Engineering Hydrology 9789352606214, 9352606213

Table of contents :
Title
Contents
1 Introduction
2 Precipitation
3 Abstractions from Precipitation
4 Runoff
5 Hydrograph Analysis
6 Hydrograph Routing
7 Groundwater
8 Irrigation and Water Resource Management
9 Statistical Methods in Hydrology
10 Measurement of Hydrologic Variables
Answers to Objective-Type Questions
References
Index

Citation preview

Engineering Hydrology

About the Authors Rajesh Srivastava is Professor at the Department of Civil Engineering, Indian Institute of Technology (IIT), Kanpur. He obtained his Bachelor’s and Master’s degrees in Civil Engineering from the University of Roorkee (now IIT Roorkee). He received his doctorate in Civil Engineering from the University of Arizona, Tucson, USA. He has taught at the University of Roorkee for four years and has been teaching at IIT Kanpur for 20 years. He also has an industrial experience of 3 years with consulting companies in the USA related to metal mining and its effect on the environment. His areas of expertise include Water Flow and Contaminant Transport in Porous Media, and Effect of Climate Change on Water Resources. He has authored more than 85 technical articles in national and international journals and conferences, 15 technical reports for consultancy projects, and two textbooks, one on open channel flow and the other on numerical methods. He has also edited a two-volume set on contemporary research on Fluid Mechanics and Fluid Power and has created a video module on Water Resources Engineering under the National Programme on Technology Enhanced Learning (NPTEL). He was awarded the G M Nawathe medal of the Indian Society for Hydraulics in 2004 and 2005, Best paper award in Asia and Pacific Division of International Association of Hydraulic Research in 2006, and Best poster award at the Annual Monsoon Workshop of Indian Institute of Tropical Meteorology in 2015. He was the first occupant of the Sir M. Visvesvaraya Chair instituted by the Ministry of Water Resources at IIT Kanpur. He is a member of the Indian Society for Hydraulics, the American Geophysical Union, and InterPore–an International Society for Porous Media. Ashu Jain is Professor at the Department of Civil Engineering, IIT Kanpur. He received his Bachelor’s degree in Civil Engineering from the Malaviya Regional Engineering College, Jaipur (now National Institute of Technology, Jaipur) and Master’s degree in Civil Engineering from IIT Bombay. He did his PhD in Civil Engineering from the University of Kentucky, Lexington, USA and then worked in two consulting firms, Michael Baker Jr., Inc., Alexandria, Virginia, USA, and The Oyekan Group USA, Inc., Dallas, Texas, USA, for a total of about 3 years in the areas of flood management and drainage design. He then joined as Assistant Professor in the Department of Civil Engineering, IIT Kanpur. He teaches undergraduate courses on Hydrology, Open Channel Flow, and Engineering Graphics; tutors undergraduate courses on Engineering Graphics, Numerical Methods, and Fluid Mechanics; and teaches postgraduate courses on Advanced Hydrology, Groundwater Hydrology, Stochastic Hydrology, and Introduction to Artificial Intelligence (AI) Techniques. He has also developed a video-based course on Advanced Hydrology for NPTEL. His research interests include Rainfallrunoff Modelling, Applications of AI Techniques for Modelling and Management of Water Resources Systems, and knowledge extraction from Trained Artificial Neural Network Hydrologic Models. He has published more than 105 articles in international and national journals and conferences, authored about 20 technical reports related to sponsored research and consultancy projects, guided seven doctoral and 37 master’s theses. He has been a peer reviewer for several journals, organized several technical sessions in international and national conferences, delivered numerous invited lectures in India and abroad, and is currently an associate editor of Hydrological Sciences Journal published by Taylor and Francis. He was awarded the Best Discussion Paper Award, 2005 by ASCE; Royal Society Fellowship, 2005-06, to visit The University of Leeds, Leeds, UK; International Centre of Excellence in Water Resources Management (ICE WaRM) Fellowship, 2008, to visit The University of Adelaide, Adelaide, Australia; and Endeavour Executive Award, 2009 by the Ministry of Education, Australia. He is a life member of the Indian Society for Hydraulics, the Indian Water Resources Society, the Indian Association of Hydrologists, and a Fellow of the Indian Water Resources Society and the Indian Society for Hydraulics.

Engineering Hydrology RAJESH SRIVASTAVA Professor, Department of Civil Engineering Indian Institute of Technology, Kanpur

ASHU JAIN Professor, Department of Civil Engineering Indian Institute of Technology, Kanpur

McGraw Hill Education (India) Private Limited CHENNAI McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai 600 116 Engineering Hydrology Copyright © 2017 by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. Print Edition ISBN (13): 978-93-5260-620-7 ISBN (10): 93-5260-620-5 e-Edition ISBN (13): 978-93-5260-621-4 ISBN (10): 93-5260-621-3 Managing Director: Kaushik Bellani Director—Science & Engineering Portfolio: Vibha Mahajan Lead—Science & Engineering Portfolio: Hemant Jha Content Development Lead: Shalini Jha Specialist—Product Development: Sachin Kumar Specialist—Product Development: Vaishali Thapliyal Production Head: Satinder S Baveja Copy Editor: Taranpreet Kaur Assistant Manager—Production: Anuj K Shriwastava General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at APS Compugraphics, 4G, PKT 2, Mayur Vihar Phase-III, Delhi 96, and printed at Cover Printer:

Visit us at: www.mheducation.co.in

Dedicated to Late Dr. A. C. Srivastava and Late Dr. P. C. Jain

Contents Preface

xv

1. Introduction

1

Learning Objectives 1 1.1 Engineering Hydrology 1 1.2 Importance of Engineering Hydrology 2 1.3 Hydrologic Cycle 5 1.3.1 Atmospheric Moisture 6 1.3.2 Precipitation 6 1.3.3 Infiltration 7 1.3.4 Groundwater 8 1.3.5 Surface Runoff 9 1.3.6 Evaporation and Transpiration 9 1.3.7 General Discussion of the Hydrologic Cycle 10 Summary 11 Exercises 12 Objective-Type Questions 12 Descriptive Questions 15 Numerical Questions 15 Useful Links 16 Glossary 16

2. Precipitation Learning Objectives 18 2.1 Introduction 18 2.2 Occurrence – Mechanism of Formation, Various Forms, Distribution 19 2.2.1 Mechanism 19 2.2.2 Forms 20 2.2.3 Intensity 21 2.2.4 Distribution 21

18

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Contents

2.3 2.4

Measurement – Rain Gauges, Remote Sensing 22 Analysis of Data – Missing Data, Consistency Check, Averaging 22 2.4.1 Completeness Check 26 2.4.2 Estimation of Missing Data 26 2.4.3 Consistency Check 29 2.4.4 Averaging of Data 33 2.5 Storm Averaging 38 Summary 44 Exercises 45 Objective-Type Questions 45 Descriptive Questions 47 Numerical Questions 48

3. Abstractions from Precipitation Learning Objectives 51 3.1 Introduction 51 3.2 Abstraction Processes 51 3.2.1 Interception 52 3.2.2 Depression Storage 52 3.2.3 Evaporation 53 3.2.4 Transpiration 53 3.2.5 Infiltration 53 3.3 Initial Abstraction 53 3.4 Evaporation 55 3.4.1 Factors Affecting Evaporation 58 3.4.2 Estimation of Evaporation 59 3.5 Evapotranspiration 75 3.5.1 Empirical Equations 75 3.5.2 Theoretical Equations 78 3.6 Infiltration 79 3.6.1 Infiltration Process 80 3.6.2 Factors Affecting Infiltration 81 3.6.3 Estimation of Infiltration Capacity 82 3.6.4 Actual Infiltration 90 Summary 95 Exercises 96 Objective-Type Questions 96 Descriptive Questions 99 Numerical Questions 99

51

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4. Runoff

ix

101

Learning Objectives 101 4.1 Runoff Generation 101 4.2 Measurement of Streamflow 102 4.2.1 Area 103 4.2.2 Velocity 103 4.3 Annual and Storm Hydrographs 104 4.3.1 Annual Hydrograph 105 4.3.2 Storm Hydrograph 105 4.4 Factors Affecting Runoff 107 4.4.1 Catchment Characteristics 107 4.4.2 Storm Characteristics 108 4.5 Rainfall-Runoff Relationships 109 4.5.1 Analytical Models 109 4.5.2 Empirical Models 110 4.6 Rational Formula 115 4.6.1 Runoff Coefficient 115 4.6.2 Intensity of Rainfall 117 4.6.3 Time of Concentration 118 4.7 SCS Method 119 4.8 Flow Duration Curve 126 4.9 Flow Mass Curve 130 Summary 135 Exercises 136 Objective-Type Questions 136 Descriptive Questions 141 Numerical Questions 142

5. Hydrograph Analysis Learning Objectives 145 5.1 Introduction 145 5.2 Unit Hydrograph 146 5.2.1 Definition 146 5.2.2 Theory and Assumptions 147 5.2.3 Application of UH 148 5.3 Derivation of Unit Hydrograph 150 5.3.1 Baseflow Separation 150 5.3.2 Effective Rainfall and Direct Runoff Hydrograph 152 5.3.3 Derivation of UH from an Isolated Storm 152 5.3.4 Derivation of UH from a Complex Storm 154 5.4 S-Hydrograph 156 5.5 Unit Hydrograph of Different Durations 158

145

x Contents

5.5.1 UH of Different Durations by Method of Superposition 159 5.5.2 UH of Different Durations by S-Hydrograph Method 161 5.6 Synthetic Unit Hydrograph 163 5.6.1 Snyder’s Synthetic Unit Hydrograph 163 5.6.2 SCS Dimensionless Synthetic Unit Hydrograph 167 5.7 Instantaneous Unit Hydrograph 170 5.8 Relationships among UH, IUH, and S-Hydrograph 171 Summary 174 Exercises 175 Objective-Type Questions 175 Descriptive Questions 182 Numerical Questions 183

6. Hydrograph Routing

187

Learning Objectives 187 6.1 Introduction 187 6.2 Hydrologic and Hydraulic Routing 188 6.2.1 Hydrologic Routing 188 6.2.2 Hydraulic Routing 189 6.3 Hydrologic Routing through a Reservoir 190 6.3.1 Modified Puls Method 191 6.3.2 Goodrich’s Method 195 6.3.3 Standard Runge–Kutta (SRK) Method 195 6.3.4 Changes in Inflow Hydrograph Characteristics 200 6.4 Hydrologic Routing through a Channel 200 6.4.1 Muskingum Method 202 6.4.2 Estimation of Parameters of Muskingum Equation 205 6.5 IUH Development 207 6.5.1 Clark’s Method 208 6.5.2 Nash’s Method 211 Summary 217 Exercises 218 Objective-Type Questions 218 Descriptive Questions 226 Numerical Questions 226 Useful Links 229

7. Groundwater Learning Objectives 230 7.1 Introduction 230 7.2 Occurrence of Groundwater 231 7.2.1 Unsaturated Zone 232

230

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7.2.2 Saturated Zone 235 7.3 Movement of Groundwater 236 7.3.1 Basic Equations 239 7.4 Flow through a Confined Aquifer 245 7.4.1 Steady One-Dimensional Flow 246 7.4.2 Steady Flow Towards a Well 248 7.4.3 Transient One-Dimensional Flow 250 7.4.4 Transient Flow Towards a Well 252 7.5 Flow through Unconfined Aquifers 257 7.5.1 Steady One-Dimensional Flow 257 7.5.2 Steady Flow Towards a Well 260 7.5.3 Transient One-Dimensional Flow 262 7.5.4 Transient Flow Towards a Well 263 7.6 Non Ideal Conditions 265 7.6.1 Well Loss and Specific Capacity 265 7.6.2 Flow through a Layered Porous Medium 267 7.6.3 Flow through Leaky Aquifers 269 7.6.4 Flow through Unsaturated Zone 271 7.6.5 Flow Near Boundaries 271 7.7 Parameter Estimation 272 7.7.1 Estimation of Flow Direction 272 7.7.2 Estimation of Recharge 273 7.7.3 Estimation of Hydraulic Conductivity 274 7.7.4 Estimation of Transmissivity and Storage Coefficient 275 Summary 281 Exercises 282 Objective-Type Questions 282 Descriptive Questions 284 Numerical Questions 285 Useful Links 287

8. Irrigation and Water Resource Management Learning Objectives 288 8.1 Introduction 288 8.2 Water Requirement of Crops 289 8.3 Canal Irrigation 292 8.4 Irrigation Methods 294 8.4.1 Basin Irrigation 296 8.4.2 Furrow Irrigation 296 8.4.3 Border Irrigation 296 8.4.4 Ring Irrigation 297 8.4.5 Sprinkler Irrigation 297

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8.4.6 Subsurface Irrigation 297 8.4.7 Drip Irrigation 297 8.5 Single-purpose and Multipurpose Projects 298 Summary 301 Exercises 302 Objective-Type Questions 302 Descriptive Questions 304 Numerical Questions 304

9. Statistical Methods in Hydrology Learning Objectives 305 9.1 Introduction 305 9.2 Basic Probabilistic and Statistical Concepts 306 9.2.1 Sample and Population 306 9.2.2 Random Variable 307 9.2.3 Probability 307 9.2.4 Probability of Discrete Random Variables 309 9.2.5 Probability of Continuous Random Variables 312 9.3 Moments and Basic Descriptive Statistics 315 9.3.1 Measures of Central Tendency 316 9.3.2 Measures of Variation 319 9.3.3 Measure of Skewness 322 9.3.4 Measures of Peakedness 323 9.4 Some Important Probability Distributions 326 9.4.1 Binomial Distribution 326 9.4.2 Normal Probability Distribution 329 9.4.3 Gumbel’s Probability Distribution 332 9.5 Frequency Analysis 334 9.5.1 Probability Plotting Method 335 9.5.2 Frequency Factor Method 338 9.6 Risk and Reliability of Water Resources Projects 349 9.6.1 Risk 350 9.6.2 Reliability 350 9.6.3 Factor of Safety 350 Summary 351 Exercises 352 Objective-Type Questions 352 Descriptive Questions 357 Numerical Questions 358

305

Contents

10. Measurement of Hydrologic Variables

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361

Learning Objectives 361 10.1 Introduction 361 10.2 Measurements – General 362 10.3 Selection of Site and Instruments 363 10.3.1 Site-selection 363 10.3.2 Instrument-selection 364 10.4 Measurement Techniques 364 10.4.1 Precipitation 364 10.4.2 Evaporation and Transpiration 370 10.4.3 Infiltration 374 10.4.4 Soil Moisture 377 10.4.5 Streamflow 379 10.4.6 Groundwater 383 10.4.7 Temperature, Pressure, Humidity, Wind, Radiation 384 10.5 Databases 385 Summary 386 Exercises 387 Objective-Type Questions 387 Descriptive Questions 389 Numerical Questions 390 Answers to Objective-Type Questions

392

References

395

Index

400

Preface Introduction to the Course Hydrology is the science dealing with the occurrence and movement of water on, above, and below the surface of the earth. The movement of water is conveniently described through a hydrologic cycle, involving evaporation of water from water bodies, return of this water to the earth in form of rain, snow, etc., transfer of rain water as runoff to surface streams and infiltration to underground water, and finally, the movement of water from surface streams and underground water to the water bodies. This book deals with basic understanding of various processes of the hydrologic cycle, with emphasis on the engineering aspect, i.e., the application for management of water resources.

Target Audience The book is primarily designed for a first course in hydrology at the undergraduate level. However, the treatment of the subject matter is such that it may also be useful in a graduate level course. Scientists and engineers dealing with water resources may use this book to get a broader perspective and learn about recent developments, particularly in the areas of statistical analysis and measurement of hydrologic variables.

Objective of this Book Several books are already available on the subject with some emphasising the basic concepts, while others concentrating on problem-solving. Following the principle of science based engineering, we believe that the students should not only know how to apply a technique, but should also know how it has been arrived at, in order to better understand its limitations. The presentation of material in this book is made with emphasis on engineering applications of hydrology, with enough fundamental concepts interwoven in between for better understanding of the basic principles.

Salient Features ∑ ∑

∑

Comprehensive coverage of the science of hydrology, incorporating the recent developments and techniques Exclusive coverage on topics like statistical methods in hydrology, estimation of evaporation and runoff, infiltration capacity models, and transient flow of groundwater, to help students develop a deeper understanding of the subject A dedicated chapter on measurement of hydrologic and climatic variables explaining the conventional and advanced methodologies, which has not been paid much attention in the existing books on hydrology

xvi

Preface

∑ ∑

∑

Summary of important concepts for a quick review before exams Over 680 chapter-end exercises and 76 solved examples to illustrate the underlying concepts and help the students prepare for first course in hydrology of their engineering degrees/diplomas and competitive examinations Online links to hydrological and meteorological databases to help students understand hydrological concepts and instructors to develop additional examples/questions

Learning Tools in the First Edition Learning Objectives Each chapter is organized into multiple learning objectives (LOs). The topics covered in the chapter closely follow these objectives with each section listing the relevant objective. This feature of tagging each section with respective learning objective will help the instructors to plan the structure of course. The tagged LO brings out the essence of every section before the readers.

Abstractions from Precipitation

3

LEARNING OBJECTIVES LO 1

Know about the various abstraction processes and initial abstraction

LO 2

Define evaporation, factors affecting it, and the methods of estimation

LO 3

Estimate potential and actual evapotranspiration

LO 4

Discuss the infiltration process and the influence of various factors on the rate of infiltration

LO 5

Summarize empirical and theoretical methods for the estimation of infiltration

3.1 INTRODUCTION Although precipitation is an important part of the hydrologic cycle, as mentioned earlier, the topic of greater interest for engineers and hydrologists is the amount of water flowing in the rivers. If we could estimate how and how much water is being abstracted from the precipitation before it runs-off to the streams, we will have

Illustrations and Diagrams It is quite essential for a book to present figures to keep the interests of the users, and hence liberal use of figures is made in the book to aid in understanding of basic concepts.

Figure 1.1 The Hydrologic Cycle

Worked-Out Examples The book contains 76 solved examples for better understanding and application of concepts. Solved example problems also help in explaining new concepts and illustrate the working of computational procedure.

EXAMPLE 2.1 The annual precipitation at the four rain gauges is shown in the table below. Determine whether more stations are needed, if the desired CV is 20%. Station

S1

S2

S3

S4

Annual Rainfall (mm)

830

1120

1000

650

Solution The mean of the station data for annual precipitation is obtained as 900 mm, and the standard deviation, 2

2

2

2

(70 220 100 250 ) 204.78 mm. The coefficient of variation is, therefore, 0.228. Since it n 1 3 is more than the desired CV of 0.2, we require more stations and the total number of stations, using Eq. (2.1), is

0.228 0.2

2

4

5.18. We should provide two more stations.

Notes: If there is another station close to the boundary, but outside the catchment area, we may expect it to be climatically similar to other stations and should include it in the analysis, even though it lies in a different catchment. Also, we should keep in mind that this technique should be applied to long-term data, e.g., annual, which tends to average out the temporal and spatial variations in rainfall.

Preface

xvii

Chapter-End Exercises OBJECTIVE-TYPE QUESTIONS 6.1 The factors affecting the size, shape, and other characteristics of an inflow hydrograph while traveling through a channel reach include (a) Storage in the river reach (b) Resistance to flow due to friction from sides and bed (c) Lateral addition or subtraction of flow within the reach (d) 6.2

DESCRIPTIVE QUESTIONS

(a) (c) 6.3 I. II. III. (a) (e)

6.1 Why do the size, shape, and characteristics of an inflow hydrograph change when it travels through a channel reach? 6.2 Explain the hydraulic and hydrologic methods for hydrograph routing with the help of equations. 6.3 Explain the Modified Puls method of hydrologic routing. 6.4 Explain the Goodrich’s method of hydrologic routing. 6.5 Explain the fourth order SRK method of hydrologic routing.

More than 680 chapter-end exercises have been carefully constructed to enhance knowledge. These are categorized into Objective-Type Questions, Descriptive Questions, and Numerical Questions, to enable the students evaluate their understanding of different concepts after end of each chapter. Answers to the Objective-type Questions are provided at the end of the book.

6.6 For hydrograph routing through a reservoir, prove that the peak of the outflow hydrograph intersects inflow hydrograph. (Hint: use continuity equation). 6.7 Discuss the importance of hydrograph routing in flood management with special emphasis on attenuation and lag 6.8

NUMERICAL QUESTIONS 6.9 6.10

7.1 Water flows through a 2 m long horizontal soil column at a constant velocity. At a section of the tube, a red dye was inserted and it was observed that it travelled a distance of 1 m in 235 seconds and the dispersion was negligible. The soil is sandy with a porosity of 0.42 and hydraulic conductivity of 1 cm/s. What would be the drop in piezometric head across the column length? 7.2 A 50 m thick confined aquifer has a porosity of 0.35. The formation compressibility is 5 × 10−8 Pa–1 and water compressibility is 1 × 10−10 Pa–1. Estimate the storage coefficient and the specific storage of the aquifer. 7.3 Two large lakes are connected by a 200 m long confined aquifer in such a way that one-dimensional flow assumption is valid. The difference in water level of the lakes is 2.5 m and it is estimated that water is being conveyed through the aquifer at a rate of 1 m3/min per meter width. Estimate the transmissivity of the aquifer. 7.4 A fully-screened well is pumping a confined aquifer at a constant rate of 1 m3/s. A prior pump test on the aquifer has provided an estimate of the transmissivity as 0.05 m2/s. Two observation wells are located at a distance of 50 m and 100 m, respectively, from the pumping well. After a long time of pumping, the water levels in the observation wells achieve a nearly constant value. If the piezometric level in the first well (at 50 m) is 120 m above mean sea level, what would be the level in the other well?

Summary A detailed chapter-end summary is provided for a quick review of the important concepts. It helps in recapitulating the ideas initiated with the outcomes achieved.

SUMMARY Presence of water and condensation nuclei in the atmosphere is required for the formation of clouds and, for precipitation to take place, a cooling mechanism is needed. The cooling may occur due to orographic, convective, frontal, or cyclonic mechanisms, and results in precipitation in any of its various forms, such as rain, snow, or hail. The severity of precipitation is expressed either in terms of its intensity, generally in mm/h, or the depth, generally over the period of a day. The variation of intensity with time is shown by a hyetograph and the variation of the depth with time is shown either through a histogram (showing daily precipitation depths) or a mass curve (showing cumulative depths). The measurement of precipitation is done by recording rain gauges, which maintain a continuous record, or non-recording gauges, which provide only the daily precipitation depths. Any missing values in a precipitation record may be estimated by utilizing the precipitation records at nearby stations. The normal ratio method, inverse distance method, quadrant method, or more advanced techniques like Kriging, may be used for this purpose. To check if the data is consistent with the general climatic conditions, a double mass curve analysis is performed. This technique also provides a method to correct the data, if found to be inconsistent. The point rainfall data measured at a rain gauge is processed to obtain meaningful quantities related to catchment area, by using the arithmetic mean method, the Thiessen polygon method, or the isohyetal method. The isohyetal method is likely to be the most accurate method

Use of Technology 1. Maps of average annual precipitation (a) India: commons.wikimedia.org/wiki/File:India_annual_rainfall_map_en.svg, www.mapsofi (b) US: (c) Global: 2. (a) (b) US: (c) Global: 3.

4.

5.

In bringing out this book, we have taken advantage of recent technological developments to create a wealth of useful information to be supplemented with the physical book. Considering the ease of internet access at most engineering institutes, useful links to literature and datasets have been provided throughout the book. Moreover, Excel based solutions are incorporated in the book.

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Preface

Approach Adopted in this Book The presentation of material in this book is made with an emphasis on engineering applications of hydrology with just enough fundamental concepts interwoven in between for better understanding of the basic principles. From an engineering viewpoint, it is the runoff component of the hydrologic cycle which is the most important and, therefore, it is only natural that this component is discussed in greater details than other components. In fact, the objective of nearly all hydrological exercises by engineers boils down to the prediction of peak flows or low flows in a stream or the amount of water available as groundwater. One could then question the utility of studying other parts of the hydrologic cycle, such as evaporation and infiltration. Due to the rapidly changing stream characteristics, in order to draw some logical conclusions about its behavior, one must have data about the streamflow covering a long period. However, the length of record of streamflow and the reliability of this data is generally not adequate. On the other hand, rainfall and other climatic data are available to us for a much longer period and are more reliable, partly because of their direct effect on our daily lives and partly because their measurement requires very little skill or training. Therefore, it is quite common to develop a relationship between rainfall and runoff for the area under consideration and then estimate the runoff on the basis of longer duration and more reliable rainfall record. The rainfall-runoff relationship will be affected by several factors which influence the evaporation, infiltration, and other losses from the precipitation.

Organization of the Book Keeping in mind these points, this book is divided into ten chapters. The chapters are organized as follows: •

Chapter 1 introduces the subject and the motivations for studying it. Chapter 2 deals with precipitation of atmospheric water onto the earth surface with emphasis on various aspects of data collection, presentation, and analysis. The measurement techniques are described very briefly since we find it more convenient to describe the measurement of all hydrologic variables together in the final chapter.

•

Chapter 3 describes various abstractions from precipitation before it runs off to the surface streams. These abstractions include interception, depression storage, evaporation and infiltration. Stress is placed on discussion of various factors affecting these abstractions and their estimation, so as to enable an engineer to derive a logical rainfall-runoff relationship.

•

Chapters 4, 5, and 6 deal with the runoff component, with Chapter 4 introducing the processes of its generation, its measurement and estimation, and the analysis of data for use in the design and operation of water systems. Chapter 5 emphasizes hydrograph analysis for estimation of runoff. It introduces the concepts of unit hydrograph, its components and estimation, S-hydrograph, instantaneous unit hydrograph, inter-relationships among them, and their application in design and operation of water systems. Chapter 6 discusses the movement of runoff water through a stream or a reservoir and how the concepts of hydrograph routing can be useful in the hydrologic design and flood control studies.

•

The subsurface reservoir of groundwater, movement of water within it, and its recharge and withdrawals are described in Chapter 7, which effectively ends with the discussion of hydrologic cycle and water supply.

Preface

xix

•

Chapter 8 then describes an aspect of water demand, related to irrigation, and application of the hydrological principles for better management of water resources of an area. Since hydrological variables vary widely in space and time, there is always an inherent uncertainty in the analysis of these variables. Any prediction we make, based on the available data and our understanding of the hydrological principles, will, therefore, not be precise.

•

Chapter 9 describes several statistical techniques to not only estimate the desired quantities but to also assign a degree of uncertainty to those. Estimation of flood magnitudes and their probability is an important component of this analysis.

•

Finally, as mentioned before, since the issues involved with measurement of most hydrological variables are similar, we have clubbed all measurement techniques in Chapter 10. Particular attention is given to more recent techniques based on remote-sensing of the data, as it is expected to become more common in future.

•

References have been provided at the end of the book and an effort has been made to include freely accessible online references, wherever possible, rather than printed books or research papers.

Online Learning Center Supplements There are a number of supplementary resources available on the book’s website: http://www.mhhe.com/srivastava_jain/eh �

For Instructors • Solutions Manual • PowerPoint Lecture Slides

�

For Students • Web links for further readings

Acknowledgements Rajesh Srivastava thanks his parents, Mrs. G. K. Srivastava and Late Dr. A. C. Srivastava, for their encouragement, wife, Jayshree, for her understanding and support, and children, Soumya and Tanu, for their delightful presence. Ashu Jain would like to express his deep sense of appreciation to his father, Late Dr. P. C. Jain, for being a constant source of inspiration not only for writing this book but for pursuing academics throughout his life. He is grateful to his mother, Mrs. Swarn Lata Jain, wife, Savita, and children, Ateendriya and Tarushi, for their support, tolerance, and understanding throughout the time spent on writing this book. He would also like to thank all other friends, relatives, and colleagues, who have helped in some form or the other. One learns so much while teaching and the acknowledgement cannot be complete without thanking all the students, teaching whom we learn the nuances of the subject matter, and for their inquisitiveness urging a teacher to delve deep into the subject matter. The authors would also like to thank the team members of McGraw Hill Education (India), especially Vibha Mahajan, Shalini Jha, Hemant Jha, Vaishali Thapliyal, Sachin Kumar, Satinder Singh Baveja, Anuj Kr. Shriwastava and Taranpreet Kaur who handled various responsibilities related to the book very patiently and effectively.

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Preface

The authors would like to welcome any constructive criticism of the book and will be grateful for any appraisal by the readers.

Rajesh Srivastava Ashu Jain

Publisher’s Note McGraw Hill Education (India) would like to acknowledge the following reviewers who have helped in improving the subject matter with their critical reviews. Himanshu Sharma Kedar Sharma Mani Kant Verma Ompal Singh Ekta Mayank Gupta Himanshu Gupta Masood Ahmad Sanchayan Mukherjee Manoj N. Langhi P. Velumani Y. Ramalinga Reddy Shivanna

National Institute of Technology Uttarakhand, Uttarakhand BML Munjal University, Gurgaon, Haryana National Institute of Technology, Raipur, Raipur, Chhattisgarh Indian Agricultural Research Institute, New Delhi, Delhi Babu Banarasi Das University, Lucknow, Uttar Pradesh CBS Group of Institutions, Belochpura, Haryana Dr. A.P.J. Abdul Kalam Technical University, Lucknow, Uttar Pradesh Maulana Azad College of Engineering and Technology, Neora, Bihar Kalyani Government Engineering College, Nadia, West Bengal Institute of Infrastructure, Technology Research and Management, Ahmedabad, Gujarat Excel Engineering College, Tiruchengode, Tamil Nadu REVA Institute of Technology and Management, Bengaluru, Karnataka Sri M. Visvesvaraya Institute of Technology, Bengaluru, Karnataka

MHE India invites suggestions and comments from the readers, all of which can be sent to [email protected] (kindly mention the title and author name in the subject line) or can be directly sent to the authors at [email protected] or [email protected] Piracy-related issues may also be reported.

1

Introduction

LEARNING OBJECTIVES LO 1

Define engineering hydrology

LO 2

Understand the importance of engineering hydrology

LO 3

Explain hydrologic cycle and its various components

1.1

ENGINEERING HYDROLOGY

The literal meaning of hydrology is water science and a complete definition LO 1 Define engineering would, therefore, involve the entire history of water on earth. There is no hydrology general accepted definition of hydrology but a workable definition could be “a science dealing with the properties, distribution, and circulation of water on and below the earth’s surface and in the atmosphere”. Although the distinction between them is sometimes not very rigid, hydrology is different from hydraulics which is the “study of liquids (generally water for most engineering applications) at rest and in motion, especially under pressure, and application of that knowledge in design and control of machines”. Similarly, hydrology, which is considered a branch of Earth Sciences, has a considerable overlap with the other earth sciences areas like geology and meteorology. It is difficult, and probably pointless, to try defining the boundaries between these branches rigidly. Depending on the focus of study, hydrology could be further classified into Theoretical, dealing with higher conceptual levels; Physical, looking at the physical processes affecting the occurrence and movement of water; and Applied (or Engineering), emphasizing the application of hydrological principles to manage the water resources. Although this book is mainly concerned with engineering hydrology, it draws heavily on physical hydrology and, to some extent, theoretical hydrology, based on the paradigm of science-based engineering.

2

Engineering Hydrology

Engineering hydrology is traditionally considered a part of civil engineering, even though hydrology is thought of as a branch of earth sciences, as it enables an engineer to regulate water for beneficial use of the society. However, considering the importance of water in almost all aspects of life, it is relevant to many other fields.

1.2

IMPORTANCE OF ENGINEERING HYDROLOGY

Engineering hydrology finds numerous applications in civil engineering (and other branches of engineering and science). Some of the applications of principles of engineering hydrology are listed below.

LO 2 Understand the importance of engineering hydrology

Note: Several terms used in this section could be unfamiliar to the reader; hence to avoid frequent breaks in the narrative we have included a glossary section comprising the definitions of such terms at the end of this chapter. •

Deciding on the need to manage the water resources of an area—The first question which a water resources engineer needs to answer is: Is there a need for “engineering” the water resources of a particular area? The answer primarily depends on the water supply and demand. For example, assuming that rainfall is the only source of water in a region, if the pattern of rainfall is such that it is just enough to satisfy the demand at all times, there is no need of interfering with the natural supply. However, it is practically impossible to have a perfect match, spatially and temporally, between the supply and the demand. The mismatch leads to a problem of either scarcity or abundance, necessitating management of the water resources. For example, in a region, we may store water during the period of abundance and use it when the supply is not enough. Or, at a certain time, we may transport water from a region having surplus water to the one with scarcity of water. The storage could also be used to protect against the damaging effects of abundance of water. Sometimes, even if enough supply is available at a point to cater to the demand, additional water may be required due to restrictions on the use of water or the poor quality of water. For example, all the water of a river passing through an area should not be used since it would be detrimental to the downstream users and to the aquatic life. Or, if water is to be used for drinking, the river water quality may not be good enough for such use. After it is decided that the water resources have to be managed, the next decision should be on the extent of this management and the techniques for the same. For this, an estimate of the availability of water and the demand is required. It is the first aspect, i.e., the availability, which is the subject matter of this book. We assume that the demand is known to us through certain sources (e.g., domestic demand from municipal sources and industrial demand from the appropriate industry); however, a brief description of the agricultural demand is provided in Chapter 8.

•

Determination of the sources of water and the available quantity/quality—The sources of water could be on the earth’s surface (e.g., rivers and lakes), below the surface (e.g., groundwater), or above the surface (e.g., rain). These should, however, not be viewed in isolation, because of their interdependence. For example, using a part of the rain may cause reduction in the river flow and groundwater, using the groundwater may reduce the river flow (if there is significant contribution to the river flow from the groundwater) and vice versa. It is also possible that a particular water source is present but is not economically viable, for example, when the groundwater is at a large depth or

Introduction

3

when the river water is so polluted that the cost of treatment is prohibitive. Our primary focus in this book is on the quantity and we leave the issues related to the quality of water to the environmental engineers. However, a hydrologist should be aware of the direct link between the quantity and quality, as dilution is often used to tackle pollution. Due to considerable spatial and temporal variations in the seemingly random hydrologic variables like precipitation and streamflow, it would be impossible to determine the exact quantity of water available through these sources at a particular place and/or at a specific time. Moreover, the physical principles governing several hydrologic processes are not yet completely understood. Therefore, most of the techniques used in estimation of the available quantity of water rely heavily on the analysis of past data and have an inherent uncertainty in the estimates. It also implies that the reliability of the estimates would be largely dependent on the quality of the data used and the spatial and temporal extent of the available data. •

Checking the completeness and consistency of the data—Much of the analysis of availability of water is based on the measured data about rainfall, river discharge, and groundwater levels. Due to the large network of measurement stations and the manpower needed to operate and maintain this network, there are bound to be errors in the data, for example, errors due to equipment malfunction, human errors in reading etc. While it is important that any data be free of errors, it is more significantly so for data-driven models. Hence, the first step after receiving a data-set should be to check whether it makes sense. One of the easiest methods to do it is by comparing the data at one station with that of some nearby stations, which not only establishes the consistency of the data but also enables us to estimate any missing values. However, one must be aware of several factors which could complicate this analysis. For example, it is possible that a storm does not occur near a rain gauge but covers other nearby rain gauges, climatic conditions at two nearby weather stations may be significantly different, and river discharge at two nearby stations could be different due to the presence of tributaries. Therefore, comparing the data of a station with a single nearby station may not be desirable. On the other hand, if several nearby stations show a consistent trend, an average trend could be used to either check the consistency of the record or to fill in the missing portion of the data at a particular station. Once we have a complete set of consistent data, we are ready to analyze it to obtain the desired quantities, e.g., precipitation, streamflow, groundwater etc. Since most data would be in the form of point values representing the measurement station, it is sometimes necessary to estimate an areal average. For example, data available at different rain gauges in a catchment may show significant variation and estimation of an average rainfall over the area may not be straightforward.

•

Estimation of average rainfall over an area—The amount of rainfall (or, more generally, precipitation which also includes snow, hail, etc.) received by an area, and its distribution over time are important parameters in determining the capacity of water storage structures. Although more recent measurement techniques provide an areal distribution of precipitation, traditionally the data has been collected using point measurements at rain gauges. Estimation of the average rainfall over an area from a few point measurements has been the subject of several studies. The easiest option would be to use the average of the point values, but it may not work well in cases where the gauges are not uniformly distributed throughout the area or where the gauges are located in regions of widely different characteristics. For example, a simple averaging is likely to result in an erroneous estimate when one gauge is in the centre of the area and the other near the boundary; or when one

4

Engineering Hydrology

gauge is in the valley and the other on a mountain. Several techniques have been suggested to account for these factors and are discussed in the next chapter. •

Estimation of flow in a stream due to rainfall in the catchment area—The response of a stream to a rainfall event in its catchment depends on several stream and catchment properties. Streamgauging stations measure the river flow which, combined with corresponding rainfall data, may be used to develop the relationship between rainfall over a catchment and the streamflow at the outlet. The unit hydrograph method, described in Chapter 5, is useful in obtaining the streamflow pattern due to a given rainfall event. And how this pattern gets modified as water travels downstream, may be analyzed using the routing methods, discussed in Chapter 6.

•

What should be the storage capacity of a reservoir—If we decide to harvest the available rainwater or streamflow by building reservoirs, we need to analyze the precipitation/streamflow data vis-à-vis the water demand, to determine the capacity of the reservoir. In simple terms, we store water during the surplus season and use it later when the demand is more than the supply. How much to store will depend on what is the deficit in supply versus demand. For example, if the demand is constant at, say, 1 unit per month, and the yearly supply is just enough (i.e., 12 units) but is spread over only 3 months (4 units per month), and assuming that we are able to use all 12 units, we would need to store 9 units to be able to supply water for the 9 deficit months. In practice, both the demand and the supply show great seasonal variations and specialized techniques are needed to estimate the storage capacity. Also, one needs to be aware of the potential water loss through evaporation, seepage, and sedimentation in these reservoirs which would result in a larger requirement for the capacity.

•

What is the safe limit for withdrawing water from the groundwater—Groundwater is a widely available and more frequently used resource than streamflow and precipitation. Major advantages in using groundwater are its better quality (due to the filtration ability of the soil) and the avoidance of an elaborate distribution system (due to local availability). However, overdrawing of the groundwater has already resulted in a lowering of the water table in several parts of the world. To find out the safe limits of withdrawal, we need to estimate the rate at which the groundwater is being replenished through infiltration of rain water. Also, we need to decide about the depth and diameter of wells to be used for extracting this water, which will require an understanding of the movement of water in the sub-surface environment.

•

What kind of extreme hydrological events could be expected in an area and what is the probability of such occurrences—Safeguarding against extreme events, like floods and drought, is as, or sometimes more, important as the proper utilization of the available water resources. An estimate of the magnitude and the associated probability of floods may be obtained by the analysis of the streamflow data. Similarly, the frequency and severity of drought could be estimated by analyzing the precipitation data. Statistical techniques are generally required for the analysis since the available data typically does not cover a sufficiently long time span. For example, one may need to estimate floods which are expected once in 100 years, but the streamflow record may be available only for 50 years. An extrapolation for estimation of a 100-year, or a 500-year flood, therefore, requires some assumptions regarding the statistical distribution of these extreme values. The results obtained from this analysis are useful for flood and drought insurance also, since they provide an idea about the risks associated with submergence of an area or damage to a crop. In addition to the extreme event analysis, the statistical techniques could also predict, for example, what amount of streamflow is likely to be available half-the-time or 90% of the time.

Introduction

5

We hope that this book will either help the reader in answering these and many more similar questions, or will provide the background which would help in searching for the answers. The first step towards this goal is to understand the occurrence and movement of water, which is achieved through a simple but powerful concept of “Hydrologic Cycle.”

1.3

HYDROLOGIC CYCLE

The water present on, above, and below the earth’s surface could be LO 3 Explain hydrologic thought of as constituting three big reservoirs: a surface reservoir (oceans cycle and its various and lakes), a subsurface reservoir (groundwater), and an atmospheric components reservoir which comprises mainly water vapour. There is a nearly continuous movement of water from one reservoir to the other through various pathways, e.g., evaporation from the oceans transfers water from the surface reservoir to the atmospheric reservoir, precipitation transfers from the atmospheric reservoir to the surface reservoir, seepage from lakes transfers from the surface to the subsurface reservoir, seepage from coastal aquifers transfers from subsurface to surface reservoir, and so on. This cycle of storage and transfer is called Hydrologic Cycle or Water Cycle and can be depicted in several forms with varying degrees of details. One such description is given in Figure 1.1 as shown below:

Atmosphere

Condensation

Ice and snow Precipitation

Evapotranspiration

Surface runoff

Snowmelt runoff Infiltration

Evaporation

Streamflow

Evaporation

Plant uptake Freshwater

Oceans

Gro und wat er fl ow Graoundwater storage

Figure 1.1 The Hydrologic Cycle We first look at the occurrence of water and then describe its movement. We also describe briefly the factors affecting the components of the water cycle, so that we are able to make an informed choice about what is important and what is not, when we analyse the hydrologic data.

6

Engineering Hydrology

The total water on earth (including the subsurface and atmospheric water) is nearly 1.4 billion km3 (we use billion as a thousand millions and not a million millions, which is also prevalent), which we will denote as 100 units to simplify the description. The surface storage is about 98.3 including about 96.5 in oceans and seas; 1.7 in icecaps, glaciers, and permanent snow; 0.013 in lakes; and less than one thousandth in rivers, swamps, and biological water. The subsurface storage is about 1.7 including 1.69 in groundwater; 0.02 as ground ice; and about one thousandth as soil moisture. Although the atmospheric storage is only about 0.001, it plays a critical role in the hydrologic cycle by accepting the evaporation from the surface and then releasing it as precipitation. It is obvious that most of the water on earth is saline and not directly usable for our purpose, unless a cost-effective method of desalination is devised. In fact, out of 0.013 units in the lakes, 0.006 is saline, and out of the 1.69 in groundwater, about 0.9 is saline. Thus, overall there are about 97.5 units of saline water and only 2.5 units of fresh water. And even out of this 2.5 units, not all is usable since it may be permanently frozen, or be very deep inside the earth. One estimate puts the amount of total usable fresh water as 200,000 km3, which is about 0.00014 units!! A note of caution is needed here that the numbers mentioned above should not be treated as exact. There have been several studies using different methodologies and these numbers represent just one of those. While some components are easier to estimate (e.g., lakes), some others may have a greater uncertainty (e.g., groundwater). We now look at the movement of water among the reservoirs. The broad components of the water cycle include three different types of transport- first, from the surface and subsurface reservoirs to the atmospheric reservoir through evaporation and transpiration; second from the atmosphere to the earth through precipitation; and third of the precipitating water to the subsurface reservoir through infiltration and to the surface reservoir through runoff. Since it is a cycle, we could start from any of these reservoirs. Traditionally, the evaporation from the surface reservoir is chosen as the start of the hydrologic cycle. However, in keeping with the order of coverage in subsequent chapters, we decide to take precipitation from the atmospheric reservoir as the start of the cycle.

1.3.1

Atmospheric Moisture

The main source of the atmospheric moisture is the evaporation from the oceans. Most of this moisture is in the form of water vapour, which is water in a gaseous state formed due to evaporation. When the water vapour condenses upon small nuclei present in the atmosphere, for example, salt, dust, etc., it takes the form of clouds or fog. It is estimated that the amount of water held in the atmosphere at any given time is about 13,000 km3, which is sufficient to produce about 2.5 cm of rain over the entire surface of the earth. Once the atmospheric conditions are conducive for precipitation, i.e., there is enough moisture, condensation nuclei are present, and the resulting water droplets grow to a sufficient size, the atmospheric moisture precipitates in the form of rain (or snow, etc.) onto the surface of the earth. However, all of this water may not reach the surface of the earth as some may evaporate while falling down.

1.3.2

Precipitation

The term precipitation is used to represent all aqueous particles which fall from the atmosphere onto the earth, mostly as rain but often in other forms like snow and hail. Since oceans cover more than three-fourths of the earth’s surface, it is not surprising that almost 80% of the precipitation occurs over the oceans. Of the total estimated annual precipitation on earth, which is about 500,000 km3 (or, in terms of the 100 units of water on earth, about 0.0004 units), nearly 400,000 km3 falls directly on the oceans. If it were to occur uniformly

Introduction

7

over the entire globe, it would represent an annual precipitation depth of about 1 m. However, there are wide variations, both spatially and temporally, in the amount of precipitation across the globe. For example, the average annual precipitation varies from around 12 m (Meghalaya, India; Cauca, Colombia) to about 0.1 mm (Antofagasta, Chile) and some parts of Antarctica receive no rain at all. Another example of the spatial heterogeneity of precipitation is that some parts of Hawaii have 350 rainy days in a year, while some areas of Chile have one rainy day in six years (some parts of Antarctica have none in a million years!). It should be kept in mind that these are average annual values, and the rainfall in any particular year may be much larger or smaller (the maximum rainfall observed in a single year is about 26 m in the state of Meghalaya, India). Even within a country there are considerable variations in precipitation. In India, the driest parts receive an annual average precipitation of 20 cm (Rajasthan) and 10 cm (Leh) while, as mentioned earlier, the wettest parts have 12 m of precipitation. Similarly, there are large temporal variations in precipitation over a significant part of the earth. The city of Kanpur in India, for example, receives nearly 90% of its annual precipitation in the four monsoon months of June, July, August, and September (and about 60% in just two months, July and August). Therefore, although the hydrologic cycle is depicted as a continuous cycle, one should be aware that it represents an average behaviour. Sometimes it would appear that the cycle has completely stopped while at some other times, it would appear to be overactive. How much precipitation occurs, and when and where it occurs, depends on several factors like humidity, air currents, nearness to water bodies, and presence of orographic barriers. The rate at which precipitation occurs is denoted by its intensity, expressed in terms of the depth of precipitation in a given time period, usually in mm/h. More details of the mechanism of precipitation and the effect of these factors are provided in the next chapter. Engineers are generally not concerned with the part of the precipitation which does not reach the earth surface. Out of the precipitation which reaches the surface, some infiltrates into the ground, some is detained in depressions or retained on the ground as a film or puddle from which it either infiltrates or evaporates, and the rest runs-off over the surface to various streams and ultimately to the surface reservoirs.

1.3.3

Infiltration

Rain and melting snow tend to moisten the surface first, and then enter the interstices of the soil under the influence of gravitational and capillary forces. The rate at which water infiltrates the soil is expressed in terms of water depth per unit time, normally in mm/h, and the maximum rate at which soil can allow water to infiltrate is known as its infiltration capacity. The infiltration capacity is affected by several factors like soil type, soil surface condition, soil moisture, vegetation, and the nature of rainfall. For example, soil surface compaction will reduce the infiltration capacity while presence of surface cracks will increase it. Similarly, dry soils will have a larger capacity than wet ones. Generally during a rain event, the infiltration capacity decreases with time due to increase in soil moisture, swelling of clay and blocking of pores by the fine particles carried by the rain water. The infiltration capacity of sandy soils is generally greater than 20 mm/h, for clayey soils it is 1–5 mm/h, and for loam it is 5–10 mm/h. The infiltrating water could stay in the upper layers of the soil as soil moisture, from where it is utilized by the vegetation and transpired back to the atmosphere, or it could percolate deeper into the subsurface reservoir. Sometimes, instead of percolating deeper, the infiltrating water encounters an impervious layer of rock close to the ground surface and flows over it to join a surface stream. This component of the runoff is called interflow and is a part of the subsurface runoff. If the precipitation intensity is more than the infiltration

8

Engineering Hydrology

capacity, water will runoff over the surface to join the stream system. This surface runoff is one of the two most important parameters (the other being the subsurface reservoir) from a water-supply perspective.

1.3.4

Groundwater

The completely saturated soil mass at some depth, which may range from zero to a few thousand meters below the ground surface is called the groundwater and is a major component of the subsurface water reservoir (the other components are soil moisture and ground ice). If this soil mass is not confined by an impervious or semipervious formation at the top, the top surface of the saturated soil will have water at atmospheric pressure and this surface is known as the water table or the phreatic surface. If it is confined, the water pressure would be more than atmospheric and water in a well tapping this soil will rise above the confining layer (Figure 1.2).

Figure 1.2 Groundwater under confined and unconfined conditions The transfer of water from/to the groundwater may take place in various ways. Infiltration is the primary source of water for the groundwater. Depending on the water level in streams and the water table height, water will flow either from the groundwater to the stream (known as the base flow of the stream) or from the stream to the groundwater (resulting in recharge of the groundwater). Water may be pumped out through tube wells or extracted from dug wells. Recharge wells are sometimes used to enhance the groundwater recharge by conveying rain water directly to the subsurface. When groundwater is under pressure, it may discharge as springs, geysers or artesian wells. In coastal areas, groundwater may discharge directly into the sea. The soil or rock porosity determines how much water could be stored in it and the permeability determines the ease with which this water can be extracted. An ideal soil formation for supplying water at a reasonable rate would, therefore, have a high porosity and large permeability, and is called an aquifer. While tapping an aquifer for water supply, one must ensure that the withdrawal rates are not significantly larger than the recharge rates; otherwise the water table will progressively go down, resulting in a possible deepening of the wells and an increase in pumping cost. The recharge rates are dependent on the precipitation magnitude and

Introduction

9

pattern, soil and vegetation type, and land cover, since, as discussed in the previous section, all these factors influence the infiltration. Recharge rates are much higher during the rainy season and rise of a few metres in the water table is commonly observed. Data collected by the Central Ground Water Board, India, in 2013 before the monsoon season, shows (compared to the data taken at the same time in 2012) a rise in water table in 44% wells and a fall in 56% wells, with most wells (77%) showing a fluctuation of less than 2 m. Nearly identical results were obtained for a decadal change from 2003 to 2013.

1.3.5 Surface Runoff The excess rain, after satisfying the infiltration capacity, first fills the surface depressions and then reaches a stream system through which it passes and discharges ultimately into the ocean or an inland water body. Some part of the water moving in the stream systems evaporates into the atmosphere and some seeps down from the channel boundaries to recharge the subsurface reservoir. It is estimated that the annual flow in the rivers of the world adds up to about 40,000 km3. As with precipitation, there are large spatial and temporal variations in the stream flows. For example, the annual flow in Amazon is about 6500 km3, while that in Ganges is about 500 km3 and in Pecos only about 0.06 km3. The flow in a river varies with location and the values listed are at the mouth of the river where it flows into the sea. It is customary to express the flow in a river in terms of the discharge in the units of m3/s (also called cumecs). For Ganges, the average discharge during a year translates into about 15,000 cumecs. Near the city of Kanpur, the average discharge in Ganges is only about 4000 cumecs, while at Allahabad located about 200 km downstream, it is about 13,000 cumecs. This increase is due to the additional catchment area because of the joining of big tributaries in this stretch of the river. An idea about the temporal variation of river flow could be obtained from the fact that in the month of May, the average discharge of Ganges at Kanpur is about 100 cumecs and at Allahabad it is about 300 cumecs. The maximum discharge observed in Ganges is about 70,000 cumecs, nearly 5 times the average value! Similarly, for the river Kosi at Baltara, the average discharge over a year is about 2200 cumecs, over the monsoon period about 5100 cumecs, and in non-monsoon period about 1200 cumecs. The maximum observed discharge over a 22-year period is 12,000 cumecs and an estimate of a 100-year flood is about 14,000 cumecs (the methodology of estimating a 100-year flood is discussed in Chapter 9). The discharge into a river at a particular point primarily depends on the area of the catchment up to that point and the climatic and physiographic characteristics of this area. For example, if the same amount of rain occurs in fewer but more intense storms, the runoff is likely to be more than that for a larger number of lower intensity rains. Similarly, runoff from a vegetated area is likely to be less than an equivalent bare area and runoff from a sandy catchment is likely to be less than that for a clayey area. A hydrograph, representing the variation of the river discharge with time, is a useful tool to analyse the flow pattern in a river to obtain peak flow, dependable flows etc. A more detailed discussion of runoff and hydrograph is provided in Chapters 4 and 5.

1.3.6

Evaporation and Transpiration

The atmospheric reservoir transfers water to the earth’s surface through precipitation, from which the subsurface reservoir is recharged through infiltration and contribution to the surface reservoir is made through runoff. To complete the water cycle, we need to look at the transfer from the surface reservoir to the atmospheric reservoir, which is effected through evaporation from water bodies and land, and transpiration from vegetation. Evaporation is the passage of water from the liquid to gaseous state when subjected to heating, generally by solar energy. Transpiration is the process of returning of the soil moisture to the

10

Engineering Hydrology

atmosphere by vegetation, which absorb water from the soil through its roots, utilize it for maintaining life and producing growth, and then discharge it through its pores as water vapour. Since the processes of evaporation and transpiration are very similar and are affected by similar factors, it is customary to combine these into a single term evapotranspiration. This is the source of all atmospheric moisture and, therefore, all precipitation. The chief contributor to evaporation are the oceans and a small portion comes from falling rain, intercepted precipitation, depression storage, ground films, rivers and lakes. Since the hydrologic cycle is in a stable state, the total global evaporation in a year is equal to the precipitation, i.e., nearly 500,000 km3 (or, in terms of water depth, about 1 m). However, if we consider the oceans and landmasses separately, the precipitation on the landmass is larger than the evaporation from it, and the precipitation on the oceans is less than the evaporation from them by an equal amount. The difference is made up by the runoff which transfers water from the landmass to the ocean. In addition to its important role in the water balance, evaporation, which requires a considerable amount of energy, plays an equally important role in the global heat transfer and the resulting circulation pattern. The two requirements for evaporation to take place are the availability of water and energy. To delink the water availability, we typically look at potential evaporation, which is the amount of evaporation that would occur if water is abundantly available. The potential evaporation depends on factors like solar radiation (more intense radiation will lead to higher evaporation), temperature, relative humidity (more humid air will result in less evaporation), and wind speed (larger wind speed leads to higher evaporation). A more detailed discussion is provided in Chapter 3. Since these factors show significant spatial and temporal variations, the potential evaporation also varies widely from one place to the other and from one time to another. For example, temperature has been observed to vary from −89°C in Antarctica to 56°C in California (for India, a range of −50°C in Ladakh to 50°C in Rajasthan has been observed), and the daily average wind speed could go as high as 175 km/h in Antarctica. The solar radiation depends on the position of the sun and the atmospheric conditions. The solar constant, which is the amount of incoming solar radiation per unit area at a distance equal to the average distance of the earth from the sun (known as one astronomical unit) on a plane perpendicular to the rays, is about 1360 W/m2 (this number varies slightly due to change in solar activity, so it is not really a constant). However, the energy per unit area on the earth’s surface, known as irradiance, is smaller due to the presence of clouds, inclination of rays etc. The average yearly sum of solar irradiance across the world generally decreases with latitude and shows a variation from about 60 W/m2 to about 400 W/m2. These variations imply that the potential evaporation also shows a wide variation from its global annual average of a depth-equivalent of about 1 m. For example, the annual potential evaporation depth in India varies from about 1.5 m to 2 m, with some places showing as high as 3.5 m. The monthly evaporation depth could be as high as 50 cm for some parts of Maharashtra in summer or as little as 5 cm or less in some parts of Assam in winter. The actual evaporation is smaller than these values due to lack of available water. Australia, for example, has annual potential evaporation ranging from 1 m to 4 m, but the actual evapotranspiration is only 20 cm to 120 cm. Globally, the actual evaporation at some places may be as low as 1 cm. The evapotranspiration transfers water from the surface and subsurface reservoirs to the atmospheric reservoir, thereby completing the hydrologic cycle.

1.3.7 General Discussion of the Hydrologic Cycle The preceding description of the Hydrologic Cycle and its various components should reinforce the fact that this cycle is not a continuous cycle, either in space or in time. However, it provides us a good starting

Introduction

11

point to study the various aspects of engineering hydrology in a systematic manner. A few notable points are summarized below: •

The cycle is not spatially uniform. Some areas may experience intense rainfall while other nearby areas may be completely dry.

•

The cycle is not temporally steady. The same area may get intense rain at times and would be completely dry at other times.

•

A general idea about the movement of water through various phases of the cycle could be obtained through an average residence time. For example, it is estimated that atmospheric reservoir has a storage volume of about 13,000 km3 and it transfers water at the rate of about 500,000 km3/year. Therefore, on an average, a water drop will stay in the atmosphere for about 9 to 10 days [~13000/ (500000/365)]. Similarly, river waters have residence time varying from about 2 to 6 months, oceans 3000 years, shallow groundwater about 100 years and deep groundwater about 10,000 years. Clearly, there is a wide variation from these average values as in the case of, say, infiltrating water percolating to the groundwater and resurfacing as interflow.

•

Changing climate of the world may have significant effect on the hydrologic cycle. For example, higher temperatures may lead to higher evaporation; some studies suggest that the precipitation pattern is likely to change to more intense storms and longer dry periods; and rise in sea levels may lead to less groundwater discharge into sea and more ingress of saline water in coastal aquifers.

SUMMARY Hydrology deals with the occurrence and movement of water on and below the earth’s surface and in the atmosphere as well. Engineering hydrology aims at applying the principles of hydrology for management of water resources. These applications include the determination of quantity and quality of available water, deciding the storage capacity of reservoirs, estimation of hydrologic extremes, for example the maximum expected flood or the minimum expected rainfall, and determination of the maximum safe withdrawal of water from the groundwater. The hydrologic cycle, which is a convenient starting point for hydrologic analysis, represents the occurrence of water in surface, subsurface, and atmospheric reservoirs, and its transfer from one reservoir to the others. Out of the total water in these reservoirs, estimated at about 1.4 × 109 km3, 97.5% is saline water and only 2.5% is freshwater. Most of this freshwater is not amenable to economic use, due to being frozen or occurring very deep into the earth, and only about 200,000 km3 freshwater is usable. The atmospheric moisture storage at any time is around 13,000 km3, with most of it being contributed by evaporation from the oceans. Only about 20% of the annual precipitation of about 500,000 km3 falls on the continents, from which some infiltrates into the ground, some evaporates, and the rest runs off to the surface stream systems. How much water infiltrates depends primarily on the type of soil and how much evaporates depends mainly on the temperature and humidity. The surface runoff is an important source of water and is estimated at about 40,000 km3 per year. The other important source of water is the groundwater, occurring below the ground surface, which is recharged by the infiltrating water. The amount of storage in the groundwater depends on the porosity and its speed of movement is governed by the permeability. There is a wide variation, in both space and time, in all the components of the hydrologic cycle. The concept of an average residence time, however, may be used to visualize the movement of water through various phases of the hydrologic cycle.

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Engineering Hydrology

OBJECTIVE-TYPE QUESTIONS

1.1 Literal meaning of hydrology is (a) Water mechanics (c) Water science

(b) Water record (d) Water budget

1.2 Hydrology is typically considered a branch of which of the following disciplines? (a) Hydraulics (b) Meteorology (c) Earth Sciences (d) Astronomy 1.3 Theoretical hydrology deals with (a) Concepts (b) Processes

(c) Application

(d) All of these

1.4 Which of the following may best be described as being a topic of study in Engineering Hydrology? (a) Pipe flow (b) Hydropower (c) Cloud formation (d) Reservoir design 1.5 Physical hydrology refers to the study of (a) Instruments (b) Processes

(c) Equations

(d) Measurement

1.6 The magnitude of possible drinking water supply from a river will depend on (a) River discharge (b) Water quality (c) Aquatic life needs (d) All of these 1.7 Will the water flow in a stream be affected by the use of groundwater for irrigation in nearby areas? (a) Never (b) Always (c) Only if groundwater use is large (d) None of these 1.8 Reliability of the estimates of water availability from collected data will depend on its (a) Quality (b) Spatial coverage (c) Temporal coverage (d) All of these 1.9 Computing areal average of rainfall over an area from the point measurements at a few rain gauges will work best in which of the following cases? (a) The area is climatically uniform (b) Rain gauges are very large in number (c) There are several hills and valleys (d) All measurements are taken at the same time 1.10 What is meant by the consistency of a hydrologic dataset obtained at a station? (a) It is spatially uniform (b) It is not changing with time (c) It shows a monotonic trend (d) It agrees well with nearby stations 1.11 The consistency of a hydrologic dataset obtained at a station is checked by (a) Ensuring that there are no missing values (b) Comparing with nearby stations (c) Ensuring that the data shows random variation (d) Analyzing the trend in data 1.12 The river discharge measured at two gauging stations situated near each other might be significantly different due to (a) Evaporation (b) Seepage (c) Tributary (d) Rainfall 1.13 The storage capacity of a reservoir will depend on (a) Supply only (b) Demand only (c) Both supply and demand (d) Neither supply nor demand

Introduction

13

1.14 Which of the following will NOT increase the required reservoir capacity? (a) Evaporation (b) Seepage (c) Sedimentation (d) Precipitation 1.15 Out of the options given below, which one is correct about the following statements? (i) The quality of groundwater is generally worse than that of surface water. (ii) Use of groundwater for water supply does not require elaborate distribution system. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 1.16 Out of the options given below, which one is correct about the following statements? (i) Streamflow record may be analyzed to obtain an estimate of flood probability. (ii) Precipitation data may be analyzed to obtain an estimate of drought frequency. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 1.17 What is the depth of rain which could be produced over the entire earth surface by the amount of water held in the atmosphere at any time? (a) 1 mm (b) 5 mm (c) 25 mm (d) 100 mm 1.18 What percentage of the total water on, above, and within the earth is freshwater? (a) 1.0% (b) 2.5% (c) 5.0% (d) 7.5% 1.19 What percentage of the total water on, above, and within the earth is surface water? (a) 96.0% (b) 97.5% (c) 98.3% (d) 99.7% 1.20 River water comprises about _____% of the surface water. (a) 1.0 (b) 0.1 (c) 0.01

(d) 0.001

1.21 What percentage of the total water on, above, and within the earth is groundwater? (a) 1.0% (b) 1.7% (c) 2.9% (d) 4.1% 1.22 The total usable fresh water on earth is about _______ km3. (a) 100,000 (b) 200,000 (c) 1,000,000

(d) 2,000,0000

1.23 The main source of atmospheric moisture is (a) Glaciers (b) Rivers (c) Oceans

(d) Groundwater

1.24 What is the order of magnitude of the largest average annual precipitation at any place on earth? (a) 1 cm (b) 1 m (c) 10 m (d) 50 m 1.25 What percentage of the global precipitation falls directly on the oceans? (a) 1% (b) 10% (c) 50% (d) 80% 1.26 What is the maximum number of rainy days in a year observed at any place on earth? (a) 200 (b) 250 (c) 300 (d) 350 1.27 What is the maximum rainfall observed at any place on earth in a single year? (a) 1 m (b) 12 m (c) 26 m (d) 37 m 1.28 Out of the options given below, which one is correct about the following statements? (i) Kanpur receives 90% of its annual precipitation in July and August. (ii) Average annual precipitation in Antofagasta, Chile, is about 1 mm. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false

14

Engineering Hydrology

1.29 The intensity of rainfall is generally expressed in the units of (a) m/s (b) m/h (c) mm/s

(d) mm/h

1.30 Out of the options given below, which one is correct about the following statements? (i) Compaction of soil generally reduces the infiltration capacity. (ii) Presence of surface cracks generally reduces the infiltration capacity. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 1.31 What happens to the infiltration capacity of soil during a precipitation event as time passes? (a) Decreases (b) Increases (c) First decreases then increases (d) First increases then decreases 1.32 What is the typical order of magnitude of the infiltration capacity of sandy soils? (a) 1 mm/h (b) 5 mm/h (c) 20 mm/h (d) 200 mm/h 1.33 The component of infiltrating water which flows over an impermeable rock near the ground surface and then joins a surface stream, is called (a) Virga (b) Transpiration (c) Interflow (d) Recharge 1.34 Out of the options given below, which one is correct about the following statements? (i) The soil or rock porosity determines how fast water will flow through it. (ii) An aquifer should have high porosity and low permeability. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 1.35 What is the average annual flow in the river Amazon? (a) 500 km3 (b) 1000 km3 (c) 3500 km3

(d) 6500 km3

1.36 What is the gage pressure at the groundwater table? (a) Negative (b) Zero (c) Positive

(d) Varies with time

1.37 The water table is also known as (a) Base flow (b) Artesian well

(d) Spring

(c) Phreatic surface

1.38 Out of the options given below, which one is correct about the following statements? (i) For the same total precipitation, fewer but more intense events are likely to produce a larger runoff compared to more, but less intense, events. (ii) Runoff from a vegetated area is likely to be more than an equivalent bare area. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 1.39 What is the total global evaporation, in 1000 km3, in a year? (a) 1 (b) 5 (c) 100

(d) 500

1.40 Out of the options given below, which one is correct about the following statements? (i) Precipitation on the landmass is larger than the evaporation from it. (ii) More humid air will result in more evaporation. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 1.41 What is the value of the solar constant? (a) 360 W/m2 (b) 1360 W/m2

(c) 2360 W/m2

(d) 3260 W/m2

1.42 How does the average yearly sum of solar irradiance generally vary with increasing latitude? (a) Decreases (b) Increases (c) Remains constant (d) Depends on humidity

Introduction

1.43 The residence time of the atmospheric moisture is about (a) 1 day (b) 10 days (c) 1 month

15

(d) 6 months

DESCRIPTIVE QUESTIONS

1.1 What is hydrology? Describe some of its applications. 1.2 What is Hydrologic cycle? Describe its components. What is meant by the statement that the hydrologic cycle is in a stable state? 1.3 Why is simple averaging of rain gauge data to obtain areal average rainfall not very accurate in some cases? 1.4 What are the factors affecting the infiltration capacity of a soil? Why does this capacity reduce with time when there is a rainfall event? 1.5 How does the rainfall pattern and presence of vegetation affect the fraction of rain which may be converted to runoff? 1.6 What is meant by the terms evaporation and transpiration? Why are evaporation and transpiration combined together in the term evapotranspiration? What is potential evaporation? 1.7 Explain the various ways in which water is transferred from/to groundwater. 1.8 What are some possible effects of climate change on the hydrologic cycle?

NUMERICAL QUESTIONS

1.1 If the latent heat of vaporization for water is 2.44 5 106 J/kg, estimate the depth of water which could be evaporated by the energy equivalent to the solar constant. 1.2 It is estimated that the average annual discharge of the river Amazon is about 200,000 m3/s and the average volume of water in the river is about 1000 km3. What is the average residence time of water particles in this river? 1.3 A lake has a surface area of 10 km2 and water surface elevation of 182.527 m above mean sea level at a certain time. A precipitation event occurred during the next four hours, with a uniform intensity of 5 mm/h and the evaporation rate during this period was estimated as 0.2 mm/h. The surface stream flowing into the lake had an average flow rate of 20 m3/s while the outflowing stream had an average flow rate of 25 m3/s during this period. If the water surface elevation at the end of the precipitation event is measured as 182.538 m, estimate the volume of water seeping into the ground during this time. (Hints: The increase of storage in the lake is equal to the difference of inflow to and outflow from the lake during the given period. The inflows are the precipitation and the inflowing streams and the outflows are the evaporation, outflowing streams and seepage. Quantities expressed in terms of depth have to be multiplied by the surface area of the lake to get the volumes). 1.4 For a catchment of area 250 km2, the estimate of annual precipitation is 900 mm, annual evapotranspiration is 120 mm, and the annual surface runoff is 0.15 km3. Assuming that 40% of the infiltration contributes to the recharge of groundwater, estimate the annual safe withdrawal volume from the groundwater. (Hints: The safe withdrawal limit is equal to the recharge of the groundwater).

16

Engineering Hydrology

1.5 There are several resources available online for finding out the precipitation and evaporation at any place (some links are provided in Chapter 10). Obtain the record of daily precipitation and actual or potential evaporation for the previous year at a station near you, plot the data, and compute the total annual precipitation and evaporation. 1.6 Streamflow data may also available for a river near your area. If so, plot the precipitation versus streamflow and see whether there is any relationship between these.

1. Maps of average annual precipitation (a) India: commons.wikimedia.org/wiki/File:India_annual_rainfall_map_en.svg, www.mapsofindia.com/maps/india/annualrainfall.htm (b) US: www.wrcc.dri.edu/pcpn/us_precip.gif (c) Global: www.whymap.org/whymap/EN/Downloads/Additional_global_maps/precipitation_pdf, www.britannica.com/science/climate-meteorology/World-distribution-of-precipitation 2. Maps of evaporation/evapotranspiration (a) India : nihroorkee.gov.in/rbis/india_information/evaporation.htm (b) US: www.extension.purdue.edu/extmedia/nch/nch-40.html, (c) Global: nelson.wisc.edu/sage/data-and-models/atlas/maps/pevapotrans/atl_pevapotrans.jpg, www.waterandclimatechange.eu/evaporation/average-monthly-1985-1999 3. Map of global surface runoff atlas.gwsp.org/atlas/img/map/a3_runoffWSAG1_0_wl.png, wwap.cesr.de/results.htm#map1 4. Map of global groundwater recharge www.whymap.org/whymap/EN/Downloads/Additional_global_maps/gw_recharge_pdf 5. Hydrologic cycle and its components scied.ucar.edu/longcontent/water-cycle

100-year-flood: The flood which is expected to occur, on an average, once in 100 years. Aquifer: A soil formation which can store enough water and can transmit it at a reasonably fast rate. Artesian well: A well from which water flows out without pumping, due to the large pressure under which water is stored in the soil. Base flow: The flow in a stream which comes from the seepage of water from the adjacent soil. Catchment: The area which catches the rain falling over it and carries it to a stream. Confined aquifer: A completely saturated aquifer bounded by impermeable layers on top and bottom and, therefore, carrying water under pressure. Depression storage: The capacity of an area to store water in pits and depressions on the land surface. Evaporation: The process by which liquid water changes into water vapour. *

A very brief and general definition is given here. More specific descriptions are provided in the relevant chapters.

Introduction

17

Evapotranspiration: It is the combined evaporation and transpiration. Groundwater: Water present below the ground surface in soil pores or rock crevices. Humidity: The amount of water vapour present in the atmosphere. Hydrograph: The graph showing variation of discharge in a stream with time. Infiltration: The process through which water enters the soil from the ground surface. Infiltration capacity: The maximum rate at which a soil can absorb infiltrating water. Interflow: The water that infiltrates into the soil and then quickly returns to the surface due to lateral movement. Permeability: Ability of a porous medium to allow passage of a fluid. Phreatic surface (or Water table): The upper surface of the saturated groundwater which is at atmospheric pressure. Porosity: A measure of the void space present in the soil. Precipitation: Water reaching the earth from the atmosphere. Rain gauge: An instrument to collect and measure the amount of rain falling over an area. Recharge: Water entering the saturated groundwater. Relative humidity: The amount of water vapour in the atmosphere in relation to that needed to saturate the air with water vapour. Routing: The technique used to estimate the change in the shape of a hydrograph as water flows through a stream or a storage structure. Runoff: The rainwater that flows off to the stream. Sedimentation: The process of settling/deposition of suspended sediments on the channel bed. Seepage: The slow passage of fluid through the pores of a porous medium. Soil moisture: Water contained in the soil pores. Stream-gauging station: An observation point at which the flow in a stream is measured. Subsurface runoff: The portion of runoff which comes from water infiltrating into the soil and then joining the river without recharging the groundwater. Surface runoff: The portion of runoff which is contributed by the water flowing on the land surface. Transpiration: Evaporation of water from the leaves, or other parts, of a plant. Unconfined aquifer: An aquifer which has an upper surface open to atmosphere, i.e., there is no confining layer at the top. Unit hydrograph: A hydrograph due to a unit rainfall occurring uniformly over the entire catchment. Water table (or Phreatic surface): The upper surface of the saturated groundwater which is at atmospheric pressure.

2

Precipitation

LEARNING OBJECTIVES LO 1

Know about precipitation, its formation and distribution

LO 2

Outline the methods of measurement of precipitation

LO 3

Summarize the methods of processing of precipitation data, estimation of missing data, completeness and consistency checks, averaging – Thiessen polygons and isohyets

LO 4

Understand presentation of precipitation data, intensity-duration curve, depth-area-duration curve

2.1

INTRODUCTION

Although evaporation is traditionally seen as the starting point of the LO 1 Know about hydrologic cycle, from an engineering point of view, precipitation is precipitation, its formation the driving force behind availability of water. The part of precipitation and distribution which infiltrates into the ground augments the groundwater and the part which runs-off over the surface contributes to the streamflow. Hence total precipitation over an area, and its spatial and temporal distribution, is probably the most significant parameter for hydrologic analysis. Moreover, due to its importance to agriculture (and, therefore the economy) and the simplicity of measuring devices, extensive measurement of precipitation has been going on for hundreds of years (the earliest measurements are thought to be around 500 B.C. and systematic measurements started in the 17th century). At times when precipitation is not the primary variable of interest (e.g., when one is interested in the streamflow), it could be beneficial to correlate the precipitation with these other variables to take advantage of the long and more reliable record of precipitation. With this in mind, we look at several aspects of precipitation in this chapter.

Precipitation

19

The description of the basic mechanism of precipitation, its various forms, and its variation in space and time is provided in the next section. The discussion is just enough to provide an understanding of the factors affecting precipitation and avoids going into greater details, which are probably more suitable for a course in meteorology.

2.2

OCCURRENCE – MECHANISM OF FORMATION, VARIOUS FORMS, DISTRIBUTION

2.2.1

LO 1

Mechanism

A brief description of the various mechanisms leading to precipitation is provided here with the intention that it would lead to a better appreciation of the various factors which affect precipitation and the extent to which they do so. For precipitation to occur there should be water in the atmosphere and the water droplets (or snow/ice crystals) should be heavy enough to overcome the upward forces of the air currents. As far as the presence of water in the atmosphere is concerned, there is almost always some water present in the air (represented as humidity); however, it is generally invisible since it is in the form of water vapour. The air has a certain capacity to hold the water vapour, which depends on its temperature: warmer air can hold more water vapour than cooler air. Hence, for the vapour to become visible in a liquid form, either more water vapour has to be added beyond this saturation point, or the air has to be cooled to a temperature beyond its dew point (the temperature at which the air becomes saturated with water vapour). The addition of water vapour may be through evaporation, transpiration, combustion etc. but is generally not enough by itself to saturate the air. Cooling is, therefore, required for the condensation of water vapour to form clouds and some mechanism is needed for fusion of the small droplets into bigger, precipitable drops. Presence of condensation nuclei is necessary for the formation of clouds. These nuclei are small particles (0.1 to 10 μm) on which the droplets or crystals form, and are abundant in the atmosphere. Condensation may occur due to adiabatic (or dynamic) cooling, radiational cooling, contact cooling, or by mixing of air masses of different temperatures. Almost all precipitation is a result of adiabatic cooling in which air mass rises to a higher altitude, expands because of lower pressure, and becomes cooler due to loss of a part of the heat energy in doing work in the process of expansion. This rise could be orographic (due to mountains), convective (due to local heating of air mass), frontal (due to a warm air mass climbing over colder air mass), or cyclonic (similar to frontal, with air masses converging into a low pressure area). Accordingly, the precipitation is classified as orographic, convective, frontal, or cyclonic.

2.2.1.1 Orographic Precipitation When an air mass encounters a topographic barrier (e.g., a mountain), it may flow up and over it. This lifting results in cooling of the air mass and sets off precipitation. The precipitation is heavier on the windward side and there is a “rain shadow” on the leeward side, i.e., there is lighter precipitation due to the air mass becoming dry and warm as it travels down from the peak. This type of precipitation is usually of low intensity and long duration. Sometimes the term relief precipitation is also used for orographic precipitation.

20

Engineering Hydrology

2.2.1.2

Convective Precipitation

On a hot day, as the ground surface gets heated, the air in contact with it also becomes warmer and therefore, lighter. This sets up a convective current in which the warm moist air rises and its place is taken by the colder (and denser) surrounding air. This colder air is then heated as it comes in contact with the ground and rises, thus continuing the convection. The rise, expansion and consequent cooling of the air mass results in convective precipitation. This type of precipitation is more common in hot and humid tropical areas and is of high intensity, generally accompanied by thunder and lightning.

2.2.1.3

Frontal Precipitation

The surface of contact between air masses of different temperatures is known as a front. If the cold air mass is moving into a warm air mass, it is called a cold front and if the warm air is moving into the colder air, it is called a warm front. In both cases, the warmer air gets lifted above the front, causing cooling and precipitation. The frontal precipitation is generally of moderate to high intensity, with the cold fronts having more intense precipitation on a relatively smaller area and the warm fronts causing less intense but widespread precipitation.

2.2.1.4 Cyclonic Precipitation A cyclone is an area of low pressure with a circulatory motion, generally inward, of air around it (the circulation is in counter-clockwise direction in the Northern Hemisphere and clockwise in the Southern Hemisphere). The air converging into this low pressure area gets lifted and causes cyclonic precipitation. The smaller diameter cyclones (up to about 1600 km) have high wind velocity and result in high precipitation, while the larger cyclones (diameter up to about 3200 km) cause widespread precipitation, similar in nature to frontal precipitation.

2.2.2

Forms

As discussed in the previous chapter, precipitation includes all aqueous particles which fall from the atmosphere onto the earth. A wider definition includes, in addition to the particles falling onto the earth, particles falling from the atmosphere but not reaching the earth due to evaporation. However, this portion of the precipitation, known as virga, is not of as much interest to engineers as it is to meteorologists. Depending on the mechanism and location of its formation and the properties of the medium through which it passes before reaching the earth, the precipitation could be in one of several possible forms, either liquid, or solid, or mixed. For example, rain and drizzle comprise liquid precipitation, with rain having drops with diameter more than 0.5 mm and drizzle having drops of smaller diameter; snow and hail are solid, with hail having balls of diameter more than 5 mm; and sleet is a mixture of liquid and solid. Sometimes, the precipitation may be in the form of liquid but becomes solid as soon as it comes in contact with a surface (freezing rain). Other forms in which the atmospheric moisture manifests itself are: Mist–small droplets suspended in air; Fog–mist with visibility of less than one kilometer; Dew–condensation of water vapour on a surface whose temperature is below the dew point of the surrounding air; Frost–when the temperature of the air near the earth’s surface is below freezing, the liquid precipitation freezes on coming in contact with the ground or other cold objects (the ice thus formed is also known as glaze). From the engineering perspective of water availability, it is common to express all forms of precipitation in terms of an equivalent water depth. Thus,

Precipitation

21

depending on its density, which in turn depends on whether the snow is fresh or compacted, 1 cm of snow depth could result in a water depth of about 1 mm to 5 mm (specific gravity of falling snow is generally in the range of 0.05 to 0.15, with most of the values closer to 0.08).

2.2.3

Intensity of precipitation

More than the form of precipitation, it is the intensity, i.e., the rate at which it is falling, which is of relevance to a hydrologist. The intensity represents the depth of precipitation accumulated per unit time and is generally expressed in mm/h. The precipitation is classified as light, medium (or moderate), and heavy, depending on the intensity. For example, on the basis of hourly intensity, rain is classified as light if the intensity is less than 2.5 mm/h, heavy if the intensity is more than 7.5 mm/h, and medium in between. Drizzle has intensity less than 1 mm/h. The India Meteorological Department classifies rain based on daily precipitation, with very light rain having an accumulated depth of precipitation in a day between 0.1 to 2.4 mm, light rain between 2.5 to 7.5 mm, moderate rain between 7.6 to 35.5 mm, rather heavy rain between 35.6 to 64.4 mm, heavy rain between 64.5 to 124.4 mm, very heavy rain between 124.5 to 244.4 mm and extremely heavy rain more than or equal to 244.5 mm. A rainy day is one which has a rainfall amount of 2.5 mm or more, i.e., a day with very light rain is not considered a rainy day. Snow is generally classified on the basis of visibility, which is an indirect measure of the intensity–light if the visibility is more than 1000 m, heavy for visibility less than 500 m, and moderate in between.

2.2.4

Distribution

As discussed in the previous subsection, precipitation is affected by several factors like presence of moisture, condensation nuclei, mountains, differential heating of air masses, etc. Therefore, it shows considerable variations in space and time. For example, if we consider the precipitation at a particular place over a year, most days may have little or no precipitation and a few days may account for more than half of the annual precipitation. Similarly, the annual precipitation may vary from year to year and it would be desirable to define an average or normal annual precipitation which could form a basis for comparing the precipitation at different places. This average is taken over a long period (usually 30 years, recomputed every 10 years; currently 1981–2010 period averages are being used and in 2021, 1991–2020 averages would be used) and clearly shows the distinct precipitation regimes at different places. For example, the normal annual precipitation averaged over the entire country is about 50 mm for Egypt and about 3000 mm for Costa Rica! Similarly, different places within a country could be compared: in USA, Hawaii receives about 1600 mm per year while Nevada only 240 mm; in India, Western Rajasthan receives 300 mm while Kerala receives 3000 mm. In addition to the large scale variations, there are small scale spatial and temporal variations which are equally important but difficult to characterize. For a single storm, i.e., strong winds with a lot of precipitation, the rainfall intensity varies widely over time, and two places within a few kilometers of each other may show considerable difference in precipitation. These variations influence the runoff amount and pattern significantly and must be considered in estimating the maximum flood in a channel or for the design of storm water drains. A proper measurement of precipitation is, therefore, necessary and is described in the next section.

22

2.3

Engineering Hydrology

MEASUREMENT – RAIN GAUGES, REMOTE SENSING

A very brief description of precipitation measurement is provided here and LO 2 Outline the methods of more details are given in Chapter 10, which deals with the measurements measurement of precipitation of several hydrologic variables. In this chapter, we assume that the precipitation is in the form of rain; else if it is in the form of snow, it should be converted to an equivalent depth of rain. The precipitation is generally measured in terms of the depth of water which will accumulate on the land surface if no part of it is allowed to infiltrate/evaporate/run-off. Traditional methods of measurement, therefore, comprise a container which collects the precipitating water/snow and which could be monitored continuously (recording gauges) or periodically (non-recording gauges). For the non-recording gauges, the precipitation depth is measured at a fixed time (in India, 8:30 am) every day and is reported as the depth of precipitation accumulated in the previous day. During intense storms, however, measurements are sometimes made every few hours or at even smaller intervals. Depending on the purpose for which the data is to be used, non-recording gauges may be adequate. However, for some cases, the more expensive recording gauges have to be used. For example, if one were interested in finding the monthly distribution of precipitation in an area, data from a non-recording gauge is sufficient. On the other hand, to estimate peak flow in a nearby river, intra-day variation of precipitation is generally needed. The flow in a river would be much higher if the same amount of rain falls in an hour compared to when it falls uniformly over the whole day. To balance the needs and the cost, typically about 10% of the gauges in an area are of the recording-type and the others are non-recording gauges which are assumed to follow a similar precipitation pattern as obtained at the recording gauges. In addition to the rain gauges, the use of radars and other remote-sensing devices is now becoming common. These devices estimate the intensity of rainfall or the velocity of raindrops by emitting an electromagnetic pulse and analyzing its reflection by the precipitating water. As mentioned earlier, Chapter 10 provides more details of these techniques of collecting the precipitation data. Here we describe the presentation of data and its analysis.

2.4

ANALYSIS OF DATA – MISSING DATA, CONSISTENCY CHECK, AVERAGING

Data from the non-recording gauges is in the form of total depth of precipitation collected at the gauge during a 24-hour period. It could be expressed as a histogram, which shows the variation of daily precipitation with time, or could be manipulated to obtain a mass curve, which shows the variation of accumulated precipitation with time, as shown below:

LO 3 Summarize the methods of processing of precipitation data, estimation of missing data, completeness and consistency checks, averaging – Thiessen polygons and isohyets

For a recording gauge, the measurement is generally of the cumulative depth and the histogram could be derived as shown below (the data corresponds exactly to that shown in Figure 2.1, but the cumulative precipitation depths are recorded every hour).

Precipitation

70

Cumulative Depth (mm)

Precipitation Depth (mm)

25

23

20 15 10 5 0

60 50 40 30 20 10 0

0

1

2

3

4

5

6

0

Time (d)

Figure 2.1

Figure 2.2

1

2

3

4

5

6

Time (d)

Histogram and mass curve for a non-recording gauge

Mass curve and histogram for a recording gauge

Precipitation Intensity (mm/h)

It is more instructive to express the data in terms of precipitation intensity, generally in mm/h. For example, we observe from Figure 2.2 that the rainfall depth increases rapidly during the second day (and most rapidly around the middle of the day) and very slowly during the 5th day. This is readily seen 1.8 from the histogram but, in order to compare 1.6 different precipitation events which may have 1.4 different recording intervals, it is more useful to 1.2 plot rainfall intensity versus time rather than the 1 precipitation depth during that interval. The plot 0.8 0.6 of rainfall intensity versus time is known as a 0.4 hyetograph, and is shown below in Figure 2.3. 0.2 (Since the observed data is at hourly intervals, the 0 hyetograph is identical to the histogram. To 0 20 40 60 80 100 120 140 Time (h) differentiate, we have plotted a smooth curve for the hyetograph, with the intensity during a time Figure 2.3 Hyetograph for a recording gauge interval assumed to correspond to the mid-point.)

24

Engineering Hydrology

The maximum intensity is at about 40 h and is equal to 1.7 mm/h, compared to an average intensity on that day of about 20 mm/d (see Figure 2.1) or about 0.85 mm/h. For a non-recording gauge, the hyetograph will be similar to the histogram since the data at a finer resolution is not available. However, data from a nearby recording gauge could be utilized to estimate the intensity variation within a day. For subsequent discussion, therefore, we will assume that the data given to us is from a recording gauge and analyze it accordingly. While some recording gauges record the time taken for a specified rainfall depth to accumulate and therefore have a discrete record, others record the continuous accumulation of rainfall over time. The continuous record could be presented as a plot but, from data storage perspective, it must be discretized. For example, we could express it as the rainfall depth accumulated in a day, an hour, or a minute. Clearly, if the data is stored as depth of rain in a day, there is no advantage gained by a recording gauge over a non-recording gauge, which also records the rainfall in a day. On the other hand, if we store the data in terms of the depth accumulated every minute, it may become voluminous. Therefore, a 15-minute interval is commonly used for reporting purposes. The analysis of precipitation data is described below by considering a hypothetical, and probably unrealistic, basin which has been idealized as a rectangle with dimensions 60 km East-West and 30 km North-South. There are four rain gauge stations as shown in Figure 2.4 given below:

Figure 2.4

Idealized basin with four rain gauges

The catchment area has 4 gauges inside it and the first question we need to address is whether these four gauges are sufficient to capture the spatial variation of rainfall over this area, or do we need to install more gauges. There are several statistical techniques which may be utilized to answer this question and we describe two of these below. (i) Based on the allowable coefficient of variation The coefficient of variation (CV) is defined as the ratio of standard deviation to the mean and is a measure of the spread of the data. We may consider a network of gauges to be dense enough if its CV is less than, say, 20%. Therefore, for the existing network if we compute the CV and find it to be less than 20%, we do not need to add more gauges. However, if the CV is more than 20%, we would need to add an adequate number of gauges to bring the CV below 20%. If we assume that addition of gauges does not significantly alter the sum of the squared deviations and the mean (in other words, the additional gauges have precipitation equal to the mean), we obtain the desired number of stations by

Precipitation

25

2

Ê CVexisting ˆ N desired = Á ˜ N existing Ë CVdesired ¯

(2.1)

The following example illustrates the procedure. � EXAMPLE 2.1 The annual precipitation at the four rain gauges is shown in the table below. Determine whether more stations are needed, if the desired CV is 20%. Station

S1

S2

S3

S4

Annual Rainfall (mm)

830

1120

1000

650

Solution The mean of the station data for annual precipitation is obtained as 900 mm, and the standard deviation, (702 + 2202 + 1002 + 2502 ) = 204.78 mm. The coefficient of variation is, therefore, 0.228. Since it 3 is more than the desired CV of 0.2, we require more stations and the total number of stations, using Eq. (2.1), s n -1 =

2

Ê 0.228 ˆ is Á ¥ 4 = 5.18. We should provide two more stations. Ë 0.2 ˜¯ Notes: If there is another station close to the boundary, but outside the catchment area, we may expect it to be climatically similar to other stations and should include it in the analysis, even though it lies in a different catchment. Also, we should keep in mind that this technique should be applied to long-term data, e.g., annual, which tends to average out the temporal and spatial variations in rainfall. (ii) Based on the allowable error in the estimation of the mean We could view the observed station data as a “sample” drawn from a large “population” of possible stations located in and around the catchment area and then estimate how close is the sample mean to the true value, i.e., the population mean (more details about sample and population characteristics are provided in Section 9.2). In statistical terms, we measure this closeness by the standard error of the mean, defined as the standard deviation of the estimates of the population mean through sample means, and given by SE =

s

, where s is the sample standard deviation n and n is the sample size. A relative standard error, RSE, is defined as the ratio of the SE and the mean, and we could stipulate that the station density is adequate if RSE is below, say, 10%. Assuming that addition of stations does not significantly change the standard deviation, the number of stations required to achieve a desired RSE is given by Ê CV ˆ N= Á ˜ Ë RSEdesired ¯

2

(2.2)

If this number comes out to be less than or equal to the number of existing stations n, the network density is adequate, otherwise we need to add N − n stations (obviously, N − n is rounded up to the next higher

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Engineering Hydrology

integer. However, considering the uncertainties involved in the analysis, we could use a threshold, e.g., 0.1, below which the value could be rounded down). � EXAMPLE 2.2 From the data of Example 2.1, determine whether more stations are needed, if the desired RSE is 10%. Solution 2 Ê 0.228 ˆ Using Eq. (2.2), the number of stations required is Á = 5.18 which indicates that we need two Ë 0.1 ˜¯ more stations. Note that if the desired RSE were larger than 0.114, the existing network would have been sufficient. There are several other more advanced techniques of not only finding the number of additional stations required, but also optimizing their locations. We have not discussed these here and the interested readers should refer to Chebbi et al., (2013) or Wadoux et al., (2016) for a recent review. Once we are satisfied with the network density, we should process the data to see if it is complete and consistent. And, if some of the data is missing or a station seems to have inconsistent data, we should look at the techniques which will address these issues.

2.4.1

Completeness Check

The precipitation record from a station comprises data at regular intervals, e.g., 1 h, 3 h, 6 h etc., representing the depth of precipitation accumulated between the two time intervals. If a station is inoperative for some time, the data will show a jump in the time values. To keep a uniform format at all stations, generally the time values are filled at regular intervals and the “missing” precipitation values are either left as blank or given a non-physical value (e.g., –99). A plot of the data will then clearly show the period during which data is not available. However, for a very long period of record, a few missing values may not be obvious from the plot, especially if blanks are used to represent the missing values. Therefore, the use of a large negative number is more common for indicating missing values. One should also be cautious about the values listed just after a missing value, to ascertain whether it was accumulated in that time step or the entire “missing” period.

2.4.2 Estimation of Missing Data The presence of missing values implies that the data will have to be adjusted before processing it to obtain, for example, the annual rainfall at that station. Sometimes, it would be a simple matter of looking at the station record and estimating the missing value. For example, if record for a single time is missing and the data shows no rain in several previous and next steps, it would be logical to assume that the missing record is also zero, especially if nearby rain gauges also show a prolonged dry period. On the other hand, if the missing step is during a rain event, we could interpolate the value from the previous and next time steps; or if the record over a longer period is missing, we could use the precipitation pattern at the nearby gauges to estimate the missing values. Due to large spatio-temporal variation of precipitation during a storm, one should be extremely cautious with such interpolation. For long-term averages, however, assumption of similar pattern at nearby stations is more justifiable.

Precipitation

27

� , at a station x, where part of the data is missing, we could take a simple To estimate the rainfall P x arithmetic mean of the rainfall at nearby stations, as follows: � = P x

 i =1,nPi n

(2.3)

where, n is the number of surrounding stations used to estimate the missing value, and Pi represents the precipitation at the ith station during the period for which the missing value is to be estimated. The selection of the n nearby stations is a little subjective. Typically, n nearest neighbours are selected, but sometimes the area around the station is divided into four quadrants (NE, SE, NW, and SW) and an appropriate number of nearest station(s) in each quadrant is selected. However, due to spatial variability, it is quite possible that the station with the missing data normally receives a different rainfall compared to other stations. Therefore, a normal ratio methodology would probably be better, in which we presume that the ratio of the rainfall at any station to the normal rainfall at that station is equal to the arithmetic mean of such ratios at the nearby stations. Therefore, � P P 1 x = Â i N x n i =1,n Ni

(2.4)

where, Ni represents the normal precipitation at the ith station corresponding to the period for which the missing value is to be estimated. The normal precipitation is generally taken as the mean value over a period of 30 years and is modified every decade. For example, currently the normal indicates a mean over the period 1981–2010, but starting in 2021 the normal will change to the mean over the period 1991–2020. Some other methods for the estimation of missing values have also been used but are not very common. For example, an inverse distance weight or inverse-distance-squared weight could be assigned to the data from nearby gauges: � = P x

 i =1,nPi di- p  i =1,ndi- p

(2.5)

where, di is the distance of the ith station from station x, and p is 1 for inverse distance weight and 2 for inverse-distance-square weight (a value of p = 3, i.e., inverse-distance-cube weight, has been found to work well in some studies). In the quadrant method mentioned earlier, the nearby gauges are chosen as the nearest one in each quadrant (i.e., NE, SE, SW, and NW) to avoid giving undue importance to stations in any one direction. Another alternative is the use of regression techniques, in which a relationship is derived between the precipitation at station x and that at other stations, based on the time periods for which data is available at all stations. This relationship is then used to estimate the missing value(s) of precipitation at the station x. Although a linear relationship is most commonly used, other forms may be used if the regression coefficient is not of acceptable accuracy. The geostatistical technique of Kriging has also been applied for spatial interpolation of data, but these are not discussed here due to their advanced nature.

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Engineering Hydrology

The following example illustrates these techniques used for the estimation of missing data. � EXAMPLE 2.3 The two additional rain gauges required for the desired accuracy, S5 (to the West of S1 and S4) and S6 (to the East of S2 and S3), were installed along the East-West centerline at a distance of 10 km from the boundaries. The following table shows the normal values at these stations and some typical values for the year 1995. Station S6 was inoperative for most months of this year. Estimate the missing values. (Note: Due to roundoff, the monthly normals may not add up to the yearly normals. Also, the differences in precipitations have been exaggerated to illustrate the method. In practice, precipitation at stations close to one another will not be so different.)

Table 2.1 The normal precipitation (in mm) at the six rain gauges and the observed values in 1995 Station

S1

S2

S3

S4

S5

S6

Jan

24.6

32.5

29.8

20.2

18.9

36.0

Feb

4.9

6.5

6.0

4.0

3.8

7.2

Mar

9.9

13.0

11.9

8.1

7.6

14.4

Apr

9.9

13.0

11.9

8.1

7.6

14.4

May

4.9

6.5

6.0

4.0

3.8

7.2

Jun

147.8

195.2

179.1

121.2

113.3

216.1

Jul

216.7

286.3

262.7

177.8

166.1

316.9

Aug

197.0

260.3

238.8

161.6

151.0

288.1

Sep

177.3

234.2

214.9

145.5

135.9

259.3

Oct

19.7

26.0

23.9

16.2

15.1

28.8

Nov

9.9

13.0

11.9

8.1

7.6

14.4

Dec

14.8

19.5

17.9

12.1

11.3

21.6

837.3

1106.1

1014.8

686.9

641.8

1224.5

1.2

1.1

0.9

0.5

0.8

?

25.1

35.0

31.6

22.3

20.1

?

848.9

1120.0

1028.0

703.0

675.0

?

Annual Jan 12, 1995 Jan 1995 1995

Solution Estimation of the missing value for a single day is not recommended due to the uncertainties involved with the storm location and pattern. For the month of January 1995, we could use the arithmetic mean method to 25.1 + 35.0 + 31.6 + 22.3 + 20.1 obtain the estimate of the rainfall at station S6 as = 26.8 mm. 5 A look at the normal rainfall for January (as well as the normal annual rainfall) clearly shows that station S6 normally receives relatively higher rainfall compared to other gauges with S2 being the most comparable. Therefore, the normal ratio method would probably be better, and results in P6 1 Ê P1 P2 P ˆ 1 Ê 25.1 35.0 31.6 22.3 20.1ˆ = Á + +º + 5 ˜ = Á + + + + = 1.064 N6 5 Ë N1 N 2 N 5 ¯ 5 Ë 24.6 32.5 29.8 20.2 18.9 ˜¯

Precipitation

29

From which we obtain the estimate of precipitation as 1.064 × 36.0 mm, i.e., 38.3 mm, quite different from the one obtained using arithmetic mean. The inverse-distance technique with linear, square, and cubic weights is shown below, with the weights as

di- p

 i =1,ndi- p

and the estimate obtained as the weighted sum of the precipitations at the stations S1 to S5.

Station

S1

S2

S3

S4

Precipitation in Jan 1995 (mm)

25.1

35.0

31.6

22.3

S5

20.1

Estimate at S6 using Eq. (2.5)

Distance from S6 (km)

30.414

10.247

10.247

30.414

40

Weight (for p = 1)

0.115

0.341

0.341

0.115

0.087

29.9

Weight (for p = 2)

0.050

0.436

0.436

0.050

0.029

32.0

Weight (for p = 3)

0.018

0.478

0.478

0.018

0.008

32.8

Note that the weights of the stations located closer to S6 tend to increase with an increase in p. In the limit when p is very large, the method becomes a nearest-neighbour method, in which the precipitation at x is assigned a value equal to that at the nearest gauge. For the estimation of the annual rainfall at S6 in 1995, using the normal ratio method, we estimate 1 Ê 848.9 1120.0 1028.0 703.0 675.0 ˆ the value as Á + + + + ¥ 1224.5 mm, i.e., 1252.6 mm. We expect these 5 Ë 837.3 1106.1 1014.8 686.9 641.8 ˜¯ estimates to be more reliable than the arithmetic mean and, as expected, these are close to the values observed at S2. If the normal precipitation values at all the stations are not significantly different from one another, the two methods will result in nearly same values. Therefore, if the normal precipitation at the other stations is within 10% of that at the “missing” station, arithmetic mean method could be used. Another point to note is that we have normalized the precipitation value with the normal rainfall for the month of January. If we use the normal annual rainfall, the estimate is not likely to change significantly, and for this example is found to be identical. Also, as we will see in the next section, the data at station S6 is not consistent, therefore the normal values of 36.0 mm for the January precipitation and 1224.5 mm for annual precipitation are incorrect, making the estimate of missing value questionable. We will come back to this issue in the next section.

2.4.3

Consistency Check

A plot of the data at a station would show any internal inconsistency in it, e.g., an unrealistically high value, a sudden change in the annual precipitation etc. A good way of assessing a change in the annual precipitation is by drawing the mass curve. If the mass curve is very close to a straight line, it indicates no change in the “average” precipitation at the station. If there is a distinct break in the slope of the mass curve (or if the curve is concave or convex), it indicates a sudden (or gradual) change in the observed precipitation, as shown in Figure 2.5. One must keep in mind that an inconsistent data set may have a perfectly logical explanation (e.g., climate change) and, by itself, does not imply that the data needs to be “corrected.” However, if there is a

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Engineering Hydrology

change in the general climate of the area, nearby stations will also be similarly affected. Hence, a double mass curve, which plots the cumulative precipitation at a station versus the average cumulative precipitation at a few nearby stations (generally 5 to 10, to minimize the effect of a possible inconsistency in one of those) would eliminate the effects of a general change in the precipitation pattern and will clearly bring out any inconsistency in the station data. The possible causes of such inconsistencies are a shift in the location, change in canopy, change in landscape etc. Figure 2.6 shows a typical double mass curve for an “inconsistent” station and indicates a sudden change in the precipitation regime in the year 1995.

Figure 2.5

Figure 2.6

Mass curve at a station with possibly inconsistent data

Typical double mass curve when the station data is inconsistent

Precipitation

31

The technique for correcting the data to make it consistent will depend on whether the older data is correct or the newer one. For example, if the older data is assumed to be correct, the data after 1995 needs to be adjusted in such a way that the double mass curve for the period after 1995 falls on the dashed line. More often, however, it is assumed that the newer data is correct (e.g., since the station was located near a tall building, it did not represent the actual precipitation, and was shifted to a more appropriate location in 1995). In this case, the double mass curve is plotted in reverse chronological order and the data prior to 1995 is modified by multiplying the annual precipitation values by the slope ratio as illustrated in the example below. � EXAMPLE 2.4 The table below shows a 30-year record for the annual precipitation at the 6 stations. Check the consistency of data for station S6, and correct the data, if needed.

Table 2.2 The annual precipitation (in mm) at the six rain gauges from 1981 to 2010 Year

2010 2009 2008 2007 2006 2005 2004 2003 2002 2001 2000 1999 1998 1997 1996 1995 1994 1993 1992 1991 1990 1989 1988 1987 1986 1985 1984 1983 1982 1981

S1

876.8 839.2 848.9 826.7 843.4 830.8 846.9 840.2 835.0 826.7 842.6 835.0 835.0 832.0 835.0 835.0 837.5 832.7 839.4 835.0 833.0 831.2 835.0 838.5 835.0 831.8 835.0 836.5 835.0 835.0

S2

S3

1158.2 1108.5 1121.4 1092.0 1114.0 1097.5 1118.8 1109.9 1103.0 1092.0 1113.0 1103.0 1103.0 1099.1 1103.0 1103.0 1106.2 1099.9 1108.8 1103.0 1100.4 1098.0 1103.0 1107.6 1103.0 1098.8 1103.0 1105.0 1103.0 1103.0

1062.6 1017.1 1028.9 1001.9 1022.1 1006.9 1026.5 1018.3 1012.0 1001.9 1021.2 1012.0 1012.0 1008.4 1012.0 1012.0 1015.0 1009.2 1017.3 1012.0 1009.6 1007.4 1012.0 1016.2 1012.0 1008.1 1012.0 1013.8 1012.0 1012.0

S4

719.3 688.4 696.4 678.2 691.9 681.6 694.8 689.3 685.0 678.2 691.2 685.0 685.0 682.6 685.0 685.0 687.0 683.1 688.6 685.0 683.4 681.9 685.0 687.9 685.0 682.4 685.0 686.2 685.0 685.0

S5

672.0 643.2 650.7 633.6 646.4 636.8 649.1 644.0 640.0 633.6 645.8 640.0 640.0 637.7 640.0 640.0 641.9 638.2 643.4 640.0 638.5 637.1 640.0 642.7 640.0 637.5 640.0 641.1 640.0 640.0

S6

1254.0 969.0 1242.6 1094.4 1026.0 1276.8 1117.2 1026.0 1254.0 1254.0 1026.0 1254.0 912.0 1254.0 1140.0 1252.6 1311.0 1311.0 1368.0 1197.0 1368.0 1333.8 1311.0 1345.2 1254.0 1288.2 1368.0 1254.0 1333.8 1368.0

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Engineering Hydrology

Solution The first thing we should note in the table above is that the value for the annual precipitation in 1995 at S6 (shown in bold) is not an observed value but is estimated in the previous example. We have deliberately taken values for the year 1996 as being identical to 1995, so that we could use the same annual value of 1140.0 mm at S6 for further analysis (assuming that the 1996 data is correct). In practical cases, there may not exist such a simple way of correcting the data beforehand and one could be forced to use the estimated value, which will lead to further propagation of error. An iterative technique could be formulated to use the estimated value, check it for consistency, correct the data to make it consistent, compute the new normal values at S6, and then go back to the estimation of missing values using the corrected normals. We leave it to the reader to try! The table below shows the computations of the mean annual precipitation for stations 1 to 5, and the preparation of the double mass curve, which is plotted in Figure 2.7. The last column is computed after plotting the mass curve and obtaining the correction factor. Year

Mean S1-5 (mm)

Cumulative Mean (m)

S6 (mm)

Cumulative at S6 (m)

2010 2009 2008 2007 2006 2005 2004 2003 2002 2001 2000 1999 1998 1997 1996 1995 1994 1993 1992 1991 1990 1989 1988 1987 1986 1985 1984 1983 1982 1981

897.8 859.3 869.3 846.5 863.6 850.7 867.2 860.3 855.0 846.5 862.8 855.0 855.0 852.0 855.0 855.0 857.5 852.6 859.5 855.0 853.0 851.1 855.0 858.6 855.0 851.7 855.0 856.5 855.0 855.0

0.898 1.757 2.626 3.473 4.336 5.187 6.054 6.915 7.770 8.616 9.479 10.334 11.189 12.041 12.896 13.751 14.608 15.461 16.320 17.175 18.028 18.880 19.735 20.593 21.448 22.300 23.155 24.011 24.866 25.721

1254.0 969.0 1242.6 1094.4 1026.0 1276.8 1117.2 1026.0 1254.0 1254.0 1026.0 1254.0 912.0 1254.0 1140.0 1140.0 1311.0 1311.0 1368.0 1197.0 1368.0 1333.8 1311.0 1345.2 1254.0 1288.2 1368.0 1254.0 1333.8 1368.0

1.254 2.223 3.466 4.560 5.586 6.863 7.980 9.006 10.260 11.514 12.540 13.794 14.706 15.960 17.100 18.240 19.551 20.862 22.230 23.427 24.795 26.129 27.440 28.785 30.039 31.327 32.695 33.949 35.283 36.651

Corrected S6 (mm)

1254.0 969.0 1242.6 1094.4 1026.0 1276.8 1117.2 1026.0 1254.0 1254.0 1026.0 1254.0 912.0 1254.0 1140.0 1140.0 1130.7 1130.7 1179.8 1032.3 1179.8 1150.3 1130.7 1160.2 1081.5 1111.0 1179.8 1081.5 1150.3 1179.8

Precipitation

Figure 2.7

33

Double mass curve for data of Table 2.2

From the figure, it is clear that the record at S6 is not consistent as there are two distinct straight lines (for real data the points will show some scatter about the straight lines. However, for the synthetic data used in the example, we purposely chose to have perfect data.). Assuming the recent records to be correct, the data corresponding to the earlier years, i.e., those before the break in slope, have to be modified in such a way that all the data plots on the solid line. For this small dataset, the year of break may be obtained by counting the points as 1995. Alternatively, we see from the figure that the break occurs at a cumulative mean of about 14, which the table shows to be occurring in 1995. The slopes of the two lines are computed as: solid line, m1 = 18.240/13.751 = 1.326; dashed line, m2 = (36.651 – 18.240)/(25.721 – 13.751) = 1.538. To get the data points prior to 1995 (i.e., the points on the dashed line) to plot on the solid line, it is apparent that the cumulative precipitation at S6 needs to be reduced in such a way that the ratio of the “difference between the corrected cumulative precipitation for any year and the cumulative precipitation for 1995” to the “difference between the uncorrected cumulative precipitation for any year and the cumulative precipitation for 1995” is equal to m1/m2, i.e., 0.862. This is achieved by multiplying the annual precipitation at S6 for all years before 1995, by this factor (0.862) as shown in the last column of the table. Another point to note is that instead of plotting the cumulative mean on the horizontal axis, we could plot the cumulative total without affecting the analysis since it only differs by a constant factor (5 in this case). However, there could be situations where the number of stations is not constant (for example, a station is added or removed during the period considered for the analysis) and the cumulative mean accounts for these. The corrected normal annual precipitation, i.e., the mean for the period 1981-2010 at S6 turns out to be 1137.3 mm, compared to 1224.5 mm obtained with uncorrected data and used in Example 2.3. We could use the ratio 1137.3/1224.5, i.e., 0.929, to obtain a better estimate of rainfall at S6 as 35.6 mm for January 1995 and 1163.7 mm for 1995.

2.4.4

Averaging of Data

Once a complete and consistent set of precipitation data is available at a network of rain gauges, we should think of how to analyze this large volume of data to obtain meaningful results. The inherent spatial and

34

Engineering Hydrology

temporal variability of precipitation patterns implies that the station data should be either analyzed statistically or should be averaged over space and/or time to reduce the uncertainty. The statistical analysis is described later in Chapter 9. Here we discuss the averaging of precipitation data.

2.4.4.1

Temporal Averaging

It is obvious that the precipitation at a place varies significantly from one season to the other within a year. Though it is not as obvious, the precipitation during a certain duration (a day, season, or a year) may also vary significantly from year to year. Therefore, in order to compare the precipitation regime at two stations, we need to define a normal precipitation (Example 2.3 had introduced the concept of normal precipitation). Generally normal precipitation is defined as the mean rainfall for the specified duration taken over a 30 year period and is updated every decade. For example, we could obtain the normal precipitation at a station for July 1 as the average of the precipitation values recorded there from 1981 to 2010, on that day. Similarly, a normal annual precipitation could be obtained by dividing the total precipitation during the years 1981 to 2010 by 30, or a normal monsoon precipitation could be obtained by averaging the monsoon-season precipitations over this period. Two different stations could then be compared to identify, for example, the wetter place; or we could utilize the normal precipitation over the entire rain gauge network over a country to prepare isopluvial maps, showing contours of equal precipitation depth, e.g., for India (saturn.20karma. com/imdguwahati/site/datafile/annual_rainfall.php) or by color-mapping, e.g., for the USA (www.wrcc.dri. edu/pcpn/us_precip.gif). In general, the temporal density of the precipitation records is much larger compared to the spatial density. This is to say that the spatial network of rain gauges could be sparse, but each gauge has a long period of daily (or hourly) data. Hence spatial averaging requires more specialized techniques, some of which are described next.

2.4.4.2

Spatial (or Areal) Averaging

Depending on the storm position and movement, rain gauges in close proximity may record widely different precipitations. For analysis of longer-term precipitation, e.g., monthly, seasonal, or annual, it may be assumed that these local variations would be averaged out. For shorter durations, e.g., hourly or daily, it may not be justifiable to average the precipitation at gauges in an area to obtain an average precipitation for the area. Let us consider the same catchment with all 6 rain gauges as shown in Figure 2.8 and assume that we are given the precipitation records for the 6 gauges and are asked to estimate the average annual precipitation for this catchment for the year 2013. The simplest method to obtain this estimate would be to take an arithmetic mean of the annual precipitation (for 2013) of the 5 stations. However, a drawback of this method is that the location of the station within the area is not given any consideration. For example, if there are only two stations in an area, one near the centre and the other near the boundary, one should expect the average precipitation to be closer to that at the central gauge. This is accounted for in the method of Thiessen Polygon, which divides the catchment area into influence areas of each gauge such that any point in the area of influence of a gauge is closer to that gauge than any other. The weight given to the precipitation at a gauge is then assigned in proportion to the area of influence. While this method takes care of the location in the horizontal plane, it does not account for other possible factors which may affect the area of influence of a gauge. For example, a gauge located on a hill may show a larger precipitation compared to the one located in a plain area and, when delineating the areas

Precipitation

Figure 2.8

35

Idealized basin with six rain gauges

of influence, it would be logical to assume that the area of influence of the gauge on the hill does not go beyond the hill. To account for these factors, we could draw isohyetal lines (i.e., lines of equal precipitation during a given time interval) using all the relevant information about the basin. Taking the same example of a station on a hill, we could draw isohyets spaced closely near that station to represent rapid variations in precipitation there. Once an isohyetal map is prepared, the average precipitation may be obtained by assigning an area of influence to each isohyet extending halfway between the neighboring isohyets. Alternatively, we could consider the area between two isohyets to have an effective rainfall equal to the average of those two isohyets. Methods based on inverse square distance weights could also be used to estimate the average depth of rainfall, by assigning weights to each rain gauge in proportion to its distance from the centroid of the catchment area. Another option could be to overlay a rectangular grid over the area, compute the precipitation depth at each grid point using, say, inverse square distance weighting, and then taking an average of all grid points lying within the catchment area. In the Index Station method, typically used in mountainous areas, the station weights are determined based on climatology. The following example illustrates the application of the arithmetic mean method, the Thiessen polygon method, and the isohyetal method. � EXAMPLE 2.5 Find the mean normal annual precipitation over the catchment area by using the arithmetic mean, Thiessen polygon, and Isohyetal methods. (Obtain the normal annual precipitation from Tables 2.2 and 2.3 with corrected data for S6.) Solution The table below shows the normal annual precipitation for the rain gauges: Station Normal Annual Precipitation (mm)

S1

S2

S3

S4

S5

S6

837.3

1106.1

1014.8

686.9

641.8

1137.3

The arithmetic mean provides the normal annual rainfall for the basin as 904.0 mm.

36

Engineering Hydrology

The figure below shows the stations joined by straight lines and perpendicular bisectors drawn on each joining line. Since these bisectors represent the points equidistant from the two stations, any point lying within the region bounded by these bisectors will be closer to one station than the other. For example, looking at the bisector of the line joining S5 and S1, any point to the left of it will be closer to S5 as compared to S1. Extending this argument, any point lying in the area ABCDE will be closer to S5 than to any other gauge and we could take this as the area of influence of S5. Because of the simple geometry of catchment, we can analytically compute this area as 375.5 km2 (in most practical cases, one would need digitization to compute this area).

Figure 2.9 Thiessen Polygons for the six rain gauges The table below shows the computation of the average precipitation: S1

S2

S3

S4

S5

S6

Area of Influence (km2)

262.3

262.3

262.3

262.3

375.5

375.5

Weight (=Area of Inf. / Total Area)

0.1457

0.1457

0.1457

0.1457

0.2086

0.2086

Station

The normal annual rainfall for the basin using the weighted mean computed from the Thiessen polygons is obtained as (0.1457 ¥ 837.3 + 0.1457 ¥ 1106.1 + 0.1457 ¥ 1014.8 + 0.1457 ¥ 686.9 + 0.2086 ¥ 641.8 + 0.2086 ¥ 1137.3) = 902.2 mm. Based on the station values, a very rough sketch of the isohyetal lines is shown in Figure 2.10 and an even more approximate estimate of the area between two isohyets is listed in the table below. The drawing of isohyetal lines is similar to the plotting of contours from the elevation values measured at some locations and is, therefore, not described in details here. Since the precipitation values are available at only six stations, there is a large uncertainty in the isohyetal pattern shown in the figure. Generally a linear variation is assumed between two stations to obtain the points on a specified isohyet. For example, to draw the 700 mm isohyet, linear interpolation between stations 1 and 4, 1 and 5, and 3 and 4 will provide three points. The extension of the isohyet beyond these points is rather arbitrary. With more stations, the preparation of the isohyetal map becomes more reliable.

Precipitation 700

800

900

1000

1100

837

600

1200

1106 1137

642 687

500

37

1015

Figure 2.10 Isohyetal map for the basin Isohyetal line range (mm)

1200

Area between lines (fraction of total area)

0.05

0.1

0.11

0.1

0.1

0.11

0.17

0.15

0.11

Effective precipitation (mm)

22.5

55

71.5

75

85

104.5

178.5

172.5

137.5

The effective precipitation is obtained as the fractional area multiplied by the average rainfall taken as the mean of the bounding isohyets (for area having less than 500 mm rainfall, the mean is considered as 450 mm and for the area with greater than 1200 mm rainfall, the mean has been taken as 1250 mm. Further fine-tuning may be done on the basis of the distance to the boundary, but may not significantly change the result). The effective precipitation essentially indicates the depth of precipitation if all the rain falling in the area enclosed by two isohyets is spread over the entire catchment area. The normal annual rainfall for the basin using the isohyetal method and obtained by adding the effective precipitation values is 902 mm. Some points to be noted about the spatial averaging of the station precipitation data: •

The average precipitation, naturally, will be between the lowest and highest observed station values. Therefore, if the station values do not differ much, any method of averaging could be used without significant error.

•

If the averaging is to be done for several different sets of data, Thiessen polygon method has an advantage over the isohyetal method since the weights assigned to different stations do not need to be recomputed but the isohyets will need to be drawn afresh for each set. The accuracy, however, is likely to be greater for the isohyetal method.

•

If a new station is added, the Thiessen polygon method will require redrawing of the polygons. The isohyetal method will also need redrawing of the isohyets but the effort may not be as significant as that for the Thiessen polygon. The arithmetic mean method, of course, does not need much additional effort even if a new set of data is considered or a new station is added.

38

Engineering Hydrology

•

2.5

The averaging process is also influenced by the intended purpose of the averaging. For example, if the purpose of the averaging is to obtain an average precipitation over a catchment during a day with a view to correlate it with the measured runoff at the outlet; one may consider the time taken by a drop of water to travel from the area near the rain gauge to the outlet. Thus, while averaging the precipitation data one could use the precipitation data of a station which is far from the outlet, from a previous day to account for the time lag. In other words, all precipitation values which are averaged need not be for the same time period. Other factors, like the direction of movement of the storm and distance of the stream from the rain gauge, may further complicate the averaging process.

STORM AVERAGING

For a particular storm event, the maximum intensity of LO 4 Understand presentation of precipitation occurs over an area near the centre of the storm and precipitation data, intensity-duration the intensity decreases as we move further away from the centre. curve, depth-area-duration curve Hence, if we compute the average depth of precipitation for the storm, it would be greatest near the centre and will generally show an exponential decrease with increasing area as we move further out. The temporal variation in the intensity of precipitation for a particular storm however may be quite irregular, with the maximum intensity occurring sometimes near the beginning of the storm and at other times around the middle of it. The temporal variation of the point rainfall values, as obtained at a rain gauge, could be analyzed in terms of an intensityduration curve or a depth-duration curve. One way of characterizing the spatial variations during a storm is through ‘Depth-Area-Duration (DAD)’ curves which depict the relationship between the area considered for analysis and the corresponding average depth of precipitation for different storm durations. Since one is generally interested in the maximum values of the precipitation, for example, to design a drainage system, both the intensity-duration curve and the depth-area-duration curve aim at obtaining the maximum possible intensity or depth of precipitation for a given duration. For example, the intensity-duration curve considers the most intense period of the storm while the depth-area-duration curve starts from the area near the stormcenter and progressively expands outwards. To illustrate, Table 2.3 shows the hourly precipitation during a 24-hour storm at a recording rain gauge (S6) in a catchment area and Figure 2.11 shows the isohyetal map of total rainfall depth drawn on the basis of ten stations (the other nine being non-recording) in and near the catchment. In absence of any other information, we assume that the rainfall pattern at all the stations is similar to that at S6. Since the total depth at the station S8 is largest, we will use it to prepare the intensityduration curve. The estimated hourly precipitation at S8 is obtained by multiplying the hourly precipitation at S6 with the ratio of the total precipitation at S8 and that at S6, as is shown in the last column of the table (although the measurements are done to the nearest tenth of mm, we have listed the estimated values up to one hundredth).

Precipitation

Table 2.3 The hourly precipitation (in mm) observed at station S6 and estimated at S8 Time (h)

Depth of precipitation at S6 (mm)

Estimated Depth of precipitation at S8 (mm)

0

0.0

0.00

1

0.1

0.10

2

0.4

0.42

3

2.1

2.20

4

4.3

4.51

5

3.7

3.88

6

5.4

5.66

7

4.8

5.03

8

7.1

7.44

9

9.8

10.27

10

5.2

5.45

11

10.7

11.22

12

13.9

14.57

13

15.2

15.93

14

16.5

17.30

15

17.0

17.82

16

11.2

11.74

17

8.9

9.33

18

9.1

9.54

19

8.2

8.60

20

7.1

7.44

21

4.5

4.72

22

0.5

0.52

23

0.2

0.21

24

0.1

0.10

Total (mm)

166.0

174.00

39

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Engineering Hydrology

(the dashed lines show the extension of isohyets beyond the catchment area for two selected lines)

Figure 2.11 Total rainfall at stations and Isohyetal map for the 24-hour storm From the table it is obvious that the maximum intensity for 1-hour duration is 17.82 mm/h in the 15th hour. It is also obvious that the maximum precipitation over a 2-hour period occurs during the 14th and 15th hours and is equal to 35.12 mm (or 17.56 mm/h). However, for some storms it may not be so obvious. In general, the maximum intensity during a 2-hour period will include the 1-hour period having the greatest intensity. But that may not always be the case, for example, when the intensity in the hour before and the hour after the most intense hour is relatively small (in case the intensity in the 14th and 16th hours had been 10 mm/h, the 12th and 13th hours would have had an average intensity of 15.25 mm/h, larger than the 13.91 mm/h in 14th and 15th or 15th and 16th hours). As we consider progressively larger durations, it would be desirable to compute the average intensity for ALL consecutive periods of same length to find out the largest average intensity. For example, to compute the maximum intensity for 3-hour duration, we would look at 0-3 hours, 1-4 hours, 2-5 hours etc. A simpler method could be to divide the 24-hour period into non-overlapping 3-hour periods, e.g., 0-3 hours, 3-6 hours and so on. However, it is possible that the maximum intensity could be missed by doing so. We have used all possible durations to generate the maximum intensity for different durations, which is shown in Table 2.4 along with the corresponding precipitation depth during that interval.

Precipitation

41

Table 2.4 Maximum intensity and precipitation depth for different duration of rainfall Duration (h)

Maximum Precipitation Depth (mm)

Period of occurrence (h)

Maximum Intensity (mm/h)

1

17.82

15

17.82

2

35.12

14-15

17.56

3

51.05

13-15

17.02

4

65.62

12-15

16.41

5

77.36

12-16

15.47

6

88.58

11-16

14.76

7

97.91

11-17

13.99

8

107.45

11-18

13.43

9

116.05

11-19

12.89

10

123.49

11-20

12.35

11

131.77

9-19

11.98

12

139.21

9-20

11.60

13

146.65

8-20

11.28

14

151.68

7-20

10.83

15

157.34

6-20

10.49

16

162.06

6-21

10.13

17

165.94

5-21

9.76

18

170.45

4-21

9.47

19

172.65

3-21

9.09

20

173.17

3-22

8.66

21

173.59

2-22

8.27

22

173.80

2-23

7.90

23

173.90

2-24

7.56

24

174.00

1-24

7.25

Figure 2.12 shows a graphical representation of the maximum intensity and rainfall depth variation for different durations. It should be noted that the maximum depth/intensity for a particular duration depends on the frequency of occurrence of the storm event. Naturally, if we want to estimate the absolute maximum values of these parameters, we should consider the most severe storm expected in the area and prepare the intensity-duration curve. A more detailed discussion appears in Chapter 4.

20

200

18

180

16

160

14

140

12

120

10

100

8

80

6

60

4

40

2

20

Depth (mm)

Engineering Hydrology

Intensity (mm/h)

42

0

0 0

4

8

12

16

20

24

Duration (h)

Figure 2.12 Maximum intensity (solid line) and depth of precipitation (dashed line) for different durations for the given 24-hour storm The analysis of spatial variations during a storm is more complex compared to the analysis of point rainfall values. We illustrate the procedure by assuming, as stated earlier, that the rainfall patterns at all stations are similar, implying that the isohyetal map for all durations are identical in shape, only the magnitude is different. Also, we consider only three isohyets (150, 160, and 170 mm shown as closed curves in Figure 2.11) since analysis of other isohyets is similar but would require more data to draw the complete (closed) contours. Note that we also include the area outside the catchment since we are analyzing the storm and not the average precipitation over the catchment. Starting from the area enclosed by the isohyet near the storm-centre, we move outward to subsequent isohyets as shown in the table below to prepare the depth-area curve for the 24-h duration. The average precipitation in the area enclosed by the 170 mm isohyets is taken as 172 mm based on the station value of 174 mm and for other areas it is taken as the mean of the surrounding isohyets. For example, the average depth of precipitation over the area enclosed by the 160 mm isohyets is computed as (20 × 172 + 140 × 165)/ (20 + 140) mm, i.e., 165.875 mm. Isohyet range (mm)

Area enclosed (km2)

Average Depth of precipitation (mm)

>170

20

172.0

160-170

140

165.9

150-160

230

159.5

Due to our assumption of same isohyetal pattern for all durations, the average depths for 1-hour duration would be equal to the values shown in the table multiplied by the ratio 17.82/174.00 since the maximum depth of rainfall during a 1-hour duration was earlier obtained as 17.82 mm. Proceeding similarly for other durations, the resulting depth-area-duration curve is shown in the figure below:

Precipitation

43

Figure 2.13 Maximum depth of precipitation for different durations over different areas These curves may be expressed in mathematical form as P=

C1 D ( A + C2 D ) n

(2.6)

where, P is the precipitation depth (generally cm), D is the relevant duration (h), A is the appropriate area (km2), and C1, C2 and n are region-dependent constants which are obtained by curve fitting. If we have hourly data available at all the stations for the total storm-duration of 24 h, there is no change in the procedure for the 24-h DAD curve. However, when we want the 12-h DAD curve, keeping in mind that we want the maximum accumulated depth, it is not immediately clear as to which 12 hour period should be used. The period of maximum rainfall may be different for different stations due to the movement of the storm. The most thorough procedure therefore would be to prepare the 12-h DAD for all the 13 possible 12-hour intervals and draw an envelope curve over the DADs so obtained. A less tedious option could be to use the 12-hour period which shows the maximum average precipitation, draw the corresponding isohyetal map and derive the DAD from it. To reduce the computational effort, sometimes Thiessen polygons are drawn to estimate the weights assigned to each station while computing the average precipitation over the area between successive isohyets. These weights will remain the same as long as the isohyetal pattern is assumed to be similar and the computation of average depth for a different duration will only require a simple calculation of the weighted mean. Since the basic procedure is the same as discussed above, we do not address these issues in detail. Most of the DAD curves follow an equation of the form: P = P0 e - KA

n

(2.7)

44

Engineering Hydrology

where, P is the average depth of precipitation (usually in mm) over an area A (usually in km2) for a d-hour storm (d is not directly included in the equation, but the parameters K and n are functions of d, in addition to being dependent on the general climate of the area under consideration. As an example, for North India, for a 48-hour storm, K is about 0.001 and n is about 0.63). Clearly, P0 in this equation signifies the precipitation depth for a zero area, implying that it corresponds to the storm centre. We could use the rainfall depth at the station with maximum precipitation as P0. However, since the storm centre does not normally coincide with a station, it is customary to obtain P0 by assuming that the station value, Ps, corresponds to an area of 25 km2, and then using the exponential equation with known values of P (=Ps), A (=25) and the given values of K and n. Precipitation is the driving force behind the run-off and, in general, higher precipitation would lead to larger run-off. However, it is not always true since the abstractions from precipitation, in the form of infiltration, evaporation, etc., may be different for the same amount of precipitation, depending on the atmospheric and soil conditions. Therefore, in the next chapter we describe various processes which abstract water from the precipitation and look at their measurement, estimation, and analysis.

SUMMARY Presence of water and condensation nuclei in the atmosphere is required for the formation of clouds and, for precipitation to take place, a cooling mechanism is needed. The cooling may occur due to orographic, convective, frontal, or cyclonic mechanisms, and results in precipitation in any of its various forms, such as rain, snow, or hail. The severity of precipitation is expressed either in terms of its intensity, generally in mm/h, or the depth, generally over the period of a day. The variation of intensity with time is shown by a hyetograph and the variation of the depth with time is shown either through a histogram (showing daily precipitation depths) or a mass curve (showing cumulative depths). The measurement of precipitation is done by recording rain gauges, which maintain a continuous record, or non-recording gauges, which provide only the daily precipitation depths. Any missing values in a precipitation record may be estimated by utilizing the precipitation records at nearby stations. The normal ratio method, inverse distance method, quadrant method, or more advanced techniques like Kriging, may be used for this purpose. To check if the data is consistent with the general climatic conditions, a double mass curve analysis is performed. This technique also provides a method to correct the data, if found to be inconsistent. The point rainfall data measured at a rain gauge is processed to obtain meaningful quantities related to catchment area, by using the arithmetic mean method, the Thiessen polygon method, or the isohyetal method. The isohyetal method is likely to be the most accurate method of estimating the average rainfall over an area, but the arithmetic mean method is the simplest and may provide reasonably accurate estimates for some catchments. The temporal variation of precipitation during an event could be analyzed using the intensity-duration curve (or the depth-duration curve), which shows the variation of the maximum intensity (or depth) observed over different durations of the precipitation event. The spatial variation of precipitation for an event is represented by a depth-area-duration curve, which shows the relation between the area analyzed and the corresponding average precipitation depth for various storm durations. Most of the depth-area-duration curves could be represented by an exponential relationship, with empirical coefficients varying from one region to the other.

Precipitation

OBJECTIVE-TYPE QUESTIONS 2.1 In which century did the systematic measurements of precipitation start? (a) 15th (b) 17th (c) 18th (d) 19th 2.2 Out of the options given below, which one is correct about the following statements? (i) Colder air can hold more water than warmer air. (ii) The size of condensation nuclei is a few mm. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 2.3 The temperature at which the air becomes saturated with the water vapour is known as the (a) Humidity (b) Dew point (c) Vapor pressure (d) Triple point 2.4 Cooling of air mass due to its rise over mountains causes _________ precipitation. (a) Orographic (b) Cyclonic (c) Frontal (d) Convective 2.5 Out of the options given below, which one is correct about the following statements? (i) When an air mass rises to a higher altitude, it expands. (ii) Convective precipitation occurs due to local heating of air mass. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 2.6 Orographic precipitation is also known as ________ precipitation. (a) Comfort (b) Relief (c) Cyclonic 2.7 A rain shadow indicates an area on the (a) Windward side of a mountain (c) Centre of a rainstorm

(d) Adiabatic

(b) Leeward side of a mountain (d) Top of the cloud

2.8 Which type of precipitation is more common in hot and humid tropical areas? (a) Orographic (b) Cyclonic (c) Frontal (d) Convective 2.9 The surface of contact between air masses of different temperatures is known as (a) Isohyet (b) Front (c) Cyclone (d) Mass curve 2.10 Out of the options given below, which one is correct about the following statements? (i) When the cold air mass moves into a warm air mass, it is called a cold front. (ii) The warm fronts generally have more intense precipitation on a relatively smaller area. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 2.11 The smaller cyclones have a diameter of up to (a) 1600 km (b) 3200 km (c) 4800 km

(d) 6400 km

2.12 Out of the options given below, which one is correct about the following statements? (i) Drizzle size is larger than the rain size. (ii) Sleet is a mixture of liquid and gas. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false

45

46

Engineering Hydrology

2.13 According to the India Meteorological Department classification, what is the daily precipitation for a moderate rain? (a) 2.5 to 7.5 mm (b) 7.6 to 35.5 mm (c) 35.6 to 64.4 mm (d) 124.5 to 244.4 mm 2.14 What is the typical value of specific gravity of falling snow? (a) 0.01 (b) 0.02 (c) 0.04

(d) 0.08

2.15 Normal precipitation values are generally based on a ____ year period. (a) 10 (b) 20 (c) 30 (d) 40 2.16 The normal annual precipitation for Egypt is about (a) 5 mm (b) 50 mm (c) 100 mm

(d) 200 mm

2.17 The normal annual precipitation for Kerala is about (a) 500 mm (b) 1000 mm (c) 2000 mm

(d) 3000 mm

2.18 What percent of rain gauges are generally kept as recording gauges? (a) 5% (b) 10% (c) 20%

(d) 30%

2.19 Out of the options given below, which one is correct about the following statements? (i) A histogram shows the variation of daily precipitation with time. (ii) The plot of rainfall intensity versus time is known as a mass curve. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 2.20 Out of the options given below, which one is correct about the following statements? (i) The coefficient of variation is the ratio of mean to standard deviation. (ii) The standard error of the mean is the standard deviation of the sample means. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 2.21 The standard error of the mean is proportional to what power of n (the number of rain gauges)? (a) −1 (b) −1/2 (c) 1/2 (d) 1 2.22 Out of the options given below, which one is correct about the following statements? (i) Assumption of similar pattern at nearby stations is justifiable for long-term averages. (ii) The quadrant method gives undue importance to stations in one direction. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 2.23 Which of the following is NOT a technique for estimation of missing rainfall values? (a) Arithmetic mean (b) Normal ratio (c) Thiessen polygon (d) Kriging 2.24 Which of the following methods is likely to be the most accurate for determining the average precipitation over an area (a) Arithmetic mean (b) Thiessen polygon (c) Isohyetal (d) Quadrant 2.25 The double mass curve technique is used to (a) Find average precipitation (b) Derive a hyetograph (c) Check the data consistency (d) Derive depth-area-duration curve 2.26 Out of the options given below, which one is correct about the following statements? (i) The arithmetic mean method accounts for the location of the rain gauge. (ii) The Thiessen polygon method accounts for the elevation of the rain gauge.

Precipitation

(a) (i) is true, (ii) is false (c) Both (i) and (ii) are true

47

(b) (i) is false, (ii) is true (d) Both (i) and (ii) are false

2.27 If spatial averaging of precipitation is to be done for the same set of rain gauges with several different sets of data, which of the following methods would be advantageous? (a) Arithmetic mean (b) Thiessen polygon (c) Isohyetal (d) Harmonic mean 2.28 Out of the options given below, which one is correct about the following statements? (i) The maximum intensity of precipitation decreases as the duration increases. (ii) The maximum depth of precipitation decreases as the duration increases. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 2.29 The station value of precipitation is assumed to represent an area of ____ km2. (a) 1 (b) 5 (c) 10 (d) 25

DESCRIPTIVE QUESTIONS 2.1 Describe the orographic, convective, frontal, and cyclonic precipitation. Which of these is likely to be of largest intensity and which is likely to be spread over a larger area? 2.2 What are various forms in which precipitation may occur? What is meant by equivalent water depth of snow? 2.3 What is the difference between recording and non-recording rain gauge? Which of these would you recommend to estimate the average annual precipitation for a catchment? 2.4 How is a histogram different from a mass curve? How will you derive the hyetograph from the mass curve? 2.5 How would you decide whether an existing rain gauge network is sufficient or more gauges are needed? What is meant by relative standard error? 2.6 What is meant by normal precipitation? How is the normal precipitation for a day, month, season, or year, computed? 2.7 Describe the normal ratio method of estimating the missing data at a station. When can the arithmetic mean method be used instead of the normal ratio method? 2.8 What is the problem in using the mass curve at a station to check its data consistency? Explain the double mass curve method for consistency check. 2.9 What is the logic behind Thiessen polygon method? Why is the isohyetal method considered to be more accurate than the Thiessen polygon method? 2.10 Explain the methodology of construction of the depth-area-duration curve.

48

Engineering Hydrology

NUMERICAL QUESTIONS 2.1 The data from weighing-type recording rain gauges shows the variation of cumulative precipitation depth with time. During a 6-hour storm, the following data was obtained. Generate the rainfall hyetograph. Time (minutes)

0

15

30

45

60

75

90

105

120

135

150

165

180

Precipitation Depth (mm)

0.00

0.01

0.03

0.09

0.21

0.42

0.73

1.14

1.63

2.17

2.72

3.26

3.74

Time (minutes)

195

210

225

240

255

270

285

300

315

330

345

360

Precipitation Depth (mm)

4.14

4.46

4.72

4.92

5.06

5.16

5.23

5.27

5.29

5.30

5.31

5.31

2.2 Another type of recording rain gauge is the tipping bucket, which records the time it takes for a specific amount of precipitation (generally 0.2 mm). During a 6-hour storm, the tipping times (the accumulated rainfall is equal to 0.2 mm between any two consecutive times) were recorded as shown in the table below. Generate the rainfall hyetograph (Hint: assume uniform intensity between two tipping times and assign it to the mid-point of the interval). Compare this hyetograph with that of the previous problem (both were obtained for the same storm). Tipping Time 58, 45 (minute, second)

73, 34 83, 43 92, 34 99, 53 106, 50 112, 58 119, 05 124, 43 130, 17 135, 49 141, 16 146, 44

Tipping Time 152, 11 157, 47 163, 20 169, 23 175, 38 182, 15 189, 45 197, 49 207, 11 218, 05 231, 00 248, 34 276, 00 (minute, second)

2.3 The annual precipitation at six rain gauges in a catchment area is shown in the table below. Determine whether more stations are needed, if the desired relative standard error is 10%. Station

1

2

3

4

5

6

Annual Rainfall (mm)

837.3

1106.1

1014.8

686.9

641.8

1137.3

2.4 A catchment may be idealized as a circle of 50 km radius, with the centre at (0, 0). Five rain gages are located within this area at S1 (0, 0), S2 (30, 0), S3 (0, 40), S4 (−40, 0), and S5 (0, −30), all distances being in km. The following table shows the annual precipitation measured at these stations for the period 1981-2015. It is known that an obstruction near S1 was removed in the beginning of the year 1990 and may have affected the observations prior to that. Check the consistency of data at S1 and correct, if needed. Year

S1

S2

S3

S4

S5

2015

-

1051.9

1066.8

879.1

910.1

2014

773.7

827.2

852.5

700.1

721.7

2013

888.9

959.1

986.4

805.2

826.8

2012

702.3

751.1

755.3

636.0

644.5

2011

757.3

823.8

828.9

693.9

702.7

2010

809.3

860.6

887.4

739.7

751.9

2009

917.0

984.9

993.4

828.9

855.8

2008

699.1

739.9

748.6

622.4

637.1

2007

961.7

1027.7

1054.7

868.5

897.2 (Contd.)

Precipitation Year

S1

S2

S3

S4

S5

2006

847.6

2005

925.6

915.1

941.1

776.9

789.7

998.3

1010.8

840.9

861.6

2004

837.4

2003

908.0

881.4

905.2

756.4

774.0

989.0

1003.8

833.7

2002

859.5

724.5

781.6

801.5

659.6

685.3

2001

939.8

1016.7

1039.2

850.3

880.8

2000

718.3

772.2

784.6

661.4

674.0

1999

953.0

1018.5

1053.7

868.6

882.6

1998

888.3

938.3

963.2

806.3

817.9

1997

945.0

1006.7

1037.9

861.6

882.4

1996

801.3

855.2

863.9

716.9

744.3

1995

814.8

886.9

902.9

750.4

764.1

1994

931.9

1000.6

1017.5

851.2

878.5

1993

705.6

762.3

774.6

644.9

661.0

1992

708.7

763.8

785.8

645.7

665.7

1991

839.4

894.6

903.3

754.9

780.4

1990

898.8

954.9

984.9

805.5

827.1

1989

727.6

858.6

892.4

728.7

758.7

1988

844.6

1006.8

1034.7

859.6

884.8

1987

899.4

1069.2

1080.1

901.2

920.4

1986

866.4

1030.7

1060.4

875.7

902.7

1985

708.5

842.9

850.6

713.8

721.2

1984

758.1

900.1

909.9

755.3

775.3

1983

856.5

1025.4

1047.8

869.6

888.4

1982

654.8

781.9

800.9

659.8

682.3

1981

788.9

923.0

958.3

785.3

801.1

49

2.5 For the above problem, estimate the annual precipitation at station S1 for the year 2015 using the arithmetic mean method, normal ratio method, and inverse distance method with cubic weight. 2.6 For the same catchment, estimate the normal annual precipitation by using the arithmetic mean, Thiessen polygon, and Isohyetal methods. 2.7 The table below shows the data about the isohyetal map prepared from the observations of a 12-hour rainfall in an area. Construct the depth-area curve and estimate the average depth of precipitation over an area equal to 1000 km2. (Note that the enclosed area represents the total area enclosed by the isohyet, not the incremental area.) Isohyetal line (mm)

20

18

16

14

12

10

8

6

4

2

Enclosed Area (km2)

35.1

135.6

300.4

843.8

1388.2

1723.3

2146.9

2762.6

3488.8

4087.7

50

Engineering Hydrology

2.8 For storms in North India, the values of K and n for a duration of 48 hours were given earlier as K = 0.001 and n = 0.63. For a 24-h storm, these values may be taken as 0.00085 and 0.66 respectively, and for a 72-h storm, as 0.0017 and 0.60 respectively. At a rain gauge, which may be assumed to be at the centre of the most intense storm, the maximum precipitation depths for various durations were observed as: 472.1 mm for 24-h, 824.6 mm for 48-h, and 1079.2 mm for 72-h. Prepare the depth-area-duration curves. 2.9 The table below shows the data of cumulative precipitation depth with time obtained from a recording rain gauge during a storm of 3-hour duration. Prepare the maximum intensity versus duration curve. Time (minutes)

0

15

30

45

60

75

90

105

120

135

150

165

180

Cumulative Precipitation Depth (mm)

0.0

1.3

2.3

3.5

5.3

6.5

8.3

10.9

12.5

13.3

14.3

14.8

14.9

3

Abstractions from Precipitation

LEARNING OBJECTIVES LO 1

Know about the various abstraction processes and initial abstraction

LO 2

Define evaporation, factors affecting it, and the methods of estimation

LO 3

Estimate potential and actual evapotranspiration

LO 4

Discuss the infiltration process and the influence of various factors on the rate of infiltration

LO 5

Summarize empirical and theoretical methods for the estimation of infiltration

3.1

INTRODUCTION

Although precipitation is an important part of the hydrologic cycle, as mentioned earlier, the topic of greater interest for engineers and hydrologists is the amount of water flowing in the rivers. If we could estimate how and how much water is being abstracted from the precipitation before it runs-off to the streams, we will have a tool to estimate the runoff from the precipitation records. In this chapter, we will describe some of these abstractions in terms of their basic mechanism, factors affecting these, and their methods of estimation. The measurement techniques are described in greater details in Chapter 10.

3.2

ABSTRACTION PROCESSES

Let us consider a stream for which we want to study the flow at a section O, as shown in Figure 3.1.

LO 1 Know about the various abstraction processes and initial abstraction

52

Engineering Hydrology

Also shown is its catchment area which may be loosely defined as the area which contributes to the flow at O (a more detailed description is provided in the next chapter). In other words, any precipitation falling on this area has the potential to reach the point O. If, or when it reaches there, is dependent on the interaction of several complex processes, e.g., the location, intensity and duration of precipitation, soil type and moisture conditions, presence and type of vegetation, catchment size and slope, etc. On one extreme, if the catchment area is completely impervious, free of obstructions and depressions, and there is negligible evaporation, the entire precipitation could run-off to the point O. On the other extreme, if the soil is dry and pervious and the precipitation intensity is small, it is likely that none of the precipitating water would reach the point O, at least Figure 3.1 A stream and its catchment area not in a short time (the water which infiltrates into the soil could show up at O after a prolonged time through underground pathways). The reality, as always, lies somewhere between these two extremes. Even for a realistic catchment, whether the storm occurs close to the outlet O or away from it may decide how much of it will run-off. For example, if the storm is close to O, more of precipitation is likely to reach O. If the storm occurs away from O, and there are unfilled depressions on its path to O, these have to be filled before the water reaches O. Or, if the storm occurs over a vegetated area, a significant amount of water may be stored on the leaves before reaching the ground. This spatial and temporal heterogeneity implies that the relationship between precipitation and the resulting runoff is not very easy to establish. To simplify the discussion, we will look at this relationship for a storm of uniform intensity, occurring over the entire catchment area and discuss the possible abstractions from it before the water reaches the outlet.

3.2.1 Interception Vegetation, buildings, and other structures intercept the precipitation before it reaches the ground. Some of this intercepted water may quickly reach the ground, e.g., through a drainpipe from the roof, and is not considered a loss. The water retained on leaves, on the other hand, could stay there for a long time and may get evaporated from there without reaching the ground. Depending on the intensity and duration of the storm and the type and areal extent of the vegetation, this loss could be a significant proportion of the precipitation, particularly for low intensity rains of short durations.

3.2.2

Depression Storage

Water running off to the outlet may have to pass through small depressions or large ponds in its path. Depending on the water level in these depressions, some of this water will be stored in them and may be lost either to the atmosphere through evaporation or to the subsurface through infiltration. The combination of interception and depression storage is termed as Initial Abstraction, since it occurs primarily during the early stages of precipitation. Once the storage capacities for interception and depressions are fulfilled, there is a very little further loss due to these, caused by evaporation and/or infiltration during the precipitation event.

Abstractions from Precipitation

3.2.3

53

Evaporation

Evaporation is the process by which water changes from liquid to gas or vapour. The reverse process of change of water vapour to liquid form is called condensation. These two processes typically occur simultaneously at the water-air interface since some water molecules evaporate into the air and some of the water vapour condenses into water. The net evaporation primarily depends on the vapour-pressure gradient between the water surface and the air slightly above it. As the air gets more saturated with water vapour due to evaporation, this gradient decreases, causing a reduction in the rate of evaporation. However, if a mechanism exists for removing this saturated air and bringing in drier air, the evaporation rate would again increase. Hence, wind speed is also an important factor affecting the rate of evaporation.

3.2.4 Transpiration Only about 1–3% of water taken up by a plant through its roots is utilized in its growth. The remaining is carried from the roots to the other parts of the plants and is lost to the atmosphere through small pores called stomata, mostly located on the underside of leaves. This is called transpiration, and it may be considered as the evaporation of water from plant leaves (or stems or flowers). Its contribution to the total moisture in the atmosphere is estimated at about 10%. In addition to the factors which affect evaporation, transpiration is affected by the type and density of vegetation as well as the availability of sunlight for photosynthesis. Since evaporation and transpiration are very similar in nature and the factors affecting both the processes are almost identical, we sometimes study these together and call it evapotranspiration or consumptive use.

3.2.5

Infiltration

The passage of water from the ground surface to the soil/rock below is called infiltration. The maximum rate at which water can infiltrate a given subsurface stratum is called its infiltration capacity, which typically decreases during a rainfall event. At the beginning, the infiltration capacity is large since the soil is dry and is able to absorb more water. As the rain continues, the soil becomes more saturated and the pore space available to accommodate the infiltrating water reduces. Also, due to the impact of raindrops, the soil gets compacted and its structure changes causing further decrease in the infiltration capacity. The actual rate of infiltration is, naturally, less than (or, at most, equal to) the infiltration capacity and depends on several factors, the most important being the rainfall intensity. Depending on the depth of the water table, the infiltrating water may remain in the unsaturated zone or it may replenish the groundwater. In the subsequent sections, we will describe each of these abstraction processes in detail.

3.3

INITIAL ABSTRACTION

LO 1

The initial abstraction Ia, comprises the interception loss and the depression storage. The interception at the beginning of a precipitation event will depend on the area available for storage on leaves and roof-tops, etc. Once this storage capacity is filled, the interception loss will depend on the additional storage capacity created due to the evaporation of water from these surfaces. Therefore, the interception loss may be written as

54

Engineering Hydrology

Interception Loss (in mm) = Interception Storage (in mm) + k E t

(3.1)

where, t is the duration of the rainfall (generally in hours), E is the rate of evaporation during this duration (in mm/h), and k is a factor representing the ratio of the area from which evaporation is taking place to the area over which precipitation is occurring. For example, if interception on a flat rooftop is considered, k would be equal to 1. However, for interception on leaves, k is generally larger than 1, since the leaves are inclined and vertically distributed and the surface area will be larger than the projected horizontal area. A leaf area index, defined as the area of the foliage (i.e., plant leaves) per unit ground surface area, multiplied with the fractional vegetation cover, defined as the fraction of area covered by vegetation, is generally used to obtain the coefficient k. The leaf area index varies from about 0.2 to 9, and is taken as about 3.5 for general crops. The fractional vegetation cover typically varies from about 0.6 to as high as 1.0. The value of E is relatively less variable, showing a range of about 0.15 to 0.30 mm/h, and may be assumed to be about 0.2 mm/h. The evaporation rate and various factors affecting it (e.g., temperature, humidity, and wind speed) are discussed in the next section. The interception storage may be obtained by multiplying the fractional vegetation cover, the leaf area index, and the maximum storage of water per unit foliage area. The last term depends on the type of vegetation, e.g., for pine tree it may be as small as 0.1 mm while for big grass it may be around 0.2 mm. The interception storage is generally around 0.25–1.25 mm and the interception loss from a structure or vegetation is typically about 10–20% of the precipitation (for a very dense forest it may be more than 25% and for a low intensity precipitation it may be as high as 100%). The depression storage is similar to the interception loss in some sense, in that there is an initial storage which has to be filled before the water is available for runoff. However, there is a major difference that the additional storage capacity is now generated through both evaporation and, more significantly, by infiltration. There is an initial depression abstraction and once the depressions on the ground surface are filled, further depression storage will depend on the rate at which water is evaporating into the atmosphere and infiltrating into the subsurface. The infiltration rate and various factors affecting it (e.g., soil type, moisture content, and ground slope) are discussed in a later section. Depression storage is generally a small portion of the total abstraction and the use of a rough value may be sufficient for practical purposes. The American Society of Civil Engineers (ASCE) recommends using a value of about 6 mm for pervious areas and 1.5 mm for impervious areas. The temporal variation of the precipitation intensity also affects the initial abstraction significantly. It is possible that during early period of a storm, all the precipitation is either intercepted or fills the depressions. With the passage of time, the abstraction will decrease. Since there are several factors which affect the interception and the depression storage and it is very difficult to estimate the effect of all these factors individually; commonly the initial abstraction is assumed to be some fraction of the precipitation. This fraction depends on several catchment and meteorological characteristics and is obtained either experimentally or empirically for a given catchment. The Soil Conservation Service (SCS) of the United States Department of Agriculture has suggested a method to estimate the initial abstraction based on an empirical parameter called the Curve Number (CN). The initial abstraction is assumed to be about 20% of the potential storage in a catchment and the potential storage depends on various factors including the soil type, land use, land cover, and the initial moisture conditions. The SCS has developed a relationship between this storage S and the Curve Number as

Abstractions from Precipitation

Ê 1000 ˆ S = 25.4 Á - 10 ˜ Ë CN ¯

55

(3.2)

where S is the potential maximum depth of retention of the catchment (mm) and CN generally ranges between about 30 to nearly 100. Smaller values of CN indicate highly pervious catchment and values of CN close to 100 indicate highly impervious catchment. More details about this methodology and its use in runoff estimation are provided in the next chapter. � EXAMPLE 3.1 A catchment is completely covered with vegetation comprising grass and the soil is sandy. The curve number for the catchment is estimated as 60. The leaf area index for grass is 3.0 and the maximum storage of water per unit leaf area is 0.2 mm. The evaporation rate from a wet surface may be assumed to be 0.2 mm/h. If rainfall at a constant intensity of 3 mm/h occurs for a day, estimate the initial abstraction. Solution We first use Eq. (3.1) to estimate the initial abstraction. The interception storage = fractional vegetation cover × leaf area index × maximum storage of water per unit foliage area = 1.0 × 3.0 × 0.2 mm = 0.6 mm k = 3.0 (i.e., the leaf area index, since the area is completely covered with grass) E = 0.2 mm/h t = 24 h Depression storage may be taken as 6 mm for pervious areas. Hence, initial abstraction = 0.6 + 3.0 × 0.2 × 24 + 6 = 21 mm Using Eq. (3.2), the potential storage, S=169.3 mm. Assuming initial abstraction to be 20%, we get 34 mm.

3.4

EVAPORATION

The molecules in a water body have a continuous random motion and LO 2 Define evaporation, are held within the water body due to cohesive forces. However, some factors affecting it, and the molecules have sufficient energy to overcome these forces and cross the methods of estimation water surface to enter the atmosphere above in the form of water vapour. Similarly, some of the water vapour molecules in the atmosphere will cross the water surface and enter the water body. The net effect of this transfer is either evaporation, when more molecules leave the water body, or condensation, when more molecules enter it. In the discussion below, we assume that evaporation is occurring. Also, we assume that evaporation is taking place from a water body such that the availability of water is not limited. In other words, we look at the potential evaporation. When we consider evaporation from a soil surface, the actual evaporation will depend on other factors as described later.

56

Engineering Hydrology

There is a thin layer at the water surface which is in equilibrium with the water body, meaning that the number of molecules entering this layer from the water body is same as that entering the water body from this layer. In other words, this thin layer is saturated with water vapour. The atmospheric layer above it is generally not saturated, and it is this water vapour gradient which drives the evaporation process. For a better understanding of the evaporation process, we define below some terms from psychrometry (which is the study of moisture in air): ∑

Dry air and moist air: The naturally occurring air is generally moist, i.e., it has some water vapour in it. If we remove all water vapour, we get dry air, which has predominance of nitrogen and oxygen (nearly 99%). The composition of dry air changes a little from place to place and from time to time. However, the amount of water vapour in the air varies widely, depending on the temperature and pressure.

∑

Saturated air: If we keep on adding water vapour to dry air at a given temperature, the air will be able to accept it only up to a limit. Any further addition of water vapour will cause condensation. The air is thus saturated with water vapour.

∑

Humidity: It is an indication of the amount of water vapour present in the air. � Absolute humidity is the mass of water vapour per unit volume of dry air (generally in g/m3). � Specific humidity or humidity ratio is the mass of water vapour per unit mass of dry air (g/kg). � Relative humidity is the ratio of mass of water vapour in a given volume of air to the mass of water vapour in the same air volume when it is saturated at same temperature and pressure.

∑

Vapour pressure: It is the partial pressure exerted by water vapour and is generally denoted by e, with units of mm of Hg, kPa, or mbar. � Saturated (or saturation) vapour pressure: It is the partial pressure exerted by the water vapour when the air is saturated. It is denoted by es, and has a strong dependence on temperature with its value changing from about 17.6 mm of Hg at 20 °C to about 31.9 mm of Hg at 30 °C. It should be noted that, using the ideal gas law, the relative humidity may also be defined in terms of the vapour pressure as e/es, with both at the same temperature.

∑

Dew point temperature: If we start reducing the temperature of an air sample, its moisture-holding capacity starts to decrease. At some temperature, the water vapour will begin to condense. This temperature is known as the dew point and may be thought of as the temperature at which the saturated vapour pressure is equal to the vapour pressure of the given air sample.

For the saturated layer above the water surface, the vapour pressure will be equal to the saturated vapour pressure es, and in the unsaturated air some distance above it, the vapour pressure ea, will be less than the saturated vapour pressure. es is the saturated vapour pressure corresponding to the water temperature Tw, and ea is the actual vapour pressure of air at temperature Ta. The evaporation rate E, generally expressed in mm/d, is proportional to the saturation deficit: E μ (es – ea)

(3.3)

This form of equation was first suggested by Dalton (1802) and is therefore known as the Dalton’s law of evaporation. The term on the right hand side of Eq. (3.3) may also be thought of as a vapour pressure gradient.

Abstractions from Precipitation

57

To estimate the evaporation rate using the Dalton’s law, we need to know the temperature of water, temperature and relative humidity of air at some distance above the water surface, a relation between the temperature and saturation vapour pressure of water, and the constant of proportionality in Eq. (3.3). Water temperature is readily measured, generally about 20-30 cm below the surface, and the relation between temperature and saturation vapour pressure of water is well established. For example, several researchers have proposed the following equations (with vapour pressure in mm Hg and temperature T in °C) Ê cT ˆ es = c1 exp Á 2 ˜ (3.4) Ë c3 + T ¯ with minor differences in the values of the coefficients (Bolton used c1 = 4.584, c2 = 17.67, c3 = 243.5; Buck used c1 = 4.584, c2 = 17.502, c3 = 240.97; WMO suggests c1 = 4.584, c2 = 17.62, c3 = 243.12). We will use the WMO formula, but written in a slightly different form (so as to make it easier to compute its derivative, which is required later). Ê 4283.8 ˆ es = exp Á 19.143 243.12 + T ˜¯ Ë D=

des 4283.8 = e dT (243.12 + T )2 s

(3.5) (3.6)

Putting T = Tw in equation (3.5), we get the value of es to be used in the Dalton’s equation. Figure 3.2 shows variation of the saturated vapour pressure and its derivative with temperature.

Figure 3.2 Saturated vapour pressure es (solid line) and its derivative des/dT (dashed line) For determination of ea, we have to decide at what elevation the air temperature and relative humidity need to be measured. Most weather stations measure the air temperature at a height of 1–2 m above the ground surface, therefore, that is the elevation used. By using T = Ta in Eq. (3.5), we get the saturation

58

Engineering Hydrology

vapour pressure and multiplying it by the relative humidity, we get the actual vapour pressure in the air ea. Alternatively, if the dew point temperature Td of air is measured, we could directly obtain ea by using Eq. (3.5) with T = Td. The most difficult and uncertain part of estimating evaporation by Dalton’s law is the determination of the proportionality constant. Since evaporation depends on several climatic factors other than the vapour pressure gradient, it is expected that this constant of proportionality would be a function of these factors. Therefore, we first look at these factors and discuss their effects on evaporation.

3.4.1 Factors Affecting Evaporation As discussed in the previous section, the primary driving force for evaporation is the vapour pressure gradient between the water body and the air above it. Since this gradient depends on the water temperature, the air temperature, and the relative humidity, these three are the most important factors affecting the rate of evaporation from a water body. However, there are several other factors which influence the rate of evaporation from a water body, some of which are described below. � Wind Speed As evaporation continues, the concentration of water vapour in the air above the water surface increases and the vapour pressure gradient decreases. If there is no wind, the evaporation rate will keep on decreasing and may become negligible after some time. However, in the presence of wind currents, the moist air above the water surface is replaced by drier air from beyond the water body boundaries, thereby restoring the vapour pressure gradient and the evaporation rate. Therefore, the rate of evaporation is likely to increase with an increase in the wind speed. However, beyond a certain wind speed, which is just sufficient to carry the moist air away from the entire water surface, the evaporation rate would stabilize. Another effect of the wind may be to create waves on the water surface, leading to an increase in evaporation due to small drops of water from the breaking waves being carried upward by the wind turbulence, and being evaporated from there (this is called mechanical evaporation). � Atmospheric Pressure Higher atmospheric pressure reduces the evaporation by making it more difficult for water molecules to cross the water surface and enter the air. Therefore, evaporation is expected to increase at higher elevations. However, since temperature decreases at higher elevations, leading to a decrease in the saturation vapour pressure, the net effect of elevation on evaporation could be either positive or negative. On the other hand, since the saturation vapour pressure is explicitly included in Dalton’s law, the constant of proportionality will increase with a decrease in the atmospheric pressure. � Depth of the water body A deep water body will have a more uniform annual distribution of evaporation, since it is able to store some heat energy in summer and use it in winter. Evaporation from a shallow water body, on the other hand, will be largely governed by the seasonal temperature patterns, and will show a wider variation from the summer to winter period. The total annual evaporation, however, is likely to be nearly same irrespective of the depth of the water body. � Presence of salt Due to presence of the salt molecules in the water, the movement of water molecules is restricted and leads to a lower saturation vapour pressure. Evaporation from saline water is, therefore, less than that from a freshwater body by about 1 percent for each 1 percent increase in salinity. For seawater the salinity is about 30 g/kg, and the evaporation is generally 2–3% less than that from a freshwater body.

Abstractions from Precipitation

3.4.2

59

Estimation of Evaporation

The most direct, and probably the most accurate, method of estimating the rate of evaporation from a water body is through measurements using evaporation pans. These pans are filled with water and the water level in the pan is periodically measured. The difference between the water levels at two observation times (corrected for any precipitation) provides the estimate of the evaporation rate during this interval. However, due to the effect of size, evaporation from a large water body is generally smaller than that measured by the pan. More details of this measurement process are given in Chapter 10. Several empirical equations, generally based on the Dalton’s law, have been developed to estimate the rate of evaporation and work well if the location at which evaporation is to be estimated has characteristics similar to the area for which the equation was developed. Due to the lack of universal applicability of the empirical equations, theoretical equations, based on water budget, energy budget, or mass transfer have been proposed. Theoretical equations should work well everywhere, provided the parameter values required in these are accurately estimated.

3.4.2.1

Empirical Equations

Empirical equations are derived by conducting extensive experiments to arrive at the coefficient k in the Dalton’s equation E = k(es – ea)

(3.7)

with k, in general, being a function of the factors affecting evaporation discussed in Section 3.4.1. Here we list some of these equations (detailed methodology of one of these, the Rohwer’s formula, is given in Box 3.1 to provide an insight into the derivation process, but may be skipped without any loss of continuity). � Meyer’s Formula Meyer’s (1915) formula provides the evaporation in a 30-day month, which may be written as daily evaporation Ê u ˆ E = C M Á 1 + 8 ˜ (es - ea ) Ë 16 ¯

(3.8)

in which, E is in mm/d, u8 is the monthly mean wind speed in km/h measured at about 8 m above the surface (ea also corresponds to the same elevation), for finding es, the water temperature is measured about 30 cm below the water surface, and CM is an empirical coefficient, with its value taken as 0.37 (11 for monthly evaporation) for a large and deep water body and 0.50 (15 for monthly evaporation) for a small and shallow water body. If the wind speed is measured at a reference height r, as ur, the assumption of a power-law velocity profile in the atmospheric boundary layer could be utilized to obtain the wind-speed at a height of h as Ê hˆ uh = ur Á ˜ Ër¯

a

(3.9)

with the value of a commonly taken as 1/7. � Rohwer’s Formula Rohwer’s (1931) equation for estimating the rate of evaporation from a large water body, is

60

Engineering Hydrology

p ˆ Ê E = 0.771 Á 1.465 - a ˜ (0.44 + 0.0733u0 )(es - ea ) Ë 1366 ¯ Ê u ˆ ª 2.48 ¥ 10 -4 (2000 - pa ) Á 1 + 0 ˜ (es - ea ) Ë 6¯

(3.10)

where E is the evaporation rate in mm/d, pa denotes the mean barometric pressure in mm of Hg, reduced to 0°C (i.e., accounting for the fact that the barometers are calibrated at 0°C, and mercury and the metal housing show thermal expansion), u0 denotes the wind speed in km/h, at the water surface (values up to about 0.6 m above the water surface could be used), and es and ea are in mm of Hg. If the wind speed is not measured at the water surface but at a height r, Eq. (3.9) could be used with h ≈ 0.6 m to estimate u0. � Harbeck and Meyers (1970) Formula The daily evaporation is given as E = C HM u2 (es - ea ) (3.11) where, E is in mm/d and u2 is the wind speed 2 m above the surface in km/h (ea also corresponds to the same elevation). The coefficient CHM was taken as 0.044, but shows a slight inverse dependence on the surface area of the water body. These, and several other empirical equations which have been proposed, require the measurement of various climatic variables. Due to easy availability and accurate measurement of the temperature data and the fact that temperature is an important indicator of the potential for evaporation, some empirical equations relate the evaporation to temperature, rather than the vapour pressure gradient. For example, Linacre (1977) simplified the Penman formula (which is based on the energy and mass-transfer concepts as described in the next section) using empirical relationships between (a) solar radiation, latitude, elevation, and temperature; and (b) diurnal temperature range and relative humidity; and expressed the evaporation from a water surface as follows: 700 (T + 0.006 z ) + 15(0.0023z + 0.37T + 0.53 R + 0.35 Ry - 10.9) 100 - f E= 80 - T where, E is the evaporation rate (mm/d) during a specific time period (day, month, year), T is the mean air temperature (°C) during that time period, f is the latitude (degrees, North or South, both taken positive), z is the elevation above mean sea level (m), R is the mean daily range of temperature (°C) during the relevant time period, and Ry is the annual range of the mean monthly temperature (°C), i.e., the difference in the mean temperature of the hottest and coldest months. Use of the formula was not recommended for places having precipitation less than 5 mm/month or having a difference of less than 4 °C between the mean air temperature and the dew point. Daily values of the maximum and minimum temperatures are widely available and the average daily temperature is generally taken as the mean of the maximum and minimum temperatures. Recent reviews of the available evaporation equations and their performance (Xu and Singh, 2002; McJannet et al., 2013) may be consulted for more detail.

Abstractions from Precipitation

BOX 3.1

ROHWER’S FORMULA

Rohwer, an engineer in the Division of Agricultural Engineering of the Bureau of Public Roads, United States, conducted experiments on evaporation at Fort Collins, Colorado, in the 1920s. Preliminary experiments were done in a laboratory under controlled conditions, where accurate measurements could be made and the effect of wind and precipitation could be controlled. The evaporation tank was made of galvanized iron and was about 90 cm × 90 cm square with a depth of 25 cm (the FPS system was used by Rohwer, we have converted the dimensions into metric units leading to some approximation). An optical evaporimeter (using a float, a mirror and a telescope) was used to accurately measure the extremely small evaporation losses. A psychrometer was used to find the vapour pressure in the air near the water surface, mercury barometer was used to measure the atmospheric pressure, and thermometers were used to measure the water and air temperatures (water temperature was taken at the water surface and air temperature 2.5 cm above it). Hourly observations of these variables were taken and it was found that the evaporation rate, in mm/d, is given by E = 0.12(Tw - Ta + 1.67)2/3 (es - ea )

(B1.1)

where, es and ea are in mm of Hg, and Tw and Ta are in °C. Observations were then made under laboratory conditions with a controlled wind generated through a fan in an air duct. The wind velocity was measured at the level of the water surface. It was thought that an additional wind factor could be included in Eq. (B1.1) to account for the wind speed. However, the water and air temperature difference was not found to have a direct effect on evaporation when there is appreciable wind. He hypothesized that the convection currents caused by this temperature difference are important in still-air conditions, but are overshadowed by the wind effects. Eq. (B1.1) was, therefore, discarded and the zero-wind equation was obtained by analyzing the evaporation data under fully-exposed outside conditions, which was given by E = 0.43(es - ea )

(B1.2)

The value of the coefficient at different wind speeds was obtained and showed a linear increase with the wind speed. The following equation was found to fit the data well E = (0.44 + 0.0733u0 )(es - ea )

(B1.3)

where, u0 denotes the wind speed, in km/h, at the water surface (values up to about 0.6 m above the water surface could be used). The next set of experiments were conducted outside under fully exposed conditions, with the evaporation tank sunk in the ground and an anemometer measuring the wind speed a little above the water surface. The wind speed was corrected for the level difference to obtain its value at the ground surface (which is same as the water surface, since the tank was sunk flush with the ground). It was seen that the formula developed for the controlled conditions in the laboratory was able to reproduce the evaporation reasonably well under outside conditions also (considering the errors introduced due to natural variability in wind, temperature, etc.).

61

62

Engineering Hydrology

The effect of atmospheric pressure was then studied by conducting experiments at locations ranging in altitude from about sea level to 4300 m (elevation of Fort Collins, where the previous experiments were performed is about 1500 m). The data fitted the following relationship well. p ˆ Ê E = Á 1.465 - a ˜ (0.44 + 0.0733u0 )(es - ea ) Ë 1366 ¯

(B1.4)

where, pa denotes the mean barometric pressure, in mm of Hg, reduced to 0°C, i.e., accounting for the fact that the barometers are calibrated at 0°C, and mercury and the metal housing show thermal expansion. If measurement of barometric pressure is not available, the value of pa at an altitude of z, above the mean sea level, could be approximated by z ˆ Ê pa = 760 Á 1 Ë 44330 ˜¯

5.2561

(B1.5)

where, z is the altitude in meters. Note that the pressure correction factor in Eq. (B1.4) for an altitude of 1500 m (which is the altitude of Fort Collins, where the equation was developed) is nearly 1. Finally, Rohwer conducted experiments at Fort Collins on a reservoir which was much larger in size (about 26 m diameter and 2 m deep) than the evaporation tanks used earlier (90 cm × 90 cm and 25 cm deep). The data showed that, except during periods of very high wind and stormy weather, the measured evaporation from the reservoir was smaller than that computed from Eq. (B1.4). This may be explained by the fact that the smaller tanks absorb heat through their metal sides also, while the reservoir receives the heat mostly through its surface. Moreover, heat penetration is more uniform in a shallow tank than in the deeper reservoir. The average ratio of the observed to computed evaporation was found to be 0.771, leading to the final form of Rohwer’s equation for estimating the rate of evaporation from a large water body, as p ˆ Ê E = 0.771 Á 1.465 - a ˜ (0.44 + 0.0733u0 )(es - ea ) Ë 1366 ¯ Ê u ˆ ª 2.48 ¥ 10 -4 (2000 - pa ) Á 1 + 0 ˜ (es - ea ) Ë 6¯

(B1.6)

with E in mm/d; pa, es, and ea in mm of Hg; and u0 in km/h. Eq. (B1.6) was then tested at the East Park reservoir in California and several other weather stations in the United States. It was found that both the pan evaporation and the reservoir evaporation could be estimated to an accuracy of about 12%, which is reasonably good considering the uncertainties involved in the measurements of several meteorological variables.

� EXAMPLE 3.2 A large lake in Srinagar (elevation 1585 m) has a surface area of about 20 km2. Daily record of water temperature 30 cm below the water surface, and air temperature, relative humidity and wind speed at a height of 8 m are available for the month of June. The average values for the month may be taken as: Water

Abstractions from Precipitation

63

temperature = 26 °C, Air temperature = 25 °C, Relative humidity = 60%, Wind speed = 8 km/h. Estimate the evaporation from the lake during this month. Solution Using Meyer’s formula, Eq. (3.8), Ê u ˆ E = C M Á 1 + 8 ˜ (es - ea ) Ë 16 ¯ Given: u8 = 8 km/h; for T = 26 °C, es = 25.2 mm Hg; for T = 25 °C and RH = 60%, ea = 0.6 × 23.7 mm Hg; for large bodies, CM = 11 Thus, E = 180 mm. Using Rohwer’s formula, Eq. (3.10), Ê u ˆ E ª 2.48 ¥ 10 -4 (2000 - pa ) Á 1 + 0 ˜ (es - ea ) Ë 6¯ Ê 0.6 ˆ Since wind speed is given at 8 m, we use Eq. (3.9) to compute u0, at 0.6 m, as u0 = 8 Á Ë 8 ˜¯ = 5.53 km/h.

1/7

km/h

pa is computed from Eq. (B1.5), with z = 1585 m, as 628 mm Hg. Therefore, E = 7.15 mm/d = 214 mm. Using Harbeck and Meyers Formula, Eq. (3.11) E = C HM u2 (es - ea ) Ê 2ˆ u2 = 8 Á ˜ Ë 8¯

1/7

km/h = 6.56 km/h. CHM is taken as 0.044. We assume that ea at a height of 2 m is same

as that at 8 m. Therefore, E = 3.16 mm/d = 95 mm. While the other two equations give reasonably close values, there seems to be a large under-prediction using the Harbeck-Meyer formula. However, the relative performance of different methods will vary from place t place. The total volume of evaporation is obtained by multiplying the monthly evaporation depth with the surface area of the lake.

3.4.2.2

Theoretical Equations

Although the empirical equations described in the previous section are based on the Dalton’s law, the use of an experimentally obtained coefficient makes them more empirical than theoretical. In this section we will discuss some methods of estimating the evaporation from a water body, which are more theoretical in nature. Dalton’s law is a “difference” or “gradient” law, which states that the rate of evaporation is proportional to the difference of water vapour concentration close to the water surface and in the air above it. Other methods, which have been proposed for estimating evaporation, are based on the concept of mass balance, energy balance, aerodynamics, or a combination of these. Note that the constant of proportionality in Dalton’s law

64

Engineering Hydrology

is a function of the wind speed, and the mass-transfer (the vertical movement of vapour) is governed by, in addition to the molecular diffusion, eddy diffusivity. Therefore, Dalton’s law could also be thought of as based on an aerodynamic or mass-transfer approach, although it was not derived using these concepts. � Mass Balance Approach Based on the fact that over a specific period of time, the net inflow into a water body is equal to the increase in storage, if we estimate all other components of the water balance, the evaporation could be estimated. The continuity equation is written as Vi – Vo = DS where, Vi represents the volume of water coming into the water body, Vo represents the volume of water going out and ΔS is the change in storage within the water body, all of them over the specified time period (generally a day, a month, or a year). The inflow comprises the direct precipitation over the lake Vp, surface flow in the form of surface runoff and streamflow Vis, and subsurface flow in the form of interflow or groundwater inflow Viss. The outflow includes evaporation Ve, surface flow Vos, subsurface flow Voss, and if there is vegetation present in the lake, transpiration Vt. If all other volumes are estimated accurately, the volume of evaporation can be obtained as: Ve = Vp + Vis + Viss - Vos - Voss - Vt - DS

(3.12)

On dividing by the surface area of the lake and the relevant time period, we will get an estimate of the evaporation rate. The inflow (and similarly, outflow) to the lake from surface water runoff and from direct precipitation is easy to measure or estimate. However, the groundwater component is rather difficult to estimate and involves a large uncertainty. For some lakes, the transpiration loss could also be significant and its estimation may not be very accurate. Since evaporation losses will generally be small compared to the other quantities in Eq. (3.12), a small error in any of these may cause large errors in the estimate of evaporation. If accurate measurements are available, the mass balance method will provide a reasonably good estimate of the evaporation rate. Also, if the time period over which evaporation is being estimated is large, the relative error in evaporation may be small. However, in most practical problems, the errors in estimate are large and, therefore, this method is not commonly used. � Energy Balance Approach Based on the fact that evaporation requires energy for converting water from a liquid phase to vapour phase, if we could estimate or measure the net energy input due to all other sources and change in energy storage in the lake, we could estimate the energy used for evaporation. For this purpose, it is useful to look at the different components of the energy balance. The primary source of energy for a lake is the solar radiation incident at that location. This radiation could be direct (i.e., a direct solar beam) or diffused (from atmospheric particles or clouds) and is mostly in the form of shortwave (wavelengths of about 1 micron) radiation. Depending on the angle of incidence and the type of surface, part of this radiation is reflected back to the atmosphere. This ratio of the reflected to incident radiation is called the reflection coefficient or albedo (generally about 0.08 for water bodies). The lake emits longwave radiation as a black body, part of which is absorbed by the atmosphere and re-emitted at longer wavelengths in all directions. Thus, the lake also receives energy as longwave radiation but, since

Abstractions from Precipitation

65

the earth’s surface is typically at a higher temperature than the atmosphere, there is a net loss in terms of longwave radiation. The net radiative flux, due to both shortwave and longwave, is balanced by the transfer of heat to the atmosphere in the form of sensible heat flux (used in raising the temperature of the air), latent heat flux (used in evaporation of water), heat transfer to the ground and through advected water (i.e., the water carried into or out of the lake), and an increase in the heat energy stored in the lake. If we estimate all other components of this energy balance, we can estimate the latent heat flux and, consequently, evaporation rate. The energy balance is written as ARis + ARil + Hig + Hiw - ( ARos + ARol + Hog + How + AHos + AHol ) =

DH s Dt

where, A is the surface area of the water body, Ri represents the incoming solar radiation to the water body (subscript s representing shortwave radiation and l, longwave), Ro represents the outgoing solar radiation, Hos is the sensible heat flux, and Hol is the latent heat flux (all in terms of energy per unit time per unit area). Hig is the heat transfer into the lake from the ground, Hog is the heat transfer from the lake to the ground, Hiw is the heat carried into the lake through incoming water (including precipitation), How is the heat carried out of the lake through outgoing water (not including evaporation) (all in terms of energy per unit time) and ΔHs is the change in heat storage (in terms of energy) within the water body over the specified time period Δt. Similar to the mass balance approach, if all other energies are estimated accurately, the energy used in evaporation can be obtained and, on dividing by the latent heat of vaporization for water (about 2.44 × 106 J/ kg at 25°C) and mass density of water, will provide an estimate of the evaporation rate. Box 3.2 provides the details of the computations of various terms in the energy equation, but may be skipped without any loss of continuity. The daily evaporation E, in mm/d, is given by E = 8.64 ¥ 107

Rn - ( H ng + H nw + DH s /Dt )/A r H L (1 + b )

(3.13)

where, Rn is the net radiation (W/m2), Hng is the net heat transfer to the ground from the lake (W), Hnw is the net heat transfer through advective water from the lake (W), r is the mass density of water (kg/m3), HL is the latent heat of vaporization (J/kg), and b is the Bowen’s ratio, representing the ratio of sensible heat flux to the latent heat flux. The coefficient 8.64 × 107 is used to convert from m/s to mm/d. Rn is given by Rn = (1 - a )

R0 p

2p ( J - 2) ˘ È Í1 + 0.0334 cos 365 ˙ (w ss sin f sin d + sin w ss cos f cos d ) Î ˚

4 4 ˆ Ê Tmax + Tmin n˘ n˘ È È a + b s Á ˜ ÈÎ ah - bh ea ˘˚ Íal + (1 - al ) ˙ s s Í ˙ N˚ 2 N˚ Î Î Ë ¯

(3.14)

where, a is the albedo of the water surface, R0 is the solar constant (1367 W/m2), J is the Julian day, f is the latitude, d is the solar declination, wss is the hour angle at sunset, n and N are the actual and potential sunshine hours respectively, s is the Stefan-Boltzmann constant (5.670 × 10−8 J/K4/m2/s), Tmax and Tmin are the maximum and minimum air temperatures on a given Julian day (in Kelvin), ea is the actual vapour pressure (mm of Hg), and the a and b are empirical constants with as and bs representing the effect of

66

Engineering Hydrology

BOX 3.2

RADIATION

The incoming solar radiation on the earth’s surface varies considerably with solar activity, distance between the earth and sun, location, time of day, and atmospheric conditions, e.g., presence of clouds and aerosol. It can be measured by using a pyranometer or a pyrheliometer (see Chapter 10). In the absence of direct measurements, various components of the radiation could be estimated as shown below. Shortwave radiation The incoming shortwave radiation could be estimated for any day of the year by the following equation: Ris =

R0 p

2p ( J - 2) ˘ n˘ È È Í1 + 0.0334 cos 365 ˙ (w ss sin f sin d + sin w ss cos f cos d ) Ías + bs N ˙ (B2.1) Î ˚ Î ˚

where, R0 is the solar constant (defined as the energy ergy incident on a un unit area of a sphere with its center at the center of the sun and radius us equal to th the he mean distance between th the centers of the sun and the earth); J is the Julian day (1 forr January 1 and d 365 5 for fo December 31); f is the latitude (positive for North and negative for South); d is the solar dec declination eclination (defi (d ned as the angle betw between the line joining the centers of the sun and earth and the equator equatori equatorial rial plane, tak ria taken aken as positive when the line passes through northern hemisphere); wss iss the th he hour angle (angle (an nglee between be n the local meridian and the meridian that is parallel to the sun-rays, taken as a zero at solar noon, negati negative ne tive before noon and pos positive after noon) at sunset; n and N are thee actual and an potential sunshine sun nshine hours, hou our urs, respectively; and as and bs are empirical constants relating the actual and potential po l shortw shortwave wave ra radiations radi diations (i.e., radiation rreceived at the top of the atmosphere, called d the extraterrestri extraterrestrial estrial radiation) radiation). ion). n)). The term t in the first squa square brackets accounts for the fact that the earth’s ’s orbit is not circular and the earth-sun earth-s th-ssun distance is largest la around January 2 and smallest around July 4.. The term in the second square brac brackets accounts for the atmospheric conditions which cause a reduction on in the shortwave radiation rea reaching the earth’s surface. All the angles in Eq. (B2.1) are in radians. R0 is usually expressed in W/m2 and a value of 1367 is generally accepted (about 1% variation is observed in solar radiation due to change in solar activity). Since the earth’s axis of rotation is tilted at an angle of about 23.45°, the solar declination varies from −23.45° at December solstice to +23.45° on June solstice, being zero at the equinoxes in March and September, and may be computed (in radians) as 2p ( J - 81) (B2.2) d = 0.409sin 365 The sunset hour angle is related to the solar declination and the latitude as w ss = arccos(- tan f tan d )

(B2.3)

The actual sunshine hours are measured using paper burn length (Chapter 10) and the potential sunshine hours on any day depend on the latitude and the time of the year. Assuming the sunlight hours to be counted from sunrise to sunset between the times when the center of the sun is at the horizon, N is given by

Abstractions from Precipitation

N=

24 w ss p

(B2.4)

A more accurate estimate, accounting for atmospheric refraction and the fact that the sunshine hours are generally based on the time when the top (not the center) of the sun is at the horizon, is given by N=

Ê 0.0145 + sin f sin d ˆ 24 arccos Á ˜¯ p cos f cos d Ë

(B2.5)

The coefficient as represents the ratio of the actual and potential radiation under completely overcast conditions (n = 0) and (as + bs) represents the ratio under clear sky conditions (n = N). These are usually obtained by a site-specific regression between the measured values of actual daylight hours and the shortwave radiation. In absence of measured data, empirical values have been suggested as as = 0.25 and bs = 0.50 (for India, as = 0.14 and bs = 0.56 have been obtained in a recent study). Some other relationships express these coefficients as functions of n/N, some express these as functions of latitude (e.g., as = 0.29 cos f, bs = 0.52) and some others use a quadratic function of n/N, as [as + bsn]/[N + cs(n/N)]2. For example, analysis of radiation and cloud cover data from 7 meteorological stations across India resulted in n Ê nˆ as = - 10.533 + 27.180 - 17.222 Á ˜ Ë N¯ N bs = 12.098 - 29.395

n Ê nˆ + 18.676 Á ˜ Ë N¯ N

2

2

(B2.6)

Considering the uncertainties involved in the data, the values as = 0.25 and bs = 0.50 may be used. These values indicate that on a clear day (n = N), 75% of the extraterrestrial radiation reaches the earth as shortwave radiation. On a completely cloudy day (n = 0), this value reduces to 25%, mostly as diffuse sky radiation due to scattering in the atmosphere. The shortwave outgoing radiation consists of the reflected radiation from the water surface and is equal to the albedo (a) times the incoming radiation, i.e., Ros = aRis

(B2.7)

The value of a for water surface is taken as around 0.05. � Longwave Radiation According to the Stefan-Boltzmann law, the longwave energy emission rate is proportional to the fourth power of the absolute temperature. Due to the absorption of radiation by the atmosphere and re-emitting of longwave radiation by the water vapour, clouds, dust, etc., the net outward longwave radiation from the lake will be smaller than the theoretical value. The outgoing longwave radiation from the lake could be written as Rol = e ws Tw4

(B2.8)

67

68

Engineering Hydrology

where, s is the Stefan-Boltzmann constant (5.670 × 10−8 J/K4/m2/s); Tw is the absolute water temperature (= 273.16 + Tw in °C); and ew is the emissivity vity of the th water surface, which is the ratio of the actual emitted energy to that by an equivalent black body. Similar Similarly, the longwave radiation from the atmosphere is written using the longwave ongwave alb albedo, lbedo edo, al, atmosphere em emissivity, ea, and absolute air temperature, Ta, as follows: Ril = ((1 1 - a l )e as Ta4

(B2.9)

The emissivity of the water ter surfac surface face is around d 0.98,, aand the emissivity of the atmosphere is affected significantly by the humidity ty and cloudiness. oudiness. Since Sin nce air a temperature is more commonly measured than the water temperature and has a linkage with the he water w surface temperature, nearly all empirical equations use the air temperature ature in finding the net out outward longwave radiation. The general form of these equations gives the net outward longwave radia radiation, in W/m2, as 4 ˆ Ê T 4 + Tmin n˘ È i Rnl = s Á max ˜ ÈÎ ah - bh ea ˘˚ Íal + (1 - al ) ˙ 2 N Î ˚ Ë ¯

(B2.10)

where, T is the absolute air temperature (in Kelvin), subscript max denoting the maximum temperature and min denoting minimum; ea is the actual vapour pressure (mm of Hg); ah and bh are the coefficients accounting for humidity; and al is a coefficient accounting for the effect of clouds on the longwave radiation. Values of the coefficients ah, bh, and al show variation from place to place and are best obtained by regression of the observed data. However, approximate values could be used as follows: ah = 0.34, bh = 0.051, and al = 0.33 for humid regions and 0.10 for arid regions.

cloudiness on shortwave radiation, ah and bh representing the effect of humidity on atmospheric emissivity, and al representing the effect of cloudiness on longwave radiation. Box 3.2 provides details of these terms and the methodology for their evaluation. The net heat transfer to the ground Hng, could be obtained from H ng = Chs

DT Aw ds Dt

where, Chs is the soil heat capacity (we will be using the volumetric heat capacity and not the specific heat), ΔT is the temperature change of the ground over the relevant period Δt, Aw is the wetted area of the lake, i.e., the area of contact between the lake and the ground, and ds is the effective soil depth. Generally, the ground temperature is assumed to follow a similar variation as the air temperature. The effective soil depth is of the order of 10 to 20 cm for a day but may be of the order of a meter if the time period is a month. In absence of measurements, the following approximation could be used to obtain the soil heat capacity: Ê r ˆ Chs = Á b + 4.18q ˜ ¥ 106 Ë 1100 ¯

Abstractions from Precipitation

69

where, Chs is in J/m3/°C, rb is the bulk density of the soil (kg/m3) and q is the volumetric water content, which may be taken equal to the porosity. The net heat transfer from the lake through the advected water Hnw, could be obtained from the temperature and volumes of the incoming (including precipitation) and outgoing (not including evaporation) water using the heat capacity of water Chw (4.18 × 106 J/m3/°C), as H nw =

Chw ÈV (T - Tp ) + Vis (Tw - Tis ) + Viss (Tw - Tiss ) - Vos (Tw - Tos ) - Voss (Tw - Toss )˘ ˚ Dt Î p w

in which the volume terms are as defined in Eq. (3.12) and the temperature terms are the corresponding average temperatures over the period of interest. The change in heat storage within the lake is estimated as DH s = Chw

DTw Vl Dt

where, ΔTw is the change in the average lake temperature during the time period Δt, and Vl is the average lake volume. The measurement of water temperatures and the lake temperature profile are generally not available. However, for the time period of a day, the net heat transfer to the ground and through the advected water are generally small, especially if the inflowing and outflowing water is nearly at the same temperature as the lake, and may be neglected from the equation. The average lake temperature is generally taken as the water surface temperature, which is a reasonable assumption for shallow lakes, but not for deeper lakes. Since the sensible heat flux is difficult to measure directly, the Bowen’s ratio b, is used to estimate it. The Bowen’s ratio is given by ∂T - ra c p Dh H os ∂z b= = ∂r v H ol - H L Dv ∂z where, ra is the air density, cp is the specific heat of the air at constant pressure (J/kg/°C), Dh is the heat diffusivity (m2/s), T is the air temperature (°C), Dv is the vapour diffusivity (m2/s), and rv is the vapour concentration in air (kg/m3). The vapour concentration is related to the vapour pressure through the ideal gas law as M r e rv = wv a (3.15) pa M da where, M represents the molecular weight (18 for water vapour and 28.9 for dry air), the atmospheric pressure pa, and the vapour pressure e, are in the same units (typically mm of Hg). Generally the two diffusivities are assumed to be equal, and the temperature and vapour pressure derivatives are obtained by assuming a linear variation. Then, b=

M da pa c p Tw - Ta T - Ta =g w M wv H L ew - ea ew - ea

(3.16)

Engineering Hydrology

70

where, the psychrometric constant g, represents a relation between the temperature and vapour pressure as obtained in a psychrometer. Using the standard values at 25°C, cp = 1000 J/kg/°C, HL = 2.44 × 106 J/kg, we get g = 6.5 ¥ 10 -4 pa per°C = 0.5 mm of Hg per °C at sea level

(3.17)

Thus, knowing the temperature and the vapour pressure at the water level and at another height, we can estimate the Bowen’s ratio and, therefore, the evaporation using the energy balance from Eq. (3.13). The equations described above provide the value for a particular day. To obtain the monthly averaged value, we could compute all the daily values and then obtain the average of the daily values. However, an easier way is to use the 15th day of the month and assume that the values on that day represent the mean for the entire month. Sometimes, for better accuracy, a representative day for the month is chosen which more closely represents the mean monthly value. Table 3.1 gives the representative day for each month along with its Julian day number.

Table 3.1 Representative day for each month for computing monthly average values Month

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

Day

17

16

16

15

15

11

17

16

15

15

14

10

Julian day

17

47

75

105

135

162

198

228

258

288

318

344

Since the potential radiation and potential sunshine hours are dependent on the latitude and Julian day only, tables for these variables could be used to help in the computation of the evaporation (Allen et al., 1998). A sample is given below for some latitudes.

Table 3.2 Monthly mean extraterrestrial radiation (as a fraction of solar constant, 1367 W/m2) Lat.

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

40° N

0.40

0.54

0.73

0.93

1.06

1.11

1.08

0.97

0.80

0.59

0.43

0.36

20° N

0.71

0.81

0.93

1.01

1.05

1.05

1.04

1.02

0.95

0.84

0.73

0.68

0°

0.96

1.00

1.01

0.98

0.92

0.89

0.90

0.95

0.99

0.99

0.97

0.95

20° S

1.11

1.06

0.97

0.83

0.70

0.64

0.67

0.77

0.91

1.03

1.10

1.12

40° S

1.15

1.01

0.81

0.59

0.41

0.34

0.37

0.51

0.72

0.95

1.12

1.19

Table 3.3 Monthly mean of potential sunshine hours in a day (hours) Lat.

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

40° N

9.6

10.7

11.9

13.2

14.4

15.0

14.7

13.8

12.5

11.2

10.0

9.4

20° N

11.1

11.5

12.0

12.6

13.1

13.3

13.2

12.8

12.3

11.7

11.2

10.9

0°

12.1

12.1

12.1

12.1

12.1

12.1

12.1

12.1

12.1

12.1

12.1

12.1

20° S

13.2

12.8

12.2

11.6

11.2

10.9

11.0

11.4

12.0

12.5

13.1

13.3

40° S

14.7

13.6

12.4

11.1

9.9

9.3

9.6

10.5

11.8

13.1

14.3

15.0

Abstractions from Precipitation

71

The energy balance method provides accurate results but requires a large number of variables to be measured. If measurements are available, the method can provide estimation of evaporation from a lake within 5% error for a weekly or longer time period, but the errors are larger for smaller time periods. � Aerodynamic Approach In order for the evaporation process to continue, it requires a steady (or periodic) removal of the saturated air mass from above the water surface. Also, the vapour flux from/to the saturated layer to/from the air will depend on, in addition to the vapour pressure gradient, the velocity profile, turbulence characteristics, and the eddy diffusivity in the planetary boundary layer. This implies that the aerodynamics will have a major impact on the evaporation rate. The vapour mass flux through a horizontal plane is proportional to the specific humidity gradient with the constant of proportionality equal to air density times the vapour eddy diffusivity. The turbulent shear stress (equivalent to momentum flux) is proportional to the velocity gradient and the constant of proportionality is equal to air density times the eddy viscosity. If it is assumed that the eddy diffusivity and eddy viscosity have similar magnitude, and the specific humidity and wind velocity profiles could be approximated by straight lines over a short distance, the vapour mass flux is given by m� = t t

s2 - s1 u2 - u1

where, tt is the turbulent shear stress, s denotes the specific humidity, u is the wind velocity, and the subscripts indicate their values measured at two different elevations, z1 and z2, which are close enough for the linear approximation to be valid. The atmospheric boundary layer may be assumed to have a logarithmic velocity profile given by u 1 z = ln u* k z0 where, u* is the shear velocity (related to the air density ra and shear stress by t t = ra u*2 ), k is the Karman’s constant (≈ 0.4), and z0 is the height of surface roughness. By applying this equation at the two elevations to obtain the value of u*, and then tt, the vapour mass flux is written as m� = rak 2

(u2 - u1 )(s2 - s1 ) Ê z2 ˆ ÁË ln z ˜¯

2

(3.18)

1

This equation is known as the Thornthwaite-Holzman equation since they first derived it in 1939. If measurements of the specific humidity and wind speed are available at two different heights (e.g., through a flux tower), the vapour mass flux could be estimated. Generally the measurements available are the temperature, relative humidity, and wind speed at a single height, and it is assumed that the other height is equal to the surface roughness height z0, where wind speed is zero and air is saturated with moisture. Also, since the specific humidity is directly proportional to the vapour pressure, and the vapour mass flux is directly proportional to the evaporation rate E, Eq. (3.18) may be written as E μ u(es - ea ) which is of the same form as the Dalton’s law.

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Engineering Hydrology

Since the application of the aerodynamic approach requires data on variables which are generally not available, and since a combination approach as described in the next section works well, we will not describe the aerodynamic approach further. � Combination Approach Since the evaporation rate is affected by both the available energy and the mass-transfer effects, a combination approach using both these concepts is likely to provide a better estimate of the potential evaporation. One of the earliest combination approaches was suggested by Penman and there have been several other modifications proposed. We describe here the Penman method and one of the more recent methods. (a) Penman method: Penman (1948) described the first approach towards estimating evaporation which combined the energy balance and the aerodynamic approach. Since the water temperature measurements are less frequently available, he eliminated it from the equations through this combination as follows: For computing evaporation over a short period, e.g., a day, or even a few weeks, the heat transferred to the ground through the advected water, and storage change will be small compared to the radiation component. Neglecting these components of the energy balance and considering only the net radiation, latent heat flux, and sensible heat flux, we can write using the Bowen’s ratio, Ê T - Ta ˆ H n = E (1 + b ) = E Á 1 + g w ew - ea ˜¯ Ë

(3.19)

with the net radiation Hn, expressed in depth of evaporable water in mm/d (using a latent heat of 2.44 × 106 J/kg and mass density of 1000 kg/m3, 1 W/m2 is equivalent to 86400 × 1000/(1000 × 2.44 × 106) mm/d, i.e., 0.035 mm/d). The aerodynamic approach results in an equation of the form E = f (u)(ew - ea )

(3.20)

with f(u) being a wind speed function. To avoid use of water temperature Tw, and the saturated vapour pressure corresponding to it, ew, Penman defined a new variable esa, representing the saturated vapour pressure corresponding to the air temperature, which could be readily obtained from the air temperature, using Eq. (3.5). He then approximated (ew – esa) by (Tw – Ta) multiplied by the slope of the (es – T) curve at Ta, denoted by Δ, i.e., ew - esa = (Tw - Ta )

des ˘ = D(Tw - Ta ) dT ˙˚T = T a

Equation (3.19) can now be written as Ê g ew - esa ˆ Ê g ew - ea + ea - esa ˆ Hn = E Á1 + = E Á1 + ˜ ˜¯ D ew - ea ¯ D ew - ea Ë Ë g Êe -e ˆ Ê gˆ Ê gˆ g = E Á 1 + ˜ - E Á sa a ˜ = E Á 1 + ˜ - f (u)(esa - ea ) Ë Ë D¯ D Ë ew - ea ¯ D¯ D

(3.21)

Abstractions from Precipitation

73

This results in the final form of the Penman equation: E=

DH n + g f (u)(esa - ea ) D+g

(3.22)

where, Hn is given by Eq. (3.14), to be multiplied by 0.035 to convert from W/m2 to mm/d. The values of the coefficients used by Penman are as = 0.18 and bs = 0.55, ah = 0.56, bh = 0.092, and al = 0.10 g is the psychrometric constant in mm of Hg/°C, and is given by Eq. (3.17) esa is the compute from Eq. (3.5) with T = Ta ea is the actual vapour pressure in the air, computed from either temperature and relative humidity or from the dew-point. f(u) = 0.35 (1 + 0.147 u2), u2 being the wind speed, in km/h, at a height of 2 m. Δ is the slope of the (es – T) curve in mm of Hg/°C and is obtained from Eq. (3.6) (see Figure 3.2) Later studies modified some of the coefficients used by Penman. For example, using a power of the wind speed in f(u), using different coefficients in the expression for Hn, etc. The currently accepted values are as = 0.25 and bs = 0.50, ah = 0.34, bh = 0.14, and al = 0.10. An empirical method proposed by Valiantzas (2006) simplifies the computations in the Penman method by empirical approximations of various terms but may need more validation studies before being used routinely. � EXAMPLE 3.3 For the lake described in Example 3.2, estimate the evaporation during June using the theoretical methods. Latitude of Srinagar is 34.1 °N. Maximum air temperature during this month may be taken as 30 °C and minimum as 15 °C. The actual number of sunny daylight hours may be assumed to be 8 hours per day. Solution Given: actual daylight hours, n = 8; Tmax = 30 °C (or 303.16 K); Tmin = 15 °C (288.16 K); latitude f = 34.1 °N (0.595 radian) Assumed: Solar constant, R0 = 1367 W/m2; albedo of water surface, a = 0.05; shortwave radiation coefficients, as = 0.25, bs = 0.50; Stefan-Boltzmann constant, s = 5.670 × 10−8 J/K4/m2/s; humidity coefficients, ah = 0.34, bh = 0.051; longwave radiation coefficient al = 0.2; latent heat of vaporization, HL = 2.44 × 106 J/kg. From Table 3.1, we will take the representative day as June 11, Julian day 162, to compute the average evaporation rate during June. Using Equations (B2.1–B2.7), Solar declination, d = 0.409sin

2p ( J - 81) = 0.403 radians (23.1 degrees) 365

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Engineering Hydrology

Sunset hour angle, w ss = arccos(- tan f tan d ) = 1.86 radians (106.8 degrees) The potential daylight hours, N =

Rn = (1 - a )

R0 p

Ê 0.0145 + sin f sin d ˆ 24 arccos Á ˜¯ = 14.4 p cos f cos d Ë

2p ( J - 2) ˘ È Í1 + 0.0334 cos 365 ˙ (w ss sin f sin d + sin w ss cos f cos d ) Î ˚

4 4 ˆ Ê Tmax + Tmin n˘ n˘ È È 2 a + b s Á ˜ ÈÎ ah - bh ea ˘˚ Íal + (1 - al ) ˙ = 199.6 W/m s Í s ˙ N˚ 2 N˚ Î Î Ë ¯ -4 Psychrometric constant g = 6.5 ¥ 10 pa per °C = 0.408 mm Hg per °C

Bowen’s ratio b = g

Tw - Ta = 0.0373 ew - ea

Using the Energy-balance approach, Eq. (3.13), and neglecting heat transferred to the ground, advected through water, and change in storage, we have E = 8.64 ¥ 107

Rn = 6.81 mm/d or 204 mm r H L (1 + b )

Using the Penman approach, we have For T = 25 °C and RH = 60%, esa = 23.7 mm Hg, ea = 0.6 × 23.7 mm Hg = 14.2 mm Hg D=

4283.8 (243.12 + Ta )2

esa =1.41 mm Hg per °C

f(u) = 0.35 (1 + 0.147 u2) = 0.688 Hn = Rn × 0.035 mm/d = 6.99 mm/d Form Eq. (3.22), E =

DH n + g f (u)(esa - ea ) = 6.88 mm/d or 206 mm D+g

(b) Actual evaporation: As discussed earlier, the formulae developed for estimating evaporation provide the value of the maximum possible, or potential evaporation under given climatic conditions. For estimating lake evaporation, therefore, these will work well. However, when we consider the evaporation from a soil surface, the actual evaporation will be limited by the availability of water within the top strata of the soil. Generally, the actual evaporation is obtained by multiplying the potential evaporation by a factor, which depends on the soil moisture content. However, since the soil will typically have some vegetation, the transpiration will also contribute to the water loss to the atmosphere. Therefore, the evaporation and transpiration are estimated together in terms of evapotranspiration or consumptive use.

Abstractions from Precipitation

3.5

75

EVAPOTRANSPIRATION

Transpiration of water from the plant leaves is, in many ways, similar to LO 3 Estimate potential and the evaporation process. The two major differences are: (i) transpiration actual evapotranspiration occurs only during day-time, and (ii) it is affected by additional factors like vegetation density, nature and stage-of-growth of the vegetation, etc. The vegetation properties are typically described through resistance terms (having dimensions of inverse of velocity), for example, surface resistance rs, the resistance of vapour flow through stomata openings, leaf area and soil surface; and aerodynamic resistance ra, the resistance to air flow due to friction over vegetative surfaces. The surface resistance is proportional to the stomatal resistance and inversely proportional to the leaf area index. The aerodynamic resistance is inversely proportional to the wind speed at a reference height (usually 2 m). First we assume that the water availability is not limited and look at the methods of estimating the potential evapotranspiration (PET). Since the PET is dependent on the crop type, stage of growth, and several other factors, the comparison of different formulae becomes very difficult. The concept of a reference crop evapotranspiration, defined as the evapotranspiration rate from a reference surface, with no shortage of water, is called the reference crop evapotranspiration or reference evapotranspiration ET0. The reference surface is taken as an extensive surface of green, well-watered grass of uniform height, actively growing and completely shading the ground which has a moderately dry soil surface. The parameters used are: height of grass equal to 12 cm, albedo of 0.23, and surface resistance of 70 s/m. Since other factors are standardized, ET0 is only affected by climatic variables and can be computed from meteorological data. The PET for other crops is obtained by applying a factor K, dependent on the crop-type and stage of growth, to ET0. For example, for natural vegetation of medium density, K is 1 indicating that the PET is close to the reference crop transpiration value. For very dense vegetation, K may be as large as 1.3 while for wheat it may be around 0.65. Measurements of evapotranspiration using a lysimeter or by remote sensing are described in Chapter 10. Empirical and theoretical methods of estimation of the PET are described below.

3.5.1

Empirical Equations

Temperature and number of daylight hours are the two most important factors affecting the PET, with the nature and stage of growth of vegetation also having some effect. Therefore, nearly all empirical equations correlate the PET with these variables on the basis of measured data. The two commonly used methods in this category are the Thornthwaite equation and the Blaney–Criddle equation.

3.5.1.1

Thornthwaite Equation

Using the estimate of PET through field measurements, using a water budget analysis, and the meteorological data across several areas in the United States, Thornthwaite (1948) observed that the monthly PET, in cm, standardized to represent a 30-day month with 12 daylight hours every day, follows a power-law with the mean monthly temperature T, (in °C) as PET = cT a

(3.23)

He also observed that although the temperature-PET plots were different for different areas, showing higher PET for the same temperature in warmer places, they all tended to converge to a single point represented by a temperature of 26.5 °C and PET of 13.5 cm. The classification of warm or cold places could not be based

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Engineering Hydrology

on the annual mean temperature since there could be sub-freezing temperatures during a part of the year. He, therefore, suggested a heat index I, based on the mean monthly temperature values as: I=

12

ÊT ˆ  ÁË 5i ˜¯ i =1

1.514

(3.24)

where, i represents the month of the year, and Ti is the monthly mean temperature (taken equal to zero, if it is negative). The coefficient c and the power a in Eq. (3.23) were then found to be related to the heat index as a = 6.75 ¥ 10 -7 I 3 - 7.71 ¥ 10 -5 I 2 + 0.01792 I + 0.49239 Ê 10 ˆ c = 1.6 Á ˜ Ë I ¯

(3.25)

a

The division of temperature by 5 in the formula for heat index makes the value of I close to 100 for a temperature of 20°C. Any other divisor may be used with the expressions for a and c modified accordingly. For example, if the heat index for a month is defined as (Ti/100)1.514, a = 0.548 I 3 - 0.671I 2 + 1.671I + 0.49239 a

Ê 0.107 ˆ and c = 1.6 Á . Ë I ˜¯ Since the standard PET value corresponded to 360 hours of daylight, a correction factor is needed to obtain the value of PET for a particular month, based on the actual daylight hours in the month. The monthly PET, in mm/month, is therefore given as Ê 10 Ta ˆ PET = 16Cdl Á Ë I ˜¯

a

(3.26)

where, Ta is the average daily temperature for that month (taken equal to zero if it is negative), and Cdl is a correction factor for the number of daylight hours in that month. Cdl depends on the latitude and the month, since the potential daily sunshine hours are a function of the latitude and the months have different number of days. Equation (B2.5) could be applied on the “average” day of the month to obtain the number of daily daylight hours in a month. Multiplying this by the number of days in a month and dividing by 360 will provide the correction factor. A sample table is given below.

Table 3.4 Monthly Correction factor for the Thronthwaite formula (February assumed to have 28 days) Lat.

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

40° N 20° N 0° 20° S 40° S

0.83 0.95 1.04 1.14 1.26

0.83 0.89 0.94 0.99 1.06

1.03 1.03 1.04 1.05 1.07

1.10 1.05 1.01 0.97 0.92

1.24 1.13 1.04 0.96 0.86

1.25 1.11 1.01 0.91 0.78

1.27 1.14 1.04 0.95 0.83

1.19 1.10 1.04 0.98 0.91

1.04 1.02 1.01 1.00 0.98

0.96 1.01 1.04 1.08 1.13

0.83 0.93 1.01 1.09

0.81 0.94 1.04 1.15

1.19

1.29

3.5.1.2

Blaney-Criddle (1962) Equation

Based on data from the Western United States, the formula for the monthly values of the reference evapotranspiration, in mm, was given as ET0 = pdl (8.13 + 0.457Ta )

(3.27)

Abstractions from Precipitation

77

where, pdl is the number of daylight hours in that month, expressed as a percent of the annual daylight hours, and Ta is the mean monthly temperature, in °C. pdl is a function of the latitude and the month and could be obtained using the formulae in Box 3.2. A sample table is shown below.

Table 3.5 Monthly potential sunshine hours as a percentage of annual potential sunshine hours Lat.

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

40° N

6.74

6.73

8.30

8.93

10.03

10.10

10.25

9.56

8.38

7.73

6.71

6.54

20° N

7.74

7.26

8.41

8.53

9.15

9.02

9.25

8.96

8.29

8.16

7.57

7.66

0°

8.49

7.67

8.49

8.22

8.49

8.22

8.49

8.49

8.22

8.49

8.22

8.49

20° S

9.26

8.08

8.58

7.92

7.85

7.43

7.75

8.02

8.12

8.79

8.85

9.34

40° S

10.29

8.62

8.69

7.52

6.99

6.37

6.75

7.40

8.00

9.18

9.70

10.48

The PET for other crops was then obtained by multiplying ET0 by the crop factor K. The formula has also been applied to estimate the daily evapotranspiration, by using the daily daylight percentage rather than the monthly value. Some other modifications have also been used, e.g., introducing another coefficient which could be a function of relative humidity, wind speed, and temperature. The temperature-based empirical methods generally underestimate the PET in dry climates and overestimate in humid regions, due to the time lag between air temperature and solar radiation. Therefore, site-specific calibration of the coefficients is desirable to improve the accuracy of prediction. � EXAMPLE 3.4 Mean monthly temperatures for Kanpur (latitude 26.5°N) are given in the table below: Month

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

Temperature (°C)

15.5

19

24.5

30

32.5

33

30.5

30

29.5

26.5

22

17.5

Find the reference crop evapotranspiration for June and also estimate the evapotranspiration from a densely vegetated field (K = 1.2). Solution For the Thornthwaite equation: 12 ÊT ˆ Heat index I = Â Á i ˜ Ë ¯ i =1 5

1.514

=147.5

Coefficient a = 6.75 ¥ 10 -7 I 3 - 7.71 ¥ 10 -5 I 2 + 0.01792 I + 0.49239 = 3.62 Using equations given in Box 3.2 (or from Table 3.4), coefficient Cdl = 1.15 a

Ê 10 Ta ˆ For the month of June, with Ta = 33 °C, PET = 16Cdl Á = 341 mm Ë I ˜¯ For the Blaney-Criddle Equation: Using equations given in Box 3.2 (or from Table 3.5), percentage pdl = 9.32

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Engineering Hydrology

For the month of June, with Ta = 33 °C, ET0 = pdl (8.13 + 0.457Ta ) = 216 mm From the densely vegetated field, PET = K × ET0 = 260 mm As seen from these values of PET, and from the next example, different methods give very different values, primarily because these are derived for different areas. As stated earlier, for a better estimate of PET, site-specific calibration of model parameters is desirable.

3.5.2 3.5.2.1

Theoretical Equations Penman-Monteith Equation

Monteith (1965) modified the Penman method to extend it to vegetated surfaces by using resistance factors and obtained the Penman-Monteith equation as (note that similar to the Penman method, here also we have ignored all heat fluxes other than radiation, evapotranspiration, and sensible heat) DRn + ET0 = 0.035

ra c p ra

(esa - ea )

Ê r ˆ D + g Á1 + s ˜ Ë r ¯

(3.28)

a

The factor 0.035 converts W/m2 to mm of evaporable water per day. The air density ra is around 1.2 kg/m3, and cp which is the specific heat of air at a constant pressure, is about 1005 J/kg/°C. The aerodynamic resistance for the reference surface is given by ra = 208/u2, with wind speed at 2 m height in m/s, and the surface resistance is 70 s/m. In the expression for Rn (see Eq. 3.14), the albedo of the soil-crop system is used with a typical value being around 0.2. Several modifications of this equation have been proposed after the initial study of Monteith in 1965. For example, the American Society of Civil Engineers has recommended the following form Cn u (e - ea ) Ta + 273 2 sa D + g (1 + Cd u2 )

DH n + g ET0 =

(3.29)

with coefficients in the numerator Cn and denominator Cd, depending on the reference crop type. Also, if hourly computations of ET0 are to be performed, Cd takes a different value during the daytime and a different (higher) value during the night-time. More details about these coefficients are available in the report (ASCE, 2005). � EXAMPLE 3.5 For the data given in Example 3.4, estimate the reference crop evapotranspiration using the Penman-Monteith method. The average wind speed at a height of 2 m is 2.5 m/s for the month of June, and the average values of Tmax and Tmin are 39 °C and 26 °C, respectively. The average actual daylight hours may be assumed to be 9 hours per day and the relative humidity 60%.

Abstractions from Precipitation

79

Solution The procedure for computing the radiation is already described in Example 3.3. Hence, here we just give the values. n = 9; Tmax = 39 °C (312.16 K); Tmin = 26 °C (299.16 K); f = 0.463 radian, albedo of reference crop, a = 0.2; as = 0.25, bs = 0.50; ah = 0.34, bh = 0.051; al = 0.2, g = 0.487 mm Hg per °C, ra = 1.15 kg/m3, cp = 1005 J/kg/°C. For Ta = 33 °C and RH = 60%, esa = 37.7 mm Hg, ea = 0.6 × 37.7 = 22.6 mm Hg, D = 2.12 mm Hg per °C. For the representative day (June 11, Julian day 162), d = 0.403 radians, wss = 1.78 radians, N = 13.77 hours, Rn = 182.5 W/m2 The aerodynamic resistance coefficient ra = 208/u2 = 83.2 s/m and the surface resistant rs = 70 s/m. Using Eq. (3.28), we get DRn + ET0 = 0.035

ra c p ra

(esa - ea )

Ê r ˆ D + g Á1 + s ˜ Ë r ¯

= 6.92 mm/d or 208 mm

a

3.5.2.2

Actual Evapotranspiration

The actual evapotranspiration (AET) will be limited by the availability of water within the root zone of the crops. Generally, it is obtained by comparing the actual versus the maximum possible availability of soil water in the root zone. The maximum possible water availability will depend on the soil-crop combination: roots of some crops may be able to apply a large suction and use water which other crops may not be able to use. Similarly, some soils may hold water with a large adhesive force and the roots may not be able to use it. Two terms which are useful in this connection are: (a) the field capacity of the soil, i.e., the moisture content of the soil which exists after flooding when the gravity drainage has become negligible, and (b) the permanent wilting point, i.e., the moisture content at which the plants wilt and do not recover even after re-wetting. The total water availability is considered to be the difference in these two moisture contents. The ratio of the actual and the maximum possible water availability could be taken as the ratio of AET and PET. Another way of obtaining the AET is to use the appropriate surface and aerodynamic resistance in the Penman-Monteith equation. However, these are difficult to estimate and the crop-factor approach is generally used to estimate AET. The crop-factor will depend on the type of crop and its growth stage, climate, and soil evaporation. Sometimes, the crop-factor is split into two factors, one accounting for the climate and crop properties and the other accounting for the soil evaporation.

3.6

INFILTRATION

Infiltration is the process of water penetrating the ground surface to reach the underlying soil. Depending on the soil type and its moisture content, sometimes a large part of the precipitation may be lost as infiltration. In this section we will look at the process of infiltration, the factors affecting it, and its estimation.

LO 4 Discuss the infiltration process and the influence of various factors on the rate of infiltration

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Engineering Hydrology

3.6.1

Infiltration Process

During a precipitation event, when rain falls on bare soil (or on vegetated areas after the interception capacity has been satisfied) part of it will cross the soil surface and enter the ground. How much water infiltrates will depend on the rainfall intensity and the infiltration capacity of the soil, which in turn, is dependent on several factors, as discussed in the next section. Once water infiltrates into the subsurface, it moves down, primarily in the vertical direction, but sometimes also in the horizontal direction. This movement is typically due to gravity but sometimes the capillary action in the soil pores also plays a major role. Whatever water is not able to infiltrate, due to the rainfall intensity being larger than the infiltration capacity, travels on the surface to fill the depressions and then contributes to the surface runoff. Some part of the infiltrated water may also return quickly to the surface and contribute to the runoff. Therefore, in order to estimate the runoff, we should have a clear understanding of the infiltration process. The infiltrating water will create a wetting front in the soil above which the water content of the soil will be higher than that before the precipitation event (one should note, however, that the wetting front does not exactly correspond to how far the infiltrating water has penetrated, since it also contains the initial pore water which is pushed down by the infiltrating water). Figure 3.3 shows a typical soil moisture profile.

Figure 3.3

A typical soil moisture profile during infiltration

The position of the wetting front will depend on the infiltration rate, which may be defined as the volumetric rate of infiltrating water per unit surface area, and is normally expressed in mm/h. The infiltration rate will depend on several factors as described in the next section. Initially, when the soil is dry, it will be able to accommodate more water and the infiltration rate would be high. However, as the soil pores fill up with the infiltrating water, the rate of infiltration will gradually decrease. This temporal variation in infiltration is an important component in the understanding of infiltration and its estimation, and there have been several models proposed to simulate it. We will look at some of these models in a subsequent section.

Abstractions from Precipitation

81

3.6.2 Factors Affecting Infiltration Infiltration is affected by several properties of the soil, topography, and the precipitation pattern. The rate at which water is being supplied at the surface, e.g., through precipitation, irrigation, etc., and the rate at which the soil can carry this water down are the two factors which influence the infiltration process. If the supply is not enough, the soil will carry all the water down and will be unsaturated. If the capacity is not enough, the soil will become saturated and some water will remain at the surface, causing ponding. The supply is governed by the precipitation pattern or the irrigation schedule. In the discussion below, we assume that sufficient water is available and look at the factors which affect the infiltration capacity, defined as the maximum possible infiltration rate at any time. ∑

Hydraulic Conductivity: The hydraulic conductivity of a soil represents the ease with which water can flow through it (discussed in more detail in Chapter 7). A larger conductivity, naturally, results in a higher infiltration capacity. The conductivity varies with the moisture content of the soil and is maximum when the soil is saturated with water. This is denoted by saturated hydraulic conductivity Ks. As the moisture content reduces, i.e., the soil gets drier, the conductivity reduces. The relative conductivity, i.e., the ratio of unsaturated hydraulic conductivity to saturated hydraulic conductivity, is a function of the degree of saturation q/qs, q denoting the actual water content (generally volumetric water content is used, but sometimes gravimetric water content is used) and qs, the saturated water content (equal to the porosity). Sandy soils typically have higher Ks but there is a rapid decrease in conductivity on drying. Clayey soils, on the other hand, have a much smaller Ks but the decrease in conductivity on drying is much more moderate. For infiltration under saturated conditions, therefore, a sandy soil will have a larger infiltration capacity.

∑

Initial moisture profile: The moisture profile in the soil at the start of the water application on the soil surface is another important factor. If the soil is initially dry, it will have a larger capacity to accommodate the infiltrating water leading to a larger infiltration capacity. If the pores are already filled with water, the soil may not be able to take much more water as the infiltration rate will be governed by the downward movement of the existing pore water.

∑

Impact of raindrops: The impact of raindrops tends to compact the surface soil and reduce its conductivity. Also, the fine particles of the soil may get displaced and clog the larger pores. Both these events lead to a reduction in the infiltration capacity.

∑

Type of surface: Since the land cover has a significant effect on the impact of raindrops, the nature of the soil surface has a bearing on the infiltration capacity. Presence of vegetation will lessen the impact of raindrops and may also cause loosening of the soil by the roots, thereby leading to an increase in the infiltration capacity.

∑

Clay content: The swelling of clay on getting wet leads to a decrease in the infiltration capacity in soils having high clay content.

∑

Water turbidity: Turbid water infiltrating into the soil may clog some of the pores with the fine particles present in the water, leading to a decrease in infiltration capacity.

∑

Soil air: The entrapped air in the soil initially reduces the infiltration capacity, by reducing the conductivity. As infiltration progresses, these bubbles may be removed and may lead to an increase in infiltration capacity.

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Engineering Hydrology

In addition to these factors affecting the infiltration capacity, there are other factors which affect the actual infiltration rate. For example, if the rainfall intensity is less than the infiltration capacity, the actual infiltration rate will be smaller. Similarly, if the ground slope is large, infiltration will be smaller than that for a gently sloping surface.

3.6.3

Estimation of Infiltration Capacity

The ‘infiltrometer’ is used to measure the infiltration rate in the laboratory or at a field scale. Measurement techniques are described in details in Chapter 10. Here we will look at several models, both empirical and theoretical, which have been proposed to characterize the variation of the infiltration capacity with time, and discuss the methods of estimating their parameters.

LO 5 Summarize empirical and theoretical methods for the estimation of infiltration

The infiltration capacity f, is expressed in the same units as the rainfall intensity, generally mm/h. It represents the volumetric flux of infiltrating water per unit area at any given time. Since most of the measurement techniques measure the volume of water infiltrated during a time period, it is useful to define a cumulative infiltration capacity F, as t

F=

Ú fdt

(3.30)

0

which represents the volume of infiltrated water per unit surface area (typically in mm). Subscripts 0, t, and ∞, are used on f and F to denote their initial values at t = 0, the value at any time t, and the ultimate values, i.e., at very large times t = ∞. Obviously, F0 = 0 and if the ultimate infiltration capacity f∞ is finite, then F∞ = ∞. It is the value of ft (or Ft, since both are interrelated) which needs to be expressed as function of time, as described in the next section.

3.6.3.1 Empirical Methods � Kostiakov Model infiltration rate as

One of the early investigators, Kostiakov (1932), proposed a power law for the ft = a K t - b K

(3.31)

where, aK and bK are empirical positive constants for the Kostiakov model, with bK less than 1. From the observed data on cumulative infiltration depth at different times, the infiltration rate at various times may be estimated by finite differences. A straight line fitted to the log-log plot of ft versus t will provide the values of the coefficients, with bK as the negative of the slope and aK as the intercept. To avoid the numerical differentiation for estimating ft, the observed cumulative infiltration data may be directly used, by writing Ft =

a K 1- b K t 1 - bK

(3.32)

and obtaining the coefficients using log-log plot of Ft versus t. However, any of the several computer-based nonlinear regression techniques readily available (e.g., in Microsoft Excel) could be used to avoid the use of graphical techniques for estimation of the coefficients.

Abstractions from Precipitation

83

Note that (a) the initial infiltration capacity is infinite, and (b) the ultimate infiltration capacity is zero, while a finite value is generally observed in experiments. It is recommended to use this equation only till the time ft becomes equal to the saturated hydraulic conductivity of the soil. After that, ft = Ks may be assumed. Another alternative, suggested by Mezencev in 1948, is to add the ultimate infiltration capacity term as ft = f• + a K t - b K

(3.33)

� Horton Model Horton (1940), proposed an exponential equation for the infiltration rate observed under conditions of ponding at the surface, as ft = f• + ( f0 - f• )e -a H t

(3.34)

where, aH is the time constant (h–1) for the Hortonian model. The model was based on the assumption of the infiltration process being an exhaustion process, in which the rate at any time is proportional to the remaining capacity. Estimating the infiltration rate at various times by finite differences, the initial infiltration rate is taken as f0. If the experiment is conducted for a long time, the ultimate infiltration rate f∞, is readily obtained as the asymptotic value. A log-log plot of (ft – f∞)/(f0 – f∞) vs t will provide the value of ah. Sometimes, the experiments are not conducted for a long time (f∞ is not known) or f0 may not be available (e.g., when using central difference to find the infiltration rate). A nonlinear regression could then be used to estimate all three parameters (f0, f∞, and Kh) of the model from the observed F versus t data, using the fact that Ft = f• t +

f0 - f• (1 - e -a H t ) aH

(3.35)

The Horton model is probably the most widely used model for infiltration capacity, although it requires more number of parameters (three) than several other models. � Holtan Model Holtan (1961), proposed an equation based on the idea that the soil water storage, surface porosity, and the plant roots have a major influence on the infiltration capacity. This equation is generally applicable to agricultural areas. Using a crop growth index icg, defined as the maturity of crop; available storage Sa, defined as the available moisture storage capacity in the surface layer; and a porosity index ip, dependent on the surface conditions and plant root density; the infiltration capacity may be written as ft = f• + icg i p Sa1.4

(3.36)

Sa is a function of time and is obtained by multiplying the depth of the surface layer with the moisture deficit, i.e., saturated water content minus the actual water content, at any time. The surface layer is usually taken to be the plow layer, i.e., the layer affected by plowing, or up to the first impeding, i.e., a less pervious, layer. � Swartzendruber-Clague Model Swartzendruber and Clague (1989), proposed a semi-empirical model which was shown to replicate all the previously proposed models very well. The infiltration capacity is given as ft = K s +

S 2 t

e -a SC

t

(3.37)

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Engineering Hydrology

and the cumulative infiltration capacity is given as Ft = K s t +

S (1 - e -a SC a SC

t

)

(3.38)

where, aSC is a coefficient of the Swartzendruber-Clague model (units 1/√time), and S is the sorptivity of the soil (length/√time) which is a measure of the capacity of the soil to absorb/desorb water. � EXAMPLE 3.6 The following table shows the observed data of cumulative infiltration depth for an infiltrometer. Estimate the parameters of the Kostiakov, Horton, and Swartzendruber-Clague models, and plot the temporal variation of the predicted cumulative infiltration capacity and compare with the observed data. 1

Time (min) Cumulative Depth (mm) Time (min) Cum. Depth (mm)

1.2

2 2.1

60

90

31.7

39.0

4 3.8 120 44.4

6 5.6 180 55.0

8 7.1 240 65.2

10 8.7 360 85.4

15 12.2

20 15.6

30 20.7

480

600

900

104.8

125.1

175.0

45 26.7 1200 225.2

Solution We first compute the infiltration rate by using finite differences and assign it to the center of the time interval. The table below shows these values (time rounded-off to 0.01 h). 0.01

Time (hours)

0.03

0.05

0.08

0.12

0.15

0.21

0.29

0.42

0.63

Infiltration rate (mm/h)

72.0

54.0

51.0

54.0

45.0

48.0

42.0

40.8

30.6

24.0

Time (h)

0.9

1.3

1.8

2.5

3.5

5.0

7.0

9.0

12.5

17.5

Inf. Rate (mm/h)

20.0

14.6

10.8

10.6

10.2

10.1

9.7

10.2

10.0

10.0

Figure 3.4 shows the plots of the infiltration rates and the cumulative infiltration depth.

Figure 3.4

Cumulative infiltration depth (solid line) and infiltration rate (dashed line) variation with time

Abstractions from Precipitation

85

It is apparent that the experiment has been conducted for a sufficiently long time for the infiltration rate to reach its asymptotic value of 10 mm/h. We could use this fact while estimating the model parameters. However, in this example we do not take advantage of this and estimate all the parameters. (a)

Kostiakov model: A plot of log Ft and log t is shown below in Figure 3.5.

Figure 3.5

Power-law fit of the Cumulative infiltration depth

The data does not follow an exact straight line, but we could fit a line and find its slope (= 1 – bK) Ê aK ˆ and the intercept at t = 1 Á = . Instead, we use Microsoft Excel to directly perform a powerË 1 - b ˜¯ K

law regression and get the power as 0.7 and intercept as 27.5 (shown in the figure). This results in bK = 0.3 and aK = 19.3 mm/h. Therefore, Ft = 27.5t 0.7 and ft = 19.3t -0.3 (b)

Horton Model: We use the solver add-in of Excel to obtain the coefficients of the model, Ft = c1t + c2 (1 - e - c3t ) , by minimizing the sum of the relative errors, as c1 = 10.0 mm/h, c2 = 25.0 mm/h, and c3 = 2.0 h–1. Therefore, from Eq. (3.35), f∞ = 10 mm/h, aH = 2.0 h–1, and f0 = 60 mm/h, resulting in -2 t ft = 10 + 50e -2 t and Ft = 10t + 25(1 - e )

Alternatively, we could use our estimate of f∞ = 10 mm/h and fit a straight line to the ln(ft – f∞) versus t plot. However, the estimates of f using finite difference tend to magnify errors in the data and it would be better to use the cumulative infiltration data directly. Also, note that small fluctuations in the values of f at large times could lead to a negative value for ft – f∞. (c)

Swartzendruber-Clague Model: The form of cumulative infiltration equation is identical to the Horton’s equation Ft = c1t + c2 (1 - e - c3t ) , therefore, c1 = 10.0 mm/h, c2 = 25.0 mm/h, and c3 = 2.0 h–1. From these values, we get Ks = 10.0 mm/h, aSC = 2.0 h–1/2 and S = 50 mm/h1/2, resulting in

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Engineering Hydrology

ft = 10 +

25 t

e -2

t

and

Ft = 10t + 25(1 - e -2 t )

Figure 3.6 shows a comparison of the cumulative infiltration and the infiltration rate from these three models with the observed data. Note that the Horton model is closest to the observed data in this case. However, in general, the ability of the models to fit the observed data is largely site-dependent.

Solid line – Horton; Dashed line – Kos akov; Do ed line – Swartzendruber-Clague

Figure 3.6

3.6.3.2

Comparison of the data (symbols) and models

Theoretical Equations

The theoretical equations are based on an application of the Darcy’s law (more detail given in Chapter 7) pertaining to flow of water through unsaturated soil. The governing equation is known as the Richards equation and is obtained by combining Darcy’s law and the continuity equation.

Abstractions from Precipitation

87

∂(y - z ) ∂z where, z is the vertical coordinate, positive downwards from the soil surface; q is the downward flux (length/ time); K is the hydraulic conductivity (length/time) which is a function of the moisture content q, under unsaturated conditions; and y is the soil water pressure head (also called the capillary pressure head or suction head with negative pressure head equivalent to positive suction head), which is dependent on q, and is negative for unsaturated conditions. q = - K (q )

Darcy’s law:

∂q ∂q + =0 ∂t ∂z

Continuity equation:

Richards equation:

∂q ∂ Ê ∂y ˆ ∂K = K ∂t ∂z ÁË ∂z ˜¯ ∂z

(3.39)

Sometimes, the soil diffusivity, D(= Kdy/dq) is used to express the Richards equation in terms of moisture-content as ∂q ∂ Ê ∂q ˆ ∂K = D ∂t ∂z ÁË ∂z ˜¯ ∂z

(3.40)

Dependence of K and y on q makes the Richards equation a nonlinear equation which is very difficult to solve. Hence various approximations have been made, resulting in different models for the infiltration capacity. � Green-Ampt Model Green and Ampt (1911), derived an expression for infiltration from a ponded surface into a deep homogeneous soil with uniform initial moisture content. The wetting front was assumed to be a sharp boundary between the saturated zone, having a uniform moisture content qw, and the unaffected zone, having a uniform moisture content qi, equal to the initial water content, as shown in the Figure 3.7 given below:

Figure 3.7

Wetting front in the Green-Ampt model

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Engineering Hydrology

Assuming the depth of ponding to be negligible, i.e., the soil surface is always saturated with a very thin layer of water existing above it, and using an effective hydraulic conductivity Ke for the wetted zone, the Darcy’s law results in ft = K e

-y + dt dt

(3.41)

where, y is the soil water pressure (negative) at the wetting front, and dt is the depth of infiltration at any time t, which is related to the cumulative infiltration Ft, as Ft = dt (q w - q i ) = dt Dq

(3.42)

A relation between the instantaneous infiltration rate and the cumulative infiltration volume is, therefore, written as ft = K e -

K eyDq b = a GA + GA Ft Ft

(3.43)

which could be used to obtain the two parameters [a GA = K e , bGA = - K eyDq ] from measured values of Ft and the derived values of ft. To avoid the use of the derivative estimation for ft, Eq. (3.43) could be written as dFt a GA Ft + bGA Ft dFt (a GA Ft + bGA - bGA )dFt = a = => = 1 => GA dt Ft a GA Ft + bGA a GA Ft + bGA On integration, with the condition that F0 = 0, we get Ft -

ˆ bGA Ê a GA ln Á Ft + 1˜ = a GA t a GA Ë bGA ¯

The ratio, b/a [= –yDq] could be treated as another parameter, gGA, to obtain Ê F ˆ Ft - g GA ln Á t + 1˜ = a GA t Ë g GA ¯

(3.44)

which is only in terms of Ft and two parameters, that could be estimated by nonlinear regression from the measured values of Ft. We could also estimate these from soil properties and moisture conditions by using the effective conductivity as some fraction (usually 0.5 to 1) of the saturated conductivity, qw as the porosity, the wetting front pressure y as a function of the soil type (e.g., around –5 cm for sand and about –30 cm for clay), and qi either from measurements or from soil properties. Several explicit approximations for Eq. (3.44) have been suggested to compute Ft for a given t. The following approximation (Parlange et al., a t F 2002) provides nearly exact values (in terms of non-dimensional parameters, Ft* = t and t * = GA ): g GA g GA

Abstractions from Precipitation

Ê ˆ Á ˜ 6 ˜ Ft* = t * + ln Á 1 + t * + 6 ˜ Á 1+ Á ˜ Ë 2t * ¯

89

(3.45)

� Philip Model Philip (1957), developed an infinite series solution of the Richards equation and found that the truncation of the series after the first two terms provided reasonably accurate values of the cumulative infiltration. The equation for cumulative infiltration is given by (3.46)

Ft = f• t + S t and, in terms of the infiltration rate, ft = f• +

S

(3.47)

2 t

with S as the sorptivity (see Eq. 3.38). The value of f∞ is generally taken in the range of Ks/3 to 2Ks/3. � Smith-Parlange Model Assuming the diffusivity to be an exponential function of the water content, Smith and Parlange (1978), obtained a solution of the Richards equation which may be written as ft = K s

ea SP Ft

(3.48)

ea SP Ft - 1

� EXAMPLE 3.7 A sandy-loam soil has saturated conductivity of 10 mm/h, wetting front pressure head of –100 mm, moisture deficit Δq = 0.3, and sorptivity of 35 mm/h1/2. Using the Green-Ampt and Philip models, generate a table of cumulative infiltration versus time. Using this table, and assuming that the soil properties are not known, estimate the parameters of the Green-Ampt model and compare with the actual values. Solution We assume that Ke in the Green-Ampt model is equal to the saturated hydraulic conductivity and f∞ in the Philip model is equal to half of the saturated hydraulic conductivity. Then, for the Green-Ampt model, a GA = K e = 10 mm/h, bGA = - K eyDq = 300 mm 2 /h, g GA = bGA /a GA = 30 mm . Using the same time values as in the previous example, the following table is generated from Equations (3.45) and (3.46). Time (min)

1

2

4

6

8

Cumulative Depth – Green Ampt (mm)

23.1

28.0

33.1

36.2

Cumulative Depth – Philip (mm)

5.2

7.5

10.7

13.1

60

90

Cum. Depth GA (mm)

59.5

66.9

73.7

86.4

Cum. Depth Philip (mm)

45.0

56.5

66.6

84.3

Time (min)

120

180

10

15

20

30

45

38.4

40.2

43.7

46.3

50.4

55.3

15.3

17.2

21.3

24.8

30.8

38.4

240

360

480

600

900

1200

98.6

122.0

144.9

167.4

222.6

277.0

100.0

128.0

153.1

176.5

229.9

278.9

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Engineering Hydrology

Figure 3.8 given below shows the plot of cumulative infiltration versus time.

Figure 3.8

Simulated infiltration depth from Green-Ampt (dashed line) and Philip (solid line) models

The Philip model fitted to the simulated data will obviously result in the exact same parameters. However, for the Green-Ampt model, estimation of the infiltration rate using finite differences may result in some error. A plot of f versus 1/Ft is shown in Figure 3.9 given below:

Figure 3.9

Straight-line fit for the Green-Ampt model

Slight error is seen in the values of aGA (9.77 vs 10) and bGA (304.6 vs 300).

3.6.4

Actual Infiltration

The previous section described the estimation of the infiltration capacity (which may also be called potential infiltration). The actual infiltration will depend on the relative values of the availability of water and the capacity of the soil to absorb it. Considering the case of rainfall, if the effective intensity (i.e., after accounting for initial losses) of the rainfall is always higher than the infiltration capacity, the actual infiltration will take place at the same rate as the infiltration capacity. However during a rainfall event, typically the intensity in the

Abstractions from Precipitation

91

beginning is less than the infiltration capacity, and with time as the infiltration capacity decreases, the rainfall intensity becomes larger than the capacity, leading to ponding and subsequent runoff. The simplest way to account for this is to assume the actual infiltration rate to be equal to the smaller of the rainfall intensity and the infiltration capacity. However, it is to be kept in mind that all infiltration capacity models described in the previous section had assumed sufficient water availability. For example, in the Hortonian model, if the origin of the time coordinate is taken at the beginning of precipitation, the infiltration capacity will show an exponential decrease with time. If the rainfall is less than the infiltration capacity for some time, the soil pores will not be filled to the extent assumed by the model. Therefore, the infiltration capacity is likely to be larger than that given by the model at any particular time. As a result, the infiltration capacity curve gets modified. In both the cases, one where the rainfall intensity is always more than f0, and the other where the intensity is always less than f∞, not much needs to be done. In the first case, the assumption of ponding is valid and the actual infiltration will follow the Hortonian model. In the second case, the actual infiltration will always be equal to the rainfall intensity, since the capacity is always larger than the available water. It is the case where the rainfall intensity is between f0 and f∞ that requires some modifications, as described next. [We consider a rainfall of uniform intensity i, such that f∞ < i < f0. Variable intensity cases are more complex and are not discussed here. Interested readers may see Assouline et al., (2007). Also, we have discussed the Hortonian model only; similar methodology will work for other models also.] If we assume that the actual infiltration rate at any time, irrespective of the surface ponding conditions, is dependent on the actual cumulative infiltration up to that time, we can combine Equations 3.34 and 3.35 as follows: f -f 1 ln 0 • aH ft - f•

From Eq. (3.34):

t=

Using this t, from Eq. (3.35): Ft =

f -f f -f f• ln 0 • + 0 t aH ft - f• aH

(3.49)

Note that Eq. (3.49) is implicit in ft and iterations are needed to obtain the value of ft for a given Ft. If we define a time of ponding tp, such that the modified infiltration capacity at that time is equal to the rainfall intensity, ponding will occur after tp. Before the time of ponding, however, the actual infiltration rate will be equal to the rainfall intensity, and the actual cumulative infiltration up to time tp will be equal to (i × tp). Putting these conditions in Eq. (3.49), we get it p =

f -f f -i f0 - f• ˘ f• 1 È ln 0 • + 0 => t p = Í f0 - i + f• ln ˙ aH i - f• aH iaH Î i - f• ˚

(3.50)

After ponding, the modified infiltration capacity ft¢, will follow the Hortonian model, but with a timeshift to account for the less-than-capacity infiltration up to that point. The time shift Δt is obtained by using the condition that ft = i at the time (tp – Δt), as Dt = t p -

f -f 1 ln 0 • aH i - f•

(3.51)

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Engineering Hydrology

The infiltration capacity and the cumulative infiltration capacity after ponding are then obtained from Equations (3.34) and (3.35) by using (t – Δt) as the time. The following Figure 3.10 shows the original and modified infiltration capacity and cumulative infiltration capacity curves. The actual infiltration rate at any time fta is given by Èi, for t £ t p fta = Í (3.52) Í f• + ( f0 - f• )e -a H (t -Dt ), for t ≥ t p Î and the cumulative infiltration Fta is given by Èit p , for t £ t p Í Fta = Í f0 - f• È1 - e -a H (t - Dt ) ˘ , for t ≥ t p Í f• (t - Dt ) + a Î ˚ H Î

(3.53)

Figure 3.10 Modifications in the infiltration capacity curves (Solid lines show the original curves and dashed lines show the modified curves.) Similarly, for the Green-Ampt model, it can be shown that (for i > aGA), t p =

Dt = t p -

bGA and, i(i - a GA )

g GA g i . The actual cumulative infiltration till the ponding time is given by + GA ln i - a GA a GA i - a GA

Ê Fta ˆ + 1˜ = a GA (t - Dt ) or its explicit form, Eq. (3.45). Fta = it and after that by (see Eq. 3.44) Fta - g GA ln Á Ë g GA ¯ Without computing Δt, we may also obtain Fta, by applying the equation at the ponding time and the current Ê F + g GA ˆ time, as Fta - it p - g GA ln Á ta ˜ = a GA (t - t p ) . Ë it p + g GA ¯

Abstractions from Precipitation

93

� EXAMPLE 3.8 For the scenario discussed in Example 3.6, assuming the Hortonian model to describe the infiltration capacity, what would be the actual infiltration profile if a rainfall of constant intensity of 40 mm/h occurs for a day? Solution Using the values of the parameters obtained in Example 3.6, i.e., f∞ = 10 mm/h, aH = 2.0 h–1, and f0 - f• ˘ 1 È f0 = 60 mm/h, and with i = 40 mm/h in Eq. (3.50), we get t p = Í f0 - i + f• ln ˙ = 0.314 h iaH Î i - f• ˚ (i.e., 18.8 minutes). The time shift, Dt = t p -

f -f 1 ln 0 • = 0.058 h, i.e., 3.5 minutes. The actual infiltration rate using aH i - f•

Eq. (3.52) is plotted below in Figure 3.11.

Figure 3.11 Actual infiltration rate using Horton’s model � SCS Model As described earlier (more detail in Chapter 4, see Eqs. 3.2 and 4.22–4.27), the SCS model relates the runoff R, to the precipitation P, and the abstractions as R=

( P - I a )2 (P - I a + S )

with the storage parameter S related to the curve number through Eq. (3.2), and the initial abstraction Ia generally taken as 20% of S. The actual infiltration is then computed as P-R. � Infiltration Index Since the computation of the actual infiltration with a variable infiltration capacity and rainfall intensity is complicated, for several engineering calculations an approximation is made that the infiltration capacity is constant. Several definitions of this constant infiltration capacity, termed as infiltration index, have been proposed. We will discuss one of these, called the f-index.

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Engineering Hydrology

If the runoff volume resulting from a given rainfall hyetograph is known, we can estimate the losses and consider these to be due to infiltration. This loss of volume divided by the catchment area will provide a depth of infiltration, using which we can estimate the constant rate of infiltration, f. Using this index, and assuming it to be invariable for all storms, the runoff resulting from any storm could be estimated. Some other infiltration indices have also been proposed but will not be discussed here (for example, the W-index accounts for Initial abstractions also). The following Figure 3.12 shows the definition of the f-index and the example shows the method of computation.

Figure 3.12 The f-index � EXAMPLE 3.9 A rainfall occurs over a catchment for a day, with its intensity varying as 1 mm/h for the first 4 hours, 3 mm/h for the next 4 hours, 5 mm/h for the next 8 hours, 4 mm/h for the next 4 hours, and 1 mm/h for the last 4 hours. What will be the f-index, if this rainfall results in a direct runoff of 44 mm? Solution Figure 3.13 given below shows the rainfall hyetograph. Total precipitation is 1 × 4 + 3 × 4 + 5 × 8 + 4 × 4 + 1 × 4 mm = 76 mm. Direct runoff is 44 mm and the losses are 32 mm. To estimate the f-index, we could start from the bottom and mark-off an area equal to the losses or we could start from the top and mark-off an area equal to the runoff. Here, starting from bottom is more convenient.

Figure 3.13 Computation of the f-index

Abstractions from Precipitation

95

If the f-index is equal to 1 mm/h, the losses will be equal to 24 mm (Figure 3.13). Since the losses are 32 mm, f-index should be larger than 1 mm/h, such that the area above the 1 mm/h intensity should be equal to 8 mm. If the f-index is less than 2 mm/h, the duration of infiltration is 16 hours (from 4 hours to 20 hours). Hence the required intensity (above 1 mm/h) would be 8/16 = 0.5 mm/h. Thus, f-index = 1.5 mm/h. After gaining an understanding of the losses, we are now ready to discuss the main variable of our interest, the runoff, in the next chapter.

SUMMARY Various abstractions take place from precipitation before it runs off to a stream. These include interception, depression storage, evaporation, transpiration, and infiltration. Interception and depression storage are combined in a single term known as initial abstraction, and evaporation and transpiration are collectively called evapotranspiration. The initial abstraction may be estimated by using the curve number methodology developed by the US Soil Conservation Service. Potential evaporation represents the rate at which evaporation will occur if there is unlimited supply of water, such as evaporation from a water body. Actual evaporation from a land surface will depend on the soil characteristics, with moisture content being the most important factor. The potential evaporation is affected by the air and water temperatures, relative humidity, wind speed, atmospheric pressure, and the size and shape of the water body. Estimation of the potential evaporation may be made using empirical or theoretical equations. The empirical equations are based on observed data and may not work well under conditions very different from those for which these are developed. The theoretical approach could be based on mass balance, energy balance, aerodynamics, or a combination thereof. Since the potential evapotranspiration is affected by the crop type and its stage of growth, the concept of a reference crop evapotranspiration is used in order to delink the crop-dependent parameters. The reference surface is specified as a standardized ground and grass surface implying that the reference evapotranspiration is a function of only climatic variables. The potential evapotranspiration is then obtained by using a crop factor. We could also use empirical or theoretical methods to estimate the potential evapotranspiration. The final component of abstraction is the infiltration of the precipitating water into the ground. The rate at which infiltration may potentially occur depends mainly on the hydraulic conductivity of the soil and its initial moisture content. The actual infiltration may be less than the potential infiltration, if the rate at which water is being supplied by the rain is smaller than the infiltration capacity. Several empirical and theoretical approaches have been suggested for the estimation of infiltration capacity. If the effective rainfall intensity is always larger than the infiltration capacity, the actual infiltration will be equal to the infiltration capacity. For other cases, techniques, based on the assumption of actual infiltration rate at any time being dependent on the actual infiltrating volume up to that time, are used to obtain the actual infiltration.

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OBJECTIVE-TYPE QUESTIONS 3.1 Which two of the following constitute the initial abstraction? (a) Infiltration (b) Interception (c) Depression Storage (d) Evaporation 3.2 Out of the options given below, which one is correct about the following statements? (i) Interception loss will be a large proportion of precipitation for low intensity and short duration rains. (ii) Initial abstraction occurs mainly during the final stages of a rainfall event. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 3.3 Change of water vapour to liquid form is called (a) Evaporation (b) Condensation (c) Boiling 3.4 The term consumptive use represents (a) Infiltration and Evaporation (c) Interception and Evaporation

(d) Freezing

(b) Evaporation and Transpiration (d) Infiltration and Runoff

3.5 Which of the following does NOT affect the interception storage? (a) Type of vegetation (b) Leaf area index (c) Type of soil (d) Vegetation cover 3.6 The leaf area index is (a) The area of a single leaf (b) The total area of all leaves (c) Ratio of single leaf area and ground surface area (d) None of these 3.7 The range of variation of the leaf area index is about (a) 0.001 to 0.09 (b) 0.02 to 0.9 (c) 0.2 to 9

(d) 20 to 100

3.8 The interception loss is generally what percent of the total precipitation? (a) 1% to 5% (b) 10% to 20% (c) 30% to 40% (d) 40% to 50% 3.9 Out of the options given below, which one is correct about the following statements? (i) The maximum storage of water per unit foliage area is more for pine tree compared to grass. (ii) ASCE recommends a value of about 1.5 mm for depression storage in pervious areas. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 3.10 Typical value of depression storage is about (a) 0.2 mm (b) 0.5 mm (c) 2 mm

(d) 20 mm

3.11 The curve number method for estimating initial abstraction was proposed by (a) Soil Conservation Service (b) NASA (c) ASCE (d) US Weather Bureau 3.12 Out of the options given below, which one is correct about the following statements? (i) The curve number of a pervious catchment is higher than that of an impervious catchment. (ii) The initial abstraction is about 50% of the potential storage in a catchment.

Abstractions from Precipitation

(a) (i) is true, (ii) is false (c) Both (i) and (ii) are true

97

(b) (i) is false, (ii) is true (d) Both (i) and (ii) are false

3.13 Out of the options given below, which one is correct about the following statements? (i) Dry air has predominance of Hydrogen and Oxygen. (ii) The amount of water vapour in air depends on the temperature and pressure. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) both (i) and (ii) are false 3.14 Out of the options given below, which one is correct about the following statements? (i) Absolute humidity is the mass of water vapour per unit volume of dry air. (ii) Saturation vapour pressure at 20 °C is about 17.6 mm of Hg. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 3.15 How does the saturation vapour pressure change with increase in temperature? (a) Increases (b) Decreases (c) Remains constant (d) Depends on the pressure 3.16 Dalton’s law relates the evaporation rate with the gradient of (a) Temperature (b) Relative humidity (c) Vapour pressure

(d) Atmospheric pressure

3.17 How does the evaporation rate change with increase in wind speed? (a) Increases continuously (b) Increases and then becomes constant (c) Decreases continuously (d) Decreases and then becomes constant 3.18 What is the primary mechanism through which wind speed affects evaporation? (a) Reducing the temperature (b) Increasing the humidity (c) Carrying saturated air away (d) Increasing the vapour pressure 3.19 What is a likely relationship between the annual evaporation from a shallow water body and that from a deep water body? (a) More in shallow water body (b) Nearly same for both (c) More in deep water body (d) Depends on the latitude 3.20 Out of the options given below, which one is correct about the following statements? (i) Higher atmospheric pressure causes an increase in evaporation. (ii) Higher salinity causes an increase in evaporation. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 3.21 The wind speed is assumed to vary as what power of height from the ground level? (a) 7th (b) 1/7th (c) 2nd (d) 1/2nd 3.22 According to the Meyer’s formula, what is the ratio of evaporation from a large and deep water body to that for a small and shallow water body? (a) 0.37 (b) 0.50 (c) 0.67 (d) 0.74 3.23 In Rohwer’s formula, if the wind speed measurement is not available at the water surface, but at some other height, then the wind speed computed at a height of ___ m is used. (a) 0.3 (b) 0.6 (c) 0.9 (d) 1.0 3.24 In the Harbeck and Meyers formula, the coefficient shows ______ dependence on the surface area of the water body. (a) Small and direct (b) Small and inverse (c) Large and direct (d) No 3.25 The contribution of transpiration to total atmospheric moisture is about (a) 1% (b) 5% (c) 10% (d) 20%

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3.26 The albedo of a water body is generally around (a) 0.02 (b) 0.08 (c) 0.20 3.27 Bowen’s ratio is the ratio of __________. (a) Evaporation and Transpiration (c) Water and air temperatures

(d) 0.90

(b) Sensible and latent heat fluxes (d) Vapour pressure and atmospheric pressure

3.28 In the mass balance approach for estimating evaporation, which of the components has the largest uncertainty? (a) Precipitation (b) Surface runoff (c) Change in storage (d) Groundwater 3.29 What rate of evaporation would result from a continuous heat energy of 1 W/m2? (a) 0.035 mm/d (b) 0.35 mm/d (c) 0.5 mm/d (d) 1 mm/d 3.30 Out of the options given below, which one is correct about the following statements? (i) Using the water surface temperature as average lake temperature is a good assumption for deeper lakes. (ii) The heat and vapour diffusivities are generally assumed to be equal. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 3.31 As we move away from the equator in the Northern hemisphere, the monthly mean extraterrestrial radiation (a) Decreases (b) Increases (c) Increases in summer (d) Decreases in summer 3.32 The monthly mean of potential sunshine hours in a day at the equator is (a) Constant (b) Largest in January (c) Largest in June (d) None of these 3.33 The Thornthwaite-Holzman equation is based on which of the following methods? (a) Aerodynamic (b) Energy balance (c) Mass balance (d) Empirical 3.34 Which of the following data is NOT required in Penman’s method? (a) Air temperature (b) Wind speed (c) Latitude

(d) Water temperature

3.35 Which of the following does NOT affect the potential evapotranspiration? (a) Temperature (b) Day length (c) Soil water content (d) Nature of vegetation 3.36 Thornthwaite observed a common value of the monthly PET equal to ____ cm at a mean monthly temperature of 26.5°C at different places. (a) 1.5 (b) 5.9 (c) 13.5 (d) 26.5 3.37 The temperature-based empirical methods generally underestimate the PET in ___ regions and overestimate in _____ regions? (a) Dry, Humid (b) Humid, Dry (c) Hilly, Plain (d) Plain, Hilly 3.38 The surface resistance of the surface used in computing reference evapotranspiration is equal to (a) 1 s/m (b) 10 s/m (c) 70 s/m (d) 130 s/m 3.39 The infiltration rate is generally expressed in units of (b) mm/h (c) m3/s (a) m3/d

(d) mm3/d

3.40 The infiltration rate ________ with an increase in the initial soil moisture content. (a) Increases (b) Decreases (c) Remains unchanged (d) Has no definite relation 3.41 Out of the options given below, which one is correct about the following statements? (i) If the soil is initially dry, it will have a smaller infiltration capacity. (ii) Soils having high clay content will have a smaller infiltration capacity.

Abstractions from Precipitation

(a) (i) is true, (ii) is false (c) Both (i) and (ii) are true

99

(b) (i) is false, (ii) is true (d) Both (i) and (ii) are false

3.42 Out of the options given below, which one is correct about the following statements? (i) Turbidity in water causes a reduction in the soil infiltration capacity. (ii) If the ground slope is large, infiltration will be smaller. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 3.43 The Kostiakov model of infiltration capacity predicts a _________ infiltration capacity at a very large time. (a) Infinite (b) Zero (c) Finite positive (d) Negative 3.44 The Horton model of infiltration may be derived by assuming that the infiltration rate is proportional to (a) Rainfall intensity (b) Remaining infiltration capacity (c) Soil moisture content (d) Soil hydraulic conductivity 3.45 The Green-Ampt model of infiltration assumes the wetting front to be (a) Sharp (b) Diffused (c) Parabolic (d) Logarithmic 3.46 The f-index model of infiltration assumes the infiltration capacity to be (a) Increasing with time (b) Decreasing with time (c) Constant (d) Equal to rainfall

DESCRIPTIVE QUESTIONS 3.1 What is meant by the term initial abstraction? How do you obtain it using the Curve Number method? 3.2 What are the factors which affect the rate of evaporation from a water body? Explain the effect of wind speed on the evaporation rate. 3.3 In addition to the factors affecting evaporation, what other factors affect the transpiration rate? 3.4 What are the various theoretical approaches of estimating evaporation from a lake? Describe the energy balance approach. How is the net heat transfer from the lake to the ground estimated? 3.5 What approximation is used in the Penman’s method to avoid the use of water temperature in the formula? Describe the Penman-Monteith method of estimating the PET. 3.6 What are the various empirical approaches of estimating PET? Describe the Thornthwaite method. 3.7 What are the factors which affect the infiltration capacity of soil? List some of the empirical models of infiltration capacity and describe the Horton model. 3.8 Why is the actual infiltration different from potential infiltration? Using the Horton model, describe how you will modify the infiltration capacity curve if the initial precipitation rate is smaller than the initial infiltration capacity? 3.9 Describe the f-index method of accounting for the infiltration.

NUMERICAL QUESTIONS 3.1 A lake has a surface area of 12.6 km2. For a particular month, the average values were measured as: Water temperature = 29.1 °C; Air temperature = 27.5 °C; Relative humidity = 58.6%; Wind speed at a height of

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2 m = 4.2 km/h. Estimate the evaporation (in m3) from the lake during this month using Meyer’s and Harbeck-Meyers Formulae. 3.2 For the lake of the previous question, the latitude is 23.3 °N, the month is May, maximum air temperature during May is 33 °C and minimum is 18 °C, and the actual number of sunny daylight hours are 10 hours per day. Estimate the monthly evapotranspiration using the Penman method. 3.3 Mean monthly temperatures for Cape Town (latitude 33.9°S) are given in the table below: Month

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

Temperature (°C)

22

23

21

19

16

14

13

14

15

17

19

21

Find the reference crop evapotranspiration for February and also estimate the evapotranspiration from a lightly vegetated field (K = 0.8). 3.4 For the data given in the previous question, estimate the reference crop evapotranspiration using the PenmanMonteith method. The average wind speed at a height of 10 m is 5 m/s for the month of February, and the average values of Tmax and Tmin are 26 °C and 16 °C, respectively. The average actual daylight hours may be assumed to be 9 hours per day and the relative humidity 70%. 3.5 An infiltrometer experiment produced the data as shown in the table below. Estimate the parameters of the Kostiakov and Horton models. Time (min)

1

2

4

6

8

10

15

20

30

Cumulative Depth (mm)

1.7

2.7

4.3

5.8

Time (min)

60

90

120

180

Cum. Depth (mm)

28.8

38.3

44.9

57.1

65.3

45

7.2

8.7

11.3

13.7

18.2

24.6

240

360

480

600

900

1200

84.0

100.5

113.7

153.0

186.6

3.6 A soil has saturated conductivity of 8 mm/h, wetting front pressure head of −70 mm, moisture deficit Δq = 0.32, and sorptivity of 30 mm/h1/2. Generate a table of cumulative infiltration similar to the one above by using the Green-Ampt and Philip models. Using this synthetic data, estimate the parameters of the GreenAmpt model. 3.7 Using the data of Exercise 3.8 with the Hortonian model, obtain the actual infiltration profile if a rainfall of constant intensity 20 mm/h occurs for a day. 3.8 A rainfall occurs over a catchment for 12 hours, with its intensity varying as 0.5 mm/h for the first 2 hours, 1.2 mm/h for the next 3 hours, 4.7 mm/h for the next 3 hours, 1.0 mm/h for the next 3 hours, and 0.2 mm/h for the last 1 hour. Estimate the f-index, if the direct run-off is 11.1 mm.

4

Runoff

LEARNING OBJECTIVES LO 1

Explain the process of runoff generation in a catchment

LO 2

Illustrate the measurement of streamflow

LO 3

Outline different types of annual hydrographs and storm hydrograph

LO 4

Summarize the impacts of various physical factors on runoff

LO 5

Develop rainfall-runoff relationships using method of least squares

LO 6

Estimate peak runoff using rational formula

LO 7

Evaluate runoff volume using SCS curve number method

LO 8

Demonstrate flow duration curve and its usefulness

LO 9

Determine the capacity of a reservoir using flow mass curve and sequent peak algorithm

4.1

RUNOFF GENERATION

Runoff represents the response of a catchment measured at the outlet LO 1 Explain the process when the catchment is subjected to rainfall. The runoff response from a of runoff generation in a catchment is an integrated effect of a wide range of physiographic and catchment climatic factors. As discussed in the previous chapter, the precipitation falling on a catchment initially gets trapped on vegetation and other obstructions or some depressions in the form of initial losses. The initial losses can be either in the form of interception or depression losses. The continuation of rainfall causes the interception and depression storage capacities of the catchment to become satisfied. Another type of abstraction from precipitation is infiltration, wherein water infiltrates into the ground leading to increased soil moisture. Once the soil just below the

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ground becomes saturated and all other abstractions (e.g., depression and interception etc.) are satisfied, the excess precipitation starts to accumulate over the land surface. During a rainfall event, as water accumulates on the surface of the Earth, it starts to travel along natural slopes. The water moving over the ground as a thin sheet is called overland flow. The overland flow travels for short distances only and then forms small channels, which combine to form bigger channels, and water ultimately runs into the main water course/river. The water moving in the channels/main water course is referred to as the channel flow. The infiltrated water percolates deep into the ground and meets the groundwater table (GWT). The groundwater in the saturated zone may enter a river, after travelling some distance, in the form of baseflow. Sometimes, due to certain local conditions, infiltrated water between the GWT and the land surface may travel laterally towards the main river and contribute to streamflow in the form of interflow (also called quickflow). Interflow occurs in steeply sloped catchments favoring lateral movement of infiltrated water between ground surface and GWT during storm events. Interflow may also occur when the land mass above GWT consists of impervious strata restricting the movement of water vertically downwards, but favors water movement laterally along the impervious strata towards the main river. Therefore, the total flow observed at the outlet of a catchment, known as streamflow, consists of three components: runoff, interflow, and baseflow. Runoff is the portion that travels on the land surface and is, therefore, also called surface runoff. The interflow and baseflow are the portions traveling below the ground and, therefore, are sometimes jointly referred to as subsurface flow. Generally, runoff travels at a fast rate, baseflow travels slowly, and the travel speed for interflow is somewhere in between. During a rainfall event, streamflow in a river is dominated by runoff and interflow, while during rainless periods contributions from baseflow are significant in a river. Unless otherwise mentioned, the term runoff refers to surface runoff or direct runoff. The direct runoff starts to occur when the soil just below the ground surface becomes saturated with water, which may happen in one of the following two ways. The first is when the soil becomes saturated by rainwater infiltrating from the top, and then any additional rain that falls becomes runoff. This type of runoff generation mechanism is known as Hortonian runoff. This runoff is simply estimated as the difference between rainfall and infiltration. On the other hand, it may be possible that the soil just below the ground surface may already be saturated due to water travelling as interflow from the upstream reaches of the catchment. The additional rain that falls on the ground, which is saturated from below, becomes runoff known as the Saturation Overland flow. The saturation overland flow occurs in hilly areas and in catchments with steep slopes that encourage interflow. The calculation of runoff under saturation overland flow conditions is much more complex than the Hortonian overland flow. In most parts of the world, including India, Hortonian overland flow concept is applicable. This chapter deals with Hortonian overland flow only and the concept of saturation overland flow is outside the scope of this book.

4.2

MEASUREMENT OF STREAMFLOW

The flow in a river at a location is the volume of water flowing across LO 2 Illustrate the measurethat location per unit of time. The unit of flow measurement is cubic ment of streamflow 3 meter per second (m /s) or more commonly called ‘cumec’. The flow expressed in the foot-pound-second (FPS) system is cubic feet per second (ft3/s) or ‘cusec’. There are a variety of methods available for the measurement of streamflow in a river. The measurement of flow is dealt with in greater details in Chapter 10. Only the most commonly used method,

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103

known as stream-gauging or the area-velocity method, is discussed here. The area-velocity method involves the use of continuity equation Q = Av, where Q is the flow in m3/s, A is the area of cross section (m2) of the river at the measurement location, and v is the average velocity of flow (m/s) at the cross section under consideration.

4.2.1

Area

At the location of flow measurement, the cross section of a river is divided into a number of segments (Figure 4.1). Each segment has an area of Ai and a velocity of vi. At the midpoint of each segment, the depth of flow is measured using either a staff-gauge or an echo-depth recorder. For shallow and small streams, staff-gauges are used while for large and deep rivers, echo-depth recorders are used. Once the depth of flow of a segment is known, its area is approximated by assuming the segmental area to be consisting of a simple geometric shape or a combination thereof. For example, corner areas may be taken as triangles and the segmental areas in the middle of a cross section may be taken as rectangles or trapezoids. The width of each segment is predecided and is known.

Figure 4.1

4.2.2

Streamflow measurements at a cross section in a river by area velocity method

Velocity

The average velocity of flow in each segmental area is needed to determine the segmental flow (Qi). The velocity at a cross section is a function of depth of flow. Figure 4.2 shows a rough approximation of the vertical distribution of velocity of flow at a cross section in a river as a function of depth below the water surface. The velocity of flow is zero at the stream bed and maximum slightly below the water surface.

Figure 4.2

Vertical distribution of velocity at a cross section in a river

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Engineering Hydrology

Let the depth of flow at a cross section be d; the average velocity at a vertical be v ; and the velocity at a distance y from water surface be vy. Then, approximating the segment as a rectangle, the average velocity at a vertical in the cross section may be calculated as follows: 1 v= d

d

Ú

v y dy

(4.1)

y=0

Mathematically, the function vy may be expressed as logarithmic, parabolic, or a power law. An accurate assessment of v requires continuous measurement of the velocity of flow along the vertical or measurement at several discrete steps along the vertical. However, observing vy at different values of y poses a lot of challenges in the field, can be extremely expensive, and may not be necessary. There are several methods available for measuring the velocity of flow in a river; however, the use of a current meter is the most common. At the vertical passing through the midpoint of each segment, the velocity is measured using a current meter. The average velocity at a vertical may be approximated using either a single point method or a two-point method depending on the type of river. In single point method, the current meter is lowered up to a depth of 0.6 d from the water surface and the velocity is measured (v0.6). In two-point method, two observations are made at 0.2 d and 0.8 d (v0.2 and v0.8). The average velocity at the vertical is then taken as follows: vi = v0.6

Single Point Method vi =

Two-Point Method

(v0.2 + v0.8 ) 2

(4.2) (4.3)

The single point method may be used for small rivers with the depth of flow limited to 2 m to 3 m. However, for large and deep rivers, the two point method is normally employed. Once the segmental areas (Ai) and velocities ( vi ) are obtained, the segmental discharge is approximated N

as their product, i.e., Qi = Ai vi and Q = ÂQi . More details about this and some other methods for streamflow i =1

measurement are provided in Chapter 10.

4.3

ANNUAL AND STORM HYDROGRAPHS

The flow measured in a river at any location is composed of several types LO 3 Outline different types of flows coming in from the catchment at that location, e.g., surface runoff, of annual hydrographs and sub-surface runoff, baseflow, etc. Baseflow or groundwater flow is the storm hydrograph portion of the flow contributed to the river from the saturated portion of the ground. A plot of total flow (or streamflow) against time is known as a hydrograph. A hydrograph can be either an annual hydrograph or a storm hydrograph. An annual hydrograph is one in which the variation of streamflow is plotted during an entire year. On the other hand, the variation of streamflow over a ‘limited time-period’ is expressed in a storm hydrograph. The ‘limited time period’ is normally due to a storm event and is taken equivalent to the ‘total time base’ of a hydrograph. Streamflow is a continuous physical variable, which may or may not be measured continuously in time. Regardless of

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105

whether measured continuously or at discrete time steps, it is usually expressed as the average value during a specified time-interval. This time-interval may be one-hour, one-day, one-week, one-month, or any other time interval depending upon the practical utility. When we need to analyze the dynamic behavior of a catchment in terms of its response to an intense rainfall event over small durations, for design purposes, then we prefer the time interval to be smaller, e.g., 1-hour, 30-minutes, 15-minutes, or even 5-minutes. On the other hand, when we are interested in the overall planning and management of water resources over a longer period of time (such as a few years), then the time interval may be taken as daily, weekly, or monthly.

4.3.1

Annual Hydrograph

An annual hydrograph is a hydrograph showing the flow in a river at a particular location throughout the year. An annual hydrograph from a river is a useful tool to obtain information about the catchment. For example, an annual hydrograph can be used to determine the total amount of runoff that is available in the catchment at that location. The total volume of runoff available is useful in various water resources planning activities. For example, allocation of water resources for different purposes, such as municipal water supply, irrigation, and hydropower, etc., would require the total supply available from year to year. An annual hydrograph also indicates the nature of the catchment. For example, a continuous annual hydrograph with its flow never becoming zero indicates a perennial river, i.e., one which does not become dry during the entire year. We know that it does not rain throughout the year, therefore the water in a perennial river during non rainfall periods must come from some other source. Typically, it comes from the baseflow and/or snowmelt in the upper reaches of the river. The annual hydrograph from a perennial river is known as a perennial hydrograph. On the other hand, we have rivers which run with water only during rain-periods and are dry at other times. Such rivers do not have any contribution from baseflow and the flow in them consists mainly of surface flow. The rivers flowing for very short periods, due to a rainfall event or snowmelt, are called ephemeral rivers and the hydrographs from them are called ephemeral hydrographs. There are rivers that have moderate amount of groundwater contributions and may become dry during some part of the year. These are called intermittent rivers and the hydrographs from them are known as intermittent hydrographs. Figure 4.3 shows the three types of annual hydrographs described here.

4.3.2

Storm Hydrograph

A storm hydrograph is a plot of streamflow versus time for limited time duration. This limited-time duration is equivalent to the time taken by the catchment in translating the inputs (rainfall) as outputs to the outlet of the catchment. A storm hydrograph (or simply hydrograph) is useful in understanding the dynamic behavior of a catchment when it is subjected to an intense short-duration rainfall event. A typical storm hydrograph is shown in Figure 4.3(d). AB and EF represent the baseflow recession, BC is the rising limb, CPD represents the crest portion, P is the peak flow, DE is the falling limb, TR is called storm duration, tpk is the time to peak flow, TB is the time base of the hydrograph, and TL is called the time lag or lag time of the catchment from which this hydrograph is produced. The time lag TL represents the maximum time taken by water to travel through the catchment (similar to time of concentration, discussed later). It is the time between the midpoint of the Effective Rainfall Hyetograph (ERH) and the point of inflexion (D) on the falling limb of the hydrograph. The point of inflexion denotes the beginning of the falling limb on which the flow is contributed from different storages of the catchment. The crest portion is the result of excess rainfall (ER). The line, or sometimes a curve, joining the two points B and E divides the hydrograph into two parts: an upper part called

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Engineering Hydrology

direct runoff hydrograph (DRH), and the lower part called the baseflow. The point B represents the beginning of DRH with the beginning of ER, while the point E denotes the end of DRH. The flow in the river is purely from baseflow contribution during the portions AB and EF, while the flow in the river during the portion BCPDE is a combined effect of baseflow and surface flow (or direct runoff).

Figure 4.3

Different types of hydrographs: (a) Perennial, (b) Intermittent, (c) Ephemeral, and (d) Storm hydrograph

� Rising Limb The rising limb of a hydrograph, also known as a concentration curve, is a result of gradual release of water from different storages of the catchment as they get filled up due to rain falling on the catchment. Initially, the catchment is on the drier side, hence there are more infiltration losses and less of runoff; therefore, the initial portions of the rising limb increase slowly or are mildly sloped. Later, as the infiltration losses decrease, there is more runoff resulting in greater concentration of flow on the rising limb, leading to a steep rise in the concentration curve. Initially, all portions of the catchment not contribute to the runoff at the outlet but with the passage of time, more and more portions of the catchment do contribute to the runoff at the outlet of the catchment. Close to the crest portion, the entire catchment contributes to the runoff at the outlet. It is important to note that the shape, size, and slope of the rising limb will be a function of both the catchment and storm characteristics. Other factors remaining same, a large sized catchment will produce flatter rising limbs as compared to small sized catchments, and a steeply sloped catchment will produce steep rising limbs with sharps peaks as compared to a mildly sloped catchment. Similarly, storms having greater rainfall intensities will produce steeply sloped rising limbs, and greater and early peak flows as compared to the DRHs resulting from the storms having lower rainfall intensities. Longer duration storms tend to produce

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107

milder rising limbs and greater time bases as compared to the case of short duration rainfall events of similar magnitudes. In addition, the nature of the rising limbs will also depend on the antecedent moisture conditions in the catchment just before the storm hits the catchment. A catchment with dry conditions will produce mildly sloped rising limbs with less magnitude and delayed peak flows as compared to the case when the catchment is close to saturation or with wet conditions. � Falling Limb The falling limb of a hydrograph is produced due to the gradual depletion of different storages of a catchment. The size, shape, and slope of the falling limb are a function of catchment characteristics only and do not depend on the storm characteristics. The falling limb, also known as recession curve, can be modeled as follows (Horton, 1933): Q(t ) = Q(t 0 ) e - K ( t - t0 )

(4.4)

where, Q(t) is the flow at time t on the falling limb, Q(t0) is the flow at time t0 on the falling limb (t > t0), and K is a recession coefficient for the catchment having dimensions of T –1. The recession coefficient can be determined using observed flow data on the falling limb. It is important to note that the above relationship is derived by assuming the catchment to behave linearly. In other words, the catchment is assumed to be a linear reservoir in which the catchment storage S, is linearly related to outflow, i.e., S = KQ. In addition to the annual and storm hydrographs, a hydrograph may also be characterized as a monthly hydrograph depicting daily mean flows over a month or a seasonal hydrograph showing daily mean flow over a season, e.g., monsoon season. While an annual hydrograph is useful in understanding the surface water potential in a catchment, reservoir planning, and in drought management studies, a storm hydrograph is useful for hydraulic design purposes. A seasonal or monthly hydrograph may be helpful in understanding and management of the monthly and seasonal water requirements in a catchment.

4.4

FACTORS AFFECTING RUNOFF

The generation of runoff in a catchment subjected to rainfall is an extremely LO 4 Summarize the complex process that gets affected by a multitude of factors. The factors impacts of various physical affecting the runoff response from a catchment include catchment, geofactors on runoff morphological, rainfall, and climatic characteristics. Depending on the practical utility, runoff may be considered over long time periods (monthly runoff, annual runoff etc.) or it may be taken at shorter time intervals (a day, an hour, or even smaller). For large time intervals, the total volume/depth of runoff is an important variable to be considered in water resources planning activities. While for shorter time intervals, the time distribution of runoff, in the form of runoff rates, peak runoff magnitude, time to peak runoff, and the shapes of the rising and falling limbs of a runoff hydrograph, are more important for hydrologic/hydraulic design. The impact of various physical factors on the runoff would be different depending on the time-scale considered. The effects of various factors on runoff are briefly discussed here.

4.4.1

Catchment Characteristics

A catchment may be characterized using several factors including size, slope, elevation, orientation, land-use/ land-cover, soil type, drainage network, etc. For the same amount of rainfall, a large sized catchment will

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result in greater magnitude of runoff as compared to a small sized catchment. In a large sized catchment, time of travel will be larger resulting in smaller and delayed peak runoffs as compared to those from a small sized catchment. A steeply sloped catchment would allow faster movement of water through it as compared to a mildly sloped catchment, leading to larger and earlier peak runoff for a rainfall event. The effect of slope on runoff volume is not as pronounced as its effect on the time distribution of runoff. The land use/land cover conditions in a catchment depend on the degree of urbanization. For example, an urbanized catchment is dominated by paved areas, in the form of buildings, pathways, parking lots, etc., that do not allow water to infiltrate into the ground. On the other hand, a rural catchment consists of a lot of vegetation, cropped areas and grasslands allowing easy infiltration. Therefore, an urbanized catchment will tend to produce hydrographs with higher peak flows, early peaks, and smaller time bases as compared to that from a rural catchment. The vertical movement of water is a function of the soil type. For example, a sandy soil allows greater and faster infiltration during a storm as compared to clayey soils. Catchments dominated by sandy soils would experience greater infiltrations leading to smaller magnitude and greater time to peak runoff as compared to a catchment dominated by clayey soils. Therefore, the shape of a hydrograph resulting from catchments consisting of different type of soils will have different characteristics. The surface runoff will also depend on the density of the channel network in a catchment. A catchment having a well-developed network of small and large channels has a greater capacity to transmit water quickly from different portions of the catchment to the outlet as compared to a catchment consisting of poorly developed channel network. Therefore, a catchment that has a well-developed network of natural channels will produce steeply sloped rising limbs in hydrographs, higher magnitudes of peak flows, and early peak flows as compared to the hydrographs produced from a catchment consisting of poorly defined network of natural channels. Lastly, it is important to note that a catchment can be thought of as consisting of several storage elements, for example, surface storage, soil storage, ground storage, etc. The runoff produced at the outlet of a catchment is essentially a function of how these different storage elements behave during a storm. As the different storages in a catchment get filled up during a storm event, they tend to release water from it depending on the type of storage. The size, shape, and nature of the runoff hydrograph will be a function of how the catchment releases water from its storages at different times.

4.4.2

Storm Characteristics

The magnitude and time distribution of runoff produced from a catchment would depend on the storm characteristics. A storm can be characterized by the total rainfall depth, intensity of rainfall, and duration of the rainfall event. The total volume of runoff produced from a catchment will obviously be higher from a rainfall event having higher total rainfall depth as compared to a rainfall event consisting of smaller total depth of rainfall. The time distribution of rainfall also plays an important role in deciding the size and shape of the runoff hydrograph produced. A storm consisting of greater rainfall intensity will produce a steep rising limb, greater peak magnitudes, and early peaks from a catchment as compared to a storm having low rainfall intensity. The duration of rainfall affects the total duration for which the runoff will be observed at the outlet of a catchment. It is important to note that while the catchment characteristics affect the shape of entire hydrograph including rising limb, crest portion, and falling limb, the storm characteristics affect only the rising limb of a hydrograph. This is due to the fact that after the rainfall has stopped, the water reaching the outlet of a catchment is essentially coming from different storage elements of the catchment.

Runoff

4.5

109

RAINFALL-RUNOFF RELATIONSHIPS

The transformation of rainfall into runoff in a catchment is an extremely LO 5 Develop rainfall-runoff complex, dynamic, and nonlinear physical process. A catchment is relationships using method of composed of several storage elements or components, e.g., interception least squares and depression storage, surface and channel storage, soil-moisture storage, and groundwater storage. As rain falls on the catchment, the water goes into these storage elements. As the rainfall continues, the capacities of these storage elements get satisfied slowly and water is gradually released from these storage elements. The spatio-temporal distribution and movement of rainwater through these storage elements dictates the shape, size, and time-distribution of the runoff observed at the outlet. There are several climatic and physiographic factors that affect the rainfall-runoff relationship, which were discussed in the previous section. Although the influence of these factors and many of their combinations in generating runoff is an extremely complex physical process and is not understood clearly, scientists have tried to capture the rainfall-runoff process in different types of mathematical models. All the rainfall-runoff models may be broadly classified into one of the following two types: (i) analytical or simulation models, and (ii) empirical or black-box models. Analytical models require the understanding of the physical processes involved. The empirical models are those in which observations on runoff and the physical variables affecting it are used to develop mathematical relationships without considering the mechanics of the physical processes involved.

4.5.1

Analytical Models

Analytical models, also known as deterministic or conceptual models, are based on the laws of physics and, therefore, are most widely accepted by traditional hydrologists. A conceptual model captures the transformation of rainfall into runoff through the use of several interconnected storage elements, each representing a certain component of the overall rainfall-runoff process. The origin of conceptual modeling of the rainfall-runoff process may be traced to the event based rainfall-runoff model developed by Mulvany (1850), called the Rational Method. Later, Sherman (1932) proposed the concept of unit hydrograph. He assumed that the runoff process was linear and time invariant. The unit hydrograph and its subsequent evolution to the instantaneous unit hydrograph provided a basis for the storm response models of Nash (1957) and Dooge (1959). Stanford Watershed Model (SWM), developed by Crawford and Linsley (1966), was probably the first conceptual catchment simulation model which was capable of producing streamflow forecasts on a continuous basis. This is a deterministic model that uses precipitation and potential evapotranspiration as meteorological inputs and produces time variant predictions of streamflow and groundwater storage as outputs. Its modified version, known as Kentucky Watershed Model (KWM), was proposed by Liou (1970) by including automatic calibration capability. The Storm Water Management Model (SWMM) developed by Metcalf and Eddy (1971), is capable of simulating both event-based and continuous hydrologic process. Hydrologic Engineering Centre (1990) developed the HEC-1 flood hydrograph model which is a very widely used rainfall-runoff model. This computer package can be used to simulate a direct runoff hydrograph from a catchment by representing the catchment with interconnected hydrologic and hydraulic components. The Rainfall-Runoff Library (RRL) (Podger, 2004) is a freely available computer model tool for rainfall-runoff modeling, consisting of five different conceptual models. The modern conceptual models come with a lot of options for modeling different components of the hydrologic cycle in terms of objective function and

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calibration methods. Detailed discussion of these computer simulation models is out of the scope for this book, and interested readers may consult the suggested references provided at the end of this book.

4.5.2

Empirical Models

Analytical models are complex, difficult to develop, require a lot of data for their development, and require a thorough understanding of the physical sub-processes involved. Alternatively, empirical models have been proposed that are easy to develop and use in the field. The empirical models may be said to have some conceptual basis in the sense that they are developed using the input data that affect the output of a physical process. The factors affecting runoff from a catchment may include a wide range of climatic and topographical factors; however, the runoff (R) from a catchment is mainly a function of the precipitation depth (P). The general form of an empirical model for representing a rainfall-runoff relationship in a catchment can be represented by the following equation: R = f (P)

(4.5)

Equation (4.5) represents a mathematical equivalent of the rainfall-relationship in a catchment. Apart from the two variables (P and R), the equation involves model coefficients depending on the structure and form of the model that need to be estimated. Data in terms of both R and P are required for the calibration of an empirical model. Since the process of transformation of precipitation into runoff is a nonlinear process, a nonlinear functional relationship ( f ) would be able to capture the complexities inherent in the data in a better manner. However, linear approximations of the rainfall-runoff relationships are quite common, as they are easy to develop, use, and provide reasonably accurate results. Several empirical models for different parts of India have been proposed in the past in the form of equations and tables, for example, Strange’s Tables, Barlow’s Tables, Binnie’s Percentages, Inglis DeSouza, and Khosla’s Formulas, etc. The details of these models can be found in Subramanya (2013). Here, we will discuss the use of regression analysis for the development of a simple empirical model, given the data in terms of rainfall and runoff during a certain time period. The time period for such models is normally a season or a year. Normally, a straight line of the following form is fitted to the data R and P: R = mP + c

(4.6)

where, m is the slope of the fitted line and c is the intercept, and these coefficients of the regression model are to be determined through calibration. Given the data on R and P, the method of least squares can be used to estimate the regression coefficients, m and c. In the method of least squares, the sum of squares of the departures between the estimated and observed values of R is minimized. Let Rˆ i be the estimated value of the runoff for known values of P, m, and c, it can be estimated as follows: Rˆ i = mPi + c

(4.7)

The departure between the estimated and observed value for the ith data point (Ri), called error in estimating runoff, can then be written as: ei = Ri - Rˆ i

(4.8)

Runoff

111

The total error from the regression model, which is to be minimized, can be written as follows: n

E=

Âei2

(4.9)

i =1

where, n is the total number of data points available and i is an index running from 1 to n. In the method of least squares, the objective is to determine the values of the regression coefficients (m and c in this case) such that the value of E is minimum for a given dataset (P, R). The minimum of a function (E) can be found by differentiating it with respect to the unknowns (m and c) and equating to zero, because at the minimum value of a function, its slope must be zero. Differentiating the global error E with respect to the unknowns and equating to zero results in what is known as the normal equations. The normal equations are linear in nature and their solution results in the estimation of the regression coefficients. There will be as many normal equations as the number of regression coefficients. In the present case, there are two regression coefficients (m and c) to be determined; hence there will be two normal equations. One normal equation will result when E is differentiated with respect to c and equated to zero, and the second normal equation will result when E is differentiated with respect to m and equated to zero. The two normal equations can be represented as follows: SR = n c + m SP

(4.10)

SPR = c SP + m SP2

(4.11)

In these equation, the summations run from 1 to n (not shown). These are two simultaneous linear equations, which can be easily solved for the two regression coefficients m and c to obtain: m=

n SPR - SP SR n SP 2 - ( SP ) 2

c=

SR - m SP n

(4.12)

(4.13)

Once the regression model has been developed, the question arises that how good or bad the model is. The quality of the regression model can be evaluated graphically using a scatter plot. If the observed runoff (R) is plotted on x-axis and the predicted runoff ( Rˆ ) is plotted on y-axis, the trend should follow a straight line at unit slope because the value of Rˆ should ideally be equal to R. Statistically, the goodness of the model is usually quantified by the coefficient of correlation (r). Coefficient of correlation is a measure of the strength of linear relationship between two variables R and Rˆ . The value of r may be calculated using the following expression: Correlation Coefficient (r):

r=

n SRRˆ - SR SRˆ Èn SR 2 - ( SR)2 ˘ Èn SRˆ 2 - ( SRˆ )2 ˘ Î ˚Î ˚

(4.14)

The value of r lies between −1.0 and 1.0, with values of r close to 1.0 indicating a very good model; the values of r close to 0.0 indicate a bad model; while the negative values of r imply that the two variables are negatively correlated, i.e., as R increases, Rˆ decreases (this is, naturally, not applicable here since the predicted value of R must increase with observed values of P or R).

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Equation (4.14) can also be used to assess the strength of a linear relationship between P and R as raw data. Should the value of r between P and R & that between R and Rˆ be same for the linear relationship such as the one represented by Eq. (4.6) is discussed later. As mentioned earlier, the rainfall-runoff relationship in a catchment is complex and nonlinear. Therefore, a nonlinear regression model would be able to capture the complexities inherent in the rainfall-runoff process in a better manner. A power regression model of the following form is normally developed as the rainfallrunoff relationship: R = e + aPb

(4.15)

The process of development of a regression model of the form represented by Eq. (4.15) is essentially the same as described above except that the structural form of the rainfall-relationship is first converted to a linear form by taking logarithm (either natural or to the base 10). ln(R – e) = ln(a) + b ln(P)

(4.16)

Letting Y = ln(R – e); c = ln(a); m = b; and X = ln(P), Eq. (4.16) essentially reduces to the form Y = mX + c, which is same as Eq. (4.6). The value of parameter e can be determined by trial and error. Alternatively, spreadsheet software (e.g., the solver module of Microsoft Excel) could be used to directly obtain the parameters of Eq. (4.15). � EXAMPLE 4.1 The seasonal precipitation and runoff data at a location in a hypothetical river basin are provided in the following table. Develop a linear and a power regression model (assume parameter e = 0.) using these data. Calculate the coefficient of correlation for the raw data (R and P). Also calculate the coefficient of correlation for the linear and nonlinear models. Which model would you recommend? Precipitation (mm)

116.1

484.0

923.1

438.8

765.7

464.5

238.0

115.5

Runoff (mm)

49.1

238.4

477.2

229.8

247.4

590.9

194.7

34.8

Solution Calculation of the coefficient of correlation for the raw data presented in the table above is shown in Table 4.1. From the Table 4.1, n = 8, SP = 3,545.7, SR = 2,062.3, SP2 = 2,164,435.3, SR2 = 789,262.0, and SPR = 1,176,690.9. Putting these values in Eq. (4.14) we get, r=

8(1,176,690.9) - (3,545.7)(2,062.3) È8(2,164, 435.3) - (3,545.7)2 ˘ È8(789,262.0) - (2,062.3)2 ˘ Î ˚Î ˚

= 0.6720

For the linear relationship, the regression coefficients m and c can be calculated from Eq. (4.12) and Eq. (4.13), respectively. m=

8(1,176,690.9) - (3545.7)(2062.3) 8(2,164, 435.3) - (3545.7)2

= 0.4430

Runoff

113

Table 4.1 Calculations for coefficient of correlation for raw data and regression coefficients for linear model in Example 4.1 S. No.

Precipitation P (mm)

Runoff R (mm)

P2

R2

PR

2410.8

5700.5

1

116.1

49.1

13,479.2

2

484.0

238.4

234,256.0

56,834.6

115,385.6

3

923.1

477.2

852,113.6

227,719.8

440,503.3

4

438.8

229.8

192,545.4

52,808.0

100,836.2

5

765.7

247.4

586,296.5

61,206.8

189,434.2

6

464.5

590.9

215,760.3

349,162.8

274,473.1

7

238.0

194.7

56,644.0

37,908.1

46,338.6

8

115.5

34.8

13,340.3

1211.0

4019.4

Sum

3545.7

2062.3

2,164,435.3

789,262.0

1,176,690.9

c=

2062.3 - (0.4430)(3545.7) = 61.5 8

The linear rainfall-runoff relationship for the example problem can thus be represented by the equation R = 61.5 + 0.443 P. Calculation of the coefficient of correlation for the linear rainfall-runoff relationship is shown in Table 4.2. From the table, n = 8, SR = 2062.3, SRˆ = 2062.3, SR2 = 789,262.0, SRˆ 2 = 647,983.1, and SRRˆ = 647,983.1. Putting these values in Eq. (4.14) we get, r=

8(647,983.1) - (2062.3)(2062.3) È8(789,262.0) - (2062.3)2 ˘ È8(647,983.1) - (2062.3)2 ˘ Î ˚Î ˚

= 0.6720

which is same as the r value obtained earlier between P and R. It is because the estimated value of runoff ( Rˆ ) is nothing but a linear combination of P and R.

Table 4.2 Calculations for coefficient of correlation for linear regression model S. No.

Precipitation P (mm)

Runoff R (mm)

Fitted Runoff R (mm)

1 2

116.1

49.1

112.9

2410.8

12,743.3

5542.7

484.0

238.4

275.9

56,834.6

76,096.1

65,763.9

3

923.1

477.2

470.4

227,719.8

221,242.1

224,457.6

4

438.8

229.8

255.8

52,808.0

65,450.5

58,790.4

5

765.7

247.4

400.6

61,206.8

160,512.5

99,118.4

6

464.5

590.9

267.2

349,162.8

71,405.1

157,898.7

7

238.0

194.7

166.9

37,908.1

27,850.4

32,492.4

8

115.5

Sum

R2

R2

R×R

34.8

112.6

1211.0

12,683.3

3919.2

2062.3

2062.3

789,262.0

647,983.2

647,983.2

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Engineering Hydrology

Calculations for the regression coefficients for the nonlinear relationship are shown in Table 4.3. From the table, n = 8, Sln(P) = 46.9, Sln(R) = 41.7, S{ln(P)}2 = 278.8, Sln(R)2 = 224.2, and Sln(P)ln(R) = 249.0. For the nonlinear relationship, the regression coefficients m and c can be calculated from Eq. (4.12) and Eq. (4.13), respectively. It must be noted that m and c refer to the log-transformed data series in this case.

Table 4.3 Calculations for regression coefficients for nonlinear model in Example 4.1 S. No.

ln(P) (X)

ln(R) (Y)

{ln(P)}2

{ln(R)}2

ln(P) × ln(R)

1

4.75

3.89

22.60

15.16

18.51

2

6.18

5.47

38.22

29.96

33.84

3

6.83

6.17

46.62

38.04

42.11

4

6.08

5.44

37.02

29.56

33.08

5

6.64

5.51

44.10

30.37

36.60

6

6.14

6.38

37.71

40.73

39.19

7

5.47

5.27

29.95

27.79

28.85

8

4.75

3.55

22.56

12.60

16.86

Sum

46.9

41.7

278.8

224.2

249.0

m= c=

8(249.0) - (46.9)(41.7) 8(278.8) - (46.9)2

= 1.118

41.7 - (1.118)(46.9) = - 1.337 8

Therefore, m = b = 1.118, c = ln(a) fi a = ec = e–1.337 = 0.267. The nonlinear relationship can be represented by R = 0.267P1.118. Calculation for the coefficient of correlation for a nonlinear rainfall-runoff relationship is shown in Table 4.4.

Table 4.4 Calculations for coefficient of correlation for a nonlinear regression model S. No.

Precipitation P (mm)

1 2 3 4 5 6 7 8

116.1 484.0 923.1 438.8 765.7 464.5 238.0 115.5

Sum

Runoff R (mm)

Fitted Runoff R (mm)

R2

R2

R×R

49.1 238.4 477.2 229.8 247.4 590.9 194.7 34.8

53.5 263.7 542.8 236.3 440.4 251.9 119.3 53.1

2410.8 56,834.6 227,719.8 52,808.0 61,206.8 349,162.8 37,908.1 1211.0

2857.2 69,548.8 294,624.7 55,857.9 193,967.4 63,438.9 14,223.4 2824.3

2624.5 62,871.1 259,021.0 54,311.5 108,959.3 148,830.5 23,220.3 1849.4

2062.3

1961.0

789,262.0

697,342.8

661,687.7

Runoff

115

From the Table 4.4, n = 8, SR = 2062.3, SRˆ = 1961.0, SR2 = 789,262.0, SRˆ 2 = 697,342.8, and ˆ SRR = 661,687.7. Putting these values in Eq. (4.14) we get, r=

8(661,687.7) - (2,062.3)(1,961.0) È8(789,262.0) - (2,062.3)2 ˘ È8(697,342.8) - (1,961.0)2 ˘ Î ˚Î ˚

= 0.6610

which is slightly less than that for the linear relationship. Therefore, based on the coefficient of correlation statistic, a linear model is recommended as the rainfall-runoff relationship for the given dataset.

4.6

RATIONAL FORMULA

While driving on a highway, we frequently come across a structure LO 6 Estimate peak runoff using below the roadway, which consists of one or more pipes laid Rational formula across the roadway to carry drainage water from one side of the road to the other. This structure is known as a culvert or pulia. The size of the pipes is such that the total area of the opening is capable of passing the peak flood of certain magnitude through the culvert. The peak flood magnitude at the location of the culvert would depend on several factors, of which the two most important ones are the area of the catchment at the culvert location (let us call it outlet) and the maximum intensity of rainfall that is expected in the catchment contributing to runoff at the outlet. As mentioned in Chapter 2 the maximum intensity of rainfall in a catchment is a function of the duration of the rainfall and the frequency of occurrence of a storm event. The expected maximum rainfall intensity increases with a decrease in the duration of rainfall and with an increase in the return period associated. The Intensity Duration Frequency (IDF) relationships, discussed later, can be used to determine the maximum rainfall intensity expected. The magnitude of peak flood for which a culvert or any other hydraulic structure needs to be designed, is calculated using the ‘Rational’ formula. The Rational formula, proposed by Mulvany in 1850, is used even today for calculating the peak discharge for drainage design purposes. It can be mathematically expressed as follows: QP = CID,T A (4.17) where, Qp is the peak flow, ID, T is the maximum rainfall intensity corresponding to a duration (D) and design return interval (T), A is the area of the catchment draining water at the outlet, and C (also called runoff coefficient) is a factor accounting for all other catchment characteristics, e.g., slope, land-use, land-cover conditions, losses due to infiltration, etc. The duration D should be at least equal to the time of concentration of the catchment at the outlet. If ID, T is in mm/hour and A is in km2, then Eq. (4.17) may be written as: QP =

1 CI A 3.6 D,T

(4.18)

where, QP is in m3/s and C is a dimensionless runoff coefficient having value less than 1.0.

4.6.1 Runoff Coefficient (C) The runoff coefficient (C) represents the ratio of runoff to rainfall in a catchment. Obviously, its value will range between 0.0 and 1.0. It depends on several physical factors including interception and depression

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storage characteristics of the catchment, infiltration, initial soil moisture conditions, soil type, land use, land cover conditions and the slope of the catchment. Therefore, it is difficult to estimate, even for a homogeneous catchment consisting of a single type of soil and land use pattern. A catchment with low infiltration and/or steep slopes will have high value of C, while a catchment predominantly consisting of vegetated or cultivated areas will have low values of C. Some representative values of runoff coefficients for different types of land use and land cover conditions are presented in Table 4.5.

Table 4.5 Values of runoff coefficients Land Use

C

Business Downtown areas Neighborhood areas

Residential Single-family areas Multi units, detached Multi units, attached Suburban

0.70 – 0.95 0.50 – 0.70

0.30 – 0.50 0.40 – 0.60 0.60 – 0.75 0.25 – 0.40

Industrial Light areas Heavy areas

0.50 – 0.80 0.60 – 0.90

Land Use

C

Lawns Sandy soil, flat: 2% Sandy soil, average: 2–7% Sandy soil, steep: 7% Heavy soil, flat: 2% Heavy soil, average: 2–7% Heavy soil, steep: 7%

0.05 – 0.10 0.10 – 0.15 0.15 – 0.20 0.13 – 0.17 0.18 – 0.22 0.25 – 0.35

Agricultural land Bare packed soil Smooth Rough Cultivated rows Heavy soil, no crop Heavy soil, with crop Sandy soil, no crop Sandy soil, with crop Pasture Heavy soil Sandy soil Woodlands

0.30 – 0.60 0.20 – 0.50 0.30 – 0.60 0.20 – 0.50 0.20 – 0.40 0.10 – 0.25 0.15 – 0.45 0.05 – 0.25 0.05 – 0.25

Streets Asphaltic Concrete Brick

0.70 – 0.95 0.80 – 0.95 0.70 – 0.85

Parks, cemeteries

0.10 – 0.25

Unimproved areas

0.10 – 0.30

Playgrounds

0.20 – 0.35

Drives and walks

0.75 – 0.85

Railroad yard areas

0.20 – 0.40

Roofs

0.75 – 0.95

In real life, no catchment will be homogeneous consisting of a single type of land use and land cover conditions. For a heterogeneous catchment consisting of different types of usage, the following weighted averaging for runoff coefficient may be employed: Ceq =

C1 A1 + C2 A2 + C3 A3 + � + Cn An A1 + A2 + A3 + � + An

(4.19)

where, C’s are the runoff coefficient values of individual sub-catchments having homogeneous characteristics and A’s are the corresponding areas.

Runoff

4.6.2

117

Intensity of Rainfall (ID, T)

There are several formulae developed for calculating maximum rainfall intensity that may occur in a specific area corresponding to a certain duration and frequency of occurrence. These relationships are called IntensityDuration-Frequency (IDF) relationships and are developed using the rainfall data taken from a specific area. The data for different durations are analyzed and some probability distributions are fitted to the data. The most commonly used probability distribution used for developing IDF relationships is the Gumbel’s distribution. Rambabu et al., (1979) have calculated the maximum rainfall intensity at different places in India for different durations and return periods and proposed the following formula: I D,T =

KTx

(4.20) ( D + a )n where, T is the return period (years), D is the duration of the storm (hours), I is the maximum rainfall intensity in an area (cm/hour), and K, x, a, n are coefficients to be determined. Rambabu et al., (1979) have employed the extensive rain gauge network of the India Meteorological Department (IMD) to estimate the values of these coefficients in different parts of the country. The representative values of these coefficients for different locations in India are presented in Table 4.6.

Table 4.6 Values of coefficients for use in Equation (4.20) Zone

Location

K

x

a

n

Central

Bhopal Nagpur Raipur Average

6.9296 11.45 4.683 7.4645

0.1892 0.1560 0.1389 0.1712

0.50 1.25 0.15 0.75

0.8767 1.0324 0.9284 0.9599

Northern

Allahabad Amritsar Dehradun Jodhpur Srinagar Average

4.911 14.41 6.00 4.098 1.503 5.914

0.1667 0.1304 0.2200 0.1677 0.2730 0.1623

0.25 1.40 0.50 0.50 0.25 0.50

0.6293 1.2963 0.8000 1.0369 1.0636 1.0127

Eastern

Agartala Kolkata Guwahati Jharsuguda Average

8.097 5.940 7.206 8.595 6.933

0.1177 0.1150 0.1157 0.1392 0.1353

0.50 0.15 0.75 0.75 0.50

0.8191 0.9241 0.9401 0.8740 0.8801

Western

Aurangabad Bhuj Veraval Average

6.081 3.823 7.787 3.974

0.1459 0.1919 0.2087 0.1647

0.50 0.25 0.50 0.15

1.0923 0.9902 0.8908 0.7327

Southern

Bangalore Hyderabad Chennai Trivandrum Average

6.275 5.250 6.126 6.762 6.311

0.1262 0.1354 0.1664 0.1536 0.1523

0.50 0.50 0.50 0.50 0.50

1.1280 1.0295 0.8027 0.8158 0.9465

[Source: Rambabu et al., 1979]

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For calculation of the peak flow by Rational formula, the value of D in Eq. (4.20) should be taken at least equal to the time of concentration of the catchment. The time of concentration of a catchment is the maximum time taken by water in traveling from the farthest location to the outlet of the catchment. As rainfall of a constant intensity falls on a catchment, initially the storage requirements of the catchment are fulfilled. Once the catchment is fully saturated, the rainfall occurring on the catchment simply travels through the catchment and reaches the outlet. In other words, the rainfall volume falling on different parts of the catchment go through pure translation and arrive at the outlet. The rainfall falling on the subareas of the catchment close to the outlet will reach the outlet earlier and the rainfall falling on subareas farther from the outlet will reach the outlet later. Therefore, the rate of runoff at the outlet will initially increase with time and as more and more portions of the catchment start contributing to the runoff at the outlet, the rate of runoff will keep on increasing till the point when whole of the catchment starts contributing to the runoff at the outlet. The catchment will produce the maximum amount of runoff when the entire catchment is contributing to runoff at the outlet. This happens when a rainfall continues for duration equal to the time of concentration of the catchment.

4.6.3 Time of Concentration The time of concentration is an important parameter in the calculation of the maximum intensity of rainfall expected at a location. It would be a function of several factors, e.g., area of the catchment, maximum length of the catchment, steepness of the catchment, surface conditions of the catchment, etc. It is extremely difficult to account for each and every factor while estimating the value of time of concentration in a catchment. Researchers have tried to develop formulae for estimating the time of concentration using the topographic, rainfall, and flow data. Although there are several methods available for estimating the time of concentration of a catchment, the one proposed by Kirpich (1940) is the most popular. The Kirpich’s formula takes into account two main factors, maximum length of travel and the slope of the catchment, and gives the time of concentration as follows: tc = 0.01947

L0.77 S 0.385

(4.21)

where, tc is the time of concentration for a catchment (minutes), L is the maximum length of travel for water (m), and S is the average slope of the catchment (dimensionless), taken as the ratio of the elevation difference between the farthest location and outlet to L. It is clear from Eq. (4.21) that the time of concentration will be higher for a catchment having greater length of travel and flatly sloped catchments and vice-versa. � EXAMPLE 4.2 A small suburban area of Allahabad is found to have the following data: Area of catchment = 240 Ha; Maximum length of travel for water = 750 m; Elevation of farthest location = 103.25 m; Elevation of outlet = 101.85 m. The catchment consists of lawns on sandy soils, flat (30%), single family residential areas (40%), light industry (20%), and roads and pavements (pakka) (10%). Calculate the peak discharge values at the outlet of the area for return periods of 1-year, 10-year, and 25-year.

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119

Solution For Allahabad region, from Table 4.6, K = 4.911, x = 0.1667, a = 0.25, and n = 0.6293. Time of Concentration The average slope of the catchment S = (103.25-101.85)/750 = 0.00187 (m/m). Using Kirpich’s equation, we obtain: tc = 0.01947

7500.77 0.001870.385

= 35.8 mins

Runoff Coefficient The values of runoff coefficients for different types of land usage in the catchment are taken from Table 4.5 and the subarea calculations are summarized in a tabular form as follows. The average runoff coefficient values of the range given in Table 4.5 are taken, except for roads and pavements for which a higher value is taken because of pakka roads. Type of Land Usage

Runoff Coefficient

Area Fraction

Sub-Area (sq.m.)

0.075

0.3

72,000

Single family residential areas (40%)

0.4

0.4

96,000

Light industry (20%)

0.65

0.2

48,000

Roads and pavements, pakka (10%)

0.9

0.1

24,000

Lawns on sandy soils, flat (30%)

Note that total catchment area is 24 hectares = 240,000 m2. The equivalent runoff coefficient is then calculated as follows: Ceq =

0.075(72,000) + 0.40(96,000) + 0.65(48,000) + 0.90(24,000) = 0.4025 240,000

Peak Discharge The value of maximum rainfall intensity is calculated from Eq. (4.20) for D = 35.8 min and the peak discharge values are calculated from Eq. (4.18) for different return periods. The calculations are summarized in a tabular form as follows: T (years) Intensity (mm/hr) Discharge (cumecs)

4.7

1 44.2 11.9

10 64.9 17.4

25 75.6 20.3

SCS METHOD

The Rational method discussed above can be used to calculate LO 7 Evaluate runoff volume using SCS the design discharge value for which the drainage system needs curve number method to be designed. However, in an urban area, it is desirable to temporarily store flood waters in flood detention basins in order to account for extremely rare rainfall events. For the construction of the flood detention basins and in other flood control and management studies, it is important to estimate the runoff volume that is expected to be generated when a design storm event occurs

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in the area under consideration. The Soil Conservation Service (SCS) of the United States has developed a method called SCS Curve Number (CN) method for estimating the runoff volume produced from a design rainfall event. The SCS-CN method can be used to estimate effective rainfall, given the total rainfall depth during a storm. Let the total rainfall depth during a storm be P and the effective rainfall be ER. The effective rainfall in the catchment represents the depth of direct runoff (DR), therefore, DR = ER. The SCS method is based on two concepts: (a) the law of conservation of mass, and (b) ratio of actual and potential runoff being equal to the ratio of actual and potential retention in the catchment. Let the potential retention (or storage) in a catchment be S, actual retention or cumulative infiltration during the storm be Fa, and initial abstraction due to interception and depression etc., be Ia. Figure 4.4 shows a rainfall hyetograph for the rainfall storm under consideration along with various variables defined above. Some of the rainfall falling in a catchment is lost in the form of interception and depression and infiltration, and the rest becomes runoff (equivalent to ER). In terms of the variables defined above, the continuity equation may be written as follows:

Figure 4.4

SCS curve number method for effective rainfall

P = Ia + ER + Fa

(4.22)

The portion of total rainfall that is abstracted as initial abstraction is not available for runoff. The portion (P – Ia) is potentially available for runoff but some of it infiltrates (Fa) and the remaining becomes runoff (ER). Therefore, the actual runoff is ER and potential runoff is (P – Ia). The amount of actual cumulative infiltration (or actual retention) is Fa and let the maximum possible retention in the catchment be S. Then based on the second concept of the SCS-CN method, Actual Ruoff Actual Retention = Potential Runoff Potential Retention

(4.23)

In terms of various variables explained above, Eq. (4.23) may be written as follows: F ER = a P - Ia S

(4.24)

The two concepts of the SCS-CN method in the form of Eq. (4.23) and Eq. (4.24) can be combined to yield the following expression for ER: ER =

( P - I a )2 P - Ia + S

(4.25)

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121

The initial abstraction ranges from 0.1S to 0.3S and is very difficult to estimate. In the absence of any information, its value may be taken as 0.2S. For Ia = 0.2S, Eq. (4.25) may be written as ER =

( P - 0.2 S )2 P + 0.8S

(4.26)

Equation (4.26) can be used to estimate runoff volume for a storm of given depth. However, the parameter S, which represents potential storage in a catchment, needs to be known. The maximum amount of water a catchment can store will depend on several factors including the soil type, land use, land cover, and the initial moisture conditions in the catchment. The SCS has defined the values of S for different types of soil, land use, land cover conditions, and the initial moisture conditions in the catchment. The SCS has developed a relationship between S and a parameter called Curve Number (CN). The original relationship developed by the SCS was in FPS system of units and its SI version can be expressed as follows: Ï 1000 ¸ S = 25.4 Ì - 10 ˝ CN Ó ˛

(4.27)

where, S is the potential maximum depth of retention of the catchment (mm) and CN is a dimensionless parameter whose value theoretically ranges between 0 and 100. The values of CN close to 0 indicate a highly pervious catchment and values of CN close to 100 indicate highly impervious catchment. It is clear from Eq. (4.27) that for CN = 100, S = 0, i.e., for a completely impervious surface there will be no retention/infiltration and the entire rainfall will become runoff. On the other hand, for CN = 0, S = ∞, i.e., for a completely pervious surface the maximum possible storage is infinity and there will be no runoff from the catchment. Obviously, for real catchments, the value of CN will be somewhere in between. In reality, its value ranges between 5 and 98 depending on the soil type and surface cover conditions in the catchment. The SCS has specified the values of CN for different types of soils and land use-land cover conditions. For the purpose of curve numbers specification, the SCS has classified different soils into four broad types: Soil Type A, Soil Type B, Soil Type C, and Soil Type D. These, along with their characteristics, are presented in the table below.

Table 4.7 Soil classification and characteristics for SCS-CM method Soil Type

Infiltration Capacity

Runoff Potential

Examples

A

High

Low

B

Moderate

Low to Moderate

Deep sands and loess, gravels, and aggregated silts Shallow loess, sandy loam, red sand/loam/sandy loam

C

Moderate to Low

Moderate

Clay loam, shallow sandy loam, soils with high clay and organic content

D

Low

High

Heavy plastic clay, deep black soils, certain saline soils

For catchments consisting of more than one type of soils and/or several surface conditions, a composite CN can be used, similar to Eq. (4.19). Table 4.8a through Table 4.8c presents the CN values for different types of soil type and land use conditions (as per TR55, 1986).

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Table 4.8a Curve numbers for urban areas1 Cover description

Curve numbers for hydrologic soil group Average percent impervious area2

Cover type and hydrologic condition

A

B

C

D

Poor condition (grass cover < 50%)

68

79

86

89

Fair condition (glass cover 50% to 75%)

49

69

79

84

Good condition (grass cover > 75%)

39

61

74

80

Paved parking lots, roofs, driveways, etc. (excluding right-of-way)

98

98

98

98

Fully developed urban areas (vegetation established) Open space (lawns, parks, golf courses, cemeteries, etc.)3:

Impervious areas:

Streets and roads: Paved; curbs and storm sewers (excluding right-of-way)

98

98

98

98

Paved; open ditches (including right-of-way)

83

89

92

93

Gravel (including right-of-way)

76

85

89

91

Dirt (including right-of-way)

72

82

87

89

Natural desert landscaping (pervious areas onlv)4

63

77

85

88

Artificial desert landscaping (impervious weed barrier, desert shrub with 1- to 2-inch sand or gravel mulch and basin borders)

96

96

96

96

Western desert urban areas:

Urban districts: Commercial and business

85

89

92

94

95

Industrial

72

81

88

91

93

Residential districts by average lot size: 1/8 acre or less (town houses)

65

77

85

90

92

1/4 acre

38

61

75

83

87

1/3 acre

30

57

72

81

86

1/2 acre

25

54

70

80

85

1 acre

20

51

68

79

84

2 acres

12

46

65

77

82

77

86

91

94

Developing urban areas Newly graded areas (pervious areas only, no vegetation)4 1

Average runoff condition, and Ia = 0.2S. The average percent impervious area shown was used to develop the composite CN’s. Other assumptions are as follows: impervious areas are directly connected to the drainage system, impervious areas have a CN of 98, and pervious areas are considered equivalent to open space in good hydrologic condition. 3 CN’s shown are equivalent to those of pasture. Composite CN’s may be computed for other combinations of open space cover type. 4 The pervious area CN’s are assumed equivalent to desert shrub in poor hydrologic condition. 2

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Table 4.8b Curve numbers for cultivated agricultural land1 Cover description 2

Cover type

Treatment

Fallow

Bare soil Crop residue cover (CR)

Row crops

Straight row (SR) SR + CR Contoured (C) C + CR Contoured & terraced (C&T) C&T+ CR

Small grain

SR SR + CR C C + CR C&T C&T+ CR

Close-seeded or broadcast legumes or rotation meadow

SR C C&T

Curve numbers for hydrologic soil group Hydrologic condition3 — Poor Good Poor Good Poor Good Poor Good Poor Good Poor

A

B

C

D

77 76 74 72 67 71 64 70 65 69 64 66

86 85 83 81 78 80 75 79 75 78 74 74

91 90 88 88 85 87 82 84 82 83 81 80

94 93 90 91 89 90 85 88 86 87 85 82

Good Poor Good Poor Good Poor Good Poor Good Poor Good Poor Good Poor Good Poor Good Poor Good Poor Good

62 65 61 65 63 64 60 63 61 62 60 61 59 60 58 66 58 64 55 63 51

71 73 70 76 75 75 72 74 73 73 72 72 70 71 69 77 72 75 69 73 67

78 79 77 84 83 83 80 82 81 81 80 79 78 78 77 85 81 83 78 80 76

81 81 80 88 87 86 84 &5 84 84 83 82 81 81 80 89 85 85 83 83 80

1

Average runoff condition, and Ia = 0.2S Crop residue cover applies only if residue is on at least 5% of the surface throughout the year. 3 Hydraulic condition is based on combination factors that affect infiltration and runoff, including (a) density and canopy of vegetative areas, (b) amount of year-round cover, (c) amount of grass or close-seeded legumes, (d) percent of residue cover on the land surface (good > 20%), and (e) degree of surface roughness. Poor: Factors impair infiltration and tend to increase runoff. 2

Good: Factors encourage average and better than average infiltration and tend to decrease runoff.

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Engineering Hydrology

Table 4.8c Curve numbers for cultivated agricultural land1 Cover description Cover type Pasture, grassland, or range—continuous forage for grazing.2 Meadow—continuous grass, protected from grazing and generally mowed for hay. Brush—brush-weed-grass mixture with brush the major element.3

Woods—grass combination (orchard or tree farm).5 Woods.6

Farmsteads—buildings, lanes, driveways, and surrounding lots.

Curve numbers for hydrologic soil group Hydrologic condition2 Poor Fair Good

A

B

C

D

68 49 39

79 69 61

86 79 74

89 84 80

—

30

58

71

78

Poor Fair Good Poor Fair Good Poor Fair Good

48 35 30 4 57 43 32 45 36 30 4

67 56 48 73 65 58 66 60 55

77 70 65 82 76 72 77 73 70

83 77 73 86 82 79 83 79 77

—

50

74

82

86

1

Average runoff condition, and Ia = 0.2S. Poor: 75% ground cover and lightly or only occasionally grazed. 3 Poor. 75% ground cover. 4 Actual curve number is less than 30; use CN = 30 for runoff computations. 5 CN’s shown were computed for areas with 50% woods and 50% grass (pasture) cover. Other combinations of conditions may be computed from the CN’s for woods and pasture. 6 Poor: Forest litter, small trees, and brush are destroyed by heavy grazing or regular blurring. Fair: Woods are grazed but not burned, and some forest litter covers the soil. Good: Woods are protected from grazing, and litter and brush adequately cover the soil. 2

The curve numbers specified in these tables take care of two factors that affect runoff from a catchment, soil type and surface conditions. Consider a catchment subjected to the same rainfall storm at two different times–in the first case, the catchment is close to saturation (say during monsoon period), and the other case when the catchment is close to dry conditions (say during peak summer heat of May/early June). The amount of runoff from the same catchment subjected to same rainfall storm at two different times will not be same, because the amount of infiltration during the dry conditions will be much more than that during the wet conditions. As a result, the amount of runoff during dry conditions will be less than that during the wet conditions. That means that the amount of runoff produced from a catchment will also be a function of its initial soil moisture conditions. The SCS-CN method accounts for the initial moisture conditions in a catchment at the beginning of the rainfall event in the form of another variable called Antecedent Moisture Condition (AMC). The SCS has classified three types of AMCs—AMC-I for dry conditions, AMC-II for normal conditions, and AMC-III for wet conditions. Although several climatic factors affect the moisture conditions in a catchment, SCS has classified the three AMC conditions based on the cumulative rainfall for

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125

the past five-day-period. This classification is given in Table 4.9. Note that water requirement of crops is higher during the growing season, hence higher amounts of rainfall for the same AMC is specified.

Table 4.9 AMC classification in SCS-CN method AMC Type

Cumulative rainfall over past five days (mm) Dormant Season

Growing Season

I

< 13

< 36

II

13 to 28

36 to 53

III

> 28

> 53

The curve numbers specified in Table 4.8a through Table 4.8c are for AMC-II. The SCS has developed correlations for calculating the CN values for AMC-I and AMC-III in terms of AMC-II. These correlations are specified as follows: CN (I) =

4.2CN (II) 10 - 0.058CN (II)

(4.28)

23CN (II) 10 + 0.13CN (II)

(4.29)

CN (III) =

The procedure of using the SCS-CN method for calculating the runoff depth is explained using an example problem below. � EXAMPLE 4.3 A small catchment has an area of 680 ha and consists of Type-C soil. The land cover is 50% good condition parks, 10% commercial and business area, 30% residential, and 10% roads/paved areas. Assuming normal moisture conditions, determine the volume of direct runoff from a design storm of total rainfall of 200 mm from this catchment. What would be the values of runoff volumes if the past 5-day total rainfalls were 20 mm and 60 mm, respectively, assuming growing season in the catchment? Solution The curve numbers for Type-C soil for different land use conditions are taken from Table 4.8. The calculations of composite CN are summarized in a tabular form given below. Based on percentages of different types of area, the composite curve number is calculated as CN = 8320/100 = 83.2.

Calculations for AMC-II Land Use

%

CN

Product

Good condition parks Commercial and business Residential Paved/Roads Sum Composite CN(II)

50 10 30 10 100 83.2

74 94 90 98

3700 940 2700 980 8320

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Engineering Hydrology

The maximum possible retention (S) is calculated by Eq. (4.27) as follows: Ï 1000 ¸ S = 25.4 Ì - 10 ˝ = 51.3 mm Ó 83.2 ˛ For P = 200 mm storm rainfall, the effective rainfall or depth of runoff can then be calculated using Eq. (4.26) as follows (assuming Ia = 0.2S): ER =

(200 - 0.2 * 51.3)2 = 149.37 mm 200 + 0.8(51.3)

The volume of runoff produced for AMC-II is calculated as follows: VR = P * A =

149.37 m2 m (680 ha)(100 ¥ 100) = 1,015,698 m 3 1000 ha

Total cumulative rainfall over the past five days of 20 mm in the growing season indicates dry conditions, i.e., AMC-I. The curve number for AMC-I is calculated by Eq. (4.28) as follows: CN(I) =

4.2(83.2) = 67.5 10 - 0.058(83.2)

By putting CN = 67.5 in Eq. (4.27), we get potential retention value for AMC-I as S = 122.1 mm. By putting S = 122.1 mm in Eq. (4.26), we get ER = 103.55 mm. For ER = 103.55 mm, the runoff volume for AMC-I condition can be calculated in a similar manner as VR = 704,168 m3. Total cumulative rainfall over the past five days of 60 mm in the growing season indicates wet conditions, i.e., AMC-III. The curve number for AMC-III is calculated by Eq. (4.29) as follows: CN(III) =

23(83.2) = 91.9 10 + 0.13(83.2)

By putting CN = 91.9 in Eq. (4.27), we get potential retention value for AMC-III as S = 22.3 mm. By putting S = 22.3 mm in Eq. (4.26), we get ER = 175.52 mm. For ER = 175.52 mm, the runoff volume for AMC-III condition can be calculated in a similar manner as VR = 1,193,560 m3.

4.8

FLOW DURATION CURVE

We have seen earlier that there are three different types of rivers, LO 8 Demonstrate flow duration curve namely, perennial rivers, intermittent rivers, and ephemeral and its usefulness rivers. The variation of runoff in different rivers will be different throughout a water year. The volume of runoff available in a river annually is an important variable that is needed in water resources planning activities in a catchment. The total runoff volume available in a catchment drained by a river may be estimated from the annual hydrograph for the river. However, in many water resources planning and development activities, the information about the total amount of runoff is not

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127

adequate and certain additional information is required. For example, data on the number of days for which a certain magnitude of flow will always be available can be extremely useful in planning for the allocation of water for various types of water demands in a catchment. Also, in hydropower generation, it is desirable to know the guaranteed amount of flow that is going to be available for most of the time during a year. This type of information can be obtained from a flow duration curve. A flow duration curve may be thought of as another type of hydrograph in which y-axis represents the flow (like in a hydrograph) but the x-axis represents the percent of time the particular flow value will be equaled or exceeded. The flow duration curve may be prepared using the plotting position method. The flow data are arranged in descending order of magnitude. Then each flow value is assigned a rank (m) starting with a rank of 1 from the top. The plotting position is then calculated using the following formula: P(%) =

m ¥ 100 n +1

(4.30)

where, m is the rank for a flow value (xm), n is the total number of data points in the flow duration analysis, and P(%) is the percent of times the xm will be equaled or exceeded. The plot between P(%) and xm is known as flow duration curve. The data employed may be daily, weekly, or monthly depending on the intended use of the flow duration analysis and the length of data available. The flow duration curve at a finer temporal resolution (say a day) will tend to be steeper than the one for the coarser temporal resolution (say a month) due to smoothening of the dynamic effects embedded in the coarser time intervals. A typical flow duration curve is shown in Figure 4.5. A flow duration curve may be plotted on arithmetic, semi-log, or log-log scale depending on the range of data available and its practical utility. Figure 4.5 shows three different flow duration curves. The curve represented by a solid line is the flow duration curve for a perennial river in which there is flow throughout the year. The minimum flow in a perennial river is available throughout the year; therefore, its probability will be 100%. From the figure, the flow value that will be equaled or exceeded 100% of the time is about 90 m3/s. The flow value with a certain

Figure 4.5

Flow duration curve

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Engineering Hydrology

probability of P(%) is also referred to as the P(%) dependable flow. For example, for the perennial river in the figure, 100% dependable flow is about 90 m3/s (or Q100 = 90 m3/s) and 50% dependable is about 215 m3/s (or Q50 = 215 m3/s). The curve represented by the dashed line is the flow duration curve for an intermittent or an ephemeral river which does not flow throughout the year. The flow duration curve for such rivers would meet the x-axis as shown. For an ephemeral or intermittent river, 100% dependable flow is zero (Q100 = 0.0 m3/s). For the example shown in Figure 4.5, 25% dependable flow for the intermittent/ephemeral river is about 200 m3/s (Q25 = 205 m3/s). The two flow duration curves discussed so far are for natural flows that are unaffected by human activities. When there is a dam across a river and the reservoir is operated to meet the water demands downstream, the nature of the flow duration curve will get modified. Such a flow duration curve is known as a regulated flow duration curve and is shown in Figure 4.5 by a dash-dot line. The magnitude of P(%) dependable flow reduces during high flow range as the flow is diverted for meeting different types of demands. However, during lean periods, minimum environmental flows are released from the reservoir causing the regulated flow duration curve to be higher than the natural flow duration curve. The flow duration curves have applications in many water resources activities. In hydropower generation, power generated is classified into two broad categories–primary and secondary. The primary power (also called firm power) supplied is the one that is available about 95% of the time during a year. This can be determined using the 95% dependable flow from the flow duration curve of the river under consideration. The secondary power (also called surplus power) is the additional power that can be generated depending upon the magnitude of flows available at other times. The hydropower generated (PH) for a given discharge value of Q can be determined using the following relation: PH =

hg Q H 1000

(4.31)

where, PH is the hydropower generated (kW), h is the efficiency of the hydropower generation system, g is the specific weight of water, Q is the discharge in the river (m3/s), and H is the head difference between upstream and downstream of the hydropower system (m). The low magnitude range of the flow duration curves are useful in drought studies. Further, the flow duration curve and sediment rating curve can be combined to determine the cumulative sediment transported through a river at different times. This provides useful information in reservoir sedimentation and other sediment management activities in a catchment. A log-log plot of the flow duration curve is useful in determining flow variation and estimating the surface and groundwater potential in the catchment. A steeply sloped flow duration curve indicates significant surface water contribution while a flatly sloped flow duration curve indicates a significant groundwater contribution in a river. Flow duration curves can also be used in extrapolation studies and assessment of runoff potential in hydrologically similar catchments having limited data availability. � EXAMPLE 4.4 The average monthly streamflow values in a river are 150, 200, 420, 400, 350, 290, 320, 230, 250, 210, 180, and 100 (in m3/s). Prepare a flow duration curve for this river using this data and answer the following:

Runoff

(a)

What is the value of firm power corresponding to 90% dependable flow?

(b)

What is the value of 75% dependable flow?

(c)

What is the dependability of 300 m3/s?

129

Assume 80% efficiency and 100 m head available for hydropower generation. Solution The calculations for the preparation of flow duration curve are presented in a tabular form below. Column 1 presents the streamflow data given, column 2 presents the streamflow data arranged in descending order of magnitude, column 3 presents the order (m) for each value of the average monthly streamflow, and column 4 represents the plotting position calculated as per Eq. (4.30) above (in %). The flow duration curve can be obtained by plotting values in column 4 (P %) on x-axis and the values in column 2 (Q m3/s) on y-axis. Monthly Flow (m3/s) (Col. 1)

Ranked Flows (m3/s) (Col. 2)

Rank (m) (Col. 3)

P (%) (Col. 4)

150

420

1

7.7

200

400

2

15.4

420

350

3

23.1

400

320

4

30.8

350

290

5

38.5

290

250

6

46.2

320

230

7

53.8

230

210

8

61.5

250

200

9

69.2

210

180

10

76.9

180

150

11

84.6

100

100

12

92.3

Firm Power: We need to calculate 90% dependable flow for estimating firm power. Using linear interpolation, it is easily calculated as Q90 = 115 m3/s. Now, putting h = 0.80, g = 9810 N/m3, Q = 115 m3/s, and H = 100 m in Eq. (4.4), we get: PH =

(0.80)(9810)(115)(100) kW = 90.25 MW 1000

Therefore, the value of firm power that will be available 90% of the time is 90.25 MW. Further, using interpolation, 75% dependable flow (Q75) = 185 m3/s, and dependability of flow of 300 m3/s = 35.9 %.

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Engineering Hydrology

FLOW MASS CURVE

In the precipitation chapter, we learnt about the concept of massLO 9 Determine the capacity of a curve. A mass-curve represents the cumulative value of a physical reservoir using flow mass curve and variable with respect to time. We learnt how rainfall mass curve sequent peak algorithm can be derived from a rainfall hyetograph. In a similar manner, a flow mass curve can be prepared by plotting cumulative runoff volume versus time. The time may be expressed as days, weeks, or months depending on the usage. The flow mass curve, when used in conjunction with a demand mass curve, can be an important tool in analyzing the wet and dry spells in a catchment. The runoff data over a year or several years can be used to determine cumulative runoff volume as a function of time and plotted accordingly. Such a plot is called flow mass curve. Figure 4.6 shows a typical flow mass curve, which consists of peaks and valleys depending on the length of the record.

Figure 4.6

Determination of reservoir capacity using flow and demand mass curves

The cumulative runoff volume (CRV) at any time ‘t’ can be represented as follows: t

CRV = ÚQ dt

(4.32)

0

where, Q is rate of flow at any time ‘t’ (m3/s). CRV is expressed in m3, million m3, hectare-meter (ha-m), m3/s-day (cumec-day) or some other suitable volumetric units. The slope of the flow mass curve at any time represents the rate of flow (Q) at that time. Now let us analyze the superposition of flow mass curve with the demand mass curve. Let us say that a water demand of D (m3/s) is required to be supplied from a river. A demand mass curve can be similarly prepared and the cumulative demand volume (CDV) at any time ‘t’ can be represented as follows: t

CDV = ÚD dt

(4.33)

0

If the demand D varies as a function of time, the demand mass curve will be a nonlinear curve but if the demand D is constant, the demand mass curve will be a straight line. In Figure 4.6, the demand mass curve

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is represented by line AB indicating that while Q varies as a function of time, D is constant with time. The flow mass curve and the demand mass curve can be analyzed together to determine the capacity of a reservoir that needs to be constructed across the river in order to meet the demand D at all times. At point A, both demand rate and flow rate are equal, and as we march ahead in time, the demand remains constant but flow rate decreases as the slope of flow mass curve flattens out. Thus, the point A represents the beginning of a dry period and as we head into the dry period, water supply available is less than the demand that needs to be met. In order to meet the demand during this period, water must come from that stored in the reservoir. The difference between demand mass curve and flow mass curve between A and B (shown by verticals at different times) represents the volume of water that should be supplied from the reservoir in order to meet the required demand D. In order to meet this demand, the reservoir must have sufficient water stored at the beginning of the dry period. Clearly, the capacity of the reservoir should be equal to (or more than) the largest of all the verticals during the period from A to B. This largest vertical is represented by x1x2 in Figure 4.6, where x1 is a point on the demand mass curve and x2 is a point on the flow mass curve. Note that from A to x2, Q < D; at x2, Q = D; and from x2 to B, Q > D. Therefore, the demand is met from the water stored in the reservoir during the phase from A to x2. The reservoir is completely emptied at a time corresponding to x2. The reservoir starts to fill up after point x2 because there is more flow than the demand. At point B, the reservoir is full again and water starts to spill over from the reservoir. Therefore, based on the analysis of the flow and demand data during A and B, the storage capacity needed is S which is equal to the difference between the ordinates of demand and flow mass curves at the largest departure. A similar analysis may be carried out for a greater length of data consisting of more numbers of peaks and valleys. For each cycle, the minimum storage capacity needed is calculated. The required storage capacity of the proposed reservoir would be the maximum of all of these values, corresponding to the various cycles in the data. Given the flow and constant demand, the storage capacity of a proposed reservoir may be determined using the graphical analysis explained above. A similar procedure can be followed if the demand is not constant but varies at different time intervals. In such a case, the demand mass curve is prepared and superimposed on the flow mass curve with proper matching of the timings. The maximum departure between the two curves will represent the minimum storage capacity needed. The calculations for determining storage capacity (for constant or variable demand) can also be carried out in a tabular form. Given the flow and storage capacity of a reservoir, it is also possible to determine the maximum constant demand that can be sustained from the reservoir during lean periods. It can be done by drawing lines at different slopes tangential to the peak around A on the flow mass curve. The slope of the line giving the storage equal to the given storage capacity will be the demand that can be sustained from the reservoir. � EXAMPLE 4.5 The monthly runoff volumes, in m3, in Mississippi River for a 2-year period are provided below. Determine the storage capacity of a proposed reservoir on the Mississippi River to meet a constant demand of 20 Mm3 per month. Month

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

Year-1

58

66

29

33

37

25

10

2

2

3

4

8

Year-2

10

31

50

53

39

39

36

12

12

14

18

20

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Solution The calculations for the determination of storage capacity of the proposed reservoir are presented in a tabular form below. Column 1 represents months and columns 2 and 3 present the flow and demand volumes (QV and DV), respectively. Column 4 presents the difference between flow and demand volumes for each month and is calculated by subtracting entries in column 3 from the entries in column 2. Notice that the entries in column 4 are both positive and negative. Positive values indicate excess flow in a month whereas, a negative entry in column 4 means that there is excess demand that cannot be met by the inflow alone. This negative demand needs to be met from the water stored in the reservoir. The negative entries start at the beginning of July, which marks the beginning of the dry season. Remember, the assumption is that the reservoir is full at the beginning of dry period. There is surplus water in February of the following year, while in August of year 2 there will be a deficit supply. In column 5, only negative values of column 4 are accumulated while in column 6 only positive values of column 4 are accumulated. Therefore, the entries in column 5 represent cumulative excess demands while column 6 represents cumulative excess flows. Note that the reservoir is full at the end of June (or at the beginning of July). In the month of July of year 1, there is excess demand of 10 Mm3 that is met by the water stored in the reservoir. Month Col. 1

Flow-Vol (QV) Col. 2 (Mm3)

Demand-Vol (DV) Col. 3 (Mm3)

QV – DV Col. 4 (Mm3)

Cum-ED Col. 5 (Mm3)

Cum-EQ Col. 6 (Mm3)

Jan

58

20

38.0

38.0

Feb

66

20

46.0

84.0

Mar

29

20

9.0

93.0

Apr

33

20

13.0

106.0

May

37

20

17.0

123.0

Jun

25

20

5.0

128.0

Jul

10

20

–10.0

–10.0

Aug

2

20

–18.0

–28.0

Sep

2

20

–18.0

–46.0

Oct

3

20

–17.0

–63.0

Nov

4

20

–16.0

–79.0

Dec

8

20

–12.0

–91.0

Jan

10

20

–10.0

–101.0

Feb

31

20

11.0

11.0

Mar

50

20

30.0

41.0

Apr

53

20

33.0

74.0

May

39

20

19.0

93.0 (Contd.)

Runoff

Jun

39

20

19.0

112.0

Jul

36

20

16.0

128.0

Aug

12

20

–8.0

–8.0

Sep

12

20

–8.0

–16.0

Oct

14

20

–6.0

–22.0

Nov

18

20

–2.0

–24.0

Dec

20

20

0.0

–24.0

133

In the month of August of year 1, there is additional excess demand of 18 Mm3, which can also be met from the reservoir. Therefore, the cumulative excess demand met from reservoir for the first two dry months of July and August is 10 Mm3 + 18 Mm3 = 28 Mm3. The reservoir will keep supplying the excess demand till the end of January of year 2 when a cumulative excess demand of 101.0 Mm3 will be met from the reservoir. Therefore, if a reservoir of minimum storage capacity of 101.0 Mm3 is constructed then a constant demand of 20 Mm3 can be met for the flow data given. It follows that the maximum (negative) entry in column 5 corresponds to the minimum storage capacity needed for the proposed reservoir, which is 101.0 Mm3 in the present case. At the end of January of year 2, the entire capacity of 101.1 Mm3 is depleted and the reservoir is empty at the end of January of year 2. Then, because of excess flow, the reservoir starts to slowly fill up again with 11 Mm3 coming in February of year 2, followed by 30 Mm3 in March, and so on. The reservoir will continue to fill up till it reaches its capacity. Notice that the cumulative excess flow is 93 Mm3 at the end of May and 112 Mm3 at the end of June of year 2. Therefore, the reservoir will get filled up (to its capacity of 101.0 Mm3) sometime in the month of June of year 2. Assuming linear variations within the month, the date on which the reservoir will fill up can also be estimated. The spill-over from the reservoir in the month of June of year 2 will be 112-101 = 11 Mm3. As seen from column 4, there would be a further spill-over of 16 Mm3 during the month of July of year 2, because the reservoir was full in June of year 2, and the excess flow will simply spill-over. In the following month of August, we would need to supply 8 Mm3 from the reservoir, as shown in column 5, since the supply is only 12 Mm3 against the demand of 20 Mm3. It is seen that the arithmetic calculations in a simple tabular form are useful in analyzing the flow and demand mass curves and determine a lot of important information about the drying up, filling up, and spillover of the reservoir etc. The example discussed above has assumed a constant demand; however, if the demand varies at each time interval, it can be easily accounted for in the arithmetic calculations. Further, there may be other physical factors, e.g., rainfall, evaporation, seepage losses, etc., that may be taken care of by either adding to or subtracting from the overall demand, as the case may be. � Sequent Peak Algorithm In using the flow mass curve analysis for the estimation of minimum storage capacity of a proposed reservoir, at least one complete cycle of peak and trough is needed. Sometimes, the trends in the data are such that a complete cycle of peak and trough is either not complete or is unclear. In such cases (and for other cases also), sequent peak algorithm can be employed. In a sequent peak algorithm, the data available is repeated in order to ensure at least one full cycle of peaks and troughs, or deficit and

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surplus. In this method, cumulative net inflow volume curve S(QVi – DVi) is prepared using the given inflow and demand data. This cumulative net inflow volume curve is scanned for various surplus and deficit periods in the form of peaks and troughs. Figure 4.7 shows a typical cumulative net inflow volume curve and the definition of sequent peaks and troughs. A peak sequent to P1 is defined as the peak (P2) that is of a greater magnitude than P1.

Figure 4.7

Sequent peaks and troughs in the cumulative net inflow volume curve

The calculation of storage capacity of a reservoir using the sequent peak algorithm is described in the following step by step procedure. 1.

From given inflow and demand data, flow volume (QVi) and demand volume (DVi) for each time interval are estimated. The total data given is repeated at least twice, i.e., if the number of data given is n, the sequence of data used is 2n with n data appended at the end.

2.

The difference between flow volume and demand volume at each time step, the net inflow volume (QVi - DVi), is calculated. Then the cumulative net inflow volume S(QVi - DVi) is calculated.

3.

Locate the first peak (P1), sequent peak (P2), and the lowest trough between the two sequent peaks (T1). Calculate the difference between first peak and trough (P1 – T1).

4.

Similarly, locate all other sequent peaks and troughs in the 2n data points and estimate their differences, i.e., (P2 – T2), (P3 – T3),…. (PK – TK), where, K is the total number of pairs of sequent peaks and troughs.

5.

The minimum storage capacity of the proposed reservoir is given by the maximum difference of the peaks and troughs, i.e., S = Max (Pk – Tk); k = 1 to K.

Like the flow mass curve method discussed in the previous section, the sequent peak algorithm also assumes the reservoir to be full at the beginning of a dry period, and the sequence of deficit and surplus water are assumed to recur in future.

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SUMMARY In this chapter, we studied two types of runoff generation mechanisms, Hortonian and Saturation overland flow. In Hortonian overland flow, the runoff can be expressed as the precipitation minus infiltration as the soil gets saturated from top. On the other hand, in Saturation overland flow, the soil may be saturated from below due to variable sources of areas in the upper catchment causing soil saturation. We learnt about the different types of flows, i.e., surface flow, base flow, groundwater flow, channel flow, interflow, and streamflow. Streamflow can be measured using the continuity equation and more details about streamflow measurement are provided in a later chapter. We also discussed about hydrograph which is a plot of flow versus time and can be one of the following four types: perennial hydrograph, intermittent hydrograph, ephemeral hydrograph, and storm hydrograph. A perennial hydrograph is the one in which there is significant groundwater flow contribution and hence a perennial river flows throughout the year. An intermittent hydrograph is derived from an intermittent river in which there is small amount of groundwater flow contribution. An ephemeral stream does not have any groundwater contribution and flows during rain periods only. The impacts of catchment characteristics (size, shape, slope, land use land cover conditions, soil type, etc.) and storm characteristics (intensity, duration, depth, etc.) on the runoff response from a catchment were discussed in great details in this chapter. The rising limb of a hydrograph is affected by both catchment and storm characteristics and the falling limb of a hydrograph is a function of catchment characteristics alone. Some of the factors affect the time distribution of runoff while some others affect the total volume of runoff. Several methods of estimating runoff were then discussed including analytical and empirical methods. The Rational formula gives peak discharge in a catchment which is very useful in the hydraulic design of culverts and storm water system. The SCS curve number method is more useful in estimating runoff depth or runoff volume from a design precipitation storm. The flow duration curve, which is a plot between magnitude of flow and the percent of time it will be equaled or exceeded, is very useful in estimating the water resources potential in a catchment and in determining the hydropower generation capacity in a river. The method to develop a flow duration curve, its characteristics, and how it gets affected by manmade regulation of storage structures were discussed in detail. The concept of flow mass curve and its use in determining the capacity of a reservoir proposed across a river, given the requirements of meeting certain water demands, was discussed next. Lastly, the steps involved in the sequent peak algorithm for determining the storage capacity of a proposed reservoir were presented. This chapter focused on the concepts of runoff magnitude over a long period of time. The response of catchment when it is subjected to intense rainfall for short spells of a few hours is dealt within the next chapter on hydrograph analysis.

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OBJECTIVE-TYPE QUESTIONS 4.1 The water moving over the ground as a thin sheet is called (a) Channel flow (b) Overland flow (c) Baseflow

(d) Streamflow

4.2 Lateral movement of water between the ground surface and groundwater table towards a river is known as (a) Streamflow (b) Baseflow (c) Interflow (d) Overland flow 4.3 Subsurface flow may consist of (a) Overland flow and interflow (c) Interflow and channel flow

(b) Baseflow and channel flow (d) Interflow and baseflow

4.4 During a rainfall event, streamflow in a river is dominated by (a) Runoff and interflow (b) Baseflow and channel flow (c) Runoff and baseflow (d) Interflow and baseflow 4.5 While measuring the streamflow in a river using continuity equation (Q = VA), the velocity should be (a) Velocity at water surface (b) Average velocity at the cross section (c) Maximum velocity (d) Velocity at stream bed 4.6 In a small river having a depth of flow of 2 m, up to what depth the current meter should be lowered to measure the average velocity? (a) 1 m (b) 0.8 m (c) 0.0 m (d) 1.2 m 4.7 In a large river having a depth of flow of 10 m, up to what depths the current meter should be lowered to measure the average velocity? (a) 2 m and 6 m (b) 4 m and 6 m (c) 2 m and 8 m (d) 4 m and 8 m 4.8 An annual hydrograph is useful for (a) Planning water resources activities (b) Allocation of water for municipal water supply, irrigation, and hydropower (c) Estimate total volume of runoff available in a catchment (d) All of these (e) None of these 4.9 Which of the following rivers has the maximum contribution from baseflow? (a) Perennial river (b) Intermittent river (c) Ephemeral river (d) All of these (e) None of these 4.10 Which of the following rivers has the least contribution from baseflow? (a) Perennial river (b) Intermittent river (c) Ephemeral river (d) All of these (e) None of these 4.11 During early portions of the rising limb of a storm hydrograph, (a) Infiltration is high and runoff is less (b) Runoff is high and infiltration is less (c) All portions of the catchment are contributing to runoff at the outlet (d) All of these (e) None of these

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4.12 The shape, size, and slope of the rising limb are a function of (a) Catchment characteristics (b) Storm characteristics (c) Both catchment and storm characteristics (d) None of these 4.13 The shape, size, and slope of the falling limb are a function of (a) Catchment characteristics (b) Storm characteristics (c) Both catchment and storm characteristics (d) None of these 4.14 Other factors remaining the same, a mildly sloped catchment (a) Will produce steep rising limbs with sharps peaks (b) Will produce flatter rising limbs with broad peaks (c) Will produce flatter rising limbs with sharp peaks (d) Shape of rising limb is independent of slope of the catchment 4.15 Other factors remaining constant, greater intensity rainfall falling on a catchment (a) Will produce steeply rising limbs, and greater and early peak flows (b) Will produce flatly rising limbs, and greater and early peak flows (c) Will produce flatly rising limbs, and smaller and early peak flows (d) Will produce flatly rising limbs, and smaller and delayed peak flows (e) Rising limb and peak characteristics are independent of intensity of rainfall 4.16 Other factors remaining constant, a catchment with dry initial conditions will produce (a) Steeply sloped rising limbs and lesser magnitude and delayed peak flows (b) Mildly sloped rising limbs and greater magnitude and early peak flows (c) Mildly sloped rising limbs and lesser magnitude and delayed peak flows (d) Rising limb and peak characteristics are independent of initial moisture conditions in a catchment 4.17 In the recession curve equation by Horton (1933), the unit of recession coefficient (K) is (c) T2 (d) T −2 (e) No units (a) T (b) T −1 4.18 As compared to a rural catchment, an urbanized catchment will tend to produce a hydrograph with (a) Higher peak flows (b) Early peaks (c) Smaller time bases (d) All of these 4.19 For a catchment consisting of sandy soil, which of the following statements are true in comparison to a catchment consisting of clayey soil? I. Will have greater infiltration losses II. Will have greater magnitude of peak runoff III. Will have smaller time of peak runoff (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 4.20 For a catchment having a fully developed channel network, which of the following statements are true in comparison to a catchment having a poorly developed channel network? I. Will have steeply sloped rising limbs in hydrographs II. Will have higher magnitudes of peak flows III. Will have smaller time of peak runoff (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false

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4.21 Catchment characteristics affect which of the following portions of a hydrograph? (a) Rising limb only (b) Falling limb only (c) Both (a) and (b) (d) None of these 4.22 Storm characteristics affect which of the following portions of a hydrograph? (a) Rising limb only (b) Falling limb only (c) Both (a) and (b) (d) None of these 4.23 The length of the main watercourse of a catchment is 2.5 km and total drop in elevation is 2 m. The time of concentration is approximately (in hours) (a) 12.6 (b) 2.1 (c) 0.2 (d) 73.9 (e) Insufficient data 4.24 For a small urbanized catchment in Bangalore, what is the maximum rainfall intensity expected for a 2-hour storm having a return period of 20 years (in cm/hour)? (a) 3.26 (b) 0.30 (c) 2.73 (d) 2.44 (e) Insufficient data 4.25 Which of the following methods you will use to calculate the peak discharge for the design of a highway culvert? (a) SCS curve number method (b) Rational method (c) Regression method (d) None of these 4.26 The maximum intensity of rainfall expected in a catchment is a function of which of the following factors? I. Duration of rainfall II. Frequency of occurrence III. Location of the catchment (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of the three statements 4.27 The runoff coefficient in a catchment is a function of which of the following physical factors? I. Interception and depression storage characteristics of the catchment II. Soil type, land use, and land cover conditions of the catchment III. Slope of the catchment (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of these physical factors 4.28 The maximum intensity of rainfall expected in a catchment (a) Increases with an increase in the duration of rainfall (b) Decreases with an increase in the duration of rainfall (c) Decreases with an increase in the return period (d) Has no relation with duration and return period 4.29 The rainfall falling on the subareas of a catchment close to the outlet will reach the outlet (as compared to subareas farther from outlet) (a) Earlier (b) Later (c) At the same time (d) Depends on other catchment characteristics 4.30 The peak runoff from a catchment will occur when the duration of the rainfall is (a) Twice the time of concentration (b) Half of the time of concentration (c) Equal to the time of concentration (d) Does not depend on the time of concentration 4.31 Which of the following statements is incorrect for the time of concentration in a catchment? (a) Depends on the area of catchment (b) Larger the maximum length of the catchment, larger it is (c) Does not depend on the steepness of the catchment

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(d) Depends on the surface conditions of the catchment 4.32 Which of the following methods you will use to calculate runoff volume to design a flood detention storage structure? (a) SCS curve number method (b) Rational method (c) Regression method (d) None of these 4.33 The SCS curve number method is based on (a) Law of conservation of mass (b) Law of conservation of momentum (c) Both mass and momentum conservation (d) None of these 4.34 Which of the following statements in incorrect for the SCS curve number method? (a) It is based on the law of conservation of mass (b) It is based on the assumption that the ratio of actual to potential runoff is equal to the ratio of actual to potential retention in the catchment (c) It ignores the initial abstraction in the form of interception and depression losses (d) All of the above statements are incorrect 4.35 Potential storage (S) during a storm in a catchment in the SCS curve number method is a function of (a) Soil type in the catchment (b) Land use and land cover conditions in the catchment (c) Initial moisture conditions in the catchment (d) All of these (e) Does not depend on any of these 4.36 The relationship between potential storage (S, mm) and curve number (CN) in SI system can be represented by which of the following?

Ï 1000 ¸ - 10 ˝ Ó CN ˛

(b)

Ï 1000 ¸ S = 25.4 Ì - 100 ˝ Ó CN ˛

Ï 100 ¸ - 10 ˝ CN Ó ˛

(d)

Ï 1000 ¸ S = 25.4 Ì - 10 ˝ CN Ó ˛

(a) S = 2.54 Ì (c) S = 25.4 Ì

4.37 Which of the following statements are true for the curve number (CN)? I. The values of CN close to 0 indicate highly pervious catchment II. The values of CN close to 100 indicate highly impervious catchment III. The value of CN is independent of initial moisture conditions of the catchment (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of the statements is true for CN 4.38 Which of the following statements is correct for the SCS curve number method? (a) For a completely pervious catchment (CN = 0, S = ∞); and there will be no runoff from the catchment (b) For a completely impervious catchment (CN = ∞, S = 0); and there will no runoff from the catchment (c) Both (a) and (b) are correct (d) None of (a) and (b) is correct 4.39 Which of the following statements is correct in SCS curve number method with reference to the initial soil moisture conditions in a catchment? (a) The volume of runoff during wet conditions will be a lot less than that during the dry conditions (b) The volume of runoff during dry conditions will be a lot less than that during the wet conditions (c) The volume of runoff during dry conditions will be same as that during the wet conditions (d) None of the above is correct

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4.40 Which of the following statements is incorrect for the antecedent moisture conditions (AMC) in the SCS curve number method? (a) AMC-I is meant for wet conditions (b) AMC-II is meant for dry conditions (c) AMC-III is meant for normal conditions (d) None of these (e) All three (a), (b), and (c) are incorrect 4.41 Which of the following statements is correct for the antecedent moisture conditions (AMC) in the SCS curve number method? (a) Amount of rainfall to classify an AMC during growing season is higher than that during a dormant season (b) Amount of rainfall to classify an AMC during dormant season is higher than that during a growing season (c) Amount of rainfall to classify an AMC during dormant season is equal to that during a growing season (d) Classification of AMC is independent of the season in a catchment 4.42 Let the cumulative rainfall over the past five days in a catchment on January 10th, June 10th, and October 10th of a year be represented as x1, x2, and x3, respectively. It has been found that the initial moisture conditions on these three days could be classified as AMC-I, AMC-II, and AMC-III, respectively. The values of x1, x2, and x3 should be such that (a) x1 = x2 = x3 (b) x1 ≥ x2 ≥ x3 (c) x1 ≤ x2 ≤ x3 (d) x1, x2, and x3 can have any values 4.43 The value of composite CN in a catchment was found to be 75.6. What is the value of potential storage (S) in the catchment (in mm)? (a) 8.2 (b) 220.4 (c) 2204.0 (d) 82.40 (e) Insufficient data 4.44 A fully developed urban area consists of 30% soil type A, 20% soil type B, 15% soil type C, and remaining area consists of soil type D. The value of composite CN for a residential district of size 0.5 acre in this area will be (a) 70.15 (b) 73.55 (c) 71.95 (d) Insufficient data 4.45 An agricultural land consists of soil type A and has a woods-grass combination in poor hydrologic condition. What is the CN for this agricultural land in the dormant season if the cumulative rainfall in the past five days is 30 mm? (a) 57 (b) 36 (c) 75 (d) Insufficient data 4.46 Which of the following statements are true for the flow duration curve? (a) It is a plot between flow value and the percent of time the particular flow value will be equaled or exceeded (b) Flow duration curve at a finer temporal resolution will be steeper than the one for the coarser temporal resolution (c) Flow duration curve for a perennial river does not cut the x-axis while that for intermittent river does (d) All of these (e) None of these 4.47 The low magnitude range of the flow duration curves are useful in (a) Flood studies (b) Drought studies (c) Both floods and drought studies (d) Not useful at all

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141

4.48 Which of the following statements are true for a flow duration curve? (a) A steeply sloped flow duration curve indicates significant groundwater contribution (b) A mildly sloped flow duration curve indicates significant surface water contribution (c) Both (a) and (b) (d) None of the above 4.49 Which of the following statements are true for a flow mass curve? (a) The graph between runoff volume and time is known as the flow mass curve (b) The second derivative of the flow mass curve at any time represents the flow at that time (c) Flow mass curve is a monotonically decreasing curve (d) All of these (e) None of these 4.50 The difference between the demand mass curve and flow mass curve at any time during a dry period represents (a) The volume of surplus water entering the reservoir at that time (b) Volume of water that can be supplied from the reservoir in order to meet the required demand (c) The rate of flow at that point of time (d) None of these 4.51 A sequent peak is defined as the peak in the cumulative net inflow volume curve that is (a) Equal to the previous peak (b) Greater than the previous peak (c) Smaller than the previous peak (d) None of these 4.52 The sequent peak algorithm is based on the following assumption(s): (a) The reservoir is full at the beginning of a dry period (b) The sequence of deficit and surplus water would recur in future (c) Both (a) and (b) (d) None of these

DESCRIPTIVE QUESTIONS 4.1 Describe what happens to the water from the time it falls on the ground to the time it reaches the outlet of a catchment in a river. 4.2 Differentiate between different types of flow, i.e., overland flow, interflow, baseflow, and streamflow. 4.3 Explain the differences between Hortonian overland flow and saturation overland flow. 4.4 Describe the different types of annual hydrographs. 4.5 Describe the different components in a storm hydrograph. 4.6 Explain how the catchment and storm characteristics affect the shape of rising limb. 4.7 Explain the SCS curve number method and derive the expression for calculating runoff volume from a catchment due to a storm. 4.8 What do you understand by a flow duration curve and how is it useful in water resources related activities? 4.9 Given demand and flow data, explain how you will determine the storage capacity of a reservoir proposed across a river using double mass curve analysis. Use rough sketch to explain your answer. 4.10 Explain the sequent peak algorithm for finding the storage capacity of a reservoir.

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NUMERICAL QUESTIONS 4.1 A catchment consists of paved area of 20 km2, residential area of 102 km2, and lawns of 15 km2. Assuming the runoff coefficient values for the three land use types as 0.9, 0.4, and 0.1 respectively, estimate the equivalent runoff coefficient for the entire catchment. 4.2 The annual rainfall (P) and runoff (R) data (both in mm) in a catchment for the period 1990–2009 are provided in the following table. Develop a linear and a power regression model (assume parameter e = 0.) using these data. Calculate the coefficient of correlation for the raw data (R and P). Also calculate the coefficient of correlation for the linear and the non-linear models. Which model would you recommend? Year

Rainfall (mm)

Runoff (mm)

Year

Rainfall (mm)

Runoff (mm)

1990

520

350

2000

560

350

1991

615

450

2001

400

270

1992

420

270

2002

520

380

1993

270

170

2003

435

302

1994

305

204

2004

395

250

1995

380

250

2005

290

190

1996

705

594

2006

430

315

1997

600

425

2007

1020

800

1998

350

210

2008

900

720

1999

550

380

2009

865

680

4.3 A small suburban catchment in Jodhpur has the following data: Area of catchment = 35 km2; Maximum length of main stream = 2.2 km; Drop in elevation along main stream = 2.5 m. The land use data of the catchment is given in the following table: Land Use/Land Cover

Area (%)

Single family residential

10

Suburban residential

10

Central Business District

20

Flat lawns with sandy soil

30

Asphaltic streets

10

Unimproved areas Total

20 100

Calculate the peak discharge values at the outlet of the catchment for return periods of 1-year, 10-year, 25year, and 50-year. 4.4 A catchment of area 10 km2 consists of 20% soil Type A, 40% soil Type B, and 40% soil Type C scattered randomly. The catchment consists of 60% fully developed urbanized area and 40% agricultural area. The distribution of land use/land cover (LU/LC) within the fully developed urbanized area and agricultural area is given in the following table.

Runoff Fully Developed Urban Area (Total 6 km2 ) Land Use, Land Cover

Agricultural Area ( Total 4 km2 ) Percent

Land Use, Land Cover

Percent

Good condition open spaces

30

Fallow bare soil

10

Parking, streets, and roads

10

Contoured small grains good hydrologic condition

30

Commercial business district

20

Woods poor condition

20

Residential district of average lot size ½ acre

10

Pastures and grasslands, fair condition

10

Residential district of average lot size 1 acre

25

Grassy Meadows

20

Residential district of average lot size 2 acre

5

Brush poor condition

Totals

143

100

10 100

Determine the runoff depth and volume from this catchment due to a severe rainfall event having a depth of 300 mm during storm on a single day. Assume normal initial moisture conditions in the catchment. 4.5 The average monthly streamflow data (in m3/s) at a certain location in a river are given in the following table.

(a) (b) (c) (d)

Month

Year 1

Year 2

January

66

95

February

48

70

March

50

58

April

45

50

May

38

40

June

30

28

July

25

30

August

40

35

September

50

45

October

80

67

November

120

80

December

90

75

Determine and plot the flow duration curve. Determine the flow that can be expected 50%, 80%, and 95% of the time. What is the percent of time a flow magnitude of 50 m3/s, 80 m3/s, and 100 m3/s can be expected? Determine the firm power when efficiency of the hydropower generation system is 85% and the head difference between upstream and downstream of the hydropower system is maintained at 10 m.

4.6 For the monthly flow data of two years in the question above, determine the storage capacity of a proposed reservoir to maintain a constant demand of 50 m3/s using (a) double mass curve method graphically, (b) double mass curve method in a tabular form, and (c) sequent peak method. Also, answer the following: (a) During which month and year the reservoir is completely empty? (b) In which month and year does the reservoir get completely filled up? (c) In which month and year there will be the first water spill-over? (d) How much is the maximum spill-over and in which month and year? 4.7 The mean monthly runoff (m3/s), pan evaporation (mm), mean monthly rainfall (mm) and monthly demand (Mm3) data at a proposed reservoir site are provided in the following table. The minimum environmental

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flow (e-flow) to be released from the proposed reservoir should be 20 Mm3 or the natural inflow, whichever is greater. The pan coefficient value is 0.65 and the average submergence area for the proposed reservoir is 8 km2. Determine the storage capacity of the proposed reservoir if the runoff coefficient for the catchment in the proposed site is estimated as 0.50. Month

Runoff (m3/s)

Evaporation (mm)

Rainfall (mm)

Demand (Mm3)

Jan

20.8

130

13

23.1

Feb

17.0

135

17

25.0

Mar

13.0

140

4

26.9

Apr

9.5

200

6

30.5

May

8.0

210

2

27.5

Jun

21.0

170

130

29.5

Jul

77.0

130

220

23.5

Aug

83.2

130

180

23.5

Sep

62.0

120

200

22.0

Oct

32.0

110

10

22.0

Nov

28.4

100

40

19.5

Dec

24.0

120

25

22.5

5

Hydrograph Analysis

LEARNING OBJECTIVES LO 1

Understand a unit hydrograph

LO 2

Develop a unit hydrograph for a catchment

LO 3

Explain S-hydrograph and instantaneous unit hydrograph

LO 4

Calculate a unit hydrograph of different durations

LO 5

Summarize different types of synthetic unit hydrographs in an ungauged catchment

LO 6

Outline the relationships among unit hydrograph, instantaneous unit hydrograph, and S-Hydrograph

5.1

INTRODUCTION

In the previous chapter, we learnt about the methods of estimation of runoff in a catchment subjected to rainfall. Those methods focused on estimating the total runoff depth or volume over a long period of time, without considering the variation of runoff over time. The process of translation of rainfall into runoff is a highly complex, nonlinear and dynamic process. From the time rain starts to fall on a catchment, the water goes through various phases or components of the hydrologic cycle, e.g., initial losses in the form of interception and depression storage, infiltration, etc., before getting converted into surface and subsurface flow. In this chapter, we will focus on the dynamic nature of the rainfall-runoff process in a catchment in which rain falling for a short duration results in a storm hydrograph at the outlet of the catchment. The methods focusing on modeling the rainfall-runoff process due to a short-duration rainfall resulting into a storm hydrograph are

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called event based rainfall-runoff models. The event based simulation of the rainfall-runoff process provides us vital information about the variation of runoff with respect to time, such as, the timing and magnitude of the peak runoff. Peak discharges are useful in hydraulic design and flood management studies. The storm hydrograph is useful in understanding the runoff-response characteristics and the physical processes in a catchment. Also, the concept of unit hydrographs is useful in developing distributed rainfall runoff models, i.e., those accounting for spatial variations in the hydrologic and sub-basin characteristics in a catchment.

5.2

UNIT HYDROGRAPH

There have been numerous methods proposed in literature to model the LO 1 Understand a Unit complex rainfall-runoff process. The first event based rainfall-runoff model Hydrograph was probably developed by Mulvany (1850) and was called the Rational Method. Sherman (1932) proposed the concept of Unit Hydrograph, which is extremely simple to use and that is probably why it is still employed in several advanced rainfallrunoff models/software.

5.2.1

Definition

A Unit Hydrograph (UH) is defined as a direct runoff hydrograph (DRH) resulting from a catchment subjected to a unit effective rainfall (ER) of 1 cm (or 1-inch, 1-mm) occurring uniformly over space and time for a specified duration (D-hours). A unit hydrograph is usually referred to as the D-hour UH, where D represent the time for which the effective rainfall occurs. Since a D-hour UH represents runoff discharge due to unit ER, the units of the ordinates of a D-hour UH are cubic meters per second per cm of ER (m3/s-cm). However, the units of a UH are normally represented as m3/s only, with the assumption that it is from a unit amount of effective rainfall. Uniformity of space means the value of rainfall is same everywhere in the catchment; while the uniformity over time means that the intensity of rainfall is same over the entire duration of D-hours. The condition of uniformity over space and time is an important condition which may be quite restrictive in reallife, particularly over large catchments subjected to storms having large values of D. While a UH is useful when rainfall is uniform over time, an instantaneous unit hydrograph (discussed later) can be useful when a catchment is subjected to varying rainfall intensity. Since 1 cm of ER occurs over D-hours, the intensity of rainfall will be equal to 1/D cm/h, which is assumed to be constant for the entire duration. It is important to note that D is the duration of effective rainfall, i.e., the catchment is assumed to be completely saturated during this period. In other words, all the abstractions in the form of interception, depression, etc., have been satisfied and the soil is saturated. Therefore, the total volume of the effective rainfall (1 cm × A, where A is the drainage area of the catchment) passes through the outlet of the catchment and is equal to the total volume of the DRH given by the area under a UH-curve. � EXAMPLE 5.1 The UH of a catchment is triangular in shape having time base of 10 hours and peak discharge of 20 m3/s. How much is the drainage area at the outlet of the catchment?

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147

Solution A UH represents runoff volume equivalent to the rainfall volume due to 1 cm of ER. The runoff volume under a UH is equal to the area under the UH. We get the drainage area by equating the rainfall and runoff volumes: UH Volume = 1 cm ER ¥ (Area) Ê m3 ˆ 1 10 (hours) ¥ 20 Á ˜ = 1 cm ER ¥ (Area) 2 Ë s ¯ Ê m3 ˆ 1 1 10 (3600)s ¥ 20 Á = m ER ¥ (Area) ˜ 2 Ë s ¯ 100 Area = 36 km2

5.2.2

Theory and Assumptions

The UH theory of modeling the effective rainfall-runoff process is based on several assumptions. One of the most important assumptions of the UH theory relates to its linear response. The runoff response from a catchment, when it is subjected to uniform ER, is assumed to be linear in nature. As mentioned earlier, the process of conversion of rainfall into runoff is highly complex and nonlinear in nature but the UH theory assumes the response to be linear. The assumption of linearity enables us to obtain reasonable results for practical use in hydrology. This is a very important assumption as it allows us to employ the principle of proportionality and the principle of superposition while using the UH in a catchment. The principle of proportionality essentially means that given the UH for a catchment, the response from the catchment subjected to x cm of ER can be obtained simply by multiplying the respective ordinates of the given UH by x. The principle of superposition allows us to combine DRHs from several different storm events in a catchment having same duration in order to obtain a DRH at the outlet. It is based on the principle that if y1 and y2 are solutions to a linear system then y1 + y2 will also be a solution. The application of the two principles is explained in greater details in the next section. Apart from the two basic assumptions of the UH theory described above, there are some other assumptions or limitations of the UH theory: ∑

The application of UH in a catchment is time invariant, i.e., the DRH response from 1 cm of ER will always be same irrespective of the time of the year.

∑

The time bases of all the DRHs resulting from different magnitudes of ERs, but having the same duration, will be same.

∑

The ordinates of different DRHs having a common time base will be directly proportional to the runoff volumes represented by the respective DRHs.

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Engineering Hydrology

∑

The DRH response from a catchment represents unchanging characteristics of the catchment, i.e., once the UH has been derived from a catchment, the changes occurring in the catchment (e.g., urbanization etc.) cannot be reflected in the UH (unless new data is obtained and a new UH is derived).

∑

The UH theory is applicable to rainfall only, and not for any other type of precipitation, e.g., snow, ice, or hail etc.

∑

The UH theory is limited to catchments with drainage areas ranging from approximately 2 km2 to 5,000 km2.

∑

Finally, the UH theory assumes that there are no storage components (either manmade, in the form of reservoirs, or natural, in the form of lakes and ponds) in the catchment. This is because of the fact that the existence of the large storage elements in a catchment makes the relationship between storage and outflow from the catchment nonlinear, which violates the assumption of linearity of the UH theory.

5.2.3

Application of UH

Assuming that a D-hour UH from a catchment is available, it can be used to determine the DRH from a complex storm consisting of two or more ERs each having duration of D-hours. In doing so, the principle of proportionality and principle of superposition are applied. This is explained with the help of an example below. � EXAMPLE 5.2 The ordinates of a 3-hour UH are given in the following table. Time (Hours)

0

3

6

9

12

15

18

21

24

27

30

3-hour UH (m3/s)

0

10

30

80

60

50

40

30

10

2

0

Determine the DRH due to (a) an ER of 2 cm, and (b) two ER pulses of 2 cm and 3 cm occurring in two successive 3-hour durations each. Solution The calculations are arranged in a tabular form below. Column [1] and column [2] represent the given 3-hour UH. Column [3] is obtained by multiplying column [2] by 2 and represents the DRH resulting from an ER of 2 cm starting at t = 0 and continuing for 3 hours. ER of 3 cm starts 3-hours later; therefore, the DRH response from this 3 cm of ER will also start with a lag of 3 hours. The entries in column [4] are obtained by multiplying ordinates in column [2] by 3 and lagging the respective ordinates by 3 hours. The combined DRH due to two ER pulses of 2 cm and 3 cm occurring in two successive 3-hour durations each is thus obtained by summing up the entries in column [3] and column [4]. The column [5] represents the combined DRH due to the complex storm having 2 cm and 3 cm ERs occurring in two consecutive 3-hour intervals. This example demonstrates the use of the principle of proportionality and superposition of the UH theory. The 3-hour UH and the DRHs are shown in Figure 5.1.

Hydrograph Analysis

Time (Hours) [1]

0 3 6 9 12 15 18 21 24 27 30 33

Given UH (m3/s) [2]

DRH 2 cm (m3/s) [3]

0 10 30 80 60 40 30 20 10 2 0

0 20 60 160 120 80 60 40 20 4 0

DRH 3 cm (m3/s) [4]

Combined DRH (m3/s) [5]

0 30 90 240 180 120 90 60 30 6 0

0 20 90 250 360 260 180 130 80 34 6 0

Figure 5.1

149

The 3-hour UH and DRHs

The above example shows the procedure to be used to calculate DRH response from a catchment subjected to a complex storm consisting of two constant-intensity rainfalls of equal duration. This procedure, in fact, can be extended to calculate DRH from a complex storm consisting of any number of equal-duration ER pulses in the ERH. This can be easily done in a tabular form as explained in the solved example above. However, a procedure is needed for computing the composite DRH response from a catchment subjected to a complex storm consisting of several ER pulses. This procedure can be easily implemented in a computer. This process of conversion of ER into runoff is known as the convolution process and since the data considered is taken at a discrete time step, the process is also known as the discrete convolution process. The process explained in Example 5.2 above to calculate DRH from a given ERH is nothing but the discrete convolution process. The above procedure of calculating the composite DRH from a complex storm can be represented by the discrete convolution process as follows: ∑

Discrete Convolution n£M

Qn =

 Pm Un - m + 1

(5.1)

m =1

where, Qn is the DRH ordinate at the outlet of the catchment at the nth time step, Pm is the depth of ER at the mth time step, and U’s are the UH ordinate at specified time (represented by n – m + 1). The time step here should be equal to the duration of the effective rainfall, D. Note that Eq. (5.1) takes care of the dynamic effects in the catchment that are responsible for converting the ER into DRH. The index m for ER runs from 1 to M, index n for DRH runs from 1 to N, and the index n – m + 1 for UH ordinates runs from 1 to N – M + 1. That is, if the maximum number of ER pulses in the complex storm is M, which result into N DRH ordinates

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then the number of ordinates of the UH from the catchment will be N – M + 1. The summation in Eq. (5.1) runs from m = 1 to n as long n is less than or equal to M. In other words, the terms in the summation for n > M are not evaluated. This is obvious because M is the maximum number of ER pulses in the ERH.

5.3

DERIVATION OF UNIT HYDROGRAPH

The derivation of a D-hour UH from a catchment requires the measured LO 2 Develop a unit streamflow hydrograph at the outlet of the catchment, the rainfall data hydrograph for a catchment in the form of hyetograph, and the drainage area at the outlet of the catchment. The storm can be either an isolated storm of D-hour duration with no rainfall before and after; or it could also be a complex storm consisting of several rainfall pulses each of D-hour duration occurring back to back with no gap in between. The amount of rainfall in each pulse of a complex storm can be different. The method of derivation of a D-hour UH would be different depending on whether it is derived from an isolated storm or a complex storm. In either case, extreme care needs to be exercised while selecting the data corresponding to a storm for deriving the D-hour UH from a catchment, in terms of the basic assumptions of the uniformity of the rainfall with respect to both space and time. The uniformity with respect to space may be ascertained by examining the rainfall magnitudes for the same storm at rain gauges located in the catchment at different locations. Storms showing spatial variations are discarded and the one having about the same magnitude of rainfall at various places (in each of the D-hour durations) is selected. Similarly, the variation of rainfall with respect to time should also be checked to be uniform. The rate of rainfall throughout each of the D-hour durations should be nearly constant in the selected storm.

5.3.1

Baseflow Separation

Once the rainfall and streamflow hydrograph data is selected for the derivation of a D-Hour UH from a catchment, the first step is separation of the baseflow. The total flow measured at a location in a catchment consists of quick flow and delayed flow. While the quick flow normally consists of surface flow and interflow, the delayed flow is essentially the baseflow. Since in the hydrograph analysis and in UH studies, we are concerned with effective rainfalls and direct runoff from a catchment, it is necessary to separate the baseflow from the total flow to obtain the DRH. Although there are several methods proposed in literature to separate the baseflow, only some commonly used methods are discussed here. Various methods of baseflow separation are depicted in Figure 5.2. These methods differ in the manner of identifying the beginning and ending points of DRH on the measured streamflow hydrograph and joining these two points. While identification of the beginning of a DRH is clearly marked by the sharp change in slope of the observed streamflow hydrograph, it is not easy to determine the end-point of DRH, which also marks the beginning of the base-flow-only portion on the streamflow hydrograph.

5.3.1.1

Straight Line Method

In this method, a horizontal line is drawn from the beginning of the DRH (point A in Figure 5.2) to intersect the streamflow hydrograph at B. The point B is taken as the point at which direct runoff ends and the baseflow starts. The horizontal line AB then separates the baseflow from the DRH and essentially means that the baseflow is assumed to be constant. This method is the simplest one and is more suitable for ephemeral or intermittent streams which have very little contribution from baseflow. This method may be modified, if required, by assuming the line AB to be inclined either slightly upwards or downwards from point A towards

Hydrograph Analysis

151

point B. A downward slope is taken if it is expected that the baseflow contribution would reduce during the storm and an upward slope may be used when it is believed that the baseflow contribution would increase during the storm. For example, in a large and/or mildly sloped catchment, the baseflow may reduce during the storm; therefore, a downward slope may be taken. On the other hand, in a small and/or steeply sloped catchment, the baseflow contribution is expected to increase during the storm; therefore, an upward slope may be taken.

Figure 5.2

5.3.1.2

Different baseflow separation methods

Fixed Base Method

In this method, the end point of DRH is estimated using the following empirical expression (Linsley et al., 1949). N = 0.83 A0.2

(5.2) 2

where, N is the time in days since the occurrence of peak flow, and A is the drainage area, in km , at the outlet of the catchment. This end point is denoted by point D in Figure 5.2. The identification of the end of DRH using Eq. (5.2) is only approximate. The baseflow curve prior to point A is extended in forward direction and its intersection with a vertical drawn from the peak flow (point P in Figure 5.2) is identified as C; therefore, AC is the extended portion of the baseflow prior to the flow hydrograph in question. Points C and D are then joined by a straight line such that the line ACD represents the baseflow separation line for the streamflow hydrograph. A variation of this method could be to join C with point B if it is believed that the contribution from baseflow is smaller, as in a large and/or flatly sloped catchment (not shown in the figure).

5.3.1.3 Variable Slope Method In this method, the end of DRH is determined graphically by using the baseflow recession curve equation (Horton, 1933): Q(t ) = Q(t 0 ) e - K ( t - t0 )

(5.3)

where, Q(t) is the flow at time t on the falling limb, Q(t0) is the flow at time t0 on the falling limb (t > t0), and K is a recession coefficient for the catchment having dimensions of T –1. Note that time t0 corresponds to the

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Engineering Hydrology

end of direct runoff on the falling limb, denoted by point F in Figure 5.2. The recession coefficient can be determined using observed streamflow data on the falling limb. Eq. (5.3) is linearized by taking log so that the curve between log [Q(t)] and time t is a straight line. The linear portion of the recession curve thus obtained is superimposed on the falling limb of the streamflow hydrograph to determine the end point of DRH, point F. Then, the baseflow curve GF is extrapolated backwards to intersect the vertical drawn from the point of inflection on the streamflow hydrograph at point E. Points C and E are then joined by a straight line. Thus, the line ACEFG represents the baseflow separation line as per the variable slope method. Clearly, this method is so named because the slope of the baseflow separation line varies several times from A to F. It is possible to join A and F directly, giving another variation of the straight line method.

5.3.2

Effective Rainfall and Direct Runoff Hydrograph

Once the baseflow has been determined, DRH from the storm is determined by subtracting the baseflow ordinates from the ordinates of total streamflow hydrograph at different time steps. The area under the DRH curve represents the volume of runoff being drained by the catchment during the storm. This volume must be equal to the volume of the effective rainfall. The effective rainfall hyetograph is determined by subtracting losses due to infiltration and other initial losses. This can be done either by using conceptual or empirical equations for infiltration, or F-index (already discussed in Chapter 3).

5.3.3

Derivation of UH from an Isolated Storm

An isolated storm is the one in which there is intense rainfall for a certain duration but no rainfall before and after the storm for considerable durations. If we are interested in finding a D-hour UH from a catchment, then the rainfall records are scanned to find out an isolated storm having duration of D-hours. The storm is selected in such a way that the magnitude of rainfall occurring in D-hours is as high as possible. The rainfall should be uniform over the entire catchment and over the entire duration of D-hours. Once the isolated storm of D-hours duration is selected, the rainfall and streamflow data is compiled. The steps involved in derivation of the D-hour UH are as follows: ∑

DRH from the storm is computed by separating baseflow using one of the methods discussed above.

∑

The volume of direct runoff is determined by estimating the area under the DRH curve.

∑

The volume of runoff is divided by the drainage area (A) to determine the depth of runoff from the catchment. This is equal to the depth of effective rainfall for the storm, say x cm.

∑

The ordinates of the DRH found in step 1 are divided by x to determine the ordinates of the D-hour UH from the catchment.

Note that the principle of proportionality is used in step 4 above to determine the D-hour UH. Often it is difficult to find an isolated storm of duration exactly equal to D-hours and satisfying the uniformity condition over space and time. Therefore, some storms are selected having durations approximately equal to D-hours. For example, if 3-hour UH is to be determined in a catchment, rainfall and streamflow data corresponding to the storms having durations of 2.8 hours to 3.2 hours may be selected. A 3-hour UH is then determined from each of such storms using the procedure outlined above. All such 3-hour UHs are then plotted together on a graph and a best fit curve is obtained to determine a representative 3-hour UH from the catchment.

Hydrograph Analysis

153

� EXAMPLE 5.3 The table below gives the ordinates of a streamflow hydrograph at the outlet of a catchment of an area 600 km2 due to a storm that is believed to have duration of 3 hours. Assuming a constant baseflow of 50 m3/s, compute the ordinates of a 3-hour UH for the catchment. Time (Hours) 3

Flow (m /s)

0

3

6

9

12

15

18

21

24

27

30

33

36

50

600

950

700

530

400

310

240

190

150

110

80

50

Solution The calculations are arranged in a tabular form below. Columns [1] and [2] present the data given. Since the baseflow is given as 50 m3/s, it is subtracted from column [2] to obtain DRH given in column [3]. The area under the DRH in column [3] is approximated by summing up the ordinates and multiplying by 3 hours. Therefore, DRH volume is = 4,00,68,000 m3. Dividing this DRH volume by the catchment area will give the ER responsible for producing the DRH. Thus, ER = 4,00,68,000 m3 / 600 km2 = 6.68 cm. The 3-hour UH ordinates are then obtained by dividing the ordinates of DRH by ER of 6.68. The ordinates of 3-hour UH (in m3/s) are given in column [4] and plotted in the Figure 5.3. Time (Hours)

Given Flow (m3/s)

DRH (m3/s)

3-Hour UH (m3/sm)

[1]

[2]

[3]

[4]

0

50

0

0.0

3

600

550

82.4

6

950

900

134.8

9

700

650

97.3

12

530

480

71.9

15

400

350

52.4

18

310

260

38.9

21

240

190

28.5

24

190

140

21.0

27

150

100

15.0

30

110

60

9.0

33

80

30

4.5

36

50

0

0.0

Figure 5.3

The ordinates of 3-hour UH (in m3/s)

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Engineering Hydrology

5.3.4

Derivation of UH from a Complex Storm

A complex storm is one in which there are several rainfall pulses each having a constant duration. If we are interested in determining a D-hour UH from the catchment, we scan the rainfall records to select a complex storm in which there are several rainfall pulses each having duration of D-hours. The condition of uniformity of rainfall over space and time should be satisfied for each of the D-hour durations of the storm as far as possible. Once a representative complex storm has been selected, ERH and DRH from the complex storm are determined as discussed earlier. Previously, we learnt about the process of convolution of ER to determine DRH from a complex storm by using the discrete form of convolution equations represented by Eq. (5.1). In the process of convolution, ER and ordinates of UH are known, and ordinates of DRH are to be determined. The discrete convolution can be used in a reverse direction to determine the ordinates of the UH when ER and ordinates of DRH are available. This is called the de-convolution process, which is used to determine a D-hour UH from a complex storm. In order to understand the procedure to determine UH using the deconvolution process, let us look at the expanded form of Eq. (5.1) given below: Q1 = P1 U1 Q2 = P1 U 2 + P2 U1 Q3 = P1 U3 + P2 U 2 + P3 U1 QM

� = P1 U M + P2 U M -1 º. + PM -1U 2 + PM U1

QM +1 = P1 U M +1 + P2 U M º. + PM U 2 QN -1 = PM -1 U N - M +1 + PM U N - M QN = PM U N - M +1

¸ Ô Ô Ô Ô Ô ˝ Ô Ô Ô Ô Ô ˛

(5.4)

In Eq. (5.4), Q’s are ordinates of DRH, U’s are ordinates of UH to be determined, and P’s are ER pulses. There are a total of M effective rainfall pulses, N DRH ordinates, and N – M +1 UH ordinates to be determined. This method is applicable when the ordinates of the DRH are available at a time interval equal to D-hours. Equation (5.4) represents simultaneous linear equations that can be easily solved to determine the unknowns, i.e., U’s. Notice that in the first equation, both Q1 and P1 are known and only U1 is unknown, which can be determined as U1 = Q1/P1. Once U1 is known, the second equation can be used to determine U2 = (Q2 – P2U1)/P1. Similarly, U3 can be determined from the third equation and this procedure can be continued until all U’s have been determined. It should be ensured that the total depth of runoff under the UH curve is equivalent to 1.0 cm. If it is not equal to 1.0 cm, then the UH ordinates are proportionately adjusted. This method of calculating the ordinates of a D-hour UH through deconvolution process is known as the method of successive substitution. This process started from the top and went downwards, to solve for the values of various U’s. Notice that the last equation also consists of only one unknown, UN–M+1, which can be

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155

calculated as UN–M+1 = QN/PM. Then starting from the bottom and moving upwards, one can solve for all U’s. If the UH volume thus obtained is not equal to 1.0 cm, then the UH ordinates are proportionately adjusted. Will the ordinates obtained by moving from bottom to top be same as those obtained by moving from top to bottom? Note that there are N equations and only N – M +1 unknown, therefore, the system of equations represented by Eq. (5.4) is over-determined. Only N – M + 1 equations are needed to solve for N – M + 1 unknowns. When we move from top to bottom, we utilize the first N – M + 1 equations and when we move from bottom to top, we utilize the last N – M + 1 equations. Since different sets of N – M + 1 equations are used to determine UH ordinates, chances are that the ordinates of the two UHs will not match. In fact, one can use any N – M + 1 equations of the N equations available to solve for U’s. However, such a procedure employs only a part of the data in the form of P’s and Q’s. This may be avoided to some extent by averaging out various UH solutions obtained by considering different sets of N – M + 1 equations. Further, in order to determine a more mathematically correct and representative UH satisfying all the N equations, the matrix method and optimization methods are employed, which are beyond the scope of this book. Regardless of the method employed, the UH obtained from a complex storm may consist of some negative values and/or fluctuating ordinates on the falling limb. A smooth curve can be drawn through the fluctuations to determine the best fit UH. �EXAMPLE 5.4 A DRH from a complex storm is given below. The DRH resulted from three 3-hour ERs of 8 cm, 3 cm, and 16 cm, respectively. Estimate the 3-hour UH from this data using the deconvolution method. Time (Hours) 3

DRH (m /s)

0

3

6

9

12

15

18

21

24

27

0

160

300

570

636

404

234

105

48

0

Solution There are three ER pulses, P1 = 8 cm, P2 = 3 cm, and P3 = 16 cm, so M = 3. The DRH consists of eight ordinates: Q1 = 160, Q2 = 300, Q3 = 570, Q4 = 636, Q5 = 404, Q6 = 234, Q7 = 105, Q8 = 48, so N = 8. The number of UH ordinates will be N – M + 1 = 8 – 3 + 1 = 6. Let the UH ordinates be U1, U2, U3, U4, U5, and U6. There will eight simultaneous linear equations as per Eq. (5.4); these are given in the table below for the example data. Equations as per Eq. (5.4)

Q1 = P1U1 Q2 = P1U2 + P2U1 Q3 = P1U3 + P2U2 + P3U1 Q4 = P1U4 + P2U3 + P3U2 Q5 = P1U5 + P2U4 + P3U3 Q6 = P1U6 + P2U5 + P3U4 Q7 = P2U6 + P3U5 Q8 = P3U6

After substituting Q and P

Eq. No.

160 = 8 U1 300 = 8 U2 + 3 U1 570 = 8 U3 + 3 U2 + 16 U1 636 = 8 U4 + 3 U3 + 16 U2 404 = 8 U5 + 3 U4 + 16 U3 234 = 8 U6 + 3 U5 + 16 U4 105 = 3 U6 + 16 U5 48 = 16 U6

1 2 3 4 5 6 7 8

156

Engineering Hydrology

These equations can be solved for the unknown U’s. Starting from the top we get U1 = Q1/P1 = 160/8 = 20 m3/s. From Eq. (2), U2 = (Q2 – P2U1)/P1 = (300 – 3 × 20)/8 = 30 m3/s. From Eq. (3), U3 = (Q3 – P2U2 – P3U1)/P1 = (570 – 3 × 30 – 16 × 20)/8 = 20 m3/s. From Eq. (4), U4 = (Q4 – P2U3 – P3U2)/ P1 = (636 – 3 × 20 – 16 × 30)/8 = 12 m3/s. From Eq. (5), U5 = (Q5 – P2U4 – P3U3)/P1 = (404 – 3 × 12 – 16 × 20)/8 = 6 m3/s. From Eq. (6), U6 = (Q6 – P2U5 – P3U4)/P1 = (234 – 3 × 6 – 16 × 12)/8 = 3 m3/s. This completes the determination of all six UH ordinates. The computed UH is presented in the following table and is plotted in Figure 5.4. Time (Hours) 3

3-Hour UH (m /s)

0

3

6

9

12

15

18

21

0

20

30

20

12

6

3

0

The computations may also be done in the reverse order starting from the bottom and moving up to obtain another UH solution. Similarly, any six equations out of the eight equations above may be picked up and solved for U’s. Then an average UH of all the UHs obtained may be determined as the representative 3-hour UH.

Figure 5.4

5.4

3-Hour UH

S-HYDROGRAPH

The S-hydrograph or S-curve is a DRH observed at the outlet of a catchment LO 3 Explain S-hydrograph when it is subjected to an effective rainfall intensity of (1/D) cm/hour for and instantaneous unit an infinite period. It can also be understood as the DRH resulting from a hydrograph catchment when it is subjected to a complex storm having infinite number of ER pulses of 1 cm depth each occurring in D-hour duration. The S-hydrograph is useful in deriving a UH for a different duration from a given UH, as we will see later. It can be obtained by adding several D-hour UHs that are occurring back to back with their starting times spaced D-hours apart. Figure 5.5 shows a D-hour UH and the resulting S-hydrograph when the ordinates of various D-hour UHs are summed up.

Hydrograph Analysis

157

The S-curve attains a maximum value known as equilibrium discharge (QS) which occurs at a time equal to the time base of the D-hour UH. As shown in Figure 5.5, the equilibrium discharge is equal to the sum of all UH ordinates spaced D-hour apart, and represents the rainfall volume due to 1 cm of ER divided by the duration D-hours. In other words, equilibrium discharge represents the maximum rate of direct runoff drained by the catchment when it is subjected to an effective rainfall intensity of (1/D) cm/hour indefinitely. The equilibrium discharge therefore can be calculated as follows: Qs =

1 cm A m3 A km 2 = 2.778 D hour D s

(5.5)

where, D is the duration of ER in hours, and A is the drainage area of the catchment at the outlet in km2. While calculating the S-hydrograph from a given D-hour UH, sometimes it may be observed that the S-curve oscillates around the equilibrium discharge close to the time base of the D-hour UH. It happens due to errors in the data and accumulation of the errors while summing up the UH ordinates. A smooth curve can be drawn through the oscillations to obtain the S-hydrograph in such situations.

Figure 5.5

S-hydrograph from a D-hour UH

�EXAMPLE 5.5 For a 3-hour UH given in the table below, find the S-hydrograph. Estimate the value of the equilibrium discharge obtained from the S-curve. Time (Hours) 3

3-Hour UH (m /s)

0

3

6

9

12

15

18

21

24

27

30

33

0

25

100

160

190

170

110

70

30

20

6

0

158

Engineering Hydrology

Solution The computations for the determination of S-hydrograph are shown in the table below. The columns [1] and [2] represent the data given. Column [3] is S-curve addition column and column [4] is S-curve ordinate column. The first entry of UH ordinate in column [2] is copied to the first entry in column [3]. The first entry in column [4] is the sum of the previous two columns, i.e., sum of 3-hour UH ordinate and the S-curve addition ordinate. The first entries are thus 0 in the first row. The entry in the S-curve ordinate thus obtained is copied in the S-curve addition column after 3-hours (as shown by the arrow). Thus 0 is copied in column [3] against a time of 3-hours. The S-curve ordinate at 3-hours is then the sum of the previous two columns, i.e., 25 + 0 = 25 m3/s. This is now transferred in column [3] against a time of 6-hours. The S-curve ordinate at 6-hours is then the sum of the previous two columns, i.e., 100 + 25 = 125 m3/s. This procedure is repeated till an equilibrium runoff of S-hydrograph is obtained. Note that the value of the equilibrium runoff is 881 m3/s. The 3-hour UH and the S-hydrograph thus obtained are plotted on the same graph and are shown in Figure 5.6.

Time (Hour)

3-Hour UH (m3/s)

S-Curve Addition (m3/s)

S-Curve Ordinate (m3/s)

[1]

[2]

[3]

[4]

0

0

0

0

3

25

0

25

6

100

25

125

9

160

125

285

12

190

285

475

15

170

475

645

18

110

645

755

21

70

755

825

24

30

825

855

27

20

855

875

30

6

875

881

33

0

881

881

36

0

881

881

Figure 5.6

The 3-hour UH and S-hydrograph

Hydrograph Analysis

5.5

159

UNIT HYDROGRAPH OF DIFFERENT DURATIONS

Till now we had focused our attention on hydrograph analysis for a storm LO 4 Calculate a unit having an effective duration of D-hours. While it is important to analyze hydrograph of different data from a catchment and determine a D-hour UH, sometimes it is durations necessary to determine a UH of a duration other than D-hours. Let us say that we have derived a D-hour UH from a catchment and need an nD-hour UH, where n is a constant which may either be an integer or a real number. There may not be sufficient data available from the catchment for storms having a duration equivalent to nD-hours in order to derive an nDhour UH. In such situations, we exploit the UH theory to calculate nD-hour UH given a D-hour UH. The nDhour UH can be determined in one of the two ways: using the method of superposition or using the method involving the use of S-hydrograph. The method of superposition is applicable when n is an integer; while the method of S-hydrograph is more general and can be employed irrespective of whether n is an integer or a real number.

5.5.1

UH of Different Durations by Method of Superposition

Let us consider that we have a D-hour UH available and an nD-hour UH is required. The two principles of the linear system theory, namely, principle of superposition and principle of proportionality are used to determine nD-hour UH from the D-hour UH. This can be done by superimposing n different D-hour UHs that are arranged back to back at D-hours apart. A D-hour UH results from 1 cm of ER starting at time t = 0; the second D-hour UH results from 1 cm ER but starts at D-hours after the first one; the third D-hour UH results from 1 cm ER and starts at D-hours after the second one; and so on. The superimposition of these n different D-hour UHs results in a DRH from n cm of ER occurring in nD-hours. Then we use the principle of proportionality and divide the resulting DRH by n to determine the nD-hour UH. The procedure is explained in the following example. �EXAMPLE 5.6 For a 2-hour UH given in the table below, find the 6-hour UH using the method of superposition. Time (Hours)

0

2

4

6

8

10

12

14

16

18

20

22

2-Hour UH (m3/s)

0

20

60

150

120

90

65

50

30

20

10

0

Solution The computations are arranged in a tabular form below. Columns [1] and [2] represent the data given. Please note that n = 6/2 = 3, so we need three columns of 2-hour UHs that are lagged by 2-hours from one another. Columns [3] and [4] contain the 2-hour UH ordinates which are lagged by 2-hours and by 4-hours, respectively. Column [5] shows the DRH due to the three ERs of 1 cm each occurring in three successive 2-hour durations (or 3 cm in 6-hours), which is essentially the sum of the previous three UHs. The 6-hour UH is then obtained by dividing the ordinates of the DRH in column [5] by 3.0. The plots of 2-hour UH and the 6-hour UH are shown in Figure 5.7. Note the difference between the 2-hour UH and 6-hour UH in the figure. The 2-hour UH has higher peak discharge that occurs earlier as compared to those for the 6-hour UH. It is because the intensity of rainfall for a 2-hour rainfall is greater (1/2 cm/h) than that for the 6-hour UH (1/6 cm/h). Also, the areas under the two UHs are same, which is equivalent to the ER depth of 1 cm.

160

Engineering Hydrology

Time (Hour)

2-Hour UH (m3/s)

2-hr UH Lagged 2-hr UH Lagged by 2-hours by 4-hours (m3/s) (m3/s)

DRH Sum [2]+[3]+[4] (m3/s)

6-Hour UH (m3/s)

[1]

[2]

[5]

[6]

0

0

0

0.0

2

20

0

20

6.7

4

60

20

0

80

26.7

6

150

60

20

230

76.7

8

120

150

60

330

110.0

10

90

120

150

360

120.0

12

65

90

120

275

91.7

14

50

65

90

205

68.3

16

30

50

65

145

48.3

18

20

30

50

100

33.3

20

10

20

30

60

20.0

22

0

10

20

30

10.0

24

0

0

10

10

3.3

26

0

0

0

0

0.0

28

0

0

0

0

0.0

Figure 5.7

[3]

[4]

Plot of 2-hour UH and 6-hour UH

Hydrograph Analysis

5.5.2

161

UH of Different Durations by S-Hydrograph Method

The method of superposition discussed above cannot be used when n is not an integer. Let us consider a case when we want to find a T-hour UH given a D-hour UH so that ratio T/D can be a fraction. We first obtain the S-hydrograph from the given D-hour UH and then use it to determine the T-hour UH. Determination of T-hour UH from a D-hour UH also uses the two principles of the linear systems theory on which UH is based, i.e., superposition and proportionality. Once the S-hydrograph has been obtained, it is lagged by T-hours and the difference between the two S-hydrographs is obtained.

Figure 5.8

Determination of T-hour UH using S-hydrograph

Figure 5.8 shows the process of determining a T-hour UH from a D-hour UH through S-hydrograph. The original S-hydrograph obtained by the given D-hour UH is denoted as SA in the figure, while the second S-hydrograph lagged by T-hours is denoted by SB. The first S-hydrograph (SA) represents the DRH from the catchment subjected to an ER intensity of (1/D) cm/hour starting at t = 0 and occurring indefinitely as shown in Figure 5.8. Similarly, the second S-hydrograph (SB) represents the DRH from the catchment subjected to an ER intensity of (1/D) cm/hour starting at t = T hours and occurring indefinitely. Therefore, the difference between the two S-hydrographs (SA – SB) represents a DRH resulting from an ER intensity of (1/D) cm/hour occurring for T hours. In other words, the difference hydrograph (SA – SB) represents a DRH resulting from an ER of (T/D) cm occurring in T-hours. Therefore, using the principle of proportionality, the T-hour UH can be determined by dividing the ordinates of (SA – SB) by (T/D). The procedure is explained in the following example. �EXAMPLE 5.7 Given a 2-hour UH, find the 3-hour UH from the catchment. Time (Hours) 3

2-Hour UH (m /s)

0

2

4

6

8

10

12

14

16

18

20

22

0

20

60

150

120

90

70

50

32

20

10

0

162

Engineering Hydrology

Solution For the determination of a 3-hour UH from a 2-hour UH, S-hydrograph method will be used. The computations are arranged in a tabular form below. Columns [1] and [2] present the 2-hour UH interpolated at every hour. Note that interpolation is necessary because the S-hydrograph will need to be lagged by 3-hours (= T) later. Columns [3] and [4] are the S-curve addition and S-curve ordinate columns. The computation of S-curve ordinates is done in the same manner as explained earlier except that transferring of the S-hydrograph ordinates to the S-curve addition column is done after 2-hours, which is the duration of the effective rainfall. The S-hydrograph is presented as SA hydrograph in column [4]. Column [5] presents the S-hydrograph lagged by 3-hours (denoted as SB-curve). Column [6] is obtained by subtracting entries in column [5] from those in column [4] to obtain the (SA-SB) hydrograph. The (SA-SB) hydrograph represents the DRH resulting from T/D = 3/2 cm of ER occurring over 3-hours. Thus, the 3-hour UH is obtained by dividing the entries in column [6] by 3/2. The 3-hour UH is presented in column [7]. Figure 5.9 shows 2-hour UH, 3-hour UH, and the two S-hydrographs. Time (Hour)

2-Hour UH (m3/s) [2]

S-Curve Addition (m3/s) [3]

SA-Curve Ordinate (m3/s) [4]

[1]

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

0 10 20 40 60 105 150 135 120 105 90 80 70 60 50 41 32 26 20 15 10 5 0 0

0 0 0 10 20 50 80 155 230 290 350 395 440 475 510 535 560 576 592 602 612 617 622 622

0 10 20 50 80 155 230 290 350 395 440 475 510 535 560 576 592 602 612 617 622 622 622 622

SB-Curve Ordinate (m3/s) [5]

SA-SB Ordinate (m3/s) [6]

3-Hour UH (m3/s) [7]

0 10 20 50 80 155 230 290 350 395 440 475 510 535 560 576 592 602 612 617 622

0 10 20 50 70 135 180 210 195 165 150 125 115 95 85 66 57 42 36 25 20 10 5 0

0.0 6.7 13.3 33.3 46.7 90.0 120.0 140.0 130.0 110.0 100.0 83.3 76.7 63.3 56.7 44.0 38.0 28.0 24.0 16.7 13.3 6.7 3.3 0.0

Hydrograph Analysis

Figure 5.9

5.6

163

2-hour UH, 3-hour UH, and the two S-hydrographs

SYNTHETIC UNIT HYDROGRAPH

The methods discussed above for derivation of a UH require rainfall LO 5 Summarize different data for the catchment and streamflow data at a particular location in the types of synthetic unit catchment and, therefore, the resulting UH is applicable at that location hydrographs in an ungauged only. In several practical situations, a UH at another location in the same catchment catchment may be needed where streamflow data is not available due to certain reasons. We may also want to calculate UH in a nearby catchment which is ungauged but is similar in hydrological characteristics. In such situations, a UH can be derived synthetically at other locations in the same catchment or a nearby catchment where the data is not available. There are several methods for estimating a Synthetic Unit Hydrograph (SUH) in a catchment, which can be broadly grouped into three categories. The first type of methods is based on developing relationships among geo-morphological characteristics of the catchment (e.g., area, length of stream, slope of catchment, etc.) and the hydrograph characteristics such as peak flow, time to peak, time base, etc. The two SUHs commonly used in this category are Snyder’s SUH (Snyder, 1938) and Gray’s SUH (Gray, 1961). The second type of methods involves developing a dimensionless UH using data from a similar catchment. One widely used method in this category is the dimensionless UH developed by the Soil Conservation Service of USA (SCS, 1972). The third type of methods to develop SUH is conceptual in nature and uses the concept of catchment storage. The two popular methods in this category are Clark’s method (1943) and Nash’s method (1957). We study the Snyder’s SUH and the SCS dimensionless SUH here and the conceptual models by Clark and Nash will be described in Chapter 6.

5.6.1

Snyder’s Synthetic Unit Hydrograph

Snyder studied many catchments of drainage areas varying from 25 km2 to 25,000 km2 in the Appalachian region in USA and developed synthetic relationships among various geo-morphological and hydrograph

164

Engineering Hydrology

characteristics of a catchment. He proposed the concept of a Standard unit hydrograph, defined as a unit hydrograph for which the time to peak (tp) is approximately equal to 5.5 times the effective storm duration (tr), tp = 5.5 tr. The two other terms used in connection with the Snyder’s method are: derived unit hydrograph and required unit hydrograph. Derived unit hydrograph is used to obtain the coefficients of the Snyder SUH based on rainfall and streamflow data from a nearby similar catchment and required unit hydrograph refers to a synthetic unit hydrograph of the required duration. The US Army Corps of Engineers (USACE, 1959) developed some additional relationships among the geo-morphological and hydrograph characteristics in a catchment for these non-standard unit hydrographs, which are used first to derive the coefficients and then to obtain the required UH, as explained below. The standard UH and the required UH are shown in Figure 5.10. The various relationships in a catchment as per Snyder (1938) and USACE (1959) in modified form are presented here. Four main characteristics of the required UH for which synthetic relationships are developed are: (a) peak runoff per unit drainage area of catchment at the outlet, qpR (m3/s-km2), (b) the basin lag defined as the time elapsed between the centroid of the ERH and the peak of the UH, tpR (hours), (c) the time base or base time of UH equal to the time elapsed between the start and end of DRH, tb (hours), and (d) widths of the SUH at 50% and 75% of peak runoff, W50 and W75 (hours). Using the values of these hydrograph characteristics, a required synthetic unit hydrograph can be drawn.

Figure 5.10 Snyder’s Standard and Required Unit Hydrographs As mentioned previously, a standard unit hydrograph is defined as one for which the following relationship holds: tp = 5.5 tr

(5.6)

The basin lag (tp, in hours) and the peak runoff for a standard UH can be calculated using the following equations as per Snyder (1938).

Hydrograph Analysis

tp = 0.75 Ct(LLC)0.3 qp = 2.78

Cp tp

165

(5.7) (5.8)

where, Ct and Cp are the coefficients that need to be derived using data from a nearby gauged catchment that is hydrologically and climatically similar to the catchment under consideration; L is the length of the main river/stream in the catchment (km); Lc is the distance between a point on the main river/stream nearest to the centroid of the catchment and the outlet of the catchment (km); and qp is the peak runoff per unit of the drainage area for the standard UH from the catchment (m3/s-km2). The two lengths are the geo-morphological parameters of a catchment that can be easily estimated using a topographic map of the area. Once determined, the basin lag and peak runoff for a standard UH can be calculated. In order to derive the estimates of coefficients Ct and Cp, the rainfall and flow data in a nearby catchment, that is climatically and hydrologically similar, are used along with its topographic map. Using the rainfall and flow data for this catchment, a unit hydrograph may be derived using the method described earlier in this chapter. Let the derived unit hydrograph be for effective storm duration of tNR with the values of basin lag equal to tNpR and peak runoff per unit drainage area equal to qNpR. Using the topographic map of this gauged catchment, the lengths L and Lc are estimated. If tNpR = 5.5 tNR, the derived UH is a standard UH and we can calculate Ct and Cp by setting tr = tNR; tp = tNpR; and qp = qNpR in Equations (5.7) and (5.8). However, if tNpR is significantly different from 5.5 tNR, then the following equation is used to define the basin lag: tp = tN pR +

tr - tN R 4

(5.9)

tN ˘ 22 È tN pR - R ˙ and tr = tp/5.5. Once the value of tp Í 21 Î 4 ˚ is known, the values of the two coefficients Ct and Cp can be easily calculated using Equations (5.7) and (5.8) with qp = qNpR. After obtaining the values of the two coefficients Ct and Cp for a gauged catchment, these values can be assumed to be applicable to the ungauged catchment for which synthetic unit hydrograph is required. The basin lag for the catchment is obtained from Eq. (5.7) and if it is equal to 5.5 times the effective rainfall duration tR of the required UH, the standard unit hydrograph is the required UH. If not, the basin lag for the required UH tpR is obtained from Eq. (5.9), using tpR in place of tNpR and tR in place of tNR. Using Equations (5.6) and (5.9), we can find, t p =

The peak runoff per unit drainage area and time base for the required synthetic unit hydrograph in the ungauged catchment can then be estimated as follows: qpR = tb =

qp t p t pR 5.56 q pR

(5.10) (5.11)

The time base is calculated using the fact that the area under a UH represents a DRH volume of 1 cm. The widths of the required synthetic UH at 50% and 75% of peak runoff can be calculated using the following equation:

166

Engineering Hydrology

W = CW(qpR)–1.08

(5.12)

where, the coefficient CW is taken as 1.22 for 75% width and 2.14 for 50% width. While sketching the synthetic UH using all the hydrograph characteristics calculated above, one-third of the widths are distributed before the peak and the remaining after the peak. �EXAMPLE 5.8 The following characteristics are determined from the topographic map of catchment A: LA = 170 km, LAc = 80 km, and drainage area AA = 4,000 km2. Using the rainfall and streamflow data, a UH was derived for catchment A consisting of the following hydrograph characteristics: tAR = 15 hours, tApR = 40 hours, and QApR = 180 m3/s. Evaluate the coefficients Ct and Cp for catchment A. Then derive a 6-hour synthetic unit hydrograph of a nearby catchment-B for which the following catchment characteristics are determined using the topographic map: LB = 120 km, LBc = 60 km, and drainage area AB = 2,500 km2. Solution Let us first calculate Ct and Cp using the data from gauged catchment A (nearby catchment). Catchment A: Note that for catchment A, tApR ≠ 5.5 tAR. So, calculate tp as follows, with tAR = 15 hours, tApR = 40 hours: tp =

tN ˘ 22 È 22 È 15 ˘ tN pR - R ˙ = 40 - ˙ = 37.98 hours. Í Í 21 Î 4 ˚ 21 Î 4˚

From Eq. (5.7), tp = 0.75Ct(LLC)0.3, we get 37.98 = 0.75 Ct (170 × 80)0.3, or Ct = 2.91. The peak discharge Cp per unit area for catchment A, qpR, is 180/4000 = 0.0450 m3/(s-km2). Therefore, from Eq. (5.8), q p = 2.78 , tp we get 0.0450 = 2.78 Cp/37.98. This gives Cp = 0.6147. Thus knowing the two coefficients, Ct and Cp, the computation of 6-hour SUH for catchment B can then be carried out as follows. Catchment B: From Eq. (5.7), tp = 0.75Ct(LLC)0.3, we get tp = 0.75 (2.91) (120 × 60)0.3 = 31.38 hours. Eq. (5.6) tp = 5.5 tr, gives tr = 31.38/5.5 = 5.71 hours. We need to find a 6-hour SUH for which tR = 6 hours. Thus Cp from Eq. (5.9) we get tpR = tp – (tr – tR)/4 = 31.38 – (5.71 - 6)/4 = 31.45 hours. From Eq. (5.8), qp = 2.78 , tp q t p p we get, qp = 2.78 × (0.6178/31.38) = 0.0545 m3/(s-km2) and Eq. (5.10) q pR = , gives t pR qpR = 0.0545 (31.38/31.45) = 0.0543 m3/(s-km2). Therefore, peak flow QpR = 0.0543 × 2500 = 135.8 m3/(s). 5.56 The time base of the SUH is given by Eq. (5.11) tb = = (5.56/0.0543) = 102.3 hours. The widths of the q pR SUH at 75% and 50% heights are given by Eq. (5.12): W = CW(qpR)–1.08. Cw = 1.22 for 75% width gives W75 = 1.22 × (0.0543)–1.08 = 28.3 hours and Cw = 2.14 for 50% width gives W50 = 2.14 × (0.0543)–1.08 = 49.7 hours. Thus knowing all the parameters of the SUH, the 6-hour SUH can be sketched. Note that 1/3rds of the widths are distributed before the peak discharge and 2/3rds after the peak discharge. The ordinates of the Snyder’s SUH can be determined using the hydrograph characteristics determined above. The values of ordinates of SUH at different times and the plotted SUH are shown in Figure 5.11.

Hydrograph Analysis

S. No.

Time

Time (Hours)

0

SUH Ordinate (m3/s)

1

Initial

2

50% Rising = (tpR – W50/3)

14.88

QpR/2 = 67.9

3

75% Rising = (tpR – W75/3)

22.00

3 × QpR/4 =101.45

4

Peak = tpR

31.45

QpR =135.8

5

75% Falling = (tpR + 2 × W75/3)

50.35

3 × QpR/4 =101.45

6

50% Falling = (tpR + 2 × W50/3)

64.60

QpR/2 = 67.9

7

End = tb

102.3

167

0.0

0.0

Figure 5.11 Synthetic Unit Hydrograph for Example 5.8

5.6.2

SCS Dimensionless Synthetic Unit Hydrograph

The Soil Conservation Service (SCS, 1972) presented a simple method of generating a synthetic unit hydrograph in an ungauged catchment. In the SCS dimensionless synthetic UH, the ordinates of the UH are expressed as the ratio of runoff and peak runoff (Q/Qp). Similarly, the times are expressed as the ratio of time t to the time to peak runoff (t/Tp). The time to peak (Tp) is also referred to as the time of rise. The coordinates of the SCS dimensionless unit hydrograph are presented in Table 5.1, which can be used to generate a synthetic unit hydrograph.

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Table 5.1 Coordinates of the SCS dimensionless synthetic unit hydrograph t/Tp

Q/Qp

t/Tp

Q/Qp

0.0

0.000

1.4

0.750

0.1

0.015

1.5

0.660

0.2

0.075

1.6

0.560

0.3

0.160

1.8

0.420

0.4

0.280

2.0

0.320

0.5

0.430

2.2

0.240

0.6

0.600

2.4

0.180

0.7

0.770

2.6

0.130

0.8

0.890

2.8

0.098

0.9

0.970

3.0

0.074

1.0

1.000

3.5

0.036

1.1

0.980

4.0

0.018

1.2

0.920

4.5

0.009

1.3

0.840

5.0

0.004

Obviously, the values of Qp and Tp for the ungauged catchment will be required to develop the SCS synthetic dimensionless UH. The SCS has suggested guidelines for selecting Qp and Tp for the ungauged catchment using the concept of a triangular unit hydrograph. The concepts of SCS synthetic dimensionless UH and the triangular UH are shown in Figure 5.12.

Figure 5.12 Concepts in SCS dimensionless unit hydrograph

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169

The volume under the triangular UH before the peak is same as that under the SCS dimensionless UH before the peak given in Table 5.1. After analyzing several catchments, SCS has suggested that the time of recession or the time between peak and end of triangular UH, is approximately equal to 1.67 times the time to peak. In other words,

Or

Tb - Tp = 1.67 Tp

(5.13)

Tb = 2.67 Tp

(5.14)

Now, as the area under the triangular UH should be equal to the direct runoff of 1 cm, therefore Q p = 2.08

A Tp

(5.15)

where, A is the drainage area in km2 and Tp is as shown (in hours). A study of several rural catchments has shown that basin lag (tp) is approximately equal to 0.6 times the time of concentration (Tc) of the catchment. Therefore, the time of rise (Tp) can be expressed as follows: Tp = 0.6 Tc +

tr 2

(5.16)

Knowing the time of concentration (Tc) and the effective rainfall duration (tr), one can calculate Tp by Eq. (5.16). Then, knowing the drainage area and Tp thus calculated, one can find Qp from Eq. (5.15). Thus, the SCS dimensionless unit hydrograph for the ungauged catchment can be easily computed using Qp and Tp in conjunction with the coordinates given in Table 5.1. �EXAMPLE 5.9 Derive a 30-minute SCS unit hydrograph for a catchment having a drainage area of 5.0 km2 and a time of concentration of 2.0 hours. Solution Given tr = 30 min = 0.5 hours; Tc = 2 hours, and drainage area A = 5.0 km2. Let us first calculate the SCS t triangular SUH parameters. Eq. (5.16) Tp = 0.6 Tc + r , gives Tp = 0.6(2.0) + (0.5/2) = 1.45 hours; Eq. (5.15) 2 A Q p = 2.08 , gives Qp = 2.08 (5.0/1.45) = 7.17 m3/s; and Eq. (5.14) Tb = 2.67 Tp, gives Tb = 2.67(1.45) = Tp 3.87 hours. Knowing the values of Qp and Tp, the SCS dimensionless SUH can be easily determined using Table 5.1. The computations are shown in a tabular form below. The SCS dimensionless SUH and the SCS triangular SUH for duration of 0.5 hours for the example problem are shown in Figure 5.13.

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Figure 5.13 SCS SUH and SCS triangular SUH for duration of 0.5 hours

5.7

INSTANTANEOUS UNIT HYDROGRAPH

LO 5

We have seen the concept of a D-hour unit hydrograph and how to derive a UH of different durations. It is possible to derive many UHs of different durations in a catchment; however good quality data for different durations may not always be available. Let us consider four different UHs from a catchment for different durations, i.e., D1-hour, D2-hour, D3-hour, and D4-hour. Figure 5.14 shows the four UHs for these four durations. Note that the duration D1 is the largest and D4 is the smallest, as intensity is reciprocal to duration (= 1/D cm/hour). Since the depth of effective rainfall for each UH is 1 cm, by definition of a UH, the intensity of effective rainfall for each UH will be increasing as we go from D1 to D4.

Figure 5.14 Concepts of instantaneous unit hydrograph

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171

The area under a UH represents DRH depth equivalent to ER depth of 1 cm. The storms with higher intensity will result in DRH response having greater and quicker peaks as shown in Figure 5.14. What happens when we keep on reducing the duration of the effective rainfall such that it approaches zero, i.e., the rainfall of 1 cm occurs instantaneously on the catchment? This will result in a DRH that will have very high and early peak as compared to the other UHs/DRHs. The DRH resulting from a catchment when it is subjected to a uniform effective rainfall of 1 cm instantaneously is known as Instantaneous Unit Hydrograph (IUH). The IUH from a catchment is independent of any duration and is a unique characteristic of the catchment in terms of its DRH response under extreme and hypothetical conditions. The concept of an IUH in a catchment is very useful in determining DRH from an ERH of variable intensity, which is difficult using the concept of UH. The concept of IUH is also useful in deriving the UH in any catchment at locations that do not have data or in an ungauged catchment. Properties of an IUH are as follows: (a) the ordinate of an IUH at any time t, u(t) is equal to zero at t = 0 and positive for t > 0, (b) u(t) Æ 0 as t Æ ∞, and (c) area under the IUH represents DRH depth equivalent of ER depth of 1 cm like a UH. While a UH is useful in finding the DRH from an ER of certain duration having a constant rainfall intensity, the IUH can be used to find DRH from variable intensity ER. We will learn how to derive a conceptual IUH from an ungauged catchment later in Chapter 6.

5.8

RELATIONSHIPS AMONG UH, IUH, AND S-HYDROGRAPH

The IUH, UH, and S-hydrograph represent the DRH response from a LO 6 Outline the relationcatchment under different inputs and conditions. Therefore, they are interships among unit hydrograph, related and this allows us to derive one from the other. Let us consider instantaneous unit hydroan S-hydrograph corresponding to a D-hour UH from a catchment. Let graph, and S-Hydrograph us denote this S-hydrograph as S1 which is the DRH response from the catchment when an ER intensity of i = 1/D cm/hour is applied to it indefinitely. Let S2 be the same S-hydrograph from the same catchment but lagged by a small time interval of dt-hours. Now, using the concept similar to determining a T-hour UH from a D-hour UH using S-hydrograph discussed earlier, it follows that a dt hour UH from this catchment can be determined as follows: dt hour UH =

S2 - S1 i dt

(5.17)

As dt Æ 0, the dt hour UH approaches towards IUH. Therefore, the ordinate of an IUH can be written as Ê S - S1 ˆ 1 dS u(t ) = lim Á 2 = dt Æ 0 Ë i dt ˜¯ i dt

(5.18)

For i = 1 cm/hour or dt = 1-hour u(t ) =

dS dt

(5.19)

where, S is an S-hydrograph from the catchment produced by an effective rainfall of intensity 1 cm/hour. Therefore, the IUH at any time t from a catchment is equal to the slope of the S-hydrograph produced from an ER intensity of 1 cm/hour. Note that this S-hydrograph can be obtained from a 1-hour UH from the catchment. Thus, knowing the ordinates of the S-hydrograph for i = 1 cm/hour, an IUH can be determined.

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Engineering Hydrology

It is possible to derive a D-hour UH given an IUH also. Integrating Eq. (5.19) between times t1 and t2, we get t2

S2 - S1 = Úu(t )dt

(5.20)

t1

If the time interval Δt = t2 – t1 is sufficiently small such that the IUH can be considered linear during this interval, then the IUH ordinate during time interval dt may be taken as the linear interpolation of the two values at the beginning and end of the time interval as follows: u(t ) = u(t ) =

1 Èu(t1 ) + u(t2 )˘˚ 2Î

(5.21)

Now combining the two Equations (5.20) and (5.21) we get, S2 - S1 1 = ÎÈu(t1 ) + u(t2 )˚˘ Dt 2

(5.22)

The term on the left hand side of Eq. (5.22) is nothing but a dt-hour UH. Letting D = dt represent the desired duration, the ordinates of a D-hour UH in a catchment can be obtained by averaging the IUH ordinates at two times such that their difference represents the D-hour duration, i.e., D = t2 – t1. Note that this method is valid only when the DRH response from the catchment during the D-hour intervals can be assumed to be linear. Therefore, the value of the effective rainfall duration should be small, preferably 1-hour or less. Once a D-hour UH has been determined, its duration can be converted to another value using one of the methods discussed earlier in this chapter. �EXAMPLE 5.10 An IUH from a catchment is given in the following table. Find: (a) 3-hour UH from the catchment, and (b) DRH due to an ERH consisting of ERs of 2 cm and 4 cm in two 3-hour intervals. Plot IUH, 1-hour UH, 3-hour UH, and the total DRH on the same graph. Time (h)

0

1

2

3

4

5

6

7

8

IUH (m3/s)

0

10

30

60

45

30

15

5

0

Solution The computations are arranged in a tabular form below. The first two columns represent the IUH data given in the example. Column [3] presents the IUH lagged by 1-hour and column [4] presents the 1-hour UH, which is calculated by taking average of IUH ordinates that are 1-hour apart. Then using the 1-hour UH, 3-hour UH is obtained by using the method of superposition in column [5] through column [8]. Column [9] is obtained by multiplying the 3-hour UH ordinates by 2 to calculate DRH due to the 2 cm ER. Then, column [10] is obtained by multiplying 3-hour UH by 4 and lagged by 3-hours to obtain the DRH from 4 cm of ER occurring after 3-hours. The last column [11] is obtained by summing up the respective entries in columns [9] and [10] to obtain the total DRH from 2 cm and 4 cm of ER. The graph shown in Figure 5.15 represents all the plots.

Hydrograph Analysis

173

Time (Hour)

IUH (m3/s)

IUH lagged by 1 hr (m3/s)

1-hour UH ([2]+[3])/2 (m3/s)

1-hr UH lagged by 1 hr (m3/s)

1-hr UH lagged by 2 hr (m3/s)

DRH due to 3 cm [4]+[5]+[6] (m3/s)

3-Hour UH [7]/3 (m3/s)

DRH due to 2 cm 2*[8] (m3/s)

DRH due to 4 cm 4*[8] (m3/s)

Total DRH [9]+[10] (m3/s)

[1]

[2]

[3]

[4]

[5]

[6]

[7]

[8]

[9]

[10]

[11]

0

0

0.0

0.0

0.0

0.0

1

10

0

5.0

0

5.0

1.7

3.3

3.3

2

30

10

20.0

5

0

25.0

8.3

16.7

16.7

3

60

30

45.0

20

5

70.0

23.3

46.7

0.0

46.7

4

45

60

52.5

45

20

117.5

39.2

78.3

6.7

85.0

5

30

50

40.0

55

45

140.0

46.7

93.3

33.3

126.7

6

15

40

27.5

45

55

127.5

42.5

85.0

93.3

178.3

7

5

30

17.5

35

45

97.5

32.5

65.0

156.7

221.7

8

0

10

5.0

20

35

60.0

20.0

40.0

186.7

226.7

0

0.0

5

20

25.0

8.3

16.7

170.0

186.7

0

5

5.0

1.7

3.3

130.0

133.3

0

0.0

0.0

0.0

80.0

80.0

12

33.3

33.3

13

6.7

6.7

14

0.0

0.0

9 10 11

0.0

Figure 5.15 IUH and various UHs for Example 5.10

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Engineering Hydrology

SUMMARY The conversion of rainfall into runoff in a catchment is a highly non-linear and complex process. The time distribution of runoff at the outlet of a catchment can be modeled using the concept of a unit hydrograph (UH) proposed by Sherman. A UH is defined as a direct runoff hydrograph (DRH) resulting from a catchment when it is subjected to a unit effective rainfall that is uniform over space and time occurring in a specified duration of D-hours. Although the rainfall-runoff process is non-linear in nature, the UH theory assumes the process to be linear. The assumption of linearity provides reasonably accurate results and allows us to employ the principles of proportionality and superposition in solving various problems involving DRH in a catchment using UH. This chapter discussed methods to determine a D-hour UH in a catchment using the rainfall and streamflow data, and the area of the catchment. The method to determine a D-hour UH is different depending upon whether the data are derived from a single storm event or a complex storm event. The first step in the determination of a UH from a catchment is to collect a representative data set satisfying the underlying assumptions of the UH theory. Once a representative data set is selected, base flow needs to be separated to obtain the DRH from the storm. This can be done by using any of the following methods discussed in this chapter: straight line method, fixed base method, variable slope method, or a variation of any of these methods. Different base flow separation methods differ in the manner of determining the end of the direct runoff during a storm. The discrete continuity equation can be used to calculate DRH knowing rainfall and UH. On the other hand, the discrete convolution equation can be used in an inverse direction to determine a D-hour UH from a complex storm, given rainfall and flow data. An S-hydrograph is a DRH observed at the outlet of a catchment when it is subjected to an effective rainfall intensity of 1/D cm/hour for an infinite period. S-hydrograph is a summation hydrograph obtained from several D-hour UHs that are arranged in series spaced D-hours apart. The S-hydrograph attains a maximum value, known as equilibrium discharge, that occurs at a time equal to the time base of the D-hour UH. While the method of superposition can be used to find an nD-hour UH from a D-hour UH when n is an integer, the S-hydrograph is useful in finding nD-hour UH when n is either an integer or a fraction; therefore, it provides a more general and robust tool for handing a variety of problems in hydrology. The rainfall and flow data are needed to obtain the UH and S-hydrograph from a catchment but sometimes the data in a catchment are not available. In such a situation, we resort to creating Synthetic UH (SUH) in a catchment using one of the following methods: (a) methods based on developing relationships among geomorphological characteristics of the catchment and the hydrograph characteristics, e.g., Snyder’s or Gray’s SUH, (b) methods based on developing a dimensionless SUH using data from a similar catchment e.g. SCS dimensionless SUH, and (c) conceptual methods based on catchment storage e.g. Clark’s method or Nash’s model (not discussed in this chapter). An Instantaneous Unit Hydrograph (IUH) is the DRH resulting from a catchment when it is subjected to a uniform effective rainfall of 1 cm occurring instantaneously and uniformly over the catchment. The concept of an IUH in a catchment is very useful in (a) determining DRH from effective rainfall of variable intensity, and (b) deriving the UH at ungauged locations in a catchment. The relationships among UH, S-hydrograph, and IUH provides important tools for hydrograph analysis in a catchment.

Hydrograph Analysis

175

OBJECTIVE-TYPE QUESTIONS 5.1 The event based simulation of the rainfall-runoff process is useful in (a) Studying the variation of runoff with respect to time (b) Hydraulic design and flood management studies (c) Developing distributed rainfall runoff models (d) All of these (e) None of (a), (b), and (c) 5.2 In a D-hour UH, D represents (a) Time base of the UH (c) Time to peak flow

(b) Time of concentration of the catchment (d) None of these

5.3 The concept of unit hydrograph was first introduced by (a) Mulvany in 1850 (b) Sherman in 1932 (c) Horton in 1933 (d) Thiessen in 1920

(e)

None of these

5.4 A unit hydrograph is a graphical representation between (a) Effective rainfall and total streamflow over time at a location in a river (b) Total rainfall and total streamflow over time at a location in a river (c) Total rainfall and direct runoff over time at a location in a river (d) Effective rainfall and direct runoff over time at a location in a river 5.5 The peak discharge of a 3-hour UH from a catchment is x m3/s. The peak discharge for a 6-hour UH from the same catchment will be (a) Equal to x (b) Less than x (c) Greater than x (d) Insufficient data to comment 5.6 The peak discharges for a 3-hour UH and a 6-hour UH occur at times tp3 and tp6, respectively. The most appropriate relation between tp3 and tp6 is (a) tp3 = tp6 (b) tp3 < tp6 (c) tp3 > tp6 (d) Insufficient data to comment 5.7 The UH of a catchment is triangular in shape having a time base of 20 hours and peak discharge of 200 m3/s. The drainage area at the outlet of the catchment (in km2) is: (a) 72 (b) 720 (c) 1440 (d) 144 (e) Insufficient data 5.8 The most important assumptions of the UH theory are (a) Time variance and nonlinear response (b) Time invariance and nonlinear response (c) Time variance and linear response (d) Time invariance and linear response (e) None of these 5.9 The assumption “ordinates of different DRHs having common time base will be directly proportional to the runoff volumes represented by the respective DRHs” represents (a) Principle of uniformity of rainfall over space and time (b) Principle of superposition (c) Principle of proportionality (d) Principle of time invariance (e) None of these

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Engineering Hydrology

5.10 The peak discharge of a 3-hour UH from a catchment is 200 m3/s. The peak discharge from the same catchment for ER = 0.5 cm occurring in 3-hours will be (in m3/s): (a) 200 (b) 400 (c) 100 (d) Insufficient data 5.11 A DRH from a catchment due to an x-hour storm resulted into y cm of direct runoff depth. The ordinates of this DRH when (a) Divided by x will give y-hour UH (b) Divided by y will give x-hour UH (c) Divided by x will give x-hour UH (d) Divided by y will give y-hour UH (e) Difficult to say 5.12 Which of the statements are true for a UH? I. The time bases of all the DRHs resulting from different magnitudes of ERs, but having the same duration, will be constant. II. The ER should be constant with respect to both space and time. III. The DRH response from a catchment represents changing characteristics of the catchment. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 5.13 A DRH, resulting from a complex storm consisting of 4 pulses of 3-hour duration each, consists of 20 ordinates. Assuming ordinates in DRH and UH to be at 3-hour interval, the number of ordinates in the 3-hour UH for this catchment will be (a) 16 (b) 21 (c) 17 (d) 24 (e) Insufficient data 5.14 What is unit in a unit hydrograph? (a) Duration of effective rainfall (c) Depth of effective rainfall

(b) Intensity of effective rainfall (d) Duration of unit hydrograph

5.15 A 6-hour storm dumped 5.8 cm rainfall on a catchment. The F-index during the storm was found to be 0.25 cm/hour. If the peak discharge of the resulting DRH was 2,460 m3/s, what is the value of the peak discharge (in m3/s) for the 6-hour UH from this catchment? (a) 424 (b) 878 (c) 487 (d) 572 (e) Insufficient data 5.16 A UH starts to rise at 8:00 am, reaches its peak at 12:00 noon and ends at 8:00 pm. Which of the following statements are incorrect? I. The time to peak is 4 hours II. The time base is 12 hours III. The duration of ER is less than 4 hours (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of the statements is incorrect 5.17 The triangular DRH from a catchment of area 389 km2 resulted in a peak runoff of 150 m3/s after 12 hours of the start of a storm. If the time base of the DRH was 72 hours, how much was the effective rainfall during the storm? (a) 0.83 cm (b) 0.50 cm (c) 5.0 cm (d) 50.0 cm (e) Insufficient data 5.18 A 6-hour UH from a catchment is triangular in shape with a peak runoff of 58 m3/s and a time base of 16 hours. What is the area of the catchment? (a) 1.67 km2 (b) 16.7 km2 (c) 167.0 km2 (d) Insufficient data

Hydrograph Analysis

177

5.19 A 6-hour UH from a catchment is triangular in shape with a peak runoff of 260 m3/s and a time base of 42 hours. The time base of a DRH from a storm having an ER of 0.50 cm occurring in 6-hours will be: (a) 42 hours (b) 21 hours (c) 84 hours (d) Insufficient data 5.20 A 6-hour UH from a catchment is triangular in shape with a peak runoff of 260 m3/s and a time base of 42 hours. The peak discharge of a DRH from a storm having an ER of 0.50 cm occurring in 6-hours will be: (a) 260 m3/s (b) 130 m3/s (c) 520 m3/s (d) Insufficient data 5.21 Which of the following is needed to derive a D-hour UH from a catchment? (a) Measured streamflow hydrograph (b) Measured streamflow hydrograph and ERH (c) Measured streamflow hydrograph, ERH, and drainage area of the catchment (d) None of the above 5.22 The total flow measured at a location in a catchment consists of (a) Direct runoff and baseflow (b) Direct runoff and delayed flow (c) Quick flow and delayed flow (d) Quick flow and baseflow (e) All of the above (f) None of (a), (b), (c), or (d) 5.23 The quick flow is the sum of (a) Overland flow and channel flow (c) Overland flow and baseflow (e) None of the above

(b) Surface flow and baseflow (d) Surface flow and interflow

5.24 Different methods of baseflow separation differ in (a) Identifying the end point of DRH on measured streamflow hydrograph (b) The manner of joining the start and end points of DRH (c) Both (a) and (b) (d) None of (a) and (b) 5.25 The ‘Straight Line’ method for baseflow separation is useful for (a) Perennial streams (b) Ephemeral streams (c) Both (a) and (b) (d) More information needed to comment 5.26 The baseflow contribution is expected to increase during a storm in a (a) Large and/or steeply sloped catchment (b) Small and/or mildly sloped catchment (c) Small and/or steep catchment (d) Large and/or mildly sloped catchment (e) Baseflow contribution does not depend on size and slope of catchment 5.27 For a catchment of size 615 km2, the end point of DRH is expected to occur after how many days of occurrence of peak flow? (a) 1 (b) 2 (c) 3 (d) 4 (e) Insufficient data 5.28 As per the Horton (1933) equation for the baseflow recession, the ratio Q(t = t0)/Q(t) can be expressed as (a) e - ( t - t0 )/ K

( t - t )/ K (b) e 0

(c)

e K ( t - t0 )

- K ( t - t0 ) (d) e

5.29 Which of the following statements is/are true for the use of deconvolution method to determine a UH from a complex storm? I. The deconvolution method does not give a unique solution. II. The number of equations is less than the number of unknowns in the method. III. The UH obtained may consist of some negative values. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of the statements is true

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5.30 A triangular DRH from a catchment resulting from a storm having 5.0 cm of effective rainfall consists of a peak runoff of 280 m3/s occurring after 15 hours from the start of rainfall. If the time base of the DRH was 72 hours, what is the area of the catchment? (a) 7.26 km2 (b) 725.8 km2 (c) 151.2 km2 (d) 806 km2 (e) Insufficient data 5.31 The triangular DRH from a catchment of area 280 km2 consists of a peak runoff of 100 m3/s and a time base of 62 hours due to a 3-hour storm. What will be the peak runoff of the 3-hour UH from this catchment? (a) 100 m3/s (b) 25 m3/s (c) 50 m3/s (d) Insufficient data 5.32 The 3-hour UHs from two catchments A and B are triangular in shape with same time base. It has been found that for catchment A: peak runoff = 50 m3/s, Area = 128 km2, and for catchment B: peak runoff = 100 m3/s. What is the area of catchment B? (a) 128 km2 (b) 256 km2 (c) 64 km2 (d) Insufficient data 5.33 The 3-hour UHs from a catchment due to 1 cm and 1 mm ER are represented by Ucm and Umm, respectively. Which of the following statements is/are true for this catchment? I. Ordinates of Ucm are 1/10th of the ordinates of Umm. II. Time base of Ucm is 1/10th of the time base of Umm. III. The peak runoff for both UHs will occur at the same time. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of the statements is true 5.34 The peak discharge of a 6-hour UH from a catchment is 50 m3/s. Later, a storm having 6-hour duration produced a total rainfall of 7 cm with a F–index of 0.50 cm/hour. If the baseflow was 10 m3/s, what is the value of peak discharge of the streamflow hydrograph from this storm? (a) 110 m3/s (b) 210 m3/s (c) 60 m3/s (d) Insufficient data 5.35 S-hydrograph is a summation of (a) Streamflow hydrograph (c) Effective rainfall hyetograph

(b) Unit hydrograph (d) Baseflow hydrograph

5.36 The S-hydrograph can be used to (a) Estimate peak flood from a given storm (b) Develop synthetic unit hydrograph (c) Develop a UH of different duration (d) Derive UH from a complex storm using deconvolution method 5.37 What is the value of the equilibrium discharge (in m3/s) resulting from a catchment when a rainfall intensity of 0.1667 cm/hour is applied on it indefinitely? (a) 108.3 (b) 301.0 (c) 350 (d) Insufficient data 5.38 When does an S-curve attain equilibrium discharge? (a) At a time equal to the time to peak of the D-hour UH (b) At a time equal to the time base of the D-hour UH (c) At a time somewhere in between (d) Not sufficient information to determine 5.39 Which of the following statements is/are true for the S-hydrograph associated to a D-hour UH in a catchment? I. The S-hydrograph is a DRH observed at the outlet of a catchment when it is subjected to an effective rainfall depth of 1 cm for an infinite period. II. The S-hydrograph is a DRH observed at the outlet of a catchment when it is subjected to an effective rainfall intensity of 1 cm/hour for an infinite period.

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179

III. The S-hydrograph is a DRH observed at the outlet of a catchment when it is subjected to an effective rainfall intensity of (1/D) cm/hour for an infinite period. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of the statements is true 5.40 Which of the following statements is/are true for the equilibrium discharge of an S-hydrograph? I. It occurs at a time equal to the time base of the D-hour UH. II. It is equal to the sum of all UH ordinates spaced D-hour apart. III. It represents the rainfall volume due to 1 cm of ER divided by the duration D-hours. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of the statements is true 5.41 Which of the following principles is/are used to determine a UH of a different duration? (a) Proportionality (b) Superposition (c) Both (a) and b (d) None of (a) and (b) 5.42 The volume of a 3-hour UH was found to be 2500 m3. What is the volume of the 6-hour UH from the same catchment? (a) 5000 (b) 2500 (c) 1250 (d) Insufficient data 5.43 Which of the following statements is/are incorrect in determining the nD-hour UH from a D-hour UH? I. Method of superposition can be used only when n is an integer. II. Method of S-hydrograph can be used when n is a fraction. III. Method of S-hydrograph can be used regardless of whether n is integer or real. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of the statements is incorrect 5.44 While calculating a 3-hour UH from a 2-hour UH, the S-hydrograph is to be lagged by (a) 2 hours (b) 3 hours (c) Either 2 hours or 3 hours (d) Depends on drainage area 5.45 While calculating a 3-hour UH from a 2-hour UH, the ordinates of the DRH represented by the difference in the lagged S-hydrographs (SA – SB) is to be divided by (a) 3 cm (b) 2 cm (c) 0.667 cm (d) 1.5 cm 5.46 Which of the following statements is/are true a synthetic unit hydrograph in a catchment? I. It is useful in the same catchment at other ungauged locations. II. It is useful in the nearby ungauged catchment. III. The nearby ungauged catchment should be hydrologically similar to the gauged catchment. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of the statements is true 5.47 Snyder’s method of finding synthetic unit hydrograph (SUH) falls under which of the following categories? (a) Those that establish relationships among geo-morphological and hydrograph characteristics (b) Dimensionless hydrograph category (c) Conceptual hydrograph category (d) All of the above (e) None of (a), (b), or (c) 5.48 Gray’s method of finding SUH falls under which of the following categories? (a) Those that establish relationships among geo-morphological and hydrograph characteristics (b) Dimensionless hydrograph category (c) Conceptual hydrograph category (d) All of the above (e) None of (a), (b), or (c)

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5.49 SCS method for finding SUH falls under which of the following categories? (a) Those that establish relationships among geo-morphological and hydrograph characteristics (b) Dimensionless hydrograph category (c) Conceptual hydrograph category (d) All of the above (e) None of (a), (b), or (c) 5.50 Snyder studied many catchments in which parts of the USA for SUH development? (a) Mount Saint Elias region (b) Mount Whitney region (c) Mount Blackburn (d) None of the above 5.51 For developing SUH, Snyder studied many catchments of drainage areas (km2) varying from? (a) 10 to 10,000 (b) 20 to 20,000 (c) 25 to 25,000 (d) 40 to 40,000 5.52 A standard unit hydrograph for a catchment is defined as a UH for which the time to peak is (a) Approximately equal to the effective storm duration (b) Approximately equal to 3.5 times the effective storm duration (c) Approximately equal to 4.5 times the effective storm duration (d) Approximately equal to 5.5 times the effective storm duration (e) Not sufficient information to comment 5.53 As per Snyder, the basin lag is defined as the time elapsed between the (a) Start of the ERH and the peak of the UH (b) Centroid of the ERH and the inflexion point on the UH (c) Start of the ERH and the inflexion point of the UH (d) Centroid of the ERH and the peak of the UH (e) None of the above 5.54 For the widths of the UH at 50% and 75% of peak runoff, W50 and W75, which of the following is correct in Snyder’s SUH? (a) W50 = W75 (b) W50 < W75 (c) W50 > W75 (d) No relation between W50 and W75 5.55 What is the unit of W50 and W75 in Snyder’s SUH? (a) m3/s (b) hour (c) m3/s/km

(d) No units

5.56 As per Snyder’s SUH, which of the following is most correct? (a) Time to peak of a UH decreases linearly with the length of the main stream in the catchment (b) Time to peak of a UH decreases nonlinearly with the length of the main stream in the catchment (c) Time to peak of a UH increases linearly with the length of the main stream in the catchment (d) Time to peak of a UH increases nonlinearly with the length of the main stream in the catchment 5.57 Which of the following statements is correct for Snyder’s SUH? (a) Higher the time to peak, higher will be the peak discharge (b) Lower the time to peak, lower will be the peak discharge (c) Higher the time to peak, lower will be the peak discharge (d) No relation between the time to peak and the peak discharge 5.58 Which of the following statements is correct for Snyder’s SUH? (a) Higher the peak discharge, higher will be the base time (b) Lower the peak discharge, lower will be the base time (c) Higher the peak discharge, lower will be the base time (d) No relation between the peak discharge and the base time

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5.59 Which of the following statements is correct for the SCS dimensionless SUH? (a) Q/Qp can be greater than 1.0 (b) Q/Qp cannot be greater than 1.0 (c) t/tp cannot be greater than 1.0 (d) Not sufficient information to comment 5.60 The SCS dimensionless SUH method assumes that the (a) Time of recession of triangular UH is approximately equal to the time of rise (b) Time of recession of triangular UH is approximately equal to 16.7 time of rise (c) time of recession of triangular UH is approximately equal to 10.67 time of rise (d) Time of recession of triangular UH is approximately equal to 1.67 time of rise (e) No such assumption is made relating the two 5.61 The time of rise was found to be 3 hours for the SCS triangular SUH. What is the time of recession (in hours)? (a) 2 (b) 3 (c) 4 (d) 5 (e) None of these 5.62 The time of rise was found to be 3 hours for the SCS triangular SUH. What is the time base (in hours)? (a) 4 (b) 6 (c) 8 (d) 10 (e) None of these 5.63 In SCS triangular SUH, which of the following is the most correct for the time to peak? (a) equal to the time of concentration (b) Less than the time of concentration (c) Greater than the time of concentration (d) Can be either less than or greater than the time of concentration depending on duration of effective rainfall (e) They have no relationship at all 5.64 For a catchment having a time of concentration of 10 hours, what will be the time to peak discharge for a storm having an effective duration of 2 hours as per the SCS triangular method? (a) 4 (b) 6 (c) 7 (d) 10 (e) Insufficient data 5.65 For a catchment having an area of 5 km2 and time to peak discharge equal to 2.08 hours, what will be the value of peak discharge (in m3/s) as per the SCS triangular method? (a) 5 (b) 6 (c) 7 (d) 8 (e) Insufficient data 5.66 Which of the following is true for a UH? (a) As the effective rainfall duration increases, the magnitude of peak discharge increases (b) As the effective rainfall duration increases, the time to peak discharge increases (c) As the effective rainfall duration increases, the time base increases (d) As the effective rainfall duration increases, the intensity of rainfall increases 5.67 An IUH is the DRH resulting from a catchment when it is subjected to an (a) Effective rainfall of intensity equal to 1 cm/hour occurring for 1-hour (b) Effective rainfall depth equal to 1 cm occurring for 1-hour (c) Effective rainfall of intensity equal to 1 cm/hour occurring instantaneously (d) Effective rainfall depth equal to 1 cm occurring instantaneously 5.68 Which of the following statements is/are true for an IUH? I. It is independent of any duration.

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II. It represents unique characteristic of a catchment in terms of its DRH response under extreme and hypothetical conditions. III. It is very useful in determining DRH from an ERH of variable intensity, which is difficult using the concept of UH. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) None of the statements is correct 5.69 The ordinate of an IUH at any time t, u(t) is (a) Equal to zero at t = 0 (b) Greater than zero at t = 0 because of the instantaneous rainfall (c) Can be negative at t = 0 (d) Depends on the catchment area (e) None of the above 5.70 The ordinate of an IUH at any time t, u(t) is (a) Equal to zero at t > 0 (b) Greater than zero at t > 0 (c) Can be negative at t > 0 (d) Depends on the catchment area (e) None of the above 5.71 Which of the following is correct for the ordinate of an IUH at a time t, u(t)? (a) u(t) Æ 0 as t Æ 0 (b) u(t) Æ ∞ as t Æ ∞ (c) u(t) Æ ∞ as t Æ 0 (d) u(t) Æ 0 as t Æ ∞ (e) None of the above 5.72 Which of the following statements is true for the relationship between an IUH and a S-hydrograph? (a) The IUH at any time t from a catchment is equal to the slope of the S-hydrograph produced from an ER depth 1 cm occurring instantaneously (b) The IUH at any time t from a catchment is equal to the slope of the S-hydrograph produced from an ER intensity of 1/D cm/hour (c) The IUH at any time t from a catchment is equal to the ordinates of the S-hydrograph produced from an ER intensity of 1 cm/hour (d) The IUH at any time t from a catchment is equal to the slope of the S-hydrograph produced from an ER intensity of 1 cm/hour (e) IUH and S-hydrograph are not related 5.73 The D-hour UH at time t can be written in terms of IUH ordinates as follows: (a)

1 [u(t ) - u(t - D)] 2

(c) [u(t) – u(t – D)]

(b)

1 [u(t ) + u(t - D)] 2

(d) [u(t) + u(t – D)]

DESCRIPTIVE QUESTIONS 5.1 Define a unit hydrograph. Discuss its different components using a neat sketch. 5.2 What are the assumptions involved in the UH theory and their implications in its application to real-life problems? 5.3 What is the discrete convolution equation and how is it useful in hydrology?

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5.4 You need to determine a 3-hour UH from a catchment. Based on the records available in the area, there are 26 different datasets of streamflow hydrograph and rainfall records having duration of rainfall equal to approximately 3-hours. How will you select a particular dataset? 5.5 Explain the three methods of baseflow separation with the help of a neat sketch. 5.6 Using the drainage area, and rainfall and flow data from an isolated storm, explain how you will determine the D-hour UH. 5.7 Explain how you will find a D-hour UH using data from a complex storm and the deconvolution method. 5.8 Given a D-hour UH, explain how you will find an nD-hour UH using the method of superposition when n is an integer. 5.9 With the help of neat sketches, explain what an S-hydrograph is and how it can be used to determine an T-hour UH given a D-hour UH. 5.10 Explain the difference between standard, derived, and required UHs in Snyder’s SUH. 5.11 Describe the Snyder’s synthetic unit hydrograph with the help of equations and sketches. 5.12 Why is the SCS triangular UH concept needed to find the SCS dimensionless SUH? 5.13 Prove that the area under the rising limb of the SCS triangular SUH is approximately equal to 37.5% of the total area under the SCS triangular SUH. 5.14 Explain the concept of an instantaneous unit hydrograph with the help of a neat sketch. Also discuss its properties. 5.15 Show that the ordinate of an IUH in a catchment is equal to the slope of the S-hydrograph produced by the catchment when it is subjected to an ER intensity of 1 cm/hour.

NUMERICAL QUESTIONS 5.1 The ordinates of a 1-hour UH are given in the following table. Time (h) 3

1-hour UH (m /s)

0

1

2

3

4

5

6

0

50

125

100

50

25

0

Determine the ordinates of a DRH resulting from a 2-hour design storm having ERs of 1.5 cm and 2.75 cm in two consecutive hours. Plot the given UH and the derived DRH on the same graph. 5.2 The value of streamflow on the falling limb on June 2 and June 20 are 120 m3/s and 100 m3/s, respectively. Calculate the coefficient of recession for the catchment. In the same catchment, what will be the value of streamflow on Jun 30? Assume no rain occurs throughout the month of June. 5.3 The streamflow hydrograph from a catchment of size 650 km2 is provided in the table below. The streamflow hydrograph is believed to have been produced by an isolated storm having duration of 2-hours. Assuming a constant baseflow of 60 m3/s, determine the 2-hour UH from the catchment. Time (Hours)

Flow (m3/s)

Time (Hours)

Flow (m3/s)

0

60

36

240

3

600

39

210

6

950

42

190

(Contd.)

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9

800

45

170

12

710

48

150

15

610

51

130

18

530

54

110

21

460

57

90

24

400

60

80

27

350

63

70

30

310

66

60

33

270

5.4 Let us assume that the streamflow hydrograph given in the Problem 3 above resulted from four successive 3-hour periods of effective rainfalls of 1.5 cm, 3.0 cm, 2.5 cm, and 2.0 cm, respectively. Write all the simultaneous linear equations in the discrete convolution equation. Determine the 3-hour UH from the catchment first starting from the top and then starting from the bottom. Then, estimate the average of the two UH ordinates thus determined. Plot the two UHs and the average UH on the same graph and comment on the results. 5.5 The residents in your neighborhood experienced a rainfall event on 19 November 2016, which started at 04:00 PM and continued until 07:00 PM, recording a total rainfall of 4.85 cm. The peak flow of 300 m3/s was recorded due to this storm in a nearby river at a location having a total drainage area of 600 km2. Determine the peak flow rate of the 3-hour UH, assuming a constant baseflow of 15 m3/s, and an average infiltration rate of 0.25 cm/hour. Also determine until what time (and date) the DRH will be observed in the nearby river due to this storm assuming the shape of the DRH to be triangular. 5.6 The ordinates of a 3-hour UH are provided in the table below. Using the method of superposition, determine the 6-hour UH and 9-hour UH from the catchment. Plot all the three UHs on the same graph and comment. 3-Hour UH (m3/s)

Time (Hours)

3-Hour UH (m3/s)

0

0

33

86

3

13

36

69

Time (Hours)

6

77

39

54

9

135

42

40

12

189

45

32

15

215

48

22

18

180

51

17

21

151

54

13

24

133

57

6

27

123

60

0

30

95

5.7 The ordinates of a 4-hour UH from a basin at 1-hour interval are 5, 30, 50, 65, 75, 63, 50, 45, 39, 30, 25, 23, 19, 15, 10, 8, 6, 4, 2, 1 (in m3/s). Determine: (a) drainage area of the catchment, (b) equilibrium discharge, and (c) S-hydrograph.

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5.8 The ordinates of a 3-hour UH are provided in the following table. Time (Hours)

3-Hour UH (m3/s)

Time (Hours)

3-Hour UH (m3/s)

0

0

8

120

1

20

9

100

2

80

10

80

3

120

11

60

4

160

12

40

5

200

13

20

6

180

14

0

7

140

Determine the S-hydrograph for the catchment using this data. 5.9 Determine a 2-hour UH for the data given in Problem-8 above. Calculate the DRH from a 4-hour storm having 2 successive 2-hour rainfall rates of 2.5 cm/hour and 4.5 cm/hour. Assume F–index for the catchment is 1.5 cm/hour. 5.10 Observed streamflow data from an ER of 3.65 cm in 12 hours is shown in the table below. If the baseflow was 375 m3/s, determine the 12-hour UH from the catchment and convert it to a 6-hour UH. Time (Hours)

Flow (m3/s)

Time (Hours)

Flow (m3/s)

0

375

42

1800

6

820

48

1300

12

2000

54

900

18

3700

60

700

24

4000

66

500

30

3250

72

420

36

2400

78

375

Then using the 6-hour UH thus computed determine the total flow hydrograph (including a baseflow of 400 m3/s) for a 24-hour storm having four 6-hour blocks of ER intensities of 1.7, 4.8, 10.3, and 2.6 cm/hour. Plot the 12-hour UH, 6-hour UH and the total flow hydrograph computed from the complex storm. 5.11 From the topographic map of a catchment, the following data was obtained: L = 200 km, Lc = 90 km, A = 4,500 km2. From the rainfall and flow data in the same catchment, the time and magnitude of peak discharge of a UH were found to be 40 hours and 267 m3/s, respectively, due to a 12-hour storm. Determine the coefficients Ct and Cp for this catchment. 5.12 Using the coefficients Ct and Cp determined above, find the Snyder’s 6-hour SUH in a nearby catchment with L = 400 km, Lc = 180 km, A = 8,500 km2.

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5.13 The data in river Narmada at a location is found as follows: A = 670 km2; L = 120 km; Lc = 80 km; Ct = 1.85; Cp = 0.565. Determine the standard SUH and the 2-hour UH for this catchment. Sketch the SUH neatly on a graph sheet and show the important hydrograph characteristics on it. 5.14 For the two hydrologically similar catchments A and B, the characteristics measured from the basin map are as follows:

A B

L = 148 km, Lc = 76 km, Area = 2,718 km2 L = 106 km, Lc = 52 km, Area = 1,400 km2

The peak discharge for a 6-h UH in the catchment A is 200 m3/s and it occurs at 17 hours from the end of the rainfall excess. Use Snyder’s method to determine the 6-h SUH for the catchment B. Sketch the SUH neatly on a graph sheet and show the important hydrograph characteristics on it. 5.15 Synthetically generate a 20-minute SCS UH for a catchment of area 5 km2 and time of concentration of 2.1 hours. Neatly sketch the SCS SUH and the triangular SUH. 5.16 An IUH from a catchment is given in the following table. Find: (a) area of the catchment, (b) 1-hour UH from the catchment, and (b) DRH due to an ERH consisting of ERs of 2.5 cm and 5.8 cm in two 2-hour intervals. Plot IUH, 1-hour UH, 2-hour UH, and the total DRH on the same graph. Time (h)

0

1

IUH (m3/s)

0

400

2

3

4

5

1080 2400 2550 1480

6

7

8

9

10

750

410

290

120

0

6

Hydrograph Routing

LEARNING OBJECTIVES LO 1

Explain hydrograph routing and its importance

LO 2

Discuss hydrologic and hydraulic routing

LO 3

Outline the methods for hydrologic routing through a reservoir

LO 4

Outline the methods for hydrologic routing through a channel

LO 5

Summarize instantaneous unit hydrograph development using routing concepts

LO 6

Discuss Clark’s method for IUH development

LO 7

Explain Nash’s method for IUH development

6.1

INTRODUCTION

In the previous chapter, we studied the concept of unit hydrograph and its LO 1 Explain hydrograph application to estimate Direct Runoff Hydrograph (DRH) from a complex routing and its importance storm in a catchment. When the base flow is added to the DRH, the total flow hydrograph at any location in a river can be obtained. The total flow hydrograph at a location in a river is often referred to as a hydrograph. Consider the reach of a river of certain length in which a hydrograph enters at the upstream end of the reach (say inflow hydrograph). The inflow hydrograph after travelling through the reach of the river comes out at the downstream end of the reach (say outflow hydrograph). Should the size, shape, peak flow, time to peak, slopes of the rising and falling limbs of the two hydrographs be the same? — the answer is no, because as the inflow hydrograph travels through a river reach, its shape and characteristics get modified due to several factors, e.g., storage in the river reach,

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resistance to flow due to friction from sides and bed, lateral addition or subtraction of flow within the reach, if any, etc. Such factors account for the spatio-temporal variation in the inflow hydrograph as it travels through a river reach. Similarly, when an inflow hydrograph travels through a storage structure (e.g., a reservoir), its characteristics change depending on the storage and outflow characteristics at the outlet of the reservoir. The study of changes in the hydrograph characteristics as it passes through a river reach or a storage structure is known as hydrograph routing or flood routing. Hydrograph routing is important in many water resources applications including flood forecasting, flood control, design of reservoirs and spillways, and catchment simulation studies. In this chapter, we will study the various methods available that can be used to determine the outflow hydrograph from a hydrologic element (a river reach or a reservoir) given the inflow hydrograph and characteristics/data pertaining to the hydrologic element.

6.2

HYDROLOGIC AND HYDRAULIC ROUTING

In hydrograph routing, we essentially try to predict the spatio-temporal LO 2 Discuss hydrologic and variation of an inflow hydrograph as it travels through a river reach or hydraulic routing a reservoir. Hydrograph routing can be one of the two types: hydrologic routing or hydraulic routing, depending on whether the method employs empirical or physics based approach, respectively. Hydrologic routing uses law of conservation of mass along with a predefined relationship between the storage and outflow within a river or a reservoir, which may be either linear or nonlinear. On the other hand, one uses both the laws of conservation of mass and momentum in hydraulic routing. The two types of hydrograph routing methods along with the basic equations used are discussed briefly here.

6.2.1 Hydrologic Routing Hydrologic routing is simpler than hydraulic routing as it employs only the continuity equation. However, it has numerous applications in real life water resources activities. For example, hydrologic routing can be used for predicting a flood at a downstream location, knowing the conditions at an upstream location. It is used in catchment simulation studies of distributed nature wherein hydrographs generated at sub-basin scales are routed to the downstream locations. National Oceanic and Atmospheric Administration (NOAA) and US Corps of Engineers employ hydrologic routing methods in their flood warning system to predict floods in advance of a severe thunderstorm. Hydrologic routing is also used for the assessment of flood control measures at a reservoir (described later) and effects of urbanization. The process of inflow hydrograph traveling through a hydrologic element (a river reach or a reservoir) is an unsteady process in which the storage in the hydrologic element and the outflow hydrograph depend on time. The inflow, outflow, and storage in the hydrologic element can be related using the law of conservation of mass as follows: I -Q=

dS dt

(6.1)

where, I is the rate of inflow (m3/s), Q is the rate of outflow (m3/s), and S is the storage in the hydrologic element (m3), all at time t (s). Eq. (6.1) is applicable in continuous time domain; however, the discharge data in a river are normally available or analyzed at discrete time steps. Let us assume that the data is available

Hydrograph Routing

189

at very small time intervals (Δt) such that the hydrologic variables (I, Q, and S) can be assumed to be linear within this small time interval. In that case, Eq. (6.1) can be rewritten by taking the average values of inflow and outflow at the beginning and end of the time interval as follows: Ê I1 + I 2 ˆ Ê Q1 + Q2 ˆ Dt = S2 - S1 ÁË 2 ˜¯ Dt - ÁË 2 ˜¯

(6.2)

where, subscript 1 refers to the beginning of the time interval and subscript 2 refers to the end of the time interval Δt. We will use Eq. (6.2) for hydrologic routing later.

6.2.2

Hydraulic Routing

Hydraulic routing is more accurate than hydrologic routing of hydrographs. Hydraulic routing is very useful in developing conceptual unit hydrographs in an ungauged catchment. It finds important applications in distributed rainfall-runoff modeling and simulation studies in a catchment. The flow in a river is an unsteady phenomenon and is governed by continuity and momentum equations in unsteady form. The continuity equation for an unsteady flow in a river (assuming no lateral inflow/outflow) can be represented as follows: ∂Q ∂y +T =0 ∂x ∂t

(6.3)

where, Q is the discharge, T is the top width of the flow cross-section, y is the depth of flow, all at time t, and x represents the direction of flow. The equation of motion of unsteady flow in a river is represented by the momentum equation as follows: ∂y V ∂V 1 ∂V + + = So - S f ∂x g ∂x g ∂t

(6.4)

where, V is the flow velocity, So is the longitudinal slope of the river (dimensionless), Sf is the friction slope of the river at the point under consideration (dimensionless), and g is the acceleration due to gravity (m/s2). The continuity and momentum equations given above were first developed by Adhémar Jean Claude Barré de Saint-Venant (1871) and they are popularly known as St. Venant equations. The St. Venant equations represent simultaneous, quasi-linear, first order partial differential equations (PDEs) of the hyperbolic type and are not easy to solve analytically in general. Analytical solutions may be possible for extremely simplified systems with several assumptions. Therefore, these are solved using numerical techniques and high-speed computers for large scale systems. A host of numerical methods are available for solving the St. Venant equations that can broadly be classified into two categories: (a) partial solution methods, and (b) complete numerical methods. In partial solution methods, either only continuity equation is used, or a highly simplified version of momentum equation along with continuity equation is solved. The hydrologic method of reservoir routing and Muskingum method of river routing (which we will study later in this chapter) and Kinematic Wave method fall under the partial solution method category. In complete numerical methods, both continuity and momentum equations are solved simultaneously using sophisticated numerical techniques. The complete numerical methods can be one of the three types: (a) finite difference (FD) methods, (b) method of characteristics (MOC), or

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(c) finite element (FE) methods. In FD methods, the partial derivatives are replaced by the finite difference approximations. This converts the St. Venant equations into algebraic equations which are solved for unknowns using any of the several methods available for solving algebraic equations. The FD methods can be further sub-divided into two kinds: explicit and implicit methods. Explicit methods are easy to use as the unknown variables at the end of each time step are explicitly determined using known quantities. On the other hand, implicit FD schemes are computationally expensive as the unknown quantities are determined through iterative procedures. In MOC, the St. Venant equations are first converted into ordinary differential equations (known as characteristic forms) and then solved using any of the FD methods. In the FE methods, the entire system is assumed to be composed of small elements and the PDEs are written and integrated at each node point to obtain the unknown quantities at each node. Hydrograph routing through natural rivers has been implemented in many computer software systems available freely and commercially. Hydrologic Engineering Center’s River Analysis System (HEC-RAS) is freely available software developed by the US Army Corps of Engineers (USACE). HEC-RAS is capable of computing water surface elevation in natural rivers using one dimensional flow analysis, carry out flood flow simulations, simple sediment transport computations, and water quality analyses. The FLDWAV, developed by the US National Weather Service (NWS) and supported by Federal Emergency Management Agency (FEMA), is another generalized flood routing computer program with the capability to model flows through a single stream or a system of interconnected waterways. Some other notable examples of popular software for hydrograph routing and analysis are FLO-2D developed by FLO-2D, Inc., USA, TUFLOW, and MIKEFLOOD. Interested readers should see the links provided at the end of this chapter for details. In this chapter, we will concentrate on the hydrologic methods of hydrograph routing. The readers interested in hydraulic methods of hydrograph routing can refer to Streeter and Wylie (1967) and Mahmood and Yevjevich (1975).

6.3

HYDROLOGIC ROUTING THROUGH A RESERVOIR

The hydrologic routing of an inflow hydrograph through a reservoir is LO 3 Outline the methods carried out using the continuity equation (Eq. 6.2) alone. The changes in for hydrologic routing through the inflow hydrograph as it passes through a reservoir are reflected in the a reservoir outflow hydrograph that are functions of the storage characteristics of the reservoir. Let I(t), Q(t) and S(t) represent the inflow, outflow, and storage in the reservoir at any time t. The storage in the reservoir will be a function of the level of water in it, i.e., S = S(h). The level of water in the reservoir will be continuously changing with time, i.e., h = h(t), as the inflow hydrograph travels through the reservoir. A reservoir normally contains an outflow structure in the form of an uncontrolled spillway. The flow over the spillway of a reservoir can be expressed using the weir equation as follows: Q=

2 Cd 2 g Le H 3/2 3

(6.5)

where, Cd is the coefficient of discharge, Le is the effective length of the crest of the spillway, and H is the height of water over the spillway. When there is another form of outlet at the end of the reservoir, e.g., a controlled spillway, sluice gates, etc., the corresponding relation for Q = Q(H) can be adopted, instead of

Hydrograph Routing

191

Eq. (6.5). Please note that h represents height of water in the reservoir above a certain datum while H represents the height of water over the spillway crest. However, they both represent the water surface elevation in the reservoir and can be transformed to a suitable common datum. The problem of hydrologic routing through the reservoir then entails determining Q(t) and S(t) given I(t), storage volume versus elevation relationship (S v/s h), outflow versus height of water over outlet structure (Q v/s H) relationship, and initial conditions, i.e., values of Q(t) and S(t) at t = 0. There are many methods available for this purpose; we will study three popular methods, namely, Modified Puls method, Goodrich’s method, and Standard fourth order Runge-Kutta (SRK) method.

6.3.1

Modified Puls Method

The Modified Puls method of hydrologic routing of an inflow hydrograph through a reservoir is a semigraphical method. Also known as level pool routing, it uses the equation of continuity (Eq. 6.2) only, which can be rearranged slightly to bring all the unknown quantities on the right hand side (RHS) as follows: Q1 Dt ˆ Ê Q2 Dt ˆ Ê I1 + I 2 ˆ Ê ÁË 2 ˜¯ Dt + ÁË S1 - 2 ˜¯ = ÁË S2 + 2 ˜¯

(6.6)

At the start of hydrograph routing, values of outflow and storage (Q1 and S1) in the reservoir are known from the initial conditions. Therefore, all the quantities on the left hand side (LHS) in the equation above are known. Therefore, the expression on the RHS of Eq. (6.6) can be estimated. The term (S + QΔt/2) is known as the indicative storage and is very useful in Modified Puls method of hydrologic routing of an inflow hydrograph through a reservoir. The indicative storage at the end of any interval Δt can be easily evaluated using Eq. (6.6) and once that is done, a semi-graphical method can be employed to determine the values of Q and S at the end of time interval Δt, (i.e., Q2 and S2). The process of determining the values of Q and S at the end of each time interval is presented in a step by step procedure below. 1.

Since the inflow and outflow relationships are assumed to be linear within the time interval Δt, a sufficiently small value of Δt is chosen. As a rough guideline, the value of Δt may be taken about one-fifth of the time to peak of the inflow hydrograph.

2.

Using the S v/s h and Q v/s h data, prepare indicative storage versus h curve, i.e., (S + QΔt/2) v/s h curve and plot it on a graph.

3.

Plot Q v/s h curve on the same graph. In this graph, the height of water in the reservoir (h) is plotted on y-axis, indicative storage is plotted on the primary x-axis, and the outflow (Q) is plotted on the secondary x-axis (see Figure 6.1).

4.

Using the known values at the start of the time interval, compute the indicative storage (S2 + Q2 Δt/2) using Eq. (6.6).

5.

From the indicative storage versus h curve, find the corresponding value of h (see Figure 6.1).

6.

From the Q v/s h curve, find the corresponding value of Q (see Figure 6.1). This is the outflow at the end of time interval Δt, i.e., Q2.

7.

Knowing Q2, S2 can be determined by subtracting Q2 Δt/2 from the indicative storage at the end of the time interval Δt.

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Engineering Hydrology

Figure 6.1

Outflow and indicative storage graph in Modified Puls method

8.

Subtract Q2 Δt from the indicative storage at the end of time interval Δt, (S2 + Q2 Δt/2), to obtain (S2 – Q2 Δt/2). This represents the second term on the LHS of Eq. (6.6) for the next time interval.

9.

Go back to step 4 and repeat the procedure of steps 4 to 8 for the next time interval and continue until Q and S have been determined for all the time intervals in the inflow hydrograph.

10.

The outflow and inflow hydrographs are then plotted on the same graph for comparison purposes.

The reduction in the peak discharge when an inflow hydrograph travels through a reservoir or a channel reach is known as attenuation. The time of occurrence of the peak discharge in the outflow hydrograph is delayed, which is known as the lag time or delay in peak during the passage of the inflow hydrograph. We will learn about these later in this chapter. � EXAMPLE 6.1 The elevation, outflow discharge, and storage data for a reservoir are given in the following table. Elevation (m)

62.00

62.50

63.00

63.50

64.00

64.50

64.75

65.00

Storage (Mm3)

3.35

3.50

3.88

4.38

4.90

5.50

5.80

5.90

0

10

20

46

72

100

116

130

3

Outflow m /s

An inflow hydrograph given below enters the reservoir when the initial water level in the reservoir was 63.00 m. Time (Hours)

Inflow (m3/s)

Time (Hours)

Inflow (m3/s)

0

10

21

36

3

20

24

28

6

55

27

20

9

80

30

15

12

73

33

13

15

58

36

11

18

46

39

10

Hydrograph Routing

193

Route the inflow hydrograph using the Modified Puls method and determine: (a) outflow hydrograph, (b) maximum reservoir elevation during the passage of the flood, (c) peak outflow discharge, (d) attenuation, and (e) time lag from the reservoir. Plot the inflow and outflow hydrographs on the same graph sheet. Solution The elevation (m), storage (Mm3), and outflow (m3/s) data given are presented in column 1, column 2, and column 3, respectively, of the table below. Since the time interval for the given inflow hydrograph is 3 hours, using Δt = (3 × 3600) = 10,800 seconds, column 4 (S + QΔt/2) is calculated using S and Q data in column 2 and column 3, respectively. Then h v/s Q and h v/s (S + QΔt/2) curves are plotted on the same graph (see Figure 6.1). This graph will be used for routing of the inflow hydrograph. Elevation (m)

Storage (Mm3)

Outflow (m3/s)

(S+QΔt/2) (Mm3)

[1]

[2]

[3]

[4]

62.00

3.35

0

3.350

62.50

3.50

10

3.554

63.00

3.88

20

3.988

63.50

4.38

46

4.631

64.00

4.90

72

5.289

64.50

5.50

100

6.040

64.75

5.80

116

6.426

65.00

5.90

130

6.602

Time (Hours)

Inflow, I (m3/s)

I_Avg (m3/s)

I_Avg*Δt (Mm3)

S – QΔt/2 (Mm3)

S + QΔt/2 (Mm3)

Elevation, h (m)

Q (m3/s)

[1]

[2]

[3]

[4]

[5]

[6]

[7]

[8]

0

10

63.000

20.0

3

20

15.0

0.162

3.772

3.934

62.965

24.9

6

55

37.5

0.405

3.665

4.070

63.040

27.6

9

80

67.5

0.729

3.772

4.501

63.393

41.7

12

73

76.5

0.826

4.050

4.877

63.686

55.7

15

58

65.5

0.707

4.275

4.983

63.767

59.9

18

46

52.0

0.562

4.336

4.898

63.702

56.5

21

36

41.0

0.443

4.287

4.730

63.575

49.9

24

28

32.0

0.346

4.191

4.537

63.422

42.9

27

20

24.0

0.259

4.074

4.333

63.255

36.2

30

15

17.5

0.189

3.942

4.131

63.090

29.6

33

13

14.0

0.151

3.811

3.962

62.937

24.0

36

11

12.0

0.130

3.703

3.833

62.798

19.5

39

10

10.5

0.113

3.622

3.735

62.694

16.2 (Contd.)

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Engineering Hydrology

42

10

10.0

0.108

3.560

3.668

62.622

13.9

45

10

10.0

0.108

3.518

3.626

62.577

12.5

48

10

10.0

0.108

3.491

3.599

62.548

11.5

51

10

10.0

0.108

3.475

3.583

62.530

11.0

54

10

10.0

0.108

3.464

3.572

62.519

10.6

57

10

10.0

0.108

3.458

3.566

62.512

10.4

60

10

10.0

0.108

3.453

3.561

62.508

10.2

63

10

10.0

0.108

3.451

3.559

62.505

10.2

66

10

10.0

0.108

3.449

3.557

62.503

10.1

69

10

10.0

0.108

3.448

3.556

62.502

10.1

72

10

10.0

0.108

3.448

3.556

62.501

10.0

The computations for the routing of inflow hydrograph are carried out in a tabular form as shown above. The first two columns of the table represent the given inflow hydrograph. Column 3 gives the average inflow value at each time interval, (I1 + I2)/2 or (Iavg), in m3/s. Column 4 is the inflow volume in each time interval, (IavgΔt) in Mm3, which is obtained by multiplying the entries in column 3 by 3 hours or 10,800 s. The first row against time t = 0 represents the initial conditions, i.e., I = 10 m3/s, h = 63.000 m, and Q = 20 m3/s. For the given initial h = 63.000 m, we have S = 3.88 Mm3. Using the values of initial S and Q, we calculate (S – QΔt/2) = 3.772 Mm3. This is entered in column 5 in the row for t = 3 hours. Now as per Eq. (6.6), the RHS is calculated as the sum of column 4 and column 5. Therefore, (S + QΔt/2) = 3.934 Mm3, this is entered in column 6 in the row for t = 3. Now using the indicative storage graph prepared (Figure 6.1), we find out the values of elevation in the reservoir (h) and outflow (Q) corresponding to the value of (S + QΔt/2) = 3.934 Mm3 just calculated, by using the arrows. Note that this step can also be easily carried out by using linear interpolation of the (S + QΔt/2) v/s h and Q v/s h data. The values of h and Q are thus found as h = 62.965 m and Q = 24.9 m3/s, respectively; and are entered in column 7 and column 8, respectively, in the row for t = 3 hours. This completes the calculations for t = 3 hours. Now, using (S + QΔt/2) = 3.934 Mm3 at the end of 3 hours, we find (S – QΔt/2) = (S + QΔt/2) – QΔt = 3.665 Mm3. With the value of (S – QΔt/2) = 3.665 Mm3 at t = 6 hours, the RHS is calculated as the sum of column 4 and column 5, as per Eq. (6.6). Therefore, (S + QΔt/2) = 4.070 Mm3, this is entered in column 6 in the row for t = 6. Now using the indicative storage graph prepared (Figure 6.1), we find out the values of h and Q corresponding to the value of (S + QΔt/2) = 4.070 Mm3. The values of h and Q are thus found as h = 63.040 m and Q = 27.6 m3/s, respectively; and are entered in column 7 and column 8, respectively, in the row for t = 6 hours. This completes the calculations for t = 6 hours. This process is then continued for each time step and the inflow hydrograph is thus routed through the reservoir to calculate the outflow hydrograph. The outflow hydrograph is presented in column 8 in the table above and the inflow and outflow hydrographs are shown in Figure 6.2 below. Note that the peak flow of 80 m3/s occurs at t = 9 hours in the inflow hydrograph, while the peak flow of 59.9 m3/s occurs at t = 15 hours in the outflow hydrograph. The maximum reservoir elevation during the passage of the flood is equal to 63.767 m occurring at t = 15 hours. The times and magnitudes of the peak flows in the inflow and outflow hydrographs are marked in bold font in the table above. It is thus easy to find that the attenuation of peak is = 80 m3/s – 59.9 m3/s = 20.1 m3/s; and delay in the peak (or lag time through the reservoir) is (15 hours – 9 hours) = 6 hours.

Hydrograph Routing

Figure 6.2

6.3.2

195

Inflow and outflow hydrographs

Goodrich’s Method

Goodrich’s method of hydrograph routing through a reservoir is also a semi-graphical method very similar to the Modified Puls method. In the rearranged continuity equation (6.6), all the quantities are in volumetric units (m3). However, it can be rearranged in a slightly different manner such that the quantities involved have units of discharge (m3/s) as follows: Ê 2S ˆ Ê 2S ˆ ( I1 + I 2 ) + Á 1 - Q1 ˜ = Á 2 + Q2 ˜ Ë Dt ¯ Ë Dt ¯

(6.7)

The steps involved in the Goodrich’s method are similar to those in the Modified Puls method. The only difference is that the indicative storage for Goodrich’s method has units of discharge and is written as (2S/Δt + Q) and 2Q2 is subtracted from the modified indicative storage in step (8) of the Modified Puls method described above.

6.3.3 Standard Runge–Kutta (SRK) Method The two methods discussed above for routing an inflow hydrograph through a reservoir are semi-graphical methods and pose difficulties in implementation on a computer. Other numerical schemes can be used for this purpose using continuity equation and the reservoir characteristics. The Runge-Kutta methods of different orders are available that can be used for hydrograph routing through a reservoir. The standard fourth order Runge-Kutta (SRK) method has been employed by hydrologists due to its accuracy and ease of implementation. As we have seen earlier, as the inflow hydrograph passes through a reservoir, its storage changes continuously with time. The change in storage in a reservoir can be expressed as follows: dS dh (6.8) = A(h) dt dt where, A is the surface area of the reservoir at height of water equal to h. Noting that the outflow from the reservoir Q is a function of the height of water in the reservoir, i.e., Q = Q(h), the continuity equation (6.1) and Eq. (6.8) may be combined as follows:

196

Engineering Hydrology

I (t ) - Q(h) = A(h)

dh dt

dh I (t ) - Q(h) = = f (t , h ) dt A(h)

Or

(6.9) (6.10)

where, f(t, h) is a function depending on values of I(t), Q(h), and A(h). I(t) is the given inflow hydrograph, Q(h) is discharge relationship available depending on the kind of outlet structure at the end of the reservoir, and A(h) is the surface area v/s depth relationship for the reservoir. The process of inflow hydrograph routing starts with the known values of initial S and Q at t = 0. The calculation of the height of water in the reservoir at the end of any time interval Δt by the fourth order SRK method can be described by the following equations: ht +Dt = ht + where

1 ( K + 2 K 2 + 2 K3 + K 4 )Dt 6 1

(6.11)

K1 = f(t, ht) Dt 1 Ê ˆ K2 = f Á t + , ht + K1 Dt ˜ Ë ¯ 2 2 Dt 1 Ê ˆ K3 = f Á t + , ht + K 2 Dt ˜ Ë ¯ 2 2

(6.12)

K 4 = f (t + Dt , ht + K3 Dt ) Given the initial conditions, I(t), A v/s h, and outflow rating curve (Q v/s h) at the outlet of the reservoir, the fourth order SRK method can be presented in a step by step procedure as follows. 1.

Choose a small value of time interval Dt such that inflow hydrograph may be assumed to be linear within the time interval Dt.

2.

For the initial value of h at t = 0, calculate outflow Q and surface area of the reservoir A from the two relationships available.

3.

Using Eq. (6.10), calculate the value of function f(t, h) for values of I, Q, and A at the start of the time interval.

4.

Calculate the SRK parameters (K1, K2, K3, and K4) using Eq. (6.12). Note that the value of inflow discharge at time step (t + Dt/2) can be linearly interpolated using the values of inflow discharges at the beginning and end of the time interval Dt, i.e., I(t) and I(t + Dt), respectively.

5.

The height of water in the reservoir at the end of the time interval Dt, ht + Dt, can then be calculated using Eq. (6.11).

6.

The outflow discharge at the end of time interval Dt, Q(t + Dt) is then calculated using the outflow rating curve (Q v/s h) available.

7.

Calculate the surface area of the reservoir at time step (t + Dt), A(t + Dt) using the available A v/s h relationship for the reservoir.

8.

Go back to step 3 and repeat the procedure of steps 3 to 7 for the next time interval and continue until the h and Q have been determined for all the time intervals in the inflow hydrograph.

Hydrograph Routing

9. 10.

197

The storage in the reservoir at each time step can also be calculated using S v/s h relationship for the reservoir. The outflow and inflow hydrographs are then plotted on the same graph for comparison purposes.

� EXAMPLE 6.2 The crest elevation of a small spillway of a reservoir is at 70.00 m having outflow discharge relationship as Q = aHb with a = b = 1.5, where H is the height of water above the spillway crest. The reservoir is rectangular in shape with vertical walls and a surface area of 0.25 km2. The following flood hydrograph enters the reservoir when the water surface elevation (WSEL) in the reservoir was 68.00 m. Time (Hours)

Inflow (m3/s)

Time (Hours)

Inflow (m3/s)

0

5.0

25

47.0

1

7.5

26

44.0

2

10.0

27

41.0

3

12.5

28

38.0

4

15.0

29

35.0

5

17.5

30

32.0

6

20.0

31

30.3

7

23.3

32

28.7

8

26.7

33

27.0

9

30.0

34

25.3

10

33.3

35

23.7

11

36.7

36

22.0

12

40.0

37

20.8

13

43.3

38

19.7

14

46.7

39

18.5

15

50.0

40

17.3

16

53.3

41

16.2

17

56.7

42

15.0

18

60.0

43

14.2

19

58.3

44

13.3

20

56.7

45

12.5

21

55.0

46

11.7

22

53.3

47

10.8

23

51.7

48

10.0

24

50.0

198

Engineering Hydrology

Route the inflow hydrograph through the reservoir using standard fourth-order Runge-Kutta method and determine the outflow hydrograph, attenuation, and delay in the peak discharge. Solution The computations are arranged in a tabular form shown below. The first two columns present the given inflow hydrograph. Column 3 is the WSEL in the reservoir; at t = 0, it is given as the initial condition as 68.00 m. Column 4 is the outflow discharge as per equation Q = 1.5h1.5 for the value of h in column 3. Note that whenever the WSEL in the reservoir is less than the spillway crest elevation of 70.00 m, the value of outflow discharge is set to 0.0. Column 5 is K1 calculated by Eq. (6.12) above. Note that since K1 (and K2, K3, K4) represent the rate of change of h, their units are m/s. Column 6 presents the interpolated inflow hydrograph ordinates at time (t + Δt/2), which will be needed to calculate the values of K2 and K3. Columns 7 is h = (ht + K1Δt/2) and column 8 is the outflow discharge from the reservoir for the WSEL in column 7 calculated using the outflow equation. Using the inflow, outflow, and area of reservoir, K2 is calculated in column 9. Columns 10 is h = (ht + K2Δt/2) and column 11 is the outflow discharge from the reservoir for h in column 10 calculated using the outflow equation. Using the inflow, outflow, and area of the reservoir, K3 is calculated in column 12 from Eq. (6.12). Similarly, computations are carried out for calculating Q(ht + K3Δt) and K4 in columns 13, 14, and 15. Then, h at time (t + Δt) is calculated using Eq. (6.11) and is represented in column 16. The outflow discharge at the next time step, Q(t + Δt) is finally calculated using the outflow equation (Q = 1.5h1.5) and is represented in column 17. This completes one cycle of calculations of WSEL and Q from the reservoir at the end of the first time interval. The values of h and Q thus obtained become the initial conditions for the next time interval and are copied in columns 3 and 4, respectively, against t = 1 hour. The calculations are then repeated, in the same manner as mentioned above, to calculate h and Q at the end of next time interval or t = 1 hour. Similar calculations are carried out to compute the entire outflow hydrograph from the reservoir, which is presented in column 17. Note that the peak discharge of 60 m3/s occurs at t = 18 hours in the inflow hydrograph. The peak discharge of 35.6 m3/s occurs at t = 28 hours in the outflow hydrograph. Therefore, the value of attenuation of peak discharge is equal to 60 m3/s – 35.6 m3/s = 24.5 m3/s and the delay in peak discharge is 10 hours. The inflow and outflow hydrographs are shown in Figure 6.3.

Figure 6.3

Inflow and outflow hydrographs

Time (Hours) [1] 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 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Inflow, I (m3/s) [2] 5.0 7.5 10.0 12.5 15.0 17.5 20.0 23.3 26.7 30.0 33.3 36.7 40.0 43.3 46.7 50.0 53.3 56.7 60.0 58.3 56.7 55.0 53.3 51.7 50.0 47.0 44.0 41.0 38.0 35.0 32.0 30.3 28.7 27.0 25.3 23.7 22.0 20.8 19.7 18.5 17.3 16.2 15.0 14.2 13.3 12.5 11.7 10.8 10.0

ht (m) [3] 68.00 68.09 68.22 68.38 68.58 68.81 69.08 69.39 69.75 70.16 70.61 71.10 71.62 72.16 72.72 73.31 73.90 74.51 75.12 75.70 76.21 76.66 77.06 77.39 77.68 77.91 78.07 78.18 78.24 78.26 78.23 78.17 78.09 78.00 77.89 77.77 77.64 77.50 77.35 77.20 77.05 76.89 76.73 76.57 76.41 76.26 76.10 75.94 75.78

Q(h) (m3/s) [4] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.1 0.7 1.7 3.1 4.8 6.7 9.0 11.6 14.4 17.4 20.4 23.2 25.8 28.1 30.1 31.9 33.3 34.4 35.1 35.5 35.6 35.4 35.0 34.5 33.9 33.3 32.5 31.7 30.8 29.9 29.0 28.1 27.1 26.2 25.3 24.4 23.5 22.6 21.7 20.9

K1 (m/s) [5] 2.00E-05 3.00E-05 4.00E-05 5.00E-05 6.00E-05 7.00E-05 8.00E-05 9.33E-05 1.07E-04 1.20E-04 1.30E-04 1.40E-04 1.48E-04 1.54E-04 1.60E-04 1.64E-04 1.67E-04 1.69E-04 1.70E-04 1.52E-04 1.34E-04 1.17E-04 1.01E-04 8.61E-05 7.24E-05 5.46E-05 3.84E-05 2.35E-05 1.00E-05 -2.35E-06 -1.36E-05 -1.88E-05 -2.35E-05 -2.78E-05 -3.17E-05 -3.54E-05 -3.87E-05 -3.99E-05 -4.10E-05 -4.20E-05 -4.30E-05 -4.39E-05 -4.48E-05 -4.44E-05 -4.41E-05 -4.39E-05 -4.37E-05 -4.35E-05 -4.35E-05

I(t+Δt/2) (m3/s) [6] 6.3 8.8 11.3 13.8 16.3 18.8 21.7 25.0 28.3 31.7 35.0 38.3 41.7 45.0 48.3 51.7 55.0 58.3 59.2 57.5 55.8 54.2 52.5 50.8 48.5 45.5 42.5 39.5 36.5 33.5 31.2 29.5 27.8 26.2 24.5 22.8 21.4 20.3 19.1 17.9 16.8 15.6 14.6 13.8 12.9 12.1 11.3 10.4 5.0

(ht+K1Δt/2) (m) [7] 68.04 68.14 68.29 68.47 68.68 68.94 69.22 69.56 69.94 70.38 70.85 71.35 71.88 72.44 73.01 73.60 74.20 74.81 75.43 75.97 76.45 76.87 77.24 77.55 77.81 78.00 78.14 78.23 78.26 78.25 78.20 78.14 78.05 77.95 77.84 77.71 77.57 77.43 77.28 77.13 76.97 76.81 76.65 76.49 76.33 76.18 76.02 75.86 75.71

Q(ht+K1Δt/2) (m3/s) [8] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.3 1.2 2.3 3.9 5.7 7.8 10.3 12.9 15.8 19.0 21.9 24.6 27.0 29.2 31.1 32.7 34.0 34.8 35.4 35.6 35.6 35.2 34.8 34.3 33.6 32.9 32.1 31.2 30.4 29.5 28.5 27.6 26.7 25.7 24.8 23.9 23.0 22.2 21.3 20.4 K2 (m/s) [9] 2.50E-05 3.50E-05 4.50E-05 5.50E-05 6.50E-05 7.50E-05 8.67E-05 1.00E-04 1.13E-04 1.25E-04 1.35E-04 1.44E-04 1.51E-04 1.57E-04 1.62E-04 1.66E-04 1.68E-04 1.70E-04 1.61E-04 1.42E-04 1.25E-04 1.09E-04 9.32E-05 7.89E-05 6.31E-05 4.61E-05 3.06E-05 1.65E-05 3.53E-06 -8.24E-06 -1.63E-05 -2.12E-05 -2.57E-05 -2.98E-05 -3.36E-05 -3.71E-05 -3.93E-05 -4.04E-05 -4.15E-05 -4.25E-05 -4.35E-05 -4.44E-05 -4.46E-05 -4.43E-05 -4.40E-05 -4.38E-05 -4.36E-05 -4.35E-05 -6.18E-05

(ht+K2Δt/2) (m) [10] 68.05 68.15 68.30 68.48 68.69 68.95 69.24 69.57 69.96 70.39 70.85 71.36 71.89 72.44 73.02 73.61 74.21 74.81 75.41 75.96 76.44 76.86 77.22 77.53 77.79 77.99 78.13 78.21 78.25 78.24 78.20 78.13 78.05 77.95 77.83 77.71 77.57 77.43 77.28 77.13 76.97 76.81 76.65 76.49 76.33 76.18 76.02 75.86 75.67

Q(ht+K2Δt/2) (m3/s) [11] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.4 1.2 2.4 3.9 5.7 7.9 10.3 12.9 15.8 18.9 21.8 24.5 26.9 29.1 31.0 32.6 33.9 34.8 35.3 35.5 35.5 35.2 34.8 34.2 33.6 32.9 32.1 31.2 30.3 29.5 28.5 27.6 26.7 25.7 24.8 23.9 23.0 22.2 21.3 20.3 K3 (m/s) [12] 2.50E-05 3.50E-05 4.50E-05 5.50E-05 6.50E-05 7.50E-05 8.67E-05 1.00E-04 1.13E-04 1.25E-04 1.35E-04 1.44E-04 1.51E-04 1.57E-04 1.62E-04 1.66E-04 1.68E-04 1.70E-04 1.61E-04 1.43E-04 1.25E-04 1.09E-04 9.35E-05 7.93E-05 6.35E-05 4.65E-05 3.10E-05 1.68E-05 3.83E-06 -7.97E-06 -1.62E-05 -2.11E-05 -2.56E-05 -2.98E-05 -3.35E-05 -3.70E-05 -3.93E-05 -4.04E-05 -4.15E-05 -4.25E-05 -4.35E-05 -4.44E-05 -4.46E-05 -4.43E-05 -4.40E-05 -4.38E-05 -4.36E-05 -4.35E-05 -6.11E-05

(ht+K3Δt) (m) [13] 68.09 68.22 68.38 68.58 68.81 69.08 69.39 69.75 70.16 70.61 71.10 71.62 72.16 72.72 73.31 73.90 74.51 75.12 75.70 76.21 76.66 77.06 77.39 77.68 77.91 78.07 78.18 78.24 78.26 78.23 78.17 78.09 78.00 77.89 77.77 77.64 77.50 77.35 77.20 77.05 76.89 76.73 76.57 76.41 76.26 76.10 75.94 75.78 75.56

Q(ht+K3Δt) (m3/s) [14] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.1 0.7 1.7 3.1 4.8 6.7 9.0 11.6 14.4 17.4 20.4 23.2 25.8 28.1 30.1 31.9 33.3 34.4 35.1 35.5 35.6 35.4 35.0 34.5 33.9 33.3 32.5 31.7 30.8 29.9 29.0 28.1 27.2 26.2 25.3 24.4 23.5 22.6 21.7 20.9 19.7 K4 (m/s) [15] 3.00E-05 4.00E-05 5.00E-05 6.00E-05 7.00E-05 8.00E-05 9.33E-05 1.07E-04 1.20E-04 1.30E-04 1.40E-04 1.48E-04 1.54E-04 1.60E-04 1.64E-04 1.67E-04 1.69E-04 1.70E-04 1.52E-04 1.34E-04 1.17E-04 1.01E-04 8.61E-05 7.24E-05 5.46E-05 3.84E-05 2.35E-05 9.99E-06 -2.36E-06 -1.36E-05 -1.88E-05 -2.35E-05 -2.78E-05 -3.17E-05 -3.54E-05 -3.87E-05 -3.99E-05 -4.10E-05 -4.20E-05 -4.30E-05 -4.39E-05 -4.48E-05 -4.44E-05 -4.41E-05 -4.39E-05 -4.37E-05 -4.35E-05 -4.35E-05 -7.88E-05

(ht+Δt) (m) [16] 68.09 68.22 68.38 68.58 68.81 69.08 69.39 69.75 70.16 70.61 71.10 71.62 72.16 72.72 73.31 73.90 74.51 75.12 75.70 76.21 76.66 77.06 77.39 77.68 77.91 78.07 78.18 78.24 78.26 78.23 78.17 78.09 78.00 77.89 77.77 77.64 77.50 77.35 77.20 77.05 76.89 76.73 76.57 76.41 76.26 76.10 75.94 75.78 75.56

Q(ht+Δt) (m3/s) [17] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.1 0.7 1.7 3.1 4.8 6.7 9.0 11.6 14.4 17.4 20.4 23.2 25.8 28.1 30.1 31.9 33.3 34.4 35.1 35.5 35.6 35.4 35.0 34.5 33.9 33.3 32.5 31.7 30.8 29.9 29.0 28.1 27.1 26.2 25.3 24.4 23.5 22.6 21.7 20.9 19.7

Hydrograph Routing 199

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6.3.4 Changes in Inflow Hydrograph Characteristics The examples for hydrograph routing from a reservoir presented above offer some insights into how the characteristics of an inflow hydrograph change when it travels through a reservoir. The study of such changes in the inflow hydrograph’s characteristics is important in effective management of the reservoir, particularly during floods. Note what happens to the peak discharge in the inflow hydrograph as it comes out from the reservoir as an outflow hydrograph. While the magnitude of the peak discharge of the outflow hydrograph is smaller than that of the inflow hydrograph, the time to peak discharge in the outflow hydrograph is delayed as compared to the time to the peak discharge of the inflow hydrograph. The reduction in magnitude of the peak discharge of an inflow hydrograph is called attenuation, and the delay in peak (time difference between occurrence of peak discharges of inflow and outflow hydrographs) is called lag. When the outflow from a reservoir is uncontrolled, then the peak of the outflow hydrograph intersects with the falling limb of the inflow hydrograph. The magnitudes of attenuation and lag depend on several factors including reservoir storage characteristics, outlet or spillway characteristics, and the initial conditions. While the reservoir and spillway characteristics are fixed for a particular site, it is possible to achieve desirable attenuation and delay of a flood by carefully choosing or managing the initial water level (hence the initial storage and outflow) in the reservoir. Therefore, the study of the subject of hydrograph routing from a reservoir forms an important aspect of flood management through a reservoir, by carefully managing the initial conditions during impending floods. Further, note that the area between the inflow and outflow hydrographs on the rising limb represents the surplus volume of water entering a reservoir when the inflow is greater than the outflow. On the other hand, the area between the outflow and inflow hydrographs on the falling limb represents the deficit storage that is released from the reservoir during a flood. The continuity equation holds good for understanding the relative magnitudes of inflow, outflow, and storage during any time interval or the entire duration of the flood. The amounts of surplus and deficit storages are functions of the reservoir and outlet characteristics and initial conditions. As such, the knowledge of varying the initial conditions and their impacts on the surplus and deficit storage volumes provide important handles or controls on the flood management activities through a reservoir.

6.4

HYDROLOGIC ROUTING THROUGH A CHANNEL

In the reservoir routing method, the storage in the reservoir is taken as a LO 4 Outline the methods unique function of the outflow in the reservoir, S = f (Q). On the other hand in channel routing, the storage in a channel depends on both the inflow and for hydrologic routing through a channel outflow. The different stages of the types of channel storages during the passage of a flood through it are shown in Figure 6.4. Figure 6.4(a) shows the channel in which there is uniform flow; hence inflow is equal to outflow, I(t) = Q(t), at all times and the water surface elevation (WSEL) is parallel to the bed of the channel. The storage in a channel thus formed is known as prism storage as shown in Figure 6.4(a). The prism storage can also be understood as the storage formed when an imaginary plane is drawn at the downstream depth of flow in the channel reach. As the flood waters enter the reach of the channel during rising flood, the WSEL at the upstream end becomes higher than that at the downstream end because inflow to the channel reach is more than the outflow from the reach. This

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201

forms an additional storage called wedge storage as shown in Figure 6.4(b). During the time when the flood has completely entered the channel reach, the WSEL can have an irregular shape depending on the relative magnitudes of I(t) and Q(t), resistance to flow, and the geometry of the channel. The irregular shaped wedge storage thus formed is shown in Figure 6.4(c).

Figure 6.4 Prism and Wedge storages during different stages of passage of a flood through a channel or a river A little later, as the flood waters start to recede from the channel reach, the WSEL at the downstream becomes higher than that at the upstream section, because outflow is greater than the inflow. The wedge storage is negative in this case as shown in Figure 6.4(d). Slowly, the flood wave passes through the channel completely, and the WSEL at the upstream and downstream sections of the channel reach attain the original positions as shown in Figure 6.4(e). The flow in the channel becomes uniform, the wedge storage becomes zero, and the total storage is equal to the prism storage. The total storage at any time in a channel reach is equal to the sum of prism and wedge storages. Note that the shape of the WESL in Figure 6.4(b) and Figure 6.4(d) can be curved also, although they are shown as straight lines. The prism storage behaves similar to a reservoir storage and is a function of the outflow while the wedge storage is a function of the inflow I(t). The total storage at time ‘t’ can thus be represented as follows (Chow, 1959): S=

b {x I m / n + (1 - x )Q m / n } a

(6.13)

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where, x is a coefficient that reflects the relative importance given to the inflow and outflow in calculating storage in the channel; coefficients a and n represent the average stage discharge characteristics of the channel reach (Q = aGn; G being the channel stage. i.e., water level with respect to a reference point, usually taken as the deepest point in the channel cross-section); and coefficients b and m reflect the average stage volume characteristics of the channel reach (S = bGm). For example, for uniform flow in a wide rectangular channel, storage varies as the first power of the stage (m = 1) and discharge varies as 5/3rd power of stage (n = 5/3 as per Manning’s formula). For natural rivers in the floodplains, the value of n is normally less than but close to 1.0.

6.4.1

Muskingum Method

The Muskingum method is one of the most popular methods of hydrograph routing through a channel reach. The Muskingum method assumes m/n = 1 and b/a = K, and putting these relations in Eq. (6.13), we obtain the Muskingum equation as follows: S = K{xI + (1 – x)Q}

(6.14)

where, K is the storage time constant for the channel reach equivalent to the time of travel of the flood wave through the channel reach, and x is a weighting factor that varies between 0 and 0.5 with average value approximately equal to 0.2. The impact of x on the shape of outflow hydrograph is depicted in Figure 6.5.

Figure 6.5

Impact of coefficient x on hydrograph routing for Muskingum method

As we have seen in case of hydrograph routing through a reservoir, the inflow hydrograph undergoes two changes, attenuation and delay. Similarly, when an inflow hydrograph travels through a channel reach, it also undergoes attenuation and delay, which are reflected in the outflow hydrograph. Figure 6.5 shows two extreme cases of outflow hydrographs for x = 0.5 and x = 0.0. The weighting factor x = 0.5, which gives equal weightage to the inflow and outflow in the channel reach, corresponds to the maximum delay and minimum attenuation (see Figure 6.5). On the other hand, the value of x = 0.0 results in maximum attenuation and minimum delay. As per Eq. (6.14) for x = 0.0, the storage in the channel reach depends on outflow just

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203

like the case for a reservoir seen earlier (S = K Q). Therefore, the routing through a channel for x = 0.0 is essentially similar to the routing through a reservoir. The calculation of an outflow hydrograph, given an inflow hydrograph and channel characteristics, in terms of coefficients K and x, is called hydrologic routing of an inflow hydrograph through a channel. It uses the continuity equation and a finite difference form of Eq. (6.14). The time domain is first discretized using a small value of time interval Δt. Expressing Eq. (6.14) in finite difference form using subscripts 1 and 2 for the beginning and end of the time interval Δt, we get, S2 - S1 = K {x( I 2 - I1 ) + (1 - x )(Q2 - Q1 )}

(6.15)

Combining Eq. (6.15) with the continuity Eq. (6.2), the following equations can be derived for hydrograph routing through a channel reach using Muskingum method. Q2 = C0I2 + C1I1 + C2Q1 where

C0 =

- Kx + 0.5Dt K - Kx + 0.5Dt

C1 =

Kx + 0.5Dt K - Kx + 0.5Dt

C2 =

(6.16)

(6.17)

K - Kx - 0.5Dt K - Kx + 0.5Dt

Equations (6.16) and (6.17) are known as Muskingum routing equations. Note that the sum of the three constants is equal to 1.0 (i.e., C0 + C1 + C2 = 1.0). The time interval should be sufficiently small such that the inflow, outflow, and storage in the channel reach can be assumed to be linear. The value of Δt is chosen such that K > Δt > 2Kx. The values of Δt < 2Kx would result in negative values of C0, which should be avoided. Some other researchers recommend the value of Δt between (K/3) and K. Given the inflow hydrograph, and the values of coefficients K and x, the process of hydrologic routing of an inflow hydrograph through a channel reach using the Muskingum method is summarized in a step by step procedure as follows: 1.

Knowing the values of coefficients K and x for the channel reach, suitable value of time interval Δt is selected, using the guidelines given above.

2.

Using Eq. (6.17), coefficients C0, C1, and C2 are calculated.

3.

Using initial conditions in terms of Q1 and known values of inflows I1 and I2 for the first time interval, Eq. (6.16) is used to compute outflow at the end of the first time interval Q2.

4.

The outflow Q2 at the end of the first time interval calculated above is set equal to the outflow at the beginning of the next time interval and the above procedure is repeated for all the time intervals in the inflow hydrograph.

5.

The outflow and inflow hydrographs are then plotted on the same graph for comparison purposes.

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� EXAMPLE 6.3 Route the following flood hydrograph using Muskingum method through a channel reach for which K = 22 hours and x = 0.25. Plot the inflow and outflow hydrographs on the same graph sheet and determine the peak lag and attenuation from the channel reach. The outflow discharge at t = 0 was 40 m3/s. Time (Hours) 3

Inflow (m /s)

0

12

24

36

48

60

72

84

96

108

120

132

144

40

65

165

250

240

205

170

130

115

85

70

60

54

Solution The first task in inflow hydrograph routing through a channel using Muskingum method is calculating the three routing coefficients C0, C1, and C2. This requires selection of the time interval Δt for routing. The value of Δt should be selected such that K > Δt > 2Kx. For the data given in this example K = 22 hours, x = 0.25, 2Kx = 2 × 22 × (0.25) = 11.0 hours. Therefore, Δt = 12 hours is chosen which is equal to the time interval of the inflow hydrograph also. Using Eq. (6.17), the routing coefficients are calculated as: C0 = 0.0222, C1 = 0.5111, C2 = 0.4667. Note that the sum of the three routing coefficients C0 + C1 + C2 = 1.000. The computations for the routing using Muskingum method are arranged in the following table. Time (Hours)

Inflow, I (m3/s)

C0*I2 (m3/s)

C1*I1 (m3/s)

C2*Q1 (m3/s)

Q (m3/s)

[1]

[2]

[3]

[4]

[5]

[6]

0

40

1.4

20.4

18.7

40

12

65

3.7

33.2

18.9

40.6

24

165

5.6

84.3

26.0

55.8

36

250

5.3

127.8

54.1

115.9

48

240

4.6

122.7

87.4

187.2

60

205

3.8

104.8

100.1

214.6

72

170

2.9

86.9

97.4

208.7

84

130

2.3

66.4

87.3

187.2

96

105

1.9

53.7

72.9

156.1

108

85

1.6

43.4

59.9

128.4

120

70

1.3

35.8

49.0

104.9

132

60

1.2

30.7

40.2

86.1

144

54

0.0

27.6

33.6

72.0

The initial condition of Q = 40 m3/s is used. Column 1 and column 2 represent the given inflow hydrograph. Columns 3, 4, and 5 represent the C0I2, C1I1, and C2Q1 components, respectively, from the routing Eq. (6.16). Column 6 shows the outflow at the time shown in column 1, with its value equal to the sum of columns 3,4, and 5, of the previous row. The inflow and outflow hydrographs are shown in Figure 6.6.

Hydrograph Routing

Figure 6.6

205

Inflow and outflow hydrographs

6.4.2 Estimation of Parameters of Muskingum Equation The coefficients K and x represent the channel characteristics and can be found using the data corresponding to the inflow and outflow hydrographs. The procedure to determine the coefficients is essentially a trial and error procedure. The coefficient K is the storage time constant for the reach representing the travel time of the flood wave through the reach. The value of K can thus be approximated using the inflow and outflow hydrograph data. The continuity equation can be used to determine the incremental storage (ΔS) in the channel reach for each time interval Δt as follows: DS =

I1 + I 2 Q + Q2 Dt - 1 Dt 2 2

(6.18)

The storage (S) versus time relationship can then be determined by calculating cumulative storage for each time step. For a trial value of coefficient x (say x = 0.2), the quantity {xI + (1 – x)Q} is calculated for each time interval, which is linearly related with the storage S in the channel reach as per the Muskingum equation (6.14). Therefore, if the chosen value of x is correct, a plot between S and the quantity {xI + (1 – x)Q} will be a straight line, otherwise the relationship will appear as a nonlinear looped one. The graph between S and {xI + (1 – x)Q} is plotted for several trial values of x and the one that gives the relationship closest to a straight line is adopted as the suitable value of x. The slope of the straight line S = K{xI + (1 – x)Q} is represented by K and can be easily calculated using this plot for the selected value of x. The procedure of estimating K and x is explained in the example below. � EXAMPLE 6.4 Using the inflow and outflow hydrographs from a channel reach given below, determine Muskingum coefficients K and x. Time (Hours)

0

4

8

12

16

20

24

28

32

36

40

44

48

52

56

Inflow (m3/s)

42

68

116

164

194

200

192

170

150

128

106

88

74

62

54

42

39

44

65

98

133

159

174

175

168

156

140

122

106

90

3

Outflow (m /s)

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206

Solution The calculation of the Muskingum coefficients (K and x) is carried out using the continuity equation (6.18). The computations are arranged in the following table. The first three columns present the inflow and outflow hydrograph data given in the example. Column 4 is the difference of inflow and outflow (I – Q) and column 5 is obtained by multiplying column 4 by time interval Δt = 4 hours. Column 5 represents incremental storage ΔS in m3/s-hour. Column 6 is storage S which is obtained by accumulating the entries in column 5. Then, the next 4 columns represent the quantity {xI + (1 – x)Q} for different trial values of x. The values tried are x = 0.2, x = 0.3, x = 0.4, and x = 0.5. The plots of storage S versus the quantity {xI + (1 – x)Q} are then drawn to ascertain if it shows up as a straight line. The value of x which results in a straight line plot will be the correct value of x. As seen from the plots, the value of x = 0.4 is the most appropriate as it gives a straight line with a very narrow scatter. The value of K is then obtained as the slope of the straight line in the S v/s {xI + (1 – x)Q} plot corresponding to x = 0.4. Selecting the storages S1 = 1,000 m3/s-hour and S2 = 1,500 m3/s-hour, the corresponding values of the quantity {xI + (1 – x)Q} are read as 125 m3/s and 168 m3/s. Then, K = (1500 – 1000)/(168 – 125) = 12.63 hours. Therefore, the values of the Muskingum coefficients for the given data are K = 12.63 hours and x = 0.4. Time (Hours)

Inflow, I (m3/s)

Outflow, Q (m3/s)

(I – Q) (m3/s)

ΔS = [4]*Δt (m3/s-h)

S = ∑ΔS (m3/s-h)

[1]

[2]

[3]

[4]

[5]

0

42

42

0

4

68

39

8

116

12

[xI + (1 – x)Q ] (m3/s) x = 0.2

x = 0.3

x = 0.4

x = 0.5

[6]

[7]

[8]

[9]

[10]

0

0

42.0

42.0

42.0

42.0

29

116

116

44.8

47.7

50.6

53.5

44

72

288

404

58.4

65.6

72.8

80.0

164

65

99

396

800

84.8

94.7

104.6

114.5

16

194

98

96

384

1184

117.2

126.8

136.4

146.0

20

200

133

67

268

1452

146.4

153.1

159.8

166.5

24

192

159

33

132

1584

165.6

168.9

172.2

175.5

28

170

174

–4

–16

1568

173.2

172.8

172.4

172.0

32

150

175

–25

–100

1468

170.0

167.5

165.0

162.5

36

128

168

–40

–160

1308

160.0

156.0

152.0

148.0

40

106

156

–50

–200

1108

146.0

141.0

136.0

131.0

44

88

140

–52

–208

900

129.6

124.4

119.2

114.0

48

74

122

–48

–192

708

112.4

107.6

102.8

98.0

52

62

106

–44

–176

532

97.2

92.8

88.4

84.0

56

54

90

–36

–144

388

82.8

79.2

75.6

72.0

Hydrograph Routing

6.5

207

IUH DEVELOPMENT

In the previous chapter, we learnt the concept of Instantaneous Unit LO 5 Summarize instanHydrographs (IUHs). An IUH is a hypothetical concept describing the taneous unit hydrograph DRH response from a catchment when it is subjected to 1 cm of effective development using routing rainfall (ER) instantaneously and uniformly over the catchment. There are concepts several catchment simulation models that have been developed to model this complex rainfall-runoff process. These catchment simulation models employ the concepts of hydrograph routing that we learnt in this chapter. In this section, we will learn about the two important conceptual models for IUH development using routing methods. The transformation of rainfall into runoff is an extremely complex, nonlinear, and dynamic process that is difficult to understand and model. The two major steps involved in the transformation of rainfall into runoff are: (i) calculation of infiltration and other losses and estimation of the effective rainfall, and (ii) subsequent transformation of the effective rainfall into runoff hydrograph through an operator, which simulates the behavior of the catchment. As rain falls on the catchment, some of it gets trapped in depressions

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and interception storages and some of it gets infiltrated. We learnt about the methods to calculate the losses in Chapter 3. The remaining or “effective” rainfall (ER) runs over the land surfaces and ultimately travels to the outlet in the form of DRH. The second step is actually responsible for modeling the nonlinear, dynamic, and complex nature of the rainfall-runoff process. During this process, the input (i.e., effective rainfall) to the system (i.e., catchment) goes through two operators: (i) ‘translation’ in time due to different areas of the catchment contributing runoff at the outlet at different times, and (ii) ‘attenuation’ due to the storage characteristics of the catchment. All the models of the rainfall-runoff process that are based on physical concepts, essentially attempt to account for these operators with varying degrees of sophistication and complexity. We will learn two methods for IUH development, namely, Clark’s method and Nash’s method in this section.

6.5.1

Clark’s Method

The Clark’s method is a conceptual method for developing IUH at the outlet LO 6 Discuss Clark’s Method of a catchment when it is subjected to 1 cm of ER occurring instantaneously for IUH development and uniformly over the catchment. As ER falls on a catchment, it undergoes two processes, translation and attenuation, which are modeled by two conceptual hydrologic elements. The translation of ER is modeled using a hypothetical hydrological element called linear channel (LC). The attenuation of the inputs is achieved by routing the translated ER through a linear reservoir (LR) to calculate the outflow hydrograph at the outlet of the catchment. Since the input consisted of 1 cm of ER occurring instantaneously, the output from the catchment is DRH in the form of IUH. The concept of IUH development by Clark’s method is illustrated in Figure 6.7.

Figure 6.7 Clark’s method for IUH development A linear channel is a fictitious hydrologic element in which the time of travel of a flood wave is constant. In Clark’s model, the linear channel is modeled using the concept of a time-area diagram (TAD) of the catchment. A time-area diagram is a histogram showing area versus time of travel from different parts to the outlet of a catchment. The maximum time of travel on a time-area diagram is the time of concentration. The time of concentration of a catchment is the time of travel of water from the most remote location to the outlet of the catchment. Time of concentration (tc) forms the backbone of the time area diagram for a catchment. In a gauged catchment, the time of concentration can be taken as the time elapsed between the time ER ends and the point of inflection of a hydrograph. For ungauged catchments, Eq. (4.28) may be adopted as per the Kirpich’s formula. In order to develop a time-area diagram for a catchment, it is sub-divided into sub-areas such that the two sub-areas are separated by a line known as an isochrone. An isochrone is a line representing equal time of travel from any point on it to the outlet of the catchment. The area between two isochrones is called inter-isochronal area and the time difference between the times of travel between two isochrones is denoted as time interval Δtc. The concept of a time area diagram is depicted in Figure 6.8. The entire catchment is sub-divided into a suitable number of sub-areas corresponding to Δtc. In Figure 6.8, Ai’s represent the inter-isochronal areas (km2).

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209

The instantaneous ER of 1 cm is assumed to fall on the catchment uniformly with respect to space. That is, amount of effective rainfall is 1 cm over each Ai; as such the volume of ER falling on each sub-area is Ai × 1 (km2-cm). The time area diagram translates this ER volume to the outlet without any attenuation as per the concept of linear channel. Therefore, the ER volume reaching the outlet of the catchment at each time interval will be equal to Ai (km2-cm). The amount of discharge at the outlet with pure translation effects only can therefore be estimated as follows: I (t ) =

Figure 6.8

Ai A m3 km 2 cm = 2.778 i Dtc Dtc s

(6.19)

Concepts of time-area diagram in Clark’s method for IUH development

The term I(t) in Eq. (6.19) represents the DRH response with pure translation effects only from the catchment when it is subjected to an instantaneous uniform ER of 1 cm. The I(t) is taken as the inflow hydrograph to a linear reservoir and the routed outflow hydrograph is calculated using Muskingum method. In other words, the outflow hydrograph at the outlet of the catchment with both translation and attenuation effects is found by routing the inflow hydrograph given by Eq. (6.19) through a linear reservoir. This routing is carried out by Muskingum method using Equations (6.18) and (6.19) with x = 0. Putting x = 0 in the Muskingum Eq. (6.14), we get S = K Q which is the storage-outflow relationship for a linear reservoir. Note that the inflow hydrograph is not a continuous function but a discretized one so that I1 = I2 for each time interval. Also, for x = 0, C0 = C1; therefore the routing equations are modified as follows: Q2 = 2 C1 I1 + C2Q1 where

C0 = C1 =

0.5Dtc K + 0.5Dtc

K - 0.5Dtc C2 = K + 0.5Dtc

(6.20)

(6.21)

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Engineering Hydrology

The inflow hydrograph I (t) is routed through the catchment using Eqs (6.20) and (6.21). The resulting outflow DRH represents the IUH from the catchment. Using the IUH, a UH of any duration (D) can be obtained using the method described in Chapter 5. The value of storage time constant (K) can be easily obtained by using the continuity equation at the point of inflection of an observed DRH in the same catchment, or a nearby catchment that is hydrologically similar, by putting I = 0 in the continuity equation as follows: -Q =

dS dQ =K dt dt

K=-

Or

(6.22)

Q dQ dt

(6.23)

� EXAMPLE 6.5 A catchment of area 244 km2 is found to have a storage constant of 4.0 hours. Given the following time-area diagram for the catchment, determine the IUH from it. Time (Hours)

0-1

1-2

2-3

3-4

4-5

5-6

Area (km2)

20

40

90

55

28

11

Solution Taking Δtc = 1 hour, K = 4.0 hours and using Eq. (6.21) above, the values of coefficients are calculated as C1 = 0.111 and C2 = 0.778. The inflow to the catchment is then calculated using the time area diagram given and the Eq. (6.19). The inflow hydrograph is then routed through the catchment by Clark’s method using Eq. (6.20). The computations are arranged in a tabular form below. In the table, the given time area diagram ordinates are presented in column 2 at 1-hour intervals. The inflow is then calculated as I = 2.778 Ai/Δtc, where Ai is the inter-isochrone area (km2) and Δtc = 1 hour. The inflow hydrograph ordinates (m3/s) with pure translation effects are presented in column 3. Column 4 and column 5 represent the two terms (2C1I1 and C2Q1) in the routing Eq. (6.20). Column 6 represents the IUH ordinate in the catchment, which is obtained as the sum of the previous two columns. Note that the IUH ordinate is taken as 0.0 at t = 0 as per the property of the IUH. The IUH thus obtained is then plotted in Figure 6.9. Time (Hours)

TAD Ordinate (km2)

Inflow Ordinate (m3/s)

2C1×I1 (m3/s)

C2×Q1 (m3/s)

IUH Ordinate (m3/s)

[1]

[2]

[3]

[4]

[5]

[6]

0

0.0

1

20

55.6

12.3

0.0

12.3

2

40

111.1

3

90

250.0

24.7

9.6

34.3

55.6

26.7

82.2 (Contd.)

Hydrograph Routing

4

55

152.8

34.0

64.0

97.9

5

28

77.8

17.3

76.2

93.4

6

11

30.6

6.8

72.7

79.5

7

61.8

61.8

8

48.1

48.1

9

37.4

37.4

10

29.1

29.1

11

22.6

22.6

12

17.6

17.6

13

13.7

13.7

14

10.6

10.6

15

8.3

8.3

16

6.4

6.4

17

5.0

5.0

18

3.9

3.9

19

3.0

3.0

20

2.4

2.4

Figure 6.9

6.5.2

211

IUH plot for Example 6.5

Nash’s Method

The Clark’s method discussed above employs two conceptual elements, LO 7 Explain Nash’s method a linear channel for pure translation and a linear reservoir for attenuation for IUH development of the input to a catchment, in order to determine the outflow hydrograph at the outlet of the catchment. Nash (1957) proposed that a catchment can be represented by a series of n identical linear reservoirs each having the same storage constant (K). The reservoirs are connected in series forming a cascade of n linear reservoirs as shown in Figure 6.10.

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Figure 6.10 Cascade of linear reservoirs for Nash model for IUH development A unit volume of rainfall input, equivalent to 1 cm of ER, is instantly applied to the first linear reservoir as shown in Figure 6.10. This is routed through the first linear reservoir and the output from the first linear reservoir is applied as an input for the second linear reservoir. The output from the second linear reservoir is applied as an input for the third linear reservoir and so on, till we receive the output from the nth linear reservoir. Each reservoir has a storage constant K and imparts both translation and attenuation effects of equal magnitude to the input it receives. Since the input applied is 1 cm of ER instantaneously, the output from the nth reservoir represents the IUH from the catchment. In addition to the storage constant K, the other parameter of the Nash’s model is n, which is a function of the catchment characteristics and represents the number of linear reservoirs needed to capture the rainfall-runoff process of the catchment. Note that the value of n need not be an integer, it could be a fraction as well, since it is a part of a hypothetical concept. The Nash’s model is based on the continuity equation: I -Q= Or

K

dS dQ =K dt dt

dQ +Q=I dt

(6.24) (6.25)

Equation (6.25) is a first order linear ordinary differential equation (ODE). Multiplying by the integrating factor et/K on either side, and simplifying we get: et / K Or

dQ 1 I + Q et / K = et / K dt K K

(6.26)

d (Qet / K ) I t / K = e dt K

(6.27)

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213

Integrating on both sides, we get: 1 (6.28) I et / K dt + C KÚ where, C is the constant of integration. In order to avoid the singularity at t = 0, we may conceptualize the instantaneous rainfall of 1 cm as occurring over a small interval, 0 to Δt, with an intensity of 1/Δt, and then let Δt approach zero. Eq. (6.28) may then be written as Qet / K =

t

Qe K =

1 K

1 = K

t

1

Ú Dt e

t/K

dt + C for t £ Dt

0

Dt

Ú 0

(6.29)

1 t/K e dt + C for t ≥ Dt Dt

Since the system starts at rest, i.e., Q = 0 at t = 0 (the outflow will show a sharp jump from t = 0 to t = Δt due to the sudden inflow), the first equation in Eq. (6.29) results in C = 0 (note that we have assumed a small but non-zero Δt). Putting C = 0 in the second equation and assuming that for very small Δt, the exponential term may be treated as a constant (we arbitrarily take this value at the midpoint of the interval), the solution can be written as t

for t ≥ Dt , Qe K =

1 K

Dt

1

Ú Dt e

Dt /2 K

dt =

0

1 Dt /2 K e K

(6.30)

And now, letting Δt go to zero, the outflow from the first reservoir becomes: Q1 =

1 -t / K e K

(6.31)

Note that the outflow has jumped at t = 0 from 0 to 1/K as a result of the instantaneous rainfall. Equation (6.31) represents the DRH from the first linear reservoir which becomes input for the second linear reservoir. Hence, Q2 =

1 -t / K t/K -t / K e ÚQ1 e dt + Ce K

(6.32)

The constant of integration, C, is again obtained by the initial condition, Q2 = 0 at t = 0, to get t

Q2 =

Or

1 -t / K 1 -t / K t / K e Ú K e e dt K 0 Q2 =

t K2

e- t / K

(6.33)

(6.34)

Equation (6.34) represents the DRH coming out of the second linear reservoir which becomes the input for the third linear reservoir. The DRH coming out of the third reservoir can be written as follows: Q3 =

t2 2K

3

e- t / K

(6.35)

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Engineering Hydrology

Similarly, it can be shown that the output from the nth linear reservoir can be written as: Qn = u(t ) =

1 (n - 1)! K n

tn -1 e

-

t K

(6.36)

Equation (6.36) represents IUH from a catchment with parameters K and n to be determined using rainfall and flow data. Note that K and t are in hours and Q, or the IUH ordinate u(t), is in cm/h in the above equation. It is easy to evaluate (n – 1)! when n is an integer as factorial of (n – 1). In order to have a generalized solution for IUH for integer as well as fractional values of n, the factorial (n – 1)! is expressed using Gamma function as follows: Qn = u(t ) =

1 G ( n) K

n

t n -1 e

1 Ê tˆ Qn = u(t ) = G (n)K ÁË K ˜¯

-

t K

n -1

e

-

t K

¸ Ô Ô ˝ Ô Ô ˛

(6.37)

Therefore, if n is an integer, G(n) = (n – 1)! and when n is a fraction then the value of G(n) can be evaluated using the Gamma function given in Table 6.1. In the table, the values of G(n) are given, for different values of n ranging from 1.00 to 2.00 in the interval of 0.02. The values of G(n) for other values of n can be found using the relation G(n + 1) = nG(n). The solved example given below illustrates the use of Gamma table and calculation of IUH using Nash’s model.

Table 6.1 Values of Gamma function G(n) for different values of n n

G(n)

n

G(n)

n

G(n)

1.00

1.000000

1.34

0.892216

1.68

0.905001

1.02

0.988844

1.36

0.890185

1.70

0.908639

1.04

0.978438

1.38

0.888537

1.72

0.912581

1 06

0.968744

1.40

0.887264

1.74

0.916826

1.08

0.959725

1.42

0.886356

1.76

0.921375

1.10

0.951351

1.44

0.885805

1.78

0.926227

1.12

0.943590

1.46

0.885604

1 80

0.931384

1.14

0.936416

1.48

0.885747

1.82

0.936845

1.16

0.929803

1.50

0.886227

1.84

0.942612

1.18

0.923728

1.52

0.887039

1.86

0.948687

1.20

0.918169

1.54

0.888178

1.88

0.955071

1.22

0.913106

1.56

0.889639

1.90

0.961766

1.24

0.908521

1.58

0.891420

1.92

0.968774

1.26

0.904397

1.60

0.893515

1 94

0.976099

1.28

0.900718

1.62

0.895924

1.96

0.983743

1.30

0.897471

1.64

0.898642

1.98

0.991708

1.32

0.894640

1.66

0.901668

2.00

1.000000

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� EXAMPLE 6.6 A catchment of area 650 km2 is found to have K = 5.0 hours and n = 4.5. Evaluate IUH and find its peak discharge and time to peak. Solution The IUH is given by Eq. (6.37). For n = 4.5, using the relation G(n + 1) = nG(n) recursively, we get G(4.5) = G(3.5) (3.5) = (3.5)(2.5) G(2.5) = (3.5)(2.5)(1.5) G(1.5). From the Table 6.1, G(1.5) = 0.886226 gives G(4.5) = 11.632. Putting these values in Eq. (6.37), we get: Qn = u(t ) =

1 Ê t ˆ (11.632)(5.0) ÁË 5.0 ˜¯

Ê t ˆ Qn = u(t ) = 0.01719 Á Ë 5.0 ˜¯

4.5 -1

e

3.5

e

-

-

t 5.0

t 5.0

Now putting the values of time (t) in the above equation, the IUH can be calculated. The computations are arranged in the table below. Column 1 is the time at 1-hour intervals. Column 2 is the time in dimensionless units (t/K) = (t/5.0). Column 3 is the IUH ordinate by putting the value of (t/K) in the equation above. The IUH ordinate in column is in cm/h. Using the catchment area A = 650 km2, the IUH ordinate is converted into m3/s. The IUH ordinate in m3/s is given in column 4 in the table below. The calculations are carried out for t = 0 to t = 68 hours. The IUH ordinates against time are plotted in Figure 6.11. Note that the IUH starts with u(t) = 0.0 m3/s at t = 0.0 hours and tends towards 0.0 m3/s for large times. The peak IUH ordinate is 75.092 m3/s, which occurs at t = 18 hours.

Figure 6.11 IUH obtained by Nash model for Example 6.6

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Engineering Hydrology

Time (Hours)

(t/K) (no units)

IUH Ordinate (cm/h)

IUH Ordinate (m3/s)

Time (Hours)

(t/K) (no units)

[1]

[2]

[3]

[4]

[1]

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 25 26 27 28 29 30 31 32 33 34

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00 2.20 2.40 2.60 2.80 3.00 3.20 3.40 3.60 3.80 4.00 4.20 4.40 4.60 4.80 5.00 5.20 5.40 5.60 5.80 6.00 6.20 6.40 6.60 6.80

0.0000 0.0001 0.0005 0.0016 0.0035 0.0063 0.0098 0.0138 0.0180 0.0222 0.0263 0.0301 0.0334 0.0362 0.0384 0.0400 0.0411 0.0416 0.0416 0.0411 0.0403 0.0391 0.0377 0.0361 0.0343 0.0324 0.0304 0.0284 0.0264 0.0245 0.0226 0.0207 0.0189 0.0173 0.0157

0.00 0.09 0.84 2.85 6.39 11.42 17.70 24.86 32.47 40.15 47.53 54.33 60.32 65.35 69.35 72.28 74.18 75.088 75.092 74.29 72.78 70.69 68.11 65.15 61.91 58.47 54.91 51.31 47.71 44.17 40.72 37.39 34.21 31.19 28.35

35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68

[2]

IUH Ordinate (cm/h) [3]

IUH Ordinate (m3/s) [4]

7.00 7.20 7.40 7.60 7.80 8.00 8.20 8.40 8.60 8.80 9.00 9.20 9.40 9.60 9.80 10.00 10.20 10.40 10.60 10.80 11.00 11.20 11.40 11.60 11.80 12.00 12.20 12.40 12.60 12.80 13.00 13.20 13.40 13.60

0.0142 0.0129 0.0116 0.0104 0.0093 0.0084 0.0075 0.0066 0.0059 0.0052 0.0046 0.0041 0.0036 0.0032 0.0028 0.0025 0.0022 0.0019 0.0017 0.0015 0.0013 0.0011 0.0010 0.0008 0.0007 0.0006 0.0005 0.0005 0.0004 0.0004 0.0003 0.0003 0.0002 0.0002

25.69 23.21 20.92 18.80 16.86 15.08 13.46 11.99 10.66 9.46 8.38 7.41 6.54 5.76 5.07 4.46 3.91 3.43 3.00 2.62 2.29 2.00 1.74 1.51 1.31 1.14 0.99 0.86 0.74 0.64 0.56 0.48 0.41 0.36

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217

There are several other conceptual models available that employ different combinations of different numbers of linear reservoirs connected in series and parallel; however, their description is beyond the scope of this book and the interested readers can find more details in Diskin and Simpson (1978) or Gupta et al. (1980).

SUMMARY When a flow hydrograph travels through a hydrologic element in a catchment (a reservoir or a channel reach), its characteristics (e.g. size, shape, peak flow, time to peak, slopes of the rising and falling limbs etc.) get modified due to several factors such as storage characteristics of the hydrologic element, resistance to flow, lateral addition/subtraction of flow etc. Such factors account for the spatio-temporal variations in the inflow hydrograph as it travels through a reservoir or a river reach; and a study of the impacts of such factors on the inflow hydrograph and calculation of the outflow hydrograph at the outlet of a hydrologic element forms the core of an important area in hydrology known as hydrograph routing. The hydrograph routing can be one of the two kinds: hydrologic routing or hydraulic routing. The hydrologic routing uses continuity equation and is easier than the hydraulic routing which uses both continuity and momentum equations. This chapter presented two methods of hydrologic routing for routing of an inflow hydrograph through a reservoir, namely, Modified Puls method and Goodrich’s method, both of which are semi-graphical in nature. The semi-graphical methods pose problems in computer simulation; hence, there are other methods available for hydrologic routing through a reservoir that can be easily implemented on a computer and incorporated in management models. The fourth order Standard Runge-Kutta method is one such method and was discussed in this chapter. When an inflow hydrograph passes through a reservoir, it undergoes two important changes, attenuation and delay. The reduction in magnitude of the peak discharge of an inflow hydrograph is called attenuation, and the time difference between occurrence of peak discharges of inflow and outflow hydrographs is called lag or delay. It is possible to achieve desirable attenuation and delay of a flood by carefully managing the initial water level in a reservoir. The area between the inflow and outflow hydrographs on the rising limb represents the surplus volume of water entering into a reservoir while the area between the outflow and inflow hydrographs on the falling limb represents the deficit storage that is released from the reservoir during a flood. The varying initial conditions, their impacts on the attenuation, delay, surplus and deficit storage volumes, and the continuity equation are useful tools for efficient operation of reservoirs during a flood. While the storage in a reservoir depends on outflow alone, the storage in a channel depends both on inflow and outflow. The total storage in a channel reach is the sum of two types of storages, a prism storage which is always positive and a wedge storage which can be either positive or negative depending on the stage of the passage of a flood through a channel reach. The prism storage behaves similar to reservoir storage and is a function of the outflow while the wedge storage is a function of the inflow. The Muskingum method of hydrologic routing is based on the continuity equation and a linear relationship between total storage in a channel, the inflow, and the outflow. Hydraulic routing is useful in distributed rainfall-runoff modeling and simulation studies in a catchment. It is based on St. Venant’s equation, which is difficult to solve analytically. Finite difference, finite element, and method of characteristics are used to obtain complete numerical solutions of the St. Venant’s equation for hydraulic routing.

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Engineering Hydrology

There are several conceptual methods for developing an instantaneous unit hydrograph from a catchment using the concept of attenuation and delay in a catchment. In Clarks’ method for conceptual IUH development, unit effective rainfall is applied to a time area diagram of the catchment first to impart pure translation effects using a linear channel element. Then the translated output from the linear channel element is routed through a linear reservoir element to impart attenuation effects and calculate the IUH. On the other hand, the Nash’s method of IUH development is based on simulating the catchment using n identical linear reservoirs and passing the input through this series of reservoirs to calculate the output from the last reservoir as the IUH.

OBJECTIVE-TYPE QUESTIONS 6.1 The factors affecting the size, shape, and other characteristics of an inflow hydrograph while traveling through a channel reach include (a) Storage in the river reach (b) Resistance to flow due to friction from sides and bed (c) Lateral addition or subtraction of flow within the reach (d) All of the above 6.2 Hydrograph routing is important in (a) Flood forecasting and flood control (c) Catchment simulation studies

(b) Design of reservoirs and spillways (d) All of the above

6.3 Which of the following statements is/are true for hydrograph routing? I. Hydraulic routing uses law of conservation of mass only. II. Hydrologic routing uses both law of conservation of mass and momentum. III. Spatio-temporal variations in an inflow hydrograph are non-existent as it travels through a river reach or a reservoir. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 6.4 Which of the following statements is true for hydrograph routing? (a) Hydrologic routing is more complex than hydraulic routing (b) Hydraulic routing is more complex than hydrologic routing (c) Both are of same difficulty (d) Insufficient information to comment 6.5 Which of the following statements is/are true for hydrograph routing? I. Hydraulic routing is more accurate than hydrologic routing. II. Hydraulic routing is very useful in developing conceptual unit hydrographs in an ungauged catchment. III. Hydrologic routing is employed by NOAA in their flood warning system. (b) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false

Hydrograph Routing

219

6.6 The passage of an inflow hydrograph traveling through a reservoir is (a) Uniform and steady flow (b) Non-uniform and steady flow (c) Non-uniform and unsteady flow (d) Uniform and unsteady flow 6.7 Which of the following finds application in distributed rainfall-runoff modeling and simulation studies in a catchment? (a) Hydrologic routing (b) Hydraulic routing (c) Both (a) and (b) (d) None of these 6.8 The continuity equation of unsteady flow in a river, with no lateral flow, can be represented as

∂Q ∂y +T =C ∂x ∂t ∂Q ∂y (c) +T =C ∂y ∂t (a)

(b)

∂y ∂y +T =0 ∂x ∂t

(d)

∂Q ∂y +T =0 ∂x ∂t

6.9 The St. Venant equations represent (a) Simultaneous, quasi-linear, first order PDEs of the hyperbolic type that are not easy to solve analytically in general (b) Simultaneous, quasi-linear, first order PDEs of the hyperbolic type that are easy to solve analytically in general (c) Simultaneous, nonlinear, first order PDEs of the hyperbolic type that are not easy to solve analytically in general (d) Simultaneous, quasi-linear, second order PDEs of the hyperbolic type that are not easy to solve analytically in general (e) None of the above 6.10 Muskingum method of river routing is an example of (a) Partial solution methods (b) Complete numerical methods (c) Both (a) and (b) (d) None of (a) and (b) 6.11 Which of the following statements is most appropriate for the partial solution methods for hydrograph routing? (a) We use either only continuity equation or highly simplified version of momentum equation and continuity equation (b) We use only continuity equation (c) We use only momentum equation (d) We use both momentum and continuity equations 6.12 The hydrologic method of reservoir routing is an example of (a) Partial solution methods (b) Complete numerical methods (c) Both (a) and (b) (d) None of (a) and (b) 6.13 The complete numerical methods employ (a) Continuity equation only (b) Momentum equation only (c) Both continuity and momentum equations (d) Method of characteristics only 6.14 The finite difference (FD) method is an example of (a) Partial solution method (b) Complete numerical method (c) Analytical method (d) None of the above

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Engineering Hydrology

6.15 The method of characteristics is an example of (a) Partial solution method (b) Complete numerical method (c) Analytical method (d) None of the above 6.16 The complete numerical method can be (a) Finite difference type (c) Method of characteristics type (e) None of (a), (b), and (c)

(b) Finite element type (d) All of the above

6.17 In finite difference methods, (a) Partial derivatives are replaced by the finite difference approximations (b) St. Venant equations are transformed into algebraic equations (c) Both (a) and (b) (d) None of (a) and (b) 6.18 Which of the following statements is true for FD methods? (a) Explicit methods are easy and more accurate (b) Implicit methods are easy and more accurate (c) Both are easy and accurate (d) None of the above 6.19 Which of the following statements is true for FD methods? (a) Explicit methods are iterative in nature and computationally more expensive (b) Implicit methods are iterative in nature and computationally more expensive (c) Both are iterative in nature (d) None of the above 6.20 In the method of characteristics (a) ODEs are first converted into PDEs and then solved using any of the FD methods (b) PDEs are first converted into ODEs and then solved using any of the FE methods (c) PDEs are first converted into ODEs and then solved using any of the FD methods (d) ODEs are first converted into PDEs and then solved using any of the FE methods 6.21 In the FE methods for hydrograph routing, the entire system is assumed to be composed of small elements and the (a) PDEs are written and integrated at each node point to obtain unknowns at each node (b) PDEs are written and differentiated at each node point to obtain unknowns at each node (c) ODEs are written and integrated at each node point to obtain the unknowns at each node (d) ODEs are written and differentiated at each node point to obtain unknowns at each node (e) None of the above 6.22 HEC-RAS is capable of computing water surface elevation in natural rivers using (a) 1-D flow analysis and 2-D sediment transport computations (b) 2-D flow analysis and 1-D sediment transport computations (c) 1-D flow analysis and 1-D sediment transport computations (d) 2-D flow analysis and 2-D sediment transport computations (e) None of the above 6.23 The computer program FLDWAV was developed by (a) Federal Emergency Management Agency (b) US National Weather Service (c) US Army Corps of Engineers (d) MIT and Stanford universities (e) None of the above

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6.24 Which of the following computer programs is useful for modeling a system of interconnected rivers? (a) HEC-RAS (b) MIKE FLOOD (c) FLDWAV (d) FLOW-2D 6.25 In addition to the inflow hydrograph, the data needed for carrying out hydrologic routing through a reservoir using semi-graphical method are (a) Storage volume versus elevation relationship and initial conditions (b) Outflow versus height of water over outlet structure relationship and initial conditions (c) Storage volume versus elevation relationship and outflow versus height of water over outlet structure relationship (d) Storage volume versus elevation relationship, outflow versus height of water over outlet structure relationship, and initial conditions (e) None of the above 6.26 The indicative storage in Modified Puls method is equivalent to (a) (Q + S t/2) (b) (S + 2Q t) (c) (S + Q t/2)

(d) (2S + Q t/2)

6.27 In the Modified Puls method, we normally plot (a) The height of water in the reservoir on x-axis, indicative storage on the primary y-axis, and the outflow on the secondary y-axis (b) The height of water in the reservoir on y-axis, indicative storage on the primary x-axis, and the outflow on the secondary x-axis (c) The height of water in the reservoir and indicative storage on y-axis, and the outflow on the x-axis (d) The height of water in the reservoir on x-axis, indicative storage on the primary y-axis, and the outflow on the secondary y-axis 6.28 As an inflow hydrograph passes through a reservoir, (a) Its peak discharge decreases and time to peak increases (b) Its peak discharge increases and time to peak increases (c) Its peak discharge decreases and time to peak decreases (d) Its peak discharge increases and time to peak decreases 6.29 The lateral inflow into a channel reach can cause (a) Decrease in the peak discharge (b) Increase in the peak discharge (c) Has no effect on peak discharge (d) Cannot comment 6.30 In the rearranged continuity equation in Modified Puls method, various variables have units of (c) m2 (d) m3/s (a) m (b) m3 6.31 In the rearranged continuity equation in Goodrich’s method, various variables have units of (a) m (b) m3 (c) m2 (d) m3/s 6.32 Which of the methods is suitable for implementation on a computer for hydrologic routing? (a) Modified Puls method (b) Goodrich’s method (c) Fourth Order SRK method (d) None of the above 6.33 In addition to the inflow hydrograph, the data needed for carrying out hydrologic routing through a reservoir using SRK method are (a) Storage volume versus elevation relationship and initial conditions (b) Outflow versus height of water over outlet structure relationship and initial conditions (c) Surface area versus elevation relationship for reservoir and outflow versus height of water over outlet structure relationship (d) Surface area versus elevation relationship for reservoir, outflow versus height of water over outlet structure relationship, and initial conditions (e) None of the above

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6.34 For a particular reservoir, which of the following is the most important aspect in flood management? (a) Reservoir characteristics (b) Spillway characteristics (c) Initial conditions (d) All of the above 6.35 Which of the following statements is/are true for hydrograph routing? I. Surplus storage enters a reservoir when the inflow is greater than the outflow. II. Deficit storage is released from a reservoir when the inflow is greater than the outflow. III. The amounts of surplus and deficit storages are functions of the reservoir and outlet characteristics, and initial conditions. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 6.36 Which of the following statements is true? (a) Surplus storage occurs on rising limb and deficit storage occurs on falling limb (b) Surplus storage occurs on falling limb and deficit storage occurs on rising limb (c) The two storages can occur on any limb depending on the storage characteristics of the reservoir (d) The two types of storages do not depend on the initial conditions 6.37 Which of the following are useful in flood management? (a) Continuity equation through the passage of the hydrograph from a reservoir (b) Impact of initial conditions on the attenuation and lag (c) Impact of initial conditions on the surplus and deficit storages (d) All of the above (e) None of (a), (b), and (c) 6.38 In channel routing, the storage in a channel depends on (a) Inflow only (b) Outflow only (c) Both inflow and outflow (d) None of (a) or (b) 6.39 Prism storage in a channel occurs when (a) Inflow is equal to outflow (b) Flow in the channel is uniform (c) Water surface elevation in the channel is parallel to the bed of the channel (d) All of the above (e) None of (a), (b), and (c) 6.40 As the flood waters enter a channel reach during rising flood, (a) WSEL at the downstream end becomes higher than that at the upstream end (b) Outflow from the channel reach is more than the inflow to the reach (c) WSEL at the upstream end becomes higher than that at the downstream end (d) Inflow to the channel reach is equal to the outflow from the reach 6.41 Wedge storage in a channel reach is positive when (a) WSEL at the downstream end is higher than that at the upstream end (b) WSEL at the upstream end is higher than that at the downstream end (c) WSEL at the upstream end is equal to that at the downstream end (d) None of the above 6.42 Wedge storage in a channel reach is negative when (a) WSEL at the downstream end is higher than that at the upstream end (b) WSEL at the upstream end is higher than that at the downstream end (c) WSEL at the upstream end is equal to that at the downstream end (d) None of the above

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6.43 During passage of a flood wave through a channel reach, the WSEL can have an irregular shape depending on the (a) Relative magnitudes of I(t) and Q(t) (b) Resistance to flow (c) Geometry of the channel (d) All of the above (e) None of (a), (b), and (c) 6.44 As the flood waters leave a channel reach during falling flood, (a) WSEL at the downstream end becomes higher than that at the upstream end (b) Inflow to the channel reach is more than the outflow from the reach (c) WSEL at the upstream end becomes higher than that at the downstream end (d) Inflow to the channel reach is equal to the outflow from the reach 6.45 The wedge storage in a channel reach is zero when (a) Upstream and downstream WSELs are equal (b) Flow in the channel is uniform (c) WSEL is parallel to channel bed (d) All of the above (e) None of the above 6.46 The total storage at any time in a channel reach is equal to the (a) Sum of prism and wedge storages (b) Average of prism and wedge storages (c) Difference of prism and wedge storages (d) None of the above 6.47 Which of the following statements is/are true for storage in a channel reach during the passage of a flood? I. The prism storage behaves similar to the reservoir storage. II. The prism storage is a function of the outflow. III. The wedge storage is a function of the inflow. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are true 6.48 As per Chow (1959), the total storage at time ‘t’ in a channel can be represented as follows:

a {x I m / n + (1 - x )Q m / n } b b (c) S = {x I n / m + (1 - x )Q n / m } a (a) S =

(b)

S=

b {x I m / n + (1 - x )Q m / n } a

(d)

S=

b {(1 - x ) I m / n + xQ m / n } a

6.49 For Muskingum method, the values of coefficients m, n, a, and b are such that (a) m/n = 1 and a/b = K (b) m/n = 1 and b/a = K (c) m/n = K and b/a = 1 (d) n/m = K and b/a = 1 6.50 The weighting factor x = 0.5 in the Muskingum method results in the (a) Minimum delay and maximum attenuation (b) Maximum delay and minimum attenuation (c) Maximum delay and maximum attenuation (d) Minimum delay and minimum attenuation 6.51 The weighting factor x = 0.0 in the Muskingum method results in the (a) Minimum delay and maximum attenuation (b) Maximum delay and minimum attenuation

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(c) Maximum delay and maximum attenuation (d) Minimum delay and minimum attenuation 6.52 Routing of a flood hydrograph through a channel reach is similar to that through a reservoir (a) for x = 0.5 (b) for x = 0.0 (c) for x = 1.0 (d) none of the above 6.53 Routing of a flood hydrograph through a channel reach uses (a) Continuity equation (b) Muskingum equation (c) Continuity and Muskingum equations (d) Continuity, momentum, and Muskingum equations 6.54 In Muskingum method of hydrologic routing through a channel, the value of Δt shall be chosen such that (a) K > t > 2Kx (b) K < t < 2Kx (c) K < t > 2Kx (d) K > t < 2Kx 6.55 Which of the following conditions would result into the negative values of C0 in Muskingum method (hence should be avoided)? (a) t > 2Kx (b) t = 2Kx (c) t < 2Kx (d) none of the above 6.56 The unit of coefficient K in Muskingum method is (a) T (b) T–1 (c) L/T

(d) T/L

6.57 The values of coefficients K and x of the Muskingum method of routing are found using (a) Continuity equation (b) Trial and error procedure (c) The fact that a plot between S and {xI + (1 – x)Q} is a straight line (d) All of the above (e) None of (a), (b), and (c) 6.58 The transformation of the effective rainfall into direct runoff hydrograph at the outlet of a catchment involves (a) Translation in time (b) Attenuation (c) Both (a) and (b) (d) None of (a) and (b) 6.59 A channel reach is found to have K = 22 hours and x = 0.25. The value of Muskingum coefficient C0 is (a) 0.022 (b) 0.511 (c) 0.467 (d) Insufficient data 6.60 A channel reach is found to have Muskingum coefficients as C0 = 0.1 and C1 = 0.6. The value of Muskingum coefficient C2 is (a) 0.70 (b) 0.30 (c) 0.40 (d) Insufficient data 6.61 A channel reach is found to have K = 6 hours and x = 0.30. For Δt = 4-hours, the value of Muskingum coefficient C1 is (a) 0.667 (b) 0.613 (c) 0.355 (d) Insufficient data 6.62 Which of the following statements is/are false for transformation of effective rainfall into DRH in a catchment? I. Storage characteristics of the catchment are responsible for translation effects. II. Different areas of the catchment contributing runoff at the outlet at different times are responsible for attenuation effects. III. While calculating effective rainfalls, the depression and interception storages are neglected. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are true

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6.63 Which of the following statements is/are true for Clark’s method of IUH development? I. Translation of ER is modeled using a linear channel (LC). II. Attenuation of translated ER is modeled using a linear reservoir (LR). III. Both LC and LR are hypothetical hydrological elements. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 6.64 A linear channel is a fictitious hydrological element in which the time of travel of a flood wave (a) Varies linearly during the passage of the flood (b) Varies nonlinearly during the passage of the flood (c) Stays constant during the passage of the flood (d) Insufficient information to comment upon 6.65 An isochrone is a line representing (a) Equal amounts of rainfall in a catchment (b) Equal amounts of runoff in a catchment (c) Equal time of travel from any point on it to the outlet of the catchment (d) None of the above 6.66 The value of storage time constant (K) in Clark’s method can be easily obtained by using the continuity equation at the (a) Point of inflection of an observed DRH (b) Point of end of DRH (c) Peak flow of the DRH (d) Start of a DRH 6.67 At a point on the falling limb of a DRH, slope of DRH was found to be –4.0 m3/s per hour and the value of DRH ordinate was 20 m3/s. The value of coefficient K in Clark’s method is (a) 80 hours (b) 8 hours (c) 4 hours (d) 5 hours (e) Insufficient data 6.68 Nash (1957) hypothesized that a catchment can be represented by (a) A series of n linear reservoirs each having a different storage constant (K) (b) A series of n nonlinear reservoirs each having the same storage constant (K) (c) A series of n linear reservoirs each having the same storage constant (K) (d) A series of n nonlinear reservoirs each having a different storage constant (K) 6.69 Which of the following statements is/are true for Nash’s method of IUH development? I. Each linear reservoir imparts some amount of translation and attenuation to the input. II. Output from one linear reservoir becomes input to the next linear reservoir. III. Input to the first linear reservoir is 1 cm ER applied instantaneously; therefore, the output from the nth reservoir represents the IUH from the catchment. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 6.70 Which of the following statements is/are true for the parameter n in Nash’s method for IUH development? I. It is a function of the catchment characteristics. II. It represents the number of linear reservoirs needed to capture the rainfall-runoff process in a catchment. III. Its value can be a fraction also. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 6.71 As per the Nash’s method for IUH development, the value of output from the first linear reservoir just at t = 0 jumps from 0.0 to a finite value. What is this finite value if the value of K was found to be 1 hour for a catchment having drainage area of 1 km2?

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(a) 1 cm/h (b) 10,000 m3/h (e) None of the (a), (b), and (c)

(c) 2.78 m3/s (f) Insufficient data

(d) All of the above

6.72 The outflow from the first linear reservoir in Nash’s method is (a) Q1 =

1 t/K e K

(b) Q1 = Ke - t / K

(c) Q1 =

1 -t / K e K

(d) Q1 = Ket / K

6.73 The number of linear reservoirs for modeling the rainfall-runoff relationship in a catchment was found to be 3.56. The value of G(n) is (a) 0.88964 (b) 2.2778 (c) 3.5529 (d) Insufficient data 6.74 If G(1.2) = 0.9182, what is the value of G(2.2)? (a) 1.2/0.9182 (b) 1.2 × 0.9182 (c) 2.2 × 0.9182

(d) 2.2/0.9182

DESCRIPTIVE QUESTIONS 6.1 Why do the size, shape, and characteristics of an inflow hydrograph change when it travels through a channel reach? 6.2 Explain the hydraulic and hydrologic methods for hydrograph routing with the help of equations. 6.3 Explain the Modified Puls method of hydrologic routing. 6.4 Explain the Goodrich’s method of hydrologic routing. 6.5 Explain the fourth order SRK method of hydrologic routing. 6.6 For hydrograph routing through a reservoir, prove that the peak of the outflow hydrograph intersects inflow hydrograph. (Hint: use continuity equation). 6.7 Discuss the importance of hydrograph routing in flood management with special emphasis on attenuation and lag effects. 6.8 Search through the internet and prepare a summary of the capabilities and applications of following computer software for hydrograph routing: HEC-RAS, MIKE FLOOD, FLDWAV, FLOW-2D, TUFLOW, and DWOPER. 6.9 Describe the concepts of prism and wedge storages in a channel when a flood wave passes through it. Use neat sketches to answer. 6.10 What do you understand by a time-area diagram? Explain with the help of sketches. How is it useful in hydrology? 6.11 Describe the Clark’s method for IUH development in a catchment. 6.12 Describe the Nash’s method for IUH development in a catchment.

NUMERICAL QUESTIONS 6.1 The storage, outflow discharge, and elevation data for a reservoir are given in the following table. The spillway crest of the outlet structure is at an elevation of 180.2 m.

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Elevation (m)

179.5

180.2

180.7

181.2

181.7

182.2

182.7

Storage (Mm3)

4.7

5.6

5.9

6.4

7.0

8.1

9.0

0

0

16

42

73

117

158

3

Outflow m /s

An inflow hydrograph given below enters the reservoir when the initial water level in the reservoir was 180.00 m. Time (Hours) 3

Inflow (m /s)

0

3

6

9

12

15

18

21

24

27

10

19

50

60

54

43

31

20

14

10

Route the inflow hydrograph using the Modified Puls method and determine: (a) outflow hydrograph, (b) maximum reservoir elevation during the passage of the flood, (c) peak outflow discharge, (d) attenuation, and (e) time lag from the reservoir. Plot the inflow and outflow hydrographs and the water surface elevation in the reservoir on the same graph sheet. 6.2 For the data in Problem 6.1 above, find out the outflow hydrograph, attenuation, and delay in peak discharge for initial water surface elevation in the reservoir of (a) 179.5 and (b) 180.5 m and comment on your results. Plot the two outflow hydrographs on same graph and find peak discharges, attenuation, and lag for each. Discuss your results obtained in this problem in conjunction with those obtained in Problem 6.1. 6.3 Solve Problem 6.1 above using Goodrich’s method and compare your results with those obtained in the solution for Problem 6.1. 6.4 The inflow hydrograph given below passes through a reservoir having an outflow given as Q = 100h1.5 m3/s and storage given as S = h + h2 Mm3, where h is the height of water above the spillway crest (in m). If the water level in the reservoir initially was at the spillway crest, route the inflow hydrograph using Modified Puls method. Plot the two hydrographs on same graph and find peak discharges, attenuation, and lag. Time (Hours) 3

Inflow (m /s)

0

4

8

12

16

20

24

28

32

36

40

44

10

28

68

78

58

36

30

23

18

14

11

10

6.5 Solve Problem 6.4 above using Goodrich’s method and compare your results with those obtained in the solution for the Problem 6.4. 6.6 A reservoir has vertical walls with a rectangular bottom of surface area of 1 km2. The flow over the outlet spillway is given as Q = 1.5h1.5 (m3/s), where h is the height of water above spillway. An inflow hydrograph given below hits the reservoir when the water surface in the reservoir was at spillway crest. Time (Hours)

Inflow (m3/s)

Time (Hours)

Inflow (m3/s)

0

0

80

280

10

60

90

240

20

120

100

200

30

180

110

160

40

240

120

120

50

300

130

80

60

360

140

40

70

320

150

0

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Engineering Hydrology

Route the inflow hydrograph using the fourth order SRK method and find the peak discharge of the outflow hydrograph, and amounts of attenuation and lag. 6.7 Route the inflow hydrograph given below from a channel element with K = 3.5 hours and x = 0.2. Use a time step of 4 hours. Plot the two hydrographs on same graph and find peak discharges, attenuation, and lag. Inflow (m3/s)

Time (Hours)

Inflow (m3/s)

0

50

28

230

4

180

32

155

Time (Hours)

8

270

36

90

12

360

40

60

16

410

44

35

20

370

48

20

24

300

6.8 Rout the inflow hydrograph given in Problem 6.7 from another channel reaches having K = 2.5 hours and K = 4.5 hours keeping the value x same. Compare the results obtained here with those in the previous problem in terms of attenuation and lag. 6.9 Determine the Muskingum routing coefficients K and x from a channel reach using the inflow and outflow hydrographs given below. Inflow (m3/s)

Outflow (m3/s)

Time (Hours)

Inflow (m3/s)

Outflow (m3/s)

0

35

35

21

235

375

3

135

57

24

140

246

6

440

200

27

95

158

Time (Hours)

9

670

455

30

60

102

12

685

612

33

42

68

15

535

617

36

35

56

18

385

510

6.10 A catchment is found to have the following data in terms of the inter-isochrone areas. If the time of concentration and storage time constant for the catchment are 24 hours and 16 hours respectively, determine and plot the IUH from the catchment using Clark’s method. Time (Hours) 2

Area (km )

0-3

3-6

10

30

6-9

9-12

12-15

15-18

18-21

21-24

75

50

15

9

5

6

2

6.11 A catchment has drainage area of 165 km , K = 12.5 hours, and tc = 7.0 hours. The time area diagram from the catchment is given in the following table. Determine and plot the IUH from the catchment using Clark’s method. Time (Hours)

0

1

2

3

4

5

6

7

Area (km2)

0

10

40

20

50

30

10

5

6.12 For a catchment of area 567 km2, the values of Nash model coefficients were found as follows: n = 5.76 and K = 4.8 hours. Determine and plot the ordinates of the IUH of the catchment. 6.13 For a catchment of area 5,237 km2, the values of Nash model coefficients were found as follows: n = 10.76 and K = 65 hours. Determine and plot the ordinates of the IUH of the catchment.

Hydrograph Routing

1. http://www.hec.usace.army.mil/software/hec-ras/ 2. https://www.fema.gov/national-weather-service-fldwav-computer-program 3. https://www.flo-2d.com/ 4. http://www.tuflow.com/ 5. https://www.mikepoweredbydhi.com/products/ mike-flood

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7

Groundwater

LEARNING OBJECTIVES LO 1

Discuss the importance of groundwater

LO 2

Discuss the occurrence and movement of groundwater and the governing equations

LO 3

Explain the flow of groundwater through confined and unconfined aquifers

LO 4

Outline the methods of estimation of aquifer parameters

7.1

INTRODUCTION

Groundwater, or subsurface water, refers to the water existing underground and the study of its occurrence and movement (and sometimes its quality) LO 1 Discuss the importance of groundwater is termed as groundwater hydrology. This water resides in the pores and fractures present in soil and rocks and its natural movement occurs primarily due to gravitational forces, but sometimes the capillary forces also become important. As mentioned in Chapter 1, the groundwater reservoir stores a significant quantity of water—about 1.7% of the entire water on, below, and above the earth surface. Of the entire groundwater, about 45% is freshwater and the rest is saline water. In terms of freshwater quantity, groundwater constitutes about 30% of all the freshwater resources, which is about 100 times larger than the water available in all the lakes and rivers (one should note that a major part, about 69%, of the freshwater is locked up in glaciers and ice caps). The groundwater storage gets recharged by infiltration of precipitation and seepage from rivers and lakes, and discharges through springs and seepage to surface water bodies. A water resources engineer should be able to evaluate the potential of this source of water and answer questions such as, how much groundwater is available (which will depend on the storage properties of the soil/rock), what is the maximum rate at which it could be withdrawn (which will depend on the transmissive properties of the subsurface), and at what rate could it be safely withdrawn (which will

Groundwater

231

depend on the recharge rate). In this chapter, we will discuss the occurrence and movement of groundwater with the ultimate goal of answering these questions. Due to reasons of accessibility, economy, and quality, groundwater is typically preferred over surface water. There is no need of a storage/conveyance system to store water and carry it from the source to the point of use, as in the case of water supply from a river. Water will generally be available at the point where it is needed (of course, since it occurs at some depth below the ground level, we will need to install pumps or use other lifting devices to bring it to the surface). The natural filtration through the soil pores generally results in good quality of water and only minimal treatment is required to make it potable. There is not as much seasonal variation in availability of water as in surface sources. Some areas of the world do not have access to a surface water source at all and groundwater provides the only source. For example, Denmark has its entire annual water demand of 1 km3 supplied by groundwater, and several Middle-East countries satisfy their entire agricultural water demand by groundwater. In India, groundwater accounts for about 65% of the irrigation water and 85% of the drinking water supply. If we look at the share of different sectors in the total groundwater withdrawal in India, it is about 89% for irrigation, 9% for domestic use, and 2% for industrial use; compared to the global values of about 60% for irrigation, 20% for domestic, and 20% for industrial use (it should be noted that the numbers mentioned here are based on different sources/years and should not be taken as exact. These are useful, nevertheless, to provide a rough idea about the groundwater usage). In quantitative terms, the annual recharge of groundwater from rainfall and other sources (including seepage from canals, ponds, etc., and irrigation return flow and subtracting natural discharge) for India is estimated at about 400 km3 and the annual groundwater consumption is estimated at about 240 km3. The fact that the consumption is only about 60% of the recharge should not give one a false sense of security, since there is tremendous spatial and temporal variation in the recharge and withdrawal rates and the safe withdrawal rate is generally smaller than the recharge rate as some of the groundwater is needed to sustain the rivers and other ecosystems. Several areas of the country have an overdraft of the groundwater, i.e., the annual withdrawal rate is higher than the recharge rate resulting in depletion of the groundwater storage. Even when an area withdraws groundwater within safe limits on an annual basis, it may have an overdraft during the dry period of the year.

7.2

OCCURRENCE OF GROUNDWATER

Figure 7.1 shows a typical soil profile and the occurrence of water within LO 2 Discuss the its pores. If we take a soil sample from close to the ground surface, it will occurrence and movement generally show very little amount of water and most of the pore spaces of groundwater and the will be occupied by air. Due to the presence of the impermeable rock at governing equations some depth below the surface, water percolating down through the soil will accumulate on top of the rock and will saturate the pore spaces of the soil above it. This saturated zone will extend above the rock to a certain height, which depends on the balance of the recharge and withdrawal rates. The top of the saturated zone is known as the groundwater table or water table, and the region between ground surface and water table is known as the unsaturated zone (also called the vadose zone or the zone of aeration), since the soil pores in this zone contain some air and are not completely saturated with water. The soil in this zone is partially saturated and is not of much concern from a water-supply perspective, since it neither holds nor is able to transmit a sizable quantity of water (this is the

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reason that often the term ‘groundwater’ is used to refer to only the water in the saturated zone, and not all the water in the subsurface). However, it is very important from an agricultural perspective, since the top portion of this zone holds the moisture necessary for plant growth.

Figure 7.1

Various zones in a typical soil profile

7.2.1 Unsaturated Zone The unsaturated zone can be subdivided into three parts: a capillary fringe, just above the water table, caused by the phenomenon of capillary rise due to the cohesive forces between water molecules and the adhesive forces between water and soil; a soil water zone just below the ground surface, from which the plant roots draw water; and an intermediate zone extending from the bottom of the soil water zone to the top of the capillary fringe. Sometimes, the entire unsaturated zone is called the soil water zone, but we will adopt the three-zone subdivision described above. Also, sometimes the water table is nearly at the ground surface and the unsaturated zone does not exist or only the capillary fringe may be present.

7.2.1.1

Soil Water Zone

The soil water zone is the region from where plants can take water and it typically extends to the depth of the plant roots, although the roots may be able to draw water from the soil below the root-zone also. For most crops, the depth of the root-zone is around 1 m, but some roots (e.g., alfalfa and date palm) may go as deep as 2.5–3 m. Most of the water, about 95%, taken up by the plant roots goes back into the atmosphere through transpiration. This zone is more important for an agricultural engineer, but not of much significance from a water resources point of view. We will, therefore, provide only a brief description of the occurrence of water within this zone. The soil water zone shows a very wide variation in its water content due to infiltration/irrigation and plant-uptake. To look at this variation, we will define some terms as given below: � Porosity The porosity of a porous media is defined as the ratio of the volume of voids Vv to total volume of the soil sample Vt (which is the sum of the volume of voids and the volume of solids Vs). It is generally denoted by h or f, and is expressed either as a fraction, between 0 and 1, or in percentages between 0% and 100%. The voids may be filled by air and water for unsaturated soils; or by water only for saturated soils. Typical porosity value for gravel is about 0.2; for sand, around 0.4; and for clay, around 0.55.

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233

Vv Vv = Vt Vv + Vs

� Water Content The water content (or, more generally, moisture content) represents the amount of water in a soil sample, either on a mass basis (knows as gravimetric water content) or on a volume basis (volumetric water content). The gravimetric water content is defined as the ratio of the mass of water in a soil sample to the total mass (sometimes, the total mass of the soil sample is obtained after drying, but mostly the wet mass is used). The volumetric water content is defined as the ratio of the volume of water in a soil sample to the total volume, i.e., volume of solids + volume of voids (the volume of voids may further be written as the sum of the volume of water, Vw, and the volume of air, Va). Clearly, the volumetric water content can be obtained from the gravimetric water content through a multiplication by the specific gravity of the soil. In hydrology, the volumetric water content is more commonly used, therefore, in this book the term water content is used to represent the volumetric water content, denoted by q. q=

Vw Vw = Vt Vw + Va + Vs

� Degree of Saturation Also called water saturation, it is defined as the ratio of the water content to the porosity, or equivalently, the ratio of the volume of water and the volume of voids, and is denoted by Sw (varying between 0 and 1) Sw =

Vw q = Vv h

� Field Capacity (FC) It is the water content of an initially saturated soil after the gravity drainage has nearly stopped. Just after irrigation or precipitation, the soil water zone may become saturated (q = h). In absence of further application of water, and under the influence of gravity, the water content reduces due to the downward percolation of the soil water. After some time, which may range from a few hours for sandy soils to a few days for clayey soils, the water content becomes nearly constant at the field capacity, typically about 20% for sand and 50% for clay. � Permanent Wilting Point (PWP) It is the water content of a soil at which the plants growing in it wilt and do not recover after re-irrigation. Since most plants show some signs of wilting even before reaching the PWP, irrigation is needed much before the soil water content reaches PWP. Typical values of PWP are about 10% for sandy soils, and 20% for clayey soils. The total available water capacity, FC−PWP, signifies the amount of water available for the plants. Comparing the typical values of FC and PWP, we see that there is about 10% water available in sandy soils and 30% for clayey soils. To avoid plant stress, irrigation is generally applied when about half of the available water has been used up by the plants. For the values used here, it implies that irrigation in a sandy soil would be needed when the water content is about 15%; and for clayey soil, when it is around 35%. � Residual Water Content We have seen above that the plant roots are not able to suck water from the soil in sufficient quantity once the water content reaches the PWP. However, if we could apply a larger suction to the soil, we will be able to take some more water out. For example, the evaporative action near the

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soil surface may be seen as applying a large suction force on the soil water. Under the action of this suction, the water content of the soil will keep on decreasing with the increase of suction. However, even at very large suction, some water will remain in the soil, e.g., in disconnected pores or as strongly attached films on soil grains. This absolute minimum water content is called the residual (or irreducible) water content qr. Typical values of qr are about 2% for sand and 10% for clay. � Specific Retention It is the water content retained in the soil after drainage under gravity and is equivalent to the field capacity. In hydrology, the term specific retention (Sr) is more frequently used, while the term field capacity is more common in agricultural science. Some hydrologists define specific retention as the water content at which moisture movement virtually ceases, but we will not use this definition. � Specific Yield This is the volume of water drained under gravity per unit volume of the formation. The specific yield, Sy, is of more relevance to us than porosity since it represents the amount of water readily available. Two soils may have the same porosity but they may have very different specific yields. For example, sand may have a porosity of about 30% and specific yield of about 25%, while clay can have a high porosity of about 50%, but the specific yield may be only 5%. From the definition of specific retention and specific yield, it is clear that these are complementary, such that Sy + Sr is equal to the porosity. The water content in the soil water zone will be equal to the porosity immediately after irrigation or heavy rains. After the irrigation/rain has stopped, or even during it, if the intensity is small, the water content will show a continuous decrease due to gravity drainage. The rate of this decrease will become smaller with time as the drainage rate reduces and the water content will reach the field capacity after sufficient time. Further decrease in water content will occur due to water uptake by plant roots, and possibly due to evaporation from the soil surface. In absence of any application of water, the water content in the soil will first reach the PWP, at which the plants will wilt, and then approach the residual water content. To sustain the plants, however, we will irrigate the soil when the water content reaches a certain predetermined value, bring its water content to the FC and the whole cycle will be repeated.

7.2.1.2 Capillary Fringe The height of the capillary rise hc, in a uniform circular tube made of a solid material (s), dipped in a liquid (l) in a gaseous medium (g) depends on the radius of the tube r, liquid-gas density difference rl − rg, surface tension at the liquid-gas interface slg, and the liquid-solid-gas contact angle qlsg. For a narrow tube, it may be approximated by hc =

2s lg cosq lsg g( rl - rg )r

(7.1)

The contact angle depends on the interfacial tension for liquid-solid, liquid-gas, and solid-gas interfaces (in fact the term slg cosqlsg is equal to ssg− ssl, but is written in this form since slg and qlsg are easy to obtain) and is close to zero for water-glass-air interaction. Therefore, for a given soil-water-air combination, assuming the soil pores to be interconnected in the form of a uniform circular tube, the capillary rise would be inversely proportional to the pore radius. In general, the soil grain size could be assumed to represent the pore size; hence the thickness of the capillary fringe will depend on the type of soil. For small grained soil, the capillary rise will be larger and for large grained soils, the rise is smaller. It should be noted, however, that the soil pores are non uniform in diameter, are not connected in the form of a straight vertical tube, and

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have large variations in the surface characteristics. Hence, Eq. (7.1) is not directly applicable and there is a wide variation in the thickness of the capillary fringe even for similar soils. Typical values of capillary rise are about 1–10 cm for gravel, 10–100 cm for sand, 1–10 m for silt, and 10–30 m for clay. The water content in the capillary fringe is at or very close to saturation and the pressure head varies from zero (i.e., atmospheric) at the lower boundary (which is the water table) to –hc at the top boundary, which is the meniscus formed at the water-air interface. The pressure then changes abruptly to zero across the meniscus as we move from the water side to the air side.

7.2.1.3

Intermediate Zone

The zone between the soil water zone and capillary fringe is the intermediate zone. Downward percolation of water from the soil water zone to the water table occurs through this zone. Depending on the capillary rise and the root depth, this zone may not even exist and sometimes it may be as thick as a few hundred metres! This zone plays an important part in the movement of water and pollutants from the ground surface to the water table.

7.2.2

Saturated Zone

Our interest is mainly in the saturated zone below the water table because this is the zone which can store and supply water in sufficient quantity when needed. Since the porous media is saturated, i.e., all the pores are filled with water, the water content will be equal to the porosity. However, we may not be able to take out all the water which is present, because some of it will be strongly attached to the grains or may be in pores which are not interconnected, and will not be available. Moreover, even when water is available in a porous medium, it may not be possible to take it out at a reasonably rapid rate. Based on the availability of water and its transmitting ability, a geologic formation is classified as either an aquifer, which has large storage and rapid transmission (e.g., sand) or a confining unit, which may or may not have large storage but is not able to transmit water at a rapid rate. The confining unit could be further classified as an aquifuge, which neither stores nor transmits water (e.g., impermeable rock); an aquiclude, which stores water but does not transmit it appreciably (e.g., clay); or an aquitard, which retards the movement of groundwater but does not completely prevent it (e.g., sedimentary rocks with high clay content). Clearly, our interest lies in the aquifers since these are the only formations which can be tapped to satisfy the water requirement of an area. Figure 7.2 shows a typical aquifer system. An unconfined aquifer has a confining unit at the bottom and a free surface, i.e., the water table, as its top boundary, where the pressure is equal to the atmospheric pressure. These are also known as water table aquifers. If a well is drilled with its screen in the unconfined aquifer, the water level in the well will rise up to the water table. A special case of an unconfined aquifer occurs at a confining unit of smaller dimension existing above the water table and is known as a perched aquifer. A confined aquifer (or artesian aquifer) has confining units both at the bottom and the top, and the water pressure is more than the atmospheric pressure. If a well is screened in the confined aquifer, water will rise above the upper confining unit to a level which depends on the water pressure. This imaginary surface, to which water will rise in wells tapping the confined aquifer, is known as the potentiometric or piezometric surface. It may happen that the piezometric surface becomes higher than the ground level at some point and a well tapping the confined aquifer will provide spontaneous discharge of water without any pumping. These wells are called flowing wells. If the confining unit of an aquifer is an aquitard, it allows for some leakage from/to the aquifer and, therefore, these are called leaky aquifers. Depending on the relative location of the

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water table in the unconfined aquifer and the piezometric level in the confined aquifer, the leakage may be downward (when water table is higher) or upward (when the piezometric level is higher). For example, if a leaky unconfined and confined aquifer system is initially in equilibrium (i.e., the water table and the piezometric level are same), pumping of the unconfined aquifer will result in lowering of the water table which will induce upward leakage from the confined aquifer to the unconfined aquifer. Similarly, pumping of the confined aquifer will induce a downward leakage. This is just one example of how the groundwater may move due to pressure difference. In order to study the general movement of groundwater and to estimate the rate of movement, we need to look at the driving and resisting forces and the fluid and material properties which influence these.

Figure 7.2

7.3

A typical aquifer system

MOVEMENT OF GROUNDWATER

LO 2

The movement of groundwater occurs from regions of higher energy to those with lower energy. The energy of a water particle could be expressed in terms of head, i.e., energy per unit weight, which comprises the p v2 potential energy head z, the kinetic energy head , and the elastic energy head (or, pressure head) . rg 2g These energy heads are called the elevation (or datum) head, velocity head, and pressure head respectively, and the sum of these three represents the total mechanical energy head.

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Since groundwater generally moves at a very small velocity, the velocity head is negligible and we define the term piezometric head (also called hydraulic head or potentiometric head) as the sum of the pressure and elevation heads, with the flow of groundwater taking place from a higher piezometric head to a lower piezometric head. The piezometric head at a point in the groundwater represents the height to which water will rise in a piezometer installed at that point. From Figure 7.2, it may be seen that the piezometric head for a point in the unconfined aquifer will be equal to the water table elevation, and that for a confined aquifer will be equal to the height of the potentiometric surface, both measured above the chosen datum. After establishing the direction of flow, we now look at the magnitude. It is the difference in the piezometric head at two points which governs the flow of water. If there is no difference in the piezometric head then there will be no movement of water. And we would expect the flow to be higher if the head difference becomes larger. Darcy, more than 150 years ago, was probably the first investigator to study the connection between the flow rate and the hydraulic head. As shown in Figure 7.3, the flow rate Q, through a vertical soil-column was adjusted by means of valves in the inlet pipe and outlet faucet, and the hydraulic head at its ends were measured using mercury manometers. Denoting the cross section area of the column by A, the head difference between the two ends as Δh, and the length of the tube by L, Darcy observed that Q was directly proportional to A and Δh, and inversely proportional to L (The term Δh/L represents the drop in hydraulic head per unit length and is called the hydraulic gradient, i.) Dh (7.2) = KAi L in which the constant of proportionality K, is called the permeability or hydraulic conductivity, and has the same units as velocity (i.e., m/s). However, since the groundwater moves at a very slow speed, generally K is expressed in m/d or mm/s. It represents the ease with which water (or another fluid) passes through soil (or another porous medium) and its typical values are about 1 m/s for gravel, 1 cm/s for sand, and about 1 mm/d for clay. The hydraulic conductivity depends on the medium properties, e.g., grain size, porosity, degree QμA

Figure 7.3

Darcy’s experimental set-up

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of compaction, etc., as well as fluid properties, viz., dynamic viscosity and mass density. Equation (7.2) is known as Darcy’s law and forms the basis for the analysis of groundwater movement. Some important points about Darcy’s law are as follows: •

Darcy’s law is valid for slow-moving laminar flow only and would not be applicable for fast-moving groundwater flow which may occur near a pumping well. Similar to the flow of liquids through closed conduits, the Reynolds number Re, is used to demarcate laminar and turbulent flows. For rQd flow through a porous medium, Re is defined as Re = , in which d is a characteristic length of Am the porous medium signifying the size of the flow tubes, with the median grain size being the most commonly used value. Darcy’s law has been found to be valid for Reynolds number values up to 1 (and approximately valid up to 10), which is rarely exceeded in groundwater flows. For very fine soils, Darcy’s law is not valid at very low Re values also, due to the effect of electrical forces.

•

Although Darcy’s law is empirical, its form may be derived using the Hagen-Poiseuille equation for 128 mQL laminar flow through circular tubes, h f = , which could be written in terms of the hydraulic pr gD 4 r gD 2 gradient as Q = Ai . This implies that the hydraulic conductivity K is a function of the fluid 32 m properties (mass density and dynamic viscosity) as well as the medium properties (diameter of pores, arrangement of the flow tubes, etc.).

•

Equation (7.2) may be written in terms of the derivative of the hydraulic head h, with respect to the flow direction s, as Q = - KA

∂h ∂s

(7.3)

The negative sign signifies that the flow is from higher to lower hydraulic head. Sometimes the porous medium may have different conductivity in different directions, e.g., for layered soils, and a tensor form of Darcy’s law has to be used. This will be discussed latter in this chapter. •

The discharge per unit area, Q/A, is denoted by q, and is called the specific discharge, Darcy velocity or apparent velocity. It is not the actual velocity with which water moves through the soil pores since the flow rate has been divided by the cross-sectional area A, which includes both the soil grain area and the pore area. To obtain the actual velocity, called the seepage velocity and denoted by v, we need to divide Q by the area of pores, which is hA (although the porosity is defined on volumetric basis, it may be shown that the ratio of the area of the pores to the total area is also equal to h). Thus, q ∂h and v = . q = -K h ∂s

•

Although for the flow of groundwater, there is not a significant variation in the fluid properties, sometimes we may encounter a different fluid, for example, a contaminant or oil. Since hydraulic conductivity is a function of both the medium properties and the fluid properties, it becomes convenient in such cases to define another property of the medium which is independent of the fluid properties. This property is known as the intrinsic permeability (k), of the porous medium, and is

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Km (see point 2 given above). The dimensions of k are rg L2 and typical values are about 10−9 m2 for sand to 10−15 m2 for clay. There are several empirical h3 equations relating k with the soil properties (e.g., the Kozeny-Carman equation, k μ d2 , (1 - h)2 with d as the particle diameter). Another unit commonly used for intrinsic permeability is the darcy, which is defined in such a way that a porous medium with an intrinsic permeability of 1 darcy will have a specific discharge of 1 cm/s under a pressure gradient of 1 atm/cm for a fluid with a dynamic viscosity of 1 centipoise (poise is the unit of dynamic viscosity in the CGS system of units and is equivalent to g/cm/s. Also note that instead of hydraulic gradient, the definition uses the pressure gradient, hence the mass density of the fluid need not be specified). One darcy is equivalent to 9.869233 × 10–13 m2. Sometimes, the intrinsic permeability is called permeability but it may cause confusion with the hydraulic conductivity, since sometimes that is also referred to as permeability. In this book, we will use the terms hydraulic conductivity (K) and permeability (k). related to the hydraulic conductivity by k =

•

In unconfined aquifers, the hydraulic gradient is equal to the slope of the water table since it represents the hydraulic head at a point. In confined aquifers, it is equal to the slope of the potentiometric surface.

� EXAMPLE 7.1 A 10 cm diameter horizontal tube contains saturated soil and carries a constant discharge of 0.4 litre per minute as measured through a pan at the outlet. There are two piezometers connected to the tube at a distance of 1 m from each other and the water level in these is measured as 102.4 cm and 94.6 cm, respectively. The porosity of the soil is 0.36. Estimate the hydraulic conductivity of the soil. How long will it take for a water particle to travel the distance between the two piezometers. Solution Dh 102.4 - 94.6 p d2 = = 0.078 , area of flow is A = = 78.54 L 100 4 cm2, and the discharge is Q = 0.41/min = 6.67 cm3/s. From Darcy’s law, Eq. (7.2), we get the hydraulic Q Ki Q conductivity, K = = 1.09 cm/s (probably a sandy soil). The seepage velocity is obtained as v = or Ai h Ah = 0.236 cm/s, implying that it would take a water particle 423 seconds to travel the distance of 1 m between the two piezometers. The hydraulic gradient is given by i =

7.3.1

Basic Equations

The three basic equations applicable to fluid flow problems are the continuity equation, momentum equation, and energy equation. Darcy’s law, though empirically obtained, can also be derived from an application of the Navier-Stokes equations, which represent the momentum equation. The energy equation needs to be used in very specific cases only, e.g., for non-isothermal flow, and will not be discussed here. The combination of the continuity and momentum equations provides the basis for solving all the groundwater flow problems

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discussed in this book. The development of this equation of groundwater motion is described in this section and its application to various cases in the next sections.

7.3.1.1

Continuity Equation

The continuity or mass conservation principle states that the mass of a system of fluid particles remains constant. Application of the Reynolds transport theorem results in the control volume based continuity equation stating that the net mass influx (i.e., the rate of inflow) of fluid in a control volume is equal to the rate of increase of mass of fluid within the control volume. Although the actual groundwater flow is threedimensional, the typical horizontal dimensions for most practical problems are much larger than the vertical dimensions. Hence, the assumption of an essentially horizontal flow is often made to simplify the analysis, and this approach of analysing the groundwater flow is known as the hydraulic approach. For the sake of completeness, though, we will first describe the three-dimensional continuity equation and then discuss its two-dimensional simplification for confined and unconfined aquifers. � Three-dimensional Form Considering the control volume to be a rectangular prism element oriented along the Cartesian coordinates, and having dimensions Δx, Δy, and Δz, as shown in Figure 7.4, we could write the net mass influx along the x-direction as ∂r q x ˆ ∂r q x Ê rq x Dy Dz - Á rq x + Dx ˜ Dy Dz = Dx Dy Dz Ë ¯ ∂x ∂x where, qx is the specific discharge in the x-direction (note that porosity is not considered in the computation of the area. If we use the seepage velocity instead of the specific discharge, the area of flow should be taken as hΔxΔy, but will result in the same equation since the seepage velocity is equal to the specific discharge multiplied by the porosity). Considering that the prism is fully saturated, the mass of water in the control ∂rhDx Dy Dz volume will be rhΔxΔyΔz, and its rate of increase will be . Therefore, considering the net ∂t

Figure 7.4

An elementary volume for mass balance analysis

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influx along the three axes, and cancelling out the volume of the element, ΔxΔyΔz, we get the continuity equation as ∂rh ∂rq x ∂rq y ∂rqz + + + =0 ∂t ∂x ∂y ∂z

(7.4)

In the above form, the continuity equation is not very useful, since the primary variable of our interest is the hydraulic head, h. Once the head is known, we could derive the flow rates using Darcy’s law. To write the first term, i.e., the rate of change of mass within the control volume, in terms of the hydraulic head, we define an aquifer property called specific storage, Ss, which represents the volume of water added to (or released from) a unit volume of a saturated aquifer per unit rise (or fall) in the hydraulic head. It is a function of the porosity, soil compressibility and water compressibility and has the dimensions of L–1. Typical values of specific storage in m–1 are about 10–2 for clay and 10–4 for sand. For the flux terms, Darcy’s law is used to express the specific discharge components in terms of the hydraulic head. However, a threedimensional counterpart of Eq. (7.3) needs to be formulated, which requires the classification of the aquifer as homogeneous or heterogeneous and isotropic or non-isotropic. Almost all aquifers are heterogeneous, i.e., their properties (we will specifically consider hydraulic conductivity here) will vary from one location to the other; and several aquifers are non-isotropic, i.e., the hydraulic conductivity at a location is different in different directions (generally non spherical particles deposited in water tend to settle with their longer side horizontal and the hydraulic conductivity in the horizontal direction is much larger than that in the vertical direction). For an isotropic aquifers, the direction of the flow vector q may not coincide with the direction of the hydraulic gradient —h and the hydraulic conductivity is written as a tensor È K xx Í K = Í K yx Í ÎÍ K zx

K xy K yy K zy

K xz ˘ ˙ K yz ˙ ˙ K zz ˚˙

with Kij relating the specific discharge in the ith direction to the hydraulic gradient in the jth direction. Darcy’s law is then written as Ê ∂h ∂h ∂h ˆ q x = - Á K xx + K xy + K xz ˜ ∂x ∂y ∂z ¯ Ë Ê ∂h ∂h ∂h ˆ q y = - Á K yx + K yy + K yz ˜ ∂x ∂y ∂z ¯ Ë

(7.5)

Ê ∂h ∂h ∂h ˆ qz = - Á K zx + K zy + K zz ˜ ∂x ∂y ∂z ¯ Ë or, more compactly, as q = - K ◊—h . The hydraulic conductivity tensor has 9 components, but it can be shown (based on Onsager’s principle of microscopic reversibility) that it is symmetric, i.e., Kij = Kji, hence only 6 components are needed to describe it. Moreover, if x, y, and z are the principal directions, the offdiagonal terms are zero, and only the 3 principal conductivities, Kx, Ky, and Kz are needed. In such cases,

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if the hydraulic gradient is in the direction of one of the principal axes, the flow would also be in the same direction. And finally, if the porous medium is isotropic, the conductivity tensor may be simplified as K = KI, where I is the identity matrix of size 3. The three-dimensional continuity equation is then written as ∂h (7.6) = — ◊ ( K ◊ —h ) ∂t by using the following assumptions: (i) the spatial variations in water density are very small implying that r could be taken out of the gradient term on the right hand side, (ii) the temporal variations in density and porosity are accounted for in the specific storage term Ss, implying that the change of mass within the control volume is given by rSsΔh per unit volume. Ss

Equation (7.6) is a partial differential equation which is first-order in time and second-order in space. Therefore, its solution requires one initial condition and two boundary conditions for each spatial variable. The solution is generally difficult, and is further complicated for an unconfined aquifer due to the fact that the top boundary of the flow domain, i.e., the water table is moving. Therefore, the governing equation is written in a two-dimensional form by using the hydraulic approach as described next. � Two-dimensional Form using Hydraulic Approach Due to the smaller vertical extent of most aquifers compared to their horizontal extent, it could be assumed that the flow is essentially horizontal. This hydraulic approach implies that the pressure distribution could be assumed to be hydrostatic and the hydraulic head variation in the vertical may be neglected, implying that the velocity component in the vertical direction is negligible. For the confined aquifers, these assumptions are quite reasonable. For unconfined aquifers, in order to obtain a solution, we frequently use the approximation put forward by Dupuit and, therefore, known as the Dupuit assumption (sometimes also called Dupuit-Forchheimer assumption, since Forchheimer first applied it to a number of seepage problems). The assumption of essentially horizontal flow employed by Dupuit resulted in two significant simplifications—(i) the hydraulic head does not vary in the vertical direction and the velocity component in the vertical direction is negligible, and (ii) the hydraulic gradient along the streamline, ∂h/∂s, could be approximated by the horizontal head gradient (note that the water table is the topmost streamline and other streamlines will have varying directions ranging from the water table at the top to the horizontal at the bottom, see Figure 7.5). The continuity equation is obtained by considering an elementary volume in the form of a prism with a rectangular base of size Δx · Δy located at the base of the aquifer and height equal to the thickness of the aquifer. For convenience, we assume that the aquifer base is horizontal and is taken as the datum. As shown in Figure 7.5, the relevant height H is then equal to the aquifer thickness B, for a confined aquifer; and the hydraulic head, h, for an unconfined aquifer. We could write the net mass influx along the x-direction as ∂ rq x H ˆ ∂ rq x H Ê rq x H Dy - Á rq x H + Dx ˜ Dy = Dx Dy Ë ¯ ∂x ∂x Additionally, for the two-dimensional case we consider an aquifer recharge, denoted by R, and representing the volume of water added to the aquifer per unit base area per unit time. The dimensions of R are LT –1, and the commonly used unit is mm/d. For unconfined aquifers, rainfall and irrigation return flow are the main sources of the recharge and for confined aquifers recharge occurs due to leakage through the

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Figure 7.5

243

Flow in confined and unconfined aquifers

confining aquitard layers. The mass of water in the control volume is rhHDxDy, and its rate of increase will be ∂rh H Dx Dy . Similar to the three-dimensional case, we define an aquifer property called storage ∂t coefficient or storativity S, which represents the volume of water added to (or released from) a unit base area of a saturated aquifer per unit rise (or fall) in the hydraulic head. For a confined aquifer, it is a function of the porosity, soil compressibility and water compressibility and typical values are about 10−5 to 10−3. For an unconfined aquifer, it is equal to the specific yield since the compressibility effects are negligible compared to the storage change, due to change of water level and also due to the smaller pressure. Considering the net influx along the two axes along with the recharge, cancelling out the base area of the element DxDy, and invoking assumptions similar to those used in the derivation of Eq. (7.6), we get the continuity equation as S

∂h = — ◊ ( HK ◊—h) + R ∂t

(7.7)

or, in expanded form, as Ê Ê ∂h ∂h ˆ ∂h ∂h ˆ ∂ H Á K xx + K xy ˜ ∂ H Á K yx + K yy ˜ ∂ x ∂ y ∂ x ∂y ¯ Ë ¯ Ë ∂h S = + +R ∂t ∂x ∂y

(7.8)

Further simplification may be made by assuming the aquifer to be homogeneous and isotropic, in which case, we get Ê ∂h ˆ ∂Á H ˜ Ë ∂x ¯ S ∂h = + K ∂t ∂x

Ê ∂h ˆ ∂Á H ˜ Ë ∂y ¯ R + ∂y K

(7.9)

This equation and its counterpart in the r-q polar coordinates, form the basis of the analysis of groundwater flow in the next sections, since we will assume the aquifer to be homogeneous and isotropic in most cases.

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The radial flow equation is useful in analysing the flow towards a pumping well and may be obtained either by a coordinate transformation (x = r cos q, y = r sin q) or from elementary mass balance principle applied to a prism as shown below in Figure 7.6: z

(r +

Dq

Figure 7.6

Dr ) Dq

H

r

Elementary prism for mass balance analysis in polar coordinates

The net mass influx along the r-direction is ∂( rqr Hr ) ˆ ∂rqr Hr Ê rqr Hr Dq - Á rqr Hr Dq + Dr ˜ Dq = Dr Dq Ë ¯ ∂r ∂r and that along the q-direction is ∂rqq H ∂rqq H Ê ˆ rqq H Dr - Á rqq H Dr + Dq ˜ Dr = Dr Dq Ë ¯ ∂q ∂q ∂h Krq ∂h The base area of the control volume is rDrDq. Using the polar form of Darcy’s law, qr = Krr + ∂r r ∂q ∂h Kqq ∂h and qq = Kq r + , the mass balance is written as ∂r r ∂q È Ê ∂h Kqq ∂h ˆ ˘ ∂h ∂h ˆ Ê ∂H Á Kq r + + Krq Í ∂H Á rKrr ˙ Ë Ë ∂r r ∂q ˜¯ ˙ ∂h 1 Í ∂r ∂q ˜¯ S = + +R ˙˚ ∂t r ÎÍ ∂r ∂q For a homogeneous and isotropic aquifer, and assuming angular symmetry (i.e., the hydraulic head is a function of r only, as is the case in flow towards a pumping well), we get ∂h ˆ Ê ∂Hr S ∂h 1 Á ∂r ˜ + R = Á (7.10) ˜ Ë K ∂t r ∂r ¯ K As discussed before, H is equal to the aquifer thickness, B for a confined aquifer and, for a constantthickness aquifer, may be calculated from the derivative. For an unconfined aquifer, it is equal to the hydraulic head and varies with time and space. Similarly, S is equal to the specific yield Sy for an unconfined aquifer, but depends on the formation and fluid compressibility for a confined aquifer. Therefore, we will discuss the

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two types of aquifers separately, starting with the simpler case of the confined aquifer. In both cases, we will start with the simplest case of steady one-dimensional flow and progress towards the more complex cases of transient two-dimensional and radial flows.

7.4

FLOW THROUGH A CONFINED AQUIFER

For a confined aquifer, we assume that there is no recharge and Eq. (7.9) is written as S ∂h ∂ 2 h ∂ 2 h = + T ∂t ∂x 2 ∂y 2

(7.11)

LO 3 Explain the flow of groundwater through confined and unconfined aquifers

where, T denotes the product of the hydraulic conductivity K, and aquifer thickness B, and is called the transmissivity. Its dimensions are L2T −1 and the typical range is from 100 m2/d to 10,000 m2/d. The storage coefficient S, as described earlier, represents the volume of water released from the saturated thickness of the aquifer per unit base area for a unit decrease in the hydraulic head. Comparing the definition with that of the specific storage Ss, it is readily seen that S = SsB (this also implies that the term S/T in the equation above could be written as Ss /K. We will, however, use the more common form with S/T). Comparing Eq. (7.11) with the diffusion equation, it is observed that T/S is analogous to the diffusivity and hence this term is sometimes known as the diffusivity of the aquifer. The water release with a drop in the hydraulic head is essentially due to two mechanisms—the expansion of water due to a drop in water pressure and the compression of the aquifer matrix due to an increase in the effective stress, i.e., the stress borne by the soil grains, which reduces the thickness and porosity. Using the definition of compressibility as the volumetric strain per unit change in pressure, we write the water compressibility b, and the formation (also called aquifer or matrix) compressibility a, as DVw b=-

Vw

Dpw

DV and a = -

V Dps

where, Vw is the volume of water, pw is the pore water pressure, V is the volume of aquifer, and ps is the effective stress (the negative sign in the equation indicates that the volume decreases with increase in pressure). Considering a rectangular prism of unit base area and height equal to the aquifer thickness B, and taking a unit drop of the potentiometric surface, the volume of water Vw, is hB and the pore water pressure change Δpw, is equal to −rg. Hence the volume change of water ΔVw, will be equal to brghB. This indicates that due to water compressibility alone, a water volume increase of brghB will occur due to a unit drop in hydraulic head from a prism of unit base area spanning the entire thickness of the aquifer. This volume is released from the aquifer and signifies the component of storativity due to water compressibility. Similarly, considering aquifer compressibility, assuming the solids to be incompressible, and assuming the total overburden pressure to be constant, the change in effective stress Δps, is equal to rg (since a decrease in pore pressure leads to an equal increase in the effective stress), the volume of aquifer (V) is equal to B, and the change in pore volume (since solids are incompressible), ΔV, is equal to −argB. This decrease in pore volume releases a water volume of argB, since the aquifer is saturated and all the pore spaces are filled with water. Combining the effects of water and aquifer compressibility, therefore, we get the relation for the storage coefficient as

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S = rgB(a + bh)

(7.12)

Typical values of compressibility (in units of Pa−1) are around 10−7 for clay and 10−8 for sand and the water compressibility is around 10−10. � EXAMPLE 7.2 A 145 m thick confined aquifer contains mostly sandy soil with a porosity of 0.4, and its compressibility is estimated to be 2.05 × 10−8 Pa−1. The water compressibility is 4.6 × 10−10 Pa−1. Find the storage coefficient and the specific storage for this aquifer. Solution Using Eq. (7.12), we get S = 1000 ¥ 9.81 ¥ 145(2.05 ¥ 10 -8 + 4.6 ¥ 10 -10 ¥ 0.4) = 2.9 × 10 –2 The specific storage is Ss =

S = 2 × 10–4 m–1. B

We are now ready to discuss the application of the continuity equation to various flow conditions encountered in a confined aquifer.

7.4.1

Steady One-Dimensional Flow

This may occur between two water bodies having different water levels and connected by a confined aquifer of uniform thickness, as shown in Figure 7.7. The water level in each of these water bodies is assumed to remain constant even though there is a transfer of water from one to the other (this assumption is justifiable if the flow rate through the aquifer is very small, or the water bodies are very large). For steady flow, the left hand side of Eq. (7.11) is zero and for one-dimensional flow (assumed to be along the x-direction), the second term on the right hand side is zero. Therefore, the governing equation reduces to d2h dx 2

Figure 7.7

=0

A confined aquifer connecting two water bodies

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This indicates that the hydraulic head varies linearly with x, as h = c0 + c1x. The two constants of integration are obtained by using the boundary conditions imposed through the water levels in the two water bodies, as h(0) = h1 and h(L) = h2, giving us the solution as h = h1 +

h2 - h1 x L

(7.13)

Some important points to be noted are: •

h - h1 ˆ Ê The head distribution is linear and the hydraulic gradient is constant Á i = 2 . Ë L ˜¯

•

The head distribution does not depend on the aquifer properties of transmissivity and storage coefficient.

•

The hydraulic gradient is negative implying that the flow is in the positive x direction.

•

The specific discharge qx, at any section is obtained using Darcy’s law and requires the value of the h -h hydraulic conductivity K. It is given by q x = K 2 1 . L K h2 - h1 The seepage velocity requires the value of the porosity h, and is given by v x = . h L

• •

Considering unit width of the aquifer perpendicular to the plane, the rate at which water is flowing h -h h -h from one water body to the other is equal to K 2 1 B = T 2 1 . L L

•

If we could somehow measure the rate of flow through the aquifer, we could estimate the transmissivity. Estimation of aquifer parameters through measurements of the water level and flow rate is an important component of groundwater flow analysis.

•

Both the confining layers have been assumed to be horizontal. If these are not parallel, but one is sloping, with a slope of S0 with respect to the other, the H term in the equation will not be constant. However, in this book we will not consider these cases.

� EXAMPLE 7.3 A 24 m thick confined aquifer connects two water bodies, which have constant water level elevations of 35.5 m and 32.1 m, respectively, above the base of the aquifer. The distance between the water bodies is 225.4 m. If the hydraulic conductivity of the aquifer is 1 cm/s, at what rate water is being transferred between the water bodies. Solution From Darcy’s law, Q = KAi = 0.01 ¥ 24 ¥

35.5 - 32.1 3 m /s per m = 13 m3/h per metre width. 225.4

The one dimensional flow problem is easy to solve but is not frequently encountered in practice. In most cases we are interested in flow towards a well, which is discussed next.

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7.4.2

Steady Flow Towards a Well

This may occur when a well which is screened through the entire depth B, is being pumped at a constant volumetric rate of Q, for a very long time as shown in Figure 7.8. Although it is a two-dimensional flow, and the hydraulic head is a function of both x and y; because of its axisymmetric nature, the governing equation could be reduced to a single dimension, i.e., the radial direction. Eq. (7.10) is written in this case as ∂h ∂r = 0 ∂r

∂r

and results in h = c1 ln r + c2. The two boundary conditions to be used to determine the constants c1 and c2 are not as obvious as in the previous case. For the time being, let us assume that the boundary conditions are obtained by observing the piezometric level in two observation wells, one located at a distance of r1 from the pumping well and the other at r2. These hydraulic heads are denoted by h1 and h2, respectively. Using h(r1) = h1 and h(r2) = h2, we obtain h = h1 +

Figure 7.8

h2 - h1 Ê r ˆ ln ln(r2 /r1 ) ÁË r1 ˜¯

(7.14)

Pumping well in a confined aquifer

Some points to be noted are as follows: •

The assumption of steady flow for the conditions considered in this problem will be valid if a well is being pumped in the centre of a circular island. The assumption of axisymmetric flow will be valid and the ocean would provide a constant supply equal to the pumping rate to maintain steady flow. For other cases, a true steady state will not be reached and the radius of influence, i.e., the distance up to which the hydraulic head is affected by pumping, will keep on increasing. However, for most practical cases, the steady state may be assumed to be reached when the rate of change of the hydraulic head in the observation wells becomes negligible.

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249

The head distribution is logarithmic and the hydraulic gradient is inversely proportional to the Ê 1 h2 - h1 ˆ distance from the well r Á i = . This indicates that the gradient is steep near the well and r ln(r /r ) ˜¯ Ë 2

1

becomes flat as we move away. •

The hydraulic gradient is positive implying that the flow is in the negative r direction, i.e., towards the well.

•

The specific discharge qr, at any section is obtained using Darcy’s law for polar coordinates. It K h2 - h1 requires the value of the hydraulic conductivity K, and is given by qr = . r ln(r2 /r1 )

•

The rate at which water is flowing towards the well at a cylindrical surface located at a distance of h -h K h2 - h1 r from the well is given by 2p rB = 2p T 2 1 , which is a constant and is naturally r ln(r2 /r1 ) ln(r2 /r1 ) equal to Q, the pumping rate. The requirement that the discharge at each section be constant explains the fact that the hydraulic gradient, and consequently the specific discharge, is inversely proportional to r, since the area of flow increases linearly with r.

•

The equation Q = 2p T

h2 - h1 ln(r2 /r1 )

(7.15)

is commonly known as the Thiem equation, since he was the first to derive it, and could be used to estimate the aquifer transmissivity by measuring the pumping rate and hydraulic head in two wells. Note that the equation requires only the difference between the two hydraulic heads and, therefore, the datum for the hydraulic heads can be arbitrary and does not need to be at the bottom of the aquifer. If we assume that the potentiometric surface before the start of pumping was horizontal and its height was measured before the start of the pumping, and define the term drawdown as the amount of lowering of the piezometric surface from its initial value, we could write the term h2 – h1 as s1 – s2, where s represents the drawdown measured in the observation well. •

The two observation wells could be located anywhere and it is tempting to use the pumping well as one of the observation wells, with r taken as the radius of the well rw, and s as the drawdown in the well sw. However, it is not a good idea since drawdown in the pumping well is affected by additional losses due to the turbulent flow close to the well.

� EXAMPLE 7.4 A confined aquifer is being pumped through a well screened over the entire aquifer depth at a constant rate of 1 m3/s. Two observation wells are located at a distance of 20 m and 50 m, respectively, from the pumping well. After a long time of pumping, the water levels in the observation wells achieve a nearly constant value, which is 99.5 m and 102.0 m, respectively, above the aquifer bed. Plot the hydraulic head profile and estimate the transmissivity of the aquifer.

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Engineering Hydrology

Solution The drawdown plot, using Eq. (7.14) is shown in Figure 7.9 corresponding to the measured heads as 99.5 m at a radial distance of 20 m and 102.0 m at 50 m distance from the well. Since it is in the shape of a cone, the drawdown profile is commonly called the cone of depression. Note that the hydraulic head has not levelled-off even at a distance of 100 m. In fact, theoretically, the hydraulic head will keep on increasing as we move away from the well. However, the rate of increase of head with distance keeps on decreasing at increasing distances. The transmissivity is estimated from the Thiem’s equation (7.15) as Q ln(r2 /r1 ) 1 ln 50 / 20 T= = m 2 /s = 210 m2/h. 2p h2 - h1 2p 102.0 - 99.5

102 100

Head (m)

104

98 96 94 92 90 100

50

Figure 7.9

0 Distance from well (m)

50

100

Drawdown in a confined aquifer due to pumping

Although the steady state solutions are easy to obtain, their practical applicability is limited since it may require a long time of pumping to reach close to the steady state conditions. Transient solutions, therefore, become very useful. We will look at the one-dimensional case first and then discuss the more complicated, but much more useful, axisymmetric case of a pumping well.

7.4.3

Transient One-Dimensional Flow

This may occur between two water bodies connected by a confined aquifer of uniform thickness, as shown in Figure 7.10, when there is a change of water level in one of the aquifers. We are interested in obtaining the temporal variation in the hydraulic head within the aquifer. We assume that initially the water level in both the water bodies is h1 and, at t = 0, the water level in one of the bodies drops down to h2. The governing equation, from Eq. (7.11), and the initial and boundary conditions for this case are S ∂h ∂ 2 h = T ∂t ∂x 2 Initial condition h(x, 0) = h1; Boundary conditions h(0, t) = h1; h(L, t > 0) = h2

(7.16)

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Figure 7.10 Transient one-dimensional flow in a confined aquifer The solution is readily obtained by the method of separation of variables after writing the desired solution h(x,t) as the sum of the steady-state solution (Eq. 7.14) and a transient component ht(x,t), such that the transient component satisfies the same governing equation but the boundary conditions become homogeneous, i.e., ht at both boundaries become zero. Then, writing ht(x,t) = c(x).t(t), the solution of •

np x Eq. (7.16) with the homogeneous boundary conditions is obtained as ht ( x, t ) = Âcn sin e L n =1 the constants cn, are obtained from the initial condition ht ( x, 0) =

n2p 2 T t L2 S

, where

h1 - h2 2(h - h )(-1)n x , as cn = - 1 2 . The L np

head distribution is, therefore, given by h -h 2(h1 - h2 ) • (-1)n np x h( x, t ) = h1 - 1 2 x sin e  L p L n =1 n

n2p 2 T t L2 S

(7.17)

The infinite series generally converges rapidly due to the fast decay of the exponential term (except at very small times, for which another form of the solution in terms of error function may be obtained. However, these are not of great practical significance and are not discussed here. For most cases, Eq. (7.17) provides reasonably accurate solutions). � EXAMPLE 7.5 A 10 m thick confined aquifer joins two water bodies which are 100 m apart and have the same water level, 16 m above the aquifer bed. The transmissivity of the aquifer is 1.2 m2/h, and the storage coefficient is 0.002. At a certain time, t = 0, the water level in one of the water bodies drops suddenly to 11 m. Obtain the variation of the hydraulic head in the aquifer. Solution Using Eq. (7.17), Figure 7.11 shows the temporal evolution of the piezometric head in the aquifer. The head distribution is plotted at different times (in hours) as shown in the legend. Note that there is very little change in the head after 5 hours, and at 10 hours the head has attained its steady state values. Also, for the earliest time considered in this example, i.e., 30 minutes, 6 terms of the infinite series were sufficient for convergence (for computation at 1 minute, it required 30 terms for convergence).

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Figure 7.11 Hydraulic head variation in a confined aquifer due to sudden change at one end

7.4.4

Transient Flow Towards a Well

The governing equation in this case is given by Ê ∂h ˆ ∂r S ∂h 1 Á ∂r ˜ = Á T ∂t r Á ∂r ˜˜ Ë ¯ However, it is more convenient to formulate the problem in terms of the drawdown, s. Assuming an initially horizontal piezometric surface at a height of h0 above the datum, we have s = h0−h, and the above equation becomes Ê ∂s ˆ ∂r S ∂s 1 Á ∂r ˜ = Á T ∂t r Á ∂r ˜˜ Ë ¯

(7.18)

The initial and boundary conditions, assuming the aquifer to be of infinite extent, are Initial condition: s(r,0) = 0; Boundary conditions: s(•, t ) = 0; -2p rT

∂s (0, t ) = Q ∂r

where the boundary condition at the well (r = 0) is obtained by applying Darcy’s law at the well (considering ∂s the well radius to be negligible) to obtain the discharge and equating it to the pumping rate Q. Note that ∂ r is negative since the drawdown decreases as we move away from the well.

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This equation could be solved by using the similarity transformation u=

r2S 4Tt

(7.19)

which converts Eq. (7.18) to an ordinary differential equation as follows: u

d2s du

2

+ (1 + u)

ds =0 du

(7.20)

Note that the form of the similarity variable u, is dictated by the well-known diffusion equation and is also evident from the exponential term in the solution of the one-dimensional problem given by Eq. (7.17). The constant “4” is arbitrary and any other number could have been used. However, use of the value 4 makes d2s ds r2S the resulting ODE simpler (for example, if we use u = , the resulting ODE is 4u 2 + (4 + u) = 0 ). du Tt du The two conditions needed for the solution of Eq. (7.20) are s(∞) = 0, which corresponds to both the initial ds Q condition and the infinite boundary condition (since u = 0 for t = 0 as well as for r = ∞) and u (0) = , du 4p T which corresponds to the condition at the well. The first step in the solution of Eq. (7.20) provides, with the use of the well boundary condition, the result ds Q e-u . The solution for the drawdown is then obtained, using the other boundary condition as =du 4p T u s=

Q 4p T

•

e- x Q Ú x dx = 4p T W (u) u

(7.21)

Equation (7.21) was first reported in the hydrology literature by Theis (1935) based on the solution sent to him by Lubin, who obtained it from the 1921 edition of the classic book of Carslaw on heat conduction. Therefore, it is commonly known as the Theis equation, and the integral, which in mathematics is known as the Exponential Integral, is called the Well Function, W(u). Several approximate expressions have been suggested for computing the values of W(u) for a given u (For example, Vatankhah (2014)). One of the approximations is based on the fact that for x < 1, the exponential function may be written as a convergent series and the integral in the above equation could be performed term-by-term. We write (for u < 1): •

W (u) =

•

e- x e- x e- x dx = dx + Ú x Ú x Ú x dx u u 1 1

=Ú u

1

•

1 e- x - 1 e- x + dx + Ú dx x x x 1 •

e- x - 1 e- x - 1 e- x dx + Ú dx + Ú dx x x x 0 0 1

u

= - ln u - Ú

1

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u

= - ln u - Ú

-x +

0

x2 x3 1 • -x +º 1 - e- x e 2! 3! dx - Ú dx + Ú dx x x x 0 1

The last two terms in the last line represent, in an integral form, the negative of the Euler’s constant g, defined by n ˆ Ê 1 g = lim Á - ln n + Â ˜ ª 0.5772 n Æ• Ë i =1 i ¯ resulting in W (u) = - 0.5772 - ln u + u -

u2 u3 + 2.2! 3.3!

(7.22)

For very small values of u (less than about 0.01), i.e., for large time periods, Cooper and Jacob (1946) suggested that the first two terms of the series provide a good approximation of the well function providing a very simple expression for the drawdown as s=

Q 4p T

Ê r2S ˆ Q 2.25Tt 0.5772 ln = ln 2 Á ˜ 4Tt ¯ 4p T r S Ë

(7.23)

This equation based on the Cooper-Jacob approximation, is useful in estimating the parameters S and T of a confined aquifer, as described later in the chapter. Figure 7.12 shows a plot of the well function and could be used to obtain the drawdown in an observation well at any time after the start of pumping, if the aquifer transmissivity and storage coefficients are known. 10

1

W(u)

0.1

0.01

0.001

0.0001

0.00001 0.1

1

10

100 1/u

Figure 7.12 Plot of the well function

1000

10000

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� EXAMPLE 7.6 A confined aquifer has a transmissivity of 1 m2/min and storage coefficient of 0.0001. The piezometric surface is initially horizontal. A fully penetrating, i.e., screened over the entire thickness of the aquifer, pumping well is then pumped at the rate of 120 m3/hr. Obtain the theoretical variation of the drawdown observed in two wells, one at 50 m from the pumping well and the other at 100 m. Solution A sample calculation is shown below for a time of 2 hours: u=

r 2 S r 2 ¥ 10 -4 = = 0.002083 for r = 100 m; 0.000521 for r = 50 m 4Tt 480

The corresponding W(u) values obtained from the table are 5.60 and 6.98, respectively. Using Eq. (7.21), the drawdowns are obtained as 0.89 m and 1.11 m, respectively, for r = 100 m and 50 m. The following table shows some values of the drawdowns and Figure 7.13 shows the complete temporal variation up to 4 hours. t (min)

0.5

1

2

5

10

50

100

120

240

s at 50 m (m)

0.26

0.36

0.46

0.61

0.72

0.97

1.08

1.11

1.22

s at 100 m (m)

0.09

0.17

0.26

0.39

0.50

0.75

0.86

0.89

1.00

Figure 7.13 Drawdown in two observation wells If we look at the form of the similarity variable u, we could see that the two separate drawdown curves shown in the figure above will plot as a single curve if the time is scaled by the factor r2. This is demonstrated in Figure 7.14 where the drawdowns in both wells are plotted against the variable t/r2.

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1.4 1.2

s (m)

1 0.8

r = 50 m

0.6

r = 100 m

0.4 0.2 0 0

0.02

0.04

0.06

0.08

0.1

t/r2(min/m2)

Figure 7.14 Drawdown in two observation wells as a function of the similarity variable After looking at combining data from two observation wells, let us see what happens if there are two (or more) pumping wells. Since the governing equation and the boundary conditions are linear, it follows that the principle of superposition is valid. Therefore, the drawdowns due to individual wells could be added to obtain the overall drawdown due to two or more wells. Similarly, if there is a single well but the pumping is not continued for a very long time, the drawdown would decrease after the stoppage of pumping. This is known as recovery since the piezometric head recovers after the reduction caused by pumping. Again, using superposition, we could view this situation as the combination of an infinite duration pumping and another infinite duration recharge, but this time with the recharge starting at t0, the time at which pumping is stopped. Therefore, we could obtain the drawdown at any time beyond t0, as s=

Ê r2S ˆ ˘ Q È Ê r2S ˆ W ÍW Á Á 4T (t - t ) ˜ ˙ 4p T ÎÍ Ë 4Tt ˜¯ Ë ˙ 0 ¯˚

(7.24)

in which the first well function corresponds to the pumping well and the second is due to the recharge well of equal magnitude, starting at t0. For time periods less than t0, only the first term is used. � EXAMPLE 7.7 For the aquifer and well system of the previous example, find the drawdown in the observation well at 50 m up to a period of 8 hours, if the pumping stops at 4 hours. Solution A sample calculation is shown below for a time of 6 hours: u=

r2S 0.25 = = 0.000694 for pumping well; 0.002083 for recharge well 4Tt 240 ¥ (6 or 2)

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The corresponding W(u) values are 6.70 and 5.60, respectively. Using Eq. (7.24), the drawdown is obtained as 0.17 m. The following table shows some values of the drawdowns and Figure 7.15 shows the complete temporal variation up to 8 hours. Note the rapid recovery immediately after the pumping is stopped at 240 minutes. t (min)

0.5

2

10

100

240

241

250

300

360

480

s at 50 m (m)

0.26

0.46

0.72

1.08

1.22

0.86

0.51

0.26

0.17

0.11

1.4

s (m)

1.2 1 0.8 0.6 0.4 0.2 0

0

120

240

360

480

t (min)

Figure 7.15 Drawdown and recovery after the stoppage of pumping

7.5

FLOW THROUGH UNCONFINED AQUIFERS

LO 3

In the previous section, the flow through confined aquifers was discussed, in which there was an impervious layer at the bottom which was assumed to be horizontal. It was also bounded by an impervious layer at the top, which was also assumed to be horizontal implying that the thickness of the aquifer remained constant, or in other words, it did not depend on the head. The piezometric head decreases towards the well when we take water out of the aquifer, but the thickness of the aquifer remains constant. On the other hand, the unconfined aquifer does not have an impervious confining layer at the top and has the water table as its upper boundary. This complicates the analysis since the aquifer thickness is now dependent on the hydraulic head. In this section we consider flow situations similar to those discussed for the confined aquifer case in the previous section. The governing equation is the same, i.e., Eq. (7.9) for two-dimensional flow and Eq. (7.10) for flow towards a well, but the difference would be that the aquifer thickness H, will now be equal to the hydraulic head h, and there may be some recharge.

7.5.1

Steady One-Dimensional Flow

This may occur between two water bodies having different water levels and connected by an unconfined aquifer, as shown in Figure 7.16. For steady flow, the left hand side of Eq. (7.9) is zero and for one-dimensional

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flow (assumed to be along the x-direction), the second term on the right hand side is zero. Therefore, the governing equation reduces to Ê dh ˆ dÁh ˜ Ë dx ¯ R + =0 dx K

i.e.,

d 2 h2 dx

2

+

2R =0 K

After applying the boundary conditions as h(0) = h1 and h(L) = h2, we get the solution as h2 = h12 -

h12 - h22 R x+ x( L - x ) L K

(7.25)

Figure 7.16 An unconfined aquifer connecting two water bodies Some points to be noted are as follows: •

In absence of recharge, the head distribution does not depend on the aquifer properties, and the Ê dh K (h12 - h22 ) ˆ discharge is independent of x Á Q = - Kh = ˜. dx 2L Ë ¯

•

When a uniform recharge occurs over the entire aquifer length, the head distribution depends on the aquifer conductivity, or to be more precise, on the ratio of the recharge rate to the hydraulic conductivity. The discharge is not constant in this case and increases with x. Using the symbol Q0 for the discharge in absence of recharge, we get ÊL ˆ Q = Q0 - R Á - x ˜ Ë2 ¯

(7.26)

indicating that the occurrence of recharge decreases the discharge in the first half of the aquifer length (x < L/2) than what it would have been without the recharge, and increases it in the latter half. RL The rate of withdrawal from the upstream water body is Q0 and the rate of discharge to the 2 RL downstream water body is Q0 + , with this additional volume of RL coming from the recharge. 2

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•

If both the water bodies have the same water level, the head distribution will be symmetric about the midpoint. Obviously, Q0 will be zero, and Q will be negative up to the midpoint and positive after that, indicating that water will flow from midpoint towards both the water bodies. In other words, the midpoint is a water divide, since the recharge before this point goes to one water body and that after it goes to the other, with the discharge at the water divide being zero. Note that at the water divide, the hydraulic head gradient will be zero and the head attains its maximum value.

•

When the water levels, h1 and h2, are different and there is a recharge, there will again be a water divide but it would not be at the midpoint. Its location x*, is obtained by finding the point where Q becomes zero. From Eq. (7.26), we have Q=

K (h12 - h22 ) L K (h12 - h22 ) ÊL ˆ - R Á - x* ˜ = 0 fi x* = Ë2 ¯ 2L 2 R 2L 2

2

K h1 - h2 towards R 2L the water body with the higher water level. As the recharge rate increases, the water divide moves closer to the midpoint, and as the recharge rate decreases the water divide moves closer to the water body, till the limiting case of the water divide reaching the water body. This happens when the second term of the expression for x* becomes equal to L/2, i.e., when the recharge is equal to K (h12 - h22 ) . For recharge rates equal to or smaller than this value, there would be no water divide L2 and the flow will be from the higher water level body to the lower body throughout the aquifer.

Therefore, the water divide will shift from the midpoint by a distance equal to

� EXAMPLE 7.8 Obtain the steady state distribution of the piezometric head in a 100 m long unconfined aquifer connecting two water bodies with the upstream head maintained at 16 m and downstream head at 11 m. Use the recharge rate to hydraulic conductivity ratio as 0.0, 0.01, 0.02, 0.05, and 0.1. Solution The head distribution is obtained from Eq. (7.25) and is plotted below for different R/K ratios as mentioned in the legend. Note that the water divide first occurs for a ratio somewhere between 0.01 and 0.02. From the h2 - h22 discussion in bullet point 4 stated above, the critical value of R/K should be 1 , i.e., 0.0135. L2 One important point to note is that for the unconfined aquifers, water table is the hydraulic head line and also a streamline. Since both the boundaries of the aquifer, which are in contact with the water bodies, are equipotential lines (potential equal to h1 and h2, respectively), the water table should intersect these boundaries at right angles. The solution does not impose this condition and from the figure also it is clear that this condition is not satisfied (except for the critical R/K ratio, for which the water table is perpendicular to the boundary at the upstream side). The solution is, therefore, not strictly applicable near the boundaries. In reality, the water table at the outflow boundary does not intersect the boundary at the water level. The point of intersection is above the water table and there is a seepage face from this point to the water level, through which water seeps out of the aquifer.

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Figure 7.17 Head variation in unconfined aquifer at different recharge rates

7.5.2

Steady Flow Towards a Well

This may occur when a fully-screened well is being pumped at a constant volumetric rate of Q, for a very long time as shown in Figure 7.18. Eq. (7.10) is written in this case as ∂h ˆ Ê ∂hr 1Á ∂r ˜ + R = 0 r ÁÁ ∂r ˜˜ K Ë ¯

or

Ê ∂h 2 ∂r 1Á ∂r Á r Á ∂r ÁË

ˆ ˜ 2R =0 ˜+ ˜ K ˜¯

R 2 r . The two boundary conditions to be used to determine the constants 2K c1 and c2 are obtained by observing the piezometric level in two observation wells, one located at a distance of r1 from the pumping well and the other at r2. These hydraulic heads are denoted by h1 and h2, respectively. Using h(r1) = h1 and h(r2) = h2, we obtain and results in h2 = c1 ln r + c2 -

2

h =

h12

+

R 2 (r - r12 ) R 2 2K 2 ln(r /r1 ) (r - r12 ) ln(r2 /r1 ) 2K

h22 - h12 +

(7.27)

Some points to be noted are as follows: •

•

In the absence of recharge, we get a logarithmic variation similar to the confined aquifer case, the only difference being that it is the square of the head which varies logarithmically. K The specific discharge qr, at any section is given by qr = 2rh

R 2 2 (r - r ) 2 K 2 1 + Rr . ln(r2 /r1 ) 2h

h22 - h12 +

Groundwater

261

•

The flow rate towards the well at a cylindrical surface located at a distance of r from the well is given R 2 2 R 2 2 È ˘ 2 2 h22 - h12 + (r - r ) Í K h2 - h1 + 2 K (r2 - r1 ) Rr ˙ 2 K 2 1 - p r 2 R , which decreases by 2p rh Í = p K ˙ ln(r2 /r1 ) 2h ˚ ln(r2 /r1 ) Î 2rh as we move away from well, since a part of the well discharge Q, is contributed by the recharge. The first term is a constant and is equal to the pumping rate, and the second term represents the contribution from recharge, which increases with the distance from the well.

•

Similar to the confined aquifer case, we would prefer to express the hydraulic head equation in terms of drawdowns. However, the nonlinear nature of the equation means that it is not possible to do it unless we make some simplifying assumptions. For example, considering that there is no recharge, we write Q = pK

h22 - h12 h -h = p K (h1 + h2 ) 2 1 ln(r2 /r1 ) ln(r2 /r1 )

(7.28)

Denoting the initial (pre-pumping) height of the water table above the aquifer bed as h0, writing the drawdown in the wells as s1 = h0 – h1 and s2 = h0 – h2; assuming that for very small drawdowns, h1 + h2 @ 2h0; and writing the transmissivity as T = Kh0, we get Q = 2p T

s1 - s2 ln(r2 /r1 )

which provides a way of estimating the transmissivity through measurement of drawdowns in two observation wells. •

If the pumping well is taken as one of the observation wells (with radius of the well rw, and drawdown sw in the well sw) and the radius of influence of the well is taken as Rw, we may write Q = 2p T . ln( Rw /rw ) However, as discussed earlier, it is not advisable to use the pumping well as an observation well due to additional losses at the well.

� EXAMPLE 7.9 An unconfined aquifer is being pumped through a fully penetrating well. Two observation wells are located at a distance of 20 m and 50 m, respectively, from the pumping well. After a long time of pumping, the water levels in the observation wells achieve a nearly constant value, which were 99.5 m and 102.0 m, respectively, above the aquifer bed. Plot the hydraulic head profile and compare with that for the confined aquifer of Example 7.4. Solution The drawdown plot, using Eq. (7.27), is shown in Figure 7.18 corresponding to the measured heads as 99.5 m at a radial distance of 20 m from the well and 102.0 m at 50 m distance, and with no recharge (R = 0). Since the drawdown is small compared to the head, the drawdown profile is almost same as that obtained for the confined case (Figure 7.9). Slight differences near the well exist but are not noticeable in the figure

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Engineering Hydrology

100

Head (m)

(e.g., drawdown at 1 m radial distance is 90.85 m for unconfined aquifer and 91.33 m for the confined aquifer; and at 200 m, it is 105.67 m for the unconfined aquifer and 105.78 m for the confined aquifer).

95

90 100

50

0 Distance from well (m)

50

100

Figure 7.18 Drawdown in an unconfined aquifer due to pumping We now look at the transient flows, first for the one-dimensional case and then for the more useful axisymmetric case of a pumping well.

7.5.3

Transient One-Dimensional Flow

This may occur between two water bodies connected by an unconfined aquifer as shown in Figure 7.16, when there is a change of water level in one of the water bodies. We are interested in obtaining the temporal variation in the hydraulic head within the aquifer. We assume that initially the water level in both the water bodies is h1 and then the water level in one of the bodies suddenly drops down to h2. Equation (7.9) and the initial and boundary conditions for this case are S ∂h = K ∂t

∂h ∂x + R ∂x K

∂h

Initial condition: h( x,0) = h1; Boundary conditions: h(0, t ) = h1 ; h( L, t > 0) = h2 The solution of this equation is complicated because of the nonlinearity in the spatial derivative term. A similarity transformation using a variable of the form x2/t could be used to convert this equation to a nonlinear ordinary differential equation. However, if we assume that the drawdown is small compared ∂h ∂s ∂s to the initial aquifer thickness h1, the term h could be written as -(h1 - s ) . Defining the @ - h1 ∂x ∂x ∂x transmissivity as T = Kh1, the governing equation and the required conditions are S ∂s ∂ 2 s R = T ∂t ∂x 2 T Initial condition: s( x,0) = 0; Boundary conditions: s(0, t ) = 0; s( L, t > 0) = h1 - h2

(7.29)

Groundwater

263

These are similar to Eq. (7.16) for the confined aquifer, except for the addition of the recharge term and may be solved using the same methodology. In the absence of recharge, the solution is given by h -h 2(h1 - h2 ) • (-1)n np x s ( x, t ) = 1 2 x + sin e  L p L n =1 n

n2p 2 T t L2 S

which is identical to that for the confined aquifer, and is therefore not illustrated here. As mentioned before, it is only an approximate solution since the drawdown is assumed to be very small compared to the aquifer thickness.

7.5.4

Transient Flow Towards a Well

The governing equation in this case is given by ∂h ˆ Ê ∂ hr S ∂h 1 Á ∂r ˜ + R = Á ˜ K ∂t r Ë ∂r ¯ K which is again a nonlinear partial differential equation and is difficult to solve. If the drawdown is small compared to the aquifer thickness, the Theis solution obtained for confined aquifers could be applied for unconfined aquifers also. However, in unconfined aquifers, it has been observed that the release of water due to the lowering of water table is not instantaneous but has some delay, thereby resulting in a delayed yield. In fact, during very early times after start of pumping, the release of water is primarily due to the elastic effects of water expansion and formation compression. The Theis solution is applicable, but with the storativity based on elastic effects only. At very large times, when the rate of decline of the water table becomes very small, the Theis solution is again applicable but with a storativity value equal to the specific yield. For intermediate times, the drawdown lies in between these two limiting cases and is a function of the relative distance of the observation well, r/h0, where h0 is the initial aquifer thickness. In terms of the well function, the solution is written as s=

Ê r 2 S r 2 Sy r 2 ˆ Ê Q Q r2 ˆ WÁ , , 2˜ = W Á uc , uu , 2 ˜ 4p T ÁË 4Tt 4Tt h0 ˜¯ 4p T Ë h0 ¯

(7.30)

where, uc is the u corresponding to the confined case, i.e., using the elastic storativity, and uu is the u corresponding to unconfined case, i.e., using the specific yield. Tables and graphs of the unconfined well function are available in several books (e.g., Mariño and Luthin (1982)) and a plot is shown in Figure 7.19 (the numbers on the curves represent the values of r/h0, the topmost curve is the Theis curve for u = uc, and the bottommost curve is the Theis curve for u = uu). To obtain the well function value, either uc or uu has to be used, but not both. For smaller time periods, i.e., prior to the horizontal portion of the relevant curve, uc axis is used, while for later time periods, i.e., beyond the horizontal portion, uu axis is used. For example, for r/h0 = 1, uc axis is used for 1/uc < 10, and uu axis is used for 1/uu > 0.1.

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1/uc 0.1

1

10

100

1000

1E4

1E5

1E6

1E7

1E8

1E+01 0.001 0.01 0.1

W(uc, uu, r/h0)

1E+00

1 2 1E-01

5

1E-02 1E-05

1E-04

1E-03

1E-02

1E-01

1E+00

1E+01

1E+02

1E+03

1E+04

1/uu

Figure 7.19 Well function for the unconfined aquifer

� EXAMPLE 7.10 An unconfined aquifer has a transmissivity of 1 m2/min, elastic storage coefficient of 0.0001, and specific yield of 0.20. The initial saturated thickness of the aquifer is 50 m and piezometric surface is horizontal. A fully penetrating pumping well is then pumped at the rate of 120 m3/hr. Obtain the variation of the drawdown observed in a well located 50 m from the pumping well. Solution Two sample calculations are shown below, one for a small time period of 15 seconds and the other for a large time period of 2 hours (note that r/h0 = 1). For t = 15 seconds, uc =

r 2 S y 502 ¥ 0.2 r 2 S 502 ¥ 10 -4 = = 0.25; uu = = = 500 4Tt 1 4Tt 1

Since 1/uc = 4, which is less than 10, the uc axis is used to read the value of the well function as 0.3. For t = 2 hours, uc =

r 2 S y 502 ¥ 0.2 r 2 S 502 ¥ 10 -4 = = 0.000521; uu = = = 1.042 4Tt 480 4Tt 480

Groundwater

265

Since 1/uu = 0.96, which is greater than 0.1, the uu axis is used to read the value of the well function as 0.54. The following table shows some values of the drawdowns and Figure 7.20 shows the complete temporal variation up to 35 days. Note that the drawdowns are much smaller than the corresponding values for the confined aquifer (Example 7.6), since the specific yield is much larger than the elastic storativity and the same amount of water can be released from the aquifer for a much smaller change in the hydraulic head. t (min)

0.25

1

30

60

120

180

240

480

50000

s (mm)

48

50

60

69

86

104

119

175

863

1000

800

s (mm)

600

400

200

0 0

5

10

15

20

25

30

35

40

t (days)

Figure 7.20 Drawdown due to pumping in an unconfined aquifer

7.6

NON IDEAL CONDITIONS

LO 3

The analysis of the flow cases discussed in the previous section was based on several assumptions. For example, Darcy’s law is assumed to be valid, aquifer is assumed to be homogeneous and isotropic and completely saturated, for unconfined aquifers the drawdown is assumed to be small, and so on. In practical cases one or more of these assumptions may not be valid. In the next few sections we discuss a few specific examples of such cases.

7.6.1

Well Loss and Specific Capacity

In all cases discussed so far, we have assumed that Darcy’s law is valid , i.e., flow is laminar and head loss is proportional to the specific discharge. In some situations, e.g., flow near a well, the flow may be turbulent and the head loss is typically proportional to a higher power (1.7 – 1.8) of q. Thus, there will be an additional loss, over the head loss obtained from Darcy’s law. Moreover, near a well, there will be some head loss when the flow passes through the well screen. The total drawdown in the well will be a combination of three terms:

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Engineering Hydrology

laminar loss (swl), turbulent loss (swt), and the screen loss (sws). The laminar head loss is called the formation loss and the sum of the other two terms is called the well loss. As we have seen, the laminar loss is proportional to the pumping rate Q. The turbulent loss is generally taken as proportional to Q2 and the screen loss is also found to be proportional to Q2, implying that the well loss is proportional to Q2, i.e., well loss = C Q2. We want the well loss to be as small as possible so that the well is more productive, i.e., produces the same discharge with a smaller drawdown. Another term, which is related to the well loss, is known as the specific capacity of the well and is defined as the discharge per unit drawdown. High specific capacity of the well indicates a more efficient well. For example, if we look at a confined aquifer being pumped at a constant rate, the specific capacity at a sufficiently large time may be written, using the Cooper-Jacob approximation, as Q = sw

1 1 2.25tT ln 2 + CQ 4p T rw S

(7.31)

We can see that the specific capacity increases with increase in transmissivity (T occurs at two places, but the log term has smaller influence) and decreases with increase in Q and t. Thus for a given aquifer, if a well is pumped at a higher rate, the specific capacity will be smaller and as the pumping time increases, the specific capacity will decrease. If the coefficient C is very large, the well may be inefficient and we may have to look for a replacement. � EXAMPLE 7.11 A 15 cm-diameter well in a confined aquifer was pumped at three different rates and the drawdown in the well was measured at 4-hour intervals, as shown below. Time (hr)

4

8

12

16

20

24

28

32

36

40

44

48

sw (m) for Q = 1 m3/min

2.148

2.208

2.236

2.258

2.275

2.292

2.301

2.318

2.324

2.337

2.343

2.353

sw (m) for Q = 2 m3/min

5.293

5.409

5.480

5.519

5.551

5.579

5.617

5.633

5.645

5.668

5.689

5.697

sw (m) for 14.602 14.824 14.961 15.037 15.110 15.160 15.229 15.282 15.296 15.321 15.370 15.409 Q = 4 m3/min

Estimate the well loss coefficient and the specific capacity of the well for these pumping rates at 1 day and 2 days. Solution To obtain the well loss coefficient, we could use sw = C1Q+ CQ2 and plot sw/Q versus Q to get a straight line, the slope of which will be equal to the well loss coefficient C. However, using Excel to obtain theses coefficients by minimising the sum of the squared errors at a time of 1 day and 2 days, we get the coefficients as shown in the table below: Time = 1 day

Time = 2 days

Formation loss coefficient, C1 (min/m2)

Well loss coefficient, C (min2/m5)

Formation loss coefficient, C1 (min/m2)

Well loss coefficient, C (min2/m5)

1.79

0.50

1.85

0.50

Groundwater

267

The well loss coefficient is, therefore, 0.50 min2/m5. As expected, the formation loss coefficient increases with time. The specific capacity of the well is directly obtained as Q/sw, and is shown below (in units of m2/min): Time = 1 day 3

3

Time = 2 days 3

3

Q = 1 m /min

Q = 2 m /min

Q = 4 m /min

Q = 1 m /min

Q = 2 m3/min

Q = 4 m3/min

0.436

0.358

0.264

0.425

0.351

0.260

The decrease in the specific capacity with increase in Q, as well as with increase in time, is clearly seen (for comparison, although we do not show the computations here, with the aquifer data used for generating the drawdowns for this example, if we decrease the transmissivity of the aquifer by a factor of 10, the specific capacity for Q = 1 m3/min at t = 1 day reduces to about 0.06 m2/min and, for a tenfold increase in transmissivity, it increases to about 1.37 m2/min).

7.6.2

Flow through a Layered Porous Medium

A natural porous medium generally has layers of different conductivities, depending on how the porous medium is formed. For examples, when soil grains are deposited in water, the larger grains will settle first and their larger dimension will be in the horizontal direction. This gives rise to a heterogeneous as well as anisotropic behaviour. In this section, we consider the flow through a layered porous media and analyse the effective hydraulic conductivity in two different directions to ascertain the anisotropy. We consider, as shown in Figure 7.21, a medium comprising n horizontal layers with the ith layer having a thickness Bi and conductivity Ki. The total thickness is B and we want to find out an effective conductivity Keff, such that replacing the layered medium with a uniform material of this conductivity will lead to the same hydraulic behaviour.

Figure 7.21 Layered porous medium with bedding in horizontal direction We first consider the flow to be in the horizontal direction, i.e., along the layers. In this case, the discharge through the medium would be the sum of the discharges in individual layers and the head loss across each layer would be same, i.e., the difference of head between the upgradient section and the downgradient section, hu – hd. Using Darcy’s law, and considering a horizontal length of the medium, L, we have Qi = Ki bi

hu - hd L

" i = 1,2,º n

and

Q = K eff B

hu - hd L

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Engineering Hydrology n

and using the fact that Q = ÂQi , we get i =1

K eff =

ÂKi bi Âbi

(7.32)

The effective conductivity is, therefore, the weighted arithmetic mean of the layer conductivities, with the layer thickness as the weight. When the flow is vertical, i.e., perpendicular to the layers, the discharge is same through all the layers, but the head loss in each layer will be different and has to be added to get the total head loss. Considering a horizontal length of the medium, L, we have Dh Qi = Q = Ki L i bi

 Dhi L i =1 n

" i = 1,2,º n

Writing the head loss in each layer as Dhi =

and

Q = K eff

B

Q bi and putting in the expression for the effective L Ki

conductivity, we get K eff =

Âbi b  Ki

(7.33)

i

The effective conductivity is, therefore, the weighted harmonic mean of the layer conductivities, with the layer thickness as the weight. Since the harmonic mean is always less than (or, at most equal to) the arithmetic mean, the effective conductivity along the layers is larger than the effective conductivity perpendicular to the layers. In fact, if we have one of the layers as impermeable, the effective conductivity for vertical flow will be zero, since the flow would be stopped by this layer. However, when the flow is taking place in the horizontal direction, the other layers will still have some flow, even when the impermeable layer does not carry any flow. � EXAMPLE 7.12 An aquifer is comprised of four distinct horizontal layers formed by the deposition of sediments in such a way that the bottommost layer is 10 m thick, the two middle layers are each 6 m thick, and the top layer is 3 m thick. The hydraulic conductivities of these layers are 0.02 cm/s for the bottom layer, 0.1 cm/s for the layer above that, 0.5 cm/s for the next layer, and 1.2 cm/s for the top layer. A natural hydraulic head gradient of 1 in 1000 exists in the aquifer. Find the effective conductivity of the aquifer and the groundwater flow rate. During a recharge event, a vertical flow rate of 10 m3/d per square meter surface area was observed. Find the effective conductivity of the aquifer and the head loss over the entire thickness of the aquifer.

Groundwater

269

Solution For the natural gradient flow (in the horizontal direction), the effective conductivity is obtained from Eq. (7.32) as

1.2 ¥ 3 + 0.5 ¥ 6 + 0.1 ¥ 6 + 0.02 ¥ 10 = 0.296 cm/s. The groundwater flow rate, per metre width 3 + 6 + 6 + 10

0.296 ¥ 86400 1 ¥ 25 ¥ = 6.39 m3/d. Most of this flow will be through the top two 100 1000 layers, since the flow is partitioned in the ratio of the transmissivities, i.e., 3.6, 3.0, 0.6, and 0.2 respectively from top to bottom. of the aquifer, is

For the recharge event (flow in vertical direction), the effective conductivity is obtained from Eq. (7.33) 10 ¥ 100 3 + 6 + 6 + 10 ¥ 25 = 6.65 m. Most of = 0.0435 cm/s and the total head loss as 0.0435 ¥ 86400 3 6 6 10 + + + 1.2 0.5 0.1 0.02 this head loss will occur in the bottommost layer, since the total head loss is partitioned in the ratio of b/K for each layer, i.e., 2.5, 12, 60, and 500 respectively from top to bottom. as

7.6.3

Flow through Leaky Aquifers

When the upper confining layer of a confined aquifer comprises an aquitard, there is a possibility of leakage from the aquifer (confined or unconfined) above this layer to the confined aquifer below it. Although water will move very slowly through the aquitard, the large areal extent may lead to a significant contribution of recharge to the confined aquifer. The flow in the aquitard will be vertical and will be governed by Darcy’s law, with the hydraulic head difference between the upper and lower aquifers driving the flow. A well function •

similar to the Theis well function was derived by Hantush for these cases as W (u, v) =

Ú u

e

Ê v2 ˆ -Á x - ˜ Ë 4x¯

x

dx , shown

in Figure 7.22, and is used to find the drawdown in leaky aquifers. This function W(u,v), is known as the Hantush well function or the leaky aquifer well function. In addition to the similarity variable u used earlier, it is now also dependent on the leakage factor, v =

r TBa Ka

, where Ba and Ka are the thickness and hydraulic

conductivity, respectively, of the aquitard (sometimes the term in the denominator,

TBa , which has the Ka

dimensions of length, is called the leakage factor, but we will call the dimensionless variable v, the leakage factor).

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Engineering Hydrology

100.00

0

10.00

0.005 0.02

0.05 0.1 0.2

1.00

0.5 1.0

1.5 2.0

0.10

2.5

0.01 1E-01

1E+00

1E+01

1E+02

1E+03

1E+04

1E+05

1E+06

Figure 7.22 Hantush well function for leaky aquifers. Numbers in the legend show the value of v, 0 for the topmost curve and 2.5 for the bottommost curve.

� EXAMPLE 7.13 A leaky aquifer has an impermeable bottom boundary and the upper confining layer is an aquitard, with thickness of 5 m and hydraulic conductivity of 0.01 m/d. The confined unit has a transmissivity of 1 m2/min and storage coefficient of 0.0001. The piezometric surface is initially horizontal. A fully penetrating pumping well is then pumped at a rate of 120 m3/hr. Obtain the theoretical variation of the drawdown in an observation well located at 100 m from the pumping well. Solution The leakage factor, v =

r TBa Ka

=

100 1 ¥ 1440 ¥ 5 0.01

= 0.118

A sample calculation is shown below for a time of 2 hours: u=

r 2 S 1002 ¥ 10 -4 = = 0.002083 4Tt 480

Groundwater

271

The corresponding W(u,v) value obtained from the figure is 4.45. The drawdown is then obtained as Q s= W (u, v) = 0.708 m (in Example 7.6, an identical confined aquifer was pumped and the drawdown at 4p T 2 hours was obtained as 0.89 m. The leakage through the aquitard leads to a reduction in drawdown of about 18 cm). The following table shows some values of the drawdowns. t (min)

0.5

1

2

5

10

50

100

120

240

s (m)

0.089

0.166

0.254

0.384

0.477

0.661

0.702

0.708

0.719

7.6.4

Flow through Unsaturated Zone

Although the unsaturated zone does not supply enough water for a sustainable water supply, it is important for the agricultural engineer, since plant roots extract water from this zone. It is also important for the environmental engineers, since most of the surface pollutants reach the groundwater only after passing through this zone. Darcy’s law is still valid but with two major changes: the pressure head is negative (also called suction), since the pressure is less than atmospheric; and the hydraulic conductivity is dependent on the pressure head. Similarly, while the porosity may be constant, the soil is not fully saturated and the moisture content is a function of the pressure head. The variation of the hydraulic conductivity and moisture content with the suction head is highly nonlinear and makes the solution of the governing equation difficult, even when using numerical methods. Several constitutive relationships have been proposed for relating the unsaturated conductivity and moisture content with the suction head, which range from a very simple exponential model of Gardner to a complex but more realistic model of van Genuchten. In this book, we will not be considering the flow through the unsaturated zone.

7.6.5

Flow Near Boundaries

The drawdown equations derived earlier assume the aquifer to be of infinite extent. When boundaries are present, the actual drawdown will be smaller than the theoretical drawdown if the boundary is a recharge boundary, e.g., a river. On the other hand, near an impermeable boundary, the drawdown will be larger. The method of images could be used to obtain the drawdown by replacing the boundary with an image well, at the same distance from the boundary as the real well but in the opposite direction. For recharge boundaries, the nature of the image well is opposite to that of the real well, i.e., a pumping well is imaged as a recharge well (pumping water into the aquifer). For impermeable boundaries, the nature of the image well is same as the real well, i.e., a pumping well is imaged as another pumping well. The drawdown at any point is then obtained by the principle of superposition as the sum of drawdowns due to both the real well and the image well. It should be noted that the drawdown due to a recharge well is negative. This implies that the net drawdown at a recharge boundary is zero and the net drawdown at an impermeable boundary is twice that due to the real well. More details can be seen in advanced texts on groundwater hydrology (e.g., Kasenow, 2001).

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7.7

PARAMETER ESTIMATION

One of the important applications of the theory of flow through porous media is in the estimation of the parameters of aquifers by observing the hydraulic head values at a few points. These parameters could be the storage coefficient, hydraulic conductivity, transmissivity, the flow direction, etc. In this section, we will look at various techniques used for this purpose.

7.7.1

LO 4 Outline the methods of estimation of aquifer parameters

Estimation of Flow Direction

For a homogeneous and isotropic aquifer, the groundwater flow direction is opposite to the gradient of the hydraulic head. Therefore, if we measure the water levels in at least three non-collinear piezometers, we could estimate the direction in which the groundwater is moving, by assuming that the hydraulic head is varying linearly. The simplest method to obtain the gradient direction is to perform a linear interpolation of heads along the line joining the largest and smallest heads to obtain the location of the point where the head is equal to that at the third point, which has an intermediate value (Figure 7.23). Joining these two points with equal heads, we get an equipotential line and the flow direction would be perpendicular to it, from higher head to lower head. Naturally, if two of the measured heads are same, we do not need to perform this step, instead we will get the equipotential line directly. And if all three heads are equal, we conclude that there is no flow near the area bounded by these three piezometers. 250 96.8

200

North (m)

98.2 (interpolated) 150

100

100.4

50 98.2 0 0

50

100

150

200

250

East (m)

Figure 7.23 Estimation of groundwater flow direction from head measurements at three wells Once we have an estimate of the direction of flow, we could obtain a rough estimate of the magnitude of the velocity by having two wells in a line along the flow direction. By introducing a tracer (e.g., dye, salt) at the upgradient well, and observing how long it takes to reach the other well through frequent sampling of the well water, we could get the seepage velocity as the distance between the wells divided by the time taken by the tracer to travel from one well to the other. However, due to natural heterogeneity of the subsurface

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and several local variations in flow directions, it is possible that the tracer does not reach the second well, especially if the distance between the wells is large. Moreover, due to diffusion and dispersion, the tracer may reach the well earlier than what would be predicted by the velocity. � EXAMPLE 7.14 Three piezometers are installed in an unconfined aquifer at coordinates (East, North) of A: (0,0), B: (150,50), and C: (50,150). The water table elevations at these wells under steady flow conditions were observed as 120.203 m, 121.189 m, and 119.678 m, respectively. Estimate the direction of the groundwater flow. Another piezometer D was then installed such that it was at a distance of 100 m upgradient of C, and it was seen that a tracer took 19 hours and 25 minutes to travel from D to C. If the aquifer porosity is 0.4, estimate its hydraulic conductivity. Solution Note that the wells are same as those shown in Figure 7.23, except that the origin of the coordinate system is placed at well A, which is the well with an intermediate head value. This makes the computations a little easier. Using linear interpolation between B and C, the coordinates of the point where the head is equal to that at A 120.203 - 119.678 120.203 - 119.678 (120.203 m) are 50 + (150 - 50) m = 84.75 m East and 150 + (50 - 150) 121.189 - 119.678 121.189 - 119.678 = 115.25 m North. The line joining A, i.e., (0,0) and (84.75,115.25), is therefore an equipotential line, which has a slope of 115.25/84.75 = 1.36 (i.e., makes an angle of 53.67° from East) and has a hydraulic head equal to 120.203. The gradient, which is perpendicular to the equipotential line is, towards the Northwest, making an angle of 36.33° from West. The seepage velocity is equal to 100/(19 × 3600 + 25 × 60) m/s = 0.00143 m/s, and Darcy velocity is 0.00143 × 0.4 = 5.72 × 10−4 m/min. The hydraulic gradient is obtained by computing the length of a flow line along the gradient between two points with known heads (if the head at D is also measured, the hydraulic gradient can be easily obtained). If we draw a perpendicular from the point C on to the 150 - 1.36 ¥ 50 equipotential line passing through A, its length will be equal to = 48.58 m. Hence, the 1 + 1.362 hydraulic gradient is

120.203 - 119.678 5.72 ¥ 10 -4 = 0.0108. The hydraulic conductivity is 48.58 0.0108

m/s

= 5.3 cm/s.

7.7.2 Estimation of Recharge Recharge is an important component of many groundwater studies but it is very difficult to measure it directly. Several indirect methods are available for the estimation of groundwater recharge. The most rigorous of these is based on the water balance, but requires the knowledge of several components of the water balance, e.g., interbasin groundwater transfer, evapotranspiration, change of storage, transfer to/from surface water bodies etc. However, several of these components are not accurately known. The simplest of the recharge estimation methods is the water table fluctuation method, in which the rise in the groundwater level Δh, observed in a well during the rainy season is used along with the estimates of specific yield Sy, to estimate the recharge as SyΔh. If reasonably accurate estimates of abstractions AG, from the groundwater during this period and the

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irrigation return flow RI are available, a better estimate of the recharge would be SyΔh + AG – RI. This is a very simplistic model since several components of the water balance are either ignored or are assumed to cancel each other out. There are several empirical methods for estimation of groundwater recharge for Indian conditions. For example, for the Indo-Gangetic plain, the Irrigation Research Institute suggested that the annual recharge, in cm, is equal to 2.15(P − 35)0.5, where P is the annual precipitation in cm.

7.7.3 Estimation of Hydraulic Conductivity If we assume ideal conditions, the method used in Example 7.14 was able to give us a value of the hydraulic gradient as well as the seepage velocity, which was used to obtain the hydraulic conductivity. However, in most practical cases, we would use other more robust methods. If we have the specific cases of one-dimensional flow occurring in a confined or unconfined aquifer, we could use the theoretical hydraulic head distributions discussed in the previous sections to obtain the hydraulic conductivity/transmissivity from a few head measurements. However, these flow situations exist very rarely and the most common method of estimating aquifer parameters is the pumping test, in which a pumping well is installed in the aquifer and a pump test is carried out. By measuring the drawdowns at a few observation wells, we aim to estimate the aquifer properties. For example, if we have a confined aquifer and a well is pumped at a constant rate for a sufficiently long time, measurement of drawdowns in two wells will give us the estimate of the transmissivity using the Thiem equation (Eq. 7.15). Similarly, for unconfined aquifers, if the drawdowns are assumed to be small, the same equation could be used to estimate the hydraulic conductivity as described in the example below. � EXAMPLE 7.15 A well in an unconfined aquifer is pumped at the rate of 120 m3/hr and the drawdown in two wells, located at 50 m and 100 m from the pumping well, were observed to become nearly constant after 4 days of pumping. The initial saturated thickness of the aquifer was 50 m and the steady-state drawdown in the observation wells were 4.234 m and 3.896 m, respectively. Estimate the hydraulic conductivity of the aquifer. Solution Using Eq. (7.28), we get K=

Q ln(r2 /r1 ) 120 ¥ 24 ln 2 = m/d = 20.5 m/d 2 2 2 p h2 - h1 p 46.104 - 45.7662

If we assume that the drawdown is small compared to the thickness and use the confined aquifer equation (Eq. 7.15), we get K=

Q ln(r2 / r1 ) 120 ¥ 24 ln 2 = m/d = 18.8 m/d 2p h0 h2 - h1 2p ¥ 50 46.104 - 45.766

This assumption, however, is more useful where the thickness of the aquifer is not known. In that case, we estimate the transmissivity Kh0, by using the fact that h2 – h1 = s1 – s2.

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The disadvantages of the steady-state tests are that it takes a long time of pumping and it cannot provide an estimate of the storage coefficient. Therefore, transient pumping tests are more common, in which we maintain a record of the temporal variation of the drawdown and estimate the aquifer parameters by analysing it, as described in the next subsection.

7.7.4

Estimation of Transmissivity and Storage Coefficient

We will discuss only the confined aquifer here but some of the methods discussed could be readily extended to analyse the unconfined aquifers also. In essence, we use the Theis equation and look at ways to obtain the values of S and T, if a set of drawdowns s, at some given times t, is known. This is an inverse problem which is not directly solvable because of the complicated nature of the well function W(u). In absence of an analytical technique, we resort to approximate analytical, graphical, or numerical methods, which will be described next. The simplest method to estimate the values of S and T is based on the Cooper-Jacob approximation. As discussed before, for small values of u, the well function may be approximated as r2S Q 2.25Tt W (u) @ - 0.5772 - ln leading to s = ln 2 . In theory, therefore, if the data exactly follows 4Tt 4p T r S this approximation, two values of drawdowns s1 and s2, observed at times t1 and t2, respectively, would be t Q sufficient to estimate the two parameters S and T. For example, s1 - s2 = ln 1 will give the value of 4p T t2 T, and then S could be obtained by applying the drawdown equation at any of the two points. However, generally the data will have observation errors or some of the assumptions made in the analysis will not be strictly valid. Therefore, it is better to use as much of the data as possible. If we plot the drawdown versus time on a semi-log paper, the data should fall on a straight line (since the approximation is valid only for u less than 0.01, some of the early drawdown data will typically not follow the straight line). The slope of this Q straight line, will give us and from there, we can obtain the value of T (note that the semi-log paper is 4p T based on the logarithm with base 10 and a factor has to be included since the equation uses natural log). To find the slope of the line, we find the change in drawdown Δs, in one log cycle of t, and then convert it into one 2.3Q natural log cycle by dividing by 2.303. We can then write T = . Further, the point on the t-axis where 4pDs the straight line intersects, i.e., the point at which the drawdown becomes zero (see Figure 7.24), is denoted by t0, and helps us obtain the storage coefficient. By using the drawdown equation, we see that s will be zero 2.25Tt 2.25Tt0 when = 1 , i.e., S = . If we have drawdown data from more than one observation well, then, 2 r S r2 as discussed earlier, instead of using the time t directly, we will use t/r2. When the Cooper-Jacob approximation is valid, the above analysis is an easy way of obtaining the aquifer parameters. However, depending on the parameter values, to get a value of u less than 0.01, we may need to continue the test for a very long time which may not be economical. Therefore, we need to look at a method that uses the entire data set rather than just the late time data. One such method is the Theis type curve method. In this method we compare two different curves on a log-log plot, one of them is known as the type curve in which we plot the well function W(u), versus 1/u, and the other is a plot of the drawdown s

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versus the time t (or, if more than one observation wells are used, t/r2). The fact that these two curves should be similar can be seen as follows: s=

Q Q W (u) fi ln s = ln + ln W (u) 4p T 4p T u=

r2S t S 1 fi ln 2 = ln + ln 4Tt 4T u r

Figure 7.24 Straight line approximation of the drawdown curve This indicates that the log-log plots of s vs t/r2 and W(u) vs 1/u would be identical, except for an axes shift which will depend on the parameters Q/4pT and S/4T. Therefore, if we plot the s vs t/r2 curve on a transparent paper, and move this curve in such a way that the data lies exactly on the type curve (the scale of both these curves should naturally be the same), the magnitude of the shift in the axes will give us an estimate of the values of S and T (see Figure 7.25, which shows the two curves separately and then the matched curves). For example, if we choose a point A, which has a value of W(u) = W* and the corresponding value of s as s*, QW * the transmissivity is obtained from Eq. (7.21) as T = . Similarly, if we choose a point B, which has 4p s * u = u* and t/r2 = t*, we get, from Eq. (7.19), S = 4Tt*u*. For convenience, point A is usually taken at

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W(u) = 1 and the point B is taken at u = 1, but these could be taken anywhere in the plot area and will result in the same values for T and S. In the figure shown, for W* = 1, we have s = 0.008 m, and for 1 @ 10 m2/min, and u* = 1, we have t/r2 = 0.000024 min/m2. If the pumping rate is 1 m3/min, T = 4p 0.008 S = 4 × 10 × 0.000024 ≈ 0.001. Although the method is straightforward, there is some subjectivity involved in matching the type curve. There are some advanced methods for parameter estimation. For example, we can use computational methods to try different values of S and T and generate the drawdown data, try to match it with the observed drawdown data, and then try to minimize the error by choosing optimum values of S and T. Or we could use slope matching or derivative matching methods which make use of the fact that the derivative of the drawdown with respect to time is of the form 1/(ueu). Computational methods are becoming more popular these days but graphical methods are still very good for a first approximate guess. We will not discuss these other methods here.

10 1

s

0.1 0.01 0.001 0.0001 0.00001 0.00001

0.0001

0.001

0.01 t/r2

0.1

1

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Figure 7.25 Theis type curve matching

� EXAMPLE 7.16 A well in a confined aquifer is pumped at a constant rate of 2 m3/min for 4 hours and the drawdown in an observation well located 50 m away is observed. The following table shows the drawdowns at different times. Obtain the transmissivity and storage coefficient of the aquifer using the Cooper-Jacob method. Time (min)

0.1

0.2

0.3

0.5

1

1.5

2

2.5

3

4

5

6

8

10

s (mm)

69

139

189

258

359

420

465

499

528

573

608

636

682

717

60

80

100

120

150

180

210

240

Time (min)

12

14

18

24

30

40

50

s (mm)

746

770

810

856

891

937

972

1001 1047 1082 1111 1147 1176 1200 1222

Solution A semi-log plot of the data is shown in Figure 7.26 given below: We fit a straight line through the data corresponding to large time and extend it to the time-axis to obtain t0 = 0.11 minutes. The value of Δs, i.e., the change in one log cycle is about 0.36 m. Using these values of Δs and t0, from Eq. (7.23), we get a transmissivity of 1.02 m2/ min and storativity of 1.01 × 10–4. Note that the plot clearly shows the applicability of the Cooper-Jacob assumption, except for a very short time, i.e., less than about 1 minute. Knowing the values of S and T, we obtain the value of u for t = 1 min as about 0.06, indicating that the method could be applied to data having u values smaller than about 0.05.

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1.5

s (m)

1

0.5

0 0.01

0.1

1

10

100

1000

t (min)

Figure 7.26 Cooper-Jacob method for estimating aquifer parameters

� EXAMPLE 7.17 For the aquifer system in the previous example, the drawdown in another observation well located 100 m away is also observed and is shown below. Obtain the transmissivity and storage coefficient of the aquifer using the Theis type curve matching. Time (min)

0.1

0.2

0.3

0.5

1

1.5

2

2.5

3

4

5

6

8

10

s (mm)

54

89

119

144

166

219

258

290

317

359

393

420

465

499

Time (min)

12

14

18

24

30

40

50

60

80

100

120

150

180

210

240

s (mm)

528

552

591

636

671

717

752

781

827

862

891

927

955

980

1001

Solution The data is plotted on log-log scale as shown in Figure 7.27. The 1/u vs W(u) curve has been shown earlier. As long as we ensure that the two plots have the same length for each log cycle, we could superimpose these two graphs and match the data with the type curve (the printed log papers come with same size but it is a little tricky when we do computer plots) keeping the axis parallel. The matched curves are shown in Figure 7.28: For 1/u = 1, the corresponding t* is about 0.000025 min/m2, and for W(u) = 1, s* is about 0.16 m. Using Equations (7.21 and 7.19), we get T = 0.995 m2/min and S = 9.95 × 10–5. These values are very close to the values obtained in the previous example. In this case, we had used synthetically generated data which resulted in nearly exact determination of the aquifer parameters. Also, due to the large diffusivity (T/S) of the aquifer, a four-hour test was sufficient for the Cooper-Jacob approximation to work. In field conditions, the data would not behave so nicely, due to observation errors, non-ideal aquifer behaviour, etc. Both the Cooper-Jacob and the Theis methods would involve some subjectivity in getting the best fit curve.

Engineering Hydrology 10

1

s (m)

280

0.1

0.01

0.001 0.00001

0.0001

0.001

0.01

t/r2 (min/m2)

Figure 7.27 Drawdown data

Figure 7.28 Theis method for estimating aquifer parameters

0.1

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SUMMARY Groundwater hydrology is the study of the occurrence and movement of the water existing underground. Use of groundwater is preferred over surface water due to its easy accessibility, lower cost, and better quality. Groundwater occurs either in saturated conditions, where all soil pores are filled with water, or in unsaturated conditions, where air is also present in the soil voids. The unsaturated zone, existing just below the ground surface, comprises the soil water zone near the surface, a capillary fringe just above the saturated zone, and an intermediate zone between these two zones. Based on the capacity of a soil or rock formation to store and transmit water, it is classified as an aquifer, aquifuge, aquitard, or aquiclude, with aquifers being the primary source of groundwater supply. Water in the aquifers may exist under unconfined conditions, with a water table at the top having atmospheric pressure, or under confined conditions, with water stored under pressure between two confining layers. Movement of groundwater is governed by the Darcy’s law which states that the velocity of flow is proportional to the hydraulic gradient with the constant of proportionality termed as hydraulic conductivity. The continuity equation is used to obtain an equation relating the temporal and spatial variations of the hydraulic head with hydraulic conductivity and specific storage, which is defined as the volume of water added to a unit volume of the aquifer per unit rise in the hydraulic head. This equation could be solved for specified initial and boundary conditions to obtain the hydraulic head distribution in an aquifer. Further simplification of the continuity equation may be obtained by using the hydraulic approach, which assumes that the flow is essentially horizontal and reduces the continuity equation to a twodimensional form. This form of the equation uses the storage coefficient, defined as the volume of water added to a unit base area of the aquifer per unit rise in the hydraulic head. For confined aquifers, instead of the hydraulic conductivity, the product of the conductivity and thickness is commonly used and is called the transmissivity. The storage coefficient for a confined aquifer depends on the water and aquifer compressibilities. Groundwater flow towards a pumping well is analyzed using the continuity equation in cylindrical coordinates. Various steady and unsteady flow cases for confined and unconfined aquifers under one-dimensional and axisymmetric have been analyzed and the resulting head distribution provides a means of estimating the aquifer properties from measurement of groundwater levels in a few wells. For example, in a confined aquifer, the steady one-dimensional flow results in a linear head variation, and steady radial flow towards a pumping well gives rise to a logarithmic variation of head. Similarly, for steady flow in an unconfined aquifer, the square of the head varies linearly in one-dimensional flow and logarithmically in radial flow. For transient flow towards a pumping well, the head variation is represented by an integral which is called the well function. The confined aquifer well function is known as the Theis well function and is readily obtained from charts, tables, or approximate equations. Theis also suggested a method to estimate the aquifer transmissivity and storage coefficient by observing the temporal variation of drawdown in an observation well and matching the plot of observed data with that of the well function. A method for estimation of the direction of flow of groundwater and the aquifer hydraulic conductivity is based on the measurement of water levels in three wells. To ascertain the effect of additional head loss near the well due to turbulent flow and/or well screen, either well-loss coefficient or specific capacity of the well may be used. The values of these parameters indicate the time when a well has become inefficient and needs to be replaced. Flow through a layered porous media is analyzed by defining an effective

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conductivity, which depends on the direction of flow vis-à-vis the layer orientation. For flow along the layers, the effective conductivity is equal to the weighted arithmetic mean of the individual layer conductivities, the weights being equal to the layer thickness. For flow perpendicular to the layers, it is equal to the weighted harmonic mean. For leaky aquifers, which are confined aquifers overlain by an aquitard, the Theis well function gets modified to account for the leakage and is known as the Hantush well function.

OBJECTIVE-TYPE QUESTIONS 7.1 Which of the options given below is correct about the following statements? (i) Of the entire global groundwater, 95% is freshwater. (ii) Groundwater shows a larger seasonal variation compared to surface water. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 7.2 Approximately what percentage of total groundwater used in India is for irrigation? (a) 10% (b) 20% (c) 50% (d) 90% 7.3 In India, the annual groundwater recharge is estimated as about _______ km3. (a) 100 (b) 200 (c) 300 (d) 400 7.4 In India, yearly groundwater consumption is ____ % of the annual groundwater recharge. (a) 10 (b) 30 (c) 60 (d) 80 7.5 Which of the options given below is correct about the following statements? (i) The saturated groundwater zone is also called the vadose zone. (ii) The capillary fringe exists just above the water table. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 7.6 What percentage of the water taken up by plants is transpired back into the atmosphere? (a) 1% (b) 25% (c) 75% (d) 95% 7.7 Which of the options given below is correct about the following statements? (i) The field capacity of sand is larger than that of clay. (ii) The permanent wilting point for sand is smaller than that of clay. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 7.8 Typical value of the residual water content for clayey soils is about (a) 1% (b) 2% (c) 10% 7.9 The sum of specific yield and specific retention is equal to the (a) Permeability (b) Porosity (c) Hydraulic conductivity (d) Transmissivity

(d) 30%

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7.10 Which of the options given below is correct about the following statements? (i) The capillary rise in sand is larger than that in clay. (ii) The water content in the saturated zone is equal to the porosity. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 7.11 The formation, which can neither store water nor transmit it, is known as (a) Aquifer (b) Aquiclude (c) Aquifuge (d) Aquitard 7.12 The water pressure in a confined aquifer is ______ the atmospheric pressure. (a) equal to (b) less than (c) greater than (d) half of 7.13 Which of the options given below is correct about the following statements? (i) The perched aquifer is a confined aquifer. (ii) Leaky aquifers have an aquitard as one of the boundaries. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 7.14 Darcy’s law is valid up to a Reynolds number equal to (a) 0.001 (b) 0.1 (c) 1

(d) 100

7.15 Which of the following statements is false? (a) Seepage velocity is larger than the Darcy Velocity. (b) Hydraulic gradient is always less than 1. (c) Intrinsic permeability does not depend on fluid properties. (d) Groundwater flow is generally laminar. 7.16 Hydraulic conductivity has the same unit as (a) Viscosity (b) Velocity (c) Discharge

(d) Hydraulic gradient

7.17 The unit darcy is equivalent to about (b) 10–12 m2 (a) 10–13 m2

(d) 10–10 m2

(c) 10–11 m2

7.18 The hydraulic approach of analyzing groundwater flow assumes (a) Approximately vertical flow (b) Unit hydraulic gradient (c) Nearly horizontal flow (d) Negligible hydraulic gradient 7.19 Which of the options given below is correct about the following statements? (i) The specific storage is a dimensionless quantity. (ii) The storativity for an unconfined aquifer is equal to the specific yield. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 7.20 For an aquifer with hydraulic conductivity K, and thickness B, the transmissivity is equal to (a) K/B (b) B/K (c) KB (d) K+B 7.21 For a steady one-dimensional flow in a confined aquifer, the hydraulic head is (a) Constant (b) Linear (c) Parabolic (d) Logarithmic 7.22 For a steady radial flow towards a pumping well in a confined aquifer, the hydraulic gradient at a distance r from the well, is proportional to (a) ln r (b) r (c) 1/r (d) 1/ ln r 7.23 The argument u, of the well function is defined as (a)

r 2t 4 ST

(b)

r 2S 2Tt

(c)

r 2S 4Tt

(d)

r2 2t

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7.24 For a pumping well in a confined aquifer, the drawdown in an observation well located at 50 m from the pumping well after 1 hour of pumping is 1 m. At what time will the drawdown in another well, which is 100 m away from the pumping well, be equal to 1 m? (a) 15 minutes (b) 1 hour (c) 2 hours (d) 4 hours 7.25 Steady one-dimensional flow occurs in an unconfined aquifer between lake A and lake B, with the water level in lake A being higher than that in B. There is a uniform recharge over the entire aquifer surface. A water divide will (a) Always form (b) Always be midway between A and B (c) Never form (d) Form if recharge rate is more than the critical value 7.26 For radial flow towards a pumping well in an unconfined aquifer, the drawdown at large times may be obtained using Theis solution, with storativity equal to (a) Porosity (b) Permeability (c) Specific Yield (d) Zero 7.27 Which of the options given below is correct about the following statements? (i) The total drawdown in a well is equal to the sum of laminar and turbulent losses. (ii) The laminar head loss is also known as the formation loss. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 7.28 How does the specific capacity of a well vary with time? (a) Increases (b) Remains Constant (c) Decreases

(d) First decreases then increases

7.29 For flow through a layered soil, the maximum effective conductivity is obtained when the flow is (a) Along the layers (b) Perpendicular to the layers (c) At 45° (d) At 30° 7.30 The minimum number of observation wells required for estimating the direction of groundwater flow is (a) 1 (b) 2 (c) 3 (d) 4 7.31 The Cooper-Jacob method of estimating aquifer parameters is valid for data observed at (a) Small times (b) Large times (c) All times (d) Exactly 1 hour 7.32 The Theis type-curve matching method of estimating parameters of a confined aquifer is valid for a drawdown data observed at (a) A single well (b) Two wells (c) Any number of equidistant wells (d) Any number of wells

DESCRIPTIVE QUESTIONS 7.1 What is meant by the unsaturated zone? Describe the subdivisions of this zone. 7.2 What is the difference between porosity and water content? Define the volumetric and gravitational water contents. 7.3 What is the difference between the specific retention and the residual water content? 7.4 What is meant by field capacity and permanent wilting point? Why is field capacity different from porosity? 7.5 Discuss various types of water storage and transmission formations. Which of these is suitable for use of groundwater as a water resource?

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285

7.6 Describe the confined and unconfined aquifers. Under what conditions will a well tapping a confined aquifer flow by itself, i.e., without any pumping? 7.7 When would you expect upward leakage in a leaky aquifer? 7.8 What are the Dupuit’s assumptions? How are they useful in analyzing groundwater flow? 7.9 Differentiate between storage coefficient and specific storage. What is meant by diffusivity of a confined aquifer? 7.10 For a confined aquifer, which fluid/formation properties affect the storage coefficient? What is the typical value of water compressibility? 7.11 In steady flow towards a pumping well in a confined aquifer, why is the hydraulic gradient inversely proportional to the radial distance from the well? 7.12 Write the integral expression for the well function and obtain the series expression for it using term-by-term integration. 7.13 What is meant by the recovery of a drawdown? How will you obtain a drawdown at any time after the stoppage of pumping? 7.14 Derive an expression for the critical recharge rate which would cause a water divide for a steady onedimensional flow in an unconfined aquifer between two water bodies. 7.15 What is meant by formation loss and well loss? How does the specific capacity of a well change with time and with pumping rate? 7.16 Describe the method of obtaining the groundwater flow direction using the data from three piezometers in an unconfined aquifer. 7.17 Explain the Theis type curve matching technique for obtaining aquifer parameters.

NUMERICAL QUESTIONS 7.1 Water flows through a 2 m long horizontal soil column at a constant velocity. At a section of the tube, a red dye was inserted and it was observed that it travelled a distance of 1 m in 235 seconds and the dispersion was negligible. The soil is sandy with a porosity of 0.42 and hydraulic conductivity of 1 cm/s. What would be the drop in piezometric head across the column length? 7.2 A 50 m thick confined aquifer has a porosity of 0.35. The formation compressibility is 5 × 10−8 Pa–1 and water compressibility is 1 × 10−10 Pa–1. Estimate the storage coefficient and the specific storage of the aquifer. 7.3 Two large lakes are connected by a 200 m long confined aquifer in such a way that one-dimensional flow assumption is valid. The difference in water level of the lakes is 2.5 m and it is estimated that water is being conveyed through the aquifer at a rate of 1 m3/min per meter width. Estimate the transmissivity of the aquifer. 7.4 A fully-screened well is pumping a confined aquifer at a constant rate of 1 m3/s. A prior pump test on the aquifer has provided an estimate of the transmissivity as 0.05 m2/s. Two observation wells are located at a distance of 50 m and 100 m, respectively, from the pumping well. After a long time of pumping, the water levels in the observation wells achieve a nearly constant value. If the piezometric level in the first well (at 50 m) is 120 m above mean sea level, what would be the level in the other well?

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7.5 Two large lakes are connected by a 200 m long confined aquifer (T = 100 m2/d, S = 0.001) in such a way that one-dimensional flow assumption is valid. The lakes have the same water level. Due to a sudden flood, the water level in one of the water bodies rises suddenly by 2 m. What would be the rise in the hydraulic head at the midpoint of the aquifer, 1 hour after the flood? 7.6 A confined aquifer (T = 1000 m2/d, S = 0.001) is being pumped at a rate of 2500 m3/d. What will be the drawdown in an observation well at 100 m from the pumping well after 2 hours of pumping? If the pumping stops at 2 hours, what would be the drawdown after another 2 hours? 7.7 A 200 m long unconfined aquifer connects two lakes, with one lake having water level 20 m above the aquifer bed, and the other, 15 m. At what minimum recharge rate to hydraulic conductivity ratio will a water divide form? If the recharge is doubled from this value, where would the water divide form? 7.8 An unconfined aquifer has an initial saturated thickness of 50 m. It is pumped at a rate of 1 m3/min for a long time and nearly constant drawdowns of 2.5 m and 0.6 m were observed in two observation wells, located at a distance of 50 m and 100 m, respectively, from the pumping well. Estimate the hydraulic conductivity of the aquifer. 7.9 An unconfined aquifer, with initial saturated thickness of 50 m, is pumped for 4 hours at a rate of 60 m3/h. Its hydraulic conductivity is 1 m/h, elastic storage coefficient is 0.0005, and specific yield is 0.3. Estimate the drawdown in a well, located 50 m from the pumping well, after 1 hour and 4 hours of pumping. 7.10 A 10 cm-diameter well in a confined aquifer was pumped at three different rates, 10 m3/h, 25 m3/h, and 50 m3/h, and the drawdown in the well was measured when it was nearly constant. These drawdown values are 0.328 m, 0.902 m, and 2.108 m, respectively. Estimate the specific capacity at these pumping rates and the well loss coefficient. 7.11 A porous medium is made of two horizontal soil layers of equal thickness, with the conductivity of the top layer being 1 cm/s. It was found that for the same discharge intensity (i.e., flow rate per unit area), the head loss for vertical flow was twice the head loss for horizontal flow. Estimate the hydraulic conductivity of the lower layer. 7.12 The aquitard above a confined aquifer (T = 100 m2/h, S = 0.0002) is 2 m thick and has a hydraulic conductivity of 1 mm/h. A pumping well is then pumped at a rate of 100 m3/h. Obtain the drawdown in an observation well, located at 100 m from the pumping well, at 1 hour and 2 hours after the start of pumping. 7.13 Three piezometers are installed in an unconfined aquifer at coordinates (East, North) of A: (50,100), B: (150,50), and C: (100,150). The water table elevations at these wells under steady flow conditions were observed as 46.345 m, 45.178 m, and 44.562 m, respectively. Find the coordinates of another piezometer D, which is to be installed 50 m upgradient of C. 7.14 A confined aquifer is pumped at a constant rate of 100 m3/h for 8 hours and the drawdown in an observation well located 50 m away was observed to be 1.302 m at 7 hours and 1.414 m at 8 hours. Estimate the transmissivity and storage coefficient of the aquifer using the Cooper-Jacob method. Also, check the suitability of the assumption that Cooper-Jacob approximation is valid. 7.15 A well in a confined aquifer is pumped at a constant rate of 30 l/s and the drawdowns in two observation wells, located 50 m and 100 m away, are observed as shown in the table below. Estimate the transmissivity and storage coefficient of the aquifer using the Theis type curve method.

Groundwater

287

Time (min)

0.1

0.2

0.3

0.5

1

1.5

2

2.5

3

4

5

6

8

10

s (mm) at 50 m

22

64

98

155

238

287

328

366

382

432

466

479

517

552

s (mm) at 100 m

0

4

11

32

82

121

153

179

201

240

271

288

333

357

Time (min)

12

14

18

24

30

40

50

60

80

100

120

150

180

210

240

s (mm) at 50 m

582

604

632

682

731

749

796

815

847

879

911

960

980

998

1008

s (mm) at 100 m

386

418

452

476

516

547

599

631

671

707

735

750

780

794

821

1. Groundwater resources of India: The Central Groundwater Board (CGWB) publishes periodic reports about the groundwater status (http://www.cgwb.gov.in/publications.html). 2. Tables of the well function are available at http://pubs.usgs.gov/wsp/wsp1536-E/pdf/wsp_1536-E_b.pdf or http://www.isws.illinois.edu/pubdoc/B/ISWSB-49.pdf

8

Irrigation and Water Resource Management

LEARNING OBJECTIVES LO 1

Understand the crop water requirements

LO 2

Define various terms related to irrigation requirements

LO 3

Discuss several aspects of canal irrigation

LO 4

Outline different types of irrigation methods

LO 5

Summarize single-purpose and multi-purpose water resource projects

8.1

INTRODUCTION

The topics discussed in this book till now mostly relate to the available water supply in terms of streamflow and precipitation. We looked at the natural demands in the form of abstractions, like interception, depression storage, evapotranspiration, and infiltration, but the primary purpose was to obtain the net available water. Once we know the amount of water available to us, we have to decide how best to utilize it, since there would be several, often conflicting, demands. In this chapter, we look at one of the major demands of water – irrigation, and then we will discuss how to supply the required water to the crops. The other demands like municipal or industrial are not described in detail, but they are equally or, in some areas, more important. Since it would generally be more cost-effective to satisfy several types of demands through the same project, we discuss the single-purpose and multipurpose projects, with a view to compare their relative merits and demerits and manage the water resources in an optimal manner.

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289

WATER REQUIREMENT OF CROPS

Water is one of the major requirements for the growth of any crop. To LO 1 Understand the crop achieve the maximum possible production with the minimum amount of water requirements water supplied to the crop, its schedule of application has to be carefully designed. How much water is needed and when, will depend on several LO 2 Define various terms factors including the type of crop, its stage of growth, and the climate related to irrigation requireof the area. For example, rice will need more water than wheat, a fullments grown crop may need more water than a just-planted crop, and crops in hot areas will require more water than those in colder areas. Since the water requirement of a crop is directly related to the evapotranspiration, the factors affecting evapotranspiration will have a significant effect on the crop water requirement. Therefore, water requirement will be more for low humidity, high wind, and less cloudy areas. Several methods for estimating the reference evapotranspiration, and the actual evapotranspiration by using a crop factor K, have been described in Chapter 3. The reference evapotranspiration, as already defined in Chapter 3, is the evapotranspiration rate from an extensive surface of green, well-watered grass of uniform height, actively growing and completely shading the ground which has a moderately dry soil surface. Since it already accounts for the climatic factors, the crop factor should only depend on the crop type and its stage of growth. However, a slight dependence on wind-speed and relative humidity has been observed, with about 5% increase in low-humidity and high-wind areas and a similar decrease for high-humidity and low-wind areas. Tables of crop factors are readily available (Brouwer and Heibloem, 1986) and some values are shown in Table 8.1. The four stages of growth of a crop are defined as: Initial stage (sowing to about 10% areal coverage), Development stage (from initial stage to nearly full areal coverage), Flowering or mid-season stage (from development stage to maturity), and Ripening or late-season stage (from maturity to harvesting).

Table 8.1 Crop factor for different stages of growth Crop Stage

Initial

Wheat

Potato

Tomato

Period (Days)

Crop Factor

Period (Days)

Crop Factor

Period (Days)

Crop Factor

15

0.35

25-30

0.45

30-35

0.45

Development

25-30

0.75

30-35

0.75

40-45

0.75

Mid-season

50-65

1.15

30-50

1.15

40-70

1.15

30-40

0.45

20-30

0.85

25-30

0.80

120-150

-

105-145

-

135-180

-

Late-season Total

For rice, the crop factor is taken as 1.1 for the initial two months after sowing and 1.0 for the last one month before harvesting. In the middle period, the crop factor may vary from 1.05 to 1.35, depending on the climatic conditions. Although the crop factor method provides a handy way of estimating the seasonal water requirements of the crops, it requires the reference evapotranspiration to be estimated for the different periods of crop growth. Moreover, for designing an irrigation system, it is sufficient to know only the total water requirement and

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the rate at which it needs to be supplied during the most critical period. We describe here the methodology commonly used in India, by first defining various terms and then describing the computations. •

Crop Period: It is the total duration from sowing to harvesting (corresponds to “Total” in the table above).

•

Base Period: It is the duration between the first and the last watering (the first watering is usually just before sowing, and is called paleo). Since it is generally not very different from the crop period, it is common to take the base period equal to the crop period.

•

Kor Period: In this period, the most intense watering is required (usually when the crop is about 2-3 cm high).

•

Delta: It defines the total water, in terms of depth (i.e., volume per unit area), required by crop over the base period.

•

Kor Depth: It defines the water required, in terms of depth, by crop during the kor period. Generally, the effective precipitation, i.e., the part of precipitation which contributes to the soil moisture, is subtracted from the total water requirement during the kor period to obtain the kor depth. (Note Earlier we had used the term effective precipitation for the part which contributes to runoff. Depending on the purpose for which it is used, the term ‘effective’ has different meanings.)

•

Command Area (or Gross Command Area): Command area of a project is defined as the area which could be benefitted from the project. For example, for an irrigation canal, all areas which could be supplied water from the canal are included in the command area.

•

Culturable Command Area (CCA): It is the portion of the command area which is fit for cultivation. The classification of irrigation schemes is done on the basis of CCA with major schemes having CCA of more than 10,000 ha, minor schemes having CCA less than 2,000 ha, and medium schemes in between these two.

•

Intensity of Irrigation: It defines the percentage of culturable command area which is proposed to be irrigated.

The water demand for irrigation will depend on the command area, the intensity of irrigation, and the type of crop. The three crop seasons generally demarcated in North India, are Kharif, Rabi, and Zayad (or Zaid), which are Arabic words. Kharif means autumn and these crops (e.g., maize, rice, peas, cotton, tobacco) are cultivated in the monsoon season, June to October. Rabi means spring, and these crops (e.g., wheat, barley, gram, peas, potatoes, tobacco) are cultivated in the winter season, October to March. Zayad means extra and these crops (e.g., watermelon, cucumber, pumpkin) are cultivated in the summer period between Rabi and Kharif, i.e., March to June. Some crops are cultivated in multiple seasons. In other parts of the country, there are different crop seasons in different regions, depending on the climatic and socio-economic factors. The same field could be used for multiple crops in different crop seasons or even in the same season. The irrigation water requirement is generally expressed in terms of the duty of canal water, D, which is the area, in hectares, which could be irrigated by a constant supply of 1 m3/s throughout the base period. In addition to the crop-related factors, the duty depends on several other factors like the type of soil (more pervious soils having less duty due to seepage from the canal and field), temperature (high temperature leads to smaller duty due to higher evapotranspiration), rainfall (higher rainfall meaning higher

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duty due to reduction in irrigation water requirement), and water-tax (no tax may lead to lower duty due to likely wastage of water). If we do not consider all these factors and denote the water requirement (delta) of the crop by Δ (in cm), and use B to denote the base period, in days, we could derive a relationship between the duty and delta by equating the volume of supply and demand as follows: D (hectare) ¥

10000 m 2 1m 86400 s ¥ D (cm) ¥ = 1 m 3 /s ¥ B (days) ¥ 1 hectare 100 cm 1 day D=

864 B D

(8.1)

Table 8.2 lists approximate values of some crop parameters, keeping in mind that there is a significant variation from one type of climate to the other.

Table 8.2 Water requirement of some crops Crop

Base Period (days)

Delta (cm)

Kor Period (days)

Kor depth (cm)

Wheat

120

40

28

12

Rice

120

125

20

30

Potato

105

75

25

25

Sugarcane

320

200

40

35

The duty of canal water for wheat is about 1800 ha/cumec while that of rice is only about half of it. This means that a continuous supply of 1 m3/s in the canal will be able to irrigate 1800 hectares of wheat but only 900 hectares of rice. � EXAMPLE 8.1 Wheat is sown in an area in the Indo-Gangetic plain in the beginning of December and is harvested after four months, in the beginning of April. The reference evapotranspiration for these months are estimated as: December – 2.3 mm/d, January – 2.5 mm/d, February – 3.0 mm/d, and March – 5.9 mm/d. Estimate the total water requirement, using the crop-factor method, and compare with the delta from Table 8.2. Also compute the duty of canal water. Solution Since the total crop period is 120 days, we will use the lower values in Table 8.1. The crop factors for wheat are, therefore, 0.35 for the first 15 days, 0.75 for the next 25 days, 1.15 for the next 50 days, and 0.45 for the last 30 days. Since the reference evapotranspiration is given month-wise, we will use weighted averages to compute the reference evapotranspiration for different growth stages. The following table shows the computations (note that an additional day has been added to the total period to get 31 days in March. It would not make a significant difference, if we use 30 days instead. All months could have been assumed to be of 30 days to simplify the calculation):

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292

Stage

Period (days)

Crop Factor

Month-wise number of days

Reference Evapotranspiration (mm)

Crop water requirement (mm)

Initial

15

0.35

15(Dec)

2.3 x 15 = 34.5

12.08

Development

25

0.75

16(Dec)+ 9(Jan)

2.3 x 16 + 2.5 x 9 = 59.3

44.48

Mid-season

50

1.15

22(Jan)+ 28(Feb)

2.5 x 22 + 3 x 28 = 139

159.85

Late-season

31

0.45

30(Mar)

5.9 x 31 = 182.9

82.31

Total

121

298.7

The total water requirement comes out to be about 30 cm. Table 8.2 shows a requirement of 40 cm on an average, but it should be noted that it will vary from place to place due to climatic factors. Using the base period of 121 days, delta of 29.87 cm, the duty of canal water is obtained from Eq. (8.1) as 864 × 121 / 29.87 = 3500 ha (per m3/s). Note that this assumes no losses and all the canal water is available to the crop. The actual duty would be smaller.

8.3

CANAL IRRIGATION

Once the water requirement of the command area is known, the canal LO 3 Discuss several aspects discharge may be estimated, either by the crop factor method, or by the of canal irrigation duty-delta method. Since the maximum discharge will be needed during the kor period, the canal design discharge is obtained considering the kor depth. However, other factors, like canal losses, precipitation, and irrigation efficiency, will also affect the design discharge. It should also be kept in mind that both the supply (river water) and the demand (irrigation requirement) are temporally variable, and the canal will not run at its design capacity for most of the time. The following example illustrates the procedure of determining the design discharge of an irrigation canal. � EXAMPLE 8.2 The command area of a canal is 5000 ha out of which 3000 ha is culturable. There are two crops grown in the area: wheat from November to February, and rice from July to October. The intensity of irrigation is 60% for wheat and 40% for rice. Using the data given in Table 8.2, obtain the design discharge of the canal. Solution The area irrigated for wheat is 0.6 × 3000 ha = 1800 ha, and that for rice is 0.4 × 3000 ha = 1200 ha. The design discharge will be based on the kor period and is obtained by dividing the irrigated area by the duty of canal water. Denoting the area (in ha) by A, the kor depth by Dk, and the kor period by Bk, the discharge should be AΔk/864Bk, which comes out to be 0.89 m3/s for wheat and 2.08 m3/s for rice. Since there is no overlap between the wheat and rice seasons, we will use the larger of these two values, i.e., 2.08 m3/s, as the canal design discharge. In case of overlapping seasons, we will have to look at the temporal variation of the demands for all crops and choose the design discharge based on the most critical period, considering all demands simultaneously.

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To carry water from the source (generally a river or reservoir) to the field where irrigation is required, an irrigation system has to be built. This system comprises an intake structure (or pumping station, if the water level in the source is lower than that required at the intake), a conveyance system (e.g., a main canal, to carry water from the source to the vicinity of the field and a branch canal or distributary, to carry water from the main canal to the field), a distribution system (e.g., a watercourse, which takes water from the distributary and carries it to various parts of the field which are to be irrigated), and an application system (e.g., furrows, which are used to apply this water to the plants). If needed, a drainage system is used to remove any excess water from the fields. Depending on the layout of the irrigation system and the irrigation requirement, some of these components may not be needed. For example, a watercourse may not be used and direct supply to the field through a branch canal may be practiced sometimes. Direct irrigation from a main canal is generally not done. In India, the branch canals and distributaries are classified according to their design discharge. The branch canal has a discharge greater than 5 m3/s and the distributaries, less than 5 m3/s. The distributaries are further sub-classified as major or minor, depending on whether the discharge is more than or less than 0.25 m3/s. Sometimes, an irrigation canal is taken out of a river without having any control structure to regulate the supply. These canals are known as inundation canals, and operate only when the flow in the river is larger than a threshold value, mostly during rainy season. The utility of these canals for irrigation is not very high, since the requirement for irrigation is generally small during rainy season. However, it does provide some utilization of the flood waters, which would otherwise go waste. We will not discuss inundation canals further in this chapter. For the controlled (or perennial) canals, a canal headwork is used to regulate the entry of river water into the main canal and an outlet is used to regulate water entry from the distributary to the watercourse. The canal headwork (or diversion headwork) comprises an obstruction across the river (a dam, barrage or weir) to raise the water level, a divide wall to create a still water pond near the head regulator at the canal offtake, scouring sluices to periodically remove the silt deposited near the canal intake, a fish ladder to create a passage of sufficiently low velocity so that fish could travel upstream against the flow, and a silt excluder/ ejector to either prevent sediments from going into the canal or remove these from the canal. The outlet from the canal to the watercourse should be strong and tamper-proof and should withdraw a proportional amount of sediment from the canal (if it withdraws less, the canal will have higher silt concentration downstream of the outlets and deposition may occur; if it withdraws more, the canal will have lower silt concentration and may lead to erosion). To reduce cost by requiring a lower water level in the canal, the outlet should be able to work under small water heads also. Once the canal takes-off from the river, it would be advantageous to align it along a ridge so that the areas on both sides could be irrigated by gravity flow alone, without the need for pumping. However, since the rivers flow through the valleys, we could either pump water to a canal intake on the nearby ridge, or we could have the intake in the valley and bring the canal to the ridge at a suitable downstream point, such that the ridge elevation is lower than the intake elevation. In the first case, we need a pumping station at the diversion headwork; and in the second case, the initial portion of the canal will not be along a ridge and will not be able to irrigate the area around it by gravity flow. Sometimes, aligning the canal along a ridge may not be cost-effective, e.g., when the ridge follows a very zigzag path. In such cases, a substantial saving in canal length may be obtained by short-circuiting the alignment. Sharp curves in the alignment are generally

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avoided since these may lead to excessive turbulence causing scouring on the outer bank and may also cause deposition of sediment due to smaller velocities on the inner bank. Irrigation canals may be constructed in natural earth but it may lead to loss of water through seepage from the boundary, bank failure due to erosion, or increased friction due to growth of plants and weeds on the canal bed. Therefore, earthen canals may be lined with concrete, brick (or rock) masonry or asphalt. Although the initial cost of lining is high, it justifies this cost by the reduction in seepage and smaller maintenance cost. Since the canal bed generally follows nearly the same longitudinal profile as the land, in steeply sloping lands, the velocity of flow could be so high as to cause excessive erosion in earthen canals or lining damage in lined canals. Canal drops are used in such cases to create a flatter bed slope in the canal, punctuated with a few vertical drops or steep chutes in the bed. Sometimes, the canal needs to cross roads, hillocks, rivers, etc. The structures built for this purpose are called crossing structures (the crossing structures at the canal-river crossing are called cross-drainage works). The crossing structures may be in form of a flume (open canal supported on pillars), culverts (a pipeline buried under a road) or inverted siphons (similar to a culvert but for roads which are at a level below the canal bottom). A drainage system may be needed in case there is excess water in the irrigated field due to rainfall, irrigation, or seepage from the distributary or water course. Shallow open drains are used to drain the excess surface water and deep open drains or underground pipes are used to remove extra groundwater. For ascertaining the effectiveness of the entire canal irrigation system, we should look at all the costs and the benefits. However, there are several factors which may complicate the analysis. For example, the seepage loss from the canal is a loss from the irrigation perspective, but it does recharge the aquifers leading to some benefits also. The canal lining, therefore, may not be as beneficial as it appears at first glance. On the other hand, if the area near the canal suffers from waterlogging due to a high groundwater level, the benefits of lining would be enhanced. Thus, the advantages of canal irrigation system, viz., the reduction in dependency on rainfall, economic development due to higher productivity, etc., and the disadvantages, viz., waterlogging and salinity, adverse environmental impacts, etc., should be carefully evaluated, before deciding on the suitability of the system.

8.4

IRRIGATION METHODS

After the water from the watercourse reaches the field, several possible LO 4 Outline different types methods of irrigation could be used to apply this water to the crop. The of irrigation methods goal of irrigation is to maintain the soil in the root zone at an optimum water content which results in a maximum yield for the given crop. Immediately after application of a sufficient quantity of water, the soil gets saturated and the moisture content is equal to the porosity. Due to gravity drainage, a portion of this water drains down below the root-zone and the soil moisture is at field capacity. The moisture content decreases with time and, if allowed to deplete without further irrigation, may go below the wilting point, at which the crop suffers a permanent damage. For minimizing the frequency of irrigation, therefore, subsequent irrigation should be done when the soil moisture reaches just above the wilting point. The other extreme would be to have very frequent irrigation, which brings the water content to just above the optimum value, allow it to deplete to just below the optimum

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value, and then irrigate again. To avoid undue stress to the plants, and to maintain the optimum water content, generally water is applied much before the wilting point. Similarly, since the soil is not able to hold water above the field capacity for a long time, the irrigation aims at increasing the moisture content to field capacity only and not to saturation. Thus, the irrigation scheduling is done in such a way that, on application of water, the soil moisture increases to the field capacity and then it is allowed to deplete to a value below the optimum water content before the next irrigation, in such a way that the average moisture content is close to optimum (Figure 8.1). The optimum moisture content varies from crop to crop and may also show a variation with time. Generally it is in the range of 50%-80% of the available water, i.e., the difference of field capacity and wilting point.

Figure 8.1

Frequency of irrigation

� EXAMPLE 8.3 A particular crop has a root-zone depth of 40 cm and water requirement of 2.5 mm/d, and is planted in a sandy loam with field capacity of 25% and permanent wilting point of 8%. The optimum soil moisture content for the crop is 20%. Assuming an irrigation efficiency of 80%, and irrigation at a moisture content of 15%, estimate the frequency of irrigation and the water depth needed at each irrigation. Solution Just after irrigation, the moisture content in the root zone depth of 40 cm will be at the field capacity of 25% and next irrigation is done when it reaches 15%. Hence the depth of irrigation required is (0.25 − 0.15) × 40 cm = 40 mm. With 80% efficiency, therefore, we will need to supply 50 mm of irrigation depth at each watering. Since the consumptive use by the crop is 2.5 mm/d, it would deplete the 40 mm of available water in the root zone in 16 days. Thus, the frequency of irrigation is 16 days and the depth required for each irrigation is 50 mm. The simplest method of irrigation is watering of plants using buckets or watering cans. But it is timeconsuming and labour-intensive and is only used for small plots, like vegetable gardens. To apply the irrigation water to the field, we could use surface methods, aerial methods, or subsurface methods. The surface irrigation uses gravity flow to supply water to the crop and is practiced by either flooding the entire plot (basin irrigation), putting water in small channels (furrow irrigation), using channels along the border of small subplots (border irrigation), or making rings around individual plants (ring irrigation). The aerial method uses water pressure to create a spray and sprinkles it on the field (sprinkler irrigation). In the subsurface method, water is applied directly to the root-zone of the plants; rather than applying it on the surface, as in the other methods, and then allowing it to seep to the root-zone. The drip irrigation method could be used with either

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surface application or subsurface application to have water drip directly to the root zone. These irrigation methods are described below.

8.4.1 Basin Irrigation Basin irrigation uses a channel to flood the plot with water and the entire land surface is covered by a sheet of water. The field does not need to be leveled and water movement is governed by the topography of the field. This leads to a large water loss since most of the applied water seeps down below the root-zone without being utilized by the plants. Moreover, some of the water remaining on the surface runs-off and some of it evaporates. The efficiency, i.e., the ratio of the water used by plants and the total water applied, is generally about 20%. Therefore, it is used in areas where water is available in abundance and at low cost. Other drawbacks are that the erosion of soil may be high due to larger velocities generated during flooding, and the depth of water is not uniform across the field. The positive aspect of this method is that it is very easy to use, requires very little initial investment and maintenance, and does not need much preparation of the land. If the land is uneven, this method ensures that water reaches all parts of the field. Basin irrigation should not be used for crops which are adversely affected by waterlogging.

8.4.2 Furrow Irrigation Furrows are narrow ditches between two rows of plants or, sometimes, with two rows of plants on either side. The crop is planted on the ridges between the furrows. In furrow irrigation, water is allowed to flow through the furrows and moves to the root zone by capillary action. If the area experiences water scarcity, alternate furrow irrigation may be practiced, in which water is put into every other furrow in one round, with the remaining furrows kept dry. Then, in the next round, these dry furrows are irrigated and water is not put in the previously irrigated furrows. The length of furrow is generally around 5 m, with smaller lengths used for permeable soils, to allow water to reach the end of the furrow. Crops planted in well-defined rows are suitable for this method of irrigation. Also, for sloping land, making furrows along contours provides an efficient way of reducing erosion and run off. The water-use efficiency is usually high for furrow irrigation. Installation and maintenance costs are low but a uniform slope in the furrow requires accurate land leveling, increasing the cost. Moreover, more skilled labour is needed and use of farm machinery on furrowed land is not very convenient. A type of furrow irrigation, known as corrugation irrigation, is also used sometimes for irregularly planted crops, in which small undulations are created on the soil surface. However, it has not been used widely.

8.4.3 Border Irrigation The border method of irrigation requires the field to be leveled and sub-divided into smaller plots bordered by embankments, about 20 cm high. Small channels are provided along the borders, through which water is supplied to the plots. The slope of these channels should be small enough to prevent erosion and large enough to provide sufficient flow, and is generally kept from 1 in 2000 to 1 in 50. The length of the plot is about 50 m (but may be as high as 500 m) with smaller lengths used for more permeable soils, and the width should be such that water flows uniformly across the entire width (generally 5 m to 20 m). The initial cost of border irrigation is high but the maintenance cost is low. It is suitable for mechanized operations since the entire sub-plot is free of obstructions.

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8.4.4 Ring Irrigation Ring irrigation is used for high value crops where individual irrigation for each tree, or for a small group of trees, may be desirable. A small basin is formed around the tree stem by providing an embankment about 20 cm high at a distance of about 50 cm. The basin may be rectangular but is generally circular, in the form of a ring, hence the name. The area enclosed by the ring is leveled except the area with the tree-base, which is slightly raised. Water may be supplied to each ring separately or furrows may be used to connect two rings. The initial cost is high but there is huge saving of water as there is very little runoff. Also, leveling is needed only within the ring and the entire area does not need to be leveled. However, it obviously rules out mechanical farming.

8.4.5 Sprinkler Irrigation Sprinkler irrigation uses pipes in which water is carried under pressure and a sprinkler at the end of a pipe creates a spray which falls on the surface, just as rain does. This method works very well in permeable, shallow, erodible, uneven, or steep lands, since the application rate is moderate and there is little possibility of high velocity sheet flow. However, it does require a pump and power supply, making the running cost higher. The main pipeline carrying water may be above the ground or it may be buried, with the sprinkler on the surface. This method results in a uniform water distribution, less obstruction to farm machinery (especially if buried pipes and hidden sprinklers are used), and easier control on the amount of water supplied. However, it has a high initial and recurring maintenance cost, and requires water to be free of silt which may clog the sprinkler nozzle. If the water quality is not good, a filter may be needed to remove the fine particles.

8.4.6 Subsurface Irrigation Subsurface irrigation (or sub-irrigation) applies water directly to the root zone of the plants. If an impervious stratum is present below the root zone, ditches could be dug up to that level and filled with water, which will then move horizontally or upward through capillary action, to supply to the root zone. Otherwise, perforated pipes are put inside the ground and water is put through these to allow seepage from the perforations into the root zone. This method has a high initial cost, but very low maintenance cost and is very efficient due to negligible evaporation. Also, it is conducive to more efficient working of farm equipment since there is no surface obstruction. Crops with shallow root system are particularly suitable for this method.

8.4.7

Drip Irrigation

In the drip irrigation system (also called trickle irrigation), water is carried to the fields through a pipe system under pressure. It then drips on the soil near the plants through outlets, called drippers or emitters, at low rate (about 10 l/h) to wet the root zone of the plant. Compared to flooding or sprinkler irrigation, which wets the entire soil surface, drip irrigation is more efficient as it wets only the area in proximity of individual plants. Therefore, it is most suitable for areas which experience water scarcity. Since the application rate is small and frequency is large, this method maintains high average moisture content in the root zone leading to higher productivity. The initial cost, however, is very high, and the method is usually used for high-value crops. Since the drippers are small in size (about 1 mm diameter), hence they may get clogged if the water contains sediments or precipitable chemicals. Filtration of the irrigation water could be used to prevent the clogging. If the water is saline, drip irrigation is better than other methods, since it applies a very small quantity of water thereby reducing the chances of land salinity.

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The irrigation method to be used for a particular application is decided by a combination of several factors, including the land characteristics, climate, quality of available water, type of crop, and availability of skilled labour and technology. Some of these are discussed here briefly. In areas subjected to strong winds, sprinkler irrigation may not be very efficient since the drop size is small and it is easily carried away. For steeply sloping areas, sprinkler or drip irrigation would be preferable, as it will reduce the potential for runoff. Similarly, for widely undulating areas, methods which do not require precise leveling, e.g., sprinkler irrigation, would be preferred. Since sand has a large permeability and smaller porosity compared to clay, it is better suited for small and frequent water applications. Sprinkler irrigation will, therefore, be better than flooding in sandy soils. If there is a large variation in soil type in an area, sprinkler or drip irrigation leads to a comparatively even distribution of applied water. If water availability is limited, sprinkler or drip irrigation, which uses less quantity of water, is preferred. If the available water carries a lot of sediments, sprinkler and drip irrigation is not suggested without the use of a filter, due to the possibility of clogging of the holes. If there is salt present in water, drip irrigation is better since it applies water at a low rate, reducing the chances of salinity of the land. Capital-intensive methods, like sprinkler or drip irrigation, are cost-effective for high-value crops only and may not be used with other crops. Furrow irrigation is suitable for row-crops while drip irrigation or ring irrigation is useful for individual plants. If skilled workers or advanced technology is not available for maintenance of the system, the sprinkler and drip irrigation systems should be avoided, since it may lead to a large down-time in case a problem develops in the system. Flooding, on the other hand, does not need skilled workers or technology for initial set-up and maintenance, and could be readily used in such cases. If the amount of water applied is small, e.g., in sandy soils, furrow irrigation would be appropriate, but if a large amount is to be applied, e.g., in clay, basin or border irrigation would be better.

8.5

SINGLE-PURPOSE AND MULTIPURPOSE PROJECTS

We have discussed the irrigation water requirements in detail. However, LO 5 Summarize singlethere are several other purposes for which water from a project may be purpose and multi-purpose used. Generally, it is more efficient to use the water for several purposes. water resource projects For example, a water-storage reservoir may be used for recreational purposes by providing swimming and boating facilities. However, several of these purposes may have conflicting requirements. For example, a reservoir for hydropower generation should be filled to the maximum possible height at all times, while a flood-control reservoir should be at its lowest possible level before the expected arrival of floods. Here we describe several purposes for which a water resource project may be used, along with the requirements for these purposes. ∑

Irrigation: Details about the benefits of irrigation and the methods have been provided earlier in this chapter. Generally when a dam is constructed, irrigation canals take off from the reservoir thus formed. To understand the importance of irrigation in water resources projects, it should be mentioned that about one-third of the irrigated land, contributing about one-sixth of the food production across the world, depends on dams.

∑

Hydroelectric Power: Electric power is generated by utilizing the potential energy of water through a hydropower plant. In addition to being a renewable source of energy, hydropower is also costeffective, with the generation cost of about Rs. 4 ($0.06) per kWh.

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∑

Flood Control: Moderation of the flood flows is achieved by temporarily storing the flood water and then releasing it in such a way as to minimize flood damage in the downstream areas.

∑

Water Supply: To provide water for domestic, industrial, and other uses in desired quantity and of desired quality. The required quality depends on the purpose for which water is being supplied and water treatment plants may have to be provided for municipal water supply.

∑

Navigation: Canals may be used for transportation of passengers and goods. Inland navigation is a very cost effective and environment-friendly method of transportation.

∑

Recreational: Although it is not a primary purpose of a project, recreational facilities for the public, e.g., swimming, boating, fishing, water sports, and walkways, add to the value of a water resources project.

∑

Fish and Wildlife Conservation: The construction of a dam generally leads to a disruption in the wildlife habitat. To minimize the impact on the habitat and prevent the loss of fish, suitable measures, e.g., providing a facility for easy migration of fish, have to be adopted.

Some other purposes of a water resource project, which are not very common, may be: prevention of waterlogging, soil conservation, sediment control, runoff reduction, land improvement, pollution control, artificial precipitation, etc. Based on the purpose(s) served, a water resource project is classified as either a single purpose project, which is meant to serve only one basic purpose or a multipurpose project, which is designed to serve two or more purposes. Generally, a project which is designed for a single purpose, but which may also serve another purpose, is not classified as a multipurpose project. For example, a reservoir meant primarily for water supply may sometimes be useful for flood mitigation. Multipurpose projects are more common since they provide a more efficient use of the available water resources. The additional cost of satisfying a second purpose is typically less than the additional benefits gained from it. Most of the large multipurpose dams combine hydropower generation with irrigation or water supply. In flood-prone areas, flood-control is included as a secondary purpose. Worldwide, however, only about one-third of the large dams are multipurpose. Most dam projects serve the single purpose of irrigation, with hydropower generation, water supply and flood control following in that order. The reason for this reluctance to develop multipurpose water resources projects is the difficulty in planning and operation to satisfy several conflicting demands. The reservoir operation is much simpler if the various purposes for which it is used are compatible with one another. This compatibility, in turn, depends on the spatial and temporal variations in water use requirements for different purposes, characteristics of the river discharge, and the ability to forecast the supply and demand variations. The availability of water, i.e., the runoff, has already been discussed in Chapters 4-6. Here we briefly discuss the water requirements for various purposes which a multipurpose project should account for in its design and operation. � Irrigation The water requirement for irrigation is seasonal and depends on the cropping pattern. In India, the maximum demand of water for irrigation is during the winter months (Rabi crops) and a small demand during the summer months (Kharif crops) before the monsoon. Obviously, the irrigation demand will be more if the precipitation is less. However, on an average, the demand per unit area remains more or less similar. A small portion of the irrigation water comes back to the surface water sources as return flow. For satisfying the irrigation demand, as much storage as possible should be reserved for it. Typically, the river-

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flow, after satisfying the downstream flow requirement and other water-supply requirements, is stored during the monsoon season, June to October, and water is supplied from storage during November to May. � Hydropower The power demand also shows a seasonal variation. However, since the power plants are normally connected to a power-grid, the seasonal variations in demand do not affect the operation of the power plant. This allows for flexibility in coordinating the hydropower requirement with other water uses. Another feature of hydropower is that the water is not consumed during the generation, only its potential energy is utilized. This means that the same water could be used for irrigation or water supply. Some hydropower plants are run-off-the-river, where there is no storage capacity and power production is dependent on the river-flow at any time. Some others are pumped-storage plants which use the less costly off-peak power to pump water back to the storage reservoir and then generate power using this stored water during times of peak demand. � Flood Control For adequate control of flood, there should be sufficient storage space available in the reservoir when the flood arrives. Therefore, as mentioned earlier, the flood-control purpose is not compatible with other purposes. This requirement is also seasonal, with maximum storage requirement during the rainy season. Also, generally the required storage is not based on the largest expected flood but on a carefully selected smaller magnitude flood. This results in lower project cost and is also able to moderate the effect of the large floods to a reasonable level. In order to use the reservoir for flood control as well as conservation, we could either reserve a specific reservoir volume at the top for flood control and the storage below it for conservation, or we could share the entire available storage for both purposes. Naturally, with shared storage, the flood storage capacity varies with time and it would be useful in such cases to have an early flood warning system in the catchment to be able to release water and create storage space for floods. � Water Supply Municipal water supply requirements do not vary much through the year, generally being a little higher in summer. However, a temporal growth due to increase in population or industry has to be accounted for. � Navigation For proper navigation, a minimum water depth is required in the river. There is a considerable seasonal variation in the demand, since the river flow and therefore the flow depth varies widely during the year. The required release of water from the reservoir is generally high during the summers. The demand also varies with the type and volume of traffic. � Recreation For recreation purposes, the reservoir should be nearly full during the period in which there is maximum demand for recreational activities like swimming and boating. There should not be abrupt changes in the water level, which may disrupt these activities. Generally, no water resources project has recreation as its main purpose, and the reservoirs meant to serve other purposes have recreation as a secondary purpose. � Fish and Wildlife For wildlife protection, there should not be sudden fluctuations in water level, the flow of water should not be completely stopped at any time, and fish ladders (or some other arrangement) should be provided to permit fish to migrate both upstream and downstream. Other considerations, e.g., environmental flows, may require adequate flow downstream of the dam and are generally compatible with other uses. Similarly, to avoid mosquito growth, typically prevalent in static water, deliberate fluctuations in the reservoir level may be caused.

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It is clear from the above description that a multipurpose project may have purposes with incompatible requirements, leading to a complexity in the operation of the project. The conflicting requirements could be of space, time, or of amount of release. For example, water conservation and flood control have requirements of full space and empty space, respectively; irrigation and power requirements both may peak at the same time in summers; the amount of release needed for hydropower generation may not match the water supply requirement. The operation of a multipurpose reservoir is, therefore, quite a complex task and is generally done according to a set of rules depending on the purpose and the priorities accorded to these. A discussion of these operating rules is, however, beyond the scope of this book. Interested readers should see Wurbs (2005) or Vogel et al., (2007).

SUMMARY We started this chapter by mentioning that we had covered the supply side of water resources, but had not discussed the demand side. Irrigation is one of the major demands of the water resources. The water needs of a crop depend on the crop type, its stage of growth, and climatic factors. The climatic factors are taken care of in the reference evapotranspiration, and the crop requirements are included in the crop factor. In North India, the three distinct crop seasons are Rabi, Kharif, and Zayad. The crop water requirements are typically specified in terms of a water depth Delta, required over the crop period. The area which could be irrigated by a unit discharge from an irrigation canal is termed as the duty of canal water. Command area refers to the area which could be irrigated by a canal, and culturable command area is the portion of it which is fit for cultivation. The intensity of irrigation is the percentage of the culturable area which is under irrigation. The design discharge of the canal depends on the period of most intense irrigation requirement, called the kor period. To carry water from a river to the field, a system of canals is used, with the canal taking off from the river known as the main canal, which is normally not used for direct irrigation. The smaller canals which carry water from the main canal to the field are known as branches/distributaries, and the water is ultimately delivered to the field through watercourses. Water is then applied to the crops either on the surface or through subsurface, with the aim of maintaining an optimal moisture content in the root zone. Surface methods include basin, furrow, border, and ring irrigation or may utilize sprinklers or drippers. The subsurface methods use either drippers or perforated pipes/tiles. The choice of the method of irrigation to be used depends on several factors related to land, climate, crop, water, labour, and technology. Finally, we looked at the various purposes which could be served by a water resources project. These include irrigation, hydroelectric power generation, water supply, navigation, recreation, wildlife conservation, soil conservation, sediment control, land improvement, and pollution control. Multipurpose projects are more cost-effective but involve complex operations due to competing objectives of the different purposes. The conflicting requirements of these purposes were described.

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OBJECTIVE-TYPE QUESTIONS 8.1 The crop period for wheat is around (a) 90 days (b) 120 days

(c) 180 days

(d) 240 days

8.2 Which of the options given below is correct about the following statements? (i) Rice needs more water than wheat. (ii) Crops in colder areas need more water than those in hot areas. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 8.3 The crop factor is generally largest at what stage of growth? (a) Initial (b) Development (c) Mid-season

(d) Late-season

8.4 Which of the following will NOT lead to lower crop water requirement? (a) High humidity (b) Low wind (c) More clouds (d) High temperature 8.5 Which of the options given below is correct about the following statements? (i) The command area of a project is always less than the culturable command area. (ii) Crop period and base period are generally assumed to be equal. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 8.6 Which of the following is NOT a Kharif crop? (a) Wheat (b) Rice (c) Cotton

(d) Peas

8.7 Rabi crops are cultivated in which season? (a) Monsoon (b) Winter

(d) Pre-monsoon

(c) Summer

8.8 The duty of a canal is generally expressed in (c) hectares (a) cm (b) m3/s

(d) days

8.9 Which of the options given below is correct about the following statements? (i) More pervious soils will have smaller duty. (ii) Higher temperature leads to larger duty. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 8.10 Which of the options given below is correct about the following statements? (i) Main canals are generally used for direct irrigation. (ii) Discharge in a minor distributary is generally more than 5 m3/s. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 8.11 The canals without any control structure at the intake for regulation of flow are called (a) Main canals (b) Branch canals (c) Inundation canals (d) Minor canals 8.12 Which of the options given below is correct about the following statements? (i) Outlets carry water from the canal to the watercourse. (ii) Silt ejector is used to create a still pond near the canal off-take.

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(b) (i) is false, (ii) is true (d) Both (i) and (ii) are false

8.13 Lining of an earthen canal is done for which of the following purposes? (a) Reducing seepage (b) Reducing maintenance cost (c) Preventing erosion (d) All of the above 8.14 Which of the options given below is correct about the following statements? (i) Canals are generally constructed along the ridge line. (ii) Culverts are used at the crossing of a canal with a river. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 8.15 The moisture content of a soil, after gravity drainage has taken place, is known as (a) Wilting point (b) Field capacity (c) Available moisture (d) Optimum moisture content 8.16 Which of the options given below is correct about the following statements? (i) Available water is the difference between the porosity and the wilting point. (ii) Water is generally applied to the field when the moisture content is at the wilting point. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 8.17 What is the typical efficiency of basin irrigation? (a) 10% (b) 20% (c) 40%

(d) 60%

8.18 Which of the options given below is correct about the following statements? (i) Initial cost of border irrigation is low but maintenance cost is high. (ii) Ring irrigation is generally used for high-value crops. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 8.19 Which of the options given below is correct about the following statements? (i) Sprinkler irrigation is suitable when the water contains sediments. (ii) Subsurface irrigation has negligible evaporation loss. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 8.20 Which of the options given below is correct about the following statements? (i) Sprinkler irrigation is not suitable for areas with strong winds. (ii) Drip irrigation is suitable when the water contains salts. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 8.21 Which of the options given below is correct about the following statements? (i) For hydropower generation projects, the reservoir should be nearly full at all times. (ii) For flood-control projects, the reservoir should be nearly full at the time of most intense rainy season. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 8.22 The maximum irrigation demand in India occurs during which season? (a) Summer (b) Winter (c) Monsoon (d) Pre-monsoon 8.23 For navigation, a minimum water _____ is needed in the channel. (a) discharge (b) depth (c) quality

(d) velocity

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DESCRIPTIVE QUESTIONS 8.1 What are the various aspects which affect the crop factor? Describe how the crop factor method is used to find the water requirement of a crop. 8.2 Describe the three crop seasons in North India. What is meant by delta of a crop? 8.3 How do you define the duty of canal water? Derive a relationship between duty, delta, and base period. 8.4 How is the design discharge for a canal obtained for an area where crops are cultivated in non-overlapping seasons? How will the method be modified to account for overlap in the cropping pattern? 8.5 What is the difference between main canal, branch, major distributary, minor distributary, and watercourse? 8.6 What is meant by canal headworks? What are its major components? 8.7 Describe various methods of surface irrigation. Under what conditions would the basin method be suitable? 8.8 Differentiate between sprinkler and drip irrigation. 8.9 List some of the purposes which could be served by a water resources project. Describe the requirements for an irrigation project and contrast these with the requirements for a flood-control project. 8.10 What are multipurpose projects? Why are they more complicated to operate?

NUMERICAL QUESTIONS 8.1 Wheat is sown in an area in the middle of November and is harvested after five months, in the middle of April. The reference evapotranspiration for these months are estimated as: November – 2.2 mm/d, December – 2.5 mm/d, January – 2.7 mm/d, February – 3.2 mm/d, March – 5.6 mm/d and April – 5.9 mm/d. Estimate the total water requirement, using the crop-factor method, and compute the duty of canal water. 8.2 A canal services a CCA of 2000 ha. There are two crops grown in the area: wheat from November to March, and rice from July to October. The intensity of irrigation is 70% for wheat and 30% for rice. What should be the design discharge of the canal? 8.3 A crop has a root-zone depth of 60 cm and water requirement of 3 mm/d, and is planted in a sandy soil with field capacity of 20% and permanent wilting point of 5%. The optimum soil moisture content for the crop is 13%. Assuming an irrigation efficiency of 70%, and irrigation at a moisture content of 8%, estimate the frequency of irrigation and the water depth needed at each irrigation.

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Statistical Methods in Hydrology

LEARNING OBJECTIVES LO 1

Discuss the basic concept of probability and statistics

LO 2

Explain random variables and probability distributions

LO 3

Analyze the moments and basic descriptive statistics and their importance in statistical hydrology

LO 4

Examine Binomial, Gaussian, and Gumbel’s Extreme Value Type-I distributions

LO 5

Demonstrate flood frequency analysis using plotting position and general frequency factor methods

LO 6

Outline the need and procedure to calculate the confidence limits and interval

LO 7

Summarize the risk and reliability concepts for water resources design

9.1

INTRODUCTION

To sustain the water resources systems, which are frequently subjected LO 1 Discuss the basic to future occurrences of extreme events, such as floods and droughts, concept of probability and proper planning is needed. The occurrence of an extreme event in future statistics will have an adverse impact on a water resources system; therefore, some information about the occurrence of such future events can be extremely useful in efficient water resources planning, operation, and management. A water resources project capable of withstanding higher magnitude floods would entail higher capital costs. On the other hand, a water resources project capable of withstanding lesser magnitude floods would entail smaller capital costs but higher costs in damages and lost lives. Such trade-offs pose challenges and difficult decision making for the hydrologists and water resources managers.

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Hydrologists are faced with several challenges in providing probabilistic statement about the future values of the hydrologic variables, for example, a hydrologist needs to answer the following questions: •

What is the probability that a hydrologic variable (e.g., streamflow) of a certain magnitude will occur in near future?

•

What is the probability that a hydrologic event having magnitude equal to or greater than a specified value will occur?

Such probabilistic statements are also useful in socio-economic evaluation of water resources projects and hence in the decision making by the water resources decision makers, or a political class, about a certain project in a specified area. Such probabilistic statements in a specific area are normally derived using the historical data available about the hydrologic variables in the study area. The hydrologic data measured over space and time may contain errors or uncertainty in them due to several reasons including instrumental limitations and measurement errors. Often the system itself is extremely complex and we assume it to be uncertain or stochastic to be able to make probabilistic statements in order to take important decisions related to a water resources project. The outcome of a physical process may depend on too many factors, information on which may be unavailable or expensive to collect. There could be fluctuations in a system due to environmental factors, climate change impacts and so on, which cannot be deterministically characterized and hence probabilistic statements are required. Further, mathematical models developed for a physical system may involve numerical errors, scale issues, and variable boundary conditions or may be based on simplistic assumptions that are not valid in real life. Therefore, we see that there are a host of factors which lead to uncertainty in the hydrologic systems. Instead of treating such hydrologic systems using the laws of physics, the remedy is to treat them as random processes or stochastic processes. The outcomes of such random or stochastic processes are governed by the laws of probability, and probabilistic and statistical analyses tools are required to deal with them. Also, most of the hydrologic analysis and designs are based on statistical data analysis methods. In this chapter, we will learn about the various probabilistic and statistical methods that are useful in the planning, design, operation and management of various water resources projects.

9.2 9.2.1

BASIC PROBABILISTIC AND STATISTICAL CONCEPTS Sample and Population

LO 2 Explain random When we talk about probability, we talk about sample and population. variables and probability Let us understand the basic difference between them and their importance distributions in probabilistic and statistical analyses. A sample is a set of observations that has been observed in the past. Let us consider that we have annual rainfall data at a location for the past 30 years, that is xi where i = 1 to 30. The set of these 30 annual rainfall values represents a sample. On the other hand, population consists of all the values of a physical variable that have been observed in the past and will be observed in the future. It is not possible to measure a variable that has already occurred in the past or which is yet to occur. Therefore, we can never know or measure the population and we rely on a sample to draw inferences about the behavior of the population. Note that a sample is a subset of the population and is drawn from a hypothetical infinite sized population

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having constant statistical properties. The statistical properties of a hydrologic variable may vary from one sample to the other. In practice, we take measurements of a hydrologic variable, collect sample data, perform probabilistic and statistical analyses, draw inferences about the population, and employ such inferences in hydrologic analysis and design. The statistical analysis allows us to make generalizations about populations using the vital information contained in random samples.

9.2.2 Random Variable A random variable is a physical variable which can assume any value within its range described by a certain probability distribution. The probability of a random variable having a value less than or equal to a specified value is represented as: P(X £ x) = p

(9.1)

where, X is a random variable, x is a specified value, and p is the probability that the value of X is less than or equal to a specified value x. The random variable can be defined in such a way that it is practically useful in hydrologic analysis and design. A random variable can be either discrete or continuous depending on its nature and the values it can take. A discrete random variable is the one that can take particular, often integer, values. For example, outcome of the roll of a dice is a discrete random variable. The outcome can be any integer from 1 to 6. Occurrence of rainfall tomorrow or a flood during a particular year are examples of a discrete random variable in hydrology having only two possible outcomes, 0 for non-occurrence and 1 for occurrence (yes or no type of outcomes). Some other examples of discrete random variables from hydrology can be number of floods in a year, number of rainy days in a month, and the number of storms in a monsoon season, etc. On the other hand, a continuous random variable is one that can take any value specified within a range including fractional values. For example, depth of monthly rainfall in a catchment, maximum discharge at a location in river in a particular year, and relative humidity, average air temperature on a given day are examples of a continuous random variable.

9.2.3

Probability

Probability can be defined as the chance or likelihood of the occurrence of an event. An event is defined as the occurrence of a specified value of a random variable. Probability is measured on a scale from 0 to 1 where the value 1 represents the certainty of the occurrence of the event while the value 0 indicates the failure of the event to occur. The probability of an event A can be estimated using the sample data and the concept of relative frequency as follows: n (9.2) N where, n is the number of occurrences of an event out of a total of N trials. The probability is normally expressed as a percentage. For example, the probability of a flood next year is 10% means that based on the past historical records, flood has occurred in 1 out of 10 years. Note that P(A) represents an estimate of the probability depending on the sample size N. As the size of the sample increases, the estimate of the probability becomes better and under the limits, a true estimate of the probability for a population is defined as follows: P( A) =

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n N Æ• N

P( A) = lim

(9.3)

Such probabilities that are based on historical data are called posterior probabilities. On the other hand, estimates of probabilities of future events based on human judgment and experience are known as prior probabilities. In this book, whenever we mention probability, it is meant to be posterior probability, unless mentioned otherwise.

9.2.3.1

Laws of Probability

The probabilities of events or random variables obey certain principles known as the laws of probability. Some of the important laws of probability are discussed here. ∑

Law of Total Probability: The sum of the probabilities of all possible individual outcomes in a single trial is equal to 1. In other words, if a sample space is divided into m different mutually exclusive sub-spaces, or events E1, E2, E3, … Em, then m

ÂP(Ei ) = 1

(9.4)

i =1

∑

Law of Complementarity: If the probability of occurrence of an event E is P(E), then the probability of non-occurrence of the event will be 1 – P(E). The non-occurrence event is denoted as E and the probability of the event E is specified as follows: P( E ) = 1 - P( E )

∑

(9.5)

Law of Intersection of Probabilities: The probability of two independent events (E1 and E2) occurring simultaneously or together is given by the product of the individual probabilities. P( E1 « E2 ) = P( E1 )* P( E2 )

(9.6)

The expression P(E1 « E2) represents the probability of the intersection of two events (also called the joint probability) and is read as “the probability of E1 and E2”. Note that the law of intersection can be extended when we deal with more than two independent events. ∑

Law of Union of Probabilities: The probability of two independent and mutually exclusive events (E1 and E2) is the sum of the probabilities of the individual probabilities of the separate events. P( E1 » E2 ) = P( E1 ) + P( E2 )

(9.7)

The expression P(E1 » E2) represents the probability of the union of two events and is read as “the probability of E1 or E2”. Note that the law of union can be extended when we deal with more than two independent and mutually exclusive events. When the two events (E1 and E2) are dependent or not mutually exclusive then the law of union is given in its more general form as follows: (9.8) P( E1 » E2 ) = P( E1 ) + P( E2 ) - P( E1 « E2 ) ∑

Law of Conditional Probabilities: The probability that an event E1 will occur given that an event E2 has already occurred is called conditional probability and is expressed as follows: P( E1 | E2 ) =

P (E1 « E2 ) P ( E2 )

(9.9)

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For example, we may be interested in finding the probability of occurrence of a flood next year given that a flood has occurred this year. Such a probability can be found using the concept of conditional probability. If the two events are independent of each other, then the LHS of equation above will be equal to P(E1). Therefore, conditional probability reduces to the law of intersection of probabilities mentioned above. � EXAMPLE 9.1 In Uttarakhand, cloudburst and landslide are the two common natural disasters that occur quite frequently. Based on the historical data, the probability of occurrence of a cloudburst given that a landslide has occurred in a particular year is 0.30 and the probability that a landslide will occur given that a cloudburst has occurred is 0.80. The joint probability of the occurrence of a cloudburst and a landslide together is 0.20. Determine the probability of occurrence of (a) a cloudburst, and (b) a landslide in a year in Uttarakhand. Solution Let us define the events as follows: Event A = Occurrence of a cloudburst; probability of occurrence of a cloudburst = P(A) Event B = Occurrence of a landslide; probability of occurrence of a landslide = P(B) The following probabilities are given: P(A | B) = 0.30; P(B | A) = 0.80; and P{A*B}= P(A « B) = 0.20. From the law of conditional probability, equation (9.9), we know that: P ( A | B) =

Or,

P ( B) =

P ( A « B) P ( B)

P ( A « B) P ( A | B)

Putting the values P(A « B) = 0.20 and P(A | B) = 0.30, we get P(B) = 0.20/0.30 = 0.667. Similarly, putting the values P(A « B) = 0.20 and P(B | A) = 0.80, we get P(A) = 0.20/0.80 = 0.250. Therefore, the probability of occurrence of a cloudburst in a year is 25% and that for a landslide is 66.7% in Uttarakhand based on historical data.

9.2.4

Probability of Discrete Random Variables

The probability of a discrete random variable is specified by probability mass function (pmf) as follows: P(X = x) = p(x)

(9.10)

where, X is a discrete random variable, x is a specified value, p(x) is the pmf and P represents the probability that the random variable X is equal to x. Since the pmf represents probability, its value for any value of X must lie between 0 and 1 inclusive as follows: 0 £ p( x ) £ 1

(9.11)

If all the possible values of X and the associated probabilities are represented by xi and p(xi), respectively, where i = 1, 2, …, N, then from the law of total probability, the sum of all the individual probabilities should be equal to 1.

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Âp( xi ) = 1

(9.12)

i =1

Sometimes it is useful to represent the cumulative probability of a discrete random variable using the concept of cumulative mass function (cmf) as follows (assuming that the xi are arranged in increasing order): i

P( X £ xi ) = F ( xi ) =

Âp( x j )

(9.13)

j =1

The pmf is useful in finding the probability of a discrete random variable X having a value less than or equal to a specified value xi. To understand the concept of probability of a discrete random variable, pmf and cmf, let us consider an example of the roll of a dice. A dice has six faces with values 1, 2, 3, 4, 5, and 6, respectively. What is the probability of getting a 2 when the dice is rolled? Will this probability be same as that for 6? The probability of getting any value of X is same and is equal to 1/6. Therefore, the individual probabilities can be specified by the pmf as follows: p( xi ) =

1 ; for i = 1, 2, 3, 4, 5, 6 6

(9.14)

Similarly, the cumulative probability can be expressed by the cmf as follows: F ( xi ) =

i ; for i = 1, 2, 3, 4, 5, 6 6

(9.15)

The pmf and cmf for this example are shown in Figure 9.1.

Figure 9.1 Probability concepts for discrete random variable: (a) pmf, and (b) cmf for the example of a roll of a dice � EXAMPLE 9.2 The probability of number of floods (N) occurring during the useful life of a bridge over Ganga river has been found as follows: p(N = 0) = 0.20, p(N = 1) = 0.25, p(N = 2) = 0.20, p(N = 3) = 0.15, p(N = 4) = 0.10, p(N = 5) = 0.05, p(N = 6) = 0.03, and p(N ≥ 7) = 0.02. Plot the pmf and cmf for this data. Also, write the probabilities for each of the following: (a) p(N > 3), (b) p(N < 4), and (c) p(3 < N < 6).

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Solution The discrete random variable here is the number of floods (N) in a given area during the useful life of a bridge over Ganga river. Table 9.1 shows the pmf and cmf for the data given in the example. The first column [1] represents the discrete random variable (N), the second column [2] presents the pmf values as the probabilities given in the example. The third column [3] is calculated as the cumulative values of the pmf column [2]. The pmf and cmf are plotted and shown in Figure 9.2.

Table 9.1 pmf and cmf values for Example 9.1 N

p(N)

F (N)

[1]

[2]

[3]

0

0.20

0.20

1

0.25

0.45

2

0.20

0.65

3

0.15

0.80

4

0.10

0.90

5

0.05

0.95

6

0.03

0.98

≥7

0.02

1.00

1.00

0.30

Cumulative Probability

Probability

0.25 0.20 0.15 0.10 0.05 0.00

0.80 0.60 0.40 0.20 0.00

0

1

2 3 4 5 Number of Floods

6

7

(a) pmf

Figure 9.2

0

1

2 3 4 5 Number of Floods

6

7

(b) cmf

Probability mass function (pmf) and cumulative mass function (cmf)

The various probabilities are calculated as follows: (a)

p(N > 3): The probability of the number of floods in the useful life of a bridge over river Ganga being greater than 3 can be found as follows: 7

P( N > 3) =

 p( N )

n=4

= p(4) + p(5) + p(6) + p(7) = 0.10 + 0.05 + 0.03 + 0.02 = 0.20

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It can also be found as follows: 3

P( N > 3) = 1 - P( N £ 3) = 1 -

 p( N ) = 1 – p(0) – p(1) – p(2) – p(3)

n=0

= 1.0 – 0.20 – 0.25 – 0.20 – 0.15 = 0.20 (b)

p(N < 4): The probability of the number of floods in the useful life of a bridge over river Ganga being less than 4 can be found as follows: 3

P( N < 4) =

 p( n )

= p(0) + p(1) + p(2) + p(3) = 0.20 + 0.25 + 0.20 + 0.15 = 0.80

n=0

(c)

p(3 < N < 6): The probability of the number of floods in the useful life of a bridge over river Ganga being between 3 and 6 (exclusive) can be found as follows: 5

P(3 < N < 6) =

 p(n) = p(4) + p(5) = 0.10 + 0.05 = 0.15

n=4

9.2.5

Probability of Continuous Random Variables

The probability of a continuous random variable is specified by the probability density function (pdf), and is denoted as f (x). The probability of occurrence of a continuous random variable is normally specified within an interval (from x1 to x2) and is given by the integral of the pdf from x1 to x2 for all the possible values in the interval inclusive. It is mathematically expressed as follows: x2

P( x1 £ X £ x2 ) =

Ú f ( x)dx

(9.16)

x1

where, X is a random variable, x1 and x2 are the lower and upper limits of the interval, and f (x) is the pdf. Notice that as x1 approaches x2, or in other words if the interval approaches zero, the probability given by the Eq. (9.16) approaches zero. This is the fundamental difference between a discrete random variable and a continuous random variable. The probability of a discrete random variable taking on a specified value is non-zero while the same for a continuous random variable is zero. Therefore, the probability of a continuous random variable is always defined for a finite interval. The sum of all probabilities over all intervals is given by the integral of the pdf over all possible values of the random variable and is equal to 1.0. This is mathematically written as follows: •

Ú f ( x)dx = 1

(9.17)

-•

The cumulative distribution function (cdf), denoted as F(x), for a continuous random variable is written as follows: x

F(x) = P (X ≤ x) =

Ú f (u)du

-•

(9.18)

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where, u is a dummy variable of integration. The cdf is monotonically increasing function with minimum and maximum values of 0 and 1, respectively. For x1 ≤ x2, P (X ≤ x1) ≤ P (X ≤ x2). The value of cdf is greater than or equal to 0.0, F(x) ≥ 0.0, for all possible values of x within the range of the random variable. Since the cdf is a cumulative function, it is related to pdf as follows: f ( x) =

dF ( x ) dx

(9.19)

If we know the cdf, pdf can be found as the first derivative of the cdf. On the other hand, if we know pdf, cdf can be found by its integration. The concepts of pdf and cdf are represented graphically in Figure 9.3. The ordinate of a cdf at any value of x is equal to the area under the pdf till that value of x, starting from the minimum value of x. The probability of a random variable falling in a certain range is equal to the area under the pdf for the range. Since the cdf is cumulative function of the pdf, the probability of a random variable falling between x1 and x2 is equal to the difference of cdf values at x2 and x1. This can be mathematically written as follows: x2

P( x1 £ X £ x2 ) =

Ú f ( x)dx = F ( x2 ) - F ( x1 )

x1

Figure 9.3

Probability concepts for continuous random variable

� EXAMPLE 9.3 The pdf of a hydrologic variable is given as follows: f(x) = kx f(x) = k(4 – x) f(x) = 0

for 0 ≤ x ≤ 2 for 2 ≤ x ≤ 4 elsewhere

(a)

Determine the value of the constant k.

(b)

Write down the complete equations of pdf and cdf.

(c)

Plot the pdf and cdf.

(d)

Determine the following: (i) P(X > 2), (ii) P(X ≤ 1), and (iii) P(1 ≤ X ≤ 3)

(9.20)

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Solution (a)

The constant k can be found out by the property of the pdf that the area under it should be equal to 1. •

Ú f ( x) dx = 1

-• 2

That is,

4

Úkxdx + Úk(4 - x) dx = 1 0

2

2 4 Ï Ê x2ˆ Ê x 2 ˆ ¸Ô Ô k Á ˜ + k Ì4( x )24 - Á ˜ ˝ = 1 Ë 2 ¯0 Ë 2 ¯2 Ô ÔÓ ˛

Ï Ê 4 - 0ˆ Ê 16 - 4 ˆ ¸ kÁ + k Ì4(4 - 2) - Á ˝ =1 ˜ Ë 2 ¯ Ë 2 ˜¯ ˛ Ó k (2) + k {8 - 6} = 1; k = (b)

The pdf is given by: 1 x 4 1 f ( x ) = (4 - x ) 4 f ( x) = 0 f ( x) =

(c)

1 4

for 0 £ x £ 2 for 2 £ x £ 4 elsewhere

The cdf can be obtained by integrating the pdf given by the above equation as follows: x

F(x) =

Ú f (u) du

-•

For the example, x

1 F ( x ) = Ú udu 4 0 1 Ê u2ˆ F ( x) = Á ˜ 4Ë2 ¯ 2

F ( x) =

x 8

x

for 0 ≤ x ≤ 2 0

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315

x

1 1 F ( x ) = Ú udu + Ú (4 - u) du 4 4 0 2 2 x Ï Ê u 2 ˆ ¸Ô 1 Ê u2ˆ 1Ô x F ( x ) = Á ˜ + Ì4(u)2 - Á ˜ ˝ 4Ë2 ¯ 4Ô Ë 2 ¯2 Ô 0 Ó ˛

and

(d)

F ( x) =

1 1 ÏÔ x 2 - 4 ¸Ô + Ì4( x - 2) ˝ 2 4 ÔÓ 2 ˛Ô

F ( x) =

1 ( x - 2) Ï ( x + 2) ¸ + Ì4 ˝ 2 4 Ó 2 ˛

F ( x) =

1 ( x - 2) + {6 - x} 2 8

for 2 ≤ x ≤ 4

Probabilities can be found out as follows: (i) P(X > 2): The probability P(X > 2) = 1 – P(X ≤ 2) = 1 – F(X = 2) = 1 – (22/8) = 1 – 0.50 = 0.50. Therefore, P(X > 2) = 0.50. Note that this can also be found out by using the functions in the range 2 to 4. (ii) P(X ≤ 1): The probability P(X ≤ 1) can be found out by putting x = 1 in the cdf in the range (0, 2). Therefore, P(X ≤ 1) = (12/8) = 1/8 = 0.1250. (iii) P(1 ≤ X ≤ 3): The probability P(1 ≤ X ≤ 3) = F(X = 3) – F(X = 1). Now, the individual probabilities can be found by putting x = 3 and x = 1 in the corresponding cdfs.

9.3

F ( X = 3) =

1 (3 - 2) 1 1 7 + {6 - 3} = + {3} = 2 8 2 8 8

and

F ( X = 1) =

12 1 = 8 8

Therefore,

P(1 £ X £ 3) =

7 1 6 - = = 0.75 8 8 8

MOMENTS AND BASIC DESCRIPTIVE STATISTICS

Each random variable is unique in its nature and characteristics. There are LO 3 Analyze the moments many probability distributions available and a given random variable may and basic descriptive statistics closely follow the characteristics of a particular probability distribution. and their importance in Therefore, it is essential to understand the ways in which we can characterize statistical hydrology a particular probability distribution. The characteristics of a pdf are normally expressed using various moments of the pdf taken either about the origin or about the mean. Once we have defined the characteristics of a population, it is easy to calculate

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the values of the moments (also called statistics) for a sample. The concept of moments also helps in fitting a particular probability distribution to a set of data. In order to completely define or characterize a probability distribution, two moments are usually sufficient. The first moment is the mean which is measured about the origin; the second moment is variance, which is measured about the mean. However, higher moments are sometimes needed. The third moment of the values measured about the mean is known as skewness, and the fourth moment of the values measured about the mean is known as kurtosis. In general, the kth moment about the origin for a random variable is defined as the product of xk and the corresponding pdf, integrated over the entire feasible range of the random variable. For a continuous random variable, it is mathematically expressed as follows: •

M k/ =

Úx

k

f ( x ) dx

(9.21)

-•

where, M k/ is the first moment about the origin, x is a continuous random variable, and f (x) is the pdf for the continuous random variable. For a discrete random variable, it is mathematically expressed as follows: n

M k / = Âxik p( xi )

(9.22)

i =1

where, p(xi) is the pmf of the discrete random variable and n is the number of data points in a sample. The kth moment about the mean for continuous and discrete random variables is defined as follows: •

Mk =

Ú (x - m)

k

f ( x )dx

(9.23)

M k = Â( xi - m )k p( xi )

(9.24)

-• n

i =1

where, μ is the mean. The moments, whether taken about the origin or the mean, represent the basic descriptive statistics for the population or its representative sample. The moments are also referred to as the measures of tendencies of the probability distributions or the random variables. The first moment about the origin measures the central tendency of a probability distribution. The second moment about the mean represents measure of variation, dispersion, or spread in the data; the third moment about the mean represents skewness or asymmetry in the data, and the fourth moment about the mean represents the sharpness in the peak or peakedness of a probability distribution.

9.3.1 Measures of Central Tendency The measures of central tendency quantify the value of a random variable around which most of the observations are expected to occur. There are three measures of central tendency normally in use: mean, median, and mode of the data or a probability distribution.

9.3.1.1

Mean

Mean is the first moment measured about the origin of the data and is also known as the average or expected value. The population mean is denoted as μ and the sample mean is denoted as x . The mean represents the

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317

expected value of the random variable X, E(X) around which the probability distribution is centered. For a continuous random variable, the mean can be expressed as follows: •

m = E( X ) =

Ú x f ( x) dx

(9.25)

-•

For a discrete random variable, the mean is expressed as follows: n

m = E ( X ) = Âxi p( xi )

(9.26)

i =1

For estimating the sample mean from a sample of size n, it is customary to give equal importance to each observation (1/n); therefore, the sample mean is written as follows: x=

1 n Âx n i =1 i

(9.27)

The mean essentially represents the value of a random variable corresponding to the centroid of the probability distribution it follows. The mean lies on the axis of symmetry for a symmetrical probability distribution. Figure 9.4 shows two symmetric pdfs, one with an axis of symmetry x = a while the other one at x = 0. If the line x = a is an axis of symmetry, then the centroid of the pdf must lie on it; therefore, mean = E(X) = μ = a. Also, if f (x) = f (-x) or x = 0 is the line of symmetry, then mean = E(X) = μ = 0.

Figure 9.4

Mean lies on the axis of symmetry

Further, the mean is a location parameter that locates the pdf along the measurement axis. Consider two random variables, X1 and X2, representing flows in a river measured at an upstream and a downstream station, respectively. Using the data observed at the two locations, a probability distribution can be fitted for flow at each location. Let f (x1) and f (x2) be the pdfs representing the two random variables, X1 and X2 having means of x1 and x2 , respectively. Figure 9.5 shows the concept of mean being a location parameter. Since the flows at a downstream station in the same river are expected to be higher than those at an upstream station most of the time, it is expected that x2 > x1 . Note that the means locate the pdf on the x-axis such that the pdf for the flow at a downstream station will be located to the right of that at an upstream location.

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318

f(x1)

f(x2)

f (x )

x1

Figure 9.5

x2

Mean as a location parameter

Note that the flows at downstream station may be smaller than the flows at an upstream station sometimes, characterized by the small overlap in the two pdfs in Figure 9.5. Also, if the two pdfs are non-overlapping, then the flow in one river having large location parameter (mean) will always be larger than the flow in the other river having small location parameter (mean). Similarly, the pdfs of two different rivers plotted on the same graph would provide information about the relative magnitudes of the flows and their means for the two rivers. �

Properties of E(X)

Some important properties of the mean or expected value are as follows:

•

Expected value of a constant is the value of the constant itself i.e., E(C) = C, where C is a constant.

•

Expected value of a random variables that is obtained by multiplying its observations by a constant is equal to the expected value of the random value multiplied by the constant i.e., E(CX) = CE(X) = Cμ

•

If Y = g(X), then the expected value of Y can be estimated as follows: E (Y ) = E [g( X )]=

•

n

-•

i =1

Ú g( x) f ( x)dx = Âg( xi ) p( xi )

(9.28)

Notice that the first expression in Eq. (9.28) is for a continuous random variable while the second expression is for a discrete random variable.

9.3.1.2

Median

Median is also a measure of central tendency in the data. It is defined as a value such that the area under the pdf to the right is equal to that at the left. In other words, it represents the mid value of the random variable within the range of values it can take. Mathematically, median is a value u such that: u

Ú f ( x) dx = 0.50

(9.29)

-•

Median is a more appropriate measure of central tendency when the pdf is asymmetric. For a symmetric pdf both mean and median will have the same value.

Statistical Methods in Hydrology

9.3.1.3

319

Mode

Mode, also a measure of central tendency of a pdf, is defined as that value of a random variable which occurs the most or has the highest probability of occurrence. Note that it is not the highest value of the random variable but the value having the highest ordinate on the pdf. The pdf having a single mode is called a unimodal distribution while the one having more than one mode is called a multi-modal distribution. For a unimodal and symmetric pdf, mean, median and mode, all will have the same value, while they will be different for a skewed pdf. Also, for a symmetric and multi-modal pdf, mean is equal to median and the mode will be the value of the random variable with the highest ordinate on the pdf. These concepts are shown in Figure 9.6.

Figure 9.6

Measures of central tendency: Mean, Median, and Mode

The units of all the three measures of central tendency, mean, median, and mode, are same as that of the random variable. For example, for a random variable representing annual rainfall in a catchment (in mm), the units of mean, median, and mode will be mm. Similarly, for a random variable representing annual peak flow in a river (in m3/s), the units of mean, median, and mode will also be m3/s.

9.3.2

Measures of Variation

The higher moments are normally taken about the mean. The second moment about the mean (denoted as s 2) represents the measure of variation or dispersion in the data. Variance and standard deviation are the measures of variation that are most commonly used to characterize a pdf. For a continuous random variable, the variance can be expressed as follows:

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Engineering Hydrology •

s 2 = E ( X - m )2 =

Ú (x - m)

2

f ( x ) dx

(9.30)

-•

For a discrete random variable, the variance is expressed as follows: n

s 2 = E ( X - m )2 = Â( xi - m )2 p( xi )

(9.31)

i =1

When each observation in a sample of size n is given equal importance (i.e., f (x) = p(xi) = 1/n), then the variance can be expressed as follows: s 2 = E ( X - m )2 =

1 n Â( xi - m )2 n i =1

(9.32)

The value of variance given by Eq. (9.32) may be biased depending on the size of the sample. An unbiased estimate of the variance is usually employed as follows: s 2 = E ( X - m )2 =

n 1 Â( xi - m )2 (n - 1) i =1

(9.33)

It is because the average value of s2 calculated from Eq. (9.33) from many samples would approach the true value of s2; however, the average value of s2 calculated from Eq. (9.32) from many samples would not approach the true value of s2. While the variance of a population is denoted as s2, its value for a sample is denoted as s2. The sample variance s2 can be calculated using the following alternative expression: 2˘ È 1 Í n 2 1Ê n ˆ ˙ s = Âxi - n ÁË Âxi ˜¯ ˙ (n - 1) Í i =1 i =1 Î ˚ 2

(9.34)

Variance is a very important parameter in statistical hydrology because it forms the basis of several advanced stochastic models for hydrological forecasting. The variance is a measure of the average distance or departure between the observations and the mean. In other words, it represents how close or far the observations are from the mean. For example, if all the observed values of a random variable are equal to the mean, then the variance of the random variable would be equal to zero. Further, higher the value of the variance, farther away are the observations from the mean and vice-versa. The pdf will be spread wider for larger variance and sharp and narrow for smaller variance or dispersion in the data. The units of the variance are the square of the units of the random variable. For example, if the random variable represents daily temperature in oC, then the units of variance would be (oC)2. Since the units of variance are not same as those of the random variable, mean, median, or mode; its use in characterizing a pdf is limited and another statistical measure that has the same unit as the random variable is normally used. This is known as standard deviation. The standard deviation is the square root of the variance. Therefore, standard deviation has the same units as the random variable and its measures of central tendency (mean, median, and mode). Standard deviation is also a measure of variance, dispersion, or spread in the data or a pdf. The standard deviation is denoted as s for a population and s for a sample. It can be calculated by taking square root of Eq. (9.34) or Eq. (9.33) with μ = x .

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The standard deviation is also known as the shape parameter of a pdf because it controls the shape of the pdf. A pdf with larger standard deviation (or variance for that matter) will be spread wider along the measurement axis as compared to a pdf with smaller standard deviation. This concept is depicted in Figure 9.7, which shows two pdfs, one having small s and the other having large s, but both having same mean μ. It is clear from this figure that the variance controls the shape of the pdf. Sometimes, it is beneficial s to express variation as a dimensionless coefficient called coefficient of variation Cv = . The coefficient of x variation helps in comparing variation in two different samples having different means.

f (x )

Small s

Same mean m

Large s

m x

Figure 9.7

9.3.2.1

pdfs with same mean but different standard deviations

Properties of Standard Deviation

Some important properties of the standard deviation are as follows: •

Standard deviation of a constant is always equal to zero, i.e., SD (C) = 0.0.

•

Standard deviation of a random variable remains unchanged when a constant is added to it or subtracted from it. If Y = X ± C, then SD (Y) = SD (X).

•

If a random variable is multiplied by a constant then its standard deviation gets multiplied by the constant i.e., if Y = CX, then SD (Y) = C SD (X).

The mean and standard deviation are important descriptors of data in stochastic hydrology. Most of the pdfs can be completely defined using these two statistical parameters. The most commonly used probability distribution is normal probability distribution that has mean and standard deviation as its parameters. While using normal distribution, it is customary to express the random variable in standardized form using the mean and standard deviation as follows: z= where, z is called the standard normal variable.

x-m s

(9.35)

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9.3.2.2 Chebyshev’s Inequality In the absence of any knowledge about the probability distribution a particular random variable follows, mean and standard deviations can be used in Chebyshev’s inequality to find the probability of occurrence of the random variable in a certain range. The inequality can be expressed as follows for any k > 1: 1ˆ Ê P[( m - ks ) £ X £ ( m + ks )] ≥ Á 1 - 2 ˜ Ë k ¯

(9.36)

That is, the probability of a random variable falling in an interval (μ ± s) must at least be equal to (1 – 1/k2) regardless of the underlying probability distribution of the random variable. � EXAMPLE 9.4 The annual rainfall in Godavari river basin is found to have a mean and standard deviation of 100 cm and 20 cm, respectively. Determine the range of annual rainfall in Godavari river basin within which it will have a probability of at least equal to 60%. Solution By Chebyshev’s inequality, the probability of at least equal to 0.60 means 1ˆ 1 1 Ê 2 ÁË 1 - 2 ˜¯ = 0.60; 2 = 0.40; k = 0.4 = 2.5; k = 2.5 = 1.58114 k k Therefore, P ÈÎ(100 - 2.5 (20)) £ X £ (100 + 2.5 (20))˘˚ ≥ 0.60 Or,

P[(68.4) £ X £ 131.6] ≥ 0.60

Therefore, the range of the annual rainfall in Godavari river basin is (68.4 cm, 131.6 cm) within which it will have a probability of at least equal to 60%.

9.3.3

Measure of Skewness

The third moment of the observations measured about the mean quantifies skewness in the data. The skew or skewness is a measure of symmetry (or rather asymmetry) in the data or a pdf. For a continuous random variable, the skewness can be expressed as follows: •

E ( X - m )3 =

Ú (x - m)

3

f ( x )dx

(9.37)

-•

For a discrete random variable, the skewness is expressed as follows: n

E ( X - m )3 = Â( xi - m )3 p( xi ) i =1

(9.38)

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There is no consensus on its notation and different books use different symbols. The units of the skewness as expressed above are the cubes of the original units of the random variable. Therefore, coefficient of skewness is expressed as the ratio of the third moment to the cube of the standard deviation as follows: g1 =

n

n

Â( xi - m )3 (n - 1)(n - 2)s 3

(9.39)

i =1

where, g1 is the coefficient of skewness. For calculating the coefficient of skewness for a sample, the population mean μ can be replaced by the sample mean x and population standard deviation s can be replaced by sample standard deviation s. Since skew measures the asymmetry of a distribution, the value of the coefficient of skewness for a symmetric pdf is zero. The pdf can be either positively skewed or negatively skewed depending on the location of the peak of the pdf/mode with respect to the mean. If the pdf has mode shifted to the left of the mean, it is called positively skewed (g1 > 0) and if the pdf has mode shifted to the right of the mean, it is called negatively skewed (g1 < 0). This concept is illustrated in Figure 9.8.

Figure 9.8

9.3.4

pdfs with different skews: (a) g1 = 0, (b) g1 > 0, (c) g1 < 0

Measures of Peakedness

The fourth moment about the mean measures the peakedness of a probability distribution, i.e., how sharp or flat the peak of the pdf is. Since it will have units of fourth power of the original units of the random variable, Kurtosis coefficient denoted as g2, is defined as the ratio of fourth to the square of the second central moments. For a sample, Kurtosis can be calculated as follows: g2 =

n(n + 1)

n

( x - x )4 4 Â i (n - 1)(n - 2)(n - 3)s

(9.40)

i =1

9.3.4.1

Coefficient of Excess

Coefficient of excess is another way of quantifying the peakedness of a pdf using the fourth moment about the mean. The coefficient of excess is defined as the difference between the Kurtosis coefficient of a random variable and that for the normal probability distribution. Since the normal distribution is the most commonly

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used probability distribution, it is taken as a standard one. It has a Kurtosis coefficient equal to 3.0. The coefficient of excess for population, denoted as E, is expressed as E = g2 – 3. For a sample data, it can be expressed as follows: E =g2 -3

(n - 1)2 (n - 2)(n - 3)

(9.41)

All pdfs can be classified as follows depending on the value of the coefficient of excess (or Kurtosis coefficient): (a) a pdf is called mesokurtic if E = 0 or g2 = 3, (b) a pdf is called leptokurtic if E > 0 or g2 > 3, and (c) a pdf is called platykurtic if E < 0 or g2 < 3. This is graphically depicted in Figure 9.9.

Figure 9.9

Types of pdfs depending on peakedness

� EXAMPLE 9.5 The following table shows data for maximum instantaneous flood peaks over a period of 25 years. 1981

1982

1983

1984

1985

1986

1987

1988

1989

1990

1991

1992

Q (m /s)

4734

5932

5212

4734

6381

5376

4545

5821

5120

4102

3554

4987

Year

1993

1994

1995

1996

1997

1998

1999

2000

2001

2002

2003

2004

2005

3801

3977

4855

4712

4476

4622

4309

4378

4256

3719

6101

6978

5555

Year 3

3

Q (m /s)

Calculate mean, median, mode, standard deviation, coefficient of variation, coefficient of skewness, kurtosis coefficient, and coefficient of excess for the data given. Solution The calculations are arranged in Table 9.2. The first two columns present the data given in the example problem. The columns [3], [4], [5], and [6] represent the first, second, third, and fourth moment terms about the mean for the data given. The last row (in bold font) represents the sum of various moments. The column [7] represents the given data sorted from the lowest to the highest value.

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325

Table 9.2 Computation of various moments for data in Example 9.5 S. No.

Flow (m3/s)

(x – μ) (m3/s)

(x – μ)2 (m3/s)2

(x – μ)3 (m3/s)3

(x – μ)4 (m3/s)4

Flow Sorted

[1]

[2]

[3]

[4]

[5]

[6]

[7]

1

4734

–155.5

24,174.0

–37,58,578.2

58,43,83,745.8

2

5932

1,042.5

10,86,848.0

1,13,30,60,725.3

11,81,23,84,67,288.7

3719

3

5212

322.5

1,04,019.2

3,35,48,256.4

10,81,99,83,649.9

3801

4

4734

–155.5

24,174.0

–37,58,578.2

58,43,83,745.8

3977

5

6381

1,491.5

22,24,631.9

3,31,80,82,987.0

49,48,98,71,36,770.0

4102

6

5376

486.5

2,36,701.7

11,51,60,116.1

56,02,76,99,706.3

4256

7

4545

–344.5

1,18,666.5

–4,08,78,225.7

14,08,17,31,197.2

4309

8

5821

931.5

8,67,729.5

80,83,07,393.5

7,52,95,45,03,219.0

4378

9

5120

230.5

53,139.5

1,22,49,710.7

2,82,38,03,314.4

4476

10

4102

–787.5

6,20,124.8

–48,83,35,838.4

3,84,55,47,06,058.7

4545

11

3554

–1,335.5

17,83,506.8

–2,38,18,37,701.9

31,80,89,66,14,083.5

4622

12

4987

97.5

9,510.2

9,27,429.9

9,04,42,960.6

4712

13

3801

–1,088.5

11,84,788.7

–1,28,96,18,815.5

14,03,72,42,88,291.3

4734

14

3977

–912.5

8,32,619.8

–75,97,48,869.8

6,93,25,56,48,756.2

4734

15

4855

–34.5

1,188.9

–40,992.3

14,13,412.8

4855

16

4712

–177.5

31,499.2

–55,90,469.2

99,21,96,475.9

4987

17

4476

–413.5

1,70,965.7

–7,06,90,901.9

29,22,92,74,132.6

5120

18

4622

–267.5

71,545.6

–1,91,37,003.8

5,11,87,65,782.0

5212

19

4309

–580.5

3,36,957.0

–19,55,96,817.0

1,13,54,00,40,336.0

5376

20

4378

–511.5

2,61,611.8

–13,38,09,198.6

68,44,07,28,876.3

5555

21

4256

–633.5

4,01,296.9

–25,42,13,566.8

1,61,03,92,10,296.6

5821

22

3719

–1,170.5

13,70,023.4

–1,60,35,85,024.8

18,76,96,41,99,845.0

5932

23

6101

1,211.5

14,67,780.7

1,77,82,45,686.3

21,54,38,02,13,822.3

6101

24

6978

2,088.5

43,61,915.8

9,10,99,48,366.6

1,90,26,30,93,62,540.9

6381

25

5555

665.5

4,42,916.9

29,47,70,035.6

1,96,17,53,54,084.9

6978

Sum =

1,22,237

0.0

1,80,88,336

9,35,37,00,125

3,62,62,81,45,52,393

The basic descriptive statistics for the given dataset are calculated as follows: Mean

x=

Standard Deviation

s=

1 n 1,22,237 m3 xi = = 4,889.5 Â n i =1 25 s n 1 1 m3 ( xi - x )2 = (1,80,88,336) = 868.1 Â (n - 1) i =1 (25 - 1) s

3554

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Engineering Hydrology

Coefficient of Variation

Cv =

Coefficient of Skewness

g1 =

Kurtosis Coefficient

g2 =

Coefficient of Excess

s 868.1 = = 0.1776 x 4,889.5 n

n

25(9,35,37,00,125)

( x - x )3 = 3 Â i (n - 1)(n - 2)s (25 - 1)(25 - 2)(868.1)3

= 0.6474

i =1

n(n + 1) (n - 1)(n - 2)(n - 3)s

E =g 2 -3

n

4

25(26)(3,62,62,81, 45,52,393)

Â( xi - x )4 = (25 - 1)(25 - 2)(25 - 3)(868.1)4

= 3.4169

i =1

(n - 1)2 (25 - 1)2 = 3.4169 - 3 = 0.0019 (n - 2)(n - 3) (25 - 2)(25 - 3)

The middle value for n = 25 will be the 13th value in the sorted data. From column [7], the 13th value is 4,734 m3/s; therefore the median of the data = 4,734 m3/s. Also, this value occurs twice, therefore the mode of the data = 4,734 m3/s. Note that mode is not really relevant for individual data, like in this example. Mode is more important when we are dealing with class intervals of data when we prepare a frequency histogram. Note that the mean, median, mode, and standard deviation have the same units as the hydrologic variable while other statistical parameters (coefficients of Variation, Skewness, Kurtosis, and Excess) are dimensionless.

9.4

SOME IMPORTANT PROBABILITY DISTRIBUTIONS

Each hydrologic variable is unique in its nature and may follow a certain LO 4 Examine Binomial, probability distribution. There are several probability distributions available Gaussian, and Gumbel’s that can be used for analyzing hydrologic variables. Some probability Extreme Value Type-I distributions are discrete and some are continuous. In this section, we will distributions study the most commonly used probability distributions in hydrology. The first one is Binomial distribution which is a discrete distribution; while the other two—Gaussian and Gumbel’s distributions, are continuous distributions. Some more probability distributions suitable for flood frequency analysis will be discussed later.

9.4.1

Binomial Distribution

Binomial probability distribution is a discrete distribution that is applicable under the following conditions/ criteria: •

There are n occurrences of a random variable or trials of an experiment.

•

Each of the n occurrences or trials is independent of each other.

•

There are only two possible outcomes of each trial (e.g., occurrence or non-occurrence).

•

The probability of occurrence of the random variable (p) and non-occurrence (q) are known and are constant from one trial to the next.

A random variable satisfying the above four assumptions/criteria can be modeled using the Binomial distribution. For example, the toss of coin n times satisfies the above four conditions because the outcome of

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327

each toss is independent of each other, there are only two possible outcomes (head or tail), and the probability of getting a head (or tail) is 0.5, which remains constant from one trial to the next. In hydrology, one may be interested in finding out the probability of occurrence of a flood in any given year such that there are only two possible outcomes (flood or no flood), occurrence of flood being independent from one year to the next, and probability of occurrence (or non-occurrence) being constant from one year to the next. One can think of many such examples of the application of Binomial distribution in hydrology. The Binomial distribution gives a special type of probability, i.e., probability of occurrence of an event exactly r times in n successive trials. Such a probability is given as follows by the Binomial distribution: P(n, r ) = rnC pr q n - r =

n! pr q n - r (n - r )! r !

(9.42)

where, P(n, r) is the probability of occurrence of an event exactly r times in n trials, rnC is the combinatorial symbol representing the number of ways in which r trials out of n can occur, p is the probability of occurrence of the event in a single trial, and q is the probability of non-occurrence of the event in a single trial. Please note that the probability of non-occurrence of an event in n trials can be found by putting r = 0 in the Eq. (9.42) as P(n, 0) = qn. Similarly, the probability of occurrence of an event exactly n times in n trials can be found by putting r = n in the Eq. (9.42) as P(n, n) = pn. The probability of occurrence of an event at least once can be found as P(at least once) = 1 – qn. The probability of occurrence of an event “r or fewer times” in n successive trials can be found out as the cumulative Binomial probability as follows: r

P( X £ r ) = Â ni C pi q n - i

(9.43)

i =0

By taking appropriate moments of the Binomial distribution, it can be shown that the descriptors of the Binomial distribution can be obtained as follows: Mean Variance Standard Deviation Coefficient of Skewness Kurtosis Coefficient

m = np

(9.44)

s2 = npq

(9.45)

s = npq

(9.46)

g1 = g1 =

(q - p ) npq

(1 - 6 pq ) npq

(9.47) (9.48)

� EXAMPLE 9.6 IIT Kanpur will be organizing an international conference during the last week of June in the coming year. Based on the historical rainfall records, it has been found that the probability of having a rainy day (on any given day) in the last week of June is 0.30. Assuming that occurrence of rain on any day of the week is

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independent of the occurrence of rain on any other day, determine the following: (a) probability that there will be no rain during the conference week, (b) probability that it will rain on all seven days of the last week of June, (c) probability of having exactly 2 rainy days in the last week of June, (d) probability that there will be at least one rainy day in the last week of June, (e) probability that there will be at least four rainy days in the last week of June. Solution There are only two possible outcomes on a given day, i.e., it will rain or it will not rain. It is given that any day being rainy is independent of what happens on any other day, so trials are independent. Therefore, Binomial probability distribution is applicable with n = 7, p = 0.30, and q = 1 – p = 1 – 0.30 = 0.70. The various probabilities are calculated as follows: (a)

Probability that there will be no rain during the conference week, P{r = 0} This probability can be found by putting r = 0 in Eq. (9.42), or P(n, 0) = qn. Thus, P(7, 0) = (0.70)7 = 0.0824.

(b)

Probability that it will rain on all seven days of the last week of June, P{r = 7} This probability can be found by putting r = 7 in Eq. (9.42), or P(7, 7) = pn. Thus, P(7, 7) = (0.30)7 = 0.0002.

(c)

Probability of having exactly 2 rainy days in the last week of June, P{r = 2} This probability can be found by putting r = 2 in Eq. (9.42), or P(7, 2). P(7, 2) =

(d)

7! (0.3)2 (0.7)7 - 2 = 0.3177 (7 - 2)!2!

Probability that there will be at least one rainy day in the last week of June, P{at least once} The probability of occurrence of an event at least once can be found as P(at least once) = 1 – qn. Therefore, P{at least one rainy day} = 1 – 0.0824 = 0.9176 [qn = 0.0824 from part (a) above].

(e)

Probability that there will be at least four rainy days in the last week of June, P{at least 4}

The probability that there will be at least four rainy days in the last week of June means that there can be either 4 rainy days, or 5 rainy days, or 6 rainy days, or 7 rainy days. Thus, this probability can be found as P{at least 4 rainy days} = P{r = 4} + P{r = 5} + P{r = 6} + P{r = 7} and the individual probabilities can be found by putting r = 4, 5, 6, and 7 in Eq. (9.42) as follows: P{r = 4}

P(7, 4) =

7! (0.3)4 (0.7)7 - 4 = 0.0972 (7 - 4)!4!

P{r = 5}

P(7, 5) =

7! (0.3)5 (0.7)7 - 5 = 0.0250 (7 - 5)!5!

Statistical Methods in Hydrology

P{r = 6}

P(7, 6) =

7! (0.3)6 (0.7)7 - 6 = 0.0036 (7 - 6)!6!

P{r = 7}

P(7, 7) =

7! (0.3)7 (0.7)7 - 7 = 0.0002 (7 - 7)!7!

329

Therefore, P{at least 4} = 0.0972 + 0.0250 + 0.0036 + 0.0002 = 0.1260. Note that the probability that there will be at least 4 rainy days in the last week of June can also be found as follows: P{r ≥ 4} = 1 – P{r < 4} = 1 – P{r ≤ 3} = 1 – P{r = 0} – P{r = 1} – P{r = 2} – P{r = 3}.

9.4.2

Normal Probability Distribution

The most commonly used probability distribution, in almost all fields of engineering and sciences, is the normal probability distribution. It is also known as Gaussian distribution and is symmetric around the mean. It has a bell-shaped appearance and its pdf can be represented by the following expression: f ( x) =

È 1 Ê x - mˆ2˘ exp Í- Á ˜ ˙ for -• £ x £ • s 2p ÍÎ 2 Ë s ¯ ˙˚ 1

(9.49)

where, μ is the location parameter and s is the scale parameter of the normal distribution. The cumulative distribution function for the normal distribution can be determined by integrating Eq. (9.49). This is, however, done for the standard normal variable defined as follows: z=

x-m s

(9.50)

For a sample, mean x and standard deviation s can be used in place of population mean μ and population s. The pdf of the standard normal variable z then can be expressed as follows: f (z) =

1 2p

e- z

2

/2

(9.51)

Note that the transformed standard normal variable z has a mean of 0.0 and a standard deviation (or variance) equal to 1.0. The area under the curve represented by Eq. (9.51) is evaluated by integrating the density of the standard normal distribution. This is normally available in standard tables for certain limits of the standard normal random variable, usually between 0.0 and 3.49. The values of cumulative probabilities for a standard normal distribution, F(z) are provided in Table 9.3. The standard normal variable ranges from -∞ to +∞; however, the values of F(z) are listed for positive values of z only (Figure 9.10). The values of F(z) for the negative values of z can be determined by exploiting the fact that the normal probability distribution is symmetric in nature. In order to use the table for z < 0, we use the relation F(z) = 1 – F(|z|), where F(|z|) is the value tabulated in Table 9.3.

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Engineering Hydrology

Table 9.3 Cumulative probabilities for standard normal distribution, F(z) .00

.01

.02

.03

.04

.05

.06

.07

08

.09

0.0 0.1 0.2 0.3 0.4

0.5000 0.5398 0.5793 0.6179 0.6554

0.5040 0.5438 0.5832 0.6217 0.6591

0.5080 0.5478 0.5871 0.6255 0.6628

0.5120 0.5517 0.5910 0.6293 0.6664

0.5160 0.5557 0.5948 0.6331 0.6700

0.5199 0.5596 0.5987 0.6368 0.6736

0.5239 0.5636 0.6026 0.6406 0.6772

0.5279 0.5675 0.6064 0.6443 0.6808

0.5319 0.5714 0.6103 0.6480 0.6844

0.5359 0.5753 0.6141 0.6517 0.6879

0.5 0.6 0.7 0.8 0.9

0.6915 0.7257 0.7580 0.7881 0.8159

0.6950 0.7291 0.7611 0.7910 0.8186

0.6985 0.7324 0.7642 0.7939 0.8212

0.7019 0.7357 0.7673 0.7967 0.8238

0.7054 0.7389 0.7704 0.7995 0.8264

0.7088 0.7422 0.7734 0.8023 0.8289

0.7123 0.7454 0.7764 0.8051 0.8315

0.7157 0.7486 0.7794 0.8078 0.8340

0.7190 0.7517 0.7823 0.8106 0.8365

0.7224 0.7549 0.7852 0.8133 0.8389

1.0 1.1 1.2 1.3 1.4

0.8413 0.8643 0.8849 0.9032 0.9192

0.8438 0.8665 0.8869 0.9049 0.9207

0.8461 0.8686 0.8888 0.9066 0.9222

0.8485 0.8708 0.8907 0.9082 0.9236

0.8508 0.8729 0.8925 0.9099 0.9251

0.8531 0.8749 0.8944 0.9115 0.9265

0.8554 0.8770 0.8962 0.9131 0.9279

0.8577 0.8790 0.8980 0.9147 0.9292

0.8599 0.8810 0.8997 0.9162 0.9306

0.8621 0.8830 0.9015 0.9177 0.9319

1.5 1.6 1.7 1.8 1.9

0.9332 0.9452 0.9554 0.9641 0.9713

0.9345 0.9463 0.9564 0.9649 0.9719

0.9357 0.9474 0.9573 0.9656 0.9726

0.9370 0.9484 0.9582 0.9664 0.9732

0.9382 0.9495 0.9591 0.9671 0.9738

0.9394 0.9505 0.9599 0.9678 0.9744

0.9406 0.9515 0.9608 0.9686 0.9750

0.9418 0.9525 0.9616 0.9693 0.9756

0.9429 0.9535 0.9625 0.9699 0.9761

0.9441 0.9545 0.9633 0.9706 0.9767

2.0 2.1 2.2 2.3 2.4

0.9772 0.9821 0.9861 0.9893 0.9918

0.9778 0.9826 0.9864 0.9896 0.9920

0.9783 0.9830 0.9868 0.9898 0.9922

0.9788 0.9834 0.9871 0.9901 0.9925

0.9793 0.9838 0.9875 0.9904 0.9927

0.9798 0.9842 0.9878 0.9906 0.9929

0.9803 0.9846 0.9881 0.9909 0.9931

0.9808 0.9850 0.9884 0.9911 0.9932

0.9812 0.9854 0.9887 0.9913 0.9934

0.9817 0.9857 0.9890 0.9916 0.9936

2.5 2.6 2.7 2.8 2.9

0.9938 0.9953 0.9965 0.9974 0.9981

0.9940 0.9955 0.9966 0.9975 0.9982

0.9941 0.9956 0.9967 0.9976 0.9982

0.9943 0.9957 0.9968 0.9977 0.9983

0.9945 0.9959 0.9969 0.9977 0.9984

0.9946 0.9960 0.9970 0.9978 0.9984

0.9948 0.9961 0.9971 0.9979 0.9985

0.9949 0.9962 0.9972 0.9979 0.9985

0.9951 0.9963 0.9973 0.9980 0.9986

0.9952 0.9964 0.9974 0.9981 0.9986

3.0 3.1 3.2 3.3 3.4

0.9987 0.9990 0.9993 0.9995 0.9997

0.9987 0.9991 0.9993 0.9995 0.9997

0.9987 0.9991 0.9994 0.9995 0.9997

0.9988 0.9991 0.9994 0.9996 0.9997

0.9988 0.9992 0.9994 0.9996 0.9997

0.9989 0.9992 0.9994 0.9996 0.9997

0.9989 0.9992 0.9994 0.99% 0.9997

0.9989 0.9992 0.9995 0.9996 0.9997

0.9990 0.9993 0.9995 0.9996 0.9997

0.9990 0.9993 0.9995 0.9997 0.9998

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331

Area

0

z

Figure 9.10 F(z) as area under pdf for Normal distribution for positive values of z As mentioned previously, the normal probability distribution is the most commonly used distribution in hydrology and other areas. For water resources management, the value of F(z) is needed repeatedly in various mathematical and management models. Therefore, the use of tables for estimating the cumulative probability of the standard normal variable poses a problem as it requires manual estimation. Alternatively, the following expressions can be used (Abramowitz and Stegun, 1965). F (z) = B = 1- B

for z < 0 for z ≥ 0

-4 1 2 3 4 B = ÈÍ1 + b1 z + b2 z + b3 z + b 4 z ˘˙ ˚ 2Î

(9.52)

where, b1 = 0.196854, b2 = 0.115194, b3 = 0.000344, and b4 = 0.019527. The values of F(z) estimated by Eq. (9.52) has an error less than 0.00025. Sometimes, we need to determine the value of a random variable (z) for a given cumulative normal probability, F(z). This can be done as follows: p = 1 – F(z) Ê 1 ˆ w = ln Á 2 ˜ for 0 ≤ p ≤ 0.50 Ëp ¯ z=w-

(9.53)

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

For p > 0.50, (1 – p) is substituted in Eq. (9.53). The error in estimating z like this is less than 0.00045. Note that the approximate methods given by Equsations (9.52) and (9.53) can be easily implemented in a computer.

Some Useful Deductions Following are some of the useful deductions related to the normal distribution.

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1.

If X is a random variable with a mean of μ and a variance of s2, i.e., X = N(μ, s2) and Y is another random variable such that Y = a + bX, then Y is also normally distributed with a mean of a + bμ and a variance of b2s2, i.e., Y = N(a + bμ, b2s2).

2.

The above can be extended if Y is a function of more than one random variable. Let Xi, i = 1, 2 2, …, n be n independent normal random variables with means of μi and variances of s i ; and Y = a + b1X1 + b2X2 + …. +bnXn. Then, Y is also normally distributed with a mean of μy and variances of s y2 , Y = N ( m y , s y2 ) , such that m y = a + b1 m1 + b2 m2 +� + bn m n s y2 = b1s 12 + b2s 22 + � + bns n2

(9.54)

In other words, any linear combination of independent normal variables will also be normally distributed. � EXAMPLE 9.7 The annual rainfall in a catchment can be modeled using the Normal distribution with a mean of 120 cm and a standard deviation of 20 cm. Find the probability that the annual rainfall will be (a) greater than 150 cm (b) between 70 cm and 130 cm. Solution Let X be the random variable representing annual rainfall such that X = N(120, 20). The standard normal variable can therefore be represented as follows: z= (a)

x - 120 = N (0, 1) 20

Probability that the annual rainfall will be greater than 150 cm, P{X > 150}: This can be evaluated as P{X > 150} = 1 – P{X ≤ 150} = 1 – F{X = 150} = 1 – F{z = (150 – 120)/20 = 1.50}. The value of cumulative probability for z = 1.50 can be found either by normal tables or by the Eq. (9.52). By putting z = 1.5 in Eq. (9.52), we get, B = 0.0667. Therefore, F(z) = 1 – B = 1 – 0.0667 = 0.9333 because z is positive. The required probability then is found as P{X > 150} = 1 – F{z = 1.50} = 1 – 0.0.9333 = 0.0667 = 6.67%.

(b)

Probability that the annual rainfall will be between 70 cm and 130 cm, P{70 ≤ X ≤ 130}: For x = 70 cm, z = (70 – 120)/20 = –2.5, B = 0.0063, F{z = –2.5} = B = 0.0063 because z is negative. Similarly, for x = 130 cm, z = (130 – 120)/20 = 0.5, B = 0.3083, F{z = 0.5} = 1 – B = 1 – 0.3083 = 0.6917 because z is positive. Therefore, the required probability is found as P{70 ≤ X ≤ 130} = F{x = 130} – F{x = 70} = F{z = 0.5} – F{x = –2.5} = 0.6917 – 0.0063 = 0.6854.

9.4.3

Gumbel’s Probability Distribution

Also known as Extreme Value Type-I (EVT-I) distribution or Double Exponential distribution, it is commonly used in hydrology for modeling the random variables of larger extremes, such as annual peak discharge,

Statistical Methods in Hydrology

333

maximum temperature in a year, maximum 24-hour rainfall in a year, etc. The density and cumulative distribution function of the Gumbel’s distribution are expressed as follows: f ( x ) = a exp[–a ( x - b ) - exp{-a ( x - b )}]

(9.55)

F ( x ) = exp [- exp{-a ( x - b )}]

(9.56)

where, x is a random variable following Gumbel’s distribution (–∞ < x < ∞), a and b are parameters of the distribution such that a > 0 and –∞ < b < ∞. Note that b is a location parameter equivalent to the mode of the distribution and a is a scale parameter. The parameters and various moments of the distributions can be calculated as follows: m=b+

Mean

s2 =

Variance Coefficient of Skewness Kurtosis Coefficient

p2 6a

2

0.5772 a =

1.645 a2

(9.57) (9.58)

g1 = 1.1396

(9.59)

g1 = 5.4

(9.60)

a=

1.28255 s

b = m - 0.45005s

(9.61) (9.62)

Letting y = a (x – b), the pdf and cdf for the Gumbel’s distribution can be rewritten as follows: f ( x ) = a exp[– y - exp{- y}]

(9.63)

F ( x ) = exp[- exp{- y}]

(9.64)

where, y is called the Gumbel’s reduced variate. � EXAMPLE 9.8 Assuming the maximum instantaneous flood peak data for a period of 25 years in Example 9.5 follow the Gumbel’s distribution, answer the following: (a)

What is the probability that the maximum instantaneous flood peak will be greater than 6000 m3/s?

(b)

What is the magnitude of the maximum instantaneous flood peak that will be exceeded with a probability of 20%?

Solution From Example 9.5, we have, mean (μ) = 4889.5 m3/s and standard deviation (s) = 868.1 m3/s. The first step is to calculate the parameters of the Gumbel’s probability distribution as follows: a=

1.28255 1.28255 = = 0.00148 s 868.1

b = m - 0.45005s = 4889.5 - 0.45005(868.1) = 4, 498.8

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Engineering Hydrology

(a)

P{X > 6,000}: P{X > 6,000} = 1 – P{X ≤ 6,000} = 1 – F(x = 6,000) Let us first calculate Gumbel’s reduced variable y for x = 6,000 m3/s as y = a(x – b) = 0.00148 (6,000 – 4498.8) = 2.22. The cumulative Gumbel’s probability for y = 2.22 is then calculated by Eq. (9.64) as follows: F ( x ) = exp[ - exp{-2.22}] = 0.8969 Therefore, P{X > 6,000} = 1 – P{X ≤ 6,000} = 1 – F(x = 6,000) = 1 – 0.8969 = 0.1031 = 10.31%.

(b)

x = ? for p = 20%: For exceedance probability p = 0.20, we have P{X > x} = 1 – P{X ≤ x} = 1 – F(x) = 0.20. Therefore, F(x) = 1 – 0.20 = 0.80 and for this value of F(x), we first find Gumbel’s reduced variable from Eq. (9.64) as follows: y = - ln[ - ln{0.80}] = 1.50

For y = 1.50, the value of x can be calculated using the relation y = a (x – b) as follows: x=

y 1.50 +b= + 4, 498.8 = 5,514.1 m 3 /s a 0.00148

Thus, the magnitude of the maximum instantaneous flood peak that will be exceeded with a probability of 20% is 5514.1 m3/s. There are several other discrete probability distributions, such as negative binomial distribution, geometric distribution, Poisson distribution, and continuous probability distributions e.g. log-normal distribution, gamma distribution, exponential distribution, beta distribution, and uniform distribution. These are beyond the scope of this book.

9.5

FREQUENCY ANALYSIS

Water resources systems are normally designed to withstand extreme LO 5 Demonstrate flood events, such as severe storm, extreme floods and droughts. Therefore, the frequency analysis using magnitude of an extreme hydrologic event needs to be calculated for a plotting position and general certain frequency of occurrence, for use in hydrologic design. The objective frequency factor methods of frequency analysis is to establish a relationship between the magnitude of a hydrologic variable and its frequency of occurrence. The frequency of occurrence is usually expressed using return period. Return period, also known as the recurrence interval, is the average time interval after which the magnitude of a hydrologic variable will be equaled or exceeded. The return period is generally expressed in years and is reciprocal of the exceedance probability. T=

1 = P[ X > x ] p

(9.65)

where, T is the recurrence interval in years and p is the exceedance probability. The recurrence interval is defined statistically and there are no guarantees that a 10-year flood will occur after 10-years if it has occurred this year. The magnitude of an extreme event is directly proportional to its frequency of occurrence in terms of return period. For example the magnitude of a 100-year flood will be much more than that of a 10-year

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335

flood. Observed hydrologic data is needed to carry out the frequency analysis. The data is assumed to be independent of each other and to follow the same probability distribution, or in other words, is identically distributed. Further, the hydrologic variable is assumed to be stochastic in nature, and spatially and temporally uncorrelated although such assumptions may not be valid in real life. There are many methods available for carrying out the frequency analysis. We consider only a few methods here.

9.5.1

Probability Plotting Method

Given sample data, probability plotting method is the easiest way to establish a relationship between the magnitude of a hydrologic variable and its frequency of occurrence. The given data is arranged in descending order of magnitude with the highest value at the top. The first observation at the top is assigned a rank of 1, the next highest value is assigned a rank of 2, and so on such that the mth value in the descending order will have a rank of m and the last value in the table will have a rank of n for sample of size n. Then the probability plotting position or exceedance probability for each magnitude of the hydrologic variable is calculated as follows: p = P ÈÎ X > xm ˘˚ =

m-b n + 1 - 2b

(9.66)

where, m is the rank of the mth observation in the descending order, n is the total number of observations in the dataset, and b is a coefficient depending on the method of calculating the plotting position. There are several formulas available for calculating the plotting position; some of them are presented in Table 9.4.

Table 9.4 Plotting position formulas S. No.

Formula

Value of b

1

Hazen

0.50

2

Chegodeyev

0.30

3

Weibull

0.00

4

Blom

3/8

5

Tukey

1/3

6

Gringorten

0.44

The best method for a particular dataset is the one that provides the plotting position with no bias and minimum variance. The Weibull’s formula is the most commonly used formula for calculating the plotting position in hydrology, which is expected to provide an unbiased estimate of exceedance probability with minimum variance. The exceedance probability or plotting position by Weibull’s method can be expressed as follows: p = P ÈÎ X > xm ˘˚ =

m n +1

(9.67)

Once the plotting position values are calculated, it is plotted on x-axis and the magnitude of the hydrologic variable is plotted on the y-axis to develop the plotting position curve for a particular dataset. Instead of

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using exceedance probability, return period can also be plotted on the x-axis. The relationship between the magnitude and exceedance probability is expected to be linear and is very useful in finding the magnitude of the hydrologic variable for any return period either within the range of the data considered (interpolation) or outside the range of the data considered (extrapolation). � EXAMPLE 9.9 The data for a 30-minute annual maximum rainfall from 1975 to 2009 is given in the table below. Calculate the plotting position (p) and corresponding return period (T) for each data point using Weibull’s, Hazen, and Chegodeyev formulas. Then plot the data and plotting position (T) for the three formulas. Year

Rainfall (mm)

Year

Rainfall (mm)

1975

15

1993

35

1976

24

1994

39

1977

11

1995

19

1978

22

1996

26

1979

20

1997

34

1980

15

1998

23

1981

27

1999

25

1982

16

2000

19

1983

27

2001

13

1984

27

2002

22

1985

24

2003

20

1986

24

2004

45

1987

21

2005

21

1988

22

2006

24

1989

43

2007

19

1990

15

2008

22

1991

12

2009

31

1992

29

Solution The plotting position computations are arranged in Table 9.5. The first two columns present the data given and the column [3] presents the rainfall data sorted in descending order. Column [4] gives the rank for each sorted rainfall value. Columns [5], [6], and [7] present the return period (=1/p) calculated as per the Weibull, Hazen, and Chigodeyev’s formulas.

Statistical Methods in Hydrology

Table 9.5 Plotting position computations for Example 9.9 Year

Rainfall

Sorted

Rank (m)

T-Weibull

T-Hazen

T-Chegodayev

[1]

[2]

[3]

[4]

[5]

[6]

[7]

1975

15

45

1

36.00

70.00

50.57

1976

24

43

2

18.00

23.33

20.82

1977

11

39

3

12.00

14.00

13.11

1978

22

35

4

9.00

10.00

9.57

1979

20

34

5

7.20

7.78

7.53

1980

15

31

6

6.00

6.36

6.21

1981

27

29

7

5.14

5.38

5.28

1982

16

27

8

4.50

4.67

4.60

1983

27

27

9

4.00

4.12

4.07

1984

27

27

10

3.60

3.68

3.65

1985

24

26

11

3.27

3.33

3.31

1986

24

25

12

3.00

3.04

3.03

1987

21

24

13

2.77

2.80

2.79

1988

22

24

14

2.57

2.59

2.58

1989

43

24

15

2.40

2.41

2.41

1990

15

24

16

2.25

2.26

2.25

1991

12

23

17

2.12

2.12

2.12

1992

29

22

18

2.00

2.00

2.00

1993

35

22

19

1.89

1.89

1.89

1994

39

22

20

1.80

1.79

1.80

1995

19

22

21

1.71

1.71

1.71

1996

26

21

22

1.64

1.63

1.63

1997

34

21

23

1.57

1.56

1.56

1998

23

20

24

1.50

1.49

1.49

1999

25

20

25

1.44

1.43

1.43

2000

19

19

26

1.38

1.37

1.38

2001

13

19

27

1.33

1.32

1.33

2002

22

19

28

1.29

1.27

1.28

2003

20

16

29

1.24

1.23

1.23

2004

45

15

30

1.20

1.19

1.19

2005

21

15

31

1.16

1.15

1.15

2006

24

15

32

1.13

1.11

1.12

2007

19

13

33

1.09

1.08

1.08

2008

22

12

34

1.06

1.04

1.05

2009

31

11

35

1.03

1.01

1.02

337

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Engineering Hydrology

The plotting positions are plotted on the same graph and shown in Figure 9.11.

Figure 9.11 Graph of plotting positions for Example 9.9

9.5.2

Frequency Factor Method

Calculation of the magnitude of a hydrologic variable for a particular return period T (xT ) requires the probability distribution to be invertible. Although it is possible for some of the probability distributions, while some other distributions are not invertible. For such probability distributions, the frequency factor method is used. However, in general the frequency factor method can be used for any probability distribution. As per the frequency factor method, the magnitude xT can be expressed as the sum of the mean (μ) and a small deviation ΔxT equal to the product of frequency factor (KT) and standard deviation (s) of the distribution. This concept is graphically depicted in Figure 9.12 and can be mathematically expressed in the following equation (Chow, 1951): xT = m + KTs

(9.68)

Equation (9.68) is for the population and for a sample of finite size, sample estimates of population mean and standard deviation can be used as follows: xT = x + KT s

(9.69)

where, xT is the magnitude of the hydrologic variable corresponding to a return period of T-years, x is the sample mean, s is the sample standard deviation, and KT is the frequency factor. Note that the frequency factor depends on the return period and the type of probability distribution used in the frequency analysis. Some of the probability distributions used in hydrology involve log, in such cases the frequency analysis using the method of frequency factor can be applied to the log transformed data y = log (x). In such cases, the general frequency equation can be written as follows: yT = y + KT s y

(9.70)

339

f (x )

Statistical Methods in Hydrology

•

P(x ≥ xT ) =

1 f (x)dx T xÚ T

KTs m

xT x

Figure 9.12 Concept of the frequency factor method where, yT is the value of the log-transformed hydrologic variable corresponding to a return period of T-years, y is the sample mean of log-transformed series, sy is the standard deviation of the log-transformed series, and KT is the frequency factor. The required value of xT is then estimated by taking antilog of yT. The values of the frequency factor (KT) for different return periods are normally available in standard statistical tables for different probability distributions. The calculation of xT by frequency factor method is very easy and involves three steps: (i) first, calculate the statistical parameters of the probability distribution using the sample data, (ii) second, calculate or read out the frequency factor KT from standard table for the particular probability distribution, and (iii) third, use Eq. (9.69) or (9.70), as the case may be, to calculate xT. The procedure of using the frequency factor method for some commonly used probability distributions is discussed here.

9.5.2.1

Normal Distribution

The frequency factor can be expressed as follows using Eq. (9.69): KT =

xT - m s

(9.71)

The right hand side of Eq. (9.71) is same as the standard normal variable (z) defined earlier. The value of z corresponding to a particular value of exceedance probability (p = 1/T) or (1 – F(z)) can be found using the Abramowitz and Stegun (1965) approximation discussed earlier and presented in Eq. (9.53). The value of z (= KT) can also be found out by using the standard normal tables for a particular value of F(z). Once the value of frequency factor is calculated for a given T, xT can be easily calculated from Eq. (9.69) knowing the sample mean and standard deviations.

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340

9.5.2.2

Gumbel’s Distribution

The frequency factor for Gumbel’s probability distribution for a sample of finite size can be expressed as follows (Chow, 1953): KT =

yT - yn sn

(9.72)

where, yT is the Gumbel’s reduced variable, yn and sn are the Gumbel parameters for finite n, called reduced mean and reduced standard deviations, respectively. The Gumbel’s parameters ( yn and sn) depend on the size of the sample and are provided in standard tables. Note that for very large value of n (tending to ∞), the value of reduced mean tends to 0.5772 and the value of reduced standard deviation tends to 1.2825. The values of yn and sn are provided in Table 9.6.

Table 9.6 Reduced mean and standard deviation for Gumbel’s distribution – (a) Reduced Mean y n in Gumbel’s Extreme Value Distribution (for sample size N) N

0

1

2

3

4

5

6

7

8

9

10

0.4952

0.4996

0.5035

0.5070

0.5100

0.5128

0.5157

0.5181

0.5202

0.5220

20

0.5236

0.5252

0.5268

0.5283

0.5296

0.5309

6.5320

0.5332

0.5343

0.5353

30

0.5362

0.5371

0.5380

0.5388

0.5396

0.5402

0.5410

0.5418

0.5424

0.5430

40

0.5436

0.5442

0.5448

0.5453

0.5458

0.5463

0.5468

0.5473

0.5477

0.5481

50

0.5485

0.5489

0.5493

0.5497

0.5501

0.5504

0.5508

0.5511

0.5515

0.5518

60

0.5521

0.5524

0.5527

0.5530

0.5533

0.5535

0.5538

0.5540

0.5543

0.5545

70

0.5548

0.5550

0.5552

0.5555

0.5557

0.5559

0.5561

0.5563

0.5565

0.5567

80

0.5569

0.5570

0.5572

0.5574

0.5576

0.5578

0.5580

0.5581

0.5583

0.5585

90

0.5586

0.5587

0.5589

0.5591

0.5592

0.5593

0.5595

0.5596

0.5598

0.5599

100

0.5600

(b) Reduced Standard Deviation Sn in Gumbel’s Extreme Value Distribution (for sample size N) N

0

1

2

3

4

5

6

7

8

9

10

0.9496

0.9676

0.9833

0.9971

1.0095

1.0206

1.0316

1.0411

1.0493

1.0565

20

1.0628

1.0696

1.0754

1.0811

1.0864

1.0915

1.0961

1.1004

1.1047

1.1086

30

1.1124

1.1159

1.1193

1.1226

1.1255

1.1285

1.1313

1.1339

1.1363

1.1388

40

1.1413

1.1436

1.1458

1.1480

1.1499

1.1519

1.1538

1.1557

1.1574

1.1590

50

1.1607

1.1623

1.1638

1.1658

1.1667

1.1681

1.1696

1.1708

1.1721

1.1734

60

1.1747

1.1759

1.1770

1.1782

1.1793

1.1803

1.1814

1.1824

1.1834

1.1844

70

1.1854

1.1863

1.1873

1.1881

1.1890

1.1898

1.1906

1.1915

1.1923

1.1930

80

1.1938

1.1945

1.1953

1.1959

1.1967

1.1973

1.1980

1.1987

1.1994

1.2001

90

1.2007

1.2013

1.2020

1.2026

1.2032

1.2038

1.2044

1.2049

1.2055

1.2060

100

1.2065

Statistical Methods in Hydrology

341

The Gumbel’s reduced variable can be calculated using the equation for the Gumbel’s cumulative distribution function. Equation (9.64) can be written as follows: F ( x ) = e- e

-y

(9.73)

Expressing F(x) = 1 – p = (1 – 1/T), taking log on both sides of Eq. (9.73), and denoting Gumbel’s reduced variable (y) for a particular exceedance probability (p) or return period (T) as yT, we can write: È Ê T ˆ˘ yT = - Íln. ln Á Ë T - 1˜¯ ˙˚ Î

(9.74)

Thus, the procedure for calculating the value of a hydrologic variable (xT) corresponding to a particular return period as per the Gumbel’s distribution for a finite sized sample can be summarized as follows: 1.

From the given sample data, calculate the mean ( x ) and standard deviation (s).

2.

For the given size of sample (n), use standard Gumbel’s tables (Table 9.6) to read out the values of Gumbel’s parameters, reduced mean ( yn ) and reduced standard deviation (sn).

3.

Using Eq. (9.74), calculate Gumbel’s reduced variable (yT) for the desired value of return period (T) or exceedance probability.

4.

Using Eq. (9.72), calculate the frequency factor KT.

5.

Using Eq. (9.69), calculate the value of hydrologic variable corresponding to a desired return period of T (xT).

The above procedure can be used to calculate xT for different values of T. Then the xT versus T data can be plotted on the semi-log, log-log, or Gumbel’s probability paper. If the data follows the Gumbel’s distribution, then the xT versus T data will follow a straight line when plotted on the Gumbel’s probability paper. This straight line is called the theoretic line. The value of xT for T = 2.33 years is known as the mean annual flood for Gumbel’s distribution. Such a straight line fit can be used to interpolate and extrapolate the values of xT for any value of T. In case Gumbel’s probability paper is not available, the theoretic line calculated above may plot as a non-linear curve when plotted on a simple graph paper (as semi-log, or log-log plot). The plotting positions are then calculated using the Weibull’s formula and plotted on the same graph paper (semi-log, or log-log). If the calculated plotting position points (by Weibull’s formula) closely follow the theoretical line, then also the data follows the Gumbel’s distribution. Although there are several statistical methods available to determine whether a given dataset follows the Gumbel’s distribution, the procedure described above offers a simple graphical method for the same purpose.

9.5.2.3

Log-Pearson Type-III Distribution

Most of the hydrologic variables for extreme values are skewed in nature. The methods discussed so far assume the data to be symmetric or do not account for the skewness in the data. The Log-Pearson Type-III distribution is useful in frequency analysis of the data for hydrologic variables that are skewed in nature. The data series is first transformed to a log series (base 10) and the frequency analysis is carried out using Eq. (9.70). The values of mean ( y ), standard deviation (sy), and coefficient of skewness (Cs) are first calculated for the log-transformed (y) series. The frequency factor for Log-Pearson Type-III distribution is a function of return period T and the coefficient of skewness Cs. The frequency factor (KT) for the Log-Pearson Type-III distribution can be approximated as per Kite (1977) as follows:

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1 1 (9.75) KT = z + ( z 2 - 1)k + ( z 3 - 6 z )k 2 - ( z 2 - 1)k 3 + zk 4 + k 5 3 3 where, k = Cs/6 and z is standard normal variable for a given return period (T) or exceedance probability p = 1/T = (1 – F(z)), which can be found by using the Abramowitz and Stegun (1965) approximation discussed earlier and presented in Eq. (9.53). The values of frequency factor KT for different T and Cs are also available in standard tables for the Log-Person Type-III distribution and can be directly read out (Table 9.7a and Table 9.7b). Once the value of KT has been determined, Eq. (9.70) can be used to calculate yT. Then, taking antilog, the value of xT can be calculated. Note that the Log-Pearson Type-III distribution reduces to Log-Normal distribution for Cs = 0.0 as the Log-Normal distribution is symmetric in nature. Therefore, the procedure outlined above for frequency analysis for Log-Pearson Type-III distribution along with Table 9.7 can be used for Log-Normal distribution also. � EXAMPLE 9.10 For the data given in Example 9.9 above, determine the 30-minute annual maximum rainfall for 50-year, 100year, and 500-year return periods using Normal, Gumbel’s Extreme Value Type-I, Log-Pearson Type-III, and Log-Normal distributions. Compare your results. Solution For the data in Example 9.9, the basic descriptive statistics is calculated as follows: x = 23.74, sx = 8.11, y = 1.3519, sy = 0.1457, and Cs–y = –0.0245 Note that the parameters with subscript y are for the log-transformed series, i.e., y = log x. The general frequency factor formula [Eq. (9.69) or Eq. (9.70)] is used for calculating the 30-minute annual maximum rainfall values by different distributions. (a) Normal Distribution For the Normal distribution, the frequency factor is same as the standard normal variable (z). The value of z corresponding to a particular value of exceedance probability (p = 1/T) or (1 – F(z)) can be found by using the Abramowitz and Stegun (1965) approximation discussed earlier and presented in Eq. (9.53). Once z is calculated for each T, general frequency factor formula is used to calculated xT for a particular return period T. The computations are shown in the following table. By Normal Distribution T (Yrs)

p

w

z

xT

50

0.0200

2.797

2.0537

40.4

100

0.0100

3.035

2.3263

42.6

500

0.0020

3.526

2.8779

47.1

(b) Gumbel’s Distribution For the size of the sample n = 35, the Gumbel’s parameters are read out from Table 9.6 as follows: yn = 0.5402 and sn = 1.1285. Then, the Gumbel’s reduced variate yT for each return period T is calculated

Statistical Methods in Hydrology

343

from Eq. (9.74) and the corresponding value of Gumbel’s frequency factor KT for each return period is calculated from Eq. (9.72). Then, general frequency factor formula is used to calculated xT for a particular return period T. The computations are shown in the following table. By Gumbel’s Distribution T (Yrs)

yT

KT

xT

50

3.9019

2.979

47.9

100

4.6001

3.598

52.9

500

6.2136

5.027

64.5

(c) Log-Pearson Type-III Distribution For k = Cs/6 = –0.0245/6 = –0.004088 and z calculated for Normal distribution for different T earlier, the frequency factor KT for the Log-Pearson Type-III distribution is calculated from Eq. (9.75). Once KT is calculated for each T, general frequency factor formula for the log-transformed data, given in Eq. (9.70), is used to calculate yT for each return period T. Then xT is calculated by taking antilog of yT for each T. The computations are shown in the following table. By Log Pearson Type III Distribution T (Yrs)

z

KT

yT

xT

50

2.0537

2.0537

1.6511

44.8

100

2.3263

2.3263

1.6908

49.1

500

2.8779

2.8779

1.7712

59.0

(d) Log Normal Distribution For Log-Normal distribution, coefficient of skew is zero. Therefore, Cs = 0.0 is taken and the remaining calculations remain the same as those for the Log-Pearson Type-III distribution presented above. The computations for finding xT for different T using Log-Normal distribution are shown in the following table. By Log-Normal Distribution T (Yrs)

p

w

z

yT

xT

50

0.0200

2.797

2.0537

1.6511

44.8

100

0.0100

3.035

2.3263

1.6908

49.1

500

0.0020

3.526

2.8779

1.7712

59.0

A comparison of the rainfall values for different return periods from Log-Pearson Type-III and LogNormal distributions indicates that the values obtained from the two distributions are the same. This is because the value of coefficient of skewness is nearly equal to zero for the data. Further, the Normal distribution gives minimum values for each return period among all the probability distributions and T. The Gumbel’s distribution gives the highest values for 30-minute annual maximum rainfall data considered. The extreme values are not properly modeled by the Normal or Log-Normal distributions. Therefore, Gumbel’s distribution is the most suitable probability distribution for the 30-minute annual maximum rainfall data.

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Engineering Hydrology

Table 9.7(a) Frequency factor (KT) for Log-Pearson Type-III distribution (positive skew) Return Period in Years Skew coefficient C

2

5

10

25

0.50

0.20

0.10

0.04

3.0

–0.396

0.420

1.180

2.9

–0.390

0.440

2.8

–0.384

0.460

2.7

–0.376

2.6 2.5

50

100

200

0.02

0.01

0.005

2.278

3.152

4.051

4.970

1.195

2.277

3.134

4.013

4.909

1.210

2.275

3.114

3.973

4.847

0.479

1.224

2.272

3.093

3.932

4.783

–0.368

0.499

1.238

2.267

3.071

3.889

4.718

–0.360

0.518

1.250

2.262

3.048

3.845

4.652

2.4

–0.351

0.537

1.262

2.256

3.023

3.800

4.584

2.3

–0.341

0.555

1.274

2.248

2.997

3.753

4.515

2.2

–0.330

0.574

1.284

2.240

2.970

3.705

4.444

2.1

–0.319

0.592

1.294

2.230

2.942

3.656

4.372

2.0

–0.307

0.609

1.302

2.219

2.912

3.605

4.298

1.9

–0.294

0.627

1.310

2.207

2.881

3.553

4.223

1.8

–0.282

0.643

1.318

2.193

2.848

3.499

4.147

1.7

–0.268

0.660

1.324

2.179

2.815

3.444

4.069

1.6

–0.254

0.675

1.329

2.163

2.780

3.388

3.990

1.5

–0.240

0.690

1.333

2.146

2.743

3.330

3.910

1.4

–0.225

0.705

1.337

2.128

2.706

3.271

3.828

1.3

–0.210

0.719

1.339

2.108

2.666

3.211

3.745

1.2

–0.195

0.732

1.340

2.087

2.626

3.149

3.661

1.1

–0.180

0.745

1.341

2.066

2.585

3.087

3.575

1.0

–0.164

0.758

1.340

2.043

2.542

3.022

3.489

0.9

–0.148

0.769

1.339

2.018

2.498

2.957

3.401

0.8

–0.132

0.780

1.336

1.993

2.453

2.891

3.312

0.7

–0.116

0.790

1.333

1.967

2.407

2.824

3.223

0.6

–0.099

0.800

1.328

1.939

2.359

2.755

3.132

0.5

–0.083

0.808

1.323

1.910

2.311

2.686

3.041

0.4

–0.066

0.816

1.317

1.880

2.261

2.615

2.949

0.3

–0.050

0.824

1.309

1.849

2.211

2.544

2.856

0.2

–0.033

0.830

1.301

1.818

2.159

2.472

2.763

0.1

–0.017

0.836

1.292

1.785

2.107

2.400

2.670

0.0

0

0.842

1.282

1.751

2.054

2.326

2.576

Exceedence probability

Statistical Methods in Hydrology

Table 9.7(b) Frequency factor (KT) for Log-Pearson Type-III distribution (negative skew) Return period in years Skew coefficient Cs

2

5

10

25

0.50

0.20

0.10

0.04

–0.1

0.017

0.846

1.270

–0.2

0.033

0.850

–0.3

0.050

0.853

–0.4

0.066

–0.5 –0.6

50

100

200

0.02

0.01

0.005

1.716

2.000

2.252

2.482

1.258

1.680

1.945

2.178

2.388

1.245

1.643

1.890

2.104

2.294

0.855

1.231

1.606

1.834

2.029

2.201

0.083

0.856

1.216

1.567

1.777

1.955

2.108

0.099

0.857

1.200

1.528

1.720

1.880

2.016

–0.7

0.116

0.857

1.183

1.488

1.663

1.806

1.926

–0.8

0.132

0.856

1.166

1.448

1.606

1.733

1.837

–0.9

0.148

0.854

1.147

1.407

1.549

1.660

1.749

–1.0

0.164

0.852

1.128

1.366

1.4M2

1.588

1.664

–1.1

0.180

0.848

1.107

1.324

1.435

1.518

1.581

–1.2

0.195

0.844

1.086

1.282

1.379

1.449

1.501

–1.3

0.210

0.838

1.064

1.240

1.324

1.383

1.424

–1.4

0.225

0.832

1.041

1.198

1.270

1.318

1.351

–1.5

0.240

0.825

1.018

1.157

1.217

1.256

1.282

–1.6

0.254

0.817

0.994

1.116

1.166

1.197

1.216

–1.7

0.268

0.808

0.970

1.075

1.1 16

1.140

1.155

–1.8

0.282

0.799

0.945

1.035

1.069

1.087

1.097

–1.9

0.294

0.788

0.920

0.996

1.023

1.037

1.044

–2.0

0.307

0.777

0.895

0.959

0.980

0.990

0.995

–2.1

0.319

0.765

0.869

0.923

0.939

0.946

0.949

–2.2

0.330

0.752

0.844

0.888

0.900

0.905

0.907

–2.3

0.341

0.739

0.819

0.855

0.864

0.867

0.869

–2.4

0.351

0.725

0.795

0.823

0.830

0.832

0.833

–2.5

0.360

0.71 1

0.771

0.793

0.798

0.799

0.800

–2.6

0 368

0.696

0.747

0.764

0.768

0.769

0.769

–2.7

0.376

0.681

0.724

0.738

0.740

0.740

0.741

–2.8

0.384

0.666

0.702

0.712

0.714

0.714

0.714

–2.9

0.390

0.651

0.681

0.683

0.689

0.690

0.690

–3.0

0.396

0.636

0.666

0.666

0.666

0.667

0.667

Exceedence probability

(Source: United States Water Resources Council (1981))

345

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Engineering Hydrology

� Confidence Limits: General The value of xT calculated by using any of the frequency analysis method described above is subjected to different types of errors, such as measurement errors, uncertainty in the system, size of the sample on which the analysis carried out being small, and the data not following the basic assumptions of frequency analysis mentioned earlier. As such, it is not possible to have 100% confidence in the frequency curve obtained by the methods outlined above. Therefore, it is customary to provide a range within which the value of xT is expected to lie rather than specifying a point value. The limits of such a range are called confidence limits; there being two limits, one upper limit and another lower limit. The width between the upper and lower limits is called the width of the confidence band or confidence interval. The width of confidence interval is affected by many factors. The first and foremost is the degree of confidence with which the width of confidence interval is specified. This degree of confidence, also called confidence level, is denoted by b, and is normally taken as 95% or 0.95. However, other values of b are also in use, such as 99% and 90% etc. It is clear that higher the confidence level, higher will be the width of the confidence interval. The second factor on which the confidence limit depends is the size of the sample n. Larger is the size of the sample, greater is the accuracy in the statistical estimates, indicating that the width of confidence interval is smaller for larger samples. Another factor on which the width of the confidence band depends is the exceedance probability or the return period itself. The width of the confidence interval will be the smallest for the mean annual flood (T~2.33 years) and will increase for larger T values. The skew in the data also affects the width of the confidence interval; the widths of confidence interval being greater for the data having greater coefficient of skewness. It is not always possible to account for all of these factors in a single method while calculating the width of the confidence interval. There are several methods available for estimating the width of the confidence interval, or the upper and lower limits of xT. We will study two methods, one applicable for Gumbel’s and normal distribution and the other for the log transformed data series, for example, Log-Pearson Type-III distribution. � Confidence Limits: Gumbel’s and Normal Distributions The upper and lower limits for the confidence interval for Gumbel’s and Normal distribution are specified around the calculated value of xT as follows:

Gumbel’s Distribution

Normal Distribution

xL /U = xT ± f (b )Se

(9.76)

È1 ˘ Se = Í (1 + 1.1396 KT + 1.1000 KT2 )˙ s n Î ˚

(9.77)

Ê 2 + z2 ˆ Se = Á ˜ s Ë n ¯

(9.78)

where, xL/U are the lower and upper limits of the confidence interval, L for the lower limit and U for the upper limit, f (b) is a function that depends on the confidence level b, Se is the standard error of estimate, s is the unbiased standard deviation of the original data, KT is the Gumbel’s frequency factor, z is the standard normal variable for a return period of T equivalent to KT that can be found by using the Abramowitz and Stegun (1965) approximation discussed earlier and presented in Eq. (9.53), and n is the size of the sample.

Statistical Methods in Hydrology

347

The values of f (b) are found using standard normal variable and are given in Table 9.8 for some commonly used values of b.

Table 9.8 Values of f(b) for commonly used values of confidence levels to determine confidence limits b

0.50

0.68

0.80

0.90

0.95

0.99

f (b)

0.674

1.00

1.282

1.645

1.960

2.80

� Confidence Limits: Log-Pearson Type-III Distribution The upper and lower limits of the confidence interval around the mean of the log-transformed data for Log-Pearson Type-III distribution can be found using the following set of equations: Upper Limit

yU = y + s y KTU,b

(9.79)

Lower Limit

yL = y + s y KTL,b

(9.80)

Frequency Factor: Lower

KTL,b =

Frequency Factor: Upper

KTU,b =

KT - KT2 - ab a KT + KT2 - ab

a =1 -

(9.81)

(9.82)

a zb2 2(n - 1)

b = KT2 -

zb2 n

(9.83)

(9.84)

where, xU and xL are the upper and lower limits of the confidence interval, y is the sample mean of the logtransformed data, sy is the sample standard deviation of the log-transformed data, KTU,b and KTL,b are the frequency factors for upper and lower limits corresponding to a confidence level of b, respectively, KT is the frequency factor corresponding to a return period of T-years as per the Log-Pearson Type-III distribution, and zb is the standard normal variable for a significance level or an exceedance probability of a (defined below). Note that in Equations (9.79) and (9.80), the lower and upper limits of confidence interval are around the mean of the log-transformed data series ( y ) and not yT. The calculation of confidence intervals is carried out for a confidence level b, which is also expressed in terms of significance level a as follows: 1- b (9.85) 2 Therefore, if b = 90%, then a = (1 – 0.9)/2 = 0.05 or 5%. It is customary to designate confidence intervals using significance levels also. A value of a = 5% or 0.05 is the most commonly used significance level in hydrological applications. a=

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Engineering Hydrology

� EXAMPLE 9.11 For the data in Example 9.10, determine the lower and upper confidence limits and the width of confidence interval for T = 100 years using Gaussian, Gumbel’s Extreme Value Type-I, and Log-Pearson Type-III probability distributions at confidence level (b) of 95%. Compare your results. Solution (a) Gaussian Distribution For b = 0.95, f (b) = 1.96. From the data in the previous example for Normal distribution (for T = 100 years), we have n = 35, xT = 42.60 mm, sx = 8.11 mm, z = 2.3263. Using these values in Eq. (9.78), we get Ê 2 + (2.3263)2 ˆ Se = Á ˜ 8.11 = 3.7307 35 ÁË ˜¯ Now from Eq. (9.76), we get: xL = xT - f (b )Se = 42.60 - (1.96)(3.7307) = 35.29 mm xU = xT + f (b )Se = 42.60 + (1.96)(3.7307) = 49.91 mm Therefore, the width of the confidence interval for Normal distribution C.I. = 49.91 – 35.29 = 14.62 mm. (b) Gumbel’s Distribution From the data in the previous example for Gumbel’s distribution (for T = 100 years), we have n = 35, xT = 52.91 mm, KT = 3.5977, sx = 8.11 mm. Putting these values in Eq. (9.77), we get È1 ˘ Se = Í (1 + 1.1396 KT + 1.1000 KT2 )˙ s În ˚ È1 ˘ = Í (1 + 1.1396(3.5977) + 1.1000(3.5977)2 )˙ (8.11) = 0.9919 35 Î ˚ Now from Eq. (9.76), we get xL = xT - f (b )Se = 52.91 - (1.96)(0.9919) = 50.97 mm xU = xT + f (b )Se = 52.91 + (1.96)(0.9919) = 54.85 mm Therefore, the width of the confidence interval for Normal distribution C.I. = 54.85 – 50.97 = 3.89 mm. (c) Log-Pearson Type-III Distribution From the data in the previous example for Log-Pearson Type-III distribution (for T = 100 years), we have y = 1.3519, sy = 0.1457, KT = 2.3263. For b = 0.95, a = (1 – 0.95)/2 = 0.025. For an exceedance probability of a = 0.025, we have cumulative probability F(z) = 1 – a = 1 – 0.025 = 0.975. From standard Normal tables for F(z) = 0.975, we get z = 1.96. Therefore, zb = 1.96. Putting these values in Eqs (9.83) and (9.84), we get

Statistical Methods in Hydrology

a =1-

zb2 2(n - 1)

b = KT2 -

zb2 n

=1-

349

(1.96)2 = 0.9435 2(35 - 1)

= (2.3263)2 -

(1.96)2 = 5.3019 35

Now using the computed values of a and b; and KT = 2.3263 in Eqs (9.81) and (9.82), the upper and lower frequency factors are calculated as follows: KTL,b = KTU,b =

KT - KT2 - ab a KT + KT2 - ab a

= =

2.3263 - (2.3263)2 - (0.9435)(5.3019) 0.9435

= 1.7875

2.3263 + (2.3263)2 - (0.9435)(5.3019) = 3.1436 0.9435

Now, the upper and lower limits of the confidence interval on the log-transformed series are determined using Eqs (9.79) and (9.80) as follows: yU = y + s y KTU,b = 1.3519 + 0.1457(3.1436) = 1.8099 yL = y + s y KTL,b = 1.3519 + 0.1457(1.7875) = 1.6123 The values of the lower and upper limits of confidence interval are calculated by taking the antilog of the respective values as follows: xL = (10)1.6123 = 40.96 mm; xU = (10)1.8099 = 64.55 mm. Therefore, the width of the confidence interval as per the Log-Pearson Type-III distribution is C.I. = 64.55 – 40.96 = 23.60 mm. Therefore, the three widths of the confidence interval by Gaussian, Gumbel, and Log-Pearson Type-III distributions for the data under consideration are 14.62 mm, 3.89 mm, and 23.60 mm, respectively. Clearly, the Gumbel’s distribution is the most suitable for the 30-minute annual maximum rainfall data as it provides a narrow band around the value of the rainfall calculated for T = 100 years at a confidence level of 95%. The Log-Pearson Type-III and Gaussian distributions are not suitable with Log-Pearson Type-III Distribution being the worst.

9.6

RISK AND RELIABILITY OF WATER RESOURCES PROJECTS

Water resources decision makers need to make an assessment of the risk associated with a large water resources project, such as construction of a LO 7 Summarize the risk and reliability concepts for water dam or a bridge across a river. These structures are designed to withstand resources design a particular value of peak discharge. However, there is no guarantee that a flood of magnitude greater than or equal to the design flood magnitude will not occur during the useful life of a hydraulic structure. Even if a flood of magnitude greater than or equal to the design flood magnitude occurs only once, there will be a finite probability that the proposed hydraulic structure may fail. The failure of a large hydraulic structure may have catastrophic effects as it may result in loss of life and property. Therefore, it is customary to ascertain the risk and reliability associated with a water resources project before being sanctioned.

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Engineering Hydrology

9.6.1

Risk

The risk associated with a water resources project is defined as the probability that a hydrologic event having magnitude greater than or equal to the design magnitude will occur at least once during the useful life time of the project. Let n be the useful life of a water resources project, T be the return period for which the project is being designed, and xT be the magnitude of the hydrologic variable (mostly peak flood). Then, the risk R can be mathematically expressed as follows: 1ˆ Ê R = 1 - (1 - p)n = 1 - Á 1 - ˜ Ë T¯

n

(9.86)

where, p is the exceedance probability, T is the return period, and n is the useful life of the project.

9.6.2

Reliability

The reliability (Re) of a water resources project is complimentary to the risk associated with the project such that: 1ˆ Ê Re = 1 - R = Á 1 - ˜ Ë T¯

n

(9.87)

Note that the risk and reliability of a water resources project depend on the return period for which the project is being planned. Therefore, the return period is decided based on the economic importance of the area in which the project is proposed and other social and political considerations. It is clear that higher the return period of a project, higher will be its reliability and lower will be the risk associated.

9.6.3

Factor of Safety

Like any other civil engineering construction project, water resources projects also involve factor of safety (FS). It is customary to design a hydraulic structure for a magnitude that is higher than the magnitude calculated based on the statistical and hydrological considerations. This is because there can be many other factors or sources of uncertainty in the calculation of the value of the design variable other than the hydrological considerations (for example, social, political, environmental, climate change, and so on.). The factor of safety (FS) can be expressed as follows: FS =

xTa

(9.88)

xTh

where, xTa is the actual value of the design variable corresponding to the design return period T, and xTh is the value of the design variable corresponding to the design return period T calculated using hydrological considerations, such as frequency analysis as discussed earlier in this chapter. The value FS is always greater than 1.0. Higher values of FS are used in areas of higher economic and social importance, such as urbanized areas, while lower values of FS may be adopted in rural areas. For example, suppose a bridge is to be designed to safely pass a 100-year flood in an area. Let the value of the 100-year flood calculated using the frequency analysis as per Gumbel’s distribution was found as 23,000 m3/s. The proposed bridge, therefore, may not be designed to safely pass a peak discharge in the river equal to 23,000 m3/s but 20% higher discharge, i.e., 23,000 × 1.2 = 27,600 m3/s. As such the factor of safety for the project will be 1.2.

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351

� EXAMPLE 9.12 A water resources project is proposed in an area and a risk and reliability analysis is to be carried out. The useful life of the project is 50 years. If a risk of more than 50% is undesirable, then determine the return period for which the project should be designed. Solution The computations are arranged in a tabular form as shown in Table 9.9. The risk and reliability for each return period are calculated using Equations (9.86) and (9.87), respectively. In the table shown, the risk and reliability are calculated for return periods of 10-, 50-, 75-, 100-, and 500-years. It can be noted from the table that the risk associated with return periods of 50-years and 75-years is about 64% and 49%, respectively. Therefore, in order to construct a water resources project for which the risk associated is not to exceed 50%, we should adopt a return period of 75-year.

Table 9.9 Computations for Example 9.12 S. No.

Return Period (T-years)

Risk (%)

Reliability R (%)

1

10

99.48

0.52

2

50

63.58

36.42

3

75

48.89

51.11

4

100

39.50

60.50

5

500

9.53

90.47

SUMMARY Most of the hydrologic variables are uncertain in nature due to several limitations in analysis; therefore, they are treated as random variables and the laws of probability and statistical methods are employed to deal with them. In this chapter, we introduced the concept of sample and population, probability, random variables, the laws of probabilities, basics of the probability density functions and cumulative distribution functions, along with their basic properties. The importance of various moments in characterizing the probability distributions was also discussed. The moments can be taken either about the origin or about the mean. Most of the probability distributions can be characterized using the first two moments, mean and variance; while some others need higher moments like skewness and Kurtosis. Some commonly used probability distributions, such as Binomial, Gaussian, and Gumbel’s Extreme Value Type-I distributions were described along with some example problems to reinforce their applications in hydrology. An approximate method for estimating standard normal variable z given F(z), and estimating F(z) given z, for Gaussian distribution was also discussed, which is handy in computer applications. Frequency analysis of hydrologic variables aims to define the relation between the magnitude of extreme events and their frequency of occurrence. The frequency of occurrence is measured using return period. Gaussian, Gumbel’s, and Log-Pearson Type-II distributions, the most commonly used distributions used for frequency analysis, were presented. Since it is

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not possible to be 100% certain while carrying out statistical analysis of hydrologic variables, the importance and methods for calculating the confidence interval was described. Finally, the concepts of risk, reliability, and factor of safety and their importance in planning of water resources projects were highlighted.

OBJECTIVE-TYPE QUESTIONS 9.1 A water resources project capable of withstanding a lesser magnitude flood would entail (a) Smaller capital costs and smaller costs in damages and lost lives (b) Higher capital costs and higher costs in damages and lost lives (c) Smaller capital costs and higher costs in damages and lost lives (d) Higher capital costs and smaller costs in damages and lost lives 9.2 Statistical treatment is needed in hydrology because of (a) Instrumental limitations and measurement errors (b) Complexity and uncertainty in the system (c) The system involves too many variables (d) Fluctuations due to environmental factors (e) All of these (f) None of (a) through (d) 9.3 Which of the following statement(s) is/are true for hydrological systems? I. Numerical errors and variable boundary conditions can cause uncertainty. II. Unrealistic assumptions made in analysis lead to uncertainty in the system. III. Fluctuations due to climate change impacts may cause uncertainty in the system. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 9.4 Which of the following statements is true for a sample and the population in statistics? (a) A population is a set of observations that has been observed in the past (b) A sample consists of all the values in the past and the future (c) A sample cannot be measured (d) Population can be measured (e) All of these (f) None of (a) through (d) 9.5 Which of the following statement(s) is/are true for statistics in general? I. We rely on a sample to draw inferences about the behavior of the population. II. The population has constant statistical properties but they may vary from one sample to the other. III. The statistical properties of a sample approach the constant statistical properties of the population as the size of the sample increases. (a) I only (b) II only (c) III only (d) I and II

Statistical Methods in Hydrology

(e) I and III

(f) II and III

(g) I, II, and III

353

(h) All three statements are false

9.6 Which of the following is an example of a discrete hydrologic variable? (a) Daily mean temperature (b) Annual peak flow (c) Wind velocity (d) Number of rainy days 9.7 In an area, flooding occurs if the flow in the river exceeds x m3/s. The flow exceeded x m3/s in each of the past 10 years. Based on this small dataset, the probability of occurrence of a flood is: (a) 0.0 (b) 1.0 (c) 0.5 (d) Insufficient information 9.8 A random experiment can be exactly divided into three mutually exclusive outcomes/events, A, B, and C. The probability of occurrences of A and B was found to be 0.5 each, then what is the probability of occurrence of event C? (a) 0.0 (b) 1.0 (c) 0.5 (d) Insufficient information 9.9 The probability of occurrence of a flood in an area is 0.05 and that for an earthquake is 0.02 in a given year. What is the probability that one of the disasters will occur next year? (Assuming the two events to be independent.) (a) 0.05 (b) 0.02 (c) 0.07 (d) 0.0014 9.10 The probability of occurrence of a flood in an area is 0.05 and that for an earthquake is 0.02 in a given year. What is the probability that both of the disasters will occur next year? (Assuming the two events to be independent.) (a) 0.05 (b) 0.02 (c) 0.07 (d) 0.001 9.11 The probability of occurrence of a flood in an area is 0.05 and that for an earthquake is 0.02 in a given year. If the probability that either one will occur in any given year is 6%, then what is the probability that both of the disasters will occur next year? (a) 0.05 (b) 0.02 (c) 0.07 (d) 0.01 9.12 The probability that it will rain on any given day is 40% and that it will rain on two consecutive days is 20%. If it has rained today, what is the probability that it will rain tomorrow? (a) 50% (b) 20% (c) 40% (d) Insufficient information 9.13 The pdf of a continuous random variable is triangular in shape. If it ranges between 0 and 200, the peak of the distribution is (a) 0.001 (b) 0.01 (c) 0.10 (d) 1.00 9.14 The cumulative distribution of a random variable is given by F(x) = kx2 for 0 < x < 1.0 and zero elsewhere. The pdf in the range (0, 1) is given by (a) f(x) = x2 (b) f(x) = kx2 (c) f(x) = 2kx (d) f(x) = kx3/3 9.15 The cumulative distribution of a random variable is given by F(x) = kx2 for 0 < x < 10 and zero elsewhere. The value of k is (a) 0.00 (b) 0.01 (c) 1.00 (d) 0.10 9.16 The cumulative distribution of a random variable is given by F(x) = kx2 for 0 < x < 10.0 and zero elsewhere. The probability that the value of x is greater than 10.0 is (a) 0.00 (b) 0.01 (c) 1.00 (d) 0.10

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9.17 The cumulative distribution of a random variable is given by F(x) = kx2 for 0 < x < 10.0 and zero elsewhere. The probability that the value of x is less than or equal to 5.0 is (a) 0.64 (b) 0.39 (c) 0.25 (d) Insufficient information 9.18 The cumulative distribution of a random variable is given by F(x) = kx2 for 0 < x < 10.0 and zero elsewhere. The probability that the value of x is less than or equal to 8.0 is (a) 0.64 (b) 0.39 (c) 0.25 (d) Insufficient information 9.19 The cumulative distribution of a random variable is given by F(x) = kx2 for 0 < x < 10.0 and zero elsewhere. The probability that the value of x is between 5.0 and 8.0 is (a) 0.64 (b) 0.39 (c) 0.25 (d) Insufficient information 9.20 Occurrence of a flood follows Binomial distribution with p = 0.10. What is the probability that there will be no flood in the useful life of a bridge of 20 years? (a) 0.0000 (b) 0.1216 (c) 0.8784 (d) Insufficient information 9.21 The kth moment about the mean for a continuous random variable (Mk) is defined as follows: •

x

(a)

Ú (x - m)

k

f ( x ) dx

(b)

•

k Ú ( x - m ) f ( x)dx

-•

k

f ( x )dx

-•

-•

(c)

Ú (x - m) x

(d)

Ú (x

k

- m ) f ( x )dx

-•

9.22 The expected value for a discrete random variable is equal to the (a) First moment about origin for the data (b) First moment about the mean for the data (c) Second moment about the origin of the data (d) Second moment about the mean of the data 9.23 The variance of a continuous distribution is equal to the (a) First moment about the origin for the distribution (b) First moment about the mean for the distribution (c) Second moment about the origin of the distribution (d) Second moment about the mean of the distribution 9.24 Which of the following statement(s) is/are true for the expected value of a random variable? I. It represents the value of a random variable corresponding to the centroid of the probability distribution it follows. II. It lies on the axis of symmetry for a symmetrical probability distribution. III. It represents the shape parameter of the probability distribution. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 9.25 The average discharge at a downstream station will be (a) Lesser than that at an upstream station (b) Greater than that at an upstream station (c) Equal to that at an upstream station (d) Insufficient information to comment 9.26 The mode of discharge at a downstream station will be (a) Lesser than that at an upstream station (b) Greater than that at an upstream station (c) Equal to that at an upstream station (d) Insufficient information to comment 9.27 The flow in a river having large location parameter will always be larger than the flow in another river having small location parameter if

Statistical Methods in Hydrology

(a) The two pdfs are identical (c) The two pdfs are non-overlapping

355

(b) The two pdfs are overlapping (d) Insufficient information to comment

9.28 The rainfall at Faridabad (x) and rainfall at Ghaziabad (y) are related as y = 2x. If the average rainfall at Faridabad was 90 cm, the expected value of rainfall at Ghaziabad is (a) 90 cm (b) 45 cm (c) 180 cm (d) Insufficient information 9.29 Which of the following statement(s) is/are true for the measures of central tendency of a random variable? I. Mode is the highest probability of occurrence of a random variable. II. Mean, median, and mode are equal for symmetric unimodal pdf. III. Mean, median, and mode are equal for symmetric multi-modal pdf. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 9.30 Standard deviation represents (a) A measure of dispersion (b) A measure of variation in the data from the mean (c) Spread in the data or the probability distribution (d) All of the above (e) None of the (a), (b), and (c) 9.31 If all the observed values of a random variable are equal to the mean, then the variance of the random variable would be equal to (a) Expected value of the random variable (b) Any observed value of the random variable (c) Zero (d) Insufficient information to comment 9.32 Which of the following statement(s) is/are true for the measures of dispersion of a random variable? I. Higher the value of the variance, closer are the observations from the mean. II. Smaller the value of the variance, further away are the observations from the mean. III. Higher the value of the variance, further away are the observations from the mean. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 9.33 A pdf with larger standard deviation will be (a) Spread narrower along the measurement axis as compared to a pdf with smaller standard deviation (b) Spread wider along the measurement axis as compared to a pdf with smaller standard deviation (c) Spread equally along the measurement axis as compared to a pdf with smaller standard deviation (d) Insufficient information to comment 9.34 The rainfall at Faridabad (x) and rainfall at Ghaziabad (y) are related as y = 20 + x. If the standard deviation in rainfall at Faridabad was found as 10 cm, the standard deviation in rainfall at Ghaziabad is (a) 10 cm (b) 20 cm (c) 30 cm (d) Insufficient information 9.35 The rainfall at Faridabad (x) and rainfall at Ghaziabad (y) are related as y = 2x. If the standard deviation in rainfall at Faridabad was found as 10 cm, the standard deviation in rainfall at Ghaziabad is (a) 10 cm (b) 20 cm (c) 40 cm (d) Insufficient information 9.36 The mean and standard deviation of a normal random variable were found as 100 and 50, respectively. The probability that the value of the random variable will be less than or equal to 200 is (a) 0.0228 (b) 0.9772 (c) 0.5324 (d) Insufficient information 9.37 The mean and standard deviation of a random variable were found as 100 and 50, respectively. The probability that the value of the random variable will be in the range (0, 200) is at least equal to (a) 0.250 (b) 0.500 (c) 0.750 (d) Insufficient information

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9.38 Which of the following statement(s) is/are true for the skewness of a random variable? I. Skew for a symmetric pdf is equal to zero. II. If the pdf has mode shifted to the left of the mean, it is called positively skewed (g1 > 0). III. If the pdf has mode shifted to the right of the mean, it is called negatively skewed (g1 < 0). (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 9.39 The coefficient of excess for normal random variable is equal to (a) 0.0 (b) 2.0 (c) 3.0

(d) Insufficient information

9.40 A pdf is called mesokurtic if (a) E = 0 or g2 = 3 (b) E > 0 or g2 > 3

(c) E < 0 or g2 < 3

(d) Insufficient information

9.41 A pdf is called leptokurtic if (a) E = 0 or g2 = 3 (b) E > 0 or g2 > 3

(c) E < 0 or g2 < 3

(d) Insufficient information

9.42 A pdf is called platykurtic if (a) E = 0 or g2 = 3 (b) E > 0 or g2 > 3

(c) E < 0 or g2 < 3

(d) Insufficient information

9.43 Occurrence of a flood follows Binomial distribution with p = 0.10. What is the probability that there will be at least one flood in the useful life of a bridge of 20 years? (a) 0.0000 (b) 0.1216 (c) 0.8784 (d) Insufficient information 9.44 Occurrence of a flood follows Binomial distribution with p = 0.10. What is the probability that there will be flood in each of the year in the useful life of a bridge of 20 years? (a) 0.0000 (b) 0.1216 (c) 0.8784 (d) Insufficient information 9.45 A hydrologic variable follows Binomial distribution. If the value of its occurrence was found as p = 0.10 using a sample of size 100, what is the value of the mean of the distribution? (a) 1 (b) 100 (c) 10 (d) Insufficient information 9.46 A hydrologic variable follows Binomial distribution. If the value of its occurrence was found as p = 0.50 using a sample of size 50, what is the value of the standard deviation of the distribution? (a) 25 (b) 12.5 (c) 3.54 (d) Insufficient information 9.47 For a Binomial distribution, p > q. Its coefficient of skewness will be (a) Positive (b) Negative (c) Zero (d) Insufficient information 9.48 For a Binomial distribution, p = q. Its coefficient of skewness will be (a) Positive (b) Negative (c) Zero (d) Insufficient information 9.49 If X = N(μ, s2) and Y = a + bX, then Y is also normally distributed such that (a) Y = N(bµ, b2s2) (b) Y = N(aµ, b2s2) 2 (d) Y = N(a + bµ, b2s2) (c) Y = N(a + bµ, bs ) 9.50 Which of the following statement(s) is/are true for the Gumbel’s distribution? I. It is also known as Extreme Value Type-I (EVT-I) distribution or Double Exponential distribution. II. It can be used in drought analysis. III. It can be used in analyzing flood situations. (b) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 9.51 The Gumbel’s distribution is usually (a) Positively skewed (c) Symmetric

(b) Negatively skewed (d) Insufficient information

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9.52 The standard deviation of a hydrologic variable was found to be 1.28255. The value of Gumbel’s parameter a is most near to (a) 0.0 (b) 1.0 (c) 1.28255 (d) Insufficient information 9.53 The cumulative distribution function, F(x) for Gumbel’s distribution can be represented as which of the following in terms of the reduced variable y? e (a) e

-y

(b) e - e

y

(c)

e- e

-y

(d)

ee

y

9.54 For a dataset having 99 points, the return period for the random variable having the highest magnitude, as per Weibull’s formula is close to (a) 10 years (b) 100 years (c) 50 years (d) Insufficient information 9.55 Which of the following probability distribution is suitable for hydrologic data containing high skew? (a) Normal (b) Log Normal (c) Gumbel’s (d) Log-Pearson Type-III 9.56 Higher the confidence level, (a) Lower will be the width of the confidence interval (b) Higher will be the width of the confidence interval (c) Width of the confidence interval does not depend on confidence level (d) Insufficient information to comment on 9.57 Which of the following statement(s) is/are true for the confidence intervals? I. Larger the size of the sample, lesser is the accuracy in estimating the width of confidence interval. II. Larger the return period, smaller is the width of confidence interval. III. Larger the skew in the data, smaller is the width of confidence interval. (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III (h) All three statements are false 9.58 The risk associated with a water resources project designed for a return period of 100-years is approximately (a) 0.20 (b) 0.30 (c) 0.40 (d) Insufficient information 9.59 The reliability of a water resources project with a useful life of 50 years and designed for a return period of 500-years is approximately (a) 0.90 (b) 0.60 (c) 0.36 (d) 0.50

DESCRIPTIVE QUESTIONS 9.1 Explain why statistical treatment of the hydrologic system in needed. 9.2 Discuss the major differences between sample and population and their relative importance. 9.3 Define a random variable, a discrete random variable, and a continuous random variable. 9.4 What do you understand by posterior and prior probabilities? 9.5 Describe the basic laws of probabilities. 9.6 Explain the pmf and cmf for discrete random variables, and pdf and cdf of continuous random variables. Give two examples each from hydrology. Discuss how different types of probabilities can be calculated from pmf, cmf, pdf and cdf. 9.7 Explain the difference between moments about origin and those about the mean. Define the four moments and the characteristics of a pdf they represent.

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9.8 The pdf of a random variable is given as follows: f(x) = ae–ax f(x) = 0

for x ≥ 0 elsewhere

where, a is the parameter of the distribution. Determine the mean, median, mode, standard deviation, coefficient of skewness, Kurtosis coefficient, and coefficient of excess for the given distribution. 9.9 Describe the properties of expected value and standard deviation of a random variable. 9.10 Explain with the help of sketches that mean is a location parameter. 9.11 Explain with the help of sketches that standard deviation is a shape parameter. 9.12 Draw pdfs with positive and negative skew and indicate mean, median, and mode on them. 9.13 Explain the peakedness of a pdf. How is it characterized? What are leptokurtic, mesokurtic and platykurtic pdfs? Explain with the help of a neat sketch. 9.14 What do you understand by Chebyshev’s inequality? Explain with the help of an example. How is it useful in hydrology? 9.15 What are the criteria/assumptions under which Binomial distribution is applicable? Write its expression and explain each term. State its corollaries. 9.16 Derive the pdf of standard normal variable z from the pdf of Normal distribution for x. 9.17 How will you use the standard normal tables to determine (a) F(z) for a given z, and (b) z for given F(z)? 9.18 Describe Gumbel’s probability distribution and its parameters. 9.19 What do you understand by frequency analysis? 9.20 Explain the plotting position method for carrying out the frequency analysis. 9.21 Describe the general frequency factor method with the help of a few probability distributions. 9.22 How will you determine if the given dataset follows Gumbel’s distribution? 9.23 What do you understand by confidence limits? Discuss the factors on which the width of confidence limits depends on. 9.24 How will you determine the confidence limits for the Normal, Gumbel’s, and Log-Pearson Type-III distribution? 9.25 Discuss the importance of risk, reliability, and factor of safety for water resources projects. Draw the risk and reliability curves for different return periods for n = 50 years and n = 100 years.

NUMERICAL QUESTIONS

f(P)

9.1 The distribution of annual maximum 1-hour rainfall in Krishna river basin (P, cm) is given in the following figure:

0

2

8

P(cm)

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359

Determine the equations of (a) pdf and cdf for the above distribution. Determine the following probabilities without using the derived pdf: (b) P{P > 3 cm}, (c) P{P < 4 cm}, (d) P{P = 2 cm}, and (e) P{1 cm ≤ P ≤ 4.5 cm}. 9.2 The probability mass function for a discrete random variable that a culvert will get flooded m times in a year is given as follows p(m) = k

for m = 0 for m = 1, 2, 3, 4, 5, 6, and 7 elsewhere

p(m) = k/m p(m) = 0

Determine the value of the constant k and plot the pmf and cmf thus obtained. Find the mean, standard deviation, and skew of the distribution thus obtained. Find the following probabilities: (a) P{m = 3}, (b) P{m ≤ 3}, (c) P{m > 4}, (d) P{4 ≤ m ≤ 6}. 9.3 The values of annual maximum discharges from 1970 to 1986 on a river are given in the following table. Year

Max Q (m3/s)

Year

Max Q (m3/s)

1970

232

1979

514

1971

974

1980

377

1972

456

1981

318

1973

289

1982

342

1974

348

1983

593

1975

564

1984

378

1976

479

1985

255

1977

303

1986

292

1978

603

Answer the following: (a) Plot the annual maximum discharge data as a time series (b) Determine the minimum, maximum, and range of the data (c) Divide the range of the data into five equal sized classes and plot the frequency histogram (d) Calculate the mean, mode, median, variance, standard deviation, and coefficients of variation, skewness, peakedness, and excess for the data. 9.4 The duration of a monsoon storm is a random variable with following pdf: f(x) = Cx2 for 0 ≤ x ≤ 4 hours f(x) = C(8 – x)2 for 4 ≤ x ≤ 8 hours f(x) = 0 elsewhere Determine the value of the constant C. What is the probability that the duration of a storm during monsoon is (a) less than 2 hours, (b) more than 7 hours, (c) between 3 hours and 6 hours, and (d) greater than 8 hours? 9.5 The probability of occurrence of a flood in Mississippi river in any year is 25%, which is independent from any other year. Water authorities are planning to build a dam on Mississippi river with a life span of 50 years. What is the probability that (a) no flood will occur in its life span, (b) flood will occur in each of the year in its life span, (c) flood will occur only once in its life span, (d) flood will occur exactly 10 times in its life span, and (e) flood will occur at least twice in its life span? 9.6 The average monthly wind speed (U) in Ghaziabad has a mean and standard deviation of 20 kmph and 10 kmph, respectively. Determine the probability that the wind speed in the month of January in a given year

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will be (a) [U ≤ 10], (b) [U ≥ 40], and (c) [15 ≤ U ≤ 35]. Solve this question by assuming that the wind speed in Ghaziabad follows Normal distribution. 9.7 Historical records in an area show that the probability of a rainy day on 1st January is 0.45. Plot the pmf and cmf of the applicable distribution. What is the probability of this day being rainy day in (a) exactly 5 years out of 11 years (2020 to 2030), (b) more than 4 years out of 11 years (2020 to 2030), (c) less than 6 years out of 11 years (2020 to 2030), (d) at least one of the years in 11 years (2020 to 2030)? 9.8 The annual rainfall in Kanpur (x) follows normal distribution with a mean of 100 cm and variance of 64 cm2. The annual rainfall in Lucknow (y) has been found to be 1.5 times the annual rainfall at Kanpur (i.e., y = 1.5 x). Find the probability that the annual rainfall at Lucknow in any year will be (a) greater than 220 cm, (b) less than 160 cm, and (c) in the range 90 cm to 310 cm. 9.9 An annual hydrologic variable follows Gumbel’s distribution with a mean of 500 and a standard deviation of 250. What is the probability that the variable will exceed a value of 750 next year? What is the value of the variable with an exceedance probability of 0.05? 9.10 The exceedance probability of a random variable for x = 250 is 0.03 and that for x = 380 is 0.01. If the variable follows Gumbel’s distribution, find its mean and standard deviation. What is the value of the random variable for an exceedance probability of 0.003? 9.11 The annual peak flood data in a river (in m3/s) are: 180, 160, 300, 140, 145, 128, 150, 120, 180, 100, 90, 147, 110, 179, 168, 280, 148, 70, 126, and 110. Determine and plot the plotting positions using any three formulas. Then calculate the annual peak flood for T = 50 years using the three formulas. Comment on the results thus obtained. 9.12 For the peak flood data of 100 years in Amazon River, the basic descriptive statistics have been calculated as follows: x = 22,000 m3/s, sx = 3,985 m3/s, y = 4.60, sy = 0.38, and Csy = 0.2354, where x = Peak Flood in Amazon river and Y = log10 (x). Determine the magnitude of peak flood in Amazon River for 50-year, 100year, and 500-year return periods using (a) Normal distribution, (b) Log Pearson Type III distribution, and (d) Gumbel’s distribution. Out of the three, which distribution is more suitable according to you and why? 9.13 For the flood data on a river for 50 years which follow Gumbel’s distribution, the following statistics were found: x = 5,000 m3/s, and sx = 1,000 m3/s. Determine the return period for flood magnitudes of 12,00 m3/s, 8,000 m3/s, 1,000 m3/s, and 800 m3/s.

10

Measurement of Hydrologic Variables

LEARNING OBJECTIVES LO 1

Outline the general requirements for locating a measuring station

LO 2

Discuss the selection criteria of a site and the instruments

LO 3

Explain the methods of measurement of precipitation, evapotranspiration, infiltration, streamflow, soil moisture, and groundwater

LO 4

Summarize the methods of measurement of temperature, pressure, humidity, wind, and radiation

LO 5

Outline the various freely available databases

10.1

INTRODUCTION

Now that we have gained an understanding of the important hydrological processes, we have a fairly good idea of which variables are important from a hydrological perspective. We include all such variables in hydrologic variables, although some of them may not appear to belong to this category (e.g., solar radiation and wind speed, which influence evaporation). Due to a lack of complete understanding of the physical processes involved in the hydrologic cycle, and natural variability of the hydrologic variables, hydrologists generally use empirical methods derived from an analysis of the available data. Therefore, it becomes very important to have good quality data with adequate spatial and temporal resolution. In this chapter, we will describe the various aspects of the measurement process, including the measurement network, instruments, and databases. Analysis of data has already been described in the relevant chapters. However, some additional discussion is included here, as needed.

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10.2

MEASUREMENTS – GENERAL

Most of the data which is useful in hydrological analysis is primarily LO 1 Outline the general meteorological data, which is collected due to its importance in weather requirements for locating a monitoring and forecasting. For example, temperature, precipitation, measuring station wind speed, relative humidity and atmospheric pressure are routinely measured at weather stations. Some other variables are measured because of their importance in other fields, e.g., soil moisture and evaporation for agriculture; and some are primarily hydrologic variables, e.g., streamflow. While deciding on a measurement network, the questions which need to be answered are: ∑

How many stations should be used (density of network)?

∑

Where should these be located (siting)?

∑

What kind of equipment should be used (instrument characteristics)?

∑

What accuracy of measurement is required (resolution)?

∑

How often should the measurement be done (frequency)?

The answers will, naturally, depend on what data is being collected and the purpose for which it is to be used. For example, evaporation measurement does not require as dense a network as required for precipitation measurement, due to a larger spatial and temporal variability in precipitation. Similarly, a shortrange weather forecast will require more frequent observations over a denser network to improve the accuracy of the forecast. For a flood-frequency analysis, only the flows of larger magnitude need to be measured, while for the design of a storage reservoir, a complete hydrograph is needed. The first step in the measurement is to set up a network of stations and specify an operating procedure, which would serve the required purpose with minimum cost. Due to inherent uncertainties in the hydrological variables, and the continually changing land-use and atmospheric conditions, the network/procedure may need to be adaptively modified. The instruments could be of recording type, which maintain a continuous record of the measured variable, or non-recording type, which are generally read once-a-day (in remote areas the reading frequency may be only once in a month or season). Based on experience, a minimum density of stations for different types of areas has been suggested by the World Meteorological Organization (WMO). For example, for measurement of evaporation, one station for each 50,000 km2 area is recommended for most areas, except for polar/arid areas where it could be one station for each 100,000 km2. Similarly, for plains, there should be one standard (i.e., non-recording) rain gauge per 575 km2 and one recording rain gauge per 5,750 km2; and for hills, there should be one standard rain gauge per 250 km2 and one recording rain gauge per 2,500 km2. For standard gauges, the measurements are taken once a day, generally in the morning at a specified time (in India, typically at 8:30 am). However, during times of rapid variation, e.g., a storm (for a rain gauge) or a flood (for a stream gauge), more frequent observations should be undertaken. For recording gauges, although the measurements are continuous, the station has to be visited for maintenance, for example, replacing the chart paper on which recording is made, emptying the container which collects the precipitation, checking for equipment malfunction, recalibration of equipment, etc. These visits could be once a week or less frequent.

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Also, to reduce the data storage requirement, a finite interval is used. The recording-interval could be fixed, say, every 5 minutes, or may be made variable-dependent, e.g., every 1-cm change in water level or every 0.1 mm of precipitation. With advances in communication technology, most recording stations are able to provide real-time data which may be viewed from anywhere in the world. In addition to the on-site measurement of the hydrologic variables at the stations, there are several methods of remote measurement, which are based on either the natural electromagnetic radiation emitted by the atmosphere and/or the earth’s surface (passive remote sensing) or their response to an artificially emitted radiation (active remote sensing). For example, the reflected radiation from the top of a cloud may help in estimating the amount of rainfall; the measured variation of natural gamma radiation may be used to estimate soil moisture since water strongly attenuates these. The sensor could be ground-based (ground radar for precipitation measurement), airborne (weather balloon for temperature, wind speed, and pressure measurement; aircrafts for turbulence, wind speed, and water vapor measurement), or space-borne (weather satellites for precipitation and near-surface soil moisture measurement).

10.3

SELECTION OF SITE AND INSTRUMENTS

10.3.1

Site-selection

LO 2 Discuss the selection criterion of a site and the instruments

For on-site measurements, the site should be such that it provides a representative value for the desired variable over the appropriate spatial and temporal scales. For example, if the precipitation data is to be used for the design of a rainwater harvesting system for a building, the rain gauge should be as close to the building as possible, and the monthly or weekly values of precipitation would be sufficient. However, if the data is to be used for estimating the runoff from a catchment, the gauges should be distributed across the catchment and daily or hourly values may be desirable. We assume that the station is a part of climatological network and the data would be used for a regional hydrological analysis. The following points should, then, be kept in mind: •

The instruments should be placed on level ground and the surrounding area also should not have very steep slopes. This will avoid local effects on the observations.

•

The ground surface should either have short grass or similar to the general surface of that area. A fence should surround the area to prevent entry of animals and unauthorized persons.

•

The site should be in an open area away from obstructions like buildings and trees. If a rain gauge is installed, the distance of any obstruction (including the fence) from the gauge should preferably be at least four times its height (measured with respect to the rim of the gauge, which is generally 30 cm above ground level). In unavoidable cases, a minimum distance equal to twice the height may be used.

•

If near-surface soil temperature is being measured, a 2m × 2m patch of bare ground is maintained and the temperature measurements are performed at depths less than 20 cm.

•

For stream-gauging sites, there should be a sufficient number of stations so that an accurate interpolation is possible to obtain the discharge at any location. The location is governed by topographic and climatic factors.

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•

There are several contradictory requirements which should be resolved in an optimum manner. For example, fencing of the area may cause deviations from the natural wind velocity and evaporation; an open area may report a smaller precipitation due to enhanced wind effects.

10.3.2

Instrument-selection

The type of instrument used at a station is governed by the data requirements as well as the desired accuracy. Recording instruments provide a complete record of the variable but are more expensive both in terms of initial cost and the operation/maintenance cost. Similarly, a more accurate instrument is likely to be more expensive. WMO has suggested accuracies for different types of measurements, which may be used as a guideline when selecting an instrument. For example, at 95% confidence interval, precipitation depth should be accurate to about 5% and intensity should be accurate to 1 mm/h. Similarly, streamflow measurement should be accurate to 5%. Some instruments may suffer a gradual decline in accuracy due to mechanical wear (e.g., in a tipping bucket rain gauge) or a change in the relationship between the measured parameter and the computed variable (e.g., a stage-discharge curve). Periodic checks of the instrument and calibration of these relationships are, therefore, desirable.

10.4

MEASUREMENT TECHNIQUES

We describe the measurement techniques in detail for precipitation, evaporation, infiltration, streamflow, soil moisture, and groundwater, which are of primary interest to a hydrologist. Measurement of other variables, which have an indirect effect on the hydrological processes, e.g., air temperature, water temperature, atmospheric pressure, wind velocity, humidity, solar radiation, sunshine duration, is described very briefly.

10.4.1

LO 3 Explain the methods of measurement of precipitation, evapotranspiration, infiltration, streamflow, soil moisture, and groundwater

Precipitation

The precipitation measured by a rain gauge should closely represent the precipitation over the nearby area. Several factors, including wind effects, evaporation, surface wetting, and splashing, may cause the measured precipitation to be different from the actual precipitation. The wind has the most dominant effect and generally reduces the precipitation collected in a rain gauge. Therefore, the rain gauge should be shielded from wind effects by using vegetation around it. For proper shielding, the height of the vegetation cover, measured with respect to the rain gauge opening, should not be less than half its distance from the gauge. To avoid obstruction to the incoming precipitation, this height should not be more than the distance from the rain gauge. To avoid obstruction from other objects, their distance from the rain gauge should be more than four times their height. The height of the rain gauge opening above the ground is normally kept at 30 cm. Heights greater than this will be subjected to stronger winds, since the wind velocity increases with height. On the other hand, smaller heights may lead to over-estimation of precipitation since some splashes from the ground may enter the rain gauge. Sometimes, the rain gauge opening is made flush with the ground level by installing it in a pit with an anti-splash sheet.

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10.4.1.1 Non-recording Rain Gauge The non-recording rain gauge uses a collector over which the precipitation falls, a funnel to channelize this precipitation, and a receiver to store it. Several sizes of the collector opening, ranging from 100 cm2 to 1000 cm2 are being used, since the actual area of collection is not important. The receiver is generally a polythene bottle of about 2-4 liter capacity and is enclosed in an outer base. The lower end of the collector and the top end of the base have rings provided for locking the collector to the base. In addition to channelizing the flow, the use of a funnel prevents evaporation by narrowing the neck. When snow is expected, the funnel and the receiving bottle are removed; snow is collected in the outer base, and is melted to find the equivalent liquid precipitation. The readings are normally taken once a day at a specific time by pouring the water from the receiver to a graduated cylinder, which is calibrated such that it directly provides the depth of precipitation. However, due to the limited storage capacity of the receiver, it needs to be emptied more frequently for intense storms, and therefore, several measurements are made and summed for the 24-hour period. Figure 10.1 shows the standard design adopted by the India Meteorological Department (IMD), with the collector area recommended as 100 or 200 cm2. Precipitation Collector

5 cm

Locking Ring

30 cm

Funnel

Base

G.L.

60 cm

Receiver

Masonry or Concrete Foundation

60 cm

Figure 10.1 IMD standard rain gauge

10.4.1.2 Recording Rain Gauge Common types of recording rain gauges are (Figure 10.2): tipping bucket, weighing, and float (or siphon). In the tipping bucket gauge, a pair of buckets is arranged in such a way that when a specific quantity (generally 0.25 mm, but sometimes 0.1 mm) of rain is collected in a bucket, it tips over, thereby bringing the other bucket in position. The tipping also causes either a record on a graph paper mounted on a drum or a signal to a remote recorder. The time between two successive tipping indicates the time taken for occurrence of, say, 0.25 mm precipitation. Clearly, more intense rain will cause more frequent tipping. Since a small but finite time is taken in the tipping process, any precipitation falling during this time is lost. To correct this, water from the bucket is collected in a can and is measured at regular intervals to verify the total rainfall recorded

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by the buckets. The discrepancy may then be distributed over the period of intense precipitation events. The other limitation of this gauge is that precipitation less than 0.25 mm is not recorded. If a precipitation event ends before the bucket tips, the collected precipitation in the bucket will be wrongly attributed to the next precipitation events. For example, if at the end of a precipitation event, 0.2 mm of rain is collected after the last tipping, it will not be registered. When the next event occurs, the bucket will tip after only 0.05 mm of rain, but it will register as 0.25 mm. Therefore, the first tip after a large time gap is not a true representation of the precipitation intensity.

Figure 10.2 Recording rain gauges (a) Tipping bucket (b) Weighing (c) Float In the weighing gauge, a bucket mounted on a weighing scale collects the precipitation. The weight may be recorded on a local chart or may be transmitted remotely. The chart is in the form of a mass curve, i.e., cumulative precipitation depth versus time. In the float gauge, the rainfall is carried to a float chamber, with the float indicating the height of water in the chamber. A pen attached to the float then records the rainfall on a graph paper. When the water level in

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the chamber reaches a specified level, generally 25 mm, a siphon empties the chamber and the pen is reset to zero level. The record, therefore, consists of a mass curve with several nearly vertical lines in it representing the siphon action over a very small time interval. Since the siphon takes about 10-15 seconds in emptying the chamber, any precipitation during this period is lost. Similar to the tipping-bucket, the total precipitation is collected and used to correct these errors. The graph paper on which recordings are made is mounted on a rotating drum, with speed ranging from one revolution in about a week to four revolutions in a day. � EXAMPLE 10.1 A float rain gauge, with a chamber capacity of 40 mm, produced the following chart during a 24-hour storm (Figure 10.3). Construct the rainfall hyetograph. 45

Cumulative Precipitation (mm)

40 35 30 25 20 15 10 5 0 0

4

8

12

16

20

24

Time (h)

Figure 10.3 Precipitation chart of float rain gauge

Solution We consider 1-hour periods and note down the cumulative precipitation depth after each hour from the Figure 10.3 given above. Note that the gauge was filled to its capacity of 40 mm between the 13th and 14th hour, therefore, the values after the emptying are increased by 40 mm. Following table shows the computations and Figure 10.4 shows the hyetograph, with the intensity assumed to be constant in each hour interval.

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Time (h)

1

2

3

4

5

6

Cum. Depth (mm)

0.2

1.0

2.2

4.0

6.2

9.0

12.2

16.0

Intensity (mm/h)

0.2

0.8

1.2

1.8

2.2

2.8

3.2

3.8

Time (h)

7

8

9

10

11

12

20.2

25.0

29.8

34.2

4.2

4.8

4.8

4.4

13

14

15

16

17

18

19

20

21

22

23

24

Cum. Depth (mm)

38.4

42.1

45.5

48.6

51.2

53.6

55.5

57.2

58.4

59.3

59.8

60.0

Intensity (mm/h)

4.2

3.7

3.4

3.1

2.6

2.4

1.9

1.7

1.2

0.9

0.5

0.2

Intensity (mm/h)

5.0 4.0 3.0

2.0 1.0 0.0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Time Interval Figure 10.4 Rainfall hyetograph A more recent development in recording rain gauges is the disdrometer (or distrometer), which measures the drop size distribution and velocity of the falling precipitation. This could be done by impact or optical technique. In the impact disdrometers, a sensor transforms the momentum of a rain drop into electrical pulse, and the size and magnitude of this pulse is used as a measure of the drop size and velocity. The transformation of the mechanical momentum to electrical signal may be done either through a displacement sensor or an acoustic sensor. The optical disdrometers use either imaging, i.e., use of high speed cameras to record every single drop of precipitation, or scattering, i.e., use of a transmitter to generate a beam of light which is converted to an electric signal by the receiver, with the changes in signal providing a way to estimate the size and velocity of the drop.

10.4.1.3

Remote Measurement

The remote measurement of precipitation may be done by ground-based radars or satellites. These are more helpful than the point measurements by rain gauges, since they provide continuous measurement in both space and time. A radar may be able to sample an area up to a radius of 200 km around it, depending on the power output and receiver sensitivity. However, since the measurements are indirect, i.e., they do not measure the precipitation directly but rely on the relationship between precipitation and some other measured characteristic (e.g., radar echo), proper calibration with on-site instruments is necessary. Moreover, there are several sources of uncertainty in the measurement due to ground clutter owing to buildings, trees, towers, etc., and shielding by hills, particularly in areas with large topographic variations.

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� Radar Measurement A radar sends out pulses of electromagnetic signal (wavelength of about 1–10 cm and duration of the order of microseconds) and then detects the return signal due to backscattering by the raindrops. The elapsed time between sending the pulse and receiving it helps in finding the distance to the target and the strength of the received signal helps in finding the radar reflectivity factor, which relates the power received and the power transmitted. Assuming the drop to be spherical and its size to be small compared to the wavelength, the Rayleigh scattering implies that the intensity of the backscattered radiation will be proportional to the 6th power of the diameter and inversely proportional to the square of the distance from the radar. The reflectivity factor z, is, therefore, defined as the sum of the 6th power of the diameter of all raindrops within a unit volume sampled by the radar beam. For a particular radar configuration, we may write: pr = C

z r2

(10.1)

where, pr is the power received by the radar (Watt), r is the distance to the target (generally in km), and z is generally expressed in mm6/m3. C is a radar-specific constant which is a function of the dielectric factor of the precipitation K, and several radar characteristics, such as, the transmitted power, horizontal and vertical beam widths, wavelength, antenna gain, and attenuation loss. Sometimes, instead of including the dielectric factor in C, it is included in the numerator of Eq. (10.1) as pr = C* K2z/r2, to have C* a function of radar characteristics only. K is defined as (e - 1)/(e + 2) where, e is the dielectric constant, about 80 for water and 3.2 for ice/snow. The multiplying factor, K2, is therefore 0.93 for water and 0.18 for ice/snow. Thus, the measurement of the power received by radar and estimation of the distance to target enables us to compute the reflectivity factor. Since the rainfall intensity i is also a function of the size of raindrops (volume is proportional to the cube of diameter and the falling speed is also a function of diameter), an empirical relationship could be derived between i and z, and is generally of the form i = azb

(10.2)

The coefficients a and b are typically obtained by calibration of the radar with the rain gauge data. With i in mm/h and z in mm6/m3, the values of a and b are suggested as 0.0365 and 0.625, respectively (for tropical convective systems, values of 0.01 and 0.833 are found to work better). Since the value of z varies over a very wide range, from about 10−3 mm6/m3 to more than 106 mm6/m3, the more commonly used variable is decibels of reflectivity, denoted by dBZ and defined as dBZ = 10 log10z

(10.3)

A dBZ value of 20 may indicate light precipitation, while more than 60 corresponds to very heavy precipitation. � EXAMPLE 10.2 The return echo for a radar from a tropical convective rainfall location is represented by a dBZ value of 40. What is the intensity of rain?

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Solution From Eq. (10.3), with dBZ = 40, we get the reflectivity factor, z = 10000 mm6/m3. Using the suggested values for a tropical convective system, a = 0.01 and b = 0.833, from Eq. (10.2), we get the intensity as 21.5 mm/h. Note If we use the suggested values of a = 0.0365 and b = 0.625, we get the intensity as 11.5 mm/h, nearly half of that obtained earlier. Sometimes, the speed of the falling rain drops is obtained by measuring the change in frequency of the received radiation. Since the radar makes use of the Doppler effect, these radars are called Doppler radars. Recent research on precipitation radar has shown that Eq. (10.2) may not be very accurate under practical conditions, due to several reasons including the varying precipitation profile, complex topography, beam blockage, etc. One of the important factors is the shape of the raindrops, which tends to be flat and not spherical. Dual polarization radars, in which both horizontal and vertical reflectivity are measured, have been found to provide a more accurate estimation of the rainfall intensity by including an additional term in Eq. (10.2) to represent the anisotropy. More details are available in Bringi and Chandrasekar (2001). � Satellite Measurement The rain gauges provide an accurate measurement of the precipitation, but are very sparsely located. The ground-based radars cover a larger area but suffer from significant gaps in spatial coverage due to mountains. Moreover, both these methods are not suitable for measurement of precipitation over oceans. Therefore, rainfall estimation using radars mounted on satellites is becoming increasingly popular. The principle of measurement using radars is similar to that for the ground radars as it emits radiation and measures the back-scattered signal. Light Detection and Ranging (Lidar) is also being used for this purpose, but is not very common yet. Frequently, passive sensing is also performed in which the reflection and scattering of natural solar radiation, or the emission of long-wave radiation, by the earth’s surface and atmosphere is measured by sensors on the satellite. Since most of the radiation does not penetrate deep into the clouds, the rainfall is estimated from the cloud-top properties. For example, on observing the reflection of radiation in the visible range of the spectrum, a brighter reflectivity indicates thicker clouds and potentially heavier rainfall. Similarly, measurement of microwave emissions from the rain drops and scattering from ice particles provides an estimate of the precipitation intensity. Use of radars has led to an improved spatial coverage. However, since a satellite samples a particular region of the earth only twice a day, the sampling error is large. Collaborative efforts between several agencies have led to measurement of the global precipitation over a fine spatial and temporal resolution. The Tropical Rainfall Measuring Mission (TRMM), by NASA and the Japan Aerospace Exploration Agency, measured precipitation data from 1997 to 2015 using microwave imager and precipitation radar; Global Precipitation Measurement (GPM) by the same two agencies provides observations of rain and snow worldwide every three hours, since February 27th, 2014. The data from these are used, in conjunction with measurements made by an international network of satellites, to quantify the amount of precipitation over the entire globe. The GPM uses a dual-frequency precipitation radar and a microwave imager to provide a three-dimensional measurement of precipitation. Radars and satellites have worked well for weather reporting and forecasting, but are not very suitable for hydrological purpose like precipitation measurement or flood forecasting, due to the large uncertainty associated with the estimates of precipitation intensity.

10.4.2

Evaporation and Transpiration

Direct measurement of evaporation/evapotranspiration from small areas is quite straightforward and is reasonably accurate. However, since direct measurement from large surfaces is not currently feasible,

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various indirect methods have been developed for estimation of evaporation from large water bodies and evapotranspiration from a large land area. Some of these methods have already been discussed in Chapter 3. Here, we discuss the direct methods using evaporation pans and lysimeters. Some other techniques have also been used, but are not very common due to accuracy or cost reasons. For example, a scintillometer is used to measures the sensible heat flux, which, combined with the measurements of net radiation and the ground heat flux, provides an estimate of the latent heat flux and, therefore, evapotranspiration. Lidar is used to measure water vapour concentration at various heights, which provides an estimate of the vapour concentration gradient and, therefore, evaporation rate. Flux towers have been used for measurement of gas flux from the surface into the atmosphere, which are converted to evapotranspiration estimates by using turbulence-based or gradient-based methods. These techniques are a little advanced for this text and are not discussed here.

10.4.2.1

Evaporation Pans

Evaporation pans are placed near a water body from which evaporation is to be estimated. The lake evaporation is then obtained by multiplying the measured pan evaporation with a pan coefficient. The pans usually have a circular or square section, and are either mounted above the ground or sunk in the ground such that the water level in the pan is approximately at the ground level. Sometimes, floating pans are used within the water body, as they are expected to provide a closer match with the actual lake evaporation. However, these are not very common and will not be discussed.

Figure 10.5 The United States Class A Evaporation Pan (dimensions in mm) Figure 10.5 shows a typical evaporation pan, known as the United States Class A pan. It is used as a standard since its performance has been extensively studied under widely varying geographic and climatic conditions. The Class A pan is made of galvanized iron and is in the form of a cylinder, with 1.2 m diameter and a depth of 25 cm, which is placed on a raised and level slatted wooden base to allow free circulation of air. A stilling well is constructed in the pan to provide a stable water level by damping the perturbations due to wind. At a pre-specified time every day, the water level in the stilling well is measured and water is added/removed to make the water level exactly 50 mm below the rim (if precipitation occurs in the preceding 24-hour period, water level may be higher than the 50 mm mark, and some water will have to be removed).

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If there is no precipitation, the depth of water which needs to be added to the pan to bring the water level 50 mm below the rim, is the pan evaporation. In case precipitation occurs, the pan evaporation will be equal to the depth of precipitation ± depth of water added/removed. However, one needs to be careful that no overflow/splash has occurred from the pan, which could happen for intense rainfalls, either due to total rainfall exceeding 50 mm or due to water splashing out of the pan due to impact of raindrops. In both cases, evaporation will be overestimated. In the United Kingdom, the Symons, or S-pan, is more commonly used while in the USSR, the GGI-3000 pan is used. The Symons pan (which is similar to the Sunken Colorado pan used in the US) is a square prism made of galvanized iron, with base size of 1830 mm and height of 610 mm. It is sunk in the ground such that the rim is 100 mm above the ground level and the water level is kept at the ground level. Its interior is painted black. Since the sides of the pan are not exposed to the sun, as is the case with the above-ground pans, the evaporation from the sunken pan is likely to be closer to the actual lake evaporation. However, due to heat transfer from the surrounding soil, the measured evaporation is still more than the lake evaporation. Other drawbacks of the sunken pans are the possibility of splashing of rain water into the pan and the difficulty in detection of any probable leakage from the pan. The Bureau of Indian Standards specifies that an evaporation pan should be cylindrical with a diameter of 1220 mm and height of 255 mm, made of copper or stainless steel, painted white on the outer surface and tinned inside, covered with a wire mesh and placed on a 1225 mm × 1225 mm white-painted square wooden platform (Figure 10.6).

Figure 10.6 The Indian Standard Evaporation Pan (dimensions in mm)

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Some general precautions to be observed when using the evaporation pans are: obstructions should not be closer than four times the height of the object above the pan, the pan should not be placed on a concrete slab or over asphalt or gravel to avoid bottom heating, other nearby instruments should not cast shadows over the pan, water level in the pan should not be allowed to decrease beyond about 75 mm below the rim, since it may lead to wind-shielding. If a mesh is used to cover the pan to prevent birds from drinking water from it, the measured evaporation may be higher due to absorption of heat by the mesh and its re-radiation. Due to side-heating in the pan, non-uniform heating in the lake, surrounding environmental conditions, and comparatively rapid rate of moisture laden air being taken away from above the water surface in the pan, the pan evaporation is generally higher than the lake evaporation. The pan coefficient, i.e., the ratio of lake evaporation to pan evaporation is therefore less than 1, and shows some variation with the season, type of lake, type of pan, surroundings of the pan, etc. For example, a deep lake tends to absorb some part of incident radiation in summer and may release this stored energy in winter, while in the pan, the absorption is more uniform across the seasons. The pan coefficient is, therefore, likely to be smaller in summer and larger in winter. Similarly, a floating pan reduces the side-heating and more closely resembles the lake conditions, thereby having a larger pan coefficient. Pans located on bare soil generally record higher evaporation than those in a grassy patch, since the air moving over the pan is drier in the former case. Commonly suggested values of the pan coefficient for annual evaporation are 0.7 for a Class A pan (with values ranging from about 0.35 to 0.85 for different wind speeds, relative humidity, and surrounding environment) and 0.8 for the Indian Standard pan and sunken and floating pans. For better accuracy, individual calibration of the pan coefficient for a particular pan-lake combination should be performed based on actual measurements. � EXAMPLE 10.3 An Indian Standard evaporation pan, with internal diameter of 1220 mm, is 255 mm deep and is filled with water up to a height of 190 mm. After 24 hours, it was observed that 2.93 liters of water has to be added to the pan to bring the water level back to its original level of 190 mm. A nearby rain gauge recorded 2.1 mm of rainfall during this 24 hour period. If the pan coefficient is 0.8, what is the lake evaporation during this period? Solution To convert the volume of water added to the depth, we divide by the internal area of the pan. Therefore, the depth of water added = 2.93 × 106 / (p × 12202/4) mm = 2.5 mm. Since the depth of rain during this period is 2.1 mm, the pan evaporation will be 2.5 + 2.1 = 4.6 mm. Using the pan coefficient of 0.8, the lake evaporation is 0.8 × 4.6 = 3.7 mm.

10.4.2.2 Lysimeters Lysimeters are instruments which measure the gains and losses of water by a column of soil. The gains are due to precipitation, irrigation, or condensation and the losses are due to evapotranspiration and, in some cases, runoff. The measurement system consists of a soil mass, generally filled in a container to prevent water loss from the sides and bottom, and a device to collect the water percolating through it. The container is put inside the ground such that the soil surface is at the same level as the natural ground and the vegetation within the lysimeter is representative of the natural vegetation of that area. By measuring all other components of

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the soil mass balance, the evapotranspiration can be estimated. Transpiration is estimated by subtracting the separately measured evaporation from the soil from the evapotranspiration. Lysimeters could be confined or unconfined, depending on whether the soil mass is contained within the walls of an impermeable material or not. In the unconfined lysimeters, there could be lateral movement of water, and the plant roots may extend beyond the soil mass. In the confined lysimeters, the confining material may interfere with the natural heat and mass transfer and could cause the estimated evapotranspiration to be different from that under natural conditions. Another classification is based on how the soil mass is placed in the field: disturbed or undisturbed. For example, if impermeable barriers are driven around an existing soil mass, it will create an undisturbed lysimeter. However, if a container is first placed in a trench and is then filled with the soil removed from the trench, it would be a disturbed lysimeter. Based on how water is collected, lysimeters could be either zerotension or tension, with the former collecting water draining freely under gravity, and the latter collecting water by applying a vacuum. Therefore, for sampling water in the unsaturated zone, the tension lysimeter is useful. Similarly, based on how the amount of water is measured, the lysimeters may be classified as weighing or non-weighing, with the entire lysimeter system resting on a weighing scale in the former. In the non-weighing lysimeters, the percolated water is collected in a vessel and measured.

10.4.2.3

Remote Sensing

Indirect estimates of evapotranspiration may be obtained by remote-sensing of parameters like solar radiation, vegetation cover, temperature, and soil moisture. This is done by measuring the electromagnetic radiation reflected or emitted from the Earth’s surface in different bands of the spectrum. For example, solar radiation and albedo is estimated from scanning the visible and infrared bands, and surface temperature may be estimated from measurements at thermal infrared wavelengths. The relationships described in Chapter 3 could then be used to estimate the evapotranspiration.

10.4.3

Infiltration

As discussed in Chapter 3, the actual infiltration rate is equal to the smaller of the precipitation rate and the infiltration capacity. The infiltration capacity of a soil decreases with time and the measurement of this temporal variation is done through an infiltration test. Since the infiltration capacity also shows a significant spatial variation due to change in soil properties like size, texture, and compaction, generally several tests are performed to obtain the representative infiltration capacity curve for any given area. The infiltration capacity at a location is measured using infiltrometers or permeameters. As the name suggests, infiltrometers measure the rate at which water infiltrates into the soil and permeameters measure the soil permeability, typically by measuring the infiltration rate. Some commonly used devices are described below. Estimation of the infiltration capacity by assuming a specific form of variation, e.g., the Horton model, has already been described in Chapter 3.

10.4.3.1

Single Ring Infiltrometer

This is the simplest arrangement for measuring the infiltration rate, in which a thin cylindrical ring, generally made of metal or hard plastic, is driven into the soil up to a certain depth and filled with water (Figure 10.7a). By measuring the variation of depth of water within the cylinder with time, the infiltration capacity curve is obtained. The readings are generally taken at 15 minute intervals for the first hour, 30 minute intervals for

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the second hour, and hourly after that. However, for soils with large hydraulic conductivity, more frequent readings should be taken. If evaporation is likely to be significant, it should be either prevented (e.g., by using an oil film on the water surface) or accounted for in the calculation. The water level in the ring could be kept constant by adding the required amount of water at specific times (constant head test) or, for less permeable soils, may be allowed to drop (falling head test). The rate of infiltration is obtained by analyzing the amount of water added in the constant head test, and the drop in water level in the falling head test. One drawback of this method is that the flow lines are not vertical as the flow expands when it seeps below the ring. This implies that the area over which infiltration is taking place is larger than the ring area and is also unknown, thus making the estimate of the infiltration rate higher than actual, hence unreliable. Another source of error is a lack of good contact between the soil and the cylinder surface, which may cause leakage of water through the inside surface of the ring and consequent overestimation of the infiltration rate. The commonly used ring-size is 3 mm thick, 100 cm diameter and 50 cm height with approximately 30 cm driven into the ground and the water level maintained at about 5 cm above the soil surface. Smaller diameter rings may also be used but tend to overestimate the infiltration rate due to the lateral flow. The bottom edge is usually beveled to allow easy driving into the soil, and the driving force should not be so excessive as to cause undue disturbance to the soil structure. Sometimes, the soil around the ring is presaturated with water to ensure vertical flow of water below the ring. A large diameter ring or greater depth of driving would reduce the error due to lateral flow. A method suggested by Bouwer et al., (1999) may be used to correct for the divergence of flow. However, the double ring infiltrometer serves the same purpose of reducing the lateral spreading and is described next.

10.4.3.2

Double Ring Infiltrometer

A double-ring infiltrometer comprises of two concentric cylinders, with a constant water level maintained in both of these (Figure 10.7b). The rate at which water is added to the inner ring (the rate of addition in the outer ring may be different due to lateral flow) provides an estimate of the infiltration capacity. Typical ring-size is 3 mm thick, 30 cm inner ring diameter, 60 cm outer diameter and 50 cm height with the outer ring driven approximately 15 cm into the ground and the inner ring about 5–10 cm into the ground. If larger diameters are used, the ratio of the diameters of the outer and inner rings is maintained at two. The water level in both rings is maintained at the same level, usually about 5–10 cm above the soil surface.

Figure 10.7 Ring infiltrometers (a) Single ring (b) Double ring (dimensions in cm)

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10.4.3.3

Other Methods of Estimating Infiltration

In addition to the ring infiltrometer, there are several other methods which are used to estimate the infiltration capacity of the soil. We provide a very brief description of some of these here. � Guelph permeameter is a constant-head permeameter for estimation of saturated hydraulic conductivity and soil sorptivity. It comprises a 10 cm deep hole, dug in the ground, in which a constant water level is maintained. Computations are based on the discharge from the hole measured for two different constant heads (generally 5 cm and 10 cm), and the corresponding water content of the soil. The measurement of discharge is done by noting down the amount of water added in the hole to maintain a constant water level. Equations available for calculation of the soil hydraulic properties involve shape factors, which account for the three-dimensional nature of flow, and are functions of the diameter of the hole and the applied heads. � Tension infiltrometer is used to estimate the unsaturated hydraulic properties of soil. It is a porous disc, 10 cm or 20 cm in diameter, placed in contact with the soil surface, with a water reservoir and a bubble tower attached to it (Figure 10.8). Sometimes a thin layer of fine sand is placed on the soil surface to improve the contact between the disc and the soil. In contrast to the ring infiltrometers, where water is supplied to the soil under ponded conditions, the tension infiltrometer supplies water at a negative pressure (suction) by supplying air through the bubble tower. By measuring the infiltration rate corresponding to different suctions, the unsaturated hydraulic conductivity of the soil is estimated.

Figure 10.8 Tension infiltrometer In Philip-Dunne permeameter, a hole is augured in the ground and a tube is inserted at its bottom. Measurement of the initial moisture content of the soil, the time taken in emptying half of the tube and the full tube, and the final water content enables one to estimate the hydraulic properties of the soil.

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� EXAMPLE 10.4 A double ring infiltrometer has an inner ring diameter of 20 cm, and outer ring diameter of 40 cm. Initially, water is filled in both the rings to a depth of 10 cm, and then the depth is brought back to 10 cm by periodic addition of water. The time durations at which water is added and the volume of water added in the inner ring to bring the water level to 10 cm, is shown in the table below. Plot the variation of the infiltration capacity with time. Time (sec)

30

60

120

240

390

600

1200

1800

2400

3600

Volume added (ml)

9.2

8.4

13.1

19.9

20.9

25.7

62.8

57.6

55.0

104.7

Solution The volume added is divided by the cross-sectional area of the inner ring to obtain the depth of infiltration during that time period. This depth divided by the elapsed time gives the infiltration capacity during the time period, which could be assumed to be representative of the capacity at the middle of the time interval. The following table shows the computations and Figure 10.9 shows a plot of the infiltration capacity versus time. 0.5

1

2

4

7

10

20

30

40

60

Depth of Infiltration (mm)

0.29

0.27

0.42

0.63

0.67

0.82

2.00

1.83

1.75

3.33

Infiltration Capacity (mm/h)

35.1

32.1

25.0

19.0

16.0

14.0

12.0

11.0

10.5

10.0

Infiltration Capacity (mm/h)

Time (min)

40 35 30 25 20 15 10 5 0 0

10

20

30

40

50

60

Time (min) Figure 10.9 Plot of infiltration capacity versus time

10.4.4 Soil Moisture The moisture content may be expressed as volumetric moisture content (ratio of the volume of water in a soil sample to the total volume) or gravimetric moisture content (ratio of the mass of water in a soil sample to the total mass). The simplest method of finding the soil moisture is the gravimetric method, which involves weighing a soil sample before and after drying it to a constant weight. However, this needs collection of several samples to account for the spatial variation in moisture content and causes disturbance of the soil. Several in-situ measurement techniques have, therefore, been developed.

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10.4.4.1

Neutron Probe

When neutrons collide with an atom, they lose a part of their energy. This loss is higher for atoms of lower atomic mass. Hydrogen is the main low-mass atom in the soil and is primarily contained in the water molecules. Therefore, if we use a radioactive source to emit fast neutrons in the soil, some of these will slow down after collision with the soil moisture. Counting the number of slow neutrons will provide an estimate of the soil moisture, with a larger number of slow neutrons representing higher moisture content.

10.4.4.2

Gamma Ray

The decrease in intensity of a gamma ray passing through a soil depends on the soil density, the coefficients of attenuation for the soil and for the water, and the moisture content. Since the other parameters are constant, measurement of intensity by a gamma-ray detector provides a direct estimate of the soil moisture. Two parallel access tubes are installed in the soil for this purpose, one housing a gamma-ray source (generally Cs-137) and the other, a gamma-ray detector.

10.4.4.3

Dielectric Constant Method

The dielectric constant of water is about 80 while that of dry soil is about 5. When high-frequency electromagnetic waves are transmitted through soil, their velocity is inversely proportional to the dielectric constant and, therefore, the moisture content of the soil. Depending on how the dielectric constant is measured, this method may be classified as Time domain reflectometry (TDR) or Frequency domain reflectometry (FDR). In the TDR method, the electromagnetic wave propagates along a cable parallel to a conducting probe that is inserted into the soil. The time lag between emission of the pulse and reception of its reflection is used to find the wave velocity. The FDR method is based on the fact that when a capacitor is connected to an oscillator to form an electrical circuit, changes in dielectric constant will lead to changes in the circuit operating frequency. Radio frequency waves are used to measure the soil capacitance, from which the volumetric water content may be estimated.

10.4.4.4

Remote Sensing

As already discussed, water has a dielectric constant which is more than an order of magnitude larger than that of the soil particles. Also, the heat capacity and thermal conductivity of water are considerably higher than those of the soil. Therefore, the remote sensing of the diurnal variation in the surface temperature or the surface emissivity and reflectivity of the soil provides a way to estimate the soil moisture. The temperature may be sensed by thermal infrared imaging, emissivity may be obtained by passive microwave response using a radiometer to measure the thermal emission from the soil surface, and reflectivity may be obtained by active microwave response using a radar to transmit the microwave and measure the back-scattering. It has been found that these techniques provide an accurate estimation of the surface soil moisture, i.e., within the top 5 cm of the soil profile. The thermal response may be affected by the presence of vegetation on the soil surface and may not be able to provide the true soil moisture. However, the microwave response has been found to be insensitive to the vegetation. � EXAMPLE 10.5 A cylindrical core sampler was used to collect a soil sample of 3 cm diameter and 5 cm depth. The sample was then put in a container of known weight and weighed to obtain the wet weight. The container was then placed in an oven and heated at 105 °C for 48 hours to remove the moisture. The container was then

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weighed again to obtain the dry weight. Estimate the volumetric and gravimetric soil moisture content for the following data: Weight of empty container: 100.000 g Weight of container + wet soil: 164.521 g Weight of container + dry soil: 148.237 g Solution Weight of the dry soil is 48.237 g and weight of wet soil is 64.521 g. Therefore, the weight of moisture within the soil sample is 16.284 g. Volume of the soil sample is π × 32/4 × 5 cm3 = 35.343 cm3. The volumetric moisture content is obtained by dividing the volume of moisture with the volume of soil. Assuming the moisture to be composed of water, its volume is equal to 16.284 cm3. Therefore, the volumetric moisture content is 16.284/35.343 = 0.461 or 46.1%. The gravimetric moisture content is obtained by dividing the mass of moisture by the total mass (which is mostly taken as the wet mass, but sometimes dry mass is used). The gravimetric moisture content is, therefore, 16.284/64.521 = 0.252 or 25.2%.

10.4.5

Streamflow

In Chapter 4, we discussed the velocity-area method of estimating the streamflow at a cross-section, by subdividing the area and measuring the velocity in each subdivision. The velocity measurements are generally done using a mechanical current meter, which consists of a rotor having a few cups, and mounted on a rod. The force of water on the cups causes the rotor to move and the speed of revolution is directly related to the flow velocity. Depending on how the rod (axis) is oriented, the current meter is called either a vertical-axis current meter or a horizontal-axis current meter. The average velocity could also be estimated by using a float to measure the surface velocity and then using a multiplying factor (generally around 0.85 to 0.9). More recent techniques for velocity measurement include the Acoustic Doppler current profiler (which sends a sound wave into the water and measures the change in its frequency on reflection by sediments or other particles being carried by the water), the electromagnetic current meter (in which the electromotive force produced by the moving water, as it passes through the magnetic field created by current passed through a coil, is measured), and the ultrasonic method (in which an ultrasonic pulse is transmitted diagonally across the stream from both banks and the difference in the time of travel is measured). � EXAMPLE 10.6 A current meter was calibrated by moving it with different speeds in a water tank and observing its number of revolutions. It was found that the equation V (m/s) = 0.7 × Revolutions per second + 0.01 provides a reasonably good estimation of the velocity V (the additive constant of 0.01 m/s implies that the threshold velocity at which the current meter starts to rotate is about 1 cm/s). The current meter was then used in a 5-m wide stream of rectangular cross-section with a flow depth of 1.25 m. The cross-section was subdivided into 10 sections of equal width (0.5 m), the current meter was put at mid-width of each subdivision, and lowered to a depth of 0.75 m (i.e., 0.6 times the flow depth) below the water surface. The numbers of

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revolutions per minute at these 10 locations were observed to be 48, 54, 68, 75, 78, 77, 73, 69, 52, and 46, respectively. Estimate the streamflow at that section. Solution Since the area of each subsection is same, we could find the average velocity for the cross-section as the average of the velocities through each subsection (in fact, since the relationship between velocity and revolutions is linear, we could compute the average number of revolutions per second to obtain the average velocity. However, we show the subsection velocities also). The velocity is obtained by the current meter equation, with the rpm divided by 60 to obtained rps. The following table shows the computations of average velocity. Section

1

2

3

4

5

6

7

8

9

10

Ave

RPM

48

54

68

75

78

77

73

69

52

46

64

V (m/s)

0.570

0.640

0.803

0.885

0.920

0.908

0.862

0.815

0.617

0.547

0.757

The discharge is then obtained as 0.757 × 5.0 × 1.25 = 4.73 m3/s For small streams, a weir (which is an obstruction constructed across the full or part-width of the channel, usually in the form of a vertical plate) could be used to estimate the streamflow by measuring the height of water surface over the weir crest h, and relating it to the discharge Q, through a discharge coefficient Cd, as Q=

2 C 2 g Bh1.5 3 d

(10.4)

where, B is the width of the weir. For weirs with thin-crest, the discharge coefficient is a function of the ratio of the water height h, and the weir height hw, and equations/graphs are available for finding its value. Construction of a weir, however, is not practical for very wide streams carrying high discharges. � EXAMPLE 10.7 A 30 cm high thin plate is used as a weir across the entire width of a 3 m wide rectangular channel. Estimate the streamflow at that section if the water depth at an upstream section, where the depth is nearly constant, is 1.23 m. The Rehbock formula for discharge coefficient, valid for h/hw < 5, is given as Cd = 0.611 + 0.08

h hw

Solution The head over the weir, h = 1.23 − 0.3 m = 0.93 m. Since h/hw is less than 5, we use the Rehbock formula to get Cd = 0.611 + 0.08 × 0.93/0.3 = 0.859. Using Eq. (10.4), with B = 3 m, we get the discharge as 6.82 m3/s. Empirical equations, e.g., Manning’s or Chezy’s equation, could be used to obtain the streamflow by measuring the water surface slope and the cross-section of the channel in a certain reach of the stream, and estimating the resistance coefficient (Manning’s n or Chezy’s C, which are primarily dependent on the grain size). However, since these empirical equations are valid under uniform flow conditions only, they may not be applicable during floods.

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The Manning’s equation relates the average velocity of flow V, with the energy slope S, and the hydraulic radius (defined as the flow area divided by the wetted perimeter) R, through the use of a roughness coefficient n, as 1 2/3 1/2 1 (10.5) R S fi Q = AR 2/3 S1/2 n n where, V is in m/s and R is in m; A is the cross-sectional flow area in m2, and the term AR2/3 is known as the section factor. If estimates of n are not available, the Strickler equation may be used to obtain n, based on the grain-size of the channel bed material, as V=

d1/6 (10.6) 24 where, d is the median diameter of the bed material in meters and the value of the coefficient in the denominator may vary from about 21 to 29, depending on the type of material and which diameter (e.g., diameter which has 90% material finer) is chosen. By measuring the water levels in a stream at two sections, which are a known distance apart and where the stream cross-section is known, we may estimate the discharge as described below. n=

The energy slope may be assumed to be equal to the water surface slope, which is computed based on the water surface elevations measured at the two sections and the distance between them. The flow area and wetted perimeter at the sections are computed based on the water depth and the known cross-section, and the hydraulic radius is calculated. Manning’s roughness may be given or it may be estimated from Eq. (10.6), if the bed material size is measured. Eq. (10.5) is then used to obtain the discharge, either by computing the discharge separately at both sections and then averaging, or by averaging the area and hydraulic radius of the two sections and then computing the discharge. Note that if the velocities at the two sections are significantly different, or if there are losses other than those due to boundary friction, the assumption of the energy slope being equal to the water surface slope will not be valid. In such cases, an iterative procedure may be required. � EXAMPLE 10.8 A certain reach of a river may be idealized as a 1000 m long straight channel with rectangular cross-sections at both ends. The upstream end has a width of 110 m and bed elevation of 123.0 m above mean sea level (msl), and the downstream section is 100 m wide and its bed is 122.5 m above msl. At a certain time, the water depth at the upstream section was observed as 2.1 m and that at the downstream section was observed as 2.5 m. The median bed material diameter is 0.2 mm. Estimate the streamflow in this reach. Solution The water elevation at the upstream section is 123.0 + 2.1 = 125.1 m, while that at the d/s section is 122.5 + 2.5 = 125.0 m. The slope of the water surface is, therefore, (125.1 – 125.0)/1000 = 0.0001. The manning’s roughness n, using Eq. (10.6) with d = 0.0002 m, is about 0.01. The flow area at the u/s section is 110 × 2.1 = 231 m2 and the wetted perimeter is 110 + 2 × 2.1 = 114.2 m.

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Ê 231 ˆ The section factor is, therefore, 231 Á Ë 114.2 ˜¯

2/3

= 369 m8/3.

Similarly, the value of the section factor at the d/s section is obtained as 446 m8/3. We may assume that the effective section factor is mean of these values, i.e., 408 m8/3, and obtain the discharge from Eq. (10.5) as 1 Q= 408 0.0001 m3/s, i.e., about 400 m3/s. If we use the mean of the areas and the hydraulic radii to 0.01 compute the section factor, we get nearly identical result. For large streams, the dilution method may work well, in which a tracer is injected in the stream and its concentration at a downstream section is measured. Any substance, which is easily measurable, dissolves readily in water, is not naturally present in the stream, does not decompose in water, does not react with sediments, and is not harmful to the environment, may be used as a tracer. Commonly used tracers are common salt, fluorescent dyes, sodium dichromate and, less commonly, radioactive elements. The tracer could be injected in the stream suddenly by dumping a known amount at a point or at a steady rate, through continuous injection of a tracer solution of known concentration. By measuring the concentration of the tracer at a downstream section, the streamflow can be computed. In the sudden injection method, a section is chosen sufficiently downstream of the section where a known mass of tracer M, is suddenly dumped, so that there is complete mixing of the tracer with the water. Assuming that there is no tracer initially in the stream, the integral of the mass flow curve at the downstream section should be equal to the total mass injected, i.e., M (in kg). Therefore, we could write •

M = ÚCQ dt fi Q = 0

M •

Ú0

(10.7)

C dt

3

where, C is the measured concentration (kg/m ), t is the time (in seconds), and Q is the streamflow (m3/s). Although the integration limits are 0 to ∞, practically the range would be much smaller, since C will be 0 till a certain time (when the tracer reaches the downstream section) and again after a certain time (when the entire tracer mass has passed through that section). Also, if the stream has a background concentration of the tracer as C0, then C in Eq. (10.7) is not the actual concentration but an increase over the background concentration. In the continuous injection method, the procedure is similar, except that instead of suddenly dumping a known mass of the tracer, a continuous injection of a tracer solution with a flow rate of Qi and tracer concentration of Ci is done. The mass balance equation, assuming the background tracer concentration to be C0, is Ci Qi + C0Q = (Q + Qi )C•

(10.8)

where, C• is the measured concentration at a large time, i.e., when the downstream concentration attains a constant value. Generally, the background concentration is zero and the injection flow rate is very small compared to the streamflow. Therefore, one may use Q=

Qi Ci C•

(10.9)

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� EXAMPLE 10.9 Two sections of a 100 m wide river, which are 1 km apart, are used for estimating the discharge through the dilution method. Initially there is no salt present in the river. At the upstream section, 500 kg of salt is dumped at a certain time and the observed concentration values at the downstream section at various times, starting just before the first non-zero concentration measurement, are shown below. Estimate the streamflow. Also, if instead of mass dumping, a continuous injection of 1 l/s of water with a salt concentration of 100 g/l is done, what would be the long-term concentration of salt observed at the downstream section? Assume that the length of 1 km is sufficient to ensure complete mixing of the tracer. Time (s)

Concentration (mg/l) Time (s)

Concentration (mg/l)

0

60

120

180

240

300

360

420

480

540

600

660

0.000 0.013 1.479 3.921 4.378 3.549 2.484 1.610 0.999 0.604 0.360 0.212 720

780

840

900

960

1020

1080

1140

1200

1260

1320

1380

0.124 0.072 0.042 0.024 0.014 0.008 0.005 0.003 0.002 0.001 0.001 0.000

Solution The concentrations are reported in mg/l, and hence we will convert it to kg/m3, on dividing by 1000. Since the measurements are equally spaced in time, we could find the integral required in Eq. (10.7) by adding up all •

the concentration values and multiplying with the time interval (60 seconds). This gives us ÚC dt = 19.905 × 0 60 / 1000 kg s/m3 = 1.194 kg s/m3. From Eq. (10.7), we get Q = 500/1.194 = 419 m3/s. For the continuous injection case, Eq. (10.9) gives us the long-term concentration as Q C 1 l/s ¥ 100 g/l C• = i i = = 0.24 mg/l. Q 419 m 3 /s Most of the stream-gauging stations routinely measure the water level instead of directly measuring the discharge, and then use a rating-curve (i.e., the depth-discharge relationship) to estimate the discharge. However, the rating-curve needs to be periodically updated since the stream cross-section keeps changing with time. The stage, i.e., the water level measured from a specified datum, may be monitored by using an automatic level sensor, such as, a float, leading to a continuous record of the streamflow. One drawback of most of the streamflow measurement methods is the assumption of uniform and steady flow. In several practical cases, e.g., during a flood, these assumptions are not satisfied and there could be significant errors in the estimation.

10.4.6

Groundwater

A technique for estimating the groundwater flow direction and speed was described in Chapter 7 by using the data on groundwater levels. Groundwater levels are measured in open dug wells, if available in the area, or with the help of piezometers, which are pipes installed in the ground with their bottom open and, sometimes, having a screen near the bottom. The measurement of water level may be made by using a measuring tape, an electric cable, an acoustic instrument (sounder), or, if a continuous record is required, a float.

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Remote Sensing The Gravity Recovery and Climate Experiment (GRACE) satellites, a joint mission of NASA and German Aerospace Center, detect changes in the gravity field caused by the redistribution of water on and below the earth’s surface. This data is used to estimate the change in groundwater storage, although at a very coarse spatial and temporal resolution. An analysis of the GRACE data from 2003 and 2013 has shown that several of world’s largest aquifers are being depleted and some of them are significantly over-exploited. For example, groundwater depletion rate over Northwest India has been estimated as about 4 cm per year.

10.4.7

Temperature, Pressure, Humidity, Wind, Radiation

These variables are not directly relevant to hydrologists. However, since these are needed in the estimation of evapotranspiration and soil moisture, we briefly describe their measurement here. More details are available in WMO (2008).

LO 4 Summarize the methods of measurement of other variables like temperature, pressure, humidity, wind, and radiation

� Temperature is measured by a thermometer, which is based on the relationship of a physical property of a substance with temperature. For example, thermal expansion of a material or change in electrical resistance with temperature is used in a majority of thermometers. Some thermometers use ultrasonic sampling to obtain the average speed of the air molecules, which is an indicator of its temperature. Remote measurement from satellites is done using radiometric thermometers, in the infrared range. Liquid-in-glass thermometers are commonly used and are based on the differential expansion of a liquid with respect to its glass container. The liquid is generally mercury but, for low temperatures, ethyl alcohol is used. Platinum resistance or thermistor elements have also been used. Manual observations are generally with a mercury-in-glass thermometer, while continuous records may be obtained with resistance or thermocouple elements. � Atmospheric Pressure is generally measured with barometers, which may be electronic, mercury, or aneroid (i.e., using no liquid). Hypsometers, which utilize the relationship between the boiling point of a liquid and the atmospheric pressure, have also been used, but not widely. � Humidity measuring instruments are called hygrometers. A gravimetric hygrometer makes use of the absorption of water vapor by a desiccant to estimate the vapor content in the air. A psychrometer (or wet- and dry-bulb hygrometer) comprises two thermometers with one covered by a thin film of water (wet bulb) and the other exposed to air (dry bulb). The wet bulb records a lower temperature due to evaporation of water and the humidity is obtained using an equation relating this temperature difference to the difference of the actual and saturation vapor pressure through a psychrometric constant. A dew point hygrometer works on the principle that when moist air is cooled, it reaches its saturation point and a deposit of dew forms, at a temperature known as the dew point temperature (or just dew point). The saturation vapor pressure corresponding to the dew point is equal to the actual vapor pressure in the air. Some other types of hygrometers are based on the fact that some materials, after interaction with water vapor, undergo a change in one of their chemical or physical properties, e.g., electrical conductivity, thermal conductivity, color, acoustic properties, emission spectrum etc. � Wind Speed is usually measured by a cup anemometer or a vane (or propeller) anemometer, which work on the principle that the revolution speed of the cup or the vane is directly proportional to the wind speed. Instantaneous wind speeds are measured and are integrated to obtain the daily average wind speed. The measurements are generally accurate to about 0.5 m/s. Other, less common, anemometers are based on the

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wind pressure (the difference of static and dynamic pressure is related to the velocity), heat radiation (cooling of a hot wire changes its electrical resistance, with the change being dependent on the wind velocity), sound propagation (change in the propagation speed of an ultrasonic wave depends on the wind velocity), and radio waves (Doppler effect, i.e., a change in the frequency of transmitted radio waves and reflected waves). � Solar (Short-Wave) Radiation is measured with a pyranometer which generally uses thermopiles (a device that converts thermal energy to electrical energy) covered by glass domes that allow penetration only over a certain wavelength range, typically 0.3–3 μm. Some other types of pyranometers use photodiodes or photovoltaic cells in place of thermopiles. Long-wave radiation is generally obtained by measuring the total radiation received from the sun and the sky with a radiometer and subtracting the solar radiation. The sunshine duration is usually defined as the duration over which the direct solar irradiance is more than 120 W/m2, and may be obtained by measuring the burn length of a paper subjected to focused direct solar radiation with a burn threshold of 120 W/m2.

10.5

DATABASES

The use of recording instruments and developments in communication LO 5 Outline the various technology have made it possible to get real-time data of various freely available databases hydrological variables at any location from anywhere in the world. However, since several quality checks are performed on the data before it is released for general use, there is usually a time lag before the data becomes available. Several national and international agencies are involved in collection of data and maintaining databases of these variables. Most of these databases are available freely, but some may be restricted due to strategic reasons (e.g., data near defense-related installations or international boundaries, data for rivers which flow through multiple countries). Sometimes, several databases are available for the same variable, e.g., precipitation over India is available with the India Meteorological Department, at the Climate Research Unit (University of East Anglia) and APHRODITE (Japan Meteorological Agency). Although these databases pertain to the same data, the values may be slightly different depending on the way the data is collected and processed. For example, the precipitation data collected at ground stations and by radar is used to interpolate the precipitation values at regularly spaced grid points. The algorithm used for interpolation will affect the values obtained at the grid points, leading to a possible difference in the gridded data. Some of the commonly used datasets are listed below: • Hydromet Division of India Meteorological Department – hydro.imd.gov.in/hydrometweb • Automatic Weather Stations of IMD – www.imdaws.com/viewawsdata.aspx • Water Resources Information System of India – www.india-wris.nrsc.gov.in • Global Precipitation Climatology Centre – gpcc.dwd.de/ • Global Runoff Data Centre – www.bafg.de/GRDC • U.S. Geological Survey – www.usgs.gov/science/mission-areas/water • Tropical Rainfall Measuring Mission - trmm.gsfc.nasa.gov • Global Summary of the Day –www7.ncdc.noaa.gov/CDO/cdoselect.cmd?datasetabbv=GSOD&cou ntryabbv=&georegionabbv • Global Historical Climatology Network – ftp://ftp.ncdc.noaa.gov/pub/data/ghcn/v3 • Giovanni (NASA) – giovanni.gsfc.nasa.gov/Giovanni • World Radiation Data Center – wrdc.mgo.rssi.ru

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SUMMARY This chapter described several aspects of the measurement processes for various hydrologic variables. These include precipitation, evapotranspiration, infiltration, streamflow, temperature, radiation, and wind speed. We looked at how to choose the locations of the measurement sites, what precautions to take at the site, and what kind of instruments to use. The density of the measurement network depends on the expected spatial variation of the variable being measured. For measurement of evaporation, a smaller density of stations is used, while for precipitation, hilly areas should have a higher station density than that in plain areas. Standard, or nonrecording, gauges measure the variables at a fixed interval, generally a day. Recording gauges provide a continuous record of the temporal variation of the variable. In addition to on-site measurement, remote methods of measurement are also gaining popularity. The measurement site should be chosen in such a way as to provide a representative value of the variable being measured. Obstructions near the site should be sufficiently distant so that they do not affect the measurements. Detailed description of the available methods of measurement was then provided for the important variables like precipitation, evapotranspiration, infiltration, soil moisture, and streamflow. For precipitation, the IMD standard rain gauge is commonly used in India, with measurement of accumulated precipitation done every day. Tipping bucket, weighing type, and float type recording rain gauges were described and the process of obtaining a hyetograph from the recorded data was also illustrated. Use of radar and satellite for precipitation measurement was briefly described. The reflected electromagnetic signal sent by a radar can be analyzed to estimate the precipitation intensity. Similarly, satellite measurement of the radiation reflected by the clouds is used to estimate the rainfall intensity. Several types of evaporation pans are used to measure the pan evaporation, which is then related to the lake evaporation through a pan coefficient. The pan evaporation is generally larger than the lake evaporation due to the effect of size, wind, and boundary heating. The Indian Standard pan, US Class A pan, and the Symons pan are some examples of evaporation pans. Measurement of evapotranspiration is done through lysimeters. The measurement of infiltration using single and double ring infiltrometers was also described. The single ring infiltrometer overestimates the infiltration capacity of a soil due to lateral expansion of water as it infiltrates below the edge of the infiltrometer. A double ring infiltrometer ensures a nearly vertical flow in the inner ring by surrounding it with an outer ring. Measurement of soil moisture is typically done by taking a soil sample and then drying it in an oven to find the amount of moisture. More recent techniques like Neutron probes and Gamma Ray attenuation are also being used extensively. We also described the slope-area method, which is based on the Manning’s equation, and the dilution method, which is based on the mass balance of a tracer introduced in the stream. The remote sensing of groundwater through satellite measurement of changes in the gravity field was briefly described. This was followed by a brief description about measurement of some other variables like temperature, pressure, wind speed, humidity, and solar radiation. Finally, various databases available for some of these variables were listed, from which data from across the globe can be accessed.

Measurement of Hydrologic Variables

387

OBJECTIVE-TYPE QUESTIONS 10.1 Most of the data useful in hydrology is primarily (a) Economic (b) Meteorological (c) Social

(d) Chemical

10.2 Which of the options given below is correct about the following statements? (i) Density of a measurement network depends on what data is being measured. (ii) Frequency of measurement of a variable does not depend on the purpose for which data is being collected. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 10.3 Which of the options given below is correct about the following statements? (i) Density of network for evaporation measurement should be larger than that for precipitation measurement. (ii) Short-range weather forecast requires more frequent measurements as compared to flood-frequency analysis. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 10.4 The reading from a non-recording gauge is generally taken once in a/an (a) Hour (b) Day (c) Week (d) Month 10.5 The recommended number of stations for evaporation measurement is 1 per (b) 20,000 km2 (c) 50,000 km2 (d) 100,000 km2 (a) 10,000 km2 10.6 The recommended number of non-recording rain gauges in a plain area is 1 per (b) 250 km2 (c) 575 km2 (d) 750 km2 (a) 100 km2 10.7 The recommended number of recording rain gauges in a hilly area is 1 per (b) 2500 km2 (c) 5750 km2 (d) 7500 km2 (a) 1000 km2 10.8 The distance of any object from a rain gauge should be at least ____ times its height above the gauge rim. (a) 1 (b) 4 (c) 10 (d) 20 10.9 At 95% confidence interval, measurement of precipitation depth should be accurate to _____ %. (a) 0.1 (b) 0.5 (c) 1 (d) 5 10.10 At 95% confidence interval, measurement of precipitation intensity should be accurate to _____ mm/h. (a) 0.1 (b) 0.5 (c) 1 (d) 5 10.11 Which of the options given below is correct about the following statements? (i) Wind generally reduces the precipitation collected in a rain gauge. (ii) Height of the rain gauge opening from the ground level is generally 60 cm. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 10.12 Which of the options given below is correct about the following statements? (i) The funnel in a rain gauge helps in reducing evaporation. (ii) The collector area for a rain gauge recommended by IMD is around 1000 cm2.

388

Engineering Hydrology

(a) (i) is true, (ii) is false (c) Both (i) and (ii) are true

(b) (i) is false, (ii) is true (d) Both (i) and (ii) are false

10.13 Which of the options given below is correct about the following statements? (i) In the tipping bucket rain gauge, more intense rainfall will cause more frequent tipping of the buckets. (ii) The emptying of chamber in a siphon rain gauge generally takes about 1 second. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 10.14 A precipitation radar is able to sample an area up to a distance of about _____ km around it. (a) 5 (b) 20 (c) 200 (d) 2000 10.15 The intensity of backscattered radiation from a raindrop is proportional to the _____ power of its diameter. (b) 4th (c) 6th (d) 8th (a) 3rd 10.16 Commonly used unit for the reflectivity factor of a radar is (c) mm3/m6 (a) mm (b) mm6/m3

(d) mm4/m3

10.17 The diameter of a US Class A evaporation pan is about (a) 30 cm (b) 60 cm (c) 1.2 m

(d) 2.4 m

10.18 In the Symons evaporation pan, the water level is kept (a) 100 mm above the ground level (b) At the ground level (c) 100 mm below the ground level (d) 50 mm above the ground level 10.19 The water level in an evaporation pan is generally not allowed to go more than 75 mm below the rim to prevent (a) Splashing (b) Wind-shielding (c) Measurement difficulty (d) All of these 10.20 The pan coefficient for an Indian Standard evaporation pan for annual evaporation is about (a) 0.6 (b) 0.7 (c) 0.8 (d) 1.0 10.21 Lysimeters are used for measurement of (a) Temperature (b) Relative humidity (c) Evapotranspiration (d) Wind speed 10.22 Which of the options given below is correct about the following statements? (i) Single ring infiltrometers are less accurate than the double ring infiltrometers. (ii) Single ring infiltrometers underestimate the infiltration capacity. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 10.23 A neutron probe is based on the principle that when neutrons collide with soil moisture, they (a) Change direction (b) Become protons (c) Lose energy (d) Emit radiation 10.24 The dielectric constant of water is typically about _____ times that of dry soil. (a) 2 (b) 8 (c) 16 (d) 96 10.25 The relation between the number of revolutions of a current meter R, and the flow velocity V, is generally of the form (a) V = R (b) V = aR (c) V = aR + b (d) V = aR − b 10.26 Which of the options given below is correct about the following statements? (i) The Rehbock formula indicates that the coefficient of discharge increases with an increase in the head over the weir. (ii) The Manning’s roughness is proportional to the 6th power of the grain size. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false

Measurement of Hydrologic Variables

389

10.27 Which of the options given below is correct about the following statements? (i) For the dilution method of streamflow measurement, the tracer used should react with the sediments. (ii) In the continuous injection method, the stream discharge is directly proportional to the asymptotic concentration. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 10.28 Which of the options given below is correct about the following statements? (i) A psychrometer has two thermometers, one covered by a thin film of water and the other exposed to air. (ii) Wind speed is measured by a pyranometer. (a) (i) is true, (ii) is false (b) (i) is false, (ii) is true (c) Both (i) and (ii) are true (d) Both (i) and (ii) are false 10.29 Humidity measuring instruments are called (a) Hummers (b) Hydrometers (c) Hygrometers

(d) Anemometers

10.30 The sunshine duration refers to a period over which the direct solar irradiance is more than ______ W/m2. (b) 1.2 (b) 12 (c) 120 (d) 1200

DESCRIPTIVE QUESTIONS 10.1 What are the precautions which need to be observed while choosing a site for meteorological observations? 10.2 What are recording and non-recording rain gauges? Describe each type of gauge. 10.3 What is the principle of radar measurement of rainfall intensity? Why is the decibel of reflectivity preferred for use over the reflectivity factor? 10.4 Why is pan evaporation different from lake evaporation? Describe the Indian Standard evaporation pan with a sketch. 10.5 How are lysimeters classified on the basis of different properties? Compare confined and unconfined lysimeters and also disturbed and undisturbed lysimeters. 10.6 Describe the single and double ring infiltrometers. List some other methods of measuring the infiltration rate in a soil. 10.7 How do you measure soil moisture using Gamma Ray attenuation? Describe the principle of remotely sensing the soil moisture in the surface layer. 10.8 For what type of stream would you use a weir to measure the streamflow? Describe the slope-area method of estimating the discharge. 10.9 What is the difference between the sudden and continuous injection methods for measurement of streamflow through dilution? Derive a relation between the long-term concentration and the injected concentration for the continuous injection method. 10.10 Using any of the relevant databases, obtain the record of daily average temperature and daily precipitation at a point close to your residence. Analyse this data to see if there is any trend in the temperature and/or precipitation with time. Also, find out if there is a correlation between these two variables.

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NUMERICAL QUESTIONS 10.1 For the data given in Example 10.1, obtain the time of tipping if a tipping bucket rain gauge is used to measure the same event. The capacity of the bucket is 0.2 mm. 10.2 The return echo for a radar, from an area over which tropical convective rainfall is taking place, is represented by a dBZ value of 30. What is the intensity of rain? If the intensity doubles, what would be the dBZ value? 10.3 A Unites States Class A evaporation pan is filled with water up to a height of 200 mm. After 24 hours, it was observed that 1.87 liters of water has to be added to the pan to bring the water level back to its original level. A nearby rain gauge recorded 2.1 mm of rainfall during this 24-hour period. If the pan coefficient is 0.7, what is the lake evaporation during this period? 10.4 A single ring infiltrometer has a ring diameter of 30 cm. Initially, water is filled to a depth of 5 cm, and then the depth is brought back to 5 cm by periodic addition of water. The times at which water is added and the volume of water, added in the ring to bring the water level to 5 cm, is shown in the table below. Plot the variation of the infiltration capacity with time. Time (sec)

30

60

120

240

390

600

1200

1800

2400

3600

Volume added (ml)

22.7

20.7

32.4

49.2

51.8

63.5

155.5

142.5

136.1

259.2

10.5 A soil sample of 5 cm diameter and 10 cm depth was obtained from the field and was weighed in a container. After removing the moisture by oven heating of the container, it was weighed again. Estimate the volumetric and gravimetric (based on both wet weight and dry weight) soil moisture content for the following data: Weight of empty container: 200.000 g Weight of container + wet soil: 353.421 g Weight of container + dry soil: 324.276 g 10.6 A current meter has the following calibration equation V (m/s) = 0.01 × RPM + 0.012 It was used in a stream with a trapezoidal cross-section with bed-width of 15.76 m, side slopes of 1H:1V, and a flow depth of 2.12 m, so that the top width is 20.00 m. The cross-section was subdivided into 10 sections with the two corner sections being 2.12 m wide, the two sections next to the corner sections being 1.88 m wide, and the central 6 sections being 2 m wide. The current meter rpm at 0.6 m depth at these 10 locations were observed to be 84, 99, 110, 116, 121, 120, 115, 111, 96, and 82, respectively. Estimate the streamflow at that section. 10.7 A 60 cm high weir across the entire width of a 5 m wide rectangular channel results in a water depth of 1.67 m, measured from the channel bed, sufficiently upstream of the weir. Estimate the streamflow at that section. 10.8 A river has a trapezoidal cross-section with bed width of 87 m and side slopes of 1H:1V. The median bed material diameter is 0.18 mm. Two sections, 500 m apart, are chosen for application of the slope-area method. The upstream section has a bed elevation of 10.92 cm above that of the downstream section. At a certain time, the water depths at both the sections were observed to be nearly same, 3.253 m at the upstream and 3.259 m at the downstream. Estimate the streamflow. 10.9 Discharge in a river is being measured through the dilution method with a tracer which is not naturally present in the river. One tonne of the tracer is dumped at a section and the observed concentration values at a sufficiently downstream section are shown below. Estimate the streamflow.

Measurement of Hydrologic Variables

391

Time (minutes)

0

1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

Concentration (mg/l)

0.001

0.015

0.821

1.674

3.078

4.439

4.789

4.955

4.528

4.018

3.335

2.812

Time (minutes)

6.5

7

8

9

10

11

12

13

15

18

25

Concentration (mg/l)

2.264

1.823

1.131

0.684

0.407

0.240

0.140

0.082

0.028

0.005

0.001

10.10 In a stream, a continuous injection of 10 l/s of water with a tracer concentration of 10 g/l is done. The background concentration of the tracer is negligible at 1 mg/m3. The long-term concentration of the tracer observed at a sufficiently downstream section is 0.287 mg/l. Estimate the streamflow.

Answers to Objective-Type Questions CHAPTER 1 1.1 (c) 1.9 (a) 1.17 (c) 1.25 (d) 1.33 (c) 1.41 (b)

1.2 (c) 1.10 (d) 1.18 (b) 1.26 (d) 1.34 (d) 1.42 (a)

1.3 (a) 1.11 (b) 1.19 (c) 1.27 (c) 1.35 (d) 1.43 (b)

1.4 (d) 1.12 (c) 1.20 (d) 1.28 (d) 1.36 (b)

1.5 (b) 1.13 (c) 1.21 (b) 1.29 (d) 1.37 (c)

1.6 (d) 1.14 (d) 1.22 (b) 1.30 (a) 1.38 (a)

1.7 (c) 1.15 (b) 1.23 (c) 1.31 (a) 1.39 (d)

1.8 (d) 1.16 (c) 1.24 (c) 1.32 (c) 1.40 (a)

2.3 (b) 2.11 (a) 2.19 (a) 2.27 (b)

2.4 (a) 2.12 (d) 2.20 (b) 2.28 (a)

2.5 (c) 2.13 (b) 2.21 (b) 2.29 (d)

2.6 (b) 2.14 (d) 2.22 (a)

2.7 (b) 2.15 (c) 2.23 (c)

2.8 (d) 2.16 (b) 2.24 (c)

3.3 (b) 3.11 (a) 3.19 (b) 3.27 (b) 3.35 (c) 3.43 (b)

3.4 (b) 3.12 (d) 3.20 (d) 3.28 (d) 3.36 (c) 3.44 (b)

3.5 (c) 3.13 (b) 3.21 (b) 3.29 (a) 3.37 (a) 3.45 (a)

3.6 (d) 3.14 (c) 3.22 (d) 3.30 (b) 3.38 (c) 3.46 (c)

3.7 (c) 3.15 (a) 3.23 (b) 3.31 (c) 3.39 (b)

3.8 (b) 3.16 (c) 3.24 (b) 3.32 (a) 3.40 (b)

4.3 (d) 4.11 (a) 4.19 (a) 4.27 (g) 4.35 (d)

4.4 (a) 4.12 (c) 4.20 (g) 4.28 (b) 4.36 (d)

4.5 (b) 4.13 (a) 4.21 (c) 4.29 (a) 4.37 (d)

4.6 (d) 4.14 (b) 4.22 (a) 4.30 (c) 4.38 (a)

4.7 (c) 4.15 (a) 4.23 (b) 4.31 (c) 4.39 (b)

4.8 (d) 4.16 (c) 4.24 (a) 4.32 (a) 4.40 (e)

CHAPTER 2 2.1 (b) 2.9 (b) 2.17 (d) 2.25 (c)

2.2 (d) 2.10 (a) 2.18 (b) 2.26 (d)

CHAPTER 3 3.1 (b),(c) 3.9 (d) 3.17 (b) 3.25 (c) 3.33 (a) 3.41 (b)

3.2 (a) 3.10 (c) 3.18 (c) 3.26 (b) 3.34 (d) 3.42 (c)

CHAPTER 4 4.1 (b) 4.9 (a) 4.17 (b) 4.25 (b) 4.33 (a)

4.2 (c) 4.10 (c) 4.18 (d) 4.26 (g) 4.34 (c)

Answers to Objective-Type Questions

4.41 (a) 4.49 (e)

4.42 (c) 4.50 (b)

393

4.43 (d) 4.51 (b)

4.44 (c) 4.52 (c)

4.45 (c)

4.46 (d)

4.47 (b)

4.48 (d)

5.3 (b) 5.11 (b) 5.19 (a) 5.27 (c) 5.35 (b) 5.43 (h) 5.51 (c) 5.59 (b) 5.67 (d)

5.4 (d) 5.12 (d) 5.20 (b) 5.28 (c) 5.36 (c) 5.44 (b) 5.52 (d) 5.60 (d) 5.68 (g)

5.5 (b) 5.13 (c) 5.21 (c) 5.29 (e) 5.37 (d) 5.45 (d) 5.53 (d) 5.61 (d) 5.69 (a)

5.6 (b) 5.14 (c) 5.22 (c) 5.30 (b) 5.38 (b) 5.46 (g) 5.54 (c) 5.62 (c) 5.70 (b)

5.7 (b) 5.15 (d) 5.23 (d) 5.31 (b) 5.39 (c) 5.47 (a) 5.55 (b) 5.63 (b) 5.71 (d)

5.8 (d) 5.16 (h) 5.24 (c) 5.32 (c) 5.40 (g) 5.48 (a) 5.56 (d) 5.64 (c) 5.72 (d)

6.3 (h) 6.11 (a) 6.19 (b) 6.27 (b) 6.35 (e) 6.43 (d) 6.51 (a) 6.59 (a) 6.67 (d)

6.4 (b) 6.12 (a) 6.20 (c) 6.28 (a) 6.36 (a) 6.44 (a) 6.52 (b) 6.60 (b) 6.68 (c)

6.5 (g) 6.13 (c) 6.21 (a) 6.29 (b) 6.37 (d) 6.45 (d) 6.53 (c) 6.61 (b) 6.69 (g)

6.6 (c) 6.14 (b) 6.22 (c) 6.30 (b) 6.38 (c) 6.46 (a) 6.54 (a) 6.62 (g) 6.70 (g)

6.7 (c) 6.15 (b) 6.23 (b) 6.31 (d) 6.39 (d) 6.47 (g) 6.55 (c) 6.63 (g) 6.71 (d)

6.8 (d) 6.16 (d) 6.24 (c) 6.32 (c) 6.40 (c) 6.48 (b) 6.56 (a) 6.64 (c) 6.72 (c)

7.3 (d) 7.11 (c) 7.19 (b) 7.27 (b)

7.4 (c) 7.12 (c) 7.20 (c) 7.28 (c)

7.5 (b) 7.13 (b) 7.21 (b) 7.29 (a)

7.6 (d) 7.14 (c) 7.22 (c) 7.30 (c)

7.7 (b) 7.15 (b) 7.23 (c) 7.31 (b)

7.8 (c) 7.16 (b) 7.24 (d) 7.32 (d)

8.3 (c) 8.11 (c) 8.19 (b)

8.4 (d) 8.12 (a) 8.20 (c)

8.5 (b) 8.13 (d) 8.21 (a)

8.6 (a) 8.14 (a) 8.22 (b)

8.7 (b) 8.15 (b) 8.23 (b)

8.8 (c) 8.16 (d)

CHAPTER 5 5.1 (d) 5.9 (c) 5.17 (c) 5.25 (b) 5.33 (c) 5.41 (c) 5.49 (b) 5.57 (c) 5.65 (a) 5.73 (b)

5.2 (d) 5.10 (c) 5.18 (c) 5.26 (c) 5.34 (b) 5.42 (b) 5.50 (d) 5.58 (c) 5.66 (c)

CHAPTER 6 6.1 (d) 6.9 (a) 6.17 (c) 6.25 (d) 6.33 (d) 6.41 (b) 6.49 (b) 6.57 (d) 6.65 (c) 6.73 (c)

6.2 (d) 6.10 (a) 6.18 (d) 6.26 (c) 6.34 (c) 6.42 (a) 6.50 (b) 6.58 (c) 6.66 (a) 6.74 (b)

CHAPTER 7 7.1 (d) 7.9 (b) 7.17 (b) 7.25 (d)

7.2 (d) 7.10 (b) 7.18 (c) 7.26 (c)

CHAPTER 8 8.1 (b) 8.9 (a) 8.17 (b)

8.2 (a) 8.10 (d) 8.18 (b)

394

Answers to Objective-Type Questions

CHAPTER 9 9.1. (c) 9.9. (c) 9.17. (c) 9.25. (b) 9.33. (b) 9.41. (b) 9.49. (d) 9.57. (h)

9.2. (e) 9.10. (d) 9.18. (a) 9.26. (b) 9.34. (a) 9.42. (c) 9.50. (e) 9.58. (d)

9.3. (g) 9.11. (d) 9.19. (b) 9.27. (c) 9.35. (b) 9.43. (c) 9.51. (a) 9.59. (b)

9.4. (f) 9.12. (a) 9.20. (b) 9.28. (c) 9.36. (d) 9.44. (a) 9.52. (b)

9.5. (g) 9.13. (b) 9.21. (b) 9.29. (b) 9.37. (c) 9.45. (c) 9.53. (c)

9.6. (d) 9.14. (c) 9.22. (a) 9.30. (d) 9.38. (g) 9.46. (c) 9.54. (b)

9.7. (b) 9.15. (b) 9.23. (d) 9.31. (c) 9.39. (a) 9.47. (b) 9.55. (d)

9.8. (a) 9.16. (a) 9.24. (d) 9.32. (c) 9.40. (a) 9.48. (c) 9.56. (b)

10.3 (b) 10.11 (a) 10.19 (b) 10.27 (d)

10.4 (b) 10.12 (a) 10.20 (c) 10.28 (a)

10.5 (c) 10.13 (a) 10.21 (c) 10.29 (c)

10.6 (c) 10.14 (c) 10.22 (a) 10.30 (c)

10.7 (b) 10.15 (c) 10.23 (c)

10.8 (b) 10.16 (b) 10.24 (c)

CHAPTER 10 10.1 (b) 10.9 (d) 10.17 (c) 10.25 (c)

10.2 (a) 10.10 (c) 10.18 (b) 10.26 (a)

References 1.

Abramowitz, M., and Stegun, I.A. (1965), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, Dover, New York, USA, p. 932.

2.

Allen, R.G., Pereira, L.S., Raes, D., and Smith, M. (1998), Crop Evapotranspiration- Guidelines for Computing Crop Water Requirements, FAO Irrigation and drainage paper 56. Annex 2. (http://www.fao.org/docrep/X0490E/x0490e0j.htm)

3.

ASCE (2005), The ASCE Standardized Reference Evapotranspiration Equation, Rep. Task Com. on Standardized Reference Evapotranspiration, EWRI-Am. Soc. Civil. Engr., Reston, VA, p. 59. (https://www.kimberly.uidaho.edu/water/asceewri/ascestzdetmain2005.pdf)

4.

Assouline, S., Selker, J. S., and Parlange, J.-Y. (2007), A Simple Accurate Method to Predict Time of Ponding Under Variable Intensity Rainfall, Water Resour. Res., 43, W03426.

5.

Blaney, H.F., and Criddle, W.D. (1962), Determining Consumptive Use and Irrigation Water Requirements, U. S. Dept. Agr., Agricultural Research Service, Tech Bull 1275, p. 59. (https://naldc.nal.usda.gov/naldc/download.xhtml?id=CAT87201264&content=PDF)

6.

Bouwer, H., Back, J.T., and Oliver, J.M. ( 1999), Predicting Infiltration and Groundwater Mounds for Artificial Recharge, J Hydro. Eng., ASCE, 4(4) pp. 350-357.

7.

Bringi, V. N., and Chandrasekar, V. (2001), Polarimetric Doppler Weather Radar: Principles and Applications, Cambridge University Press, p. 636.

8.

Brouwer, C. and Heibloem, M. (1986), Irrigation Water Management: Irrigation Water Needs, Training manual no. 3, Food and Agriculture Organization. (http://www.fao.org/docrep/s2022e/s2022e07.htm#3.2.4%20determination%20of%20crop%20 factors)

9.

Chebbi, A., Bargaoui, Z. K., and da Conceição Cunha, M. (2013), Development of a Method of Robust Rain Gauge Network Optimization Based on Intensity-Duration-Frequency Results, Hydrol. Earth Syst. Sci., 17, pp. 4259–4268. (www.hydrol-earth-syst-sci.net/17/4259/2013/hess-17-4259-2013.pdf)

10.

Chow, V.T. (1951), A General Formula for Hydrologic Frequency Analysis, EOS, Transactions American Geophysical Union, 32(2) pp. 231-237.

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References

11.

Chow, V.T. (1959), Open Channel Hydraulics, McGraw Hill Book Company, Inc, New York, New York, USA, p. 680.

12.

Clark, C.O. (1943), Storage and the Unit Hydrograph, In Proceedings of the American Society of Civil Engineers, 9 pp. 1333-1360.

13.

Cooper, H.H., Jr., and Jacob, C.E. (1946), A Generalized Graphical Method for Evaluating Formation Constants and Summarizing Well-Field History, Transactions of the American Geophysical Union, 27(4) pp. 526-534.

14.

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Index Symbol f-index 93 A Abramowitz and Stegun 331 Absolute humidity 56 Acoustic doppler current profiler 379 Active remote sensing 363 Actual evaporation 55, 74 Actual evapotranspiration 79 Actual infiltration 90 Aerial irrigation 295 Aerodynamic approach 71 Aerodynamic resistance 75 Albedo 64 AMC classification 125 Anisotropic aquifers 241 Annual hydrograph 105 Antecedent moisture condition 124 Apparent velocity 238 Aquiclude 235 Aquifer 8 Aquifer compressibility 245 Aquifuge 235 Aquitard 235 Area-velocity method 103 Artesian aquifer 235 Assumptions of the UH theory 147

Asymmetry 322 Atmospheric boundary layer 71 Atmospheric emissivity 68 Atmospheric moisture 6 Attenuation 200, 208 Automatic weather stations 385 Available water 295 Available water capacity 233 B Barometers 384 Baseflow 102 Baseflow separation 150 Base period 290 Basin irrigation 296 Basin lag 164 Binomial distribution 327 Blaney-Criddle equation 76 Border irrigation 296 Bowen’s ratio 65 Branch canal 293 C Canal headwork 293 Canal water 290 Capillary fringe 232, 234 Catchment 9 Catchment area 52 Channel flow 102

Index

Chebyshev’s inequality 322 Chezy’s equation 380 Chow 201 Clark’s method 208 Coefficient of excess 324 Coefficient of skewness 323 Coefficient of variation 24 Cold front 20 Command area 290 Complex storm 150 Condensation 53 Condensation nuclei 19 Conditional probabilities 308 Confidence interval 346 Confidence level 346 Confidence limits 346 Confined aquifer 235 Confining unit 235 Consistency check 29 Constant demand 131 Constant head test 375 Consumptive use 53 Continuous injection method 382 Convective precipitation 20 Convolution 149 Cooper-jacob approximation 254 Cooper-jacob method 278 Correlation coefficient 111 Crop factor 77, 289 Crop growth index 83 Crop period 290 Crop water requirement 289 Culturable command area 290 Culvert 115 Cumulative distribution function 312 Cumulative infiltration capacity 82 Cumulative mass function 310 Cup anemometer 384 Current meter 379

Curve number 54, 122–124 Cyclonic precipitation 20 D Dalton’s law 56 Darcy’s law 87, 238 Darcy velocity 238 Databases 385 Decibels of reflectivity 369 De-convolution process 154 Degree of saturation 81, 233 Delay 200 Delayed yield 263 Delta 290 Demand mass curve 130 Density of network 362 Density of stations 362 Dependable flow 128 Depression storage 52 Depth-area-duration 38 Depth-duration curve 38 Derived unit hydrograph 164 Dew 20 Dew point 19, 56 Dew point hygrometer 384 Diffusivity of the aquifer 245 Dilution method 382 Discharge coefficient 380 Discrete convolution 149 Disdrometer 368 Dispersion 320 Distributary 293 Diversion headwork 293 Divide wall 293 Doppler radars 370 Double mass curve 30 Double-ring infiltrometer 375 Drawdown 249 Drawdown recovery 256

401

402

Index

Drip irrigation 297 Drizzle 20 Dupuit assumption 242 Dupuit-Forchheimer assumption 242 Duration 146 Duty 290 E Eddy viscosity 71 Effective conductivity 267 Elastic storativity 263 Electromagnetic current meter 379 Empirical models 110 Energy balance approach 64 Engineering hydrology 1 Environmental flows 300 Ephemeral hydrographs 105 Ephemeral rivers 105 Equilibrium discharge 157 Equivalent water depth 20 Estimation of evaporation 59 Estimation of infiltration capacity 82 Evaporation 9 pans 59, 371 rate 56 Evapotranspiration 10, 75 Expected value 317 Extreme Value Type-I (EVT-I) distribution 332 F Factor of safety 350 Factors affecting evaporation 58 Factors affecting infiltration 81 Factors affecting the runoff 107 Falling head test 375 Falling limb 107 Field capacity 79, 233 Finite Difference (FD) methods 189 Finite Element (FE) methods 190

Firm power 128 Fish and wildlife conservation 299 Fish ladder 293 Fixed base method 151 Float gauge 366 Flood control 299 Flow duration curve 127 Flow mass curve 130 Fog 20 Formation compressibility 245 Formation loss 266 Fractional vegetation 54 Freezing rain 20 Frequency analysis 334 Frequency domain reflectometry 378 Frequency factor 339 Frequency factor method 338 Frontal precipitation 20 Frost 20 Furrow irrigation 296 G Gamma function 214 Gamma ray 378 Gaussian distribution 329 Giovanni (NASA) 385 Global historical climatology network 385 Global precipitation climatology centre 385 Global precipitation measurement 370 Global runoff data centre 385 Global summary of the day 385 Goodrich’s method 195 Gravimetric hygrometer 384 Gravimetric method 377 Gravimetric water content 233 Gravity recovery and climate experiment 384 Green-Ampt Model 87 Gross command area 290 Groundwater 8

Index

Guelph permeameter 376 Gumbel’s distribution 333 Gumbel’s reduced mean and standard deviation 340 Gumbel’s reduced variable 341 H Hail 20 Hantush well function 269 Harbeck and Meyers (1970) formula 60 Heat index 76 HEC-1 109 Histogram 22 Holtan model 83 Hortonian overland flow 102 Horton model for infiltration 83 Horton model for recession 107 Hydraulic approach 240 Hydraulic conductivity 81, 237 Hydraulic gradient 237 Hydraulic head 237 Hydraulic routing 189 Hydroelectric power 298 Hydrograph 104 Hydrograph routing 188 Hydrologic cycle 5 Hydrologic routing 188 Hydrology 1 Hydromet division 385 Hyetograph 23 Hygrometers 384 I Image well 271 IMD standard rain gauge 365 Impermeable boundary 271 Index station method 35 Indian standard evaporation pan 372

403

Infiltration 7 capacity 7 index 93 rate 80 Infiltrometer 82 Initial abstraction 52 Instantaneous unit hydrograph 170–171, 207 Instrument-selection 364 Intensity-duration curve 38 Intensity-Duration-Frequency (IDF) relationships 117 Intensity of irrigation 290 Intensity of precipitation 21 Interception 52 Interception storage 54 Interflow 7, 102 Inter-isochronal area 208 Intermediate zone 235 Intermittent hydrographs 105 Intermittent rivers 105 Intrinsic permeability 238 Inundation canals 293 Inverse-distance-squared weight 27 Irrigation efficiency 292 Isochrone 208 Isohyetal method 35 Isolated storm 150 Isopluvial maps 34 K Kentucky Watershed Model 109 Kharif 290 Kirpich 118 Kor depth 290 Kor period 290 Kostiakov model 82 Kozeny-Carman equation 239 Kurtosis coefficient 323, 324

404

Index

L Lag 200 Lake evaporation 371 Laminar loss 266 Latent heat flux 65 Latent heat of vaporization 65 Laws of probability 308 Layered porous medium 267 Leaf area index 54 Leakage factor 269 Leaky aquifers 235 Leaky aquifer well function 269 Linear channel 208 Linear reservoir 107, 208 Linsley method for time after peak 151 Location parameter 318 Log-Person Type-III distribution 342 Longwave radiation 64 Lysimeters 373 M Main canal 293 Manning’s equation 381 Mass balance approach 64 Mass curve 22 Matrix compressibility 245 Mean 316 Mean annual flood 341 Measurement network 362 Measurement of streamflow 102 Measurement techniques 364 Mechanical evaporation 58 Mechanism of precipitation 19 Median 318 Method of characteristics 189 Method of superposition 159 Meyer’s formula 59 Missing data 26 Mist 20

Mode 319 Modified infiltration capacity 91 Modified Puls method 191 Moisture deficit 83 Moisture storage capacity 83 Moments 316 Monthly hydrograph 107 Monthly mean extraterrestrial radiation 70 Monthly mean of potential sunshine hours 70 Movement of groundwater 236 Movement of water 6 Multipurpose project 299 Muskingum method 202 N Nash model 212 Navigation 299 Neutron probe 378 Non-recording gauges 22 Non-recording rain gauge 365 Normal annual precipitation 21 Normal distribution 329 Normal equations 111 Normal precipitation 34 Normal ratio method 27 O Observation wells 248 Occurrence of groundwater 231 Occurrence of water 5 Operating rules 301 Optimum water content 295 Orographic precipitation 19 Outlet 293 Overland flow 102 P Paleo 290 Pan coefficient 371

Index

Pan evaporation 371 Passive remote sensing 363 Peak discharge 115 Peakedness 323 Peak runoff 164 Penman method 72 Penman-Monteith equation 78 Perched aquifer 235 Perennial canals 293 Perennial hydrograph 105 Perennial river 105 Permanent wilting point 79, 233 Permeameters 374 Philip-Dunne permeameter 376 Philip model 89 Phreatic surface 8 Physical hydrology 1 Piezometric surface 235 Plotting position 127 Point of inflexion 105 Point rainfall 38 Ponding 81 Population 306 Porosity 232 Porosity index 83 Posterior probabilities 308 Potential evaporation 10 Potential evapotranspiration 75 Potentiometric head 237 Power-law velocity profile 59 Power regression model 112 Precipitation 6 Precipitation, relief 19 Primary power 128 Prior probabilities 308 Prism storage 200 Probability 307 Probability density function 312 Probability mass function 309

Probability plotting 335 Psychrometer 384 Psychrometric constant 70 Pumping well 248 Pure translation 209 Pyranometer 385 Q Quadrant method 27 Quickflow 102 R Rabi 290 Radar measurement 369 Radar reflectivity factor 369 Radiometer 385 Radius of influence 248 Rainfall hyetograph 368 Rainfall-runoff library 109 Rainfall-runoff relationship 110 Rain gauges 22 Rain shadow 19 Rainy day 21 Random process 306 Random variable 307 Rational formula 115 Recession coefficient 107 Recession curve 107 Recharge boundary 271 Recording gauges 22 Recording rain gauge 365 Recreational 299 Reference crop evapotranspiration 75 Regulated flow duration curve 128 Relative conductivity 81 Relative standard error 25 Reliability 350 Remote measurement of precipitation 368 Representative day for the month 70

405

406

Index

Reservoir operation 299 Reservoir routing 190 Residence time 11 Residual water content 233 Return period 334 Richards equation 87 Ring irrigation 297 Rising limb 106 Risk 350 Rohwer’s formula 59 RRL 109 Runoff 101 Runoff coefficient 115, 116 Run-off-the-river 300 S Sample 306 Satellite measurement 370 Saturated hydraulic conductivity 81 Saturated vapour pressure 56 Saturated zone 231 Saturation deficit 56 Saturation overland flow 102 Scouring sluices 293 Screen loss 266 SCS Curve Number (CN) method 120 SCS dimensionless synthetic unit hydrograph 168 SCS model 93 S-curve 157 Seasonal hydrograph 107 Secondary power 128 Seepage face 259 Seepage loss 294 Seepage velocity 238 Sensible heat flux 65 Sequent peak algorithm 133 Shape parameter 321 S-curve 157 S-hydrograph 156

S-hydrograph method 161 Significance level 347 Silt excluder 293 Single point method 104 Single purpose project 299 Single ring infiltrometer 374 Site-selection 363 Skewness 322 Sleet 20 Smith-Parlange model 89 Snyder’s synthetic unit hydrograph 163 Soil classification 121 Soil compressibility 241 Soil diffusivity 87 Soil moisture 377 Soil water zone 232 Solar constant 10 Solar declination 65 Solar irradiance 10 Sorptivity 89 Spatial averaging 34 Specific capacity 265 Specific discharge 238 Specific humidity 56 Specific humidity gradient 71 Specific retention 234 Specific storage 241 Specific yield 234 Spread 320 Sprinkler irrigation 297 Stages of growth of a crop 289 Standard deviation 320 Standard error of the mean 25 Standard normal variable 321, 329 Standard Runge–Kutta (SRK) method 195 Standard unit hydrograph 164 Stanford Watershed Model 109 Stefan-Boltzmann constant 65 Stochastic process 306

Index

Storage capacity 131 Storage coefficient 243 Storage time constant 210 Storativity 243 Storm 21 Storm averaging 38 Storm hydrograph 105 Storm Water Management Model 109 Straight line method 150 Streamflow 102 St. Venant equations 189 Subsurface irrigation 297 Suction 271 Sudden injection method 382 Sunken Colorado pan 372 Sunshine duration 385 Sunshine hours 65 Surface irrigation 295 Surface resistance 75 Surface Runoff 9 Surplus power 128 Swartzendruber-Clague model 83 Symons pan 372 Synthetic unit hydrograph 163

Time lag 105 Time of concentration 118 Time of ponding 91 Time of rise 167 Tipping bucket 365 Total probability 308 Tracer 272 Translation 208 Transmissivity 245 Transpiration 9 Triangular unit hydrograph 168 Tropical rainfall measuring mission 370, 385 Turbulent loss 266 Two-point method 104

T

V

Temporal averaging 34 Tension infiltrometer 376 Theis equation 253 Theis type curve method 275 Theoretical hydrology 1 Thermometer 384 Thiem equation 249 Thiessen polygon 34 Thornthwaite equation 75 Thornthwaite-Holzman equation 71 Time-area diagram 209 Time base 164 Time domain reflectometry 378

Vadose zone 231 Vane (or propeller) anemometer 384 Vapour diffusivity 69 Vapour eddy diffusivity 71 Vapour pressure 56 Vapour pressure gradient 56 Variable demand 131 Variable slope method 151 Variance 320 Virga 20 Volumetric heat capacity 68 Volumetric water content 233

U Ultrasonic method 379 Unconfined aquifer 235 Unconfined well function 263 United States Class A pan 371 Unit hydrograph 146 Unsaturated hydraulic conductivity 81 Unsaturated zone 231 U.S. Geological Survey 385

407

408

Index

W Warm front 20 Water compressibility 241, 245 Water content 233 Watercourse 293 Water divide 259 Waterlogging 294 Water resources information system 385 Water supply 299 Water table 8 Water table aquifers 235 Water table fluctuation method 273 Water vapour 19 Wedge storage 201

Weibull’s method 335 Weighing gauge 366 Weir 380 Well function 253 Well loss 265 Wetting front 80 Widths of the SUH 164 Wilting point 294 World radiation data center 385 Z Zayad 290 Zone of aeration 231