the distance from the catchment boundary to the drain is 30 m. The drainage channel is 60 m long, 20 cm wide, 25 cm deep, and has a slope of 0.6%. For

= 1.7 * 10−5 kg/m3 = 0.017 mg/L = 17 µg/L

ine eri n

an effective rainfall rate of 50 mm/h, the flow rate in the channel is estimated to be 0.02 m3 /s. Estimate the time of concentration of the catchment. 10.6. Storm-event data for a 121-ha watershed are given in Table 10.19. Field experiments within the watershed have shown that losses are primarily by soil infiltration, which can be characterized by a constant INFILTRATION capacity. Use the data in Table 10.19 with any appropriate method of base-flow separation to estimate the following: (a) the average (constant) infiltration capacity and hence the predominant soil group within the watershed; (b) the average curve number of the watershed; (c) the time of concentration; and (d) describe practical uses of the parameters calculated in this problem. 10.7. The surface of a 2-ha catchment is characterized by a runoff coefficient of 0.5, a Manning’s n for overland flow of 0.25, an average overland-flow length of 60 m, and an average slope of 0.5%. Calculate the time of concentration using the kinematic-wave equation. The drainage channel is to be sized for the peak runoff rate resulting from a 10-year rainfall event, and the 10-year IDF curve is given by 150 i= (t + 8.96)0.78

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TABLE 10.19: Rainfall/Runoff Measurements

Hyetograph Time of day

Intensity (mm/h)

0857 74 0903 102 0906 19 0910 5 0916 122 0920

ww

49

0925 0930 0935

Hydrograph

21

Time of day

Flow (m3 /s)

0902 0912 0914 0915 0916 0918 0922 0926 0928 0929 0930 0931 0936 0937 0940 0942 0944 0945 0946 0947 0948 0949 0950 0954 1000 1006 1012 1018 1024 1030 1036 1048 1100

0.009 0.093 0.355 0.370 0.311 0.415 0.612 0.910 1.119 1.295 1.488 1.739 2.647 2.801 3.032 3.187 3.486 3.752 3.812 3.778 3.778 3.752 3.598 3.212 2.476 1.897 1.465 1.119 0.869 0.677 0.553 0.376 0.276

error is included, and which gives the lowest peak runoff rate when the error is included. If the 10% error occurs simultaneously in all the drainage parameters, what would be the design peak discharge at the inlet? 10.9. Explain why higher runoff coefficients should be used for storms with longer return periods. 10.10. An exfiltration trench serves a 2-ha residential catchment area that is 40% impervious, and the 10-year IDF curve for the region is given by i=

where i is the average rainfall intensity in mm/h and t is the storm duration in minutes. The 2-year 24-h rainfall amount for the area is 12 cm. Based on a site development map, the overland-flow length to the trench is approximately 100 m over asphalt pavement with an average slope of 0.7%. Estimate the peak flow rate to be accommodated by the exfiltration trench. If the exfiltration trench is designed for this capacity, what is the probability of the catchment area being flooded in any given year? 10.11. Suppose that the catchment described in Problem 10.7 contains 0.5 ha of impervious area that is directly connected to the storm sewer. If the runoff coefficient of the impervious area is 0.9, Manning’s n for overland flow on the impervious surface is 0.035, the average flow length is 30 m, and the average slope is 0.5%, estimate the peak runoff rate from the catchment. 10.12. The 10-year IDF curve for a 10-ha site in Cleveland, Ohio, is given by

w .E asy En g 3

where i is the average rainfall intensity in cm/h and t is the duration in minutes. The minimum time of concentration is 5 min. Determine the peak runoff rate. 10.8. A 20-ha townhouse development is to be drained by a single drainage inlet. The average runoff coefficient for the site is estimated to be 0.7, Manning’s n for overland flow is 0.25, the average overland-flow length is 100 m, and the average slope of the site toward the inlet is 0.6%. The site is located in an area with an IDF curve given by i=

1020 (t + 8.7)0.75

where i is the rainfall intensity in mm/h and t is the duration in minutes. What is the peak runoff rate expected at the drainage inlet? If an error of 10% is possible in each of the assumed site parameters, determine which parameter gives the highest peak runoff rate when the 10%

2029 (t + 7.24)0.73

ine eri n i=

1209 (t + 8.86)0.79

g .n

where i is the average rainfall intensity in mm/h and t is the rainfall duration in minutes. (a) Determine the peak runoff rate to a drainage structure located in a residential area with single-family housing where the catchment area is 0.5 ha and the time of concentration is 10 min. (b) If the available soil storage on the site is 8 cm, determine the curve number and the volume of runoff (in m3 ) for a 10-year 24-h storm. 10.13. A 4.2-km2 catchment with 0.5% pond area has a curve number of 79, a time of concentration of 3 h, and a 24-h Type II precipitation of 10 cm. Estimate the peak runoff rate. 10.14. A 1-km2 catchment with 3% pond area has a curve number of 70, a time of concentration of 1.7 h, and a 24-h Type I precipitation of 13 cm. Estimate the peak runoff rate. 10.15. A suburban residential catchment covers 1 ha in Atlanta, Georgia, and consists mostly of 0.4-ha lots. The catchment has an average land slope of 0.5%, a runoff distance of slightly less than 100 m, and consists mostly of sandy loam soil. (a) What range of times of concentration would you expect? (b) Contrast the 10-year

et

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Downloaded From : www.EasyEngineering.net Problems peak runoff rates predicted using the rational and TR-55 methods and assess the implication of this result on drainage design. 10.16. A 10-ha single-family residential development in Atlanta is estimated to have a curve number of 70 and a time of concentration of 15 min. There are no ponds or swamps on the site, and the 10-year IDF curve is given by 2029 i= (t + 7.24)0.73 where i is the rainfall intensity in mm/h and t is the storm duration in minutes. Estimate the peak runoff rate from the site, and determine the equivalent runoff coefficient that could be used with the rational method to estimate the peak runoff rate. 10.17. You are designing a drainage system for a residential site in Atlanta using a 10-year return period. The site has an area of 1 ha, a characteristic flow length of 100 m, an average slope of 1%, a characteristic Manning’s n of 0.25, and Type B soil. Drainage regulations require that the kinematic-wave equation be used to calculate the time of concentration, and you are allowed to use either the rational method or the TR-55 method to calculate the peak runoff rate. It is suggested that you use the rational method with a runoff coefficient of 0.4, which is typical of residential sites. What curve number would be required for both the rational method and the TR-55 method to give the same peak runoff rate? 10.18. A 2.79-km2 watershed located just west of Boston, Massachusetts, is estimated to have a time of concentration of 96 minutes. (a) If a previous investigation has estimated the curve number of the watershed as 55, estimate the peak runoff rate from a 10-year 24-h storm; and (b) comment on how the peak runoff rate from a shorter-duration storm with a 10-year return period might compare with your calculated value. 10.19. The 15-min unit hydrograph for a 2.1-km2 urban catchment is given by

ww

10.20. The 15-min unit hydrograph for a 2.1-km2 catchment is given in Problem 10.19. Determine the S-hydrograph, calculate the 50-min unit hydrograph for the catchment, and verify that your calculated unit hydrograph corresponds to a 1-cm rainfall excess. 10.21. The 15-min unit hydrograph for a 2.1-km2 catchment is given in Problem 10.19. Estimate the runoff resulting from the following 120-min storm:

Runoff (m3 /s)

Time (min)

Runoff (m3 /s)

0 30 60 90 120 150 180

0 1.4 3.2 1.5 1.2 1.1 1.0

210 240 270 300 330 360 390

0.66 0.49 0.36 0.28 0.25 0.17 0

(a) Verify that the unit hydrograph is consistent with a 1-cm rainfall excess; (b) estimate the runoff hydrograph for a 15-min rainfall excess of 2.8 cm; and (c) estimate the runoff hydrograph for a 30-min rainfall excess of 10.3 cm.

Time (min)

Rainfall (cm)

0–30 30–60 60–90 90–120

2.4 4.5 2.1 0.8

10.22. In a 315-km2 watershed the measured rainfall and resulting runoff in a small river leaving the catchment are as given in Table 10.20. Explain why the flow in the drainage channel is nonzero even when there is no rain. What is the name given to this no-rain flow? Separate the flow in the drainage channel into the component that results from rainfall (i.e., direct runoff) and the component that does not result from rainfall. Determine the unit hydrograph based on the given data. Compare the

w .E asy En g

Time (min)

541

TABLE 10.20: Measured Rainfall and Runoff

ine eri n Time (h)

Rainfall (cm)

0

River flow (m3 /s) 100

0.5

1

2.5

2

2.5

3

100

g .n 300 700

0.5 4

1000

5

800

6

600

7

400

8

300

9

200

10

100

11

100

et

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Fundamentals of Surface-Water Hydrology II: Runoff

total rainfall depth with the total (direct) runoff depth to determine the total losses during the storm. If this loss is distributed uniformly over the duration of the storm event (4 h), estimate the duration of rainfall excess to be associated with the unit hydrograph. Use this derived unit hydrograph to estimate the runoff hydrograph from the following storm event:

Time (h)

Effective rainfall (cm/h)

0–2 2–4 4–6 6–8

0.5 1.5 2.0 1.0

ww

10.23. Derive the 3-h Snyder unit hydrograph for a 100-km2 watershed where the main stream is 15 km long and the distance from the watershed outlet to the point on the stream nearest to the centroid of the watershed is 7 km. Assume Cp = 0.7 and Ct = 1.8.

and the time of concentration of the site is estimated to be 30 min. (a) From the design hyetograph, identify the 30-min interval with the highest rainfall intensity. State the maximum rainfall intensity. (b) Estimate the incremental runoff hydrograph produced by rainfall during the interval identified in Part (a). (c) Describe how you would calculate the runoff hydrograph for the entire storm. 10.29. A 75-ha catchment has the following time-area relation:

w .E asy En g

10.24. A 3-km2 catchment has an average curve number of 60, an average slope of 0.8%, and a flow length from the catchment boundary to the outlet of 1000 m. If the catchment is 40% impervious, determine the NRCS unit hydrograph for a 20-min storm. 10.25. A 16-km2 rural catchment is estimated to have a time of concentration of 14.4 h. Estimate the runoff hydrograph resulting from a storm with the following rainfall excess: Time (min)

Rainfall excess (cm)

1–30 30–60

2.9 1.5

10.26. Determine the approximate NRCS triangular unit hydrograph for the catchment described in Problem 10.24, and verify that it corresponds to a rainfall excess of 1 cm. 10.27. Show that any triangular unit hydrograph that has a peak runoff rate, Qp , in (m3 /s)/cm, given by Qp = 2.08

A Tp

must necessarily have a runoff duration equal to 2.67Tp , where Tp is the time to peak in hours, and A is the catchment area in km2 . 10.28. A 80-ha residential development is to contain 0.4-ha (1-ac) lots on 80% of the development, and surface runoff from the entire development is to be routed to a detention pond located in a park which takes up the remaining 20% of the development. The 25-year 24-h rainfall for the area is 25 cm, the area typically experiences Type II rainfall, the native soils are Type B,

Time (min)

Contributing area (ha)

0 5 10 15 20 25 30

0 2 7 19 38 68 75

Estimate the runoff hydrograph using the time-area model for a 30-min rainfall-excess distribution given by Time (min)

Average intensity (mm/h)

0–5 5–10 10–15 15–20 20–25 25–30

120 70 50 30 20 10

ine eri n

10.30. A catchment has an area of 80 km2 , a time of concentration of 7.0 h, and the Clark storage coefficient is estimated as 3.0 h. Estimate the 1-h unit hydrograph for the catchment. 10.31. A 2-ha catchment consists of pastures with an average slope of 0.5%. The width of the catchment is approximately 250 m and the depression storage is estimated to be 10 mm. Use the nonlinear-reservoir model to calculate the runoff from the following 20-min rainfall excess:

g .n

Time (min)

Effective rainfall (mm/h)

0–5 5–10 10–15 15–20

110 50 40 10

et

10.32. A 1-km2 catchment has a time of concentration of 25 min and is 45% impervious. Use the Santa Barbara Urban Hydrograph method to estimate the runoff hydrograph for the following rainfall event:

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Time (min)

Rainfall (mm/h)

Rainfall excess (mm/h)

0 10 20 30 40

0 50 200 103 52

0 0 130 91 11

Stage (m)

Storage (m3 )

8.0 8.5 9.0 9.5 10.0 10.5 11.0

0 1041 2288 3761 5478 7460 9726

543

Use a time increment of 10 min to calculate the runoff hydrograph. 10.33. A 3-km2 catchment has a time of concentration of 50 min and is 30% impervious.

The discharge weir from the detention basin has a crest elevation of 8.0 m, and the weir discharge, Q ( m3 /s), is given by

(a) Calculate the 10-min NRCS triangular unit hydrograph and use this result to estimate the 1-h unit hydrograph for the area. (b) The design rainfall distribution is a 1-h storm with an average intensity of 6 cm/h, the pervious area has a constant infiltration capacity of 1 cm/h and a depression storage of 4 mm, and the impervious area has a depression storage of 2 mm.

where h is the height of the water surface above the crest of the weir in meters. The catchment runoff hydrograph is given by:

ww

3

Q = 3.29 h 2

w .E asy En g

(i) What is the depth of runoff? (ii) Use the 1-h unit hydrograph calculated in Part (a) to estimate the runoff hydrograph. (iii) Estimate the runoff hydrograph using the Santa Barbara Urban Hydrograph method with 10-min time increments. (iv) Compare your results from Part (ii) and (iii) and state which hydrograph you would use to design the detention pond for the catchment.

10.34. A 5-km2 area is to be developed, and the development is to be 35% impervious and have a time of concentration of 1 h. The impervious area is to be distributed throughout the catchment and will not be directly connected to the outlet. The pervious area is estimated to have an infiltration capacity of 200 mm/h and the 10-year IDF curve of the area is

i=

3000 (t + 9.30)0.80

where i is the average rainfall intensity in mm/h and t is the storm duration in minutes. For a storm duration of 1 h, compare the 10-year runoff hydrographs calculated using the NRCS unit hydrograph and Santa Barbara Urban Hydrograph methods. Use time increments of 10 minutes. 10.35. A stormwater detention basin has the following storage characteristics:

Time (min)

Runoff (m3 /s)

Time (min)

Runoff (m3 /s)

0 30 60 90 120 150 180

0 3.6 8.4 5.1 4.2 3.6 3.3

210 240 270 300 330 360 390

2.7 2.3 1.8 1.5 0.84 0.51 0

ine eri n

If the prestorm stage in the detention basin is 8.0 m, estimate the discharge hydrograph from the detention basin. 10.36. The flow hydrograph at a channel section is given by: Time (min)

Flow (m3 /s)

0 30 60 90 120 150 180

0.0 12.5 22.1 15.4 13.6 12.4 11.7

g .n Time (min)

Flow (m3 /s)

210 240 270 300 330 360 390

10.8 9.9 8.4 8.1 7.5 4.2 0.0

et

Use the Muskingum method to estimate the hydrograph 1000 m downstream from the channel section. Assume that X = 0.3 and K = 35 min. 10.37. Measured flows at an upstream and downstream section of a river are given in Table 10.21: Estimate the Muskingum coefficients X and K that should be used in routing flows through this section of the river. 10.38. Write a simple finite-difference model to solve the kinematic-wave equation for channel routing. [Hint: The kinematic-wave equation is obtained by combining Equations 10.137 and 10.144.]

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Fundamentals of Surface-Water Hydrology II: Runoff TABLE 10.21

Time (min)

ww

0 30 60 90 120 150 180 210 240 270 300

Upstream flow (m3 /s)

Downstream flow (m3 /s)

Time (min)

Upstream flow (m3 /s)

Downstream flow (m3 /s)

10.0 10.0 25.0 45.0 31.3 27.5 25.0 23.8 21.3 19.4 17.5

10.0 10.0 12.2 23.4 35.1 32.1 28.8 26.2 24.3 22.1 20.1

330 360 390 420 450 480 510 540 570 600 —

16.3 13.5 12.1 10.0 10.0 10.0 10.0 10.0 10.0 10.0 —

18.3 16.6 14.4 12.6 11.0 10.3 10.1 10.1 10.0 10.0 —

10.39. Plans are being considered to develop a 60-ha area for commercial and residential use. Annual rainfall in the area is 120 cm, the existing (undeveloped) land is approximately 3% impervious, and the developed land is expected to be 30% impervious. Estimate the current and future amounts of lead in the runoff.

annual rainfall is 98 cm, estimate the annual load of phosphate (PO4 ) expected in the runoff. 10.43. A stormwater-management system is to be designed for a new 10-ha residential development. The site is located in a region where there are 62 storms per year with at least 1.3 mm of rain, and the IDF curve is given by

w .E asy En g

10.40. A 70-ha catchment is 50% impervious, with 60% commercial and industrial use. The site is located in a city where there are typically 84 storms per year, the mean annual rainfall is 95 cm, and the mean minimum January temperature is 8.1◦ C. Estimate the annual load of suspended solids contained in the runoff. 10.41. Calculate the annual load of lead in the runoff for the site described in Problem 10.40.

10.42. A 80-ha residential development has a population density of 15 persons per hectare, streets are swept every 2 weeks, and the area is 25% impervious. If the average

i=

6000 mm/h t + 20

where i is the average rainfall intensity and t is the duration of the storm in minutes. The mean annual rainfall in the region is 98.5 cm, and the mean January temperature is 9.6◦ C. The developed site is to have an impervious area of 65%, an estimated depression storage of 1.5 mm, an estimated population density of 25 persons/hectare, and no street sweeping. Estimate the average concentration in mg/L of suspended sediments in the runoff using: (a) the USGS model, and (b) the EPA model.

ine eri n

g .n

et

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C H A P T E R

11

Design of Stormwater-Collection Systems 11.1

ww

Introduction

Roadways and parking lots are usually the largest components of impervious areas in urbanized watersheds, and the adequacy of drainage systems serving these areas are particularly important for traffic safety and flood control. Streets are either curbed or uncurbed, and runoff is typically handled differently in these two cases. In curbed streets, roadway runoff is typically directed toward gutters that channel the runoff to inlets which lead to subsurface drainage conduits that transport the runoff away from the roadway. In uncurbed streets, roadway runoff is typically directed toward roadside swales that transport the runoff to a safe discharge location. The principal hydraulic elements of a typical drainage system are shown in Figure 11.1. Runoff from (curbed) roadways and parking-lots, illustrated on the left in Figure 11.1, typically enters pipelines called storm sewers via inlet structures, and the collected runoff is transported in the storm-sewer system to a treatment unit such as a detention basin or an infiltration basin. Effluent from the treatment unit is usually discharged into a receiving water such as a river or a lake. Collected runoff in excess of the capacity of the treatment unit typically bypasses the treatment unit via a weir-type structure and is discharged directly into the receiving water without treatment. Runoff from surfaces other than roadways can also be collected by inlet structures and directed to a treatment unit prior to discharge into a receiving water. Figure 11.1 (on the right) illustrates the case where both collected runoff from storm sewers and direct surface runoff flow into a retention or detention basin, and any overflow from this basin is discharged into the receiving water. Open-channel drainage pathways that transport stormwater to receiving areas are generically called drainage ways, which can range from a landscaped swales to concrete paved channels. In some older U.S. and European cities, storm and sanitary sewers are combined into a single system; these are called combined-sewer systems.

w .E asy En g 11.2

Street Gutters

ine eri n

g .n

et

Surface runoff from urban streets are typically routed to sewer pipes through street gutters and inlets. To facilitate drainage, urban roadways are designed with both cross slopes and longitudinal slopes. The cross slope directs the surface runoff to the sides of the roadway, where the pavement intersects the curb and forms an open channel called a gutter. Longitudinal slopes direct the flow in the gutters to stormwater inlets that direct the flow into sewer pipes. Typical cross slopes on urban roadways are in the range of 1.5%–2%, with slopes of 2%–6% on shoulders. Typical longitudinal slopes are in the range of 0.5%–5%, depending on the topography, although it is generally recommended that longitudinal slopes exceed 0.3% except within 15 m (50 ft) of sag points (USFHWA, 2009). Curbs and gutters are typically restricted to use at the outside edge of pavements for low-speed roadways, and in some instances adjacent to shoulders on moderate to high-speed roadways. Standard paved gutters have a triangular shape and typically extend a distance of 0.3–1 m (1–3 ft) from the curb. Flow in a triangular curb gutter is illustrated in Figure 11.2. The incremental flow, dQ [L3 T−1 ], through any gutter width dx [L] can be estimated using the relation dQ = Vy dx (11.1)

545

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FIGURE 11.1: Principal hydraulic elements in a stormwater-management system

Storm sewer

Roadway

Parking lot Stormwater inlet

Detention/retention storage

Outlet structure

ww

Drainage channel

Weir Overflow

w .E asy En g Treatment

Receiving water body

FIGURE 11.2: Triangular curb gutter

T

dx

y

ine eri n

d Sx 1

g .n

et

where V is the average velocity [LT−1 ] and y is the depth of flow [L] within an elemental flow area of width dx. Using the Manning equation, the average velocity within the elemental flow area can be estimated by V=

1 2 21 y 3 S0 n

(11.2)

where n is the Manning roughness coefficient [dimensionless], and S0 is the longitudinal slope of the gutter [dimensionless]. If Sx is the cross slope of the gutter [dimensionless], then dy = Sx dx

(11.3)

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Street Gutters

547

Combining Equations 11.1 to 11.3 gives 1

1 5 S2 dQ = y 3 0 dy n Sx

(11.4)

Since the flow depth, y, varies from 0 to d across the gutter, the total flow, Q, in the gutter is given by Q=



d

0

1

1 5 S02 y3 dy n Sx

(11.5)

which yields

ww

Q = 0.375



1 nSx



8

1

d 3 S02

(11.6)

Equation 11.6 is viewed as preferable to the direct application of the Manning equation to the gutter, since the hydraulic radius does not adequately describe the gutter cross section, particularly when the top width, T [L], exceeds 40 times the depth at the curb (ASCE, 1992). A limitation of Equation 11.6 is that it neglects the resistance of the curb face, which is negligible if the cross slope is 10% or less (Johnson and Chang, 1984). The depth and top width of gutter flow are related by

w .E asy En g

(11.7)

d = TSx

Combining Equations 11.7 and 11.6, the Manning equation can also be expressed in terms of the spread, T, in the gutter as Q=

ine eri n 0.375 35 12 8 Sx S0 T 3 n

(11.8)

Typical Manning’s n values for street and pavement gutters are given in Table 11.1 (USFHWA, 2009a), and a Manning’s n of 0.016 is recommended for most applications (Guo, 2000). A gutter cross slope may be the same as that of the adjacent pavement or may be designed to be steeper. The gutter should normally be at least 15 cm (6 in.) deep and 60 cm (2 ft) wide, with the deepest portion adjacent to the curb. When flow spreads outside the gutter section onto the adjacent pavement (which might be a shoulder or driving lane) the entire flow section is considered to be the gutter. In such cases, the gutter section can become a composite section which has flow areas with different cross slopes and surface textures.

g .n

TABLE 11.1: Typical Manning’s n Values for Street and Pavement Gutters

Type of gutter or pavement

et

Manning n

Concrete gutter, troweled finish Asphalt pavement Smooth texture Rough texture Concrete gutter with asphalt pavement Smooth Rough Concrete pavement Float finish Broom finish

0.012 0.013 0.016 0.013 0.015 0.014 0.016

Source: USFHWA (2009).

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Design of Stormwater-Collection Systems TABLE 11.2: Minimum Recommended Design Frequency and Spread

Roadway type

ww

Speed limit

Design frequency (year)

Allowable Spread

High volume, divided, or bidirectional

15 cm (6 in.) between the pipe inverts, and the absence of a semicircular channel called a bench at the bottom of the manhole (ASCE, 1992). The head loss, hL [L], in manholes can be estimated using the equation

w .E asy En g

hL = K

V2 2g

(11.48)

where K is a head-loss coefficient [dimensionless] and V is the average velocity in the inflow pipe [LT−1 ]. Head-loss coefficients for manholes with single inflow and outflow pipes aligned opposite to each other vary in the range of 0.12–0.32, while the loss coefficients vary in the range of 1.0–1.8 for inflow and outflow pipes at 90◦ to each other (ASCE, 1992). To prevent backwater effects, a minimum of 3–6 cm (1–2 in.) drop in the sewer invert at manholes is recommended, although the minimum required drop, z [L], can be calculated using the energy equation as   V12 V22 − z = (y2 − y1 ) + (11.49) + hL 2g 2g

ine eri n

g .n

were y1 and y2 are the normal flow depths in the conduits upstream and downstream of the manhole [L], respectively, and V1 and V2 are the velocities in the conduits upstream and downstream of the manhole [LT−1 ], respectively. Under no conditions should the crown of the upstream pipe be lower than the crown of the downstream pipe. In cases where the pipe diameter increases, it is recommended that the crowns of the upstream and downstream pipes be aligned, or, if more of a drop is required, the crown of the outlet pipe can be below the crown of the inlet pipe. In cases where two or more large storm drain conduits intersect, junction chambers are sometimes provided. These types of structures are usually required where storm drains are

et

TABLE 11.6: Typical Manhole Spacings

Pipe size (mm) (in.) 300–600 700–900 1000–1400 Ú1500

Maximum spacing (m) (ft)

12–24 27–36 42–54 Ú60

100 125 150 300

300 400 500 1000

Source: USFHWA (2009).

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Storm Sewers

577

larger than the size that can be accommodated by standard manholes. Junction chambers do not need to extend to the ground surface and can be completely buried; however, it is usually recommended that riser structures be used to provide for surface access (USFHWA, 2009). EXAMPLE 11.14 A manhole is located at a change of grade where a 610-mm sewer pipe on a slope of 0.8% enters the manhole and a 685-mm-diameter sewer pipe on a slope of 0.5% exits the manhole directly opposite to the inflow pipe. Both sewer pipes are RCP, and it can be assumed that Manning’s n is 0.013 and the head-loss coefficient at the manhole is 0.30. If the design flow rate is 0.5 m3 /s, determine the required invert drop at the manhole. Solution From the given data: D1 = 0.610 m, S1 = 0.008, D2 = 0.685 m, S2 = 0.005, n = 0.013, K = 0.30, and Q = 0.5 m3 /s. Solving for the normal depths of flow in the upstream and downstream pipes using the Manning equation yields

ww

y1 = 0.441 m V1 = 2.21 m/s y2 = 0.467 m

w .E asy En g

V2 = 1.87 m/s

V2 2.212 hL = K 1 = 0.30 = 0.0747 m 2g 2(9.81)

Substituting these values into Equation 11.49 gives the required drop, z, as ⎡ ⎤   V12 V22 2.212 1.872 ⎣ ⎦ − − z = (y2 − y1 ) + + 0.0747 = 0.03 m + hL = (0.467 − 0.441) + 2g 2g 2(9.81) 2(9.81)

ine eri n

Therefore, the minimum required drop in the sewer invert is 0.03 m. However, since the pipes change diameter at the manhole, alignment of the crowns of the incoming and outgoing pipes produces a drop of 0.685 m − 0.610 m = 0.075 m. Hence the invert drop of 0.075 m associated with aligning the crowns of the incoming and outgoing pipes is more than sufficient to accommodate the head loss at the manhole and prevent backwater effects that could be caused by the change of grade.

11.5.4

Determination of Impervious Area

g .n

et

The minimization of directly connected impervious area (DCIA) is by far the most effective method of controlling the quantity and quality of surface runoff and, in many cases, DCIA is a key indicator of urbanization’s effect on receiving waters (Lee and Heany, 2003; Jones et al., 2005). By directing roof-drainage downspouts onto pervious areas (e.g., lawns) significant attenuation of roof runoff can be attained (Mueller and Thompson, 2009). Typically, runoff from non-DCIA areas occurs only for larger storms, and the accurate estimation of DCIA is an important component of the cost-effective design and adequacy of roadway drainage systems (Aronica and Lanza, 2005b). Portions of roadways contributing to DCIA generally have curbs and gutters, while portions drained by roadside swales are usually not associated with DCIA. A variety of methods can be used for estimating impervious area, with varying levels of accuracy (Chabaeva et al., 2009). The most common contemporary practice is to correlate impervious surface to the area covered by a series of land-use classes (Brabec, 2009), although these methods mostly focus on total impervious area (TIA) rather than DCIA. Automated methods have been proposed for estimating DCIA using satellite imagery and geographic information systems (e.g., Han and Burian, 2009); however, these methods are not in widespread use and are significantly less accurate than manual methods (Beighley et al., 2009). Manual digitization of impervious surfaces from aerial photos has been shown to provide accurate measures of TIA; however, DCIA can only be accurately measured by

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field assessments (Roy and Shuster, 2009). Several empirical formulae have been derived for estimating DCIA from TIA, such as ⎧ ⎪ 0.15 * TIA1.41 (Alley and Veenhuis, 1983) ⎪ ⎪ ⎪ ⎨ DCIA = (1.046 * TIA) − 6.23% (11.50) (Wenger et al., 2008) ⎪ ⎪ ⎪ ⎪ ⎩(0.627 * TIA) − 1.86% (Roy and Shuster, 2009)

where TIA is the total impervious area [%]. The formulae given in Equation 11.50 have coefficients of determination (r2 ) on the order of 0.55 when applied to individual land parcels, and are more appropriate for use on catchment scales, typically having areas greater than 25 ha (60 ac) (Roy and Shuster, 2009). In hydrologic applications, the importance of errors in estimating TIA and DCIA will ultimately depend on the sensitivity of predicted runoff rate to these parameters.

ww

11.5.5

System-Design Computations

Design of storm drains can be facilitated using the computation worksheet shown in Figure 11.17, and the computations guided by Figure 11.17 must be followed up by calculation of the energy and hydraulic grade lines. The sequence of computations to be followed in filling out each column of the design spreadsheet are as follows:

w .E asy En g

Columns 1 and 2: Reference station numbers upstream and downstream of the segment, respectively. Column 3: Length of the segment, L.

Column 4: Incremental drainage area, Ai .

ine eri n

! Column 5: Total drainage area, A, contributing to the segment, where A = N i=1 Ai and N is the total number of incremental contributing areas above the segment. Column 6: Runoff coefficient of the incremental drainage area, Ci .

Column 7: Ci * Ai # !N " Column 8: i=1 Ci * Ai

g .n

Column 9: Time of concentration for incremental drainage area, tci .

Column 10: Time of concentration for total drainage area, tcA .

et

Column 11: Using a time of concentration, tc = max(tci , tcA ), determine the rainfall intensity, I, from the intensity-duration-frequency (IDF) curve. Column 12: Peak flow rate, Q, determined by ⎧ ! " ⎨I N Ci * Ai # tc = t (Column 8 * Column 11) cA i=1 Q= " # ⎩I Ci * Ai tc = tci (Column 7 * Column 11)

Column 13: Pipe diameter, D, corresponding to pipe slope, S0 , in Column 21. Various (D,S0 ) combinations can be determined using the method described in Section 11.5.2. Round D up to the nearest commercial size. Column 14: Full-flow capacity of selected pipe, QFULL . Column 15: Velocity when Q = QFULL ; this is VFULL . Column 16: Velocity when Q = QDESIGN ; this is VDESIGN .

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ww

w .E asy En g A×C

C

Inc.

tc Total Inlet System

(–) (6)

(ha) (7)

(ha) (min) (8) (9)

Area, A Pipeline

L

Inc. Total

From To (m) (ha) (1) (2) (3) (4)

(ha) (5)

(min) (10)

V

Flow

I Q D Q FULL Full Design time (mm/h) (m3/s) (mm) (m3/s) (m/s) (m/s) (min) (17) (11) (12) (13) (14) (15) (16)

Invert elevation

Invert

U/S

D/S

drop

Slope

(m) (18)

(m) (19)

(m) (20)

(–) (21)

ine eri n

FIGURE 11.17: Storm-Sewer computation worksheet

g .n

et

579

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Chapter 11

Design of Stormwater-Collection Systems

Column 17: Travel time in pipe segment, ti = L/VDESIGN . Column 18: Invert elevation at upstream (U/S) end of segment, zU , must be consistent with invert elevation in upstream segment and invert drop given in Column 20 Column 19: Invert elevation at downstream (D/S) end of segment, zD , must be consistent with zU in Column 18 and S0 in Column 21 such that zD = zU − S0 L. Column 20: Invert drop, z, at upstream inlet structure to offset energy losses, at least equal to z given by Equation 11.49. Column 21: Pipe slope, S0

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Each line of the computation worksheet (Columns 1 to 21) is completed for each segment of the storm-sewer system, starting from the upstream end and terminating at the downstream end. After this is completed, the next step is to compute the energy grade line (EGL) and hydraulic grade line (HGL) to determine the state of the sewer system under specified conditions at the outfall. The objective here is to ensure that the performance of the system will not be compromised by the EGL being above the ground surface. These computations are equivalent to performing a backwater computation in the sewer system.

w .E asy En g EXAMPLE 11.15

A stormwater drainage system is to be designed for a new development, and the proposed pipeline segments and manholes are shown in Figure 11.18. Runoff contributions from individual subcatchments are added at the manhole locations, and the catchment and pipeline characteristics are as follows:

Location

Elevation (m)

1 2 3 4 5 6 7

30.064 29.263 29.655 29.705 28.328 28.558 27.653

Catchment characteristics Inlet Runoff Area time coefficient (ha) (min) (−) 2.00 1.50 0.71 0.86 1.20 0.30 —

8.0 7.4 6.5 6.8 7.0 5.0 —

Pipeline

Length (m)

1–2 3–2 4–2 2–5 6–5 5–7

100 55 73 110 30 90

ine eri n 0.70 0.85 0.55 0.62 0.90 0.75 —

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Local regulations require that all pipelines be RCP with a design Manning’s n of 0.013, the minimum and maximum velocities under design conditions are 0.90 m/s and 4.5 m/s, respectively, the minimum pipe diameter is 455 mm, and the minimum cover is 1.0 m. The project is located in a region where the IDF curve is given by 2020 (11.51) i= (t + 7.24)0.73 where i is the average rainfall intensity [mm/h] and t is the duration [min]. Design the drainage system between Manholes 1 and 7. FIGURE 11.18: Drainage system

4

1

2

5

7

6 3

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Solution The objectives of the design are to determine the required pipe diameters, and the upstream and downstream invert elevations of each pipeline. The calculations are described below and are summarized in Figure 11.19. Line 1−2: The design flow rate is the peak runoff rate to Manhole 1. The characteristics of the contributing area are A = 2.00 ha, C = 0.70, and tc = 8.0 min. Substituting t = tc in Equation 11.51 yields i = 277 mm/h. The design flow rate, Q, is therefore given by (with appropriate unit conversion) Q = CiA = 0.7 * 277 mm/h * 2.00 ha = 1.08 m3 /s

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Several combinations of diameter, D, and pipe slope, S0 , can be used. With D = 760 mm and S0 = 0.00875 the pipe flows full, the slope is only slightly greater than the ground slope (to maintain minimum cover), and the velocity criteria are met (VFULL = VDESIGN = 2.38 m/s). Since the upstream (U/S) ground elevation is 30.064 m, with minimum cover (1.0 m) the U/S invert elevation is 30.064 m − 1.0 m − 0.760 m = 28.304 m. Since the slope of the pipe is 0.00875 and the length of the pipe is 100 m, the downstream (D/S) invert elevation is 28.304 m − (0.00875)(100) = 27.429 m. Line 3−2: Similar calculation procedure as for Line 1–2. For this line, A = 0.71 ha, C = 0.55, and tc = 6.5 min which yields i = 298 mm/h and Q = 0.32 m3 /s. With D = 455 mm and S0 = 0.01190 the pipe flows full, the slope is only slightly greater than the ground slope, and the velocity criteria are met. Since the upstream (U/S) ground elevation is 29.655 m, with minimum cover (1.0 m) the U/S invert elevation is 29.655 m − 1.0 m − 0.455 m = 28.200 m. Since the length of the pipe is 55 m, the downstream (D/S) invert elevation is 28.200 m − (0.01190)(55) = 27.546 m.

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Line 4−2: For this line, A = 0.86 ha, C = 0.62, and tc = 6.8 min which yields i = 294 mm/h and Q = 0.43 m3 /s. With D = 535 mm and S0 = 0.00902 the pipe flows full, the slope is only slightly greater than the ground slope, and the velocity criteria are met. Since the upstream (U/S) ground elevation is 29.705 m, with minimum cover (1.0 m) the U/S invert elevation is 29.705 m − 1.0 m − 0.535 m = 28.170 m. Since the length of the pipe is 73 m, the downstream (D/S) invert elevation is 28.170 m − (0.00902)(73) = 27.512 m. ! Line 2−5: For this line A = 1.50 ha, C = 0.85, and CA = 3.60 ha for this and all upstream catchment areas contributing to Line 2–5. For all paths leading to Line 2–5 the maximum time of concentration is from Line 1–2, and the time of concentration of the downstream end of Line 1–2 is 8.0 min + 0.7 min = 8.7 min. Using t = tc = 8.7 min in Equation 11.51 yields i = 268 mm/h, and hence the peak runoff rate, Q, is given by (with appropriate unit conversion) $ CA = 277 mm/h * 3.60 ha = 2.68 m3 /s Q = CiA = i

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The minimum diameter and corresponding slope that meets the maximum-velocity criterion under full-flow conditions is D = 915 mm and S0 = 0.02003. Determination of the upstream invert elevation of Line 2–5 requires an analysis of the pipelines entering Manhole (MH) 2 as shown below:

Pipeline

Type

Diameter (mm)

1–2 3–2 4–2

Inflow Inflow Inflow

760 455 535

27.429 27.546 27.512

28.189 28.001 28.047

2–5

Outflow

915

27.086

28.001

et

Elevation (m) Invert Crown

It is clear that the minimum crown elevation of all pipes entering MH 2 is 28.001 m and so the crown elevation of the exiting pipe (Line 2–5) should be set at 28.001 m and so the invert elevation of the U/S end of Line 2–5 is 28.001 m − 0.915 m = 27.086 m. The minimum invert elevation of pipes entering MH 2 is 27.429 m and so the minimum invert drop at MH 2 is 27.429 m − 27.086 m = 0.343 m. Since the length of the pipe is 110 m, the downstream (D/S) invert elevation is 27.086 m − (0.02003)(110) = 24.883 m.

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ww Pipeline

L

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Area, A Inc. Total

C

A×C tc Inc. Total Inlet System

From (1)

To (m) (ha) (2) (3) (4)

(ha) (5)

(–) (6)

(ha) (7)

(ha) (min) (8) (9)

1 3 4 2 6 5

2 100 2.00 2 55 0.71 2 73 0.86 5 110 1.50 5 30 0.30 7 90 1.20

2.00 0.71 0.86 5.07 0.30 6.57

0.70 0.55 0.62 0.85 0.75 0.90

1.40 0.39 0.53 1.28 0.23 1.08

1.40 0.39 0.53 3.60 0.23 4.90

8.0 6.5 6.8 7.4 5.0 7.0

(min) (10) 8.0 6.5 6.8 8.7 5.0 9.1

V Flow Q D Q FULL Full Design Time (mm/h) (m3/s) (mm) (m3/s) (m/s) (m/s) (min) (14) (15) (16) (17) (11) (12) (13) I

277 298 294 268 325 262

1.08 0.32 0.43 2.68 0.20 3.57

760 455 535 915 455 1065

Invert elevation U/S D/S (m) (18)

ine eri n 1.08 0.32 0.43 2.68 0.20 3.57

2.38 1.97 1.91 4.10 1.58 4.01

FIGURE 11.19: Storm-Sewer design calculations

2.38 1.97 1.91 4.10 1.75 4.01

0.7 0.5 0.6 0.4 0.3 0.4

28.304 28.200 28.170 27.086 27.103 24.733

Invert drop

Slope

(m) (19)

(m) (20)

(–) (21)

27.429 27.546 27.512 24.883 26.873 23.310

0.000 0.000 0.000 0.343 0.000 0.150

0.00875 0.01190 0.00902 0.02003 0.00767 0.01581

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Line 6−5: For this line, A = 0.30 ha, C = 0.75, and tc = 5.0 min which yields i = 325 mm/h and Q = 0.20 m3 /s. With D = 455 mm and S0 = 0.00767 the slope is equal to the ground slope, and the velocity criteria are met. The pipe does not flow full. Since the upstream (U/S) ground elevation is 28.558 m, with minimum cover (1.0 m) the U/S invert elevation is 28.558 m − 1.0 m − 0.455 m = 27.103 m. Since the length of the pipe is 30 m, the downstream (D/S) invert elevation is 27.103 m − (0.00767)(30) = 26.873 m. ! Line 5−7: For this line A = 1.20 ha, C = 0.90, and CA = 4.90 ha. For all paths leading to Line 5–7 the maximum time of concentration is from Line 2–5, and the time of concentration of the downstream end of Line 2–5 is 8.7 min + 0.4 min = 9.1 min. Using t = tc = 9.1 min in Equation 11.51 yields i = 262 mm/h, and hence the peak runoff rate, Q, is given by $ CA = 262 mm/h * 4.90 ha = 3.57 m3 /s Q = CiA = i

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The minimum diameter and corresponding slope that meets the maximum-velocity criterion under full-flow conditions is D = 1065 mm and S0 = 0.01581. Determination of the upstream invert elevation of Line 5–7 requires an analysis of the pipelines entering MH 5 as shown below: Diameter

Elevation (m)

Pipeline

Type

(mm)

Invert

Crown

2–5 6–5

Inflow Inflow

915 455

24.883 26.873

25.798 27.328

5–7

Outflow

1065

24.733

25.798

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It is clear that the minimum crown elevation of all pipes entering MH 2 is 25.798 m and so the crown elevation of the exiting pipe (Line 5–7) should be set at 25.798 m and so the invert elevation of the U/S end of Line 5–7 is 25.798 m − 1.065 m = 24.733 m. The minimum invert elevation of pipes entering MH 5 is 24.883 m and so the minimum invert drop at MH 5 is 24.883 m − 24.733 m = 0.150 m. Since the length of the pipe is 90 m, the downstream (D/S) invert elevation is 24.733 m − (0.01581)(90) = 23.3101 m. MH 5 will require a special drop-manhole structure since Line 6–5 is entering the manhole at a crown elevation 1.53 m above Line 5–7.

11.5.6

Other Design Considerations

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Other key constraints that must generally be taken into account in designing storm-sewer systems are:

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⊲ Storm sewers are typically located a short distance behind the curb, or in the roadway near the curb. ⊲ Storm sewers should be straight between manholes (where possible); where curves are necessary to conform to street layout, the radius of curvature should not be less than 30 m (100 ft). Pipe sizes smaller than 1220 mm (48 in.) should not be designed with curves. ⊲ Pipe centerlines should generally be set below the frost line (if any) and there should typically be at least 0.9 m (3 ft) of cover over the crowns of the sewer pipes to prevent excessive loading on the pipe. However, smaller diameter reinforced concrete pipes, up to 1220 mm (48 in.), do not have structural loading problems as long as the cover exceeds around 30 cm (1 ft); a common rule of thumb among light construction contractors is to set the minimum cover to 30 cm (1 ft) plus one-fourth the pipe diameter (Seybert, 2006). Where heavy vehicles and shallow cover are encountered, structural integrity might be a concern and a detailed analysis of the loading should be performed. ⊲ Crossings with underground utilities should be avoided whenever possible, but, if necessary, should be at an angle greater than 45◦ . Local regulatory guidelines must generally be followed if they exist, and various manuals of practice (e.g., USFHWA, 2009) provide further details of various practical considerations relevant to designing storm-sewer systems.

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Chapter 11

Design of Stormwater-Collection Systems

Problems 11.1. A four-lane collector roadway is to be constructed with 3.66-m (12-ft) lanes, a cross slope of 1.5%, a longitudinal slope of 0.8%, and pavement made of smooth asphalt. The roadway-drainage system consists of curbs and gutters and is to be designed for a rainfall intensity of 120 mm/h. Local drainage regulations require that at least one lane in each direction must be free of water under design conditions. Determine the spacing of the inlets. 11.2. A roadway with a rough-asphalt pavement has a cross slope of 2.5%, a longitudinal slope of 1.5%, a curb height of 15 cm, and a 90-cm-wide concrete gutter. If the flow rate in the gutter is 0.09 m3 /s, determine the length of an 11-cm-high curb inlet that is required to remove all the gutter flow. Consider the cases where: (a) there is an inlet depression of 30 mm over a width of 70 cm; and (b) there is no inlet depression. What inlet length would reduce the gutter flow rate by 70%? 11.3. A roadway has a flow depth at the curb of 9 cm and a corresponding flow rate in the gutter of 0.1 m3 /s. If the gutter flow is to be removed in a vertical sag by a 15-cmhigh curb inlet, determine the minimum required length of the inlet. Consider the cases where: (a) the width of the inlet depression is 0.3 m, and (b) there is no inlet depression. 11.4. A sump curb inlet is 1.5 m long and has a vertical opening of 15 cm. (a) Use Equation 11.17 to estimate the capacity of the inlet for a ponded water depth of 5 cm, and (2) compare your result with the inlet capacity estimated using the appropriate conventional USFHWA (2009) equation. 11.5. Use the form of the Manning equation given by Equation 11.6 to show that the ratio, Rw , of the flow rate over width W0 to the flow rate over the top width T in a triangular curb gutter is given by

ww

inflow to the unclogged inlet when the ponding depth is 5 cm; and (b) compare your result with the inlet capacity estimated using the appropriate conventional USFHWA (2009) equation. 11.9. Explain why the interception capacity of a combination inlet with a grate placed alongside a curb opening on grade does not differ materially from the interception capacity of the grate only. 11.10. Show that the ratio, Rw , of the flow rate over width W0 to the flow rate over the top width T in a depressed triangular curb gutter is given by

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Rw = 1 −



1 −

W0 T

8

11.11.

11.12.

3

Explain the importance of this equation in the design of grate inlets. 11.6. A roadway with a rough-asphalt pavement has a cross slope of 2%, a longitudinal slope of 2.5%, a curb height of 8 cm, and a 90-cm-wide concrete gutter. If the flow rate in the gutter is 0.07 m3 /s, determine the size and interception capacity of a reticuline grate that should be used to intercept as much of the flow as possible. 11.7. A roadway has a cross slope of 1.5%, a flow depth at the curb of 9 cm, and a corresponding flow rate in the gutter of 0.1 m3 /s. The gutter flow is to be removed in a vertical sag by a grate inlet that is mounted flush with the curb. Calculate the minimum dimensions of the grate inlet. 11.8. A 914 mm * 457 mm vane grate inlet is located in a roadway sag. (a) Use Equation 11.24 to estimate the

11.13.

11.14.

⎫−1 ⎧ ⎪ ⎪ S /S w x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 8 ⎪ ⎪ ⎡ ⎤ ⎪ ⎪ ⎪ ⎪ 3 ⎬ ⎨ ⎢ ⎥ Rw = 1 + ⎢ S /S w x ⎥ ⎪ ⎪ ⎪ ⎢1 + ⎥ − 1 ⎪ ⎪ ⎪ T ⎪ ⎪ ⎣ ⎦ ⎪ ⎪ ⎪ ⎪ − 1 ⎪ ⎪ ⎭ ⎩ W0

where W0 is the top width of the flow in the depressed gutter, Sw is the slope of the depressed gutter, and Sx is the slope of the undepressed gutter. Explain the importance of this equation in the design of combination inlets. A combination inlet consists of a 3-m-long curb inlet and a 0.6-m by 1.2-m reticuline grate inlet adjacent to the downstream 1.2 m of the curb opening. The gutter section has a width of 0.6 m, a longitudinal slope of 1.5%, a cross slope of 3.5%, and a gutter depression of 30 mm. The roadway pavement consists of rough asphalt. For a flow rate of 0.09 m3 /s in the gutter, calculate the interception capacity of the inlet. A combination inlet in a sag location of a roadway is to accommodate a flow rate of 0.20 m3 /s. The inlet consists of a 12-cm-high by 3-m-long curb opening with a 0.6-m by 0.9-m P-30 grate centered in front of the curb opening. If the roadway has a cross slope of 3.5% and the perimeter of the grate is 20% clogged, determine the spread of water on the roadway at the sag location. A subdivision is to have two-lane roads that are 7.62 m wide and are to be drained by curb inlets, grate inlets, and combination inlets. Consider a street where the longitudinal slope is 1%, the cross slope is 2%, and the encroachment onto the pavement is to be limited to 1.524 m in each of the two lanes of the road. Identify the alternative inlet designs (curb, grate, combination) that will intercept 90% of the maximum flow rate in the gutter. Make a tentative recommendation on the preferred inlet design. A roadway has a cross slope of 1.5%, a longitudinal slope of 0.5%, and a flow rate in the gutter of 0.2 m3 /s. The flow is to be removed by a slotted drain with a slot width of 5 cm, and the Manning’s n of the roadway is estimated as 0.017. Estimate the minimum length of slotted drain that should be used.

ine eri n

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Downloaded From : www.EasyEngineering.net Problems 11.15. A storm drainage system is to be designed for an urban development having a total area of 4.96 ha. The average site slope is to be 0.1%, and the site is to be made up of both pervious and impervious area. All of the impervious area will be directly connected to the drainage system. The IDF curve for the region is given by

concentration of the catchments contributing to Pipes I and II are:

7836 i= 40.7 + 0.930t

ww

Surface

C

L (m)

n

S0

A

Pervious Impervious Pervious Impervious

0.2 0.9 0.2 0.9

80 60 140 65

0.2 0.1 0.2 0.1

0.01 0.01 0.01 0.01

11.18.

11.19.

11.20.

11.21. The IDF curve of the design storm is given by i=

Area

tc (min)

I

All DCIA∗ All DCIA∗

25 12 30 —

Note: *DCIA means directly connected impervious area.

w .E asy En g

Catchment

B

Pipe

II

where i is the rainfall intensity in mm/h and t is the duration in minutes. Determine the limiting fraction of DCIA for the DCIA not to be the controlling factor in the design of the drainage system. Do you see any implication of your results on roadway drainage in the development? State your assumptions. [Hint: You can estimate the overland flow length as the square root of the contributing area.] 11.16. Consider the two inlets and two pipes shown in Figure 11.16. Catchment A has an area of 0.5 ha and is 60% impervious, and catchment B has an area of 1 ha and is 15% impervious. The runoff coefficient, C; length of overland flow, L; roughness coefficient, n; and average slope, S0 , of the pervious and impervious surfaces in both catchments are given in the following table:

8000 t + 40

where i is the average rainfall intensity in mm/h and t is the duration of the storm in minutes. Calculate the peak flow rates to the inlets and in the pipes. 11.17. Within the storm-sewer system for the site described in Problem 10.43, Pipe I and Pipe II intersect (at a manhole) and flow into Pipe III. All pipes have a slope of 2%. Pipe I and Pipe II both drain 1-ha areas that are 65% impervious, with the impervious area directly connected to the inlet to Pipe I, but with the inlet to Pipe II not directly connected to any impervious area. The runoff coefficients can be taken as 0.3 and 0.9 for the pervious and impervious areas, respectively. The times of

585

11.22.

11.23.

If the flow times in Pipe I and Pipe II are 3 min (in each pipe), estimate the design flow rates in Pipes I, II, and III. What diameter of concrete pipe would you use for Pipe III? A concrete sewer pipe is to be laid on a slope of 0.90% and is to be designed to carry 0.50 m3 /s of stormwater runoff. Estimate the required pipe diameter using the Manning equation. Repeat Problem 11.18 using the Darcy–Weisbach equation. A section of a storm sewer is to be designed where the length of the segment is 65 m, the peak flow rate is 0.5 m3 /s, the upstream diameter is 610 mm, and the slope to maintain minimum cover is 0.001. The minimum and maximum allowable velocities under design conditions are 0.9 m/s and 4.0 m/s, respectively, and reinforced concrete pipe is to be used. Determine the diameters and corresponding slopes that could be used in the new pipe segment. If service manholes are placed along the pipeline in Problem 11.18, estimate the head loss at each manhole. A manhole is located at a change of grade where a 610mm RCP enters the manhole on a slope of 0.8% and exits the manhole on a slope of 0.5%. The head-loss coefficient at the manhole is 0.35. If the design flow rate is 0.4 m3 /s, determine the required invert drop at the manhole. An identical drainage system to that described in Example 11.15 is to be located in a region where the IDF curve is given by 818 i= (t + 8.54)0.76

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Design the drainage system. [Note: The IDF curve in Example 11.15 is characteristic of Miami, FL, while the IDF curve given here is characteristic of Santa Fe, NM.]

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C H A P T E R

12

Design of StormwaterManagement Systems 12.1

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Introduction

Urban stormwater-management systems are designed to control both the quantity and quality of stormwater runoff. In the absence of quantity and quality controls, increased urbanization would inevitably lead to increased flooding of downstream areas and deterioration of the quality of receiving waters. The conventional approach to designing a stormwatermanagement system is to first identify the performance goals for the system, and then determine the size, configuration, and location of the system components that will meet the required performance goals. Components of stormwater-management systems are commonly referred to as best-management practices (BMPs) or simply as stormwater control measures (SCMs), with the former term being more common and the latter term being more appropriate (NRC, 2009). Engineers typically select system components from a suite of proven SCMs, or from a restricted set of SCMs acceptable to the local jurisdiction. The challenge for design engineers is to select SCMs that meet the performance goals of the system with due consideration to cost and environmental impact.

w .E asy En g 12.2

Performance Goals

The performance goals of stormwater-management systems are typically set by local regulations, and these regulations must be followed where applicable. The performance goals that have been adopted in most jurisdictions are fairly similar to each other and the most common performance goals will be covered here. Generally, stormwater-management systems have both quantity-control and quality-control components, and there are different performance goals for these two components. 12.2.1

Quantity Control

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The typical goal for the quantity-control component of a stormwater-management system is that the postdevelopment peak runoff rate from the site be less than or equal to the predevelopment peak runoff rate. Jurisdictions will typically specify the return periods and durations of the design rainfall events that must be used in making this comparison. Some jurisdictions specify a single design event, while others specify multiple design rainfall events. For example, a jurisdiction specifying a single event might specify that a 25-year 24-h duration rainfall event with a Type II NRCS rainfall distribution be used as the basis for ensuring that the postdevelopment peak offsite discharge be less than the predevelopment peak offsite discharge. 12.2.2

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Quality Control

Goals for the quality-control component of a stormwater-management system are of three different types: Type 1: The runoff from a specified depth of rainfall must be treated onsite; Type 2: A specified depth of runoff must be treated onsite; or Type 3: A specified fraction of a certain pollutant must be removed prior to offsite discharge. Type 1 performance goals specify that the runoff from a water-quality rainfall depth be treated onsite. The water-quality rainfall depth is usually determined from an analysis of 586

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Design of Stormwater Control Measures

587

the local rainfall characteristics, where a characteristic local rainfall record is first separated into events, the depth of rainfall in each event is determined, and then the water-quality rainfall is the rainfall depth corresponding to a given percentile. The 90-percentile rainfall depth is typically used in determining the water-quality rainfall depth (ASCE, 2012). Waterquality rainfall depths can vary significantly between areas with different climates, with 90-percentile water-quality rainfall depths typically in the range of 10–40 mm (0.5–1.5 in.). The water-quality volume (WQV) is the runoff corresponding to the water-quality rainfall, and the WQV will generally depend on the catchment characteristics (primarily the catchment imperviousness). The water-quality control component of the stormwater-management system must then be designed to treat the WQV within the catchment. Clearly the WQV will increase with the imperviousness of the catchment. Type 2 performance goals specify a fixed WQV, regardless of the catchment characteristics. This fixed WQV is typically on the order of 13 mm (0.5 in.). The basis of this approach is the assumption that most of the pollutants in stormwater runoff are contained in the so-called first flush of runoff contained in the fixed WQV. Field measurements have shown that the first-flush phenomenon exists in some (not all) urban catchments, and is not found in natural catchments (Yoon and Stein, 2008; Francey et al., 2010). The limitation of the fixed WQV approach is that the level of service provided decreases with increased imperviousness, since greater imperviousness corresponds to greater runoff volumes and less fractions of runoff being treated within the catchment. Type 3 performance goals require that a minimum fraction of the annual-average load of specified pollutants be removed from the runoff prior to discharge from the catchment, where the required removal fraction is commonly called the required removal efficiency. The pollutants of concern in stormwater runoff commonly include the nutrients nitrogen (N) and phosphorus (P). The required removal efficiencies are typically either the removal efficiencies required to reduce the postdevelopment loads to being less than or equal to the predevelopment loads, or the removal efficiencies required to reduce the postdevelopment loads to a baseline condition. A typical baseline condition is a predevelopment catchment with 15% imperviousness. Attainment of a desired performance goal can be achieved by selecting SCMs in which there are validated relationships between size and performance, and then sizing the SCM based on required performance.

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Design of Stormwater Control Measures

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Stormwater control measures constitute the basic elements of stormwater-management systems. To achieve the performance goals for a particular catchment, it is incumbent on the design engineer to choose the SCMs that are most suitable for the conditions and constraints in a particular catchment. The design procedures for several widely used SCMs are covered in the following sections. 12.3.1

Storage Impoundments

et

Storage impoundments are used in both the quantity-control and quality-control components of stormwater-management systems, and so it is important to understand their configurations and performances. Impoundments are defined as depressed storage areas (basins) that collect surface runoff and release it either at a reduced rate through an outlet or by infiltration into the ground. The two major types of storage areas are detention basins and retention basins. Detention basins are storage areas where the stored water is released gradually through an uncontrolled outlet, and retention basins are storage areas where there is either no outlet or the impounded water is stored for a prolonged period. Typically, detention basins are used when there is minimal infiltration from the storage basin into the underlying soil, and retention basins are used otherwise. Storage impoundments that permanently contain water are called wet basins, and impoundments without a permanent body of water are called dry basins.

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Chapter 12

Design of Stormwater-Management Systems

12.3.1.1

Detention basins—design parameters

Detention basins must have at least one outflow structure and an emergency spillway. In some cases, multiple outlets at different elevations are used to facilitate the discharge from the basin under multiple design storms with return periods between 20 and 50 years (ASCE, 1992). An emergency spillway provides for controlled overflow during large storm events, typically the 100-year storm (Yen and Akan, 1999). When detention basins are designed for water-quality control, the bases for design are typically: the water-quality volume (WQV), the evacuation time, Te , and, for wet detention basins, the detention time. Water-quality volume. The water-quality volume, WQV, is the initial volume of runoff that must be stored and treated prior to offsite discharge. The WQV should ideally be determined from local rainfall characteristics and is commonly taken as the 90-percentile rainfall-event depth.

ww

Evacuation time. The evacuation time, Te [T], is the time for the WQV [L3 ] to be discharged from the detention basin, as given by the relation  Te O(t)dt = WQV (12.1)

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0

where O(t) is the discharge hydrograph [L3 T−1 ], and t is time [T], where the basin has WQV stored at t = 0. Other definitions of the evacuation time are also used, such as: (1) the time lag between the centroid of the outflow hydrograph and the centroid of the corresponding inflow hydrograph under design conditions (e.g., Hancock et al., 2010); and (2) the time from the peak of the outflow hydrograph, O(t) (when the storage is a maximum), to the time that 95% of the surcharge storage has been evacuated (Wurbs and James, 2002). Regardless of which definition of evacuation time is used, the evacuation time is a measure of the time required for the detention basin to return to pre-storm conditions and be ready to accommodate the next storm. Ideally, the evacuation time should be less than the average interval between storms, which varies with the local climate. Regulatory maximum evacuation times are typically on the order of 3 d (72 h).

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Detention time. The detention time is a parameter that is only applicable to wet detention basins (i.e., detention ponds), and measures the average time that surface runoff remains in the detention pond. The detention time, Td [T], is usually estimated by the relation Vpond Td = (12.2) Q

g .n

et

where Vpond is the volume of the detention pond [L3 ], and Q is the average runoff rate [L3 T−1 ]. Typically, Q is the average annual runoff determined by multiplying the average annual rainfall by a runoff coefficient. Ideally, the detention time Td should be greater than the detention time to provide adequate contaminant removal in the detention pond. Regulatory minimum detention times are typically on the order of 14 d. EXAMPLE 12.1 A 5-ha site is to be developed with a 35 m * 112 m detention pond that is 2.5 m deep under pre-storm (control) conditions. The regulatory water-quality volume to be passed through the detention pond is 2.5 cm, and this volume is to be discharged offsite via a triangular weir whose vertex is at the control elevation of the pond. Flow through the weir, Q [m3 /s], can be estimated by the relation 5

Q = 0.101H 2

(12.3)

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where H is the height of the water surface above the vertex of the weir [m]. The average annual rainfall at the site is 150 cm, and the average runoff coefficient is 0.7. Determine the evacuation time for the water-quality volume and the average detention time in the pond. Solution From the given data: A = 5 ha = 5 * 104 m2 , L = 112 m, W = 35 m, and d = 2.5 m. The water-quality volume, WQV, area of the lake, Alake , and ponded height, H0 , are given by WQV = 0.025A = 0.025(5 * 104 ) = 1250 m3 Alake = L * W = (112)(35) = 3920 m2 H0 =

WQV 1250 = = 0.318 m Alake 3920

To route the WQV through the lake it is typically assumed that the WQV is added instantaneously to the lake, in which case the outflow hydrograph, Q(t), satisfies the continuity equation:

ww

Q = Alake

dH dt

(12.4)

where H is the height of the water surface above the control elevation. Combining Equations 12.3 and 12.4 and substituting Alake = 3920 m2 gives the following equation describing the stage hydrograph,

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5

0.101H 2 = 3920

dH dt

Solving this equation with the initial condition that H = H0 = 0.318 m when t = 0 sec yields H=



2613 14570 − 0.101t

2

3

and combining this result with Equation 12.3 gives the outflow hydrograph as Q(t) =

ine eri n 50,067

(12.5)

5

(14,570 − 0.101t) 3

where Q is in m3 /s and t is in seconds. The evacuation time is calculated using Equation 12.1 which requires that  Te O(t)dt = WQV 0

 Te 0

50,067 5

dt = 1250 m3

(14,570 − 0.101t) 3

g .n

et

which yields Te = 93,370 s = 25.9 h. Therefore, the evacuation time of the WQV in the detention pond is 25.9 h. From the given data, the average annual rainfall, drain , is 1.50 m and the runoff coefficient, C, is 0.70; the average runoff, Q, and the volume of the pond, Vpond , are given by (1.50)(5 * 104 )(0.7) d AC = = 143.8 m3 /d Q = rain 365 365 Vpond = LWd = (112)(35)(2.5) = 9800 m3 Therefore the detention time, Td , is given by Equation 12.2 as Td =

Vpond Q

=

9800 = 68 d 143.8

In summary, for a water-quality volume corresponding to 2.5 cm of runoff, the detention pond will have an evacuation time of approximately 26 h and a detention time of approximately 68 d.

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12.3.1.2

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Wet detention basins

Wet detention basins, commonly known as detention ponds, are the most widely used type of detention basin in stormwater management. Detention ponds are used for both quantity control and quality control. The key feature of the detention pond is the permanent pool of water, where the design water-surface elevation is at the control elevation. A typical cross section of a detention pond is shown in Figure 12.1. Detention ponds are used when native soils have low infiltration capacity and/or the water table is near the ground surface. In areas of high water table, detention ponds might be the only option for attenuating the peak runoff rate from a catchment, and in these cases the control elevation is normally taken to be at the seasonal-high water table elevation. When the water table is not near the ground surface and the infiltration capacity of the native soils are high, detention ponds must be lined with an impermeable geofabric. Inflows to detention ponds must be first passed through catch basins where coarse materials such as sand can settle and not cause excessive sediment build-up in the pond. The lowest opening in the outflow structure is located at the control elevation such that any surcharge of the permanent pool of water is gradually discharged through the outflow structure, and there is no outflow from the structure when the pool of water is below the control elevation. An example of a wet detention basin (i.e., detention pond) adjacent to a shopping center is shown in Figure 12.2, where 12.2(a) shows inflow (mitered) pipes and 12.2(b) shows the outflow structure, which in this case consists of a compound weir (facing left) protected by a metal baffle to restrain floating debris from entering the structure. A wide view of the detention pond is shown in Figure 12.2(c), with the shopping center in the background.

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Design for water-quality control. When used for water-quality control, detention ponds primarily remove pollutants by sedimentation, uptake of dissolved contaminants by vegetation, and adsorption of colloidal particles on plants and bottom sediments. The three most important factors in determining the pollutant removal efficiency of detention ponds are: (1) the volume of the permanent pool; (2) the depth of the permanent pool; and (3) the presence of a shallow littoral zone. Specific design guidelines are as follows (from ASCE 2012, unless otherwise indicated):

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⊲ The volume of the permanent pool should be sufficient to provide 60–120 days (2–4 weeks) of detention time so that algae can grow. ⊲ For nutrient removal, significantly longer detention times (such as 200 d) are necessary (Harper and Baker, 2007). ⊲ The ratio of the volume of the detention pond to the water-quality volume should be at least 4 to achieve total suspended-sediment removal rates of 80%–90%. ⊲ The depth of the permanent pool should be greater than 1–2 m (3–6 ft), to prevent wind-generated waves from resuspending accumulated bottom sediments and to reduce bottom-weed growth by minimizing sunlight penetration to the bottom of the pond.

g .n

et

FIGURE 12.1: Detention pond

Flood-control volume Outflow Water-quality volume

Control elevation

Inflow Permanent storage volume Drainage channel Outflow structure

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FIGURE 12.2: Wet detention

(a) Inflow structure

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(b) Outflow structure

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(c) Wide view

⊲ The depth of the pond should be less than 3–5 m (10–16 ft) so that the water remains well mixed and the bottom sediment remains aerobic. An anaerobic condition in the bottom of the pond will mobilize nutrients and metals into the water column and significantly reduce the effectiveness of the detention pond. ⊲ The presence of a littoral zone is essential to the proper performance of a detention pond, since the aquatic plants in the littoral zone provide much of the biological assimilation of the dissolved stormwater pollutants. The littoral zone should cover 25%–50% of the surface area of the detention pond (ASCE, 2012) and have a slope of 6:1 (H:V) or less to a depth of 60 cm (2 ft) below the permanent-pond elevation. This is sometimes called an aquatic bench (Paine and Akan, 2001). Small side slopes provide a measure of public safety, especially for children. ⊲ The flow length in the detention pond should be extended as much as possible between the inlet and outlet structures and the outlet structure should be designed such that the water-quality volume is drained in approximately 72 h, or whatever is the local interstorm duration. ⊲ A minimum length-to-width ratio of 3:1 is recommended to minimize short circuiting, enhance sedimentation, and prevent vertical stratification within the permanent pool (Hartigan, 1989; ASCE, 2012).

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g .n

et

An example of regulatory design requirements for wet detention basins is as follows: ⊲ (Florida) It is recommended that a storage volume equal to 25 mm (1 in.) of runoff be provided above the control elevation, with no more than half of this volume released in 60 h and all released in 120 h. The detention time in the permanent pool should be at least 14 d. Wet detention basins can achieve high removal rates of sediment, BOD, nutrients, and trace metals. Environmental problems frequently encountered in detention ponds include eutrophication, excessive presence of mosquitoes, and associated problems of mosquitoborne diseases. Particular environmental concerns that can be minimized with proper pond design include excessive sedimentation, areas with stagnant water, and anaerobic conditions

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within the pond. Normal sediment accumulation rates in detention ponds are in the range of 0.5–2 cm/yr (0.2–0.8 in/yr), and accumulated sediments should typically be removed when the effective volume of the pond has been reduced by around 10%. The sediment removed is potentially hazardous. Desirable aerobic conditions typically require that dissolved oxygen concentrations be maintained above 4 mg/L, and under no circumstances go lower than 2 mg/L. This level of dissolved oxygen is typically sufficient for retaining phosphorus and metals bound in the sediment, and avoiding obnoxious smells associated with the reducing conditions of low dissolved oxygen (Hvitved-Jacobsen et al., 2010). 12.3.1.3

ww

Dry detention basins

Dry detention basins are areas that are normally dry, but function as detention reservoirs during runoff events. A schematic diagram of a dry detention basin is shown in Figure 12.3. The key feature of a dry detention basin is an open area characterized by low-level outlets that can discharge any accumulated basin inflows. Dry detention basins generally empty after a storm. In many cases, dry detention basins have a dual purpose in both quality and quantity (peak flow rate) control. For dual-purpose dry detention basins, the lower portion of the basin is designed with its own low-level outlet structure for water-quality control, while the basin as a whole is designed to control the peak discharge. The lower portion of the basin used for water-quality control typically has a detention time much longer than the basin as a whole and this lower portion of the basin is sometimes called an extended dry-detention basin or simply extended detention basin. The flow normally enters the basin through a sediment forebay that is intended to capture incoming sediment, and then a meandering path is built to slow down the inflow. The runoff exits through an outflow control structure built to retard the flow and discharge the water-quality volume over a specified time interval that is characteristic of the local or regulatory interstorm period. The combination of the meandering path and the low-level outlet that regulates the outflow serves the purpose of extending the time that surface runoff spends in the detention basin, hence the name “extended detention basin.” Some dry detention basins are hardly noticed by the public, since they may be implemented as multiple-use facilities that function as parks and recreation areas, and are submerged only during exceptional storms. Maintenance of dry detention basins is both essential and costly, with the general objectives being to prevent clogging, prevent standing water, and prevent the growth of weeds and wetland plants. Maintenance items include removal of built-up sediment from the sediment forebay, harvesting of grasses to remove accumulated nutrients, and repair of berms and structures after storm events.

w .E asy En g

ine eri n

g .n

Design for Water-Quality Control. Dry detention basins remove pollutants primarily by sedimentation. In designing dry detention basins for water-quality control, the volume of the basin is typically sized to accommodate a specified WQV (water-quality volume) and release this volume in no less than a specified period of time. Sizing the basin to store the WQV is usually adequate if the catchment area is less than 100 ha (250 ac). If the catchment area is larger than 100 ha (250 ac), reservoir routing is necessary to determine the required volume of the basin. Specific design guidelines are as follows (from ASCE 2012, unless otherwise indicated): FIGURE 12.3: Dry detention basin

Maximum storage level

et

Emergency spillway

Water-quality volume Low-level outlet Inflow

Bottom of detention basin

Outflow

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⊲ The required volume (= WQV) of the detention basin should be increased by 20% to account for sediment accumulation. ⊲ The outlet structure, such as a V-notch weir or perforated riser, should be designed to drain the basin in the design evacuation time. To ensure that small-runoff events will be adequately detained, the outlet should empty less than 50% of the design volume in the first one-third of the design evacuation time. In the United States, typical design evacuation times are 24–48 h (Barrett, 2008), although some engineers recommend that evacuation times greater than 40 h be used (Urbonas and Stahre, 1993). Dry detention basins with long evacuation times tend to be breeding grounds for mosquitoes, have “boggy” bottoms with wetland vegetation, and are usually difficult to maintain and clean. ⊲ The shape of dry detention basins should be such that they gradually expand from the inlet and contract toward the outlet to reduce short circuiting. ⊲ Riprap lining or other methods of stabilization should be provided within a low-flow channel and at the outflow channel to resist erodible velocities. ⊲ A length-to-width ratio of 2:1 or greater, preferably up to a ratio of 4:1 should be used. ⊲ Basin side slopes of 4:1 (H:V) or less provide for facility maintenance and safety. ⊲ A forebay with a volume equal to approximately 10% of the total design volume can help with the maintenance of the basin by facilitating sediment deposition near the inflow, thereby extending the service life of the remainder of the basin. ⊲ Embankments for small on-site basins should be protected from at least the 100-year flood, whenever possible.

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The removal efficiency of dry detention basins is regarded as poor for evacuation times shorter than 12 h, good for evacuation times longer than 24 h, and excellent for evacuation times greater than 48 h (USFHWA, 2009). Removal of as much as 90% of particulates is possible if stormwater is retained for 24 h or more. However, dry detention basins only slightly reduce the levels of soluble phosphorus and nitrogen found in urban runoff (USFHWA, 2009). The design of dry detention basins for pollutant removal is much less scientific than for wet detention basins, and basin performance is derived mostly from empirical results. An example of a dry detention basin adjacent to a small commercial establishment is shown in Figure 12.4, where Figure 12.4(a) shows the inflow structure that directs surface runoff from the parking lot into the dry detention basin, Figure 12.4(b) shows the outflow structure consisting of a circular orifice below a rectangular weir, and Figure 12.4(c) shows a wide view of the dry detention basin, with the inflow structures on the left and the outflow structure in the background. In cases where there is not ample surface area available to meet storage requirements, underground detention might be necessary. Underground detention usually consists of a series of large pipes or prefabricated custom chambers manufactured specifically for underground detention. Examples of regulatory design requirements for dry detention basins are as follows:

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g .n

et

⊲ (Maryland) Dry detention basins must accommodate runoff from the 1-year 24-h storm to be released over a minimum of 24 h. ⊲ (Washington) Dry detention basins must accommodate runoff from the 2-year 24-h storm to be released over a minimum of 40 h. In both of the above examples, the runoff described is, by definition, equal to the waterquality volume. 12.3.1.4

Design of outlet structures

The most common types of outlet structures from detention ponds are weir-type and orificetype outlets, and these outlets are typically part of a single-stage riser as shown in Figure 12.5. Other types of outlet structures can also be used, but are less common (e.g., Tullis et al., 2008). Both of the outlets shown in Figure 12.5 are appropriate for use in a detention pond where the control elevation is equal to Z0 . When the water level in the detention pond is

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FIGURE 12.4: Dry detention

(b) Outflow structure

(a) Inflow structure

ww

w .E asy En g

(c) Wide view

FIGURE 12.5: Single-stage riser

Inflow/overflow

Lw

Z1

Weir

Z0 Inflow

Orifice

Hw D

Riser

ine eri n Z0

Outflow

Inflow Riser

(a) Weir outlet

Hw

(b) Orifice outlet

Inflow/overflow

Z2 Z1

g .n

Outflow

et

below Z0 , there is no outflow from the structure. When the water level in the detention pond is above Z0 , water flows over the rectangular weir or through the orifice into the riser and out of the pipe leading from the riser to the offsite discharge location. For the weir outlet shown in Figure 12.5(a), the top of the weir is at Z1 , which is usually designed to match the maximum water level in the detention pond for the design flood. When the water level in the detention pond exceeds Z1 , water flows over the top of the outlet structure into the riser, with the top of the outlet structure functioning like a weir. For the orifice outlet shown in Figure 12.5(b), the opening in the side of the riser functions like a weir for water levels between Z0 and Z1 , and functions like an orifice for water levels greater than Z1 . The elevation of the top of the riser is Z2 , which usually coincides with maximum water level for the design flood, and the top of the riser functions like a weir for water levels exceeding Z2 .

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The discharge capacity of outlet structures are mostly derived from weir-flow and orifice-flow equations. The discharge over a weir, Qw [m3 /s], is given by 3

(12.6)

Qw = Cw Lw h 2

where Cw is the weir coefficient [dimensionless], Lw is the length of the weir [m], and h is the elevation of the water surface above the crest of the weir [m]. The weir coefficient, Cw , depends on the height of the weir crest above the bottom of the reservoir, Hw [m], and can be expressed as h Cw = 1.81 + 0.22 (12.7) Hw with Cw taken as 1.84 when h/Hw … 0.3. In lieu of explicitly recognizing the h/Hw dependence of Cw , a typical value of Cw = 1.83 is commonly assumed. For orifice flow, the discharge through the orifice, Q0 [L3 T−1 ], is given by

ww

 Q0 = Cd A0 2gh

w .E asy En g

(12.8)

where Cd is a discharge coefficient [dimensionless], A0 is the cross-sectional area of the orifice [L2 ], and h is the elevation of the water surface above the center of the orifice (for free discharges) or h is the difference between the headwater and tailwater elevations (for submerged discharges) [L]. Typical values of Cd are 0.6 for square-edge uniform-entrance conditions, and 0.4 for ragged-edge orifices (USFHWA, 2009). Laboratory results have indicated that, in reality, Cd depends on the head over the orifice, the height of the orifice above the bottom of the detention pond, and the ratio of the orifice diameter to riser-pipe diameter (Prohaska et al., 2010). In spite of these dependencies, the aforementioned constant values of Cd are commonly used. For outlet structures with multiple orifices, the total discharge is the sum of the discharge through all of the orifices. For single-orifice outlets from dry detention basins, the minimum allowable orifice dimension is usually taken as 300 mm (12 in.) to avoid problems with clogging (ASCE, 2012). Where drainage regulations require the control of flow rates of two exceedance frequencies, the two-stage riser is an alternative for control. Its structure is similar to that of the single-stage riser, except that it includes either a weir and an orifice or two weirs as shown in Figure 12.6. Typically, the lower outlet is used to control the more frequent event, and the larger event is controlled using both outlets. The runoff from the smaller and larger events are sometimes referred to as the low-stage and high-stage events. There have been few theoretical or empirical studies on the hydraulics of two-stage risers, and a number of procedures have been proposed for routing flows through these structures. With regard to the weir/orifice outlet, some proposed procedures have assumed that the flow through the orifice ceases when weir flow begins, while other proposed procedures have assumed that

FIGURE 12.6: Two-stage riser

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Inflow

Inflow

Z2 Inflow

Z1

H0 Z0 Inflow

et

Inflow

Z2 Z1

g .n

Outflow Hw

(a) Weir/orifice outlet

Z0

Outflow Inflow

Hw

(b) Weir/weir outlet

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FIGURE 12.7: Orifice–weir discharge structure

the flow through the orifice is independent of the flow over the weir. It appears to be more realistic to assume that the two flows are not independent and that the interdependence between weir flow and orifice flow increases as the elevation of the water surface above the weir increases. The general form of the stage-discharge relationship for a weir/orifice outlet is given by ⎧ Z … Z0 ⎪ ⎪0, ⎪ ⎪ ⎪ ⎪ ⎪ 1.5 ⎪ Z0 … Z … Z0 + H0 ⎪ ⎨Cw Lw (Z − Z0 ) ,  Q= (12.9) ⎪ 1 ⎪ H ), Z + H … Z … Z 2g(Z − Z − C A ⎪ d 0 0 0 0 1 ⎪ 2 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩C L (Z − Z )1.5 + C A 2g(Z − Z ), Z1 < Z w w d 0 1 1 where Z is the water-surface elevation in the pond surrounding the riser [L], Z0 is the elevation of the bottom of the orifice [L], H0 is the height of the orifice [L], and Z1 is the elevation of the weir [L]. If the riser is large enough to assume that the orifice  and weir flows are inde-

pendent, then the orifice part of Equation 12.9 would be Cd A0 2g(Z − Z0 − 12 H0 ) rather  than Cd A0 2g(Z − Z1 ). A discharge structure with a low-level orifice and a high-level weir is shown in Figure 12.7, where the orifice discharge is through a low-level PVC pipe, and a metal baffle is placed upstream of the discharge structure to prevent clogging by floating debris. An outlet riser with orifices at multiple levels can be designed to control multiple runoff events. Orifice and weir equations are used to calculate the discharge from the detention pond into the riser as a function of the water-surface elevation in the detention pond, and the relationship between the flow into the riser and the water-surface elevation in the detention pond is called the rating curve. In calculating the rating curve of an outlet riser, care must be taken to ensure that the pipe discharging flow from the riser to a downstream drainage channel has sufficient capacity, and that the headwater elevation (in the riser) on the discharge pipe for any given discharge is taken into account when applying the (freeflow or submerged-flow) orifice equations for flow into the riser. Discharge structures can be designed to handle flows up to the 100-year flow. In retention areas designed to retain a fixed volume of runoff (i.e., the water-quality volume, WQV), the outlet structure must be such that no offsite discharge occurs when water is ponded at or below the level corresponding to the storage of the WQV within the retention area. The water level corresponding to storage of the WQV is sometimes called the control elevation. In these cases, the lowest elevation of the discharge opening is at the

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et

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FIGURE 12.8: Discharge structure from retention area

ww

control elevation. Figure 12.8 is an example of this type structure, which shows a view of the structure from a downstream location. In this case, discharge from the retention area is via a culvert; however, a concrete weir covering the culvert exit does not allow any discharge until the water upstream of the culvert is ponded to the height of the weir crest. A riprap apron downstream of the discharge structure protects the receiving area from erosion. Outlet structures should be readily accessible for maintenance, and should ideally be equipped with trash racks or baffles, or have openings small enough to intercept debris that might block the outlet pipe.

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Emergency spillway. The primary purpose of an emergency spillway is to provide a way for excess water in the impoundment to exit safely, thus preventing the overtopping of the berm surrounding the detention basin. If the berm is overtopped it could be breached, allowing impounded water to flood areas downstream of the basin. If the detention basin does not have a berm, the need for an emergency spillway is lessened. Most detention basin designs incorporate one of two general types of emergency spillway: (1) broad-crested weir leading to a grassed channel or waterway located separate from the outflow structure; or (2) allowing water to flow into the top of the outflow structure and through the outflow pipe. Emergency spillways of the broad-crest weir type and vertical-pipe (riser) type are shown in Figures 12.9(a) and (b), respectively. The entrance to the vertical-pipe spillway is covered with a trash-screen cover, as is usually the case. It is a common design practice to specify the crest elevation of the emergency spillway as equal to the maximum water elevation in

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g .n

FIGURE 12.9: Emergency spillways

(a) Broad-crested spillway

et

(b) Outflow riser

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the detention pond when the discharge structure is discharging the runoff from the 100-year 24-hour rainfall event. Emergency spillways are typically designed to function when the discharge structure is clogged, in which case the emergency spillway is designed to pass the 100-year discharge from the pond. Where concrete broad-crested weirs are used for emergency spillways, shallow flow depths, typically on the order of 15 cm (6 in.), are specified (Seybert, 2006). EXAMPLE 12.2

ww

An outlet structure from a detention pond is to be designed for both water-quality and water-quantity control purposes. For water-quality control, the outlet structure is to be sized such that the first 2.5 cm of runoff from the catchment is discharged in not less than 24 h. For water-quantity control, the outlet structure is to be sized such that the peak runoff rate from the site is less than or equal to the predevelopment peak runoff rate of 0.92 m3 /s. The catchment area is 46 ha, and the average wet-season pool elevation in the detention pond is 102.50 m. Storage of the water-quality volume causes the water-surface elevation in the detention pond to rise to 102.85 m, and under design-flood conditions the maximum allowable pool elevation in the detention pond is 104.00 m. The outlet structure is to consist of a 1.21-m diameter vertical circular pipe (riser), with the water-quality volume discharged through an orifice in the riser, and the flood flow discharged over a high-level weir at the top of the riser. The high-level discharge weir must have a crest length not greater than 50% of the riser diameter. Determine the required dimensions and elevations of the orifice and weir components of the two-stage riser outlet.

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Solution The water-quality volume will be discharged through an unsubmerged orifice (which acts like a weir), where the bottom side of the orifice is at the average wet-season pool elevation of 102.50 m and the top of the orifice is at the pool elevation when the water-quality volume is stored, which is 102.85 m. If the catchment area is 46 ha = 4.6 * 105 m2 and the water-quality depth is 2.5 cm = 0.025 m, then water-quality volume = 4.6 * 105 m2 * 0.025 m = 11, 500 m3

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If this water-quality volume is to be detained in the pond for 1 d (= 86,400 seconds), then the average discharge rate through the orifice/weir, Q0 , is given by

11, 500 m3 water-quality volume = = 0.133 m3 /s 1d 86, 400 s Specifying the maximum orifice discharge rate as equal to the average discharge rate for 24-h detention (which guarantees a detention time greater than 24 h) yields, according to Equation 12.9, Q0 =

Q0 = Cw Lw (Z − Z0 )1.5

g .n

et

where Cw can be taken as 1.83, Z = 102.85 m, and Z0 = 102.50 m. Substituting into the above equation yields 0.133 = 1.83Lw (102.85 − 102.50)1.5 which gives Lw = 0.35 m The required height of the orifice, H0 , is 102.85 − 102.50 m = 0.35 m, and therefore the dimensions of the orifice required to detain the water-quality volume for at least 24 h (Lw * H0 ) is 0.35 m * 0.35 m. The riser diameter, D, is equal to 1.21 m and the maximum length of the flood-discharge weir is 0.5D = 0.5(1.21 m) = 0.605 m. The discharge from the outlet structure when the pool elevation exceeds the crest elevation of the flood-discharge weir is given by Equation 12.9 as  Q = Cw Lw (Z − Z1 )1.5 + Cd A0 2g(Z − Z1 ) When the pool elevation is at its maximum allowable level, Q = 0.92 m3 /s, Cw = 1.83, Lw = 0.605 m, Z = 104.00 m, Cd = 0.6, and A0 = 0.35 m * 0.35 m = 0.1225 m2 . Substituting into the above equation yields  0.92 = 1.83(0.605)(104.00 − Z1 )1.5 + 0.6(0.1225) 2(9.81)(104.00 − Z1 )

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which gives Z1 = 103.30 m Therefore, the bottom of the flood-discharge weir should be at elevation 103.30 m to give a design-flood discharge equal to the predevelopment value of 0.92 m3 /s. In summary, the required two-stage riser should have a 0.35-m * 0.35-m orifice with top and bottom elevations of 102.85 m and 102.50 m, and a 0.605-m long flood-discharge weir with a crest elevation of 103.30 m. The riser will meet both the water-quality and water-quantity (flood-discharge) objectives of the detention pond.

ww

By their very nature, flood-control basins tend to be wet and soggy. It should never be assumed that a concrete outlet structure can be adequately supported in this environment without piles or other geotechnical treatment. Without proper support, the outlet structure can settle and separate from the outfall (Paine and Akan, 2001). As the height of the outlet riser increases, conditions become more unstable, and, in order to increase stability, the outlet structure should be located in the embankment rather than in the open bottom of the basin. 12.3.1.5

Design for flood control

w .E asy En g

Detention basins are commonly used for both water-quality control and flood (waterquantity) control. In such cases, the performance of the detention basin should be evaluated separately for both functions. The design of detention basins for flood control is described below.

Step 1: Make a preliminary selection of a drainage basin. A preliminary estimate of the required drainage-basin volume can be obtained by subtracting the predevelopment runoff volume from the postdevelopment runoff volume. This volume represents the approximate storage requirement to ensure that the postdevelopment peak discharge rate is less than or equal to the predevelopment peak runoff rate. Step 2: Make a preliminary selection of an outlet structure. Use the required volume of the detention basin estimated in Step 1 with the storage-elevation function for the site to estimate the maximum headwater elevation at the outlet location. Select an outlet structure that will pass the maximum allowable outflow at this headwater elevation. Step 3: Route the runoff hydrograph through the detention basin. The runoff hydrograph is routed through the detention basin using the modified Puls method described in Section 10.6.1.1. This procedure yields the discharge hydrograph from the detention basin. Step 4: Assess the performance of the detention basin. Determine the peak discharge rate from the detention basin. If the calculated postdevelopment peak discharge rate is not equal to the predevelopment peak discharge rate, adjust the outlet structure and repeat Step 3 until these discharge rates are equal. The result of these iterative calculations yield both the required structure dimensions and the maximum water-level elevation in the storage basin. If the duration of the design storm is not specified by regulations, the performance of the detention basin should be calculated for a range of storm durations (for a given return period) to ensure that the predevelopment peak discharge rate is not exceeded for any storm duration. Many drainage ordinances require that postdevelopment peak discharge rates not exceed predevelopment peak discharge rates for multiple events, such as the 2-year, 10-year, 25-year, and 50-year storms (ASCE, 1996a). If applicable, these requirements should be checked.

ine eri n

g .n

et

Detention facilities should be designed to drain within the typical interstorm period, usually on the order of 72 h (Debo and Reese, 1995).

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EXAMPLE 12.3 The estimated runoff hydrographs from a site before and after development are as follows: Time (min)

0 30

Before (m3 /s)

60

90

120 150

180

210

240

270

300

330

0 1.2 1.7 2.8 1.4

1.2

After (m3 /s)

0 2.2 7.7 1.9 1.1

0.80 0.70 0.58 0.38 0.22 0.11 0

360

1.1

0.91 0.74 0.61 0.50 0.28 0.17 0 0

390

0

The postdevelopment detention basin is to be a detention pond drained by an outflow weir. The elevation versus storage in the detention pond is

ww

Elevation (m)

Storage (m3 )

0 0.5 1.0 1.5

0 5544 12,200 20,056

w .E asy En g

where the weir crest is at elevation 0 m, which is also the initial elevation of the water in the detention pond prior to runoff. The performance of the weir is given by 3

Q = 1.83bh 2

where Q is the overflow rate (m3 /s), b is the crest length (m), and h is the head on the weir (m). Determine the required crest length of the weir for the detention pond to perform its desired function. What is the maximum water-surface elevation expected in the detention pond?

ine eri n

Solution The required detention-pond volume is first estimated by subtracting the predevelopment runoff volume, V1 , from the postdevelopment runoff volume, V2 . From the given hydrographs: V1 = (30)(60)[1.2 + 1.7 + 2.8 + 1.4 + 1.2 + 1.1 + 0.91 + 0.74 + 0.61 + 0.50 + 0.28 + 0.17] = 22,698 m3

g .n

V2 = (30)(60)[2.2 + 7.7 + 1.9 + 1.1 + 0.80 + 0.70 + 0.58 + 0.38 + 0.22 + 0.11] = 28,242 m3 A preliminary estimate of the required volume, V, of the detention pond above the normal pool elevation is V = V2 − V1 = 28, 242 − 22, 698 = 5544 m3

et

From the storage-elevation function, the head h corresponding to a storage volume of 5544 m3 is 0.50 m. The maximum predevelopment runoff rate, Q, is 2.8 m3 /s, and the weir equation gives 3

Q = 1.83bh 2 or b=

Q 3 1.83h 2

=

2.8

= 4.33 m

3

1.83(0.50) 2

Based on this preliminary estimate of the crest length, use a trial length of 4.25 m. The corresponding weir-discharge equation is 3

3

3

Q = 1.83bh 2 = 1.83(4.25)h 2 = 7.78h 2 The postdevelopment runoff hydrograph can be routed through the detention pond using t = 30 min. The storage and outflow characteristics of the detention pond can be put in the following form:

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Elevation (m)

Storage, S (m3 )

Outflow, O (m3 /s)

2S/t + O (m3 /s)

0 0.5 1.0 1.5

0 5544 12,200 20,056

0 2.75 7.78 14.29

0 8.91 21.34 36.58

601

The routing computations (using the modified Puls method) are summarized in the following table:

ww

Time (min)

Inflow, I (m3 /s)

2S/t − O (m3 /s)

2S/t + O (m3 /s)

O (m3 /s)

0 30 60 90 120

0 2.2 7.7 1.9 1.1

0 0.84 3.76 4.26 2.78

0 2.2 10.74 13.36 7.26

0 0.68 3.49 4.55 2.24

w .E asy En g

From these results, it is already clear that the maximum outflow from the detention pond is 4.55 m3 /s, which is higher than the predevelopment peak of 2.8 m3 /s and is therefore unacceptable. Decreasing the crest length (= 4.25 m) of the weir by the factor 2.8/4.55 = 0.62 gives a new crest length, b, of 0.62(4.25 m) = 2.64 m. Using a rounded number of 2.50 m, the revised weir equation is 3

3

3

Q = 1.83bh 2 = 1.83(2.5)h 2 = 4.58h 2

The revised storage-outflow characteristics of the detention basin are:

ine eri n

Elevation (m)

Storage, S (m3 )

Outflow, O (m3 /s)

2S/t + O (m3 /s)

0 0.5 1.0 1.5

0 5544 12,200 20,056

0 1.62 4.58 8.41

0 7.78 18.14 30.70

The routing computations are summarized in the following table:

g .n

Time (min)

Inflow, I (m3 /s)

2S/t − O (m3 /s)

2S/t + O (m3 /s)

O (m3 /s)

0 30 60 90 120

0 2.2 7.7 1.9 1.1

0 1.28 6.00 7.90 5.88

0 2.2 11.18 15.60 10.90

0 0.46 2.59 3.85 2.51

et

From these results, it is already clear that the maximum outflow from the detention basin is 3.85 m3 /s, which is higher than the predevelopment peak of 2.8 m3 /s and is therefore unacceptable. The crest length must be further decreased until the maximum weir discharge is less than or equal to 2.8 m3 /s. This occurs when the crest length, b, is decreased to 1.30 m. The revised weir-discharge equation is then given by 3

3

3

Q = 1.83bh 2 = 1.83(1.30)h 2 = 2.38h 2

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Storage, S (m3 /s)

Outflow, O (m3 /s)

2S/t + O (m3 /s)

0 0.5 1.0 1.5

0 5544 12,200 20,056

0 0.84 2.38 4.37

0 7.00 15.94 26.66

The routing computations are summarized in the following table:

ww

Time (min)

Inflow, I (m3 /s)

2S/t − O (m3 /s)

2S/t + O (m3 /s)

O (m3 /s)

0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 630 660 690 720

0 2.2 7.7 1.9 1.1 0.80 0.70 0.58 0.38 0.22 0.11 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 1.68 8.32 12.42 10.84 9.08 7.66 6.60 5.68 4.78 3.89 3.04 2.32 1.76 1.34 1.02 0.78 0.60 0.46 0.34 0.26 0.20 0.16 0.12 0.10

0 2.2 11.58 17.92 15.42 12.74 10.58 8.94 7.56 6.28 5.11 4.00 3.04 2.32 1.76 1.34 1.02 0.78 0.60 0.46 0.34 0.26 0.20 0.16 0.12

0 0.26 1.63 2.75 2.29 1.83 1.46 1.17 0.94 0.75 0.61 0.48 0.36 0.28 0.21 0.16 0.12 0.09 0.07 0.06 0.04 0.03 0.02 0.02 0.01

w .E asy En g

ine eri n

g .n

et

For a crest length of 1.30 m, the maximum postdevelopment discharge rate is 2.75 m3 /s, which is slightly less than the target of 2.8 m3 /s and is therefore acceptable. The maximum water level in the detention pond corresponds to a weir overflow rate of 2.75 m3 /s; from the weir-discharge equation, this corresponds to h = 1.10 m. If this water elevation is excessive, the engineer could consider expanding the proposed detention basin. It is interesting to note that for any pond with an uncontrolled outlet the peak water-surface elevation in the pond will always occur at the time when the outflow hydrograph intersects the receding limb of the inflow hydrograph. Prior to this intersection, inflow exceeds outflow and the water level is rising, and beyond this intersection outflow exceeds inflow and the water level is falling.

Flood-control systems are primarily designed to ensure that postdevelopment peakdischarge rates do not exceed predevelopment peak-discharge rates. Using detention basins to accomplish this goal generally results in a postdevelopment-discharge hydrograph that is

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603

shifted in time and has an overall greater volume compared to the predevelopment-discharge hydrograph. Consequently, the postdevelopment-discharge hydrograph generally has higher off-peak discharge rates than the predevelopment hydrograph. Effects of increased runoff volumes from developed areas include: (1) prolonged rise in the water surface downstream of the development, which might affect the slope and stability of channels; (2) decreased onsite groundwater recharge; and (3) increased volume of water released into downstream detention ponds. 12.3.2

ww FIGURE 12.10: Schematic diagram of an infiltration basin

Infiltration Basins

Infiltration basins are natural or excavated areas that impound stormwater runoff, which then infiltrates into the ground. Infiltration basins, also called dry retention basins, are similar in appearance and construction to dry detention basins, except that the detained stormwater runoff is exfiltrated through permeable soils underlying the basin rather than being discharged through an outlet structure. Infiltration basins may or may not be lined with vegetation, but require permeable soils in order to be effective. A schematic diagram of a typical infiltration basin is illustrated in Figure 12.10. Infiltration basins generally have overflow structures to safely discharge water in excess of the storage volume provided by the basin. These outflow structures can be of the weir type illustrated in Figure 12.10, or can be (broad-crested) spillways. An actual infiltration basin is shown in Figure 12.11(a), with the accompanying spillway lined with riprap shown in Figure 12.11(b). The spillway shown in Figure 12.11(b) is located just to the right of the scene shown in Figure 12.11(a). To remove pollutants more effectively and to facilitate maintenance, infiltration basins are sometimes lined with 15–30 cm (6–12 in.) of sand, which serves to trap sediments in runoff and can be readily replaced when the infiltration capacity is reduced below an acceptable level. Infiltration basins typically serve catchment areas ranging from front yards to 20 ha (50 ac) (ASCE, 1992). Infiltration basins are classified as either on-line or off-line. On-line infiltration basins retain a specified water-quality volume; when a larger runoff occurs, it overflows the basin,

w .E asy En g Grass buffer strip Inflow Inflow

ine eri n

Overflow outlet

Storage volume Pipe from catch basin

g .n

Infiltration/outflow

Outflow

et

FIGURE 12.11: Actual infiltration basin with overflow spillway

(a) Infiltration basin

(b) Overflow spillway

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which then acts as a detention pond for the larger event. Some drainage systems divert the water-quality volume out of the normal drainage path and into off-line infiltration basins that provide storage for the treatment volume only. Off-line systems typically utilize a weir structure to divert runoff into the retention area. As a practical matter, to avoid the loss of infiltration capacity due to compaction, excavation of the basin should be performed from the perimeter of the basin, and heavy machinery should be prevented from entering the basin. Design guidelines. Infiltration basins are typically designed to capture and infiltrate a designated water-quality volume, and are usually on the order of 60 cm (24 in.) deep. Typical design guidelines are as follows:

ww

⊲ It is a common design practice to size infiltration basins with sufficient volume to store the water-quality volume. Such an approach is generally conservative since it neglects infiltration during the time in which the WQV accumulates. ⊲ Infiltration basins must be located in soils that allow the design runoff to infiltrate within 72 h, or within 24–36 h for infiltration areas that are planted with grass. ⊲ Soils with saturated infiltration rates less than 8 mm/h (0.3 in/h) are not suitable for infiltration basins (ASCE, 2012). A minimum infiltration rate of 13 mm/h (0.5 in/h) is accepted by many jurisdictions, but since infiltration capacity tends to decrease over time a minimum infiltration capacity of 25 mm/h (1 in/h) is recommended by some engineers (e.g., Pazwash, 2011). The design infiltration rate should be limited to a maximum of 50 mm/h (2 in/h) to account for clogging (ASCE, 2012). Infiltration capacities sometimes vary seasonally with lower capacities observed in lower temperatures (Horst et al., 2011). ⊲ Water ponding in infiltration basins should be less than 30 cm (12 in.) during the design storm (ASCE, 2012). When water is above the design ponding level, an emergency outlet should be provided. ⊲ The bottom of an infiltration basin should be graded as flat as possible to allow for uniform ponding and infiltration, and the side slope of the basin should be less than 3:1 (H:V) to allow for easier mowing and better bank stabilization. ⊲ The seasonal-high water table should be at least 0.6–1.2 m (2–4 ft) below the ground surface in the infiltration basin to assure that the pollutants in the runoff are removed by the vegetation and soil before reaching the water table (Urbonas and Stahre, 1993; Shammas and Wang, 2011). ⊲ In areas where the water table is shallow, particular care should be taken to assess the mounding of the water table below the basin. In cases where the infiltrated water causes the water table to rise to the ground elevation, ponding will occur and the infiltration basin will not function properly. ⊲ Water-tolerant turf such as bahia grass, reed canary grass, or tall fescue should be used in the basin to promote better infiltration and pollutant filtering.

w .E asy En g

ine eri n

g .n

et

In urban settings, infiltration basins are commonly integrated into recreational areas, greenbelts, neighborhood parks, and open spaces, while in highway drainage they may be located in rights-of-way or in open space within freeway interchange loops. An example of an infiltration basin integrated into the greenbelt surrounding a townhouse development is shown in Figure 12.12, where ponding from a recent storm is apparent. Some infiltration basins are susceptible to clogging and sedimentation, and can require large land areas. In some cases, lost infiltration capacity can be restored by scraping the soil from the bottom of the basin or by replacing the sand layer. Although the two main problems typically associated with infiltration basins are clogging and contamination of underlying soil and groundwater, data collected at several infiltration basins have indicated minimal clogging and soil-contamination depths less than 50 cm (20 in.) after about 20 years of operation (Dechesne et al., 2005). During routine operation, long-term standing water in infiltration basins can create problems of security and insect breeding.

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605

FIGURE 12.12: Infiltration basin integrated into greenbelt

ww

w .E asy En g EXAMPLE 12.4

An infiltration basin is to be designed to retain the first 1.3 cm of runoff from a 10-ha catchment. The area to be used for the infiltration basin is turfed, and field measurements indicate that the native soil has a minimum infiltration rate of 150 mm/h. If the retained runoff is to infiltrate within 24 h, determine the surface area that must be set aside for the basin. Solution The volume, V, of the runoff corresponding to a depth of 1.3 cm = 0.013 m on an area of 10 ha = 105 m2 is given by V = (0.013)(105 ) = 1300 m3

The native soil has a minimum infiltration rate of 150 mm/h. To account for clogging, however, the design infiltration rate will be taken as 50 mm/h = 0.05 m/h. The area, A, of infiltration basin required to infiltrate 1300 m3 in 24 h is given by

ine eri n

1300 = 1080 m2 = 0.11 ha A= (0.05)(24)

g .n

If runoff is not to pond to more than 0.3 m, then the maximum volume that can be handled by a 0.11-ha infiltration basin is 0.3 m * 0.11 ha = 330 m3 . Since the runoff volume to be handled is 1300 m3 , the area of the infiltration basin should be at least 1300/330 * 0.11 ha = 0.43 ha.

et

Infiltration basins can be integrated with flood-control detention basins, in which case the control elevation of the discharge structure is at or above the level corresponding to the required retention volume to be infiltrated in the basin. When the performance of this combined retention-detention system is assessed for flood control, infiltration is usually neglected. 12.3.3

Swales

Swales are shallow vegetated (grass) open channels with small longitudinal and side slopes that transport runoff from adjacent land areas. These are sometimes referred to as vegetated swales. Swales differ from regular grass-lined open channels in that they are shallow and have small side slopes, with an operational definition of a swale being a channel that has a depth of 45 cm (1 ft) or less and side slopes of 3:1 (H:V) or less (Pazwash, 2011). Swales are desirable compared to drainage pipes since they can provide a level of treatment that pipes are not able to provide, particularly at low flows. Swales are commonly used in highway medians and for roadside drainage on rural roads. Typically, swales are designed as free-flowing grasslined open channels with inclined slopes. In these contexts, swales are primarily designed for their runoff transport capabilities, with their water-quality benefits being secondary and not

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counted toward the treatment of surface runoff. In arid areas, grass linings are usually not sustainable and stone lining is more practical. To be effective measures for treating surface runoff, swales can be designed to be nonflowing, with all the design runoff retained and infiltrated into the ground; nonflowing swales are sometimes called retention swales (Field et al., 2000). Alternatively, for waterquality control purposes, swales can be designed for flow depths that are less than the height of the vegetation contained in the swale, and such swales are called biofiltration swales. The designs of both retention swales and biofiltration swales are described in the following sections. 12.3.3.1

ww

Retention swales

Retention swales can be designed either to retain a fixed volume of runoff or to retain the entire runoff from a design storm. Retention swales used for water-quality control are typically designed to retain a specified water-quality volume and are sized based on their stage-storage relation; elevated outlets, such as those shown in Figure 12.13, are used for excess runoff amounts above the water-quality volume. Retention swales used for waterquantity control are designed to retain the entire runoff from a design storm, and are sized such that the volumetric infiltration rate is equal to the peak runoff rate. In this case, if the peak runoff rate is Qp [L3 T−1 ], then the length, L [L], of swale required for the infiltration rate to be equal to the runoff rate is given by

w .E asy En g

L=

Qp fP

(12.10)

where f is equal to the infiltration capacity of the soil [LT−1 ], and P is the wetted perimeter of the swale [L]. For any cross-sectional shape, the wetted perimeter can be expressed in terms of the runoff rate via the Manning equation. For triangular-shaped swales, Equation 12.10 can be combined with the Manning equation to yield (Wanielista et al. 1997)

L=

ine eri n 5

3

5

151, 400Qp8 m 8 S 16 3

(12.11)

5

n 8 (1 + m2 ) 8 f

g .n

where L is the swale length [m], Qp is the peak runoff rate [m3 /s], m is the side slope [dimensionless], S is the longitudinal slope [dimensionless], n is the Manning roughness coefficient, and f is the infiltration capacity [cm/h]. Typical values of the Manning roughness coefficient are n = 0.20 for routinely mowed swales and n = 0.24 for infrequently mowed swales (ASCE, 2012). In the case of trapezoidal sections

et

FIGURE 12.13: Overflow weirs from retention areas

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360, 000Qp ⎫ L= ⎧ ⎡ ⎤3 ⎪ ⎪ 8 ⎪ ⎪ ⎪ ⎪ ⎬ ⎨  Qp n ⎢ ⎥ 2 % & b + 2.38 ⎣  1 + m f ⎦ 1 ⎪ ⎪ ⎪ ⎪ 2 1 + m2 − m S 2 ⎪ ⎪ ⎭ ⎩

607

(12.12)

where b is the bottom width of the swale [m]. Equation 12.12 applies to the best trapezoidal section, where the perimeter, P [L], is related to the flow depth, y [L], and the side slope m [dimensionless] by  P = 4y 1 + m2 − 2my (12.13) Design guidelines for retention swales are as follows:

ww

⊲ The infiltration rate of the underlying soil should be greater than 13 mm/h (0.5 in/h). ⊲ The longitudinal grade should be set as flat as possible to promote infiltration, never steeper than 3%–5% (Urbonas and Stahre, 1993; Yu et al., 1994). ⊲ Side slopes should be flatter than 4:1 (H:V) to maximize the contact area. ⊲ It is sometimes recommended that swales be designed for 2-year storms and that the runoff volume in nonflowing or slow-moving swales be infiltrated within 36 h (Urbonas and Stahre, 1993). ⊲ In cases where the length of the swale required to infiltrate the design peak runoff rate is excessive, a check dam can be used to store the runoff within the swale, in which case the depth in the swale should not exceed 0.5 m (1.6 ft). Check dams usually consist of crushed rock, gabions, or pretreated timber up to 0.6 m (2 ft) in height. Rocks used in check dams should have a median diameter of 25–75 mm (1–3 in.), and the spacing of dams should not exceed the horizontal distance from the toe of the upstream dam to the same elevation on the top of the downstream dam (Nicklow et al., 2006). ⊲ Grassed swales should be mowed to stimulate vegetative growth, control weeds, and maintain the capacity of the system.

w .E asy En g EXAMPLE 12.5

ine eri n

A triangular-shaped swale is to retain the runoff from a catchment with design peak runoff rate of 0.02 m3 /s. The longitudinal slope of the swale is to be 3% with side slopes of 4:1 (H:V). If the grassed swale has a Manning’s n of 0.24 (infrequently mowed grass) and a minimum infiltration rate of 150 mm/h, determine the length of swale required.

g .n

et

Solution From the given data: Qp = 0.02 m3 /s, m = 4, S = 0.03, n = 0.24, and f = 150 mm/h = 15 cm/h. Equation 12.11 gives the required swale length, L, as 5 8

L = 151, 400

3 5 Qp m 8 S 16 3

5

n 8 (1 + m2 ) 8 f

= 151, 400

3 5 5 (0.02) 8 (4) 8 (0.03) 16 3

5

(0.24) 8 (1 + 42 ) 8 15

= 314 m

(12.14)

This length of swale (314 m) is quite long. A ponded infiltration basin would require less area and would probably be more cost-effective.

12.3.3.2 Biofiltration swales In cases where the swale is to provide treatment of the surface runoff primarily through trapping a portion of the sediment and organic biosolids in the vegetative cover, the swale is called a biofiltration swale. In contrast to retention swales, biofiltration swales do not rely on infiltration for their effectiveness. A biofiltration swale is designed as a biofilter for which following design criteria are recommended (ASCE, 2012): ⊲ Minimum hydraulic residence time of 5 min ⊲ Maximum flow velocity of 0.3 m/s (1 ft/s)

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⊲ Maximum bottom width of 2.4 m (8 ft) ⊲ Minimum bottom width of 0.6 m (2 ft) ⊲ Maximum depth of flow no greater than one-third of the gross or emergent vegetation height for infrequently mowed swales, or no greater than one-half of the vegetation height for regularly mowed swales, up to a maximum height of approximately 75 mm (3 in.) for grass and approximately 50 mm (2 in.) below the normal height of the shortest plant species ⊲ Minimum length of 30 m (100 ft) ⊲ Longitudinal slope preferably in the range of 1%–2.5%, with a minimum of 0.5% and a maximum of 5%. When the longitudinal slope is less than 1%–2%, perforated underdrains should be installed or, if there is adequate moisture, wetland species should be established. If the slope is greater than 5%, check dams should be used to reduce the effective slope to approximately 2%–2.5%.

ww

Using these guidelines, the following design procedure is proposed for biofiltration swales (updated from ASCE, 2012): Step 1: Estimate the peak runoff rate for the design event and limit the discharge to approximately 0.03 m3 /s by dividing the flow among several swales, installing upstream detention to control release rates, or reducing the developed surface area to reduce the runoff coefficient and gain space for biofiltration. Step 2: Establish the slope of the swale. Step 3: Select a vegetation cover suitable for the site. Step 4: Estimate the height of vegetation that is expected to occur during the storm runoff season. The design flow depth should be at least 50 mm (2 in.) less than this vegetation height and a maximum of approximately 75 mm (3 in.) in biofiltration swales and 25 mm (1 in.) in filter strips. Step 5: Typically, biofiltration swales are designed as trapezoidal channels (skip this step for filter-strip design). When using a rectangular section, provide reinforced vertical walls. Step 6: For a trapezoidal cross section, select a side slope that is no steeper than 3:1 (H:V), with 4:1 (H:V) or flatter preferred. Step 7: Compute the bottom width and flow velocity. Limit the design velocity to less than 0.3 m/s (1 ft/s). Step 8: Compute the swale length using the design velocity from Step 7 and an assumed hydraulic detention time, preferably greater than 5 min. If the computed swale length is less than 30 m (100 ft), increase the swale length to 30 m (100 ft) and adjust the bottom width.

w .E asy En g

ine eri n

g .n

et

Biofiltration swales are low-cost stormwater control measures that have proven effective for controlling runoff pollution from land surfaces, especially highways and agricultural lands. Biofiltration swales are attractive options for agencies such as departments of transportation, since they are easily incorporated into the landscape, such as highway medians. The primary mechanisms for pollutant removal in swales are filtration by vegetation, settling of particulates, and infiltration into the subsurface zone. In general, biofiltration swales show good performance for removal of suspended solids, but are not considered efficient for removal of nutrients (Yu et al., 2001). Some studies have indicated that the most effective biofiltration swales have a minimum length of 75 m and a maximum longitudinal slope of 3% (e.g., Yu et al., 2001). EXAMPLE 12.6 A biofiltration swale is to be constructed on a 1% slope to handle a design runoff rate of 0.03 m3 /s. During the wet season, the swale is expected to be covered with grass having an average height of 130 mm with Class E retardance. Design the biofiltration swale.

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Solution From the given data: Q = 0.03 m3 /s, S0 = 0.01, and the average height of the vegetation is 130 mm. This given data covers the specifications in Steps 1 to 3 of the design procedure. Step 4: The design depth in the biofiltration swale should be at least 50 mm below the height of the vegetation (130 mm − 50 mm = 80 mm), with a maximum height of 75 mm. Therefore, in this case, the design flow depth is taken as 75 mm. Step 5: Use a trapezoidal section for the swale. Step 6: Use side slopes of 4:1 (m = 4), a bottom width b, and a flow depth y = 75 mm (= 0.075 m). The flow area, A, wetted perimeter, P, and hydraulic radius, R, are given by A = by + my2 = b(0.075) + (4)(0.075)2 = 0.075b + 0.0225   P = b + 2 1 + m2 y = b + 2 1 + 42 (0.075) = b + 0.618 R=

ww

0.075b + 0.0225 A = P b + 0.618

(12.15)

where 0.6 m < b < 2.4 m. Step 7: The Manning equation requires that

w .E asy En g

5

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

In this case,

0.03 =

5

1 1 (0.075b + 0.0225) 3 (0.01) 2 2 n (b + 0.618) 3

or

1 (0.075b + 0.0225)5 = 0.027 n3 (b + 0.618)2 For Class E retardance, Manning’s n can be estimated using Equation 5.55 as n=

ine eri n

(12.16)

1

1.22R 6

(12.17)

52.1 + 19.97 log(R1.4 S00.4 )

Simultaneous solution of Equations 12.15, 12.16, and 12.17 with S0 = 0.01 gives b = 4.44 m

g .n

et

which exceeds the maximum bottom width (for a uniform flow distribution) of 2.4 m. Repeated solution of Equations 12.15, 12.16, and 12.17 shows that a flow of 0.014 m3 /s will require a bottom width, b of 2.4 m, with a flow velocity (= Q/A) of 0.067 m/s. Since the flow velocity is less than the maximum velocity of 0.3 m/s, two swales each having a bottom width of 2.4 m and handling half the flow can be used. Step 8: Using a detention time of 5 min with the design velocity of 0.067 m/s gives the length, L, of the swale as L = Vt = (0.067)(5 * 60) = 20.1 m which is shorter than the minimum length of 30 m. Therefore, use a length of 30 m. In summary, two parallel biofiltration swales are required, each with a trapezoidal cross section with a bottom width of 2.4 m, side slopes of 4 : 1, and 30 m long.

Biofiltration systems sometimes operate by filtering diverted runoff through dense vegetation followed by vertical filtration through soil filter media. Surface treatment in these systems is by sedimentation, fine filtration, sorption, and biological uptake, and water collected in underdrains at the base of the filter media can be either discharged to receiving waters or stored for reuse. These types of systems are capable of high pollutant removal efficiencies (e.g., Hatt et al., 2009).

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12.3.4

ww

Vegetated Filter Strips

Vegetated filter strips, which are also called buffer strips or simply filter strips, are mildly sloping vegetated surfaces, usually grass, that are located between impervious surfaces (pollutant sources) and drainage channels. The typical layout of a filter strip is illustrated in Figure 12.14. Vegetated filter strips are designed to slow the runoff velocity from the impervious area, thereby reducing the peak runoff rate and increasing the opportunities for infiltration, sedimentation, and trapping of the pollutants. Vegetated filter strips are often used as pretreatment (to remove sediments) for other structural practices such as dry detention basins and exfiltration trenches. These areas are designed to receive overland sheet flow, and they provide little treatment for concentrated flows. Level spreaders are commonly used at the upstream end of vegetated filter strips to prevent concentrated flows. The design procedure for filter strips is the same as that for biofiltration swales, with the additional constraints that the average depth of flow be no more than 50 mm (2 in.) and that the hydraulic radius be taken as equal to the flow depth (ASCE, 2012). The width of the filter strip should be sufficiently limited to achieve a uniform-flow distribution. Grassed filter strips sometimes develop a berm of sediment at the upper edge that must be periodically removed. Mowing will maintain a thicker vegetative cover, thereby providing better sediment retention. Recommended areas for use are in agriculture, low-density developments, parking lots, and alongside roadways. Filter strips can be natural or manmade. Although studies indicate highly varying effectiveness, trees in filter strips can be more effective than grass filter strips alone because of their greater uptake and long-term retention of plant nutrients. Properly constructed forested and grass filter strips can be expected to remove more than 60% of the particulates and perhaps as much as 40% of plant nutrients in urban runoff (Dodson, 1998). In arid and semiarid climates, grass buffer strips need to be irrigated (Field et al., 2000).

w .E asy En g 12.3.5

Bioretention Systems

Bioretention systems, also referred to as rain gardens, bioinfiltration systems, and other names, are among the most versatile and widely used SCMs in the United States and many parts of the world. A bioretention cell, which is essentially a vegetated infiltration basin with a subsurface collection pipe or natural groundwater recharge, is identified as a preferred site practice for green building design and leadership in energy and environmental design (LEED) certification. A schematic diagram of a typical bioretention cell with a subsurface drainage system is shown in Figure 12.15. In cases where there is no subsurface drainage, the unit is sometimes defined as a bioinfiltration cell. Bioretention cells are typically located at the upper end of a stormwater-management system and are designed for smaller storms, which minimizes their footprint and allows for distributed treatment of the smaller volumes and distributed volume reduction. Bioretention cells are commonly found in parking-lot islands, roadway medians, and lawns. In cases where the drainage water is used for groundwater recharge, the biofiltration cell is located in a ground depression into which surface runoff flows, infiltrates, and percolates to the water table. This arrangement is typical of tropical rain gardens. The design criteria for bioretention cells are evolving and can vary with the design objectives (Hunt et al., 2012). Design objectives for bioretention cells can be grouped into hydrologic objectives and water-quality objectives. Hydrologic objectives include mitigation of peak runoff rates and storage of runoff volume,while water-quality objectives include

FIGURE 12.14: Filter strip

ine eri n

g .n

et

Surface runoff Impervious surface

Grass Vegetated filter strip

Drainage channel

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Design of Stormwater Control Measures

Vegetation and mulch

611

Runoff

Runoff

Filter layer (sand, fines, organics) Gravel layer Underdrain (some cases) Groundwater recharge (some cases) Water table

ww

w .E asy En g

reduction in the concentrations of particular pollutants. Commonly cited design criteria that are consistent with meeting most design objectives are as follows: ⊲ Bioretention cells should have sufficient volume to accommodate a specified waterquality volume. Some jurisdictions require that the water-quality volume be stored above ground, while other jurisdictions allow the water-quality volume to be stored below ground (within the bioretention system). The above-ground storage volume in a bioretention cell is called the bowl, and a bowl depth of 30 cm (1 ft) is commonly recommended. If the water-quality volume is to be stored in the bowl, then the surface area of the bowl can be adjusted accordingly. ⊲ Side slopes of the bowl should be no steeper than 4:1 (H:V). Side slopes should include a vegetative strip to capture coarse sediments, thereby reducing sediment load and the potential for clogging. ⊲ The filter medium should extend to a depth of 0.6–1.2 m (2–4 ft) below the land surface and be covered with a 75–100 mm (3–4 in.) layer of mulch. The layer containing the filter medium is commonly called the filter layer. The filter medium typically consists primarily of sand, with soil fines and organic matter (e.g., leaf compost) added to manipulate the infiltration capacity and organic content. The fines fraction in the filter medium should be in the range of 8%–12%. The filter medium should be surrounded by a filter fabric to prevent the intrusion of fines from the surrounding soil. The infiltration capacity of the filter medium should normally be in the range of 25–50 mm/h (1–2 in./h). The mulch layer above the filter layer attenuates heavy metals, reduces weed establishment, regulates soil temperature and moisture, and adds organic matter to the soil. ⊲ The filter medium must have sufficient capacity to drain the cell within 2–3 d. ⊲ Bioretention areas are typically planted with pollution- and water-tolerant trees, shrubs, and herbaceous species. These types of vegetation provide a carbon source, help facilitate microbial activity, and improve infiltration rates. ⊲ The bottom of the bioretention cell should be at least 30 cm (1 ft) above the seasonalhigh water table. ⊲ Some bioretention cells have underdrains while others do not. Underdrains typically consist of small-diameter plastic perforated pipes with diameters in the range of 100–150 mm (4–6 in.) and surrounded by coarse stones and wrapped in filter fabric to prevent clogging by fines. Underdrains should have sufficient capacity to convey the filtered water. In cases where water storage in the filter media is desirable for water-quality purposes, the drainage outlet can be configured such that outflow from the drainage system only occurs after water ponds to a certain depth within the filter medium.

ine eri n

g .n

et

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ww

The key parameter in predicting the performance of a bioretention cell in reducing the volume of runoff is the bioretention abstraction volume (BAV), which is defined as the storage volume in a bioretention cell that will fill before underdrain or overflow discharge from the system will occur (Davis et al., 2012). For bioretention cells without underdrains, the average BAV can be taken as the available storage in the vegetation root zone plus the bowl storage, while for bioretention cells with underdrains the average BAV can be taken as available storage in the entire filter medium. An operational bioretention cell located in the parking lot of a commercial area (in Nashville, NC) is illustrated in Figure 12.16. Runoff from the parking lot enters the bioretention cell (= rain garden) via curb openings on the lefthand and righthand sides of the bioretention cell shown in Figure 12.16. Erosion at curb openings is minimized by stone linings, as is apparent on the lefthand side of Figure 12.16. Bioretention cells must be maintained to ensure continuous effective operation. Routine visual inspection, infiltration testing (using an infiltrometer), and measurement of drain time are all measures that should be considered to identify malfunctioning cells (Asleson et al., 2009). Field measurements to assess the performance of some bioretention cells have shown that they can perform well in removing some contaminants, such as some metals (e.g., Zn), suspended solids, nutrients, and pathogenic microorganisms, while performing poorly in removing other contaminants, such as Cu (Trowsdale and Simcock, 2011; Zhang et al., 2011). In the case of nutrients (TN and TP), some bioretention systems are characterized by an average effluent concentration that is not significantly correlated to the influent concentrations (McNett et al., 2011).

w .E asy En g 12.3.6

Exfiltration Trenches

Exfiltration trenches, also called infiltration trenches, percolation trenches, or French drains, are common in urban areas with large impervious areas and high land costs. An exfiltration trench typically consists of a long narrow excavation, with a depth in the range of 1–4 m (3–12 ft), backfilled with gravel aggregate 2.5–7.6 cm (1.5–3 in.) in diameter and surrounded by a filter fabric to prevent the migration of fine soil particles into the trench, which can cause clogging of the gravel aggregate. Exfiltration trenches can be used for water-quality

FIGURE 12.16: Bioretention cell Source: Bill Hunt, North Carolina State University.

ine eri n

g .n

et

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control and for flood control. When an exfiltration trench is used for water-quality control, the volume of the trench is usually sized to accommodate the water-quality volume and the exfiltration rate from the trench must be sufficient to evacuate the trench within the typical storm interevent time, usually on the order of 3 d (72 h). When an exfiltration trench is used for flood control, the exfiltration capacity of the trench (when it is full) must be greater than or equal to the peak rate of runoff entering the trench. In many cases, an exfiltration trench designed for flood control will also satisfy the water-quality control requirement. In coastal areas with high sand dunes, exfiltration trenches can be a viable alternative to discharging surface runoff directly into coastal waters, since high dunes can provide the necessary separation between the bottom of the trench and the local water-table elevation, which is typically slightly higher than local sea level. Such systems are sometimes called dune infiltration systems (e.g., Bright et al., 2011). These systems can be particularly effective in reducing the levels of pathogenic microorganisms in coastal waters.

ww

12.3.6.1

General design guidelines

Exfiltration trenches can have their top elevation either at the ground surface or below ground. Surface trenches have been designed to underlay grass swales (e.g., Abida and Sabourin, 2006) and/or to receive sheet flow runoff directly from adjacent areas after it has been filtered by a grass buffer. In subsurface exfiltration trenches, runoff is collected by a stormwater inlet on top of a catch basin, and water collected in the catch basin is delivered to the below-ground exfiltration trench by a perforated pipe. In some cases, open-bottomed, high density polyethylene (HDPE) chambers have been used to distribute influent water within a trench. A subsurface exfiltration trench under a paved area is shown in Figure 12.17, where Figure 12.17(a) shows the trench under construction and Figure 12.17(b) shows the paved area above the trench after construction. The exfiltration trench consists of a catch basin in the center, with perforated pipe extending from both sides of the catch basin into the trench which is filled with rock and surrounded by a filter fabric. Exfiltration trenches can be used only at sites with porous soils, favorable site geology, and proper groundwater conditions. Site conditions that are favorable to exfiltration trenches are (Harrington, 1989; Stahre and Urbonas, 1989; ASCE, 2012):

w .E asy En g

FIGURE 12.17: Infiltration trench under paved area

(a) During construction

ine eri n

g .n

et

(b) After construction

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⊲ The hydraulic conductivity surrounding the trench exceeds 2 m/d (6 ft/d). ⊲ The distance between the bottom of the trench and the seasonal-high water table or impervious soil layer exceeds 0.6–1.2 m (2–4 ft). ⊲ Water-supply wells are more than 30 m (100 ft) from the trench (to prevent possible contamination). ⊲ The trench is located at least 6 m (20 ft) from building foundations (to avoid possible hydrostatic pressures on foundations or basements). ⊲ The ground slope downstream of the trench does not exceed 20%, which would increase the chance of downstream seepage and slope failure. ⊲ Exfiltration trenches 1 m (3 ft) wide and 1–2 m (3–6 ft) deep are usually the most efficient (ASCE, 2012). 12.3.6.2

ww

Design for flood control

When designed for flood control, exfiltration trenches are sized to exfiltrate water at the peak runoff rate when they are filled. The conventional design approach assumes that the water in the trench percolates through one-half of the trench height, saturated-flow conditions exist between the trench and the water table, and there is negligible outflow from the bottom of the trench (due to clogging). Using these assumptions, the total outflow rate, Qout [L3 T−1 ], from the two (long) sides of the trench is given by Darcy’s law (Section 14.2) as   H Qout = 2 Kt L = Kt HL (12.18) 2

w .E asy En g

where Kt is the trench hydraulic conductivity [LT−1 ], H is the height of the trench [L], L is the length of the trench [L], and the hydraulic gradient is assumed equal to unity. In cases where saturated-flow conditions do not exist between the trench and the water table, the assumption of a unit hydraulic gradient gives a conservative estimate of the trench outflow (Duchene et al., 1994). Taking t as the duration of the storm [T], the volume exfiltrated in time t, Vout [L3 ], is given by

ine eri n

(12.19)

Vout = Qout t = Kt HLt

The runoff volume entering the trench during time t, Vin [L3 ], can be estimated by the rational formula (see Section 10.4.1) as Vin = CiAt

g .n

(12.20)

et

where C is the runoff coefficient [dimensionless], i is the average intensity [LT−1 ] of a storm with duration t [T], and A is the area of the catchment contributing flow to the exfiltration trench [L2 ]. The storage capacity of the trench, Vstor [L3 ], is given by Vstor = nWHL

(12.21)

where n is the porosity in the trench [dimensionless], typically taken as 0.40 (ASCE, 2012). The trench dimensions must be such that Vin = Vstor + Vout

(12.22)

Combining Equations 12.19 to 12.22 yields CiAt = nWHL + Kt HLt

(12.23)

Solving for the trench length, L, gives L=

CiAt (nW + Kt t)H

(12.24)

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The rainfall intensity, i [LT−1 ], is related to the storm duration, t [T], by an intensity– duration–frequency (IDF) curve that typically has the form i=

a (t + b1 )c1

(12.25)

where a, b1 , and c1 are constants (see Section 9.2.2.2). The combination of Equations 12.24 and 12.25 indicates that the required trench length, L, varies as a function of the storm duration, t, and the design length should be chosen as the maximum length required for any storm duration (with a given return period). It is commonly recommended that a factor of safety of 2 or 3 be applied to the trench hydraulic conductivity, K, to account for soil clogging with time. EXAMPLE 12.7

ww

An exfiltration trench is to be designed to accommodate the runoff from a 1-ha commercial area with an average runoff coefficient of 0.7. The IDF curve for the design rainfall is i=

w .E asy En g

548 (t + 7.24)0.73

where i is the average rainfall intensity in mm/h and t is the storm duration in minutes. The trench hydraulic conductivity estimated from field tests is 15 m/d, the seasonal-high water table is 5 m below the ground surface, and local regulations require that a safety factor of 2 be applied to the trench hydraulic conductivity to account for clogging. Design the exfiltration trench.

Solution From the given data: C = 0.7, A = 1 ha = 10,000 m2 , and Kt = 15/2 = 7.5 m/d = 0.31 m/h (using a safety factor of 2). According to ASCE (2012) guidelines, the porosity, n, of the gravel fill in the trench can be taken as 40% (n = 0.4), and a trench width, W, of 1 m and height, H, of 2 m can be expected to perform efficiently. Substituting these values into Equation 12.24 gives L=

ine eri n

(0.7)i(10, 000)t 7000it CiAt = = (nW + Kt t)H (0.4 * 1 + 0.31t)(2) 0.8 + 0.62t

where i in m/h, and t in hours are related by i=

0.548 m/h (60t + 7.24)0.73

g .n

(12.26)

(12.27)

Combining Equations 12.26 and 12.27 gives the required trench length, L, as a function of the storm duration, t, as 3836t (12.28) L= (0.8 + 0.62t)(60t + 7.24)0.73 Taking the derivative with respect to t gives

et

[(0.8 + 0.62t)(60t + 7.24)0.73 ]3836 − 3836t[(0.8 + 0.62t)0.73(60t dL = dt

+ 7.24)−0.27 60 + (60t + 7.24)0.73 0.62] [(0.8 + 0.62t)(60t + 7.24)0.73 ]2 3836(0.8 + 0.62t)(60t + 7.24)0.73 − 168, 000t(0.8 + 0.62t)(60t

=

+ 7.24)−0.27 − 2378t(60t + 7.24)0.73 [(0.8 + 0.62t)(60t + 7.24)0.73 ]2

and the maximum-value criterion, dL/dt = 0, yields 3836(0.8 + 0.62t)(60t + 7.24)0.73 − 168, 000t(0.8 + 0.62t)(60t + 7.24)−0.27 − 2378t(60t + 7.24)0.73 = 0 which gives t = 0.759 h, and substituting into Equation 12.28 yields L = 127 m. On the basis of this result, a trench length of 127 m gives the trench volume required to handle the design storm without

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Design of Stormwater-Management Systems causing surface ponding. Since the seasonal-high water table is 5 m below the ground surface and the minimum allowable spacing between the bottom of the trench and the water table is 1.2 m, a (maximum) trench height of 5 m − 1.2 m = 3.8 m would still be adequate and would yield the shortest possible trench length (for the specified trench width of 1 m). Since the trench length is inversely proportional to the trench height (see Equation 12.24), using a trench height, H, of 3.8 m would give a required trench length, L, of 2 = 67 m L = 127 m * 3.8 Hence, a trench capable of handling the design storm is 1 m wide, 3.8 m high, and 67 m long. If the (vertical) side slopes are not stable for a trench of this depth, then the trench height and length can be adjusted according to their inverse proportionality.

ww

In designing an exfiltration trench, there must be reasonable assurance that the geologic formation beneath the trench can conduct water away from the trench as fast as water exfiltrates from the trench to the water table. Under normal circumstances, the water table beneath the trench will mound until there is sufficient induced lateral hydraulic gradient to conduct the exfiltrated water away. The trench will fail if the water-table mound rises above the trench bottom to interfere with the operation of the trench. A method to predict the mounding depth under exfiltration trenches is presented in Section 16.7.

w .E asy En g

12.3.6.3 Design for water-quality control When designed for water-quality control, exfiltration trenches must be sized to store the WQV and to be able to exfiltrate the WQV within a regulatory maximum evacuation time. Sizing for the storage requirement requires that n * L * W * H Ú WQV

(12.29)

where n is the porosity of the trench fill, L, W, and H are the length, width, and height of the trench, respectively [L], and WQV is the water-quality volume [L3 ]. To satisfy the evacuation time requirement, the following approximation is commonly used,

ine eri n

nLWH … Te Kt HL

(12.30)

g .n

where Kt is the trench hydraulic conductivity [LT−1 ] and Te is the required evacuation time [T], which is typically on the order of 3 d. On the left-hand side of Equation 12.30, the numerator is the storage volume in the exfiltration trench and the denominator is the estimated exfiltration rate when the trench is full. The trench dimensions L and H cancel out in Equation 12.30, which becomes nW (12.31) … Te Kt

et

where it is apparent that the evacuation time of the trench depends on only n, W, and Kt . EXAMPLE 12.8 An exfiltration trench captures the initial runoff from a 0.2-ha catchment. The regulatory water-quality depth is 2.5 cm and the hydraulic conductivity of the surrounding geologic formation is estimated as 4 m/d. It is proposed to use a trench that is 1 m wide and 2 m deep, and the porosity of the fill material is estimated to be 0.4. Determine the minimum required trench length to store the water-quality volume, and verify that the evacuation time of the trench meets the regulatory requirement of a maximum of 3 d. Solution From the given data: A = 0.2 ha = 2000 m3 , dWQV = 2.5 cm = 0.025 m, Kt = 4 m/d, W = 1 m, H = 2 m, and n = 0.4. The water-quality volume, WQV, is calculated as WQV = A * dWQV = (2000)(0.025) = 50 m3

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If L is the length of the trench required to store the WQV, then Equation 12.29 requires that nLWH = WQV (0.4)L(1)(2) = 50 m3 which yields L = 63 m. The evacuation time, Te , of the runoff stored in the trench is derived from Equation 12.31 as Te =

nW (0.4)(1) = = 0.1 d = 2.4 h Kt (4)

Therefore, a trench length of 63 m will provide adequate retention storage, and the evacuation time of 0.1 d is much less than the regulatory limit of 3 d.

ww

As indicated previously, an exfiltration trench sized for flood control usually meets the design requirements for water-quality control. However, a trench sized for water-quality control does not usually meet the requirements for flood control. The practicality of the chosen design goal will depend on space availability and subsurface conditions, which must be evaluated on a case-by-case basis. Accurate estimation of the drainage time for exfiltration trenches requires numerical simulation of subsurface flow in the unsaturated zone between the trench and the water table (e.g., Chahar et al., 2012); however, such numerical simulations are not commonly performed in the design of exfiltration trenches.

w .E asy En g 12.3.7

Subsurface Exfiltration Galleries

Underground exfiltration galleries, also called infiltration galleries, are similar in form and function to exfiltration trenches, with the main difference being that exfiltration galleries are installed in larger excavated areas rather than in narrow trenches. Surface runoff is distributed within exfiltration galleries via either a network of perforated pipes or by the installation of high-strength modular units with open bottoms that allow for exfiltration into the surrounding soil. The modular water-distribution units that are sometimes used are typically made of high-density polyethylene (HDPE), and an illustration of such a system being installed is shown in Figure 12.18. The distribution system within the subsurface infiltration gallery (either perforated pipes of modular open-bottom units) are typically embedded in gravel aggregate and surrounded by a filter fabric to prevent the migration of fine soil particles into the subsurface volume containing the exfiltration gallery. The specifications for the gravel aggregate in exfiltration galleries is the same as for exfiltration trenches. The volume of the exfiltration gallery is typically sized to accommodate the water-quality volume, and the exfiltration capacity is calculated using the same equations as used for exfiltration trenches.

ine eri n

g .n

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FIGURE 12.18: Exfiltration gallery

(a) During construction

(b) After construction

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12.4

Selection of SCMs for Water-Quality Control

Methods of controlling quality of stormwater runoff are categorized as nonstructural or structural stormwater control measures (SCMs). Nonstructural SCMs are practices that improve water quality by reducing the accumulation and generation of potential pollutants at or near their source. Structural SCMs involve building structures to control the quantity and quality of surface runoff. A comprehensive stormwater-management system should include a combination of structural and nonstructural components that are properly selected, designed, implemented, inspected, and maintained. 12.4.1

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Nonstructural SCMs

Nonstructural SCMs generally fall into the following four categories: planning and regulatory tools; source controls; maintenance and operational procedures; and educational and outreach programs. Planning and regulatory tools include hazardous materials codes, zoning, land-development and land-use regulations, water-shortage and conservation policies, and controls on the types of flow allowed to drain into municipal storm-sewer systems. Source controls include erosion and sediment control during construction, collection and proper disposal of waste materials, and controlled use of chemicals such as fertilizers, pesticides, and herbicides. Maintenance and operational procedures include turf and landscape management, street cleaning, catch-basin cleaning, road maintenance, and canal/ditch maintenance. Educational and outreach programs include distributing toxic checklists for meeting household hazardous-waste disposal regulations, producing displays and exhibits for school programs, distributing free seedlings for erosion control, and creating volunteer opportunities such as water-quality monitoring. Prevention practices such as planning and zoning tools to ensure setback, buffers, and open-space requirements can usually be implemented with ease at the planning stage of any development with a high degree of success.

w .E asy En g 12.4.2

Structural SCMs

ine eri n

Structural SCMs for controlling stormwater runoff generally fall into two main categories: retention systems and detention systems. Retention SCMs include exfiltration trenches, infiltration basins, grassed swales, vegetated filter strips, and retention ponds. Detention SCMs include dry and wet detention basins. When there is sufficient infiltration capacity, retention systems are generally more effective than detention systems in controlling the quality of surface runoff, since retention systems remove 100% of the pollutants contained in the retention volume while the removal efficiencies are generally less than 100% in detention systems. Although some water-quality control is provided by structures designed to control the peak runoff rate (e.g., Marttila and Kløve, 2009), additional structural controls are usually required to provide an adequate level of water-quality control. Important factors to be considered in selecting a SCM are the amount of runoff to be handled by the system and the soil type. Guidance for selecting SCMs is given in Table 12.1. Generally, well-designed SCMs listed in Table 12.1 can provide high pollutant removal for nonsoluble particulate pollutants, such as suspended sediment and sorbed trace metals.

g .n

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TABLE 12.1: Water-Quality Control System Selection Criteria

Stormwater control measures Exfiltration trenches Infiltration basins Grassed swales Filter strips Retention ponds Detention ponds

0–2 • • •

Contributing area (ha) 2–4 4–12 12–20 >20 • • •

• •

•

• •

• •

• •

Acceptable NRCS soil types A, B A, B A, B, C A, B, C B, C, D A, B, C, D

Source: USFHWA (2009).

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Much lower removal rates are achieved for soluble pollutants that include phosphorus and nitrogen. Each site has natural attributes that influence the type and configuration of the preferred stormwater-management system. For example, a site with sandy soils would suggest the use of infiltration practices such as retention areas integrated into a development’s open space and landscaping, while natural low areas offer opportunities for detention systems. The low-impact development (LID) approach to stormwater management has the goal of adapting postdevelopment hydrology to more closely mimic the predevelopment hydrology. In the LID approach, bioretention systems, surface sand filters, subsurface gravel wetlands, street trees, and porous asphalt systems have shown comparable performance to conventional systems (Roseen et al., 2009), and design guidelines for these systems are continually evolving as more experience is gained (e.g., Davis et al., 2009; Li et al., 2009; Roy-Poirier et al., 2010). 12.4.3

ww

Other Considerations

Stormwater-management systems typically consist of SCM components arranged in a sequence such that the “upstream” components maximize infiltration of surface runoff and the “downstream” components attenuate the peak discharge rate from the site. The minimization of directly connected impervious areas remains one of the most effective source controls that can be implemented to reduce the quantity and improve the quality of surface runoff at the source, and it can significantly reduce the capacity requirements in other runoff controls (e.g., Thorndahl et al., 2008). Under ideal conditions, the minimization of directly connected impervious areas can virtually eliminate surface runoff from storms with less than 13 mm (0.5 in.) of precipitation (Urbonas and Stahre, 1993). In deciding whether to use a wet or dry detention basin for water-quality control, an important consideration is whether nutrient removal (nitrogen and phosphorus) is an important requirement. This is particularly the case when the quality of the receiving water is sensitive to nutrient loadings. Properly designed wet detention basins (detention ponds) generally provide much better nutrient removal than dry detention basins, since many of the nutrients in surface runoff are in dissolved form and are not significantly affected by the sedimentation process in dry detention basins (Hartigan, 1989). This functional advantage of wet detention basins must be balanced against their greater land requirements. For example, the permanent pool of a wet detention basin can require anywhere from two to seven times more storage than the alternative dry detention basin.

w .E asy En g 12.5

Major Drainage System

ine eri n

g .n

Stormwater-management systems for quantity and quality control are typically designed for rainfall events with return periods of 25 years or less, and such systems are commonly referred to as minor drainage systems. These minor drainage systems will be overwhelmed by rainfall events with larger return periods, such as 100-year events, and the drainage pathways for larger rainfall events are accommodated in the major drainage system. The major drainage system includes features such as natural and constructed open channels and drainage easements such as floodplains. Major drainage systems must be planned concurrently with the design of the minor drainage system. In major urban drainage systems, concrete- and grass-lined channels are the most common, with grass lining usually preferred for aesthetic reasons. Concrete-lined channels have smaller roughness coefficients, require smaller flow areas, and are used when hydraulic, topographic, and right-of-way constraints are important considerations.

et

Problems 12.1. A 100 m * 25 m detention pond is proposed for a 7.5-ha site. The detention pond is 2.0 m deep under control (pre-storm) conditions and the water-quality depth is 2.0 cm. Storage in the detention pond is to be discharged offsite via a triangular weir whose vertex is at the control elevation, and the flow rate over the weir, Q [m3 /s], is given by

5

Q = 0.08H 2

where H is the height of the water surface above the vertex of the weir [m]. The average annual rainfall at the site is 120 cm, and the average runoff coefficient is 0.75. Determine the evacuation time and the average detention time in the pond.

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Chapter 12

Design of Stormwater-Management Systems The postdevelopment detention basin is to be a wet detention reservoir drained by an outflow weir. The elevation versus storage in the detention basin is

12.2. An outlet from a detention pond consists of a 1.50-m diameter riser with a 0.40 m * 0.40 m square orifice between elevations 1.00 m and 1.40 m. Flood flows discharge over the top of the riser (as weir flow), and the elevation of the top of the riser is 2.00 m. Determine the outlet discharges for pond elevations between 1.00 m and 2.40 m in 0.20-m intervals. 12.3. Show that if the bottom of a storage reservoir is a rectangle of dimensions L * W, the longitudinal cross section is a trapezoid with base W and side slope angle α, and the transverse cross section is rectangular with base L, then the stage-storage relation is given by

where S is the storage, and h is the depth. 12.4. A 5-ha residential development with 0.4-ha lots is to be drained such that the first 25 mm of runoff is retained onsite and the postdevelopment peak discharge rate from the site is not to exceed 1 m3 /s for the 25-year 1-d rain event. The natural soil on the site is sandy loam. The IDF curve for the region is given by

ww

7836 34.11 + 0.7052t

where i is the average rainfall intensity in mm/h, and t is the duration in minutes. The retention and detention areas are to have square bases with side slopes not greater than 3:1 (H:V) and regulations require that retention areas drain in less than 72 h. Determine the percentage of the site area that should be set aside for retention and describe how this portion of the site should be developed. If the retention area is also to be used for the detention basin, make a preliminary estimate of the length and elevation of the weir crest in the discharge structure. Describe the physical appearance of the discharge structure. 12.5. The runoff hydrographs from a site before and after development are as follows: Time (min)

Before (m3 /s)

After (m3 /s)

0 30 60 90 120 150 180 210 240 270 300 330 360 390

0 2.0 7.5 1.7 0.90 0.75 0.62 0.49 0.30 0.18 0.50 0 0 0

0 3.5 10.6 7.5 5.1 3.0 1.5 0.98 0.75 0.62 0.51 0.25 0.12 0

0.0 0.5 1.0 1.5

0 11,022 24,683 41,522

3

Q = 1.83bh 2 where Q is the overflow rate in m3 /s, b is the crest length in m, and h is the head on the weir in m. Determine the crest length of the weir for the detention basin to perform its desired function. What is the maximum watersurface elevation expected in the detention basin? 12.6. The postdevelopment runoff hydrograph from a 5-km2 development is approximated by an NRCS triangular hydrograph with a base of 120 min, a peak of 14.8 m3 /s, and the peak occurs at a time of 40 min. The water-quality volume corresponding to the first 2.5 cm of runoff is to be retained in a dry retention area, and the maximum allowable discharge rate is 8 m3 /s. The park area that will be used for retention has an area of 0.5 km2 and the berm surrounding the park has approximately vertical side slopes. Determine the geometry and specifications of the discharge structure in the retention area. For the design storm, what is the maximum depth of ponding expected in the retention area? 12.7. The runoff from the site described in Problem 10.43 is to be routed to a dual-purpose dry detention reservoir for both flood control and water-quality control. The predevelopment and postdevelopment runoff volumes can be estimated using the USEPA empirical formula (Equation 10.150), where the predevelopment imperviousness and depression storage are 10% and 2.5 mm, respectively. The peak predevelopment runoff rate is 3 m3 /s, and the stage-storage relation of the detention reservoir is:

w .E asy En g

i=

Storage (m3 )

where the weir crest is at elevation 0.0 m, which is also the initial elevation of the water surface in the detention basin prior to runoff. The performance of the weir is given by

L 2 h + (LW)h tan α

S=

Elevation (m)

ine eri n

g .n

Elevation (m)

Storage (m3 )

0.0 0.5 1.0 1.5

0 300 600 900

et

The basin is to be drained by a circular orifice, where the discharge, Q, is related to the area of the opening, A, and the stage, h, by

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(12.32)

Make a preliminary estimate of the required diameter of the orifice opening. 12.8. A 5-ha development contains 2.5 ha of open area and 2.5 ha occupied by buildings, where all roof drainage is onto the pervious areas of the site. The site is to be graded such that the lowest ground elevation is 6.200 m and the highest ground elevation is 6.300 m. A berm is to be placed around the development such that the runoff from the 25-year 3-d storm is entirely retained onsite. The 25-year 3-d rainfall amount for the area is 25.4 cm and the average onsite soil storage is 7 cm. (a) Determine the required elevation of the top of the berm. (b) Consider an alternative design in which all of the runoff is routed to a detention area having the same stage-storage characteristics as the graded site. The discharge from the detention area is to be made via a compound weir, where the bottom elevation of the V notch is 6.200 m and the top elevation of the V notch corresponds to a storage equal to 1.3 cm of runoff from the entire site. If the storage corresponding to 1.3 cm of runoff is to be discharged over at least 24 h, determine the required V-notch angle in the discharge structure. 12.9. A retention area is to be designed to accommodate the runoff from a 10-year 45-min storm. The average rainfall intensities in 5-min intervals are as follows: 24.1, 30.4, 41.6, 67.4, 169, 97.3, 51.6, 36.2, and 26.8 mm/h. If the catchment area is 5 ha, and the infiltration capacity of the soil remains constant at 50 mm/h during the storm, estimate the storage volume in the retention area required to store the excess runoff. If the retention area covers 700 m2 , what depth of ponding would you expect to occur during the design storm? 12.10. The IDF curve for Miami, Florida, is given by the relation 7836 i= 48.6T −0.11 + t(0.5895 + T −0.67 )

ww

(lake), where the control elevation in the lake is 4.00 m NGVD, the water surface in the lake is at the control elevation immediately prior to the occurrence of the design storm, the dimensions of the lake at the control elevation are 15 m * 15 m, and the side slopes are 6 : 1. Determine the lake setback distance required to completely retain the 25-year 3-d runoff in the lake. 12.11. An infiltration basin is to retain the first 2.5 cm of runoff from a 20-ha catchment. The area to be used for the infiltration basin is turfed, and the soil has a minimum infiltration rate of 100 mm/h. If the retained runoff is to infiltrate within 36 h, determine the surface area to be set aside for the basin. 12.12. The IDF curve for rainfall in Miami-Dade County (Florida) is given by i=

w .E asy En g

where i is the average rainfall intensity in mm/h, T is the return period in years, and t is the rainfall duration in minutes. A 4-ha residential development consists primarily of single-family residences on 0.2-ha (1/2-acre) lots, the native soil is classified as Group B, the maximum distance from the catchment boundary to the drainage outlet is 200 m, and the average slope along the drainage path is 0.5%. The drainage outlet for the development is located adjacent to the roadway and has 0.8-ha of directly connected impervious area, consisting of asphalt pavement with a maximum drainage length of 80 m and an average slope of 0.7%. If the minimum allowable time of concentration is 5 min, determine the peak runoff rate from the catchment for a 25-year return period. Local drainage regulations require that the stormwater-management system be designed for a 25-year 3-d rainfall event. Estimate the volume of runofffrom the site to be stored in a retention pond

621

12.13.

12.14.

12.15.

12.16. 12.17.

12.18.

12.19.

7836 48.6T −0.11 + t(0.5895 + T −0.67 )

where i is the rainfall intensity in mm/h, T is the return period in years, and t is the rainfall duration in minutes. Estimate the maximum rainfall amount for all storms with a return period of 10 years. If a development site is characterized by a runoff coefficient of 0.4 and has a retention capacity for 50 mm of runoff, will any 10-year runoff be discharged from the site? Derive Equation 12.11 for a swale with a triangular cross section. Derive Equation 12.12 for a swale with a trapezoidal cross section. A triangular-shaped swale is to retain the runoff from a catchment with a design peak runoff rate of 0.01 m3 /s. The longitudinal slope of the swale is to be 1.5% with side slopes of 5 : 1 (H : V). If the grassed swale has a Manning’s n of 0.030 and a minimum infiltration rate of 200 mm/h, determine the length of swale required. Repeat Problem 12.15 for a trapezoidal swale with a bottom width of 1 m. A biofiltration swale is to be laid on a 2% slope to handle a design peak runoff rate equal to 0.002 m3 /s. During the wet season, the swale is expected to be covered with grass having an average height of 100 mm with Class E retardance. Design the biofiltration swale. An exfiltration trench is to be designed to accommodate a peak runoff rate of 0.03175 m3 /s. Percolation tests indicate that the trench hydraulic conductivity is approximately 10 m/d and the required factor of safety is 2. Calculate the required trench dimensions. An exfiltration trench is to be designed to handle the runoff rate from a 3-ha residential area with a runoff coefficient of 0.5. The IDF curve for the design rainfall is given by 403 i= (t + 8.16)0.69

ine eri n

g .n

et

where i is the average rainfall intensity in mm/h, and t is the storm duration in minutes. The trench hydraulic

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Chapter 12

Design of Stormwater-Management Systems

conductivity is 35 m/d, and the seasonal-high water table is 4.6 m below the ground surface. Assuming a safety factor of 3 for the trench hydraulic conductivity, design the exfiltration trench. 12.20. An exfiltration trench is to be used to drain a residential area in Coral Gables, Florida. The trench will be adjacent to a main roadway and is designed to remove runoff from all 10-year storms, for which the IDF curve is given by the local Public Works Department as 7836 37.7 + 0.803t where i is the average rainfall intensity in mm/h, and t is the rainfall duration in minutes. The trench has a catchment area of 0.03 ha, of which 35% is directly connected impervious area (C = 0.9), and 65% is effectively pervious area (C = 0.4). If the trench hydraulic conductivity is measured at 5 m/d, the width of the trench is 1 m, and the height of the trench is 3 m, determine the required length of the trench. What is the total runoff depth that will be retained by the trench? 12.21. An exfiltration trench is to be designed to handle runoff from a 0.5-ha residential area with an average runoff coefficient of 0.6. The IDF curve for the trench location is given by i=

ww i=

w .E asy En g 7836

i=

Q K(2H2 Du − D2u + 2H2 Ds )

(12.33)

where Q is the exfiltration rate (L3 T−1 ), K is the (modified) trench hydraulic conductivity (T−1 ), H2 is the vertical distance between the water table and the ground surface (L), Du is the unsaturated depth of the trench below the backfill (L), and Ds is the saturated depth of the trench.

308.5 48.6T −0.11 + t(0.5895 + T −0.67 )

where i is the average intensity in inches per hour, T is the return period in years, and t is the rainfall duration in minutes. Consider a site where the ground elevation is at 2.088 m NGVD, the wetseason water table is at elevation 1.250 m NGVD, the bottom of the trench is at elevation −2.484 m NGVD, and there is 0.305 m of backfill used at the top of the trench. The width of the trench is to be 1 m, slug tests indicate that the trench hydraulic conductivity is 1.74 * 10−3 s−1 , the runoff coefficient is 0.6, and the catchment area is 0.8 ha. Design the exfiltration trench for a return period of 5 years. (c) An alternative way to estimate the length of an exfiltration trench is to optimize the length according to the relation

48.6T −0.11 + t(0.5895 + T −0.67 )

where i is the average rainfall intensity in mm/h, T is the return period in years, and t is the rainfall duration in minutes. Slug tests indicate that the trench hydraulic conductivity is 8 m/d, the seasonal-high water table is 3.5 m below the ground surface, and a safety factor of 2 is to be used in the design. If the trench is required to prevent flooding caused by any 10-year storm, and is required to retain the first 25 mm of runoff, design the trench. [Hint: In cases where a specified depth of runoff is to be retained by a trench, ASCE (2012) recommends that the trench be designed using the IDF curve with at least a return period in which the corresponding rainfall depth occurs in 1 h.] 12.22. Review the design of exfiltration trenches in MiamiDade County as described by Chin (2004).∗ A typical exfiltration trench in South Florida is illustrated in Figure 12.19, and the required length, L, of the trench is locally estimated using the relation L=

(a) Derive Equation 12.33 and state the assumptions associated with using this equation. (b) The exfiltration rate, Q, in Equation 12.33 is routinely assumed to be equal to the peak runoff rate corresponding to an assumed time of concentration of 10 min. The IDF curve for Miami-Dade County is given by

L=

at∗

(12.34)

(b + t∗ )(c + dt∗ )

ine eri n where

a=

b=

0.130CA

0.5895 + T −0.67 2916T −0.11

0.5895 + T −0.67

c = 0.5WDu

g .n

d = 2K[Du (H2 − 0.5Du ) + Ds H2 ] ' bc ∗ t = d Derive Equation 12.34 and state the associated assumptions. This approach is identical to the “mass-curve” approach used by the regulatory agency in Miami-Dade County. (d) For the site conditions given in Part (b), design the trench using Equation 12.34. Compare the trench length with that determined using Equation 12.33 and state which trench length you would use in the final design. Explain why. (e) For a return period of 5 years and a time of concentration of 10 min, determine the trench conditions under which it would always be preferable to use Equation 12.34 rather than Equation 12.33 in the final design.

et

∗ This reference can be found on the web at: http://pubs.usgs.gov/of/2004/1346/pdf/ofr20041346.pdf.

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623

Base Exfiltration trench

Select backfill Pea gravel

Ds

Backfill

15 cm (6 in.)

Gravel

15 cm (6 in.) minimum

Pipe cover

Pipe

30 cm (12 in.) minimum

Perforated pipe diameter

Coarse rock

30 cm (12 in.) minimum

Pipe bed

H2 Du

30 cm (12 in.)

Seasonal-high water table

Trench width W

ww

FIGURE 12.19: Exfiltration trench Source: South Florida Water Management District (2000).

w .E asy En g

12.23. An exfiltration trench is to retain the initial runoff rate from a 0.5-ha catchment where the regulatory waterquality depth is 2.5 cm. The hydraulic conductivity of the surrounding sandy aquifer is 10 m/d, and the trench is to

be 1 m wide and 1.5 m deep. The porosity of fill material is estimated as 0.4. What is the minimum required trench length? Does the evacuation time meet the regulatory requirement of a maximum of 3 d?

ine eri n

g .n

et

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C H A P T E R

13

Estimation of Evapotranspiration 13.1

ww

Introduction

On an annual basis, evapotranspiration is usually the largest consumptive use of water and is usually the second most important quantity in regional water budgets, second only to rainfall. It is not unusual for evapotranspiration to consume around 70% of rainfall on an annual basis. Evaporation is the process by which water is transformed from the liquid phase to the vapor phase, and transpiration is the process by which water moves through plants and evaporates through leaf stomatae. In cases where the ground surface is covered by vegetation, it is usually difficult to differentiate between evaporation from the ground surface and transpiration through plants. The combined process of evaporation and transpiration is called evapotranspiration and is commonly denoted by the acronym “ET.” Three standard ET rates are commonly used: (1) potential ET, (2) reference-crop ET, and (3) actual ET. These are defined as follows:

w .E asy En g

Potential ET: Potential evapotranspiration (PET) is used synonymously with the term potential evaporation, which is defined as the quantity of water evaporated per unit area, per unit time, from an idealized, extensive free water surface under existing atmospheric conditions. Potential evaporation is typically used as a measure of the meteorological control on evaporation from an open water surface (e.g., lake, reservoir) or bare saturated soil. Penman (1956) originally defined potential evapotranspiration as the amount of water transpired per unit time by a short green crop, completely shading the ground, of uniform height and never short of water.

ine eri n

Reference-Crop ET: Reference-crop evapotranspiration is the rate of evapotranspiration from an area planted with a specific (reference) crop, where water availability is not a limiting factor. Reference-crop evapotranspiration is used as a measure of evapotranspiration from a standard vegetated surface. Grass and alfalfa are by far the most commonly used reference crops in hydrologic practice. Grass is characteristic of short vegetation that is 8–15 cm (3–6 in.) in height, and alfalfa is characteristic of tall vegetation that is greater than 50 cm (20 in.) in height.

g .n

et

Actual ET: The actual evapotranspiration is the evapotranspiration that occurs under actual soil, ground-cover, and water-availability conditions. Actual evapotranspiration is sometimes referred to as crop evapotranspiration. Typically, the actual evapotranspiration is taken as the reference-crop evapotranspiration multiplied by a crop coefficient. Numerous empirical and semi-empirical methods of estimating standard ET rates have evolved, and these can be broadly classified as radiation, temperature, combination, and evaporation-pan methods. Radiation and temperature methods relate ET solely to net radiation (solar plus longwave) and air temperature, respectively; combination methods account for both energy utilization and the processes required to remove the water vapor once it has evaporated; and evaporation-pan methods relate reference-crop ET to measured water evaporation from standardized open pans. Combination methods are widely regarded as the most accurate in estimating actual and reference-crop ET, and there is a broad consensus among hydrologists that the physically based Penman–Monteith (PM) equation provides the best description of the evapotranspiration process. 13.2

Penman–Monteith Equation

The basic hypothesis of the Penman–Monteith equation (Monteith, 1965; 1981) is that transpiration of water through leaves is composed of three serial processes: the transport of 624

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Penman–Monteith Equation

625

water through the surface of the leaves against a surface or canopy resistance, rs ; molecular diffusion against a molecular boundary-layer resistance, rb ; and turbulent transport against an aerodynamic resistance, ra , between the layer in the immediate vicinity of the canopy surface and the planetary boundary layer. The boundary-layer resistance, rb , for water vapor is usually much smaller than the aerodynamic resistance, ra , or the surface resistance, rs , and can be ignored in comparison with both other resistances. The Penman–Monteith equation for estimating the evapotranspiration rate, ET [LT−1 ], from vegetated surfaces where availability of water is not a limiting factor is given by ⎡ ⎤ es − ea (Rn − G) + ρa cp ⎥ 1 ⎢ r ⎢ ⎥   a ET = (13.1) ⎥ ⎢ rs ⎦ ρw λ ⎣  + γ 1 + ra

ww

where ρw is the density of water [ML−3 ], λ is the latent heat of vaporization of water [EM−1 ],∗  is the gradient of the saturated vapor pressure versus temperature curve [FL−2 Θ−1 ];† Rn is the net radiation flux (solar plus long wave) [EL−2 T−1 ], G is the soil heat flux [EL−2 T−1 ], ρa is the density of moist air [ML−3 ], cp is the specific heat of moist air [= 1.013 kJ/(kg◦ C)]; es is the saturation vapor pressure [FL−2 ], ea is the ambient vapor pressure [FL−2 ], ra is the aerodynamic resistance to vapor and heat diffusion [TL−1 ], γ is the psychrometric constant [FL−2 Θ−1 ], and rs is the bulk surface resistance [TL−1 ]. Equation 13.1 is dimensionally homogeneous, and any consistent set of units can be used. The Penman–Monteith equation is commonly associated with a standardized methodology used to estimate the parameters in the equation. The Penman–Monteith equation is applicable only when the availability of water is not a limiting factor; hence, the estimated ET depends on weather-related and crop parameters only. Recommended methods to be used in estimating the parameters in the Penman– Monteith equation are given in the following sections.

w .E asy En g 13.2.1

Aerodynamic Resistance

ine eri n

The aerodynamic resistance, ra , measures the resistance to vapor flow from air flowing over vegetated surfaces. In the absence of significant thermal stratification, ra [TL−1 ] can be estimated by (Pereira et al., 1999) 



zh − d zm − d ln ln z0m z0h ra = (13.2) 2 k uz

g .n

et

where zm is the height of wind measurements [L], d is the zero-plane displacement height [L], zh is the height of air temperature and humidity measurements [L], z0m is the roughness length governing momentum transfer [L], z0h is the roughness length governing heat and ´ an ´ constant [dimensionless], and uz is the wind speed vapor transfers [L], k is the von Karm measured at height zm [LT−1 ]. Typically, for full-cover uniform crops, d and z0m are related to the crop height, h [L], by (Brutsaert, 1982; ASCE, 1990) d=

2 h 3

and

z0m = 0.123h

(13.3)

Since momentum transfer governs the heat and vapor transfer, the roughness height z0h is assumed to be a function of z0m , where z0h = az0m

(13.4)

∗ “E” refers to the standard unit of energy in the system of units being used (e.g., joules in SI system).

† “Θ” refers to the standard unit of temperature in the system of units being used (e.g., ◦ C or K in SI system).

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Chapter 13

Estimation of Evapotranspiration

For tall and partially covering crops a = 1 and for fully covering crops a = 0.1 (Monteith, 1973; Campbell, 1977; Thom and Oliver, 1977; Brutsaert, 1982; Allen et al., 1989; ASCE, 1990). Three-cup and propeller anemometers are both widely used to measure wind speed, and measurements by both types are reliable, provided that the mechanical parts are functioning properly. Anemometers are typically placed at a standard height of either 5 m (16 ft) or 10 m (32 ft) above ground in meteorologic applications and 2 m (6.6 ft) or 3 m (10 ft) in agrometeorologic applications; for ET applications, wind speeds at 2 m (6.6 ft) above the ground surface are generally required. To adjust wind speed collected at elevations other than the standard height of 2 m (6.6 ft), the following logarithmic wind-speed profile can be used above a short grassed surface to adjust measurements (Allen et al., 1998) u2 =

ww

4.87 uz ln (67.8z − 5.42)

(13.5)

where u2 is the wind speed 2 m (6.6 ft) above the ground surface, and uz is the wind speed measured z m above the ground surface. Care should be taken that measured wind speeds are characteristic of the local area. In cases where the wind speed is not measured over a short grassed surface, the following equation should be used to adjust the wind speed:   2 − d ln z u2 =  0m  uz (13.6) z − d ln z0m

w .E asy En g

where d and z0m are given by Equation 13.3. It is important to note that Equation 13.5 is derived from Equation 13.6 by assuming a crop height, h, of 0.12 m (4.7 in.). The standard grass reference crop has a height, h, of 0.12 m (4.7 in.). Using the standardized height for wind speed, temperature, and humidity as 2 m (zm = zh = 2 m), assuming a fully covering ´ an ´ constant, k, as 0.41, Equations 13.2 to 13.4 give crop (a = 0.1), and taking the von Karm



2 − 23 (0.12) 2 − 32 (0.12) ln ln 0.123(0.12) (0.1)0.123(0.12) 208 ra = (13.7) = 2 u2 (0.41) u2

ine eri n

g .n

This equation is commonly used as a starting point to estimate the aerodynamic resistance of a standard grass reference crop. The aerodynamic resistance of open water can be estimated using the equation (Thom and Oliver, 1977) 

zm 2 4.72 ln z0 ra = (13.8) s/m 1 + 0.536u2

et

where zm is the measurement height of meteorological variables above the water surface, and z0 is the aerodynamic roughness of the surface. Taking zm = 2 m and z0 = 1.37 mm (Thom and Oliver, 1977), the aerodynamic resistance is determined by the wind speed, u2 . The aerodynamic resistance of open water is generally higher than found for either short vegetation or forest, and reflects a greater resistance to vapor transport due to lower levels of turbulence. 13.2.2

Surface Resistance

The surface resistance, rs , describes the resistance to vapor flow through leaf stomatae, and is sometimes referred to as the canopy resistance or stomatal resistance. The surface resistance, rs , varies as leaf stomatae open and close in response to micrometeorological conditions and is dependent on the particular plant species. It is generally assumed that the surface resistances of trees are greater than those of shorter vegetation, since trees tend to have stomatal

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Penman–Monteith Equation

627

control, while shorter vegetation does not. An acceptable approximation for estimating the surface resistance, rs [TL−1 ], of dense full-cover vegetation is (Allen et al., 1998) rs =

ww

rl LAIactive

(13.9)

where rl is the bulk stomatal resistance of a well-illuminated leaf [TL−1 ], and LAIactive is the active (sunlit) leaf-area index (dimensionless). The bulk stomatal resistance, rl , of a single well-illuminated leaf typically has a value on the order of 100 s/m, and the leaf area index, LAI, is defined as the ratio of the surface area of the leaves (upper side only) to the projection of the vegetation on the ground surface. The active leaf area index includes only the leaf area that actively contributes to the surface heat and vapor transfer, typically the upper, sunlit-portion of a dense canopy. LAI values for various crops differ widely, but values of 3–5 are common for many mature crops. The LAI for grass and alfalfa reference crops can be estimated using the following relations (Allen et al., 1998):  24h, for clipped grass LAI = (13.10) 5.5 + 1.5 ln h, for alfalfa

w .E asy En g

where h is the height of the vegetation in meters. For various reference crops (grass, alfalfa), a general equation for estimating LAIactive is (Allen et al., 1998) LAIactive = 0.5LAI

(13.11)

For a standard grass reference surface with a height of 0.12 m and a stomatal resistance of 100 s/m, the bulk surface resistance, rs , is given by rs =

100 = 70 s/m 0.5(24)(0.12)

ine eri n

(13.12)

The bulk surface resistance, rs , for large open-water bodies such as lakes is zero by definition, since there are no restrictions to water moving between the liquid and vapor phases at the surface of the lake. 13.2.3

Net Radiation

g .n

The net radiation, Rn , is approximately equal to the net solar (shortwave) radiation, Sn , plus the net longwave radiation, Ln , hence Rn = Sn + Ln

et

(13.13)

The net radiation, Rn , can be measured directly using a net radiometer, which senses both short- and longwave radiation using upward- and downward-facing sensors. Net radiation is difficult to measure accurately because net radiometers are hard to maintain and calibrate, resulting in systematic biases (EWRI, 2005). As a consequence, the net radiation, Rn , is often predicted using empirical equations which are usually highly accurate (EWRI, 2005). The net radiation, Rn , is normally positive during the day and negative during the night, and the total daily value of Rn is almost always positive, except in extreme conditions at high latitudes (Allen et al., 1998). 13.2.3.1 Shortwave radiation The net shortwave radiation, Sn , is equal to the fraction of the incoming solar radiation that is not reflected by the ground cover and is given by Sn = (1 − α)Rs

(13.14)

where α is the albedo or canopy reflection coefficient, defined as the fraction of shortwave radiation reflected at the surface, and Rs is the total incoming solar radiation, also known

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Estimation of Evapotranspiration TABLE 13.1: Typical Albedos

Albedo, α Land cover Open water Tall forest Tall farm crops Cereal crops Short farm crops Grass and pasture Bare soil

ww

Minimum

Maximum

0.08 0.11 0.15 0.20 0.20 0.20 0.10

0.08 0.16 0.20 0.26 0.26 0.26 0.35

as the insolation. Typical albedos for various surfaces are given in Table 13.1. The albedo of an open water body is sometimes taken as a function of Rs [MJ m−2 d−1 ] using the relation (Koberg, 1964; Ojha et al., 2011) αopen water = 0.127 exp(−0.0258Rs )

w .E asy En g

(13.15)

Total incoming solar radiation, Rs , is commonly measured using a pyranometer, which measures the incoming shortwave radiation to a solid angle in the shape of a hemisphere oriented upward. When a pyranometer is oriented downward it measures the reflected shortwave radiation and is called an albedometer. When two pyranometers are used together, one oriented upward and the other downward, the net shortwave radiation, Sn , is measured directly and the instrument is called a net pyranometer. Fairly accurate estimates of Rs can be derived from geostationary operational environmental satellites (GOES), which provide an alternative data source for Rs when ground-based measurements are limited and high accuracy in ET estimates is not required (e.g., Jacobs et al., 2009). Estimates of Rs derived from GOES measurements have been shown to be within 10% of ground measurements (Paech et al., 2009). Empirical formulae relating Rs directly to measurements of maximum and minimum temperature, Tmax and Tmin , are very convenient in areas where only Tmax , Tmin , and daily rainfall are recorded on a regular basis. Empirical estimates of Rs can be obtained using the relation (Hargreaves and Samani, 1982)

ine eri n

Rs = (KT)(Tmax − Tmin )0.5 S0

g .n

(13.16)

et

where KT is an empirical coefficient, S0 is the extraterrestrial radiation (MJ m−2 d−1 ), Tmax is the maximum daily temperature (◦ C), and Tmin is the minimum daily temperature (◦ C). Hargreaves (1994) recommended using KT = 0.19 for coastal regions, and KT = 0.162 for interior regions. Several other methods for estimating KT have been proposed (Allen, 1997; Samani, 2000; Samani et al., 2011; Thepadia and Martinez, 2012); however, these methods typically give similar estimates of KT at lower land elevations. The conventional method for estimating the total incoming solar radiation, Rs , in ET applications is the Angstrom formula, which relates the solar radiation to extraterrestrial radiation and relative sunshine duration by   n Rs = as + bs S0 (13.17) N where n is the number of bright-sunshine hours per day (h), N is the total number of daylight hours in the day (h), S0 is the extraterrestrial radiation (MJ m−2 d−1 ), as is a regression constant expressing the fraction of extraterrestrial radiation reaching the earth on overcast days (n = 0), and as + bs is the fraction of extraterrestrial radiation reaching the earth on clear days (n = N). Depending on atmospheric conditions (humidity, dust) and solar declination (latitude and month), the Angstrom values as and bs will vary. Where no actual solar radiation data are available and no calibration has been carried out to estimate as and bs ,

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629

the values as = 0.25 and bs = 0.50 are recommended. Sunshine duration, n, is commonly recorded using a Campbell–Stokes heliograph in which a glass globe focuses the radiation beam to a special recording paper and a trace is burned on the paper as the sun is moving. No trace is recorded in the absence of bright sunshine. The extraterrestrial solar radiation, S0 , in Equation 13.17 can be estimated using the following equation (Duffie and Beckman, 1980): S0 =

ww

24(60) Gsc dr (ωs sin φ sin δ + cos φ cos δ sin ωs ) π

(13.18)

where S0 is in MJ/(m2 ·d), Gsc is the solar constant equal to 0.0820 MJ/(m2 ·min), dr is the inverse relative distance between the earth and the sun (dimensionless), ωs is the sunset-hour angle in radians, φ is the latitude in radians, and δ is the solar declination in radians. Variables in Equation 13.18 can be estimated using the following relations:   2π dr = 1 + 0.033 cos J (13.19) 365   2π δ = 0.4093 sin J − 1.405 (13.20) 365

w .E asy En g

ωs = cos−1 (− tan φ tan δ)

(13.21)

where J is the number of the day in the year between 1 (1 January) and 365 or 366 (31 December), and J is commonly referred to as the Julian day. Equation 13.18 is appropriate for calculating the extraterrestrial solar radiation, S0 , over daily time scales; however, for hourly or shorter durations the following modified equation is recommended (Allen et al., 1998): S0 =

ine eri n

24(60) Gsc dr [(ω2 − ω1 ) sin φ sin δ + cos φ cos δ(sin ω2 − sin ω1 )] π

(13.22)

where ω1 and ω2 are the solar angles at the beginning and end of the time interval. The number of daylight hours, N, is directly related to the sunset-hour angle, ωs , and is given by (Duffie and Beckman, 1980) N=

24 ωs π

g .n

(13.23)

et

Cloudy skies generally reduce the number of hours of bright sunshine from N to n. Combining Equations 13.14, 13.17, and 13.18 gives the following combined relation for estimating the daily-averaged net shortwave radiation: 

 n 1440 Gsc dr (ωs sin φ sin δ + cos φ cos δ sin ωs ) Sn = (1 − α) as + bs N π 13.2.3.2

(13.24)

Longwave radiation

In addition to the shortwave (0.3–3 μm) solar energy that is added to the surface vegetation, there is also longwave radiation (3–100 μm) that is emitted by both the atmosphere and the ground. The rate of longwave emission from any body is proportional to the absolute temperature of the surface of the body raised to the fourth power. This relation is expressed quantitatively by the Stefan–Boltzmann law. The net flux of longwave radiation leaving the earth’s surface is less than that given by the Stefan–Boltzmann law due to the absorption and downward radiation from the sky. Water vapor, clouds, carbon dioxide, and dust in the atmosphere are absorbers and emitters of longwave radiation. As humidity and cloudiness play an important role, the Stefan–Boltzmann law is corrected by these two factors when

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estimating the net flux of longwave radiation at the ground surface. The net incoming longwave radiation, Ln [MJ m−2 d−1 ], can be estimated using the equation ⎛ ⎞ 4 4 Tmax, + Tmin, K K ⎠ ǫ′f Ln = −σ ⎝ (13.25) 2

where σ is the Stefan–Boltzmann constant [= 4.903 * 10−9 MJ m−2 K−4 d−1 ], Tmax, K is the daily maximum absolute temperature [K], and Tmin, K is the daily minimum absolute temperature [K], ǫ ′ is the net emissivity between the atmosphere and the ground [dimensionless], and f is the cloudiness factor [dimensionless]. The net emissivity, ǫ ′ , can be estimated using the equation (Brunt, 1932; Doorenbos and Pruitt, 1975; 1977; Allen et al., 1989) √ ǫ ′ = 0.34 − 0.14 ea (13.26)

ww

where ea is the vapor pressure [kPa] of water in the atmosphere. The cloudiness factor, f , can be estimated using the relation Rs f = 1.35 − 0.35 (13.27) Rs0

w .E asy En g

where Rs0 is the clear-sky solar radiation that can be related to the extraterrestrial radiation, S0 , and the elevation, z, using the relation (Doorenbos and Pruitt, 1977) Rs0 = (0.75 + 2 * 10−5 z)S0

(13.28)

where z is the land-surface elevation [m]. In cases where measurements of Rs are not available, the cloudiness factor, f , can be estimated by the equation f = 0.22

n N

ine eri n

(13.29)

Combining Equations 13.25, 13.26, and 13.27 gives the following relation to be used in estimating the net longwave radiation: ⎛

Ln = −σ ⎝

4 4 Tmax, K + Tmin, K

2

⎞

   √  ⎠ 0.34 − 0.14 ea 1.35 Rs − 0.35 Rs0

g .n

(13.30)

The relative solar radiation, Rs /Rs0 , indicates the relative cloudiness and must be limited to 0.25 … Rs /Rs0 … 1.0.

et

Summary. In summary, the recommended method for determining Rn in the Penman– Monteith equation is to use the measured solar radiation, Rs , with Equations 13.13, 13.14, 13.25, 13.26, 13.27, and 13.28. This is the standardized approach when using the PM equation. Using empirical equations in lieu of direct measurements of Rs can lead to substantial deviations in the estimated ET compared to using direct measurements (Irmak et al., 2011). 13.2.4

Soil Heat Flux

The soil heat flux, G, is the energy utilized in heating the soil, and is positive when the soil is warming and negative when the soil is cooling. The soil heat flux, G [EL−2 T−1 ], can be estimated using the relation Ti − Ti−1 G = cs (13.31) z t where cs is the soil heat capacity [EL−3 −1 ], Ti−1 and Ti are the soil temperatures [ ] at the beginning and end of time step t [T], respectively, and z is the effective soil depth [L]. It is commonly assumed that the soil temperature is equal to the air temperature; however,

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Penman–Monteith Equation

631

the soil temperature generally lags the air temperature and therefore time intervals, t, of at least 1 d are recommended (Allen et al., 1998). The penetration depth of the temperature wave is determined by the length of the time interval. The effective soil depth, z, is only 0.10–0.20 m (4–8 in.) for a time interval of one or a few days but might be 2 m (6.6 ft) or more for monthly periods. The soil heat capacity, cs , is related to the mineral composition and water content of the soil. Averaged over 1 d, G is typically small, but becomes more significant for hourly or monthly time periods. For hourly (or shorter) time intervals, G beneath a dense cover of grass or alfalfa does not correlate well with air temperature; however, hourly G does correlate well with net radiation and can be approximated as a fraction of Rn according to the following relations: Short reference crop (grass): Ghour =

ww



0.1Rn 0.5Rn

(day) (night)

0.04Rn 0.2Rn

(day) (night)

Tall reference crop (alfalfa): 

w .E asy En g Ghour =

For daily time intervals beneath a grass reference surface, it can be assumed that Gday = 0

(13.32)

For monthly time intervals, assuming a constant heat capacity of 2.1 MJ/(m3◦ C) and an effective soil depth of about 2 m, Equation 13.31 can be used to derive G for monthly time intervals as Gmonth = 0.07(Ti + 1 − Ti−1 ) (13.33)

ine eri n

where Ti−1 and Ti+1 are the mean temperatures in the previous and following months, respectively. If Ti+ 1 is unknown, then the following equation can be used in lieu of Equation 13.33: Gmonth = 0.14(Ti − Ti−1 ) (13.34)

g .n

where Ti is the mean temperature in the current month. The soil heat flux, G, can be measured directly using soil heat flux plates and thermocouples or thermistors. 13.2.5

Latent Heat of Vaporization

et

The latent heat of vaporization, λ [MJ/kg], is the energy required to change a unit mass of water from liquid to vapor at a constant temperature and pressure. The latent heat of vaporization, λ, can be expressed as a function of the water-surface temperature, Ts [◦ C], using the empirical equation (Harrison, 1963) λ = 2.501 − 0.002361Ts

(13.35)

Since λ varies only slightly over normal temperature ranges, a constant value of 2.45 MJ/kg corresponding to a temperature of 20◦ C is commonly assumed. 13.2.6

Psychrometric Constant

The psychrometric constant, γ [kPa/◦ C], depends on the atmospheric pressure, p [kPa], and the latent heat of vaporization, λ [MJ/kg], and is defined as γ =

cp p p = 0.0016286 ǫλ λ

(13.36)

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where the specific heat of moist air, cp , is taken as 1.013 kJ/(kg◦ C), and ǫ is the ratio of the molecular weight of water vapor to the molecular weight of dry air (= 0.622). The atmospheric pressure, p [kPa], at an elevation z [m] above sea level can be estimated using the relation (Allen et al., 1998)   293 − 0.0065z 5.26 p = 101.3 (13.37) 293 Assuming an atmospheric pressure of 101.32 kPa, and a latent heat of vaporization of 2.444 MJ/kg, the psychrometric constant, γ , given by Equation 13.36 is equal to 0.06752 kPa/◦ C. 13.2.7

ww

Saturation Vapor Pressure

The saturation vapor pressure, es [kPa], can be estimated from the air temperature, T [◦ C], using the relation 

17.27T es (T) = 0.6108 exp (13.38) T + 237.3

For hourly time intervals, T can be taken as the mean air temperature during the hourly period, while for estimates of daily-averaged ET, the saturation vapor pressure, es [kPa], should be estimated using the relation

w .E asy En g

es (Tmax ) + es (Tmin ) (13.39) 2 Combining Equations 13.38 and 13.39 gives the following expression for estimating the daily average saturation vapor pressure from temperature measurements: es =



es = 0.3054 exp

13.2.8



17.27Tmax Tmax + 237.3

Vapor-Pressure Gradient



+ exp



17.27Tmin Tmin + 237.3

ine eri n



(13.40)

The gradient of the saturation vapor pressure versus temperature curve,  [kPa/◦ C], can be estimated using the equation 4098es (13.41) = (T + 237.3)2

g .n

et

where es is the saturation vapor pressure [kPa], and T is the air temperature [◦ C]. Since es is determined by the air temperature, T, via Equation 13.38,  can be calculated directly from the air temperature using the relation   17.27T 4098 0.6108 exp T + 237.3 = (13.42) (T + 237.3)2

For daily ET estimates, the air temperature, T, is taken as the average of the maximum and minimum air temperatures. 13.2.9

Actual Vapor Pressure

The actual vapor pressure, ea [kPa], can be estimated directly from the measured dew-point temperature, Tdew [◦ C], using Equation 13.38, which yields 

17.27Tdew ea = es (Tdew ) = 0.6108 exp (13.43) Tdew + 237.3

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633

Equation 13.43 follows directly from the definition of the dew-point temperature, which is the temperature to which air needs to be cooled to make the air saturated. Hence, the actual vapor pressure is equal to the saturation vapor pressure at dew-point temperature. The dew-point temperature, Tdew , is often measured with a mirror-like metallic surface that is artificially cooled, and when dew forms on the surface, its temperature is sensed as Tdew . Other dew-sensor systems use chemical or electric properties of certain materials that are altered when absorbing water vapor. Instruments for measuring dew-point temperature require careful operation and maintenance and are seldom available at weather stations. As an alternative to Equation 13.43, the actual vapor pressure, ea , can be estimated using measured temperatures and relative humidities using the relation es (Tmax ) ea =

ww

RHmax RHmin + es (Tmin ) 100 100 2

(13.44)

where es (Tmax ) and es (Tmin ) are the saturation vapor pressures at the maximum and minimum temperatures, respectively, and RHmax and RHmin are the maximum and minimum relative humidities, respectively, over the course of 1 d. Relative humidity is measured using a hygrometer. Modern hygrometers use a film from a dielectric polymer that changes its dielectric constant with changes in surface moisture, thus inducing a variation of the capacity of a condenser using that dielectric. Equation 13.44 can also be used to estimate the average vapor pressure over periods longer than 1 d. For periods of a week, 10 d, or a month, RHmax and RHmin can be taken as the average of daily maxima and minima (respectively) during the averaging period. When using equipment where errors in estimating RHmin can be large, or when RH data integrity is in doubt, it is recommended to use only RHmax in the following equation (Allen et al., 1998): RHmax ea = es (Tmin ) (13.45) 100

w .E asy En g

ine eri n

In the absence of RHmax and RHmin , the following equation can be used:

RHmean es (Tmax ) + es (Tmin ) ea = 100 2

g .n

(13.46)

where RHmean is the mean relative humidity, defined as the average between RHmax and RHmin . Equation 13.46 is less desirable than Equation 13.44 or 13.45. Field measurements under semiarid tropical climatic conditions have shown that Equation 13.44 yields more accurate estimates of ea compared to using either Equation 13.45 or 13.46 (Ojha et al., 2011). 13.2.10

Air Density

et

The density of moist air, ρa [kg/m3 ], is a function of the temperature and pressure of the air, and can be estimated using the ideal gas law ρa =

p TKv R

(13.47)

where p is atmospheric pressure [kPa], TKv is the virtual temperature [K] given by (Allen et al., 1998) TKv = 1.01(T + 273) (13.48) where T is the air temperature [◦ C], and R is the specific gas constant for air [0.287 kJ/(kg·K)]. Combining Equations 13.47 and 13.48 and taking R equal to 0.287 kJ/(kg·K) yields ρa = 3.450

p T + 273

(13.49)

where the atmospheric pressure, p, can be assumed to equal 101.32 kPa.

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13.3

Application of the PM Equation

The Penman–Monteith (PM) equation is given by Equation 13.1, and the preferred methods to estimate the parameters in the PM equation are given in Sections 13.2.1 to 13.2.10. Application of the PM equation is illustrated by the following example. EXAMPLE 13.1

ww

A constructed cattail (Typha latifolia) marsh is located in South Florida at 23◦ 38′ N and 80◦ 25′ W, and measurements during the month of February indicate the following conditions (Abtew, 1996): Maximum temperature = 26.0◦ C, minimum temperature = 16.9◦ C, mean temperature = 21.3◦ C, maximum humidity = 98.2%, minimum humidity = 63.2%, and average wind speed at 10 m height = 3.96 m/s. The temperature and humidity measurements were collected at 4 m above the ground surface, typical cattail height in the marsh is 0.6 m, the cattails only partially cover the wetland, the surface resistance of cattails has been shown to be approximately 90 s/m (Abtew and Obeysekera, 1995), the average amount of sunshine at this location is typically 69% of the possible amount (Winsberg, 1990), and the mean temperatures in previous and following months (January and March) are 18.7◦ C and 21.3◦ C, respectively. Assuming an albedo of 0.20, use these data to estimate the evapotranspiration of cattails during the month of February. Subsequent direct measurements reported by Abtew (1996) indicate an average vapor-pressure deficit of 0.67 kPa, average incoming solar radiation of 12.95 MJ/(m2 ·d), and average net radiation of 8.83 MJ/(m2 ·d) for February. How would these measurements affect your predictions?

w .E asy En g

Solution From the given data: φ = 23◦ 38′ = 23.63◦ = 0.412 rad, Tmax = 26.0◦ C, Tmin = 16.9◦ C, RHmax = 98.2%, RHmin = 63.2%, zm = 10 m, u10 = 3.96 m/s, zh = 4 m, h = 0.6 m, rs = 90 s/m = 1.04 * 10−3 d/m, sunshine fraction (n/N) = 0.69, Ti−1 = 18.7◦ C, Ti + 1 = 21.3◦ C, and α = 0.20. The parameters in the Penman–Monteith equation are estimated using the conventional procedure as follows: ra : The aerodynamic resistance, ra , is given by Equation 13.2 as



zm − d zh − d ln ln z0m z0h ra = k2 u z

ine eri n

where d and z0m are given by Equation 13.3 as

2 2 h = (0.6) = 0.4 m 3 3 z0m = 0.123h = 0.123(0.6) = 0.0738 m d=

g .n

and, since the cattails only partially cover the wetland, Equation 13.4 gives z0h = (1)z0m = (1)(0.0738) = 0.0738 m

et

´ an ´ constant, k, as 0.41 and substituting the given and derived data into the Taking the von Karm expression for ra yields 



4 − 0.4 10 − 0.4 ln ln 0.0738 0.0738 ra = = 28.4 s/m = 3.29 * 10−4 d/m (0.41)2 (3.96) : The slope of the vapor pressure versus temperature curve, , is given by Equation 13.42 as   17.27T 4098 0.6108 exp T + 237.3 = (T + 237.3)2

where  is in kPa/◦ C, and T is the average temperature in ◦ C. When estimating daily-averaged evapotranspiration rates, T is taken as the average of the maximum and minimum daily temperature; hence 26.0 + 16.9 Tmax + Tmin = = 21.5◦ C T= 2 2

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635

Substituting into the expression for  yields ⎡ ⎤  17.27(21.5) ⎦ 4098 ⎣0.6108 exp 21.5 + 237.3 = 0.157 kPa/◦ C = (21.5 + 237.3)2 Rn : The net radiation, Rn , is equal to the sum of the net shortwave and longwave radiation according to Equation 13.13, where Rn = Sn + Ln The net shortwave radiation can be estimated using Equation 13.24 as

 

1440 n (13.50) Sn = (1 − α) as + bs Gsc dr (ωs sin φ sin δ + cos φ cos δ sin ωs ) N π

ww

where as and bs can be taken as 0.25 and 0.50, respectively (since there are no calibrated values); the solar constant, Gsc , is equal to 0.0820 MJ/(m2 ·min); the Julian day, J, for mid-February is 45, dr is, the relative distance between the earth and the sun, given by Equation 13.19 as     2π 2π J = 1 + 0.033 cos 45 = 1.024 dr = 1 + 0.033 cos 365 365

w .E asy En g

the solar declination, δ, is given by Equation 13.20 as     2π 2π δ = 0.4093 sin J − 1.405 = 0.4093 sin 45 − 1.405 = −0.241 radians 365 365 and the sunset-hour angle, ωs , is given by Equation 13.21 as

ωs = cos−1 [−tan φ tan δ] = cos−1 [−tan(0.412) tan(−0.241)] = 1.463 radians

Substituting the given and derived values of α, as , bs , n/N (= sunshine fraction), Gsc , dr , ωs , φ, and δ into Equation 13.50 yields    1440 Sn = (1 − 0.20) 0.25 + 0.50(0.69) (0.0820)(1.024)[1.463 sin(0.412) sin(−0.241) π  + cos(0.412) cos(−0.241) sin(1.463)] = 13.6 MJ/(m2 ·d)

ine eri n

g .n

The net longwave radiation, Ln , is given by Equation 13.30 as ⎛ ⎞   4 4 Tmax, + T √  Rs K min, K ⎠  ⎝ − 0.35 Ln = −σ 0.34 − 0.14 ea 1.35 2 Rs0

(13.51)

et

where σ = 4.903 * 10−9 MJ m−2 K−4 d−1 ; Tmax, K and Tmin, K are the maximum and minimum daily temperatures given as 299.2 K (= 26.0◦ C) and 290.1 K (= 16.9◦ C), respectively; ea is the actual vapor pressure given by Equation 13.44 es (Tmax ) ea =

RHmin RHmax + es (Tmin ) 100 100 2

where ⎡

⎤ 17.27(26.0) ⎦ = 3.36 kPa 26.0 + 237.3





17.27Tmax Tmax + 237.3

⎡





⎡  ⎤ ⎤ 17.27Tmin 17.27(16.9) ⎦ = 0.6108 ⎣exp ⎦ = 1.93 kPa Tmin + 237.3 16.9 + 237.3



17.27T T + 237.3

es (Tmax ) = 0.6108 exp

es (Tmin ) = 0.6108 ⎣exp es (T) = 0.6108 exp





= 0.6108 ⎣exp

⎡

= 0.6108 ⎣exp





⎤ 17.27(21.5) ⎦ = 2.64 kPa 21.5 + 237.3

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Chapter 13

Estimation of Evapotranspiration Substituting the given and derived values of es (Tmax ), RHmax , es (Tmin ), and RHmin into the expression for ea yields 63.2 98.2 + 1.93 3.36 100 100 = 2.26 kPa ea = 2 According to Equation 13.17, the incoming shortwave (solar) radiation, Rs , can be estimated by the relation     n S0 = 0.25 + 0.50 * 0.69 S0 = 0.595S0 Rs = as + bs N where S0 is the extraterrestrial radiation, and the clear-sky solar radiation, Rs0 , can be estimated by Equation 13.28 (taking z = 0) as Rs0 = [0.75 + 2 * 10−5 (0)]S0 = 0.75S0

ww

Substituting the given and derived values of σ , Tmax, K , Tmin, K , ea , Rs , and Rs0 into Equation 13.51 yields      √ 299.24 + 290.14  0.595S0 −9 0.34 − 0.14 2.26 * 1.35 Ln = −4.903 * 10 − 0.35 2 0.75S0 = −3.46 MJ/(m2 ·d)

w .E asy En g

where the negative value indicates that the net longwave radiation in February is away from the earth. The total available energy, Rn , in February is then equal to the sum of Sn and Ln and is given by Rn = Sn + Ln = 13.6 − 3.46 = 10.1 MJ/(m2 ·d)

G: According to Equation 13.33,

Gmonth = 0.07(Ti + 1 − Ti−1 ) = 0.07(21.3 − 18.7) = 0.182 MJ/(m2 ·d)

γ : The psychrometric constant, γ , can be estimated using Equation 13.36 as p γ = 0.0016286 kPa/◦ C λ where the atmospheric pressure, p, can be taken as 101.32 kPa, and the latent heat of vaporization, λ, can be taken as 2.45 MJ/kg. Therefore, γ is given by

ine eri n

101.32 = 0.0674 kPa/◦ C 2.45 Since the pressure, p, and latent heat of vaporization, λ, remain approximately constant throughout the year, γ remains approximately constant. ρa : The density of air, ρa , can be estimated using Equation 13.49 as p 101.32 = 3.450 = 1.187 kg/m3 ρa = 3.450 T + 273 21.5 + 273 γ = 0.0016286

g .n

et

Taking the density of water, ρw , as 998.2 kg/m3 ; the latent heat of vaporization, λ, as 2.45 MJ/kg; the specific heat of moist air, cp , as 1.013 kJ/(kg◦ C) = 1.013 * 10−3 MJ/(kg◦ C), and substituting into the Penman–Monteith equation, Equation 13.1 yields ⎤ ⎡ es − ea (R − G) + ρ c ⎥ n a p 1 ⎢ r ⎥ ⎢  a  ET = ⎢ ⎥ rs ⎦ ρw λ ⎣  + γ 1 + ra ⎤ ⎡ 2.64 − 2.26 ⎢ ⎢ 0.157(10.1 − 0.182) + (1.187)(1.013 * 10−3 ) 1 ⎢ 3.29 * 10−4   = ⎢ −3 (998.2)(2.45) ⎢ 1.04 * 10 ⎣ 0.157 + 0.0674 1 + 3.29 * 10−4

= 0.0028 m/d = 2.8 mm/d

⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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Potential Evapotranspiration

637

Therefore, the average evapotranspiration during February is estimated to be 2.8 mm/d. Subsequent direct measurements indicate that es − ea = 0.67 kPa and Rn = 8.83 MJ/(m2 ·d), and using these values in the Penman–Monteith equation gives ⎤ ⎡ 0.67 ⎢ ⎢ 0.157(8.83 − 0.182) + (1.187)(1.013 * 10−3 ) 1 ⎢ 3.29 * 10−4   ET = ⎢ −3 (998.2)(2.45) ⎢ 1.04 * 10 ⎣ 0.157 + 0.0674 1 + 3.29 * 10−4 = 0.0036 m/d = 3.6 mm/d

⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Therefore, using direct measurements of net radiation and vapor pressure deficit increases the estimated evaporation rate by 29% (3.6 mm/d vs. 2.8 mm/d). A close inspection of this result indicates that the increased evapotranspiration is primarily caused by the measured vapor-pressure deficit being much higher than the estimated vapor-pressure deficit.

ww

13.4

Potential Evapotranspiration

The potential evapotranspiration rate, originally, introduced by Penman (1948), is defined as “the amount of water transpired in a given time by a short green crop, completely shading the ground, of uniform height and with adequate water status in the soil profile.” According to this definition, the potential evapotranspiration rate is not related to a specific type of vegetation, and many types of vegetation can be described as a “short green crop.” Potential evaporation is a term used synonymously with potential evapotranspiration and has been defined as “the maximum evaporation rate when the moisture content of the soil and vegetation conditions do not limit evaporation” (Hendriks, 2010). Several equations have been proposed for estimating the potential evapotranspiration. In many cases, the preferred method is that suggested by Penman (1956), which can be derived from the Penman–Monteith equation, Equation 13.1, repeated here for convenience as ⎤ ⎡ es − ea (Rn − G) + ρa cp ⎥ 1 ⎢ r ⎥ ⎢  a  ET = (13.52) ⎥ ⎢ r ⎦ ⎣ ρw λ s  + γ 1 + ra

w .E asy En g

ine eri n

g .n

The Penman (1956) equation for estimating the potential evapotranspiration, ETp , is derived directly from the Penman–Monteith equation (Equation 13.52) by taking rs = 0, which gives the Penman equation as ⎤ es − ea (Rn − G) + ρa cp ⎥ 1 ⎢ ra ⎢ ⎥ ETp = ⎣ ⎦ ρw λ  + γ ⎡

et

(13.53)

This equation has been used in several regional studies to estimate the distribution of potential evapotranspiration (e.g., Bidlake et al., 1996), with open water and wetland systems commonly assumed to evaporate at the potential rate (Abtew et al., 2003). EXAMPLE 13.2 Estimate the potential evapotranspiration at the location of the cattail marsh described in the previous example. Comment on the difference between the actual and potential evapotranspiration at the site. Solution Potential evapotranspiration is calculated using the same Penman–Monteith equation as for actual evapotranspiration, with the exception that rs = 0 and ra is the aerodynamic resistance of open water, which is given by Equation 13.8 as

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4.72 ln ra =

zm z0

2

1 + 0.536u2

s/m

where zm = 2 m (standard for open water), z0 = 1.37 mm = 0.00137 m, and the wind speed at 2 m height is derived from the given wind speed at 10 m height (u10 = 3.96 m/s) using Equation 13.5, which yields 4.87 4.87 u2 = uz = (3.96) = 2.96 m/s ln[67.8z − 5.42] ln[67.8(10) − 5.42] Substituting the known and derived values of zm , z0 , and u2 into the expression for ra gives 2

2 4.72 ln 0.00137 = 97 s/m = 1.12 * 10−3 d/m ra = 1 + 0.536(2.96)

ww

For open water, the albedo is 0.08, and since the albedo used in the previous example is 0.20, the previously calculated net solar radiation, Sn [= 13.6 MJ/(m2 ·d)], becomes Sn = 13.6

(1 − 0.08) = 15.6 MJ/(m2 ·d) (1 − 0.20)

w .E asy En g

Since Ln = −3.46 MJ/(m2 ·d), the net radiation, Rn , over open water is given by Rn = Sn + Ln = 15.6 + (−3.46) = 12.1 MJ/(m2 ·d)

Substituting into the Penman–Monteith equation, using applicable data from the previous example, yields ⎡ ⎤ es − ea (R − G) + ρ c ⎥ n a p 1 ⎢ r ⎢ ⎥  a  ET = ⎢ ⎥ rs ⎦ ρw λ ⎣  + γ 1 + ra ⎤ ⎡ 2.64 − 2.26 −3 ⎥ ⎢ 0.157(12.1 − 0.182) + (1.187)(1.013 * 10 ) 1 ⎢ 1.12 * 10−3 ⎥   = ⎥ ⎢ 0 ⎦ (998.2)(2.45) ⎣ 0.157 + 0.0674 1 + −3 1.12 * 10 = 0.0042 m/d = 4.2 mm/d

ine eri n

g .n

Therefore, the (theoretical) potential evapotranspiration of 4.2 mm/d at the cattail marsh is significantly higher than the (theoretical) actual transpiration of 2.8 mm/d. This difference is primarily due to the reduced surface resistance and increased absorption of solar energy associated with open water versus cattail vegetation, with the increased aerodynamic resistance over open water not being sufficient to offset these factors.

13.5

et

Reference Evapotranspiration

The reference evapotranspiration concept was introduced between the late 1970s and early 1980s to avoid the ambiguity in surface vegetation associated with the definition of potential evapotranspiration. The reference evapotranspiration is frequently defined as “the rate of evapotranspiration from a hypothetical reference crop with an assumed crop height of 0.12 m, a fixed surface resistance of 70 s/m, and an albedo of 0.23, closely resembling the evapotranspiration from an extensive surface of green grass of uniform height, actively growing, well-watered, and completely shading the ground.” In the literature, the terms reference evapotranspiration and reference-crop evapotranspiration are used synonymously. The most widely used and recommended methods for estimating reference evapotranspiration are the FAO56 and ASCE Penman–Monteith equations, with evaporation pans and empirical methods also being used. In California, the California Irrigation Management Information System (CIMIS) Penman equation is also widely used. All Penman equations give similar estimates of the reference evapotranspiration (Temesgen et al., 2005).

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13.5.1

Reference Evapotranspiration

639

FAO56-Penman–Monteith Method

The International Commission for Irrigation and Drainage (ICID) and the Food and Agriculture Organization of the United Nations (FAO) have recommended that the FAO56Penman–Monteith (FAO56-PM) method be used as the standard method to estimate reference-crop evapotranspiration. For grass and alfalfa reference crops, ra [s/m] can be approximated by ⎧ ⎪ ⎪ 208 , for grass reference crop ⎪ ⎨u 2 ra = (13.54) ⎪ 110 ⎪ ⎪ , for alfalfa reference crop ⎩ u2

where u2 is the wind speed [m/s] at a standardized height of 2 m above the ground (Allen et al., 1989). The surface resistance, rs [s/m], of grass and alfalfa reference crops are given by

ww

rs =

⎧ ⎨70 s/m

⎩45 s/m

w .E asy En g

for grass reference crop (13.55) for alfalfa reference crop

According to the Penman–Monteith equation (Equation 13.1), the evapotranspiration rate of any vegetated surface is given by ⎡ ⎤ es − ea (Rn − G) + ρa cp ⎥ 1 ⎢ r ⎢ ⎥  a  (13.56) ET = ⎢ ⎥ rs ⎦ ρw λ ⎣  + γ 1 + ra

ine eri n

The FAO56-PM method assumes that the density of water is equal to 1000 kg/m3 , and since values of λ vary only slightly over normal temperature ranges, a constant value of 2.45 MJ/kg corresponding to a temperature of 20◦ C is assumed. Equation 13.36 can be used to estimate the specific heat of water, cp , and Equation 13.49 can be used to estimate the density of air. Grass is the most widely used reference crop, and substituting ra and rs for grass (Equations 13.54 and 13.55) along with the aforementioned values of ρw , λ, and cp yields the conventional FAO56-PM equation given by

g .n

900 u2 (es − ea ) 0.408(Rn − G) + γ Tmean + 273 ET0 =  + γ (1 + 0.34u2 )

et

(13.57)

where ET0 is the grass reference evapotranspiration [mm/d] and the other variables and units are:  [kPa/◦ C], Rn [MJ/m2 ·d], G [MJ/m2 ·d], γ [kPa/◦ C], Tmean [◦ C], u2 [m/s], es [kPa], and ea [kPa]. Equation 13.57 is appropriate for use with 24-h time steps. For hourly estimates of ET0 , 900 is replaced by 37 in Equation 13.57. It is generally recommended that the climate data used to estimate reference evapotranspiration be collected in an environment similar to that of the reference crop (grass). Therefore, weather stations used to collect data to estimate reference evapotranspiration should be located in well-irrigated and well-maintained grass areas, with the surrounding grass area exceeding 2 ha (5 ac). In many cases, especially in developing countries, the quality of data and the difficulties in gathering all the necessary weather parameters can present serious limitations to using the FAO56-PM method. In particular, wind speed and humidity data measured at airports and hill locations can introduce errors in ET0 estimates, since these locations have different microclimates than irrigated agricultural areas for which ET0 estimations are made. Evapotranspiration estimates using the FAO56-PM method are typically most sensitive to relative humidity, followed by shortwave radiation, air temperature, and

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Estimation of Evapotranspiration

ww

wind speed (Gong et al., 2006). The FAO56-PM method recommends using the equations described in Section 13.2 to estimate the weather parameters in the FAO56-PM equation when direct measurements are not available—the minimum data requirements in this case are minimum and maximum temperature and wind speed. Although alternative equations are available in the technical literature to estimate weather parameters that are not directly measured, strict adherence to the FAO56-recommended equations should be followed, since alternative equations can lead to a lack of consistency in estimating the FAO56-PM reference evapotranspiration. It has been shown that using the FAO56-PM recommended equations with the minimum required weather parameters in a humid climate produces monthly ET0 estimates that are more exact than any other method (Trajkovic and Kolakovic, 2009). The accuracy of ET0 estimation models should ideally be measured relative to data derived from lysimeter measurements. However, environmental and management requirements for lysimeter experiments are very demanding, and it has been suggested that the FAO56-PM equation be considered superior to most lysimeter-measured ET0 , and that the FAO56-PM equation can be used in calibrating other ET0 models (Allen et al., 1994). In areas where substantial changes in wind speed, dew point, or cloudiness occur during the day, calculation of ET0 using hourly time steps is generally better than using 24-h time steps. However, the hourly averaged surface resistance, rs , of grass can deviate significantly from the daily-averaged value of 70 s/m. Under most conditions, application of the FAO56PM equation with 24-h time steps produces accurate results.

w .E asy En g EXAMPLE 13.3

Weather conditions on typical days in Miami during the months of January, April, July, and October are shown in the following table (Winsberg, 1990):

Month

January April July October

Minimum humidity (%)

Maximum humidity (%)

Minimum temperature (◦ C)

Maximum temperature (◦ C)

Wind speed (m/s)

Daylight fraction

59 54 63 64

84 79 85 86

14.4 20.0 24.4 21.7

23.9 28.3 31.7 28.9

4.5 4.9 3.6 4.0

0.69 0.78 0.76 0.72

ine eri n

g .n

Given that Miami is located at approximately 26◦ N latitude, compute the grass reference evapotranspiration for the given months using the FAO56-PM method. Typical monthly rainfall amounts for January, April, July, and October are 5.0 cm, 8.2 cm, 15.2 cm, and 19.1 cm, respectively. Compare the estimated monthly reference evapotranspiration with the monthly rainfall, and assess the implications of this differential on available water resources.

et

Solution The FAO56-PM equation is given by Equation 13.57, and the parameters of this equation either are known or can be estimated from the given data. The computational procedure will be illustrated for the month of January. The FAO56-PM equation gives the grass reference evapotranspiration, ET0 , as 900 u (es − ea ) 0.408(Rn − G) + γ T + 273 2 ET0 =  + γ (1 + 0.34u2 ) and the computation of the parameters in this equation for January is as follows: : The slope of the vapor pressure versus temperature curve, , is given by Equation 13.42 as

4098 0.6108 exp =



17.27T T + 237.3

(T + 237.3)2



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641

where  is in kPa/◦ C, and T is the average temperature in ◦ C. When estimating daily-averaged evapotranspiration rates, T is taken as the average of the maximum and minimum daily temperature, and for a typical day in January the given data indicate that T= and therefore

14.4 + 23.9 = 19.2◦ C 2

⎡

=

4098 ⎣0.6108 exp



⎤ 17.27(19.2) ⎦ 19.2 + 237.3

(19.2 + 237.3)2

= 0.139 kPa/◦ C

Rn : The net radiation, Rn , is equal to the sum of the net shortwave and longwave radiation according to Equation 13.13, where Rn = Sn + Ln

ww

The net shortwave radiation can be estimated using Equation 13.24 as 



1440 n Gsc dr (ωs sin φ sin δ + cos φ cos δ sin ωs ) Sn = (1 − α) as + bs N π

w .E asy En g

(13.58)

where the albedo, α, is equal to 0.23 for the grass reference crop; as and bs can be taken as 0.25 and 0.50, respectively (since there are no calibrated values); the fraction of bright sunshine during daylight hours for January is given as 0.69; the solar constant, Gsc , is equal to 0.0820 MJ/(m2 ·min); dr is the relative distance between the earth and the sun, given by Equation 13.19 as     2π 2π dr = 1 + 0.033 cos J = 1 + 0.033 cos 15 = 1.032 365 365

where the mean Julian day, J, for January is taken as 15; the solar declination, δ, is given by Equation 13.20 as     2π 2π J − 1.405 = 0.4093 sin 15 − 1.405 = −0.373 radians δ = 0.4093 sin 365 365

ine eri n

and the sunset-hour angle, ωs , is given by Equation 13.21 as

ωs = cos−1 [−tan φ tan δ] = cos−1 [−tan(0.454) tan(−0.373)] = 1.38 radians

g .n

where the latitude, φ, is given as 0.454 rad (= 26◦ ). Substituting the values of α, as , bs , n/N, Gsc , dr , ωs , φ, and δ into Equation 13.58 yields    1440 (0.0820)(1.032)[1.38 sin(0.454) sin(−0.373) Sn = (1 − 0.23) 0.25 + 0.50(0.69) π  + cos(0.454) cos(−0.373) sin(1.38)]

et

= 10.68 MJ/(m2 ·d) The net longwave radiation, Ln , is given by Equation 13.30 as ⎛ ⎞   4 4 Tmax, + Tmin,  √  K K ⎠ 0.34 − 0.14 ea 1.35 Rs − 0.35 Ln = −σ ⎝ 2 Rso

(13.59)

where σ = 4.903 * 10−9 MJ m−2 K−4 d−1 ; Tmax, K and Tmin, K are the maximum and minimum daily temperatures given as 297.1 K (= 23.9◦ C) and 287.6 K (= 14.4◦ C), respectively; ea is the actual vapor pressure given by Equation 13.44 es (Tmax ) ea =

RHmax RHmin + es (Tmin ) 100 100 2

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Chapter 13

Estimation of Evapotranspiration where RHmax and RHmin are the maximum and minimum relative humidities given as 84% and 59%, respectively, and ⎡ ⎤    17.27(23.9) 17.27Tmax ⎦ = 2.97 kPa es (Tmax ) = 0.6108 exp = 0.6108 ⎣exp Tmax + 237.3 23.9 + 237.3 ⎡





⎡ ⎤ ⎤  17.27Tmin 17.27(14.4) ⎦ = 0.6108 ⎣exp ⎦ = 1.64 kPa Tmin + 237.3 14.4 + 237.3



17.27T T + 237.3

es (Tmin ) = 0.6108⎣exp es (T) = 0.6108 exp



which yields

⎡

= 0.6108 ⎣exp



⎤ 17.27(19.2) ⎦ = 2.22 kPa 19.2 + 237.3

59 84 + 1.64 100 100 = 1.73 kPa ea = 2 According to Equation 13.17, the net shortwave (solar) radiation, Rs , can be estimated by the relation     n S0 = 0.25 + 0.50 * 0.69 S0 = 0.595So Rs = as + bs N and the clear-sky solar radiation, Rs0 , can be estimated by Equation 13.28 (taking z = 0) as 2.97

ww

w .E asy En g

Rs0 = [0.75 + 2 * 10−5 (0)]S0 = 0.75S0

Substituting the known values of σ , Tmax, K , Tmin, K , ea , Rs , and Rs0 into Equation 13.59 yields      √ 297.14 + 287.64  0.595So −9 Ln = −4.903 * 10 0.34 − 0.14 1.73 * 1.35 − 0.35 2 0.75So = −4.03 MJ/(m2 ·d)

ine eri n

where the negative value indicates that the net longwave radiation in January is away from the earth. The total available energy, Rn , in January is then equal to the sum of Sn and Ln and is given by Rn = Sn + Ln = 10.68 − 4.03 = 6.65 MJ/(m2 ·d)

g .n

G: According to Equation 13.32, G = 0 MJ/(m2 ·d) when averaged over 1 d. γ : The psychrometric constant, γ , can be estimated using Equation 13.36 as p γ = 0.0016286 kPa/◦ C λ where the atmospheric pressure, p, can be taken as 101.32 kPa, and the latent heat of vaporization, λ, can be taken as 2.45 MJ/kg. Therefore, γ is given by

et

101.32 = 0.0674 kPa/◦ C 2.45 Since the pressure, p, and latent heat of vaporization, λ, remain approximately constant throughout the year, γ remains approximately constant. γ = 0.0016286

Based on the given and calculated parameters for the month of January, the grass reference-crop evapotranspiration estimated by the FAO56 Penman–Monteith equation is given by

ET0 =

=

900 u (es − ea ) T + 273 2  + γ (1 + 0.34u2 )

0.408(Rn − G) + γ

900 (4.5)(2.22 − 1.73) 19.2 + 273 0.139 + 0.0674(1 + 0.34 * 4.5)

0.408(0.139)(6.65 − 0) + (0.0674)

= 2.7 mm/d

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643

The computations illustrated here for January are repeated for April, July, and October, and the results are tabulated below:

Month

 (kPa/◦ C)

Rn (MJ/m2 ·d)

ET (mm/d)

ET (mm)

Rain (mm)

0.138 0.181 0.221 0.192

6.7 14.3 16.4 10.4

2.7 5.3 5.6 3.9

83 159 174 121

50 82 152 191

January April July October

These results indicate that the grass reference evapotranspiration exceeds rainfall in January, April, and July, which indicates that grass irrigation could be necessary during these months.

ww

There have been several applications of the FAO56-PM method to study the spatial and temporal trends of ET0 within various countries. For example, a study of the temporal trends of ET0 in India between 1971 and 2002 showed a significant decreasing trend in ET0 in all Indian agricultural regions during the study period, which was mainly caused by a significant increase in the relative humidity and a consistent significant decrease in the wind speed throughout the country (Bandyopadhyay et al., 2009). A study of the spatial and temporal trends of ET0 in Iran between 1965 and 2005 showed both positive and negative trends in annual and monthly ET0 , with increasing ET0 trends in the agricultural areas in northwest and northeast of Iran and decreasing trends in nonproductive areas in the center of Iran (Dinpashoh et al., 2011). In this latter study, wind speed was reported to be the most influential variable responsible for decreasing and increasing trends in ET0 at most locations. A study of the spatial and temporal trends of ET0 in the 750,000-km2 (290,000-mi2 ) Yellow River Basin in China between 1957 and 2008 showed negative trends in some parts of the basin and positive trends in other parts of the basin (Wang et al., 2012). Warming trends in temperature and decreasing trends in wind speed and sunshine duration were reported as the principal climatic factors driving the trends in ET0 .

w .E asy En g 13.5.2

ine eri n

ASCE Penman–Monteith Method

The American Society of Civil Engineers (ASCE) has developed equations to estimate reference evapotranspiration for two types of vegetated surfaces: a short crop with an approximate height of 0.12 m (4.7 in.) (similar to clipped grass), and a tall crop with an approximate height of 0.50 m (20 in.) (similar to full-cover alfalfa). Both estimation equations are derived from the Penman–Monteith equation (Equation 13.1) and are expressed in the generalized form (Walter et al., 2000)

ETref =

g .n

Cn u2 (es − ea ) T + 273  + γ (1 + Cd u2 )

0.408(Rn − G) + γ

et

(13.60)

where ETref is the standardized reference-crop evapotranspiration for short (ET0 ) or tall (ETr ) vegetation, Cn is a numerator constant that changes with reference-crop type and calculation time step, and Cd is a denominator constant that also changes with reference-crop type and calculation time step. Values of Cn and Cd used in the ASCE Penman–Monteith (ASCE-PM) equation are given in Table 13.2. The ASCE-PM equation (Equation 13.60) is derived from the Penman–Monteith equation (Equation 13.1) and is applied by substituting the parameter values shown in Table 13.3. The ASCE-PM equation accounts for surface resistance being significantly higher during the nighttime than during the daytime, which is due to stomatal closure during nighttime hours; nighttime is considered to occur whenever the net radiation, Rn , is negative. On reviewing the assumed parameter values in Table 13.3, it is apparent that for daily time intervals for short reference crops the assumed parameter values are the same as those for the FAO56-PM equation, and the ASCE-PM equation is the same as the FAO56-PM equation in this case. The ASCE-PM equation goes beyond the

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Chapter 13

Estimation of Evapotranspiration TABLE 13.2: Values of Cn and Cd in ASCE Standardized Method

Short reference, ETo

Tall reference, ETr

Units for

Units for

Calculation time step

Cn

Cd

Cn

Cd

ETo , ETr

Rn , G

Daily Hourly (in daytime) Hourly (in nighttime)

900 37 37

0.34 0.24 0.96

1600 66 66

0.38 0.25 1.7

mm/d mm/h mm/h

MJ/(m2 ·d) MJ/(m2 ·h) MJ/(m2 ·h)

TABLE 13.3: Assumed Parameter Values in the ASCE Standardized Method

Parameter

ww

Reference vegetation height, h Height of air temperature and humidity measurements, zh Height corresponding to wind speed, zw Zero plane displacement height Latent heat of vaporization, λ Surface resistance, rs (daily) Surface resistance, rs (daytime) Surface resistance, rs (nighttime)

w .E asy En g

ETo

ETr

0.12 m 1.5–2.5 m 2.0 m 0.08 m 2.45 MJ/kg 70 s/m 50 s/m 200 s/m

0.50 m 1.5–2.5 m 2.0 m 0.08 m 2.45 MJ/kg 45 s/m 30 s/m 200 s/m

FAO56-PM equation, which applies only to a short (grass) reference crop, in that it is applied also to a tall reference crop (alfalfa), and also the ASCE-PM equation uses different surface resistances for day and night, compared to the FAO56-PM equation which uses the same surface resistance for day and night. 13.5.3

Evaporation Pans

ine eri n

Evaporation pans provide a direct measure of evaporation in the field and are sometimes used to estimate the reference-crop evapotranspiration, ET0 , by multiplying pan-evaporation measurements by pan coefficients. Evaporation pans measure the integrated effect of radiation, wind, temperature, and humidity on the evaporation from an open water surface. The most commonly used pan design in the United States is the U.S. Weather Bureau Class A pan, which is 120.7 cm (4 ft) in diameter, 25 cm (10 in.) deep, and made of either galvanized iron (22 gauge) or MonelTM metal (0.8 mm). If galvanized, the pan should be painted annually with aluminum paint. The pan is mounted on a wooden frame (slatted platform) 15 cm (6 in.) above ground level, the soil is built up to within 5 cm (2 in.) of the bottom of the pan, and the pan should be surrounded by short grass turf 4–10 cm (1.5–4 in.) high that is well irrigated during the dry season. The pan is filled to a depth of 20 cm (8 in.), which is within 5 cm (2 in.) of the rim. The water level should not fall more than 7.5 cm (3 in.) below the rim, and the water should be regularly renewed, at least weekly, to eliminate excessive turbidity. A typical Class A pan station is shown in Figure 13.1. Inside the pan, a small stilling well is used to facilitate measurement in the presence of ripples generated under adverse wind conditions. Water can be automatically added or removed from the pan based on the reference level of water in the stilling well using an automatically controlled water adding and removing system (e.g., Xing et al., 2008). A Class A pan station generally includes an anemometer mounted 15 cm (6 in.) above the pan rim and, in some cases, a mesh cover to prevent animals from entering the pan or drinking the water. These mesh screens are sometimes called “bird guards.” A 12.5-mm (0.5-in.) mesh screen is commonly used; however, the mesh lowers the measured evaporation by 5%–20% (Stanhill, 1962; Dagg, 1968); an adjustment of 7% is recommended for use in Australia (van Dijk, 1985). Screens over the pan are not standard equipment and preferably should not be used if at all possible. Pans should be protected by fences to keep ground animals away from the pan. Pan-evaporation rates

g .n

et

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Reference Evapotranspiration

645

FIGURE 13.1: Class A evaporation pan

ww

w .E asy En g

are sensitive to the size of the pan, with pans smaller than the standard Class A pan yielding higher evaporation rates (Ekwue and Stone, 2004). In cases where there are gaps in pan data or pan data need to be extrapolated, pan-evaporation models have been developed to relate pan evaporation to local meteorological conditions, typically characterized by incident radiation, wind speed, relative humidity, and saturation vapor-pressure deficit (e.g., Piri et al., 2009). Pan-evaporation rates from Class A pans are generally greater than evaporation rates from large bodies of water. Numerous studies have shown that large water bodies have evaporation rates far higher near the edge of the water than toward the center, where the air is more saturated and able to absorb less water vapor. The small size of an evaporation pan means that the whole pan is effectively an ‘edge’ and will have higher evaporation rates than much larger bodies of water (Davie, 2002). A second, smaller, problem is that the sides of the pan and the water inside the pan will absorb radiation and warm up quicker than a much larger body of water, providing an extra energy source and greater evaporation rate. Ideally, evaporation pans used to estimate evaporation from large bodies of water should be located adjacent to the water body, since extrapolation from remote evaporation pans can be a dominant source of error (e.g., Lowe et al., 2009). Experimental results have shown that the Class A pan-evaporation rate will be influenced by the water depth when the water depth is much smaller than the standard water depth of 20 cm (8 in.) due to the shelter effect of the rim of the pan (Chu et al., 2010). In addition, wind blows water over the edges of the pan when the water depth in the pan is 20 cm (8 in.) and wind speeds exceed 7.0 m/s (16 mph). Pan-evaporation estimates can be poor and erratic over daily time scales, and are better suited to longer-term estimates, such as weekly and monthly. Class A pans constructed to float in lakes and ponds have been proposed as giving more realistic measures of evaporation from surface-water bodies (e.g., Masoner et al., 2008); however, such applications are nonstandard and are not commonly used. Reference-crop evapotranspiration, ET0 [LT−1 ], is typically estimated from standard Class A pan measurements using multiplicative factors called pan coefficients. Therefore

ine eri n

ET0 = kp Ep

g .n

et

(13.61)

where kp is the pan coefficient [dimensionless], and Ep is the measured pan evaporation [LT−1 ]. Pan coefficients vary seasonally, are typically in the range of 0.35–0.85 (Doorenbos and Pruitt, 1975; 1977), and representative pan coefficients for various site conditions are given in Table 13.4. Key factors affecting the pan coefficient are the average wind speed,

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Chapter 13

Estimation of Evapotranspiration TABLE 13.4: Pan Coefficients for Various Site Conditions

Case A: Pan surrounded by short green crop

Case B: Pan surrounded by dry, bare (fallow) area

Mean relative humidity, % Upwind fetch of green crop, m

Low < 40

Medium 40–70

Light (< 2 m/s)

1 10 100 1000

0.55 0.65 0.70 0.75

Moderate (2–5 m/s)

1 10 100 1000

Strong (5–8 m/s)

Very strong (> 8 m/s)

Wind

ww

Mean relative humidity, %

High > 70

Upwind fetch of dry fallow, m

Low < 40

Medium 40–70

High > 70

0.65 0.75 0.80 0.85

0.75 0.85 0.85 0.85

1 10 100 1000

0.70 0.60 0.55 0.50

0.80 0.70 0.65 0.60

0.85 0.80 0.75 0.70

0.50 0.60 0.65 0.70

0.60 0.70 0.75 0.80

0.65 0.75 0.80 0.80

1 10 100 1000

0.65 0.55 0.50 0.45

0.75 0.65 0.60 0.55

0.80 0.70 0.65 0.60

1 10 100 1000

0.45 0.55 0.60 0.65

0.50 0.60 0.65 0.70

0.60 0.65 0.70 0.75

1 10 100 1000

0.60 0.50 0.45 0.40

0.65 0.55 0.50 0.45

0.70 0.65 0.60 0.55

1 10 100 1000

0.40 0.45 0.50 0.55

0.45 0.55 0.60 0.60

0.50 0.60 0.65 0.65

1 10 100 1000

0.50 0.45 0.40 0.35

0.60 0.50 0.45 0.40

0.65 0.55 0.50 0.45

w .E asy En g Source: Doorenbos and Pruitt (1977).

ine eri n

upwind fetch characteristics, and ambient humidity. The estimation of evapotranspiration using the pan coefficients in Table 13.4 is commonly called the FAO-24 pan evaporation method. This method was published by the Food and Agriculture Organization (FAO) in paper number 24, hence the name. The pan coefficients given in Table 13.4 can be described by the following regression equations (Allen and Pruitt, 1991; Allen et al., 1988):

g .n

Case A: Surrounded by short green crop.

et

kp = 0.108 − 0.0286u2 + 0.0422 ln(FET) + 0.1434 ln(RHmean ) − 0.000631[ln(FET)]2 ln(RHmean )

(13.62)

Case B: Surrounded by dry, bare area. kp = 0.61 + 0.00341RHmean − 0.000162u2 RHmean − 0.00000959u2 FET + 0.00327u2 ln(FET) − 0.00289u2 ln(86.4u2 ) − 0.0106 ln(86.4u2 ) ln(FET) + 0.00063[ln(FET)]2 ln(86.4u2 )

(13.63)

where u2 is the average daily wind speed at 2 m height [m/s], RHmean is the average daily relative humidity [%] calculated by averaging the maximum and minimum relative humidity

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Reference Evapotranspiration

647

over the course of a day, and FET is the upwind fetch [m]. In applying Equations 13.62 and 13.63, the following data ranges must be strictly observed (Allen et al., 1998):

ww

1 m … FET … 1000 m

(13.64)

30% … RHmean … 84%

(13.65)

1 m/s … u2 … 8 m/s

(13.66)

The pan coefficients given in Table 13.4 are applicable to short, irrigated grass turf. For taller and aerodynamically rougher crops, the values of kp would be higher and vary less with differences in weather conditions (ASCE, 1990). It is recommended that the evaporation pan be installed inside a short green cropped area with a size of at least 15 m * 15 m (50 ft * 50 ft), and at least 10 m (30 ft) from the green crop edge in the general upwind direction (Allen et al., 1998). Empirical equations other than Equations 13.62 and 13.63 have been proposed for estimating pan coefficients. In the case of an evaporation pan surrounded by short green vegetation (Case A), Cuenca (1989) suggested the relation kp = 0.475 − 2.4 * 10−4 u2 + 5.16 * 10−3 RHmean + 1.18 * 10−3 FET

w .E asy En g

− 1.6 * 10−5 RHmean 2 − 1.01 * 10−6 FET2 − 8.1 * 10−9 RHmean 2 u2

− 1.0 * 10−8 RHmean 2 FET

(13.67)

where FET is the upwind distance fetch of low-growing vegetation [m], RHmean is the average daily relative humidity [%], and u2 is the wind speed at 2 m above ground [km/day]. Also for the case of an evaporation pan surrounded by short green vegetation (Case A), Snyder (1992) suggested the relation kp = 0.482 − 0.000376u2 + 0.024 ln(FET) + 0.0045RHmean

ine eri n

(13.68)

where the variables and units are the same as in Equation 13.67. A comparative analysis in a maritime region of Canada showed that both the Cuenca and the Snyder equations performed adequately, with the Snyder equation performing better than the Cuenca equation (Xing et al., 2008). For the case of an evaporation pan surrounded by a fallow area (Case B), Abdel-Wahed and Snyder (2008) proposed the following relation

g .n

kp = 0.62407 − 0.02660 ln FET − 0.00028u2 + 0.00226RHmean

(13.69)

et

where it was also demonstrated that Equation 13.69 is in better agreement with Table 13.4 than Equation 13.63, with the added advantage that it is simpler and less prone to calculation errors. In addition to Equations 13.67 to 13.69 which use FET, u2 , and RHmean as the independent variables, other region-specific expressions for estimating kp from meteorological measurements have also been developed. For example, Lee and Cho (2012) showed that the following relationship adequately represents pan coefficients in the Korean peninsula, kp = 0.092u2 + 0.011T2 + 0.476

(13.70)

where u2 and T2 are the wind speed [m/s] and temperature [◦ C] at 2 m above the ground, respectively. Where a standard pan environment is maintained and where strong dry-wind conditions occur only occasionally, mean monthly evapotranspiration for a well-watered short grass should be predictable to within ;10% or better for most climates. Daily pan-based estimates of ET0 tend to be noisy and can sometimes be improved by taking 2d averages of the measured data (Xing et al., 2008). The accuracy of the pan-evaporation method is dependent on the quality of station maintenance and is susceptible to the microclimatic conditions under which the pan operates.

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Chapter 13

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EXAMPLE 13.4 Evaporation-pan measurements indicate daily evaporation depths over the course of 1 week during the summer as: 8.2 mm, 7.5 mm, 7.6 mm, 6.8 mm, 7.6 mm, 8.9 mm, and 8.5 mm. During this period the average wind speed is 1.9 m/s and the average daily relative humidity is 70%. If the evaporation pan is installed in a green area with an upstream fetch of up to 1 km, determine the 7-d average reference evapotranspiration. Solution From the given data: u2 = 1.9 m/s, RHmean = 70%, and FET = 1000 m. Based on these data, Table 13.4 and Equation 13.62 give kp = 0.85. The 7-d average pan-evaporation rate, Ep , is given by 8.2 + 7.5 + 7.6 + 6.8 + 7.6 + 8.9 + 8.5 = 7.9 mm/d Ep = 7 and hence the reference evapotranspiration, ET0 , is given by Equation 13.61 as ET0 = kp Ep = (0.85)(7.9) = 6.7 mm/d

ww

Therefore, a 7-d average grass-reference evapotranspiration of 6.7 mm/d is indicated by the pan data.

w .E asy En g 13.5.4

Empirical Methods

The preferred FAO56-PM and ASCE-PM equations require solar radiation, wind speed, air temperature, and humidity data to calculate ET0 . However, all of these input variables are not always available for a given location. Although temperature and humidity data are routinely measured, and solar radiation data can be estimated with sufficient accuracy, windspeed data are rarely available, and there are no widely accepted and reliable methods to predict wind speeds with sufficient accuracy. In such cases, simplified approximate methods which require fewer input parameters can be used to estimate the grass-reference evapotranspiration (ET0 ), alfalfa-reference evapotranspiration (ETr ), or potential evapotranspiration (PET). Empirical equations that have been successfully used to approximate either the FAO56PM equation or the ASCE-PM equation are shown in Table 13.5, with the included variables defined in Table 13.6. In most cases, empirical equations that include solar radiation as a predictor variable have been found to provide better approximations to the FAO56-PM and ASCE-PM equations than purely temperature-based methods. The success of temperaturebased methods is mostly dependent of there being a strong correlation between solar radiation and temperature, and such correlations can be particularly weak in high-elevation areas which have lower temperatures and higher solar radiation (e.g., Juday et al., 2011). The relative performance of empirical equations generally vary by geographical location, and in many cases the standard coefficients in the empirical equations are adjusted to provide a better fit for a particular location. For example, in the Hargreaves equation, coefficients of 0.0028 and 0.0031 have been found to provide improved estimates of ET0 in the arid and colder climatic regions of Iran, respectively, compared to using the standard coefficient of 0.0023 (Tabari and Talaee, 2011b). Also, adjusted coefficients for using the Turc equation in Florida have been developed (Thepadia and Martinez, 2012), where it has been shown that the conventional coefficients in the Turc equation generally provide underestimates of PET in Florida. Irmak et al. (2003a) evaluated the performance of 21 ET0 and ETr empirical methods in Florida (humid climate) and reported that all methods produced significantly different daily ET0 values than the FAO56-PM method. To accommodate limitations on the availability of weather data and the limitations of several empirical ET0 estimation methods, Irmak et al. (2003b) developed a set of empirical equations that are in closer agreement with the FAO56PM equation than any of the other commonly used empirical methods to estimate ET0 in humid climates in the southeastern United States. These equations have been calibrated to match the FAO56-PM equation and are given by

ine eri n

g .n

et

ET0 = −0.611 + 0.149Rs + 0.079Tmean

(13.71)

ET0 = 0.489 + 0.289Rn + 0.023Tmean

(13.72)

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ww Name

Blaney–Criddle Hansen

Hargreaves Priestly-Taylor

Turc

w .E asy En g

TABLE 13.5: Empirical Equations for Estimating Evapotranspiration

Equation

References

ET0 = p(0.46Tavg + 8.13)

PET = 0.7



  + γ



Rs λ

Blaney and Criddle (1962)



Hansen (1984) Xu and Singh (2002)

1

ET0 = (0.408)(0.0023)Ra (Tmax − Tmin ) 2 (Tmean + 17.8)

ine eri n

 (Rn − G)  + γ



  Tmean 50 − RH PET = 0.013 (23.9Rs + 50) 1 + Tmean + 15 70 

Tmean (23.9Rs + 50) PET = 0.013 Tmean + 15

ET0 = 1.26

Hargreaves and Samani (1985) Priestly and Taylor (1972)

(RH < 50%)

(RH > 50%)

Turc (1961)

g .n

Note: (1) units of ET0 and PET are in mm/d; (2) conversion factor, 1 MJ/m2 ·d = 0.408 mm/d; and (3) see Table 13.6 for definitions and units of independent variables.

et

649

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ww

Symbol G

Soil heat flux

w .E asy En g

TABLE 13.6: Parameters in Empirical Evapotranspiration Equations

Definition

Units MJ m−2 d−1

Measured

Calculation —

Reference

Ra

Extraterrestrial radiation

MJ m−2 d−1

Equation 13.18∗

Allen et al. (1998)

Rn

Net solar radiation

MJ m−2 d−1

Measured

—

Rs

Total incoming solar radiation

MJ m−2 d−1

Measured

—

RH

Relative air humidity

%

Measured

—

p

Percentage of annual daylight hours (monthly)

%

Measured

—

Maximum temperature (daily)

◦C

Measured

—

Minimum temperature (daily)

◦C

Measured

—

Mean air temperature (daily)

◦C

Measured

—

Average of daily maximum and minimum temperatures (monthly)

◦C

—

Slope of saturation vapor pressure curve

kPa/◦ C

1 !N (T T avg = 2N i=1 max,i + Tmin,i )

γ

Psychrometric constant

kPa/◦ C

Equation 13.36

λ

Latent heat of vaporization

MJ/kg

Equation 13.35

Tmax Tmin Tmean Tavg 

Note: ∗ Apply appropriate unit conversion to the dependent variable in the indicated equation.

ine eri n Equation 13.42

Allen et al. (1998)

g .n

Allen et al. (1998)

Allen et al. (1998)

et

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Actual Evapotranspiration

651

where ET0 is the reference evapotranspiration [mm/d], Rs is the incoming solar radiation [MJ/(m2 ·d)], Tmean is the mean daily air temperature [◦ C] computed as the average of daily maximum and minimum air temperature, and Rn is the net radiation [MJ/(m2 ·d)]. Equation 13.71 is recommended for use when measurements or estimates of Rs and Tmean are available, and Equation 13.72 when Rn and Tmean are available. The formulation of empirical ET models generally involves the application of linear regression techniques between ET and several (climatic) predictor variables. In developing such regression models, it must be recognized that many climatic variables that affect ET are not independent of each other, and the most efficient regression model uses the least number of climatic parameters (Kovoor and Nandagiri, 2007). 13.6

ww

Actual Evapotranspiration

Precision weighing lysimeters are the standard for the direct measurement of ET. Lysimeters consist of isolated volumes of soil and vegetation in which the water balance is monitored by measuring the inputs (e.g., rainfall, irrigation), outputs (e.g., drainage), and changes in storage (e.g., moisture content) of water within the control volume. Two different applications of lysimeters are shown in Figure 13.2. In both cases the study area is isolated within a tank, with Figure 13.2(a) showing a lysimeter in a cattail (Typha domingensis) marsh and Figure 13.2(b) showing a lysimeter in an open-water wetland. If lysimeters are not available, the Bowen ratio energy balance method is often used as an alternative to lysimeter measurements of actual ET (e.g., Irmak and Irmak, 2008). However, care must be taken in using the Bowen ratio method, which can yield erroneous data for a variety of reasons, including limitations in the accuracy of the instruments, limitations of the method itself under certain conditions, configuration of the instrumentation, and the intrinsic numerical instability of the method (Romano and Giudici, 2009). Lysimeters and Bowen-ratio measurements of actual ET are mostly used in research applications. In areas with shallow water tables, the ET of surface vegetation can sometimes be estimated from measurements of diurnal fluctuations in the water table (e.g., Gribovszki et al., 2010). Actual ET can be estimated directly from meteorological and vegetation data using the general Penman–Monteith equation (Equation 13.1); however, this approach is seldom taken because of the difficulty in estimating such vegetation-specific parameters as albedo, α, aerodynamic resistance, ra , and surface resistance, rs .

w .E asy En g 13.6.1

Index-of-Dryness Method

ine eri n

g .n

A widely accepted hypothesis that was originally proposed by Budyko (1974), and is commonly referred to as the Budyko hypothesis, is that the ratio of mean annual evaporation to mean annual precipitation, ET/P, is primarily controlled by the ratio of mean annual potential evaporation to mean annual precipitation ETp /P. The ratio (ETp /P) has been assigned a variety of names, such as the index of dryness, aridity index (Oudin et al., 2008), and climate dryness index (Wang and Hejazi, 2011); the ratio ET/P is commonly called the evaporation ratio. The Budyko hypothesis asserts that ET approaches P in regions where the index of dryness is much greater than one, and ET approaches the ETp where the index of dryness is much less than one. A variety of empirical relationships (called Budyko-type curves) have been developed to describe this relationship, such as (Fu, 1981; Zhang et al., 2004)

et

FIGURE 13.2: Two lysimeters Source: Wossenu Abtew, South Florida Water Management District.

Vegetation Tank

(a) Cattail marsh

Tank

(b) Open-water wetland

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Chapter 13

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ETp ET =1 + − P P



1 +



ETp P

w w1

(13.73)

where the parameter w represents the integrated effects of catchment characteristics on annual evapotranspiration. For forested catchments w can be taken as 2.84, and for grassed catchments as 2.55. These relative values of w indicate higher evapotranspiration from forested areas than grassed areas under the same climatic conditions. EXAMPLE 13.5 In western Kansas the mean annual potential evapotranspiration is 125 cm, and the mean annual precipitation is 80 cm. Estimate the index of dryness and the actual evapotranspiration for grassed catchments.

ww

Solution From the given data: ETp = 125 cm, P = 80 cm, and the index of dryness is given by index of dryness =

w .E asy En g

ETp 125 = = 1.56 P 80

Using Equation 13.73 with w = 2.55 (for grassed catchments) gives ETp ET =1 + − P P



1 +



 1 " # 1 ETp w w = 1 + 1.56 − 1 + 1.562.55 2.55 = 0.820 P

Therefore, the annual evapotranspiration, ET, in western Kansas can be estimated by ET = 0.820P = 0.820(80 cm) = 66 cm

The estimated annual evapotranspiration (66 cm) is much closer to the annual precipitation (80 cm) compared with the potential evapotranspiration (125 cm). It is also notable that the estimated annual evapotranspiration is less than the annual precipitation, while the annual potential evapotranspiration is greater than the annual precipitation.

ine eri n

Other estimates of annual ET/P in terms of ETp /P have also been proposed. Gardner (2009) analyzed the rainfall and runoff in many watersheds in the United States, and by equating the annual ET to the measured annual precipitation minus the measured annual runoff found that the following equation provided a good fit to the estimated annual ET   ETp ET = 1 − exp − P P

g .n

et

(13.74)

where the annual potential evapotranspiration, ETp [mm], is estimated by the Holland (1978) formula given by   4620 ETp = 1.2 * 1010 exp − (13.75) TK and TK is the mean annual temperature [K]. Turc (1954) and Pike (1964) proposed the following functional form of the Budyko curve  − 1  ETp −ν ν ET = 1 + P P

(13.76)

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Actual Evapotranspiration

653

where ν is a locally fitted constant. A study of annual ET, ETp , and P data in 547 catchments across the United States over the 24-year period from 1983 to 2006 showed that ET/P could be linearly related to ETp /P with an average goodness of fit of 0.928 (Cheng et al., 2011). Although the fit was generally found to be linear, the parameters of the linear fit (i.e., slope and intercept) typically varied between catchments. 13.6.2

Crop-Coefficient Method

In the crop-coefficient approach, the actual ET is estimated by the relation ET = kc ET0

ww

(13.77)

where kc is a crop coefficient, which varies primarily with the type of vegetation and only to a limited extent with climate. The crop-coefficient approach enables the transfer of standard values of kc between locations and between climates. The crop-coefficient method can be used to predict ET where no limitations are placed on vegetation growth or evapotranspiration due to water shortage, vegetation density, disease, weed, insect, or salinity pressures. The crop coefficient, kc , represents an integration of the effects of four primary characteristics that distinguish the vegetation from reference grass: vegetation height, albedo, surface resistance, and evaporation from the underlying exposed soil. Changes in these characteristics over the growing season generally affect the crop coefficient. The crop-coefficient approach is particularly useful when short-term estimates of ET are needed, such as in irrigation scheduling for individual agricultural fields (Allen et al., 2005). Time-varying crop coefficients are also commonly used to describe seasonal variation in ET for natural vegetation (e.g., Hutchinson et al., 2008). Crop coefficients can be taken as independent of location only to the extent that they are based on the same equation used to estimate the reference evapotranspiration. In an effort to put all crop coefficients on a common basis, there is a movement in the United States to refer all crop coefficients to the ASCE-PM equation for either grass or alfalfa, where it is noted that the ASCE-PM equation is the same as the FAO56-PM equation for a grass reference crop using daily time steps. Regional cropping patterns can have a significant effect on ET and hence significantly influence regional water yields. For example, over the last century, land use and land cover in the United States Corn Belt region shifted from mixed perennial and annual cropping systems to primarily annual crops. This change in cropping patterns impacted the annual water balance in many Midwestern basins by decreasing annual ET and increasing streamflow and base flow (Schilling et al., 2008).

w .E asy En g 13.6.3

Remote Sensing

ine eri n

g .n

et

Remote sensing data collected by satellites are sometimes used to estimate regional ET, while recognizing that estimates of ET using ground-based measurements are generally preferable and more accurate at local scales. The spatial and temporal resolution of remote sensing data from the existing set of earth-observing satellites are not sufficient to estimate spatially distributed ET for on-farm irrigation scheduling purposes, especially at field scales on the order of 10–200 ha (25–500 ac) (Gowda et al., 2007). However, ET estimates derived from remote sensing can be used to improve the calibration of large-scale regional distributed hydrologic models (Immerzeel and Droogers, 2008). A variety of algorithms have been proposed for estimating ET from satellite-measured data. Techniques such as the surface energy balance algorithm for land (SEBAL), and the mapping evapotranspiration at high resolution and with internalized calibration (METRIC) algorithm, use surface energy balance algorithms and thermal remote-sensing data to estimate ET directly for a given image time and location. Application of these algorithms have been demonstrated in many areas of the United States (e.g., Bastiaanssen et al., 2005; 1998; Allen et al., 2007a; 2007b; Ambast et al., 2008). Alternative and simpler estimates of ET based on remote-sensing data have also been derived from vegetation coverage. For example, Groeneveld et al. (2007) proposed a method that combines satellite-measured vegetation coverage with ground-measured annual precipitation and reference ET to yield

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first-order estimates of actual ET—an approach that is particularly applicable under shallow ground water conditions. Rafn et al. (2008) demonstrated a fairly robust method in which crop coefficients can be estimated from satellite-based measurements of the normalized difference vegetation index (NDVI); these crop coefficients are then combined with groundbased reference ET estimates to estimate the actual ET. Singh and Irmak (2009) have also shown high correlations between crop coefficients and NDVI. Brunner et al. (2008) proposed and demonstrated a methodology to extract direct groundwater evaporation from remotely sensed maps of evapotranspiration. A common problem with using remote sensing to estimate ET is that remotely sensed data are commonly once-a-day measurements, and scaling methods must be used to approximate daily total ET (Colaizzi et al., 2006). 13.7

ww

Selection of ET Estimation Method

The distinction between actual ET, reference ET, and potential ET should always be kept in mind. The preferred method of estimating actual ET is either using the Penman–Monteith equation (Equation 13.1) directly or using crop coefficients combined with a reference ET calculated using either the FAO56-PM equation (Equation 13.57) or the ASCE-PM equation (Equation 13.60), with the latter preferred in the United States. The Penman equation (Equation 13.53) is the preferred method of estimating potential ET, although utilization of the potential ET is discouraged in favor of the reference ET as a measure of meteorological effects on ET. Considerations in using Penman-type equations to estimate reference ET are the complexity of the model, lack of sufficient meteorological data to estimate the model parameters, and a recognition that in some cases reference ET is dominated by solar radiation and therefore the use of simpler methods with fewer parameters is justified. For example, Abtew (1996) has noted that in South Florida most of the variance in daily reference ET is explained by solar radiation, with the effect of humidity and wind speed relatively minimal. In contrast, Irmak et al. (2008) have shown that wind speed and humidity significantly influence reference ET in Nebraska and radiation-based methods are far less preferable to combination methods, such as the ASCE-PM and FAO56-PM equations, in estimating reference ET. In general, the adequacy of alternative reference ET estimation methods depends on local climatic conditions; however, when supporting data are available the ASCE-PM or FAO56-PM methods are always preferable.

w .E asy En g

Problems 13.1. Show that the wind-speed profile over a grassed surface (Equation 13.5) can be derived from the general windspeed profile (Equation 13.6) by assuming a vegetation height of 0.12 meters. 13.2. Determine the soil heat flux in April in Algiers (Algeria). The mean monthly temperatures in March, April, and May are 14.1◦ C, 16.1◦ C, and 18.8◦ C. What is the equivalent evaporation rate? 13.3. The Greeley, Colorado, weather station is located at latitude 40.41◦ N and longitude 104.78◦ W at an elevation of 1462.4 meters. The anemometer is located 3 m above ground, and instruments measuring air temperature and relative humidity are located 1.68 m above ground. The weather station is surrounded by irrigated grass with a height of 0.12 meters. Daily measurements during a 1-week period in July are given in the following table:

ine eri n

g .n

et

Day

Tmax (◦ C)

Tmin (◦ C)

ea (kPa)

Rs (MJ/m2 · d)

u3 (m/s)

1 2 3 4 5 6 7

32.4 33.6 32.6 33.8 32.7 36.3 35.5

10.9 12.2 14.8 11.8 15.9 15.8 16.7

1.27 1.19 1.40 1.18 1.59 1.58 1.13

22.4 26.8 23.3 29.0 27.9 29.2 23.2

1.94 2.14 2.06 1.97 2.98 2.37 2.43

where Tmax and Tmin are the maximum and minimum temperatures in the day, ea is the actual vapor pressure, Rs is the incoming solar radiation, and u3 is the wind speed 3 m above the ground. Determine the grass and

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655

TABLE 13.7: Measured Meteorological Parameters

Month Tmean (◦ C) Tmax (◦ C) Tmin (◦ C) RHmin (%) RHmax (%) f (%) u10 (m/s) Rs (MJ m−2 d−1 ) Rain (cm)

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

17.3 23.2 8.5 70.9 96.1 69 3.52 12.51 3.40

19.1 23.4 12.1 71.9 94.8 68 3.60 14.60 4.77

20.9 25.0 15.1 71.3 96.0 77 3.84 18.18 8.86

22.5 25.8 17.8 67.2 94.0 78 3.45 20.03 6.05

25.1 27.3 22.0 74.0 94.1 71 2.91 21.70 7.95

26.4 28.3 24.0 79.5 95.0 74 2.66 19.54 18.50

27.2 28.8 24.9 80.6 93.9 76 2.49 19.95 16.42

27.2 28.9 24.8 81.1 95.8 75 2.44 18.33 18.13

26.7 28.4 24.2 80.8 96.9 72 2.56 15.97 18.01

24.8 27.4 21.0 75.8 95.2 72 3.11 16.11 14.03

21.4 25.3 16.1 76.0 96.9 67 3.39 13.69 7.87

18.8 23.4 10.3 74.3 97.1 65 3.35 11.59 6.01

alfalfa reference evapotranspiration rates for each day of the week, and evaluate the degree of variability. 13.4. The mean monthly weather parameters in vicinity of a constructed wetland in South Florida are given in Table 13.7, where Tmean , Tmax , and Tmin are the daily mean, maximum, and minimum temperatures respectively, RHmin and RHmax are the daily minimum and maximum relative humidities, f is the average amount of sunshine (% of possible amount), u10 is the wind speed at 10 m, and Rs is the solar radiation. It has been shown that the ET for shallow wetlands in South Florida can be estimated by the simple relation (Abtew, 2004)

ww

w .E asy En g

Rs (13.78) λ where ET is in mm/d, Rs is the incident solar radiation (MJ m−2 d−1 ), and λ is the latent heat of vaporization of water (MJ kg−1 ). ET = 0.53

(a) Compare the monthly ET estimates given by Equation 13.78 with estimates given by the Penman–Monteith equation and determine the monthly variations in albedo that would provide the best agreement between the two methods.

(b) By comparing monthly ET with rainfall, identify the wet season and the dry season, and assess the relationship, if any, between the albedo and season. (c) Equation 13.78 assumes that the wind speed and humidity have a minimal effect on ET in South Florida. Use the PM equation to assess this assumption and discuss the implication of this result on using Equation 13.78.

13.5. An evaporation pan measures a monthly evaporation of 152 mm. The pan is surrounded by a short green crop for approximately 300 m in the upwind direction, the average wind speed is 4 m/s, and the mean relative humidity is 80%. Estimate the reference evapotranspiration for short irrigated grass turf. 13.6. Show that Equations 13.62 and 13.63 are adequate representations of the data in Table 13.4. 13.7. In central Florida the mean annual potential evapotranspiration is 125 cm, and the mean annual precipitation is 150 cm. Estimate the index of dryness and the annual evapotranspiration for grassed catchments in central Florida.

ine eri n

g .n

et

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C H A P T E R

14

Fundamentals of Groundwater Hydrology I: Governing Equations 14.1

ww

Introduction

Groundwater hydrology is the science dealing with the quantity, quality, movement, and distribution of water below the surface of the earth. The field of groundwater hydrology is sometimes called geohydrology or hydrogeology, where the former term is used in the context of hydrologic practice and the latter term in the context of geologic practice. A major engineering application of the principles of groundwater hydrology is in the development of water supplies by means of wells and infiltration galleries. Other important engineering applications include the evaluation, mitigation, and remediation of contaminated groundwater; the temporary storage of freshwater in underground reservoirs; the control of groundwater levels to permit crop growth; and the temporary lowering of groundwater levels to facilitate subsurface construction. Groundwater accounts for approximately 30% of all the freshwater on earth, which is second only to the polar ice (69%), and two orders of magnitude greater than the amount of freshwater in lakes and rivers (0.3%). The subsurface environment is a porous medium in which the void spaces have varying degrees of water saturation. The region where void spaces are completely filled with water is called the saturation zone, and the region where void spaces are not completely filled with water is called the aeration zone. Water in the aeration zone is sometimes called vadose∗ water, and the aeration zone is sometimes called the vadose zone. Typically, the aeration zone (vadose zone) lies above the saturation zone, and the upper boundary of the saturation zone is called the phreatic surface† or water table. At the water table, the pressure is equal to atmospheric pressure, and this condition is illustrated in Figure 14.1. Within the aeration zone are three subzones: the soil-water, intermediate, and capillary zones. The soil-water zone is the region containing the roots of surface vegetation, voids left by decayed roots of earlier vegetation, and animal and worm burrows. The maximum moisture content in the soil-water zone corresponds to the maximum moisture that can be held by the soil against the force of gravity, regardless of the depth of the water table below ground surface. The maximum moisture content in the soil-water zone is called the field capacity, and the thickness of the soil-water zone is typically on the order of 1–3 m (3–10 ft). Beneath the soil-water zone is the intermediate zone, which extends from the bottom of the soil-water zone to the upper limit of the capillary zone. The capillary zone extends from the water table up to the limit of the capillary rise of the water from the saturation zone. The thickness of the capillary zone depends on the pore sizes in the material above the water table, and can vary from around 1 cm (0.4 in.) for gravels to several meters for clays. An aquifer‡ is a geologic formation containing water that can be withdrawn in significant amounts. A common misconception is that an aquifer is always completely saturated with water; however, an aquifer actually consists of the entire water-bearing geologic unit (or entire group of water-bearing units) and not just its saturated portion. Aquicludes contain water but are incapable of transmitting it in significant quantities, and aquifuges neither contain nor transmit water. A clay layer is an example of an aquiclude; solid rock is an example of an aquifuge; and, for most practical purposes, aquicludes can be taken as impervious

w .E asy En g

ine eri n

g .n

et

∗ Vadose is a derivative of the Latin word vadosus, which means “shallow.”

† Phreatic is a derivative of the Greek word phreatos, which means “well.” In this context, a saturated zone is

encountered when a well is dug. ‡ Aquifer is a derivative of the Latin words aqua (“water”) and ferre (“to bear”).

656

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Introduction

657

Surface vegetation Ground surface Soil-water zone Zone of aeration

Intermediate zone Capillary zone

Water table

Zone of saturation

ww

Aquifer

Impervious or semipervious formation

w .E asy En g

formations. Aquifers are classified as either unconfined or confined. Unconfined aquifers are open to the atmosphere, as illustrated in Figure 14.1, and are also called phreatic aquifers or water-table aquifers. In some cases, water accumulates above an elevated stratum of low permeability, and water above this stratum spills over into an unconfined aquifer. In such cases, the water table above the low-permeability stratum is called a perched water table. Clay lenses in sedimentary deposits often have shallow perched water bodies overlying them. In confined aquifers, water in the saturated zone is bounded above by either impervious or semipervious formations, and water at the top of a confined aquifer is normally at a pressure greater than atmospheric pressure. A typical configuration of a confined aquifer is shown in Figure 14.2, where the water in the confined aquifer is recharged by inflows at A (usually from rainfall). Land surfaces that supply water to aquifers are called recharge areas, and maintaining an adequate recharge area (and recharge-water supply) is particularly important in urban areas where groundwater is a major source of drinking water. The primary groundwater recharge mechanism is the infiltration of rainfall. Piezometers are observation wells with very short screened openings that are used to measure the piezometric head, φ, at the screened opening, which for an incompressible fluid is given by

ine eri n

φ=

FIGURE 14.2: Confined, unconfined, and artesian aquifers

p + z γ

g .n

et

(14.1)

Recharge ⫽ Impervious or semipervious formation

A Unconfined aquifer

Ground surface

Unconfined aquifer D

B

C Piezometers

Confined aquifer

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Chapter 14

Fundamentals of Groundwater Hydrology I: Governing Equations

ww

where p and z are the pressure and elevation, respectively, at the opening of the observation well (piezometer), and γ is the specific weight of the groundwater. If the confined aquifer shown in Figure 14.2 is penetrated by piezometers at B and C, then the water levels in these piezometers rise to levels equal to the piezometric heads at B and C, respectively. At B, the water in the piezometer rises above the top confining layer of the (confined) aquifer, indicating that the water pressure at the top of the confined aquifer is greater than atmospheric pressure. This condition can be contrasted with the case of a piezometer in an unconfined aquifer, such as at D, where the water level in the piezometer rises to the water table. The piezometer at C behaves similarly to the piezometer at B, the difference being that the water level in the piezometer at C rises above the ground surface. In practical terms, this means that if a well were extended from the ground surface down into the confined aquifer at C, then groundwater would flow continuously from the well until the piezometric head was reduced to the elevation at the top of the well. An aquifer that produces flowing water when penetrated by a well from the ground surface is called an artesian aquifer.∗ As indicated in Figure 14.2, a confined aquifer can be an artesian aquifer at some locations, such as at C, and nonartesian at other locations, such as at B. The presence of artesian aquifers and the distribution of piezometric heads in aquifers is frequently identified by plotting the areal distribution of piezometric head, and such plots are commonly referred to as piezometric surfaces or potentiometric surfaces. In hydrologic practice, the term potentiometric surface is preferred. In cases where artesian aquifers intersect the ground surface, concentrated flows of groundwater called springs are formed. Groundwater inflows into surface-water channels are a common source of perennial discharge in streams, which is commonly referred to as the base flow of the stream. Stream flows mostly consist of the base flow plus the flow resulting from stormwater runoff. Both unconfined and confined aquifers can be bounded by semipervious formations called aquitards, which are significantly less permeable than the aquifer but are not impervious. Of course, unconfined aquifers can only be bounded by impervious or semipervious layers on the bottom, while confined aquifers are bounded on both the top and bottom by either impervious or semipervious layers. Aquifers bounded by semipervious formations are called leaky aquifers, and terms such as leaky-unconfined aquifer and leaky-confined aquifer are commonly used. A microscopic view of the flow through a porous medium is illustrated in Figure 14.3, where water flows through the void space and around the solid matrix within the porous medium. It is difficult to describe the details of the flow field within the void spaces, since this would necessarily require a detailed knowledge of the geometry of the void space within the porous medium. To deal with this problem, it is convenient to work with spatially averaged variables rather than variables at a point (which is itself a spatial average over a very small volume). Referring to Figure 14.3, a property of the porous medium at P can be taken as the average value of that property within a volume centered at P. The scale of the averaging volume is called the support scale. Almost all properties of a porous medium that are relevant

w .E asy En g

FIGURE 14.3: Microscopic view of flow through a porous medium

ine eri n

Void space

g .n

Solid matrix

et

Averaging volume centered at P

P

Support scale

∗ The name “artesian” is derived from the name of the northern French city of Artois, where wells penetrating artesian aquifers are common.

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Introduction

659

to groundwater engineering are associated with a support scale, and in most cases the value of the averaged quantity is independent of the size of the support scale. A case in point is the porosity, n, of a porous medium, which is defined by the relation n=

ww

volume of voids sample volume

(14.2)

where the sample volume can be taken as the spherical volume with diameter equal to the support scale. A typical relationship between the porosity and support scale is shown in Figure 14.4. When the support scale is very small, the porosity is sensitive to the location and size of the sample volume. Clearly, for sample volumes of the order of the size of the void space, it makes a significant difference whether the sample volume is located within a void space or within the solid matrix. As the support scale gets larger, the porosity becomes less sensitive to the location and size of the sample volume and approaches a constant value that is independent of the support scale. The porosity remains independent of the support scale until the averaging volume becomes so large that it encompasses portions of the porous medium that have significantly different characteristics; under these circumstances, the porosity again becomes dependent on the size of the support scale. The range within which the porosity is independent of the support scale is given by the interval between the scales L0 and L1 shown in Figure 14.4. Therefore, as long as the support scale is between L0 and L1 , the porosity need not be associated with any particular support scale. The sample volume associated with the support scale L0 is commonly referred to as the representative elementary volume (REV). For sandy soils, an averaging volume with a diameter on the order of 10–20 grain diameters appears to be adequate for obtaining a stable average (Charbeneau, 2000). The relationship between the porosity and support scale is typical of the relationship between many other hydrogeologic parameters and support scales, although the REV of other parameters may be different. In general, there is no guarantee that a REV exists for any hydrogeologic parameter, and in the absence of a REV the value of the parameter must be associated with a support scale. All earth materials are collectively known as rocks; and the three main categories of rocks are igneous, sedimentary, and metamorphic rocks. Igneous rocks, such as basalt and granite, are formed from molten or partially molten rock (magma) formed deep within the earth; sedimentary rocks, such as sand, gravel, sandstone, and limestone, are formed by the erosion of previously existing rocks and/or the deposition of marine sediment; and metamorphic rocks, such as schist and shale, are formed through the alteration of igneous or sedimentary rock by extreme heat or pressure or both. Typically, igneous and metamorphic rocks have less pore space and fewer passageways for water than sedimentary deposits. Rocks are further classified as either consolidated or unconsolidated. Solid masses of rock are referred to as consolidated, while rocks consisting of loose granular material are termed unconsolidated. In consolidated formations, original porosity or primary porosity is associated with pore spaces created during the formation of the rock, while secondary porosity is associated with pore spaces created after rock formation. Examples of secondary porosity in consolidated formations include fractures in granite, and solution cavities in limestone. Secondary porosity in limestone is illustrated in Figure 14.5, where the scale indicates the size

w .E asy En g

FIGURE 14.4: Porosity versus support scale

ine eri n

g .n

et

Porosity, n

1.0

0.5

0

L0 L1 Support scale

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Chapter 14

Fundamentals of Groundwater Hydrology I: Governing Equations

FIGURE 14.5: Limestone with solution cavities

TABLE 14.1: Representative Hydrologic Properties in Consolidated Formations

ww

Material

Porosity

Specific yield

Hydraulic conductivity (m/d)

Sandstone Limestone Schist Siltstone Claystone Shale Till Basalt Pumice Tuff

0.05–0.50 0–0.56 0.01–0.50 0.20–0.48 0.41–0.45 0–0.10 0.22–0.45 0.01–0.50 0.80–0.90 0.10–0.55

0.01–0.41 0–0.36 0.20–0.35 0.01–0.35 — 0.01–0.05 0.01–0.34 — — 0.01–0.47

10−5 –4 10−4 –2000 10−4 –0.2 10−6 –0.001 — 10−8 –0.04 10−5 –30 10−6 –2000 — —

w .E asy En g

ine eri n

of the cavities in centimeters (left) and inches (right). This sample was taken from the surficial aquifer system in South Florida, which is one of the most productive aquifer systems in the world. Representative values of porosity in consolidated formations are given in Table 14.1. Between 60% and 90% of all developed aquifers consist of granular unconsolidated rocks (Todd and Mays, 2005), where the porosities are associated with the intergranular spaces determined by the particle-size distribution. Granular material is classified by particlesize distribution, and many different organizations have established classification standards for use in various disciplines. The United States Department of Agriculture (USDA) soil classification system is one of the most widely used in water-resources engineering and is given in Table 14.2, along with corresponding values of porosity. The porosities of granular materials tend to decrease with increasing particle size; however, this does not mean that water flows with more resistance through aquifers composed of larger particle sizes. In fact, the opposite is true. Porosities are considered small when n < 0.05, medium when 0.05 … n … 0.20, and large when n > 0.20 (Kashef, 1986). The most common aquifer materials are unconsolidated sands and gravels, which occur in alluvial valleys, coastal plains, dunes, and glacial deposits. Consolidated formations that make good aquifers are sandstones, limestones with solution channels, and heavily fractured volcanic and crystalline rocks. Clays, shales, and dense crystalline rocks are the most common materials found in aquitards. Aquifers range in thickness from less than 1 m (3 ft) to several hundred meters. They may be long and narrow, as in small alluvial valleys, or they may extend over millions of square kilometers. The depth from the ground surface to the top of the saturated zone of an aquifer may range from 1 m (3 ft) to more than several hundred meters.

g .n

et

Geologic perspective. Geologic characterization of the subsurface environment is an essential component of most groundwater studies. Although geologic characterizations are typically performed by professional geologists, water-resources engineers are expected to have sufficient understanding of geologic principles and nomenclature to collaborate with

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Introduction

661

TABLE 14.2: USDA Classification and Representative Hydrologic Properties in Unconsolidated Formations

Classification*

ww

Very coarse gravel Coarse gravel Medium gravel Fine gravel Very fine gravel Very coarse sand Coarse sand Medium sand Fine sand Very fine sand Silt Clay

Particle size* (mm) 32.0–64.0 16.0–32.0 8.0–16.0 4.0–8.0 2.0–4.0 1.0–2.0 0.5–1.0 0.25–0.5 0.10–0.25 0.05–0.10 0.002–0.05 100. Nonlinear flow conditions are routinely found in the immediate

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Darcy’s Law

665

vicinity of large water-supply wells, in fractured rock formations, and in karst formations (Worthington and Gunn, 2009). A variety of empirical and semiempirical equations have been suggested for describing nonlinear flows (Bear, 1972), with the Forchheimer equation or Dupuit–Forchheimer equation being commonly used. For isotropic porous media, the Forchheimer equation is given by (14.12)

−§φ = aq + bq|q|

ww

where § is the gradient operator, φ is the piezometric head, q is the specific discharge vector, and a and b are constants (Forchheimer, 1901). Although groundwater flow is adequately described by the Darcy equation in most practical cases, some groundwater flows have been more appropriately described by the Forchheimer equation or some other similar nonlinear formulation (e.g., Moutsopoulos and Tsihrintzis, 2005; Mathias et al., 2008; Wen et al., 2009). Flows that obey Darcy’s law are commonly classified as linear flows or Darcian flows, while nonlinear flows that obey the Forchheimer equation are sometimes classified as Forchheimer flows. Experimental results validating Equation 14.12 can be found in Chin et al. (2009) and Moutsopoulos et al. (2009). In cases of large water-supply wells located in karst formations, turbulent flow can predominate over distances on the order of several kilometers surrounding the wells (Shoemaker et al., 2008). In modeling turbulent flow in karst formations, it has been asserted that the piezometric head gradient can be taken as proportional to the square of the specific discharge, in which case the hydraulic conductivity is taken as (Reimann et al., 2011) ⎧ $ ⎨ Klam Recrit Re > Recrit Re Kturb = (14.13) ⎩Klam Re … Recrit

w .E asy En g

where Kturb and Klam are the hydraulic conductivities under turbulent and laminar flow conditions, respectively, Recrit is the limiting (critical) Reynolds number for laminar-flow conditions, and Re is the actual Reynolds number. Equation 14.13 has been validated using permeameter data with samples extracted from a karst formation in South Florida (Kuniansky et al., 2008). However, another investigation using these same data has shown that direct application of the Dupuit–Forcheimer equation provides a better description of turbulent flows than Equation 14.13 (Chin et al., 2009). In some karst formations, the ability to accurately account for turbulence has been found to be of secondary importance to accounting for the interchange between the rock matrix and the primary flow conduits (Hill et al., 2010).

ine eri n

EXAMPLE 14.2

g .n

et

A sand aquifer has a 10-percentile particle size of 0.4 mm and an effective porosity of 0.3. If the temperature of the water in the aquifer is 20◦ C, estimate the range of seepage velocities for which Darcy’s law is valid. Solution Darcy’s law can be taken to be valid when Re < 10, qd < 10 ν which can be put in the form q < or

10ν d

10ν ne d where the specific discharge, q, is related to the seepage velocity, v, and the effective porosity, ne , by v = q/ne . From the given data, ne = 0.3, d L d10 = 0.4 mm = 4 * 10−4 m, and at 20◦ C, ν = 1.00 * 10−6 m2 /s. Hence v
1000

Very high

Clean gravel

Vesicular and scoriaceous basalt and cavernous limestone and dolomite

High

Clean sand, and sand and gravel

Clean sandstone and fractured igneous and metamorphic rocks

Moderate

Fine sand

Laminated sandstone, shale, mudstone

Low

Silt, clay, and mixtures of sand, silt, and clay

Massive igneous and metamorphic rocks

Very low

Massive clay

10–1000

0.01–10 0.0001–0.01

ww

8, where Q is the rate of flow of freshwater per unit breadth of the aquifer. The Ghyben–Herzberg approximation generally assumes steady-state conditions and ignores the time required for steady-state conditions to be attained. In this respect, it is generally recognized that the Ghyben–Herzberg approximation can only provide an estimate of the extent of saltwater intrusion. Numerical experiments have shown that for a steady rise in sea level the actual extent of saltwater intrusion can be either greater than or less than the estimate based on Ghyben–Herzberg approximation (Watson et al., 2010). Besides saltwater intrusion caused by the density difference between saltwater and freshwater, a second important mechanism for saltwater intrusion is associated with the construction of unregulated coastal drainage canals. These canals allow the inland penetration of saltwater via tidal inflow and subsequent leakage of saltwater from the canals into the aquifer. To prevent saltwater intrusion in coastal drainage canals, salinity-control gates are typically placed at the downstream end of the canal to maintain a freshwater head (on the upstream side of the gate) over the sea elevation (on the downstream side of the gate). The freshwater head should be sufficient to prevent saltwater intrusion in accordance with the Ghyben–Herzberg equation. During periods of high runoff and when the stages in the canals are above a prescribed level, the canal gates are opened to permit drainage while maintaining a freshwater head that is sufficient to prevent saltwater intrusion.

w .E asy En g EXAMPLE 15.20

Consider the gated canal in a coastal aquifer illustrated in Figure 15.20. If the aquifer thickness below the canal is 30 m, and at high tide the depth of seawater on the downstream side of the gate is 3 m, find the minimum depth of freshwater on the upstream side of the gate that must be maintained to prevent saltwater intrusion.

FIGURE 15.20: Gated canal

Gate

Canal Freshwater

h 3m

Aquifer

z

30 m

ine eri n

Saltwater

g .n

et

Solution The minimum elevation of the freshwater surface at the upstream side of the gate must be sufficient to maintain the saltwater interface at a depth of 33 m below sea level. According to the Ghyben–Herzberg equation (Equation 15.247), the height of the freshwater surface above sea level, h, is given by h = ǫz where ǫ is the buoyancy factor and z is the depth of the interface below sea level. Taking ǫ = 0.025 and z = 33 m yields h = (0.025)(33) = 0.83 m Therefore, the freshwater on the upstream side of the gate must be at least 0.83 m above the sea level on the downstream side of the gate. Under this condition, the total depth of freshwater in the canal is 3 m + 0.83 m = 3.83 m.

In addition to salinity-control gates in coastal drainage channels, other methods of controlling saltwater intrusion include modification of pumping patterns, creation of freshwater

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Saltwater Intrusion

757

recharge areas, and installation of extraction and injection barriers. Extraction barriers are created by maintaining a continuous pumping trough with a line of wells adjacent to the sea, and injection barriers are created by injecting high-quality freshwater into a line of recharge wells to create a high-pressure ridge. In extraction barriers, seawater flows inland toward the extraction wells and freshwater flows seaward toward the extraction wells. The pumped water is brackish and is normally discharged to the sea. Whenever water-supply wells are installed above the saltwater interface, the pumping rate from the wells must be controlled so as not to pull the saltwater up into the well. The process by which the saltwater interface rises in response to pumping is called upconing. This phenomenon is illustrated in Figure 15.21. Schmorak and Mercado (1969) used equations developed by Dagan and Bear (1968) to propose the following approximation of the rise height, z, of the saltwater interface in response to pumping: z=

ww

Qw 2π dKx ǫ

(15.258)

where Qw is the pumping rate, d is the depth of the saltwater interface below the well before pumping, and Kx is the horizontal hydraulic conductivity of the aquifer. Equation 15.258 incorporates both the Dupuit and Ghyben–Herzberg approximations, and therefore care should be taken in cases where significant deviations from these approximations occur (Nordbotten and Celia, 2006). Equation 15.258 also assumes that the thickness of the saltwater–freshwater interface is small relative to the thickness of the aquifer. In cases where these approximations are valid, experimental data indicate that Equation 15.258 provides a reasonable approximation to the height of upconing (Werner et al., 2009). Experiments have shown that whenever the rise height, z, exceeds a critical value, the saltwater interface accelerates upward toward the well. This critical rise height has been estimated to be in the range 0.3d–0.5d. Taking the maximum allowable rise height to be 0.3d in Equation 15.258 corresponds to a pumping rate, Qmax , given by

w .E asy En g

ine eri n

Qmax = 0.6π d2 Kx ǫ

(15.259)

Therefore, as long as the pumping rate is less than or equal to Qmax , pumping of freshwater above a saltwater interface remains viable, although pumping rates must remain steady to avoid blurring the interface. For anisotropic aquifers in which the vertical component of the hydraulic conductivity is less than the horizontal component, a maximum well discharge greater than that given by Equation 15.259 is possible (Chandler and McWhorter, 1975).

FIGURE 15.21: Upconing under a partially penetrating well

Qw

g .n

et

Phreatic surface

Freshwater

Saltwater interface with pumping d

Saltwater

z

Saltwater interface without pumping Base of aquifer

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Chapter 15

Fundamentals of Groundwater Hydrology II: Applications

EXAMPLE 15.21 A well pumps at 5 L/s in a 30-m thick coastal aquifer that has a hydraulic conductivity of 100 m/d. How close can the saltwater wedge approach the well before the quality of the pumped water is affected? Solution From the given data: Qw = 5 L/s = 432 m3 /d, Kx = 100 m/d, and it can be assumed that ǫ = 0.025. Equation 15.259 gives the minimum allowable distance of the saltwater wedge from the well as Qmax 432 = = 9.6 m d= 0.6π Kx ǫ 0.6π (100)(0.025) Therefore, the quality of pumped water will be impacted when the saltwater interface less than or equal to 9.6 m below the pumping well.

ww FIGURE 15.22: Saltwater intrusion in a confined aquifer

The Ghyben–Herzberg approximation as given by Equation 15.247 can be applied to confined aquifers as illustrated in Figure 15.22(a). In this case, h represents the height of the piezometric surface above sea level, z is the depth of the saltwater interface below sea level, and these are related by h z= ǫ

w .E asy En g

where ǫ is the buoyancy factor. In confined aquifers where the pumping well fully penetrates the aquifer, the pumping rate must be limited to ensure that the toe of the saltwater wedge does not intersect the well (Mantoglou, 2003). In fact, the toe of a saltwater wedge should not even be allowed to come near a pumping well because the toe will accelerate and be sucked into the well as it approaches the well. The case of a pumping well in a confined aquifer Piezometric surface

h

z Qf

b

ine eri n Sea level

Seawater

Saltwater interface

(a) Confined aquifer without pumping

g .n

et

Qw Piezometric surface Sea level Seawater Qf

Saltwater interface

L (b) Confined aquifer with pumping

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with saltwater intrusion is illustrated in Figure 15.22(b). The limitation on pumping rate was investigated by Strack (1976) who showed that the limiting pumping condition when the toe of the saltwater wedge intersects the pumping well is given by the implicit relation ⎡ ⎤ 1  ∗ 2 Q ⎣1 − 1 − ⎦ 1  π ∗ ∗ Q Q 2 ⎤ ln ⎡ + λ∗ = 2 1 − (15.260) 1  π π ∗ 2 Q ⎣1 + 1 − ⎦ π where λ∗ and Q∗ are nondimensional variables defined by

ww

λ∗ =

Kbǫ Lqf

and

Q∗ =

Qmax bLqf

(15.261)

where K is the aquifer hydraulic conductivity (LT−1 ), b is the aquifer thickness (L), ǫ is the buoyancy factor (dimensionless), qf is the specific freshwater discharge from inland (LT−1 ), L is the distance of the well from the saltwater boundary (L), and Qmax is the pumping rate when the toe of the saltwater interface intersects the pumping well (L3 T−1 ). A shortcoming of Equation 15.260 is that it assumes a sharp saltwater interface and neglects mixing at the interface. As a consequence, Equation 15.260 associates the maximum pumping rate with the saltwater interface being at the well, whereas the maximum pumping rate will occur when a portion of the mixed zone intersects the pumping well. If mixing of the saltwater interface due to aquifer dispersivity is taken into account and the maximum pumping rate is associated with a salinity of 0.1% at the pumping well, then the maximum pumping rate can still be calculated using Equation 15.260; however, the buoyancy factor, ǫ, in λ∗ must be replaced by a modified buoyancy factor, ǫ ∗ , where ⎡ ⎤  1 6 α T ⎦ ǫ ∗ = ǫ ⎣1 − (15.262) b

w .E asy En g

ine eri n

and αT is the transverse dispersivity (Pool and Carrera, 2011). It is interesting to note that the modified buoyancy factor, ǫ ∗ , can also be used in the regular Ghyben–Herzberg equation (Equation 15.247) to estimate the location of the saltwater interface as h z= ∗ ǫ

g .n

et

(15.263)

where z is the depth below sea level to the most saline portion of the mixing zone (mixing ratios between 50% and 75% seawater), and h represents either the elevation of the water table above sea level (unconfined aquifer) or the elevation of the piezometric surface above sea level (confined aquifer). EXAMPLE 15.22 A coastal community is planning to develop a new wellfield in an area that is approximately 500 m from the coastline. A geological investigation indicates that the region is underlain by a confined aquifer of thickness 160 m, hydraulic conductivity of 80 m/d, transverse dispersivity of 1 m, and the regional freshwater flow toward the coast is approximately 0.5 m/d. It is expected that the water-supply wells will fully penetrate the aquifer. Estimate the maximum allowable pumping rate with and without taking dispersion into account. Solution From the given data: L = 500 m, b = 160 m, K = 80 m/d, αT = 1 m, and qf = 0.5 m/d. It will be assumed that ǫ = 0.025.

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Chapter 15

Fundamentals of Groundwater Hydrology II: Applications Without Dispersion: From the given data: λ∗ =

Kbǫ (80)(160)(0.025) = = 1.28 Lqf (500)(0.5)

Substituting λ∗ into Equation 15.260 requires that



1.28 = 2 1 −

ww

Q∗ π

1

⎡

2

+

⎢ ⎣1 −



1 −

Q∗ π

1 2

⎤ ⎥ ⎦

Q∗ ⎤ ln ⎡

1 π 2 ∗ Q ⎥ ⎢ ⎦ ⎣1 + 1 − π

which yields Q∗ = 0.555, and hence the maximum allowable flow rate, Qmax , is given by Qmax = Q∗ bLqf = (0.555)(160)(500)(0.5) = 22, 200 m3 /d = 257 L/s With Dispersion: The modified buoyancy factor, ǫ ∗ , is given by Equation 15.262 as ⎡ ⎡ ⎤ ⎤ 1 1   6 6 α 1 T ∗ ⎦ = (0.025) ⎣1 − ⎦ = 0.0143 ǫ = ǫ ⎣1 − b 160

w .E asy En g and hence λ∗ is taken as

λ∗ =

(80)(160)(0.0143) Kbǫ ∗ = 0.732 = Lqf (500)(0.5)

Substituting λ∗ into Equation 15.260 yields Q∗ = 1.24, and hence the maximum allowable flow rate, Qmax , is given by

ine eri n

Qmax = Q∗ bLqf = (1.24)(160)(500)(0.5) = 49, 600 m3 /d = 574 L/s

Taking dispersion into account increases the maximum allowable pumping rate from 257 L/s to 574 L/s. It is apparent that neglecting dispersion of the saltwater interface can lead to very conservative estimates of the maximum allowable pumping rate.

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et

Besides single-well cases, models have been developed to control saltwater intrusion in cases where multiple wells are used in coastal aquifers (e.g., Mantoglou and Papantoniou, 2008). The interconnectedness of water-supply wells in coastal aquifers allows pumping rates to be redistributed to account for seasonal and aperiodic shifts in water availability, demand, and saltwater intrusion (e.g., St. Germain et al., 2008). Classification of saline groundwater. Saline groundwater is a general term used to describe groundwater containing more than 1000 mg/L of total dissolved solids. There are several classification schemes for groundwater based on total dissolved solids, and a widely cited one, initially proposed by Carroll (1962), is given in Table 15.8. TABLE 15.8: Classification of Saline Groundwater

Classification

Total dissolved solids (mg/L)

Freshwater Brackish water Saline water Brine

0–1000 1000–10,000 10,000–100,000 >100,000

Source: Carroll (1962).

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Seawater has a total dissolved solids concentration of approximately 35,000 mg/L. Other forms of saline groundwater include connate water∗ that was originally buried along with the aquifer material, water salinized by contact with soluble salts in the porous formation where it is situated, and water in regions with shallow water tables where evapotranspiration concentrates the salts in solution. Problems 15.1. Consider a two-layer stratified aquifer between two reservoirs. The water surfaces in the reservoirs are at elevations 5 m and 4 m NGVD, respectively; the ground surface between the aquifers is at elevation 10 m NGVD; the top layer of the aquifer extends from ground surface down to elevation −10 m NGVD, and the base of the aquifer (and reservoirs) is at elevation −20 m NGVD. The hydraulic conductivity of the top layer of the aquifer is 50 m/d and that of the bottom layer is 100 m/d. If the reservoirs are 2 km apart, find the equation of the phreatic surface and the flow rate between the reservoirs. Neglect surface recharge. 15.2. Show that the flow rate, Q, between two reservoirs separated by a two-layer aquifer can be expressed as  b K  Q = 1 h2L − h2R + (K1 − K2 ) 2 (hR − hL ) 2L L

ww

where C1 and C2 are constants given by

C1 = and

C2 =

Nx2 K1 2 h + (K2 − K1 )b2 h + = C1 x + C2 2 2

L

sL

ine eri n L

Canal

H

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Ground surface

Water table

Water table Aquifer

K1 2 h + (K2 − K1 )b2 hL 2 L

where N is the recharge rate between the two reservoirs. This equation describes a mounded phreatic surface, with flow to the left of the mound peak going toward the left-hand reservoir, and flow to the right of the mound peak going toward the right-hand reservoir. Derive an expression for the location of the mound peak. It has been stated that the mound peak will always be located between the two reservoirs. Use your derived expression to determine whether this statement is true or false. 15.6. Derive the general equation for the phreatic surface in a three-layer aquifer between two reservoirs. Neglect surface recharge. 15.7. A well pumps at 400 L/s from a confined aquifer whose thickness is 24 m. If the drawdown 50 m from the well is 1 m and the drawdown 100 m from the well is 0.5 m, then calculate the hydraulic conductivity and transmissivity of the aquifer. Do you expect the drawdowns at 50 m and 100 m from the well to approach a steady state? Explain your answer. If the radius of the pumping well is 0.5 m and the drawdown at the pumping well is measured as 4 m, then calculate the radial distance to where the drawdown is equal to zero. Why is the steady-state drawdown equation not valid beyond this distance?

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15.3. Consider the case of a fully penetrating canal shown in Figure 15.23 in which the drawdowns at a distance L from the sides of the canal are sL and sR on the lefthand and right-hand sides of the canal, respectively, and the effective hydraulic conductivity of the aquifer is K. Derive an expression for the leakage out of the canal per unit length of canal. Calculate the leakage when K = 30 m/d, H = 20 m, L = 70 m, and sL = sR = 5 cm. 15.4. Derive the general equation for the phreatic surface in a two-layer aquifer between two reservoirs when the recharge, N(x), is not equal to zero. [Hint: An equation similar to Equation 15.8, but with an additional term to account for recharge.] 15.5. The equation describing the phreatic surface in a twolayer aquifer between two reservoirs has been shown to be

 (h − hL ) NL K1  2 hR − h2L + (K2 − K1 )b2 R + 2L L 2

sR

Aquifer Bottom of aquifer

FIGURE 15.23: Leakage from a fully penetrating canal ∗ The word connate is derived from the latin word connatus, which means “born together.”

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15.8. (a) For a confined aquifer, the Thiem equation can be used to express the drawdown surrounding a well in terms of well pumping rate, Qw , transmissivity, T, and radius of influence, R. Explain why this equation does not accurately account for the drawdown at the well. Show that by selecting the radius of influence as   2π Tsw R = rw exp Qw the Thiem equation can be made to accurately account for the drawdown at the well. What are the limitations of the Thiem equation and why does it have these limitations? Under what circumstances would you use the Thiem equation? (b) In deriving the Thiem equation, it is assumed that Darcy’s law is applicable in the aquifer immediately surrounding the well intake. If a well has a pumping rate of 30 L/s, diameter of 10 cm, and a 1-m long intake, estimate the range of hydraulic conductivities for which the Thiem equation could be applied without violating Darcy’s law. 15.9. A confined aquifer is 20 m thick and a 0.70-m diameter water-supply well is to be installed and pumped at a design rate of 40.5 L/s. A geologic investigation reveals that the confined aquifer is stratified and contains a high hydraulic conductivity layer of thickness 3 m. It is further estimated that the hydraulic conductivity of this layer is five times the hydraulic conductivity of the other layers in the aquifer, where the other layers have approximately the same hydraulic conductivity. The aquifer porosity is found to be 0.18 in all layers. Using the pilot hole for the water-supply well, a pump test is conducted at 4.05 L/s and the steadystate piezometric heads at 30 m and 60 m from the well are found to be 42.015 m and 42.030 m. Prior to pumping, the piezometric head was uniform at 42.301 m. After the water-supply well is installed and is pumping at its design rate, it is noted that the piezometric head at the well is 41.103 m. (a) Estimate the hydraulic conductivity of the aquifer layers. (b) Estimate the radius of influence of the well. (c) Estimate the variation of seepage velocity with depth at a location 100 m away from the well. 15.10. Repeat Problem 15.7 for an unconfined aquifer, assuming that the saturated thickness is 24 m prior to pumping. If actual drawdowns, rather than the modified drawdowns, are used to calculate the transmissivity of the aquifer, then what would be the percentage difference in the calculated transmissivity? 15.11. A water-supply well is located at the center of a small circular island that has a radius of 1 km. The well has a diameter of 1 m and a pumping rate of 333 L/s, and the (phreatic) aquifer has a hydraulic conductivity of 40 m/d, a specific yield of 0.15, and a porosity of 0.2. If the water surrounding the island is at mean sea level and the base of the aquifer is 45 m below sea level,

ww

15.12.

15.13.

15.14.

w .E asy En g 15.15.

15.16.

15.17.

estimate the water-table elevation at the well intake. Use the seepage velocity halfway between the well and the coastline to estimate how long a contaminant would take to travel from the perimeter of the island to the well. A pumping well has a radius of 0.5 m and extracts water at a rate of 400 L/s from a semiconfined aquifer whose thickness is 24 m. The drawdown 50 m from the well is 1 m, and the drawdown 100 m from the well is 0.5 m. Use Equation 15.71 to calculate the leakage factor, transmissivity, and hydraulic conductivity of the aquifer. Compare the transmissivity with that obtained by neglecting leakage. (See Problem 15.7.) Repeat Problem 15.12 using the approximate relation given by Equation 15.74. Verify that it is appropriate to use Equation 15.74. If the piezometric surface in Problem 15.12 was initially 40 m above the base of the aquifer, then plot the leakage rate into the semiconfined aquifer as a function of the radial distance from the pumping well. A semiconfined aquifer is recharged from an overlying unconfined aquifer through an aquitard, and the estimated recharge rate is 2.92 cm/yr. Observations indicate that the average piezometric surface in the semiconfined aquifer is 10 m below the water table in the unconfined aquifer and the aquitard is 1 m thick with a porosity of 0.2. What is the hydraulic conductivity of the aquitard? If a well in the semiconfined aquifer pumps 57.9 L/s, how much recharge area would be required to sustain the flow to the well? An aquifer test is conducted in an unconfined aquifer whose saturated thickness is 25 m prior to pumping. Steady-state drawdowns 100 m and 150 m from the 0.5-m diameter pumping well are 0.98 m and 0.82 m, respectively, when the pumping rate is 34.7 L/s. Estimate the average hydraulic conductivity of the formation. After the aquifer test, it is discovered that a low-hydraulic conductivity layer of thickness 1.5 m is located above the intakes of the pumping and monitoring wells, with 12 m of aquifer below the low-hydraulic conductivity layer. Would this finding change your estimate of the hydraulic conductivity? If so, what is your new estimate? A small 10-cm diameter well extracts water at a rate of 4.05 L/s from a subsurface formation that is characterized as being a leaky confined aquifer. Observed drawdowns at several distances from the well are given below. What is the theoretical equation for the drawdown versus distance from the well under steady-state conditions. Estimate the transmissivity and leakage factor of the aquifer.

ine eri n

g .n

et

Distance from 8.0 20.0 50.0 100.0 125.0 350.0 well, (m) Drawdown, (m) 0.220 0.172 0.125 0.106 0.094 0.047

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Downloaded From : www.EasyEngineering.net Problems 15.18. The transmissivity in a confined aquifer is estimated from steady-state drawdown measurements using Equation 15.34, and in a semiconfined aquifer using Equations 15.72 and 15.73. Identify the range of leakage factors for which the leakage can be neglected in the analysis of the drawdown data. Assume that leakage can be neglected whenever the neglect of leakage results in less than 0.1% error in the calculated transmissivity. 15.19. A well of radius 0.5 m in an unconfined aquifer pumps at a rate of 400 L/s. If the hydraulic conductivity of the aquifer is 30 m/d, the saturated thickness of the aquifer at the well is 24 m, and the recharge rate to the aquifer is 500 mm/yr, find an expression for the distribution of saturated thickness surrounding the well. Determine the distribution of seepage velocity surrounding the well. 15.20. Consider the fully penetrating well described in Problem 15.7. What will be the drawdown in the well if only the top half of the aquifer is penetrated? Express the drawdown in the well as a function of the penetration factor. Can the drawdowns at 50 m and 100 m be reasonably estimated by assuming full penetration of the well, even if the well penetrates only the top half of the aquifer? Explain. 15.21. Consider the fully penetrating well described in Problem 15.7. Compare the additional drawdown in the well expected if the well penetrates only the top 40% of the aquifer with the additional drawdown expected if the well screen is centered in the aquifer and has a length equal to 40% of the aquifer thickness. 15.22. A 200-mm diameter water-supply well is to be installed in a confined aquifer of thickness 30 m, and the radius of influence of the well is expected to be 250 m. Estimate the screen length that would be required to limit the effect of partial penetration on drawdown at the well to less than 10 %. Use both the Kozeny (1933) and DeGlee (1930) equations in your analysis. 15.23. Derive Equations 15.104 to 15.106 using the chain rule and Equation 15.103. 15.24. A well pumps water from a confined aquifer at a rate of 400 L/s, where the radius of the well is 0.5 m and the thickness of the aquifer is 24 m. If the aquifer’s storage coefficient is 0.0012 and its hydraulic conductivity is estimated to be 300 m/d, then calculate the drawdown at distances of 0.5 m, 50 m, and 100 m from the well as a function of time. 15.25. It has been widely asserted that the steady-state equation for estimating the transmissivity in confined aquifers given by Equation 15.34 can be used to estimate the transmissivity in spite of the unsteadiness in the measured drawdowns. Use the drawdown results derived in Problem 15.24 to estimate the time required for the transmissivity to be estimated within 1% accuracy by the steady-state equation.

ww

763

15.26. A fully penetrating well pumps 200 L/s from an unconfined aquifer where the specific yield is 0.15 and the hydraulic conductivity is 100 m/d. Prior to pumping, the saturated thickness was uniformly equal to 28 m. Calculate the drawdowns at 50 m and 100 m from the well after 1 second, 1 min, 1 h, 1 d, 1 month, and 6 months. At each time, use the steady-state (Thiem) equation to estimate the transmissivity of the aquifer based on the calculated drawdowns at 50 m and 100 m. On the basis of these results, what can you say about using a steady-state equation to estimate the transmissivity from pairs of drawdowns during the transient aquifer pump test? 15.27. A confined aquifer of thickness 24 m is pumped at a rate of 400 L/s, and the recorded drawdowns in a monitoring well located 50 m from the pumping well are given in the table below. Estimate the hydraulic conductivity, transmissivity, and storage coefficient of the aquifer.

w .E asy En g

Time (s)

Drawdown (cm)

1 10 100 1000 10,000 100,000

0.00 9.54 76.39 150.03 256.85 319.72

ine eri n

15.28. Explain clearly how equations for isotropic formations can be utilized in anisotropic formations. Explain why some aquifers are anisotropic in the horizontal plane. Consider a case in which a confined aquifer is anisotropic with hydraulic conductivities given approximately by Kxx = 30 m/d, Kyy = 15 m/d, and Kxy = 10 m/d. The thickness of the aquifer is 20 m, the storage coefficient is 0.005, and the porosity of the aquifer is 0.15. An aquifer pump test is to be conducted in the aquifer to confirm the hydraulic conductivities. If the pumping rate is 40 L/s, find the location of a monitoring well such that a drawdown of 1 m would be expected after 1 d of pumping. How many possible locations for the monitoring well are there? 15.29. (a) Consider the case of an aquifer pump test in a confined homogeneous anisotropic aquifer. Starting with the Theis equation, express the drawdown at location (x,y) relative to the pumping well as a function of time, well pumpage, transmissivity components, and storage coefficient. (b) During an aquifer pump test in a confined aquifer, drawdowns measured at two locations (1 and 2) are as follows:

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Time (h)

Drawdown at 1 (cm)

Drawdown at 2 (cm)

1 2 3 4 5 6 7 8 9 10

2.62 3.37 3.80 4.11 4.35 4.54 4.70 4.85 4.97 5.09

2.50 3.24 3.67 3.98 4.22 4.41 4.58 4.72 4.85 4.96

The well pumping rate is 3.47 L/s, and the (N,S) coordinates of the measurement locations relative to the well location are (86.60 m, 56.00 m) and (75.00 m, 129.90 m). Use the given measurements to estimate the transmissivity components and storage coefficient of the aquifer. [Hint: Observe that the difference in drawdown at locations 1 and 2 stabilizes quickly.]

ww

Use the Cooper–Jacob approximation of the well function to estimate the hydraulic conductivity and the storage coefficient of the aquifer. At what time does the Cooper–Jacob approximation become reasonable? 15.32. An aquifer pump test is conducted at a pumping rate of 22 L/s in an unconfined aquifer that has a saturated thickness of 30 m. Observed drawdowns 70 m from the well are as follows: DrawTime down (min) (m)

w .E asy En g

15.30. A well of radius 0.5 m pumps water out of an unconfined aquifer at a rate of 400 L/s. The saturated thickness prior to pumping is 24 m, and the measured drawdowns 50 m from the pumping well as a function of time are given below. Determine the hydraulic conductivity, transmissivity, and specific yield of the aquifer. Time (s)

Drawdown (cm)

1 10 100 1000 10,000 100,000

0.00 0.00 0.12 12.21 36.89 68.64

15.31. A confined aquifer of thickness 24 m is pumped at 270 L/s and the recorded drawdowns in a monitoring well 50 m from the pumping well are given below. Time (s)

Drawdown (cm)

1 10 100 1000 10,000 100,000

0.00 6.39 51.18 100.52 172.09 214.21

0.01 0.01 0.02 0.03 0.06 0.11 0.19

0.001 0.008 0.013 0.023 0.034 0.036 0.036

Time (min)

Drawdown (m)

1.88 18.75 59.40 187.50 333.00 594.00 1053.00

0.036 0.036 0.046 0.058 0.067 0.091 0.122

Time (min)

Drawdown (m)

1875.00 3330.00 5940.00 10,530.00 18,750.00

0.157 0.187 0.248 0.267 0.312

Determine the aquifer properties using the Neuman type-curve method. 15.33. An aquifer pump test is conducted in an unconfined aquifer with a saturated thickness of 25 m. The pumping well is pumped at 40 L/s, and the drawdowns are measured at a monitoring well located 40 m away from the pumping well. The measured drawdowns as a function of time are given in the following table:

ine eri n Time

18 s 37 s 1.08 min 1.85 min 3.07 min 6.15 min 10.8 min 18.4 min 1.54 h 2.69 h 4.61 h 7.68 h 15.4 h 1.12 d 1.92 d

Drawdown (cm) 0.8 4.4 10.2 17.7 25.2 33.7 37.8 39.3 43.3 45.5 49.5 55.1 67.9 82.8 98.4

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Use a delayed-yield analysis to determine the storage coefficient, specific yield, horizontal hydraulic conductivity, and vertical hydraulic conductivity of the aquifer. 15.34. An aquifer pump test was conducted in an unconfined aquifer near Fairborn, Ohio (Lohman, 1979). The fully penetrating well was pumped at a constant rate of

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765

TABLE 15.9

Time Drawdown Time Drawdown Time Drawdown (min) (cm) (min) (cm) (min) (cm)

ww

0.165 0.25 0.34 0.42 0.50 0.58 0.66 0.75 0.83 0.92 1.00 1.08 1.16 1.24 1.33 1.42 1.50 1.68 1.85 2.00 2.15 2.35

3.66 5.94 7.77 10.06 11.89 13.11 14.93 16.15 17.37 18.59 19.51 20.42 21.33 21.94 22.55 23.16 23.77 24.99 25.60 26.21 26.52 27.43

2.5 2.65 2.8 3.0 3.5 4.0 4.5 5.0 6.0 7.0 8.0 9.0 10.0 12.0 15.0 18.0 20.0 25.0 30.0 35.0 40.0 50.0

27.74 28.04 28.35 28.65 28.95 29.56 29.72 29.87 30.17 30.48 30.78 30.94 31.09 31.39 31.70 32.00 32.31 32.92 34.44 35.05 35.66 36.27

w .E asy En g

68.14 L/s, and the drawdowns measured at a monitoring well 22.25 m away are given in Table 15.9. The saturated thickness of the aquifer under static (no pumping) conditions is 23.77 m. Calculate the hydraulic parameters of the aquifer using a delayed-yield analysis. 15.35. A confined anisotropic aquifer has a hydraulic conductivity of Kxx = 45 m/d, Kyy = 15 m/d, and Kxy = 0 m/d; a thickness of 14 m; and a storage coefficient of 10−4 . If a well is installed in the aquifer and pumped at 16.7 L/s, estimate the drawdown at x = 100 m, y = 100 m after 1 week of pumping. How would this result change if the confining layer had a leakage factor of 2000 m? 15.36. A fully penetrating well of radius 0.5 m pumps water at 400 L/s from a semiconfined aquifer of thickness 24 m. If the hydraulic conductivity of the aquifer is 300 m/d and the storage coefficient is 0.0012, estimate the drawdown as a function of time at a distance of 50 m from the well for leakage factors of 1 m, 10 m, and 100 m. Explain your results. If the thickness of the semiconfining layer is 5 m, then what are the hydraulic conductivities of the semiconfining layers corresponding to leakage factors of 10 m and 100 m? 15.37. A rapid aquifer pump test is performed in an unconfined aquifer and the drawdown measurements 50 m from the pumping well are shown below.

60 70 80 90 100 120 150 200 250 300 400 500 600 700 800 900 1000 1200 1500 2000 2500 3000

37.18 38.10 39.01 39.32 39.93 41.45 44.19 46.33 48.46 50.29 51.81 56.39 59.43 61.26 63.70 65.53 67.05 69.19 71.62 75.89 78.94 81.07

Drawdown (m)

ine eri n Time

0.16 s 1.6 s 16 s 2.7 min 27 min

0.0698 0.576 1.24 1.66 1.68

g .n

Water is pumped from the aquifer at a rate of 2310 L/s during the test. Using these data, estimate the storage coefficient (not specific yield!) and transmissivity of the aquifer using a delayed-yield analysis. A closer look at the geological samples collected during installation of the well indicates that the subsurface formation might actually consist of a 50-m deep homogeneous high hydraulic conductivity formation overlain by a 5-m deep low hydraulic conductivity formation. The pumping-well and monitoring-well intakes are located in the high hydraulic conductivity formation. Using this information, reanalyze the measurements to estimate the leakage factor and the hydraulic conductivity of the low hydraulic conductivity formation. If possible, use your overall results to estimate the vertical hydraulic conductivity in the high hydraulic conductivity formation.

et

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15.38. An aquifer pump test is conducted in a subsurface formation that is characterized as being a leaky confined aquifer. Geological logs indicate that aquifer thickness is 80 m and the thickness of the semiconfining layer is 28 m. The pumping rate is 7.23 L/s and the measured drawdowns 105 m from the well are given below. Time (days)

Drawdown (m)

Time (days)

Drawdown (m)

0.0 0.0340 0.0428 0.0525 0.0702 0.1010 0.1340

0.0 0.055 0.062 0.066 0.073 0.090 0.098

0.188 0.251 0.312 0.400 0.453 0.533 0.612

0.108 0.116 0.122 0.135 0.136 0.137 0.138

ww

Estimate the hydraulic conductivities of the aquifer and semiconfining layer, and the storage coefficient of the aquifer. 15.39. An aquifer pump test was conducted in a 5-m thick shallow aquifer confined above by 4.5 m of clay. The 200-mm diameter well was pumped at the rate of 1.58 L/s for 20 h, and the drawdown was measured in a monitoring well located 20 m west of the pumping well. The drawdown data are shown in the table below. Estimate the transmissivity and storage coefficient of the aquifer, and the hydraulic conductivity of the confining layer.

w .E asy En g

Time Drawdown Time Drawdown Time Drawdown (min) (m) (min) (m) (min) (m) 5 28 41 60

0.23 1.01 1.09 1.24

1200 m. If the planar coordinates of the well are (0 m, 0 m), and a fully penetrating river runs along the line x = 500 m, then calculate the steady-state drawdowns at (100 m, 0 m) and (−100 m, 0 m). 15.43. A second well located at (0 m, 200 m) is added to the wellfield described in Problem 15.42. If this well also pumps at 400 L/s and the radius of influence is 1200 m, then calculate the drawdowns at (100 m, 0 m) and (−100 m, 0 m) and compare with those calculated in Problem 15.42. If this second well is to be placed parallel to the river and contribute no more than 1% of the total drawdown at the designated points, then determine the coordinates of the second well. 15.44. A well with an effective radius of 0.8 m pumps water at 500 L/s from an unconfined aquifer of saturated thickness 24 m, hydraulic conductivity 250 m/d, and specific yield 0.2. A canal that feeds the aquifer is located 500 m east of the well, and the canal runs in a north–south direction. Determine the fraction of the pumped water that originates in the canal. You may need to use the following integral relation to solve this problem:    x 1 dx = tan−1 2 2 a a a + x

75 244 493 669

1.34 1.67 1.82 1.86

958 1129 1185

15.45.

15.46.

15.47.

1.91 1.95 1.96

15.40. In the aquifer described in Problem 15.39, the thickness and hydraulic conductivity of the overlying aquifer is 20 m and 25 m/d, respectively, and the semiconfining layer has a storage coefficient of 0.0015. Assess whether neglecting storage in the semiconfining layer, neglecting drawdown in the overlying aquifer, and neglecting storage in the pumping well are justified. 15.41. Three wells are located in an infinite unconfined aquifer of saturated thickness 20 m, specific yield 0.2, and hydraulic conductivity 40 m/d. The planar coordinate locations of the wells are: Point A (0 m, 0 m), Point B (200 m, 200 m), and Point C (200 m, −200 m). If all wells begin pumping at the same time and at a rate of 200 L/s, then calculate the drawdown as a function of time at a location (100 m, 100 m). 15.42. A well of radius 0.5 m pumps water at 400 L/s from a confined aquifer of thickness 24 m. The hydraulic conductivity of the aquifer is 300 m/d, the storage coefficient is 0.012, and the radius of influence can be taken as

15.48.

where a is any constant. Repeat Problem 15.42, but with the fully penetrating river replaced by an impermeable barrier. Repeat Problem 15.43, with the fully penetrating river replaced by an impermeable barrier. A well of radius 0.5 m pumps water at 400 L/s from a confined aquifer of thickness 24 m, hydraulic conductivity 300 m/d, and storage coefficient 0.012. If a fully penetrating river is located 1 km east of the well and an impermeable boundary is located 1 km west of the well, then calculate the drawdown in the aquifer at points 200 m east of the well and 200 m west of the well. Assume that the radius of influence of the well is 1200 m. Show that the drawdown distribution caused by a well in an infinite strip between two fully penetrating streams is given by

ine eri n s=

g .n

et

q (x + x0 − 2nd)2 + y2 Qw  ln 4π T (x − x0 − 2nd)2 + y2 n=−q

(15.264)

where Qw is the pumping rate, T is the aquifer transmissivity, x0 is the distance of the well from one stream, and d is the distance between the streams. 15.49. Show the arrangement of image and real wells that is required to calculate the drawdown induced by a well in an aquifer quadrant, where the other three quadrants contain impermeable formations. 15.50. Show that the buildup caused by injecting water at a rate Qw into a fully penetrating well in an unconfined infinite aquifer is exactly the same as the drawdown caused by withdrawing water at a rate Qw from the same aquifer.

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Downloaded From : www.EasyEngineering.net Problems 15.51. The water table at a given location in a coastal aquifer is 1 m above mean sea level, and the saturated zone in the aquifer is 24 m thick. Do you expect saltwater intrusion to be a problem at this location? 15.52. In a coastal aquifer, the saltwater interface intersects the bottom of the aquifer approximately 3 km from the

shoreline. If the hydraulic conductivity of the aquifer is 40 m/d and the bottom of the aquifer is 50 m below sea level, estimate the freshwater discharge per kilometer of shoreline. 15.53. The average yearly high and low water-table elevations in Miami-Dade County, Florida, are shown in

80° 30′

80° 45′

80° 15′

Snake Creek C an al

BROWARD CO

w .E asy En g

an

SR-826

al

6

Mi

am

i

Turnpike

NS

ER

Levee 30

VA T 3B ION

AR

Dade-Broward Levee

EA

Le

iam

iC

CO

ww

M

5

ve

COLLIER CO

Le ve e6 7A e6 7C

4

4 Rive

r

Levee 31 N SR-27

US-41

4 3

US-

1

SR

-8

74

r Creek

ine eri n

y Ba ne

cay

Perrine Ca na l

Bis

5

pe

ek

6

ap

Cre

Krome Ave

7

Sn

Black

8

Miami Beach

4

10

9

3

3

Tamiami Trail Tamiami Canal

45′

US-1

26°

NT

IC

OCE

AN

g .n

et

US-1

Levee 31 W

Lev

ee 3

1E

Florida City

LA

Homestead

AT

MONROE CO

30′

4

C-1

11

3 EXPLANATION Contours of water table elevation. Local depressions in the water table caused by wellfields are not included.

25° 15′

0 0

767

5 5

10 MILES 10 KILOMETERS

FIGURE 15.24: Average annual high water-table elevations in Miami-Dade County, Florida Source: Lietz et al. (2002).

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Figures 15.24 and 15.25 respectively. High water-table conditions are characteristic of wet-season conditions and low water-table conditions are characteristic of dryseason conditions. The transmissivity contours of the surficial aquifer system are shown in Figure 15.26, and the bottom elevations of the surficial aquifer system

80° 30′

80° 45′

80° 15′

Snake Creek C an al M

i Ca

na

l

8

SR-826

2 SR

-8

74

Creek

ine eri n l

Ba

1

na

ne

Ca

y

Perrine

2

cay

3

E

g .n

et

US-1

Levee 31 W

1

AT

LA

NT

Leve e 31

Florida City

IC

Homestead

2

er

Bis

Krome Ave

30′

Miami Beach

pp

ek

3

r

Cre

5

4

1 Rive

1

Sna

Black

SR-27

6

i

US-41

Levee 31 N

7

am

1

w .E asy En g

Tamiamii Trail Tamiami Canal

Mi

US-

CONSERVATION AREA 3B

Turnpike

Levee 30 Dade-Broward Levee

Le

ww 45′

US-1

iam

ve

COLLIER CO

Le ve e6 7A e6 7C

BROWARD CO

AN

26°

are shown in Figure 15.27. (a) Estimate the range of saltwater intrusion between high and low water-table conditions; and (b) estimate the range of freshwater inflows to Florida Bay between the wet and dry seasons. 15.54. A drainage canal in a coastal area terminates in a gated structure, and the thickness of the aquifer beneath the

OCE

Chapter 15

MONROE CO

768

C-1

11

EXPLANATION Contours of water table elevation. Local depressions in the water table caused by wellfields are not included.

25° 15′

0 0

5 5

10 MILES 10 KILOMETERS

FIGURE 15.25: Average annual low water-table elevations in Miami-Dade County, Florida Source: Lietz et al. (2002).

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80° 45′

26°

80° 15′

Snake Creek C an al US-1

Le ve e6 7A e6 7C

BROWARD CO

M

iam

an

300

al Mi

SR-826

Turnpike

Levee 30

300

75

CONSERVATION AREA 3B

iC

1,000

Dade-Broward Levee

Le

ve

COLLIER CO

769

am

i

Rive

r

Miami Beach

Tamiami Trail Tamiami Canal US-41

74 -8

US-

1

SR

0 00

1,

Levee 31 W

y

11

AN OCE

AT

LA

NT

IC

1E

US-1

C-1

Ba Bis

ine eri n Florida City

Lev

Homestead

300

ee 3

TE ES W E AT EVERGLADES M NATIONAL PARK XI

cay

Krome Ave

Perrine Ca na l

ne

75

IFE R

AQ U

CA YN E

BI S

O F

IT

LI M

MONROE CO

l na

0

30

Ca

Creek

ek

O PR AP

er

Cre

w .E asy En g 1,000

RN

30′

Snapp

Black

ww

Levee 31 N SR-27

45′

g .n

et

EXPLANATION Contours of transmissivity in 1,000’s of feet squared per day. Dashed where approximately located. Hachures indicate depression.

25° 15′

0 0

5 5

10 MILES 10 KILOMETERS

FIGURE 15.26: Transmissivity distribution of surficial aquifer system in Miami-Dade County, Florida Source: Fish and Stewart (1991).

canal is 24 m. On the downstream side of the gate, the seawater fluctuates between 30 cm above and below mean sea level. If the elevation of the bottom of the canal is 3 m below mean sea level, determine the minimum fresh water elevation on the upstream side of the gate to prevent saltwater intrusion.

15.55. A water-supply well is being threatened by saltwater intrusion. The well is pumping 40 L/s from an aquifer that has a hydraulic conductivity of 500 m/d, saturated thickness of 50 m, porosity of 0.2, and specific yield of 0.15. If the well is screened to within 15 m of the bottom of the aquifer, what will be the thickness of the saltwater

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Chapter 15

Fundamentals of Groundwater Hydrology II: Applications 30′

80° 45′

80° 15′

26°

−160

Snake Creek

−200

BROWARD CO

n Ca

−260

−

US-1 i Bea ch

Miami am i Rive

Mi

Miam

Florida’s Turnpike

20

−2

r

US-41

Snapp

er Creek Canal

y Ba ne cay

80

AN

Bis

−1

Krome Ave

Perrine Ca na l

Levee 31 W

0

5

111

NT LA

g .n

EXPLANATION

et

STRUCTURE CONTOUR: Shows altitude below sea level of base of surficial aquifer system. Dashed where approximately located. Contour interval 20 feet. Hachures indicate depression.

25° 15′

0

IC

1E

Lev

ee 3

0

−16

AT

0

US-1

C-

OCE

Cit

y

ine eri n

−14

rida

Ho

80

−1

Flo

me

ste

ad

−220

ek

00

EVERGLADES NATIONAL PARK

−200 Cre

0

−2

−20

w .E asy En g

US-

1

SR

-8

74

Levee 31 N SR-27

Levee 67

Extension

Tamiami Canal Tamiami Trail 180

Black

ww

Dade-Broward Levee

−16 0

Le

Le ve e6 7A e6 7C

−200

Le

ve

Levee 28

ve

e

al

−220

CONSERVATION AREA 3B

45′

MONROE CO

an

−200

CONSERVATION AREA 3A

−240

iC

30

−180

SR-826

80

iam

−180

COLLIER CO

M

Levee 29

30′

al

−1

5

10 MILES 10 KILOMETERS

FIGURE 15.27: Elevation distribution of bottom of surficial aquifer system in Miami-Dade County, Florida Source: Fish and Stewart (1991).

wedge under the well when the well is likely to become contaminated with saltwater? 15.56. A water-supply well is planned for installation in a confined aquifer that has a hydraulic conductivity of 75 m/d, a thickness of 100 m, an estimated transverse dispersivity

of 2 m, and freshwater flow toward the coast of approximately 0.1 m/d. If the well is to be located 1 km from the coastline and fully penetrate the aquifer, estimate the maximum allowable pumping rate with and without accounting for saltwater dispersion.

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C H A P T E R

16

Design of Groundwater Systems 16.1

ww

Introduction

In areas where groundwater is a source of public water supply, wellfields must be designed with a variety of constraints, such as pumpage from individual wells must not produce excessive drawdowns, the rates of pumpage must be sustainable, and the wellfield must be adequately protected from contaminants released into the groundwater from surrounding areas. The impacts of wellfields on the environment, such as reduced water availability to plants caused by lower water tables and decreased surface-water flows caused by well capture must generally be assessed. Many surface-water management systems and water-treatment effluent-discharge systems can also be considered as groundwater systems since they are designed to provide groundwater recharge via infiltration (e.g., infiltration basins), and the performance of such systems must typically be assessed in terms of the capacity of the subsurface matrix to percolate the design infiltration amounts, and the effect that the induced groundwater mounding will have on the performance of these systems.

w .E asy En g 16.2

Design of Wellfields

The objective of wellfield design is to determine the number and locations of production wells required to supply water at a specified rate. For large municipal and agricultural production wells, pumping-rate requirements are typically in the range of 30–250 L/s (500–4000 gpm). Small- and medium-size community water systems may depend on water wells that produce 5–30 L/s (100–500 gpm). Individual homes’ domestic wells usually meet their needs with as few as 0.07–0.3 L/s (1–5 gpm), depending on local regulations. A primary constraint in wellfield design is the maximum allowable drawdown in the aquifer caused by wellfield operation. To accommodate this constraint, the following design procedure is recommended:

ine eri n

Step 1: Assume that each well in the wellfield causes a drawdown (at the well) equal to onehalf the allowable drawdown, with the other half of the allowable drawdown caused by interference from other wells, boundary effects, and well losses. Determine the pumping rate, Qw , at each well that would cause a drawdown (at the well) equal to one-half the allowable drawdown. This can be done using the Theis equation, with a pumping time equal to 1 year. Step 2: The number of wells required in the wellfield is equal to the required water-supply rate, Q, divided by the pumping rate from each well, Qw . This ratio is rounded upward to the nearest integer. Step 3: Arrange the wells such that the total drawdown at each well does not exceed the allowable drawdown. Selecting the relative locations of the wells will depend on the individual site characteristics, such as property boundaries and existing pipe networks, but the wells should generally be aligned parallel to and as close as possible to surface-water recharge boundaries, such as rivers and canals, and as far away as possible from impermeable boundaries. Typically, production wells are spaced at least 75 m (250 ft) apart (Walton, 1991).

g .n

et

A wellfield design is illustrated in the following example.

771

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Chapter 16

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EXAMPLE 16.1 A small municipal wellfield is to be developed in an unconfined sand aquifer with a hydraulic conductivity of 50 m/d, saturated thickness of 30 m, and specific yield of 0.2. A service demand of 81.0 L/s is required from the wellfield, and the diameter of each well is to be 60 cm. If the drawdown in the aquifer is not to exceed 3 m when the wellfield is operational, develop a proposed layout for the wells. There are no nearby surface-water bodies. Solution Step 1: Estimate the pumping rate from a single well that would cause a drawdown of 1.5 m, equal to one-half of the allowable drawdown of 3 m. The drawdown, s, resulting from the operation of each well can be estimated using the Theis equation, s=

ww

Qw W(u) 4π T

where Qw is the pumping rate from the well, T is the transmissivity of the aquifer, W(u) is the well function, and u is defined by the relation

w .E asy En g

u=

r 2 Sy 4Tt

where r is the distance from the well, Sy is the specific yield of the aquifer, and t is the time since the beginning of pumping. Since the drawdowns in the wellfield will vary between zero and 3 m, the average transmissivity, T, to be used in the wellfield design is given by   3 = 1425 m2 /d T = KH = 50 * 30 − 2

For a single well with t = 365 d,

r 2 Sy (0.3)2 (0.2) = = 8.652 * 10−9 uw = w 4Tt (4)(1425)(365)

ine eri n

Since uw … 0.004, with less than 0.1% error

W(uw ) = −0.5772 − ln uw and therefore

g .n

W(uw ) = −0.5772 − ln (8.652 * 10−9 ) = 17.99

According to the Theis equation, the pumping rate, Qw , required to produce a drawdown, sw , is given by 4π Tsw Qw = W(uw ) For sw = 1.5 m (one-half the allowable drawdown), Qw =

et

4π (1425)(1.5) = 1493 m3 /d 17.99

Step 2: Since the service demand is 81.0 L/s = 7000 m3 /d, the number of wells required is given by number of wells =

7000 = 4.69 L 5 wells 1493

Step 3: Consider the five-well wellfield with the arrangement shown in Figure 16.1. Each of the 5 wells is to operate at the same pumping rate, Qw , given by Qw =

7000 = 1400 m3 /d = 16.2 L/s 5

and the arrangement consists of a central well (No. 1) surrounded by four wells (Nos. 2, 3, 4, 5) at a distance R away from the central well. Clearly, when all five wells are in operation, the

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Design of Wellfields

773

Qw 3

Qw 2

Qw 1

Qw 4

R Qw 5

ww

drawdown will be greatest at the central well (No. 1). Denoting the drawdown at well No. 1 by s1 , according to the principle of superposition

w .E asy En g

5 Qw  W(ui ) 4π T

s1 =

(16.1)

i=1

where ui are given by

r 2 Sy (0.3)2 (0.2) = = 8.652 * 10−9 u1 = w 4Tt 4(1425)(365)

and

u2 = u3 = u4 = u5 =

and therefore,

R2 Sy R2 (0.2) = = 9.613 * 10−8 R2 4Tt 4(1425)(365)

ine eri n

(16.2)

W(u1 ) = 17.99

W(u2 ) = W(u3 ) = W(u4 ) = W(u5 )

g .n

Equating the drawdown at well No. 1 to the maximum allowable value of 3 m, Equation 16.1 yields ! 1400 17.99 + 4W(u2 ) 3= 4π (1425) which leads to

W(u2 ) = 5.096

et

which, by the definition of the well function leads to u2 = 0.003437 Therefore, on the basis of Equation 16.2, 0.003437 = 9.613 * 10−8 R2 which leads to R = 189 m Therefore, a wellfield consisting of 5 wells arranged as shown in Figure 16.1, with each well pumping 16.2 L/s and spaced 189 m apart will yield the required amount of water and produce drawdowns that will not exceed the specified maximum of 3 m.

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Design of Groundwater Systems

16.3

Wellhead Protection

Public water-supply wells must generally be protected from contaminants introduced into the aquifer from areas surrounding the wells. The most effective way of protecting wells from contamination is to control land uses in the surrounding areas by zoning regulations. In the United States, all states are required by law to develop comprehensive wellhead-protection (WHP) plans, and such plans must contain the following elements:

ww

1. Specify the roles and duties of state agencies, local government entities, and public water suppliers in the development and implementation of WHP programs. 2. Delineate a wellhead protection area (WHPA) for each public water-supply well based on reasonably available hydrogeologic information on groundwater flow, recharge and discharge, and other information necessary to adequately determine the WHPA. 3. Identify sources of contaminants within each WHPA, including all potential anthropogenic sources that may have an adverse effect on health. 4. Develop management approaches that include, as appropriate, technical assistance, financial assistance, implementation of control measures, education, training, and demonstration projects that are used to protect the water supply within the WHPA from such contaminants. 5. Develop contingency plans indicating the location and provision of alternate drinking water supplies for each public water-supply system in the event of well contamination. 6. Site new wells properly to maximize yield and minimize potential contamination. 7. Ensure public participation by establishing procedures encouraging the public to participate in developing the WHP program elements.

w .E asy En g

There is no universal approach to wellhead protection, given the variety of political, social, economic, and technical constraints that are encountered in practice. A WHP plan should generally contain the seven key elements listed above, be tailored to local conditions, and afford a defined level of protection to the public water supply. 16.3.1

ine eri n

Delineation of Wellhead Protection Areas

The most important and challenging component of any WHP plan is the delineation of the wellhead protection area. A wellhead protection area is defined as the surface and subsurface area surrounding a well or wellfield that supplies a public water system through which contaminants are likely to pass and eventually reach the water well or wellfield. A pumping well produces drawdowns in the aquifer surrounding the well, and the portion of the aquifer within which groundwater moves toward the pumping well is called the zone of contribution (ZOC) of the well. Only pollutants introduced within the ZOC can contribute to contamination of the pumped water. Although engineers must be concerned with all pollutant sources within the ZOC, the required level of protection necessarily varies with the distance from the well. In the immediate vicinity of the well, within tens of meters, protection is most stringent and all pollutant sources are typically prohibited; it is not unusual for this area to be fenced and have controlled access. This area is sometimes called the remedial action zone. Beyond this area, land uses that generate a limited amount of groundwater contamination, or potential for groundwater contamination, are typically permitted; however, sufficient distance to the well is provided such that the pollutant concentrations are attenuated to acceptably low levels by the time they reach the well. This area, where contaminant sources are controlled by regulatory permits, is commonly called the attenuation zone. Beyond the attenuation zone, but still within the ZOC, contaminant sources are unlikely to have any significant impact on the pumped water and regulatory burdens are usually reduced. This outer zone is commonly called the wellfield management zone. The primary factors to be considered in delineating WHPAs are the types and amounts of contaminants that could possibly be introduced into the ZOC and the attenuation characteristics of the contaminants in the subsurface environment. The types of contaminants that are of most concern are organic chemicals and viruses. Organic chemicals are used as industrial solvents and pesticides, and they are the major component of gasoline, which is usually stored in underground storage tanks at gas stations.

g .n

et

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Wellhead Protection

775

TABLE 16.1: WHPA Delineation Criteria

ww

Criteria

Assessment

Distance

Does not directly incorporate contaminant fate and transport processes. Commonly an arbitrary policy decision.

0.3–3 km (0.2–2 mi)

Drawdown

Does not directly incorporate contaminant fate and transport processes. Defines areas where the influence of pumping is the same.

3–30 cm (1–10 in.)

Time of travel

Incorporates advection and decay processes, neglects dispersion. Widely used criterion.

5–50 y

Flow boundaries

Highest level of protection, boundary of WHPA is taken as the ZOC boundary. Not practical in most cases.

Physical and hydrologic

Assimilative capacity

Incorporates all significant contaminant fate and transport processes. A rational approach to wellhead protection. Requires much technical expertise and is rarely used.

Requires drinking water standards to be met at well

w .E asy En g

Typical thresholds

Source: USEPA (1987).

Viruses in groundwater originate primarily from septic-tank effluent, onsite domestic wastewater treatment systems, and broken sewer pipes. The attenuation of various contaminants in groundwater occurs primarily via the processes of dispersion, decay, sorption, and filtration. Criteria commonly used for delineating WHPAs are: (1) distance, (2) drawdown, (3) time of travel, (4) flow boundaries, and (5) assimilative capacity. An assessment of these criteria, and typical threshold values, are given in Table 16.1. The specific criteria selected to delineate a WHPA depend on a variety of considerations, including overall protection goals, technical considerations, and policy considerations. 16.3.2

Time-of-Travel Approach

ine eri n

g .n

The most commonly used approach is the time of travel (TOT) approach, which assigns to each location within the ZOC the shortest travel time from that location to any wellhead within the wellfield. Travel times are commonly determined by using a groundwater flow model with a particle-tracking algorithm (e.g., McMahon et al., 2008; Tosco et al., 2008; Visser et al., 2009; Chin et al., 2010). In some cases, the travel-time determinations take into account uncertainties in the hydraulic properties of the aquifer, in which case a stochastic wellhead protection zone or probability wellhead protection zone is determined (Kunstmann and Kastens, 2006; Esling et al., 2008; Guha, 2008). To measure the level of protection of pumped water from a particular contaminant, the time of travel from the contaminant source to a production well can be contrasted with the time for the contaminant to undergo an acceptable amount of attenuation. The time of travel is sometimes associated with the time that is available to clean up a spill within the WHPA. In cases where decay or biotransformation is the dominant mode of attenuation, reductions in contaminant concentration are commonly described by the first-order process dc = −λc dt

et

(16.3)

where c is the contaminant concentration in groundwater, t is the time since release of the contaminant, and λ is the decay parameter. Equation 16.3 can be solved to yield c = e−λt c0

(16.4)

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Chapter 16

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where c0 is the initial concentration at t = 0. Equation 16.4 gives the attenuation of contaminant concentration as a function of travel time. Therefore, if a pollutant source is expected to contaminate the groundwater at a concentration c0 and the allowable concentration at the well contributed by this source is c, the minimum allowable time of travel to the well can be calculated using Equation 16.4. A useful parameter to characterize the time scale of decay is the half-life, T50 , which is the time for the concentration to decay to one-half of its original value. Equation 16.4 gives T50 =

0.693 λ

(16.5)

Time-of-travel contours surrounding single wells can be derived using the continuity relation Qw t = π r2 bn

ww

(16.6)

where Qw is the pumping rate, t is time, r is the distance traveled in time t, b is the saturated thickness, and n is the porosity. Equation 16.6 can be put in the form

w .E asy En g

r=



Qw t π bn

(16.7)

which gives the radial distance from the well that has a travel time t. Equation 16.7 assumes that the aquifer hydraulic properties are radially symmetric around the pumping well, a condition that is usually difficult to determine and subject to uncertainty. In the case of nonuniform aquifers and more complex multiwell scenarios, numerical groundwater models and probabilistic approaches can be used to calculate travel times. Sources of uncertainty in delineating capture areas for specified travel times usually include the spatial distribution of hydraulic conductivity, aquifer thickness, porosity, and recharge. In a study of the capture areas surrounding a 9.3 L/s (150 gpm) water-supply well in the Chicot aquifer in southwestern Louisiana, the capture area was found to be significantly sensitive to the hydraulic conductivity and aquifer thickness, and least sensitive to the porosity field (Rahman et al., 2008). In karstic aquifers, wellhead protection can be particularly challenging since the contamination pathways in these cases are sometimes difficult to identify. In such cases, the use of natural age-dating tracers, such as SF6 , 3 H, and 3 He, and/or installation of monitoring wells to detect anthropogenic contaminants and possibly identify their sources might provide useful information in developing a wellhead protection plan (e.g., Katz et al., 2009).

ine eri n

EXAMPLE 16.2

g .n

et

A WHPA is to be delineated around a municipal water-supply well that pumps 350 L/s from an aquifer with a saturated thickness of 25 m and a porosity of 0.2. The contaminant of concern is viruses from residential septic tanks. Residential development is to be permitted on 0.2-ha lots, and the viral concentration under each lot resulting from septic-tank effluent is expected to be 50/L. A risk analysis indicates that the maximum-allowable viral concentration in the pumped water is 0.01/L, and the decay constant, λ, for viruses can be taken as 0.5 d−1 (Yates, 1987). If there is to be 1 km2 of residential development surrounding the water-supply well, estimate the boundary of the WHPA within which no residential development should be allowed. Solution The allowable viral concentration in the pumped water is 0.01/L. Since the lots are uniformly distributed around the water-supply well, when each lot contributes a concentration of 0.01/L to the pumped water, the average concentration in the pumped water is 0.01/L. Since c0 = 50/L and λ = 0.5 d−1 , Equation 16.4 requires that c = e−λt c0 0.01 = e−0.5t 50

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Design and Construction of Water-Supply Wells

777

which gives t = 17.0 d. The radial distance, r, corresponding to a travel time of 17.0 d (= 1.47 * 106 seconds) and a pumping rate of 350 L/s = 0.35 m3 /s is given by Equation 16.7 as

r=



Qw t = π bn



(0.35)(1.47 * 106 ) = 181 m π (25)(0.2)

Therefore, prohibiting residential development within a radius of 181 m of the well will provide a satisfactory level of protection from viral contamination. Groundwater monitoring is recommended to verify assumptions regarding viral concentrations in the groundwater under septic tanks.

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The delineation of WHPAs based on travel time is appropriate when decay is the dominant attenuation process, contaminant dispersion is negligible, and the aquifer and pollutant source parameters are known with a reasonable degree of certainty. In cases where these conditions are not met, more comprehensive WHP models are recommended. 16.4

Design and Construction of Water-Supply Wells

Water-supply wells can be broadly classified as either public supply wells or domestic wells. Public supply wells deliver water at large flow rates to public water-distribution systems, while domestic wells deliver water at relatively low flow rates to individual homes. The focus of this section is on public supply wells; however, most of the concepts and methods apply equally well to domestic wells.

w .E asy En g 16.4.1

Types of Wells

Wells used for water supply are typically classified according to their method of construction and include: driven wells, drilled wells, and radial wells. These types of wells are described briefly as follows:

ine eri n

Driven well. A driven well consists of a pointed screen, called a drive point or well point, and lengths of pipe attached to the top of the drive point. The drive point is a perforated pipe covered with woven wire mesh or a tubular brass jacket. A pointed steel tip at the base of the drive point breaks through pebbles and thin layers of hard material and opens a passageway for the point. The diameter of a driven well is typically in the range of 32–100 mm (1 1/4–4 in.), and a driven well can typically be installed to a maximum depth of 9–12 m (30–40 ft) (AWWA, 2003b). For municipal water supplies, a driven well is used where thin deposits of sand and gravel are found at shallow depths.

g .n

et

Drilled well. Major water-supply wells are typically constructed using rotary drilling methods. A drilled well is constructed using a drill rig to excavate or drill a hole, into which a casing is forced or placed to prevent it from collapsing. When a water-bearing formation of sufficient capacity is reached, a screen is set in place that allows water to flow into the casing and holds back fine materials in the formation. When the drilled well passes through rock, a screen is usually not used, unless the formation is fractured. A drilled well is commonly used for municipal water supply, with pipe diameters typically in the range of 50–1210 mm (2–48 in.). Radial well. A radial well is a combination of a large-diameter central well and a series of horizontally driven wells projecting out of the vertical walls of the central well. The main well, or central caisson, serves as a collector for the water produced from the individual horizontal wells, called laterals. The laterals are installed in coarse formations, often in more than one layer or tier. The radial well, also called a horizontal collector well, is widely used because it can produce very large quantities of water. Horizontal wells are commonly used in bank filtration, where surface water is extracted via recharge through river bed sediments into horizontal wells located underneath or next to a stream. Horizontal wells can be particularly effective when the saturated

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Chapter 16

Design of Groundwater Systems

thickness of the aquifer is small but the hydraulic conductivity of the aquifer is high. Prediction of drawdowns resulting from pumping horizontal wells typically requires the use of numerical groundwater flow models, although some analytic approximations are available (e.g., Haitjema et al., 2010). Drilled wells are the most commonly used type of water-supply wells, and guidelines for their design and construction are given below. 16.4.2

ww

Design of Well Components

A generally accepted, all-inclusive standard for designing water-supply wells is not available; however, the American Water Works Association (AWWA, 1990) and the National Water Well Association (Lehr et al., 1988) describe a number of commonly used design standards. A schematic diagram of a typical water-supply well is illustrated in Figure 16.2, and the design of the components shown in Figure 16.2 are encompassed by the well-design process. The geology of the aquifer surrounding the well has a dominant influence on the well design, and the greater the certainty with which the geology surrounding the well is known, the better is the well design. Wells drilled in consolidated formations such as sandstone or limestone generally do not require any casing, screen, or gravel pack, and these wells are commonly referred to as open-hole wells. Wells drilled in unconsolidated formations such as sand or gravel will generally require a casing and screen and may require a gravel pack. For major water-supply wells with yields greater than 8 L/s (150 gpm), it is usually advisable to first drill a pilot hole or test hole at the proposed well location to collect the relevant geologic data; if this location proves to be acceptable, the expanded pilot hole is then used as the water-supply well. Information that can be derived from a pilot hole includes the depth of water-producing zones, confining beds, well-production capabilities, water levels, and groundwater quality. If the pilot hole indicates that the proposed site is inadequate, the site can be abandoned without the major cost of drilling a production well. For minor wells, well-drilling costs may be about the same as for a pilot hole. Consequently, smaller wells are drilled with a preliminary design developed on the basis of readily available information, and necessary adjustments and refinements in design are made during drilling and construction, as appropriate to maximize well yield (AWWA, 2001). Several criteria for determining the optimal withdrawal rate from a well have been suggested. One of the most popular rules of thumb is that wells should be pumped at a rate

w .E asy En g

FIGURE 16.2: Typical water-supply well

ine eri n

Pumped water Sanitary seal

Borehole Grout

P

Static water table

g .n

et

Submersible pump

Pump intake Casing

Gravel pack Screen/intake

Sand trap

Aquifer

Base of aquifer

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that maintains a maximum allowable drawdown. A variety of criteria are used to establish the maximum allowable drawdown, including consideration of existing pump-intake elevations, allowable stage fluctuations in connected surface-water bodies, and the sometimes-used criteria that the maximum allowable drawdown should not exceed 50–60% of an unconfined aquifer’s saturated thickness. Once the design pumping rate has been specified, with due consideration to the allowable drawdown, the structural components of the water-supply well are designed. These structural components include the casing, screen, pump, and gravel pack. 16.4.2.1

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Casing

Well casing refers to the tubular components used in the construction of water wells. There are typically two functional types of casing. Surface casing is used to line the borehole from the ground surface to the first impermeable strata, or a minimum of 6 m (20 ft) according to many codes (Rafferty, 2001). Pump-housing casing contains the pump that is used to extract water from the well. The pump-housing casing is connected to the top of a screened interval in unconsolidated formations or to the top of the producing interval in consolidated formations. The pump-housing casing is commonly referred to as the well “casing.” In production wells, the diameter of the (pump-housing) casing should be at least 5 cm (2 in.) larger than the nominal diameter of the pump bowls and the flow velocity in the casing should be less than 1 m/s (3 ft/s) to prevent excessive friction losses (Lehr et al., 1988; Todd and Mays, 2005). Recommended casing diameters based on pump sizes used in the water-well industry are given in Table 16.2 (Johnson Division, 1966; Lehr et al., 1988); however, in final design the manufacturer of the selected pump system should be consulted for advice. It is generally desirable to have a casing diameter that is two nominal pipe sizes larger than the pump to be installed (Rafferty, 2001). If the casing diameter based on pump size is less than the diameter required for a casing flow velocity of 1 m/s (3 ft/s), then a casing diameter based on a 1 m/s (3 ft/s) flow velocity should be selected. Casing materials commonly used in practice include alloyed or unalloyed steel, fiberglass, and polyvinyl chloride (PVC). Fiberglass and PVC casings are used primarily in shallow, small-diameter wells where corrosion might be an issue, while steel casing is used in the majority of water-supply wells. Well casings should generally extend above the ground surface to prevent surface water from running down the hole and contaminating the well water. Joints for permanent steel casings should have threaded couplings or be welded to ensure watertightness from the bottom of the casing to above grade. Thermoplastic casing typically is bell-and-socket construction, joined by cementing, or else is joined with spline-lock fittings.

w .E asy En g 16.4.2.2

Screen intake

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Wells completed in unconsolidated formations (such as sands and gravels) generally have screen intakes which keep the pump from being clogged with aquifer material. Wells can be screened continuously along the borehole or at specific depth intervals. The latter is

et

TABLE 16.2: Typical Casing Diameters Based on Pump Size

Pumping rate (L/s) (gpm) 820 250 820 200 650 160 520 120 390 100 330 80 260 60 200 40 130 20 70 0.25 mm are good candidates

et

TABLE 16.4: Minimum Screen Diameters

Pumping rate (L/s) (gpm) 0.25 mm d10 > 0.25 mm d10 > 0.25 mm

Screen slot size d40 –d60 d40 –d70 d50 –d70

for being developed naturally, meaning that initial pumping of the well will remove the fines from the surrounding aquifer material, leaving the screen surrounded by a very permeable coarse-grained annular region. In cases where the well can be developed naturally, the criteria shown in Table 16.5 should be followed for selecting the screen slot size (Lehr et al., 1988; Todd and Mays, 2005). Within the range of slot sizes indicated in Table 16.5, sizes at the low end of the range are selected whenever the well is not overlaid with a firm layer of soil, such as clay or shale, and slot sizes at the high end of the range are selected whenever firm soil layers are present (Lehr et al., 1988). In cases where d10 < 0.25 mm, an artificial gravel pack should be used. By designing a screen slot size based on criteria in Table 16.5, the finer material in the aquifer surrounding the screen is removed during well development, which includes any mechanism that removes silt, fine sand, or other such material from the zone immediately surrounding a well intake. High-rate pumping and surge plungers are common methods used in well development. After well development, the screen intake is surrounded by a coarse layer with a hydraulic conductivity significantly higher than that of the undisturbed aquifer matrix. The coarse layer surrounding the screen is typically about 50-cm (20-in.) thick (Boonstra, 1998). If a screen is not available in the required slot size, the next smaller standard size should be selected. Slot openings in commercially available well screens are typically in the range of 1–6 mm, and slot sizes are commonly specified as multipliers of 0.025 mm (0.001 in.). Hence, a 40-slot screen has 1.0-mm (0.040-in.) slot widths. In cases where the grain-size distribution of the aquifer matrix varies over the depth of the aquifer, the specification of screen sections with different slot sizes is common.

w .E asy En g

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Screen length. To maximize the pumping rate for a given drawdown, it is recommended that wells in confined aquifers be screened through the entire thickness of the aquifer (USBR, 1995), or at least through 70%–90% of the aquifer thickness (Wurbs and James, 2002; Karamouz et al., 2011). If the aquifer is homogeneous, the well screen is centered in the aquifer; otherwise, the screen is set in the coarser part of the aquifer. To maximize the pumping rate for a given drawdown in unconfined aquifers, wells should be screened in the lower one-third to one-half of the saturated thickness, with the upper two-thirds to one-half of the saturated thickness reserved for drawdown. The screen-length specifications for confined and unconfined aquifers are not economically feasible in very deep and thick aquifers, and in these cases the usual practice is to penetrate a sufficient thickness of the aquifer to achieve the required discharge capacity at an acceptable drawdown. In cases where the screen does not fully penetrate the aquifer, drawdowns will be greater than estimated by assuming full penetration. This represents a decrease in well yield. A minimum of 1–2 m (3–6 ft) of screened intake is usually required for domestic wells (Fetter, 2001). Longer screen lengths are preferable and generally result in higher pumping rates per unit drawdown.

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Screen depth. The depth of the screen in unconsolidated formations is designed to protect the intake from surface contamination, and the minimum screen depth will vary with soil formations and surrounding conditions. In unconsolidated formations, screen depths of 8–9 m (25–30 ft) or more provide a reasonable level of protection from surface contamination (AWWA, 2003b). The top of the screen should generally be within the saturated zone of the aquifer at all times. Screen materials. Screen sections are typically joined together by welding or couplings to give almost any length of screen. The least expensive and most commonly available screens are made of low-carbon steel. Screens made of nonferrous metals and alloys, plastics, and fiberglass are used in areas of aggressive corrosion and encrustation. To reduce the possibility

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of corrosion, the well screen and its fittings should be made of the same material. Slotted, louvered, and bridge-slotted screens and continuous wire-wrap screens are the most common types of screens. Slotted screens provide poor open area and are not well suited for proper well development and maintenance. Wire-wrap screens or pipe-based wire-wrap screens provide a higher degree of open area and give the best performance. Water-supply wells typically have stainless-steel screens with continuous slots that are made by winding wire around vertical rods. 16.4.2.3

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Gravel pack

In some cases, the aquifer material is so fine that selection of screen openings on the basis of its size and uniformity coefficient would yield openings so small that the entrance velocity would be unacceptably high. Under these circumstances, gravel packs or (equivalently) filter packs are placed in the annular region between the screen and the perimeter of the borehole. In this context, the term gravel pack refers to any filtering media that is placed around the well screen and is not limited to a coarse gravel material as the name implies. Fine to medium sand is commonly used as gravel-pack material. Gravel packs in unconsolidated formations are usually justified when Uc < 3 and d10 < 0.25 mm (0.010 in). The gravel pack does not exclude fine silt and clay particles, and where these occur in significant amounts in a formation it is usually best to use blank casing sections (i.e., no screen or gravel pack).

w .E asy En g

Gravel-pack material. The specifications of the gravel pack are determined primarily by the aquifer matrix, and criteria for selecting the gravel pack and corresponding screen slot size are given in Table 16.6. Typically, a gravel pack should have a uniformity coefficient between 1 and 2.5, with a median grain size between six and nine times the median grain TABLE 16.6: Criteria for Gravel-Pack Selection

Uniformity coefficient of aquifer matrix, Uc 5

Multiply the 30% passing size of the aquifer by 6 and 9 and locate the points on the grain-size distribution graph on the same horizontal line. Through these points draw two parallel lines representing materials with Uc … 2.5. Select gravel pack material that falls between the two lines.

5%–10% passing size of the gravel pack

Source: USBR (1995).

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size of the aquifer material. The corresponding slot size of the screen opening should be between the 5- and 10-percentile grain size of the gravel pack, with the 10-percentile size usually being preferable to minimize entrance losses. The maximum grain size of a gravel pack should generally be around 10 mm, but less than 9 mm if placed through a nominal 100 mm (4 in.) tremie pipe (USBR, 1995). Gravel packs usually consist of quartz-grained material that is sieved and washed to remove the finer material such as silt and clay. The grains of the gravel-pack material must be well rounded, to maximize the porosity; angular grains will tend to lock into adjacent grains, reducing the openings through which water and fine aquifer material can move. Placement of the gravel pack is generally accomplished by either pouring from the surface (in shallow wells) or by placement through a tremie in wells deeper than 300 m (1000 ft).

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Thickness of gravel pack. The thickness of the gravel pack is typically in the range of 8–23 cm (3–9 in.) (Boonstra, 1998; Todd and Mays, 2005), with ideal thicknesses in the range of 10–15 cm (4–6 in.) (Lehr et al., 1988). It is difficult to develop a well through a gravel pack much thicker than 20 cm (8 in.), since the seepage velocities induced during the development procedure must be able to penetrate the gravel pack to repair the damage done by drilling, break down any residual drilling fluid on the borehole wall, and remove finer particles near the borehole (USBR, 1995). The thickness of the gravel pack should not be less than 8 cm (3 in.) to ensure that a continuous pack will surround the entire screen (Boonstra, 1998).

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Length of gravel pack. The gravel pack should extend at least 1 m (3 ft) above the screen. For domestic wells (i.e., those that serve individual households), it is recommended that gravel packs not extend beyond this minimum distance above the screen, since this causes the well capture zone to increase with associated increased risk of contamination of the pumped water (Horn and Harter, 2009).

Other considerations. Gravel packs are usually considered essential in sandy aquifers. In unconsolidated formations, whether a gravel pack is used or not, a properly developed well is surrounded by a coarse annular region that has a hydraulic conductivity much higher than the surrounding aquifer. This condition serves to increase the effective radius of the well, which is roughly equal to the radial extent of the coarse annular region surrounding the well screen, and drawdown calculations must be based on the effective radius. 16.4.2.4

Pump

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g .n

Pumps are installed in water wells to lift the water in the well to the ground surface and deliver it to the point of use. The types of pumps used to remove water from wells are: vertical turbine pumps, submersible pumps, jet pumps, pneumatic pumps, airlift pumps, positive displacement pumps, and suction pumps. These types of pumps are described in Table 16.7. The vertical turbine pump is often the most suitable pump for groundwater applications, especially for moderate to large discharge rates. Submersible pumps are also used frequently and have the advantage of being able to lift water from deep wells where long shafts in crooked casings might prohibit the use of vertical turbine pumps. Both vertical turbine and submersible pumps are categorized as centrifugal pumps, since the rotating impellers induce water flow through centrifugal force. Typical vertical turbine and submersible pumps are illustrated in Figure 16.4. The sizes and performance characteristics of these pumps must meet the design/operational requirements of head (pressure) and flow rate to be delivered by the pump. The performance characteristics (head versus discharge) provided by manufacturers are generally given per stage in the pump. For multiple-stage pumps, the performance characteristics are derived by adding all the heads (added at each stage) for a given discharge. Well-design specifications typically caution against installing the intake of a pump in the well screen, since such designs are assumed to increase screen entrance velocities, thereby decreasing well efficiencies (Driscoll, 1986; Roscoe Moss Company, 1990; Bailey and Wicklein, 2007). However, there is some evidence that placing the pump intake within the

et

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TABLE 16.7: Water-Well Pumps

Pump type

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Description

Vertical turbine pumps

Vertical turbine pumps have the motor located on the discharge side at the ground surface and require a drive shaft extending down the well to the pump located below the water surface.

Submersible pumps

The term submersible is applied to turbine pumps when the motor is close-coupled beneath the bowl assembly of the pump and both are installed under water. This type of construction eliminates the surface motor, long drive shaft, shaft bearings, and lubrication system of the conventional turbine pump; however, the electrical connections are submerged. Submersible pumps are especially useful for high-head low-capacity applications such as domestic water supply.

Jet pumps

The jet pump combines two principles of pumping, that of the injector (jet) and that of the centrifugal pump. Jet pumps are inefficient when compared to ordinary centrifugal pumps, but are used in domestic wells because of other favorable characteristics.

w .E asy En g Pneumatic pumps

Pneumatic pumps operate using air pressure and are generally used under special conditions such as contaminant cleanup and monitoring. They are used for purging, sampling, product-only pumping, and low- to moderate-flow pumping.

Airlift pumps

Airlift pumps release compressed air into the discharge pipe, and the reduced specific weight of the water column lifts the water to the surface. Airlift pumps are inefficient and expensive in comparison to other pumping methods and are rarely used.

Positive displacement pumps

Positive displacement pumps displace water through a pumping mechanism. There are several types of positive displacement pumps including: piston pumps, rotor peristaltic pumps, and Lemoineau-type pumps.

Suction pumps

Suction pumps are typically centrifugal pumps located above the ground surface, and their use is limited by the suction lift that can be developed. In practice, the suction lift is usually limited to where the water table is about 7 m (23 ft) below the pump intake.

ine eri n

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well screen actually increases well efficiencies (e.g., Von Hofe and Helweg, 1998; Korom et al., 2003). As mentioned in the context of screen design, the pump intake diameter should be approximately 60% of the screen diameter for efficient operation. If the pump is located within the well casing, a minimum clearance of 25 mm (1 in.) around the pump bowls is recommended (USBR, 1995). The pump intake should remain below the water level in the well during the lifetime of the well. 16.4.2.5

Other considerations

Besides the main structural components of water-supply wells described in the previous sections, the following components of wells must also be specified. Grouting. Well grouting consists of placing a cement slurry (called grout) between the well casing and the borehole; grouting is generally necessary to prevent polluted surface water or low-quality water in overlying aquifers from seeping along the outside of the casing and

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FIGURE 16.4: Water-supply pumps Courtesy Goulds Water Technology, a Xylem brand.

Motor Pump bowls

Intake Outflow

Motor

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Pump bowls

w .E asy En g Intake

(a) Vertical turbine pump

(b) Submersible pump

contaminating the well or aquifer. Grouting also serves to anchor the casing inside the borehole, protect the casing from corrosion, and prevent the caving in of aquifer material surrounding the well. Depending on the method used for grouting, the borehole is drilled 50–150 mm (2–6 in.) larger than the outside diameter of the casing. The cement slurry used in grouting is usually a mix of about 50 L of water per 100 kg of cement, with additives used to provide special properties such as preventing the grout from cracking and accelerating or retarding the time of setting. Additives that are commonly used include bentonite clay, pozzolans, and perlite (Bouwer, 1978). Grout should be placed by pumping it through a pipe or hose (called a tremie) to the bottom of the casing and above the gravel pack. The tremie pipe is placed between the casing and the surrounding soil or rock. Pumping the grout from the bottom up ensures that no gaps or voids are left between the casing and the soil or rock. To prevent the penetration of the cement slurry into the gravel pack, a layer of clean sand (particle size 0.3–0.6 mm) is placed on top of the gravel pack. This layer of sand is commonly referred to as a sand bridge, and cement slurries generally do not penetrate the sand layer by more than a few centimeters. A minimum grout thickness of 50 mm (2 in.) is recommended and might even be required by some regulatory agencies.

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Sand trap. A sand trap is a section of well casing installed below the screened intake section. Its function is to store sand and silt entering the well during pumping. The length of the sand trap is usually 2–6 m (6–20 ft), and the diameter is typically the same as the screen. Sanitary seal. At the surface, all wells should have a sanitary seal, which prevents contamination from entering the well casing. This seal is in addition to the grout that is placed between the borehole wall and the well casing to prevent surface contamination of water entering the well intake. The sanitary seal for a small well having a submersible pump is illustrated in Figure 16.5 and consists of a metal plate with a rubberized gasket around its perimeter that fits snugly into the top of the well casing. The sanitary seal has openings into the well for the discharge pipe and the pump power cable, and an air vent to let air into the casing as the water level drops.

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FIGURE 16.5: Sanitary seal in water-supply well Source: AWWA (2003a).

Air vent Discharge line

Pump power cable Sanitary seal

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w .E asy En g EXAMPLE 16.3

Well casing

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A water-supply well is to be installed in an unconfined aquifer which has a saturated thickness of 25 m and a hydraulic conductivity of 30 m/d. A grain-size analysis of the aquifer indicates a uniformity coefficient of 2.7 and a 50-percentile grain size of 1 mm. If the well is to be pumped at 20 L/s, design the well screen and the gravel pack.

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Solution According to Table 16.6, since the uniformity coefficient (Uc ) of the aquifer matrix is 2.7, the gravel pack should have a Uc between 1 and 2.5 and a 50% size not greater than nine times the 50-percentile size (d50 ) of the aquifer. Hence, d50 for the gravel pack should be less than 9 * 1 mm = 9 mm. A check with a sand and gravel company will generally yield a commercial gravel-pack material that satisfies these criteria. For example, you may find “5 mm * 9 mm” gravel, which contains only grain sizes between 5 mm and 9 mm. Assuming that the particle sizes are uniformly distributed between 5 mm and 9 mm, d10 L 5.4 mm, d60 L 7.4 mm, and Uc of the gravel is approximately 7.4/5.4 = 1.4. Hence, the 5 mm * 9 mm gravel is acceptable. Since gravel packs should be between 8 cm and 23 cm thick, a reasonable thickness would be 10 cm. According to Table 16.6, the screen slot size should be between the 5% and 10% passing size of the gravel pack. In this case, the 10% passing size is estimated to be 5.4 mm, and it is reasonable to select a slot size of 5 mm. A review of manufacturers’ literature on well screens will give the commercially available slot widths and associated fractions, P, of open area. Typically, P = 0.10. Since the screen length, Ls , in unconfined aquifers should be between 0.3 and 0.5 times the saturated thickness, select a screen length of 0.5 times the saturated thickness to maximize the pumping rate per unit drawdown, in which case Ls = 0.5 * 25 m = 12.5 m. For a pumping rate of 20 L/s, Table 16.2 indicates an optimum casing diameter of 250 mm, and Table 16.4 indicates a minimum screen diameter of 150 mm. Taking the screen and casing diameter as 250 mm, and assuming that 50% of the open area is clogged by the aquifer material, the screen entrance velocity, vs , is estimated by Equation 16.8 as vs =

et

Qw cπ ds Ls P

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1.2 = 2.4 m/min (0.5)π (0.25)(12.5)(0.1)

According to Table 16.3, this screen entrance velocity is too high for an aquifer with a hydraulic conductivity of 30 m/d, where the desirable screen size would produce an entrance velocity on the order of 1.05 m/min. Taking vs = 1.05 m/min, the corresponding screen diameter, ds , is given by ds =

Qw 1.2 = = 0.583 m L 600 mm cπ vs Ls P (0.5)π (1.05)(12.5)(0.1)

In summary, a screen length of 12.5 m, diameter of 600 mm, and slot size of 5 mm is acceptable. This screen should be surrounded by a 5 mm * 9 mm gravel pack with a thickness of about 10 cm; the diameter of the well casing should be 600 mm so that it can be easily joined to the screen.

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16.4.3

Performance Assessment

The pumping rate corresponding to the maximum allowable drawdown can be estimated from the specific capacity of a well, which is defined as the well pumping rate per unit drawdown, where the drawdown is equal to the drawdown in the aquifer (at the boundary of the borehole) plus the well (head) loss resulting from flow through the gravel pack and well screen. The specific capacity is sometimes considered to be the single most informative factor in well performance (AWWA, 2003b). The specific capacity in the case of a fully penetrating well in a homogeneous isotropic confined aquifer with negligible head losses in the developed region surrounding the intake can be derived directly from the Thiem steady-state equation (Equation 15.26) as Qw 2π T Specific capacity = (16.10) = sw ln (R/rw )

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or from the Theis transient equation (Equation 15.123) as Specific capacity =

Qw 4π T = 2 S/4Tt) sw W(rw

(16.11)

where Qw is the pumping rate, sw is the drawdown at the well, T is the transmissivity of the aquifer, R is the radius of influence, rw is the effective radius of the well, W(u) is the well function, S is the storage coefficient, and t is the time since the beginning of pumping. Although both Equations 16.10 and 16.11 are used to estimate the specific capacity of a well, Equation 16.11 is more realistic and indicates that the specific capacity gradually decreases with time. Indeed, a reduction of up to 40% in the specific capacity has been observed in 1 year in wells deriving water entirely from storage (Gupta, 2001). Classifications of well productivity in terms of specific capacity as suggested by S¸en (1995) are given in Table 16.8. The specific capacity of water-supply wells is typically monitored over time to detect whether the well is becoming clogged. The specific capacity should be determined at least annually, and the analysis should occur only after a well is allowed to fully recover. Then, the well

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TABLE 16.8: Classification of Well Productivity

Specific capacity, Qw /sw [(L/s)/m]

[gpm/ft]

Productivity

5

20

Negligible Very low Low Moderate High

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should be pumped for an hour to determine the specific capacity. This result is then compared to the original data and plotted to show trends (AWWA, 2003b). An aquifer’s natural microbiology usually proliferates when a well is installed and operating. The available supply of food or dissolved minerals greatly increases around the borehole of the well and screen interval because of increased flow velocity and turbulence. This allows more rapid growth of biological deposits or formation of mineral encrustations that can plug screen openings and pore spaces near the screen and borehole. Well operation also pulls fines in from surrounding formations, which plug the pore spaces and eventually reduce the open area for water to enter the well. Rehabilitation of clogged wells should begin before the specific capacity falls below 85%–90% of the original value (Thomas, 2002; Bailey and Wicklein, 2007). The maximum rate at which water can be extracted from a well is defined as the well yield, which is equal to the specific capacity multiplied by the allowable drawdown. The well yield should not be confused with the safe aquifer yield, which is the maximum rate at which water can be withdrawn from an aquifer without depleting the water supply. Water managers are keenly aware that increases in groundwater withdrawals must generally be balanced by an increase in aquifer recharge, a decrease in aquifer discharge, and/or a loss of groundwater storage.

w .E asy En g EXAMPLE 16.4

A water-supply well is to be constructed in a surficial aquifer where the saturated thickness is 25 m, and the maximum allowable drawdown in the surficial aquifer is 2 m. The well is to have an effective radius of 400 mm, and the hydraulic conductivity and specific yield of the aquifer are estimated as 20 m/d and 0.15, respectively. Estimate the specific capacity and the well yield after 3 years of service. Classify the productivity of the well.

Solution From the given data: H = 25 m, K = 20 m/d, T = KH = (20)(25) = 500 m2 /d, rw = 400 mm = 0.4 m, Sy = 0.15, and t = 3 years = 3(365) d = 1095 d. Substituting these data into Equation 16.11 (using the small-drawdown assumption for surficial aquifers) gives Specific capacity =

4π T

2 S /4Tt) W(rw y

=

ine eri n

4π (500) = 355 m2 /d = 4.12 (L/s)/m W(0.42 * 0.15/4 * 500 * 1095)

Since the maximum allowable drawdown is 2 m, the well yield is given by Well yield = 4.12 (L/s)/m * 2 m = 8.24 L/s

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and therefore the maximum allowable pumping rate is 8.24 L/s. According to Table 16.8, a well with a specific capacity of 4.12 (L/s)/m has a moderate productivity.

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Head losses resulting from turbulent flow through the well screen and the coarsegrained material surrounding the well are collectively called well losses, and these well losses cause the water level inside the well to be less than the water level in the aquifer adjacent to the well. This is illustrated in Figure 16.6. Well losses are added to the formation losses to estimate the total drawdown. In some cases, there is an imperfect hydraulic connection between the well bore and the aquifer formation caused by residual well construction effects; this is called the wellbore skin. The added head loss associated with the wellbore skin is called a skin effect and is usually included in the well loss (e.g., Barrash et al., 2006), although sometimes the wellbore skin is explicitly modeled (e.g., Chen and Yeh, 2009). The well loss, hw , is typically estimated by the relation hw = αQnw

(16.12)

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FIGURE 16.6: Well losses and formation losses

Well centerline Aquifer material

Effective radius

Coarse material

Well screen Water level in aquifer

Formation loss, βQw

Total drawdown, βQw ⫹ αQwn

Well loss, αQwn

ww

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TABLE 16.9: Well-Loss Coefficients

Well-loss coefficient, α (s2 /m5 )

(s2 /ft5 )

Well condition

14,400

37.9

Properly designed and developed Mild deterioration or clogging Severe deterioration or clogging Difficult to restore well to original capacity

Source: Walton (1962).

ine eri n

where α is a constant called the well-loss coefficient, Qw is the pumping rate, and n is an exponent that typically varies in the range of 1.5–3 (Lennox, 1966), but is commonly assumed to be equal to 2. The assumption of n = 2 was originally proposed by Jacob (1947) and is still widely accepted. Values of α corresponding to various states of a well were suggested by Walton (1962) and are given in Table 16.9. These data indicate that for a properly designed and functional well, α should be less than 1800 s2 /m5 (4.7 s2 /ft5 ). Since formation losses are typically proportional to the well pumping rate, Qw , and well losses are typically proportional to Q2w , the drawdown in a well, sw , can typically be expressed in the form sw = βQw + αQ2w

g .n

et

(16.13)

where β is the formation-loss coefficient. Dividing both sides of Equation 16.13 by Qw yields sw = β + αQw Qw

(16.14)

which indicates a linear relationship between sw /Qw and Qw . The coefficients α and β can be determined in the field using a step-drawdown pumping test, where the well pumping rate is increased in a series of discrete steps, with drawdown measurements taken at fixed time intervals. Typically, a step-drawdown test includes five to eight pumping steps, each lasting from 1 to 2 h, with the highest pumping rate being 25%–50% higher than the design pumping rate, if possible. The resulting data yield a relationship between sw /Qw and Qw , which is fitted to the linear relationship given by Equation 16.14, and the slope and intercept of the best-fit line are taken as α and β, respectively. This method for estimating the well-loss and

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formation-loss coefficients, α and β, is commonly referred to as Jacob’s straight-line method, and is a standard method for estimating the hydraulic parameters of production wells (Jha et al., 2006). Approximate methods have also been suggested for estimating well losses from early-drawdown data (Singh, 2008a). The well efficiency, ew , is defined as the percentage of the total drawdown in the well caused by formation losses, and based on Equation 16.13 the well efficiency can be put in the form ew =

βQw * 100 sw

(16.15)

In using values of β derived in the step-drawdown test care must be taken to maintain each stepped pumping rate for a sufficient amount of time for the response to (quasi-) stabilize. Otherwise, transient values of β will be obtained that cannot be used to adequately characterize the well loss (Mathias and Todman, 2010).

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EXAMPLE 16.5 A step-drawdown test is conducted at a well where a pumping rate of 82.3 L/s yields a drawdown in the well of 1.54 m, and a pumping rate of 123.5 L/s yields a drawdown of 2.36 m. Assess the condition of the well, and estimate the well efficiency and specific capacity at a pumping rate of 190 L/s.

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Solution The flow rates used in the step-drawdown test are 82.3 L/s = 864 m3 /d and 123.5 L/s = 1296 m3 /d, and the given data can be put in the following tabular form:

According to Equation 16.14,

Qw (m3 /d)

sw (m)

sw /Qw (d/m2 )

864 1296

1.54 2.36

0.00178 0.00181

ine eri n

sw = β + αQw Qw and combining this equation with the step-drawdown data gives 0.00178 = β + 864α

0.00181 = β + 1296α Solving these equations simultaneously yields

α = 6.9 * 10−8 d2 /m5 = 515 s2 /m5

and

g .n

β = 0.00172 d/m2

et

Comparing α = 515 s2 /m5 with the guidelines in Table 16.9 indicates that the well is in good condition. At Qw = 190 L/s = 2000 m3 /d, the expected drawdown at the well, sw , is given by Equation 16.13 as sw = βQw + αQ2w = (0.00172)(2000) + (6.9 * 10−8 )(2000)2 = 3.72 m The drawdown associated with formation losses is βQw = 0.00172(2000) = 3.44 m, and hence the well efficiency, ew , is given by Equation 16.15 as ew =

βQw 3.44 * 100 = 92% * 100 = sw 3.72

At Qw = 190 L/s = 2000 m3 /d, the specific capacity of the well is given by Specific capacity = =

Qw 1 Qw = = 2 sw β + αQw βQw + αQw 1 = 538 m2 /d = 6.2 (L/s)/m 0.00172 + (6.9 * 10−8 )(2000)

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Design of Groundwater Systems Comparing this result with the classification system in Table 16.8 indicates that the well is highly productive at a pumping rate of 190 L/s. The results of this example should be viewed cautiously since they are based on only two stepped drawdowns. Normally, a minimum of five to eight stepped drawdowns would be used.

This analysis assumes a priori that the well losses are proportional to Q2w , and this assumption is justified by the fact that the well-loss coefficient used to assess the condition of the well in Table 16.9 is based on this assumption. In the more general case, sw = βQw + αQnw

ww

(16.16)

which can be put in the form   sw − β = log α + (n − 1) log Qw log Qw

(16.17)

which shows that the relation between log(sw /Qw − β) and log Qw is linear with a slope of n − 1 and an intercept of log α. To apply Equation 16.17 to step-drawdown data, various values of β are tried until the relation between log(sw /Qw − β) and log Qw becomes approximately linear, and values of n and α are obtained directly from the slope and intercept of the best-fit straight line. This approach, originally proposed by Rorabaugh (1953), is sometimes referred to as the Rorabaugh method.

w .E asy En g EXAMPLE 16.6

The drawdowns, sw (m), and corresponding pumping rates, Qw (m3 /d), at a well during a stepdrawdown test were plotted on a graph of log (sw /Qw − β) versus log Qw for various values of β. The plotted curve is most linear when β = 6.36 * 10−4 d/m2 , and the corresponding best-fit line through the data has a slope of 1.14 and an intercept of −6.72. Determine the formation and well-loss coefficients, and estimate the efficiency and productivity of the well when the pumping rate is 85.8 L/s

ine eri n

Solution From the given data, β = 6.36 * 10−4 d/m2 ; comparing the slope and intercept of the best-fit line with Equation 16.17 indicates that log α = −6.72, which gives

(n − 1) = 1.14

α = 1.91 * 10−7 d2 /m5 ,

n = 2.14

g .n

et

Hence, the formation and well-loss coefficients are β = 6.36 * 10−4 d/m2 and α = 1.91 * 10−7 d2 /m5 (= 1430 s2 /m5 ), respectively, and the drawdown, sw , is given by sw = 6.36 * 10−4 Qw + 1.91 * 10−7 Q2.14 w The well efficiency, ew , is equal to the percentage of total drawdown caused by formation losses and can be expressed in the form ew =

β βQw * 100 * 100 = βQw + αQnw β + αQn−1 w

When Qw = 85.8 L/s = 900 m3 /d, ew =

6.36 * 10−4 6.36 * 10−4 + (1.91 * 10−7 )(900)2.14−1

* 100 = 59%

Since the well-loss coefficient, α = 1430 s2 /m5 , indicates that the annular region surrounding the well is in good condition (see Table 16.9), the low well efficiency of 59% can be attributed to formation losses

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793

being of comparable magnitude to the well losses. At Qw = 85.8 L/s = 900 m3 /d, the productivity of the well is measured by the specific capacity as Specific capacity =

1 Qw 1 Qw = = = −4 + (1.91 * 10−7 )(900)1.14 sw βQw + αQnw 6.36 * 10 β + αQn−1 w

= 925 m2 /d = 10.7 (L/s)/m Comparing this result with the classification system in Table 16.8 indicates that the well is highly productive at a pumping rate of 85.8 L/s.

Step-drawdown tests are commonly performed after well construction to determine the well efficiency and the effectiveness of the well development. In addition, the results of step-drawdown tests can be used to estimate the transmissivity of the formation surrounding the well (e.g., Dufresne, 2011). This is done by using the specific drawdown based only on the formation loss, equal to 1/β in Equation 16.14, in Equation 16.11 to yield

ww

4π T 1 = 2 S/4Tt) β W(rw

w .E asy En g

(16.18)

where the value of rw is determined from the physical characteristics of the well, and t is typically taken as 1 d. Using Equation 16.18, values for the transmissivity, T, can then be estimated for various probable values of the storage coefficient, S. 16.4.4 Well Drilling

Numerous techniques and a wide variety of specialized equipment are used to drill water wells. These techniques can be broadly categorized as either percussion or rotary methods. A drill rig mounted on a truck, along with a large gasoline or diesel engine to drive the system, is used for each of these methods, and the appropriate drilling method is selected based on the local geology and the size and depth of the well to be installed.

ine eri n

Percussion methods. Percussion drilling uses pounding or jetting to drill a hole in the ground. Three common percussion methods are: cable-tool drilling, hammer drilling, and jetting. In cable-tool drilling, a heavy drill string, supporting a sharpened tool or bit, is repeatedly lifted and dropped into the hole by a cable. The mixture of cuttings and water is periodically removed from the hole using a sand pump or bailer. The cable-tool method is mostly used in rock to depths of 600 m (2000 ft). In the hammer method, compressed air is cycled to repeatedly pound the bit downward, a process much like that used in a jack-hammer. Cuttings are removed from the hole using compressed air. Jetting is similar to cable-tool drilling, with pressurized water on either side of a chisel-like bit, and water and cuttings flowing up the annulus between the drill string and the wall of the open hole.

g .n

et

Rotary methods. Rotary-drilling methods use a rotating bit to drill a hole in the ground. The rotating drill bit has cutting teeth generally made of steel. For very hard geologic material, the teeth may be tipped with titanium or diamond. Rotary methods differ primarily in the type of drilling fluid used and the direction of circulation. In conventional mudor air-rotary methods, fluid is pumped down the inside of the drill pipe down to the drill bit, and this fluid comes back up to the surface through the annulus between the pipe and the wall of the open hole. This fluid, sometimes called drilling mud, is recirculated during the drilling operation and, in addition to cooling the drill bit and forcing the tailings (geologic material) from the borehole, it also exerts pressure on the walls of the borehole and forms a clay lining on the wall of the well, preventing it from caving (in unconsolidated formations) during the drilling process. As the drill bit rotates and moves downward, creating a borehole, the drill crew adds additional lengths of 6-m (20-ft) pipe, and this string of pipes is increased in length until the desired well depth is reached. Drilling mud consists of a suspension of water, bentonite clay, and various organic additives.

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16.5

ww

Design of Aquifer Pumping Tests

Aquifer pumping tests or aquifer tests∗ are used to determine the hydraulic properties of aquifers. The methodology consists of matching field measurements of drawdowns caused by pumping wells with the corresponding theoretical drawdowns and determining the hydraulic properties of the aquifer that produce the best fit. The details of several analytic models for analyzing aquifer-test data in a variety of geologic formations have been presented in Sections 15.2 and 15.3. Aside from analytic techniques for processing the measured data, several operational issues must be addressed to properly conduct aquifer tests. Aquifer tests are usually conducted using a pumping well and at least one observation well. The key elements in designing an aquifer test at any site are: (1) number and location of observation wells, (2) design of observation wells, (3) approximate duration of the test, and (4) discharge rate from the pumping well. Aquifer tests are usually quite expensive, and the installation of a pumping well and surrounding observation wells is typically justified only in cases where exploitation of the aquifer by water-supply production wells at the site is being contemplated. In most cases, the pumping well is subsequently utilized as a production well. Conventional aquifer tests involve a single pumping well and several monitoring wells. In some instances, it might be possible to use several existing production wells and strategically placed monitoring wells to extract the aquifer properties (e.g., Harp and Vasselinov, 2011); however, this approach is not commonly used. In some cases, the results of aquifer tests can be correlated to surficial geophysical measurements, such as vertical electrical sounding, and the results used to estimate aquifer hydraulic properties at locations where only geophysical measurements are available (e.g., Soupios et al., 2007). Prior to conducting an aquifer test, basic data on the aquifer must be collected. These data must include, if possible, the depth, thickness, areal extent, and lithology of the aquifer, the locations of aquifer discontinuities caused by changes in lithology, the locations of surface-water bodies, preliminary estimates of the transmissivity and storage coefficient of the aquifer, and preliminary estimates of leakage coefficients if semiconfining layers are present. These data facilitate the design of the discharge rate from the pumping well as well as aid in the location of observation wells. Preliminary values of the transmissivity and storage coefficient of an aquifer can be estimated by conducting slug tests (see Section 16.6) on wells near the site, but such tests have no more than order-of-magnitude accuracy. It is advisable to use existing wells to conduct aquifer tests whenever possible; however, care should be taken that existing wells are properly constructed and developed and that these wells are screened in the same aquifer zone as the one being investigated.

w .E asy En g 16.5.1

Pumping Well

ine eri n

g .n

The design of the pumping well includes consideration of: (1) well construction, (2) well development procedure, (3) well access for water-level measurements, (4) a reliable power source, (5) type of pump, (6) discharge-control and measurement equipment, and (7) method of waste disposal.

et

Well construction. The diameter of the pumping well must be large enough to accommodate both the test pump and space for water-level measurement. Guidelines for casing diameter given in Table 16.2 are recommended. The well screen must have sufficient open area to minimize local well losses; guidelines given in Table 16.4 are recommended. If the well is located in an unconsolidated aquifer, a gravel pack should be placed in the annular region between the well screen and the perimeter of the borehole. The gravel pack should extend at least 30 cm (12 in.) above the top of the well screen and be designed in accordance with the guidelines given in Section 16.4. A seal of bentonite pellets should be placed on top of the gravel pack. A minimum of 1 m (3 ft) of pellets should be used, and an annulus seal of cement and/or bentonite grout should be placed on top of the bentonite pellets. The well

∗ “Aquifer test” is the preferred terminology used by the United States Geological Survey (USGS).

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casing should be protected at the surface with a concrete pad around the well to isolate the well bore from surface runoff. Well development. Pumping wells should be adequately developed to ensure that well losses are minimized. See Section 16.4 for a thorough discussion of well losses. If the well is suspected to be poorly developed or nothing is known, it is advisable to conduct a stepdrawdown test to determine the magnitude of the well losses. Water-level measurement access. It must be possible to measure the depth of water in the pumping well before, during, and after pumping. Usually, electric-sounder or pressuretransducer systems are used to measure drawdown in the well.

ww

Reliable power source. Power must be continuously available to the pump during the test. A power failure during the test usually requires that the test be terminated and sufficient time permitted for water levels to stabilize. This can cause expensive and time-consuming delays. Pump selection. Electrically powered pumps produce the most constant discharge and are recommended for use during an aquifer test. The discharge of gasoline- or diesel-powered pumps may vary greatly over a 24-h period and requires more frequent monitoring during the aquifer test. A diesel-powered turbine pump may have more than a 10% variation in discharge as a result of daily variations in temperature (Osbourne, 1993).

w .E asy En g

Discharge-control and measurement equipment. Common methods of measuring well discharge include orifice plates and manometers, inline flow meters, inline calibrated pitot tubes, weirs or flumes, and (for low discharge rates) measuring the time taken to discharge a measured volume. The discharge of wells yielding less than 7 L/s (100 gpm) can be readily measured with sufficient accuracy using a calibrated bucket or drum and a stopwatch (USBR, 1995). An important pump characteristic is that as the pump lift increases, the discharge decreases for a pump running at a constant speed. Therefore, the pump speed will usually need to be increased during the test to maintain a constant discharge.

ine eri n

Water disposal. The volume of pumped water produced, the storage requirements, disposal alternatives, and any treatment needs must be assessed during the planning phase. The pumped water cannot be allowed to infiltrate back into the aquifer during the aquifer test. If the aquifer is unconfined and the unsaturated zone overlying the aquifer is relatively permeable, the pumped water should be transported by pipeline to a location beyond the area of influence that will develop during the aquifer test. If the water table is more than 30 m (100 ft) below the ground surface and the unsaturated zone has a low hydraulic conductivity, an open ditch may be used to transport the pumped water (USBR, 1995). 16.5.2

Observation Wells

g .n

et

If existing wells are to be used as observation wells, then it should be verified that these wells are screened in the aquifer being investigated and that the screens are not clogged due to the buildup of iron compounds, carbonate compounds, sulfate compounds, or bacterial growth. The response test (Stallman, 1971; Black and Kipp, 1977) is recommended if existing wells are to be used as observation wells. Well diameter. The well casing should be large enough to allow for accurate, rapid waterlevel measurements. Well casings 50 mm (2 in.) in diameter are usually adequate in aquifers less than 30 m deep, but they are often difficult to develop. Well casings 100–150 mm (4–6 in.) in diameter are easier to develop and should have a better aquifer response. If a water-depth recorder is to be used, then well casings of 100–150 mm (4–6 in.) will usually be required. Difficulties in drilling a straight hole usually dictate that a well over 60 m (200 ft) deep must be at least 100 mm (4 in.) in diameter. Wells with diameters larger than 150 mm (4 in.) may

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cause a lag in response time due to water held in casing storage, so smaller diameters are usually preferable (Lehr et al., 1988). The importance of this time lag can be measured by the time constant, Tc , where r2 (16.19) Tc = c KL where rc is the radius of the casing, K is the hydraulic conductivity of the aquifer material or gravel pack surrounding the observation well, and L is the length of the screened interval. In an ideal monitoring well, the maximum deviation between the aquifer head and the water level in the well should not exceed the resolution of the pressure transducer, which is typically around 0.5 cm (0.2 in.). This typically requires a time constant, Tc , of less than 10 seconds (Gardner, 2009).

ww

Well construction. Observation wells should ideally have 1.5–6 m (5–20 ft) of perforated casing or screening near the bottom of the well. In addition, the placement of the gravel pack, bentonite pellets, cement or bentonite grout, and a concrete pad should follow the same guidelines as for the pumping well. After installation, observation wells should be developed by surging with a block, and/or submersible pump for a sufficient period (usually several hours) to meet a predetermined level of turbidity (Campbell and Lehr, 1972; Driscoll, 1986).

w .E asy En g

Distance from pumping well. Single observation wells are usually located about three to five times the saturated thickness away from the pumping well, with observation wells in unconfined aquifers located closer to the pumping well than observation wells in confined aquifers (Lehr et al., 1988). This usually works out to a distance of 10–100 m (30–300 ft) (Boonstra, 1998). If the pumping well is partially penetrating, observation wells should be located at a minimum distance equal to one-and-a-half to two times the saturated thickness from the pumping well (USBR, 1995). At least three observation wells at different distances from the pumping well are desirable, so that results can be averaged and obviously erroneous data can be disregarded. Whenever multiple observation wells are used, they are typically placed in a straight line or along perpendicular rays originating at the pumping well. If aquifer anisotropy is expected, then observation wells should be located in a pattern based on the suspected or known anisotropic conditions at the site. If the principal directions of anisotropy are not known, then at least three wells on different rays are required. If the aquifer is vertically anisotropic, then the minimum distance, rmin , of an observation well from the pumping well should be taken as (USBR, 1995) rmin

ine eri n 

Kh = 1.5b Kv

1 2

g .n

(16.20)

et

where b is the saturated thickness of the aquifer, Kh is the horizontal hydraulic conductivity, and Kv is the vertical hydraulic conductivity. Observation wells should be located far enough away from geologic and hydraulic boundaries to permit recognition of drawdown trends before the boundary conditions influence the drawdown readings. 16.5.3

Field Procedures

Aside from the design of the pumping and observation wells, there are several operational guidelines to be considered. Establishment of baseline conditions. Prior to the initiation of an aquifer test, it is essential to monitor the water levels in the pumping and observation wells in addition to wells adjacent to the site. These measurements will indicate whether a measurable trend exists in the groundwater levels. In addition, the influence and scheduling of offsite pumping on the aquifer test must be assessed, and controlled if necessary. As a general rule, at least 1 week of observations must be available prior to the initiation of the aquifer test. It is advisable to record barometric pressure, rainfall, and water levels in surface-water bodies within the anticipated cone of depression of the test well during the period of the aquifer test. It is

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797

relevant to note that a barometric pressure increase of 1 cm (0.4 in.) of mercury may result in a fall of about 8 cm (3 in.) in the water level in an observation well in a confined aquifer (Prakash, 2004).

ww

Water-level measurements during aquifer test. Immediately before pumping is to begin, static water levels in all test wells should be recorded. The recommended time intervals for recording water levels during an aquifer test are given in Table 16.10 (after ASTM Committee D-18, D 4050). After pumping is terminated, recovery measurements should be taken at the same time intervals as listed in Table 16.10, where the times are measured from the instant the pump is turned off. A check valve should be used to prevent backflow after pumping is terminated, and recovery measurements should continue until the prepumping state is recovered. The drawdown, s′ , after pumping is terminated in homogeneous isotropic aquifers can be approximated by (temporal) superposition of the Theis equation (Equation 15.123) as ! Qw W(u) − W(u′ ) s′ = (16.21) 4π T where Qw is the constant pumping rate during the aquifer test, T is the aquifer transmissivity, W(u) is the well function, and

w .E asy En g u=

r2 S 4Tt

and

r2 S′ 4Tt′

u′ =

(16.22)

where r is the distance of the observation well from the pumping well, S is the storage coefficient during drawdown, S′ is the storage coefficient during recovery, t is the time since pumping began, and t′ is the time since pumping stopped. For small values of u and u′ (u, u′ < 0.004), the Cooper–Jacob approximation to the well function (Equation 15.134) can be applied to Equation 16.21, yielding

or

S′ t Qw ln ′ 4π T St

ine eri n

s′ =

Qw s′ = 4π T



S′ t ln ′ + ln t S



(16.23)

(16.24)

g .n

This equation is a linear relationship between s′ and ln t/t′ . Matching Equation 16.24 to the recovery data yields a slope of Qw /4π T and an intercept of Qw /4π T ln S′/S. Knowing the pumping rate, Qw , and the storage coefficient during the drawdown test, S, the aquifer transmissivity, T, and recovery storage coefficient, S′ , can be estimated from the slope and intercept of recovery data. The estimated value of T can be used to confirm the transmissivity determined from the pump-drawdown measurements; however, no independent confirmation of the drawdown storage coefficient is obtained from this analysis. In some cases, S′ is assumed to be equal to S (e.g., de Marsily, 1986), in which case the transmissivity of the

et

TABLE 16.10: Recommended Time Intervals for Water-Level Measurement

Time since beginning of test 0–3 min 3–15 min 15–60 min 60–120 min 120 min–10 h 10 h–48 h 48 h–shutdown

Measurement interval Every 30 s Every 1 min Every 5 min Every 10 min Every 30 min Every 4 h Every 24 h

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aquifer is all that is extracted from the recovery data. Equations 16.21 and 16.24 are applicable to the common case in which the pumping rate remains constant during the pumping phase. In cases where the pumping rate varies during the pumping phase, alternate analytical approaches are available (e.g., Singh, 2009b). Discharge-rate measurements. During the initial hour of the aquifer test, the discharge from the pumping well should be measured as frequently as practical, and it is important to bring the discharge rate up to the design rate as quickly as possible. Because of the variety of environmental factors that can affect the discharge rate from diesel- and gasolinedriven pumps, the discharge should be checked four times per day, preferably early morning, mid-morning, mid-afternoon, and early evening. Osbourne (1993) has indicated that a 10% variation in discharge can result in a 100% variation in the estimate of aquifer transmissivity, and it is recommended that the discharge should never be allowed to vary by more than 5% during an aquifer test.

ww

Length of test. The test should be of sufficient length to accurately identify the shape of the drawdown versus time curve from which the aquifer hydraulic properties are extracted. The drawdown curve should be plotted during the aquifer test. As a general guideline, when three or more drawdown readings taken at 1 h intervals at the most distant well fall on a straight line (on the log drawdown versus log time curve), the aquifer test can be terminated (Lehr et al., 1988). Aquifer tests are typically conducted for around 24 h in confined aquifers and around 72 h in unconfined aquifers.

w .E asy En g 16.6

Design of Slug Tests

In some field situations, the hydraulic conductivity of the porous medium may be too small or the diameter, depth, or yield of the well may be too small to conduct an aquifer pumping test, or the scope of the investigation may not warrant an aquifer pumping test. In such cases, a slug test might be useful as a relatively quick and cost-effective method to estimate the aquifer hydraulic conductivity in the immediate vicinity of the well. This test is applicable to fully or partially penetrating wells in unconfined aquifers and fully penetrating wells in confined aquifers. The slug test is sometimes called a bail-down test (Fetter, 2001). A typical field setup for a slug test is illustrated in Figure 16.7, where a well with casing radius rc extends to a depth Lw below the water table in an unconfined aquifer of saturated thickness H. The well intake (screen) has a length Le and effective radius rw .

FIGURE 16.7: Field setup for a slug test Source: Bouwer and Rice (1976).

ine eri n

2rc

Water table yt

g .n

et

Lw 2rw

Le

H

Impermeable

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Design of Slug Tests

799

The effective radius is generally greater than the actual radius of the intake and includes the high-permeability region surrounding the well caused by either well development or the placement of a gravel pack. The slug test is performed by instantaneously removing a volume V of water from the well (a “slug” of water), and relating the recovery of the water level in the well to the hydraulic conductivity of the surrounding aquifer. The instantaneous removal of water can be accomplished using a bailer (a type of bucket), or by submerging a closed cylinder in the well, letting the water reach equilibrium, and then quickly pulling the cylinder out. In some cases, compressed air (or partial vacuum) can be used to displace a slug of water in the well. A variety of conventional methods can be used to relate aquifer properties to the recovery data (Hvorslev, 1951; Cooper et al., 1967; Bouwer and Rice, 1976; Van der Kamp, 1976; Nguyen and Pinder, 1984; Bouwer and Rice, 1989), and an enhancement to the Cooper et al. (1967) method can be found in Swamee and Singh (2007). The appropriate method of analysis depends primarily on the response characteristics of the water in the well. If the water level recovers to the initial static level in a smooth manner, then the response is called overdamped, while in cases where the water level in the well oscillates about the static water level, the response is called underdamped. An overdamped response is typical of most cases where the test well is screened in aquifers of moderate hydraulic conductivity, and methods proposed by Cooper et al. (1967), Hvorslev (1951), Bouwer and Rice (1976), and Bouwer (1989) are appropriate in these cases. Underdamped responses typically occur in wells that are screened in high hydraulic conductivity aquifers, and methods proposed by Van der Kamp (1976), Zurbuchen et al. (2002), Butler et al. (2003), and Audouin and Bodin (2007) are appropriate for these cases. The overdamped representation should be reasonable for most partially penetrating wells in high hydraulic conductivity isotropic formations when the screen length is less than 130 times the radius of the well (Butler and Zhan, 2004). The most widely used slug-test procedure, applicable to underdamped responses in fully or partially penetrating wells in unconfined aquifers, was developed by Bouwer and Rice (1976) and subsequently updated by Bouwer (1989). The basis of the Bouwer and Rice method is the Thiem equation (Equation 15.26), which can be put in the form   Qw Re y= (16.25) ln 2π KLe rw

w .E asy En g

ine eri n

where y is the drawdown in the well, Qw is the rate at which groundwater is removed from the aquifer, K is the hydraulic conductivity of the aquifer, and Re is the effective radial distance over which the head y is dissipated. Equation 16.25 neglects well losses. Continuity requires that the flow into the well, Qw , be equal to the rate at which the volume of water in the well increases, therefore Qw dy =− 2 (16.26) dt π rc

g .n

et

Combining Equations 16.25 and 16.26 (eliminating Qw ), integrating, and solving for K yields K=

rc2 ln (Re /rw ) 1 y0 ln 2Le t yt

(16.27)

where y0 is the initial drawdown in the well and yt is the drawdown at time t. Using Equation 16.27 to estimate the hydraulic conductivity, K, requires that rc , rw , and Le be known from the dimensions of the well, and t, y0 , and yt be measured during the recovery of the water level in the well. The effective radial distance, Re , can be estimated using the following empirical relation (Bouwer and Rice, 1976): Re ln = rw

1

1.1 A + B ln[(H − Lw )/rw ] + ln (Lw /rw ) (Le /rw )

2−1

(partially penetrating well) (16.28)

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FIGURE 16.8: Parameters in slug-test analysis Source: Bouwer and Rice (1976).

14 C 12

A and C

10

A

8

4 B

6

3 B

ww

4

2

2

1

0 1

5

10

50 100

500 1000

5000

0

Le rw

w .E asy En g

where A and B are dimensionless parameters that are related to Le /rw , as shown in Figure 16.8. Analyses by Bouwer and Rice (1976) have indicated that if ln[(H − Lw )/rw ] > 6, a value of 6 should be used for this term in Equation 16.28. If H = Lw , the well is fully penetrating, the term ln[(H − Lw )/rw ] given in Equation 16.28 is indeterminate, and the following equation should be used to estimate Re Re ln = rw

1

ine eri n

1.1 C + ln (Lw /rw ) (Le /rw )

2−1

(fully penetrating well)

(16.29)

where C is a dimensionless parameter related to Le /rw , as shown in Figure 16.8. Values of ln (Re /rw ) are within 10% of experimental values when Le /Lw > 0.4, and within 25% of experimental values when Le /Lw < 0.2 (Bouwer, 1978). It should be noted that the analysis presented here applies equally well to cases in which a “slug” of water is instantaneously added to the well at the beginning of the test (Batu, 1998). Values of K derived from a slug test primarily measure the average horizontal component of the hydraulic conductivity over a cylindrical volume of aquifer with inner radius rw , outer radius Re , and height slightly larger than Le . Because of the relatively small portion of the aquifer sampled by slug tests, it is not unusual for the hydraulic conductivity values derived from slug tests at closely spaced wells to differ by several orders of magnitude. Part of the reason is that hydraulic conductivities derived from slug tests are highly sensitive to minor variations in well-construction details, such as the screen position and the dimensions of the gravel pack. Averaging the hydraulic conductivities derived from spatially distributed slug tests to get a representative areal average should be done with caution, since significant underestimates of the areal average are frequently obtained (Bouwer, 1996). Slug tests are typically limited to aquifers with relatively low hydraulic conductivities, since rapid recovery in the test well yields poor data resolution (Lee, 1999). Slug tests are particularly useful in investigating localized flow conditions, which have an important effect on contaminant transport in groundwater.

g .n

et

Data analysis. Data collected during slug tests consist primarily of the drawdown, yt , as a function of time, t, and a methodology for combining these data with Equation 16.27 to estimate the hydraulic conductivity is as follows:

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ww

Design of Slug Tests

801

Step 1: Plot the measured data of ln yt versus t. Step 2: According to Equation 16.27, the plot in Step 1 should be a straight line. If some portion of the data does not show a linear relation, then only the linear portion is used in the analysis. In some cases, the plot of ln yt versus t will yield a curve with two straight-line segments. Such a situation occurs when the water in the gravel pack drains rapidly into the well. Once the water level in the gravel pack equals the water level in the well, the second straight-line segment forms. This reflects the hydraulic conductivity of the undisturbed aquifer. If a double straight line forms, the second segment should be used in the analysis. If the water level in the well is not lowered to the level of the gravel pack or if the gravel pack is so permeable that it will drain at the same rate that the water table is lowered, then the double-straight-line effect should not develop. If the gravel pack is surrounded by a less permeable zone, the data may fit three straight lines, one at very small values of t, the second at intermediate values of t, and the third at large values of t (Prakash, 2004). Again, the last straight line is representative of the hydraulic conductivity of the undisturbed aquifer. Step 3: Fit a straight line through the linear portion of the curve identified in Step 2. The equation of this line can be written as

w .E asy En g

ln yt = mt + ln y0

(16.30)

Determine the slope, m, of the fitted line. Equation 16.30 also indicates that this slope is given by 1 y0 (16.31) m = − ln t yt

Step 4: Based on the known values of Lw and H, select the equation to be used for ln (Re /rw ) from Equations 16.28 and 16.29. Step 5: If Equation 16.28 is to be used, determine the values of A and B from Figure 16.8. If Equation 16.29 is to be used, determine the value of C from Figure 16.8. Step 6: Calculate the value of ln (Re /rw ) using the results of Step 5. Step 7: Calculate the hydraulic conductivity, K, using Equation 16.27 with the values of m, ln (Re /rw ), rc , and Le , where

ine eri n

K=−

rc2 ln (Re /rw ) m 2Le

g .n

The application of this analysis is illustrated in the following example. EXAMPLE 16.7

(16.32)

et

A slug test is conducted in a monitoring well that penetrates 5 m below the water table in an unconfined aquifer with a saturated thickness of 15 m. The well casing has a 100-mm diameter, and the bottom 3 m of the well is screened and surrounded by a 100-mm thick gravel pack. The water level in the well is instantaneously drawn down 1500 mm, and the observed time-drawdown relation during recovery is given in the following table: Time (s) Drawdown (mm)

0 1500

2 645

4 372

6 195

8 130

10 67

12 44

14 23

16 15

Estimate the hydraulic conductivity and transmissivity of the aquifer. Solution The measured data of ln yt versus t is plotted in Figure 16.9. These data indicate that except for the first data point (t = 0, yt = 1500 mm), the relationship between ln yt and t appears quite linear and can be fitted to the regression equation ln yt = −0.270t + 6.976

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Chapter 16

Design of Groundwater Systems

FIGURE 16.9: Analysis of slug-test data

8 7 ln yt (mm)

6 5 4 3 2 1 0 0

ww

5

10 t (s)

15

20

which has a slope, m, of −0.270. From the given dimensions of the well and aquifer, Lw = 5 m, H = 15 m, rc = 50 mm, Le = 3 m, rw = rc + 100 mm = 150 mm, and Le /rw = 3/0.15 = 20. Equation 16.28 is appropriate for calculating ln (Re /rw ) (since the well is partially penetrating), and the dimensionless parameters A and B are given by Figure 16.8 as A = 2.1,

w .E asy En g

B = 0.25

Substituting these data into Equation 16.28 yields 2−1

Re = ln rw

1

1.1 A + B ln[(H − Lw )/rw ] + ln (Lw /rw ) (Le /rw )

=

1

1.1 2.1 + 0.25 ln[(15 − 5)/0.15] + ln (5/0.15) (3/0.15)

2−1

ine eri n

= 2.12

and putting this result into Equation 16.32 gives

r2 ln (Re /rw ) (0.05)2 (2.12) (−0.270) = 2.39 * 10−4 m/s = 20.6 m/d K=− c m=− 2Le 2(3)

This result indicates an average (horizontal) hydraulic conductivity of 20.6 m/d in the immediate vicinity of the well. The transmissivity, T, of the aquifer can be estimated by T = KH = (20.6)(15) = 309 m2 /d

g .n

et

Although the Bouwer and Rice (1976) slug-test analysis was originally designed for unconfined aquifers, the test can also be used in confined or stratified aquifers if the top of the screen is some distance below the upper confining layer (Bedient et al., 1999). Specialized data-analysis techniques for fully penetrating slug tests in confined aquifers can be found in Cooper et al. (1967) and Singh (2007b). Guidelines. The Kansas Geological Survey (Butler et al., 1996) and Fetter (2001) have reviewed several slug-test methodologies and suggested the following guidelines for conducting slug tests: ⊲ Three or more slug tests should be performed at a given well. If the well has not been properly developed prior to slug testing, the slug test itself might cause additional development. This can be detected by a shift in the calculated transmissivity during the repeated slug tests. Also the slug test can mobilize fine material, which can result in a decrease in the estimated formation transmissivity. ⊲ Two or more different initial displacements should be used during testing of a given well. A series of tests with initial head displacement, y0 , which varies by a factor of at least 2, should be employed; however, the first and last tests should have the same y0 so

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Design of Exfiltration Trenches

803

that a dynamic skin effect can be detected. The smearing of clay and silt on the surface of boreholes by augers sometimes results in a borehole skin that has a lower hydraulic conductivity than the surrounding aquifer. This skin effect is usually removed by well development. ⊲ The slug should be introduced in a near-instantaneous fashion and a good estimate of the initial displacement should be obtained. This is a basic assumption of the slug-test methodology. While it is easy to accomplish in a low-permeability system, it can be difficult to do in a system that has a very rapid response. ⊲ Appropriate data-acquisition equipment should be used. Manual methods of measurement might be satisfactory for a well in a low-permeability sediment that responds over a period of many minutes to hours. For more permeable formations, where the total period of response can be less than 1 minute a data logger with a pressure transducer is mandatory. The pressure transducer should have the proper sensitivity for the planned head displacement.

ww

It is important to keep in mind that slug tests measure the hydraulic conductivity in the immediate vicinity of the well, and therefore these tests may not be useful in estimating larger-scale aquifer hydraulic conductivities, such as those measured by aquifer pump tests. Comparison of hydraulic conductivities estimated by slug tests with values estimated by aquifer pump tests indicate that slug-test results are generally low (Bromley et al., 2004; Prakash, 2004). The likely reason for this discrepancy between small-scale and large-scale measurements of hydraulic conductivity is that slug tests do not account for preferential flow paths within aquifers. Another important consideration in interpreting slug-test data is the possible presence of a lower-hydraulic conductivity region surrounding the borehole that is caused by disturbances during well installation. This disturbed area surrounding the borehole is commonly called the wellbore skin or simply the skin of the well. Neglecting the existence of a skin in analyzing slug-test data will generally lead to a lower-bound estimate of the aquifer hydraulic conductivity (Cardiff et al., 2011). In cases where it is desirable to estimate the hydraulic characteristics of both the aquifer and the skin, more advanced slug-test analysis methods must be used (e.g., Malama et al., 2011); however, such advanced approaches might not be able to accurately and uniquely identify the aquifer and skin characteristics. In characterizing the hydraulic properties of contaminated aquifers where disposal of water extracted during aquifer pump tests might be an issue, slug tests have the advantage that no water is extracted. This consideration could have a significant impact on characterization costs.

w .E asy En g 16.7

Design of Exfiltration Trenches

ine eri n

g .n

et

Exfiltration trenches, sometimes called French drains or percolation trenches, are commonly used to discharge stormwater runoff into the subsurface and to return treated water from pump-and-treat systems into surficial aquifers. A typical exfiltration trench is illustrated in Figure 16.10. The operational characteristics of exfiltration trenches are influenced by the hydraulic conductivity of the aquifer material immediately surrounding the trench. This trench hydraulic conductivity is commonly determined from slug tests. The functional characteristics of an exfiltration trench depend on whether the trench is discharging water at a constant rate, such as when used in pump-and-treat systems, or discharging surface-water runoff from a design storm, as in the case of stormwater-management applications. The design of a constant-flow exfiltration trench is described below. Constant-flow design. Constant-flow exfiltration trenches are commonly designed based on the assumption that water in the trench is exfiltrated through both the bottom and the long sides of the trench. The outflow from the bottom of the trench, Qb , is given by Darcy’s law as Qb = Kt WL

(16.33)

where Kt is the trench hydraulic conductivity, W is the width of the trench, L is the length of the trench, and the hydraulic gradient is assumed equal to unity, as would be expected

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Chapter 16

Design of Groundwater Systems

FIGURE 16.10: Typical exfiltration trench

Ground surface Select backfill

Backfill

Pea gravel

H

15 cm

Pipe Coarse rock

Gravel

15 cm minimum

Pipe cover

30 cm minimum

Perforated pipe diameter

30 cm minimum

Pipe bed

Trench width W

ww

1.2 m minimum

w .E asy En g Water table

for vertical flow above the water table (Hart et al., 2008). The outflow from each side of the trench, Qs , is given by Darcy’s law as (16.34)

Qs = Kt Aperc

ine eri n

where Aperc is the side area through which water exfiltrates and the hydraulic gradient is taken as unity. It is common to assume that water exfiltrates through one-half of the trench height, H, in which case 1 Aperc = LH (16.35) 2 and Equations 16.34 and 16.35 combine to give Qs =

1 Kt HL 2

g .n

et

(16.36)

The total outflow rate from the exfiltration trench, Q, is equal to the sum of the flows out of the two (long) sides plus the flow out of the bottom. Therefore,   1 Kt HL = Kt L(W + H) Q = Qb + 2Qs = Kt WL + 2 (16.37) 2 Solving this equation for the trench length, L, yields L=

Q Kt (W + H)

(16.38)

The objective of the trench design is to select a length, L, width, W, and height, H, of the exfiltration trench, where the trench hydraulic conductivity, Kt , and the outflow rate, Q, are fixed by the site and design conditions, respectively. The maximum trench depth is limited by trench-wall stability, seasonal-high water-table elevation, and the depth to any impervious

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ww

Design of Exfiltration Trenches

805

soil layer. Exfiltration trenches 1 m (3 ft) wide and 1–2 m (3–6 ft) deep are considered to be most efficient (ASCE, 2012). Adequate groundwater (quality) protection is generally obtained by providing at least 1.2 m (4 ft) separation between the bottom of the trench and the seasonal-high water table. After selecting the geometry of the exfiltration trench, the aggregate mix to be placed in the trench surrounding the perforated discharge pipe must be selected such that the hydraulic conductivity of the aggregate mix significantly exceeds the hydraulic conductivity of the porous matrix surrounding the trench. The hydraulic conductivities of various aggregate mixes have been measured by Cedergren et al. (1972), who showed that the hydraulic conductivity of any gravel mix can be expected to exceed 4000 m/d (13,000 ft/d), which should be adequate for practically all exfiltration-trench designs. Consequently, any commercially available gradation of gravel can be used to surround the perforated discharge pipe in the exfiltration trench. The perforated discharge pipe used in exfiltration trenches usually consists of corrugated steel pipe with approximately 320 perforations per square meter; the diameter of the perforations is typically 0.95 mm (American Iron and Steel Institute, 1995). After designing an exfiltration trench capable of discharging the design flow, Q, into the subsurface, the final step is to verify that the aquifer is capable of transporting the recharge water away from the trench at a sufficient rate to prevent the groundwater from mounding to within a specified depth below the trench. The mounding of groundwater under a rectangular recharge area of length L and width W has been investigated by Hantush (1967), who derived the following expression for the maximum saturated thickness, hm , under a rectangular recharge area as a function of time:

w .E asy En g

2N νtS∗ h2m (t) = h2i + K



W L √ , √ 4 νt 4 νt



(16.39)

where hi is the (initial) saturated thickness without recharge; N is the recharge rate, which can be expressed in terms of the design flow, Q, by the relation

ine eri n N=

Q LW

(16.40)

and K is the hydraulic conductivity of the aquifer; ν is a combination of terms given by ν=

Kb Sy

g .n

(16.41)

et

where b is a mean saturated thickness, typically taken to be the average of hi and hm (t); Sy is the specific yield of the aquifer; t is the time since the beginning of recharge; and S∗ (α, β) is a function defined by ∗

S (α, β) =



1 0



α erf √ τ





β erf √ τ



dτ

(16.42)

where erf(ξ ) is the error function. General evaluation of the function S∗ (α, β) requires numerical integration of Equation 16.42; however, values of the function for a range of α and β are presented in Table 16.11. For values of α and β beyond the range of Table 16.11, the following relations and approximations for S∗ (α, β) are useful for practical computations (Hantush, 1967): S∗ (α, β) = S∗ (β, α) ∗

∗

S (0, β) = S (α, 0) = 0

(16.43) (16.44)

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Chapter 16

Design of Groundwater Systems TABLE 16.11: S∗ Function

β

ww

α

0.0000

0.30

0.60

0.90

1.2

1.5

1.8

2.1

2.4

2.7

3.0

0.00 0.10 0.30 0.60 0.90 1.20 1.50 1.80 2.10 2.40 2.70 3.00

0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

0.0000 0.1290 0.3009 0.4314 0.4860 0.5070 0.5142 0.5164 0.5170 0.5171 0.5172 0.5172

0.0000 0.1764 0.4314 0.6426 0.7360 0.7729 0.7857 0.7897 0.7907 0.7909 0.7910 0.7910

0.0000 0.1957 0.4860 0.7360 0.8504 0.8966 0.9129 0.9180 0.9193 0.9196 0.9197 0.9197

0.0000 0.2030 0.5070 0.7729 0.8966 0.9472 0.9653 0.9709 0.9724 0.9728 0.9728 0.9728

0.0000 0.2052 0.5142 0.7857 0.9129 0.9653 0.9841 0.9900 0.9915 0.9919 0.9920 0.9920

0.0000 0.2062 0.5164 0.7897 0.9180 0.9709 0.9900 0.9959 0.9975 0.9979 0.9979 0.9979

0.0000 0.2065 0.5170 0.7907 0.9193 0.9724 0.9915 0.9975 0.9991 0.9995 0.9995 0.9995

0.0000 0.2065 0.5171 0.7909 0.9196 0.9728 0.9919 0.9979 0.9995 0.9998 0.9999 0.9999

0.0000 0.2065 0.5172 0.7910 0.9197 0.9728 0.9920 0.9979 0.9995 0.9999 1.0000 1.0000

0.0000 0.2065 0.5172 0.7910 0.9197 0.9728 0.9920 0.9979 0.9995 0.9999 1.0000 1.0000

⎧ ⎪ 1 − 4i2 erfc(β), ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎨1 − 4i erfc(α), ∗ S (α, β) M 1 1 2  ⎪ ⎪ ⎪ 4 β β α α ⎪ 2 2 −1 −1 ⎪ ⎪ ⎩ π αβ 3 + W(α + β ) − β tan α + α tan β

w .E asy En g

α Ú 3.0 β Ú 3.0 α Ú 3.0, β Ú 3.0 α 2 + β 2 … 0.10 (16.45)

√

where erfc(x) is the complementary error function, i = −1, and W(u) is the Theis well function. Equation 16.39 gives the maximum height of the groundwater mound under the trench as a function of time. Typically, operators are concerned with the long-term effect of recharge on groundwater, such as the maximum height of the groundwater mound 20, 50, or 100 years from initial operation (Bouwer, 1999; Bouwer et al., 2002). Groundwater levels that intersect the bottom of the exfiltration trench will cause the trench to malfunction. The design of an exfiltration trench is illustrated by the following example.

EXAMPLE 16.8

ine eri n

g .n

et

Design an exfiltration trench to inject 2.40 L/s from a pump-and-treat system. Several slug tests at the site have indicated a minimum trench hydraulic conductivity of 24 m/d. The depth from the ground surface to the seasonal-high water table is 4.35 m, and the aquifer is estimated to have a hydraulic conductivity of 107 m/d, a saturated thickness of 10.7 m, and a specific yield of 0.2. Local regulations require a factor of safety of 2 in the assumed trench hydraulic conductivity and a minimum backfill of 50 cm. The pump-and-treat system is expected to operate for a maximum of 20 years. Solution The length, L, of the trench is given by Equation 16.38 as Q Kt (W + H)

(16.46)

207 Q = = 5.75 m Kt (W + H) 12(1 + 2)

(16.47)

L=

where the dimensions H and W are illustrated in Figure 16.10. Following ASCE (2012) guidelines, specify W = 1 m (a typical backhoe dimension) and H = 2 m (a typical depth for vertical slope stability). The design injection rate, Q, is 2.40 L/s = 207 m3 /d, and the design trench hydraulic conductivity, Kt , is 12 m/d, equal to the minimum measured value of 24 m/d divided by the factor of safety of 2. Substituting these values for Q, Kt , W, and H into Equation 16.46 yields L=

A trench dimension of 6 m long by 1 m wide by 2 m deep, when filled with gravel, will be capable of discharging water into the aquifer at a rate of 2.40 L/s. Since the seasonal-high water table is 4.35 m below the ground surface, there is sufficient room to install a 2-m deep trench covered with 0.5 m

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Design of Exfiltration Trenches

807

(= 50 cm) of backfill and still maintain a distance of at least 1.2 m between the bottom of the trench and the seasonal-high water table. The next question to be addressed is whether the aquifer will be able to transport the effluent away from the trench as fast as it is supplied, without causing the water table to rise to within 1.2 m of the bottom of the trench. The height of the water table above the base of the aquifer as a function of time is given by

 W L 2N 2 2 ∗ νtS (16.48) hm (t) = hi + √ , √ K 4 νt 4 νt where hi = 10.7 m and N is given by N=

ww

207 Q = = 34.5 m/d LW (6)(1)

The hydraulic conductivity of the aquifer, K, is 107 m/d, and the parameter ν in Equation 16.48 is given by Kb K(hi + hm )/2 107(10.7 + hm )/2 = = ν= = 267.5(10.7 + hm ) Sy Sy 0.2 Substituting the trench dimensions (L = 6 m, W = 1 m) and aquifer properties into Equation 16.48 yields

 6 2(34.5) 1 2 2 ∗

[267.5(10.7 + hm )]tS , hm = 10.7 + 107 4 [267.5(10.7 + hm )]t 4 [267.5(10.7 + hm )]t

 0.0917 0.0153 ∗

, (16.49) = 114 + 172.5(10.7 + hm )tS (10.7 + hm ) (10.7 + hm )

w .E asy En g

This equation relates the maximum saturated thickness, hm , below the trench to the time, t, since the trench began operation. Values of hm for several values of t are shown in Table 16.12, where S∗ (α, β) has been estimated using Equation 16.45. The values shown in Table 16.12 indicate that the saturated thickness under the trench steadily increases, and after 10,000 days (27 years) the saturated thickness will be 10.94 m. This indicates that the water table below the exfiltration trench will rise (mound) by about 10.94 m −10.70 m = 0.24 m after 27 years and will remain at an acceptable depth (>1.2 m) below the trench. The final trench design is illustrated in Figure 16.11.

ine eri n

TABLE 16.12: hm versus t

FIGURE 16.11: Trench design

t (days)

hm (m)

1 10 100 1000 10,000

10.81 10.84 10.87 10.90 10.94

Backfill

g .n

et

50 cm

Gravel-filled trench

2m

1m Length ⫽ 6 m

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Chapter 16

Design of Groundwater Systems

The above example illustrates the application of the Hantush (1967) solution to an exfiltration trench of dimensions L * W discharging water at a constant rate, Q, into the saturated zone beneath the trench. This solution can also be applied to estimate the mounding beneath infiltration basins of dimensions L * W. In cases where the exfiltration rate, Q, does not remain constant over time, the Hantush (1967) solution is not appropriate since it does not account for storage in the vadose zone, which causes the exfiltration rate to not equal the recharge rate to the saturated zone. Under these circumstances, a vadose-zone flow model should be coupled with a saturated-zone flow model to more accurately describe the mounding effect (e.g., Nimmer et al., 2010). 16.8

ww

Seepage Meters

A conventional seepage meter is illustrated in Figure 16.12 and consists of a pan, formed from an inverted bucket or drum, connected to a collection bag made from thin plastic film. The pan is pushed into the bed of a lake or stream and the collection bag is weighed and attached to the pan. The bag is weighed again after some time has elapsed, and the flux of water into or out of the bed is readily calculated using the change in weight of the bag, the elapsed time, and the area of the pan. Interpretation of this measurement assumes that water flows into the pan at the same rate as it would if the seepage meter were absent. This assumption is intuitively appealing, because head losses in the pan and collection bag are expected to be negligible, although this is not always the case (Murdoch and Kelly, 2003). Some seepage meters use electronic flow meters to avoid the need for a collection bag. The seepage meter was initially developed to measure losses from irrigation canals (Israelsen and Reeve, 1944), and its application has been expanded to measure groundwater seepage into lakes, ponds, and estuaries (Martin et al., 2007), determine water budgets, obtain water samples for chemical analyses, study groundwater interactions with streams, and measure water fluxes across sediment-water interfaces in streams (Rosenberry, 2008). Tests by Lee (1977), Erickson (1981), and Belanger and Montgomery (1992) suggest that seepage meters can yield measurements that are repeatable within 10%, while other studies (Isiorho and Meyer, 1999) are less encouraging, and show that repeat measurements in the laboratory can vary by more than a factor of 10. Seepage flux measured in the laboratory is typically less than the actual seepage rate, according to Erickson (1981) and Belanger and Montgomery (1992), who found the ratio of measured to actual seepage in the range of 0.66–0.77. Special low-profile seepage meters are recommended for use in streams where the seepage meter itself interferes with flow and significantly influences seepage (Rosenberry, 2008).

w .E asy En g

FIGURE 16.12: Seepage meter

ine eri n

Water body

Tubing and fittings Pan

g .n

et

Collection bag Top of sediment

Seepage

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809

Problems 16.1. A wellfield is to be developed to produce 4.44 m3 /s from an unconfined aquifer with a saturated thickness of 35 m, a hydraulic conductivity of 500 m/d, and a specific yield of 0.2. If the radius of each well is to be 0.5 m, the drawdown is not to exceed 2 m, and the wells are to be arranged along a straight line, then determine the number of wells, the pumping rate from each well, and the spacing between wells. 16.2. A new wellfield is to be developed within a 500-m * 500-m block of land bounded on all sides by canals that are intended to feed the wellfield. The stages in all canals are to be maintained at 1.000 m NGVD, and the drawdown in the wellfield is not to exceed 2 m. The diameter of each well is to be 800 mm, and the wells are to be a minimum of 50 m apart and a minimum of 100 m from any canal. Field tests indicate that the aquifer has a saturated thickness of 20 m, a hydraulic conductivity of 75 m/d, and a specific yield of 0.26. Determine the number and location of the wells required to supply 69.4 L/s. 16.3. A wellfield is to be developed such that the wells are all parallel to a stream, with each well having a diameter of 100 mm. The unconfined aquifer has a saturated thickness of 20 m, an effective porosity of 0.15, and a hydraulic conductivity of 60 m/d. If the wellfield is expected to deliver 83.3 L/s, each well is to be 100 m from the stream, and the maximum allowable drawdown is 2 m, estimate the number of wells required, the pumping rate from each well, and the spacing between the wells. 16.4. An existing wellfield consists of four wells at the corners of a rectangular area 100 m long by 75 m wide. Each well has a design capacity of 467 L/s and the observed drawdown in the center of the rectangular area is 2 m when all wells are pumping at their capacity. The specific yield of the aquifer is estimated as 0.2. Consideration is being given to the addition of a fifth well to the wellfield such that the new well is no more than 75 m from any of the existing wells, and the drawdown in the center of the rectangular area is not to exceed the current maximum value of 2 m. Taking the pumping capacity of all wells in the expanded wellfield to be the same, determine the increase in total pumping capacity of the wellfield that can be achieved. 16.5. A 80 m * 80 m site is to be dewatered to provide for construction 1.5 m below the initial water-surface elevation of 90 m. Four pumps are to be used in 500-mm diameter wells at the four corners of the site. The transmissivity and storage coefficient of the aquifer are 1600 m2 /d and 0.16 respectively. The site needs to be ready after 1 month of pumping. Determine the required pumping rate. Assume fully penetrating wells, but assess the effect that using partially penetrating wells would have on your final specification.

ww

16.6. Show that the half-life of a contaminant in groundwater can be described by Equation 16.5. 16.7. The risk, R, of illness from viruses in pumped water is related to the viral concentration, c, by the relation R = 1 − (1 + 4.76c)−94.9

w .E asy En g 16.8.

16.9.

16.10.

16.11.

where c is in #/L. If an illness rate of 1 in 10,000 people is acceptable (R = 10−4 ), determine the allowable viral concentration in the pumped water. A municipal water-supply well is to be installed 100 m from a school that uses a septic tank for on-site wastewater disposal. The well is to have a rated capacity of 400 L/s, the saturated thickness of the aquifer is 20 m, and the porosity of the aquifer is 0.17. If the allowable risk of viral infection from the pumped water is 10−4 and the decay constant for viruses in groundwater is 0.3 d−1 , estimate the maximum allowable virus concentration in the groundwater below the school. A 30-cm diameter water-supply well is to pump 50 L/s and be located 150 m from a fully penetrating river in an aquifer with a saturated thickness of 25 m. The hydraulic conductivity of the aquifer is 30 m/d, the porosity is 0.2, and the radius of influence of the well is 1 km. Calculate the shortest time of travel of the groundwater from the river to the well intake. If the decay constant of a contaminant is 0.01 d−1 , and the allowable concentration of the contaminant in the pumped water is 1 μg/L, estimate the maximum allowable concentration of the contaminant in the river by assuming the shortest time of travel. Assess the validity of this assumption. A square-shaped 50-ha parcel of land has been identified as a possible site for a wellfield to supply a population of 50,000 people. The saturated thickness of the aquifer is 35 m, the hydraulic conductivity is 85 m/d, the porosity is 0.2, and the specific yield is 0.15. The per-capita demand of the population is 580 L/person/d. Design a wellfield that can be accommodated on the parcel of land. The radius of each well can be taken as 500 mm, and the drawdown must not exceed 6 m. If the wells are screened over the bottom 20 m of the aquifer and the screen has a diameter of 500 mm with 50% open area, assess whether the screen is adequate. A well of radius 500 mm is to be installed in an unconfined aquifer of hydraulic conductivity 300 m/d. What is the maximum allowable screen entrance velocity, and what is the corresponding maximum pumping rate? If the uniformity coefficient of the aquifer matrix is 3.2, then write the specifications for the gravel pack and the screen slot size. An aquifer is approximately 40 m thick, has a reported hydraulic conductivity of 1803 m/d, and the aquifer material consists entirely of medium sand with particle

ine eri n

g .n

et

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16.12.

16.13.

16.14.

16.15.

Chapter 16

Design of Groundwater Systems

sizes uniformly distributed between 0.25 mm and 0.5 mm. A water-supply well in the aquifer is to be pumped at 384 L/s, and two alternatives for the well screen are being considered. For the alternative in which a gravel pack is not to be used, design the well screen. For the alternative in which a gravel pack is to be used, design the well screen and gravel pack. A water-supply well is to be installed in a 30-m thick sand aquifer with a uniform grain-size distribution between 0.04 mm and 2.2 mm. The hydraulic conductivity of the aquifer is estimated to be 50 m/d, the porosity is 0.15, and the specific yield is 0.2. If the pumping rate is to be 13.3 L/s, design the casing, screen, and gravel pack. Sand gradations sold by Ricci Bros. Sand Co. Inc. (Port Norris, New Jersey) are shown in Table 16.13. Determine the appropriate range of screen slot sizes and aquifer properties that could be used with each gradation. Screen slot sizes are available from Ricci Bros. Sand Co. in the following sizes: 0.25 mm, 0.30 mm, 0.42 mm, 0.60 mm, 0.85 mm, 1.18 mm, and 1.70 mm. Exploration of an unconfined aquifer at the site of a proposed water-supply well indicates a transmissivity of 4200 m2 /d and a specific yield of 0.2. Drilling a pilot hole at the proposed well location indicates that the aquifer has a saturated thickness of 30 m, and a sieve analysis of the borehole cuttings indicates 10-percentile and 60-percentile particle sizes of 0.5 mm and 2.1 mm, respectively. The well is expected to yield 41.7 L/s, and the turbine pump to be used has a bowl diameter of 300 mm. Available screens have slot sizes in the range of 30 to 100 in increments of 5, and all have 15% open area. Specify the following well dimensions: (1) casing diameter; (2) borehole diameter; (3) screen diameter; (4) screen length; (5) screen slot size; (6) gravel-pack specifications (if used); and (7) location of pump intake. A 300-mm diameter well is to be installed in a 20-m thick confined aquifer in which the allowable drawdown of the potentiometric surface at the well is 5 m. The hydraulic

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conductivity and storage coefficient of the aquifer are estimated as 30 m/d and 10−4 , respectively. Estimate the specific capacity and the well yield after 1 year, and classify the productivity of the well. 16.16. Estimates of the transmissivity in the surficial aquifer system in Miami-Dade County, Florida, have been related to the observed specific capacities of wells by (Fish and Stewart, 1991) T = 270Sc

where T is the transmissivity in ft2 /d and Sc is the specific capacity in gpm/ft. Data from several watersupply wells in Miami-Dade County are shown in Table 16.14 and the specific yield of the aquifer system in Miami-Dade County is estimated to be in the range of 0.16–0.3 (Chin and Patterson, 2005). Use the data in Table 16.14 to assess the reliability of using Equation 16.50 to estimate the transmissivity from specific capacity measurements. If the Fish and Stewart (1991) relationship is inadequate, propose an improved relationship. 16.17. A step-drawdown test is conducted in an aquifer and yields the results shown in Table 16.15. Determine the well-loss coefficient, formation-loss coefficient, and the specific capacity of the well. Assess the condition and productivity of the well at a pumping rate of 57.9 L/s. 16.18. The drawdowns, sw (cm), and corresponding pumping rates, Qw (L/min), collected at a well during a step-drawdown test were plotted on a graph of log(sw /Qw − β) versus log Qw for various values of β. The plotted curve is most linear when β = 55.2 s/m2 , and the corresponding best-fit line through the data has a slope of 1.22. If the well efficiency is known to be 32% when the pumping rate is 10.3 L/s, estimate the wellloss coefficient. Assess the productivity of the well when pumping at 16.7 L/s.

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TABLE 16.13: Gravel Pack Gradations

Sieve No. 4 6 8 10 16 20 30 40 50 60 70 100

Size (mm) (in.) 4.75 0.188 3.35 0.132 2.36 0.094 1.70 0.067 1.18 0.045 0.85 0.033 0.60 0.023 0.42 0.016 0.30 0.012 0.25 0.010 0.21 0.008 0.15 0.008

OOO

100 95 75 45 30 5

P40

100 95 80 60 35 10 5 2 1

(16.50)

Gradation (% passing) OO OON O 1

100 95 40 5 2 1

100 99 35 5 1

100 99 50 5 1

100 95 45 5 1

2 100 95 55 10 1

3 100 99 55 5 1

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4 99 50 10 1

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TABLE 16.14: Hydraulic Data for Water-Supply Wells in Miami-Dade County, Florida

Site 1

2

3

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4

Well S-3011 S-3012 S-3013 S-3014 S-3005 S-3006 S-3007 S-3008 S-3009 S-3010 — — — — — — S-981 S-983 S-3065 S-3066 S-3045

Diameter (cm) 110 110 110 110 110 110 110 110 110 110 20 20 20 51 51 51 30 30 30 30 20

Production zone* (m ) 18.41–40.54 17.37–40.23 18.87–40.23 18.56–40.23 17.98–40.23 16.46–39.93 20.09–40.23 16.61–39.17 16.61–39.11 16.76–39.93 13.72–14.33 13.72–14.33 20.73–21.03 9.14–17.98 9.14–20.12 9.14–16.46 12.19–16.15 11.58–16.76 31.39–35.05 31.39–35.05 33.83–35.97

Discharge (L/s) 442 442 442 442 175 175 175 175 175 175 33 33 43 245 245 245 74 69 76 76 60

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Drawdown (m) 4.60 5.27 4.15 4.30 0.518 1.067 0.610 1.219 0.610 0.610 0.305 0.305 0.457 1.280 1.219 0.975 1.280 1.280 1.219 1.219 0.671

Pumping period (h) 8 8 8 8 2 2 2 2 2 2 2 2 2 0.25 0.5 0.5 2.25 2 8 8 8

Note: *Production zone given in meters below land surface.

TABLE 16.15: Step-Drawdown Results

Pumping rate (m3 /s)

Drawdown (cm)

0 0.05 0.10 0.15 0.20 0.25 0.30

0 20 53 86 147 210 283

16.19. A step-drawdown test yields the following results: Pumping rate (L/min)

Drawdown (cm)

40 60 80 100 120 140 160

15 23 33 41 53 62 73

Use the Rorabaugh method to analyze the measurements and determine the efficiency and specific capacity

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of the well when the pumping rate is 8.33 L/s. Compare your result with the conventional approximation that well losses are proportional to the square of the pumping rate. 16.20. The following results are derived from a step-drawdown test in a deep aquifer: Pumping rate (m3 /d) 500 1000 1500 2000 2500 3000 3500

g .n Drawdown (m) 2.40 5.38 9.28 14.36 20.82 28.87 38.70

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Use the Rorabaugh method to estimate an empirical relation between pumping rate and drawdown. Determine the well loss as a percentage of total drawdown for the pumping rates shown in the above table. [Adapted from Todd and Mays (2005).] 16.21. Give an example of how you would use estimates of aquifer properties prior to conducting an aquifer pump test to aid in the design of the pump test.

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16.22. Use the principle of superposition to show that Equation 16.21 can be used to describe the recovery data in a pump test. 16.23. Apply the Cooper–Jacob approximation to Equation 16.21 to show that the recovery data in a pump test can be described by Equation 16.24. 16.24. An aquifer pump test is conducted at a rate of 33.3 L/s, and pumping is terminated after 4 h. Measurements of drawdown versus time after pumping is terminated are given in the following table:

Drawdown (m)

Time since end of pumping (min)

1.01 0.90 0.83 0.75 0.70 0.61 0.55 0.60 0.42 0.37 0.31 0.26 0.23 0.19 0.15

1 2 3 5 7 10 15 20 30 40 60 80 100 140 180

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If the pump-drawdown test indicates a storage coefficient of 0.0001, estimate the aquifer transmissivity and the recovery storage coefficient. 16.25. It has been stated that a 10% variation in discharge during an aquifer pump test can result in a 100% variation in the estimate of aquifer transmissivity (Osbourne, 1993). Assess the validity of this statement. 16.26. The following slug-test data were measured at a test hole that penetrates the entire thickness of a confined aquifer. Time Drawdown Time Drawdown Time Drawdown (sec) (m) (sec) (m) (sec) (m) 0 3 6 9 12 15 18 21

0.560 0.457 0.392 0.345 0.308 0.280 0.252 0.224

scale would you associate with your estimated hydraulic conductivity? 16.27. A slug test was performed in alluvial deposits of the Salt River bed west of Phoenix, Arizona. The geometry of the aquifer is shown in Figure 16.13. A solid cylinder with a volume equivalent to a 0.32-m change in water level in the well was also placed below the water level. When the water level had returned to equilibrium, the cylinder was quickly removed. The measured values of yt versus time (t) are given in the following table:

24 27 30 33 36 39 42 45

0.205 0.187 0.168 0.149 0.140 0.131 0.112 0.108

48 51 54 57 60 63

0.093 0.089 0.082 0.075 0.071 0.065

The radius of the test hole is 760 mm and the thickness of the confined aquifer is 98 m. Estimate the hydraulic conductivity of the confined formation. What support

t (s)

yt (m)

yt /y0

0.0 0.7 1.4 2.8 3.7 6.0 8.8 12.8 18.6 28.6 38.6

0.290 0.238 0.190 0.150 0.117 0.072 0.030 0.012 0.005 0.002 0.001

1.000 0.821 0.655 0.517 0.403 0.248 0.103 0.041 0.017 0.007 0.003

The theoretical value of y0 calculated from the displacement of the submerged cylinder is 0.32 m. Applying the Bouwer and Rice slug-test data analysis method, determine the horizontal hydraulic conductivity of the aquifer. 16.28. A monitoring well that penetrates 4 m below the water table in an unconfined aquifer with a saturated thickness of 12 m is used to conduct a slug test. The well casing has a 150 mm diameter, and the bottom 2 m of the well is screened and surrounded by a 110-mm thick gravel pack. The observed time-drawdown relation during the slug test is given in the following table:

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Time (s)

Drawdown (mm)

0 3 6 9 12 15 18 21 24

700 392 260 137 91 47 31 16 11

et

Estimate the hydraulic conductivity and transmissivity of the aquifer. 16.29. Repeat Problem 16.28 for the case in which the monitoring well fully penetrates the aquifer.

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Ground surface

3m Water table

0.94 m

5.5 m 80 m

4.56 m

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0.24 m

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Aquifer bottom

FIGURE 16.13: Slug test

16.30. An exfiltration trench is to be designed to discharge 5.79 L/s from a pump-and-treat system. The trench hydraulic conductivity is 35 m/d, and the depth from the ground surface to the seasonal-high water table is 5.22 m. The aquifer is estimated to have a hydraulic conductivity of 70 m/d, a saturated thickness of 15 m, and a specific yield of 0.22. Design the exfiltration trench assuming a factor of safety of 2.5 in the trench hydraulic conductivity. The pump-and-treat system is expected to operate for a maximum of 25 years. 16.31. Repeat Problem 16.30 assuming 1 m of backfill over the trench (so that surface vegetation can be planted). 16.32. An exfiltration trench is to be designed such that the water table remains at least 1.2 m below the trench after 1 year of operation. The trench hydraulic conductivity is 10 m/d, the width of the trench is to be 1 m, the depth is to be 1.5 m, and space limitations will allow a maximum trench length of 10 m. The wet-season water table is 3.1 m below ground surface, the hydraulic conductivity of the aquifer is 15 m/d, the porosity is 0.15, and the saturated thickness of the aquifer is 7 m. What is the maximum allowable flow rate that can be exfiltrated through the trench? 16.33. An exfiltration trench is being considered as a means to discharge 5.79 L/s continuously for 20 years into an aquifer having a saturated thickness of 17 m, a hydraulic conductivity of 7 m/d, and a specific yield of 0.14. The trench hydraulic conductivity is estimated to be 20 m/d, and the depth from the ground surface to the water table is 4 m. A factor of safety of 2, and at least 30 cm of backfill above the trench must be used. Determine the

required trench dimensions, and assess the practicality of the trench. 16.34. A 100-m * 100-m pond is to be used as a recharge basin in which water is expected to infiltrate at the rate of 5 meters per day. The unconfined aquifer beneath the recharge basin has a hydraulic conductivity of 90 m/d, a specific yield of 0.2, and a wet-season saturated thickness of 35 meters. At the start of recharge operations, the wet-season water table is 20 m below the bottom of the pond. Based on a consideration of the mounding effect, determine whether the pond can be successfully operated for 20 years. 16.35. A planned infiltration basin of dimension 10 m * 200 m is to intermittently infiltrate water at a rate of 0.2 m/d into an aquifer that has a hydraulic conductivity of 2 m/d, a specific yield of 0.15, and a saturated thickness of 5.333 m.

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(a) Estimate the maximum mounding height under the basin after 6 d of continuous operation. (b) Swamee and Ojha (1997) have proposed the following algebraic expression to approximate the Hantush mounding function,      0.5785 2.75 0.5785 2.75 + S∗ (α, β) L 1 + α β +



 −0.325 0.36 2.4 αβ

Repeat Part (a) using the Swamee and Ojha (1997) approximation.

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16.36. A large artificial lake in support of groundwater recharge is excavated in Tempe, Arizona, and the average leakage from the lake is observed to be 45.7 cm/d. The depth to bedrock in the area is about 45.7 m and the regional aquifer has a saturated thickness of around 30.5 m. The average transmissivity in the area is

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929 m2 /d and the average porosity is 0.2. If the lake is 305 m wide and 3.22 km long, estimate how long it takes for the area to become waterlogged. If the time for the area to become waterlogged is to be no less than 1 year, what lake width would be required?

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C H A P T E R

17

Water-Resources Planning 17.1

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Introduction

Water-resources planning can be categorized into two levels: comprehensive planning and functional planning. Comprehensive planning coordinates a wide range of activities, sets overall direction, identifies priorities, and provides a basis for managing conflicts (Dzurik, 2003). Comprehensive planning generally considers interrelationships between water, society, and economic development. In contrast to comprehensive planning, functional planning is confined to one major area or resource of primary interest, such as water-resources, transportation, or land-use planning. Functional planning is typically more detailed and technical than comprehensive planning and tends to be the only type of planning that remains politically palatable over time. Water-resources planning and management is closely related to water-resources systems engineering, which is defined as the formulation and evaluation of alternative plans to determine the system configuration or set of actions that will best accomplish the project objectives within the constraints of natural laws, engineering principles, economics, environmental protection, social and political pressures, legal restrictions, and institutional and financial capabilities (Wurbs and James, 2002). The primary factors and constraints to be considered in planning the development of water resources are: sustainability, technical feasibility, public policy, regulations, political realities, and economic factors. The benefits obtained from good planning include improved quality of decisions, better understanding of problems and issues, more consistent policy decisions with respect to longterm goals, more focus on priority issues, consensus building, and more rationality in public decision making.

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Planning Process

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The planning process as applied to water resources is comparable to other types of planning. It is a logical series of steps, beginning with identification of needs, proceeding to recommendations for action, and culminating in implementation and monitoring. Descriptions of all steps in the planning process are given below.

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Step 1: Problem identification. Problem identification is derived from the expressed needs and concerns relating to the water resources of an area. Problem identification should reflect the concerns of different groups, both public and private, and should be described in sufficient detail to allow for adequate attention in the planning process. Public involvement as well as coordination of various agencies and groups should begin at this stage. Specific needs usually result either from a problem that has already been experienced, such as flooding, or from an anticipated problem, such as inadequate water supply to meet future needs. In some instances, water-resources planning studies may be more generalized and not deal with any single problem or need. River-basin studies and areawide water-quality studies are typically of this type. Step 2: Data collection and analysis. Following problem identification, the study area should be defined and existing information for the study area analyzed for its relevance to the problem under consideration. This analysis should include identification of relevant geophysical and biological features; social, demographic, and cultural characteristics; land use; and economic activity such as manufacturing, commerce, and agriculture. It is also important to determine anticipated future conditions as shown by existing planning documents, which should be reviewed and modified as needed.

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Water-Resources Planning

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In the United States, extensive data on the water resources of an area is typically available from the U.S. Geological Survey (USGS), while state and local governments can also provide relevant information. As part of the data-collection process, forecasts should be made of appropriate variables to determine likely future conditions of the problem under investigation. Step 3: Goals and objectives. Specifying relevant planning goals and objectives and defining their relative importance is one of the most difficult tasks in water-resources planning, since there are many divergent interests that often generate conflicting objectives. The most common conflict in water-resources planning is between economic and environmental objectives. Statements of goals and objectives indicate what the planning effort hopes to accomplish. Goals are broad and general, such as the attainment of clean water or provision of adequate water supplies. Although they are stated in general terms, goal statements relate human values to natural resources and the environment. Objectives are more specific statements of plans to be developed, and often several objectives must be accomplished to attain a goal. Objectives provide the basis for evaluating alternatives. Examples of typical water-resources planning objectives are: • • • • •

Prevent continued water degradation by waterborne wastes. Prevent or reduce flood damages. Provide water-based recreation. Provide for efficient reuse of treated wastewater. Provide for efficient development and management of water supplies.

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It is difficult to plan for vague or unrealistic objectives. Therefore, objectives should be stated as clearly and specifically as possible in order to facilitate their achievement. Step 4: Problem diagnosis. Problem diagnosis involves formulating a clear understanding of the problem so that alternative solutions will effectively address it. Most commonly, this analysis will indicate that there are numerous alternative solutions. For example, flooding problems may be solved by either structural or nonstructural solutions. Step 5: Formulation of alternatives. Alternative plans must be formulated to help the decision maker identify how they relate to the objectives and understand the trade-offs among them. Formulation of alternatives begins with identification of measures that will address the defined problems. Public and interagency participation is important at this point to ensure that the full range of measures is considered. Whenever possible, the alternatives should range from capital-intensive structural measures to nonstructural management or policy solutions. Some objectives have economic efficiency as their primary goal, while other frequently cited objectives focus on preservation or restoration of the environment. There is no standard number of alternatives to be developed. Judgment must be exercised to determine which plans are appropriate and to decide which alternatives to carry forward for more detailed study. Step 6: Analysis of alternatives. The two primary aspects of analyzing alternatives are economic evaluation and impact assessment. In both cases, measures are developed which quantify the changes resulting from the alternative plans. It is always useful to have a no-action plan as a baseline for comparison. Economic evaluation usually requires a benefit-cost analysis, for which several methods have been developed. The simplest form of benefit-cost analysis compares the present value of all the costs with the present value of all the benefits. If the benefits exceed the costs, then the plan is economically justified. The present-value approach is appropriate if the economic lives of the alternatives are the same. If they differ, it is more appropriate to convert each time stream of net benefits to an equivalent average-annual benefit for purposes of comparison (Dzurik, 2003). Impact assessment includes environmental, social, and economic impacts, but the major concern is usually the environmental impact. Impact assessment is an analysis

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Planning Process

817

of the potential positive and negative impacts, and significant changes that might result from selecting an alternative. Impacts are identified by comparing inputs, outputs, and facility requirements of an alternative to the base condition in the absence of the alternative. Although no single measure of environmental impact can be obtained (such as the benefit–cost ratio for economic analysis), environmental assessment of each alternative allows for a detailed comparison of potential impacts, which allows for more informed decision making. Step 7: Evaluation and recommendations. Evaluation is the process of analyzing alternative plans and comparing their beneficial and adverse contributions for the purpose of recommending a plan. A simple approach to evaluation proceeds with the following steps:

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1. Identify the issues and objectives to which each alternative is directed. 2. Determine the positive and negative impacts of the alternatives, using public input as well as professional judgment. 3. Display the results of the evaluation so that decision makers know how each alternative relates to local, regional, state, and national issues and policies. This display could include trade-offs and choices for each alternative.

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Specific criteria used by the U.S. Army Corps of Engineers that are useful in evaluating plans and reducing the number of alternatives are: ⊲ Acceptability: Assess the workability and viability of a plan in terms of its acceptance by affected parties and its accommodation of known institutional constraints. ⊲ Effectiveness: Appraise a plan’s technical performance and contribution to planning objectives. ⊲ Efficiency: Assess the plan’s ability to meet objectives functionally and in the least costly way. ⊲ Completeness: Assess whether all necessary investments to fully attain a plan are included. ⊲ Certainty: Analyze the likelihood of the plan meeting planning objectives. ⊲ Geographic Scope: Determine if the area is large enough to fully address the problem. ⊲ Benefit–Cost Ratio: Determine the economic effectiveness of the plan. ⊲ Reversibility: Measure the capability to restore a complete project to original condition. ⊲ Stability: Analyze the sensitivity of the plan to potential future developments.

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The significance of each of the above tests in comparing plans is a matter of judgment and will vary with the type of plan being developed. Final plan selection is done by those decision makers with legal authority, based on comparisons and recommendations set forth by planners. The selection should be based on the best use of resources considering all effects, monetary and nonmonetary. Usually only one plan is selected, although the planning process may be repeated to develop new alternatives or combinations of existing alternatives. Step 8: Implementation. Implementation means carrying out a selected plan or set of recommendations. At this stage, the plan is adopted and put forward for design and construction, or for adoption of laws, policies, and management procedures. Often considerable effort goes into developing plans that are never adopted. In some cases, a plan is approved and adopted but never carried out, or set aside for years. In the latter case, care must be taken not to implement a plan that was adopted years ago unless a thorough reevaluation is done. In like fashion, a recently adopted plan that is being implemented should be continuously reviewed.

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Step 9: Surveillance and monitoring. After a water-resources plan is implemented, the project should be monitored to see how well it meets the original goals and objectives. Many water-resources projects require long-term investments, so it is likely that modifications will be required as conditions change. It is common for such modifications to be needed long before the useful life of the investment is completed. 17.3

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Economic Feasibility

Economic analysis is an essential component of evaluating alternatives in water-resources planning. In many cases, a proposed action may be politically attractive, technologically feasible, and environmentally acceptable, while economic analysis shows the action to cost more than it provides in benefits. Often complicating the economic analysis of a project, and particularly of water-resources projects, is determining the proper perspective from which to view the costs and benefits. For example, a project to develop water supplies may be viewed from the utility perspective, the ratepayer perspective, or the society perspective, and determining the benefits and costs for projects will vary according to the perspective from which the analysis is done. The perspective that is appropriate for evaluation must be determined for each project, and no one perspective is appropriate for all cases. In water-supply projects, the society perspective provides the broadest coverage of costs, including those related to the environment, but the utility perspective should always be examined to make sure the program is affordable to the utility and its ratepayers. These types of projects should be evaluated from each perspective to assure the decision makers that significant, relevant impacts have not been overlooked. Benefit–cost analysis is the commonly used procedure for economic evaluation of public projects, and its success depends on the ability to assign monetary values to social and environmental costs and benefits. Benefit–cost analysis is most suitable for ranking or comparing alternatives designed to attain the same ends, rather than for testing the absolute desirability of one project. The boundaries of a benefit–cost analysis must always be specified at the outset; these are usually political boundaries (county, state), but may also be those of a corporate entity.

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Costs. Costs to an entity developing a water-resources project are generally classified as either direct or indirect. Direct costs are those borne by the entity itself, and typically include: design costs, construction costs, real-estate costs, rights-of-way costs, capital costs for equipment, and labor costs for operating and maintaining the system. Expenditures by those who receive direct benefits from a project in order to utilize it are included as direct costs. Indirect costs are those borne by parties not directly related to the project. Costs imposed on society by environmental degradation are prime examples of indirect costs. For overall economic efficiency, all costs should be accounted for, as under the society perspective, and this requires that both direct and indirect costs be included to the fullest extent possible.

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Benefits. Benefits are classified as direct (internal), indirect (external), or intangible. Direct or internal benefits include additional revenue from such sources as water sales, power generation, recreation-use fees, impact fees from growth supported by a new water supply, additional tax revenues from growth in the service area, and direct savings of reduced potential flood damages to residences, businesses, and public infrastructure. Direct benefits may also include increases in land value because of the increase in market value of property protected from flood damages. Avoided costs coming about as a result of not having to use resources in some other way are generally counted as direct benefits. Indirect or external benefits accrue to parties not directly associated with entities responsible for developing water-resources projects; for example, creating a lake for water-supply purposes may also provide recreation benefits, habitat for wildlife, and flood-control benefits, and new businesses or residences may be established in direct response to a newly completed project. However, if a new activity is a transfer of an existing activity from another location, it should not be counted as a benefit from a national point of view except for any value added to the new location. Intangible benefits are those that cannot be quantified and are not included directly in the benefit–cost analysis. These benefits must be incorporated in some other way if they are to be part of the overall study.

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Economic Feasibility

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A concern in using benefits and costs to quantitatively assess the economics of a project is that not all of them can be assigned monetary values. For example, aesthetic values vary widely among people and cannot be assigned a true monetary value. The value of fish and wildlife, aquatic ecosystems, and vegetation cannot be assigned true monetary values, and estimates are frequently made using surrogate measures such as the commercial value of fish or trees or the value of fishing and hunting experiences. Spending on water-resources projects often consists of multiple-year capital programs, with operating and maintenance costs spread out over time, and benefits that may be even further spread out over time. To provide a common basis for cost and benefit comparisons, financial analysts commonly use the concept of present value which adjusts future amounts of costs or savings to equivalent values today. To make this conversion, a planner needs to know the earnings that are possible for investing available funds. The interest rate used is called the discount rate, and is defined as the highest rate of return that could be earned by investing available funds with the same level of risk. The discount rate should reflect either the best alternative use of the funds available or the cost of capital, which for most public agencies is equivalent to the interest rate on long-term debt. For U.S. federal government projects, the Office of Management and Budget (OMB) provides guidelines for choosing the discount rates, based on interest rates for treasury notes and bonds, with maturities that can be matched to project lives. From a water-supply utility perspective, the discount rate would likely be based on long-term tax-exempt bonds of the utility. An additional consideration in choosing a discount rate is whether it should be the nominal or real discount rate. The nominal discount rate reflects current market conditions, taking into account current inflationary impacts of money. Market rates are nominal rates. The real discount rate is the nominal rate less expected inflation, which erodes the purchasing power of money over time. Since the real discount rate is an inflationary adjusted value, it is sometimes referred to as the constant-dollar rate. The choice of real or nominal discount rates depends on whether costs and benefits are evaluated based on a real (constant-dollar) or nominal basis. Many federal water-resources project planners, the U.S. Army Corps of Engineers, and the U.S. Bureau of Reclamation have been criticized for using low discount rates. The actual figure is set by legislation enacted by the U.S. Congress; therefore, all U.S. agencies are consistent in their selection of a discount rate. The economic life of a project ends when the incremental benefit from continued use no longer exceeds the cost of continued operation. The period of economic (benefit–cost) analysis should not exceed the economic life, and the same period of analysis must be adopted for all alternatives, even if their economic lives differ.

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Compound-Interest Factors

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In order to use one of the discounting techniques, alternative discounting factors must be considered to convert cash flows to a single number. Variables used in these analyses include the present value (P), future value (F), uniform annual value (A), interest rate (i), and number of payment periods (typically years or months) denoted by n. 17.3.1.1

Single-payment factors

The single-payment compound amount factor gives the amount that will have accumulated after n years per unit initial investment at a return rate of i percent per year. Common notation for this factor is (F/P, i, n), which indicates a future value, F, of a present amount, P, invested at i percent for n years or payment periods. Based on this definition, F = P(1 + i)n

(17.1)

and 

 F F , i, n = (1 + i)n = P P

(17.2)

The single-payment present-worth factor is the inverse of the single-payment compound amount factor (Equation 17.2), and gives the amount that must be invested initially at i

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Chapter 17

Water-Resources Planning

percent in order to have unit (e.g., one dollar) return at the end of n years or other payment periods. The notation for this factor is (P/F, i, n), which indicates a present value, P, which must be invested at i percent for n years or time periods in order to yield a future value, F. Based on this definition, Equation 17.1 gives 

 1 P P , i, n = = n F (1 + i) F

(17.3)

This factor is important when finding the present value of future costs or benefits that appear as discrete values. 17.3.1.2 Uniform-series factors

ww

In many cases, costs and benefits occur in uniform amounts year after year, and a better alternative is to use uniform annual series factors to show the present or future values of equal annual costs and/or benefits. The sinking-fund factor gives the uniform amount, A, that must be invested at i percent at the end of each of n years to accumulate a unit (e.g., one dollar) return. Based on this definition, the sinking-fund factor (A/F, i, n) is given by

w .E asy En g 

 i A A , i, n = = n F (1 + i) − 1 F

(17.4)

The uniform-series compound-amount factor, which is the reciprocal of the sinking-fund factor, gives the amount that will accumulate from unit (e.g., one dollar) investments at the end of each of n years at i percent. Based on this definition, the uniform-series compound-amount factor (F/A, i, n) is given by 

ine eri n

 F (1 + i)n − 1 F , i, n = = A i A

(17.5)

g .n

The capital-recovery factor is the annual value that, after n years, will yield the equivalent of a unit (e.g., one dollar) initial investment at i percent. Based on this definition, the capitalrecovery factor (A/P, i, n) is given by 

 A i(1 + i)n A , i, n = = P (1 + i)n − 1 P

et

(17.6)

The uniform-series present-worth factor, which is the reciprocal of the capital-recovery factor, gives the present value (or the amount that must be invested initially) at i percent to yield a unit (e.g., one dollar) return at the end of each of n years. Based on this definition, the uniform-series present-worth factor (P/A, i, n) is given by 

 P P (1 + i)n − 1 = , i, n = n A i(1 + i) A

(17.7)

17.3.1.3 Arithmetic-gradient factors The arithmetic-gradient present-worth factor gives the amount that must be invested initially at i percent in order to obtain an incremental unit (e.g., one dollar) return at the end of the second year, an incremental two-unit (e.g., two dollars) return at the end of the third year,

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and continuing to an incremental n − 1 unit (e.g., n − 1 dollars) at the end of the nth year. Based on this definition, the arithmetic-gradient present-worth factor (P/G, i, n) is given by 

 (1 + i)n − in − 1 P P , i, n = = 2 n G G i (1 + i)

(17.8)

where G is the increment in investment return from one year to the next. To determine the present worth of a uniformly decreasing series, subtract a uniformly increasing series from a uniform annual series. The arithmetic-gradient uniform-series factor (A/G, i, n) can be derived by multiplying the A/P and P/G factors and is given by 

ww

 1 n A A , i, n = − = G i (1 + i)n − 1 G

(17.9)

The arithmetic-gradient future-worth factor (F/G, i, n) can be derived by multiplying the P/G and F/P factors and is given by 

   F 1 (1 + i)n − 1 A , i, n = − n = G i i G

w .E asy En g

(17.10)

In cases where the initial return at the end of the first year is A0 and the annual return increases by an increment G in subsequent years, the present worth of these is calculated by adding the present worth of the annual series A0 over n years to the present worth of the uniform-gradient series with annual increments of G. 17.3.1.4 Geometric-gradient factors

ine eri n

In cases where investment returns or expenditures change from period to period by a constant percentage, this defines a geometric-gradient series. To characterize a geometricgradient series, it is convenient to define the parameter, g, as the constant rate of change, in decimal form, by which amounts increase or decrease from one time period to the next. Defining P as the present worth of the entire cash-flow series, and A0 as the initial cost or revenue, the geometric-gradient present-worth factor (P/A0 , g, i, n) is given by (Blank and Tarquin, 2002) ⎧   ⎪ 1+g n ⎪ 1 − ⎪ ⎪   1+i ⎨ P P , g Z i (17.11) , g, i, n = = i − g ⎪ A0 A0 ⎪ n ⎪ ⎪ , g=i ⎩ 1 + i

g .n

et

Corresponding factors for equivalent A and F values can be derived; however, it is easier to determine P and then multiply by A/P or F/P factors to determine the required information. EXAMPLE 17.1

A project to develop the water-supply infrastructure in western Montana is being considered, and four alternatives have been proposed. All alternatives have a 20-year design life, and projected economic conditions indicate a 6% interest rate should be used in comparing alternatives. The first alternative will lead to a lump-sum return of $100,000 at the end of the first 10 years, and a lump-sum return of $200,000 at the end of the second 10 years. The second alternative yields annual returns of $15,000 for all 20 years of the project. The third alternative yields a return of $6000 at the end of the first year, and the return increases by $1000 per year in subsequent years. The fourth alternative yields a return of $6000 at the end of the first year, and the returns are projected to increase by 8% per year in subsequent years. Determine the equivalent present-worth return for each alternative. If all projects have approximately the same present-worth cost, which of the four alternatives provides the greatest return on investment?

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Water-Resources Planning Solution Alternative 1: This alternative produces a lump-sum return of $100,000 at the end of year 10, and $200,000 at the end of year 20. The present worth of each of these lump-sum returns can be determined using the single-payment present-worth factor (Equation 17.3), which can be put in the form   1 F P= (1 + i)n For the lump-sum return of $100,000 at the end of year 10: F = $100,000, i = 0.06, n = 10, and   1 P= ($100,000) = $55,839 (1 + 0.06)10

ww

For the lump-sum return of $200,000 at the end of year 20: F = $200,000, i = 0.06, n = 20, and   1 ($200,000) = $62,361 P= (1 + 0.06)20 Therefore, the total present worth of the lump-sum returns at the end of years 10 and 20 is $55,839 + $62,361 = $118,200.

w .E asy En g

Alternative 2: This alternative produces annual returns of $15,000 per year for all 20 years of the project. The present worth of these returns can be determined using the uniform-series presentworth factor (Equation 17.7), which can be put in the form   (1 + i)n − 1 A P= i(1 + i)n In this case, A = $15,000, i = 0.06, n = 20, and   (1 + 0.06)20 − 1 ($15,000) = $172,049 P= 0.06(1 + 0.06)20

ine eri n

Therefore, the present worth of uniform annual returns is $172,049.

Alternative 3: This alternative produces a return of $6000 at the end of the first year, with returns increasing by $1000 per year thereafter. The present worth of the annual incremental returns can be determined using the arithmetic-gradient present-worth factor (Equation 17.8), which can be put in the form   (1 + i)n − in − 1 G P= i2 (1 + i)n

g .n

In this case, G = $1000, i = 0.06, and n = 20, and   (1 + 0.06)n − (0.06)(20) − 1 ($1000) = $87,230 P= (0.06)2 (1 + 0.06)n

et

The incremental return of $1000 per year (starting in year 2) is superimposed on a uniform annual return of $6000 per year beginning at the end of year 1. The present worth of the uniform annual series is given by the uniform-series present-worth factor (Equation 17.7), which can be put in the form   (1 + i)n − 1 A P= i(1 + i)n In this case, A = $6000, i = 0.06, n = 20, and   (1 + 0.06)n − 1 ($6000) = $68,820 P= 0.06(1 + 0.06)20 Therefore, the total present worth of the uniform-gradient annual returns is $87,230 + $68,820 = $156,050.

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Alternative 4: This alternative produces a return of $6000 at the end of the first year and increases by 8% each year for the duration of the 20-year design life. The present worth of these returns can be determined using the geometric-gradient present-worth factor (Equation 17.11), which can be put in the form ⎡  ⎤  1 −

⎢ P=⎣

1+g n 1+i ⎥

i − g

⎦ A1

In this case, A1 = $6000, g = 0.08, i = 0.06, n = 20, and ⎡  ⎤  1 + 0.08 ⎢ 1 − 1 + 0.06 P=⎢ ⎣ 0.06 − 0.08

20

⎥ ⎥ ($6000) = $135,993 ⎦

Therefore, the present worth of the geometric-gradient annual returns is $135,993.

ww

Comparison of Alternatives: The present worth of the returns of the four alternative proposals are as follows: Alternative Present Worth 1 $118,200 2 $172,049 3 $156,050 4 $135,993

w .E asy En g

Based on these results, and the fact that all proposed alternatives have approximately the same cost, Alternative 2 produces the greatest return on investment.

17.3.2

Evaluating Alternatives

Four different approaches can be used to deal with comparisons of time and value in considering the economic merits of alternative plans. These approaches are: present-worth analysis, annual-worth analysis, rate-of-return analysis, and benefit–cost analysis. They are different ways of analyzing the same information. Each emphasizes a particular economic aspect, and each has its own advantages and disadvantages. However, each approach should identify the same alternative as the best choice from an economic perspective. 17.3.2.1 Present-worth analysis

ine eri n

g .n

A present-worth analysis uses the net present value (NPV) of benefits minus costs as the basis for comparing alternatives, and the alternative with the highest NPV is ranked highest, as it contributes the greatest amount to net benefits. In using present-worth analysis, annual net benefits (benefits minus costs) are calculated for each year of the project, the present value of each is determined, and the net sum of all present values is calculated. Suggested rules in performing present-worth analyses are as follows:

et

⊲ Use the same time period and discount rates for all alternatives. ⊲ Calculate the present worth of each alternative. Choose all alternatives having a positive present worth. Reject the rest. ⊲ In a set of mutually exclusive alternatives, choose the one having the greatest present worth. ⊲ If the alternatives in the set of mutually exclusive alternatives have benefits which cannot be quantified but are approximately equal, choose the alternative having the least cost. Individual future amounts are discounted using the present-worth factor (P/F, i, n); series of equal amounts over n years (periods) are discounted using the uniform-series present-worth factor (P/A, i, n); and for a nonuniform series of payments, present worth can be determined by calculating all individual present worths, or by using an arithmetic-gradient present-worth factor, if appropriate. By using net present worth, comparisons can be made of costs and benefits on an equivalent basis throughout the life of the project.

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EXAMPLE 17.2 Three alternative proposals for developing the water resources of a region are being evaluated for funding. The design life of all alternatives is 30 years and market conditions indicate an interest rate of 5%. The first alternative requires an initial investment of $100,000, produces annual revenues beginning with $25,000 in year 1 and increasing by $1000 per year, operating costs begin with $3000 in the first year and increase by $2000 per year, and the salvage value of the capital investment is $10,000 at the end of 30 years. The second alternative requires an initial investment of $50,000, produces annual revenues beginning with $10,000 in year 1 and increasing by 7% per year, operating costs begin with $5500 in the first year and increase by 8% per year, and the salvage value of the capital investment is $5000 at the end of 30 years. The third alternative requires an initial investment of $65,000, produces annual revenues of $10,000 per year and operating costs of $6000 per year, and the salvage value of the capital investment is $12,000 at the end of 30 years. Compare these alternatives using the present-worth method and identify any that are not economically feasible.

ww

Solution For each alternative, the present worth of each cost and revenue item is calculated individually, and the net present value (NPV) is calculated as the sum of the present values of the cost and revenue items as follows: Alternative 1:

w .E asy En g Item

Initial investment, I Base revenue, Ar

Formula

P=I  P=

Parameters I = −$100,000



(1 + i)n − 1 Ar i(1 + i)n



 (1 + i)n − in − 1 Revenue gradient, P= Gr i2 (1 + i)n Gr   (1 + i)n − 1 Base operating cost, P = Ac i(1 + i)n Ac   (1 + i)n − in − 1 Operating cost gra- P = Gc i2 (1 + i)n dient, Gc   1 Salvage value, F P= F (1 + i)n

−$100,000

i = 5%, n = 30, Ar = $25,000

$384,250

i = 5%, n = 30, Gr = $1000

$168,623

ine eri n

i = 5%, n = 30, Ac = −$3000

−$46,110

i = 5%, n = 30, Gc = −$2000

−$337,240

g .n

i = 5%, n = 30, F = $10,000

Net present value Alternative 2: Item Initial investment, I Revenue, Ar , g

Cost, Ac , g

Salvage value, F Net present value

Formula P=I ⎡

1 −

⎡

1 −

⎢ P=⎣ ⎢ P=⎣

Parameters

et

$71,837

Present worth −$50,000

⎥ ⎦ Ar

i = 5%, n = 30, Ar = $10,000, g = 7%

$380,652

⎦ Ac

i = 5%, n = 30, Ac = −$5500, g = 8%

−$243,540

1+g 1+i



 ⎤ 1+g n 1+i ⎥

i − g   1 P= F (1 + i)n

$2314

I = −$50,000 n ⎤



i − g

Present worth

i = 5%, n = 30, F = $5000

$1155 $88,267

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Alternative 3: Item Initial investment, I

Parameters

P=I

Present worth

I = −$65,000

−$65,000

Revenue, Ar

P=



Cost, Ac

P=



 (1 + i)n − 1 Ac i(1 + i)n

i = 5%, n = 30, Ar = −$6000

P=



 1 F (1 + i)n

i = 5%, n = 30, F = $12,000

Salvage value, F

ww

Formula

 (1 + i)n − 1 Ar i(1 + i)n

i = 5%, n = 30, Ar = $10,000

$153,724

−$92,235

Net present value

$2772 −$739

Based on these results, Alternative 2 has the greatest net present value ($88,267), and Alternative 3 is not economically feasible.

w .E asy En g 17.3.2.2 Annual-worth analysis

This method is similar to the present-worth method, but it converts present worths to equivalent uniform annual values. Annual net benefits are first calculated for each year of the project, the present value of each is determined, and the net sum of all present values is obtained. Once this value is obtained, the appropriate capital recovery factor (A/P, i, n) is applied to determine equivalent annual figures. The appropriate decision rule in this method is to select the alternative having the greatest annual benefit. Although the annual-cost method is much like the present-worth method, it is often preferred by people more accustomed to thinking of annual costs rather than present worth. EXAMPLE 17.3

ine eri n

g .n

Two proposed alternatives have present worths of $72,000 and $88,000, respectively. If the design life of these alternatives is 30 years and the interest rate is 5%, express these alternatives in terms of annual worths.

et

Solution The present worth, P, is related to annual worth, A, by the capital recovery factor (Equation 17.6), which can be put in the form   i(1 + i)n P A= (1 + i)n − 1 In this case, i = 5% = 0.05, and n = 30, which yields   0.05(1 + 0.05)30 P = 0.0651P A= (1 + 0.05)30 − 1 For P = $72,000, A = 0.0651($72,000) = $4684; and for P = $88,000, A = 0.0651($88,000) = $5725. Hence, the annual worths (= annual net benefits) of the proposed alternatives are $4684 and $5725, respectively.

17.3.2.3 Rate-of-return analysis This approach identifies the rate of return at which the net present worth of all benefits and costs over the project life is equal to zero. The rate of return is also called the internal rate of return, return on investment, and the profitability index. Rate-of-return analysis provides

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Water-Resources Planning

for a direct comparison between the earning power, or return from the proposed investment, and alternative forms of investment. Suggested rules in rate-of-return analysis are as follows: • Compare all alternatives over the same period of analysis. • Calculate the rate of return for each alternative. Choose all alternatives having a rate of return exceeding the minimum acceptable value. Reject the rest. • Rank the alternatives in the set of mutually exclusive alternatives in order of increasing cost. Calculate the rate of return on the incremental cost and incremental benefits of the next alternative above the least costly alternative. Choose the more costly alternative if the incremental rate of return exceeds the minimum acceptable discount rate. Otherwise choose the less costly alternative. Continue the analysis by considering the alternatives in order of increased costliness, the alternative on the less costly side of each increment being the most costly project chosen thus far.

ww

The incremental procedure described above must be used in place of choosing the mutually exclusive alternative having the highest rate of return in order to have the same decisions as provided by a present-worth analysis. If the rate of return of a proposed investment exceeds the minimum attractive rate of return, the project is considered to be economically feasible. Typically, a target rate of return is determined based on some alternative investment opportunity. Rate-of-return analysis is used less than other methods because it requires prior calculation of net present worth, and it must be used with caution in comparing alternatives. Rate-of-return analysis provides a means of screening projects for economic feasibility but should not be used to rank projects for implementation.

w .E asy En g EXAMPLE 17.4

Three alternative proposals to develop the water resources of a region are being reviewed. All proposals require an initial capital investment, generate revenues that increase at a constant rate, have operating costs that increase at a constant rate, and have a residual value at the end of the 20-year design life. These costs and revenues are summarized in the following table

ine eri n

Alternative

Initial cost

First-year revenue

Revenue growth rate

First-year operating cost

Operating-cost growth rate

Residual value

1 2 3

$100,000 $50,000 $75,000

$15,000 $10,000 $12,000

5% 4% 3%

$5000 $3000 $7500

g .n

$5000 $1000 $1500

6% 5% 5%

et

If the minimum attractive rate of return on the investment is 8%, identify the alternatives that are economically justified. What is the preferred alternative? Solution All proposed alternatives are described by an initial cost, I, first-year revenue, R0 , revenue growth rate, gr , first-year cost, C0 , cost growth rate, gc , residual value, F, and design life, n, equal to 20 years. The return rate, i∗ , for each alternative satisfies the relation

−I +





   P P P ∗ ∗ ∗ ,i ,n F = 0 , gr , i , n R0 − , gc , i , n C0 + R0 C0 F

(17.12)

where (P/A0 , g, i, n) is the geometric-gradient present-worth factor given by Equation 17.11, and (P/F, i, n) is the single-payment present-worth factor given by Equation 17.3. Combining Equations 17.3, 17.11, and 17.12 gives ⎡

⎢ −I + ⎣

1 −



 ⎤ 1 + gr n ∗ 1+i ⎥

i∗ − gr

⎡

⎢ ⎦ R0 − ⎣

1 −



 ⎤ 1 + gc n ∗ 1+i ⎥

i∗ − gc

⎦ Co +



 1 F=0 (1 + i∗ )n

(17.13)

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Substituting given values of I, R0 , gr , C0 , gc , and F for each of the proposed alternatives yields the following results:

ww

Alternative

I

R0

gr

C0

gc

F

i∗

1 2 3

$100,000 $50,000 $75,000

$15,000 $10,000 $12,000

5% 4% 3%

$5000 $3000 $7500

6% 5% 5%

$5000 $1000 $1500

12% 16% 0.1%

Since the minimum attractive rate of return is 8%, only alternatives 1 and 2 are economically feasible and deserve further consideration. To compare alternatives, they must first be ranked by cost, starting with the lower-cost alternative. Alternative 2 has the lower cost of the two viable alternatives and will be ranked number one, and Alternative 1 has the next-lowest cost and will be ranked number two. The next step is to calculate the incremental cash flow between the number-two ranked alternative (Alternative 1) and the number-one ranked alternative (Alternative 2), and these increments are given in the following table: Year

Alternative 2

Alternative 1

Increment

0 i = 1, 20 i = 1, 20 20

−$50,000 R02 (1 + gr2 )i−1 −C02 (1 + gc2 )i−1 $1000

−$100,000 R01 (1 + gr1 )i−1 −C01 (1 + gc1 )i−1 $5000

−$50,000 R01 (1 + gr1 )i−1 − R02 (1 + gr2 )i−1 −C01 (1 + gc1 )i−1 + C02 (1 + gc2 )i−1 $4000

w .E asy En g

where R01 and R02 are the initial revenues for Alternatives 1 and 2, respectively, and gr1 and gr2 are the corresponding growth rates; C01 and C02 are the initial operating costs for Alternatives 1 and 2, respectively, and gc1 and gc2 are the corresponding growth rates. The next step is to calculate the rate of return, i∗ , of the incremental costs, which requires that 





P P P ∗ ∗ ∗ , g , i , n R01 − , g , i , n R02 − , g , i , n C01 −50,000 + R01 r1 R02 r2 C01 c1

   P P ∗ , i , n (4000) = 0 + , gc2 , i∗ , n C02 + C02 F

ine eri n

g .n

which can be put in the form ⎡ ⎡ ⎡    n ⎤ n ⎤ n ⎤ 1+g 1+g 1+g 1 − 1 + ir1 1 − 1 + ic1 1 − 1 + ir2 ∗ ∗ ∗ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ −50,000 + ⎣ ⎦ R02 − ⎣ ⎦ R01 − ⎣ ⎦ C01 i∗ − gr1 i∗ − gr2 i∗ − gc1 ⎡

⎢ +⎣

1 −



 ⎤ 1 + gc2 n ∗ 1+i ⎥

i∗ − gc2

⎦ C02 +



et

 1 (4000) = 0 (1 + i∗ )n

Substituting gr1 = 0.05, R01 = $15,000, gr2 = 0.04, R02 = $10,000, gc1 = 0.06, C01 = $5000, gc2 = 0.05, and C02 = $3000, and n = 20 gives ⎡ ⎡ ⎡  ⎤  ⎤  ⎤    1 + 0.04 20 1 + 0.06 20 1 + 0.05 20 1 − 1 − ∗ ∗ ∗ ⎢ ⎢ ⎥ ⎥ ⎥ ⎢1 − 1+i 1+i 1+i ⎥ (15,000) − ⎢ ⎥ (10,000) − ⎢ ⎥ (5000) −50,000 + ⎢ ⎣ ⎣ ⎦ ⎦ ⎦ ⎣ ∗ ∗ ∗ i − 0.05 i − 0.04 i − 0.06 ⎡

 ⎤    1 + 0.05 20 1 − ⎢ ⎥ 1 1 + i∗ ⎢ ⎥ +⎣ ⎦ (3000) + (1 + i∗ )20 (4000) = 0 i∗ − 0.05

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Water-Resources Planning which yields i∗ = 0.078 Since the rate of return of the incremental costs (7.8%) is less than the minimum attractive rate of return (8%), the additional investment returns generated by selecting Alternative 1 over Alternative 2 are not justified. Therefore, Alternative 2 is preferred. If more alternatives are to be considered, then the incremental costs and associated rate of return relative to the currently preferred alternative (Alternative 2) must be considered.

17.3.2.4

ww

Benefit–cost analysis

This approach is based on the premise that the ratio of benefits to costs must exceed 1.0 for a project to be considered economically feasible. Public-investment analysis is commonly done using the benefit–cost ratio method, particularly by U.S. government agencies. All benefits and costs are identified individually, with appropriate monetary values assigned and time periods determined, then all benefits and costs are brought to a comparable time (using present-worth analysis), and the benefit–cost ratio computed at this time. Benefits are viewed as all the positive returns from a project, regardless of who benefits, while costs are measured as the outlays made by the project sponsors as well as losses suffered by groups directly affected by the project. The benefit–cost ratio can have either of the following forms:

w .E asy En g B − D C

or

B C + D

(17.14)

where B is the benefit ($), C is the cost ($), and D measures the disbenefits, hardships, and losses caused by the project ($). The alternative benefit–cost ratios in Equation 17.14 give slightly different results for the same set of numbers. Benefit–cost ratios can be based on present values (AWWA, 2001), annual values (Prakash, 2004), or future values (Blank and Tarquin, 2002); the present-value approach is more common. It is very important to note that the benefit–cost ratio provides information about the economic feasibility of the project and the efficiency of allocating monetary resources, but benefit–cost ratios should not be used to develop a ranking of feasible projects; for that purpose the net-present-value approach is preferable. EXAMPLE 17.5

ine eri n

g .n

Two proposals are being considered for funding a public-works project with a design life of 20 years. The first proposal requires an initial capital cost of $400,000 and will produce annual net benefits of $35,000 for the life of the project. The second proposal will require an initial capital cost of $600,000 and produce annual net benefits of $60,000, but funding it will require that spending on other public-works projects be reduced by $45,000 per year for the first 5 years of the project. Using an interest rate of 4%, assess the economic feasibility of each of the proposed projects.

et

Solution From the given data: the design life, n, is 20 years, and the interest rate, i, to be used in the analysis is 4%. Considering the first proposal, the present value of the cost, C, is $400,000, the annual benefit, A, is $35,000, and the present value of the benefits, B, is given by       (1 + i)n − 1 (1 + 0.04)20 − 1 P , i, n A = A= ($35,000) = $475,661 B= A i(1 + i)n 0.04(1 + 0.04)20 Therefore the benefit–cost ratio, B/C, is given by $475,661 B = = 1.19 C $400,000 Since the benefit–cost ratio (1.19) is greater than one, the proposed project is economically feasible. Considering the second proposal, the present value of the cost, C, is $600,000, the annual benefit, A, is $60,000, and the present value of the benefits, B, is given by       (1 + i)n − 1 (1 + 0.04)20 − 1 P , i, n A = B= A= ($60,000) = $815,419 A i(1 + i)n 0.04(1 + 0.04)20

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829

The disbenefit to the second proposal is that annual spending on other projects, AD , is reduced by $45,000 for the first 5 years of the project, so the present value of the disbenefit, D, is given by       (1 + 0.04)5 − 1 (1 + i)n − 1 P D= ($45,000) = $200,332 AD = , i, n AD = A i(1 + i)n 0.04(1 + 0.04)5 The benefit–cost ratio of the second proposal, taking the disbenefit into account, is given by $815,419 − $200,332 B − D = = 1.03 C $600,000

or

B $815,419 = = 1.02 C + D $600,000 + $200,332 Since the benefit–cost ratio (1.02) is greater than one, this second proposal is economically feasible. Both proposed projects are economically feasible, and present-worth analysis must be done to determine their ranking.

ww

Problems

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17.1. Derive the expression for the sinking-fund factor given by Equation 17.4. 17.2. Derive the expression for the capital-recovery factor given by Equation 17.6. 17.3. Derive the expression for the arithmetic-gradient present-worth factor given by Equation 17.8. 17.4. Derive the expression for the geometric-gradient present-worth factor given by Equation 17.11. 17.5. Two alternatives for developing the water resources of a region are being considered. The design life of both alternatives is 30 years, and economic conditions indicate a 4% interest rate should be used in comparing alternatives. The first alternative will yield annual returns of $50,000 for all 30 years of the project, while the second alternative yields a return of $17,000 at the end of the first year, and the returns are projected to increase by 6% per year in subsequent years. Determine the equivalent present-worth return for each alternative. 17.6. Two alternative proposals are being evaluated. The design life of both alternatives is 20 years, and an interest rate of 4% can be assumed. The first alternative requires an initial investment of $200,000 and produces annual revenues beginning with $50,000 in year 1 and increasing by $2000 per year, operating costs begin with $6000 in the first year and increase by $4000 per year; and the salvage value of the capital investment is $20,000 at the end of 20 years. The second alternative requires an initial investment of $100,000 and produces annual revenues beginning with $20,000 in year 1 and increasing by 6% per year; operating costs begin with $11,000 in the

first year and increase by 6% per year; and the salvage value of the capital investment is $10,000 at the end of 20 years. Compare these alternatives using the presentworth method. 17.7. Two alternative proposals to develop the water resources of a region are being reviewed. Both require an initial capital investment, generate revenues that increase at a constant rate, have operating costs that increase at a constant rate, and have a design life of 15 years. These projects have no residual value at the end of their design life, and the alternatives are summarized in the following table:

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Initial cost

First- OperatingFirst- Revenue year cost year growth operating growth revenue rate cost rate

$130,000 $16,000 $75,000 $12,000

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4.5% 3.5%

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$6000 $3200

5.0% 3.5%

If the minimum attractive rate of return on the investment is 6%, what is the preferred alternative? 17.8. A proposal for a project with a 25-year design life will require an initial capital cost of $800,000 and produce annual net benefits of $80,000. Funding this project will require that spending on other projects be reduced by $55,000 per year for the first 8 years of the project. Using an interest rate of 5%, determine the benefit–cost ratio of the project.

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A P P E N D I X

A

Units and Conversion Factors A.1 Units ` ´ (International System of Units or SI system) was adopted The Systeme International d’Unites by the 11th General Conference on Weights and Measures (CGPM) in 1960 and is now used by almost the entire world. In the SI system, all quantities are expressed in terms of seven base (fundamental) units. These base units and their standard abbreviations are as follows:

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Meter (m): Distance light travels in a vacuum during 1/299 792 458 of a second.∗ Kilogram (kg): Mass of a cylinder of platinum–iridium alloy kept in Paris.

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Second (s): Duration of 9 192 631 770 cycles of the radiation corresponding to the transition between two hyper fine levels of the ground state of the cesium-133 atom. Ampere (A): Magnitude of the current that, when flowing through each of two long parallel wires of negligible cross section separated by one meter in a vacuum, results in a force between the two wires of 2 * 10−7 newtons for each meter of length. Kelvin (K): Defined in the thermodynamic scale by assigning 273.16 K to the triple point of water (freezing point, 273.16K = 0◦ C).

Candela (cd): Luminous intensity of 1/600 000 of a square meter of a radiating cavity at the temperature of freezing platinum (2042 K).

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Mole (mol): Amount of substance which contains as many specified entities (molecules, atoms, ions, electrons, photons, etc.) as there are atoms in exactly 0.012 kg of carbon-12. In addition to the seven base units of the SI system, there are two supplementary SI units: the radian and the steradian. The radian (rad) is defined as the angle at the center of a circle subtended by an arc equal in length to the radius, and the steradian (sr) is defined as the solid angle with its vertex at the center of a sphere that is subtended by an area of the spherical surface equal to the radius squared. The SI units should not be confused with the now obsolete metric units, which were developed in Napoleonic France approximately 200 years ago. The primary difference between metric and SI units is that the former uses centimeters and grams to measure length and mass, while these quantities are measured in meters and kilograms in SI units. The United States is gradually moving toward the use of SI units; however there is still widespread use of the “English” system of units, which are more appropriately referred to as “U.S. Customary” units. In addition to the seven SI base units, there are several derived units that are given special names. These derived units are used for convenience rather than necessity, and are listed in Table A.1. When units are named after people, like the newton (N), joule (J), and pascal (Pa), they are capitalized when abbreviated but not capitalized when spelled out. The abbreviation capital L for liter is a special case, used to avoid confusion with one (1). In the SI system the unit of absolute temperature is the degree kelvin, which is abbreviated K without the degree symbol. The units of second, minute, hour, day, and year are correctly abbreviated as s, min, h, d, and y.

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∗ “Meter” is the accepted spelling in the United States of America; the rest of the world uses the spelling “Metre.”

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Appendix A

Units and Conversion Factors TABLE A.1: SI Derived Units

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Unit name

Quantity

becquerel coulomb degree Celsius farad gray henry hertz joule lumen lux newton ohm pascal siemens∗ sievert telsa volt watt weber

Activity of a radionuclide Quantity of electricity, electric charge Celsius temperature Capacitance Absorbed dose of ionizing radiation Inductance Frequency Energy, work, quantity of heat Luminous flux Illuminance Force Electric resistance Pressure, stress Conductance Dose equivalent of ionizing radiation Magnetic flux density Electric potential, potential difference Power, radiant flux Magnetic flux

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Symbol

In terms of base units

Bq C ◦C F Gy H Hz J lm lx N  Pa S Sv T V W Wb

s−1 A·s K C·V−1 J·kg−1 Wb·A−1 s−1 N·m cd·sr lm·m−2 kg·m·s−2 V·A−1 N·m−2 A·V−1 J·kg−1 Wb·m−2 W·A−1 J·s−1 V·s

∗ The siemens was previously called the mho.

In using prefixes with SI units, multiples of 103 are preferred in engineering usage, with other multiples such as cm avoided if possible. It is conventional practice to separate sequences of digits into groups of three by spaces rather than commas. A.2 Conversion Factors

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In most cases, application of unit conversion factors results in converted numbers that have more significant digits than the original numbers. In these cases, the converted number should be rounded off such that rounding error is consistent with the rounding error of the converted number.

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EXAMPLE A.1

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The height of a water-control structure is reported as 19.3 feet. Convert this dimension to meters. Solution The conversion factor is given in Table A.2 as 1 foot = 0.3048 meters, hence, 19.3 ft = 19.3 * 0.3048 = 5.88264 m Since 19.3 ft could have resulted in rounding any number between 19.25 ft and 19.35 ft, the maximum possible rounding error is ;0.05/19.3 = ;0.26%. Similarly, rounding 5.88264 m to 5.88 m gives a maximum rounding error of ;0.005/5.88 = ;0.085%, and rounding to 5.9 m gives an error of ;0.05/5.9 = ;0.85%. Hence, accuracy is lost by taking 19.3 ft as 5.9 m, while 5.88 m is more accurate than indicated by 19.3 ft. It is usually prudent not to discard accuracy, so take 19.3 ft = 5.88 m.

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Conversion Factors

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TABLE A.2: Multiplicative Factors for Unit Conversion

Quantity

Convert From

Convert To

Area

ac mi2

ha km2

Energy

Btu cal

J J

1054.350264 4.184∗

Energy/Area

ly†

kJ/m2

41.84∗

cfs gpm mgd†

m3 /s L/s m3 /s m3 /d L/s

lbf

N

Flow rate

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Force

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Mass

Permeability

Power

Pressure

Speed

Multiply By 0.404687 2.59000

0.02831685 0.06309 0.04381 3785.412 43.81 4.4482216152605∗

ft in. mi (U.S. statute) mi (U.S. nautical) yd

m m km km m

0.3048∗ 0.0254∗ 1.609344∗ 1.852000∗ 0.9144∗

g lbm

kg kg

0.001∗ 0.45359237∗

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darcy

m2

hp

W

745.69987

atm bar mm Hg (at 0◦ C) psi torr

kPa kPa kPa kPa kPa

101.325∗ 100.000∗ 0.133322 6.894757 0.133322

knot

m/s

0.514444444 0.001∗

Viscosity (dynamic)

cp

Pa·s

Viscosity (kinematic)

cs

m2 /s

Volume

gal

L

0.987 * 10−12

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10−6∗ 3.785411784∗

∗ Exact conversion

† ly K langley, mgd K million gallons per day.

A good rule of thumb is that the converted number should have the same number of significant digits as the original number, assuming that the conversion factor is always more accurate than the original number.

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A P P E N D I X

B

Fluid Properties B.1 Water TABLE B.1: Properties of Water

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Temperature (◦ C) 0 5 10 15 20 25 30 40 50 60 70 80 90 100

Density (kg/m3 )

Dynamic viscosity (mPa·s)

Heat of vaporization (MJ/kg)

Saturation vapor pressure (kPa)

Surface tension (mN/m)

999.8 1000.0 999.7 999.1 998.2 997.0 995.7 992.2 988.0 983.2 977.8 971.8 965.3 958.4

1.781 1.518 1.307 1.139 1.002 0.890 0.798 0.653 0.547 0.466 0.404 0.354 0.315 0.282

2.499 2.487 2.476 2.464 2.452 2.440 2.428 2.405 2.381 2.356 2.332 2.307 2.282 2.256

0.611 0.872 1.227 1.704 2.337 3.167 4.243 7.378 12.340 19.926 31.169 47.367 70.113 101.325

75.6 74.9 74.2 73.5 72.8 72.0 71.2 69.6 67.9 66.2 64.4 62.6 60.8 58.9

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Bulk modulus (106 kPa) 2.02 2.06 2.10 2.14 2.18 2.22 2.25 2.28 2.29 2.28 2.25 2.20 2.14 2.07

The properties given in Table B.1 are for pure water. Pure water seldom exists in nature, where the density of water can be significantly influenced by salinity, temperature, and possibly other properties through an equation of state. The general dependence of water density on temperature has been found to be approximately parabolic, with a maximum at 4◦ C. However, the temperature corresponding to the maximum density of water changes with increasing salinity, decreasing to about 0◦ C for highly saline systems. To a first-order approximation, density is linearly dependent on salinity over much of the normal range of interest. B.2 Organic Compounds Found in Water

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The properties of several organic compounds commonly found in contaminated waters are given in Table B.2. Substances with boiling points at or below 100◦ C are classified as volatile organic compounds (VOCs). The chemical properties given in Table B.2 are those most frequently used in the quantitative analysis of the fate of organic contaminants in the environment.

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Compound

TABLE B.2: Properties of Organic Compounds Commonly Found in Water

Formula

Molecular weight (g/mol)

Boiling point (◦ C)

Density @ 20◦ C (kg/m3 )

Dynamic viscosity @ 20◦ C (mPa·s)

Solubility in water @ 20◦ C (mg/L)

Sorption coefficient, log Koc (log mL/g)

Saturation vapor pressure @ 20◦ C (kPa)

Henry’s law constant @ 20◦ C (Pa m3 /mol)

CH3 COCH3 C6 H6 C24 H38 O4 C6 H5 Cl CH3 CH2 Cl CHCl3 C2 H4 Cl2 C2 H4 Cl2 C8 H10 — (CH3 )COCH3 CH2 Cl2 C10 H8 C6 H6 O CCl2 CCl2 C6 H5 CH3 CCl3 CH3 C2 HCl3 CH2 CHCl C6 H4 (CH3 )2 C6 H4 (CH3 )2

58.08 78.11 390.57 112.56 64.5 119.4 99.0 99.0 106.2 — 88.2 84.9 128.2 94.1 165.8 92.1 133.4 131.5 62.5 106.2 106.2

56.1j –56.2b 80.1b 230r –387r 132l — 62l 57.3l 83.5l — — — — — — 121.4l — — 86.7j — — —

790b 870h –880b 985b 1106b 897b –920i 1480c –1490h 1174i –1180c 1235b –1253b 867b –870c 680s –750f 740k 1327b –1336b 1030c –1162b 1058b –1070b 1623b –1630m 867b –870c 1339b –1350i 1460c –1460m 910c –912b 880b 861b –881b

— — — 0.8m — 0.56m 0.5m 0.84m — 0.29s –0.31f — — — — 0.9m — 0.84m 0.57m — — —

440,600a 1710b –1796b 0.041b –0.285h 466h –500m 5710b –5740d 8000i –8200b 5100m –5500b 8000d –8800b 150i –152c — 48,000h 13,000c –20,000h 31.7h –38b 93,000d 149b –200d 500i –535d 480b –4400d 1000c –1100m 90c –267h 152b –175c 160b

–0.59b –0.34h 1.39b –2.95b 3.77h –5.15b 1.92b –2.63d 0.51b –1.57d 1.46b –1.94b 1.48b –1.80d 1.15h –1.57d 1.98b –3.13d — 0.55k –1.05h 0.94b 2.62b –5.00b 1.15h –3.49b 2.25b –2.99b 1.57b –3.05b 2.02b –3.40b 1.66b –2.83b 0.39b –1.76h 1.92b –2.92h 2.31b –2.72b

3.07g –35.97h 1.20g –12.67h 1.09*10−9b –26.7*10−9b 1.17l –1.2b 133.3i –133.7h 20.2h –25.9m 23.9l –29.5m 8.13i –10.9m 0.911h –1.28g 55.2s 32.7h 48.2h 0.0113g –0.0307h 0.027p –0.045h 1.87q –2.53g 2.93p –3.75h 13.3i –16.6m 7.70h –10.0m 355h –400g 0.667p –0.912h —

3.3a –10t 458b –557p 0.0011a 263p –288b 1030n 278l –486p 435l –550m 92p –152m 559n –882p — 64.3o 229b 56.0o 0.030 1327l –1763m 529n –578p 1337l –1824p 722l –1013p 2200n 409n –537p 555n

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Acetone (dimethyl ketone) Benzene Bis(2-ethylhexyl)phthalate (DEHP) Chlorobenzene Chloroethane Chloroform (Trichloromethane) 1,1-Dichloroethane 1,2-Dichloroethane Ethlybenzene Gasolinee Methyl tertiary-butyl ether (MTBE) Methylene chloride (Dichloromethane) Naphthalene Phenol Tetrachloroethene (PCE) Toluene (Methylbenzene) 1,1,1-Trichloroethane Trichloroethene (TCE) Vinyl chloride (Chloroethene) o-Xylene (1,2 Dimethylbenzene) p-Xylene (1,4 Dimethylbenzene)

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Sources: a Montgomery (2000) at 25◦ C; b Montgomery (2000); c Hemond and Fechner (1993); d Schnoor (1996); e Typical properties at 15.6◦ C. Properties of petroleum products vary; f Munson et al. (1990); g USEPA (1996); h Charbeneau (2000); i Fetter (1999); j Sage and Sage (2000); k U.S. Environmental Protection Agency (1998); l Pankow and Cherry (1996); m Pankow and Cherry (1996) at 25◦ C; n Rathbun (1998); o Rathbun (1998) at 25◦ C; p Jackson et al. (1985); q Nyer et al. (1991); r Montgomery (2000) at 5 mm Hg; s Finnemore and Franzini (2002) at 20◦ C; t Davis and Masten (2004).

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Appendix B

Fluid Properties

B.3 Air at Standard Atmospheric Pressure TABLE B.3: Properties of Air at Standard Atmospheric Pressure

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Temperature (◦ C)

Density (kg/m3 )

Dynamic viscosity (mPa · s)

Specific heat ratio

Speed of sound (m/s)

−40 −20 0 5 10 15 20 25 30 40 50 60 70 80 90 100 200 300 400 500 1000

1.514 1.395 1.292 1.269 1.247 1.225 1.204 1.184 1.165 1.127 1.109 1.060 1.029 0.9996 0.9721 0.9461 0.7461 0.6159 0.5243 0.4565 0.2772

0.0157 0.0163 0.0171 0.0173 0.0176 0.0180 0.0182 0.0185 0.0186 0.0187 0.0195 0.0197 0.0203 0.0207 0.0214 0.0217 0.0253 0.0298 0.0332 0.0364 0.0504

1.401 1.401 1.401 1.401 1.401 1.401 1.401 1.401 1.400 1.400 1.400 1.399 1.399 1.399 1.398 1.397 1.390 1.379 1.368 1.357 1.321

306.2 319.1 331.4 334.4 337.4 340.4 343.3 346.3 349.1 354.7 360.3 365.7 371.2 376.6 381.7 386.9 434.5 476.3 514.1 548.8 694.8

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A P P E N D I X

C

Statistical Tables C.1 Areas Under Standard Normal Curve TABLE C.1: Areas Under Standard Normal Curve

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z

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

−4.0 −3.9 −3.8 −3.7 −3.6 −3.5 −3.4 −3.3 −3.2 −3.1 −3.0 −2.9 −2.8 −2.7 −2.6 −2.5 −2.4 −2.3 −2.2 −2.1 −2.0 −1.9 −1.8 −1.7 −1.6 −1.5 −1.4 −1.3 −1.2 −1.1 −1.0 −0.9 −0.8 −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 0.0 0.1 0.2

0.0000 0.0000 0.0001 0.0001 0.0002 0.0002 0.0003 0.0005 0.0007 0.0010 0.0013 0.0019 0.0026 0.0035 0.0047 0.0062 0.0082 0.0107 0.0139 0.0179 0.0228 0.0287 0.0359 0.0446 0.0548 0.0668 0.0808 0.0968 0.1151 0.1357 0.1587 0.1841 0.2119 0.2420 0.2743 0.3085 0.3446 0.3821 0.4207 0.4602 0.5000 0.5398 0.5793

0.0000 0.0000 0.0001 0.0001 0.0002 0.0002 0.0003 0.0005 0.0007 0.0009 0.0013 0.0018 0.0025 0.0034 0.0045 0.0060 0.0080 0.0104 0.0136 0.0174 0.0222 0.0281 0.0351 0.0436 0.0537 0.0655 0.0793 0.0951 0.1131 0.1335 0.1562 0.1814 0.2090 0.2389 0.2709 0.3050 0.3409 0.3783 0.4168 0.4562 0.5040 0.5438 0.5832

0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0003 0.0005 0.0006 0.0009 0.0013 0.0018 0.0024 0.0033 0.0044 0.0059 0.0078 0.0102 0.0132 0.0170 0.0217 0.0274 0.0344 0.0427 0.0526 0.0643 0.0778 0.0934 0.1112 0.1314 0.1539 0.1788 0.2061 0.2358 0.2676 0.3015 0.3372 0.3745 0.4129 0.4522 0.5080 0.5478 0.5871

0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0003 0.0004 0.0006 0.0009 0.0012 0.0017 0.0023 0.0032 0.0043 0.0057 0.0075 0.0099 0.0129 0.0166 0.0212 0.0268 0.0336 0.0418 0.0516 0.0630 0.0764 0.0918 0.1093 0.1292 0.1515 0.1762 0.2033 0.2327 0.2643 0.2981 0.3336 0.3707 0.4090 0.4483 0.5120 0.5517 0.5910

0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0003 0.0004 0.0006 0.0008 0.0012 0.0016 0.0023 0.0031 0.0041 0.0055 0.0073 0.0096 0.0125 0.0162 0.0207 0.0262 0.0329 0.0409 0.0505 0.0618 0.0749 0.0901 0.1075 0.1271 0.1492 0.1736 0.2005 0.2297 0.2611 0.2946 0.3300 0.3669 0.4052 0.4443 0.5160 0.5557 0.5948

0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0003 0.0004 0.0006 0.0008 0.0011 0.0016 0.0022 0.0030 0.0040 0.0054 0.0071 0.0094 0.0122 0.0158 0.0202 0.0256 0.0322 0.0401 0.0495 0.0606 0.0735 0.0885 0.1056 0.1251 0.1469 0.1711 0.1977 0.2266 0.2578 0.2912 0.3264 0.3632 0.4013 0.4404 0.5199 0.5596 0.5987

0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0003 0.0004 0.0006 0.0008 0.0011 0.0015 0.0021 0.0029 0.0039 0.0052 0.0069 0.0091 0.0119 0.0154 0.0197 0.0250 0.0314 0.0392 0.0485 0.0594 0.0721 0.0869 0.1038 0.1230 0.1446 0.1685 0.1949 0.2236 0.2546 0.2877 0.3228 0.3594 0.3974 0.4364 0.5239 0.5636 0.6026

0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0003 0.0004 0.0005 0.0008 0.0011 0.0015 0.0021 0.0028 0.0038 0.0051 0.0068 0.0089 0.0116 0.0150 0.0192 0.0244 0.0307 0.0384 0.0475 0.0582 0.0708 0.0853 0.1020 0.1210 0.1423 0.1660 0.1922 0.2206 0.2514 0.2843 0.3192 0.3557 0.3936 0.4325 0.5279 0.5675 0.6064

0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0003 0.0004 0.0005 0.0007 0.0010 0.0014 0.0020 0.0027 0.0037 0.0049 0.0066 0.0087 0.0113 0.0146 0.0188 0.0239 0.0301 0.0375 0.0465 0.0571 0.0694 0.0838 0.1003 0.1190 0.1401 0.1635 0.1894 0.2177 0.2483 0.2810 0.3156 0.3520 0.3897 0.4286 0.5319 0.5714 0.6103

0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0002 0.0003 0.0005 0.0007 0.0010 0.0014 0.0019 0.0026 0.0036 0.0048 0.0064 0.0084 0.0110 0.0143 0.0183 0.0233 0.0294 0.0367 0.0455 0.0559 0.0681 0.0823 0.0985 0.1170 0.1379 0.1611 0.1867 0.2148 0.2451 0.2776 0.3121 0.3483 0.3859 0.4247 0.5359 0.5753 0.6141

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Appendix C

Statistical Tables TABLE C.1: Areas Under Standard Normal Curve (Cont’d)

z

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0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.6179 0.6554 0.6915 0.7257 0.7580 0.7881 0.8159 0.8413 0.8643 0.8849 0.9032 0.9192 0.9332 0.9452 0.9554 0.9641 0.9713 0.9772 0.9821 0.9861 0.9893 0.9918 0.9938 0.9953 0.9965 0.9974 0.9981 0.9987 0.9990 0.9993 0.9995 0.9997 0.9998 0.9998 0.9999 0.9999 1.0000 1.0000

0.6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.8438 0.8665 0.8869 0.9049 0.9207 0.9345 0.9463 0.9564 0.9649 0.9719 0.9778 0.9826 0.9864 0.9896 0.9920 0.9940 0.9955 0.9966 0.9975 0.9982 0.9987 0.9991 0.9993 0.9995 0.9997 0.9998 0.9998 0.9999 0.9999 1.0000 1.0000

0.6255 0.6628 0.6985 0.7324 0.7642 0.7939 0.8212 0.8461 0.8686 0.8888 0.9066 0.9222 0.9357 0.9474 0.9573 0.9656 0.9726 0.9783 0.9830 0.9868 0.9898 0.9922 0.9941 0.9956 0.9967 0.9976 0.9982 0.9987 0.9991 0.9994 0.9995 0.9997 0.9998 0.9999 0.9999 0.9999 1.0000 1.0000

0.6293 0.6664 0.7019 0.7357 0.7673 0.7967 0.8238 0.8485 0.8708 0.8907 0.9082 0.9236 0.9370 0.9484 0.9582 0.9664 0.9732 0.9788 0.9834 0.9871 0.9901 0.9925 0.9943 0.9957 0.9968 0.9977 0.9983 0.9988 0.9991 0.9994 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1.0000 1.0000

0.6331 0.6700 0.7054 0.7389 0.7704 0.7995 0.8264 0.8508 0.8729 0.8925 0.9099 0.9251 0.9382 0.9495 0.9591 0.9671 0.9738 0.9793 0.9838 0.9875 0.9904 0.9927 0.9945 0.9959 0.9969 0.9977 0.9984 0.9988 0.9992 0.9994 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1.0000 1.0000

0.6368 0.6736 0.7088 0.7422 0.7734 0.8023 0.8289 0.8531 0.8749 0.8944 0.9115 0.9265 0.9394 0.9505 0.9599 0.9678 0.9744 0.9798 0.9842 0.9878 0.9906 0.9929 0.9946 0.9960 0.9970 0.9978 0.9984 0.9989 0.9992 0.9994 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1.0000 1.0000

0.6406 0.6772 0.7123 0.7454 0.7764 0.8051 0.8315 0.8554 0.8770 0.8962 0.9131 0.9279 0.9406 0.9515 0.9608 0.9686 0.9750 0.9803 0.9846 0.9881 0.9909 0.9931 0.9948 0.9961 0.9971 0.9979 0.9985 0.9989 0.9992 0.9994 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1.0000 1.0000

0.6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.8577 0.8790 0.8980 0.9147 0.9292 0.9418 0.9525 0.9616 0.9693 0.9756 0.9808 0.9850 0.9884 0.9911 0.9932 0.9949 0.9962 0.9972 0.9979 0.9985 0.9989 0.9992 0.9995 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1.0000 1.0000

0.6480 0.6844 0.7190 0.7517 0.7823 0.8106 0.8365 0.8599 0.8810 0.8997 0.9162 0.9306 0.9429 0.9535 0.9625 0.9699 0.9761 0.9812 0.9854 0.9887 0.9913 0.9934 0.9951 0.9963 0.9973 0.9980 0.9986 0.9990 0.9993 0.9995 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1.0000 1.0000

0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621 0.8830 0.9015 0.9177 0.9319 0.9441 0.9545 0.9633 0.9706 0.9767 0.9817 0.9857 0.9890 0.9916 0.9936 0.9952 0.9964 0.9974 0.9981 0.9986 0.9990 0.9993 0.9995 0.9997 0.9998 0.9998 0.9999 0.9999 0.9999 1.0000 1.0000

w .E asy En g

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Downloaded From : www.EasyEngineering.net Section C.2

Frequency Factors for Pearson Type III Distribution

C.2 Frequency Factors for Pearson Type III Distribution TABLE C.2: Pearson Type III Frequency Factors

ww

Skew

2

5

3.0 2.9 2.8 2.7 2.6 2.5 2.4 2.3 2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 −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 −1.4 −1.5 −1.6 −1.7 −1.8 −1.9

−0.396 −0.390 −0.384 −0.376 −0.368 −0.360 −0.351 −0.341 −0.330 −0.319 −0.307 −0.294 −0.282 −0.268 −0.254 −0.240 −0.225 −0.210 −0.195 −0.180 −0.164 −0.148 −0.132 −0.116 −0.099 −0.083 −0.066 −0.050 −0.033 −0.017 −0.000 0.017 0.033 0.050 0.066 0.083 0.099 0.116 0.132 0.148 0.164 0.180 0.195 0.210 0.225 0.240 0.254 0.268 0.282 0.294

0.420 0.440 0.460 0.479 0.499 0.518 0.537 0.555 0.574 0.592 0.609 0.627 0.643 0.660 0.675 0.690 0.705 0.719 0.732 0.745 0.758 0.769 0.780 0.790 0.800 0.808 0.816 0.824 0.830 0.836 0.842 0.846 0.850 0.853 0.855 0.856 0.857 0.857 0.856 0.854 0.852 0.848 0.844 0.838 0.832 0.825 0.817 0.808 0.799 0.788

Return period, T 10 25 50 1.180 1.195 1.210 1.224 1.238 1.250 1.262 1.274 1.284 1.294 1.302 1.310 1.318 1.324 1.329 1.333 1.337 1.339 1.340 1.341 1.340 1.339 1.336 1.333 1.328 1.323 1.317 1.309 1.301 1.292 1.282 1.270 1.258 1.245 1.231 1.216 1.200 1.183 1.166 1.147 1.128 1.107 1.086 1.064 1.041 1.018 0.994 0.970 0.945 0.920

w .E asy En g

2.278 2.277 2.275 2.272 2.267 2.262 2.256 2.248 2.240 2.230 2.219 2.207 2.193 2.179 2.163 2.146 2.128 2.108 2.087 2.066 2.043 2.018 1.993 1.967 1.939 1.910 1.880 1.849 1.818 1.785 1.751 1.716 1.680 1.643 1.606 1.567 1.528 1.488 1.448 1.407 1.366 1.324 1.282 1.240 1.198 1.157 1.116 1.075 1.035 0.996

3.152 3.134 3.114 3.093 3.071 3.048 3.023 2.997 2.970 2.942 2.912 2.881 2.848 2.815 2.780 2.743 2.706 2.666 2.626 2.585 2.542 2.498 2.453 2.407 2.359 2.311 2.261 2.211 2.159 2.107 2.054 2.000 1.945 1.890 1.834 1.777 1.720 1.663 1.606 1.549 1.492 1.435 1.379 1.324 1.270 1.217 1.166 1.116 1.069 1.023

100

200

4.051 4.013 3.973 3.932 3.889 3.845 3.800 3.753 3.705 3.656 3.605 3.553 3.499 3.444 3.388 3.330 3.271 3.211 3.149 3.087 3.022 2.957 2.891 2.824 2.755 2.686 2.615 2.544 2.472 2.400 2.326 2.252 2.178 2.104 2.029 1.955 1.880 1.806 1.733 1.660 1.588 1.518 1.449 1.383 1.318 1.256 1.197 1.140 1.087 1.037

4.970 4.909 4.847 4.783 4.718 4.652 4.584 4.515 4.444 4.372 4.398 4.223 4.147 4.069 3.990 3.910 3.828 3.745 3.661 3.575 3.489 3.401 3.312 3.223 3.132 3.041 2.949 2.856 2.763 2.670 2.576 2.482 2.388 2.294 2.201 2.108 2.016 1.926 1.837 1.749 1.664 1.581 1.501 1.424 1.351 1.282 1.216 1.155 1.097 1.044

ine eri n

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Appendix C

Statistical Tables TABLE C.2: Pearson Type III Frequency Factors (Cont’d)

Skew −2.0 −2.1 −2.2 −2.3 −2.4 −2.5 −2.6 −2.7 −2.8 −2.9 −3.0

ww

2 0.307 0.319 0.330 0.341 0.351 0.360 0.368 0.376 0.384 0.390 0.396

5 0.777 0.765 0.752 0.739 0.725 0.711 0.696 0.681 0.666 0.651 0.636

10

Return period, T 25 50

0.895 0.869 0.844 0.819 0.795 0.771 0.747 0.724 0.702 0.681 0.660

w .E asy En g

0.959 0.923 0.888 0.855 0.823 0.793 0.764 0.738 0.712 0.683 0.666

0.980 0.939 0.900 0.864 0.830 0.798 0.768 0.740 0.714 0.689 0.666

100

200

0.990 0.946 0.905 0.867 0.832 0.799 0.769 0.740 0.714 0.690 0.667

0.995 0.949 0.907 0.869 0.833 0.800 0.769 0.741 0.714 0.690 0.667

ine eri n

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Downloaded From : www.EasyEngineering.net Section C.3

Critical Values of the Chi-Square Distribution

C.3 Critical Values of the Chi-Square Distribution TABLE C.3: Critical Values of the Chi-Square Distribution

α

ww

ν

0.995

0.990

0.975

0.950

0.050

0.025

0.010

0.005

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 50 60 70 80 90 100

0.000 0.010 0.072 0.207 0.412 0.676 0.989 1.344 1.735 2.156 2.603 3.074 3.565 4.075 4.601 5.142 5.697 6.265 6.844 7.434 8.034 8.643 9.260 9.886 10.520 11.160 11.808 12.461 13.121 13.787 20.707 27.991 35.534 43.275 51.172 59.196 67.328

0.000 0.020 0.115 0.297 0.554 0.872 1.239 1.646 2.088 2.558 3.053 3.571 4.107 4.660 5.229 5.812 6.408 7.015 7.633 8.260 8.897 9.542 10.196 10.856 11.524 12.198 12.879 13.565 14.256 14.953 22.164 29.707 37.485 45.442 53.540 61.754 70.065

0.001 0.051 0.216 0.484 0.831 1.237 1.690 2.180 2.700 3.247 3.816 4.404 5.009 5.629 6.262 6.908 7.564 8.231 8.907 9.591 10.283 10.982 11.689 12.401 13.120 13.844 14.573 15.308 16.047 16.791 24.433 32.357 40.482 48.758 57.153 65.647 74.222

0.004 0.103 0.352 0.711 1.145 1.635 2.167 2.733 3.325 3.940 4.575 5.226 5.892 6.571 7.261 7.962 8.672 9.390 10.117 10.851 11.591 12.338 13.091 13.848 14.611 15.379 16.151 16.928 17.708 18.493 26.509 34.764 43.188 51.739 60.391 69.126 77.929

3.841 5.991 7.815 9.488 11.071 12.592 14.067 15.507 16.919 18.307 19.675 21.026 22.362 23.685 24.996 26.296 27.587 28.869 30.144 31.410 32.671 33.924 35.172 36.415 37.652 38.885 40.113 41.337 42.557 43.773 55.758 67.505 79.082 90.531 101.879 113.145 124.342

5.024 7.378 9.348 11.143 12.833 14.449 16.013 17.535 19.023 20.483 21.920 23.337 24.736 26.119 27.488 28.845 30.191 31.526 32.852 34.170 35.479 36.781 38.076 39.364 40.646 41.923 43.195 44.461 45.722 46.979 59.342 71.420 83.298 95.023 106.629 118.136 129.561

6.635 9.210 11.345 13.277 15.086 16.812 18.476 20.090 21.666 23.210 24.725 26.217 27.688 29.141 30.578 32.000 33.409 34.805 36.191 37.566 38.932 40.289 41.638 42.980 44.314 45.642 46.963 48.278 49.588 50.892 63.691 76.154 88.379 100.425 112.329 124.116 135.807

7.880 10.597 12.838 14.861 16.750 18.548 20.279 21.956 23.590 25.189 26.757 28.300 29.819 31.319 32.801 34.267 35.718 37.156 38.582 39.997 41.401 42.796 44.181 45.559 46.928 48.290 49.645 50.993 52.336 53.672 66.766 79.490 91.952 104.215 116.321 128.299 140.170

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Appendix C

Statistical Tables

C.4 Critical Values for the Kolmogorov–Smirnov Test Statistic TABLE C.4: Critical Values for the Kolmogorov–Smirnov Test Statistic

Significance level

ww

Sample size (n) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 25 30 35 40 50 60 70 80 90 100

0.20 0.900 0.684 0.585 0.494 0.446 0.410 0.381 0.358 0.339 0.322 0.307 0.295 0.284 0.274 0.266 0.258 0.250 0.244 0.237 0.231 0.210 0.190 0.180 · · · · · · ·

0.15 0.925 0.726 0.597 0.525 0.474 0.436 0.405 0.381 0.360 0.342 0.326 0.313 0.302 0.292 0.283 0.274 0.266 0.259 0.252 0.246 0.220 0.200 0.190 · · · · · · ·

0.10 0.950 0.776 0.642 0.564 0.510 0.470 0.438 0.411 0.388 0.368 0.352 0.338 0.325 0.314 0.304 0.295 0.286 0.278 0.272 0.264 0.240 0.220 0.210 · · · · · · ·

0.05 0.975 0.842 0.708 0.624 0.563 0.521 0.486 0.457 0.432 0.409 0.391 0.375 0.361 0.349 0.338 0.328 0.318 0.309 0.301 0.294 0.264 0.242 0.230 0.210 0.190 0.170 0.160 0.150 0.140 0.140

1.07 √ n

1.14 √ n

1.22 √ n

1.36 √ n

w .E asy En g Asymptotic Formula

ine eri n

0.01 0.995 0.929 0.829 0.734 0.669 0.618 0.577 0.543 0.514 0.486 0.468 0.450 0.433 0.418 0.404 0.391 0.380 0.370 0.361 0.352 0.320 0.290 0.270 0.250 0.230 0.210 0.190 0.180 · ·

g .n 1.63 √ n

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A P P E N D I X

D

Special Functions D.1 Error Function The error function, erf(z), is defined by the relation 2 erf(z) = √ π

ww



z

2

e−x dx

(D.1)

0

The error function is closely related to the cumulative distribution function of a normal probability distribution and is defined for −q … z … q. The error function is antisymmetric, such that

w .E asy En g

erf(−z) = −erf(z)

(D.2)

The constant before the integral sign in Equation D.1 is simply a normalizing constant such that erf(z) approaches 1 as z approaches infinity. For small values of z, it is convenient to use 2 the series expansion for e−x to obtain (Carslaw and Jaeger, 1959) 2 erf(z) = √ π



z

⎡

q  (−1)n x2n

⎤

⎣

n=0

n!

(D.3)

⎦ dx

ine eri n 0

Since the series is uniformly convergent, it can be integrated term by term to yield ⎡

⎤ q n 2n+1  (−1) z 2 ⎦ erf(z) = √ ⎣ (2n + 1)n! π n=0

which can also be written in the form 2 erf(z) = √ π



g .n

z5 z7 z3 + − + ... z − 3 2!5 3!7



(D.4)

et

(D.5)

This relationship is particularly useful in estimating erf(z) for small values of z. A closely related function to the error function is the complementary error function, erfc(z), which is defined by erfc(z) = 1 − erf(z)

(D.6)

Values of the error function, erf(z), are tabulated below.

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Appendix D

Special Functions TABLE D.1: Error Function

z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

erf(z) 0.00000 0.11246 0.22270 0.32863 0.42839 0.52050 0.60386 0.67780

z 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5

erf(z) 0.74210 0.79691 0.84270 0.88021 0.91031 0.93401 0.95229 0.96611

z 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3

erf(z) 0.97635 0.98379 0.98909 0.99279 0.99532 0.99702 0.99814 0.99886

z 2.4 2.5 2.6 2.7 2.8 2.9 3.0 q

erf(z) 0.99931 0.99959 0.99976 0.99987 0.99992 0.99996 0.99998 1.00000

D.2 Bessel Functions D.2.1 Definition

ww

A second-order linear homogeneous differential equation of the form x2

dy d2 y + x + (x2 − n2 )y = 0, dx dx2

w .E asy En g

n Ú 0

(D.7)

is called Bessel’s equation. The general solutions of Bessel’s equation are y = AJn (x) + BJ−n (x), y = AJn (x) + BYn (x),

n Z 1, 2, . . .

(D.8)

all n

(D.9)

where Jn (x) is called the Bessel function of the first kind of order n, and Yn (x) is called the Bessel function of the second kind of order n. If Bessel’s equation (Equation D.7) is slightly modified and written in the form x2

ine eri n

d2 y dy − (x2 + n2 )y = 0, + x 2 dx dx

n Ú 0

(D.10)

then this equation is called the modified Bessel’s equation. The general solutions of the modified Bessel’s equation are y = AIn (x) + BI−n (x),

n Z 1, 2, . . .

y = AIn (x) + BKn (x),

all n

g .n

(D.11) (D.12)

et

where In (x) is called the modified Bessel function of the first kind of order n, and Kn (x) is called the modified Bessel function of the second kind of order n. D.2.2 Evaluation of Bessel Functions The Bessel functions cannot generally be expressed in closed form, and are usually presented as infinite series. The Bessel functions are usually standard functions in electronic spreadsheets and commercial numerical analysis programs. D.2.2.1

Bessel function of the first kind of order n

This function is given by 

x4 xn x2 Jn (x) = n + − ··· 1 − 2 Ŵ(n + 1) 2(2n + 2) 2 · 4(2n + 2)(2n + 4) =

q  (−1)k (x/2)n+2k k!Ŵ(n + k + 1)

(D.13)

(D.14)

k=0

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Bessel Functions

845

The Bessel function J−n (x) can be derived from Equation D.14 by replacing n by −n in the formula. A convenient relationship to note is J−n (x) = (−1)n Jn (x), D.2.2.2

n = 0, 1, 2, . . .

(D.15)

Bessel function of the second kind of order n

This function is given by

Yn (x) =

D.2.2.3

ww

⎧ ⎪ Jn (x) cos nπ − J−n (x) ⎪ ⎪ , ⎪ ⎨ sin nπ

n Z 0, 1, 2, . . . (D.16)

⎪ Jp (x) cos pπ − J−p (x) ⎪ ⎪ , lim ⎪ ⎩ p→n sin pπ

n = 0, 1, 2, . . .

Modified Bessel function of the first kind of order n

This function is given by 

x4 xn x2 + + ··· In (x) = n 1 + 2 Ŵ(n + 1) 2(2n + 2) 2 · 4(2n + 2)(2n + 4)

w .E asy En g =

q  k=0

(x/2)n+2k k!Ŵ(n + k + 1)

(D.17)

(D.18)

The Bessel function I−n (x) can be derived from Equation D.18 by replacing n by −n in the formula. A convenient relationship to note is I−n (x) = In (x),

D.2.2.4

ine eri n

(D.19)

Modified Bessel function of the second kind of order n

This function is given by

D.2.2.5

n = 0, 1, 2, . . .

⎧ π ⎪ n Z 0, 1, 2, . . . ⎪ ⎨ 2 sin nπ [I−n (x) − In (x)], Kn (x) = π ⎪ ⎪ [I−p (x) − Ip (x)], n = 0, 1, 2, . . . ⎩ lim p→n 2 sin pπ

Tabulated values of useful Bessel functions

TABLE D.2: Useful Bessel Functions

x

I0 (x)

K0 (x)

I1 (x)

K1 (x)

0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010 0.020 0.030 0.040

1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0001 1.0002 1.0004

7.0237 6.3305 5.9251 5.6374 5.4143 5.2320 5.0779 4.9443 4.8266 4.7212 4.0285 3.6235 3.3365

0.0005 0.0010 0.0015 0.0020 0.0025 0.0030 0.0035 0.0040 0.0045 0.0050 0.0100 0.0150 0.0200

999.9962 499.9932 333.3237 249.9877 199.9852 166.6495 142.8376 124.9782 111.0871 99.9739 49.9547 33.2715 24.9233

g .n

(D.20)

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Appendix D

Special Functions TABLE D.2: Useful Bessel Functions (Cont’d)

ww

x

I0 (x)

K0 (x)

I1 (x)

K1 (x)

0.050 0.060 0.070 0.080 0.090 0.100 0.110 0.120 0.130 0.140 0.150 0.160 0.170 0.180 0.190 0.200 0.210 0.220 0.230 0.240 0.250 0.260 0.270 0.280 0.290 0.300 0.310 0.320 0.330 0.340 0.350 0.360 0.370 0.380 0.390 0.400 0.410 0.420 0.430 0.440 0.450 0.460 0.470 0.480 0.490 0.500 0.510 0.520 0.530 0.540 0.550 0.560 0.570

1.0006 1.0009 1.0012 1.0016 1.0020 1.0025 1.0030 1.0036 1.0042 1.0049 1.0056 1.0064 1.0072 1.0081 1.0090 1.0100 1.0111 1.0121 1.0133 1.0145 1.0157 1.0170 1.0183 1.0197 1.0211 1.0226 1.0242 1.0258 1.0274 1.0291 1.0309 1.0327 1.0345 1.0364 1.0384 1.0404 1.0425 1.0446 1.0468 1.0490 1.0513 1.0536 1.0560 1.0584 1.0609 1.0635 1.0661 1.0688 1.0715 1.0742 1.0771 1.0800 1.0829

3.1142 2.9329 2.7798 2.6475 2.5310 2.4271 2.3333 2.2479 2.1695 2.0972 2.0300 1.9674 1.9088 1.8537 1.8018 1.7527 1.7062 1.6620 1.6199 1.5798 1.5415 1.5048 1.4697 1.4360 1.4036 1.3725 1.3425 1.3136 1.2857 1.2587 1.2327 1.2075 1.1832 1.1596 1.1367 1.1145 1.0930 1.0721 1.0518 1.0321 1.0129 0.9943 0.9761 0.9584 0.9412 0.9244 0.9081 0.8921 0.8766 0.8614 0.8466 0.8321 0.8180

0.0250 0.0300 0.0350 0.0400 0.0450 0.0501 0.0551 0.0601 0.0651 0.0702 0.0752 0.0803 0.0853 0.0904 0.0954 0.1005 0.1056 0.1107 0.1158 0.1209 0.1260 0.1311 0.1362 0.1414 0.1465 0.1517 0.1569 0.1621 0.1673 0.1725 0.1777 0.1829 0.1882 0.1935 0.1987 0.2040 0.2093 0.2147 0.2200 0.2254 0.2307 0.2361 0.2415 0.2470 0.2524 0.2579 0.2634 0.2689 0.2744 0.2800 0.2855 0.2911 0.2967

19.9097 16.5637 14.1710 12.3742 10.9749 9.8538 8.9353 8.1688 7.5192 6.9615 6.4775 6.0533 5.6784 5.3447 5.0456 4.7760 4.5317 4.3092 4.1058 3.9191 3.7470 3.5880 3.4405 3.3033 3.1755 3.0560 2.9441 2.8390 2.7402 2.6470 2.5591 2.4760 2.3973 2.3227 2.2518 2.1844 2.1202 2.0590 2.0006 1.9449 1.8915 1.8405 1.7916 1.7447 1.6997 1.6564 1.6149 1.5749 1.5364 1.4994 1.4637 1.4292 1.3960

w .E asy En g

ine eri n

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Bessel Functions

TABLE D.2: Useful Bessel Functions (Cont’d)

ww

x

I0 (x)

K0 (x)

I1 (x)

0.580 0.590 0.600 0.610 0.620 0.630 0.640 0.650 0.660 0.670 0.680 0.690 0.700 0.710 0.720 0.730 0.740 0.750 0.760 0.770 0.780 0.790 0.800 0.810 0.820 0.830 0.840 0.850 0.860 0.870 0.880 0.890 0.900 0.910 0.920 0.930 0.940 0.950 0.960 0.970 0.980 0.990 1.000 1.100 1.200 1.300 1.400 1.500 1.600 1.700 1.800 1.900 2.000

1.0859 1.0889 1.0920 1.0952 1.0984 1.1017 1.1051 1.1084 1.1119 1.1154 1.1190 1.1226 1.1263 1.1301 1.1339 1.1377 1.1417 1.1456 1.1497 1.1538 1.1580 1.1622 1.1665 1.1709 1.1753 1.1798 1.1843 1.1889 1.1936 1.1984 1.2032 1.2080 1.2130 1.2180 1.2231 1.2282 1.2334 1.2387 1.2440 1.2494 1.2549 1.2604 1.2661 1.3262 1.3937 1.4693 1.5534 1.6467 1.7500 1.8640 1.9896 2.1277 2.2796

0.8042 0.7907 0.7775 0.7646 0.7520 0.7397 0.7277 0.7159 0.7043 0.6930 0.6820 0.6711 0.6605 0.6501 0.6399 0.6300 0.6202 0.6106 0.6012 0.5920 0.5829 0.5740 0.5653 0.5568 0.5484 0.5402 0.5321 0.5242 0.5165 0.5088 0.5013 0.4940 0.4867 0.4796 0.4727 0.4658 0.4591 0.4524 0.4459 0.4396 0.4333 0.4271 0.4210 0.3656 0.3185 0.2782 0.2437 0.2138 0.1880 0.1655 0.1459 0.1288 0.1139

0.3024 0.3080 0.3137 0.3194 0.3251 0.3309 0.3367 0.3425 0.3483 0.3542 0.3600 0.3659 0.3719 0.3778 0.3838 0.3899 0.3959 0.4020 0.4081 0.4142 0.4204 0.4266 0.4329 0.4391 0.4454 0.4518 0.4581 0.4646 0.4710 0.4775 0.4840 0.4905 0.4971 0.5038 0.5104 0.5171 0.5239 0.5306 0.5375 0.5443 0.5512 0.5582 0.5652 0.6375 0.7147 0.7973 0.8861 0.9817 1.0848 1.1963 1.3172 1.4482 1.5906

w .E asy En g

K1 (x) 1.3638 1.3328 1.3028 1.2738 1.2458 1.2186 1.1923 1.1668 1.1420 1.1181 1.0948 1.0722 1.0503 1.0290 1.0083 0.9882 0.9686 0.9496 0.9311 0.9130 0.8955 0.8784 0.8618 0.8456 0.8298 0.8144 0.7993 0.7847 0.7704 0.7564 0.7428 0.7295 0.7165 0.7039 0.6915 0.6794 0.6675 0.6560 0.6447 0.6336 0.6228 0.6122 0.6019 0.5098 0.4346 0.3725 0.3208 0.2774 0.2406 0.2094 0.1826 0.1597 0.1399

ine eri n

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Appendix D

Special Functions TABLE D.2: Useful Bessel Functions (Cont’d)

ww

x

I0 (x)

K0 (x)

I1 (x)

K1 (x)

2.100 2.200 2.300 2.400 2.500 2.600 2.700 2.800 2.900 3.000 3.100 3.200 3.300 3.400 3.500 3.600 3.700 3.800 3.900 4.000 5.000 6.000 7.000 8.000 9.000 10.000

2.4463 2.6291 2.8296 3.0493 3.2898 3.5533 3.8417 4.1573 4.5027 4.8808 5.2945 5.7472 6.2426 6.7848 7.3782 8.0277 8.7386 9.5169 10.3690 11.3019 27.2399 67.2344 168.5939 427.5641 1093.5880 2815.7170

0.1008 0.0893 0.0791 0.0702 0.0623 0.0554 0.0493 0.0438 0.0390 0.0347 0.0310 0.0276 0.0246 0.0220 0.0196 0.0175 0.0156 0.0140 0.0125 0.0112 0.0037 0.0012 0.0004 0.0001 0.0001 0.0000

1.7455 1.9141 2.0978 2.2981 2.5167 2.7554 3.0161 3.3011 3.6126 3.9534 4.3262 4.7343 5.1810 5.6701 6.2058 6.7927 7.4357 8.1404 8.9128 9.7595 24.3356 61.3419 156.0391 399.8731 1030.9150 2670.9880

0.1227 0.1079 0.0950 0.0837 0.0739 0.0653 0.0577 0.0511 0.0453 0.0402 0.0356 0.0316 0.0281 0.0250 0.0222 0.0198 0.0176 0.0157 0.0140 0.0125 0.0040 0.0013 0.0005 0.0002 0.0001 0.0000

w .E asy En g D.3 Gamma Function

ine eri n

TABLE D.3: Gamma Function

g .n

x

Ŵ(x)

x

Ŵ(x)

x

Ŵ(x)

1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20

1.00000 0.99433 0.98884 0.98355 0.97844 0.97350 0.96874 0.96415 0.95973 0.95546 0.95135 0.94740 0.94359 0.93993 0.93642 0.93304 0.92980 0.92670 0.92373 0.92089 0.91817

1.34 1.35 1.36 1.37 1.38 1.39 1.40 1.41 1.42 1.43 1.44 1.45 1.46 1.47 1.48 1.49 1.50 1.51 1.52 1.53 1.54

0.89222 0.89115 0.89018 0.88931 0.88854 0.88785 0.88726 0.88676 0.88636 0.88604 0.88581 0.88566 0.88560 0.88563 0.88575 0.88595 0.88623 0.88659 0.88704 0.88757 0.88818

1.68 1.69 1.70 1.71 1.72 1.73 1.74 1.75 1.76 1.77 1.78 1.79 1.80 1.81 1.82 1.83 1.84 1.85 1.86 1.87 1.88

0.90500 0.90678 0.90864 0.91057 0.91258 0.91467 0.91683 0.91906 0.92137 0.92376 0.92623 0.92877 0.93138 0.93408 0.93685 0.93969 0.94261 0.94561 0.94869 0.95184 0.95507

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Exponential Integral

849

TABLE D.3: Gamma Function (Cont’d)

ww

x

Ŵ(x)

x

Ŵ(x)

x

Ŵ(x)

1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.30 1.31 1.32 1.33

0.91558 0.91311 0.91075 0.90852 0.90640 0.90440 0.90250 0.90072 0.89904 0.89747 0.89600 0.89464 0.89338

1.55 1.56 1.57 1.58 1.59 1.60 1.61 1.62 1.63 1.64 1.65 1.66 1.67

0.88887 0.88964 0.89049 0.89142 0.89243 0.89352 0.89468 0.89592 0.89724 0.89864 0.90012 0.90167 0.90330

1.89 1.90 1.91 1.92 1.93 1.94 1.95 1.96 1.97 1.98 1.99 2.00 –

0.95838 0.96177 0.96523 0.96877 0.97240 0.97610 0.97988 0.98374 0.98768 0.99171 0.99581 1.00000 –

D.4 Exponential Integral

w .E asy En g

The standard definition of the exponential integral is  q −xt e dt, x > 0, En (x) = tn 1

n = 0, 1, . . .

The function defined by the principal value of the integral  x t  q −t e e dt = dt, Ei(x) = − −q t −x t

x > 0

ine eri n

(D.21)

(D.22)

is also called an exponential integral. Note that Ei(x) is related to E1 (x) by analytic continuation, where Ei(−x) = −E1 (x) (D.23)

g .n

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A P P E N D I X

E

Pipe Specifications E.1 PVC Pipe

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Pipe dimensions of interest to engineers are usually the diameter and wall thickness. The pipe diameter is typically specified by the nominal pipe size and the wall thickness is usually specified by the schedule. Nominal pipe sizes are typically given in either inches or millimeters, and represent rounded approximations to the inside diameter of the pipe. The schedule of a pipe is a number that approximates the value of the expression 1000P/S, where P is the service pressure and S is the allowable stress. Higher schedule numbers correspond to thicker pipes, and schedule numbers in common use are 5, 5S, 10, 10S, 20, 20S, 30, 40, 40S, 60, 80, 80S, 100, 120, 140, and 160. The schedule numbers followed by the letter “S” are primarily intended for use with stainless steel pipe (ASME B36.19M).

w .E asy En g Nominal pipe size (mm)

Outside diameter (mm)

50 80 100 125 150

60 90 114 141 168

Schedule 5 Wall Inside thickness diameter (mm) (mm) 1.7 2.1 2.1 2.8 2.8

57 85 110 136 163

Source: Fetter (1999).

Nominal pipe size (in.)

Outside diameter (in.)

2 3 4 5 6

2.375 3.500 4.500 5.563 6.625

Schedule 5 Wall Inside thickness diameter (in.) (in.) 0.065 0.083 0.083 0.109 0.109

2.245 3.334 4.334 5.345 6.407

Schedule 10 Wall Inside thickness diameter (mm) (mm) 2.8 3.0 3.0 3.4 3.4

55 83 108 134 161

Schedule 40 Wall Inside thickness diameter (mm) (mm) 3.9 5.5 6.0 6.6 7.1

ine eri n

Schedule 10 Wall Inside thickness diameter (in.) (in.) 0.109 0.120 0.120 0.134 0.134

2.157 3.260 4.260 5.295 6.357

53 78 102 128 154

5.5 7.6 8.6 9.5 11.0

g .n

Schedule 40 Wall Inside thickness diameter (in.) (in.) 0.154 0.216 0.237 0.258 0.280

Schedule 80 Wall Inside thickness diameter (mm) (mm)

2.067 3.068 4.026 5.047 6.065

49 74 97 122 146

Schedule 80 Wall Inside thickness diameter (in.) (in.)

et

0.218 0.300 0.337 0.375 0.432

1.939 2.900 3.826 4.813 5.761

Source: Fetter (1999).

E.2 Ductile-Iron Pipe Ductile-iron pipe is manufactured in diameters in the range of 100–1200 mm (4–48 in.), and 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 ductile-iron pipe are 5.5 m (18 ft) and 6.1 m (20 ft).

850

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Physical Properties of Common Pipe Materials

E.3 Concrete Pipe TABLE E.1: Commercially Available Sizes of Concrete Pipe

Non-reinforced pipe Diameter Diameter (mm) (in.)

ww

100 150 205 255 305 380 455 535 610 685 760 840 915 — — — — — — — — — — — —

4 6 8 10 12 15 18 21 24 27 30 33 36 — — — — — — — — — — — —

w .E asy En g

Reinforced pipe Diameter Diameter (mm) (in.) — — — — 305 380 455 535 610 685 760 840 915 1065 1220 1370 1525 1675 1830 1980 2135 2285 2440 2590 2745

ine eri n

E.4 Physical Properties of Common Pipe Materials

Material Concrete Concrete (reinforced) Ductile iron PVC Steel

— — — — 12 15 18 21 24 27 30 33 36 42 48 54 60 66 72 78 84 90 96 102 108

g .n

Young’s modulus, E (GPa)

Poisson’s ratio

14–30 30–60 165–172 2.4–3.5 200–207

0.10–0.15 — 0.28–0.30 0.45–0.46 0.30

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A P P E N D I X

F

Unified Soil Classification System F.1 Definition of Soil Groups The soil groups in the Unified Soil Classification (USC) System are defined as follows: Group symbol

Major divisions

ww

Coarse-grained soils Gravels More than 50% 50% or more of retained on the coarse fraction 0.075-mm retained on the (No. 200) sieve 4.75-mm (No. 4)

Clean Gravels

w .E asy En g Sands 50% or more of coarse fraction passes the 4.75-mm (No. 4) sieve

Fine-grained soils More than 50% passes the 0.075-mm (No. 200) sieve

Silts and clays Liquid limit 50% or less

GW GP

Gravels with Fines

GM

Clean Sands

SW

GC

SP

Silty gravels, gravel-sand-silt mixtures Clayey gravels, gravel-sand-clay mixtures Well-graded sands and gravelly sands, little or no fines Poorly graded sands and gravelly sands, little or no fines

SM SC

Silty sands, sand-silt mixtures Clayey sands, sand-clay mixtures

ML

Inorganic silts, very fine sands, rock four, silty or clayey fine sands Inorganic clays of low to medium plasticity, gravelly/sandy/silty/lean clays Organic silts and organic silty clays of low plasticity

CL

MH

CH OH Highly organic soils

Well-graded gravels and gravel-sand mixtures, little or no fines Poorly graded gravels and gravelsand mixtures, little or no fines

ine eri n

Sands with Fines

OL Silts and clays Liquid limit greater than 50%

Typical names

PT

g .n

et

Inorganic silts, micaceous or diatomaceous fine sands or silts, elastic silts Inorganic clays or high plasticity, fat clays Organic clays of medium to high plasticity Peat, muck, and other highly organic soils

Prefix: G = Gravel; S = Sand; M = Silt; C = Clay; O = Organic Suffix: W = Well graded; P = Poorly graded; M = Silty; L = Clay, LL < 50%; H = Clay, LL > 50%

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Terminology

F.2 Terminology Terminology used in the USC system in the context of classifying soils is as follows: Term

ww

Meaning

Coarse-grained soils

More than 50% retained on a 0.075-mm (No. 200) sieve

Fine-grained soils

50% or more passes a 0.075-mm (No. 200) sieve

Gravel

Material passing a 75-mm (3-in.) sieve and retained on a 4.75-mm (No. 4) sieve

Coarse gravel

Material passing a 75-mm (3-in.) sieve and retained on a 19-mm (3/4-in.) sieve

Fine gravel

Material passing a 19-mm (3/4-in.) sieve and retained on a 4.75-mm (No. 4) sieve

Sand

Material passing a 4.75-mm (No. 4) sieve and retained on a 0.075-mm (No. 200) sieve

w .E asy En g Coarse sand

Material passing a 4.75-mm (No. 4) sieve and retained on a 2.00-mm (No. 10) sieve

Medium sand

Material passing a 2.00-mm (No. 10) sieve and retained on a 0.475-mm (No. 40) sieve

Medium sand

Material passing a 0.475-mm (No. 40) sieve and retained on a 0.075-mm (No. 200) sieve

Clay

Material passing a 0.075-mm (No. 200) sieve that exhibits plasticity and strength when dry (PI Ú 4)

Silt

Material passing a 0.075-mm (No. 200) sieve that is nonplastic and has little strength when dry (PI < 4)

Peat

Soil of vegetable matter

ine eri n

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Index Note; The letters ‘fn’ followed by locators refers to footnotes cited in the text

A

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Acceptability criterion, plan evaluation, 817 Access holes, storm sewers, 576 Accuracy of estimates, channel flow, 486–487 Accuracy of unit-hydrograph models, 509 Active conservation pool, 322 Active leaf area index, 627 Active storage, 322 Actual evapotranspiration, 624, 651–654 crop-coefficient method, 653 index-of-dryness method, 651–653 remote sensing, 653–654 Actual vapor pressure, 632–633 Aeration, zones of, 656 Aerodynamic resistance, 625–626 Aerosols, 402 Affinity laws for homologous pumps, 51–52 A-Horizon, 438 Air chambers, 39 Air density, 633 Air-relief valves, 85 Air-and-vacuum relief valves, 85 Albedo, 627–628 Albedometer, 628 Allievi equation, 36 Allowable velocities, pipeline systems, 91 Alternating-block method, temporal distribution, 423–424 American Society of Civil Engineers (ASCE), 319, 360 American Society of Testing Materials (ASTM), 191 American WaterWorks Association (AWWA), 70, 77–78, 83, 778, 780, 788 Anemometers, 626 Angle of repose, 174 Anisotropy, 672, See also Darcy’s Law equivalent anisotropic/isotropic media, 679–680 hydraulic conductivity, 670–673 primary cause of, 672 ratio, 672 Annual flood series, 360 Annual rainfall averages, 4 Annual series, IDF curves, 408 Annual-worth analysis, 825 Antecedent runoff condition (ARC), 455–457 Apron, 274 Aquatic bench, 591

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Aquicludes, 656 Aquifer pumping tests (aquifer tests), 794 data collection, 797–798 design, 794–801 field procedures, 796–797 discharge-rate measurements, 798 establishment of baseline conditions, 796–797 length of test, 798 water-level measurements during aquifer test, 797–798 observation well design, 795–796 distance from pumping well, 796 well construction, 796 well diameter, 795–796 pumping well design, 794–795 discharge-control and measurement equipment, 795 pump selection, 795 water disposal, 795 water-level measurement access, 795 well construction, 794–795 well development, 795 Aquifer tests, 794fn, See also Aquifer pumping tests (aquifer tests) Aquifers, 656–658 artesian, 658 derivation of names of, 658–659 leaky-confined aquifer, 658 leaky-unconfined aquifer, 658 phreatic, 657 poorly graded aquifer material, 781 safe yield, 322 specific storage coefficients in, 677 thickness, 661 unconfined, 681–684 water table, 681 well-graded aquifer material, 781 Aquifuges, 656 Aquitards, 658 common materials found in, 660 Area under the standard normal curve, 361 Areal porosity, 663 Areal-reduction factors (ARFs), 428–429 Areawide water-quality studies, 815 Arithmetic average, 347 Arithmetic-gradient future-worth factor, 821 Arithmetic-gradient present-worth factor, 820–821

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912

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Arterial mains, 81 Artesian aquifers, 658 ASCE Penman-Monteith (ASCE-PM) equation, 643–644 actual vapor pressure, 632–633 aerodynamic resistance, 625–626 air density, 633 bulk surface resistance, 625, 627 latent heat of vaporization, 631 net radiation, 627 psychrometric constant, 631–632 saturation vapor pressure, 632 soil heat flux, 630–631 vapor-pressure gradient, 632 Aspect ratio, 31 Atmospheric aerosols, 402 Atmospheric water, typical residence time for, 4 Attenuation zone, 774 Automatic-recording NWS gages, 403–404 Available net positive suction head, NPSHA, 55 Average conveyance method, 144 Average friction slope method, 144 Average daily demand, 76 Average per-capita demand, 71 Average recurrence interval (ARI), 344 Axial flow pumps, 46–51 efficiency of, 49–50

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Backflow-prevention devices, 85 Bail-down test, 798 Bank filtration, 777 Bare soil, See Vegetative linings Base flow, 530, 653 Baseflow index, 468 Base time, 498 Basic principles drainage channels, 167–180 open channels, 103–132 Basins, 4 Bazin weirs, 282 Bedrock, 439 ´ Belanger’s equation, 140 Benches, 231 Bends, 177–178 Benefit-cost analysis, 818–819, 828 benefits, 818–819 constant-dollar rate, 819 costs, 818 defined, 818 discount rate, 819 present value, 819 real discount rate, 819 Bergeron-Findeisen process, 402

913

Bernoulli equation, 25, 125 Bernoulli probability distribution, 352 Bernoulli process, 352, 354 Bernoulli trials, 352, 354 Best hydraulic section, 167–170 geometric characteristics of, 169 Best trapezoidal section, 607 Best-efficiency point (BEP), pumps, 49 Best-management practices (BMPs), 586 B-Horizon, 438 Binomial coefficient, 352 Binomial distribution, 351–352 Biochemical oxygen demand (BOD), 235 Biofilter, 607 Biofiltration swale, 607–608 Bioretention systems, 610–612 Bird guards, 644 Bivariate probability distributions, 345 Bleeder, 292 Blowoff valves, 85 Boltzman variable, 719 Booster pumps, 85 Boulder Zone (Floridan Aquifer System), 662 Boundary shear stress, 13 Bounded exponential distribution, 368 Boussinesq coefficient, 105 Boussinesq equation, 684 Bowl, 611 Brake horsepower (BHP), 54 Bresse’s equation, 140 Broad-crested weirs, 294–299 flow conditions over, 160 rectangular weirs, 294–297 Buckingham pi theorem, 13, 49, 666 Budyko-type curves, 651 Buffer strips, 610 Building, water-supply systems, 93 design procedure, 93–94 minimum pressures, specification, 94–96 pipe diameters, 96–101 specification, 94 Bulletin 17B (USIAC), 391–392 ´ Bureau de Recherches Geologiques et ` Minieres ant Saint Pardon de Conques, 735 Buttress dam, 320–321 Buttresses, defined, 320 Buoyancy factor, 753 Bypass flow (carryover flow), 549

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C Cable-tool drilling, 793 California Irrigation Management Information System (CIMIS) Penman equation, 638 Campbell-Stokes heliograph, 629 Canopy reflection coefficient, 627–628

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Canopy resistance, 625 Capillary fringe, 692 dynamics of water movement in, 692–693 Capillary potential, 439 Capillary rise in unconsolidated materials, 692–693 Capillary water, 694 Capillary zone, 656, 692 Capital-recovery factor, 820 Capture zone, 745 Carryover flow, 549 Casing, 779 Catchment, 401 scale of, 474 Catchment-averaged rainfall, calculation of, 420 Catchments, shape factor, 503 Cavitation, 55 Cavitation number, 56 Cells, 377 Central caisson, 777 Central limit theorem, 361–362 Centrifugal pumps, 46 specific speed, 50 Certainty criterion, plan evaluation, 817 Channel flow, 484–486 Channel slopes, 178 Check dams, 607 Check valve, centrifugal pumps, 46 Chen and Wong formula, 422, 477 ´ Antoine de, 111 Chezy, ´ coefficient, 111 Chezy Chi-square distribution, 371–372 degrees of freedom, 371 Chi-square test, 376–378 Chlorination, in sanitary sewers, 235 Choked flows, 130 C-Horizon, 439 Chute, 301 Chute spillways, 322 Cipolletti weir, 286 Clark method, 512 Classification hydraulic conductivities, 670–671 hydropower capacity, 335 saline groundwater, 760–761 USDA soil system, 438, 661 water surface profiles, 134–139 well productivity, 788 Clean-out structures, storm sewers, 576 Climate dryness index, 651 Climatic spectrum, 4–5 Clogged wells, rehabilitation of, 789 Clogging/sedimentation, susceptibility, 604 Closed-conduit flow, 9–62 computer models, 46 pipe networks, 39–46

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continuity equation, 39 energy equation, 39 loop method, 42–46 nodal method, 40–46 pumps, 46–64 single pipelines diameter for a given flow rate and head loss, 19–21 empirical friction-loss formulae, 32–35 energy and hydraulic grade lines, 25–26 flow rate for given head loss, 19–20 head losses in noncircular conduits, 31–32 head losses in transitions and fittings, 27–30 steady-state continuity equation, 9–10 steady-state energy equation, 22–25 steady-state momentum equation, 10–25 velocity profile, 27 water hammer, 35–39 Closest-depth rule, 129 Cloud droplets, formation of, 402 Cloudiness factor, 630 Coefficient of contraction, 277 Coefficient of permeability, 663, 667 Coefficient of variation, 362 Cohesive versus noncohesive materials, 172–176 Cold front, 402–403 Colebrook equation, 15–17, 108 Combination inlets, 560–564 interception capacity of, 561–566 sweeper configuration, 560 sweeper inlet, 560 Combined sewers, 211, 236, 545 Commercially available pumps, 53–54 Common probability functions binomial distribution, 351–353 central limit theorem, 361–362 chi-square distribution, 371–372 exponential distribution, 356–357 extreme-value distributions, 364–371 Gamma/Pearson Type III distribution, 357–360 geometric distribution, 353–354 log-normal distribution, 362–363 normal distribution, 360–361 Poisson distribution, 354–355 uniform distribution, 363–364 Communication factor, 77 Completeness criterion, plan evaluation, 817 Complete hydraulic jump, 141 Components, water-distribution systems minimum size, 82 pipelines, 81–82

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Components, water-distribution (continued) pipe materials, 83–85 asbestos cement, 84 concrete, 85 ductile iron, 83 fiberglass, 84–85 plastic, 84 service lines, 83 steel, 84 Composite channels, 115 Composite linings, 205–206 Composite roughness, 115–116 Compound channels, 115–118 Compound-composite channels, 115 Compound weirs, 291–293 Compound-interest factors, 819–826 arithmetic-gradient future-worth factor, 821 arithmetic-gradient present-worth factor, 820–821 capital-recovery factor, 820 geometric-gradient factors, 821 single-payment compound amount factor, 819–820 single-payment present-worth factor, 819–820 sinking-fund factor, 820 uniform annual series factors, 820 uniform-series compound-amount factor, 820 uniform-series present-worth factor, 820 Comprehensive planning, defined, 815 Compressibility coefficient, 677fn Computation, water-surface profiles, 143–159 average conveyance method, 144 average friction slope method, 144 compound channels, 152–154 control sections, 150–151 direct-integration method, 145–147 direct-step method, 147–148 geometric mean friction slope method, 144 harmonic mean friction slope method, 144–145 length of backwater profile, 151 natural channels, 151–152 profiles across bridges, 154–159 standard-step method, 148–150 steady-state assumption, 151 Computer models closed-conduit flow, 103 open-channel flow, 103 HEC-RAS (U.S. Army Corps of Engineers), 256 NRCS-TR55 method, 492–494 WSP2, 253 pipe networks, 46

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915

surface-water hydrology, 492–494 Concrete arch dam, 320 Cone of depression, 705 Confined aquifers, 657, 687–691 storage coefficient, 687–688 transmissivity of, 689 Conjugate depths, 140 Connate water, 761 Conservation storage, 322 Consolidated rocks, 660 Constant-baseflow method, 465 Constant-discharge baseflow separation method, 465 Constant-dollar rate, 819 Constant-flow design, 803–808 Constant-flow exfiltration trenches, 803–806 Constant-head boundary, 746–749 Construction factor, 77 Contact angle, 692 Continuity equation, pipe networks, 39 Continuous probability distributions, 346–347 defined, 345 Continuous sample space, 345 Continuous-runoff models, 495–520 extreme runoff events, 519–520 probable maximum flood (PMF), 519 standard project flood (SPF), 519–520 instantaneous unit hydrograph (IUH), 501–502 kinematic-wave model, 514–515 nonlinear-reservoir model, 515–517 Santa Barbara Urban Hydrograph (SBUH) model, 517–519 time-area models, 509–517 Unit-hydrograph model, 495, 502–509 accuracy of, 509 NRCS dimensionless unit hydrograph, 506–509 Snyder unit-hydrograph model, 503–505 unit-hydrograph theory, 495–501 Continuous transverse grates, 566 Controlled (gated) spillways, 307–312 Control elevation, 596 Control sections, 150 Control structures, 134, 203 Convective lifting, 402–403 Convective precipitation, 403 Convective storms (thunderstorms), 403 Conveyance, 144 Cooper-Jacob approximation, 723, 725, 797 Coriolis coefficient, 123 Corporation stop, 83 Correlation length scales, 674 Crest broad-crested weirs, 294–299 dams, 318–319

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Crest (continued) height, 284 sharp-crested weirs, 282–287 Critical duration, drainage basins, 433 Critical fixture, 96 Critical flow conditions, 127–128 Critical flow depth, 127 Critical-event design, water-resource systems, 5 Critical shear stress, 171 Crop coefficient method, ET estimation, 653–654 Crop evapotranspiration, 625–626 calculating, 638 Cross-drainage structures, 250 Cross slopes, 545–549 Crown, 218 Crust, 439 Culverts, 250–275 capacity determination of, 251 submerged entrance, 252 unsubmerged entrance, 259 cross-drainage structures, 250 defined, 250 design, 250–251, 262–264 allowable velocities, 262 debris control, 264 flow, 262 headwater elevation, 262 inlet, 263–264 material, 263 outlet, 264 size, 264 fixed-flow method, 269 fixed-headwater method, 265 minimum-performance method, 271 slope, 262 fall, 264 free-entrance conditions, 252 headwater depth, 252 hydraulic analysis of, 250–252 hydraulically long, 255–256 hydraulically short, 255–256 performance curves, 265 riprap/outlet protection, 274–275 roadway overtopping, 271–273 submerged-entrance conditions, 252 tailwater depth, 252 unsubmerged entrances, 259–262 Types 1 and 2 flow, 252–255 Type 3 flow, 255–259 Type 4 flow, 260 Type 5 flow, 260–261 Type 6 flow, 261–262 Culvert performance curves, 265 Cumulative distribution function, 345

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Curb inlets, 550–554 Curve-number model, 453 Cyclonic storms, 402

D Dams, 318–335, See also Reservoirs abutments, 319 buttress dam, 320–321 concrete arch dam, 320 crest, 319 earthen embankment dam (earthfill dam), 321–322 face, 319 gravity concrete dam, 319 hydropower, 328–335 feasibility, 334–335 turbines, 328–334 locks, 319 nonoverflow dams, 322 overflow dams, 322 parapet wall, 319 principal parts of, 319 toe, 319 Darcy, Henry, 14fn Darcy’s law, 447, 614, 662–676, 688, 701, See also Hydraulic conductivity defined, 662–663 Darcy’s velocity, 663 Darcy-Weisbach equation, 14, 17–19, 32, 34, 42, 96–97, 106–111, 143 and flow in sanitary sewers, 216 and open-channel flow, 106–110 Data collection and analysis, 815–816 Dead/inactive storage, 322 Decay, and groundwater contamination, 776 Declining-growth phase, 73 Degrees of freedom, 371 Delayed yield, 730 Delineation, WHP areas, 774–775 Delta function model, See Green-Ampt model Demand factors, 76 Density of the fluid, 13 Depression storage, 437 typical values of, 437 Depth-duration-frequency curve, 466 Depth, sanitary sewer, 231 Deriaz turbine, 331 Design aids, hydraulic design computation, 237 Design, aquifer pumping tests, 795–798 field procedures, 796–798 observation wells, 795–796 pumping well, 794–795 Design computations sanitary sewers, 224–227, 237–239 diameter and slope, 224–227

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Design computations (continued) Design aids, 237 Manning’s n, 237 Minimum slope for self-cleansing, 237–239 Design constraints, culverts, 262–264 Design/construction, water-supply wells, 777–778 Design of channels flexible linings, 182–183 rigid lining, 180–182 Design, exfiltration trenches, 803–808 Design flood, 520 Design, flood control stormwater control measures, 599–603 stormwater management system, 614–617 Design flow rate, calculation, 568–571 Design flows, water-supply system, 79–81 Design, groundwater systems, 771–808 Design, hydraulic structures, 250–335 Design, outlet structures, 593–599 Design point, pumps, 49 Design procedure, hydraulic structure, 314–318 Design rainfall, 421–433 alternating-block method, 423–425 depth, 422 duration, 422 NRCS 24-hour hyetographs, 425–427 return period, 421 spatial distribution, 428–429 temporal distribution, 422–423 alternating-block method, 423–425 NRCS 24-hour hyetographs, 425–427 triangular method, 422–423 Design, sanitary sewers, 211–247 Design, slug tests, 798–803 data analysis, 800–802 guidelines, 802–803 Design storms, synthetic, 421 Design, stormwater-collection systems, 545–583 Design of stormwater control measures, 587–589 Design, stormwater-management systems, 586–691 Design, water-distribution systems, 70–101 Design, water-quality control, 590–592, 616–617 Design, well components, 778–788 Design, wellfields, 771–773 Detention basins, 588–590, 592 design criteria, 592–593 Detention BMP systems, 588 Detention ponds, 588, 590–591 aquatic bench, 591

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917

littoral zone, presence of, 591 overflow from, 594 Detention systems, structural SCMs, 618–619 Determination impervious area, 577–578 pipe diameters, 96–101 storage requirements, 326–327 Deterministic processes, 344 Diameter slope of pipes and, 231 Diffusion model, hydraulic-routing models, 531 Direct benefit (internal), 818 Direct reuse, 816 Direct runoff, 464 Direct-integration method, water-surface profiles, 145–147 Directly connected impervious area (DCIA), 568 Direct-step method, water-surface profiles, 145, 147 Discharge coefficient, 49, 272 Discount rate, 819 Discrete probability distributions, 345–346 defined, 345 Discrete sample space, 345 Dispersion, and groundwater contamination, 775 Distinguishing condition, 278 Distributed-parameter models, 473 Distribution mains, 81 Distribution-system reservoirs, 87 Domestic wells, 777 Double-exponential distribution, 365 Double-overhung installation, 328 Draft tube, 331 Drain valves, 85 Drainage basins (areas), 401 Drainage channels 166–207 bare soil, 187–196 basic principles, 167 bends, 177–178 best hydraulic section, 167–170 boundary shear stress, 170–172 cobble linings, 199–203 cohesive versus non cohesive materials, 172–176 composite linings, 205–206 concrete-lined channels, 181 density-stiffness coefficient, 189 earthen channels, 166 flexible linings, 182–183 flexible-boundary channels, 166 freeboard, 179–180 gabions, 203 general design procedure, 183–187 gravel linings, 199–203

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Drainage channels (continued) lined channels, 166 RECP Linings, 197–199 retardance, grass-lined channels, 187–188 rigid-boundary channels, 166 rigid linings, 166, 180–182 riprap linings, 199–203 rolled erosion control products (RECPs), 183 Sheilds diagram, 172 slopes, 178 unlined channels, 166 vegetative linings, 187–196 Drainage ways, 545 Drawdown, 704 Drilled well, 777 Drilling percussion methods, 793 rotary methods, 793 wells, 793 Drilling mud, 793 Drive point, 777 Driven well, 777 Drop inlets, 566 Drop manholes, 232 Droughts, 4 Dry basins, 587 Dry-detention basins, 592–595 defined, 592 extended, 592 maintenance of, 592 removal rates in, 591–592 use of, 592 wet-detention basins vs., 591–592 Dry-retention basins, See Infiltration basins Dry well, pump stations, 233 Dual water system(s), 82 Dual-purpose basins, 592 Dune infiltration systems, 613 Dupuit approximation, 681–684 ` Dupuit, Arsene, 682fn Dupuit equation, 707 Dupuit-Forchheimer approximation, 665, 682 Dynamic model, hydraulic-routing models, 531 Dynamic viscosity, 13

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E Earthen embankment dam (earthfill dam), 321–322 Economically justified benefit, 816 Economic evaluation, 816 Economic feasibility, 818–829 arithmetic-gradient factors, 820–821 benefit-cost analysis, 818–819 compound-interest factors, 819–821

evaluating alternatives, 823–829 geometric-gradient factors, 821–823 uniform-series factors, 820 Effective biochemical oxygen demand, 235 Effective grain diameter, 664 Effective impervious area (EIA), 568 Effective porosity, 663 Effective precipitable water, 430 Effective rainfall, 473 Effectiveness criterion, plan evaluation, 817 Efficiency criterion, plan evaluation, 817 E-Horizon, 438 Elevated storage tanks, 87 Emergency spillway, 323 Empirical equations, gates, 281–282 Empirical formulae, hydraulic conductivity, 666–670 Empirical friction-loss formulae, 32–34 Empirical methods ASCE-PM equations, 648–651 FAO56-PM equations, 648–651 Energy coefficient, 123 Energy equation, pipe networks, 39 Energy grade line, 25–26, 125 Engineered systems groundwater hydrology, 771–808 aquifer pumping test design, 794–803 exfiltration trench design, 803–808 seepage meters, 808 slug test, 798–803 water-supply well, design/construction, 778–788 water-supply well performance assessment, 788–793 well drilling, 793 wellfield design, 771–773 wellhead protection, 774–777 Engineering hydrologists, 1 Environmental assessments (EAs), 817 Environmental impact documents environmental assessments (EAs), 817 environmental impacts, 816 Environmental Protection Agency (EPA), 7–8, 84, 537–538, 835 EPANET, 46 Epistemic uncertainty, 344 Equivalent anisotropic/isotropic media, 679–680 Equivalent cross slope, 551 Equivalent porous medium, 666 Equivalent sand roughness, 15–18 Estimated limiting value (ELV), 5 Estimation, evapotranspiration, 624–654 Estimation, per-capita demand, 71 Estimation, population, 72–76

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Estimation, population distribution Hypothesis tests, 375–378 Model selection criteria, 379 Probability distribution, 372–375 Estimation, population parameters maximum-likelihood method, 382–383 method of L-moments, 383–387 method of moments, 379–382 Evaluating alternatives. See Economic feasibility Evaluation of alternatives, water resources planning annual-worth analysis, 825 benefit-cost analysis, 828–829 present-worth analysis, 823–825 rate-of-return analysis, 825–828 Evaporation pans, 644–648 Evaporation ratio, 651 Evapotranspiration, 6, 624–654 actual, 624, 651–654 ASCE Penman-Monteith (ASCE-PM) equation, 643–644 actual vapor pressure, 632–633 aerodynamic resistance, 625–626 air density, 633 bulk surface resistance, 626–627 latent heat of vaporization, 631 net radiation, 627–629 psychrometric constant, 631–632 saturation vapor pressure, 632 soil heat flux, 630–631 vapor-pressure gradient, 632 combination methods, 624 crop, 625–626 calculating, 638 defined, 624 evaporation-pan method, 624 methods for estimating, 624 potential, 624, 637 radiation methods, 624 reference, 624 reference-crop, 624, 638 selection of estimation method, 654 temperature methods, 624 Event-mean concentrations (EMCs), 533–535 Exceedance probabilities of hydrologic events, 344 Exfiltration galleries, 617 Exfiltration trenches, 612–613 constant-flow design, 803–808 design of, 613–617, 803–808 as off-line systems, 603–604 operational characteristics, 803 trench hydraulic conductivity, 803 use of, 612–613 Experimental semivariogram, 417 Exponential correlation function, 674

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919

Exponential distribution, 356–357 Exponential integral, 721 Exposure factor, 77 Extended dry-detention basins, 592 Extended-period simulations, 92 Extraction barriers, 757 Extrapolation, 73 Extrapolation models, 70, 72 Extreme rainfall, 429–433 probable maximum precipitation (PMP), 429 rational estimation method, 429 statistical estimation method, 430 world-record precipitation amounts, 432 probable maximum storm (PMS), 432–433 standard project storm (SPS), 519 Extreme runoff events, 519–520 probable maximum flood (PMF), 519 standard project flood (SPF), 519–520 Extreme-value distributions, 364 extreme-value Type I (Gumbel) distribution, 364–366 Extreme-value Type I distribution, 393 frequency analysis, 393

F Fall, 242 FAO56-Penman-Monteith method, 639–643 data required for, 639–640 FAO-24 pan evaporation method, 646 Farm ditch, 618 Feasibility, hydropower, 334–335 Federal Emergency Management Agency (FEMA), 6 Fiberglass, pipe material, 84–85 Field capacity, 693 Field procedures, aquifer pumping tests, 796–798 baseline conditions, 796–797 discharge-rate measurement, 798 length, 798 water-level measurement, 797–798 Filter layer, 611 Filter packs, See Gravel packs Filtration velocity, 663 Fire demand, 77–79 fire hydrants, 79 Fire engines, 79 Fire hydrants, 86–87 guidelines for the placement of, 86 riser, 86 Fire-service pumps, 85 Firm yield, 322 First flush, 587 Fish and Wildlife preservation, 7, 819

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Fisher-Tippett Type I distribution, 365 Fixed-flow method, culverts, 269–270 Fixed-headwater method, culverts, 265–269 Flexible linings, 567 channel design, 182–183 general design procedure, 183–187 Flexible pipes, 229 Flexible-boundary channels, 166–167 bends, 177–178 composite linings, 205–207 design procedure for, 182–187 grass lining, 189 Manning roughness coefficient for nonvegetative flexible-lining materials, 195 retardance, of linings, 194 riprap linings, 175, 199, 203 Float-type recording gages, 404 Flood control, 599–603 system design, 599 underground detention, 593 Flood, defined, 6 Flood stage, 6 Flood storage, 323 Flood-control pool, 323 Flood-control structures, 5–6 Floodplains, 6, 117–119 defined, 6 Manning roughness coefficient for, 118 Flood-risk zones, 118 Flow coefficient, 49 Flow in open channels, See Open-channel flow Flow routing (flood routing), 520 Flow, unsaturated zone, 691–696 Food and Agriculture Organization of the United Nations (FAO), 639, 646 Foot valve, centrifugal pumps, 46 Force mains, 229, 233–234 Forchheimer, Philipp, 682fn Forchheimer-Dupuit equation, 665 Formation losses, 789 Formation-loss coefficient, 790 Formations, 657 Formulation of alternatives, 816 Francis, James B., 330 Francis turbine, 330–332 Franklin D. Roosevelt reservoir (Lake Roosevelt), 319 Frechet distributions, 364 Freeboard, 178–179, 323 Free discharge, gates, 276–279 Free-surface flow, See Open-channel flow French drains, 612, 803, See also Exfiltration trenches Frequency analysis, 387–394 defined, 387

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extreme-value Type I distribution, 393–394 frequency factor, 388 gamma/Pearson Type III distribution, 390 general extreme-value (GEV) distribution, 394 log-normal distribution, 389 normal distribution, 388–389 Frequency factor, 388 Frequency-based design, water-resource systems, 5 Free-surface hydraulic jump, 141 Frontal lifting, 402 Frontal precipitation, 402 Frontal storms, 402 Frontal-flow interception efficiency, 555, 558 Fully turbulent flow, 15 Functional planning, defined, 815 Fundamentals flow in open channels, 103–159 groundwater hydrology I and II, 656–696, 700–761 surface-water hydrology I and II, 401–468, 473–539

G Gabions, 203–206 Gabion weirs, 298–299 Gamma distribution, 357–361 defined, 357 log-Pearson Type III distribution, 360 one-parameter, 359 Pearson Type III distribution, 359–360 three-parameter, 359 Gamma/Pearson Type III distribution, 357–359, 390–393 frequency analysis, 391–393 Gate valves, 85 Gated spillways, 307 Gates, 275–282 discharge coefficient, 277 empirical equation, 281–282 free discharge, 276–279 radial gates, 275 sluice coefficient, 277 submerged discharge, 279–281 Tainter gates, 275 vena contracta, 277 vertical gates, 275 Gates seat, spillway crest, 308–310 downstream, 309–310 Gaussian distribution, 360–361 General design guidelines, exfiltration trenches, 613–614 procedure, 183–187

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General extreme-value (GEV) distribution, 370, 394 frequency analysis, 387–388 General flow equation, control volume, 676–681 Geographic Information Systems (GIS), 577 Geographic scope criterion, plan evaluation, 817 Geohydrology, 656 Geologic units, 661 Geometric controls, 549 Geometric distribution, 353–354 Geometric mean, 674 Geometric mean friction slope method, 144–145 Geometric-gradient factors, 821–823 Geometric-gradient present-worth factor, 821–823 Geothermal gradient, 670 Ghyben-Herzberg approximation, 752–755, 756 Ghyben-Herzberg equation, 753, 756 Global hydrologic cycle, fluxes in, 4 Goals, 586 Gradient algorithm, 46 Gradually varied flow (GVF), 143 Grand Coulee Dam, 319–320 Granular materials, porosities of, 660 Graphical peak-discharge method, 492 Grass lining, 189–190 Grate inlets, 554–560 defined, 554–555 design procedures, 555 frontal-flow interception efficiency, 555, 558 in sag vertical curves, 553 side-flow interception efficiency, 555 splash-over, 555, 557 splash-over velocity, 555 Gravel packs, 783–784 length, 784 material, 783–784 thickness, 784 Gravitational water, 694 Gravity concrete dam, 319 Gravity drainage, 442 Great Lakes-Upper Mississippi River Board of State and Provincial Public Health and Environmental Managers (GLUMRB), 88–89, 91 Green-Ampt model, 447–453, 460–461, 495 computation of rainfall excess using, 451 typical values of Green-Ampt parameters, 449 Greenland Ranch, Death Valley, California, annual rainfall, 4 Ground storage reservoirs, 87–88

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921

Groundwater, recharge mechanism, 658 Ground-level tanks, 87 Groundwater drought, 326 Groundwater flow equation, solutions of, 741 hydraulic approach, 682 partially penetrating wells, 714–718 steady flow to a well in a confined aquifer, 702–705 in a leaky confined aquifer, 709–711 in an unconfined aquifer, 706–708 in an unconfined aquifer with recharge, 713–714 steady unconfined flow between two reservoirs, 700–702 in the unsaturated zone, 691–696 unsteady flow to a well in a confined aquifer, 728–729 in a leaky confined aquifer, 736–741 in an unconfined aquifer, 728–736 Groundwater hydrology, 656–696 aquicludes, 656 aquifers, 657 aquifuges, 65 aquitards, 658 artesian aquifers, 658 base flow, 658 capillary zone, 692 confined aquifers, 948 Darcy’s law, 662–666 defined, 656 engineered systems, 771–808 aquifer pumping test design, 794–798 exfiltration trench design, 803–808 seepage meters, 808 slug test, 798–803 water-supply well design/construction, 777–788 water-supply well performance assessment, 788–793 well drilling, 793 wellfield design, 771–773 wellhead protection, 774–777 general flow equation, 676–681 intermediate zone, 692 leaky-confined aquifer, 658 leaky-unconfined aquifer, 658 major application of the principles of, 56 method of images, 746–752 constant-head boundary, 746–749 impermeable boundary, 750–752 phreatic aquifers, 657 phreatic surface, 656 piezometers, 657–658 piezometric surfaces, 658 porosity, 659–660

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Groundwater hydrology (continued) potentiometric surfaces, 658 recharge areas, 657 representative elementary volume (REV), 659 rocks, 659 saltwater intrusion, 752–761 soil-water zone, 656 solid matrix, 658 springs, 658 subsurface environment, 660–661 geologic characterization of, 660–661 superposition principle, 741–742 multiple wells, 742–744 well in uniform flow, 744–746 support scale, 674–675 two-dimensional approximations, 681–691 confined aquifers, 687–691 unconfined aquifers, 681–686 unconfined aquifers, 657 vadose water, 656 vadose zone, 656 void space, 656 water table, 656 water-table aquifers, 657 zones of aeration, 657 zones of saturation, 657 Groundwater hydrology, applications, 700–760 method of images, 746–752 constant-head boundary, 746–749 image well, 747 impermeable boundary, 750–752 other applications, 752 other solutions, 741 hydraulic tomography, 741 principle of superposition, 741–746 capture zone, 745 multiple wells, 742–744 well in uniform flow, 744–746 saltwater intrusion, 752–760 buoyancy factor, 753 classification, saline groundwater, 760 Ghyben-Herzberg equation, 753 saltwater wedge, 752 upconing, 757 steady-state solutions, 700–718 leakage factor, 710 partially penetrating wells, 714–718 unconfined flow between two reservoirs, 700–702 well in a confined aquifer, 702–705 well in an unconfined aquifer, 706–708 well in a leaky confined aquifer, 709–713 well in an unconfined aquifer with recharge, 713–714 unsteady-state solutions, 718–741

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Boltzman variable, 719 Hantush well function, 737 Theis equation, 721 unconfined well function, 730 well in a confined aquifer, 718–728 well in a leaky confined aquifer, 736–741 well in an unconfined aquifer, 728–736 Grout, defined, 785–786 Grouting, water-supply well, 785–786 Groundwater systems, designs, 771–808 aquifer pumping tests, 794–798 exfiltration trenches, 803–808 seepage meters, 808 slug test, 798–803 well components, 778–783 well construction, 777–793 well drilling, 793 wellfields, 771–773 wellhead protection, 774–777 Gumbel extreme-value distribution, 364–366 Gutters, 545–548

H Hagen-Poiseuille flow, 664 Half-life, 776 Hammer drilling, 793 Hantush and Jacob well function for leaky aquifers, 736–738 Hardrock wells, 780 Hardy Cross method, 42–46 Harmonic mean friction slope method, 144–145 Hazen-Williams formula, 32–35 Head, 123 Head coefficient, 49 Head loss, 12–15, 96–97 in noncircular conduits, 31–32 in transitions and fittings, 27–30 Headwater depth, 252 elevation, 260–261 Heat from industry, as water contaminant, 3 High spillways, 299 High-lift pumps, 79 High-stage events, 595 HMRs, 430 Homogeneous, porous formation, 672 Homologous series of pumps, 49 Hoover Dam, 320 Horizons, vertical soil profile, 438–439 Horizontal collector well, 777–778 Horton model, 442–444, 460–461 Hortonian overland flow, 473 Hunter curves, 94 Hydraulically short jump, 138 Hydraulically long jump, 138

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Hydraulic approach, 682 Hydraulic conductivity, 666–676, 678 anisotropic properties, 670–673 classification of, 670 empirical formula, 666–670 statistical characterization of, 674–675 stochastic properties, 674–676 Hydraulic criteria, 231 Hydraulic depth, 127 Hydraulic grade line, 25–26 Hydraulic gradient, 663 Hydraulic head, 663 Hydraulic jump, 139–143 conjugate depths of, 143 Hydraulic radius, 13, 106 Hydraulic routing, 531–533 Hydraulic structures, design of, 250–335 culverts, 250–275 dams/reservoirs, 318–327 gates, 275–276 hydropower, 328–335 spillways, 299–312 stilling basins, 312–318 weirs, 282–299 Hydraulic tomography, 741 Hydraulics, manholes, 227–228 Hydraulics, sewers, 216–227 Hydraulically long culverts, 256 Hydraulically short culverts, 256 Hydraulic-routing models, 520–533 diffusion model, 531–532 dynamic model, 531 inertial terms, 531 kinematic model, 532 Saint-Venant equations, 531 Hydraulics of sewers, 216–228 Hydroelectric power generation, 7 Hydroelectric power (hydropower), 328–335 estimation of annual hydropower generation potential of a facility, 334 extraction of, from surface runoff, 323 impulse turbines, 328–330 double-overhung installation, 328 feasibility, 334–335 forebay, 592 hydraulic efficiency, 329–330 net head (effective head), 329 nozzle loss, 329 overall efficiency, 329–330 penstock, 328 schematic layout of installation, 328 single-overhung installation, 328 static head, 328–329 power plant classifications, 328 pumped-storage plant, 332 reaction turbines, 330–334 run-of-river plants, 328

w .E asy En g

923

Hydrogen sulfide control, 234–236 Hydrogeology, 656, 661 Hydrograph, 421 Hydrologic cycle, 1–5 Hydrologic data, 372–394 estimation of population distribution, 372–375 estimation of population parameters, 379–387 frequency analysis, 387–394 defined, 387 extreme-value Type I distribution, 393–394 frequency factor, 388 gamma/Pearson Type III distribution, 390–393 general extreme-value (GEV) distribution, 394 log-normal distribution, 389–390 normal distribution, 388–389 hypothesis tests, 376–378 L-moments, method of, 383–387 maximum likelihood method, 382–383 method of moments, 379–382 probability distribution of observed data, 372–375 Hydrologic risk, 524 Hydrologic-routing models, 520–530 modified Puls method, 520–524 Muskingum method, 524–530 Hydrologic units, 401 Hydrology, 1 subsurface, 1–2 Hydrometeorological reports (HMRs), 430 Hydropower, 328–335 feasibility, 334–335 turbines, 328 impulse, 328–330 performance, 333–334 reaction, 330–333 Hydropower generation potential, 334 Hydrostratigraphic units, 661 HYDRO-35, 410 Hyetograph, 422 Hygrometers, 633 Hygroscopic water, 694 Hypothesis tests, 376–378 chi-square test, 376–377, 387 Kolmogorov-Smirnov test, 377–378, 387

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I Ice-crystal process, 402 Igneous rocks, 659 Image well, 747 Images, method of, 746–752 constant-head boundary, 746–749 impermeable boundary, 750–752

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Impact assessment, 816–817 Impeller, 22 Impermeable boundary, 750–752 Implicit equation, 16 Impulse turbines hydraulic efficiency, 329–330 net head (effective head), 329 nozzle loss, 329 overall efficiency, 329–330 penstock, 328–329 static head, 328–329 Impulse wheel, 328 Incomplete hydraulic jump, 141 Index-of-dryness method, 651–653 Indirect benefit (external), 818 Indirect cost, 818 Industrial use, of water, 211 Ineffective storage, 89 Inertial terms, 531 Infectious agents, as water contaminant, 3 Infiltration, 437–464 capacity, 437 capillary potential, 439–440 comparison of models of, 460–461 defined, 213, 437 Green-Ampt model, 447–453 Horton model, 442–447 matric potential, 439 NRCS curve-number model, 453–460 percolation, 439–440 potential infiltration rate, 442 process, 439–442 soil texture, 438 specific moisture capacity, 440 specific water capacity, 440 suction gradient, 440 wetness gradient, 440 Infiltration and inflow (I/I), 213–214 Infiltration basins, 603–605 clogging/sedimentation, susceptibility to, 604 location of, 603–604 off-line, 603 on-line, 603 Infiltration excess, 473 Infiltration galleries, 617 Infiltration process, 439–440 Infiltration trenches, 612–613 Inflow, defined, 213–214 Initial depth, 140 Injection barriers, 757 Inlet depth factor, 280 Inlet efficiency, 549 Inlets, 487–509 bypass flow (carryover flow), 549 combination, 560–564 curb, 550–554

w .E asy En g

grate, 554–560 slotted-drain, 565–567 Inorganic chemicals, as water contaminant, 3 Installed capacity, 334 Instantaneous unit hydrograph (IUH), 501–502 Intangible benefit, 818 Intensity-duration-area-frequency (IDAF) curves, 429 Intensity-duration-frequency (IDF) curves, 405, 422, 431 Intensity-frequency-duration (IFD) curves, 405 Intercept coefficient, 480 Interception, 433–437 percentages, in selected studies, 434 Interflow, 464 Intermediate zone, 656, 692 Internal rate of return, 825–826 International Commission for Irrigation and Drainage (ICID), 639 Intrinsic permeability, 667 equations for estimating, 667 Inverse-distance-squared method, 418 Invert, 218 Irrigation systems, design of, 6 Izzard equation, 481–482, 483, 486

J

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Jain equation, 16–17, 32 Jetting, 793–735 John Day lock (Columbia River), 319 Joint probability density function, 345 Joint probability distribution function, 346 Joukowsky-Frizell equation, 36 Julian day, 629 Junctions, 576

K

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Kansas Geological Survey, 802 Kaplan turbine, 331 Kerby equation, 482–484, 486 Kerby-Hathaway equation, 482 Kinematic model, hydraulic-routing models, 532 Kinematic-wave equation, 474–477, 483–484 Kinematic-wave model, 475–476 Kinetic energy correction factors, 24 Kirpich formula\equation, 481 Kolmogorov-Smirnov test, 377–378, 387 Kriging, 417

L Land-use models, 71 Large catchments, 428 Latent heat, vaporization, 631 Laterals, 222, 777 Leaf area index (LAI), 435, 627

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Leakage factor, 710–711 Leaky aquifers, 658 Leaky well function, 737 Leaky-confined aquifer, 658 Leaky-unconfined aquifer, 658 Lift stations, 233, See also Pump stations Likelihood function, 382 Limited-service spillways, 332 Limits, pump location, 55–58 Linear-reservoir assumption, 512 Lithology of a stratigraphic unit, 661 LMNO Engineering and Research Software (LMNO), 290 L-moments, method of, 383–387 Local head losses, 29 Local rainfall, 410–416 Locks, dams, 319 Logistic curve, 73 Log-normal distribution, 362–363, 383 frequency analysis, 387–388 Log-Pearson Type III distribution, 360, 391 Long-based weirs, See Broad-crested weirs Longwave radiation, 629–630 Longitudinal slopes, 178, 545, 551 Long-term projections, 73 Loop method, 42–46 Hardy Cross method, 42–46 Low spillways, 299 Low-lift pumps, 79 Low-stage events, 595 Lumped-parameter models, 473

w .E asy En g M

Maximum permissible shear stress, 171 Mean areal precipitation, 416 Mean flow velocity, 13 Measurement, rainfall, 403–405 Mechanisms, surface runoff, 473–474 Median channels, 566–567 Metamorphic rocks, 659 Meteorological tables, 430 Meters, 85–86, 831fn Method of images, 746–752 constant-head boundary, 746–749 impermeable boundary, 750–752 Method of L-moments, 383–387 Method of moments, 379–382 Midsize catchments, 474 Minimum permissible velocity, 224 Minimum-performance method, 271 Minimum size, pipelines, 82 Minimum slope, self-cleansing, 237–239 Minor drainage systems, 619 Minor head losses, 29 Mixed flow, inlets, 559 Mixed-flow pumps, 46 Mixed-flow turbines, 330 Model selection criteria, 379 Modified drawdowns, 707–708 Modified Puls method, 520–524 Modular limit, 297 Modular submergence, 308 Moisture content, 693 Momentum coefficient, 105 Momentum correction coefficient, 11, 105 Moody diagram, 15–17 Mount Wai’ale’ale (Kauai, Hawaii) annual rainfall, 4 Multiple-pump systems, 58–60 Multiple-regression models, 70–71 Multiple wells, 742–744 Multislit weirs, 286 Multistage pumps, 47, 59 Multivariate probability distribution, 345 Municipal water-distribution systems, design protocol for, 70 Muskingum method, 524–530 defined, 524 hydrologic routing, 520 negative wedge, 524 objective function, 527 prism storage, 524 wedge storage, 524 Muskingum-Cunge method, 529–530

ine eri n

Macroscale roughness, 108 Macroscopic hydraulic conductivities, 675 Main sewer, 229 Major drainage system, 619 Manholes, 231–233, 576–577 drop, 576 Manning equation, 110–120, 167 with constant n, 218–220 and flow in sanitary sewers, 219 with variable n, 220–223 Manning formula, 32–34 Manning roughness coefficient, 117, 180, 253, 476 in closed conduits, 571–572 Manning’s n, 199–203 Marginal probability distribution function, 346 Mass-curve analysis, 326 Master plan, 73 Mathematical expectation and moments, 347–350 Matric potential, 439 Maximum daily demand, 76 Maximum hourly demand, 76 Maximum likelihood method, 382–383

925

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N Nameplate point, 49 Nappe, 283 National Climatic Data Center (NCDC), 7

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National Flood Frequency (NFF) Program, 77–79 National Water Well Association, 778 National Weather Service (NWS), 7, 366, 403–404, 410, 421, 430 National Weather Service River Forecast System (NWSRFS), 421 Natural Resources Conservation Service (NRCS), 7–8, 425, 442, 453fn, 492, 506 NRCS 24-hour hyetographs, 425–428 NRCS curve-number model, 453–460 antecedent runoff condition (ARC), 455–457 curve numbers, 453–460 initial abstraction, 453–454 potential maximum retention, 454 rainfall excess, 453 retention, 453–454 treatment, in agricultural practice, 455 USDA Natural Resources Conservation Service Soil Survey Geographic (SSURGO) database, 459 NRCS dimensionless unit hydrograph, 506–509 NRCS-TR55 method, peak-runoff models, 492–494 Negative wedge, 524 Net radiation, Penman-Monteith Equation, 627–630 Network analysis, water distribution system, 92–94 NRCS method, overland flow, 474 Natural uncertainty, 344 Navigation planning, 7 Needed fire flow, 77–79 Net positive suction head (NPSH), 54–58 Net pyranometer, 628 Net radiometers, 627 Neuman’s well function, 730 Newton-Raphson method, 46 Next-generation weather radar (NEXRAD), 404 Nodal method, pipe networks, 40–42 Nonhomogeneous, porous formation, 672 Nonlinear-reservoir model, 515–517 Nonoverflow dams, 322 Nonparametric test, 377 Nonprismatic channels, 103 Nonreinforced concrete culverts, 263 Nonresidential sources, wastewater, 212–213 Nonstructural SCMs, 618 Nonstructural management, 816 Nonuniform flow, 103 Normal depth of flow, 111 Normal distribution, 360–361, 388

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frequency analysis, 388 Nozzle loss, 329 n-year event, 262fn

O Objective function, 527 Objectives, 816 Observation well design, 795–796 distance from pumping well, 796 well construction, 796 well diameter, 795–796 Occupancy factor, 77 Office of Management and Budget (OMB), 819 Off-line infiltration basins, 603 Ogallala aquifer, 661 Ogee, defined, 299fn O-Horizon, 438 One-parameter gamma distribution, 359 On-line infiltration basins, 603 On-site runoff controls, 593 Open-channel flow, 103–159, See also Hydraulic structures, design of basic principles, 103 computation, water-surface profiles, 143 direct-integration method, 145 direct-step method, 147 hydraulic jump, 139 practical considerations, 150 profile equation, 132 standard-step method, 148 Darcy-Weisbach equation, 106–110 defined, 103 energy grade line, 125 Manning equation, 110–119 nonprismatic channels, 103 other equations, 119–120 prismatic channels, 103 sanitary sewers, design, 211–247 average per-capita wastewater domestic flow rates, 212 chlorination, 235 commercial and industrial wastewater flow rates, 211 depth of sewers, 231 design computations, 224–227 force mains, 214, 233–234 hydraulics of sewers, 216–228 hydrogen sulfide control, 234–236 inflow and infiltration (I/I), 213–214 main sewer, 229 manholes, 227–228 peak-flow factors, 214–216 pump stations, 233 saturation density, 212 service flow, 211–212

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Open-channel flow (continued) sewer-pipe material, 230 system layout, 229 trunk sewer, 215 specific energy, 125–132 steady-state continuity equation, 103–104 steady-state energy equation, 121–132 energy grade line, 125 specific energy, 125–132 steady-state momentum equation, 104–106 Darcy-Weisbach equation, 106–110 Manning equation, 110–119 velocity distribution in open channels, 120–121 water-surface profiles, 132–159 classification of, 134–139 computation of, 143–145 hydraulic jump, 139–143 profile equation, 132–134 Open-hole wells, 778 Operating point, pump, 54 Organic chemicals, as water contaminant, 3 Original porosity, 659 Orographic lifting, 403 Orographic storms, 403 Outlet works, 323 Overdamped response, 799 Overflow dams, 322 Overflow spillways, 299–302 Overland flow, 474–484 Izzard equation, 481–482 Kerby equation, 482–484 kinematic-wave equation, 474–477 Kirpich formula, 481 NRCS method, 478–481 time of concentration, 474–487 Izzard equation, 481–482 Kerby equation, 482–484 kinematic-wave equation, 474–477 Kirpich formula, 481 NRCS method, 478–481 Oxygen-demanding wastes, as water contaminant, 3

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P Pan coefficients, 644–647 Parent distribution, 364–365 Partial-duration series, IDF curves, 408–410 Partially penetrating wells, 714–718 Pavement encroachment, regulatory requirements for, 548 Peaking factors, 214–216 Peak-runoff models, 487–494 NRCS-TR55 method, 492–494 rational method, 487–492 Pearson Type III distribution, 357–360, 390

927

Pelton, Lester A., 328 Pelton wheel, 328 Pendular water, 693 Penetration fraction, 715–716 Penman-Monteith equation, 624–637 actual vapor pressure, 632–633 aerodynamic resistance, 625–626 air density, 633 application, 634–637 latent heat of vaporization, 631 net radiation, 627–630 psychrometric constant, 631–632 soil heat flux, 630–631 surface resistance, 626–627 vapor-pressure gradient, 632 Penstock, 328–329 Per-capita water use, distribution of, 70, 71–72 Perched water table, 657 Percolation, 439 Percolation trenches, 612, See also Exfiltration trenches Percussion drilling, 793 Performance assessment, pumping rate, 788–793 Performance criteria, water-distribution systems, 90–93 allowable velocities, 91 network analysis, 92–93 service pressures, 91 water quality, 91–92 Performance goals, stormwater-management systems, 586 Permissible shear stress, 171 Phreatic aquifers, 657 Phreatic surface, 656 Piano key weirs, 294 Piezometers, 657–658 Piezometric surfaces, 658, 681 Pilot hole, 778 Pipelines, 81–82 Pipe materials, 83–85, 229–231 asbestos cement, 84 concrete, 85 ductile iron, 83 fiberglass, 84–85 plastic, 84 service lines, 83 sewers, 229–231 steel, 84 Pipe networks, 39–48 computer programs, 46 continuity equation, 39 energy equation, 39 loop method, 42–46 nodal method, 40–42 Pipe roughness coefficients, 33

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Pipelines, 81–85 arterial mains, 81 depth, 82 distribution mains, 83 material, 83–85 minimum size, 82 service lines, 83 transmission lines, 81 trench width, 82 Pipe, sizing and selection, 571–575 Planning process, 815–778 analysis of alternatives, 816–817 data collection and analysis, 815–816 defined, 815 economic evaluation, 816–817 evaluation, 817 formulation of alternatives, 816 goals, 816 impact assessment, 816–817 implementation, 817 objectives, 816 problem diagnosis, 816 problem identification, 815 recommendations, 817 surveillance and monitoring, 818 Plant nutrients, as water contaminant, 3 Poisson distribution, 354–355 Poisson process, 354, 357 Poisson rectangular pulse (PRP) model, 215 Polyvinyl chloride (PVC) casing, 779 Poorly graded aquifer material, 781 Population distribution, estimation of, 372–394 chi-square test, 376–377 extreme-value Type I distribution, 393–394 frequency analysis, 387–388 Gamma/Pearson Type III distribution, 390 general extreme-value (GEV) distribution, 394 hypothesis tests, 376 Kolmogorov-Smirnov test, 377–378 log-normal distribution, 389–390 log-Pearson Type III distribution, 391–393 maximum-likelihood method, 382–383 method of L-moments, 383–387 method of moments, 379–382 model selection criteria, 379 normal distribution, 388–389 parameters, 379 probability distribution, 372–374 Population forecasting declining-growth phase, 73 logistic curve, 73 long-term projections, 73

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short-term projections, 73 and water demand, 70 Population parameters, estimation of, 379 Porosity, 659 Positive displacement pumps, 46 Potential evaporation, 624 Potential evapotranspiration, 624, 637–638 Potential infiltration rate, 442 Potentiometric surfaces, 658, 681 Pour point, 401 Power law equation, 27 Present value, 819 Present-worth analysis, 823–825 Pressure-relief valves, 39 Pressure sewers, 234 Pretreatment, stormwater, 533 Primary factors, WHPAs, 774–775 Primary porosity, 659 Principal axes, 671 Principle of invariance, 498 Principle of superposition, 741–746 multiple wells, 742–744 well in uniform flow, 744–746 Prismatic channels, 103 Prism storage, 524 Probability density function, 346–347 Probability distribution function, See Probability functions Probability distribution of observed data, 372–375 Probability distributions, 344–372, See also Hydrologic data continuous probability distributions, 346–347 defined, 345 continuous sample space, 345 discrete probability distributions, 345–346 defined, 345 discrete sample space, 345 mathematical expectation and moments, 347–350 probability functions, 346 common, 351–353 return period, 350–351 Probability functions, 351–372 binomial distribution, 351–353 central limit theorem, 361–362 chi-square distribution, 371–372 common, 351 exponential distribution, 356–357 extreme-value distributions, 364 extreme-value Type I (Gumbel) distribution, 364–366 extreme-value Type III (Weibull) distribution, 366–368 Frechet distributions, 295

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Probability functions (continued) generalized extreme-value (GEV) distribution, 369–371 gamma distribution, 357–365 defined, 359 log-Pearson Type III distribution, 360 one-parameter, 359 Pearson Type III distribution, 359–360 three-parameter, 359 geometric distribution, 353–354 log-normal distribution, 362–363 normal distribution, 360–361 Poisson distribution, 354–355 uniform distribution, 363–364 Probability theory, in water-resources engineering, common application of, 344 Probable maximum flood (PMF), 519–520 Probable maximum precipitation (PMP), 429–432 rational estimation method, 429 statistical estimation method, 430–432 world-record precipitation amounts, 432 Probable Maximum Storm (PMS), 432–433 Problem diagnosis, 816 Problem identification, 815 Procedure for system design, sanitary sewers, 240–247 Profile equation, water surface, 132–134 Profiles, bridges, 154–159 Profitability index, 825 Propeller turbine, 330 Public-supply wells, 77 Pump-housing casing, 779 Pump stations, 233 design of, 233 dry well, 233 lift station, 233 wet well, 233 Pump systems, 58–59 Pump turbines, 332 Pumped-storage facilities, 332 Pumper trucks, 79 Pumping well design, 794–795 discharge-control and measurement equipment, 795 pump selection, 795 reliable power source, 795 water disposal, 795 water-level measurement access, 795 well construction, 794–795 well development, 795 Pumps, 46–62, 70 affinity laws, 51–53 airlift pumps, 784–785 axial flow pumps, 46, 50 best-efficiency point (BEP), 49

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929

categories of, 46 centrifugal pumps, 50 design point, 49 homologous series of, 49 specific speed, 50 jet pumps, 784–785 mixed-flow pumps, 46 multiple-stage pumps, 784 multistage pumps, 59 nameplate point, 49 pneumatic pumps, 784–785 positive displacement pumps, 46, 784–785 rotodynamic (kinetic) pumps, 46 selection, 53–58 cavitation, 55 characteristic curve, 53 commercially available pumps, 53–54 limits on pump location, 55–58 operating point, 54 performance curve, 53 pitting, 55 required net positive suction head, 54 system characteristics, 54 system curve, 54 selection guidelines, 51 selection of, multiple-pump systems, 58–60 and service pressures, 85 single-stage pumps, 47 sizes and performance characteristics, 784–785 submersible pumps, 785 suction pumps, 785 three-stage pumps, 47 vertical turbine pumps, 784, 785 water-supply well, 777–778 Pump stations, design criteria, 233 Pyranometers, 628 Psychrometric constant, 631–632

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Quality control stormwater-management systems, 586–587 Quantity control stormwater-management systems, 586 Quantity, wastewater, 211–216 Quickflow, 468

R Radial gates, 275 Radial well, 777–778 Radial-flow pumps, efficiency of, 50 Radioactive substances, as water contaminant, 3

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Radius of influence, 704 Rainfall, 8, 401–433 aerosols, 402 Bergeron–Findeisen process, 402 cloud droplets, formation of, 402 convective lifting, 402 convective storms (thunderstorms), 403 design, 421–433 alternating-block method, 423–425 depth, 422 duration, 422 NRCS 24-hour hyetographs, 425–428 return period, 421 spatial distribution, 428–429 temporal distribution, 422 extreme, 429–433 probable maximum precipitation (PMP), 429–432 probable maximum storm (PMS), 432–433 standard project storm (SPS), 519 formation of precipitation, 401 frontal lifting, 402 ice-crystal process, 402 orographic lifting, 402–403 orographic storms, 403 spatially averaged, 416–421 Rainfall abstractions, 433–464 defined, 375 depression storage, 437 infiltration, 437–442 capillary potential, 382 comparison of models of, 400–401 Green-Ampt model, 387–393 Horton model, 442–447 matric potential, 439 NRCS curve-number model, 453–460 percolation, 439–441 potential infiltration rate, 442 process, 439–442 specific moisture capacity, 440 specific water capacity, 440 suction gradient, 440 wetness gradient, 382 interception, 433–437 percentages, in selected studies, 433 rainfall excess on composite areas, 461–464 Rainfall depth, 422 Rainfall duration, 422 Rainfall excess, 451, 461 Rainfall excess on composite areas, 461–464 Rainfall Frequency Atlas (U.S. Weather Bureau), 410 Rainfall measurements, 405–408 Rain shadow effect, 403 Random space function (RSF), 674

w .E asy En g

Rainfall statistics, United States, 410 Random variables, 344 variance, 347 Rapidly varied flow (RVF), 143 Rapid valve closures, 38 Rate of return, defined, 825–826 Rate-of-return analysis, 825–826 Rating curve, 265, 596 Rational estimation method, 430–432 probable maximum precipitation (PMP), 429–430 Rational method, 474, 487–492 Reaction turbines, 330–335 axial-flow turbine (propeller turbine), 330 defined, 330 draft tube, 331 effective head, 329 Francis turbine, 330–332, 333 net head (effective head), 329 operation of, 331 Real discount rate, 819 Recharge areas, 657 Reciprocal-distance approach, 418 Recommendations, in planning, 817 RECP linings channel design, 197 Manning’s n, 184 Rectangular weirs, 282–287, 294–297 Recurrence interval, 344 Reference evapotranspiration, 638–651 ASCE Penman-Monteith method, 643–644 defined, 638–639 empirical methods, 648–651 evaporation pans, 644–648 FAO56-Penman-Monteith method, 639–643 Reference-crop evapotranspiration, 624, 638 Regional runoff controls, 619 Regression equations, 535–539 Reinforced concrete sewer pipe, 231 Reliability, 344, 353 Remedial action zone, 774 Remote sensing, 653–654 Removal efficiency, 587 Representative elementary volume (REV), 659 Required fire-flow durations, 79 Required net positive suction head (NPSHR), 54, 58 Reservoirs, 318–326 capacity, probabilistic analysis of, 319–320 emergency spillway, 323 firm yield, 322 outlet works, 323

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Reservoirs (continued) pumped-storage reservoir, 332 secondary yield, 322 sedimentation amounts, prediction of, 323 service spillway, 323 storage, 322–327 active storage, 322–323 dead/inactive storage, 322 flood storage, 323 freeboard, 323 storage calculations, 326 storage zones, 323–326 trap efficiency, 323 water demand, 323 Residential sources, wastewater, 211–212 Retardance, of linings, 194 Retention curve, 693 application, 693–694 Retention ponds, 618 Retention swales, 606 Retention systems, 610–617 exfiltration trenches, 612–617 infiltration basins, 603–605 swales, 605–609 vegetated filter strips, 610 Return on investment, 825–826 Return period, 350–351 design rainfall, 421 Reversibility criterion, plan evaluation, 817 Reynolds number, 13–17, 27, 29, 664–665 shear velocity, 108 Reynolds, Osbourne, 14fn R-Horizon, 439 Rigid linings, 166 channels design, 180–182 roughness coefficients, 180 Rigid pipes, 229 Rigid-boundary channels, 166, See also Rigid linings Rippl analysis, 326 Riprap linings, 175, 183 outlet protection, 274–275 Riser, 86 Risk, in design of water-resource systems, 5–6 Risk of failure, 344 Risk-based design, water-resource systems, 5 River-basin studies, 815 Roadside channels, 566–567 Roadway overtopping, 271–273 Rocks, 659–660 formations, 660 Rorabaugh method, 792 Rotary drilling, 793 Rotodynamic (kinetic) pumps, 46–47 Rotor, 22 Rough pipes, 14

w .E asy En g

931

Rough turbulent flow, 34–35 Routing models, 520–533 flow routing (flood routing), 520 hydraulic-routing models, 520–533 diffusion model, 531 dynamic model, 531 inertial terms, 531 kinematic model, 532 Saint-Venant equations, 531–532 hydrologic-routing models, 520 modified Puls method, 520–524 Muskingum method, 524–530 routing, defined, 529 Runner, 22, 328 Runoff models, 473–538 base flow, 530 continuous-runoff models, 495–509 extreme runoff events, 519–520 instantaneous unit hydrograph kinematic-wave model, 474–477 nonlinear-reservoir model, 515–517 Santa Barbara Urban Hydrograph (SBUH) model, 517–519 time-area models, 509–514 unit-hydrograph model, 502–509 unit-hydrograph theory, 495–501 defined, 473 direct runoff, 464 distributed-parameter models, 473 effective rainfall, 473 Hortonian overland flow, 473 infiltration excess, 473 interflow, 464 lumped-parameter models, 473 peak-runoff models, 487–494 NRCS-TR55 method, 492–494 rational method, 487–492 USGS regional regression equations, 535–537 quickflow, 468 rainfall excess, 461–464 rational method, 474 saturation excess, 473 scale of catchment, 474 surface mechanisms, 473–474 time of concentration, 474–487 accuracy of estimates, 486–487 channel flow, 486–487 overland flow, 486–487 unit hydrograph models, 474 variable source-area hydrology, 473 Run-of-river plants, 328

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S Safe aquifer yield, 789 Safe yield, 322 Sag curb inlets, 553

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Saint-Venant equations, 531–532 Saline groundwater, 760–761 classification of, 760–761 Salinity intrusion, 752 Saltwater intrusion, 752–761 caused by the density difference between saltwater and fresh water, 756 and coastal drainage canals, 756 upconing, 757 Saltwater wedge, 752 Sand bridge, 786 Sand drive, 780 Sand trap, 786 Sanitary sewers, design, 211–247 average per-capita wastewater domestic flow rates, 212 chlorination, 235 commercial and industrial wastewater flow rates, 211 depth of sewers, 231 design computations, 224–227 force mains, 214, 233–234 hydraulics of sewers, 216–228 hydrogen sulfide control, 234–236 inflow and infiltration (I/I), 213–214 main sewer, 229 manholes, 227–228 peak-flow factors, 214–216 pump stations, 233 saturation density, 212 service flow, 211–212 sewer-pipe material, 230 system layout, 229 trunk sewer, 215 Santa Barbara Urban Hydrograph (SBUH) model, 517–520 Saturation density, 243 Saturation excess, 473 Saturation, vapor pressure, 632 Scale effect, 675 Scour prevention, 224 Screen intake, 779–783 depth of screen in unconsolidated formations, 782 design constraint in sizing, 781–782 diameter constraints, 780–781 entrance-velocity constraints, 780 length, 782 materials, 782–783 minimum screen diameters, 781 screen sections, 782–783 slot size, 781–782 uniformity coefficient, 781 well screen diameters, 780–781 Scroll case, 330 Secondary estimation, IDF curves, 410–416 Secondary porosity, 659

w .E asy En g

Secondary yield, 322 Sediment accumulation, 323–326 Sediment from land erosion, as water contaminant, 3 Sedimentary rocks, 659 Seepage meters, 808 Seepage velocity, 663 Segment, storm sewers, 568 Selection, ET estimation method, 654 Selection of SCMs, water-quality control, 618–619 nonstructural SCMs, 618 other considerations, 619 structural SCMs, 618–619 Self-cleansing velocities, 223–224 Semivariogram, 417 Sequent depth, 140 Sequent depth ratio, 140 Service flow, sanitary sewers, 211–212 Service lines, 83 Service pressures, water-distribution system, 91 Service spillway, 322 Sewage treatment plant design, 211 Sewer-pipe material, 229–231 Sewers, See Sanitary sewers, design Shaft spillways, 322 Shaft work, 22 Shallow concentrated flow, 478 Shape factor, catchments, 503 Shape number, 50 Sharp-crested weirs, 282–294 as control structure, 286 crest height, 284 Shear velocity Reynolds number, 108 Sheet flow, 478 Sheilds diagram, 172 “Short green crop,” 624 Short-term projections, 73 Shortwave radiation, 627–639 Shoulder, weir, 298 Shutoff valves, 85 S-hydrograph, 497–499 Side slopes of excavated channels, 178 Side-channel spillways, 322 Side-flow interception efficiency, 555 Sill, 285 Single-overhung installation, 328 Single-payment compound amount factor, 819–820 Single-payment present-worth factor, 819–823 Single pipelines closed-conduit flow diameter for a given flow rate and head loss, 24–26 empirical friction-loss formulae, 32–35 energy and hydraulic grade lines, 25–26

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Single pipelines (continued) flow rate for given head loss, 19–20 head losses in noncircular conduits, 31–32 head losses in transitions and fittings, 27–30 steady-state continuity equation, 9–10 steady-state energy equation, 22–32 steady-state momentum equation, 10–19 velocity profile, 27 Single-stage pumps, 47 Single-stage riser, 593–595 Singular open-channel section, 127 Sinking-fund factor, 820 Sizing calculations, culverts, 264–271 Skeletonization, 92 Skewness, 348 Skewness coefficient, 348 Skin effect, 789 Slit weirs, 287 Slope stability, exfiltration trench, 806 Slot size, 781–782 Slotted-drain inlets, 565–566 installed perpendicular to the flow direction, 566 in sag locations, 566 slotted drains, main advantage of, 566 Slow valve closures, 38 Slug test, 798–803 Bouwer and Rice method, 799 data analysis, 800–802 defined, 798 guidelines, 802–803 uses of, 799–800 Slugging, 220 Sluice coefficient, 277 Sluice, defined, 275–276 Sluice gates, 275–276 Small catchments, 474 Smooth pipes, 14 Snowpack Telemetry (SNOTEL) system, 8 Snyder unit-hydrograph model, 503–505 variability in parameters, 503–504 Soil Conservation Service (SCS), 453fn Soil heat flux, 630–631 Soil horizons, 438–439 Soil texture, 438 Soil water, 656 Soil-water zone, 656, 692 Solid matrix, 658 Sorption, and groundwater contamination, 775 Spatial distribution of storms, 428–429 Spatially averaged rainfall, 416–421 Specification design flows, 94–101 minimum pressure, 94–96

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933

Specific discharge, 447, 670 Specific energy, 125–132, 294 Specific moisture capacity, 440 Specific momentum, 140 Specific speed, 50–51 Specific storage, 677 Specific water capacity, 440 Specific yield, 683 Spillways, 299–312 chute, 301, 312 gated, 307–311 high, 299 limited-service, 322 low, 299 overflow, 299 shaft, 322 side-channel, 322 trough, 322 uncontrolled, 307 Splash-over, 555 grate inlets, 554–555, 559 Splash-over velocity, 555 grate inlets, 555 Springs, 658 St. Venant equations See Saint-Venant equations Stability criterion, plan evaluation, 817 Stage-discharge relationship, general form of, 596 Standard deviation, 360 Standard error, 381 Standard fire stream, 79 Standard normal deviate, 361 Standard project flood (SPF), 519–520 Standard project storm (SPS), 519 Standard-step method, water-surface profiles, 145, 148–150 Standpipes, 88–90 Stanton diagram, See Moody diagram Static head, 55 Static suction head, 55 Statistical estimation method, 430–432 probable maximum precipitation (PMP), 432–433 Statistics, rainfall data, 405–416 Steady nonuniform flows, 103 Steady open-channel flow, 103 Steady-state continuity equation, 103–104 open-channel flow, 103–104 single pipelines, 9–10 Steady-state energy equation, 121–132 energy grade line, 125 open-channel flow, 121–132 energy grade line, 125 specific energy, 125–132 single pipelines, 22–35

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Steady-state momentum equation, 104–121 Darcy-Weisbach equation, 106–110 Manning equation, 110–119 open-channel flow, 104–121 Manning channel flow, 111–120 velocity distribution in open channels, 120–121 single pipelines, 10–35 velocity distribution in open channels, 120–121 Steady-state solutions partially penetrating wells, 714–718 unconfined flow between two reservoirs, 700–702 well in a confined aquifer, 702–704 well in a leaky confined aquifer, 709–713 well in an unconfined aquifer with recharge, 713–714 well in an unconfined aquifer, 705–708 Stefan-Boltzmann constant, 629 Step-drawdown pumping test, 790 Stilling basins, 143, 250, 312–318 chute blocks, 312 matching the tailwater and conjugate depth, 316 setting the floor elevation of, 314–315 Type I, 312 Type II, 314 Type III, 314 Type IV, 312–313 Stochastic processes, defined, 344 Stochastic properties, hydraulic conductivity, 674–676 Stomatal resistance, 626 Storage coefficient, confined aquifers, 687 Storage impoundments, 587–603 detention basins-design parameters, 588–589 dry detention basins, 592–593 emergency spillways, 597–599 flood control, design, 599–603 outlet structures, design, 593–597 wet detention basins, 590–592 Storage-indication method, 521 Storativity, 687 Storm drains, 567 Storm sewers, 545–549 access holes, 576 clean-out structures, 576 combined-sewer systems, 545 defined, 545 Stormwater best-management practices (BMPs), 586 Stormwater-collection systems 545–567 design combined-sewer systems, 545 drainage ways, 545

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inlets, 549–550 combination, 560–564 curb, 550–554 grate, 554–560 slotted, 565 street gutters, 545–549 roadside and median channels, 566–567 Stormwater control measures, 589–617 design biofiltration swales, 607–609 bioretention systems, 610–612 detention basins, parameters, 588–589 dry detention basins, 592–593 exfiltration trenches, 612–613 flood control, 599, 603, 614–616 general guidelines, 613–614 infiltration basins, 603–605 outlet structures, 593–599 retention swales, 606–607 storage impoundments, 587 subsurface exfiltration galleries, 617 swales, 605–606 vegetated filter strips, 610 water-quality control, 616–617 wet detention basins, 590 Stormwater impoundments, 587–592 detention basins, 592 detention ponds, 590 dry basins, 587 retention basins, 587 single-stage riser, 593–595 two-stage riser, 595 Stormwater inlets, 549–566 Stormwater-management systems, See also Stormwater-collection systems; Stormwater control measures design of, 586–619 best-management practices (BMPs), 586 stormwater control measures (SCMs), 586 performance goals, 586 quality control, 586–587 quantity control, 586 Storm sewers, 567–583 design flows, calculation of, 568–571 pipe sizing and selection, 571–575 manholes, 576–578 impervious area, determination of, 578 system-design computations, 578–583 other design considerations, 583 Stratigraphic unit, 661 Stratigraphy, 661 Stream flows, 658 Street gutters and inlets, 545–566 Street hydraulic conveyance capacity, 548 Street-sweeping wastes, disposal of, 538

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Strickler coefficient, 111 Strickler equation, 111 Structural measures, planning process, 816 Structural SCMs, 618–619 Subcritical flows, 127 Submerged discharge, gates, 279–281 Submerged entrances, culverts, 252–259 Submerged weirs, 286 Submersible pumps, 784–785 Subsurface environment, 656 geologic characterization of, 661–662 Subsurface exfiltration galleries, 617 Subsurface hydrology, 1–2 Suction gradient, 440 Suction lift, 55 Sump curb inlets, 553 Sunshine duration, and Campbell-Stokes heliograph, 629 Supercritical flows, 127 Superposition principle, 741–746 multiple wells, 742–744 well in uniform flow, 744–746 Supervisory control and data acquisition (SCADA) systems, 90 Support scale, 658–659, 674 Surcharge capacity, 323 Surface casing, 779 Surface Energy Balance Algorithm for Land (SEBAL), 653 Surface resistance, Penman-Monteith Equation, 626–627 Surface water, 2 Surface-water hydrology catchment, 428–429 defined, 401 drainage basins (areas), 519 evapotranspiration, 624–654 hydrograph, 429 hyetograph, 422 rainfall, 401–433 rainfall abstractions, 433–464 routing models, 520–533 flow routing (flood routing), 520 hydraulic-routing models, 520 hydrologic-routing models, 520–524 routing, defined, 520 runoff models, 495–520 stormwater-management systems, design of, 586–619 water-quality models, 533–539 EPA model, 537–539 event-mean concentrations (EMCs), 533–535 USGS model, 535–537 watershed, 401 Surge-relief valves, 39 Surge tanks, 39

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935

Surveillance and monitoring, 818 Susceptibility assessments, 85 Swales, 605–609 biofiltration, 607–609 retention, 606–607 Swamee-Jain equation, 16 Sweeper configuration, 560 Sweeper inlet, 560 Synthetic design storms, 421 Synthetic unit hydrograph, 502–503 System characteristics, pump selection, 54–55 System-Design Computations, storm drain, 578–583 System design criteria, sanitary sewer combined sewers, 236 depth, 231 diameter and slope of pipes, 231 force mains, 233–234 hydraulic criteria, 231 hydrogen-sulfide control, 234–236 manholes, 231–233 pipe material, 229–231 pump stations, 233 system layout, 229 System curve, 54, 57

T Tailrace, 331 Tailwater, 252 Tainter gates, 275–276 Tainter, Jeremiah B., 275fn Technical Memorandum Number NWS HYDRO-35 (NOAA), 410 Temporal distribution alternating-block method, 423–425 NRCS 24-hour hyetographs, 425–428 triangular method, 422–423 Temporal variations, water demand, 76–77 Test hole, 778 Theis equation, 721–722, 724 Thermoplastic casing, 779 Thiem equation, 704, 705fn Thiessen polygon approach, 417 Thin-plate weirs, See Sharp-crested weirs Thomson weirs, 289 Three Gorges Dam (China), 334 Three-parameter gamma distribution, 359, 390 Three-parameter Weibull distribution, 368 Three-stage pumps, 47 Thunderstorms, 403 Time of concentration, 474–487 accuracy of estimates, 486–487 channel flow, 484–486 defined, 474 design storms, 421 overland flow, 474–484 Izzard equation, 481–482, 487

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Time of concentration (continued) Kerby equation, 482–484 kinematic-wave equation, 478, 480, 485 Kirpich formula, 481, 483–484 NRCS method, 478–481, 486 Time of travel (TOT) approach, 775–777 Time-area models, 509–514 Time-series analysis, 372 Toe, 319 Topsoil, 438 Total dissolved solids (TDS), 760 Total drawdown, 748 Total dynamic head, 47 Total head, 12 at cross section of a pipe, 25–26 Total suspended solids (TSS), 534 TP-40, 410, 414 TP-42, 410 TP-43, 410 TR 20, 492 Tractive force ratio, 174 Tractive force method, 223 Transition turbulent flow, 107 Transmission lines, 81 Transmissivity, of confined aquifers, 689–690 Transpiration, defined, 624 Trap efficiency, 323 Tremie, 786 Trench hydraulic conductivity, 803 Triangular method, temporal distribution, 422–423 Triangular weirs, 288 Trough spillways, 322 Trunk sewer, 215fn Trunnion, 276 TR-55 method, for estimating peak runoff, 492 Tube turbine, 331 Turbines impulse, 328–330 performance, 333–334 reaction, 330–333 Turbulent pipe flows, empirical formulae for estimating the friction factor in, 14 Two-Dimensional Approximations, 681–691 confined aquifers, 687–691 unconfined aquifers, 681–687 Two-stage riser, 595 Type curve, 722 Type number, 50

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U Uncertainty analysis, 395–397 Unconfined aquifers, 657, 681–686 Unconfined flow between two reservoirs, 700–702

Unconfined well function, 730 Unconsolidated rocks, 660 Uncontrolled spillways, 299–307 Underdamped response, 799 Underground detention, 593 Underground exfiltration galleries, 617 Uniform annual series factors, 820 Uniform distribution, 363–364 Uniform flow, 103 Uniformity coefficient, 667 Uniform-series compound-amount factor, 820 Uniform-series present-worth factor, 820 Unit hydrograph models, 474, 502–509 defined, 502 instantaneous, 501–502 Unit tractive force, 170 Unit-hydrograph model, 495, 502–509 accuracy of, 509 NRCS dimensionless unit hydrograph, 506–509 Snyder unit-hydrograph model, 503–505 Unit-hydrograph theory, 495–501 Unsaturated zone capillary zone, 692 defined, 691 distribution of moisture capacity within, 692–693 gravitational water extracted from, 694 groundwater flow in, 778–783 intermediate zone, 692 soil-water zone, 692 Unsteady nonuniform flows, 103 Unsteady open-channel flow, 103 Unsteady-state solutions others, 741 well in a confined aquifer, 718–728 well in a leaky confined aquifer, 736–741 well in an unconfined aquifer, 728–736 Unsubmerged entrances, 260–262 Unsuppressed (contracted) weirs, 282 Upconing, 757 Urban roadway design, 545 Urban runoff, quality of, 533 Urban stormwater control systems, 6 design of, 533 U.S. Army Corps of Engineers (USACE), 7–8, 127, 132, 144, 416, 418, 425, 504, 519, 817, 819 HEC-RAS, 256 Hydrologic Engineering Center (HEC), 511–512, 526 U.S. Bureau of Reclamation (USBR), 7, 178, 290–291, 299, 312–315, 319–321, 780, 819

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U.S. Department of Agriculture (USDA), 132, 437 Natural Resources Conservation Service Soil Survey Geographic (SSURGO) database, 459 soil classification system, 438 soil texture triangle, 438 U.S. Environmental Protection Agency (USEPA), See Environmental Protection Agency (EPA) U.S. Federal Emergency Management Agency (FEMA), 6 U.S. Federal Highway Administration (USFHWA), 171, 178 U.S. Geological Survey (USGS), 7, 120, 255, 533, 816 water quality model, 533–539 Water Resources Division, 7 U.S. National Weather Service (NWS), 7, 366, 430 Weather Surveillance Radar 1988 Doppler (WSR-88D), 404 U.S. Natural Resources Conservation Service (NRCS), 7–8 U.S. water-resources agencies, 7 U.S. Weather Bureau, Technical Paper Number 40 (Rainfall Frequency Atlas), 410

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Vadose water, 656 Vadose zone, 656, See also Unsaturated zone Valves, 85 air-and-vacuum relief, 85 air-relief, 85 blowoff, 85 drain, 85 gate, 85 shutoff, 85 Vapor-pressure gradient, 632 Variable source-area hydrology, 473 Variable-speed pumps, 60–62 Variance, random variable, 347 Varied flow, 103 Vegetated filter strips, 610 Vegetative linings, 187–196 alternative retardance-based design, 195–196 channel shapes, 188–189 effective shear stress, 189–194 retardance-based design, 194–195 Velocity distribution in open channels, 120–121 Velocity profile, 27 Vena contracta, 277 Vertical lift gates, 275 Vertical sluice gates, 275–276

937

Vertical soil profile, horizons, 438 Villemonte’s formula, 286, 291 V-notch weirs, 288–299 broad-crested weirs, 294–297 compound weirs, 291–293, 297–299 defined, 288 Kindsvater and Shen equation, 290 open-channel flow measurement guidelines (USBR), 156 sharp-crested weirs, 293–294 Void space, 658 ´ an ´ constant, 108 von Karm

W Warm front, 402 Wastewater flows, 211 Wastewater-treatment plants, location of, 235 Water consumption, 70 Water contaminants, classes of, 3 Water-control systems, 5 Water demand, 64–75 average daily demand, 76 demand factors, 76 design flows, 79–81 fire demand, 77–79 maximum daily demand, 76 maximum hourly demand, 76 peaking factors, 76 per-capita, distribution of, 71 population forecasting, 72–76 declining-growth phase, 73 logistic curve, 73 long-term projections, 73 short-term projections, 72–73 typical distribution of, 72 variations in, 76–77 Water environment, negative impact on, 298 Water hammer, 35–39 Water-hammer pressure, 36 Water mains, trenches for, 81 Water meters, 71 Water quality distribution system, 91–92 hydrologic cycle, 3 models event-mean concentrations, 533–535 regression equations, 535–539 Water resources, distribution of, 3–5 Water Resources Division, U.S. Geological Survey (USGS), 7 Water table, 617 Water use systems, 6–7 domestic supply, 6 domestic wastewater-collection, 6 hydroelectric power, 7 irrigation, 6

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Water yield, 653 Water-control systems, 5–6 Water-distribution systems components, 81–85 design of, 70–93 design periods/capacities in, 80 fire hydrants, 86–87 meters, 85–86 minimum acceptable pressures in, 91 network analysis, 92–93 performance criteria for, 90–93 pipelines, 81–85 pumps, 85 skeletonizing, 92–93 supervisory control and data acquisition (SCADA) systems, 90 valves, 85 water demand, 70–81 water quality, 91–92 water-storage reservoirs, 87–90 Water-quality control detention systems, 618–619 dry-detention basins, 592–599 exfiltration trenches, 612–617 infiltration basins, 603–605 swales, 605–609 vegetated filter strips, 610 water-quality volume, 588 wet-detention basins, 590–592 Water-quality models, 533–538 EPA model, 537–538 event-mean concentrations (EMCs), 533–535 USGS model, 535–537 Water-quality volume, 587 Water-resource systems, 6 critical-event design, 5 design of, 1, 5–6 examples of, 1 failure of, 5 frequency-based design, 5 risk-based design, 5 water-control systems, 5–6 water-use systems, 6–7 Water-resources agencies, U.S., 7 Water-resources engineering core science of, 1 defined, 1 hydrologic cycle, 1–5 Internet sites relevant to, 8 probability/statistics in, 344–397 technical areas fundamental to, 1 Water-resources planning and management benefits of, 815 comprehensive planning, defined, 815 dams and reservoirs, 318–327 economic feasibility, 818–828

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benefit-cost analysis, 818–819 compound-interest factors, 819–823 evaluating alternatives, 823–828 functional planning, defined, 815 hydroelectric power (hydropower), 328–330 impulse turbines, 857–860 reaction turbines, 330–333 Water resources, planning process, 815–818 analysis of alternatives, 816–817 data collection and analysis, 815 defined, 815 economic evaluation, 818–819 evaluation, 817 formulation of alternatives, 816 goals, 816 impact assessment, 816–817 implementation, 817 objectives, 816 problem diagnosis, 816 problem identification, 815 recommendations, 817 surveillance and monitoring, 818 water-supply projects, 818 Water-resources systems, designs federal agencies (supporting in the U.S.), 7–8 water control system, 5–6 water-use systems, 6–7 Water-resources systems engineering, defined, 815 Watersheds, 4, 401 Water-storage reservoirs, 87–90 elevated storage tanks, 87 ground storage reservoirs, 87–88 standpipes, 87 Water-supply projects, 818 Water-supply wells design/construction, 777–793 drilled well, 777 driven well, 777 gravel packs, 783–784 grouting, 785–786 performance assessment of, 788–793 pumps, 784–785 radial well, 777–778 sand trap, 786 sanitary seal, 786–788 screen intake, 779–783 well casing, 779 Water-surface profiles, 132–159 classification of, 134–139 computation of, 143–145 direct-integration method, 145–147 direct-step method, 147 standard-step method, 148–150, 155

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Water-surface profiles (continued) hydraulic jump, 139–143 profile equation, 132–134 Water table aquifers, 657 Water-treatment systems, 6 Water-use systems, 6–7 design capacity of, 6 Wave speed, 475 Weather Surveillance Radar 1988 Doppler (WSR-88D), 404 Weber number, 148 Wedge storage, 524 Weibull distribution, 366–368 three-parameter, 368 Weibull formula, 372 Weighing-type recording gage, 404 Weir coefficient, 285 Weirs, 282–299, 520 Cipolletti weir, 286 long-based, 294 sharp-crested weirs, 282–287 submerged, 286–287 suppressed (uncontracted) rectangular weirs, 282–283 thin-plate weirs, See Sharp-crested weirs triangular, 293 unsuppressed (contracted) weirs, 282 V-notch weirs, 288–299 Weisbach, Julius, 14fn Well casings, water-supply well, 779 Well development, 782 Well drilling, 793 Well function, 721, 731–732 leaky, 737 Well intake, 782 Well losses, 789 Well point, 777 Wellbore skin, 803 Wellfield management zone, 774 Wellfields design, 771–773 objective of, 771 primary constraint in, 771 Wellhead protection, 774–775 Wellhead protection area (WHPA), 774–775 attenuation zone, 774 delineation of, 774–775 criteria for, 775 remedial action zone, 774 wellfield management zone, 774 zone of contribution (ZOC), 774 Wellhead-protection (WHP) plans, 774 Well-loss coefficients, 790

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Wells, See also Water-supply wells clogged, rehabilitation of, 789 drilling, 793 percussion methods, 793 rotary methods, 793 effective radius of, 784 steady-state solutions partially penetrating wells, 714–718 unconfined flow between two reservoirs, 700–702 well in a confined aquifer, 702–704 well in a leaky confined aquifer, 709–713 well in an unconfined aquifer with recharge, 713–714 well in an unconfined aquifer, 705–708 productivity, classification of, 788 unsteady-state solutions others, 741 well in a confined aquifer, 718–728 well in a leaky confined aquifer, 736–741 well in an unconfined aquifer, 728–736 well efficiency, 791 well-loss coefficients, 790 Wells, observation construction, 796 diameter, 795–796 distance from pumping well, 796 Wells, structural components grouting, 785–786 sand trap, 786 sanitary seal, 786–788 Wet basins, 587 Wet well, pump stations, 233 Wet-detention basins, 590–592, 618 Wetness gradient, 440 Wetting front, 447 Wicket gates, 330 Wide channels, lateral boundaries in, 120 Wilting point, 450 World Meteorological Organization (WMO), 428 World water quantities (estimated), 3 World-record precipitation amounts, 432 Wren, Sir Christopher, 403

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Y Yellow River, temporal trends ET, 643

Z Z formula, 235 Zone of aeration, See Unsaturated zone Zone of contribution (ZOC), 774 Zones of aeration, 691 Zones of saturation, 692

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